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+[
+  {
+    "id": "unit1_fill_1_1",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "3/4 的 ( ) 是 3/8",
+    "options": [],
+    "answer": "1/2",
+    "solution": "根据倒数的意义，3/4 × ( ) = 3/8，将 3/8 除以 3/4 即可得到 1/2。计算：3/8 ÷ 3/4 = 3/8 × 4/3 = 1/2",
+    "knowledge_points": [
+      "分数除法",
+      "倒数"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_001",
+    "original_id": "unit1_fill_1_1"
+  },
+  {
+    "id": "unit1_fill_1_2",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "( ) 的 5/6 是 10/12",
+    "options": [],
+    "answer": "1",
+    "solution": "( ) × 5/6 = 10/12，将 10/12 除以 5/6 即可得到 2/5。计算：10/12 ÷ 5/6 = 10/12 × 6/5 = 2/5",
+    "knowledge_points": [
+      "分数除法",
+      "倒数"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_002",
+    "original_id": "unit1_fill_1_2"
+  },
+  {
+    "id": "unit1_fill_2_1",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "简单",
+    "question": "9/10 乘 4 的结果是 ( )",
+    "options": [],
+    "answer": "36/10 或 3.6",
+    "solution": "分数乘整数的意义是求一个数的几分之几是多少，用分数乘整数。9/10 × 4 = 36/10 = 3.6",
+    "knowledge_points": [
+      "分数乘整数"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_003",
+    "original_id": "unit1_fill_2_1"
+  },
+  {
+    "id": "unit1_fill_2_2",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "简单",
+    "question": "6 乘 3/7 的结果是 ( )",
+    "options": [],
+    "answer": "18/7",
+    "solution": "整数乘分数的意义同分数乘整数，用整数乘分数的分子，分母不变。6 × 3/7 = 18/7",
+    "knowledge_points": [
+      "整数乘分数"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_004",
+    "original_id": "unit1_fill_2_2"
+  },
+  {
+    "id": "unit1_fill_3_1",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "简单",
+    "question": "3/5 × 5 = ( )",
+    "options": [],
+    "answer": "3",
+    "solution": "3/5 × 5 = 15/5 = 3。这里是将分数与整数相乘，整数移到分子上进行运算",
+    "knowledge_points": [
+      "分数乘整数",
+      "约分"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_005",
+    "original_id": "unit1_fill_3_1"
+  },
+  {
+    "id": "unit1_fill_3_2",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "简单",
+    "question": "3/5 × 5 = 3，这里运用了 ( ) 律",
+    "options": [],
+    "answer": "乘法交换",
+    "solution": "运用了乘法交换律，即 a × b = b × a。3/5 × 5 可以看作 5 × 3/5",
+    "knowledge_points": [
+      "乘法交换律"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_006",
+    "original_id": "unit1_fill_3_2"
+  },
+  {
+    "id": "unit1_fill_4",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "简单",
+    "question": "1/3 × 4/5 = ( )",
+    "options": [],
+    "answer": "4/15",
+    "solution": "分数乘分数，用分子相乘的积作分子，分母相乘的积作分母。1/3 × 4/5 = (1×4) / (3×5) = 4/15",
+    "knowledge_points": [
+      "分数乘分数"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_007",
+    "original_id": "unit1_fill_4"
+  },
+  {
+    "id": "unit1_fill_5_1",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "简单",
+    "question": "5/12 的倒数是 ( )",
+    "options": [],
+    "answer": "12/5 或 2.4",
+    "solution": "求一个分数的倒数，把分子和分母调换位置。5/12 的倒数是 12/5",
+    "knowledge_points": [
+      "倒数"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_008",
+    "original_id": "unit1_fill_5_1"
+  },
+  {
+    "id": "unit1_fill_5_2",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "简单",
+    "question": "( ) 没有倒数",
+    "options": [],
+    "answer": "0",
+    "solution": "0 没有倒数，因为任何数乘以 0 都等于 0，不等于 1",
+    "knowledge_points": [
+      "倒数"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_009",
+    "original_id": "unit1_fill_5_2"
+  },
+  {
+    "id": "unit1_fill_6_1",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "简便计算：5/9 × 18 = ( )",
+    "options": [],
+    "answer": "10",
+    "solution": "5/9 × 18 = 5 × (18/9) = 5 × 2 = 10。运用了乘法结合律和分数的基本性质",
+    "knowledge_points": [
+      "简便计算",
+      "乘法结合律"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_010",
+    "original_id": "unit1_fill_6_1"
+  },
+  {
+    "id": "unit1_fill_6_2",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "简便计算：1/4 × 202 = ( )",
+    "options": [],
+    "answer": "50.5",
+    "solution": "1/4 × 202 = 1/4 × (200 + 2) = 1/4 × 200 + 1/4 × 2 = 50 + 0.5 = 50.5",
+    "knowledge_points": [
+      "简便计算",
+      "乘法分配律"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_011",
+    "original_id": "unit1_fill_6_2"
+  },
+  {
+    "id": "unit1_fill_7",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "小明吃了蛋糕的 2/5，小华吃了蛋糕的 3/8，谁吃的多？ ( ) 吃的多",
+    "options": [],
+    "answer": "小明",
+    "solution": "比较分数大小：2/5 = 16/40，3/8 = 15/40。因为 16/40 > 15/40，所以 2/5 > 3/8，小明吃的多",
+    "knowledge_points": [
+      "分数比较大小",
+      "通分"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_012",
+    "original_id": "unit1_fill_7"
+  },
+  {
+    "id": "unit1_fill_8_1",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一根绳子长 1 米，剪去它的 3/4，剪去 ( ) 米",
+    "options": [],
+    "answer": "3/4",
+    "solution": "剪去绳子的 3/4，是把 1 米（单位\"1\"）平均分成 4 份，剪去了其中的 3 份。所以剪去 3/4 米",
+    "knowledge_points": [
+      "分数的意义",
+      "求一个数的几分之几"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_013",
+    "original_id": "unit1_fill_8_1"
+  },
+  {
+    "id": "unit1_fill_8_2",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一根绳子长 1 米，剪去它的 3/4，还剩下 ( ) 米",
+    "options": [],
+    "answer": "1/4",
+    "solution": "还剩下的占全长的 1 - 3/4 = 1/4。所以还剩下 1/4 米",
+    "knowledge_points": [
+      "分数减法",
+      "求剩余部分"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_014",
+    "original_id": "unit1_fill_8_2"
+  },
+  {
+    "id": "unit1_judge_1",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "两个分数相乘，积一定比这两个分数都大",
+    "options": [
+      "正确",
+      "错误"
+    ],
+    "answer": "错误",
+    "solution": "例如，1/2 × 1/2 = 1/4，积 1/4 比原来的两个分数 1/2 都小。所以积不一定比这两个分数都大",
+    "knowledge_points": [
+      "分数乘法性质"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_015",
+    "original_id": "unit1_judge_1"
+  },
+  {
+    "id": "unit1_judge_2",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "判断题",
+    "difficulty": "简单",
+    "question": "互为倒数的两个数的乘积是 1",
+    "options": [
+      "正确",
+      "错误"
+    ],
+    "answer": "正确",
+    "solution": "这是倒数的定义。如果 a × b = 1，且 a、b 都不为 0，根据乘法的交换律，b × a = 1，所以 a 和 b 互为倒数",
+    "knowledge_points": [
+      "倒数的定义"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_016",
+    "original_id": "unit1_judge_2"
+  },
+  {
+    "id": "unit1_judge_3",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "判断题",
+    "difficulty": "简单",
+    "question": "3/4 × 2/3 和 2/3 × 3/4 的结果相同",
+    "options": [
+      "正确",
+      "错误"
+    ],
+    "answer": "正确",
+    "solution": "分数乘法满足交换律，即 a × b = b × a。所以 3/4 × 2/3 和 2/3 × 3/4 的结果相同",
+    "knowledge_points": [
+      "乘法交换律"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_017",
+    "original_id": "unit1_judge_3"
+  },
+  {
+    "id": "unit1_judge_4",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "任何不等于 0 的数都有倒数",
+    "options": [
+      "正确",
+      "错误"
+    ],
+    "answer": "正确",
+    "solution": "0 没有倒数，因为任何数乘以 0 都等于 0，不等于 1。但题目说的是\"任何不等于 0 的数\"，这个说法是正确的。实际上这道题的答案应该是正确",
+    "knowledge_points": [
+      "倒数"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_018",
+    "original_id": "unit1_judge_4"
+  },
+  {
+    "id": "unit1_judge_5",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "判断题",
+    "difficulty": "简单",
+    "question": "如果 a × b = 1，那么 a 和 b 互为倒数",
+    "options": [
+      "正确",
+      "错误"
+    ],
+    "answer": "正确",
+    "solution": "根据倒数的定义，如果 a × b = 1，且 a、b 都不为 0，那么 a 和 b 互为倒数",
+    "knowledge_points": [
+      "倒数的定义"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_019",
+    "original_id": "unit1_judge_5"
+  },
+  {
+    "id": "unit1_choice_1",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "下面算式中，运用了乘法交换律的是",
+    "options": [
+      "A. 1/2 × 3/4 = 3/8",
+      "B. 5/6 × 0 = 0",
+      "C. 2/7 × 4 = 4 × 2/7",
+      "D. 3/5 × 1 = 3/5"
+    ],
+    "answer": "C",
+    "solution": "A 是分数乘分数。B 是 0 乘任何数都得 0。C 是 2/7 × 4 和 4 × 2/7 的位置互换了，运用了乘法交换律。D 是分数乘 1",
+    "knowledge_points": [
+      "乘法交换律"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_020",
+    "original_id": "unit1_choice_1"
+  },
+  {
+    "id": "unit1_choice_2",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "计算 2/9 × 15 的最简结果是",
+    "options": [
+      "A. 5/3",
+      "B. 3/5",
+      "C. 30/9",
+      "D. 10/3"
+    ],
+    "answer": "D",
+    "solution": "2/9 × 15 = (2 × 15) / 9 = 30 / 9。分子分母都有公因数 3，约分得 10 / 3",
+    "knowledge_points": [
+      "分数乘整数",
+      "约分"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_021",
+    "original_id": "unit1_choice_2"
+  },
+  {
+    "id": "unit1_choice_3",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一个数的倒数是 5/7，这个数是",
+    "options": [
+      "A. 5/7",
+      "B. 7/5",
+      "C. 1",
+      "D. 0"
+    ],
+    "answer": "B",
+    "solution": "一个数的倒数是 5/7，说明这个数乘以 5/7 等于 1。这个数就是 1 除以 5/7，即 1 ÷ (5/7) = 7/5",
+    "knowledge_points": [
+      "倒数",
+      "分数除法"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_022",
+    "original_id": "unit1_choice_3"
+  },
+  {
+    "id": "unit1_choice_4",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "简单",
+    "question": "某班有学生 40 人，其中男生占 5/8，男生有多少人",
+    "options": [
+      "A. 25",
+      "B. 15",
+      "C. 32",
+      "D. 50"
+    ],
+    "answer": "A",
+    "solution": "男生占全班的 5/8，全班有 40 人。男生人数 = 40 × 5/8 = (40 ÷ 8) × 5 = 5 × 5 = 25 人",
+    "knowledge_points": [
+      "求一个数的几分之几"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_023",
+    "original_id": "unit1_choice_4"
+  },
+  {
+    "id": "unit1_choice_5",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "简单",
+    "question": "甲数是 15，乙数是甲数的 3/5，乙数是",
+    "options": [
+      "A. 3/5",
+      "B. 5/3",
+      "C. 9",
+      "D. 24"
+    ],
+    "answer": "C",
+    "solution": "乙数是甲数的 3/5，甲数是 15。乙数 = 15 × 3/5 = (15 ÷ 5) × 3 = 3 × 3 = 9",
+    "knowledge_points": [
+      "求一个数的几分之几"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_024",
+    "original_id": "unit1_choice_5"
+  },
+  {
+    "id": "unit1_calc_1_1",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "简单",
+    "question": "1/3 × 6 =",
+    "options": [],
+    "answer": "2",
+    "solution": "1/3 × 6 = 6/3 = 2",
+    "knowledge_points": [
+      "分数乘整数"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_025",
+    "original_id": "unit1_calc_1_1"
+  },
+  {
+    "id": "unit1_calc_1_2",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "简单",
+    "question": "3/4 × 8 =",
+    "options": [],
+    "answer": "6",
+    "solution": "3/4 × 8 = 24/4 = 6",
+    "knowledge_points": [
+      "分数乘整数"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_026",
+    "original_id": "unit1_calc_1_2"
+  },
+  {
+    "id": "unit1_calc_1_3",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "简单",
+    "question": "2/3 × 9 =",
+    "options": [],
+    "answer": "6",
+    "solution": "2/3 × 9 = 18/3 = 6",
+    "knowledge_points": [
+      "分数乘整数"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_027",
+    "original_id": "unit1_calc_1_3"
+  },
+  {
+    "id": "unit1_calc_1_4",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "简单",
+    "question": "1 × 7/10 =",
+    "options": [],
+    "answer": "7/10",
+    "solution": "1 × 7/10 = 7/10",
+    "knowledge_points": [
+      "1乘任何数"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_028",
+    "original_id": "unit1_calc_1_4"
+  },
+  {
+    "id": "unit1_calc_1_5",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "简单",
+    "question": "5 × 2/9 =",
+    "options": [],
+    "answer": "10/9",
+    "solution": "5 × 2/9 = 10/9",
+    "knowledge_points": [
+      "整数乘分数"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_029",
+    "original_id": "unit1_calc_1_5"
+  },
+  {
+    "id": "unit1_calc_1_6",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "简单",
+    "question": "1/2 × 1/4 =",
+    "options": [],
+    "answer": "1/8",
+    "solution": "1/2 × 1/4 = 1/8",
+    "knowledge_points": [
+      "分数乘分数"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_030",
+    "original_id": "unit1_calc_1_6"
+  },
+  {
+    "id": "unit1_calc_2_1",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "1/5 × 50 =",
+    "options": [],
+    "answer": "10",
+    "solution": "1/5 × 50 = 50 × 1/5 = 50/5 = 10。运用乘法交换律进行简便计算",
+    "knowledge_points": [
+      "简便计算",
+      "乘法交换律"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_031",
+    "original_id": "unit1_calc_2_1"
+  },
+  {
+    "id": "unit1_calc_2_2",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "3/7 × 14 =",
+    "options": [],
+    "answer": "6",
+    "solution": "3/7 × 14 = 14 × 3/7 = 14/7 × 3 = 2 × 3 = 6",
+    "knowledge_points": [
+      "简便计算",
+      "乘法交换律"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_032",
+    "original_id": "unit1_calc_2_2"
+  },
+  {
+    "id": "unit1_calc_2_3",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "6/11 × 33 =",
+    "options": [],
+    "answer": "18",
+    "solution": "6/11 × 33 = 6 × (33/11) = 6 × 3 = 18",
+    "knowledge_points": [
+      "简便计算",
+      "乘法结合律"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_033",
+    "original_id": "unit1_calc_2_3"
+  },
+  {
+    "id": "unit1_calc_2_4",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "9/20 × 100 =",
+    "options": [],
+    "answer": "45",
+    "solution": "9/20 × 100 = 100 × 9/20 = (100/20) × 9 = 5 × 9 = 45",
+    "knowledge_points": [
+      "简便计算",
+      "乘法交换律"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_034",
+    "original_id": "unit1_calc_2_4"
+  },
+  {
+    "id": "unit1_calc_3_1",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "5/12 × 8/15 =",
+    "options": [],
+    "answer": "2/9",
+    "solution": "5/12 × 8/15 = (5×8)/(12×15) = 40/180。分子分母同时除以 20 得 2/9",
+    "knowledge_points": [
+      "分数乘分数",
+      "约分"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_035",
+    "original_id": "unit1_calc_3_1"
+  },
+  {
+    "id": "unit1_calc_3_2",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "2/5 × 3/7 × 5 =",
+    "options": [],
+    "answer": "6/7",
+    "solution": "2/5 × 3/7 × 5 = 2/5 × 5 × 3/7 = (2×5/5) × 3/7 = 2 × 3/7 = 6/7。运用乘法交换律和结合律",
+    "knowledge_points": [
+      "分数连乘",
+      "简便计算"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_036",
+    "original_id": "unit1_calc_3_2"
+  },
+  {
+    "id": "unit1_calc_3_3",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "1/4 × 3/8 + 1/4 × 5/8 =",
+    "options": [],
+    "answer": "1/4",
+    "solution": "1/4 × 3/8 + 1/4 × 5/8 = 1/4 × (3/8 + 5/8) = 1/4 × 8/8 = 1/4 × 1 = 1/4。运用乘法分配律",
+    "knowledge_points": [
+      "乘法分配律",
+      "分数加法"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_037",
+    "original_id": "unit1_calc_3_3"
+  },
+  {
+    "id": "unit1_app_1",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "简单",
+    "question": "一块地总面积是 8 公顷，其中 5/8 种粮食，种粮食的面积是多少公顷？",
+    "options": [],
+    "answer": "5公顷",
+    "solution": "种粮食的面积占总面积的 5/8。用总面积乘以这个分率即可求出种粮食的面积。8 × 5/8 = (8/8) × 5 = 1 × 5 = 5 公顷",
+    "knowledge_points": [
+      "求一个数的几分之几",
+      "分数乘法应用"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_038",
+    "original_id": "unit1_app_1"
+  },
+  {
+    "id": "unit1_app_2",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "简单",
+    "question": "一件衣服原价 200 元，打八折出售，现价是多少元？",
+    "options": [],
+    "answer": "160元",
+    "solution": "打八折意味着现价是原价的 80%。用原价乘以 80%（或 0.8）即可求出现价。200 × 0.8 = 160 元",
+    "knowledge_points": [
+      "百分数与分数",
+      "打折问题"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_039",
+    "original_id": "unit1_app_2"
+  },
+  {
+    "id": "unit1_app_3",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "一本书共 120 页，小华第一天看了全书的 1/3，第二天看了全书的 1/4，两天一共看了多少页？",
+    "options": [],
+    "answer": "70页",
+    "solution": "两天一共看了全书的 (1/3 + 1/4) 的数量。先计算 1/3 + 1/4 的和，再乘以总页数。先通分得 4/12 + 3/12 = 7/12。然后 120 × 7/12 = (120 ÷ 12) × 7 = 10 × 7 = 70 页",
+    "knowledge_points": [
+      "分数加法",
+      "求一个数的几分之几"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_040",
+    "original_id": "unit1_app_3"
+  },
+  {
+    "id": "unit1_app_4",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "困难",
+    "question": "一根绳子，第一次用去全长的 2/5，第二次用去剩下部分的 3/4，还剩下全长的几分之几？",
+    "options": [],
+    "answer": "3/20",
+    "solution": "先求出第一次用去占全长的几分之几 (2/5)，再求出剩下的几分之几 (1 - 2/5 = 3/5)，然后用剩下的几分之几乘以第二次用去占剩下部分的分率 (3/4)，求出第二次用去占全长的几分之几 (3/5 × 3/4 = 9/20)，最后用 1 减去第一次和第二次用去的分率之和 (1 - 2/5 - 9/20 = 3/20)",
+    "knowledge_points": [
+      "分数减法",
+      "分数乘法",
+      "复合应用"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_041",
+    "original_id": "unit1_app_4"
+  },
+  {
+    "id": "unit1_app_5",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "简单",
+    "question": "学校图书馆有故事书 1500 本，科技书是故事书的 2/5，科技书有多少本？",
+    "options": [],
+    "answer": "600本",
+    "solution": "科技书的本数占故事书本数的 2/5。用故事书的本数乘以这个分率即可求出科技书的本数。1500 × 2/5 = (1500 ÷ 5) × 2 = 300 × 2 = 600 本",
+    "knowledge_points": [
+      "求一个数的几分之几",
+      "分数乘法应用"
+    ],
+    "source": "人教版六年级上册第一单元测试题",
+    "create_time": "2024-09-24 15:30:00",
+    "merged_id": "unit1_042",
+    "original_id": "unit1_app_5"
+  },
+  {
+    "id": "test2025_fill_1",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "填空题",
+    "difficulty": "简单",
+    "question": "3/8 + 3/8 + 3/8 + 3/8 = ( ) × ( ) = ( )",
+    "options": [],
+    "answer": "3/8, 4, 3/2",
+    "solution": "4个3/8相加，根据乘法的意义可转化为乘法算式3/8×4，计算得3/2",
+    "knowledge_points": [
+      "分数加法转乘法",
+      "乘法意义"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_043",
+    "original_id": "test2025_fill_1"
+  },
+  {
+    "id": "test2025_fill_2_1",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "填空题",
+    "difficulty": "简单",
+    "question": "12个5/6是( )",
+    "options": [],
+    "answer": "10",
+    "solution": "求12个5/6用乘法12×5/6=10",
+    "knowledge_points": [
+      "分数乘整数"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_044",
+    "original_id": "test2025_fill_2_1"
+  },
+  {
+    "id": "test2025_fill_2_2",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "填空题",
+    "difficulty": "简单",
+    "question": "24的2/3是( )",
+    "options": [],
+    "answer": "16",
+    "solution": "求24的2/3用24×2/3=16",
+    "knowledge_points": [
+      "求一个数的几分之几"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_045",
+    "original_id": "test2025_fill_2_2"
+  },
+  {
+    "id": "test2025_fill_3_1",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "填空题",
+    "difficulty": "简单",
+    "question": "3/5时 = ( )分",
+    "options": [],
+    "answer": "36",
+    "solution": "1时=60分，3/5×60=36分",
+    "knowledge_points": [
+      "时间单位换算",
+      "分数乘法应用"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_046",
+    "original_id": "test2025_fill_3_1"
+  },
+  {
+    "id": "test2025_fill_3_2",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "填空题",
+    "difficulty": "简单",
+    "question": "5/8千克 = ( )克",
+    "options": [],
+    "answer": "625",
+    "solution": "1千克=1000克，5/8×1000=625克",
+    "knowledge_points": [
+      "重量单位换算",
+      "分数乘法应用"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_047",
+    "original_id": "test2025_fill_3_2"
+  },
+  {
+    "id": "test2025_fill_4",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "填空题",
+    "difficulty": "简单",
+    "question": "一个正方形的边长是3/4分米，它的周长是( )分米",
+    "options": [],
+    "answer": "3",
+    "solution": "正方形周长=边长×4，3/4×4=3分米",
+    "knowledge_points": [
+      "正方形周长",
+      "分数乘法应用"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_048",
+    "original_id": "test2025_fill_4"
+  },
+  {
+    "id": "test2025_fill_5_1",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一堆煤有15吨，运走它的1/3，还余下( )吨",
+    "options": [],
+    "answer": "10",
+    "solution": "运走15吨的1/3，余下15×(1-1/3)=10吨",
+    "knowledge_points": [
+      "求剩余部分",
+      "分数减法"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_049",
+    "original_id": "test2025_fill_5_1"
+  },
+  {
+    "id": "test2025_fill_5_2",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一堆煤有15吨，运走它的1/3，再运走3/5吨，还余下( )吨",
+    "options": [],
+    "answer": "9又2/5",
+    "solution": "先运走15×1/3=5吨，余下10吨；再运走3/5吨，余下10-3/5=9又2/5吨",
+    "knowledge_points": [
+      "分步计算",
+      "分数减法"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_050",
+    "original_id": "test2025_fill_5_2"
+  },
+  {
+    "id": "test2025_fill_6_1",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "在〇里填上\">\"\"<\"或\"=\"：6/7×5/9〇6/7",
+    "options": [],
+    "answer": "<",
+    "solution": "一个数乘小于1的数，积小于原数",
+    "knowledge_points": [
+      "分数乘法性质",
+      "大小比较"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_051",
+    "original_id": "test2025_fill_6_1"
+  },
+  {
+    "id": "test2025_fill_6_2",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "在〇里填上\">\"\"<\"或\"=\"：5/8×7/5〇5/8",
+    "options": [],
+    "answer": ">",
+    "solution": "一个数乘大于1的数，积大于原数",
+    "knowledge_points": [
+      "分数乘法性质",
+      "大小比较"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_052",
+    "original_id": "test2025_fill_6_2"
+  },
+  {
+    "id": "test2025_fill_6_3",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "在〇里填上\">\"\"<\"或\"=\"：6/5〇3/4×6/5",
+    "options": [],
+    "answer": ">",
+    "solution": "6/5乘以小于1的数3/4，积小于6/5",
+    "knowledge_points": [
+      "分数乘法性质",
+      "大小比较"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_053",
+    "original_id": "test2025_fill_6_3"
+  },
+  {
+    "id": "test2025_fill_6_4",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "在〇里填上\">\"\"<\"或\"=\"：9/10×10/9〇1/6×6",
+    "options": [],
+    "answer": "=",
+    "solution": "9/10×10/9=1，1/6×6=1，所以相等",
+    "knowledge_points": [
+      "分数乘法计算",
+      "互为倒数"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_054",
+    "original_id": "test2025_fill_6_4"
+  },
+  {
+    "id": "test2025_fill_7",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一根绳子长7/8米，第一次用去1/4，第二次用去1/4米，第( )次用去的多",
+    "options": [],
+    "answer": "二",
+    "solution": "第一次用去7/8×1/4=7/32米，7/32<1/4（即8/32），第二次用去的多",
+    "knowledge_points": [
+      "分数乘法应用",
+      "分数比较"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_055",
+    "original_id": "test2025_fill_7"
+  },
+  {
+    "id": "test2025_fill_8_1",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "填空题",
+    "difficulty": "简单",
+    "question": "六（1）班有50人，女生占全班人数的2/5，女生有( )人",
+    "options": [],
+    "answer": "20",
+    "solution": "女生50×2/5=20人",
+    "knowledge_points": [
+      "求一个数的几分之几"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_056",
+    "original_id": "test2025_fill_8_1"
+  },
+  {
+    "id": "test2025_fill_8_2",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "填空题",
+    "difficulty": "简单",
+    "question": "六（1）班有50人，女生占全班人数的2/5，男生有( )人",
+    "options": [],
+    "answer": "30",
+    "solution": "男生50-20=30人",
+    "knowledge_points": [
+      "求剩余部分"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_057",
+    "original_id": "test2025_fill_8_2"
+  },
+  {
+    "id": "test2025_fill_9_1",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "填空题",
+    "difficulty": "简单",
+    "question": "一袋大米25kg，已经吃了它的2/5，吃了( )kg",
+    "options": [],
+    "answer": "10",
+    "solution": "吃了25×2/5=10kg",
+    "knowledge_points": [
+      "求一个数的几分之几"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_058",
+    "original_id": "test2025_fill_9_1"
+  },
+  {
+    "id": "test2025_fill_9_2",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "填空题",
+    "difficulty": "简单",
+    "question": "一袋大米25kg，已经吃了它的2/5，还剩( )kg",
+    "options": [],
+    "answer": "15",
+    "solution": "剩25-10=15kg",
+    "knowledge_points": [
+      "求剩余部分"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_059",
+    "original_id": "test2025_fill_9_2"
+  },
+  {
+    "id": "test2025_fill_10_1",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "比30多1/6的数是( )",
+    "options": [],
+    "answer": "35",
+    "solution": "比30多1/6是30×(1+1/6)=35",
+    "knowledge_points": [
+      "分数加法应用",
+      "求比一个数多几分之几的数"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_060",
+    "original_id": "test2025_fill_10_1"
+  },
+  {
+    "id": "test2025_fill_10_2",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "比36少3/4的数是( )",
+    "options": [],
+    "answer": "9",
+    "solution": "比36少3/4是36×(1-3/4)=9",
+    "knowledge_points": [
+      "分数减法应用",
+      "求比一个数少几分之几的数"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_061",
+    "original_id": "test2025_fill_10_2"
+  },
+  {
+    "id": "test2025_judge_1",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "一个数（0除外）乘分数，积小于这个数",
+    "options": [
+      "正确",
+      "错误"
+    ],
+    "answer": "错误",
+    "solution": "一个数（0除外）乘大于1的分数，积大于这个数，例如2×3/2=3>2",
+    "knowledge_points": [
+      "分数乘法性质"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_062",
+    "original_id": "test2025_judge_1"
+  },
+  {
+    "id": "test2025_judge_2",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "判断题",
+    "difficulty": "简单",
+    "question": "1吨的4/5和4吨的1/5一样重",
+    "options": [
+      "正确",
+      "错误"
+    ],
+    "answer": "正确",
+    "solution": "1吨的4/5是4/5吨，4吨的1/5是4/5吨，重量相等",
+    "knowledge_points": [
+      "分数乘法应用",
+      "重量比较"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_063",
+    "original_id": "test2025_judge_2"
+  },
+  {
+    "id": "test2025_judge_3",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "一根电线长3米，用去2/5米后，还剩下3/5米",
+    "options": [
+      "正确",
+      "错误"
+    ],
+    "answer": "错误",
+    "solution": "剩下3-2/5=2又3/5米，而非3/5米",
+    "knowledge_points": [
+      "分数减法",
+      "长度计算"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_064",
+    "original_id": "test2025_judge_3"
+  },
+  {
+    "id": "test2025_judge_4",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "60的2/5相当于80的3/10",
+    "options": [
+      "正确",
+      "错误"
+    ],
+    "answer": "正确",
+    "solution": "60的2/5=24，80的3/10=24，结果相等",
+    "knowledge_points": [
+      "分数乘法计算",
+      "数值比较"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_065",
+    "original_id": "test2025_judge_4"
+  },
+  {
+    "id": "test2025_judge_5",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "冰箱的数量相当于电视机的7/8，冰箱的数量比电视机少1/8",
+    "options": [
+      "正确",
+      "错误"
+    ],
+    "answer": "正确",
+    "solution": "设电视机为1，则冰箱为7/8，少1-7/8=1/8",
+    "knowledge_points": [
+      "分数减法",
+      "数量关系"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_066",
+    "original_id": "test2025_judge_5"
+  },
+  {
+    "id": "test2025_choice_1",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "选择题",
+    "difficulty": "简单",
+    "question": "求3个2/11的和，应该怎样列式简便",
+    "options": [
+      "A. 3＋2/11",
+      "B. 2/11×3",
+      "C. 2/11+2/11+2/11"
+    ],
+    "answer": "B",
+    "solution": "求几个相同加数的和用乘法简便，即2/11×3",
+    "knowledge_points": [
+      "乘法意义",
+      "简便计算"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_067",
+    "original_id": "test2025_choice_1"
+  },
+  {
+    "id": "test2025_choice_2",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "选择题",
+    "difficulty": "简单",
+    "question": "一瓶油重8千克，吃掉1/2千克，还剩多少千克",
+    "options": [
+      "A. 4",
+      "B. 7又1/2",
+      "C. 1/2"
+    ],
+    "answer": "B",
+    "solution": "剩下8-1/2=7又1/2千克",
+    "knowledge_points": [
+      "分数减法",
+      "重量计算"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_068",
+    "original_id": "test2025_choice_2"
+  },
+  {
+    "id": "test2025_choice_3",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "下列算式中，结果最大的是",
+    "options": [
+      "A. 15×3/5",
+      "B. 15×5/3",
+      "C. 15＋3/5",
+      "D. 15－3/5"
+    ],
+    "answer": "B",
+    "solution": "15×3/5=9，15×5/3=25，15+3/5=15.6，15-3/5=14.4，最大的是25",
+    "knowledge_points": [
+      "分数乘法计算",
+      "数值比较"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_069",
+    "original_id": "test2025_choice_3"
+  },
+  {
+    "id": "test2025_choice_4",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "下面哪个的积在1/5和7/10之间",
+    "options": [
+      "A. 1/5×1/2",
+      "B. 2/3×2/5",
+      "C. 3/8×5"
+    ],
+    "answer": "B",
+    "solution": "A. 1/5×1/2=1/10（小于1/5）；B. 2/3×2/5=4/15≈0.27（在1/5=0.2和7/10=0.7之间）；C. 3/8×5=15/8=1.875（大于7/10）",
+    "knowledge_points": [
+      "分数乘法计算",
+      "数值范围判断"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_070",
+    "original_id": "test2025_choice_4"
+  },
+  {
+    "id": "test2025_choice_5",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "有两袋3千克的大米，第一袋大米取出1/3千克，第二袋中取出1/3，剩下的大米",
+    "options": [
+      "A. 第一袋重",
+      "B. 第二袋重",
+      "C. 一样重",
+      "D. 无法确定"
+    ],
+    "answer": "A",
+    "solution": "第一袋剩3-1/3=2又2/3千克；第二袋剩3×(1-1/3)=2千克，2又2/3>2",
+    "knowledge_points": [
+      "分数减法",
+      "数量比较"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_071",
+    "original_id": "test2025_choice_5"
+  },
+  {
+    "id": "test2025_calc_1_1",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "计算题",
+    "difficulty": "简单",
+    "question": "3/4×16=",
+    "options": [],
+    "answer": "12",
+    "solution": "3/4×16=12",
+    "knowledge_points": [
+      "分数乘整数"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_072",
+    "original_id": "test2025_calc_1_1"
+  },
+  {
+    "id": "test2025_calc_1_2",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "计算题",
+    "difficulty": "简单",
+    "question": "12×5/3=",
+    "options": [],
+    "answer": "20",
+    "solution": "12×5/3=20",
+    "knowledge_points": [
+      "整数乘分数"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_073",
+    "original_id": "test2025_calc_1_2"
+  },
+  {
+    "id": "test2025_calc_1_3",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "计算题",
+    "difficulty": "简单",
+    "question": "0.3×5/6=",
+    "options": [],
+    "answer": "0.25",
+    "solution": "0.3×5/6=0.25",
+    "knowledge_points": [
+      "小数乘分数"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_074",
+    "original_id": "test2025_calc_1_3"
+  },
+  {
+    "id": "test2025_calc_1_4",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "计算题",
+    "difficulty": "简单",
+    "question": "3/5×10/9=",
+    "options": [],
+    "answer": "2/3",
+    "solution": "3/5×10/9=2/3",
+    "knowledge_points": [
+      "分数乘分数"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_075",
+    "original_id": "test2025_calc_1_4"
+  },
+  {
+    "id": "test2025_calc_1_5",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "计算题",
+    "difficulty": "简单",
+    "question": "1×8/7=",
+    "options": [],
+    "answer": "8/7",
+    "solution": "1×8/7=8/7",
+    "knowledge_points": [
+      "1乘任何数"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_076",
+    "original_id": "test2025_calc_1_5"
+  },
+  {
+    "id": "test2025_calc_1_6",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "5/7×2/3×7/5=",
+    "options": [],
+    "answer": "2/3",
+    "solution": "5/7×2/3×7/5=2/3",
+    "knowledge_points": [
+      "分数连乘",
+      "约分"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_077",
+    "original_id": "test2025_calc_1_6"
+  },
+  {
+    "id": "test2025_calc_1_7",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "1/2×1/3×2×1/3=",
+    "options": [],
+    "answer": "1/9",
+    "solution": "1/2×1/3×2×1/3=1/9",
+    "knowledge_points": [
+      "分数连乘"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_078",
+    "original_id": "test2025_calc_1_7"
+  },
+  {
+    "id": "test2025_calc_1_8",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "计算题",
+    "difficulty": "简单",
+    "question": "0×7/12+1/8=",
+    "options": [],
+    "answer": "1/8",
+    "solution": "0×7/12+1/8=1/8",
+    "knowledge_points": [
+      "0乘任何数",
+      "分数加法"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_079",
+    "original_id": "test2025_calc_1_8"
+  },
+  {
+    "id": "test2025_calc_2_1",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "7/18×1/4+3/4×7/18",
+    "options": [],
+    "answer": "7/18",
+    "solution": "7/18×(1/4+3/4)=7/18×1=7/18。运用乘法分配律，提取公因式7/18",
+    "knowledge_points": [
+      "乘法分配律",
+      "简便计算"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_080",
+    "original_id": "test2025_calc_2_1"
+  },
+  {
+    "id": "test2025_calc_2_2",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "(3/4-5/8)×32",
+    "options": [],
+    "answer": "4",
+    "solution": "3/4×32-5/8×32=24-20=4。运用乘法分配律",
+    "knowledge_points": [
+      "乘法分配律",
+      "分数减法"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_081",
+    "original_id": "test2025_calc_2_2"
+  },
+  {
+    "id": "test2025_calc_2_3",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "5/9×10-5/9",
+    "options": [],
+    "answer": "5",
+    "solution": "5/9×(10-1)=5/9×9=5。运用乘法分配律",
+    "knowledge_points": [
+      "乘法分配律",
+      "简便计算"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_082",
+    "original_id": "test2025_calc_2_3"
+  },
+  {
+    "id": "test2025_calc_2_4",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "17×9/16",
+    "options": [],
+    "answer": "9又9/16",
+    "solution": "(16+1)×9/16=16×9/16+1×9/16=9+9/16=9又9/16。将17拆分为16+1",
+    "knowledge_points": [
+      "乘法分配律",
+      "拆分计算"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_083",
+    "original_id": "test2025_calc_2_4"
+  },
+  {
+    "id": "test2025_calc_2_5",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "3/4×2/9×8",
+    "options": [],
+    "answer": "4/3",
+    "solution": "3/4×8×2/9=6×2/9=4/3。运用乘法交换律",
+    "knowledge_points": [
+      "乘法交换律",
+      "简便计算"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_084",
+    "original_id": "test2025_calc_2_5"
+  },
+  {
+    "id": "test2025_calc_2_6",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "5/7×4/15×7/8",
+    "options": [],
+    "answer": "1/6",
+    "solution": "5/7×7/8×4/15=5/8×4/15=1/6。运用乘法交换律，逐步约分",
+    "knowledge_points": [
+      "乘法交换律",
+      "约分",
+      "分数连乘"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_085",
+    "original_id": "test2025_calc_2_6"
+  },
+  {
+    "id": "test2025_calc_3_1",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "2/3与4/5的积加上6乘1/45的积，和是多少？",
+    "options": [],
+    "answer": "2/3",
+    "solution": "2/3×4/5+6×1/45=8/15+2/15=10/15=2/3",
+    "knowledge_points": [
+      "分数乘法",
+      "分数加法",
+      "列式计算"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_086",
+    "original_id": "test2025_calc_3_1"
+  },
+  {
+    "id": "test2025_calc_3_2",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "一个数的5/6加上2/5，和是3/5，这个数是多少？",
+    "options": [],
+    "answer": "6/25",
+    "solution": "设这个数为x，则5/6·x+2/5=3/5，5/6·x=1/5，x=1/5×6/5=6/25",
+    "knowledge_points": [
+      "方程解法",
+      "分数运算"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_087",
+    "original_id": "test2025_calc_3_2"
+  },
+  {
+    "id": "test2025_app_1_1",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "某工厂一月份用电4800度，二月份比一月份节约用电1/10，二月份比一月份节约用电多少度？",
+    "options": [],
+    "answer": "480度",
+    "solution": "节约用电4800×1/10=480度",
+    "knowledge_points": [
+      "求一个数的几分之几",
+      "分数乘法应用"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_088",
+    "original_id": "test2025_app_1_1"
+  },
+  {
+    "id": "test2025_app_1_2",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "某工厂一月份用电4800度，二月份比一月份节约用电1/10，二月份实际用电多少度？",
+    "options": [],
+    "answer": "4320度",
+    "solution": "实际用电4800-480=4320度",
+    "knowledge_points": [
+      "求剩余部分",
+      "分数减法应用"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_089",
+    "original_id": "test2025_app_1_2"
+  },
+  {
+    "id": "test2025_app_2",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "应用题",
+    "difficulty": "简单",
+    "question": "一辆汽车每小时行驶80千米，从甲地到乙地共行驶了4/5小时。甲、乙两地相距多少千米？",
+    "options": [],
+    "answer": "64千米",
+    "solution": "距离=速度×时间，80×4/5=64千米",
+    "knowledge_points": [
+      "路程公式",
+      "分数乘法应用"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_090",
+    "original_id": "test2025_app_2"
+  },
+  {
+    "id": "test2025_app_3_1",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "应用题",
+    "difficulty": "简单",
+    "question": "一个果园占地20公顷，其中的2/5种苹果树，苹果树种了多少公顷？",
+    "options": [],
+    "answer": "8公顷",
+    "solution": "苹果树20×2/5=8公顷",
+    "knowledge_points": [
+      "求一个数的几分之几"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_091",
+    "original_id": "test2025_app_3_1"
+  },
+  {
+    "id": "test2025_app_3_2",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "应用题",
+    "difficulty": "简单",
+    "question": "一个果园占地20公顷，其中的1/4种梨树，梨树种了多少公顷？",
+    "options": [],
+    "answer": "5公顷",
+    "solution": "梨树20×1/4=5公顷",
+    "knowledge_points": [
+      "求一个数的几分之几"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_092",
+    "original_id": "test2025_app_3_2"
+  },
+  {
+    "id": "test2025_app_4",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "修一条长12千米的公路，开工3天修了1.5千米。照这样计算，修完这条路一共需要多少天？",
+    "options": [],
+    "answer": "24天",
+    "solution": "每天修1.5÷3=0.5千米，总天数12÷0.5=24天",
+    "knowledge_points": [
+      "工程问题",
+      "除法运算"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_093",
+    "original_id": "test2025_app_4"
+  },
+  {
+    "id": "test2025_app_5",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "应用题",
+    "difficulty": "简单",
+    "question": "商店运来一批水果，其中苹果有180kg，梨比苹果多1/9，梨比苹果多多少千克？",
+    "options": [],
+    "answer": "20千克",
+    "solution": "多180×1/9=20千克",
+    "knowledge_points": [
+      "求一个数的几分之几",
+      "比较问题"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_094",
+    "original_id": "test2025_app_5"
+  },
+  {
+    "id": "test2025_app_6",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "六年级同学给灾区的小朋友捐款。六（1）班捐了500元，六（2）班捐的是六（1）班的4/5，六（3）班捐的是六（2）班的9/8。六（3）班捐款多少元？",
+    "options": [],
+    "answer": "450元",
+    "solution": "六（2）班500×4/5=400元，六（3）班400×9/8=450元",
+    "knowledge_points": [
+      "分步计算",
+      "分数乘法应用"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_095",
+    "original_id": "test2025_app_6"
+  },
+  {
+    "id": "test2025_app_7",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法检测卷",
+    "question_type": "应用题",
+    "difficulty": "困难",
+    "question": "一本故事书有140页，小红第一天读了全书的1/7，第二天读了余下的1/6，第三天应从第几页读起？",
+    "options": [],
+    "answer": "第41页",
+    "solution": "第一天读140×1/7=20页，余下120页；第二天读120×1/6=20页；两天共读40页，第三天从第41页读起",
+    "knowledge_points": [
+      "分步计算",
+      "余下部分计算"
+    ],
+    "source": "2025年人教版六年级上册第一单元检测卷",
+    "create_time": "2024-09-24 16:00:00",
+    "merged_id": "unit1_096",
+    "original_id": "test2025_app_7"
+  },
+  {
+    "id": "readme_choice_1",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "(9/13 × 4/35) × 35 = 9/13 × (4/35 × 35)，这一步运算运用了什么律？",
+    "options": [
+      "A. 乘法交换律",
+      "B. 乘法结合律",
+      "C. 乘法分配律",
+      "D. 加法结合律"
+    ],
+    "answer": "B",
+    "solution": "乘法结合律的形式是 (a × b) × c = a × (b × c)，即三个数相乘，先把前两个数相乘，再和第三个数相乘，或者先把后两个数相乘，再和第一个数相乘，积不变。题目中等式左边是先把 9/13 和 4/35 相乘，再与 35 相乘；等式右边是先把 4/35 和 35 相乘，再与 9/13 相乘，符合乘法结合律的形式。",
+    "knowledge_points": [
+      "分数的连乘运算",
+      "整数乘法运算定律推广到分数乘法",
+      "乘法结合律"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-24 21:28:32",
+    "merged_id": "unit1_097",
+    "original_id": "readme_choice_1"
+  },
+  {
+    "id": "readme_choice_2",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "简单",
+    "question": "小冬每分钟能打80个字，他5/6分钟打的字与80个相比",
+    "options": [
+      "A. 多于",
+      "B. 少于",
+      "C. 等于"
+    ],
+    "answer": "B",
+    "solution": "80 × 5/6 = 66又2/3个。因为66又2/3 < 80，所以他5/6分钟打的字少于80个。",
+    "knowledge_points": [
+      "整数乘分数",
+      "分数乘法应用"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-24 21:28:32",
+    "merged_id": "unit1_098",
+    "original_id": "readme_choice_2"
+  },
+  {
+    "id": "readme_fill_3_1",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "简单",
+    "question": "9/20吨 = ( )千克",
+    "options": [],
+    "answer": "450",
+    "solution": "9/20吨 = 9/20 × 1000 = 450千克",
+    "knowledge_points": [
+      "吨、千克之间的换算与比较",
+      "分数乘整数"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-24 21:28:32",
+    "merged_id": "unit1_099",
+    "original_id": "readme_fill_3_1"
+  },
+  {
+    "id": "readme_fill_3_2",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "简单",
+    "question": "1/8米 = ( )厘米",
+    "options": [],
+    "answer": "12.5或12又1/2",
+    "solution": "1/8米 = 1/8 × 100 = 25/2 = 12又1/2厘米",
+    "knowledge_points": [
+      "毫米、厘米、分米、米之间的进率与换算",
+      "分数乘整数"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-24 21:28:32",
+    "merged_id": "unit1_100",
+    "original_id": "readme_fill_3_2"
+  },
+  {
+    "id": "readme_fill_3_3",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "简单",
+    "question": "4/15时 = ( )分",
+    "options": [],
+    "answer": "16",
+    "solution": "4/15时 = 4/15 × 60 = 16分",
+    "knowledge_points": [
+      "时、分的认识及换算",
+      "分数乘整数"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-24 21:28:32",
+    "merged_id": "unit1_101",
+    "original_id": "readme_fill_3_3"
+  },
+  {
+    "id": "readme_fill_4",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个乒乓球从50米高的地方下落，每次弹起的高度是下落时高度的2/5，这个乒乓球第三次下落时的高度是( )米",
+    "options": [],
+    "answer": "8",
+    "solution": "50 × 2/5 × 2/5 = 20 × 2/5 = 8米。第一次弹起高度是50 × 2/5 = 20米，第二次弹起高度（即第三次下落高度）是20 × 2/5 = 8米。",
+    "knowledge_points": [
+      "分数的连乘运算",
+      "连续求一个数的几分之几是多少的问题"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-24 21:28:32",
+    "merged_id": "unit1_102",
+    "original_id": "readme_fill_4"
+  },
+  {
+    "id": "readme_calc_5_1",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "简单",
+    "question": "35 × 1/7 =",
+    "options": [],
+    "answer": "5",
+    "solution": "35 × 1/7 = 5",
+    "knowledge_points": [
+      "分数乘整数",
+      "整数乘分数",
+      "分数乘小数"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-24 21:30:41",
+    "merged_id": "unit1_103",
+    "original_id": "readme_calc_5_1"
+  },
+  {
+    "id": "readme_calc_5_2",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "简单",
+    "question": "1/2 - 2/5 =",
+    "options": [],
+    "answer": "1/10",
+    "solution": "1/2 - 2/5 = 5/10 - 4/10 = 1/10",
+    "knowledge_points": [
+      "分数乘整数",
+      "整数乘分数",
+      "分数乘小数"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-24 21:30:41",
+    "merged_id": "unit1_104",
+    "original_id": "readme_calc_5_2"
+  },
+  {
+    "id": "readme_calc_5_3",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "简单",
+    "question": "0.36 × 1/9 =",
+    "options": [],
+    "answer": "0.04",
+    "solution": "0.36 × 1/9 = 0.04",
+    "knowledge_points": [
+      "分数乘整数",
+      "整数乘分数",
+      "分数乘小数"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-24 21:30:41",
+    "merged_id": "unit1_105",
+    "original_id": "readme_calc_5_3"
+  },
+  {
+    "id": "readme_calc_5_4",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "简单",
+    "question": "8/9 × 9/16 × 0 =",
+    "options": [],
+    "answer": "0",
+    "solution": "任何数乘以0都等于0",
+    "knowledge_points": [
+      "分数乘整数",
+      "整数乘分数",
+      "分数乘小数"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-24 21:30:41",
+    "merged_id": "unit1_106",
+    "original_id": "readme_calc_5_4"
+  },
+  {
+    "id": "readme_calc_5_5",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "简单",
+    "question": "3/4 × 20 =",
+    "options": [],
+    "answer": "15",
+    "solution": "3/4 × 20 = 15",
+    "knowledge_points": [
+      "分数乘整数",
+      "整数乘分数",
+      "分数乘小数"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-24 21:30:41",
+    "merged_id": "unit1_107",
+    "original_id": "readme_calc_5_5"
+  },
+  {
+    "id": "readme_calc_5_6",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "简单",
+    "question": "2/3 × 1.5 =",
+    "options": [],
+    "answer": "1",
+    "solution": "2/3 × 1.5 = 2/3 × 3/2 = 1",
+    "knowledge_points": [
+      "分数乘整数",
+      "整数乘分数",
+      "分数乘小数"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-24 21:30:41",
+    "merged_id": "unit1_108",
+    "original_id": "readme_calc_5_6"
+  },
+  {
+    "id": "readme_calc_5_7",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "简单",
+    "question": "7/15 × 30 =",
+    "options": [],
+    "answer": "14",
+    "solution": "7/15 × 30 = 14",
+    "knowledge_points": [
+      "分数乘整数",
+      "整数乘分数",
+      "分数乘小数"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-24 21:30:41",
+    "merged_id": "unit1_109",
+    "original_id": "readme_calc_5_7"
+  },
+  {
+    "id": "readme_calc_5_8",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "简单",
+    "question": "2/3 × 9 =",
+    "options": [],
+    "answer": "6",
+    "solution": "2/3 × 9 = 6",
+    "knowledge_points": [
+      "分数乘整数",
+      "整数乘分数",
+      "分数乘小数"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-24 21:30:41",
+    "merged_id": "unit1_110",
+    "original_id": "readme_calc_5_8"
+  },
+  {
+    "id": "readme_calc_6_1",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "47.62 - (7.23 + 5.89)",
+    "options": [],
+    "answer": "34.5",
+    "solution": "47.62 - (7.23 + 5.89) = 47.62 - 13.12 = 34.5",
+    "knowledge_points": [
+      "小数的加、减法混合运算"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-24 21:30:41",
+    "merged_id": "unit1_111",
+    "original_id": "readme_calc_6_1"
+  },
+  {
+    "id": "readme_calc_6_2",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "205 × 99 + 205",
+    "options": [],
+    "answer": "20500",
+    "solution": "205 × 99 + 205 = 205 × 99 + 205 × 1 = 205 × (99 + 1) = 205 × 100 = 20500。运用乘法分配律",
+    "knowledge_points": [
+      "整数乘法分配律"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-24 21:30:41",
+    "merged_id": "unit1_112",
+    "original_id": "readme_calc_6_2"
+  },
+  {
+    "id": "readme_calc_6_3",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "(1/4 - 1/6 + 5/12) × 36",
+    "options": [],
+    "answer": "18",
+    "solution": "(1/4 - 1/6 + 5/12) × 36 = 1/4 × 36 - 1/6 × 36 + 5/12 × 36 = 9 - 6 + 15 = 18。运用乘法分配律",
+    "knowledge_points": [
+      "整数乘法运算定律推广到分数乘法",
+      "乘法分配律"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-24 21:30:41",
+    "merged_id": "unit1_113",
+    "original_id": "readme_calc_6_3"
+  },
+  {
+    "id": "readme_calc_6_4",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "32 × 1.25 × 2.5",
+    "options": [],
+    "answer": "100",
+    "solution": "32 × 1.25 × 2.5 = 8 × 4 × 1.25 × 2.5 = 8 × 1.25 × 4 × 2.5 = (8 × 1.25) × (4 × 2.5) = 10 × 10 = 100。运用乘法交换律和结合律",
+    "knowledge_points": [
+      "整数乘法运算定律推广到小数乘法",
+      "乘法交换律",
+      "乘法结合律"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-24 21:30:41",
+    "merged_id": "unit1_114",
+    "original_id": "readme_calc_6_4"
+  },
+  {
+    "id": "readme_app_7",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "一个正方形的边长是9/16分米，它的周长是多少分米？",
+    "options": [],
+    "answer": "9/4分米",
+    "solution": "正方形周长 = 边长 × 4，9/16 × 4 = 9/4分米",
+    "knowledge_points": [
+      "正方形的周长",
+      "分数乘整数"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-24 21:30:41",
+    "merged_id": "unit1_115",
+    "original_id": "readme_app_7"
+  },
+  {
+    "id": "readme_app_8",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "某工厂有A、B、C三个车间，已知A车间人数是B车间人数的3/7，C车间人数又是A车间人数的2/5，已知B车间人数有140人，C车间有多少人？",
+    "options": [],
+    "answer": "24人",
+    "solution": "140 × 3/7 × 2/5 = 60 × 2/5 = 24人。先求A车间人数，再求C车间人数",
+    "knowledge_points": [
+      "连续求一个数的几分之几是多少的问题"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-24 21:30:41",
+    "merged_id": "unit1_116",
+    "original_id": "readme_app_8"
+  },
+  {
+    "id": "readme_app_9",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "聪聪看一本300页的书，第一天看了全书的1/5，第二天看的页数是第一天的3/4，聪聪第二天看了多少页？",
+    "options": [],
+    "answer": "45页",
+    "solution": "300 × 1/5 × 3/4 = 60 × 3/4 = 45页。先求第一天看的页数，再求第二天看的页数",
+    "knowledge_points": [
+      "连续求一个数的几分之几是多少的问题"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-24 21:30:41",
+    "merged_id": "unit1_117",
+    "original_id": "readme_app_9"
+  },
+  {
+    "id": "readme_app_10",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "上半年计划生产560万吨石头纸，实际比计划多生产1/4，上半年实际生产了多少万吨？",
+    "options": [],
+    "answer": "700万吨",
+    "solution": "560 × (1 + 1/4) = 560 × 5/4 = 700万吨。实际生产是计划的(1 + 1/4)倍",
+    "knowledge_points": [
+      "求比一个数多/少几分之几的数是多少"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-24 21:30:41",
+    "merged_id": "unit1_118",
+    "original_id": "readme_app_10"
+  },
+  {
+    "id": "readme_choice_1_corrected",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "(9/13 × 4/35) × 35 = 9/13 × (4/35 × 35)，这一步运算运用了什么律？",
+    "options": [
+      "A. 乘法交换律",
+      "B. 乘法结合律",
+      "C. 乘法分配律",
+      "D. 加法结合律"
+    ],
+    "answer": "B",
+    "solution": "乘法结合律的形式是 (a × b) × c = a × (b × c)，即三个数相乘，先把前两个数相乘，再和第三个数相乘，或者先把后两个数相乘，再和第一个数相乘，积不变。题目中等式左边是先把 9/13 和 4/35 相乘，再与 35 相乘；等式右边是先把 4/35 和 35 相乘，再与 9/13 相乘，符合乘法结合律的形式。",
+    "knowledge_points": [
+      "分数的连乘运算",
+      "整数乘法运算定律推广到分数乘法",
+      "乘法结合律"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-24 21:45:00",
+    "merged_id": "unit1_119",
+    "original_id": "readme_choice_1_corrected"
+  },
+  {
+    "id": "readme_choice_2_corrected",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "简单",
+    "question": "小冬每分钟能打80个字，他5/6分钟打的字与80个相比",
+    "options": [
+      "A. 多于",
+      "B. 少于",
+      "C. 等于"
+    ],
+    "answer": "B",
+    "solution": "80 × 5/6 = 66又2/3个。因为66又2/3 < 80，所以他5/6分钟打的字少于80个。",
+    "knowledge_points": [
+      "整数乘分数",
+      "分数乘法应用"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-24 21:45:00",
+    "merged_id": "unit1_120",
+    "original_id": "readme_choice_2_corrected"
+  },
+  {
+    "id": "readme_fill_unit_conversion_1",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "简单",
+    "question": "9/20吨 = ( )千克",
+    "options": [],
+    "answer": "450",
+    "solution": "9/20吨 = 9/20 × 1000 = 450千克。根据单位换算，1吨 = 1000千克",
+    "knowledge_points": [
+      "吨、千克之间的换算与比较",
+      "分数乘整数"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-24 21:45:00",
+    "merged_id": "unit1_121",
+    "original_id": "readme_fill_unit_conversion_1"
+  },
+  {
+    "id": "readme_fill_unit_conversion_2",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "简单",
+    "question": "1/8米 = ( )厘米",
+    "options": [],
+    "answer": "12.5或12又1/2",
+    "solution": "1/8米 = 1/8 × 100 = 25/2 = 12又1/2厘米。根据单位换算，1米 = 100厘米",
+    "knowledge_points": [
+      "毫米、厘米、分米、米之间的进率与换算",
+      "分数乘整数"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-24 21:45:00",
+    "merged_id": "unit1_122",
+    "original_id": "readme_fill_unit_conversion_2"
+  },
+  {
+    "id": "readme_fill_unit_conversion_3",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "简单",
+    "question": "4/15时 = ( )分",
+    "options": [],
+    "answer": "16",
+    "solution": "4/15时 = 4/15 × 60 = 16分。根据单位换算，1时 = 60分",
+    "knowledge_points": [
+      "时、分的认识及换算",
+      "分数乘整数"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-24 21:45:00",
+    "merged_id": "unit1_123",
+    "original_id": "readme_fill_unit_conversion_3"
+  },
+  {
+    "id": "readme_fill_ball_bounce",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个乒乓球从50米高的地方下落，每次弹起的高度是下落时高度的2/5，这个乒乓球第三次下落时的高度是( )米",
+    "options": [],
+    "answer": "8",
+    "solution": "50 × 2/5 × 2/5 = 20 × 2/5 = 8米。第一次弹起高度是50 × 2/5 = 20米，第二次弹起高度（即第三次下落高度）是20 × 2/5 = 8米。",
+    "knowledge_points": [
+      "分数的连乘运算",
+      "连续求一个数的几分之几是多少的问题"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-24 21:45:00",
+    "merged_id": "unit1_124",
+    "original_id": "readme_fill_ball_bounce"
+  },
+  {
+    "id": "readme_app_square_perimeter",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "一个正方形的边长是9/16分米，它的周长是多少分米？",
+    "options": [],
+    "answer": "9/4分米",
+    "solution": "正方形周长 = 边长 × 4，9/16 × 4 = 9/4分米",
+    "knowledge_points": [
+      "正方形的周长",
+      "分数乘整数"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-24 21:45:00",
+    "merged_id": "unit1_125",
+    "original_id": "readme_app_square_perimeter"
+  },
+  {
+    "id": "readme_app_factory_workers",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "某工厂有A、B、C三个车间，已知A车间人数是B车间人数的3/7，C车间人数又是A车间人数的2/5，已知B车间人数有140人，C车间有多少人？",
+    "options": [],
+    "answer": "24人",
+    "solution": "140 × 3/7 × 2/5 = 60 × 2/5 = 24人。先求A车间人数：140 × 3/7 = 60人，再求C车间人数：60 × 2/5 = 24人",
+    "knowledge_points": [
+      "连续求一个数的几分之几是多少的问题"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-24 21:45:00",
+    "merged_id": "unit1_126",
+    "original_id": "readme_app_factory_workers"
+  },
+  {
+    "id": "readme_app_reading_book",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "聪聪看一本300页的书，第一天看了全书的1/5，第二天看的页数是第一天的3/4，聪聪第二天看了多少页？",
+    "options": [],
+    "answer": "45页",
+    "solution": "300 × 1/5 × 3/4 = 60 × 3/4 = 45页。先求第一天看的页数：300 × 1/5 = 60页，再求第二天看的页数：60 × 3/4 = 45页",
+    "knowledge_points": [
+      "连续求一个数的几分之几是多少的问题"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-24 21:45:00",
+    "merged_id": "unit1_127",
+    "original_id": "readme_app_reading_book"
+  },
+  {
+    "id": "readme_app_stone_paper",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "上半年计划生产560万吨石头纸，实际比计划多生产1/4，上半年实际生产了多少万吨？",
+    "options": [],
+    "answer": "700万吨",
+    "solution": "560 × (1 + 1/4) = 560 × 5/4 = 700万吨。实际生产是计划的(1 + 1/4)倍，即计划加上多生产的部分",
+    "knowledge_points": [
+      "求比一个数多/少几分之几的数是多少"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-24 21:45:00",
+    "merged_id": "unit1_128",
+    "original_id": "readme_app_stone_paper"
+  },
+  {
+    "id": "readme_q_002",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "下列算式中,计算结果最小的是( )。",
+    "options": [
+      "A. 4/5 × 8/7 B. 4/5 × 3/7 C. 4/5 × 3/5",
+      "B. 3/7 < 1,4/5 × 3/7 < 4/5 ;",
+      "C. 3/5 < 1,4/5 × 3/5 < 4/5 ;"
+    ],
+    "answer": "B",
+    "solution": "A. 8/7 > 1,4/5 × 8/7 > 4/5 ,排除； B. 3/7 < 1,4/5 × 3/7 < 4/5 ; C. 3/5 < 1,4/5 × 3/5 < 4/5 ; 3/7 < 3/5 计算结果最小的是 4/5 × 3/7 。 故答案为：B",
+    "knowledge_points": [
+      "因数和积的大小关系(分数乘法)"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_004",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "课间时,甲、乙、丙、丁四个学习小组在讨论算式 1/2 × 3/4 × 5/6 × ... × 97/98 × 99/100 的结果,正确的( )。",
+    "options": [
+      "A. 甲组说: 大于 2 B. 乙组说: 大于 99/100",
+      "C. 丙组说: 大于 1/2 且小于 99/100 D. 丁组说: 大于 0 且小于 1/2",
+      "B. 算式 1/2 × 3/4 × 5/6 × ... × 97/98 × 99/100 中, 99/100 乘的数都小于 1,所以积都小于 99/100 ,不可能大于 99/100 ,所以该选项说法错误;",
+      "C. 算式 1/2 × 3/4 × 5/6 × ... × 97/98 × 99/100 中, 1/2 乘的数都小于 1 ,所以积都比 1/2 小,而非大于 1/2 ,所以该选项说法错误；",
+      "D. 算式 1/2 × 3/4 × 5/6 × ... × 97/98 × 99/100 中,所有分数都大于 0,所以积大于 0,同时 1/2 乘的数都小于 1,所以积一定比 1/2 小,因此结果大于 0 小于 1/2 ,该选项说法正确。"
+    ],
+    "answer": "D",
+    "solution": "A. 算式 1/2 × 3/4 × 5/6 × ... × 97/98 × 99/100 中, 1/2 乘的数都小于 1,所以积一定比 1/2 小,不可能大于 2,所以该选项说法错误； B. 算式 1/2 × 3/4 × 5/6 × ... × 97/98 × 99/100 中, 99/100 乘的数都小于 1,所以积都小于 99/100 ,不可能大于 99/100 ,所以该选项说法错误; C. 算式 1/2 × 3/4 × 5/6 × ... × 97/98 × 99/100 中, 1/2 乘的数都小于 1 ,所以积都比 1/2 小,而非大于 1/2 ,所以该选项说法错误； D. 算式 1/2 × 3/4 × 5/6 × ... × 97/98 × 99/100 中,所有分数都大于 0,所以积大于 0,同时 1/2 乘的数都小于 1,所以积一定比 1/2 小,因此结果大于 0 小于 1/2 ,该选项说法正确。 故答案为：D",
+    "knowledge_points": [
+      "因数和积的大小关系(分数乘法)"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_007",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "下面的算式中,( )与 1/7 × 3 的结果不同。",
+    "options": [
+      "A. 1/7 + 1/7 + 1/7 B. 1/7 + 1/7 × 2 C. ( {1/3 + 1/7}) × 3 D. 1/7 × 4 - 1/7",
+      "B. 根据乘法分配律, 1/7 + 1/7 × 2 = 1/7 × 1 + 1/7 × 2 = 1/7 × ( {1 + 2}) = 1/7 × 3 ,所以该选项与 1/7 × 3 结果相同。",
+      "C. 根据乘法分配律,计算 ( {1/3 + 1/7}) × 3 = 1/3 × 3 + 1/7 × 3 = 1 + 3/7 = 13/7 ,而 1/7 × 3 = 3/7 ,两者结果不同,所以该选项符合要求。",
+      "D. 根据乘法分配律, 1/7 × 4 - 1/7 = 1/7 × 4 - 1/7 × 1 = 1/7 × ( {4 - 1}) = 1/7 × 3 ,所以该选项与 1/7 × 3 结果相同。 故答案为: C"
+    ],
+    "answer": "C",
+    "solution": "A. 根据乘法的意义, 1/7 × 3 表示 3 个 1/7 相加,也就是 1/7 + 1/7 + 1/7 ,所以该选项与 1/7 × 3 结果相同。 B. 根据乘法分配律, 1/7 + 1/7 × 2 = 1/7 × 1 + 1/7 × 2 = 1/7 × ( {1 + 2}) = 1/7 × 3 ,所以该选项与 1/7 × 3 结果相同。 C. 根据乘法分配律,计算 ( {1/3 + 1/7}) × 3 = 1/3 × 3 + 1/7 × 3 = 1 + 3/7 = 13/7 ,而 1/7 × 3 = 3/7 ,两者结果不同,所以该选项符合要求。 D. 根据乘法分配律, 1/7 × 4 - 1/7 = 1/7 × 4 - 1/7 × 1 = 1/7 × ( {4 - 1}) = 1/7 × 3 ,所以该选项与 1/7 × 3 结果相同。 故答案为: C",
+    "knowledge_points": [
+      "整数乘法运算定律推广到分数乘法、分数乘整数"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_012",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一个饲养场,养鸭 1200 只,养的鸡比鸭多 3/5 ,养的鸡比鸭多多少只? 正确的列式是( )。",
+    "options": [
+      "A. {1200} + {1200} × 3/5 B. {1200} - {1200} × 3/5",
+      "C. {1200} × 3/5 D. {1200} ( {1 - 3/5})"
+    ],
+    "answer": "C",
+    "solution": "{1200} × 3/5 = {720} (只) 养的鸡比鸭多 720 只。 求养的鸡比鸭多多少只,列式为: {1200} × 3/5 。 故答案为： C",
+    "knowledge_points": [
+      "整数乘分数、求一个数的几分之几的问题"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_018",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "小林说: “一个数乘一个分数所得的积一定比这个数小。”下面算式中, 能证明他的说法错误的算式是 ( )。",
+    "options": [
+      "A. 3 × 2/3 B. 1/3 × 2/3 C. 4 × 1/2 D. {0.4} × 3/2",
+      "B. 1/3 × 2/3 ,因为为 2/3 < 1 ,所以 1/3 × 2/3 < 1/3 ,不符合题意；",
+      "C. 4 × 1/2 ,因为 1/2 < 1 ,所以 4 × 1/2 < 4 ,不符合题意；",
+      "D. {0.4} × 3/2 ,因为 3/2 > 1 ,所以 {0.4} × 3/2 > {0.4} ,符合题意。"
+    ],
+    "answer": "D",
+    "solution": "A. 3 × 2/3 ；因为 2/3 < 1 ,所以 3 × 2/3 < 3 ,不符合题意； B. 1/3 × 2/3 ,因为为 2/3 < 1 ,所以 1/3 × 2/3 < 1/3 ,不符合题意； C. 4 × 1/2 ,因为 1/2 < 1 ,所以 4 × 1/2 < 4 ,不符合题意； D. {0.4} × 3/2 ,因为 3/2 > 1 ,所以 {0.4} × 3/2 > {0.4} ,符合题意。 小林说: “一个数乘一个分数所得的积一定比这个数小。”下面算式中,能证明他的说法错误的算式是 {0.4} × 3/2 。 故答案为：D",
+    "knowledge_points": [
+      "因数和积的大小关系(分数乘法)"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_021",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "简单",
+    "question": "一瓶花生油有 9/2 升,半瓶有多少升？列式正确的是( )。",
+    "options": [
+      "A. 9/2 × 1/2 B. 9/2 + 1/2 C. 9/2 × 2",
+      "B. 表示一瓶花生油的体积再加上 1/2 升,不符合题意;",
+      "C. 表示一瓶花生油的体积的 2 倍, 不符合题意;"
+    ],
+    "answer": "A",
+    "solution": "A. 表示一瓶花生油体积的一半,列式正确； B. 表示一瓶花生油的体积再加上 1/2 升,不符合题意; C. 表示一瓶花生油的体积的 2 倍, 不符合题意; 故答案为：A ## 二、填空题",
+    "knowledge_points": [
+      "分数乘分数"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_022",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一根电线长 24 米,如果用去它的 5/8 ,那么还剩下( )米；如果用去 5/8 米,那么剩下( )米。",
+    "options": [],
+    "answer": "9",
+    "solution": "将这根电线看作单位 “1”,用去 5/8 后,还剩下： {24} × ( {1 - 5/8}) = {24} × 3/8 = 9 (米) 用去 5/8 米,那么剩下: {24} - 5/8 = {23}3/8 (米)。",
+    "knowledge_points": [
+      "同分母分数加、减法、求一个数的几分之几的问题"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_023",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "成人的脚长约是身高的 1/7 ,爸爸身高是 175 厘米,他的脚长是 ( )厘米。",
+    "options": [],
+    "answer": "25",
+    "solution": "{175} × 1/7 = {25} (厘米) 所以他的脚长是 25 厘米。 ---",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_026",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "{32} {kg} 的 3/8 是 ( ) {kg} ；6 个 3/8 相加的和是( )。",
+    "options": [],
+    "answer": "12",
+    "solution": "{32} × 3/8 = {12} (千克); 3/8 × 6 = 9/4 。 即 {32} {kg} 的 3/8 是 {12} {kg};6 个 3/8 相加的和是 9/4 。",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_027",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "简单",
+    "question": "像 1/5 + 1/5 + 1/5 + ... + 1/5 这样 a 个 1/5 相加改成乘法算式是( ),当 a = 35 时,结果等于( )。",
+    "options": [],
+    "answer": "1/5 × a/a × 1/5 7",
+    "solution": "把 1/5 + 1/5 + 1/5 + ... + 1/5 这样 a 个 1/5 相加改成乘法算式是 1/5 × a 。 当 a = {35} 时,代入 1/5 × a,1/5 × {35} = 7 。 所以改成乘法算式是 1/5 × a ,当 a = {35} 时,结果等于 7 。",
+    "knowledge_points": [
+      "用字母表示数、数量关系、分数乘整数"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_028",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "( )吨比 40 吨少 1/4;{2024} 年的第一季度共有( )天。",
+    "options": [],
+    "answer": "30 91",
+    "solution": "{40} × ( {1 - 1/4}) = {40} × 3/4 <!-- Meanless: 试卷第 14 页,共 44 页--> --- = {30} (吨) 所以,30 吨比 40 吨少 1/4 。 {2024} 4 = {506} 所以, 2024 年是润年, 2 月有 29 天。 {31} + {29} + {31} = {60} + {31} = {91} (天) 所以, 2024 年的第一季度共有 91 天。 ---",
+    "knowledge_points": [
+      "平年、闰年的认识及判定方法、年、月、日的认识及换算、求比一个数多/少几分之几的数是多少"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_029",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一幅长方形书法作品长 7/8 米,宽比长短 1/4 。这幅作品的面积是( )平方米。",
+    "options": [],
+    "answer": "147/256",
+    "solution": "7/8 × ( {1 - 1/4}) = 7/8 × 3/4 = 21/32 (米) 7/8 × 21/32 = 147/256 (平方米) 故这幅作品的面积是 147/256 平方米。",
+    "knowledge_points": [
+      "分数乘分数、求比一个数多/少几分之几的数是多少"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_030",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "甲数是 40,乙数是甲数的 3/5 ,丙数是乙数的 1/6 。丙数是( )。",
+    "options": [],
+    "answer": "4",
+    "solution": "{40} × 3/5 × 1/6 = {24} × 1/6 = 4 丙数是 4。",
+    "knowledge_points": [
+      "连续求一个数的几分之几是多少的问题"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_031",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "有红、黑两种颜色的球共 40 个,其中红球个数占 5/8 。红球比黑球多( )个。",
+    "options": [],
+    "answer": "10",
+    "solution": "红球: {40} × 5/8 = {25} (个) 黑球: {40} - {25} = {15} (个) {25} - {15} = {10} (个) 有红、黑两种颜色的球共 40 个,其中红球个数占 5/8 。红球比黑球多 10 个。",
+    "knowledge_points": [
+      "整数乘分数"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_033",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "20 米的 3/4 是( )米； 7/12 千克的 3/14 是( )千克。",
+    "options": [],
+    "answer": "15 1/8",
+    "solution": "{20} × 3/4 = {15} (米), 7/12 × 3/14 = 21/168 = 1/8 (千克) <!-- Meanless: 试卷第 16 页,共 44 页--> 故 20 米的 3/4 是 15 米; 7/12 千克的 3/14 是 1/8 千克。",
+    "knowledge_points": [
+      "整数乘分数、求一个数的几分之几的问题、分数乘分数"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_034",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "5/9 + 5/9 + 5/9 + 5/9 = ( ) × ( ) = ( )。",
+    "options": [],
+    "answer": "5/9 4 20/9",
+    "solution": "5/9 + 5/9 + 5/9 + 5/9 = 5/9 × 4 = 5 × 4/9 = 20/9 所以, 5/9 + 5/9 + 5/9 + 5/9 = 5/9 × 4 = 20/9 。 35. {2.4} × 1/2 = ( ) /5 × 1/2 = ( ) {2.4} × 1/2 = {2.4} × ( )(填小数) = ( ) 11/15 × {0.45} = 11/( matrix {15 \\\\ {15} matrix) } × ( ) /20 = ( ) /( matrix & {15 \\\\ {15} & matrix) }11/15 × {0.45} = 11/( matrix {15 \\\\ {15} matrix) } × {0.45} = ( )",
+    "knowledge_points": [
+      "分数乘整数"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_037",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "我们知道 1/2 × 2/3 = 1/3,1/2 × 2/3 × 3/4 = 1/4 ,那么 1/2 × 2/3 × 3/4 × ... × n{n + 1} = ( )。",
+    "options": [],
+    "answer": "1/n + 1",
+    "solution": "我们知道 1/2 × 2/3 = 1/3,1/2 × 2/3 × 3/4 = 1/4 ,那么 1/2 × 2/3 × 3/4 × ... × n{n + 1} = 1/n + 1 。",
+    "knowledge_points": [
+      "算式的规律(分数)、用字母表示数、数量关系、分数乘分数"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_038",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "六年级有 a 名男生,女生人数是男生人数的 3/4 ,六年级一共有_____人。",
+    "options": [],
+    "answer": "7/4a",
+    "solution": "a + a × 3/4 = a + 3/4a <!-- Meanless: 试卷第 18 页,共 44 页--> = 7/4a (人) 六年级有 a 名男生,女生人数是男生人数的 3/4 ,六年级一共有 7/4a 人。",
+    "knowledge_points": [
+      "用字母表示数、数量关系、求一个数的几分之几的问题、含有字母式子的化简与求值"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_039",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "周角的 1/2 = _____ {}^{ } ,直角的 1/5 = _____ {}^{ } ,平角的 3/4 = _____ {}^{ } 。",
+    "options": [],
+    "answer": "",
+    "solution": "{360}^{ } × 1/2 = {180}^{ } {90}^{ } × 1/5 = {18}^{ } {180}^{ } × 3/4 = {135}^{ } 周角的 1/2 = {180}^{ } ,直角的 1/5 = {18}^{ } ,平角的 3/4 = {135}^{ } 。",
+    "knowledge_points": [
+      "直角、钝角、锐角的认识及特征、整数乘分数、求一个数的几分之几的问题、平角、周角的认识及特征"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_040",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "1 年 = _____日(平年) 1 年 = _____日(闰年)",
+    "options": [],
+    "answer": "365 366 28 29 15 90",
+    "solution": "1/4 × {60} = {15},1/4 时 = {15} 分； 1分 = {60} 秒, {60} + {30} = {90},1 分30秒 = {90} 秒。 1 年 = {365} 日(平年)；1 年 = {366} 日(闰年)； 平年 2 月 = {28} 天；闰年 2 月 = {29} 天； 1/4 时 = {15} 分； 1 分 30 秒 = {90} 秒。",
+    "knowledge_points": [
+      "平年、闰年的认识及判定方法、年、月、日的认识及换算、时、分、秒之间的换算与比较、分数乘整数"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_041",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一瓶 4/5 升的饮料,小明喝了 3 瓶。求小明喝了多少升,列式为 4/5 × 3 ,这个算式表示( )；小刚喝了",
+    "options": [],
+    "answer": "4/5 的 3 倍是多少 2/3 的 4/5 是多少",
+    "solution": "据分析可知,一瓶 4/5 升的饮料,小明喝了 3 瓶。求小明喝了多少升,列式为 4/5 × 3 ,这个算式表示 4/5 的 3 倍是多少; 小刚喝了 2/3 瓶,求小刚喝了多少升,列式为 4/5 × 2/3 ,这个算式表示 2/3 的 4/5 是多少。",
+    "knowledge_points": [
+      "求一个数的几分之几的问题、分数乘整数、分数乘分数"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_042",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "简单",
+    "question": "东东看一本书,共有 160 页,已看的页数占全书的 2/5 ,接下来东东从第 ( ) 页看起。",
+    "options": [],
+    "answer": "65",
+    "solution": "{160} × 2/5 + 1 = {64} + 1 = {65} (页) 东东看一本书,共有 160 页,已看的页数占全书的 2/5 ,接下来东东从第 65 页看起。",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_043",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "简单",
+    "question": "一袋 {20}{kg} 的大米,吃了它的 1/5 ,吃了( ) {kg} 。",
+    "options": [],
+    "answer": "4",
+    "solution": "{20} × 1/5 = 4( {kg}) 吃了 4kg。",
+    "knowledge_points": [
+      "整数乘分数、求一个数的几分之几的问题"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_044",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "简单",
+    "question": "3/4 吨的 3/5 是( )吨, 3/8 米的 1/3 是 ( )米。",
+    "options": [],
+    "answer": "9/20/{0.45} 1/8/{0.125}",
+    "solution": "3/4 × 3/5 = 9/20 (吨) 3/8 × 1/3 = 1/8 (米) 3/4 吨的 3/5 是 9/20 吨(或 0.45 吨), 3/8 米的 1/3 是 1/8 米(或 0.125 吨)。",
+    "knowledge_points": [
+      "求一个数的几分之几的问题、分数乘分数"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_045",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "简单",
+    "question": "1/12 + 1/12 + 1/12 + 1/12 + 1/12 + 1/12 + 1/12 = ( ) × ( ) = ( )。",
+    "options": [],
+    "answer": "7 1/12 7/12",
+    "solution": "由分析可知: 1/12 + 1/12 + 1/12 + 1/12 + 1/12 + 1/12 + 1/12 = 7 × 1/12 = 7/12 。",
+    "knowledge_points": [
+      "初步认识乘法意义、分数乘整数"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_046",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "5 米长的绳子先剪去 1/5 ,再接上 1/5 米,结果和原来一样长。( )",
+    "options": [],
+    "answer": "× 5 × ( {1 - 1/5}) + 1/5 = 5 × 4/5 + 1/5 = 4 + 1/5 = 41/5 (米)； 41/5 米 5 米",
+    "solution": "5 × ( {1 - 1/5}) + 1/5 = 5 × 4/5 + 1/5 = 4 + 1/5 = 41/5 (米) 41/5 米 5 米。 5 米长的绳子先剪去 1/5 ,再接上 1/5 米,结果和原来一样长。(✗) 理由: 5 × ( {1 - 1/5}) + 1/5 = 5 × 4/5 + 1/5 = 4 + 1/5 = 41/5 (米); 41/5 米 5 米。",
+    "knowledge_points": [
+      "整数乘分数、求比一个数多/少几分之几的数是多少、同分母分数的大小比较"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_047",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "将一根丝带对折 3 次后量得长为 2/5 米,这根丝带原来长( )米。",
+    "options": [],
+    "answer": "16/5",
+    "solution": "2/5 × 8 = 16/5 (米) 这根丝带原来长 16/5 米。",
+    "knowledge_points": [
+      "分数乘整数"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_048",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "简单",
+    "question": "5/7 × {21} 表示 21 个 5/7 相( )。",
+    "options": [],
+    "answer": "加",
+    "solution": "根据分析可知, 5/7 × {21} 表示 21 个 5/7 相加。",
+    "knowledge_points": [
+      "分数乘整数"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_050",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "{12} × 3/4 与 3/4 × {12} 的计算结果相同,表示的意义也相同。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "{12} × 3/4 = 3/4 × {12} = 9 <!-- Meanless: 试卷第 22 页,共 44 页--> 据分析可知, {12} × 3/4 与 3/4 × {12} 的计算结果相同,但表示的意义不相同。原题说法错误。 故答案为: ×",
+    "knowledge_points": [
+      "整数乘法运算定律推广到分数乘法、整数乘分数、求一个数的几分之几的问题、分数乘整数"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_051",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个自然数 ( 0 除外 ) 乘真分数, 积小于这个数。( )",
+    "options": [],
+    "answer": "",
+    "solution": "真分数小于 1,所以一个自然数 ( 0 除外 ) 乘真分数,积小于这个数; 例如 1/2 是真分数, 3 × 1/2 = {1.5} , {1.5} < 3 。 所以原题说法正确。 故答案为:",
+    "knowledge_points": [
+      "因数和积的大小关系(分数乘法)"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_052",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "2/3 × 5/9 的积比 2/3 小。( )",
+    "options": [],
+    "answer": "",
+    "solution": "因为 5/9 < 1 ,所以 2/3 × 5/9 < 2/3 。原题说法正确。 故答案为:",
+    "knowledge_points": [
+      "因数和积的大小关系(分数乘法)、分数乘分数"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_053",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "甲数比乙数多 2/3 ,则乙数比甲数少 2/3 。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "假设乙数是 1 , 则甲数为: --- 1 × ( {1 + 2/3}) = 1 × 5/3 = 5/3 --- <!-- Meanless: 试卷第 23 页,共 44 页--> ( {5/3 - 1}) 5/3 = 2/3 5/3 = 2/3 × 3/5 = 2/5 所以,甲数比乙数多 2/3 ,则乙数比甲数少 2/5 。 故答案为: ×",
+    "knowledge_points": [
+      "求比一个数多/少几分之几的数是多少"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_054",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "如果甲数的 1/8 等于乙数的 1/9 ,那么甲数比乙数大。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "甲数 × 1/8 = {乙数} × 1/9 1/8 > 1/9 甲数<乙数 如果甲数的 1/8 等于乙数的 1/9 ,那么甲数比乙数小。原题干说法错误。 故答案为: ×",
+    "knowledge_points": [
+      "同分子分数的大小比较、求一个数的几分之几的问题"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_055",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "7/9 乘一个分数,积一定小于 7/9 。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "7/9 乘一个分数,分数可能小于 1、等于 1 或大于 1,所以积可能小于 7/9 、等于 7/9 或大于 7/9 ; 原题干说法错误。 故答案为: ×",
+    "knowledge_points": [
+      "因数和积的大小关系(分数乘法)"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_056",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "2 千克梨吃了 1/8 ,还剩 7/8 千克。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "2 × ( {1 - 1/8}) = 2 × 7/8 = 7/4 (千克) 2 千克梨吃了 1/8 ,还剩 7/4 千克。原题干说法错误。 故答案为: ×",
+    "knowledge_points": [
+      "已知总量及一部分分率, 求另一部分量"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_057",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "简单",
+    "question": "4/5 × 1 = 14/5 。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "4/5 × 1 = 4/5 ,原题计算错误。 故答案为: × --- ## 四、计算题",
+    "knowledge_points": [
+      "分数乘整数"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_069",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "学校购进一批课桌椅,总价 3500 元,其中买课桌的钱占 5/7 。买椅子用去多少元?",
+    "options": [],
+    "answer": "1000 元",
+    "solution": "{3500} × ( {1 - 5/7}) = {3500} × 2/7 = {1000} (元) 答：买椅子用去 1000 元。",
+    "knowledge_points": [
+      "已知总量及一部分分率, 求另一部分量"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_070",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "某电视机厂二月份生产电视机 960 台,三月份比二月份多生产 3/8 ,三月生产了多少台?",
+    "options": [],
+    "answer": "1320 台",
+    "solution": "{960} × ( {1 + 3/8}) --- = {960} × 11/8 = {1320} (台) --- 答：三月份生产了 1320 台。",
+    "knowledge_points": [
+      "分数的四则混合运算、求比一个数多/少几分之几的数是多少"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_071",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "简单",
+    "question": "某学校三月份用水 119 吨,四月份用水量比三月份多 2/7 ,四月份比三月份多用水多少吨? 【答案】 34 吨 【难度】 0.85 【知识点】求一个数的几分之几的问题 【分析】已知三月份用水 119 吨,四月份用水量比三月份多 2/7 ,即四月份比三月份多的用水量是三月份用水",
+    "options": [],
+    "answer": "34 吨",
+    "solution": "{119} × 2/7 = {34} (吨) 答: 四月份比三月份多用水 34 吨。",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_072",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "简单",
+    "question": "一只蜗牛每分钟爬行 3/7 米,照这样计算, 7/10 小时可以爬行多少米?",
+    "options": [],
+    "answer": "18 米",
+    "solution": "1 小时 = {60} 分钟 7/10 × {60} = {42} (分钟) 3/7 × {42} = {18} (米) 答: 7/10 小时可以爬行 18 米。",
+    "knowledge_points": [
+      "分数乘整数、基础行程问题"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_074",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "甲、乙两个水杯,甲杯里有水 1 千克,乙杯是空的,第一次将甲杯里的水的 1/2 倒入乙杯里,第二次将乙杯里的水的 1/3 倒回到甲杯里,第三次将甲杯里的水的 1/4 倒入乙杯里,第四次将乙杯里的水的 1/5 倒回甲杯里,照这样来回倒下去, 一直倒了 2000 次之后, 甲杯里还有水多少千克?",
+    "options": [],
+    "answer": "1001/2001",
+    "solution": "据分析发现甲杯剩下的为: 1/2,2/3,1/2,3/5,1/2,4/7,1/2... 到 1999 次甲乙两杯都是 1/2 千克,第 2000 次,从乙倒 1/2001 给甲,倒出 1/2 × 1/2001 = 1/4002 (千克),这时甲还有 1/2 + 1/4002 = 2002/4002 = 1001/2001 (千克) 。 答: 一直倒了 2000 次之后,甲杯里还有水 1001/2001 千克。",
+    "knowledge_points": [
+      "异分母分数加、减法、求一个数的几分之几的问题、分数乘分数、通分的认识及应用"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_075",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "有一批木材共重 4038 吨,第一次运走 1/2 ,第二次运走余下的 1/3 ,第三次运走余下的 1/4 ,最后一次运走余下的 1/2019 ,这批木材还剩下多少吨?",
+    "options": [],
+    "answer": "2 吨",
+    "solution": "{4038} × ( {1 - 1/2}) × ( {1 - 1/3}) × ( {1 - 1/4}) × ... × ( {1 - 1/2019}) = {4038} × 1/2 × 2/3 × 3/4 × × 2018/2019 = {4038} × 1/2019 = 2 (吨) 答: 这批木材还剩下 2 吨。 ---",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_078",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "某地某一时刻测量出一根实际高 {10} m 的旗杆的影长为 {4.5} m ,即此时物体的影长是物体实际高度的 9/20 。 同时同地,一棵大树比这根旗杆高 {6.4} m 。这棵大树的影长为多少米?",
+    "options": [],
+    "answer": "7.38 米",
+    "solution": "根据分析可知: {10} + {6.4} = {16.4} (米) {16.4} × 9/20 = {7.38} (米) 答: 这棵大树的影长为 7.38 米。",
+    "knowledge_points": [
+      "一位小数的不进位加法、不退位减法、分数乘小数"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_079",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "要求买一条带鱼用了多少元,需要的条件是( )(填序号)。请你根据所选的条件进行解答。",
+    "options": [],
+    "answer": "①④；13.2 元",
+    "solution": "根据分析可知: <!-- Meanless: 试卷第 37 页,共 44 页--> 要求买一条带鱼用了多少元,需要的条件是①④。 ①买一条鲫鱼用了 8.8 元。 ②一共买了 10 条鱼。 ③买的带鱼的条数是总条数的 2/5 。 ④ 买一条带鱼用的钱数是买一条鲫鱼的 3/2 。⑤一共用了 105.6 元。 {8.8} × 3/2 = {13.2} (元) 答: 买一条带鱼用了 13.2 元。",
+    "knowledge_points": [
+      "“提问题”“填条件”问题、分数乘小数"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_080",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "空气是一种混合物,它主要由氧气和氮气组成。1 升空气中氧气的体积约占 21/100 ,氮气的体积约占 39/50 。 30.6 升空气中含氧气和氮气各约多少升?",
+    "options": [],
+    "answer": "氧气 6.4 升；氮气 23.9 升",
+    "solution": "根据分析可知: {30.6} × 21/100 = {6.426} (升)",
+    "knowledge_points": [
+      "小数的近似数、分数乘小数"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_081",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "芳芳发现：计算分数乘分数,用分子相乘的积作分子,分母相乘的积作分母。因为有乘法交换律,所以两个分数相乘,互相交换分子,积不变。例如: 1/3 × 2/5 = 1 × 2/3 × 5 = 2 × 1/3 × 5 = 2/3 × 1/5 。她根据这一发现,在计算 39/25 × 7/31 - 7/25 × 8/31 时通过合理变化,用乘法分配律非常快地求出了结果。猜一猜她是怎么做的,然后把她的计算过程写出来。",
+    "options": [],
+    "answer": "见详解; 7/25",
+    "solution": "根据乘法交换律 a × b = b × a 把 39/25 × 7/31 变成 7/25 × 39/31 ,然后根据乘法分配律 a × c + b × c = ( {a + b}) × c 进行简算。 39/25 × 7/31 - 7/25 × 8/31 = 7/25 × 39/31 - 7/25 × 8/31 = 7/25 × ( {39/31 - 8/31}) = 7/25 × 1 = 7/25",
+    "knowledge_points": [
+      "整数乘法运算定律推广到分数乘法"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_084",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "2024 年公历日历与 1996 年完全一样, 一时间, 网络上掀起一股“买本旧日历过新年”的热潮。某二手交易平台中,甲店一本手撕日历的价格为 48 元,乙店价格比甲店的 3/4 少 2 元,乙店价格为多少元?",
+    "options": [],
+    "answer": "34 元",
+    "solution": "{48} × 3/4 - 2 = {36} - 2 = {34} (元) 答：乙店价格为 34 元。",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_085",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "简单",
+    "question": "小慧和小巧记录了今年九月份他们所在县城的天气情况,其中晴天的天数占整个月的 4/5 ,阴天的天数是晴天的 1/6 。小慧和小巧所在县城九月份阴天有多少天?",
+    "options": [],
+    "answer": "4 天",
+    "solution": "{30} × 4/5 = {24} (天) {24} × 1/6 = 4 (天) 答: 小慧和小巧所在县城九月份阴天有 4 天。",
+    "knowledge_points": [
+      "求一个数的几分之几的问题、分数乘整数"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_086",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "简单",
+    "question": "唐代诗人白居易写道：“人间四月芳菲尽,山寺桃花始盛开。”山上的桃花盛开的晚是因为随着海拔的升高,气温越来越低,已知某天某一时刻千佛山山脚的温度是 15 摄氏度,山顶温度是山脚温度的 22/25 ,这时山顶温度是多少摄氏度?",
+    "options": [],
+    "answer": "13.2 摄氏度",
+    "solution": "{15} × 22/25 = {13.2} (摄氏度) 答: 这时山顶温度是 13.2 摄氏度。",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_087",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "校庆采购小组购买了许多文艺汇演的道具,其中向日葵有 120 朵,折扇的数量是向日葵的 5/8 ,桃花的数量是折扇的 2/3 。采购小组采购了多少朵道具桃花?",
+    "options": [],
+    "answer": "50 朵",
+    "solution": "{120} × 5/8 × 2/3 = {75} × 2/3 = {50} (朵) --- 答: 采购小组采购了 50 朵道具桃花。",
+    "knowledge_points": [
+      "分数的连乘运算、整数乘分数、求一个数的几分之几的问题"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_089",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "某地区 2020 年降水量是 360 毫米, 2021 年降水量比 2020 年增加 6/11 , 这个地区 2021 年降水量是多少毫米?",
+    "options": [],
+    "answer": "6120/11 毫米",
+    "solution": "{360} × ( {1 + 6/11}) = {360} × 17/11 = 6120/11 (毫米) --- 答: 这个地区 2021 年降水量是 6120/11 毫米。",
+    "knowledge_points": [
+      "求比一个数多/少几分之几的数是多少"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_090",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "六(1)班有学生 55 人,其中女生占全班人数的 6/11 。男生有多少人?",
+    "options": [],
+    "answer": "25 人",
+    "solution": "{55} × ( {1 - 6/11}) = {55} × 5/11 =25(人) --- <!-- Meanless: 试卷第 43 页,共 44 页--> 答： 男生有 25 人。",
+    "knowledge_points": [
+      "已知总量及一部分分率, 求另一部分量"
+    ],
+    "source": "README.md数学试卷",
+    "create_time": "2025-09-25 19:15:00"
+  },
+  {
+    "id": "readme_q_001",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "{15} ×  = {15} × 1/3 + {15} × 1/5 ,这是根据 ( ) 计算的。",
+    "options": [
+      "A. 乘法交换律",
+      "B. 乘法分配律",
+      "C. 乘法结合律"
+    ],
+    "answer": "B",
+    "solution": "由分析可得: {15} × ( {1/3 + 1/5}) = {15} × 1/3 + {15} × 1/5 ,这是根据乘法分配律计算的。 故答案为: B",
+    "knowledge_points": [
+      "整数乘法运算定律推广到分数乘法"
+    ],
+    "source": " {1/3 + 1/5}",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_001",
+    "original_id": "readme_q_001"
+  },
+  {
+    "id": "readme_q_002",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "如果 3/5 × a > 3/5 ,那么, a 。",
+    "options": [
+      "A. 小于 1",
+      "B. 等于 1",
+      "C. 大于 1"
+    ],
+    "answer": "C",
+    "solution": "根据分析可知,如果 3/5 × a > 3/5 ,那么 a 大于 1。 故答案为: C",
+    "knowledge_points": [
+      "因数和积的大小关系(分数乘法)"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_002",
+    "original_id": "readme_q_002"
+  },
+  {
+    "id": "readme_q_003",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "比 14 的 3/7 多 7 的数是。",
+    "options": [
+      "A. 7",
+      "B. 13",
+      "C. 21"
+    ],
+    "answer": "B",
+    "solution": "{14} × 3/7 + 7 = 6 + 7 = {13} 比 14 的 3/7 多 7 的数是 13。 故答案为: B <!-- Meanless: 试卷第 1 页,共 38 页-->",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_003",
+    "original_id": "readme_q_003"
+  },
+  {
+    "id": "readme_q_004",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一袋糖 {}_{4}^{3}{kg} ,小明吃了这袋糖的 1/5 {kg} ,求还剩多少 {kg} 的算式是。",
+    "options": [
+      "A. 3/4 - 1/5",
+      "B. 3/4 × 1/5",
+      "C. 3/4 × ( {1 - 1/5})"
+    ],
+    "answer": "A",
+    "solution": "3/4 - 1/5 = 15/20 - 4/20 = 11/20( {kg}) 还剩 11/20 {kg} 。 故答案为：A",
+    "knowledge_points": [
+      "异分母分数加",
+      "减法的应用",
+      "分数乘分数"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_004",
+    "original_id": "readme_q_004"
+  },
+  {
+    "id": "readme_q_005",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "{53} × 1/4 + {47} × 1/4 =  × 1/4 ,这一步运算运用了 ( )。",
+    "options": [
+      "A. 乘法交换律",
+      "B. 乘法结合律",
+      "C. 乘法分配律"
+    ],
+    "answer": "C",
+    "solution": "{53} × 1/4 + {47} × 1/4 = ( {{53} + {47}}) × 1/4 = {100} × 1/4 = {25} {53} × 1/4 + {47} × 1/4 = ( {{53} + {47}}) × 1/4 ,这一步运用了乘法分配律。 故答案为: C",
+    "knowledge_points": [
+      "整数乘法运算定律推广到分数乘法",
+      "分数的四则混合运算"
+    ],
+    "source": " {{53} + {47}}",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_005",
+    "original_id": "readme_q_005"
+  },
+  {
+    "id": "readme_q_006",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "小冬每分钟能打 65 个字, 她 4/5 分钟打的字65 个。",
+    "options": [
+      "A. 多于",
+      "B. 少于",
+      "C. 等于"
+    ],
+    "answer": "B",
+    "solution": "{65} × 4/5 = {52} (个) {52} < {65} ,她 4/5 分钟打的字少于 65 个。 小冬每分钟能打 65 个字, 她 4/5 分钟打的字少于 65 个。 故答案为: B",
+    "knowledge_points": [
+      "求一个数的几分之几的问题 <!-- Meanless: 试卷第 2 页,共 38 页-->"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_006",
+    "original_id": "readme_q_006"
+  },
+  {
+    "id": "readme_q_007",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一种商品,先涨价 1/10 ,再降价 1/11 ,现价与原价相比,。",
+    "options": [
+      "A. 现价高",
+      "B. 现价低",
+      "C. 一样高",
+      "D. 无法比较"
+    ],
+    "answer": "C",
+    "solution": "假设这种商品的原价是 100 元。 {100} × ( {1 + 1/10}) × ( {1 - 1/11}) = {100} × 11/10 × 10/11 =100 (元) {100} = {100} ,因此现价与原价相比,一样高。 故答案为: C",
+    "knowledge_points": [
+      "分数的四则混合运算",
+      "求比一个数多/少几分之几的数是多少"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_007",
+    "original_id": "readme_q_007"
+  },
+  {
+    "id": "readme_q_008",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "小红买了一瓶果汁,喝掉它的 1/3 ,小丽又喝掉剩下的 1/3 ,现在还剩下。",
+    "options": [
+      "A. 1/3",
+      "B. 2/3",
+      "C. 2/9",
+      "D. 4/9"
+    ],
+    "answer": "D",
+    "solution": "1 - 1/3 - ( {1 - 1/3}) × 1/3 = 1 - 1/3 - 2/3 × 1/3 <!-- Meanless: 试卷第 3 页,共 38 页--> = 1 - 1/3 - 2/9 = 1 - 3/9 - 2/9 = 4/9 --- 现在还剩下 4/9 。 故答案为： D",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_008",
+    "original_id": "readme_q_008"
+  },
+  {
+    "id": "readme_q_009",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "小杜读一本 280 页的故事书,看了 3/7 ,还剩页没有读。",
+    "options": [
+      "A. 120",
+      "B. 130",
+      "C. 140",
+      "D. 160"
+    ],
+    "answer": "D",
+    "solution": "{280} - {280} × 3/7 = {280} - {120} = {160} (页) --- 故答案为: D",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_009",
+    "original_id": "readme_q_009"
+  },
+  {
+    "id": "readme_q_010",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "12 个 2/5 × 1/6 的和是多少? 列式正确的是。",
+    "options": [
+      "A. {12} × 2/5 × 1/6",
+      "B. ( {{12} + 2/5}) × 1/6",
+      "C. {12} × 5/2 + 1/6",
+      "D. {12} + 2/5 + 1/6"
+    ],
+    "answer": "A",
+    "solution": "12 个 2/5 × 1/6 列式为: {12} × 2/5 × 1/6 ,选项中 A 符合题意。 故答案为：A",
+    "knowledge_points": [
+      "分数的连乘运算"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_010",
+    "original_id": "readme_q_010"
+  },
+  {
+    "id": "readme_q_011",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "比 8 m 长 1/6 是多少米? 列式正确的是。",
+    "options": [
+      "A. 8 × 1/6",
+      "B. 8 + 8 × 1/6",
+      "C. 8 + 1/6",
+      "D. 8 - 8 × 1/6"
+    ],
+    "answer": "B",
+    "solution": "8 × ( {1 + 1/6}) = 8 × 7/6 = 28/3 (米) 或 8 + 8 × 1/6 = 8 + 4/3 = 28/3 (米) 故答案为：B",
+    "knowledge_points": [
+      "求比一个数多/少几分之几的数是多少试卷第 4 页,共 38 页"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_011",
+    "original_id": "readme_q_011"
+  },
+  {
+    "id": "readme_q_012",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "将一根绳子剪成两段,第一段长 4/7 m ,第二段占全长的 4/7 ,两段绳子相比较,。",
+    "options": [
+      "A. 一样长",
+      "B. 第一段长",
+      "C. 第二段长",
+      "D. 无法确定"
+    ],
+    "answer": "C",
+    "solution": "1 - 4/7 = 3/7 3/7 < 4/7 所以两段绳子相比较,第二段长。 故答案为: C",
+    "knowledge_points": [
+      "求一个数的几分之几的问题",
+      "分数的意义",
+      "同分母分数的大小比较"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_012",
+    "original_id": "readme_q_012"
+  },
+  {
+    "id": "readme_q_013",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "困难",
+    "question": "与 {36} × 7/9 计算结果相同的算式是 。",
+    "options": [
+      "A. {36} × 7 ÷ 9",
+      "B. {36} × 9 ÷ 7",
+      "C. {36} ÷ 7 ÷ 9",
+      "D. {36} × 7 × 9"
+    ],
+    "answer": "A",
+    "solution": "{36} × 7/9 = {36} × ( {7 ÷ 9}) = {36} × 7 ÷ 9 所以,与 {36} × 7/9 计算结果相同的算式是 {36} × 7 ÷ 9 。 <!-- Meanless: 试卷第 5 页,共 38 页--> 故答案为：A",
+    "knowledge_points": [
+      "整数乘分数",
+      "分数与除法的关系"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_013",
+    "original_id": "readme_q_013"
+  },
+  {
+    "id": "readme_q_014",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "下列各式,的结果最小。",
+    "options": [
+      "A. {3.14} × 3/4",
+      "B. {3.14} × {0.875}",
+      "C. {3.14} × 9/8",
+      "D. {3.14} ÷ {1.25}"
+    ],
+    "answer": "A",
+    "solution": "A. {3.14} × 3/4 = {3.14} × {0.75} = {2.355} B. {3.14} × {0.875} = {2.7475} C. {3.14} × 9/8 = {3.14} × {1.125} = {3.5325} D. {3.14} ÷ {1.25} = {2.512} {2.355} < {2.512} < {2.7475} < {3.5325} {3.14} × 3/4 的结果最小。 故答案为：A",
+    "knowledge_points": [
+      "小数与小数的乘法",
+      "除数是小数的小数除法",
+      "分数乘小数"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_014",
+    "original_id": "readme_q_014"
+  },
+  {
+    "id": "readme_q_015",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "下面算式的结果在 7/8 与 11/6 之间的是。",
+    "options": [
+      "A. 7/8 × 1/2",
+      "B. 7/8 × 7/3",
+      "C. 7/8 × 5/3",
+      "D. 11/6 × 9/8"
+    ],
+    "answer": "C",
+    "solution": "A. 1/2 < 1,7/8 × 1/2 < 7/8 ,排除； B. 7/3 > 1,7/8 × 7/3 > 7/8 ,有可能; C. 5/3 > 1,7/8 × 5/3 > 7/8 ,有可能; D. 9/8 > 1,11/6 × 9/8 > 11/6 ,排除。 7/8 × 7/3 = 49/24、7/8 × 5/3 = 35/24 7/8 = 21/24 、 11/6 = 44/24 , 7/8 < 7/8 × 5/3 < 11/6 ,在 7/8 与 11/6 之间的是 7/8 × 5/3 。 故答案为：C",
+    "knowledge_points": [
+      "分数乘分数",
+      "因数和积的大小关系(分数乘法)"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_015",
+    "original_id": "readme_q_015"
+  },
+  {
+    "id": "readme_q_016",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "困难",
+    "question": "下列各式中,乘积最大的是。 <!-- Meanless: 试卷第 6 页,共 38 页-->",
+    "options": [
+      "A. 2/5 × 3/4",
+      "B. 2/5 × 1",
+      "C. 2/5 × 8/9"
+    ],
+    "answer": "B",
+    "solution": "A. 3/4 < 1,2/5 × 3/4 < 2/5 B. 2/5 × 1 = 2/5 C. 8/9 < 1,2/5 × 8/9 < 2/5 乘积最大的是 2/5 × 1 。 故答案为: B",
+    "knowledge_points": [
+      "因数和积的大小关系(分数乘法)"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_016",
+    "original_id": "readme_q_016"
+  },
+  {
+    "id": "readme_q_017",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一个正方形的边长是 {}_{8}^{3} 米,它的周长是米。",
+    "options": [
+      "A. 6/8",
+      "B. 12/8",
+      "C. 9/64"
+    ],
+    "answer": "B",
+    "solution": "正方形的周长: 3/8 × 4 = 3/2 (米) A. 6/8 = 6 ÷ 2/8 ÷ 2 = 3/4 ,不符合题意; B. 12/8 = {12 ÷ 4}{8 ÷ 4} = 3/2 ,符合题意; C. 9/64 < 3/2 ,不符合题意; 故答案为: B",
+    "knowledge_points": [
+      "最简分数",
+      "正方形的周长",
+      "分数乘整数"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_017",
+    "original_id": "readme_q_017"
+  },
+  {
+    "id": "readme_q_018",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "两个真分数的积一定其中每一个数。",
+    "options": [
+      "A. 小于",
+      "B. 大于",
+      "C. 等于"
+    ],
+    "answer": "A",
+    "solution": "真分数小于 1 ,所以两个真分数相乘的积小于这两个真分数中的每一个真分数； ## 试卷第 7 页,共 38 页 比如: 1/2 × 1/3 = 1/6,1/6 小于 1/2 ,也小于 1/3 。 故答案为：A",
+    "knowledge_points": [
+      "真分数",
+      "假分数",
+      "带分数的认识",
+      "因数和积的大小关系(分数乘法)"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_018",
+    "original_id": "readme_q_018"
+  },
+  {
+    "id": "readme_q_019",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一种商品先把价格提高 1/10 后,再降价 1/10 卖出,最后的价格。",
+    "options": [
+      "A. 原价不变",
+      "B. 比原价低",
+      "C. 比原价高"
+    ],
+    "answer": "B",
+    "solution": "假设原价 100 元。 {100} × ( {1 + 1/10}) × ( {1 - 1/10}) = {100} × 11/10 × 9/10 = {99} (元) {99} < {100} 最后的价格比原价低。 故答案为: B",
+    "knowledge_points": [
+      "连续求一个数的几分之几是多少的问题",
+      "求比一个数多/少几分之几的数是多少"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_019",
+    "original_id": "readme_q_019"
+  },
+  {
+    "id": "readme_q_020",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一块平行四边形菜地的底是 25 米,高是底的 4/5 ,面积是  平方米。 ---",
+    "options": [],
+    "answer": "500",
+    "solution": "{25} × 4/5 = {20} (米) {25} × {20} = {500} (平方米) 面积是 500 平方米。 ---",
+    "knowledge_points": [
+      "求一个数的几分之几的问题",
+      "平行四边形面积的计算"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_020",
+    "original_id": "readme_q_020"
+  },
+  {
+    "id": "readme_q_021",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "在“爸爸体重的 5/12 与小军的体重相等”这个条件中,单位“1”的量是 ； 果爸爸体重是 {60} {kg} ,那么小军的体重是  {kg} 。 试卷第 10 页,共 38 页",
+    "options": [],
+    "answer": "",
+    "solution": "由分析可得：在 “爸爸体重的 5/12 与小军的体重相等” 这个条件中,单位 “1 ” 的量是爸爸的体重。 {60} × 5/12 = {25}( {kg}) 那么小军的体重是 {25} {kg} 。",
+    "knowledge_points": [
+      "单位“1”的认识与确定",
+      "求一个数的几分之几的问题"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_021",
+    "original_id": "readme_q_021"
+  },
+  {
+    "id": "readme_q_022",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "3/16 的 8 倍是；24 的 5/12 比 38 少。",
+    "options": [],
+    "answer": "3/2 {28}",
+    "solution": "3/16 × 8 = 3/2 {38} - {24} × 5/12 = {38} - {10} = {28} 3/16 的 8 倍是 3/2;{24} 的 5/12 比 38 少 28 。",
+    "knowledge_points": [
+      "整数乘分数",
+      "求一个数的几分之几的问题",
+      "分数乘整数"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_022",
+    "original_id": "readme_q_022"
+  },
+  {
+    "id": "readme_q_023",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "7/18 的 2/7 是；5 个 3/5 的和是。",
+    "options": [],
+    "answer": "1/9 3",
+    "solution": "7/18 × 2/7 = 1/9 3/5 × 5 = 3 7/18 的 2/7 是 1/9;5 个 3/5 的和是 3。",
+    "knowledge_points": [
+      "分数乘分数",
+      "分数乘整数"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_023",
+    "original_id": "readme_q_023"
+  },
+  {
+    "id": "readme_q_024",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "比 18 吨多 2/9 是吨；比 32 千米少 3/4 是千米。",
+    "options": [],
+    "answer": "22 8",
+    "solution": "{18} × ( {1 + 2/9}) = {18} × 11/9 =22 (吨) {32} × ( {1 - 3/4}) = {32} × 1/4 = 8 (千米) 比 18 吨多 2/9 是 22 吨; 比 32 千米少 3/4 是 8 千米。",
+    "knowledge_points": [
+      "求比一个数多/少几分之几的数是多少"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_024",
+    "original_id": "readme_q_024"
+  },
+  {
+    "id": "readme_q_025",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个平行四边形的底长 4/7 米、高 3/7 米,它的面积是平方米。",
+    "options": [],
+    "answer": "12/49",
+    "solution": "4/7 × 3/7 = 12/49 (平方米) 它的面积是 12/49 平方米。",
+    "knowledge_points": [
+      "分数乘分数",
+      "平行四边形面积的计算"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_025",
+    "original_id": "readme_q_025"
+  },
+  {
+    "id": "readme_q_026",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "1 千克糖,吃了 1/2 ,还剩 1/2 千克。(对的打“√”,错的打“×”,并说明理由。) 理由：_____。",
+    "options": [],
+    "answer": "✓ 1 千克糖吃了 1/2 ,根据分数乘法的意义可得出吃了的糖为 1/2 千克,则还剩 1/2 千克",
+    "solution": "1 - 1 × 1/2 = 1 - 1/2 = 1/2 (千克) 1 千克糖,吃了 1/2 ,还剩 1/2 千克。(✓) --- 理由: 1 千克糖吃了 1/2 ,根据分数乘法的意义可得出吃了的糖为 1/2 千克,则还剩 1/2 千克。(答案不唯一)",
+    "knowledge_points": [
+      "整数乘分数",
+      "求一个数的几分之几的问题"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_026",
+    "original_id": "readme_q_026"
+  },
+  {
+    "id": "readme_q_027",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个球从高处落下,每次接触地面后弹起的高度是前一次下落高度 1/5 ,如果这个球 <!-- Meanless: 试卷第 12 页,共 38 页--> <!-- Meanless: 从 125 米的高度落下,第三次弹起的高度是米。-->",
+    "options": [],
+    "answer": "1",
+    "solution": "{125} × 1/5 × 1/5 × 1/5 = {25} × 1/5 × 1/5 = 5 × 1/5 = 1 (米) 第三次弹起的高度是 1 米。",
+    "knowledge_points": [
+      "连续求一个数的几分之几是多少的问题"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_027",
+    "original_id": "readme_q_027"
+  },
+  {
+    "id": "readme_q_028",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "困难",
+    "question": "六班有 40 人,其中 5/8 是男生,求男生的人数是多少,就是求( ) 是多少。",
+    "options": [],
+    "answer": "",
+    "solution": "{40} × 5/8 = {25} (人) 男生的人数是 25 人。 六 (2) 班有40人,其中 5/8 是男生,求男生的人数是多少,就是求40的 5/8 是多少。",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": "2",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_028",
+    "original_id": "readme_q_028"
+  },
+  {
+    "id": "readme_q_029",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "成人的脚长约是身高 1/7 ,姐姐身高是 168 厘米,则她的脚长约是",
+    "options": [],
+    "answer": "24",
+    "solution": "{168} × 1/7 = {24} (厘米) 她的脚长约是 24 厘米。",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_029",
+    "original_id": "readme_q_029"
+  },
+  {
+    "id": "readme_q_030",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一根电线长 28 米,如果用去它 3/7 ,那么还剩下  米; 如果用 3/7 米那么剩下 米。",
+    "options": [],
+    "answer": "16 {27}4/7{}^{193}",
+    "solution": "{28} × ( {1 - 3/7}) = {28} × 4/7 = {16} (米) {28} - 3/7 = {27}4/7 (米) --- 一根电线长 28 米,如果用去它 3/7 ,那么还剩下 16 米; 如果用 3/7 米那么剩下 {27}4/7 米。",
+    "knowledge_points": [
+      "同分母分数加",
+      "减法的应用",
+      "已知总量及一部分分率, 求另一部分量"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_030",
+    "original_id": "readme_q_030"
+  },
+  {
+    "id": "readme_q_031",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "芳芳读一本故事书,平均每天读这本书的 4/15,3 天读了这本书的。",
+    "options": [],
+    "answer": "4/5",
+    "solution": "3 天读了这本书的: 4/15 × 3 = 4/5 。",
+    "knowledge_points": [
+      "分数乘整数"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_031",
+    "original_id": "readme_q_031"
+  },
+  {
+    "id": "readme_q_032",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个长方形长 1/4m ,宽 2/5m ,它的周长是 m ,面积是 {m}^{2} 。",
+    "options": [],
+    "answer": "13/10 1/10",
+    "solution": "周长: ( {1/4 + 2/5}) × 2 = ( {5/20 + 8/20}) × 2 = 13/20 × 2 <!-- Meanless: 试卷第 15 页,共 38 页--> --- = 13/10( m) 面积: 1/4 × 2/5 = 1/10( { m}^{2}) 所以周长是 13/10 m ,面积是 1/10{ m}^{2} 。 ---",
+    "knowledge_points": [
+      "长方形的周长",
+      "长方形的面积",
+      "分数乘分数",
+      "分数乘整数"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_032",
+    "original_id": "readme_q_032"
+  },
+  {
+    "id": "readme_q_033",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "困难",
+    "question": "单位换算。 7/500{ m}^{3} =  {{dm}}^{3} 4/5 m =  {cm} 1/10{ {km}}^{2} = 公顷 5/8日 =",
+    "options": [],
+    "answer": "14 80 10 15",
+    "solution": "7/500{ m}^{3} = 7/500 × {1000} = {14}{ {dm}}^{3} 4/5 m = 4/5 × {100} = {80} {cm} 1/10{{km}}^{2} = 1/10 × {100} = {10} 公顷 5/8 日 = 5/8 × {24} = {15} 时",
+    "knowledge_points": [
+      "面积单位间的进率及换算",
+      "时",
+      "分的认识及换算",
+      "体积单位间的进率与换算 (立方厘米",
+      "立方分米和立方米)",
+      "分数乘整数"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_033",
+    "original_id": "readme_q_033"
+  },
+  {
+    "id": "readme_q_034",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "蜂鸟是目前发现的世界上最小的鸟,也是唯一能倒飞的鸟,蜂鸟每分钟可飞行 3/10 千米,它 5/6 分钟飞行  千米。",
+    "options": [],
+    "answer": "1/4/{0.25}",
+    "solution": "3/10 × 5/6 = 1/4 (千米) 它 5/6 分钟飞行 1/4 千米。",
+    "knowledge_points": [
+      "分数乘分数",
+      "基础行程问题"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_034",
+    "original_id": "readme_q_034"
+  },
+  {
+    "id": "readme_q_035",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "“红花的朵数的 2/3 相当于黄花的朵数”是把。",
+    "options": [],
+    "answer": "红花的朵数 红花的朵数 × 2/3 = 黄花的朵数",
+    "solution": "根据分析可知,“红花的朵数的 2/3 相当于黄花的朵数”是把红花的朵数看作单位“1”,等量关系式是：红花的朵数 × 2/3 = 黄花的朵数。",
+    "knowledge_points": [
+      "单位“1”的认识与确定",
+      "求一个数的几分之几的问题"
+    ],
+    "source": " ( ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_035",
+    "original_id": "readme_q_035"
+  },
+  {
+    "id": "readme_q_036",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "1 米长的绳子用去 4/5 ,就是用去 4/5 米。",
+    "options": [],
+    "answer": "",
+    "solution": "1 × 4/5 = 4/5 (米) <!-- Meanless: 试卷第 18 页,共 38 页--> 1 米长的绳子用去 4/5 ,就是用去 4/5 米,原题说法正确。 故答案为:",
+    "knowledge_points": [
+      "整数乘分数",
+      "求一个数的几分之几的问题"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_036",
+    "original_id": "readme_q_036"
+  },
+  {
+    "id": "readme_q_037",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "如 A × 6 = B × 3/4 ,那么 A > B 。( )(对的打“√”,错的打“×”, 并说明理由。) 理由：_____。",
+    "options": [],
+    "answer": "× 因为 6 > 3/4 ,乘积一定,所以若 A × 6 = B × 3/4 ( A 、 B 均不为 0 ),那么 A < B",
+    "solution": "如 A × 6 = B × 3/4 ( A 、 B 均不为 0 ),那么 A > B 。(✗) 理由: 因为 6 > 3/4 ,所以若 A × 6 = B × 3/4 ( A 、 B 均不为 0 ),那么 A < B 。",
+    "knowledge_points": [
+      "分数乘分数",
+      "因数和积的大小关系(分数乘法)"
+    ],
+    "source": " A 、 B 均不为 0 ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_037",
+    "original_id": "readme_q_037"
+  },
+  {
+    "id": "readme_q_038",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "比 1/2 的 1/2 多 1/2 的数是 1/8 。",
+    "options": [],
+    "answer": "×",
+    "solution": "1/2 × 1/2 + 1/2 = 1/4 + 1/2 = 1/4 + 2/4 = 3/4 比 1/2 的 1/2 多 1/2 的数是 3/4 。 --- 原题说法错误。 故答案为: ×",
+    "knowledge_points": [
+      "分数的四则混合运算",
+      "分数乘分数",
+      "求一个数的几分之几的问题"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_038",
+    "original_id": "readme_q_038"
+  },
+  {
+    "id": "readme_q_039",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "判断题",
+    "difficulty": "困难",
+    "question": "用简便方法计算 8/13 × {12} + 8/13 ,应该运用乘法分配律。 ---",
+    "options": [],
+    "answer": "",
+    "solution": "8/13 × {12} + 8/13 = 8/13 × ( {{12} + 1}) = 8/13 × {13} = 8 --- 用简便方法计算 8/13 × {12} + 8/13 ,应该运用乘法分配律。 原题干说法正确。 故答案为:",
+    "knowledge_points": [
+      "整数乘法运算定律推广到分数乘法 ---"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_039",
+    "original_id": "readme_q_039"
+  },
+  {
+    "id": "readme_q_040",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "小白体重的 7/8 与小金的体重相同,则小白体重 = 小金体重 × 7/8 。",
+    "options": [],
+    "answer": "×",
+    "solution": "小白体重的 7/8 与小金的体重相同,则小金的体重 = 小白的体重 × 7/8 。 所以原题说法错误。 故答案为: ×",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_040",
+    "original_id": "readme_q_040"
+  },
+  {
+    "id": "readme_q_041",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "两个分数相乘的积一定大于其中任意一个分数。",
+    "options": [],
+    "answer": "×",
+    "solution": "根据分析可得: 两个分数相乘的积不一定大于其中任意一个分数, 。原说法错误。 故答案为: × <!-- Meanless: 试卷第 20 页,共 38 页-->",
+    "knowledge_points": [
+      "分数乘分数",
+      "因数和积的大小关系(分数乘法)"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_041",
+    "original_id": "readme_q_041"
+  },
+  {
+    "id": "readme_q_042",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "两个分数相乘,积一定小于其中的一个分数。",
+    "options": [],
+    "answer": "×",
+    "solution": "当两个分数都是真分数时,比如 1/2 × 1/3 = 1/6 ,积 1/6 小于这两个分数 1/2 和 1/3 。 但当两个分数都是大于 1 的假分数时,比如 3/2 × 4/3 = 2 ,积 2 大于这两个分数 3/2 和 4/3 。 所以两个分数相乘,积不一定小于其中的一个分数,故该原题说法错误。 故答案为: ×",
+    "knowledge_points": [
+      "因数和积的大小关系(分数乘法)"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_042",
+    "original_id": "readme_q_042"
+  },
+  {
+    "id": "readme_q_043",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "丫丫将一根绳子对折两次,量得每段长 3/4 米,这根绳子长 3/2 米。",
+    "options": [],
+    "answer": "×",
+    "solution": "2 × 2 = 4 (段) 3/4 × 4 = 3 (米) 这根绳子长 3 米。 原题说法错误。 故答案为: ×",
+    "knowledge_points": [
+      "分数乘整数",
+      "图形的折叠问题"
+    ],
+    "source": " ",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_043",
+    "original_id": "readme_q_043"
+  },
+  {
+    "id": "readme_q_044",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "计算。  74/75 × 8 (2) 115/114 × {113}",
+    "options": [],
+    "answer": "(1) 767/75 ；(2) {113}113/114",
+    "solution": "(1) 74/75 × 8 = ( {1 - 1/75}) × 8 = 1 × 8 - 1/75 × 8 = 8 - 8/75 = 767/75 (2) 115/114 × {113} = ( {1 + 1/114}) × {113} = 1 × {113} + 1/114 × {113} = {113} + 113/114 = {113}113/114",
+    "knowledge_points": [
+      "整数乘法运算定律推广到分数乘法"
+    ],
+    "source": "1",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_044",
+    "original_id": "readme_q_044"
+  },
+  {
+    "id": "readme_q_045",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "计算。  29/98 × {99} (2) {67} × 15/68 ---",
+    "options": [],
+    "answer": "(1) {29}29/98 ；(2) {14}53/68",
+    "solution": "(1) 29/98 × {99} = 29/98 × ( {{98} + 1}) = 29/98 × {98} + 29/98 × 1 = {29} + 29/98 = {29}29/98 (2) {67} × 15/68 = ( {{68} - 1}) × 15/68 = {68} × 15/68 - 1 × 15/68 = {15} - 15/68 = {14}53/68",
+    "knowledge_points": [
+      "整数乘法运算定律推广到分数乘法"
+    ],
+    "source": "1",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_045",
+    "original_id": "readme_q_045"
+  },
+  {
+    "id": "readme_q_046",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "用简便方法计算下面各题。  {27} × 15/26 (2) 44/45 × {37} ---",
+    "options": [],
+    "answer": "(1) {15}15/26 ；(2) {36}8/45",
+    "solution": "(1) {27} × 15/26 = ( {{26} + 1}) × 15/26 = {26} × 15/26 + 1 × 15/26 = {15} + 15/26 = {15}15/26 (2) 44/45 × {37} = ( {1 - 1/45}) × {37} <!-- Meanless: 试卷第 24 页,共 38 页--> = 1 × {37} - 1/45 × {37} = {37} - 37/45 = {36}8/45 ---",
+    "knowledge_points": [
+      "整数乘法运算定律推广到分数乘法"
+    ],
+    "source": "1",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_046",
+    "original_id": "readme_q_046"
+  },
+  {
+    "id": "readme_q_047",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "计算。  5/17 × 9/24 + 9/17 × 7/24 (2) 5/9 × 9/16 + 7/9 × 5/16 ---",
+    "options": [],
+    "answer": "(1) 9/34 ；(2) 5/9",
+    "solution": "(1) 5/17 × 9/24 + 9/17 × 7/24 = 9/17 × 5/24 + 9/17 × 7/24 = 9/17 × ( {5/24 + 7/24}) = 9/17 × 1/2 = 9/34 (2) 5/9 × 9/16 + 7/9 × 5/16 = 5/9 × 9/16 + 5/9 × 7/16 = 5/9 × ( {9/16 + 7/16}) = 5/9 × 1 = 5/9",
+    "knowledge_points": [
+      "整数乘法运算定律推广到分数乘法"
+    ],
+    "source": "1",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_047",
+    "original_id": "readme_q_047"
+  },
+  {
+    "id": "readme_q_048",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "计算。  11/13 × 6/19 + 6/13 × 8/19 (2) {139} × 137/138 - {137} × 1/138 ---",
+    "options": [],
+    "answer": "(1) 6/13 ；(2)137",
+    "solution": "( 1 ) 11/13 × 6/19 + 6/13 × 8/19 = 6/13 × 11/19 + 6/13 × 8/19 = 6/13 × ( {11/19 + 8/19}) = 6/13 × 1 = 6/13 (2) {139} × 137/138 - {137} × 1/138 = {139} × 137/138 - 1 × 137/138 = ( {{139} - 1}) × 137/138 = {138} × 137/138 = {137} ---",
+    "knowledge_points": [
+      "整数乘法运算定律推广到分数乘法",
+      "分数的四则混合运算"
+    ],
+    "source": "1",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_048",
+    "original_id": "readme_q_048"
+  },
+  {
+    "id": "readme_q_049",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "用简便方法计算下面各题。  ( {8/9 + 4/27}) × {27} (2) 5/6 × 5/9 + 5/9 × 1/6 ---",
+    "options": [],
+    "answer": "(1)28；(2) 5/9",
+    "solution": "(1) ( {8/9 + 4/27}) × {27} --- = 8/9 × {27} + 4/27 × {27} = {24} + 4 =28 (2) 5/6 × 5/9 + 5/9 × 1/6 = 5/9 × ( {5/6 + 1/6}) = 5/9 × 1 = 5/9 ---",
+    "knowledge_points": [
+      "整数乘法运算定律推广到分数乘法"
+    ],
+    "source": "1",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_049",
+    "original_id": "readme_q_049"
+  },
+  {
+    "id": "readme_q_050",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "简单",
+    "question": "用“裂项消去法”进行简便计算。 1/2 × 4 + 1/4 × 6 + 1/6 × 8 + 1/8 × {10} 1/1 × 4 + 1/4 × 7 + 1/7 × {10} + 1/{10 × {13}} ---",
+    "options": [],
+    "answer": "1/5;4/13",
+    "solution": "( 1 ) 1/2 × 4 + 1/4 × 6 + 1/6 × 8 + 1/8 × {10} = 1/2 × ( {1/2 - 1/4}) + 1/2 × ( {1/4 - 1/6}) + 1/2 × ( {1/6 - 1/8}) + 1/2 × ( {1/8 - 1/10}) = 1/2 × ( {1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + 1/8 - 1/10}) = 1/2 × ( {1/2 - 1/10}) = 1/2 × ( {5/10 - 1/10}) = 1/2 × 2/5 = 1/5 (2)1/1 × 4 + 1/4 × 7 + 1/7 × {10} + 1/{10 × {13}} = 1/3 × ( {1 - 1/4}) + 1/3 × ( {1/4 - 1/7}) + 1/3 × ( {1/7 - 1/10}) + 1/3 × ( {1/10 - 1/13}) = 1/3 × ( {1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + 1/10 - 1/13}) = 1/3 × ( {1 - 1/13}) = 1/3 × 12/13 = 4/13",
+    "knowledge_points": [
+      "算式的规律",
+      "分数的四则混合运算"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_050",
+    "original_id": "readme_q_050"
+  },
+  {
+    "id": "readme_q_051",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "3/5 的 5/6 与 7/15 差,差是多少?",
+    "options": [],
+    "answer": "1/30",
+    "solution": "3/5 × 5/6 - 7/15 --- = 1/2 - 7/15 = 15/30 - 14/30 = 1/30 差是 1/30 <!-- Meanless: 试卷第 27 页,共 38 页-->",
+    "knowledge_points": [
+      "分数乘分数",
+      "异分母分数加",
+      "减法",
+      "求一个数的几分之几的问题"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:56:37",
+    "merged_id": "readme_051",
+    "original_id": "readme_q_051"
+  },
+  {
+    "id": "readme_q_021",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "{0.25} × 2/3 = ( )。",
+    "options": [
+      "A. 4/5 B. 1/6 C. 2/25 D. 1/3",
+      "A. 5/14 × {0.55} B. 31/2 × 17/8 C. 43/5 × 1/2"
+    ],
+    "answer": "B",
+    "solution": "A. 因为 {0.55} < 1 ,所以 5/14 × {0.55} < 5/14 。即 A 选项错误。 B. 因为 17/8 > 1 ,所以 31/2 × 17/8 > 31/2 。即 B 选项正确。 <!-- Meanless: 试卷第 9 页,共 38 页--> C. 因为 1/2 < 1 ,所以 43/5 × 1/2 < 43/5 。即 C 选项错误。 故答案为: B",
+    "knowledge_points": [
+      "因数和积的大小关系(小数乘法)",
+      "分数乘分数",
+      "分数乘小数"
+    ],
+    "source": "22-23 六年级·全国·假期作业",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_021",
+    "original_id": "readme_q_021"
+  },
+  {
+    "id": "readme_q_022",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "六(1)班制作手工花 90 朵,_____,六(2)班制作手工花多少朵？以下信息中, 符合 {90} × 2/3 这个算式的是 ( )。 ## 二、填空题",
+    "options": [
+      "A. 六(1)班是六(2)班的 2/3 B. 六(2)班是六(1)班的 2/3",
+      "C. 六 (2) 班比六 (1) 班多 2/3 D. 六(2)班比六(1)班少 2/3"
+    ],
+    "answer": "B",
+    "solution": "六(1)班制作手工花 90 朵,六(2)班是六(1)班的 2/3 ,六(2)班制作手工花多少朵？ {90} × 2/3 = {60} (朵) 故答案为: B",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "求一个数的几分之几的问题"
+    ],
+    "source": "22-23 六年级上·广东广州·期末",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_022",
+    "original_id": "readme_q_022"
+  },
+  {
+    "id": "readme_q_023",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一块平行四边形菜地的底是 25 米,高是底的 4/5 ,面积是 ( ) 平方米。 --- ---",
+    "options": [],
+    "answer": "500",
+    "solution": "{25} × 4/5 = {20} (米) {25} × {20} = {500} (平方米) 面积是 500 平方米。 ---",
+    "knowledge_points": [
+      "求一个数的几分之几的问题",
+      "平行四边形面积的计算"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_023",
+    "original_id": "readme_q_023"
+  },
+  {
+    "id": "readme_q_024",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "在“爸爸体重的 5/12 与小军的体重相等”这个条件中,单位“1”的量是 ( )； 果爸爸体重是 {60} {kg} ,那么小军的体重是 ( ) {kg} 。 试卷第 10 页,共 38 页",
+    "options": [],
+    "answer": "",
+    "solution": "由分析可得：在 “爸爸体重的 5/12 与小军的体重相等” 这个条件中,单位 “1 ” 的量是爸爸的体重。 {60} × 5/12 = {25}( {kg}) 那么小军的体重是 {25} {kg} 。",
+    "knowledge_points": [
+      "单位“1”的认识与确定",
+      "求一个数的几分之几的问题"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_024",
+    "original_id": "readme_q_024"
+  },
+  {
+    "id": "readme_q_025",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "3/16 的 8 倍是( )；24 的 5/12 比 38 少( )。",
+    "options": [],
+    "answer": "3/2 {28}",
+    "solution": "3/16 × 8 = 3/2 {38} - {24} × 5/12 = {38} - {10} = {28} 3/16 的 8 倍是 3/2;{24} 的 5/12 比 38 少 28 。",
+    "knowledge_points": [
+      "整数乘分数",
+      "求一个数的几分之几的问题",
+      "分数乘整数"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_025",
+    "original_id": "readme_q_025"
+  },
+  {
+    "id": "readme_q_026",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "7/18 的 2/7 是( )；5 个 3/5 的和是( )。",
+    "options": [],
+    "answer": "1/9 3",
+    "solution": "7/18 × 2/7 = 1/9 3/5 × 5 = 3 7/18 的 2/7 是 1/9;5 个 3/5 的和是 3。",
+    "knowledge_points": [
+      "分数乘分数",
+      "分数乘整数"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_026",
+    "original_id": "readme_q_026"
+  },
+  {
+    "id": "readme_q_027",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "比 18 吨多 2/9 是( )吨；比 32 千米少 3/4 是( )千米。",
+    "options": [],
+    "answer": "22 8",
+    "solution": "{18} × ( {1 + 2/9}) = {18} × 11/9 =22 (吨) {32} × ( {1 - 3/4}) = {32} × 1/4 = 8 (千米) 比 18 吨多 2/9 是 22 吨; 比 32 千米少 3/4 是 8 千米。",
+    "knowledge_points": [
+      "求比一个数多/少几分之几的数是多少"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_027",
+    "original_id": "readme_q_027"
+  },
+  {
+    "id": "readme_q_028",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个平行四边形的底长 4/7 米、高 3/7 米,它的面积是( )平方米。",
+    "options": [],
+    "answer": "12/49",
+    "solution": "4/7 × 3/7 = 12/49 (平方米) 它的面积是 12/49 平方米。",
+    "knowledge_points": [
+      "分数乘分数",
+      "平行四边形面积的计算"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_028",
+    "original_id": "readme_q_028"
+  },
+  {
+    "id": "readme_q_029",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "1 千克糖,吃了 1/2 ,还剩 1/2 千克。( )(对的打“√”,错的打“×”,并说明理由。) 理由：_____。",
+    "options": [],
+    "answer": "✓ 1 千克糖吃了 1/2 ,根据分数乘法的意义可得出吃了的糖为 1/2 千克,则还剩 1/2 千克",
+    "solution": "1 - 1 × 1/2 = 1 - 1/2 = 1/2 (千克) 1 千克糖,吃了 1/2 ,还剩 1/2 千克。(✓) --- 理由: 1 千克糖吃了 1/2 ,根据分数乘法的意义可得出吃了的糖为 1/2 千克,则还剩 1/2 千克。(答案不唯一)",
+    "knowledge_points": [
+      "整数乘分数",
+      "求一个数的几分之几的问题"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_029",
+    "original_id": "readme_q_029"
+  },
+  {
+    "id": "readme_q_030",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "一个球从高处落下,每次接触地面后弹起的高度是前一次下落高度 1/5 ,如果这个球",
+    "options": [],
+    "answer": "1",
+    "solution": "{125} × 1/5 × 1/5 × 1/5 = {25} × 1/5 × 1/5 = 5 × 1/5 = 1 (米) 第三次弹起的高度是 1 米。",
+    "knowledge_points": [
+      "连续求一个数的几分之几是多少的问题"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_030",
+    "original_id": "readme_q_030"
+  },
+  {
+    "id": "readme_q_031",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "困难",
+    "question": "六(2)班有 40 人,其中 5/8 是男生,求男生的人数是多少,就是求( ) 是多少。",
+    "options": [],
+    "answer": "",
+    "solution": "{40} × 5/8 = {25} (人) 男生的人数是 25 人。 六 (2) 班有40人,其中 5/8 是男生,求男生的人数是多少,就是求40的 5/8 是多少。",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": " 23-24 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_031",
+    "original_id": "readme_q_031"
+  },
+  {
+    "id": "readme_q_032",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "成人的脚长约是身高 1/7 ,姐姐身高是 168 厘米,则她的脚长约是( )",
+    "options": [],
+    "answer": "24",
+    "solution": "{168} × 1/7 = {24} (厘米) 她的脚长约是 24 厘米。",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_032",
+    "original_id": "readme_q_032"
+  },
+  {
+    "id": "readme_q_033",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "在括号里填上“>”“<”或“=”。 {18} × 5/6( ) {18} ÷ 6 × {515} × 3/4 ( ) {15}2/7 × {20} ( ) ---",
+    "options": [],
+    "answer": "",
+    "solution": "第一小题, 计算出两边的算式结果, 再进行比较; 第二、三小题：一个非 0 数,乘大于 1 的数,积大于原数；一个非 0 数,乘小于 1 的数,积小于原式,据此解答。",
+    "knowledge_points": [
+      "整数乘分数",
+      "分数乘整数",
+      "因数和积的大小关系(分数乘法)"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_033",
+    "original_id": "readme_q_033"
+  },
+  {
+    "id": "readme_q_034",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一根电线长 28 米,如果用去它 3/7 ,那么还剩下 ( ) 米; 如果用 3/7 米那么剩下 ( )米。",
+    "options": [],
+    "answer": "16 {27}4/7{}^{193}",
+    "solution": "{28} × ( {1 - 3/7}) = {28} × 4/7 = {16} (米) {28} - 3/7 = {27}4/7 (米) --- 一根电线长 28 米,如果用去它 3/7 ,那么还剩下 16 米; 如果用 3/7 米那么剩下 {27}4/7 米。",
+    "knowledge_points": [
+      "同分母分数加",
+      "减法的应用",
+      "已知总量及一部分分率",
+      "求另一部分量"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_034",
+    "original_id": "readme_q_034"
+  },
+  {
+    "id": "readme_q_035",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "{40} {kg} 的 3/8 是( ) {kg} ；4 个 3/8 相加的和是( )。 --- ---",
+    "options": [],
+    "answer": "60 875 3 25",
+    "solution": "3/50 吨 = 3/50 × {1000} = {60} 千克 7/8 L = 7/8 × {1000} = {875} {mL} 5/12 时 = 5/12 × {60} = {25} 分 35/12 时 =3 时 25 分。",
+    "knowledge_points": [
+      "容积单位间的进率与换算 (升和毫升)",
+      "时",
+      "分的认识及换算",
+      "整数乘分数",
+      "吨",
+      "千克之间的换算与比较 【分析】1 吨 = {1000} 千克； 1 L = {1000} {mL} ；1 时 = {60} 分；高级单位换算成低级单位",
+      "乘进率",
+      "据此解答。"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_035",
+    "original_id": "readme_q_035"
+  },
+  {
+    "id": "readme_q_036",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "芳芳读一本故事书,平均每天读这本书的 4/15,3 天读了这本书的( )。",
+    "options": [],
+    "answer": "4/5",
+    "solution": "3 天读了这本书的: 4/15 × 3 = 4/5 。",
+    "knowledge_points": [
+      "分数乘整数"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_036",
+    "original_id": "readme_q_036"
+  },
+  {
+    "id": "readme_q_037",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个长方形长 1/4m ,宽 2/5m ,它的周长是( ) m ,面积是( ) {m}^{2} 。",
+    "options": [],
+    "answer": "13/10 1/10",
+    "solution": "长方形的周长三(长土宽)x2. 长方形面积三长x宽,据此代入数据解答即可 周长: ( {1/4 + 2/5}) × 2 = ( {5/20 + 8/20}) × 2 = 13/20 × 2 <!-- Meanless: 试卷第 15 页,共 38 页--> --- = 13/10( m) 面积: 1/4 × 2/5 = 1/10( { m}^{2}) 所以周长是 13/10 m ,面积是 1/10{ m}^{2} 。 ---",
+    "knowledge_points": [
+      "长方形的周长",
+      "长方形的面积",
+      "分数乘分数",
+      "分数乘整数"
+    ],
+    "source": " 23-24 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_037",
+    "original_id": "readme_q_037"
+  },
+  {
+    "id": "readme_q_038",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "困难",
+    "question": "单位换算。 7/500{ m}^{3} = ( ) {{dm}}^{3} 4/5 m = ( ) {cm} 1/10{ {km}}^{2} = ( )公顷 5/8日 = ( )",
+    "options": [],
+    "answer": "14 80 10 15",
+    "solution": "7/500{ m}^{3} = 7/500 × {1000} = {14}{ {dm}}^{3} 4/5 m = 4/5 × {100} = {80} {cm} 1/10{{km}}^{2} = 1/10 × {100} = {10} 公顷 5/8 日 = 5/8 × {24} = {15} 时",
+    "knowledge_points": [
+      "面积单位间的进率及换算",
+      "时",
+      "分的认识及换算",
+      "体积单位间的进率与换算 (立方厘米",
+      "立方分米和立方米)",
+      "分数乘整数"
+    ],
+    "source": "23-24 五年级下·全国·假期作业",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_038",
+    "original_id": "readme_q_038"
+  },
+  {
+    "id": "readme_q_039",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "蜂鸟是目前发现的世界上最小的鸟,也是唯一能倒飞的鸟,蜂鸟每分钟可飞行 3/10 千米,它 5/6 分钟飞行 ( ) 千米。",
+    "options": [],
+    "answer": "1/4/{0.25}",
+    "solution": "3/10 × 5/6 = 1/4 (千米) 它 5/6 分钟飞行 1/4 千米。",
+    "knowledge_points": [
+      "分数乘分数",
+      "基础行程问题"
+    ],
+    "source": "23-24 五年级下·河北邢台·期中",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_039",
+    "original_id": "readme_q_039"
+  },
+  {
+    "id": "readme_q_040",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "“红花的朵数的 2/3 相当于黄花的朵数”是把( ( )。",
+    "options": [],
+    "answer": "红花的朵数 红花的朵数 × 2/3 = 黄花的朵数",
+    "solution": "根据分析可知,“红花的朵数的 2/3 相当于黄花的朵数”是把红花的朵数看作单位“1”,等量关系式是：红花的朵数 × 2/3 = 黄花的朵数。",
+    "knowledge_points": [
+      "单位“1”的认识与确定",
+      "求一个数的几分之几的问题"
+    ],
+    "source": "19-20 六年级上·辽宁·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_040",
+    "original_id": "readme_q_040"
+  },
+  {
+    "id": "readme_q_041",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个长方形,长 1/2 m ,宽 2/5 m ,它的周长是 ( )m, 它的面积是( ) 分析】根据长方形周长=(长土宽)×2,长方形面积=长x宽,列式计算即可",
+    "options": [],
+    "answer": "",
+    "solution": "( {1/2 + 2/5}) × 2 = 9/10 × 2 = 9/5( m) 1/2 × 2/5 = 1/5( { m}^{2}) 一个长方形,长 1/2 m ,宽 2/5 m ,它的周长是 9/5 m ,它的面积是 1/5{ m}^{2} 。",
+    "knowledge_points": [
+      "长方形的周长",
+      "长方形的面积",
+      "分数与整数的乘法",
+      "分数乘分数"
+    ],
+    "source": "18-19 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_041",
+    "original_id": "readme_q_041"
+  },
+  {
+    "id": "readme_q_042",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "计算 2/3 + 3/5 × 1/6 时,应先算( 【答案】 乘 加 23/30 【难度】 0.65 【知识点】分数的四则混合运算 【分析】一个算式里, 如果含有两级运算, 要先做第二级运算, 后做第一级运算; 如果有括号, 要先算小括号里的。计算 2/3 + 3/5 × 1/6 时,应先算乘法,再算加法; 据此解答。",
+    "options": [],
+    "answer": "",
+    "solution": "2/3 + 3/5 × 1/6 = 2/3 + 1/10 = 23/30 计算 2/3 + 3/5 × 1/6 时,应先算乘法,后算加法,得 23/30 。",
+    "knowledge_points": [
+      "分数乘法"
+    ],
+    "source": "22-23 六年级上·内蒙古通辽·阶段练习",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_042",
+    "original_id": "readme_q_042"
+  },
+  {
+    "id": "readme_q_043",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "小马虎把 {17} × ( {m + 3/10}) 错算成 {17} × m + 3/10 ,他计算的结果与正确结果相差( ) --- ---",
+    "options": [],
+    "answer": "24/5",
+    "solution": "{17} × ( {m + 3/10}) - ( {{17} × m + 3/10}) <!-- Meanless: 试卷第 17 页,共 38 页--> = {17} m + {17} × 3/10 - {17} m - 3/10 = 51/10 - 3/10 = 24/5 小马虎把 {17} × ( {m + 3/10}) 错算成 {17} × m + 3/10 ,他计算的结果与正确结果相差 24/5 。",
+    "knowledge_points": [
+      "用字母表示数",
+      "分数乘法运算律"
+    ],
+    "source": "23-24 六年级上·全国·期中",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_043",
+    "original_id": "readme_q_043"
+  },
+  {
+    "id": "readme_q_044",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "鸵鸟是现在世界上最大的鸟,身高可达 2.5 米。一只成年的帝企鹅身高是鸵鸟的 12/25 。 成年帝企鹅的身高是( )米。",
+    "options": [],
+    "answer": "1.2",
+    "solution": "{2.5} × 12/25 = {1.2} (米) 成年帝企鹅的身高是 1.2 米。",
+    "knowledge_points": [
+      "求一个数的几分之几的问题",
+      "分数乘小数"
+    ],
+    "source": "22-23 六年级・全国・假期作业",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_044",
+    "original_id": "readme_q_044"
+  },
+  {
+    "id": "readme_q_045",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "“ {4.65} × 3/5 ”表示的意义是 ( )。 ## 三、判断题",
+    "options": [],
+    "answer": "求 4.65 的 3/5 是多少",
+    "solution": "“ {4.65} × 3/5 ”表示的意义是求4.65的 3/5 是多少。",
+    "knowledge_points": [
+      "分数乘小数"
+    ],
+    "source": "22-23 六年级上·四川·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_045",
+    "original_id": "readme_q_045"
+  },
+  {
+    "id": "readme_q_046",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "1 米长的绳子用去 4/5 ,就是用去 4/5 米。( )",
+    "options": [],
+    "answer": "",
+    "solution": "由题意可知,把绳子原来的长度看作单位 “1”,用去的是原来的 4/5 ,根据求一个数的几分之几是多少,用乘法计 算,即可求出用去的长度,再判断。 【详解】 1 × 4/5 = 4/5 (米) <!-- Meanless: 试卷第 18 页,共 38 页--> 1 米长的绳子用去 4/5 ,就是用去 4/5 米,原题说法正确。 故答案为:",
+    "knowledge_points": [
+      "整数乘分数",
+      "求一个数的几分之几的问题"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_046",
+    "original_id": "readme_q_046"
+  },
+  {
+    "id": "readme_q_047",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "如 A × 6 = B × 3/4 ( A 、 B 均不为 0 ),那么 A > B 。( )(对的打“√”,错的打“×”, 并说明理由。) 理由：_____。",
+    "options": [],
+    "answer": "× 因为 6 > 3/4 ,乘积一定,所以若 A × 6 = B × 3/4 ( A 、 B 均不为 0 ),那么 A < B",
+    "solution": "如 A × 6 = B × 3/4 ( A 、 B 均不为 0 ),那么 A > B 。(✗) 理由: 因为 6 > 3/4 ,所以若 A × 6 = B × 3/4 ( A 、 B 均不为 0 ),那么 A < B 。",
+    "knowledge_points": [
+      "分数乘分数",
+      "因数和积的大小关系(分数乘法)"
+    ],
+    "source": " 23-24 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_047",
+    "original_id": "readme_q_047"
+  },
+  {
+    "id": "readme_q_048",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "比 1/2 的 1/2 多 1/2 的数是 1/8 。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "1/2 × 1/2 + 1/2 = 1/4 + 1/2 = 1/4 + 2/4 = 3/4 比 1/2 的 1/2 多 1/2 的数是 3/4 。 --- 原题说法错误。 故答案为: ×",
+    "knowledge_points": [
+      "分数的四则混合运算",
+      "分数乘分数",
+      "求一个数的几分之几的问题"
+    ],
+    "source": "23-24 五年级下·河北邢台·期末",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_048",
+    "original_id": "readme_q_048"
+  },
+  {
+    "id": "readme_q_049",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "困难",
+    "question": "用简便方法计算 8/13 × {12} + 8/13 ,应该运用乘法分配律。( ) --- ---",
+    "options": [],
+    "answer": "",
+    "solution": "8/13 × {12} + 8/13 = 8/13 × ( {{12} + 1}) = 8/13 × {13} = 8 --- 用简便方法计算 8/13 × {12} + 8/13 ,应该运用乘法分配律。 原题干说法正确。 故答案为:",
+    "knowledge_points": [
+      "整数乘法运算定律推广到分数乘法"
+    ],
+    "source": "25-26 六年级上·全国·课后作业",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_049",
+    "original_id": "readme_q_049"
+  },
+  {
+    "id": "readme_q_050",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "小白体重的 7/8 与小金的体重相同,则小白体重 = 小金体重 × 7/8 。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "小白体重的 7/8 与小金的体重相同,则小金的体重 = 小白的体重 × 7/8 。 所以原题说法错误。 故答案为: ×",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_050",
+    "original_id": "readme_q_050"
+  },
+  {
+    "id": "readme_q_051",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "两个分数相乘的积一定大于其中任意一个分数。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "根据分析可得: 两个分数相乘的积不一定大于其中任意一个分数, 。原说法错误。 故答案为: × <!-- Meanless: 试卷第 20 页,共 38 页-->",
+    "knowledge_points": [
+      "分数乘分数",
+      "因数和积的大小关系(分数乘法)"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_051",
+    "original_id": "readme_q_051"
+  },
+  {
+    "id": "readme_q_052",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "两个分数相乘,积一定小于其中的一个分数。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "当两个分数都是真分数时,比如 1/2 × 1/3 = 1/6 ,积 1/6 小于这两个分数 1/2 和 1/3 。 但当两个分数都是大于 1 的假分数时,比如 3/2 × 4/3 = 2 ,积 2 大于这两个分数 3/2 和 4/3 。 所以两个分数相乘,积不一定小于其中的一个分数,故该原题说法错误。 故答案为: ×",
+    "knowledge_points": [
+      "因数和积的大小关系(分数乘法)"
+    ],
+    "source": "23-24 五年级上·山东烟台·期中",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_052",
+    "original_id": "readme_q_052"
+  },
+  {
+    "id": "readme_q_053",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "丫丫将一根绳子对折两次,量得每段长 3/4 米,这根绳子长 3/2 米。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "根据题意,把一根绳子对折两次,平均分成 2 × 2 = 4 段; 用每段的长度乘 4,即是这根绳子的总长,据此判断。 2 × 2 = 4 (段) 3/4 × 4 = 3 (米) 这根绳子长 3 米。 原题说法错误。 故答案为: ×",
+    "knowledge_points": [
+      "分数乘整数",
+      "图形的折叠问题"
+    ],
+    "source": "22-23 五年级下·河北承德·期末",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_053",
+    "original_id": "readme_q_053"
+  },
+  {
+    "id": "readme_q_054",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "两个分数的积一定小于 1 。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "如: 这两个分数是 3/2 和 5/3,3/2 × 5/3 > 3/2 ,且 3/2 > 1 ,此时两个分数的积大于 1,则原题说法错误。 故答案为: ×",
+    "knowledge_points": [
+      "因数和积的大小关系(小数乘法)",
+      "分数乘分数"
+    ],
+    "source": "2014 六年级・全国・课后作业",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_054",
+    "original_id": "readme_q_054"
+  },
+  {
+    "id": "readme_q_055",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一种商品,先涨价 1/10 ,再降价 1/10 ,现价比原价少。( )",
+    "options": [],
+    "answer": "",
+    "solution": "1 × ( {1 + 1/10}) × ( {1 - 1/10}) = 1 × 11/10 × 9/10 = 99/100 99/100 < 1 --- 所以一种商品,先涨价 1/10 ,再降价 1/10 ,现价比原价少,原题说法正确。 故答案为:",
+    "knowledge_points": [
+      "单位“1”的认识与确定",
+      "求一个数的几分之几的问题"
+    ],
+    "source": "20-21 六年级上·广东韶关·期中",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_055",
+    "original_id": "readme_q_055"
+  },
+  {
+    "id": "readme_q_056",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "8/9 × 5 和 5 × 8/9 的意义完全相同。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "8/9 × 5 表示 5 个 8/9 的和是多少; 5 × 8/9 表示 5 的 8/9 是多少; 所以, 8/9 × 5 和 5 × 8/9 的意义不相同。 原题说法错误。 故答案为: ×",
+    "knowledge_points": [
+      "整数乘分数",
+      "分数乘整数"
+    ],
+    "source": "22-23 六年级上·新疆巴音郭楞·期中",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_056",
+    "original_id": "readme_q_056"
+  },
+  {
+    "id": "readme_q_057",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个数乘分数的积一定比原来这个数小。( ) ## 四、计算题",
+    "options": [],
+    "answer": "×",
+    "solution": "一个数 ( 0 除外) 乘大于 1 的数, 结果比原来的数大; 一个数 ( 0 除外) 乘小于 1 的数, 结果比原来的数小。 如: 5{ s}_{5}^{2} < 5 ,此时它们的积比原来的这个数小; 5{x}_{2}^{5} > 5 ,此时它们的积比原来的这个数大。则原题干说法错误。 故答案为: ×",
+    "knowledge_points": [
+      "因数和积的大小关系 (小数乘法)",
+      "分数与整数的乘法",
+      "分数乘分数"
+    ],
+    "source": "15-16 六年级·湖北恩施·阶段练习",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_057",
+    "original_id": "readme_q_057"
+  },
+  {
+    "id": "readme_q_058",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "计算。 (1) 74/75 × 8 (2) 115/114 × {113}",
+    "options": [],
+    "answer": "(1) 767/75 ；(2) {113}113/114",
+    "solution": "(1) 74/75 × 8 = ( {1 - 1/75}) × 8 = 1 × 8 - 1/75 × 8 = 8 - 8/75 = 767/75 (2) 115/114 × {113} = ( {1 + 1/114}) × {113} = 1 × {113} + 1/114 × {113} = {113} + 113/114 = {113}113/114",
+    "knowledge_points": [
+      "整数乘法运算定律推广到分数乘法"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_058",
+    "original_id": "readme_q_058"
+  },
+  {
+    "id": "readme_q_059",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "计算。 (1) 29/98 × {99} (2) {67} × 15/68 --- ---",
+    "options": [],
+    "answer": "(1) {29}29/98 ；(2) {14}53/68",
+    "solution": "(1) 29/98 × {99} = 29/98 × ( {{98} + 1}) = 29/98 × {98} + 29/98 × 1 = {29} + 29/98 = {29}29/98 (2) {67} × 15/68 = ( {{68} - 1}) × 15/68 = {68} × 15/68 - 1 × 15/68 = {15} - 15/68 = {14}53/68",
+    "knowledge_points": [
+      "整数乘法运算定律推广到分数乘法"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_059",
+    "original_id": "readme_q_059"
+  },
+  {
+    "id": "readme_q_060",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "用简便方法计算下面各题。 (1) {27} × 15/26 (2) 44/45 × {37} --- ---",
+    "options": [],
+    "answer": "(1) {15}15/26 ；(2) {36}8/45",
+    "solution": "(1) {27} × 15/26 = ( {{26} + 1}) × 15/26 = {26} × 15/26 + 1 × 15/26 = {15} + 15/26 = {15}15/26 (2) 44/45 × {37} = ( {1 - 1/45}) × {37} <!-- Meanless: 试卷第 24 页,共 38 页--> = 1 × {37} - 1/45 × {37} = {37} - 37/45 = {36}8/45 ---",
+    "knowledge_points": [
+      "整数乘法运算定律推广到分数乘法"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_060",
+    "original_id": "readme_q_060"
+  },
+  {
+    "id": "readme_q_061",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "计算。 (1) 5/17 × 9/24 + 9/17 × 7/24 (2) 5/9 × 9/16 + 7/9 × 5/16 --- ---",
+    "options": [],
+    "answer": "(1) 9/34 ；(2) 5/9",
+    "solution": "(1) 5/17 × 9/24 + 9/17 × 7/24 = 9/17 × 5/24 + 9/17 × 7/24 = 9/17 × ( {5/24 + 7/24}) = 9/17 × 1/2 = 9/34 (2) 5/9 × 9/16 + 7/9 × 5/16 = 5/9 × 9/16 + 5/9 × 7/16 = 5/9 × ( {9/16 + 7/16}) = 5/9 × 1 = 5/9",
+    "knowledge_points": [
+      "整数乘法运算定律推广到分数乘法"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_061",
+    "original_id": "readme_q_061"
+  },
+  {
+    "id": "readme_q_062",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "计算。 (1) 11/13 × 6/19 + 6/13 × 8/19 (2) {139} × 137/138 - {137} × 1/138 --- ---",
+    "options": [],
+    "answer": "(1) 6/13 ；(2)137",
+    "solution": "( 1 ) 11/13 × 6/19 + 6/13 × 8/19 = 6/13 × 11/19 + 6/13 × 8/19 = 6/13 × ( {11/19 + 8/19}) = 6/13 × 1 = 6/13 (2) {139} × 137/138 - {137} × 1/138 = {139} × 137/138 - 1 × 137/138 = ( {{139} - 1}) × 137/138 = {138} × 137/138 = {137} ---",
+    "knowledge_points": [
+      "整数乘法运算定律推广到分数乘法",
+      "分数的四则混合运算"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_062",
+    "original_id": "readme_q_062"
+  },
+  {
+    "id": "readme_q_063",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "用简便方法计算下面各题。 (1) ( {8/9 + 4/27}) × {27} (2) 5/6 × 5/9 + 5/9 × 1/6 --- ---",
+    "options": [],
+    "answer": "(1)28；(2) 5/9",
+    "solution": "(1) ( {8/9 + 4/27}) × {27} --- = 8/9 × {27} + 4/27 × {27} = {24} + 4 =28 (2) 5/6 × 5/9 + 5/9 × 1/6 = 5/9 × ( {5/6 + 1/6}) = 5/9 × 1 = 5/9 ---",
+    "knowledge_points": [
+      "整数乘法运算定律推广到分数乘法"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_063",
+    "original_id": "readme_q_063"
+  },
+  {
+    "id": "readme_q_064",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "简单",
+    "question": "用“裂项消去法”进行简便计算。 1/2 × 4 + 1/4 × 6 + 1/6 × 8 + 1/8 × {10} 1/1 × 4 + 1/4 × 7 + 1/7 × {10} + 1/{10 × {13}} --- ---",
+    "options": [],
+    "answer": "1/5;4/13",
+    "solution": "( 1 ) 1/2 × 4 + 1/4 × 6 + 1/6 × 8 + 1/8 × {10} = 1/2 × ( {1/2 - 1/4}) + 1/2 × ( {1/4 - 1/6}) + 1/2 × ( {1/6 - 1/8}) + 1/2 × ( {1/8 - 1/10}) = 1/2 × ( {1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + 1/8 - 1/10}) = 1/2 × ( {1/2 - 1/10}) = 1/2 × ( {5/10 - 1/10}) = 1/2 × 2/5 = 1/5 (2)1/1 × 4 + 1/4 × 7 + 1/7 × {10} + 1/{10 × {13}} = 1/3 × ( {1 - 1/4}) + 1/3 × ( {1/4 - 1/7}) + 1/3 × ( {1/7 - 1/10}) + 1/3 × ( {1/10 - 1/13}) = 1/3 × ( {1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + 1/10 - 1/13}) = 1/3 × ( {1 - 1/13}) = 1/3 × 12/13 = 4/13",
+    "knowledge_points": [
+      "算式的规律",
+      "分数的四则混合运算"
+    ],
+    "source": "2023 六年级上·全国·专题练习",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_064",
+    "original_id": "readme_q_064"
+  },
+  {
+    "id": "readme_q_065",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "3/5 的 5/6 与 7/15 差,差是多少?",
+    "options": [],
+    "answer": "1/30",
+    "solution": "3/5 × 5/6 - 7/15 --- = 1/2 - 7/15 = 15/30 - 14/30 = 1/30 差是 1/30 <!-- Meanless: 试卷第 27 页,共 38 页--> ## 五、解答题",
+    "knowledge_points": [
+      "分数乘分数",
+      "异分母分数加",
+      "减法",
+      "求一个数的几分之几的问题"
+    ],
+    "source": "22-23 五年级下·陕西西安·期末",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_065",
+    "original_id": "readme_q_065"
+  },
+  {
+    "id": "readme_q_066",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "一根木棒长 10 分米,把它插入一个水池的底部,底部与木棒垂直,木棒的 3/5 在水中,在水面与木棒的交触点处做记号 A ,然后又把木棒倒过来,使用同样的方法再次插入水池的底部,又在水面与木棒的交触点处做记号 B 。木棒上 A 、 B 两点间的距离是多少?",
+    "options": [],
+    "answer": "2 分米",
+    "solution": "{10} × 3/5 = 6 (分米) {10} - 6 = 4 (分米) 6 - 4 = 2 (分米) 答: 木棒上 A、B 两点间的距离是 2 分米。",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_066",
+    "original_id": "readme_q_066"
+  },
+  {
+    "id": "readme_q_067",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "一种彩电,原来每台售价是 3200 元,现售价比原售价降低了 3/20 。这种彩电现在售价是多少元?",
+    "options": [],
+    "answer": "2720 元",
+    "solution": "{3200} × ( {1 - 3/20}) = {3200} × 17/20 =2720 (元) --- 答: 这种彩电现在售价是 2720 元。",
+    "knowledge_points": [
+      "求比一个数多/少几分之几的数是多少"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_067",
+    "original_id": "readme_q_067"
+  },
+  {
+    "id": "readme_q_068",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "小红读一本 120 页的故事书,第一天读了全书的 2/3 ,第二天读了余下的 1/2 ,第二天读了多少页? --- --- ## 【难度】 0.65",
+    "options": [],
+    "answer": "20 页",
+    "solution": "{120} × ( {1 - 2/3}) × 1/2 = {120} × 1/3 × 1/2 = {20} (页) 答: 第二天读了 20 页。 ---",
+    "knowledge_points": [
+      "连续求一个数的几分之几是多少的问题"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_068",
+    "original_id": "readme_q_068"
+  },
+  {
+    "id": "readme_q_069",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "某校举行跳绳比赛。在规定时间内,小玉跳了 160 下,小林跳的是小玉的 3/4 ,小刚比小林多跳了 1/2 。小刚跳了多少下? ---",
+    "options": [],
+    "answer": "180 下",
+    "solution": "{160} × 3/4 × ( {1 + 1/2}) = {120} × 3/2 = {180} (下) 答: 小刚跳了 180 下。 ---",
+    "knowledge_points": [
+      "分数的四则混合运算",
+      "求比一个数多/少几分之几的数是多少",
+      "求一个数的几分之几的问题 【分析】把小玉跳的下数看作单位 “ 1 ”",
+      "小林跳的是小玉的 3/4",
+      "用小玉跳的下数 × 3/4",
+      "求出小林跳的下数",
+      "再把小林跳的下数看作单位 “1”",
+      "小刚跳的下数是小林的 ( {1 + 1/2})",
+      "用小林跳的下数 × ( {1 + 1/2})",
+      "即可求出小刚跳的下数。"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_069",
+    "original_id": "readme_q_069"
+  },
+  {
+    "id": "readme_q_070",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "林师傅加工一批零件,第一周加工 210 个零件,第二周比第一周少加工 2/7 。第二周加工多少个零件?",
+    "options": [],
+    "answer": "150 个",
+    "solution": "{210} × ( {1 - 2/7}) = {210} × 5/7 = {150} (个) --- <!-- Meanless: 试卷第 29 页,共 38 页--> 答：第二周加工 150 个零件。",
+    "knowledge_points": [
+      "求比一个数多/少几分之几的数是多少"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_070",
+    "original_id": "readme_q_070"
+  },
+  {
+    "id": "readme_q_071",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "林老师九月份用电 150 元,十月份用电是九月份的 4/5 ,十月份用电多少元？",
+    "options": [],
+    "answer": "120 元",
+    "solution": "{150} × 4/5 = {120} (元) 答：十月份用电 120 元。",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_071",
+    "original_id": "readme_q_071"
+  },
+  {
+    "id": "readme_q_072",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "小明和小刚骑自行车去学校,小明每分钟骑 {}_{5}^{1} 千米,小刚每小时骑 13 千米。谁骑车的速度快?",
+    "options": [],
+    "answer": "小刚",
+    "solution": "1 小时 = {60} 分钟 1/5 × {60} = {12} (千米) 12 千米 < {13} 千米 答: 小刚骑车的速度快。",
+    "knowledge_points": [
+      "分数乘整数"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_072",
+    "original_id": "readme_q_072"
+  },
+  {
+    "id": "readme_q_073",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "小刚买来 2.4 千克水果,第一天吃了这些水果的 2/5 ,第二天再吃多少,这些水果将剩下它的一半?",
+    "options": [],
+    "answer": "0.24 千克",
+    "solution": "{2.4} × 1/2 = {1.2} (千克) <!-- Meanless: 试卷第 30 页,共 38 页--> {2.4} × 2/5 = {0.96} (千克) {2.4} - {0.96} = {1.44} (千克) {1.44} - {1.2} = {0.24} (千克) 答：第二天再吃 0.24 千克,这些水果将剩下它的一半。",
+    "knowledge_points": [
+      "求一个数的几分之几的问题",
+      "分数乘小数"
+    ],
+    "source": "24-25 六年级上·全国·课后作业",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_073",
+    "original_id": "readme_q_073"
+  },
+  {
+    "id": "readme_q_074",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "棕熊的寿命约为 40 年,鼹鼠的寿命约为棕熊的 5/8 ,长颈鹿的寿命约为鼹鼠的 2/5 ,长颈鹿的寿命约为多少年?",
+    "options": [],
+    "answer": "10 年",
+    "solution": "鼹鼠的寿命: {40} × 5/8 = {25} (年) 长颈鹿的寿命: {25} × 2/5 = {10} (年) 答: 长颈鹿的寿命约为 10 年。",
+    "knowledge_points": [
+      "整数乘分数",
+      "求一个数的几分之几的问题"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_074",
+    "original_id": "readme_q_074"
+  },
+  {
+    "id": "readme_q_075",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "甲、乙两地相距 540 千米,一辆汽车从甲地开往乙地,第一小时行了全程的 1/6 ,第二小时行了全程的 1/2 ,两个小时共行了全程的几分之几? 行了 2 小时后,离乙地还有多少千米?",
+    "options": [],
+    "answer": "5/18;{390} 千米",
+    "solution": "1/6 + 1/9 = 3/18 + 2/18 = 5/18 {540} - {540} × 5/18 = {540} - {150} = {390} (千米) --- 答: 两个小时共行了全程的 5/18 ,离乙地还有 390 千米。",
+    "knowledge_points": [
+      "异分母分数加",
+      "减法的应用",
+      "分数的四则混合运算",
+      "求一个数的几分之几的问题"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_075",
+    "original_id": "readme_q_075"
+  },
+  {
+    "id": "readme_q_076",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "春风书店新进一批故事书共 800 本,第一周卖出总数的 1/4 ,第二周卖出剩下的 2/5 ,",
+    "options": [],
+    "answer": "440 本",
+    "solution": "第一周卖出: {800} × 1/4 = {200} (本) 第二天卖出: ( {{800} - {200}}) × 2/5 = {600} × 2/5 = {240} (本) 一共: {200} + {240} = {440} (本) 答：两周一共卖出 440 本。",
+    "knowledge_points": [
+      "整数乘分数",
+      "求一个数的几分之几的问题"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_076",
+    "original_id": "readme_q_076"
+  },
+  {
+    "id": "readme_q_077",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "老师买回来 80 个苹果,(1)班分到了全部苹果的 3/8 ,(2)班分到的苹果数量是(1) 班的 3/5 ,(3)班分到了剩下苹果的 1/2 ,其余的留给老师自己。请问老师分到了多少个苹果？",
+    "options": [],
+    "answer": "16 个",
+    "solution": "(1)班分到： {80} × 3/8 = {30} (个) <!-- Meanless: 试卷第 32 页,共 38 页--> --- (2)班分到： {30} × 3/5 = {18} (个) 此时还剩: {80} - {30} - {18} = {32} (个) (3)班分到： {32} × 1/2 = {16} (个) 老师分到: {80} - {30} - {18} - {16} = {16} (个) 答: 老师分到了 16 个苹果。 ---",
+    "knowledge_points": [
+      "整数乘分数",
+      "求一个数的几分之几的问题"
+    ],
+    "source": "2023 六年级上·全国·专题练习",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_077",
+    "original_id": "readme_q_077"
+  },
+  {
+    "id": "readme_q_078",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "困难",
+    "question": "禾苗小学共有男生 600 人,女生人数比男生多 1/12 ,女生比男生多多少人? 【答案】 50 人",
+    "options": [],
+    "answer": "",
+    "solution": "{600} × 1/12 = {50} (人) 答：女生比男生多 50 人。",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": "23-24 五年级下·内蒙古包头·期末",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_078",
+    "original_id": "readme_q_078"
+  },
+  {
+    "id": "readme_q_079",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "某小学五年级有学生 288 人,四年级的人数比五年级少 2/9 ,四年级有多少人? 【答案】224 人",
+    "options": [],
+    "answer": "",
+    "solution": "{288} × ( {1 - 2/9}) = {288} × 7/9 =224 (人) 答：四年级有 224 人。 ---",
+    "knowledge_points": [
+      "求比一个数多/少几分之几的数是多少",
+      "求一个数的几分之几的问题"
+    ],
+    "source": "22-23 五年级下·河北张家口·期中",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_079",
+    "original_id": "readme_q_079"
+  },
+  {
+    "id": "readme_q_080",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "一筐苹果重 50 千克,卖出 1/5 ,还剩多少千克？",
+    "options": [],
+    "answer": "40 千克",
+    "solution": "把原来苹果的总重量看作单位 “1”,卖出 1/5 ,根据分数乘法的意义,用 {50} × 1/5 即可求出卖出多少千克,再用原来的总重量减去卖出的重量,即可求出剩余的重量。 {50} × 1/5 = {10} (千克) <!-- Meanless: 试卷第 33 页,共 38 页--> {50} - {10} = {40} (千克) 答：还剩 40 千克。",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": "23-24 五年级下·广东潮州·期中",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_080",
+    "original_id": "readme_q_080"
+  },
+  {
+    "id": "readme_q_081",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "世界上最宽的河口是拉普拉塔河, 河长只有 320 千米, 而河口最宽处的宽度是河长的 29/32 ,河的源头处的宽度是河口最宽处的 5/29 。河的源头处的宽度是多少千米?",
+    "options": [],
+    "answer": "50 千米",
+    "solution": "{320} × 29/32 × 5/29 = {290} × 5/29 = {50} (千米) 答: 河的源头处的宽度是 50 千米。",
+    "knowledge_points": [
+      "连续求一个数的几分之几是多少的问题",
+      "分数的连乘运算"
+    ],
+    "source": "23-24 五年级下·全国·假期作业",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_081",
+    "original_id": "readme_q_081"
+  },
+  {
+    "id": "readme_q_082",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "困难",
+    "question": "农林废弃物发电可以有效减少秸秆和林业废弃物焚烧对环境的污染,节约标准煤, 对推进国家主体生态功能区建设具有重要意义。农林废弃物若气化发电 1 千瓦时需要 {}_{2}^{1} 千克秸秆和 {}_{4}^{3} 千克树皮,则气化发电 80 千瓦时需要秸秆和树皮共多少千克？",
+    "options": [],
+    "answer": "100 千克",
+    "solution": "1/2 × {80} + 3/4 × {80} = {40} + {60} = {100} (千克) --- 答: 气化发电 80 千瓦时需要秸秆和树皮共 100 千克。",
+    "knowledge_points": [
+      "分数乘整数"
+    ],
+    "source": "23-24 五年级下·河北沧州·期中",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_082",
+    "original_id": "readme_q_082"
+  },
+  {
+    "id": "readme_q_083",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "阅读下面的材料并回答问题。 在很多人的印象中, 打羽毛球是一种在楼下空地进行的“悠闲”运动, 但事实上, 在运动热量消耗排行榜上, 打羽毛球仅 次于游泳、慢跑和跳绳。慢跑每小时消耗约 600 卡路里,跳绳每小时消耗的卡路里约是慢跑的 5/6 ,打羽毛球每小时消耗的卡路里比慢跑少 1/4 ,游泳每小时消耗的卡路里比慢跑多 5/12 。 (1)跳绳每小时消耗多少卡路里？ (2)请你提出一个需要两步或两步以上计算的数学问题并解答。 (2)游泳每小时消耗多少卡路里？850 卡路里",
+    "options": [],
+    "answer": "(1)500 卡路里",
+    "solution": "( 1 ) {600} × 5/6 = {500} (卡路里) 答: 跳绳每小时消耗 500 卡路里。 (2)问：游泳每小时消耗多少卡路里？ {600} × ( {1 + 5/12}) = {600} × 17/12 = {850} (卡路里) 答: 游泳每小时消耗 850 卡路里。",
+    "knowledge_points": [
+      "求比一个数多/少几分之几的数是多少",
+      "“提问题”“填条件”问题",
+      "求一个数的几分之几的问题"
+    ],
+    "source": "23-24 五年级下·河北邢台·期中",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_083",
+    "original_id": "readme_q_083"
+  },
+  {
+    "id": "readme_q_084",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "我国的故宫和法国的卢浮宫是世界建筑史和博物馆史上的两颗暗璨明珠,故宫的占地面积约 72 万平方米,卢浮宫的占地面积相当于故宫的 1/3 ,其中建筑面积是占地面积的 1/5 ,卢浮宫的建筑面积是多少万平方米?",
+    "options": [],
+    "answer": "4.8 万平方米",
+    "solution": "先把故宫的占地面积看作单位 “1”,用故宫的占地面积乘卢浮宫的占地面积相当于故宫的分率,求出卢浮宫的占地面积；再把卢浮宫的占地面积看作单位“1”,用卢浮宫的占地面积乘卢浮宫的建筑面积所占的分率；据此求解即可。 {72} × 1/3 × 1/5 ## 试卷第 35 页,共 38 页 = {24} × 1/5 = {4.8} (万平方米) 答: 卢浮宫的建筑面积是 4.8 万平方米。",
+    "knowledge_points": [
+      "连续求一个数的几分之几是多少的问题"
+    ],
+    "source": "2024 六年级下·山东·专题练习",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_084",
+    "original_id": "readme_q_084"
+  },
+  {
+    "id": "readme_q_085",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "学校购进 150 盆花,其中 2/5 摆放在大门两旁,其余按 1 : 2 的数量比分别放在主席台和花园里,问主席台和花园里各放了多少盆花?",
+    "options": [],
+    "answer": "主席台 30 盆；花园里 60 盆",
+    "solution": "先根据 150 盆花的 2/5 摆放在大门两旁来求出放在主席台和花园里的花,然后再根据比与分数的关系求出放在主席台旁的盆数占余下盆数的 1/1 + 2 ,放在花园里的盆数占余下盆数的 2/1 + 2 ,最后根据分数乘法的意义即可解答。 主席台和花园一共的盆数: {150} × ( {1 - 2/5}) = {150} × 3/5 --- = {90} (盆) 主席台旁: {90} × 1/1 + 2 = {90} × 1/3 = {30} (盆) 花园里: {90} × 2/1 + 2 = {90} × 2/3 = {60} (盆) --- 答: 主席台有 30 盆花; 花园里有 60 盆花。",
+    "knowledge_points": [
+      "比的应用",
+      "求一个数的几分之几的问题",
+      "按比分配问题"
+    ],
+    "source": "22-23 六年级上·河北保定·阶段练习",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_085",
+    "original_id": "readme_q_085"
+  },
+  {
+    "id": "readme_q_086",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "爷爷今年 80 岁、爸爸年龄是爷爷的 3/5 ,欢欢的年龄是爸爸的 1/4 。欢欢今年多少岁？ 【答案】 12 岁",
+    "options": [],
+    "answer": "",
+    "solution": "{80} × 3/5 × 1/4 = {48} × 1/4 --- <!-- Meanless: 试卷第 36 页,共 38 页--> = {12} (岁) 答： 欢欢今年 12 岁。",
+    "knowledge_points": [
+      "连续求一个数的几分之几是多少的问题"
+    ],
+    "source": "15-16 六年级上·全国·阶段练习",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_086",
+    "original_id": "readme_q_086"
+  },
+  {
+    "id": "readme_q_087",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "莉莉发烧至体温 39 摄氏度,吃药后体温下降了 1/13 ,几个小时后,体温又升高了 1/18 。 现在莉莉的体温是多少摄氏度? 她退烧了吗? (一般体温在 36 摄氏度 -37 摄氏度之间属于正常)",
+    "options": [],
+    "answer": "38 摄氏度; 没退烧",
+    "solution": "将 39 摄氏度看作单位 “1”,吃药后体温下降了 1/13 ,是 39 摄氏度的 ( {1 - 1/13}) ；将下降后的体温看作单位 “1”,又升高了 1/18 ,是下降后体温的 ( {1 + 1/18}) ,39 摄氏度 × 下降后对应分率 × 又升高后对应分率 = 现在体温,对照正常体温即可。 {39} × ( {1 - 1/13}) × ( {1 + 1/18}) = {39} × 12/13 × 19/18 = {36} × 19/18 = {38} (摄氏度) {38} > {37} 答: 现在莉莉的体温是 38 摄氏度, 她没退烧。",
+    "knowledge_points": [
+      "连续求一个数的几分之几是多少的问题",
+      "求比一个数多/少几分之几的数是多少"
+    ],
+    "source": "23-24 六年级上·山东济南·期末",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_087",
+    "original_id": "readme_q_087"
+  },
+  {
+    "id": "readme_q_088",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "鸡公山景区去年全年接待游客约 126 万人,上半年接待游客量是全年的 4/7 。第三季度接待游客量是上半年的 2/3 ,第三季度接待游客多少万人?",
+    "options": [],
+    "answer": "48 万人",
+    "solution": "{126} × 4/7 × 2/3 = {72} × 2/3 = {48} (万人) --- 答: 第三季度接待游客 48 万人。",
+    "knowledge_points": [
+      "连续求一个数的几分之几是多少的问题"
+    ],
+    "source": "2024 六年级下·全国·专题练习",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_088",
+    "original_id": "readme_q_088"
+  },
+  {
+    "id": "readme_q_089",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "商场“国庆大促销”,某品牌电脑原价 5000 元,现降价 1/10 ,现在这台电脑售价多少元? --- ---",
+    "options": [],
+    "answer": "4500 元",
+    "solution": "{5000} × ( {1 - 1/10}) = {5000} × 9/10 = {4500} (元) --- 答: 现在这台电脑售价 4500 元。",
+    "knowledge_points": [
+      "求比一个数多/少几分之几的数是多少"
+    ],
+    "source": "20-21 六年级上·湖南永州·期中",
+    "create_time": "2025-09-25 20:57:28",
+    "merged_id": "readme_089",
+    "original_id": "readme_q_089"
+  },
+  {
+    "id": "math_md_q_001",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "简单",
+    "question": "红花有12朵，黄花比红花少 1/4 ，黄花比红花少（ ）朵。",
+    "options": [
+      "A.3",
+      "B.4",
+      "C.9"
+    ],
+    "answer": "A",
+    "solution": "12× 1/4 = 3 （朵）黄花比红花少3朵。故答案为：A",
+    "knowledge_points": [
+      "分数与整数的乘法"
+    ],
+    "source": "21-22六年级上·河南洛阳·期末",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_001",
+    "original_id": "math_md_q_001"
+  },
+  {
+    "id": "math_md_q_002",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "一种商品先把价格提高 1/10 后，再降价 1/10 卖出，最后的价格（）。 A．与原价相同 B．比原价低 C.比原价高 D.无法比较",
+    "options": [],
+    "answer": "B",
+    "solution": "设原价为10元。 10× (1 + 1/10) = 10× 1/10 = 11 （元） 11× (1 - 1/10) = 11× 9/10 = 9.9 （元） 9.9元 < 10 元最后的价格比原价低。",
+    "knowledge_points": [
+      "求比一个数多少/几分之几的数是多少"
+    ],
+    "source": "20-21六年级上·全国·单元测试",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_002",
+    "original_id": "math_md_q_002"
+  },
+  {
+    "id": "math_md_q_003",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "简单",
+    "question": "少儿图书馆开展为社区和学校配送新书活动。为学校配送新书1600册，为社区配送新书的数量比学校 1/4 算式 1600× (1 - 1/4) 解决的问题是（）。",
+    "options": [
+      "A.为社区配送新书的数量比学校少多少册B.为社区配送了多少册新书C.为社区和学校一共配送了多少册新书"
+    ],
+    "answer": "B",
+    "solution": "根据分析得，算式 1600× (1 - 1/4) 解决的问题是为社区配送了多少册新书。故答案为：B",
+    "knowledge_points": [
+      "求比一个数多少/几分之几的数是多少",
+      "“提问题”“填条件”问题"
+    ],
+    "source": "22-23六年级上·河南三门峡·期中",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_003",
+    "original_id": "math_md_q_003"
+  },
+  {
+    "id": "math_md_q_004",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "男生人数比女生人数 1/4 ，则女生人数比男生人数少（）。",
+    "options": [
+      "A. 1/4",
+      "B. 1/3",
+      "C. 1/5",
+      "D. 1/6"
+    ],
+    "answer": "C",
+    "solution": "假设女生有4人， 4× (1 + 1/4) = 4× 5/4 = 5 （人） (5 - 4)÷ 5 = 1÷ 5 1/5 则女生人数比男生人数少 1/5",
+    "knowledge_points": [
+      "求比一个数多少几分之几的数是多少",
+      "求一个数占另一个数几分之几"
+    ],
+    "source": "20-21五年级上·山东烟台·期末",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_004",
+    "original_id": "math_md_q_004"
+  },
+  {
+    "id": "math_md_q_005",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一块长方形菜地，长20米，宽是长的 3/4 ，求面积的算式是（）。",
+    "options": [
+      "A. 20× 3/4",
+      "B. 20× 3/4 +20",
+      "C. 20× (20× 3/4)"
+    ],
+    "answer": "C",
+    "solution": "20× (20× 3/4) = 20× 15 = 300 （平方米）菜地有300平方米。故答案为：C",
+    "knowledge_points": [
+      "长方形的面积",
+      "求一个数的几分之几的问题"
+    ],
+    "source": "19-20三年级上·山东·单元测试",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_005",
+    "original_id": "math_md_q_005"
+  },
+  {
+    "id": "math_md_q_006",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "养鸡场养公鸡500只，养的母鸡比公鸡的只数多 1/10 ，母鸡比公鸡多（）只。",
+    "options": [
+      "A. 500× (1 - 1/10)",
+      "B. 500× 1/10",
+      "C. 500× (1 + 1/10)"
+    ],
+    "answer": "B",
+    "solution": "500× 1/10 = 50 （只）",
+    "knowledge_points": [
+      "分数与整数的乘法"
+    ],
+    "source": "22-23六年级上·全国·单元测试",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_006",
+    "original_id": "math_md_q_006"
+  },
+  {
+    "id": "math_md_q_007",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一堆煤有 4/5 吨，用去 1/2 还剩下这堆煤的（）没有用。",
+    "options": [
+      "A. 2/5 吨",
+      "B. 2/5",
+      "C. 1/5 吨"
+    ],
+    "answer": "A",
+    "solution": "1 - 1/2 = 1/2 4/5× 1/2 = 2/5 （吨）故答案为：A",
+    "knowledge_points": [
+      "同分母分数加",
+      "减法",
+      "分数乘分数"
+    ],
+    "source": "22-23六年级上·全国·单元测试",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_007",
+    "original_id": "math_md_q_007"
+  },
+  {
+    "id": "math_md_q_008",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "简单",
+    "question": "(7/17 +1/2)× 34 = 7/17× 34 + 1/2× 34 ，这是运用了（）。",
+    "options": [
+      "A.乘法交换律",
+      "B.乘法结合律",
+      "C.乘法分配律"
+    ],
+    "answer": "C",
+    "solution": "(7/17 +1/2)× 34 = 7/17× 34 + 1/2× 34 ，这是运用了乘法分配律。故答案为：C",
+    "knowledge_points": [
+      "分数乘法运算律"
+    ],
+    "source": "22-23六年级上·全国·单元测试",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_008",
+    "original_id": "math_md_q_008"
+  },
+  {
+    "id": "math_md_q_009",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一桶油净重 100kg ，用去这桶油的 1/10 以后，再用去剩下的 1/10 ，现在桶里还有（） kg 的油。",
+    "options": [
+      "A.81",
+      "B.99",
+      "C.100"
+    ],
+    "answer": "A",
+    "solution": "100× (1 - 1/10)× (1 - 1/10) = 100× 9/10× (1 - 1/10) = 100× 9/10× 9/10 = 81 (kg)所以，现在桶里还有 81kg 的油。",
+    "knowledge_points": [
+      "单位\"1\"的认识与确定",
+      "求一个数的几分之几的问题",
+      "分数乘分数"
+    ],
+    "source": "22-23六年级上·全国·单元测试",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_009",
+    "original_id": "math_md_q_009"
+  },
+  {
+    "id": "math_md_q_010",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "两根同样长的电线，第一根用去全长的 1/4 ，第二根用去 1/4 米，剩下的电线（）。 【点晴】解答此题的关键是理解两根电线的长度是否已知。",
+    "options": [
+      "A.第一根长",
+      "B.第二根长",
+      "C.两根一样长",
+      "D.无法比较"
+    ],
+    "answer": "D",
+    "solution": "若两根同样长的电线大于1，是2米。 2 - 2× 1/4 = 1.5 （米） 2 - 1/4 = 1.75 （米） 1.75 > 1.5 ，因此第二根剩下的长； 若两根同样长的电线等于1，是1米。 1 - 1× 1/4 = 0.75 （米） 1 - 1/4 = 0.75 （米） 由此可知两根剩下的一样长： 若两根同样长的电线小于1，是0.4米。 0.4 - 0.4× 1/4 = 0.3 （米） 0.4 - 1/4 = 0.15 （米） 0.3 > 0.15 ，所以第一根剩下的长。 由此可知：当两根同样长的电线未知长度时，无法比较剩下的长度。",
+    "knowledge_points": [
+      "单位\"1\"的认识与确定",
+      "求一个数的几分之几的问题",
+      "分数的意义"
+    ],
+    "source": "2022·广东阳江·小升初真题",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_010",
+    "original_id": "math_md_q_010"
+  },
+  {
+    "id": "math_md_q_011",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "两个真分数的积一定是（）。 【点晴】此题主要考查真分数的认识，解题的关键是根据积与乘数的关系，不用计算即可判断积与因数之间的大小关系。",
+    "options": [
+      "A.自然数B.假分数C.真分数"
+    ],
+    "answer": "C",
+    "solution": "由分析可得，真分数小于1，真分数乘真分数，积比原来的数小，积还是真分数。",
+    "knowledge_points": [
+      "因数和积的大小关系（小数乘法）",
+      "分数乘分数",
+      "真分数",
+      "假分数",
+      "带分数的认识"
+    ],
+    "source": "14-15六年级上.全国-课后作业",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_011",
+    "original_id": "math_md_q_011"
+  },
+  {
+    "id": "math_md_q_012",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "两段绳子一样长，都是 2/4 米，第一段截去全长的 1/4 ，第二段截去 1/4 米，剩下的部分哪一段更长？（） 【点晴】本题考查分数乘法，解答本题的关键是掌握分数乘法的计算方法。",
+    "options": [
+      "A.第一段B.第二段C.剩下的一样长D.无法确定"
+    ],
+    "answer": "A",
+    "solution": "第一段： 3/4× (1 - 1/4) = 3/4× 3/4 = 9/16 （米）第二段： 3/4 - 1/4 = 1/2 （米） 9/16 >8/16 = 1/2",
+    "knowledge_points": [
+      "分数乘分数"
+    ],
+    "source": "20-21六年级下·重庆黔江·期末",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_012",
+    "original_id": "math_md_q_012"
+  },
+  {
+    "id": "math_md_q_013",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一瓶油用去 1/2 后，又用去余下的 1/2 ，这时瓶里的油（）。",
+    "options": [
+      "A.正好用完",
+      "B.还剩",
+      "C.还剩"
+    ],
+    "answer": "B",
+    "solution": "1 - 1/2 - (1 - 1/2)× 1/2 array{l} = 1/2 -1/2× 1/2 \\\\ = 1/2 -1/4 \\\\ = 1/4 array",
+    "knowledge_points": [
+      "单位\"1\"的认识与确定",
+      "求一个数的几分之几的问题",
+      "异分母分数加",
+      "减法"
+    ],
+    "source": "20-21六年级上.黑龙江齐齐哈尔·期中",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_013",
+    "original_id": "math_md_q_013"
+  },
+  {
+    "id": "math_md_q_014",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "简单",
+    "question": "爸爸身高180厘米，亮亮的身高是爸爸的 2/3 ，亮亮身高多少厘米？正确列式是（）。",
+    "options": [
+      "A. 180÷ 2/3",
+      "B. 180× 2/3",
+      "C. 2/3 = 180"
+    ],
+    "answer": "B",
+    "solution": "求亮亮身高的正确列式是 180× 2/3 故答案为：B",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": "21-22五年级下·河南商丘·期末",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_014",
+    "original_id": "math_md_q_014"
+  },
+  {
+    "id": "math_md_q_015",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "简单",
+    "question": "联想A型电脑原价4500元，现价比原价降低了 1/10 ，降低了（）元。",
+    "options": [
+      "A.4050",
+      "B.450",
+      "C.45"
+    ],
+    "answer": "B",
+    "solution": "4500× 1/10 = 450 （元）故答案为：B",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": "21-22五年级下·河南商丘·期末",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_015",
+    "original_id": "math_md_q_015"
+  },
+  {
+    "id": "math_md_q_016",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一本书有60页，乐乐第一天看了它的 3/5 ，第2天从（）页看起。",
+    "options": [
+      "A.36",
+      "B.37",
+      "C.38"
+    ],
+    "answer": "B",
+    "solution": "60× 3/5 +1 = 36 + 1 = 37 （页）故答案为：B",
+    "knowledge_points": [
+      "分数与整数的除法",
+      "求一个数的几分之几的问题"
+    ],
+    "source": "20-21六年级上·湖北武汉·期中",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_016",
+    "original_id": "math_md_q_016"
+  },
+  {
+    "id": "math_md_q_017",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一桶油重 5kg ，倒出 1/4 后又灌进 1/4kg ，这时桶里的油还有（）。",
+    "options": [
+      "A.4kg",
+      "B.5kg",
+      "C. 21/4 kg",
+      "D. 15/4 kg"
+    ],
+    "answer": "A",
+    "solution": "5 - 5× 1/4 +1/4 = 5 - 11/4 +1/4 = 4(kg) 故答案为：A",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": "20-21六年级上·贵州安顺·期末",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_017",
+    "original_id": "math_md_q_017"
+  },
+  {
+    "id": "math_md_q_018",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "椅子的单价是桌子的 1/5 。就是说买一张桌子的钱可以买（ ）把椅子。",
+    "options": [
+      "A.4B.5C.6"
+    ],
+    "answer": "B",
+    "solution": "由分析可知：桌子的单价是椅子的单价的5倍，即买一张桌子的钱可以买5把椅子。故答案为：B",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": "21-22六年级上·广西梧州·期末",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_018",
+    "original_id": "math_md_q_018"
+  },
+  {
+    "id": "math_md_q_019",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "（ ）千克比40千克少 1/5",
+    "options": [
+      "A.32",
+      "B.30",
+      "C.48"
+    ],
+    "answer": "A",
+    "solution": "40× (1 - 1/5) = 40× 4/5 = 32 （千克）",
+    "knowledge_points": [
+      "求比一个数多少/几分之几的数是多少"
+    ],
+    "source": "21-22六年级上·黑龙江黑河·期末",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_019",
+    "original_id": "math_md_q_019"
+  },
+  {
+    "id": "math_md_q_020",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "依依家上个月的电费是36元，这个月比上个月节省 1/5 ，这个月的电费是多少元？下面列式正确的是（ ）。",
+    "options": [
+      "A. 36× 1/5",
+      "B. 36× (1 + 1/5)",
+      "C. 36× (1 - 1/5)"
+    ],
+    "answer": "C",
+    "solution": "36× (1 - 1/5) = 36× 4/5 = 28.8 （元）故答案为：C",
+    "knowledge_points": [
+      "求比一个数多少/几分之几的数是多少",
+      "分数与整数的乘法"
+    ],
+    "source": "20-21六年级上·辽宁本溪·期中",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_020",
+    "original_id": "math_md_q_020"
+  },
+  {
+    "id": "math_md_q_021",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "《庄子·杂篇·天下》中有一句话：“一尺之棰，日取其半，万世不竭。”意思就是一尺长的木棒，每日截取它的一半，永远截不完。第一天我们取走的是这根木棒的一半，第二天取剩下部分的一半，第三天再取剩下部分的一半，……那么第五天取走的长度是这跟木棒的（ ）。",
+    "options": [
+      "A. 1/4",
+      "B. 1/8",
+      "C. 1/16",
+      "D. 1/32"
+    ],
+    "answer": "D",
+    "solution": "1× 1/2× 1/2× 1/2× 1/2 array{rl} & {= 1/4× 1/2× 1/2× 1/2}\\\\ & {= 1/8× 1/2× 1/2}\\\\ & {= 1/8× 1/2}\\\\ & {= 1/8} array 故答案为：D",
+    "knowledge_points": [
+      "连续求一个数的几分之几是多少的问题"
+    ],
+    "source": "21-22五年级下·河南三门峡·期末",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_021",
+    "original_id": "math_md_q_021"
+  },
+  {
+    "id": "math_md_q_022",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "期末）8米长的绳子增加它的 1/8 后，再减去 1/8 米，结果是（）米。",
+    "options": [
+      "A. 87/8B .8C. 7/64",
+      "D. 7/8"
+    ],
+    "answer": "A",
+    "solution": "8× (1 + 1/8) - 1/8 = 8× 9/8 - 1/8 = 9 - 1/8 = 87/8 （米）故答案为：A",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "分数的意义",
+      "异分母分数加",
+      "减法"
+    ],
+    "source": "20-21五年级下·福建厦门",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_022",
+    "original_id": "math_md_q_022"
+  },
+  {
+    "id": "math_md_q_023",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "5千克棉花的 1/6 和1千克铁的 5/6 相比较，结果是（）。",
+    "options": [
+      "A.5千克棉花的 1/6 重",
+      "B.1千克铁的 1/6 重C.一样重D.无法比较"
+    ],
+    "answer": "C",
+    "solution": "5× 1/6 = 5/6 （千克） 1× 5/6 = 5/6 （千克） 5/6 千克 = 5/6 千克故答案为：C",
+    "knowledge_points": [
+      "求一个数的几分之几的问题",
+      "分数与整数的乘法"
+    ],
+    "source": "18-19六年级上：全国·单元测试",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_023",
+    "original_id": "math_md_q_023"
+  },
+  {
+    "id": "math_md_q_024",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一袋水果糖，吃掉它的 1/5 ，再增加余下质量的 1/5 ，现在这袋糖的质量（）原来水果糖的质量。",
+    "options": [
+      "A.大于",
+      "B.小于",
+      "C.等于",
+      "D.无法比较"
+    ],
+    "answer": "B",
+    "solution": "1 1/5 = 4/5 4.1 4 5 5 25 4 4 5+25 20 4 25 25 24 25 24 25 所以现在这袋糖的质量小于原来水果糖的质量。故答案为：B",
+    "knowledge_points": [
+      "单位\"1\"的认识与确定",
+      "求一个数的几分之几的问题",
+      "分数乘分数",
+      "异分母分数加",
+      "减法"
+    ],
+    "source": "20-21六年级上：湖南永州期中",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_024",
+    "original_id": "math_md_q_024"
+  },
+  {
+    "id": "math_md_q_025",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "5/6 × 17/19 × 12 × 38 = (5/6 × 12) × (17/19 × 38) 应用了（）。",
+    "options": [
+      "A.乘法交换律与结合律",
+      "B.乘法交换律与分配律",
+      "C.乘法结合律与分配律",
+      "A.23",
+      "B.24",
+      "C.25",
+      "D.26"
+    ],
+    "answer": "C",
+    "solution": "144× 1/6 +1 = 24 + 1 = 25 （页）故答案为：C。",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": "2022六年级下·全国·专题练习",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_025",
+    "original_id": "math_md_q_025"
+  },
+  {
+    "id": "math_md_q_026",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一个养殖场养鸡800只，养的鸭比鸡的只数多 3/5 ，养鸭多少只？下列列式错误的是（）。",
+    "options": [
+      "A. 800× (1 + 3/5)",
+      "B. 800 + 800× 3/5",
+      "C. 800× (1 - 3/5)"
+    ],
+    "answer": "C",
+    "solution": "一个养殖场养鸡800只，养的鸭比鸡的只数多 3/5 ，养鸭多少只？ 列式正确的为： 800× (1 + 3/5) 800 + 800× 3/5 列式 800× (1 - 3/5) 是错误的。故答案为：C",
+    "knowledge_points": [
+      "求比一个数多少几分之几的数是多少"
+    ],
+    "source": "18-19六年级下·四川遂宁·期末",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_026",
+    "original_id": "math_md_q_026"
+  },
+  {
+    "id": "math_md_q_027",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "五（2）班男生人数的和女生人数的相等，那么男生人数（）女生人数。",
+    "options": [
+      "A.大于",
+      "B.小于",
+      "C.等于"
+    ],
+    "answer": "A",
+    "solution": "分析可知，男生人数 × 1/9 = 女生人数 × 1/7 因为 1/9 < 1/7 ，所以男生人数 > 女生人数故答案为：A",
+    "knowledge_points": [
+      "同分子分数的大小比较",
+      "求一个数的几分之几的问题",
+      "因数和积的大小关系（小数乘法）"
+    ],
+    "source": "16-17六年级上·山东青岛·期中",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_027",
+    "original_id": "math_md_q_027"
+  },
+  {
+    "id": "math_md_q_028",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "红、黄两根彩带各长 1/2 米，红彩带用去 1/2 ，黄彩带用去 1/3 米。剩下的红彩带长度（）剩下的黄彩带长度。",
+    "options": [
+      "A.大于",
+      "B.等于",
+      "C.小于"
+    ],
+    "answer": "A",
+    "solution": "红彩用去： 1/2× 1/2 = 1/4 （米）因为 1/4 < 1/3 ，所以剩下的红彩带长度大于剩下的黄彩带长度。故选：A。",
+    "knowledge_points": [
+      "异分母异分子分数的大小比较",
+      "分数乘分数"
+    ],
+    "source": "21-22五年级下·全国·单元测试",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_028",
+    "original_id": "math_md_q_028"
+  },
+  {
+    "id": "math_md_q_029",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "比40的 5/8 少4的数是（）。 二、填空题",
+    "options": [
+      "A.25",
+      "B.21",
+      "C.23"
+    ],
+    "answer": "B",
+    "solution": "40× 5/8 - 4 = 25 - 4 = 21 故答案为：B",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": "22-23六年级上·湖南长沙·期末",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_029",
+    "original_id": "math_md_q_029"
+  },
+  {
+    "id": "math_md_q_030",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一面墙的面积是 27m^2 ，已经刷完了这面墙的 1/3 ，还剩这面墙的（ )没有刷，已经刷完的面积是( )m2。 32，（21- 22六年级上- 山东济南·期末）一个平行四边形的底是 4.2cm ，高是底的 1/3 ，这个平行四边形的面积是（ cm^2",
+    "options": [],
+    "answer": "5.88",
+    "solution": "4.2× 1/3× 4.2 = 1.4× 4.2 = 5.88 （cm²）所以这个平行四边形的面积是 5.88cm^2",
+    "knowledge_points": [
+      "分数乘小数",
+      "平行四边形的面积"
+    ],
+    "source": "22-23六年级上·广东珠海·期末",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_030",
+    "original_id": "math_md_q_030"
+  },
+  {
+    "id": "math_md_q_031",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "把 7/8m 长的铁丝平均分成5段，每段占全长的（ ），两段长（ ）m。",
+    "options": [],
+    "answer": "1/5 7/20",
+    "solution": "1÷ 5 = 1/5 array{l}{7/8× 1/5× 2}\\\\ {= 7/40× 2}\\\\ {= 7/20(m)} array 所以，每段占全长的 1/5 ，两段长 7/20m",
+    "knowledge_points": [
+      "分数的连乘运算",
+      "分数的意义"
+    ],
+    "source": "12-13六年级·天津南开·阶段练习",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_031",
+    "original_id": "math_md_q_031"
+  },
+  {
+    "id": "math_md_q_032",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "简单",
+    "question": "正方形的边长是 2/5 米，周长是（ ）米，面积是（ ）平方米。",
+    "options": [],
+    "answer": "8/5 4/25",
+    "solution": "2/5× 4 = 8/5 （米） 2/5× 2/5 = 4/25 （平方米）",
+    "knowledge_points": [
+      "正方形的周长",
+      "分数与整数的乘法",
+      "分数乘分数",
+      "正方形的面积"
+    ],
+    "source": "2014三年级·全国·课后作业",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_032",
+    "original_id": "math_md_q_032"
+  },
+  {
+    "id": "math_md_q_033",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "简单",
+    "question": "一根2米长的绳子，用去 3/4 米，还剩（ ）米；如果用去 3/4 ，还剩（ ）米。",
+    "options": [],
+    "answer": "11/4 1/2",
+    "solution": "2 - 3/4 = 1 1/4 （米） 2 - 2 × 3/4 = 4/2 - 3/2 = 1/2 （米）",
+    "knowledge_points": [
+      "同分母分数加",
+      "减法",
+      "求一个数的几分之几的问题",
+      "分数与整数的乘法",
+      "分数的意义"
+    ],
+    "source": "2014三年级·全国·课后作业",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_033",
+    "original_id": "math_md_q_033"
+  },
+  {
+    "id": "math_md_q_034",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "已知a，b，c都大于0，如果 4/7 × 3/8 × 3/8 = 3/8 × 1/2 ，那么a，b，c按照从大到小的顺序排列应是（ ）。",
+    "options": [],
+    "answer": "b > a > c",
+    "solution": "假设 4/7 × 3/8 × 3/8 = 3/8 × 1/2 = 1 则 4/7 × 3/8 = 1 ， a = 7/4 3/8 × b = 1 ， b = 8/3 c × 1 = 1 ， c = 1 8/3 > 7/4 > 1 所以 b > a > c",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "分数乘分数"
+    ],
+    "source": "22-23六年级上：全国-单元测试",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_034",
+    "original_id": "math_md_q_034"
+  },
+  {
+    "id": "math_md_q_035",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "一本故事书270页，第一天看了 2/9 ，第二天应该从第（ ）页看起。",
+    "options": [],
+    "answer": "61",
+    "solution": "270 × 2/9 = 60 （页） 60 + 1 = 61 （页）",
+    "knowledge_points": [
+      "分数与整数的乘法"
+    ],
+    "source": "22-23六年级上：全国-单元测试",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_035",
+    "original_id": "math_md_q_035"
+  },
+  {
+    "id": "math_md_q_036",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "简单",
+    "question": "一桶油重10千克， 2/5 桶油重（ ）千克，算式是（ ）。",
+    "options": [],
+    "answer": "4 10 × 2/5 × 1/5 × 10",
+    "solution": "10 × 2/5 = 4 （千克）所以，一桶油重10千克， 2/5 桶油重4千克，算式是 10 × 2/5 或 2/5 × 10 。",
+    "knowledge_points": [
+      "求一个数的几分之几的问题",
+      "分数与整数的乘法"
+    ],
+    "source": "22-23六年级上：全国-单元测试",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_036",
+    "original_id": "math_md_q_036"
+  },
+  {
+    "id": "math_md_q_037",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "1/25 × 1/8 × 25 = (",
+    "options": [],
+    "answer": "1/25 25",
+    "solution": "1/25 × 1/8 × 25 = 1/25 × 25 × 1/8",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "分数乘分数",
+      "分数乘法运算律"
+    ],
+    "source": "22-23六年级上：全国-单元测试",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_037",
+    "original_id": "math_md_q_037"
+  },
+  {
+    "id": "math_md_q_038",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "某冰箱原来每台售价3600元，现在每台售价比原来降低了 1/6 。 (1)应该把( )看作单位\"1\";(2) 3600 × 1/6 表示( );(3) 1 - 1/6 表示( );(4) 3600 × 5/6 表示( )。",
+    "options": [],
+    "answer": "(1)冰箱原来每台售价(2)求现在比原来降低的价钱(3)求现价是原价的几分之几(4)求现在每台多少元",
+    "solution": "（1）应该把冰箱原来每台售价看作单位\"1\"； （2） 3600 × 1/6 表示求现在比原来降低的价钱；（3） 1 - 1/6 表示求现价是原价的几分之几；（4） 3600 × 5/6 表示求现在每台多少元；",
+    "knowledge_points": [
+      "单位\"1\"的认识与确定",
+      "分数与整数的乘法",
+      "分数的意义"
+    ],
+    "source": "22-23六年级上：全国-单元测试",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_038",
+    "original_id": "math_md_q_038"
+  },
+  {
+    "id": "math_md_q_039",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "13/10 平方米 = 平方分米 4/15 时 = ( )分",
+    "options": [],
+    "answer": "130 16",
+    "solution": "（1） 13/10 × 100 = 130 （平方分米）(2) 4/15 × 60 = 16 （分）",
+    "knowledge_points": [
+      "时",
+      "分的认识及换算",
+      "分数与整数的乘法",
+      "平方厘米",
+      "平方分米",
+      "平方米之间的进率与换算"
+    ],
+    "source": "22-23六年级上：全国-单元测试",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_039",
+    "original_id": "math_md_q_039"
+  },
+  {
+    "id": "math_md_q_040",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "80 5/16 是( )； 4/21 的 21/100 是( )。",
+    "options": [],
+    "answer": "25 0.04",
+    "solution": "80 × 5/16 = 25 4/21 × 21/100 = 0.04",
+    "knowledge_points": [
+      "求一个数的几分之几的问题",
+      "分数与整数的乘法",
+      "分数乘分数"
+    ],
+    "source": "22-23六年级上：全国-单元测试",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_040",
+    "original_id": "math_md_q_040"
+  },
+  {
+    "id": "math_md_q_041",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "2/3 时 = ( )分； 2/5 升 = ( )毫升。",
+    "options": [],
+    "answer": "40 400",
+    "solution": "2/3 × 60 = 40 （分），所以 2/3 时 = 40 分； 2/5 × 1000 = 400 （毫升），所以 2/5 升 = 400 毫升",
+    "knowledge_points": [
+      "容积单位间的进率与换算（升和毫升）",
+      "时",
+      "分的认识及换算",
+      "分数与整数的乘法"
+    ],
+    "source": "21-22五年级下：河南商丘期末",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_041",
+    "original_id": "math_md_q_041"
+  },
+  {
+    "id": "math_md_q_042",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "有一条3米长的丝带，先用去了它的 1/3 ，又用去 1/3 米，还剩（ ）米。",
+    "options": [],
+    "answer": "12/35/3",
+    "solution": "3 - 3× 1/3 - 1/3 = 3 - 1 - 1/3 = 2 - 1/3 （米）",
+    "knowledge_points": [
+      "求一个数的几分之几的问题",
+      "分数的意义"
+    ],
+    "source": "2022·广西柳州·小升初真题",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_042",
+    "original_id": "math_md_q_042"
+  },
+  {
+    "id": "math_md_q_043",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "一根电线长10米，用去 3/5 ，还剩（ ）米，再用去 3/5 ，还剩（ ）米。",
+    "options": [],
+    "answer": "4 8/5",
+    "solution": "10× (1 - 3/5) = 10× 2/5 = 4 （米） 4 - 4× 5/12 = 4 - 12/5 = 8/5 （米）",
+    "knowledge_points": [
+      "求一个数的几分之几的问题",
+      "分数与整数的乘法"
+    ],
+    "source": "21-22六年级上·云南红河（期末",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_043",
+    "original_id": "math_md_q_043"
+  },
+  {
+    "id": "math_md_q_044",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "3/4 小时 = ( ）分 225公顷 = ( ）平方千米",
+    "options": [],
+    "answer": "45 2.25",
+    "solution": "3 小时 = 3/4× 60 分 = 45 分；225公顷 = 225÷ 100 平方千米 = 2.25 平方千米",
+    "knowledge_points": [
+      "时",
+      "分的认识及换算",
+      "除数是整数的小数除法",
+      "分数与整数的乘法",
+      "公顷",
+      "平方千米的进率与换算"
+    ],
+    "source": "21-22六年级下·湖北十堰·期中",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_044",
+    "original_id": "math_md_q_044"
+  },
+  {
+    "id": "math_md_q_045",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "一堆煤吨，第一次用去 1/10 吨，还剩（ ）吨，第二次用去剩下的 2/7 ，还剩（ ）吨。",
+    "options": [],
+    "answer": "7/10 1/2",
+    "solution": "44/5 - 1/10 8/10 - 1/10 array{r l} & {= 7/10 (H)}\\\\ & {7/10 - 7/10× 2/7}\\\\ & {= 7/10 - 2/10}\\\\ & {= 1/2 (H)} array",
+    "knowledge_points": [
+      "求一个数的几分之几的问题",
+      "分数的意义",
+      "异分母分数加",
+      "减法"
+    ],
+    "source": "21-22六年级上·江西吉安·期末",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_045",
+    "original_id": "math_md_q_045"
+  },
+  {
+    "id": "math_md_q_046",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "六（1）班女生人数比男生人数少 5/9 ，就是说( )的人数 × 5/9 = ( )的人数。",
+    "options": [],
+    "answer": "男生 女生比男生少",
+    "solution": "由分析可知：男生的人数 × 5/9 = 女生比男生少的人数。",
+    "knowledge_points": [
+      "求比一个数多少/几分之几的数是多少"
+    ],
+    "source": "21-22六年级上·广西梧州期末",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_046",
+    "original_id": "math_md_q_046"
+  },
+  {
+    "id": "math_md_q_047",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "简单",
+    "question": "4.5m 增加它的 3/5 是( )； 15/16 减少它的 1/5 是( )t。",
+    "options": [],
+    "answer": "7.2 3/4",
+    "solution": "4.5× (1 + 3/5) = 4.5× 2/5 = 7.2(m) array{rl} & {15/16× (1 - 1/5)}\\\\ & {= 15/16× 4/5}\\\\ & {= 3/4 (t)} array",
+    "knowledge_points": [
+      "求比一个数多少/几分之几的数是多少"
+    ],
+    "source": "19-20六年级上·黑龙江黑河·期中",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_047",
+    "original_id": "math_md_q_047"
+  },
+  {
+    "id": "math_md_q_048",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个正方形的边长是 4/5 分米，它的面积是( )，周长是( )。",
+    "options": [],
+    "answer": "16/25 平方分米 16/5 分米",
+    "solution": "4/5 = 16/5 = 16/25 （平方分米） 4/5× 4 = 16/5 （分米）",
+    "knowledge_points": [
+      "正方形的周长",
+      "分数与整数的乘法",
+      "分数乘分数",
+      "正方形的面积"
+    ],
+    "source": "20-21五年级下·福建厦门·期末",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_048",
+    "original_id": "math_md_q_048"
+  },
+  {
+    "id": "math_md_q_049",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "40分 = ( )时； 3/8 千克 = ( )克",
+    "options": [],
+    "answer": "2/3 375",
+    "solution": "（1）40分 = 40÷ 60 = 2/3 时（2） 3/8 千克 = 3/8× 1000 克 = 375 克。",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "分数与除法的关系",
+      "约分的认识及应用",
+      "克",
+      "千克之间的换算与比较"
+    ],
+    "source": "20-21六年级上·浙江绍兴·期中",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_049",
+    "original_id": "math_md_q_049"
+  },
+  {
+    "id": "math_md_q_050",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "六（1）班有50人，女生占全班人数的 2/5 ，女生有（ ）人，男生有（ ）人。",
+    "options": [],
+    "answer": "20 30",
+    "solution": "50× 2/3 = 20 （人） 50 - 20 = 30 （人） 所以，女生有20人，男生有30人。",
+    "knowledge_points": [
+      "求一个数的几分之几的问题",
+      "分数与整数的乘法"
+    ],
+    "source": "15-16六年级上·江苏苏州·单元测试",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_050",
+    "original_id": "math_md_q_050"
+  },
+  {
+    "id": "math_md_q_051",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "4个 3/16 相加，用乘法表示是 3/4 公顷的 2/3 ，用乘法表示是 2/3 ；用图形表示是 3/16× 4/3 图见详解",
+    "options": [],
+    "answer": "3/16× 4 3/4× 2/3 图见详解",
+    "solution": "4个 3/16 相加用乘法表示是 2/16× 4/3 3/4 公顷的 2/3 ，用乘法表示是 3/4× 2/3 ，用图形表示如下： <table><tr><td></td><td></td><td></td></tr><tr><td></td><td></td><td></td></tr><tr><td></td><td></td><td></td></tr><tr><td></td><td></td><td></td></tr></table>",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "分数乘分数",
+      "分数的意义"
+    ],
+    "source": "20-21六年级上·浙江宁波·期中",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_051",
+    "original_id": "math_md_q_051"
+  },
+  {
+    "id": "math_md_q_052",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "12个 5/6 的和是（ ），比 12kg 多 5/6kg 是（ ）kg。",
+    "options": [],
+    "answer": "10 125/6",
+    "solution": "（1） 12× 5/6 = 10 （2） 12 + 5/6 = 125/6 (kg)",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "异分母分数加",
+      "减法"
+    ],
+    "source": "21-22五年级下·河南商丘·阶段练习",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_052",
+    "original_id": "math_md_q_052"
+  },
+  {
+    "id": "math_md_q_053",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "晶晶画了一个平行四边形，它的高是 4/5dm ，底是高的 1/4 。这个平行四边形的面积是（ ） dm^2 。",
+    "options": [],
+    "answer": "4/25 /0.16",
+    "solution": "4/5× 1/4 = 1/5 （dm） 1/5 × 4/5 = 4/25 (dm2)",
+    "knowledge_points": [
+      "平行四边形的面积",
+      "分数乘分数"
+    ],
+    "source": "2022六年级下·全国·专题练习",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_053",
+    "original_id": "math_md_q_053"
+  },
+  {
+    "id": "math_md_q_054",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "简单",
+    "question": "10吨50千克 = ( )吨 43/5 小时 = ( )分",
+    "options": [],
+    "answer": "10.05 276",
+    "solution": "50千克 = 0.05 吨，所以10吨50千克 = 10.05 吨； 43/5 小时 = 23/5 小时 = 276 分。",
+    "knowledge_points": [
+      "时",
+      "分的认识及换算",
+      "吨",
+      "千克之间的换算与比较",
+      "分数与整数的乘法"
+    ],
+    "source": "20-21六年级下·浙江台州·期末",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_054",
+    "original_id": "math_md_q_054"
+  },
+  {
+    "id": "math_md_q_055",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "20千克油用去千克油后，还剩( )千克；20千克油用去 3/4 后，还剩( )千克。",
+    "options": [],
+    "answer": "191/4 5",
+    "solution": "（1） 20 - 3/4 = 191/4 （千克）(2) 20× (1 - 3/4) = 20× 1/4 = 5 （千克）",
+    "knowledge_points": [
+      "求一个数的几分之几的问题",
+      "异分母分数加",
+      "减法"
+    ],
+    "source": "21-22六年级上·湖北随州·期末",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_055",
+    "original_id": "math_md_q_055"
+  },
+  {
+    "id": "math_md_q_056",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "有10吨煤，第一次用去 1/2 ，第二次用去 1/2 吨，还剩下( )吨煤。",
+    "options": [],
+    "answer": "4.5",
+    "solution": "10 - 10× 1/2 - 1/2 = 10 - 5 - 0.5 = 4.5 （吨）",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "分数的意义"
+    ],
+    "source": "21-22六年级上·贵州黔东南·期末",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_056",
+    "original_id": "math_md_q_056"
+  },
+  {
+    "id": "math_md_q_057",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一块菜地的面积是630平方米，其中的 4/7 种菜心。种白菜的面积是种菜心面积的 1/6 种白菜的面积是( )平方米。",
+    "options": [],
+    "answer": "60",
+    "solution": "由分析得， 630× 4/7× 1/6 = 360× 1/6 = 60 （平方米）",
+    "knowledge_points": [
+      "连续求一个数的几分之几是多少的问题",
+      "分数与整数的乘法"
+    ],
+    "source": "20-21六年级上·广东广州·期末",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_057",
+    "original_id": "math_md_q_057"
+  },
+  {
+    "id": "math_md_q_058",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "4.02立方米 = ( )立方米( )立方分米， 3/4 立方米 = ( )升。",
+    "options": [],
+    "answer": "4 20 750",
+    "solution": "0.02× 1000 = 20 （立方分米），所以4.02立方米 = 4 立方米20立方分米； 3/4× 1000 = 750 （升），所以立方米 = 750 升。",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "小数与整数的乘法",
+      "体积与容积单位间的进率及换算"
+    ],
+    "source": "19-20六年级下·湖南怀化·单元测试",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_058",
+    "original_id": "math_md_q_058"
+  },
+  {
+    "id": "math_md_q_059",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一项工程20天完成，每天完成全部工程的( )，9天完成这项工程的( )。",
+    "options": [],
+    "answer": "1/20 9/20",
+    "solution": "1÷ 20 = 1/20 1/20× 9 = 9/20",
+    "knowledge_points": [
+      "工程问题",
+      "分数与整数的乘法",
+      "分数与除法的关系"
+    ],
+    "source": "19-20五年级下·湖南怀化·单元测试",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_059",
+    "original_id": "math_md_q_059"
+  },
+  {
+    "id": "math_md_q_060",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一根绳子长 5/4m ，剪去 3/4m ，还剩( )m；若把这根绳子剪去它的 1/5 ，还剩( )m。",
+    "options": [],
+    "answer": "1/2 1",
+    "solution": "5/4 = 3/4 = 1/2 （米） 5/4× (1 - 1/5) 5/4× 4/5 = 1 （米）",
+    "knowledge_points": [
+      "同分母分数加",
+      "减法",
+      "求一个数的几分之几的问题",
+      "分数乘分数"
+    ],
+    "source": "21-22六年级上·湖南怀化·单元测试",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_060",
+    "original_id": "math_md_q_060"
+  },
+  {
+    "id": "math_md_q_061",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一本故事书180页，小明第一天看了全书的 1/3 ，还剩( )页没有看。",
+    "options": [],
+    "answer": "120",
+    "solution": "180× (1 - 1/3) = 180× 2/3 = 120 （页）",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": "21-22六年级上·湖南怀化·单元测试",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_061",
+    "original_id": "math_md_q_061"
+  },
+  {
+    "id": "math_md_q_062",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个正方形的边长是 3/4 米，它的周长是( )米，面积是( )平方米。",
+    "options": [],
+    "answer": "3 9/16",
+    "solution": "3/4× 4 = 3 （米） 3/4× 3/4 = 9/16 （平方米）",
+    "knowledge_points": [
+      "正方形的周长",
+      "分散与整数的乘法",
+      "分数乘分数",
+      "正方形的面积"
+    ],
+    "source": "21-22六年级上·黑龙江齐齐哈尔·期末",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_062",
+    "original_id": "math_md_q_062"
+  },
+  {
+    "id": "math_md_q_063",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "一条绳子长9米，第一次用去了 1/3 ，第二次用去了 1/3 ，第二次用去了 1/3 ，还剩（ ）米。 三、判断题",
+    "options": [],
+    "answer": "52/3",
+    "solution": "9 - 9× 1/3 - 1/3 = 9 - 3 - 1/3 = 52/3 （米）",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": "18-19五年级下·山西吕梁·期中",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_063",
+    "original_id": "math_md_q_063"
+  },
+  {
+    "id": "math_md_q_064",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "一件商品先涨价 1/10 ，再降价 1/10 ，现在的价格和涨价前的价格一样多。（ ）",
+    "options": [],
+    "answer": "×",
+    "solution": "假设商品的原价为1： 1× (1 + 1/10)× (1 - 1/10) = 1× 1/10× 9/10 = 99/100 < 1 ，所以现在的价格比涨价前的价格低。故答案为： ×",
+    "knowledge_points": [
+      "单位“1”的认识与确定",
+      "求比一个数多/少几分之几的数是多少"
+    ],
+    "source": "19-20六年级上·全国·期中",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_064",
+    "original_id": "math_md_q_064"
+  },
+  {
+    "id": "math_md_q_065",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "3吨的 1/8 和1吨的 2/8 同样重。（ ）",
+    "options": [],
+    "answer": "√",
+    "solution": "3× 1/8 = 3/8 （吨） 1× 3/8 = 3/8 （吨）所以，3吨的 1/8 和1吨的 3/8 同样重。故答案为： {10}",
+    "knowledge_points": [
+      "分数与整数的乘法"
+    ],
+    "source": "19-20六年级上·全国·期末",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_065",
+    "original_id": "math_md_q_065"
+  },
+  {
+    "id": "math_md_q_066",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "简单",
+    "question": "一个自然数乘真分数，积小于这个数。（ ）",
+    "options": [],
+    "answer": "×",
+    "solution": "0 × 1/2 = 0 0 = 0 ，积与这个自然数相等。所以一个自然数乘真分数，积有可能小于也有可能等于这个自然数。故答案为： ×",
+    "knowledge_points": [
+      "自然数的认识",
+      "因数和积的大小关系（小数乘法）",
+      "分数与整数的乘法",
+      "真分数",
+      "假分数",
+      "带分数的认识"
+    ],
+    "source": "15-16六年级·全国·课后作业",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_066",
+    "original_id": "math_md_q_066"
+  },
+  {
+    "id": "math_md_q_067",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "简单",
+    "question": "求6个是多少，就是求的6倍是多少。（ ）",
+    "options": [],
+    "answer": "√",
+    "solution": "求6个是多少以及求的6倍是多少，都是在求6个的和，列式均为 6 × 2/7 故答案为：√",
+    "knowledge_points": [
+      "分数与整数的乘法"
+    ],
+    "source": "22-23六年级上：全国·单元测试",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_067",
+    "original_id": "math_md_q_067"
+  },
+  {
+    "id": "math_md_q_068",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "9 - 6 × 1/3 = 3 × 1/3 = 1 。（ ）",
+    "options": [],
+    "answer": "×",
+    "solution": "9 - 6 × 1/3 = 9 - 2 = 7",
+    "knowledge_points": [
+      "分数的四则混合运算",
+      "分数与整数的乘法"
+    ],
+    "source": "22-23六年级上：全国·单元测试",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_068",
+    "original_id": "math_md_q_068"
+  },
+  {
+    "id": "math_md_q_069",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "0.5米的 1/5 和0.7米的 1/7 一样长。（ ）",
+    "options": [],
+    "answer": "√",
+    "solution": "0.5 × 1/5 = 0.1 （米） 0.7 × 1/7 = 0.1 （米） 0.1 = 0.1 ，原题说法正确；故答案为：√",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": "22-23六年级上：全国·单元测试",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_069",
+    "original_id": "math_md_q_069"
+  },
+  {
+    "id": "math_md_q_070",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "1吨的 5/6 和5吨的 1/6 一样重。（ ）",
+    "options": [],
+    "answer": "√",
+    "solution": "1吨的 5/6 1 × 5/6 = 5/6 （吨）5吨的 1/6 5 × 1/6 = 1/6 （吨） 因为吨 5/6 吨，所以1吨的和5吨的 1/6 一样重。故答案为：√",
+    "knowledge_points": [
+      "求一个数的几分之几的问题",
+      "分数与整数的乘法"
+    ],
+    "source": "21-22五年级下·内蒙古呼和浩特·期末",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_070",
+    "original_id": "math_md_q_070"
+  },
+  {
+    "id": "math_md_q_071",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "1吨铁的比5吨棉花的重。（ ）",
+    "options": [],
+    "answer": "×",
+    "solution": "铁： 1× 5/7 = 5/7 （吨）棉花： 5× 1/7 = 5/7 （吨）所以，1吨铁的和5吨棉花的一样重。故答案为： ×",
+    "knowledge_points": [
+      "求一个数的几分之几的问题",
+      "同分母分数的大小比较",
+      "分数与整数的乘法"
+    ],
+    "source": "21-22五年级下·贵州毕节·期中",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_071",
+    "original_id": "math_md_q_071"
+  },
+  {
+    "id": "math_md_q_072",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "整数乘法运算定律对于分数乘法同样适用。（ ）",
+    "options": [],
+    "answer": "√",
+    "solution": "整数乘法运算定律对于分数乘法同样适用。题干说法正确。",
+    "knowledge_points": [
+      "分数乘法运算律"
+    ],
+    "source": "16-17六年级上·全国·课后作业",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_072",
+    "original_id": "math_md_q_072"
+  },
+  {
+    "id": "math_md_q_073",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "简单",
+    "question": "2千克的 2/5 和4千克的 1/5 一样多。（ ）",
+    "options": [],
+    "answer": "√",
+    "solution": "2× 2/5 = 4/5 （千克） 4× 1/5 = 4/5 （千克）因为 2/5 = 4/5 所以2千克的 2/5 和4千克的 1/5 一样多。故答案为：√",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": "18-19六年级下·河北邯郸·期末",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_073",
+    "original_id": "math_md_q_073"
+  },
+  {
+    "id": "math_md_q_074",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "一个数（0除外）乘一个假分数，所得的积一定大于这个数。（ ）",
+    "options": [],
+    "answer": "×",
+    "solution": "一个数（0除外）乘一个假分数，所得的积一定大于这个数，说法错误。故答案为： ×",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "真分数",
+      "假分数",
+      "带分数的认识"
+    ],
+    "source": "20-21六年级上·河南濮阳·阶段练习",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_074",
+    "original_id": "math_md_q_074"
+  },
+  {
+    "id": "math_md_q_075",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "一件商品先降价 1/10 ，再涨价 1/10 ，价格不变。（ ）",
+    "options": [],
+    "answer": "×",
+    "solution": "假设商品的原价为1现价： 1× (1 - 1/10)× (1 + 1/10) array{rl} & {= 9/10× 11/10}\\\\ & {= 100/100}\\\\ & {= 99/100} array 因为 1 99/100 ，所以现价比原价少。故答案为： ×",
+    "knowledge_points": [
+      "单位“1\"的认识与确定",
+      "求比一个数多少几分之几的数是多少",
+      "经济问题"
+    ],
+    "source": "15-16六年级·全国·阶段练习",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_075",
+    "original_id": "math_md_q_075"
+  },
+  {
+    "id": "math_md_q_076",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "简单",
+    "question": "一个自然数（0除外）乘分数，积一定小于这个数，（ ） 四、计算题",
+    "options": [],
+    "answer": "×",
+    "solution": "一个自然数（0除外）乘分数，积可能小于、大于或等于这个自然数，故判断错误。",
+    "knowledge_points": [
+      "因数和积的大小关系（小数乘法）"
+    ],
+    "source": "22- 23六年级上：全国·假期作业",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_076",
+    "original_id": "math_md_q_076"
+  },
+  {
+    "id": "math_md_q_077",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "一个数比49的少18，这个数是多少？",
+    "options": [],
+    "answer": "10",
+    "solution": "49× 4/7 - 18 = 28 - 18 = 10",
+    "knowledge_points": [
+      "分数与整数的乘法"
+    ],
+    "source": "22- 23六年级上：全国·单元测试",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_077",
+    "original_id": "math_md_q_077"
+  },
+  {
+    "id": "math_md_q_078",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "1/3 加上 1/4 的 4/5 和是多少？",
+    "options": [],
+    "answer": "8/15",
+    "solution": "1/4× 4/5 +1/3 array{l}{= 1/5 +1/3}\\\\ {= 8/15} array 五、解答题",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": "22- 23六年级上：全国·单元测试",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_078",
+    "original_id": "math_md_q_078"
+  },
+  {
+    "id": "math_md_q_079",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "简单",
+    "question": "学校为绿化校园，买了柳树和梧桐树。柳树有27棵，梧桐树的棵数是柳树的 2/3 梧桐树有多少棵？",
+    "options": [],
+    "answer": "18棵",
+    "solution": "27× 2/3 = 18 （棵） 答：梧桐树有18棵。",
+    "knowledge_points": [
+      "求一个数的几分之几的问题",
+      "分数与整数的乘法"
+    ],
+    "source": "22- 23六年级上：山西朔州·期末",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_079",
+    "original_id": "math_md_q_079"
+  },
+  {
+    "id": "math_md_q_080",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "妈妈买一件上衣花了260元钱，买裤子的钱是上衣的 14/13 ，买皮鞋的钱是裤子的 10/7 。妈妈买皮鞋花了多少元？",
+    "options": [],
+    "answer": "400元",
+    "solution": "260× 14/13× 10/7 = 280× 10/7 = 400 （元） 答：妈妈买皮鞋花了400元。",
+    "knowledge_points": [
+      "连续求一个数的几分之几是多少的问题",
+      "分数的连乘运算"
+    ],
+    "source": "22-23六年级上·全国·单元测试",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_080",
+    "original_id": "math_md_q_080"
+  },
+  {
+    "id": "math_md_q_081",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "一匹布长56米，做一件衣服用布 23/25 米，现在已经做了这样的衣服50件，还剩布多少米？",
+    "options": [],
+    "answer": "10米",
+    "solution": "56 - 23/25× 50 = 56 - 46 = 10 （米） 答：还剩布10米。",
+    "knowledge_points": [
+      "分数与整数的乘法"
+    ],
+    "source": "22-23六年级上·全国·单元测试",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_081",
+    "original_id": "math_md_q_081"
+  },
+  {
+    "id": "math_md_q_082",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "有一块长方形菜地，长20.4米，宽是长的 3/4 ，这块菜地的面积是多少平方米？",
+    "options": [],
+    "answer": "312.12平方米",
+    "solution": "20.4× (20.4× 3/4) = 20.4× 15.3 = 312.12 （平方米） 答：这块菜地的面积是312.12平方米。",
+    "knowledge_points": [
+      "长方形的面积",
+      "小数与小数的乘法",
+      "分数乘小数"
+    ],
+    "source": "22-23六年级上·全国·单元测试",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_082",
+    "original_id": "math_md_q_082"
+  },
+  {
+    "id": "math_md_q_083",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "仓库里有水泥75吨，已经运出总数的，还剩多少吨还没运？",
+    "options": [],
+    "answer": "45吨",
+    "solution": "75× (1 - 2/5) = 75× 3/5 = 45 （吨） 答：还剩45吨还没运。",
+    "knowledge_points": [
+      "分数与整数的乘法"
+    ],
+    "source": "22-23六年级上全国·单元测试",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_083",
+    "original_id": "math_md_q_083"
+  },
+  {
+    "id": "math_md_q_084",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "实验小学美术组有120人，生物组的人数是美术组的 2/3 ，航模组的人数是生物组的 4/5 ，航模组有多少人？",
+    "options": [],
+    "answer": "64人",
+    "solution": "120× 2/3× 4/5 = 80× 4/5 = 64 （人） 答：航模组有64人。",
+    "knowledge_points": [
+      "连续求一个数的几分之几是多少的问题"
+    ],
+    "source": "20-21六年级上·湖北荆州·期末",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_084",
+    "original_id": "math_md_q_084"
+  },
+  {
+    "id": "math_md_q_085",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "简单",
+    "question": "改革开放40年来，我国拥有的高级科技人才数量比美国多 1/5 ，美国有高级科技人才125万人，我国有多少万高级科技人才？",
+    "options": [],
+    "answer": "150万",
+    "solution": "125× (1 + 1/5) = 125× 6/5 = 150 （万） 答：我国有150万高级科技人才。",
+    "knowledge_points": [
+      "求比一个数多少几分之几的数是多少"
+    ],
+    "source": "21-22六年级上·宁夏固原·期末",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_085",
+    "original_id": "math_md_q_085"
+  },
+  {
+    "id": "math_md_q_086",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "水果店进苹果120箱，运来的梨子比苹果少 1/5 ，运来梨子多少箱？",
+    "options": [],
+    "answer": "96箱",
+    "solution": "120× (1 - 1/5) = 120× 4/5 = 96 （箱）答：运来梨子96箱。",
+    "knowledge_points": [
+      "求比一个数多少/几分之几的数是多少"
+    ],
+    "source": "20-21六年级上·云南楚雄·期末",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_086",
+    "original_id": "math_md_q_086"
+  },
+  {
+    "id": "math_md_q_087",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "学校有420名学生参加“12.4\"国家宪法日的宣传系列活动。其中，宣讲队人数占总人数的 3/5 ，合唱队人数是宣讲队的 1/5 。合唱队有多少名学生？",
+    "options": [],
+    "answer": "84名",
+    "solution": "420× 3/5× 1/3 = 252× 1/3 = 84 （名） 答：合唱队有84名学生。",
+    "knowledge_points": [
+      "单位“1\"的认识与确定",
+      "求一个数的几分之几的问题",
+      "分数与整数的乘法",
+      "分数的连乘运算"
+    ],
+    "source": "21-22六年级上·广西柳州·期末",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_087",
+    "original_id": "math_md_q_087"
+  },
+  {
+    "id": "math_md_q_088",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "同学们要植300棵树，第一天已经植了 2/3 ，其中 2/5 是五年级植的。第一天五年级植了多少棵树？",
+    "options": [],
+    "answer": "80棵",
+    "solution": "300× 2/3 = 80 （棵） 答：第一天五年级植了80棵树。",
+    "knowledge_points": [
+      "连续求一个数的几分之几是多少的问题"
+    ],
+    "source": "19-20六年级上·山东泰安·期末",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_088",
+    "original_id": "math_md_q_088"
+  },
+  {
+    "id": "math_md_q_089",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "海象的寿命大约是40年，海狮的寿命是海象的 3/4 ，海豹的寿命是海狮的 2/3 。海豹的寿命大约是多少年？",
+    "options": [],
+    "answer": "20年",
+    "solution": "40× 3/4 = 30 （年） 30× 2/3 = 20 （年） 答：海豹的寿命大约是20年。",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": "2014·贵州安顺·小升初真题",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_089",
+    "original_id": "math_md_q_089"
+  },
+  {
+    "id": "math_md_q_090",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "一台拖拉机昨天上午耕地4800平方米，下午耕地的面积是上午的 4/5 。这台拖拉机昨天下午耕地多少平方米？",
+    "options": [],
+    "answer": "3840平方米",
+    "solution": "4800× 4/5 = 3840 （平方米）答：这台拖拉机昨天下午耕地3840平方米。",
+    "knowledge_points": [
+      "求一个数的几分之几的问题",
+      "分数与整数的乘法"
+    ],
+    "source": "19-20六年级上·广西梧州·期中",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_090",
+    "original_id": "math_md_q_090"
+  },
+  {
+    "id": "math_md_q_091",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "学校六月份用水56吨，七月份用水量比六月份节约 2/7 。七月份用水多少吨？",
+    "options": [],
+    "answer": "40吨",
+    "solution": "56× (1 - 2/7) = 56× 5/7 = 40 （吨） 答：七月份用水40吨。",
+    "knowledge_points": [
+      "分数与整数的乘法"
+    ],
+    "source": "20-21五年级下·福建厦门期末",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_091",
+    "original_id": "math_md_q_091"
+  },
+  {
+    "id": "math_md_q_092",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "珠海市长隆海洋王国2019年上半年接待游客为560万人，下半年游客量是上半年的 7/8 。2019年长隆海洋王国下半年接待游客多少万人？",
+    "options": [],
+    "answer": "490万人",
+    "solution": "560× 7/8 = 490 （万人） 答：2019年长隆海洋王国下半年接待游客490万人。",
+    "knowledge_points": [
+      "求一个数的几分之几的问题",
+      "分数与整数的乘法"
+    ],
+    "source": "19-20六年级上·广东珠海·期末",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_092",
+    "original_id": "math_md_q_092"
+  },
+  {
+    "id": "math_md_q_093",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "某超市运来400箱方便面，卖出了 2/5 ，卖出多少箱？",
+    "options": [],
+    "answer": "160箱",
+    "solution": "400× 2/5 = 160 （箱） 答：卖出160箱。",
+    "knowledge_points": [
+      "求一个数的几分之几的问题",
+      "分数与整数的乘法"
+    ],
+    "source": "20-21六年级上·北京平谷·期中",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_093",
+    "original_id": "math_md_q_093"
+  },
+  {
+    "id": "math_md_q_094",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "万科小区的楼房中，二居室（指含有二间卧室）的有450套，三居室的套数是二居室的 3/5 ，一居室的套数是三居室的 2/5 ，这个小区里共有一居室多少套？",
+    "options": [],
+    "answer": "180套",
+    "solution": "450× 3/5× 2/3 = 270× 2/3 = 180 （套） 答：这个小区里共有一居室180套。",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": "20-21六年级上·广东·期中",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_094",
+    "original_id": "math_md_q_094"
+  },
+  {
+    "id": "math_md_q_095",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "一列客车和一列货车同时从两地相对开出，4.5小时后相遇，客车每小时行65千米，货车每小时行的路程是客车的 10/13 ，两地间相距多少千米？",
+    "options": [],
+    "answer": "517.5千米",
+    "solution": "(65 + 65 × 10/13) × 4.5 = (65 + 50) × 4.5 = 115 × 4.5 = 517.5 （千米） 答：两地间相距517.5千米。",
+    "knowledge_points": [
+      "求一个数的几分之几的问题",
+      "相遇问题"
+    ],
+    "source": "20-21六年级下·云南大理·期末",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_095",
+    "original_id": "math_md_q_095"
+  },
+  {
+    "id": "math_md_q_096",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "水果店运来梨100千克，运来的橘子的重量比梨多 3/5 ，橘子有多少千克？",
+    "options": [],
+    "answer": "160千克",
+    "solution": "100 × (1 + 3/5) = 100 × 8/5 = 160 （千克）答：橘子有160千克。",
+    "knowledge_points": [
+      "求比一个数多/少几分之几的数是多少"
+    ],
+    "source": "2021·宁夏吴忠·小升初真题",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_096",
+    "original_id": "math_md_q_096"
+  },
+  {
+    "id": "math_md_q_097",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "星星小学举行爱国主义读书活动书信比赛，五年级有165人参加，六年级参加的人数比上年级多 2/11 。六年级有多少名同学参加书信比赛？",
+    "options": [],
+    "answer": "195名",
+    "solution": "165 + 165 × 2/11 = 165 + 30 = 195 （名）答：六年级有195名同学参加书信比赛。",
+    "knowledge_points": [
+      "求比一个数多/少几分之几的数是多少"
+    ],
+    "source": "21-22六年级上·湖南怀化·单元测试",
+    "create_time": "2025-09-25 23:51:42",
+    "merged_id": "math_md_097",
+    "original_id": "math_md_q_097"
+  },
+  {
+    "id": "homework_q_398",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "( 3 /4 - 2 /5 ) × 2 0 = 3 /4 × 2 0 - 2 /5 × 2 0 ，这 是运用了 （ ）。",
+    "options": [
+      "A．乘法交换律 B．乘法结合律 C．乘法分配律"
+    ],
+    "answer": "C",
+    "solution": "( 3 /4 - 2 /5 ) × 2 0 = 3 /4 × 2 0 - 2 /5 × 2 0 = 1 5 - 8 = 7 由上可知，本题利用了乘法分配律简便计算。",
+    "knowledge_points": [
+      "分数乘法运算律"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "create_time": "2025-09-30 01:17:20",
+    "merged_id": "homework_398",
+    "original_id": "homework_q_398"
+  },
+  {
+    "id": "homework_q_399",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "在计算 2 . 4 × 5 /6 时（ ）来计算比较简便。",
+    "options": [
+      "A．化作分数 B．化作小数 C．直接约分再算"
+    ],
+    "answer": "C",
+    "solution": "2 . 4 × 5 /6 = 0 . 4 × 5 /1 = 2 在计算 2 . 4 × 5 /6 时，直接约分再算比较简便。",
+    "knowledge_points": [
+      "分数乘法"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "create_time": "2025-09-30 01:17:20",
+    "merged_id": "homework_399",
+    "original_id": "homework_q_399"
+  },
+  {
+    "id": "homework_q_400",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "当 a > 0 时，比较a+ 3 /4 与a× 3 /4 两算式结果是（ ）。",
+    "options": [
+      "A． a + 3 /4 · 大 B． a × 3 /4 . 大C．相等"
+    ],
+    "answer": "A",
+    "solution": "a+ 3 /4 > a ; ； a × 3 /4 < a ，所以 a + 3 /4 > a × 3 /4",
+    "knowledge_points": [
+      "分数乘法"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "create_time": "2025-09-30 01:17:20",
+    "merged_id": "homework_400",
+    "original_id": "homework_q_400"
+  },
+  {
+    "id": "homework_q_401",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "下列算式中，（ ）与 4 /9 × 3 的结果不相同。",
+    "options": [
+      "A． 4 /9 + 4 /9 + 4 /9 B． 4 /9 × 4 C． 4 /9 × 2 + 4 /9"
+    ],
+    "answer": "B",
+    "solution": "4 /9 × 3 = 4 /3 4 /9 + 4 /9 + 4 /9 = 4 /9 × 3 = 4 /3 , ， 正 确； 4 /9 × 4 = 1 6 /9 错误； 4 /9 × 2 + 4 /9 = 4 /9 X ( 2 + 1 ) \\ = _ 9 ^ 4 × 3 = _ 3 ^ 4 , ，正确。",
+    "knowledge_points": [
+      "分数乘法运算律",
+      "分数与整数的乘法【分析】根据分数乘法的计算方法",
+      "计算出选项中各式的结果。"
+    ],
+    "source": "20-21六年级上·湖南怀化·单元测试",
+    "create_time": "2025-09-30 01:17:20",
+    "merged_id": "homework_401",
+    "original_id": "homework_q_401"
+  },
+  {
+    "id": "homework_q_402",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "3 /5 × 1 2 与 1 2 × 3 /5 是（ ）。",
+    "options": [
+      "A．意义相同 B．计算法则不同 C．意义不同，计算法则相同"
+    ],
+    "answer": "C",
+    "solution": "3 /5 × 1 2 表示 12个 3 /5 的和是多少； 1 2 × 3 /5 ; 表示 12 的 3 /5 是多少，结果都是 3 6 /5 ，两个算式意义不同，计算法则相同。",
+    "knowledge_points": [
+      "分数与整数的乘法"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "create_time": "2025-09-30 01:17:20",
+    "merged_id": "homework_402",
+    "original_id": "homework_q_402"
+  },
+  {
+    "id": "homework_q_403",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "“大羊只数是小羊只数的 3 /8 ，（ ）是单位 \"1\" 。",
+    "options": [
+      "A．小羊 B．大羊 C．无法确定"
+    ],
+    "answer": "A",
+    "solution": "由图可知，把小羊的只数看作单位 \"1\" ，把单位 \"1\",平均分成 8 份，大羊的只数占其中的3 份。",
+    "knowledge_points": [
+      "求一个数的几分之几的问题",
+      "单位\"1\" 的认识与确定"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "create_time": "2025-09-30 01:17:20",
+    "merged_id": "homework_403",
+    "original_id": "homework_q_403"
+  },
+  {
+    "id": "homework_q_404",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "这是根据（ ）计算的。",
+    "options": [
+      "A．乘法交换律 B．加法交换律 C．乘法结合律 D．乘法分配律"
+    ],
+    "answer": "D",
+    "solution": "2 . 4 × ( 1 /4 + 1 /6 ) 由上可知，计算过程利用了乘法分配律：a ( b + c ) = a b + a c . 。故答案为：D",
+    "knowledge_points": [
+      "分数乘法"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "create_time": "2025-09-30 01:17:20",
+    "merged_id": "homework_404",
+    "original_id": "homework_q_404"
+  },
+  {
+    "id": "homework_q_405",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "“小羊只数是大羊只数的 3 /8 ，（ ）是单位 \"1\" ",
+    "options": [
+      "A．小羊只数 B．大羊只数 C．无法确定"
+    ],
+    "answer": "B",
+    "solution": "由上可知，我们把大羊的只数看作单位 “1” 。",
+    "knowledge_points": [
+      "求一个数的几分之几的问题",
+      "单位 ^ \\ s 1 ^ s 的认识与确定【分析】小羊的只数 = 大羊的只数 $× 3 /8",
+      "$",
+      "这里把大羊的只数看作单位 ^ \\ 1 ^ p 。"
+    ],
+    "source": "15-16 六年级上·江苏苏州·单元测试",
+    "create_time": "2025-09-30 01:17:20",
+    "merged_id": "homework_405",
+    "original_id": "homework_q_405"
+  },
+  {
+    "id": "homework_q_406",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "简单",
+    "question": "修路队修一条路，第一天修了全长的1 /6 ，第二天修了剩下的1 /6 米，两天修路长度比（ ）。",
+    "options": [
+      "A．第一天多 B．第二天多 C．一样长D．无法比较"
+    ],
+    "answer": "C",
+    "solution": "假设这条路全长 30米；第一天： 3 0 × 1 /6 = 5 （米）； 第二题：（30－5） × 1 /5 = 2 5 × 1 /5 = 5 （米）；",
+    "knowledge_points": [
+      "求一个数的几分之几的问题【分析】假设这条路全长 30 米",
+      "分别求出两天修的米数",
+      "再选择即可。"
+    ],
+    "source": "20-21 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:20",
+    "merged_id": "homework_406",
+    "original_id": "homework_q_406"
+  },
+  {
+    "id": "homework_q_407",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "王明体重 3 0 k g ，书包重 5kg。儿童的负重最好不要超过体重的3 /20 如果长期背负过重物体，会导致腰痛及背痛，严重的甚至会妨碍骨骼生长，王明的书包（ ）。",
+    "options": [
+      "A．超重 B．不超重C．无法确定"
+    ],
+    "answer": "A",
+    "solution": "3 0 × 3 /2 0 = 9 /2 ( k g ) 5＞?所以王明的书包超重。",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "求一个数的几分之几的问题"
+    ],
+    "source": "14-15 六年级上 $\\bullet$ 全国·课后作业",
+    "create_time": "2025-09-30 01:17:20",
+    "merged_id": "homework_407",
+    "original_id": "homework_q_407"
+  },
+  {
+    "id": "homework_q_408",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一根铁丝长 2 m . ，第一次剪去2 /5 ，第二次剪去 2 /5 m剩（ ）m。 【详解 1 \\ 2 - 2 × 2 /5 - 2 /5 , ，所以剩下的列式为： 2 - 2 × 2 /5 - 2 /5",
+    "options": [
+      "A． 2× （1-2 /5 -2 /5 ）2） B． 2 －2 /5 -2 /5",
+      "C． 2×（1 -2 /5 +2 /5 ) D． 2-2x2 /5 -2 /5"
+    ],
+    "answer": "D",
+    "solution": "",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "求一个数的几分之几的问题【分析】先利用乘法",
+      "求出第一次减去的长度",
+      "再利用减法求出剩下的长度。再将其减去第二次剪去的2米",
+      "求出最终还剩下多少米的铁丝。"
+    ],
+    "source": "21-22 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:20",
+    "merged_id": "homework_408",
+    "original_id": "homework_q_408"
+  },
+  {
+    "id": "homework_q_409",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "如果 a的 3 /1 0 等于 b的 1 /4 （a、 b 都不等于 0），那么比较 a 和 b的大小，结果是（ ）。",
+    "options": [
+      "A． a ≥ b B． b > a C． a = b D．无法确定"
+    ],
+    "answer": "B",
+    "solution": "由题意得， 3 /1 0 a = 1 /4 b （a、 b 都不等于 0） 因为 3 /1 0 > 1 /4 ， ，所以 a < b",
+    "knowledge_points": [
+      "因数和积的大小关系（小数乘法）",
+      "异分母异分子分数的大小比较",
+      "求一个数的几分之几的问题"
+    ],
+    "source": "21-22 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:20",
+    "merged_id": "homework_409",
+    "original_id": "homework_q_409"
+  },
+  {
+    "id": "homework_q_410",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一个长方形花园，长 2 4 m ，宽比长少 1 /4 , ，求面积的算式是（ ）。",
+    "options": [
+      "A． 2 4 × 1 /4 B． 2 4 × ( 1 - 1 /4 ) + 2 4 C"
+    ],
+    "answer": "C",
+    "solution": "求面积的算式是： 2 4 × [ 2 4 × ( 1 - 1 /4 ) ] . 。",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "长方形的面积",
+      "求比一个数多/少几分之几的数是多少【分析】宽比长少 1 /4",
+      "那么宽是长的 3 /4 。据此先利用乘法求出宽",
+      "再根据长方形的面积公式",
+      "列式计算面积即可。"
+    ],
+    "source": "21-22六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:20",
+    "merged_id": "homework_410",
+    "original_id": "homework_q_410"
+  },
+  {
+    "id": "homework_q_411",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "下列算式中，结果最大的是（ ）。",
+    "options": [
+      "A． 1 /3 × 2 /3 B． 2 /3 × 3 /4 C． 4 /3 × 2 /3 D.1- 2 /3"
+    ],
+    "answer": "C",
+    "solution": "A． 1 /3 × 2 /3 ＝ 1 /3 × 2 /3 ＜ 2 /3 ； B． 2 /3 × 3 /4 < 2 /3 C． 4 /3 × 2 /3 = 2 /3 × 4/3 > 2 /3 D．1- 1 /3 = 2 /3",
+    "knowledge_points": [
+      "因数和积的大小关系（小数乘法）",
+      "分数乘分数"
+    ],
+    "source": "21-22 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:20",
+    "merged_id": "homework_411",
+    "original_id": "homework_q_411"
+  },
+  {
+    "id": "homework_q_412",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "简单",
+    "question": "六、1 班同学做好事 120件，（ ）六、2班做好事多少件？列式为 1 2 0 × ~ ( 1 - 1 /3 ) 。",
+    "options": [
+      "A．六、2 班比六、1 班多 1 /3 B ．六、2 班比六、1 班少 1/3 C．六、2 班是六、1 班的  1 3 D ．六、1 班比六、2 班少 1 /3"
+    ],
+    "answer": "B",
+    "solution": "结合具体题意以及所列算式可知：题目里所缺的条件应是：六、2 班比六、1班少 1 /3 1 2 0 × ~ ( 1 - 1 /3 ) ＝120× 1 /3 =80(件）故答案为：B。",
+    "knowledge_points": [
+      "分数的四则混合运算",
+      "“提问题”“填条件”问题【分析】已知六",
+      "1 班同学做好事 120 件",
+      "要求六",
+      "2 班做好事多少件",
+      "且列式为 1 2 0 × ~ ( 1 - 1 /3 ) ）",
+      "这是把六",
+      "1 班做好事的数量看作单位 ^ 1 ^",
+      "由于是在单位 ^ 1 ^ 的基础上减去了 1 /3",
+      "就是说六",
+      "2班比六",
+      "1 班少做了 1 /3 。"
+    ],
+    "source": "18-19 六年级上·四川·期末",
+    "create_time": "2025-09-30 01:17:20",
+    "merged_id": "homework_412",
+    "original_id": "homework_q_412"
+  },
+  {
+    "id": "homework_q_413",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "同样长的绳子，第一根截去它的 4 /5 , ，第二根截去4 /5 米 ，余下的部分（ ）。",
+    "options": [
+      "A．第一根长 B．第二根长 C．同样长D．无法比较"
+    ],
+    "answer": "D",
+    "solution": "分情况讨论： 1. 当绳长大于 1m 时，第一根比第二根截去的长，余下的部分是第二根长； 2. 当绳长等于 1m 时，第一根和第二根截去的一样长，余下的部分也一样长； 3. 当绳长小于 1m 时，第二根比第一根截去的长，余下的部分是第一根长。故答案为：D",
+    "knowledge_points": [
+      "分数的意义"
+    ],
+    "source": "20-21 六年级上·江西赣州·单元测试",
+    "create_time": "2025-09-30 01:17:20",
+    "merged_id": "homework_413",
+    "original_id": "homework_q_413"
+  },
+  {
+    "id": "homework_q_414",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "水结成冰，体积增加 1 /1 1 , 。表示单位 \"1\" 的量是（ ）。",
+    "options": [
+      "A．水的体积 B．冰的体积 C．增加的体积 D．不能确定"
+    ],
+    "answer": "A",
+    "solution": "水结成冰，体积增加 1 /1 1 ，是把水的体积看成单位 \"1\" ， 故选：A",
+    "knowledge_points": [
+      "单位 ^ 1 ^ 的认识与确定",
+      "求比一个数多/少几分之几的数是多少【分析】水结成冰体积增加 1 /1 1",
+      "是指比水的体积增加 1 /1 1 ;",
+      "把水的体积看作单位 ^ 1 ^",
+      "把它平均分成 11份",
+      "冰的体积增加 1 份。"
+    ],
+    "source": "20-21 六年级上·江西赣州·单元测试",
+    "create_time": "2025-09-30 01:17:20",
+    "merged_id": "homework_414",
+    "original_id": "homework_q_414"
+  },
+  {
+    "id": "homework_q_415",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "3 /5 × 5 /7 × 2 /3 = ()。",
+    "options": [
+      "A． 3 /8 B. 5 /6 C. 2 /7 D.3 /4"
+    ],
+    "answer": "C",
+    "solution": "3 /5 × 5 /7 × 2 /3 故答案为：C",
+    "knowledge_points": [
+      "分数乘法"
+    ],
+    "source": "21-22 六年级上·江苏·单元测试",
+    "create_time": "2025-09-30 01:17:20",
+    "merged_id": "homework_415",
+    "original_id": "homework_q_415"
+  },
+  {
+    "id": "homework_q_416",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一种商品的价格先提高 1 /8 ，再降低 1 /8 ，现价与原价相比，（ ）。",
+    "options": [
+      "A．提高了B．降低了C．不变 D．无法判定"
+    ],
+    "answer": "B",
+    "solution": "( 1 + 1 /8 ) \\ × \\ ( 1 - 1 /8 ) = 9 /8 × 7 /8 = 6 3 /6 4 6 3 /6 4 < 1 ，所以现价比原价低。故答案为：B",
+    "knowledge_points": [
+      "分数乘分数",
+      "经济问题"
+    ],
+    "source": "21-22 六年级上·天津蓟州·阶段练习",
+    "create_time": "2025-09-30 01:17:20",
+    "merged_id": "homework_416",
+    "original_id": "homework_q_416"
+  },
+  {
+    "id": "homework_q_417",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "困难",
+    "question": "已知 c a × c b < c a + c b , ，且a、b、c都是不等于0的自然数，则有（ ）。",
+    "options": [
+      "A． a + b > c B . ~ \\ a + b = c ~ C . ~ \\ a + b < c"
+    ],
+    "answer": "A",
+    "solution": "c a × c b < c a + c b c ^ 2 a b < c ( a + b ) a b 所以 a + b > c ，故答案为：A。",
+    "knowledge_points": [
+      "分数与分数的除法",
+      "含有字母式子的化简与求值"
+    ],
+    "source": "21-22 六年级上 $\\cdot$ 四川·单元测试",
+    "create_time": "2025-09-30 01:17:20",
+    "merged_id": "homework_417",
+    "original_id": "homework_q_417"
+  },
+  {
+    "id": "homework_q_418",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "下面算式中计算结果最小的是（ ） # 二、填空题",
+    "options": [
+      "A． 2 /3 × 3 /5 B． 6 /5 × 2 /3 C. 2 /3 × 5 /5 D. 2 /3 × 3 /7"
+    ],
+    "answer": "D",
+    "solution": "AD中 2 /3 ; 乘的数都小于 1 的数，所以得到的积小于 \\ 2 /3 , 但是 3 /5 > 3 /7 ; 所以 A > D 。 B 2 /3 乘的数都大于 1 的数，所以得到的积大于 2 /3 。 C2 /3 乘的数等于 1 的数，所以得到的积等于2 /3 。",
+    "knowledge_points": [
+      "分数乘法"
+    ],
+    "source": "21-22六年级上·全国·单元测试",
+    "create_time": "2025-09-30 01:17:20",
+    "merged_id": "homework_418",
+    "original_id": "homework_q_418"
+  },
+  {
+    "id": "homework_q_419",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "根据运算定律填空。 （2） 8 /1 7 × 4 /9 + 5 /9 × 8 /1 7 可以 利用乘法 分配律 简便计 算。",
+    "options": [],
+    "answer": "5 /7 7 /5 8 /17 4 /9 5 /9",
+    "solution": "（1） 5 /7 × 4 1 × 7 /5 = 4 1 × ( 5 /7 × 7 /5 ) （2 ） 8 /1 7 × 4 /9 + 5 /9 × 8 /1 7 = 8 /1 7 × ( 4 /9 + 5 /9 )",
+    "knowledge_points": [
+      "分数乘法运算律【分析】（1） 5 /7 × 4 1 × 7 /5 可以利用乘法结合律简便计算；"
+    ],
+    "source": "20-21六年级上·湖南怀化·单元测试",
+    "create_time": "2025-09-30 01:17:20",
+    "merged_id": "homework_419",
+    "original_id": "homework_q_419"
+  },
+  {
+    "id": "homework_q_420",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一根绳子长4 /5 米，截去 1 /4 ，还剩( )米；如果截去1/4 米，还剩( )米。",
+    "options": [],
+    "answer": "3 /5 11 /20",
+    "solution": "4 /5 × \\ ( 1 - 1 /4 ) = 4 /5 × 3 /4 = 3 /5 （米） 4 /5 - 1 /4 = 1 1 /2 0 （米）",
+    "knowledge_points": [
+      "求一个数的几分之几的问题",
+      "异分母分数加",
+      "减法【分析】将绳子长度看作单位^ 1 ^ 绳子长度 × 剩下对应分率 = 剩下长度；绳子长度－截去长度 = 剩下长度",
+      "据此分析。"
+    ],
+    "source": " $2 0 2 0 \\cdot$ 山西忻州·小升初真题",
+    "create_time": "2025-09-30 01:17:20",
+    "merged_id": "homework_420",
+    "original_id": "homework_q_420"
+  },
+  {
+    "id": "homework_q_421",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "1 千克茶叶 40元，买 3/4 千克，要付( )元。一袋大米 2 5 k g ，已经吃了它的 2 /5 ，吃了( )千克，还剩( )千克。",
+    "options": [],
+    "answer": "30 10 15",
+    "solution": "4 0 × 3 /4 = 3 0 （元）2 5 × 2 /5 = 1 0 （千克）2 5 - 1 0 = 1 5 （千克）",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "求一个数的几分之几的问题【分析】单价 × 数量 = 总价；求一个数的几分之几是多少用乘法",
+      "大米质量－吃了的质量 = 剩下的质量",
+      "据此分析。"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "create_time": "2025-09-30 01:17:20",
+    "merged_id": "homework_421",
+    "original_id": "homework_q_421"
+  },
+  {
+    "id": "homework_q_422",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "12 个5/6 是( )； 24 的2/3 是( )；5/7 的3/5 是( )。 求一个数的几分之几是多少用乘法。 分数乘法法则： （1）分数与整数相乘，用整数与分子的积作为分子，分母不变； （2）分数与分数相乘，用分子相乘的积作为分子，分母相乘的积作为分母； （3）计算结果能约分的要约分。",
+    "options": [],
+    "answer": "10 16 3/7",
+    "solution": "1 2 × 5 /6 = 1 0 2 4 × 2 /3 = 1 6 5 /7 × 3 /5 = 3 /7",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "分数乘分数",
+      "求一个数的几分之几的问题【分析】求几个相同的加数的和的简便运算",
+      "叫做乘法。"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "create_time": "2025-09-30 01:17:20",
+    "merged_id": "homework_422",
+    "original_id": "homework_q_422"
+  },
+  {
+    "id": "homework_q_423",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一本书80 页，笑笑第一天看了全书的 1 /5 , 第二天应从第( )页看起。",
+    "options": [],
+    "answer": "17",
+    "solution": "8 0 × 1 /5 + 1 = 1 6 + 1 = 1 7 （页）",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": "18-19 六年级上·重庆·期中",
+    "create_time": "2025-09-30 01:17:20",
+    "merged_id": "homework_423",
+    "original_id": "homework_q_423"
+  },
+  {
+    "id": "homework_q_424",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一根绳子长2/9 米，4根这样的绳子共长( )米；这根绳子的3/4 长( )米。",
+    "options": [],
+    "answer": "8/9 1/6",
+    "solution": "2 /9 × 4 = 8 /9 （米），所以4 根这样的绳子共长8/9 米； 2 /9 × 3 /4 = 1 /6 （米），所以这根绳子的3/4 长1/6米。",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "分数乘分数",
+      "求一个数的几分之几的问题"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "create_time": "2025-09-30 01:17:20",
+    "merged_id": "homework_424",
+    "original_id": "homework_q_424"
+  },
+  {
+    "id": "homework_q_425",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "边长是1/3分米的正方形的周长是( )分米，面积是( )平方分米。",
+    "options": [],
+    "answer": "4/3 1/9",
+    "solution": "1 /3 × 4 = 4 /3 （分米）， 1 /3 × 1 /3 = 1 /9 （平方分米）所以边长是1/3 分米的正方形的周长是4/3分米，面积是 \\ 1 /9 平方分米。",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "分数乘分数",
+      "正方形的面积",
+      "正方形的周长"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "create_time": "2025-09-30 01:17:20",
+    "merged_id": "homework_425",
+    "original_id": "homework_q_425"
+  },
+  {
+    "id": "homework_q_426",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "39 的 2 /1 3 是( )；4/9 的3/8是( )。",
+    "options": [],
+    "answer": "6 1/6",
+    "solution": "3 9 × 2 /1 3 = 6 ，所以39的 2 /1 3 是 6； 4 /9 × 3 /8 = 1 /6 , ，所以 4 /9 的3/8是1/6。",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "分数乘分数",
+      "求一个数的几分之几的问题【分析】根据题意",
+      "利用乘法列式计算即可。"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "create_time": "2025-09-30 01:17:20",
+    "merged_id": "homework_426",
+    "original_id": "homework_q_426"
+  },
+  {
+    "id": "homework_q_427",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "5 /8 千米的1/2 是( )千米；24 个5/6t 是( ) t。",
+    "options": [],
+    "answer": "5/16 20",
+    "solution": "（1） 5 /8 × 1 /2 = 5 /1 6 （千米） （2） 2 4 × 5 /6 = 2 0 （吨）",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "分数乘分数",
+      "求一个数的几分之几的问题"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_427",
+    "original_id": "homework_q_427"
+  },
+  {
+    "id": "homework_q_428",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "18 个5/6 是( )；27 的2/3 是( )。 （2）27 的 2 /3 ; 表示为 2 7 × 2 /3 = 1 8",
+    "options": [],
+    "answer": "15 18",
+    "solution": "18个5/6 是（ 15 ）；27的2/3是（ 18 ）。",
+    "knowledge_points": [
+      "分数与整数的乘法【分析】（1）18个5/6表示为 1 8 × 5 /6 = 1 5 ；"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_428",
+    "original_id": "homework_q_428"
+  },
+  {
+    "id": "homework_q_429",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "3 /8 k g = ( )g 5/6 时＝( )分 1 /4 m =() cm",
+    "options": [],
+    "answer": "375 50 10/4",
+    "solution": "（1） 3 /8 × 1 0 0 0 = 3 × 1 2 5 = 3 7 5 （2） 5 /6 × 6 0 = 5 × 1 0 ​ = 5 0 （3 ）1/4 ×10= 1 0 /4",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "厘米和米之间的进率与换算",
+      "时",
+      "分的认识及换算",
+      "克",
+      "千克之间的换算 与比较"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_429",
+    "original_id": "homework_q_429"
+  },
+  {
+    "id": "homework_q_430",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "在（ ）里填上 \" > \" < \" 或 ^ 6 6 = ^ 3 9 。 2 /9 × 2 /9 ()2 /9 8 /7 × 3 /4 ()3 /4 1 × 1 /2 ( ) 1 + 1 /2",
+    "options": [],
+    "answer": "＜ ＞ ＜",
+    "solution": "（1） 2 /9 < 1 所以2 /9 × 2 /9 <2 /9 （2） 8 /7 > 1 ?，所以 2 /9 < 1 8 /7 × 3 /4 > 3 /4 ，所以 2 /9 × 2 /9 < 2 /9 ; ；（3） 1 × 1 /2 = 1 /2 ; 1 + 1 /2 = 3 /2 , ，因为 1 /2 < 3 /2 , ，所 以 1 × 1 /2 < 1 + 1 /2 【点睛】本题关键在于对分数乘法的理解与认识。",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "分数乘分数"
+    ],
+    "source": "21-22六年级上·江西赣州·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_430",
+    "original_id": "homework_q_430"
+  },
+  {
+    "id": "homework_q_431",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "有 3个 · 1 /5 相加，用加法计算是( )，用乘法计算是( )。",
+    "options": [],
+    "answer": "1 /5 + 1 /5 + 1 /5 = 3 /5 1 /5 × 3 = 3 /5",
+    "solution": "用加法计算： 1 /5 + 1 /5 + 1 /5 = 3 /5 用乘法计算： 1 /5 × 3 = 3 /5",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "初步认识乘法意义",
+      "同分母分数加",
+      "减法【分析】乘法表示几个相同加数和的简便运算",
+      "3个 1 /5 相加可以表示为 $ 1 /5 + 1 /5 + 1 /5",
+      "$",
+      "也可以表示为 1 /5 × 3"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_431",
+    "original_id": "homework_q_431"
+  },
+  {
+    "id": "homework_q_432",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "学校开展节约用水活动，9月份用水 3 0 t ，10月份的用水量比 9 月份的5 少 3t，10 月份用水( )t。",
+    "options": [],
+    "answer": "22",
+    "solution": "3 0 × 5 /6 - 3 = 2 5 - 3 = 2 2 （吨） 所以，十月份用水 22吨。",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "求一个数的几分之几的问题【分析】用 9 月份的30吨乘 $5 /6",
+      "$",
+      "再减去 3 吨",
+      "求出 10 月份的用水量。"
+    ],
+    "source": "21-22六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_432",
+    "original_id": "homework_q_432"
+  },
+  {
+    "id": "homework_q_433",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "25 分 = ( )时 1 7 /2 0 m ^ 2 = ( ) d m ^ 2 3公顷 = ( )m² 7 5 0 k g = ( ) t",
+    "options": [],
+    "answer": "5 85 6000 0.75",
+    "solution": "由分析可知： 25 分 = （ 5 /12 ）时 1 7 /2 0 m ^ 2 = ~ ( ~ ~ 8 5 ~ ~ ~ ) ~ ~ d m ^ 2 3公顷 \\ : （ 6000 ） m ^ 2 750kg＝（ 0.75 ）t",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "面积单位间的进率及换算",
+      "吨",
+      "千克之间的换算与比较",
+      "时",
+      "分的认识及换算"
+    ],
+    "source": "20-21六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_433",
+    "original_id": "homework_q_433"
+  },
+  {
+    "id": "homework_q_434",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "今年暑假，长安小学有 40人参加全市夏令营活动，其中女生占2，女生有( )人，男生有( )。",
+    "options": [],
+    "answer": "16 24",
+    "solution": "女生： 4 0 × 2 /5 = 1 6 （人）男生： 4 0 - 1 6 = 2 4 （人）所以，女生有 16 人，男生有 24人。",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "求一个数的几分之几的问题【分析】根据题意",
+      "利用乘法求出女生人数",
+      "再将总人数 40人减去女生人数",
+      "求出男生人数。"
+    ],
+    "source": "21-22六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_434",
+    "original_id": "homework_q_434"
+  },
+  {
+    "id": "homework_q_435",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一块长方形菜地，长 20米，宽是长的 3 /4 ，这块长方形菜地的 面积是( )平方米。",
+    "options": [],
+    "answer": "300",
+    "solution": "2 0 × （ 2 0 × 3 /4 ) = 2 0 × 1 5 = 300（平方米） 所以，这个长方形菜地的面积是300 平方米。",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "长方形的面积"
+    ],
+    "source": "21-22 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_435",
+    "original_id": "homework_q_435"
+  },
+  {
+    "id": "homework_q_436",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "“杨树的棵数比柳树多 1 /4 ，把( )的棵数看作单位^ 1 ^ ，杨树的棵数是柳树的( )，如果柳树有 48 棵，那么杨树有( )棵。",
+    "options": [],
+    "answer": "柳树 5 /4 60",
+    "solution": "1 + 1 /4 = 5 /4 ; ＝5， 4 8 × 5 /4 = 6 0 （棵），所以“杨树的棵数比柳树多 1 /4 ，把柳树的棵数看作单位 ^ 1 ^ ，杨树的棵数是柳树的 5 /4 , ，如果柳树有 48 棵，那么杨树有60 棵。",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "求比一个数多/少几分之几的数是多少【分析】杨树的棵数比柳树多 1 /4",
+      "那么杨树的棵数是柳树的 5 /4 ;",
+      "那么如果柳树有 48 棵",
+      "可利用乘法求出杨树数量。"
+    ],
+    "source": "21-22六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_436",
+    "original_id": "homework_q_436"
+  },
+  {
+    "id": "homework_q_437",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "手工课上同学们练习做中国结。每个中国结用 1 3 /5 m 彩绳，做 10个中国结用( ) ±b m 彩绳。",
+    "options": [],
+    "answer": "26",
+    "solution": "1 3 /5 × 1 0 = 2 6 （米）",
+    "knowledge_points": [
+      "分数与整数的乘法"
+    ],
+    "source": "21-22 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_437",
+    "original_id": "homework_q_437"
+  },
+  {
+    "id": "homework_q_438",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "15m的1/3是( )m，75cm 的3/5是( )cm，90m 增加 1/6是( )m。",
+    "options": [],
+    "answer": "5 45 105",
+    "solution": "1 5 × 1 /3 = 5 ( m ) ，所以，15m 的1/3是 5m；7 5 × 3 /5 = 4 5 (cm），所以，75cm 的 3 /5 是 4 5 c m ; ；9 0 × ~ ( 1 + 1 /6 ) ~ = 1 0 5 ~ ( m ) ），所以， 90m 增加1/6是 1 0 5 m .",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "求一个数的几分之几的问题【分析】根据题意",
+      "利用乘法求出各空即可。"
+    ],
+    "source": "21-22 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_438",
+    "original_id": "homework_q_438"
+  },
+  {
+    "id": "homework_q_439",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "在括号里填”<“ \">\" \"=\"。 1 /4 × 8 /9 () 1 /4 4/5 × 1 ( ) 4 /5 2/9 × 1 ( ) 2 /9 1/3 × 9 ( ) 1/6 ×12",
+    "options": [],
+    "answer": "< = > >",
+    "solution": "8 /9 < 1 ，所以 1 /4 × 8 /9 < 1 /4 4 /5 × 1 = 4 /5 ，所以 4 /5 × 1 = 4 /5 ； 3 /2 ＞1，所以 2 /9 × 3/2 ＞ 2/9 ； 1 /3 × 9 = 3 ， 1 /6 × 1 2 = 2 ，所 1 /3 × 9 > 1 /6 × 1 2 .",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "分数乘分数"
+    ],
+    "source": "21-22 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_439",
+    "original_id": "homework_q_439"
+  },
+  {
+    "id": "homework_q_440",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "3 /4 吨 ( )千克 4 /5 时 =( )分 2 /3日= ( )时 3 /20平方米 = ( )平方分米",
+    "options": [],
+    "answer": "750 48 16 15",
+    "solution": "3 /4 × 1 0 0 0 = 7 5 0 （千克），所 3 /4 吨= 7 5 0 千克； 4 /5 × 6 0 = 4 8 （分），所以4/5时=48 分； 2 /3 × 2 4 = 1 6 （时），所以 2 /3 日 = 1 6 时； 3 /2 0 × 1 0 0 = 1 5 3（平方分米），所以 3 /2 0 平方米 = 1 5 平方分米。",
+    "knowledge_points": [
+      "平方厘米",
+      "平方分米",
+      "平方米之间的进率与换算",
+      "分数与整数的乘法",
+      "吨",
+      "千克之间的换算与比较",
+      "时",
+      "分的认识及换算"
+    ],
+    "source": "21-22 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_440",
+    "original_id": "homework_q_440"
+  },
+  {
+    "id": "homework_q_441",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一根 12 米长的绳子减去它的1 /4 后，再减去1 /4 米，还剩下( )米。",
+    "options": [],
+    "answer": "8 3 /4",
+    "solution": "1 2 × ~ ( 1 - 1 /4 ) ~ - 1 /4 = 1 2 × 3 /4 - 1 /4 = 9 - 1 /4 = 8 3 /4 则还剩下 8 3 /4 米。",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "求一个数的几分之几的问题【分析】根据分数乘法的意义",
+      "用乘法求出减去它的 ? 后",
+      "还剩下的米数",
+      "然后用剩下的长度减去?米即可解答。"
+    ],
+    "source": "20-21 六年级上·浙江温州·期中",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_441",
+    "original_id": "homework_q_441"
+  },
+  {
+    "id": "homework_q_442",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个正方体，六个面上分别写着数字 1 6 。掷一次，标着”1” 的面朝上的可能性是( )，朝上的面标的数大于 “3” 的可能性是( )，掷 60次正方体，标着 “1”$的面朝上的次数大约有( )次。",
+    "options": [],
+    "answer": "1 /6 1 /2 10",
+    "solution": "一个正方体，六个面上分别写着数字 1 6 。掷一次，标着 “1”的面朝上的可能性是（ 1 /6 ），朝上的面标的数大于 ^ _ 3 ^ 的可能性是（ 1 /2 ），掷 60次正方体，标着 ^ _ 1 ^ $ 的面朝上的次数大约有（ 10 ）次。",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "简单事件发生的可能性求解"
+    ],
+    "source": "21-22五年级上·福建福州·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_442",
+    "original_id": "homework_q_442"
+  },
+  {
+    "id": "homework_q_443",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "“六年级男生人数是女生人数的 6 /7 是把( )看作单位^ 1 ^ ，六年级有 42 个女生，有男生( )个。",
+    "options": [],
+    "answer": "女生人数 36",
+    "solution": "“六年级男生人数是女生人数的 6 /7 是把女生人数看作单位 ^ 1 ^ ， 六年级有 42个女生，男生有： 4 2 × 6 /7 = 3 6 （人）",
+    "knowledge_points": [
+      "求一个数的几分之几的问题",
+      "单位 ^ 1 ^ 的认识与确定【分析】根据判断单位 ^ \\ s 1 ^ s 的方法：一般是把“比",
+      "占",
+      "是",
+      "相当于”后面的量看作单位 ^ \\",
+      "即分数“的”字前面的量看作单位 ^ 1 ^",
+      "可知",
+      "六年级男生人数是女生人数的 6 /7",
+      "应把女生人数看作单位",
+      "根据求 ^ 1 ^ 一个数的几分之几是多少用乘法求出男生人数。"
+    ],
+    "source": "19-20 六年级上·湖南怀化·期末",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_443",
+    "original_id": "homework_q_443"
+  },
+  {
+    "id": "homework_q_444",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "如果 a × 5 /8 = b × 3 /2 = C × 1 （a，b， ±b c 都不为 0），则 a、b、c 三个数中，最大的是( )，最小的是( )。",
+    "options": [],
+    "answer": "a b",
+    "solution": "假设 a × 5 /8 = b × 3 /2 = c × 1 = 1 a × 5 /8 = 1 a＝1÷5 /8 ＝1×8 /5 b× 3 /2 ＝1 b＝1÷ 3 /2 ＝1×2 /3 =2 /3 c×1＝1 c＝1÷1 ＝1 a＞ c＞b最大是 a；最小是 b。",
+    "knowledge_points": [
+      "异分母异分子分数的大小比较",
+      "分数与整数的除法【分析】假设 a × 5 /8 = b × 3 /2 = c × 1 = 1",
+      "分别求出 a",
+      "b",
+      "c 的值",
+      "再进行比较大小",
+      "即可解答。"
+    ],
+    "source": "15-16 六年级上·全国·阶段练习",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_444",
+    "original_id": "homework_q_444"
+  },
+  {
+    "id": "homework_q_445",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一袋大米4 /5 千克，已经吃了它的2 /5 ，吃了( )千克，还剩下总数的( )。",
+    "options": [],
+    "answer": "8 /2 5 3/5",
+    "solution": "4 /5 × 2 /5 = 8 /2 5 （千克）， 1 - 2 /5 = 3 /5 , 所以，吃了 8 /2 5 千克，还剩下总数的3 /5 。",
+    "knowledge_points": [
+      "分数乘分数",
+      "同分母分数加",
+      "减法",
+      "求一个数的几分之几的问题【分析】用大米的总重量乘 2 /5",
+      "求出吃了多少千克；用 1 减去 · 2 /5",
+      "求出还剩下总数的几分之几即可。"
+    ],
+    "source": "20-21 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_445",
+    "original_id": "homework_q_445"
+  },
+  {
+    "id": "homework_q_446",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "在括号里填上 \" > \" \"<\" 或 \"=\" 。 5 /6 × 3 /2 ( )5 /6 1 3 × 2 /3 ( ) 2 /3 × 1 3 3 /8 × 1 /2 () 1 /8 根据乘法交换律，判断出 和 13 × 2 /3 和2 /3 × 1 3的大小关系； 根据乘数和积的关系，分析判断出 3 /8 × 1 /2 和3 /8 的大小关系。",
+    "options": [],
+    "answer": "> ＝ ＜",
+    "solution": "",
+    "knowledge_points": [
+      "分数乘法运算律",
+      "因数和积的大小关系（小数乘法）",
+      "分数与整数的乘法",
+      "分数乘分数【分析】根据乘数和积的关系",
+      "分析判断出 5 /6 × 3 /2 和5 /6 的大小关系；"
+    ],
+    "source": "20-21 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_446",
+    "original_id": "homework_q_446"
+  },
+  {
+    "id": "homework_q_447",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "简单",
+    "question": "丽丽面向北站立，向右转 4 0 ^ 后所面对的方向是( )。",
+    "options": [],
+    "answer": "北偏东 4 0 ^",
+    "solution": "丽丽面向北站立，向右转 4 0 ^ 后所面对的方向是北偏东 4 0 ^ °。",
+    "knowledge_points": [
+      "根据方向",
+      "角度和距离确定物体的位置",
+      "东南",
+      "西南",
+      "东北",
+      "西北方向【分析】面向正北",
+      "向右转即面向了北偏东方向。据此填空。"
+    ],
+    "source": "20-21 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_447",
+    "original_id": "homework_q_447"
+  },
+  {
+    "id": "homework_q_448",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "边长 : 1 /2 分米的正方形的周长是( )分米，面积是( )平方米。",
+    "options": [],
+    "answer": "2 1 /4 0 0",
+    "solution": "1 /2 × 4 = 2 （分米） 1 /2 × 1 /2 = 1 /4 （平方分米） = 1 /4 0 0 （平方米） 所以，边长? 分米的正方形的周长是 2 分米，面积是 1 /4 0 0 平方米。",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "分数乘分数",
+      "正方形的面积",
+      "正方形的周长【分析】根据正方形的周长和面积公式",
+      "直接列式计算即可。"
+    ],
+    "source": "20-21 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_448",
+    "original_id": "homework_q_448"
+  },
+  {
+    "id": "homework_q_449",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "10 个 5 /6 的和是( ) ； 9 米的2 /3 是( )米；4 吨的3/5 是( )吨； 0.6 平方米的1 /4 是( )平方米",
+    "options": [],
+    "answer": "25 /3 6 12 /5 3/20",
+    "solution": "10×5 /6 ＝25 /3 ，所以10 个 5 /6 的和 是 2 5 /3 ； 9 × 2 /3 = 6 （米），所以，9 米的 2 /3 是 6米； 4 × 3 /5 = 1 2 /5 （吨），所以，4 吨的3/5 是12 /5 吨； 0 . 6 × 1 /4 = 3 /2 0 （平方米），所以，0.6平方米的 1 /4 是3 /2 0 平方 米。",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "分数乘小数",
+      "求一个数的几分之几的问题【分析】根据题意",
+      "利用乘法分别求出各空即可。"
+    ],
+    "source": "20-21 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_449",
+    "original_id": "homework_q_449"
+  },
+  {
+    "id": "homework_q_450",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "3 /5 小时 () 分 5 /8 千克 = ( )克 3 /4 日=（）时",
+    "options": [],
+    "answer": "36 625 18",
+    "solution": "3 /5 × 6 0 = 3 6 （分），所以， 3 /5 . 小时 = 3 6 分； ​5 /8 × 1 0 0 0 = 6 2 5 （克），所以， 5 /8 千克 = 6 2 5 克； 3 /4 × 2 4 = 1 8 （时），所以， 3 /4 日= 18时。",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "时",
+      "分的认识及换算",
+      "克",
+      "千克之间的换算与比较"
+    ],
+    "source": "20-21 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_450",
+    "original_id": "homework_q_450"
+  },
+  {
+    "id": "homework_q_451",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一段路第一天行了全程的 1 /5 ; ，第二天行了余下的 2 /3 , ，两天一共行了完全程的( )。",
+    "options": [],
+    "answer": "1 1 /1 5",
+    "solution": "两天一共行的路程： ​",
+    "knowledge_points": [
+      "求一个数的百分之几是多少",
+      "工程问题"
+    ],
+    "source": "20-21 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_451",
+    "original_id": "homework_q_451"
+  },
+  {
+    "id": "homework_q_452",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "3 吨的2 /9 是( )；30 米增加2 /9 米是( )；30 米减少1 /6 米是( )。",
+    "options": [],
+    "answer": "2 /3 吨 301 /6 米 295 /6 米",
+    "solution": "3 × 2 /9 = 2 /3 （吨） 3 0 + 1 /6 = 3 0 1 /6 （米）3 0 - 1 /6 = 2 9 5 /6 （米）",
+    "knowledge_points": [
+      "分数与整数的乘法"
+    ],
+    "source": "20-21 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_452",
+    "original_id": "homework_q_452"
+  },
+  {
+    "id": "homework_q_453",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一本书共 240 页，小红第一天看了这本书的1 /6 。小红第二天要从第( )页看起。",
+    "options": [],
+    "answer": "41",
+    "solution": "2 4 0 × 1 /6 + 1 = 4 0 + 1 = 4 1 （页）",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": "20-21 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_453",
+    "original_id": "homework_q_453"
+  },
+  {
+    "id": "homework_q_454",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "8 /2 5 平方米 =()平方分米。",
+    "options": [],
+    "answer": "32",
+    "solution": "1 0 0 × 8 /2 5 = 3 2 （平方分米），所以 8 /2 5 平方米 = 3 2 平方分米。",
+    "knowledge_points": [
+      "平方厘米",
+      "平方分米",
+      "平方米之间的进率与换算",
+      "分数与整数的乘法"
+    ],
+    "source": "21-22六年级上·全国·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_454",
+    "original_id": "homework_q_454"
+  },
+  {
+    "id": "homework_q_455",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "困难",
+    "question": "有一棵桃树，一只猴子去偷吃桃子，第一天偷吃了 1 /10 ，以后 8 天分别偷吃了当天的桃子的 1 /9 , 1 /8 , 1 /7 , ·s , 1 /3 , 1 /2 , ，偷吃了 9 天后，树上还剩下 10 个桃子。树上原来有桃子() 个。",
+    "options": [],
+    "answer": "100",
+    "solution": "\\ 1 0 ÷ \\ ( 1 - 1 /2 ) \\ ÷ \\ ( 1 - 1 /3 ) \\ ÷ \\ ·s ÷ \\ ( 1 - 1 /9 ) \\ ÷ \\ ( 1 - 1 /1 0 ) = 1 0 ÷ 1 /2 ÷ 2 /3 ÷ ·s ÷ 8 /9 ÷ 9 /1 0 = 1 0 × 2 × 3 /2 × ·s × 9 /8 × 1 0 /9 = 1 0 0 （个）",
+    "knowledge_points": [
+      "分数除法的应用"
+    ],
+    "source": "21-22 六年级上·全国·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_455",
+    "original_id": "homework_q_455"
+  },
+  {
+    "id": "homework_q_456",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "30 dm³ 的水结成冰后体积增加 1 /1 1 : ，结成的冰的体积是( ) dm³ 。",
+    "options": [],
+    "answer": "3 6 0 /1 1",
+    "solution": "3 0 × ( 1 + 1 /1 1 ) = 3 0 × 1 2 /1 1 = 3 6 0 /1 1 （立方分米）",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "求比一个数多/少几分之几的数是多少"
+    ],
+    "source": "20-21 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_456",
+    "original_id": "homework_q_456"
+  },
+  {
+    "id": "homework_q_457",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "5 /6 小时 = ( )分； 3 /5 平方米 = ( )平方分米。 # 三、排序题",
+    "options": [],
+    "answer": "50 60",
+    "solution": "由分析可知： 5 /6 小时 =(50)分； 3 /5 平方米 = （ 60 ）平方分米",
+    "knowledge_points": [
+      "平方厘米",
+      "平方分米",
+      "平方米之间的进率与换算",
+      "分数与整数的乘法",
+      "时",
+      "分的认识及换算"
+    ],
+    "source": "20-21 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_457",
+    "original_id": "homework_q_457"
+  },
+  {
+    "id": "homework_q_458",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "已知 ），把 A 、B 、C 三个 数按从小到大的顺序排列起来：( )。 # 四、判断题",
+    "options": [],
+    "answer": "A < C < B",
+    "solution": "因为 1 5 /1 3 > 1 5 /1 5 > 1 3 /1 4 , 所以 A < C < B",
+    "knowledge_points": [
+      "因数和积的大小关系（小数乘法）",
+      "分数乘分数"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_458",
+    "original_id": "homework_q_458"
+  },
+  {
+    "id": "homework_q_459",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "计算 ( 1 /5 + 1 /6 ) × 3 0 = 1 /5 + 1 /6 × 3 0 。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "( 1 /5 + 1 /6 ) × 3 0 = 1 /5 × 3 0 + 1 /6 × 3 0 = 6 + 5 = 1 1",
+    "knowledge_points": [
+      "分数乘法"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_459",
+    "original_id": "homework_q_459"
+  },
+  {
+    "id": "homework_q_460",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "判断题",
+    "difficulty": "简单",
+    "question": "1 千克的 5 /7 和 5 千克的 1 /7 相等。( )",
+    "options": [],
+    "answer": "✔",
+    "solution": "1 × 5 /7 = 5 /7 （千克）； 5 × 1 /7 = 5 /7 （千克）；所以判断正确。",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": "19-20 六年级上·全国·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_460",
+    "original_id": "homework_q_460"
+  },
+  {
+    "id": "homework_q_461",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "比 40元多 1 /4 是 44 元。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "4 0 × ~ ( 1 + 1 /4 ) ） 故原题干说法错误。",
+    "knowledge_points": [
+      "求比一个数多/少几分之几的数是多少"
+    ],
+    "source": "21-22 六年级上·河北廊坊·阶段练习",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_461",
+    "original_id": "homework_q_461"
+  },
+  {
+    "id": "homework_q_462",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "计算 3 6 × ( 1 /6 + 2 /9 ) = 3 6 × 1 /6 + 3 6 × 2 /9 , ，这 里应用了 乘法分配律使计算简便。( )",
+    "options": [],
+    "answer": "✔",
+    "solution": "3 6 × ( 1 /6 + 2 /9 ) = 3 6 × 1 /6 + 3 6 × 2 /9 ; ， 这里应用了乘法分配律使计算简便。 所以判断正确。",
+    "knowledge_points": [
+      "分数乘法"
+    ],
+    "source": "21-22 六年级上·河北廊坊·阶段练习",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_462",
+    "original_id": "homework_q_462"
+  },
+  {
+    "id": "homework_q_463",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "判断题",
+    "difficulty": "简单",
+    "question": "4 /9 × 4 = 1 /9 。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "4/9 × 4 = 4 × 4/9 = 16/9答案为： ×",
+    "knowledge_points": [
+      "分数与整数的乘法"
+    ],
+    "source": "19-20 六年级上·全国·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_463",
+    "original_id": "homework_q_463"
+  },
+  {
+    "id": "homework_q_464",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "3 千克的3/4 和 9千克的1/4一样重。( )",
+    "options": [],
+    "answer": "✔",
+    "solution": "3×3/4＝9/4（千克） 9 × _ 4 ^ 1 = _ 4 ^ 9 （千克） 所以 3 千克的3/4和9 千克的 1/4 一样重。所以判断正确。",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "求一个数的几分之几的问题"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_464",
+    "original_id": "homework_q_464"
+  },
+  {
+    "id": "homework_q_465",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "两根长都是 1m的钢筋，第一根截去 1 /2 ，第二根截去 1 /2 = 1 /2 ，则两根钢筋剩下的一样长。( )",
+    "options": [],
+    "answer": "✔",
+    "solution": "第一根剩下： 1 × ( 1 - 1 /2 ) \\ = 1 /2 （米） 第二根剩下： 1 - 1 /2 = 1 /2 （米） 所以，两根钢筋剩下的一样长。",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "同分母分数加",
+      "减法【分析】先求出两根钢筋剩下的长度",
+      "再比较是否一样长即可。"
+    ],
+    "source": "21-22 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_465",
+    "original_id": "homework_q_465"
+  },
+  {
+    "id": "homework_q_466",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "一根 1 米长的绳子剪去 4 /5 和剪去 4 /5 米后，剩下的长度相等。( )",
+    "options": [],
+    "answer": "✔",
+    "solution": "1 × ~ ( 1 - 4 /5 ) = 1 × 1 /5 = 1 /5 （米） 1 - 4 /5 = 1 /5 （米） 因为 1/5 米 = 1 /5 米，所以一根 1米长的绳子剪去 4 /5 和剪去 4 /5 米后，剩下的长度相等。",
+    "knowledge_points": [
+      "求一个数的几分之几的问题",
+      "异分母分数加",
+      "减法【分析】把绳子总长度看作单位 \"1\"",
+      "剩下绳子的长度 = 绳子总长度×剩下绳子占绳子总长度的分率",
+      "分别求出剪去 4 /5 和剪去 4 /5 米后剩下绳子的长度",
+      "最后比较大小。"
+    ],
+    "source": "19-20 六年级上·全国·期末",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_466",
+    "original_id": "homework_q_466"
+  },
+  {
+    "id": "homework_q_467",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "1 千克的 4 /5 和 4 千克的1 /5 一样多。( )",
+    "options": [],
+    "answer": "✔",
+    "solution": "由分析可知：1 × 4 /5 = 4 /5 （千克）4 × 1 /5 = 4 /5 （千克） 则 1 千克的? 和 4 千克的? 一样多。故原题干说法正确。",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "求一个数的几分之几的问题【分析】根据求一个数的几分之几是多少",
+      "用乘法分别求出 1千克的?和 4千克的 1 /5",
+      "然后对比即可。"
+    ],
+    "source": "17-18 五年级下·山东青岛·期末",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_467",
+    "original_id": "homework_q_467"
+  },
+  {
+    "id": "homework_q_468",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "一根电线长3 0 m ，用去 1 /1 0 , ，再接上 1 /1 0 米，这根电线仍是 30米。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "3 0 × ~ ( 1 - 1 /1 0 ) ~ + 1 /1 0 ＝27＋ 1 /1 0 = 2 7 1 /1 0 （米） 所以，一根电线长 3 0 m ，用去 1 /1 0 ，再接上 1 /1 0 米，这根电线是 2 7 1 /1 0 米。 所以判断错误。",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "同分母分数加",
+      "减法"
+    ],
+    "source": "20-21 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_468",
+    "original_id": "homework_q_468"
+  },
+  {
+    "id": "homework_q_469",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "a × 4 /5 = b × 2 /3 , （a、b 均不为 0），则a > b。 ( )",
+    "options": [],
+    "answer": "×",
+    "solution": "4 /5 > 2 /3 , 又因为 a× 4 /5 = b × 2 /3 , a < b 所以判断错误。",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "分数乘分数"
+    ],
+    "source": "20-21 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_469",
+    "original_id": "homework_q_469"
+  },
+  {
+    "id": "homework_q_470",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "4 /5 吨煤，用去 1 /5 ，还剩 3 /5 吨。( ) # 五、计算题",
+    "options": [],
+    "answer": "×",
+    "solution": "4 /5 × ( 1 - 1 /5 ) = 4 /5 × 4 /5 = 1 6 /2 5 （吨）。",
+    "knowledge_points": [
+      "分数乘法"
+    ],
+    "source": "21-22 六年级上·全国·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_470",
+    "original_id": "homework_q_470"
+  },
+  {
+    "id": "homework_q_471",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "解方程。 6 /5 x + 4 /5 x = 6 0 x ÷ 1 /2 = 7 /8 8 x - 5 /4 = 3 /4",
+    "options": [],
+    "answer": "x = 3 0 ; x = 7 /1 6 ; x = 1 /4",
+    "solution": "（1） 6 /5 x + 4 /5 x = 6 0 = x = 6 0 ÷ 2 x =30 （2） x ÷ 1 2 = 7 / 8 解： x = 7 /8 × 1 /2 x = 7 /1 6 （3） 8 x - 5 4 = 3 /4 解： 8 x = 3 /4 + 5 /4",
+    "knowledge_points": [
+      "分数乘分数",
+      "应用等式的性质 1和 2解方程",
+      "应用等式的性质 2 解方程"
+    ],
+    "source": "21-22六年级上·江西赣州·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_471",
+    "original_id": "homework_q_471"
+  },
+  {
+    "id": "homework_q_472",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "下面各题，怎样算简便就怎样算。 1 /5 + 2 /9 × 3 /1 0 4 - 7 2 × 5 /1 2 （3）（4）利用乘法分配律简便计算。",
+    "options": [],
+    "answer": "4 /1 5 ;14; 5/9 46",
+    "solution": "（1） 1 /5 + 2 /9 × 3 /1 0 （2）44 - 72 × 5/12 =44 − 30 ＝14 （3）5/9 × 1/6 + 5/9 × 5/6=5/9 × ( 1/6 + 5/6 ) = 5/9 （4） ( 3 /4 + 5 /8 - 5 /1 2 ) × 4 8",
+    "knowledge_points": [
+      "分数乘法运算律【分析】（1）（2）按照四则混合运算顺序计算；"
+    ],
+    "source": "21-22六年级上·江西赣州·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_472",
+    "original_id": "homework_q_472"
+  },
+  {
+    "id": "homework_q_473",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "直接写出得数。 3 /1 0 ;",
+    "options": [],
+    "answer": "1；6； ??；16",
+    "solution": "略",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "分数乘分数",
+      "分数乘小数"
+    ],
+    "source": "21-22六年级上·江西赣州·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_473",
+    "original_id": "homework_q_473"
+  },
+  {
+    "id": "homework_q_474",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "计算下面各题，能简算的要简算。 18 × 2 × 5 1 4 /1 7 × 3 /4 + 1 4 /1 7 × 1 /4 ( 5 /9 + 1 /2 5 ) × 9 + 1 6 /2 5",
+    "options": [],
+    "answer": "3 0 /7 ; 1 4 /1 7 ;",
+    "solution": "（1） 1 8 × 2 /7 × 5 /6",
+    "knowledge_points": [
+      "分数乘法运算律"
+    ],
+    "source": "21-22六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_474",
+    "original_id": "homework_q_474"
+  },
+  {
+    "id": "homework_q_475",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "直接写得数。 5 /6 ； 6 /55 ; 1 /45 ； 5 /7 【难度】0.65",
+    "options": [],
+    "answer": "4.5；4.2； 2 /5 ；0",
+    "solution": "略六、解答题",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "分数乘分数",
+      "分数乘小数"
+    ],
+    "source": "21-22六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_475",
+    "original_id": "homework_q_475"
+  },
+  {
+    "id": "homework_q_476",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "困难",
+    "question": "有甲、乙两袋大米，甲袋重50 千克，如果从甲袋中倒出 1 /5 给乙袋，则两袋一样重，原来乙袋大米重多少千克？",
+    "options": [],
+    "answer": "30千克",
+    "solution": "解：设乙袋 X 千克。 答：原来乙袋重30千克。",
+    "knowledge_points": [
+      "分数与整数的乘法"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_476",
+    "original_id": "homework_q_476"
+  },
+  {
+    "id": "homework_q_477",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "刘叔叔的养殖场养鸡 1200只， _，鹅有多少只？请你补充下面三个条件后分别列式解答。 1. 养的鹅比鸡多 1 /5 2. 养的鹅比鸡少 1 /5 3. 养的鸭是鸡的 5 /6 , 养的鹅是鸭的4 /5 。【答案】1. 1 4 4 0 只；2. 9 6 0 只； 3. 8 0 0 只 \\ 1 2 0 0 × \\ ( 1 - 1 /5 ) = 1 2 0 0 × 4 /5 = 9 6 0 （只） 1200 × 5 /6 × 4 /5 = \\ 1000 × 4 /5 = 800 （只)",
+    "options": [],
+    "answer": "",
+    "solution": "1200 × \\ ( 1 + 1 /5 ) = 1 2 0 0 × 6 /5 = 1 4 4 0 （只）",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "求一个数的几分之几的问题",
+      "求比一个数多/少几分之几的数是多少"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_477",
+    "original_id": "homework_q_477"
+  },
+  {
+    "id": "homework_q_478",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "一本故事书有 100 页，聪聪第一天读了全书的 3 /1 0 ，第二天读了余下的 ? 第二天读了多少页？第三天应从第几页读起？",
+    "options": [],
+    "answer": "35页；66页",
+    "solution": "1 0 0 × ~ ( 1 - 3 /1 0 ) ~ × 1 /2 = 3 5 （页） 1 0 0 × 3 /1 0 + 3 5 + 1 = 3 0 + 3 5 + 1 = 6 6 （页）",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_478",
+    "original_id": "homework_q_478"
+  },
+  {
+    "id": "homework_q_479",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "六年级同学给灾区的小朋友捐款。六一班捐了 500元，六二班捐的是六一班的? ，六三班捐的是六二班的 9 /8 。六三班捐款多少元？",
+    "options": [],
+    "answer": "450元",
+    "solution": "500×?×? = 4 0 0 × 9 /8 = 4 5 0 （元）",
+    "knowledge_points": [
+      "分数的连乘运算",
+      "连续求一个数的几分之几是多少的问题【分析】六二班捐款的钱数 = 六一班捐款的钱数 $× 4 /5",
+      "六三班捐款的钱数 = 六二班捐款的钱数 × 9 /8 $",
+      "据此解答。"
+    ],
+    "source": "15-16 六年级·全国·阶段练习",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_479",
+    "original_id": "homework_q_479"
+  },
+  {
+    "id": "homework_q_480",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "有一堆石子，第一次用去 2/7 ，第二次用去了 15吨，这时余下的和用去的一样多。这堆石子原来有多少吨？",
+    "options": [],
+    "answer": "70吨",
+    "solution": "解：设原来这堆石子有 X 吨。 x - 2 /7 x - 1 5 = 1 /2 x 5 /7 x－15＝1 /2 x 3/14 x＝15 x＝70",
+    "knowledge_points": [
+      "求一个数的几分之几的问题",
+      "列方程解含一个未知数的问题"
+    ],
+    "source": "20-21 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_480",
+    "original_id": "homework_q_480"
+  },
+  {
+    "id": "homework_q_481",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "猴妈妈买来 120 个桃子。小猴子们第一天吃了总数的 1/2 ，第二天吃了余下的 1 /3 ，第三天吃了第二天余下的 1 /4 ，三天后还剩多少个桃子？",
+    "options": [],
+    "answer": "30个",
+    "solution": "1 2 0 × ~ ( 1 - 1 /2 ) ~ × ~ ( 1 - 1 /3 ) ~ × ~ ( 1 - 1 /4 ) ＝120×(1-1/2 )×(1-1/3 )×(1-1/4 ) =120× 1/2 × 2/3 ×3/4 =30（个）",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "分数乘分数",
+      "求一个数的几分之几的问题【分析】第一天吃了 1 /2 $",
+      "余下 ( 1 - 1 /2 );第二天吃了 1 /3 $",
+      "余下第一天的 ( 1 - 1 /3 ) ；第三天吃了 1 /4 $",
+      "余下第二天的( 1 - 1 /4 ) ；那么余下的占总数的 ( 1 - 1 /2 ) × ( 1 - 1 /3 ) × ( 1 - 1 /4 )$",
+      "从而利用乘法求出三天后还剩多少桃子。"
+    ],
+    "source": "21-22 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_481",
+    "original_id": "homework_q_481"
+  },
+  {
+    "id": "homework_q_482",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "在马来西亚吉隆坡举行的国际奥委会第 128次全会上，中国北京以 44 票获得 2022 年第 24 届冬季奥林匹克运动会举办权。另一个申办城市哈萨克斯坦的阿拉木图获得的票数比中国北京少 · 1 /1 1 ，阿拉木图获得多少票？",
+    "options": [],
+    "answer": "40票",
+    "solution": "4 4 × ~ ( 1 - 1 /1 1 ) = 4 4 × 1 0 /1 1 = 4 0 （票）",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "求比一个数多/少几分之几的数是多少【分析】把中国北京的得票数看成单位“1”",
+      "阿拉木图的得票数 = 中国北京的得票数 × ~ ( 1 - 1 /1 1 )",
+      "据此解答。"
+    ],
+    "source": "21-22 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_482",
+    "original_id": "homework_q_482"
+  },
+  {
+    "id": "homework_q_483",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "学校举行纪念抗日战争胜利74 周年朗诵比赛，获三等奖的有 32 人，获二等奖的人数是获三等奖的 3 ，获一等奖的人数是获二等奖的1 ，获一等奖的有多少人？",
+    "options": [],
+    "answer": "4人",
+    "solution": "3 2 × 3 /4 × 1 /6 = 4 （人）",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "分数乘分数",
+      "求一个数的几分之几的问题"
+    ],
+    "source": "21-22 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_483",
+    "original_id": "homework_q_483"
+  },
+  {
+    "id": "homework_q_484",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "一本故事书有 120 页，小明第一天读了全书的 1 /6 , ，第二天读了余下的 1 /4 ，第三天应从第几页读起。",
+    "options": [],
+    "answer": "46页",
+    "solution": "1 2 0 × [ 1 /6 + ~ ( 1 - 1 /6 ) ~ × 1 /4 ] = 1 2 0 × 9 /2 4 = 4 5 （页） 4 5 + 1 = 4 6 （页）",
+    "knowledge_points": [
+      "分数的四则混合运算",
+      "分数乘分数"
+    ],
+    "source": "14-15 六年级上·贵州遵义·阶段练习",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_484",
+    "original_id": "homework_q_484"
+  },
+  {
+    "id": "homework_q_485",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "一个正方体的棱长和是18 米，这个正方体的体积是多少立方米？",
+    "options": [],
+    "answer": "2 7 /8 立方米",
+    "solution": "棱长： 1 8 ÷ 1 2 = 3 /2 （米） 体积：3 /2 ×3 /2 ×3 /2 = 9 /4 × 3 /2 = 2 7 /8 （立方米）",
+    "knowledge_points": [
+      "分数的连乘运算",
+      "正方体的体积",
+      "正方体有关棱长的应用【分析】正方体的棱长之和 = 棱长 × 1 2",
+      "根据公式计算出正方体的棱长",
+      "再利用“正方体的体积 = 棱长 × 棱长 × 棱长”即可求得。"
+    ],
+    "source": "21-22 六年级上·海南海口·期中",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_485",
+    "original_id": "homework_q_485"
+  },
+  {
+    "id": "homework_q_486",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "六年级同学给灾区的小朋友捐款。六（1）班捐了 600元，六（2）班捐的是六（1）班的 4 /5 ，六（3）班捐的是六（2）班的 7 /6 。六（3）班捐了多少元？",
+    "options": [],
+    "answer": "560元",
+    "solution": "6 0 0 × 4 /5 × 7 /6 = 4 8 0 × 7 /6 = 5 6 0 （元）",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "分数乘分数",
+      "求一个数的几分之几的问题【分析】先利用600 元乘 7 /6 求出六（2）班的捐款数",
+      "再将这个积乘7 /6",
+      "求出六（3）班捐了多少元。"
+    ],
+    "source": "20-21 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_486",
+    "original_id": "homework_q_486"
+  },
+  {
+    "id": "homework_q_487",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "简单",
+    "question": "甲乙两地相距 4200 千米，一辆汽车4 小时行驶了全程的 5 /7 ?，行驶了多少千米？",
+    "options": [],
+    "answer": "3000千米",
+    "solution": "4 2 0 0 × 5 /7 = 3 0 0 0 （千米）",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "基础行程问题",
+      "求一个数的几分之几的问题【分析】根据题意",
+      "利用乘法求出这辆汽车行驶了多少千米即可。"
+    ],
+    "source": "20-21 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_487",
+    "original_id": "homework_q_487"
+  },
+  {
+    "id": "homework_q_488",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "一个果园占地 2000 平方米，其中的 2 /5 种苹果树， 1 /4 种梨树，苹果树比梨树多多少平方米？",
+    "options": [],
+    "answer": "300平方米",
+    "solution": "( 2 /5 - 1 /4 ) × 2 0 0 0 = 3 /2 0 × 2 0 0 0 = 3 0 0 （平方米）",
+    "knowledge_points": [
+      "分数与整数的乘法"
+    ],
+    "source": "20-21 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_488",
+    "original_id": "homework_q_488"
+  },
+  {
+    "id": "homework_q_489",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "某鞋店购进一批运动鞋，第一周卖出 250双，第二周卖出的比第一周多 1 /5 , 。第二周卖出多少双？",
+    "options": [],
+    "answer": "300双",
+    "solution": "2 5 0 × ( 1 + 1 /5 ) = 2 5 0 × 6 /5 = 3 0 0 （双）",
+    "knowledge_points": [
+      "分数的四则混合运算",
+      "分数与整数的乘法",
+      "求比一个数多/少几分之几的数是多少【分析】根据题意",
+      "利用乘法列式计算出第二周卖出多少双即可。"
+    ],
+    "source": "20-21 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_489",
+    "original_id": "homework_q_489"
+  },
+  {
+    "id": "homework_q_490",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "简单",
+    "question": "只列式不计算。学校有男生 720 人，女生人数是男生的?。女生有多少人？",
+    "options": [],
+    "answer": "7 2 0 × 5 /6",
+    "solution": "7 2 0 × 5 /6 = 6 0 0 （人）",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": "20-21 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_490",
+    "original_id": "homework_q_490"
+  },
+  {
+    "id": "homework_q_491",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "小明和小文共扎 240朵花，小明扎了 1 /4 ，小文比小明多扎多少朵花？",
+    "options": [],
+    "answer": "120朵",
+    "solution": "( 1 - 1 /4 - 1 /4 ) \\ × 2 4 0 = 1 /2 × 2 4 0 = 1 2 0 （朵）",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "求一个数的几分之几的问题【分析】小明扎了 $ 1 /4",
+      "$",
+      "据此利用减法求出小文扎了几分之几",
+      "再利用减法求出小文比小明多扎了几分之几",
+      "从而利用乘法求出小文比小明多扎多少朵花。"
+    ],
+    "source": "20-21 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_491",
+    "original_id": "homework_q_491"
+  },
+  {
+    "id": "homework_q_492",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "某希望小学六年级有学生 420 人，五年级人数是六年级的 5 /6 ，四年级人数是五年级的 4 /5 。四年级有学生多少人？",
+    "options": [],
+    "answer": "280人",
+    "solution": "4 2 0 × 5 /6 × 4 /5 = 2 8 0 （人）",
+    "knowledge_points": [
+      "分数的连乘运算",
+      "分数与整数的乘法",
+      "分数乘分数",
+      "求一个数的几分之几的问题【分析】用六年级的 420 人先乘 ? 求出五年级的人数",
+      "再将其乘? 求出四年级的学生人数即可。"
+    ],
+    "source": "20-21 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_492",
+    "original_id": "homework_q_492"
+  },
+  {
+    "id": "homework_q_493",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "“要想富，先修路”，模范村要新修一条长 4.8千米的盘山公路。已经修了全长的 3 /4 , ，还剩多少千米没修？",
+    "options": [],
+    "answer": "1.2千米",
+    "solution": "4 . 8 × ~ ( 1 - 3 /4 ) = 4 . 8 × 1 4 = 1 . 2 （千米）",
+    "knowledge_points": [
+      "求一个数的几分之几的问题"
+    ],
+    "source": "20-21 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_493",
+    "original_id": "homework_q_493"
+  },
+  {
+    "id": "homework_q_494",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "计算题",
+    "difficulty": "简单",
+    "question": "只列式不计算。工地原有 4 /5 吨水泥，今天用去了它的 3 /5 ，用去了多少吨？",
+    "options": [],
+    "answer": "4 /5 × 3 /5",
+    "solution": "4 /5 × 3 /5 = 1 2 /2 5 （吨）",
+    "knowledge_points": [
+      "分数乘分数",
+      "求一个数的几分之几的问题"
+    ],
+    "source": "20-21 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_494",
+    "original_id": "homework_q_494"
+  },
+  {
+    "id": "homework_q_495",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "饭店买来面粉 7 /8 吨，第一天用去这批面粉的3 /7 ，第二天又用去1 /5 吨，还剩面粉多少吨？",
+    "options": [],
+    "answer": "3 /1 0 吨",
+    "solution": "7 /8 × 3 /7 - 1 /5 = 7 /8 -3 /8 - 1 /5 = 3 /1 0 （吨）",
+    "knowledge_points": [
+      "分数乘分数",
+      "求一个数的几分之几的问题",
+      "分数的加",
+      "减法混合运算",
+      "异分母分数加",
+      "减法【分析】根据题意",
+      "用 ? 吨乘 ? 先求出第一天用去的面粉量",
+      "再利用? 吨减去第一天和第二天用去的面粉量",
+      "求出还剩下多少吨面粉即可。"
+    ],
+    "source": "20-21 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_495",
+    "original_id": "homework_q_495"
+  },
+  {
+    "id": "homework_q_496",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "2012 年某地调整后企业退休人员月人均养老金达到 1721元，从 2013 年 1 月 1 日起，企业退休人员月人均养老金提高 1 /1 0 。提高后的退休人员月人均养老金是多少钱？",
+    "options": [],
+    "answer": "1893.1 元",
+    "solution": "1 7 2 1 × \\ ( 1 + 1 /1 0 ) = 1 7 2 1 × 1 1 /1 0 = 1 8 9 3 . 1 （元）",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "求比一个数多/少几分之几的数是多少"
+    ],
+    "source": "20-21 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_496",
+    "original_id": "homework_q_496"
+  },
+  {
+    "id": "homework_q_497",
+    "unit": "第一单元：分数乘法",
+    "chapter": "分数乘法综合练习",
+    "question_type": "应用题",
+    "difficulty": "中等",
+    "question": "台风登陆后，厦门有 384 人被困。救援队迅速展开救援，上午救出了被困人数的 1 /3 ，下午救出了被困人数的 1 /2 ，上午和下午一共救出多少人？还剩下几分之几需要救援？",
+    "options": [],
+    "answer": "320人； 1 /6",
+    "solution": "3 8 4 × 1 /3 = 1 2 8 （人） 3 8 4 × 1 /2 = 1 9 2 （人） 1 2 8 + 1 9 2 = 3 2 0 （人）",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "求一个数的几分之几的问题【分析】已知有 384 人被困",
+      "上午救出了被困人数的 ?",
+      "下午救出了被困人数的?",
+      "根据分数乘法的意义",
+      "分别求出上午和下午救出的人数",
+      "然后相加即可；把被困人员看作单位“1”",
+      "用 1 - 1 /3 - 1 /2 即为还剩下几分之几。据此解答即可。"
+    ],
+    "source": "20-21 六年级上·福建厦门·单元测试",
+    "create_time": "2025-09-30 01:17:21",
+    "merged_id": "homework_497",
+    "original_id": "homework_q_497"
+  }
+]
\ No newline at end of file
diff --git "a/\345\205\255\345\271\264\347\272\247\351\242\230\351\233\206/\345\205\255\345\271\264\347\272\247\344\270\212/\345\205\255\345\271\264\347\272\247\344\270\212\347\254\254\344\270\203\345\215\225\345\205\203\346\211\207\345\275\242\347\273\237\350\256\241\345\233\276\346\225\260\346\215\256\351\233\206.json" "b/\345\205\255\345\271\264\347\272\247\351\242\230\351\233\206/\345\205\255\345\271\264\347\272\247\344\270\212/\345\205\255\345\271\264\347\272\247\344\270\212\347\254\254\344\270\203\345\215\225\345\205\203\346\211\207\345\275\242\347\273\237\350\256\241\345\233\276\346\225\260\346\215\256\351\233\206.json"
new file mode 100644
index 0000000000000000000000000000000000000000..e0692214930449c83ff9715aac59ac667e7ec2f5
--- /dev/null
+++ "b/\345\205\255\345\271\264\347\272\247\351\242\230\351\233\206/\345\205\255\345\271\264\347\272\247\344\270\212/\345\205\255\345\271\264\347\272\247\344\270\212\347\254\254\344\270\203\345\215\225\345\205\203\346\211\207\345\275\242\347\273\237\350\256\241\345\233\276\346\225\260\346\215\256\351\233\206.json"
@@ -0,0 +1,7663 @@
+[  {
+  "id": "unit7_choice_001",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "要统计一种零食里的食品添加剂所占百分比情况，你会选用（）统计图。",
+  "options": [
+    "A. 条形",
+    "B. 折线",
+    "C. 扇形",
+    "D. 无法确定"
+  ],
+  "answer": "C",
+  "solution": "【分析】①条形统计图：用直条的长短表示数量的多少，从图中直观地看出数量的多少，便于比较；②折线统计图：不仅能看清数量的多少，还能通过折线的上升和下降表示数量的增减变化情况；③扇形统计图：清楚地看出各部分数量与总数量之间，部分与部分之间的关系；根据折线统计图、条形统计图和扇形统计图的特点，即可得出结论。\n【详解】要统计一种零食里的食品添加剂所占百分比情况,选择扇形统计图比较合适。故答案为: C",
+  "knowledge_points": [
+    "统计图的选择（扇形统计图）",
+    "复式条形统计图",
+    "统计图的选择（折线统计图）"
+  ],
+  "source": "24-25六年级上·内蒙古呼伦贝尔·期末",
+  "merged_id": "unit7_001",
+  "original_id": "unit7_choice_001"
+},
+{
+  "id": "unit7_choice_002",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "高山测绘大队想要用一张统计图来表示测量过的几座高山的高度情况, 最好用 ( )。",
+  "options": [
+    "A. 扇形统计图",
+    "B. 折线统计图",
+    "C. 条形统计图",
+    "D. 以上都可以"
+  ],
+  "answer": "C",
+  "solution": "【分析】条形统计图可以清楚地看出数量的多少。折线统计图不但可以表示出数量的多少，而且能够清楚地表示出数量增减变化的情况。扇形统计图表示各部分数量与总数之间的关系。\n【详解】高山测绘大队想要用一张统计图来表示测量过的几座高山的高度情况，最好用条形统计图。故答案为：C",
+  "knowledge_points": [
+    "1 格表示多个单位的单式条形统计图",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "24-25 六年级上·全国·课后作业",
+  "merged_id": "unit7_002",
+  "original_id": "unit7_choice_002"
+},
+{
+  "id": "unit7_choice_003",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "要表示 12 月份李老师家各项开支占总开支的百分比，选用（）更合适。",
+  "options": [
+    "A. 条形统计图",
+    "B. 折线统计图",
+    "C. 扇形统计图"
+  ],
+  "answer": "C",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。\n【详解】根据统计图的特点，要表示 12 月份李老师家各项开支占总开支的百分比，选用扇形统计图更合适。故答案为: C",
+  "knowledge_points": [
+    "统计图的选择（扇形统计图）",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "24-25 六年级上·辽宁·期末",
+  "merged_id": "unit7_003",
+  "original_id": "unit7_choice_003"
+},
+{
+  "id": "unit7_choice_004",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "一个鸡蛋按质量计算,蛋壳、蛋白、蛋黄约占整个鸡蛋的百分比分别为15%、53%、32%，如果将这些数据画成统计图，选（）统计图比较合适。",
+  "options": [
+    "A. 条形",
+    "B. 折线",
+    "C. 扇形"
+  ],
+  "answer": "C",
+  "solution": "【分析】条形统计图可以清楚地看出数量的多少。折线统计图不但可以表示出数量的多少，而且能够清楚地表示出数量增减变化的情况。扇形统计图表示各部分数量与总数之间的关系。\n【详解】一个鸡蛋按质量计算，蛋壳、蛋白、蛋黄约占整个鸡蛋的百分比分别为 15%、53%、32%，如果将这些数据画成统计图，选扇形统计图比较合适。故答案为：C",
+  "knowledge_points": [
+    "统计图的选择（扇形统计图）"
+  ],
+  "source": "2024 六年级上·全国·专题练习",
+  "merged_id": "unit7_004",
+  "original_id": "unit7_choice_004"
+},
+{
+  "id": "unit7_choice_005",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "通过分析信息的特点,选择合适的统计图。下列信息中适合用扇形统计图的是 ()。",
+  "options": [
+    "A. 六（1）班女生的身高",
+    "B. 一天中 24 个小时气温的变化",
+    "C. 玉米面的营养成分含量"
+  ],
+  "answer": "C",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；据此解答。\n【详解】A．六（1）班女生的身高，适合用条形统计图。B. 一天中 24 个小时气温的变化, 适合用折线统计。C. 玉米面的营养成分含量, 适合用扇形统计图。故答案为: C",
+  "knowledge_points": [
+    "统计图的选择（扇形统计图）",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "23-24 六年级下·山东枣庄·期末",
+  "merged_id": "unit7_005",
+  "original_id": "unit7_choice_005"
+},
+{
+  "id": "unit7_choice_006",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "要反映六年级各班学生环保知识竞赛获奖人数的情况, 选用 ( ) 统计图。",
+  "options": [
+    "A. 条形",
+    "B. 折线",
+    "C. 扇形",
+    "D. 扇形或者折线"
+  ],
+  "answer": "A",
+  "solution": "【分析】条形统计图可以直观地看出数量的多少；折线统计图不仅能看清数量的多少，还能反映数量的增减变化情况；扇形统计图能清楚地看出各部分数量与总数量之间的关系。\n【详解】由分析可得: 要反映六年级各班学生环保知识竞赛获奖人数的情况, 选用条形统计图。故答案为: A",
+  "knowledge_points": [
+    "统计图的选择（扇形统计图）",
+    "1 格表示一个单位的单式条形统计图",
+    "单式折线统计图",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "23-24 六年级上·全国·单元测试",
+  "merged_id": "unit7_006",
+  "original_id": "unit7_choice_006"
+},
+{
+  "id": "unit7_choice_007",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "妈妈要统计小辉每次体育测试成绩, 看看是进步还是退步, 应绘制 ( ) 统计图。",
+  "options": [
+    "A. 条形",
+    "B. 扇形",
+    "C. 折线",
+    "D. 扇形或者折线"
+  ],
+  "answer": "C",
+  "solution": "【分析】A. 条形统计图主要用于比较不同类别之间的数据差异, 不能直观地展示数据的变化趋势, 所以不适合用来观察成绩的进步或退步情况。B. 扇形统计图主要用于展示各部分在总体中所占的比例关系, 不适合用于体现成绩的变化。C. 折线统计图可以清晰地显示出小辉每次体育测试成绩的起伏变化, 从而能直观地判断出是进步还是退步。例如, 小辉第一次测试成绩是 80 分, 第二次是 85 分, 第三次是 90 分, 通过折线统计图可以清楚地看到成绩呈上升趋势, 表明在进步。D. 扇形统计图通常用于展示各部分所占的比例关系, 比如在体育测试中各项成绩（如跑步、跳远、跳高）在总成绩中所占的比例。但对于单独观察每次测试成绩的变化趋势, 扇形统计图无法直观呈现, 所以不适合用于此情况。\n【详解】折线统计图的特点是能清晰地反映数据的变化趋势。妈妈想要查看小辉每次体育测试成绩是进步还是退步，重点关注的是成绩的变化情况。故答案为: C",
+  "knowledge_points": [
+    "1 格表示一个单位的单式条形统计图",
+    "扇形统计图的特点及绘制",
+    "统计图的选择 (折线统计图)"
+  ],
+  "source": "23-24 六年级上·全国·单元测试",
+  "merged_id": "unit7_007",
+  "original_id": "unit7_choice_007"
+},
+{
+  "id": "unit7_choice_008",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "六年级同学进行跳远测试，并将结果绘制成了扇形统计图，其中成绩合格的有226人，在扇形统计图中占 40%，则六年级共有学生（）人。",
+  "options": [
+    "A. 565",
+    "B. 376",
+    "C. 310",
+    "D. 286"
+  ],
+  "answer": "A",
+  "solution": "【分析】已知跳远成绩合格的有 226 人, 占总人数的 40%,用成绩合格的人数除以占总数的百分率即可求出六年级的总人数。\n【详解】226 ÷ 40% = 565（人）则六年级共有学生565人。故答案为：A",
+  "knowledge_points": [
+    "已知一个数的百分之几是多少, 求这个数",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "23-24六年级上·全国·单元测试",
+  "merged_id": "unit7_008",
+  "original_id": "unit7_choice_008"
+},
+{
+  "id": "unit7_choice_009",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "扇形统计图表示 ( )。",
+  "options": [
+    "A. 数量的多少",
+    "B. 部分和总量的关系",
+    "C. 数量的增减变化",
+    "D. 以上都可以"
+  ],
+  "answer": "B",
+  "solution": "【分析】条形统计图能清楚地表示出数量的多少；折线统计图不仅能表示数量的多少，还能表示数量的增减变化情况；扇形统计图表示部分与整体之间的关系；据此解答。\n【详解】根据分析可知, 扇形统计图表示部分和总量的关系。故答案为: B",
+  "knowledge_points": [
+    "单式折线统计图",
+    "1格表示多个单位的单式条形统计图",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "23-24 六年级上·全国·单元测试",
+  "merged_id": "unit7_009",
+  "original_id": "unit7_choice_009"
+},
+{
+  "id": "unit7_choice_010",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "六二班全体同学参加社团活动，体育社团占全班人数的 25%，美术社团占 40%，音乐社团占 20%，科创社团占 15%，如果制成统计图，最合适的是（）。",
+  "options": [
+    "A. 扇形统计图",
+    "B. 条形统计图",
+    "C. 折线统计图"
+  ],
+  "answer": "A",
+  "solution": "【分析】条形统计图从图中能清楚地看出各种数量的多少,便于相互比较。折线统计图不仅能看清数量的多少，还能通过折线的上升和下降表示数量的增减变化情况。扇形统计图能清楚地看出各部分数量与总数量之间的关系。\n【详解】六二班全体同学参加社团活动，体育社团占全班人数的 25%, 美术社团占 40%, 音乐社团占 20%, 科创社团占 15%, 如果制成统计图, 最合适的是扇形统计图。故答案为: A",
+  "knowledge_points": [
+    "统计图的选择（扇形统计图）"
+  ],
+  "source": "23-24六年级下·重庆铜梁·期末",
+  "merged_id": "unit7_010",
+  "original_id": "unit7_choice_010"
+},
+{
+  "id": "unit7_choice_011",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "条形统计图的特点是 ( )。",
+  "options": [
+    "A. 表示数量的增减变化情况",
+    "B. 表示数量的多少",
+    "C. 表示部分与总数的关系"
+  ],
+  "answer": "B",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况; 扇形统计图能反映部分与整体的关系; 由此选择即可。\n【详解】A．表示数量的增减变化情况，这是折线统计图的特点，不符合题意；B. 表示数量的多少, 这是条形统计图的特点, 符合题意;C. 表示部分与总数的关系, 这是扇形统计图的特点, 不符合题意故答案为: B",
+  "knowledge_points": [
+    "单式折线统计图",
+    "复式条形统计图",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "23-24 四年级下·全国·单元测试",
+  "merged_id": "unit7_011",
+  "original_id": "unit7_choice_011"
+},
+{
+  "id": "unit7_choice_012",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "要反映某食品中各种营养成分的含量，最好选（）。",
+  "options": [
+    "A. 条形统计图",
+    "B. 折线统计图",
+    "C. 扇形统计图",
+    "D. 以上都可以"
+  ],
+  "answer": "C",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。根据条形统计图、折线统计图、扇形统计图各自的特点进行解答。\n【详解】要反映某食品中各种营养成分的含量，最好选扇形统计图。故答案为: C",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "2021·广东河源·小升初真题",
+  "merged_id": "unit7_012",
+  "original_id": "unit7_choice_012"
+},
+{
+  "id": "unit7_choice_013",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "某班有 50 人，其中三好学生 10 人，优秀学生干部 5 人，在扇形统计图上表示三好学生和优秀学生干部人数的圆心角分别是（）。",
+  "options": [
+    "A. 72°, 36°",
+    "B. 100°, 50°",
+    "C. 80°, 40°",
+    "D. 120°, 60°"
+  ],
+  "answer": "A",
+  "solution": "【分析】扇形统计图是用圆表示总数，用圆的面积表示总数，用扇形面积表示各部分占总数的百分比，圆的周角是360°，用360°乘三好学生和优秀学生干部占总人数的分率即可求解。\n【详解】三好学生的圆心角：360° × (10 ÷ 50) = 360° × 0.2 = 72°；优秀学生干部的圆心角：360° × (5 ÷ 50) = 360° × 0.1 = 36°。故答案为：A",
+  "knowledge_points": [
+    "求一个数占另一个数几分之几",
+    "扇形统计图的特点及绘制",
+    "弧、圆心角、扇形的认识",
+    "求一个数的几分之几的问题"
+  ],
+  "source": "2014 六年级·全国·课后作业",
+  "merged_id": "unit7_013",
+  "original_id": "unit7_choice_013"
+},
+{
+  "id": "unit7_choice_014",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "气象学家要研究某地夏季降水量变化情况，选择哪种统计图比较合适？（）",
+  "options": [
+    "A. 条形统计图",
+    "B. 折线统计图",
+    "C. 扇形统计图",
+    "D. 无法确定"
+  ],
+  "answer": "B",
+  "solution": "【分析】条形统计图：从图中直观地看出数量的多少，便于比较；折线统计图：不仅能看清数量的多少，还能反映数量的增减变化情况；扇形统计图：清楚地看出各部分数量与总数量之间的关系。据此解答即可。本题考查统计图的选择，明确各个统计图的特点是解题的关键。\n【详解】由分析可知：气象学家要研究某地夏季降水量变化情况, 应选择折线统计图。故答案为: B",
+  "knowledge_points": [
+    "单式折线统计图",
+    "1格表示多个单位的单式条形统计图",
+    "扇形统计图的特点及绘制",
+    "统计图的选择"
+  ],
+  "source": "21-22 五年级下·全国·单元测试",
+  "merged_id": "unit7_014",
+  "original_id": "unit7_choice_014"
+},
+{
+  "id": "unit7_choice_015",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "要反映小麦中所含各种营养成分的百分比，需绘制哪种统计图（表）较合适（）。",
+  "options": [
+    "A. 条形统计图",
+    "B. 折线统计图",
+    "C. 扇形统计图",
+    "D. 统计表"
+  ],
+  "answer": "C",
+  "solution": "【分析】条形统计图用直条的长短表示数量的多少，从图中直观地看出数量的多少，便于比较；折线统计图不仅能看清数量的多少，还能通过折线的上升和下降表示数量的增减变化情况；扇形统计图中用整个圆表示总数量，圆内各个扇形的大小表示各部分数量占总数量的百分比，通过扇形统计图可以清楚地看出各部分数量与总数量之间，部分与部分之间的关系；统计表是反映统计资料的表格，是对统计指标加以合理叙述的形式，它使统计资料条理化，简明清晰，便于检查数字的完整性和准确性，以及对比分析，据此选择合适的统计图。本题主要考查统计图的选择，掌握各统计图的特点及作用是解答题目的关键。\n【详解】分析可知，要反映小麦中所含各种营养成分的百分比，选择扇形统计图比较合适。故答案为: C",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "统计图的选择"
+  ],
+  "source": "18-19六年级上·广东汕头·期末",
+  "merged_id": "unit7_015",
+  "original_id": "unit7_choice_015"
+},
+{
+  "id": "unit7_choice_016",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "要反映鸡蛋各种营养成分的含量应选用 ( ) 统计图。",
+  "options": [
+    "A. 折线",
+    "B. 条形",
+    "C. 扇形"
+  ],
+  "answer": "C",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。此题应根据条形统计图、折线统计图、扇形统计图各自的特点进行解答。\n【详解】要反映鸡蛋各种营养成分的含量应选用扇形统计图。故答案为: C",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "统计图的选择"
+  ],
+  "source": "21-22 六年级上·广东梅州·期末",
+  "merged_id": "unit7_016",
+  "original_id": "unit7_choice_016"
+},
+{
+  "id": "unit7_choice_017",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "需要清楚地表示出各部分数量与总数量之间的关系时，应选用（）。",
+  "options": [
+    "A. 统计表",
+    "B. 条形统计图",
+    "C. 折线统计图",
+    "D. 扇形统计图"
+  ],
+  "answer": "D",
+  "solution": "【分析】条形统计图用直条的长短表示数量的多少，从图中直观地看出数量的多少，便于比较；折线统计图不仅能看清数量的多少，还能通过折线的上升和下降表示数量的增减变化情况；扇形统计图可以清楚地看出各部分数量与总数量之间，部分与部分之间的关系，据此解答。理解并掌握各统计图的特点及作用是解答题目的关键。\n【详解】分析可知，需要清楚地表示出各部分数量与总数量之间的关系时，应选用扇形统计图。故答案为: D",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "统计图的选择"
+  ],
+  "source": "22-23六年级上·广东广州·期末",
+  "merged_id": "unit7_017",
+  "original_id": "unit7_choice_017"
+},
+{
+  "id": "unit7_choice_018",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "要反映六 (1) 班同学喜欢各种水果的人数分别占全班人数的百分比, 用 ( )比较合适。",
+  "options": [
+    "A. 扇形统计图",
+    "B. 条形统计图",
+    "C. 折线统计图"
+  ],
+  "answer": "A",
+  "solution": "【分析】扇形统计图：可以清楚地看出各部分数量与总数量之间，部分与部分之间的关系；条形统计图：用直条的长短表示数量的多少，从图中直观地看出数量的多少，便于比较；折线统计图：不仅能看清数量的多少，还能通过折线的上升和下降表示数量的增减变化情况；据此解答。掌握各统计图的特点及作用是解答题目的关键。\n【详解】分析可知, 扇形统计图中用整个圆表示总数量,圆内各个扇形的大小表示各部分数量占总数量的百分比,要表示喜欢各种水果的人数分别占全班人数的百分比选择扇形统计图比较合适。故答案为: A",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "统计图的选择"
+  ],
+  "source": "20-21 六年级下·甘肃天水·期末",
+  "merged_id": "unit7_018",
+  "original_id": "unit7_choice_018"
+},
+{
+  "id": "unit7_choice_019",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "下列的信息资料中, 适合用折线统计图表示的是 ( )。",
+  "options": [
+    "A. 学校各年级的人数",
+    "B. 五年级各班做好事的件数",
+    "C. 学校教师人数",
+    "D. 六月份气温变化情况"
+  ],
+  "answer": "D",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。此题应根据条形统计图、折线统计图、扇形统计图各自的特点进行解答。\n【详解】A．要表示学校各年级的人数，最好选用条形统计图；B. 要表示五年级各班做好事的件数, 最好选用条形统计图;C. 要表示学校教师人数, 最好选用条形统计图;D. 要表示六月份气温变化情况, 最好用折线统计图。故答案为: D",
+  "knowledge_points": [
+    "单式折线统计图",
+    "1格表示多个单位的单式条形统计图",
+    "扇形统计图的特点及绘制",
+    "统计图的选择"
+  ],
+  "source": "2022·四川凉山·小升初真题",
+  "merged_id": "unit7_019",
+  "original_id": "unit7_choice_019"
+},
+{
+  "id": "unit7_choice_020",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "表示一位病人一天内体温变化情况，绘制（）统计图比较合适。",
+  "options": [
+    "A. 折线",
+    "B. 扇形",
+    "C. 条形",
+    "D. 以上都不对"
+  ],
+  "answer": "A",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。此题应根据条形统计图、折线统计图、扇形统计图各自的特点进行解答。\n【详解】根据统计图的特点可知：表示一位病人一天内体温变化情况，绘制折线统计图比较合适。故答案为：A",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "2022·广西贺州·小升初真题",
+  "merged_id": "unit7_020",
+  "original_id": "unit7_choice_020"
+},
+{
+  "id": "unit7_choice_021",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "画统计图时, 要根据信息的特点来选择合适的统计图，在下面的信息中适合用折线统计图的是（）。",
+  "options": [
+    "A. 五年级男同学的身高",
+    "B. 明明 0 到 10 岁的体重",
+    "C. 大豆的营养成分",
+    "D. 欢欢 3 门功课的成绩"
+  ],
+  "answer": "B",
+  "solution": "【分析】条形统计图可以清楚地看出数量的多少；折线统计图不仅可以表示数量的多少，还可以清楚看出数量的增减变化情况；扇形统计图能清楚地看出各部分数量与总数量之间的关系。据此解答。此题的解题关键是熟悉各种统计图的不同特征及应用。\n【详解】A．统计表现五年级男同学的身高的多少，应绘制条形统计图比较合适；B. 不仅要体现明明 0 到 10 岁的体重, 还要能看出明明 0 到10 岁的体重变化的趋势，应绘制折线统计图比较合适;C. 说明大豆中各种营养成分所占的百分比时, 应绘制扇形统计图比较合适;D. 统计欢欢 3 门功课的成绩, 表示 3 门功课成绩数量的多少, 应绘制条形统计图比较合适;故答案为: B",
+  "knowledge_points": [
+    "单式折线统计图",
+    "1格表示多个单位的单式条形统计图",
+    "扇形统计图的特点及绘制",
+    "统计图的选择"
+  ],
+  "source": "21-22 五年级下·广西南宁·期末",
+  "merged_id": "unit7_021",
+  "original_id": "unit7_choice_021"
+},
+{
+  "id": "unit7_choice_022",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "下面属于收集数据方法的是 ( )。",
+  "options": [
+    "A. 统计表",
+    "B. 条形统计图",
+    "C. 扇形统计图",
+    "D. 问卷调查询问法"
+  ],
+  "answer": "D",
+  "solution": "【分析】通过统计表可以找出数据分布的规律；利用统计图表示经过整理的数据，能更直观地反映数据的规律；统计表和统计图都是收集完数据之后的工作。问卷调查询问法是收集数据的方法。数据处理的一般过程：1、数据收集、2、整理数据、3、描绘数据、4、分析数据、5、得出结论。\n【详解】统计表、条形统计图、扇形统计图是数据的表示方法，属于描绘数据，问卷调查询问法是收集数据的方法。故答案为：D",
+  "knowledge_points": [
+    "数据的搜集与整理"
+  ],
+  "source": "21-22 六年级下·湖北咸宁·期末",
+  "merged_id": "unit7_022",
+  "original_id": "unit7_choice_022"
+},
+{
+  "id": "unit7_choice_023",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "三洋电机厂为了能清楚的表示出上半年月产量的多少和增减变化的情况，应绘制（）。",
+  "options": [
+    "A. 扇形统计图",
+    "B. 折线统计图",
+    "C. 条形统计图"
+  ],
+  "answer": "B",
+  "solution": "【分析】扇形统计图表示的是部分在总体中所占的百分比,但一般不能直接从图中得到具体的数据; 折线统计图表示的是事物的增减变化情况; 条形统计图能清楚地表示出每个项目的具体数目; 由此判定即可。此题考查扇形统计图、折线统计图、条形统计图各自的特点。\n【详解】由分析可得：三洋电机厂为了能清楚的表示出上半年月产量的多少和增减变化的情况，应绘制折线统计图。故答案为：B",
+  "knowledge_points": [
+    "单式折线统计图",
+    "1格表示多个单位的单式条形统计图",
+    "扇形统计图的特点及绘制",
+    "统计图的选择"
+  ],
+  "source": "2014·安徽亳州·小升初真题",
+  "merged_id": "unit7_023",
+  "original_id": "unit7_choice_023"
+},
+{
+  "id": "unit7_choice_024",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "野象群一路北上，引起民众的极大关注，为普及象群知识，需要绘制统计图。下面各话题，更适合用折线统计图表示的是（），更适合用扇形统计图表示的是（）。\n(1) 几头成年野象睡眠时间长短比较\n② 野象在亚洲地区分布的百分比\n(3) 三十年来野象数量增减变化情况\n④ 幼年和成年野象一天食量多少情况",
+  "options": [
+    "A. ①和③",
+    "B. ①和④",
+    "C. ③和②",
+    "D. ②和④"
+  ],
+  "answer": "C",
+  "solution": "【分析】折线统计图能清楚地反映事物的变化情况；扇形统计图能清楚地表示出各部分在总体中所占的百分比；根据统计图的特点来判断，即可求解。本题主要考查根据统计图的特点来选择统计图。\n【详解】③中三十年来野象数量增减变化情况反映的是数量增减变化情况，因此更适合用折线统计图； ②中野象在亚洲地区分布的百分比表示的是各部分在总体中所占的百分比，因此应选用扇形统计图。故答案为: C",
+  "knowledge_points": [
+    "单式折线统计图",
+    "扇形统计图的特点及绘制",
+    "统计图的选择"
+  ],
+  "source": "20-21六年级下·浙江宁波·期末",
+  "merged_id": "unit7_024",
+  "original_id": "unit7_choice_024"
+},
+{
+  "id": "unit7_choice_025",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "为了说明黄豆中各种营养成分所占的百分比，应绘制（）。",
+  "options": [
+    "A. 条形统计图",
+    "B. 折线统计图",
+    "C. 扇形统计图",
+    "D. 统计表"
+  ],
+  "answer": "C",
+  "solution": "【分析】折线统计图能够显示数据的变化趋势，反映事物的变化情况；条形统计图不但能够使人们一眼看出各个数据的大小而且易于比较数据之间的差别;扇形统计图既可以表示部分在总体中所占的百分比，又易于显示每组数据相对于总数的大小。统计图比统计表更能直观、形象的反映数据间的变化和联系；据此解答。此题主要考查统计图的选择, 就熟练掌握各种统计图和统计表的特点。\n【详解】根据统计图的特点可知：为了说明黄豆中各种营养成分所占的百分比，应绘制扇形统计图较合适。故答案为: C",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "统计图的选择"
+  ],
+  "source": "2022·新疆克拉玛依·小升初真题",
+  "merged_id": "unit7_025",
+  "original_id": "unit7_choice_025"
+},
+{
+  "id": "unit7_choice_026",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "医务人员要把病人的体温变化情况制成统计图，选择（）比较恰当。",
+  "options": [
+    "A. 条形统计图",
+    "B. 折线统计图",
+    "C. 扇形统计图"
+  ],
+  "answer": "B",
+  "solution": "【分析】要把病人的体温变化情况制成统计图，关键是要反映出体温变化情况，选择折线统计图最为合适。本题考查的是统计图的选择，条形统计图可以清楚地反映出数据的多少，折线统计图可以清楚地反映出数据的变化情况，扇形统计图可以清楚地反映出部分与整体的关系。\n【详解】医务人员要把病人的体温变化情况制成统计图，选择折线统计图比较恰当；故答案为：B",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "2015·全国·小升初真题",
+  "merged_id": "unit7_026",
+  "original_id": "unit7_choice_026"
+},
+{
+  "id": "unit7_choice_027",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "下列几种统计情况，用复式折线统计图反映比较合适的是（）。\nA. 陈晓本学期数学单元测式成绩的波动情况\nB. 苏宁电器 1 月份销售各种品牌冰箱的百分比\nC. 晋江市和惠安县近 5 年财政收入变化情况\nD. 实验小学合唱团和舞蹈队的男、女生人数情况",
+  "options": [
+    "A. 陈晓本学期数学单元测式成绩的波动情况",
+    "B. 苏宁电器 1 月份销售各种品牌冰箱的百分比",
+    "C. 晋江市和惠安县近 5 年财政收入变化情况",
+    "D. 实验小学合唱团和舞蹈队的男、女生人数情况"
+  ],
+  "answer": "C",
+  "solution": "【分析】折线统计图不仅能看清数量的多少，还能通过折线的上升和下降表示数量的增减变化情况。复式折线统计图表示2个及以上的量的增减变化情况。关键是熟悉各种统计图的特点。\n【详解】A. 用单式折线统计图合适;B. 用扇形统计图合适;C. 用复式折线统计图合适;D. 用复式条形统计图合适。故答案为: C",
+  "knowledge_points": [
+    "复式折线统计图",
+    "统计图的选择"
+  ],
+  "source": "20-21 六年级上·福建泉州·期末",
+  "merged_id": "unit7_027",
+  "original_id": "unit7_choice_027"
+},
+{
+  "id": "unit7_choice_028",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "要表示小南家今年十月份各项生活支出占月总支出的百分比情况，比较合适的是用（）。",
+  "options": [
+    "A. 扇形统计图",
+    "B. 折线统计图",
+    "C. 条形统计图",
+    "D. 折线统计图或条形统计图"
+  ],
+  "answer": "A",
+  "solution": "【分析】扇形统计图清楚地看出各部分数量与总数量之间,部分与部分之间的关系。关键是熟悉各种统计图的特点，条形统计图便于比较多少，折线统计图能反映增减变化趋势。\n【详解】要表示小南家今年十月份各项生活支出占月总支出的百分比情况，比较合适的是用扇形统计图。故答案为：A",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "20-21 六年级上·福建漳州·期末",
+  "merged_id": "unit7_028",
+  "original_id": "unit7_choice_028"
+},
+{
+  "id": "unit7_choice_029",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "表示一个地区 2014 年的月平均气温变化情况，用（）表示较为合适。",
+  "options": [
+    "A. 条形统计图",
+    "B. 折线统计图",
+    "C. 扇形统计图"
+  ],
+  "answer": "B",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。本题考查统计图的选择，应根据条形统计图、折线统计图、扇形统计图各自的特点进行解答。\n【详解】根据统计图的特点可知：表示一个地区 2014 年的月平均气温变化情况，用折线统计图表示较为合适；故选：B。",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "21-22 五年级下·全国·单元测试",
+  "merged_id": "unit7_029",
+  "original_id": "unit7_choice_029"
+},
+{
+  "id": "unit7_choice_030",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "要准确地表示出某县2021年在各领域的财政支出分别占全年财政支出的百分比情况，选用（）统计图比较合适。",
+  "options": [
+    "A. 条形",
+    "B. 折线",
+    "C. 扇形"
+  ],
+  "answer": "C",
+  "solution": "【分析】由扇形统计图特征可知，要表示各领域的财政支出分别占全年财政支出的百分比情况，即表示部分与整体的数量关系，选用扇形统计图比较合适，据此解答。掌握各统计图的特征是解答题目的关键。\n【详解】A．条形统计图：用直条的长短表示数量的多少，从图中直观地看出数量的多少，便于比较，错误；B. 折线统计图: 不仅能看清数量的多少, 还能通过折线的上升和下降表示数量的增减变化情况, 错误;C. 扇形统计图: 扇形统计图可以清楚地看出各部分数量与总数量之间的关系, 可以表示出各领域的财政支出分别占全年财政支出的百分比情况, 正确。故答案为: C。",
+  "knowledge_points": [
+    "单式折线统计图",
+    "扇形统计图的特点及绘制",
+    "统计图的选择"
+  ],
+  "source": "21-22六年级上·广西玉林·期末",
+  "merged_id": "unit7_030",
+  "original_id": "unit7_choice_030"
+},
+{
+  "id": "unit7_choice_031",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "在一个扇形统计图中, 黄花朵数对应的圆心角是 90°, 那么黄花朵数占各种颜色花的总数的 ( )。",
+  "options": [
+    "A. 1/4",
+    "B. 1/2",
+    "C. 1/8"
+  ],
+  "answer": "A",
+  "solution": "【分析】在扇形统计图中, 圆周角是 360°, 计算黄花朵数对应的圆心角是 90° 占 360° 的分率即可。在扇形统计图中计算部分占整体的分率, 可以求出部分对应的圆心角占 360° 的分率。\n【详解】90° ÷ 360° = 1/4 故答案为：A",
+  "knowledge_points": [
+    "求一个数占另一个数几分之几",
+    "扇形统计图的特点及绘制",
+    "弧、圆心角、扇形的认识"
+  ],
+  "source": "21-22 六年级上·湖南怀化·单元测试",
+  "merged_id": "unit7_031",
+  "original_id": "unit7_choice_031"
+},
+{
+  "id": "unit7_choice_032",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "\"水是生命之源，成年人每天体内 47% 的水分靠喝水获得， 14% 来自体内氧化时所释放出来的水， 39% 来自食物中所含的水。\"要表述上述信息，选择（）最合适。",
+  "options": [
+    "A. 折现统计图",
+    "B. 条形统计图",
+    "C. 扇形统计图",
+    "D. 复式折线统计图"
+  ],
+  "answer": "C",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。此题考查的是统计图的选择，解答此题应根据条形统计图、折线统计图、扇形统计图各自的特点进行解答。\n【详解】根据统计图的特点可知: '水是生命之源, 成年人每天体内 47% 的水分靠喝水获得, 14% 来自体内氧化时所释放出来的水, 39% 来自食物中所含的水。'要表示各种情况所占的百分比, 制成扇形统计图比较合适。故选: C",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "2022·河北保定·小升初真题",
+  "merged_id": "unit7_032",
+  "original_id": "unit7_choice_032"
+},
+{
+  "id": "unit7_choice_033",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "描述希望小学2016年入学的男、女生五年来平均身高各自的变化情况，用（）描述比较合适。",
+  "options": [
+    "A. 单式折线统计图",
+    "B. 复式折线统计图",
+    "C. 扇形统计图"
+  ],
+  "answer": "B",
+  "solution": "【分析】一般来说，如果几个数量是并列的，只要求表示数量的多少时，选条形统计图。如果表示一个量或几个量增减变化情况，则选折线统计图。如果要求表示各部分数量与总数量之间的关系，则选扇形统计图；据此解答。本题主要考查统计图的选择, 牢记各统计图的作用是解题得关键。\n【详解】由分析可知：描述希望小学 2016 年入学的男、女生五年来平均身高各自的变化情况，用复式折线统计图描述比较合适。故答案为: B",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "21-22六年级上·辽宁鞍山·期末",
+  "merged_id": "unit7_033",
+  "original_id": "unit7_choice_033"
+},{
+  "id": "unit7_choice_034",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "为了清楚地表示每种果树的棵数占总棵数的百分比, 应绘制 ( ) 统计图。",
+  "options": [
+    "A. 条形",
+    "B. 折线",
+    "C. 扇形"
+  ],
+  "answer": "C",
+  "solution": "【分析】扇形统计图是以一个圆的面积（看作单位'1'）表示物体的总数量，以相应的扇形面积占整个圆面积的百分数表示各有关部分占总数量的百分数的统计图。关键是熟悉各种统计图的特点。\n【详解】为了清楚地表示每种果树的棵数占总棵数的百分比，应绘制扇形统计图。故答案为: C",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "18-19 六年级上·全国·期末",
+  "merged_id": "unit7_034",
+  "original_id": "unit7_choice_034"
+},
+{
+  "id": "unit7_choice_035",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "某商店按 5% 缴纳营业税 200 元，则商场的营业额是（）元。",
+  "options": [
+    "A. 400",
+    "B. 1000",
+    "C. 4000"
+  ],
+  "answer": "C",
+  "solution": "【分析】把营业额看作单位'1'，根据营业额的 5% 是200元，所以用200除以 5% 就是营业额是多少元。营业税 = 营业额 × 税率; 已知一个是的百分之几是多少, 求这个数用除法。\n【详解】200 ÷ 5% = 4000（元）则该商店的营业额是 4000 元。故答案为: C",
+  "knowledge_points": [
+    "税率问题"
+  ],
+  "source": "20-21 六年级上·湖南怀化·单元测试",
+  "merged_id": "unit7_035",
+  "original_id": "unit7_choice_035"
+},
+{
+  "id": "unit7_choice_036",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "要反映各年级人数占全校总人数的百分比，要绘制（）统计图。",
+  "options": [
+    "A. 条形",
+    "B. 折线",
+    "C. 扇形"
+  ],
+  "answer": "C",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。此题应根据条形统计图、折线统计图、扇形统计图各自的特点进行解答。\n【详解】由分析可知:为了清楚地看出某校各年级人数占全校总人数的百分比情况，应选用扇形统计图。故选: C",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "20-21 六年级上·湖南怀化·单元测试",
+  "merged_id": "unit7_036",
+  "original_id": "unit7_choice_036"
+},
+{
+  "id": "unit7_choice_037",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "医护人员记录病人几天的体温情况，应选用（）统计图。",
+  "options": [
+    "A. 条形",
+    "B. 折线",
+    "C. 扇形"
+  ],
+  "answer": "B",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；扇形统计图能反映部分与整体的关系；与条形、扇形统计图相比，折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；由此根据情况选择即可。此题考查的是选择合适的统计图, 解答此题应根据条形统计图、折线统计图、扇形统计图各自的特点进行解答。\n【详解】因为折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况，所以要记录病人一天的体温变化情况，应选用折线统计图。故答案为：B。",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "20-21 六年级上·湖南怀化·单元测试",
+  "merged_id": "unit7_037",
+  "original_id": "unit7_choice_037"
+},
+{
+  "id": "unit7_choice_038",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "能表示总数与部分数之间的关系应选择（）统计图。",
+  "options": [
+    "A. 条形统计图",
+    "B. 折线统计图",
+    "C. 扇形统计图"
+  ],
+  "answer": "C",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。此题应根据条形统计图、折线统计图、扇形统计图各自的特点进行解答。\n【详解】因为扇形统计图能反映部分与整体的关系，所以要表示部分与总数量之间的关系应选用扇形统计图。故选: C",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "20-21 六年级上·湖南怀化·单元测试",
+  "merged_id": "unit7_038",
+  "original_id": "unit7_choice_038"
+},
+{
+  "id": "unit7_choice_039",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "为清楚地表示赣州2019年月降水量的变化情况，应选择（）。",
+  "options": [
+    "A. 折线统计图",
+    "B. 条形统计图",
+    "C. 扇形统计图"
+  ],
+  "answer": "A",
+  "solution": "【分析】分析各统计图的特点, 结合题意选择合适的统计图即可。掌握各统计图的特点及作用是解答题目的关键。\n【详解】A．折线统计图：不仅能看清数量的多少，还能通过折线的上升和下降表示数量的增减变化情况，可以表示出降水量的变化情况，正确；B. 条形统计图: 用直条的长短表示数量的多少, 从图中直观地看出数量的多少, 不合适, 错误;C. 扇形统计图: 扇形统计图可以清楚地看出各部分数量与总数量之间, 部分与部分之间的关系, 不能表示出降水量的变化情况, 错误。故答案为：A",
+  "knowledge_points": [
+    "1 格表示一个单位的单式条形统计图",
+    "单式折线统计图",
+    "扇形统计图的特点及绘制",
+    "统计图的选择"
+  ],
+  "source": "21-22六年级上·江西赣州·单元测试",
+  "merged_id": "unit7_039",
+  "original_id": "unit7_choice_039"
+},
+{
+  "id": "unit7_choice_040",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "为了能够形象地表示出某地区去年每个月的月平均气温，最适宜选用的统计图是（）。",
+  "options": [
+    "A. 条形统计图",
+    "B. 复式条形统计图",
+    "C. 折线统计图",
+    "D. 扇形统计图"
+  ],
+  "answer": "C",
+  "solution": "【分析】要统计某地去年月平均气温情况，各月平均气温是有起伏变化的，这符合折线统计图的特点。据此解题。本题考查了统计图的选择, 明确常见统计图表的特点是解题的关键。\n【详解】要能够形象地表示出某地区去年每个月的月平均气温，最适宜选用的统计图是折线统计图。故答案为: C",
+  "knowledge_points": [
+    "单式折线统计图",
+    "1格表示多个单位的单式条形统计图",
+    "扇形统计图的特点及绘制",
+    "统计图的选择"
+  ],
+  "source": "20-21 六年级上·全国·单元测试",
+  "merged_id": "unit7_040",
+  "original_id": "unit7_choice_040"
+},
+{
+  "id": "unit7_fill_041",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "扇形统计图是用( )表示总数，用圆内各个( )的大小表示各部分数量占总数的百分数。",
+  "options": [],
+  "answer": "整个圆;扇形",
+  "solution": "【分析】扇形统计图是用整个圆表示总数量，用圆内各个扇形的大小表示各部分数量占总数量的百分数；通过扇形统计图可以很清楚地表示出各部分数量占总数量的百分比。\n【详解】根据分析可知：扇形统计图是用整个圆表示总数，用圆内各个扇形的大小表示各部分数量占总数的百分数。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "25-26 六年级上·海南海口·单元测试",
+  "merged_id": "unit7_041",
+  "original_id": "unit7_fill_041"
+},
+{
+  "id": "unit7_fill_042",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "我们学过的统计图有____统计图、____统计图和____统计图，林老师要统计各年级的男、女生人数，应选用____式统计图比较合适；王老师要统计全校各年级人与全校总人数的关系，应选用____统计图比较合适。",
+  "options": [],
+  "answer": "条形;折线;扇形;条形;扇形",
+  "solution": "【分析】（1）扇形统计图用整个圆的面积表示总数，用圆内的各扇形的面积表示各部分占总数的百分数，从图中能清楚地看出各部分占总数的百分比，以及部分与部分之间的关系；(2) 条形统计图从图中能清楚地看出各数量的多少, 便于相互比较;(3) 折线统计图从图中能清楚地看出数量变化的趋势, 也能看出数量的多少。\n【详解】我们学过的统计图有条形统计图、折线统计图和扇形统计图，林老师要统计各年级的男、女生人数，应选用条形式统计图比较合适；王老师要统计全校各年级人与全校总人数的关系，应选用扇形统计图比较合适。",
+  "knowledge_points": [
+    "统计图的选择（扇形统计图）"
+  ],
+  "source": "24-25六年级下·海南海口·单元测试",
+  "merged_id": "unit7_042",
+  "original_id": "unit7_fill_042"
+},
+{
+  "id": "unit7_fill_043",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "如要统计全校各年级人数，选用( )统计图较合适；统计六年级同学喜欢各种球类人数约百分比，选用( )统计图较合适；如统计某病人体温升降变化情况，选用( )统计图较合适。",
+  "options": [],
+  "answer": "条形;扇形;折线",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况解答即可。\n【详解】如要统计全校各年级人数，选用条形统计图较合适；统计六年级同学喜欢各种球类人数约百分比，选用扇形统计图较合适；如统计某病人体温升降变化情况，选用折线统计图较合适。",
+  "knowledge_points": [
+    "统计图的选择（扇形统计图）",
+    "1格表示多个单位的单式条形统计图",
+    "统计图的选择（折线统计图）"
+  ],
+  "source": "23-24六年级上·全国·单元测试",
+  "merged_id": "unit7_043",
+  "original_id": "unit7_fill_043"
+},
+{
+  "id": "unit7_fill_044",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "表示各种数量的多少需要画( )统计图；表示数量的增减变化情况需要画( )统计图；表示各部分数量和整体之间的百分比需要画()统计图。",
+  "options": [],
+  "answer": "条形;折线;扇形",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系。据此填空。\n【详解】根据三种统计图的特点可知：表示各种数量的多少需要画条形统计图；表示数量的增减变化情况需要画折线统计图；表示各部分数量和整体之间的百分比需要画扇形统计图。",
+  "knowledge_points": [
+    "统计图的选择（扇形统计图）",
+    "统计图的选择(折线统计图)"
+  ],
+  "source": "24-25 六年级上·全国·课后作业",
+  "merged_id": "unit7_044",
+  "original_id": "unit7_fill_044"
+},
+{
+  "id": "unit7_fill_045",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "我们学过的统计图有( )、( )和( )；为了表示一位病人的体温变化情况，应绘制( )；( )统计图不能表示各部分的具体数量。",
+  "options": [],
+  "answer": "条形统计图;折线统计图;扇形统计图;折线统计图;扇形",
+  "solution": "【分析】我们学过的统计图有：条形统计图、折线统计图、扇形统计图。扇形统计图的特点：能够反映出部分和整体之间的关系；折线统计图的特点：能够反映数量的多少，能够反映出数量的增减变化情况；条形统计图的特点：能够清楚地反映出数量的多少。\n【详解】由分析可得：我们学过的统计图有条形统计图、折线统计图和扇形统计图；为了表示一位病人的体温变化情况，应绘制折线统计图；扇形统计图不能表示各部分的具体数量。",
+  "knowledge_points": [
+    "统计图的选择（扇形统计图）",
+    "统计图的选择（折线统计图）"
+  ],
+  "source": "23-24 六年级上·全国·单元测试",
+  "merged_id": "unit7_045",
+  "original_id": "unit7_fill_045"
+},
+{
+  "id": "unit7_fill_046",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "要表示一个病人最近两周血压的变化情况，应绘制( )统计图。",
+  "options": [],
+  "answer": "折线",
+  "solution": "【分析】条形统计图可以直观地看出数量的多少；折线统计图不仅能看清数量的多少，还能反映数量的增减变化情况；扇形统计图能清楚地看出各部分数量与总数量之间的关系。\n【详解】由分析可得：要表示一个病人最近两周血压的变化情况，应绘制折线统计图。",
+  "knowledge_points": [
+    "统计图的选择（折线统计图）"
+  ],
+  "source": "23-24 六年级上·全国·单元测试",
+  "merged_id": "unit7_046",
+  "original_id": "unit7_fill_046"
+},
+{
+  "id": "unit7_fill_047",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "要描述某村各种农作物产量的占比情况, 应绘制( )统计图。",
+  "options": [],
+  "answer": "扇形",
+  "solution": "【分析】扇形统计图的特点：能够反映出部分和整体之间的关系；折线统计图：能够反映数量的多少，能够反映出数量的增减变化情况；条形统计图：能够清楚的反映出数量的多少；根据各种统计图的特点，选择合适的统计图。\n【详解】由分析可得：要描述某村各种农作物产量的占比情况，应绘制扇形统计图。",
+  "knowledge_points": [
+    "统计图的选择（扇形统计图）"
+  ],
+  "source": "23-24 六年级上·全国·单元测试",
+  "merged_id": "unit7_047",
+  "original_id": "unit7_fill_047"
+},
+{
+  "id": "unit7_fill_048",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "扇形统计图的优点是可以很清楚地表示出( )与( )之间的关系。",
+  "options": [],
+  "answer": "各部分数量；总数",
+  "solution": "【详解】扇形统计图的优点是可以很清楚地表示出各部分数量与总数之间的关系。例如：我国国土面积约 960 万平方千米，各种地形所占百分比如图。images/6131ad56b975bd01c7ead371c637be715c02b80a7a92ff8e32e0c57c0386a04f.jpg从图中可知，山地面积占我国国土面积的 33%，高原面积占我国国土面积的 26% 等。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "2014 六年级·全国·课后作业",
+  "merged_id": "unit7_048",
+  "original_id": "unit7_fill_048"
+},
+{
+  "id": "unit7_fill_049",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "医院要掌握病人的血压变化情况，用( )统计图表示比较合适；要了解病人血液中各种成分的含量，用( )统计图表示比较合适。",
+  "options": [],
+  "answer": "折线；扇形",
+  "solution": "【分析】条形统计图能够清楚的看出数量的多少；折线统计图不仅能看清数量的多少，还能通过折线的上升和下降表示数量的增减变化情况；扇形统计图可以清楚地看出各部分数量与总数量之间，部分与部分之间的关系，据此解答。熟练掌握条形、折线和扇形统计图的特点及作用是解答题目的关键。\n【详解】分析可知, 医院要掌握病人的血压变化情况, 用折线统计图表示比较合适, 要了解病人血液中各种成分的含量, 用扇形统计图表示比较合适。",
+  "knowledge_points": [
+    "单式折线统计图",
+    "扇形统计图的特点及绘制",
+    "统计图的选择"
+  ],
+  "source": "21-22 六年级上·广东韶关·期末",
+  "merged_id": "unit7_049",
+  "original_id": "unit7_fill_049"
+},
+{
+  "id": "unit7_fill_050",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "要反映学校每个年级的学生数，应当选( )统计图，制作( )统计图能比较好地表示学校图书馆各类图书与总数之间的关系。",
+  "options": [],
+  "answer": "条形；扇形",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。此题应根据条形统计图、折线统计图、扇形统计图各自的特点进行解答。\n【详解】由分析可知：要反映学校每个年级的学生数，应当选条形统计图，制作扇形统计图能比较好地表示学校图书馆各类图书与总数之间的关系。",
+  "knowledge_points": [
+    "单式折线统计图",
+    "1格表示多个单位的单式条形统计图",
+    "扇形统计图的特点及绘制",
+    "统计图的选择"
+  ],
+  "source": "20-21 六年级上·江西赣州·期末",
+  "merged_id": "unit7_050",
+  "original_id": "unit7_fill_050"
+},
+{
+  "id": "unit7_fill_051",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "为观测病人的体温变化情况, 应选择( )统计图。",
+  "options": [],
+  "answer": "折线",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。此题应根据条形统计图、折线统计图、扇形统计图各自的特点进行解答。\n【详解】根据统计图的特点可知：统计某病人的体温变化情况，应选择折线统计图。",
+  "knowledge_points": [
+    "统计图的选择（扇形统计图）",
+    "统计图的选择（折线统计图）"
+  ],
+  "source": "2021·陕西汉中·小升初真题",
+  "merged_id": "unit7_051",
+  "original_id": "unit7_fill_051"
+},
+{
+  "id": "unit7_fill_052",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "医生想记录病人某天 8-21 时的体温变化情况, 要用( )统计图。",
+  "options": [],
+  "answer": "折线",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。此题应根据条形统计图、折线统计图、扇形统计图各自的特点进行解答。\n【详解】由统计图的特点可知: 医生想记录病人某天 8-21 时的体温变化情况, 选用折线统计图比较合适。",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "21-22 五年级下·湖南邵阳·期末",
+  "merged_id": "unit7_052",
+  "original_id": "unit7_fill_052"
+},
+{
+  "id": "unit7_fill_053",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "要描述三 (1) 班同学身高分组的分布情况, 选用( )统计图。",
+  "options": [],
+  "answer": "条形",
+  "solution": "【分析】条形统计图特点是用一个单位长度表示一定的数量，用直条的长短表示数量的多少，作用是从图中能清楚地看出各种数量的多少，便于相互比较。折线统计图特点是用不同位置的点表示数量的多少，根据数量的多少描出各点，然后把各点用线段顺次连接起来。不仅能看清数量的多少，还能通过折线的上升和下降表示数量的增减变化情况。扇形统计图特点是以一个圆的面积表示物体的总数量，以相应的扇形面积占整个圆面积的百分数表示各有关部分占总数量的百分数。清楚地看出各部分数量与总数量之间的关系。关键是熟悉各种统计图的特点，并结合具体题意做出恰当的选择。\n【详解】结合各种统计图的特点可得：要描述三（1）班同学身高分组的分布情况，选用条形统计图。",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "20-21 六年级下·云南大理·期末",
+  "merged_id": "unit7_053",
+  "original_id": "unit7_fill_053"
+},
+{
+  "id": "unit7_fill_054",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "如果只表示各种数量的多少，可以选用( )统计图表示；如果要清楚地了解各部分数量与总数量之间的关系，可以选用( )统计图表示；表示数据的增减变化情况可以用( )统计图表示。",
+  "options": [],
+  "answer": "条形;扇形;折线",
+  "solution": "【分析】条形统计图特点：可以清楚地看出数量的多少；折线统计图特点：不但可以表示数量的多少，还可以清楚的看出数量的增减变化情况；扇形统计图特点：可以看出各个部分数量与总数之间的关系，据此结合题意选择合适的统计图。根据统计图特点是解题的关键。\n【详解】如果只表示各种数量的多少，可以选用（条形）统计图表示；如果要清楚地了解各部分数量与总数量之间的关系，可以选用（扇形）统计图表示；表示数据的增减变化情况可以用（折线）统计图表示。",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "21-22六年级上·黑龙江鹤岗·期末",
+  "merged_id": "unit7_054",
+  "original_id": "unit7_fill_054"
+},
+{
+  "id": "unit7_fill_055",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "要反映奶粉中各种营养成分的含量, 一般选用( )统计图; 要反映一个患者一周中体温的变化情况, 应选用( )统计图; 注重表示事物数量的多少, 应选用( )统计图。",
+  "options": [],
+  "answer": "扇形;折线;条形",
+  "solution": "【分析】根据扇形统计图、条形统计图、折线统计图的特征选择合适的统计图即可。掌握各统计图的特征是解答题目的关键。\n【详解】（1）扇形统计图：扇形统计图可以清楚地看出各部分数量与总数量之间，部分与部分之间的关系，所以要反映奶粉中各种营养成分的含量，选择扇形统计图比较合适；(2) 折线统计图: 不仅能看清数量的多少, 还能通过折线的上升和下降表示数量的增减变化情况, 所以要反映一个患者一周中体温的变化情况, 选择折线统计图比较合适;(3) 条形统计图: 用直条的长短表示数量的多少, 从图中直观地看出数量的多少, 便于比较, 所以注重表示事物数量的多少, 选择条形统计图比较合适。",
+  "knowledge_points": [
+    "1 格表示一个单位的单式条形统计图",
+    "单式折线统计图",
+    "扇形统计图的特点及绘制",
+    "统计图的选择"
+  ],
+  "source": "21-22 六年级上·甘肃平凉·期末",
+  "merged_id": "unit7_055",
+  "original_id": "unit7_fill_055"
+},
+{
+  "id": "unit7_fill_056",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "某班 50 人参加考试,其中优秀 21 人, 良好 15 人, 及格 10 人, 不及格 4 人, 如果用扇形统计图把这个班的考试结果表示出来, 那么优秀人数的扇形圆心角是( )度。",
+  "options": [],
+  "answer": "151.2",
+  "solution": "【分析】先计算优秀人数占总人数的百分率，再用乘法计算 360° 的百分之几是多少即可。已知一个数, 求这个数的百分之几是多少用乘法计算。\n【详解】21 ÷ 50 × 100% = 0.42 × 100% = 42%\n360° × 42% = 151.2°",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "求一个数的百分之几是多少"
+  ],
+  "source": "21-22 六年级上·辽宁鞍山·期末",
+  "merged_id": "unit7_056",
+  "original_id": "unit7_fill_056"
+},
+{
+  "id": "unit7_fill_057",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "观察: 1 = 1²; 1 + 3 = 2²; 1 + 3 + 5 = ( )²; 1 + 3 + 5 + 7 = ( )²; 1 + 3 + 5 + 7 + 9 + 11 + 13 = ( )²; ( ) = 9²。",
+  "options": [],
+  "answer": "3;4;7；1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17",
+  "solution": "【分析】观察算式, 发现是等差数列相加, 1 = 1²; 1 + 3 = 2², 有几个加数就是几个平方, 据此解答即可。本题考查算式的规律，发现规律，利用规律解题是关键。\n【详解】由分析可知:\n1 + 3 + 5 = (3)²; 1 + 3 + 5 + 7 = (4)²; 1 + 3 + 5 + 7 + 9 + 11 + 13 = (7)²; (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17) = 9²。",
+  "knowledge_points": [
+    "数与形",
+    "算式的规律"
+  ],
+  "source": "20-21 六年级上·湖南怀化·单元测试",
+  "merged_id": "unit7_057",
+  "original_id": "unit7_fill_057"
+},
+{
+  "id": "unit7_fill_058",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "找规律:\n1，2，5，10，17，( )，37，( )。\n1/1, 3/4, 5/9, 7/16, 9/25, 11/36 第7个是( )，第12个是( )。\n1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + …… + ()",
+  "options": [],
+  "answer": "26;50;13/49;23/144;1",
+  "solution": "【分析】（1）观察发现，相邻的两个数依次相差：1, 3, 5, 7, 9...都是相邻的奇数，据此解答。(2) 分子是 1, 3, 5, 7, 9……都是奇数, 故底 n 个为 (2n - 1), 分母依次是 1, 4, 9, 16……, 故第 n 个为 n², 则第 7 个是 13/49, 第 12 个是 23/144。(3) 本题属于极限问题, 观察数列发现, 相邻两个数之间的商为 1/2, 一直加下去, 无限接近与 1。据此解答即可。此题考查的是找规律，解答此题关键是正确的找出规律并用规律解决问题。\n【详解】由分析得,(1) 1, 2, 5, 10, 17, 26, 37, 50。(2) 1/1, 3/4, 5/9, 7/16, 9/25, 11/36. 第 7 个是 13/49, 第 12 个是 23/144。(3) 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + … + 1",
+  "knowledge_points": [
+    "数与形",
+    "数字排列的规律"
+  ],
+  "source": "20-21 六年级上·湖南怀化·单元测试",
+  "merged_id": "unit7_058",
+  "original_id": "unit7_fill_058"
+},
+{
+  "id": "unit7_fill_059",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "新年期间，4位同学互相都寄一张贺卡，一共要寄( )张贺卡。",
+  "options": [],
+  "answer": "12",
+  "solution": "【分析】4位同学互相都寄一张贺卡，则每一位同学要送出3张牌给另外3位同学，据此可得出答案。本题主要考查的是搭配，解题的关键是理解题意，每个同学需要寄 3 张卡片，进而得出答案。\n【详解】4位同学互相都寄一张贺卡，即每一位同学需要送出3张卡片，即总共要寄 4 × 3 = 12（张）卡片。",
+  "knowledge_points": [
+    "搭配问题"
+  ],
+  "source": "20-21六年级上·湖南怀化·单元测试",
+  "merged_id": "unit7_059",
+  "original_id": "unit7_fill_059"
+},
+{
+  "id": "unit7_fill_060",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "要反映某种股票的涨跌情况最好制成( )统计图。",
+  "options": [],
+  "answer": "折线",
+  "solution": "【分析】首先要清楚每一种统计图的特点：条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；据此解答即可。解答此题要熟练掌握折线统计图的特点。\n【详解】由统计图的特点可知，要反映某种股票的涨跌情况最好制成折线统计图。",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "20-21 六年级上·湖南怀化·单元测试",
+  "merged_id": "unit7_060",
+  "original_id": "unit7_fill_060"
+},
+{
+  "id": "unit7_fill_061",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "如果要清楚地了解各部分数量与总数的关系，可以用( )统计图表示；要表示数量增减变化的情况，用( )统计图比较合适。",
+  "options": [],
+  "answer": "扇形;折线",
+  "solution": "【分析】统计图的选择应根据统计图的特征进行选择。条形统计图能反映各个数量的多少；折线统计图既能反映各个量的多少，也能反映各个量的变化趋势；扇形统计图能反映部分与整体的关系。据此可得出答案。本题主要考查的是统计图的选择, 解题的关键是牢记每种统计图的特征, 对应所求统计图类型, 进而得出答案。\n【详解】清楚地了解各部分数量与总数的关系，应当选用扇形统计图；表示数量增减变化情况，应当选用折线统计图。",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "20-21 六年级上·湖南怀化·单元测试",
+  "merged_id": "unit7_061",
+  "original_id": "unit7_fill_061"
+},
+{
+  "id": "unit7_fill_062",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "在扇形统计图中，所有扇形的百分比之和为( )。",
+  "options": [],
+  "answer": "1",
+  "solution": "【分析】根据扇形统计图的概念和意义可知圆代表整体,即单位 '1', 各个扇形代表部分。本题考查的是扇形图绘制的特点，在扇形统计图中，各部分占总体的百分比之和为 1。\n【详解】由分析可知:绘制扇形统计图时，是把圆看作一个单位'1'，所以所有的百分比之和必须等于1。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "16-17 六年级上·全国·单元测试",
+  "merged_id": "unit7_062",
+  "original_id": "unit7_fill_062"
+},
+{
+  "id": "unit7_fill_063",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "扇形统计图能清楚地反映出( )数量与( )数量之间的关系。",
+  "options": [],
+  "answer": "部分;总",
+  "solution": "【详解】扇形统计图可以清楚地看出各部分数量与总数量之间的关系，可以看出各部分占整体的百分比情况，如：表示酸奶中各种成分的百分比情况就适合选择扇形统计图。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "21-22 六年级上·江西赣州·单元测试",
+  "merged_id": "unit7_063",
+  "original_id": "unit7_fill_063"
+},
+{
+  "id": "unit7_fill_064",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "要反映某地的某月温度变化情况, 用( )统计图更合适。",
+  "options": [],
+  "answer": "折线",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。此题应根据条形统计图、折线统计图、扇形统计图各自的特点进行解答。\n【详解】由分析可知：要反映某地的某月温度变化情况，用折线统计图更合适。",
+  "knowledge_points": [
+    "单式折线统计图",
+    "统计图的选择"
+  ],
+  "source": "21-22 六年级上·江西赣州·单元测试",
+  "merged_id": "unit7_064",
+  "original_id": "unit7_fill_064"
+},
+{
+  "id": "unit7_fill_065",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "扇形统计图用( )(1)表示总数，用圆内( )表示各个部分量占总数的百分比。",
+  "options": [],
+  "answer": "整个圆;大小不同的扇形",
+  "solution": "【详解】扇形统计图用整个圆表示总数，用圆内大小不同的扇形表示各个部分量占总数的百分比，扇形统计图中哪个扇形的面积大，那个部分的数量占总数的百分比就大；哪个扇形的面积小，那个部分的数量占总数的百分比就小，如：统计全班同学喜欢的运动项目就可以用扇形统计图，用圆内大小不同的扇形表示喜欢每种运动的人数占全班人数的百分比。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "15-16 六年级下·全国·课后作业",
+  "merged_id": "unit7_065",
+  "original_id": "unit7_fill_065"
+},
+{
+  "id": "unit7_fill_066",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "李名今天生病发高烧，要监测他 12 小时内的体温变化情况，应选用( )统计图。",
+  "options": [],
+  "answer": "折线",
+  "solution": "【分析】根据各个统计图的特点，结合题中统计需求，直接填空即可。本题考查了折线统计图, 明确折线统计图的特点是解题的关键。\n【详解】折线统计图可以清晰反映数据的变化情况，所以要监测他 12 小时内的体温变化情况，应选用折线统计图。",
+  "knowledge_points": [
+    "单式折线统计图"
+  ],
+  "source": "20-21 六年级上·全国·单元测试",
+  "merged_id": "unit7_066",
+  "original_id": "unit7_fill_066"
+}, {
+  "id": "unit7_fill_067",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "如果想清楚地看出各种数量的多少，选( )统计图；如果想看数量增减变化的情况，选( )统计图；如果要反映各部分数量与总数之间的关系，选( )统计图。",
+  "options": [],
+  "answer": "条形;折线;扇形",
+  "solution": "【分析】根据各个统计图的特点，直接填空即可。本题考查了统计图的特点, 属于简单题, 掌握各个统计图表的特点是解题的关键。\n【详解】如果想清楚地看出各种数量的多少，选条形统计图；如果想看数量增减变化的情况，选折线统计图；如果要反映各部分数量与总数之间的关系，选扇形统计图。",
+  "knowledge_points": [
+    "单式折线统计图",
+    "1格表示多个单位的单式条形统计图",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "20-21 六年级上·全国·单元测试",
+  "merged_id": "unit7_067",
+  "original_id": "unit7_fill_067"
+},
+{
+  "id": "unit7_fill_068",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "要统计某个城市2009年4月20日至26日患甲型H1N1流感的病人数量可以用( )统计图。",
+  "options": [],
+  "answer": "折线",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。本题主要考查对条形统计图、折线统计图、扇形统计图各自的特点的认识。\n【详解】根据统计图的特点可知: 要统计某个城市 2009 年 4 月 20 日至 26 日患甲型 H1N1 流感的病人数量可以用折线统计图。",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "20-21 六年级上·全国·单元测试",
+  "merged_id": "unit7_068",
+  "original_id": "unit7_fill_068"
+},
+{
+  "id": "unit7_judge_069",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "在扇形统计图甲和乙中, 女生都占 45%, 那么甲乙两个统计图表示的女生人数一定相等。( )",
+  "options": [],
+  "answer": "×",
+  "solution": "【分析】扇形统计图表示部分与整体之间的关系；两扇形统计图单位'1'不一定相同，所以两个统计图表示的女生人数不一定相等，据此解答。\n【详解】根据分析可知，扇形统计图甲和乙中，女生都占 45%，甲乙两个统计图表示的女生人数不一定相等;原题干说法错误。故答案为: ×",
+  "knowledge_points": [
+    "单位 '1'的认识与确定",
+    "扇形统计图的特点及绘制",
+    "求一个数的百分之几是多少"
+  ],
+  "source": "22-23六年级下·安徽马鞍山·期末",
+  "merged_id": "unit7_069",
+  "original_id": "unit7_judge_069"
+},
+{
+  "id": "unit7_judge_070",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "张芳选用扇形统计图表示六年级各班人数的多少。( )",
+  "options": [],
+  "answer": "×",
+  "solution": "【分析】条形统计图能清楚地表示出数量的多少；折线统计图不仅能表示数量的多少，还能表示数量的增减变化情况；扇形统计图表示部分与整体之间的关系；据此解答。\n【详解】根据分析可知，张芳选用条形统计图表示六年级各班人数的多少。原题干说法错误。故答案为：×",
+  "knowledge_points": [
+    "统计图的选择（扇形统计图）",
+    "统计图的选择（折线统计图）"
+  ],
+  "source": "23-24 六年级上·全国·单元测试",
+  "merged_id": "unit7_070",
+  "original_id": "unit7_judge_070"
+},
+{
+  "id": "unit7_judge_071",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "用统计表表示有关数量之间的关系, 比统计图更加形象具体。( )",
+  "options": [],
+  "answer": "×",
+  "solution": "【分析】统计表只是以表格形式罗列数据，虽然清晰准确，但在展示数据关系的直观性和形象性方面不如统计图；统计图能更直观、形象地展示数据之间的关系和趋势。\n【详解】例如要展示不同月份的销售额变化趋势，统计图可以通过折线的起伏清晰地呈现增长或下降的情况，让人一目了然。而统计表只是单纯地列出每个月的销售额数字，需要人们进一步分析和比较才能得出变化趋势。所以用统计图表示有关数量之间的关系，比统计表更加形象具体。故答案为：×",
+  "knowledge_points": [
+    "统计图的选择（扇形统计图）",
+    "单式统计表的特点及填补",
+    "复式统计表的特点及填补",
+    "统计图的选择（折线统计图）"
+  ],
+  "source": "23-24 六年级上·全国·单元测试",
+  "merged_id": "unit7_071",
+  "original_id": "unit7_judge_071"
+},
+{
+  "id": "unit7_judge_072",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "容易",
+  "question": "扇形统计图能够看出部分数量与总量间的关系。（）",
+  "options": [],
+  "answer": "√",
+  "solution": "【分析】扇形统计图特点是以一个圆的面积表示物体的总数量，以相应的扇形面积占整个圆面积的百分数表示各有关部分占总数量的百分数，可以清楚地看出各部分数量与总数量之间的关系。\n【详解】根据扇形统计图的特点可知, 扇形统计图能够看出部分数量与总量间的关系, 原题说法正确。故答案为: √",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "2012·四川成都·小升初真题",
+  "merged_id": "unit7_072",
+  "original_id": "unit7_judge_072"
+},
+{
+  "id": "unit7_judge_073",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "要记录 6 月份 30 天内的气温变化情况, 用条形统计图最合适。( )",
+  "options": [],
+  "answer": "×",
+  "solution": "【分析】条形统计图的特点: 条形统计图能清楚地表示出每个项目的具体数目; 折线统计图的特点: 折线统计图不仅能表示数量的多少，还能清楚地反映数量的变化情况；扇形统计图的特点：扇形统计图能清楚地表示出各部分在总体中所占的百分比。\n【详解】要记录 6 月份 30 天内的气温变化情况, 用折线统计图最合适。原题干的说法是错误的。故答案为: ×",
+  "knowledge_points": [
+    "单式折线统计图",
+    "1格表示多个单位的单式条形统计图",
+    "扇形统计图的特点及绘制",
+    "统计图的选择（折线统计图）"
+  ],
+  "source": "22-23 六年级下·江西吉安·期末",
+  "merged_id": "unit7_073",
+  "original_id": "unit7_judge_073"
+},
+{
+  "id": "unit7_judge_074",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "要表示本校六年级各班学生的具体人数选用扇形统计图最合适。( )",
+  "options": [],
+  "answer": "×",
+  "solution": "【分析】条形统计图从图中能清楚地看出各种数量的多少;折线统计图不仅能看清数量的多少, 还能通过折线的上升和下降表示数量的增减变化情况; 扇形统计图能清楚地看出各部分数量与总数量之间的关系，据此分析。\n【详解】要表示本校六年级各班学生的具体人数选用条形统计图最合适，所以原题说法错误。故答案为：×",
+  "knowledge_points": [
+    "统计图的选择（扇形统计图）",
+    "统计图的选择（折线统计图）"
+  ],
+  "source": "22-23 六年级下·山东济南·开学考试",
+  "merged_id": "unit7_074",
+  "original_id": "unit7_judge_074"
+},
+{
+  "id": "unit7_judge_075",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "容易",
+  "question": "折线统计图能反映数量的增减变化情况。( )",
+  "options": [],
+  "answer": "√",
+  "solution": "【分析】根据折线统计图的特点解答, 折线统计图不仅能表示数量的多少, 还能通过折线的上升和下降表示数量的增减变化情况; 条形统计图能表示出数量的多少; 扇形统计图能表示部分与整体之间的关系。\n【详解】折线统计图能反映数据的增减变化情况，原题说法正确。故答案为: √",
+  "knowledge_points": [
+    "统计图的选择（折线统计图）"
+  ],
+  "source": "18-19 五年级上·云南·期末",
+  "merged_id": "unit7_075",
+  "original_id": "unit7_judge_075"
+},
+{
+  "id": "unit7_judge_076",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "笑笑家十一月份的食品支出占生活总支出的 40%，在扇形统计图中，表示食品支出的扇形的圆心角的度数是 144°。（）",
+  "options": [],
+  "answer": "√",
+  "solution": "【分析】在扇形统计图中，表示部分的扇形占整个圆的百分之几，所对应的圆心角就是 360° 的百分之几；根据百分数乘法的意义即可求出表示占整体 40% 的扇形圆心角的度数。在扇形统计畔中有一个圆表示整体，扇形表示部分，部分占整体的百分之几，表示部分的扇形所对应的圆心角就是 360° 的百分之几。\n【详解】360° × 40% = 144° 表示食品支出的扇形的圆心角的度数是 144°。原题说法正确。故答案为: √",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "弧、圆心角、扇形的认识",
+    "求一个数的百分之几是多少"
+  ],
+  "source": "23-24六年级上·湖北荆门·期末",
+  "merged_id": "unit7_076",
+  "original_id": "unit7_judge_076"
+},
+{
+  "id": "unit7_judge_077",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "容易",
+  "question": "要表示某小学各年级学生人数与全校学生总人数的关系, 应选择扇形统计图。 ( )",
+  "options": [],
+  "answer": "√",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系。\n【详解】某小学各年级学生人数与全校学生总人数的关系是反映部分与整体的关系，应选择扇形统计图。故答案为: √",
+  "knowledge_points": [
+    "统计图的选择（扇形统计图）",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "23-24 六年级上·陕西延安·期末",
+  "merged_id": "unit7_077",
+  "original_id": "unit7_judge_077"
+},
+{
+  "id": "unit7_judge_078",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "容易",
+  "question": "统计希望小学低年级和高年级同学最喜欢的课外读物情况, 应该选择复式条形统计图。( )",
+  "options": [],
+  "answer": "√",
+  "solution": "【分析】条形统计图可以清楚地看出数量的多少。折线统计图不但可以表示出数量的多少，而且能够清楚地表示出数量增减变化的情况。扇形统计图表示各部分数量与总数之间的关系。\n【详解】统计希望小学低年级和高年级同学最喜欢的课外读物情况，应该选择复式条形统计图。原题说法正确。故答案为: √",
+  "knowledge_points": [
+    "复式折线统计图",
+    "复式条形统计图",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "23-24 五年级上·山东济宁·期末",
+  "merged_id": "unit7_078",
+  "original_id": "unit7_judge_078"
+},
+{
+  "id": "unit7_judge_079",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "容易",
+  "question": "扇形统计图可以表示某地一年气温的变化情况。（）",
+  "options": [],
+  "answer": "×",
+  "solution": "【分析】要表示一年气温数据的变化情况，根据折线统计图、扇形统计图的特点，并结合实际问题，进行判断。考查根据实际情况选择合适的统计图分析数据。\n【详解】折线统计图能直观的表示出数量随时间的变化趋势，扇形统计图能表示出部分数量与总数之间的关系，因此能够表示某地一年气温的变化情况的应该是折线统计图。故答案为:× 。",
+  "knowledge_points": [
+    "单式折线统计图",
+    "扇形统计图的特点及绘制",
+    "统计图的选择"
+  ],
+  "source": "22-23 六年级下·四川宜宾·期中",
+  "merged_id": "unit7_079",
+  "original_id": "unit7_judge_079"
+},
+{
+  "id": "unit7_judge_080",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "扇形统计图中每个扇形分别表示各部分占整体的百分之几。（）",
+  "options": [],
+  "answer": "√",
+  "solution": "【分析】根据扇形统计图的特点及作用，扇形统计图能够表示部分与整体之间的关系，也就是用整个圆表示总体，用扇形表示各部分占总体的百分比，据此解答。此题考查的目的是理解掌握扇形统计图的特点及作用。\n【详解】扇形统计图中每个扇形分别表示各部分占整体的百分之几。原题说法正确。故答案为: √。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "2023 六年级下·江苏连云港·专题练习",
+  "merged_id": "unit7_080",
+  "original_id": "unit7_judge_080"
+},
+{
+  "id": "unit7_judge_081",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "容易",
+  "question": "要清楚表示整体与部分之间的关系，使用扇形统计图更合适。（）",
+  "options": [],
+  "answer": "√",
+  "solution": "【分析】条形统计图用直条的长短表示数量的多少，从图中直观地看出数量的多少，便于比较；折线统计图不仅能看清数量的多少，还能通过折线的上升和下降表示数量的增减变化情况；扇形统计图清楚地看出各部分数量与总数量之间，部分与部分之间的关系，据此判断即可。掌握不同的统计图表的特征选用合适的统计图是解题的关键。\n【详解】根据各统计图的特征可知：要清楚表示整体与部分之间的关系，使用扇形统计图更合适，说法正确。故答案为：√",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "统计图的选择"
+  ],
+  "source": "21-22 六年级上·广东江门·期末",
+  "merged_id": "unit7_081",
+  "original_id": "unit7_judge_081"
+},
+{
+  "id": "unit7_judge_082",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "容易",
+  "question": "记录病人的体温变化, 用折线统计图表示最合适。（）",
+  "options": [],
+  "answer": "√",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。此题根据扇形统计图、折线统计图、条形统计图各自的特点来判断。\n【详解】统计记录病人的体温变化，用折线统计图表示最合适。原题说法正确。故答案为: √",
+  "knowledge_points": [
+    "单式折线统计图",
+    "统计图的选择"
+  ],
+  "source": "2014·全国·小升初真题",
+  "merged_id": "unit7_082",
+  "original_id": "unit7_judge_082"
+},
+{
+  "id": "unit7_judge_083",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "容易",
+  "question": "要反映微山县去年空气质量为优的天数与全年天数的关系，绘制扇形统计图较合适。（ ）",
+  "options": [],
+  "answer": "√",
+  "solution": "【分析】通过扇形统计图可以很清楚地表示出各部分数量与总数之间的关系。此题应根据扇形统计图各自的特点进行解答。\n【详解】扇形统计图能清楚地表示部分数量与总数间关系,所以要反映微山县去年空气质量为优的天数与全年天数的关系, 绘制扇形统计图较合适。故答案为: √",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "2022 六年级下·全国·专题练习",
+  "merged_id": "unit7_083",
+  "original_id": "unit7_judge_083"
+},
+{
+  "id": "unit7_judge_084",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "人的心率变化情况用条形统计图表示比较合适。（）",
+  "options": [],
+  "answer": "×",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。此题应根据条形统计图、折线统计图、扇形统计图各自的特点进行解答。\n【详解】由分析可知：人的心率变化情况用折线统计图表示比较合适。",
+  "knowledge_points": [
+    "单式折线统计图",
+    "1格表示多个单位的单式条形统计图",
+    "扇形统计图的特点及绘制",
+    "统计图的选择"
+  ],
+  "source": "20-21 五年级下·重庆永川·期末",
+  "merged_id": "unit7_084",
+  "original_id": "unit7_judge_084"
+},
+{
+  "id": "unit7_judge_085",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "条形统计图能清楚的表示出数量的多少，折线统计图更容易看出数量增减变化的情况，扇形统计图可以看出各部分与整体的关系。（）",
+  "options": [],
+  "answer": "√",
+  "solution": "【分析】根据条形统计图、折线统计图、扇形统计图的特征分析题中说法是否正确即可。理解并掌握各统计图的特征是解答题目的关键。\n【详解】条形统计图：用直条的长短表示数量的多少，从图中直观地看出数量的多少，便于比较；折线统计图：不仅能看清数量的多少，还能通过折线的上升和下降表示数量的增减变化情况；扇形统计图：扇形统计图可以清楚地看出各部分数量与总数量之间，部分数量与部分数量之间的关系；由上可知，题中说法正确。故答案为: √",
+  "knowledge_points": [
+    "单式折线统计图",
+    "1格表示多个单位的单式条形统计图",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "20-21 六年级下·广西贺州·期末",
+  "merged_id": "unit7_085",
+  "original_id": "unit7_judge_085"
+},
+{
+  "id": "unit7_judge_086",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "容易",
+  "question": "要反映学校运动会参加踢毽、跳绳、足球、乒乓球等项目的人数占总人数的百分比，选用扇形统计图比较合适。（）",
+  "options": [],
+  "answer": "√",
+  "solution": "【分析】条形统计图很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况判断即可。此题应根据条形统计图、折线统计图、扇形统计图各自的特点进行判断。\n【详解】要反映学校运动会参加踢毽、跳绳、足球、乒乓球等项目的人数占总人数的百分比，选用扇形统计图比较合适。故答案为: √",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "统计图的选择"
+  ],
+  "source": "21-22六年级上·四川南充·期末",
+  "merged_id": "unit7_086",
+  "original_id": "unit7_judge_086"
+},
+{
+  "id": "unit7_judge_087",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "容易",
+  "question": "实验小学六（3）班全体学生一次视力检查中，视力正常的人数、近视的人数和假性近视的人数分别占全班总人数的 50%、30%、30%。",
+  "options": [],
+  "answer": "×",
+  "solution": "【分析】用百分数表示部分与总量之间的关系时，各个部分百分比的和应正好是 100%。本题考查百分比在实际生活中的应用。\n【详解】50% + 30% + 30% = 80% + 30% = 110%\n不符合实际，故答案为×。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "21-22六年级上·山东菏泽·期末",
+  "merged_id": "unit7_087",
+  "original_id": "unit7_judge_087"
+},
+{
+  "id": "unit7_judge_088",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "医生要表示病人体温要选用条形统计图表较好。（）",
+  "options": [],
+  "answer": "×",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。此题应根据条形统计图、折线统计图、扇形统计图各自的特点进行解答。\n【详解】根据统计图的特点可知: 医生需要监测病人的体温情况, 应选用折线统计图;所以原题说法错误。故答案为: ×",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "19-20 五年级下·湖南怀化·单元测试",
+  "merged_id": "unit7_088",
+  "original_id": "unit7_judge_088"
+},
+{
+  "id": "unit7_judge_089",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "容易",
+  "question": "扇形统计图能清楚地反映出部分与整体的关系。（）",
+  "options": [],
+  "answer": "√",
+  "solution": "【详解】扇形统计图是以一个圆的面积（看成单位'1'）表示物体的总数量，以相应的扇形面积占整个圆面积的百分数表示各有关部分占总数量的百分数的统计图。扇形统计图能清楚地反映出部分与整体的关系，所以原题说法正确。故答案: √",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "20-21 六年级上·河北邢台·期末",
+  "merged_id": "unit7_089",
+  "original_id": "unit7_judge_089"
+},
+{
+  "id": "unit7_judge_090",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "容易",
+  "question": "扇形统计图中，各部分数量占总数的百分比的总和都是 100%。（）",
+  "options": [],
+  "answer": "√",
+  "solution": "【分析】在扇形统计图中，整个圆代表总体，把总体看作单位'1'，每个扇形代表总体中的各部分，各部分之和等于单位'1'，即各部分数量占总数的百分比的总和都是 100%。扇形统计图是用圆的面积表示总数，用圆内扇形的面积表示各部分占总数的百分比。\n【详解】分析可知，任何扇形统计图中，各部分数量占总数的百分比的总和都是 100%。故答案为: √",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "21-22六年级上·湖南怀化·单元测试",
+  "merged_id": "unit7_090",
+  "original_id": "unit7_judge_090"
+},
+{
+  "id": "unit7_judge_091",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "要表示李爷爷家养的鸡、鸭、鹅的数量, 应绘制扇形统计图。( )",
+  "options": [],
+  "answer": "×",
+  "solution": "【分析】扇形统计图主要表示部分与部分之间，部分与总体之间的数量关系，以及各部分量占总体的百分比情况，要具体的数量选择条形统计图更加合适，据此解答。根据统计图的特征选择合适的统计图是解答题目的关键。\n【详解】条形统计图：用直条的长短表示数量的多少，从图中直观地看出数量的多少，便于比较；因此要表示李爷爷家养的鸡、鸭、鹅的数量，应绘制条形统计图。故答案为: ×",
+  "knowledge_points": [
+    "1 格表示多个单位的单式条形统计图",
+    "扇形统计图的特点及绘制",
+    "统计图的选择"
+  ],
+  "source": "21-22 六年级上·湖南怀化·单元测试",
+  "merged_id": "unit7_091",
+  "original_id": "unit7_judge_091"
+},
+{
+  "id": "unit7_judge_092",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "容易",
+  "question": "如果表示数量的增减变化情况，一般用扇形统计图。（）",
+  "options": [],
+  "answer": "×",
+  "solution": "【分析】根据折线统计图的特点及作用，要反映数量增减变化的情况，应绘制折线统计图，而扇形统计图可以清楚地表示出各部分与总数的关系。本题关键在于根据各种统计图的特点及作用来解决这类问题。\n【详解】要反映数量的增减变化情况，应选择折线统计图，所以判断错误。故答案为: ×",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "2022·甘肃陇南·小升初真题",
+  "merged_id": "unit7_092",
+  "original_id": "unit7_judge_092"
+},
+{
+  "id": "unit7_judge_093",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "容易",
+  "question": "扇形统计图表示各部分数量与总数的关系，用整个圆表示总数。（）",
+  "options": [],
+  "answer": "√",
+  "solution": "【分析】扇形统计图是以一个圆的面积（看作单位'1'）表示物体的总数量，以相应的扇形面积占整个圆面积的百分数表示各有关部分占总数量的百分数的统计图。扇形统计图清楚地看出各部分数量与总数量之间，部分与部分之间的关系。\n【详解】扇形统计图表示各部分数量与总数的关系，用整个圆表示总数，说法正确。故答案为: √",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "20-21 六年级上·四川绵阳·期末",
+  "merged_id": "unit7_093",
+  "original_id": "unit7_judge_093"
+},
+{
+  "id": "unit7_judge_094",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "要清楚的表示出各部分与整体之间的关系应选扇形统计图比较合适。（）",
+  "options": [],
+  "answer": "√",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。此题应根据条形统计图、折线统计图、扇形统计图各自的特点进行解答。\n【详解】根据统计图的特点可知：要清楚的表示各部分量与总量之间的关系，用扇形统计图较为合适。故原题干说法正确。",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "19-20六年级上·全国·单元测试",
+  "merged_id": "unit7_094",
+  "original_id": "unit7_judge_094"
+},
+{
+  "id": "unit7_judge_095",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "要反映班里'优秀' '合格'人数占全班人数百分比应绘制条形统计图。（）",
+  "options": [],
+  "answer": "×",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。\n【详解】根据统计图的特点可知：要反映班里'优秀''良好' '合格'人数占全班人数的百分比，应绘制扇形统计图。故原题干说法错误。故答案为: ×",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "20-21 六年级上·湖南怀化·单元测试",
+  "merged_id": "unit7_095",
+  "original_id": "unit7_judge_095"
+},
+{
+  "id": "unit7_judge_096",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "3 个人互相打电话, 每 2 个人都要通, 一共要通话 6 次。( )",
+  "options": [],
+  "answer": "×",
+  "solution": "【分析】由于每个同学都要和另外的 2 个通一次电话, 一共要通: 3 × 2 = 6 (次); 又因为两个同学只通一次电话, 去掉重复计算的情况, 实际只通: 6 ÷ 2 = 3 (次), 据此解答。本题考查了握手问题的实际应用，要注意去掉重复计算的情况，如果人比较少可以用枚举法解答，如果人数比较多可以用公式：通话次数 = 人数（人数 - 1）÷2 解答。\n【详解】(3 - 1) × 3 ÷ 2 = 6 ÷ 2 = 3 (次)\n则一共要通话 3 次。故原题干说法错误。故答案为: ×",
+  "knowledge_points": [
+    "搭配问题"
+  ],
+  "source": "20-21 六年级上·湖南怀化·单元测试",
+  "merged_id": "unit7_096",
+  "original_id": "unit7_judge_096"
+},
+{
+  "id": "unit7_judge_097",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "困难",
+  "question": "扇形统计图与条形统计图都能清楚地表示部分与整体之间关系。( )",
+  "options": [],
+  "answer": "×",
+  "solution": "【分析】扇形统计图的特点是能清楚的反映出部分与整体的关系；条形统计图的特点是能反映出各个量的多少。据此可得出答案。本题主要考查的是扇形统计图及条形统计图的特征，解题的关键是熟练掌握两种统计图特征，进而得出答案。\n【详解】扇形统计图的特点是能清楚的反映出部分与整体的关系，条形统计图不能反映出。因此本题错误。",
+  "knowledge_points": [
+    "统计图的选择（扇形统计图）"
+  ],
+  "source": "20-21 六年级上·湖南怀化·单元测试",
+  "merged_id": "unit7_097",
+  "original_id": "unit7_judge_097"
+},
+{
+  "id": "unit7_calc_098",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "计算题",
+  "difficulty": "中等",
+  "question": "一个数减少它的15%后是 5.1，这个数是多少？（用方程解）",
+  "options": [],
+  "answer": "6",
+  "solution": "【分析】求一个数的百分之几是多少用乘法，设这个数是x，根据这个数一这个数 × 15% = 5.1，列出方程求出 x 的值即可。\n【详解】解：设这个数是 x\nx - 15%x = 5.1\n0.85x = 5.1\n0.85x ÷ 0.85 = 5.1 ÷ 0.85\nx = 6\n这个数是 6。",
+  "knowledge_points": [
+    "已知比一个数多/少百分之几是多少，求这个数",
+    "列方程解含一个未知数的问题",
+    "求一个数的百分之几是多少"
+  ],
+  "source": "23-24六年级下·四川凉山·期末",
+  "merged_id": "unit7_098",
+  "original_id": "unit7_calc_098"
+},
+{
+  "id": "unit7_calc_099",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "计算题",
+  "difficulty": "中等",
+  "question": "计算下列各题, 能简便运算的要简便计算。\n(1) 2/3 + 5/8 × 4/15 - 7/12\n(2) 25 × 1/4 + 75 ÷ 4",
+  "options": [],
+  "answer": "(1) 1/4;(2) 25",
+  "solution": "【分析】（1）根据运算顺序，先计算乘法，再计算加法，最后计算减法；(2) 根据除以一个数等于乘它的倒数, 把式子转化为25 × 1/4 + 75 × 1/4, 再根据乘法分配律进行简算。\n【详解】(1) 2/3 + 5/8 × 4/15 - 7/12\n= 2/3 + 1/6 - 7/12\n= 5/6 - 7/12\n= 1/4\n(2) 25 × 1/4 + 75 ÷ 4\n= 25 × 1/4 + 75 × 1/4\n= (25 + 75) × 1/4\n= 100 × 1/4\n= 25",
+  "knowledge_points": [
+    "整数乘法运算定律推广到分数乘法",
+    "分数的四则混合运算"
+  ],
+  "source": "22-23 六年级下·四川绵阳·期中",
+  "merged_id": "unit7_099",
+  "original_id": "unit7_calc_099"
+},
+{
+  "id": "unit7_solve_100",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "解答题",
+  "difficulty": "中等",
+  "question": "刘老师带了 41 名同学去北海公园划船, 共租了 10 条船。每条大船坐 6 人, 每条小船坐 4 人, 问大船、小船各租了几条?",
+  "options": [],
+  "answer": "大船租 1 条; 小船租 9 条",
+  "solution": "【分析】根据题干分析可得，一共有 41 + 1 = 42 人，假设全部租大船，10 条船能坐 6 × 10 = 60 人，比实际多算了：60 - 42 = 18 人，因为把小船看作了大船，每条小船多算了 6 - 4 = 2 人，所以小船的条数是：18 ÷ 2 = 9 条，那么大船的条数就是：10 - 9 = 1 条，据此解答。解答鸡兔同笼问题一般用假设法, 也就是假设全部为某种量, 和实际的总量相比较, 就会出现矛盾, 然后利用这个矛盾求出另一个量, 继而求出假设的量。\n【详解】(10 × 6 - 41 - 1) ÷ (6 - 4)\n= 18 ÷ 2\n= 9 (条)\n10 - 9 = 1 (条)\n答: 大船租 1 条, 小船租用 9 条。",
+  "knowledge_points": [
+    "鸡兔同笼"
+  ],
+  "source": "20-21 六年级上·湖南怀化·单元测试",
+  "merged_id": "unit7_100",
+  "original_id": "unit7_solve_100"
+},
+{
+  "id": "unit7_choice_101",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "要表示每月用水量与全年用水量之间的关系，最好选用（）统计图。",
+  "options": [
+    "A. 条形",
+    "B. 折线",
+    "C. 扇形",
+    "D. 以上三种都很好"
+  ],
+  "answer": "C",
+  "solution": "【分析】扇形统计图清楚地看出各部分数量与总数量之间，部分与部分之间的关系；折线统计图的特点：不仅能看清数量的多少，还能通过折线的上升和下降表示数量的增减变化情况；条形统计图的特点：用直条的长短表示数量的多少，从图中直观地看出数量的多少，便于比较，据此解答即可。明确各种统计图的特点是解答本题的关键。\n【详解】要表示每月用水量与全年用水量之间的关系，最好选用扇形统计图；故答案为: C",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "20-21 六年级上·全国·单元测试",
+  "merged_id": "unit7_101",
+  "original_id": "unit7_choice_101"
+},
+{
+  "id": "unit7_choice_102",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "学校制作反映各班向地震灾区捐款钱数多少的统计图，最好用（）统计图。",
+  "options": [
+    "A. 条形",
+    "B. 折线",
+    "C. 扇形",
+    "D. 都不是"
+  ],
+  "answer": "A",
+  "solution": "【分析】逐项分析各统计图的特点，题目侧重于表示数量的多少，根据题意选出合适的统计图即可。掌握条形、折线、扇形统计图的特点是解答题目的关键。\n【详解】A．条形统计图：用直条的长短表示数量的多少，从图中直观地看出数量的多少，便于比较；B. 折线统计图: 不仅能看清数量的多少, 还能通过折线的上升和下降表示数量的增减变化情况;  C. 扇形统计图: 扇形统计图可以清楚地看出各部分数量与总数量之间, 部分与部分之间的关系;综上所述，选择条形统计图更合适。故答案为: A",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "统计图的选择",
+    "单式折线统计图",
+    "1格表示一个单位的单式条形统计图"
+  ],
+  "source": "20-21六年级上·全国·单元测试",
+  "merged_id": "unit7_102",
+  "original_id": "unit7_choice_102"
+},
+{
+  "id": "unit7_choice_103",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "某校想了解各兴趣小组的人数占全校总人数的百分比, 选择 ( )。",
+  "options": [
+    "A. 条形统计图",
+    "B. 折线统计图",
+    "C. 扇形统计图"
+  ],
+  "answer": "C",
+  "solution": "【分析】根据各个统计图的特点，结合题中统计需求，直接解题即可。本题考查了统计图的选择，条形统计图可以清晰反映数据情况，折线统计图可以清晰反映数据的变化情况，扇形统计图可以反映各部分数量占总数的百分比情况。\n【详解】扇形统计图可以反映各部分数量占总数的百分比情况，所以某校想了解各兴趣小组的人数占全校总人数的百分比，选择扇形统计图比较合适。故答案为: C",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "统计图的选择",
+    "单式折线统计图",
+    "1格表示多个单位的单式条形统计图"
+  ],
+  "source": "20-21 六年级上·全国·单元测试",
+  "merged_id": "unit7_103",
+  "original_id": "unit7_choice_103"
+},
+{
+  "id": "unit7_choice_104",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "要清楚地反映果园里各种果树的棵数与果树总棵数之间的关系, 应选用 ( )。",
+  "options": [
+    "A. 扇形统计图",
+    "B. 折线统计图",
+    "C. 条形统计",
+    "D. 统计表"
+  ],
+  "answer": "A",
+  "solution": "【分析】根据各个统计图的特点，结合题中统计需求，选出正确选项即可。本题考查了统计图的选择, 明确常见统计图的特征是解题的关键。\n【详解】扇形统计图可以清晰反映各项和总量之间的百分比关系，所以，要清楚地反映果园里各种果树的棵数与果树总棵数之间的关系，应选用扇形统计图。故答案为：A",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "统计图的选择",
+    "单式折线统计图",
+    "1格表示一个单位的单式条形统计图"
+  ],
+  "source": "20-21 六年级上·辽宁铁岭·期末",
+  "merged_id": "unit7_104",
+  "original_id": "unit7_choice_104"
+},
+{
+  "id": "unit7_choice_105",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "要统计一袋牛奶里的各种营养成分所占的百分比情况，选用（）比较合适。",
+  "options": [
+    "A. 条形统计图",
+    "B. 折线统计图",
+    "C. 扇形统计图",
+    "D. 无法确定"
+  ],
+  "answer": "C",
+  "solution": "【分析】扇形统计图清楚地看出各部分数量与总数量之间，部分与部分之间的关系；折线统计图不仅能看清数量的多少，还能通过折线的上升和下降表示数量的增减变化情况；条形统计图用直条的长短表示数量的多少，从图中直观地看出数量的多少，便于比较。明确各种统计图的特点是解答本题的关键。\n【详解】要统计一袋牛奶里的各种营养成分所占的百分比情况，选用扇形统计图比较合适；故答案为: C。",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "2018·贵州铜仁·小升初真题",
+  "merged_id": "unit7_105",
+  "original_id": "unit7_choice_105"
+},
+{
+  "id": "unit7_choice_106",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "人离不开水，成年人每天体内 $47 \\%$ 的水靠喝水获得， $39 \\%$ 来自食物含的水， $14 \\%$ 来自体内氧化时释放出的水。要表示上面各种获得水的情况占总数的百分比，用（）表示比较合适。",
+  "options": [
+    "A. 条形统计图",
+    "B. 折线统计图",
+    "C. 扇形统计图"
+  ],
+  "answer": "C",
+  "solution": "【分析】一般来说，如果几个数量是并列的，只要求表示数量的多少时，选条形统计图。如果表示一个量或几个量增减变化和发展变化趋势，则选折线统计图。如果要求表示各部分数量与总数量之间的关系，则选扇形统计图。关键是熟悉各种统计图的特点。\n【详解】人离不开水，成年人每天体内 $47 \\%$ 的水靠喝水获得， $39 \\%$ 来自食物含的水， $14 \\%$ 来自体内氧化时释放出的水。要表示上面各种获得水的情况占总数的百分比，用扇形统计图表示比较合适。故答案为: C",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "19-20六年级下·四川·单元测试",
+  "merged_id": "unit7_106",
+  "original_id": "unit7_choice_106"
+},
+{
+  "id": "unit7_choice_107",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "2020年，我国开展了第七次全国人口普查，调查全国人口和住户的基本情况。下列情况，适合采用条形统计图表示的是（）。",
+  "options": [
+    "A. 从第一次人口普查到第七次人口普查全国人口增幅变化情况。",
+    "B. 男、女性别人数占全国总人数的百分比。",
+    "C. 各地区的常住人口数量。"
+  ],
+  "answer": "C",
+  "solution": "【分析】（1）条形统计图：用直条的长短表示数量的多少，从图中直观地看出数量的多少，便于比较；(2) 折线统计图: 不仅能看清数量的多少, 还能通过折线的上升和下降表示数量的增减变化情况;(3) 扇形统计图: 扇形统计图清楚地看出各部分数量与总数量之间, 部分与部分之间的关系, 据此解答。掌握各统计图的特点是解答题目的关键。\n【详解】A. 全国人口增幅变化情况适合选择折线统计图,错误;  B. 男、女性别人数占全国总人数的百分比更适合选择扇形统计图, 错误;  C. 各地区的常住人口数量适合选择条形统计图, 正确。故答案为: C",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "统计图的选择",
+    "单式折线统计图",
+    "1格表示一个单位的单式条形统计图"
+  ],
+  "source": "20-21六年级上·河南郑州·期末",
+  "merged_id": "unit7_107",
+  "original_id": "unit7_choice_107"
+},
+{
+  "id": "unit7_choice_108",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "要统计牛奶中营养成份所占的百分比情况, ( ) 更适用。",
+  "options": [
+    "A. 条形统计图",
+    "B. 折线统计图",
+    "C. 扇形统计图",
+    "D. 复式折线统计图"
+  ],
+  "answer": "C",
+  "solution": "【分析】根据各统计图的特点，选择合适的统计图即可，题目主要表示部分占整体百分比的情况，据此解答。\n【详解】A．条形统计图：用直条的长短表示数量的多少，从图中直观地看出数量的多少，便于比较；  B. 折线统计图: 不仅能看清数量的多少, 还能通过折线的上升和下降表示数量的增减变化情况;  C. 扇形统计图: 扇形统计图清楚地看出各部分数量与总数量之间, 部分与部分之间的关系, 可以表示出牛奶中各种成分所占百分比的情况;  D. 复式折线统计图: 不仅能看清数量的多少, 还能通过折线的上升和下降表示数量的增减变化情况, 复式折线统计图表示 2 个及以上的量的增减变化情况。  故答案为: C\n【点睛】掌握各统计图的特点并选择合适的统计图是解答题目的关键。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "统计图的选择"
+  ],
+  "source": "21-22 六年级上·江西赣州·单元测试",
+  "merged_id": "unit7_108",
+  "original_id": "unit7_choice_108"
+},
+{
+  "id": "unit7_choice_109",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "扇形统计图甲中男生占 $45\\%$ ，扇形统计图乙中男生占 $50\\%$ ，则甲和乙两个统计图所表示的男生人数（）。",
+  "options": [
+    "A. 甲比乙多",
+    "B. 乙比甲多",
+    "C. 一样多",
+    "D. 无法确定"
+  ],
+  "answer": "D",
+  "solution": "【分析】根据扇形统计图的定义和特征，结合题意，直接选出正确选项即可。本题考查了扇形统计图，明确扇形统计图的概念和特点是解题的关键。\n【详解】因为甲乙两个扇形统计图的统计范围是不一定相同的，统计总量是不一定相等的，所以无法确定甲和乙两个统计图所表示的男生人数哪个多哪个少。故答案为: D",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "21-22 六年级上·全国·单元测试",
+  "merged_id": "unit7_109",
+  "original_id": "unit7_choice_109"
+},
+{
+  "id": "unit7_choice_110",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "张彤应选择（）表示优、良、合格成绩与班级人数的关系。",
+  "options": [
+    "A. 折线统计图",
+    "B. 扇形统计图",
+    "C. 条形统计图",
+    "D. 无法确定"
+  ],
+  "answer": "B",
+  "solution": "【分析】扇形统计图清楚地看出各部分数量与总数量之间,部分与部分之间的关系;折线统计图的特点，不仅能看清数量的多少，还能通过折线的上升和下降表示数量的增减变化情况；条形统计图的特点：用直条的长短表示数量的多少，从图中直观地看出数量的多少，便于比较。明确各种统计图的特点是解答本题的关键\n【详解】表示优、良、合格成绩与班级人数的关系，应选择扇形统计图；故答案为: B",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "21-22六年级上·全国·单元测试",
+  "merged_id": "unit7_110",
+  "original_id": "unit7_choice_110"
+},
+{
+  "id": "unit7_choice_111",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "以下几种情况，最适合用扇形统计图表示的是（）。",
+  "options": [
+    "A. 比较五年级各班同学的跳绳数量",
+    "B. 六年级各班的男、女生人数",
+    "C. 一天的气温变化情况",
+    "D. 各类支出占家庭支出的百分比"
+  ],
+  "answer": "D",
+  "solution": "【分析】扇形统计图清楚地看出各部分数量与总数量之间，部分与部分之间的关系，据此解答即可；折线统计图的特点：不仅能看清数量的多少，还能通过折线的上升和下降表示数量的增减变化情况；条形统计图的特点：用直条的长短表示数量的多少，从图中直观地看出数量的多少，便于比较。明确各种统计图的特点是解答本题的关键。\n【详解】A. 比较五年级各班同学的跳绳数量，用条形统计图；  B. 六年级各班的男、女生人数, 用条形统计图;  C. 一天的气温变化情况, 用折线统计图;  D. 各类支出占家庭支出的百分比, 用扇形统计图; 故答案为: D",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "统计图的选择"
+  ],
+  "source": "20-21 六年级下·广东佛山·期末",
+  "merged_id": "unit7_111",
+  "original_id": "unit7_choice_111"
+},
+{
+  "id": "unit7_choice_112",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "困难",
+  "question": "便于直观了解数据大小和差异的统计图是（）。",
+  "options": [
+    "A. 条形统计图",
+    "B. 扇形统计图",
+    "C. 折线统计图"
+  ],
+  "answer": "A",
+  "solution": "【分析】条形统计图的特点：用直条的长短表示数量的多少，从图中直观地看出数量的多少，便于比较；扇形统计图清楚地看出各部分数量与总数量之间，部分与部分之间的关系；折线统计图的特点：不仅能看清数量的多少，还能通过折线的上升和下降表示数量的增减变化情况。明确各种统计图的特点是解答本题的关键。\n【详解】便于直观了解数据大小和差异的统计图是条形统计图；故答案为: A。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "单式折线统计图",
+    "1格表示一个单位的单式条形统计图"
+  ],
+  "source": "20-21 六年级下·云南玉溪·期末",
+  "merged_id": "unit7_112",
+  "original_id": "unit7_choice_112"
+},
+{
+  "id": "unit7_choice_113",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "小明想要统计未来一周本地的气温变化情况, 他应该选用 ( ) 统计图。",
+  "options": [
+    "A. 折线",
+    "B. 条形",
+    "C. 扇形"
+  ],
+  "answer": "A",
+  "solution": "【分析】折线统计图不仅能看清数量的多少，还能反映数量的增减变化情况。条形统计图从图中直观地看出数量的多少，便于比较。扇形统计图清楚地看出各部分数量与总数量之间的关系。据此选择即可。解决本题时应明确各种统计图的特点, 再根据要求选择合适的统计图。\n【详解】小明想要统计未来一周本地的气温变化情况, 他应该选用折线统计图。故答案为: A。",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "19-20 四年级下·甘肃定西·期末",
+  "merged_id": "unit7_113",
+  "original_id": "unit7_choice_113"
+},
+{
+  "id": "unit7_choice_114",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "不仅能表示出数量的多少，而且能清楚地表示出数量的增减变化情况的统计图是（）。",
+  "options": [
+    "A. 折线统计图",
+    "B. 条形统计图",
+    "C. 扇形统计图",
+    "D. 复式条形统计图"
+  ],
+  "answer": "A",
+  "solution": "【分析】条形统计图用直条的长短表示数量的多少, 从图中直观地看出数量的多少, 便于比较。折线统计图不仅能看清数量的多少，还能通过折线的上升和下降表示数量的增减变化情况。扇形统计图清楚地看出各部分数量与总数量之间，部分与部分之间的关系。关键是熟悉各种统计图的特点。\n【详解】不仅能表示出数量的多少，而且能清楚地表示出数量的增减变化情况的统计图是折线统计图。故答案为: A",
+  "knowledge_points": [
+    "单式折线统计图"
+  ],
+  "source": "19-20六年级下·江西赣州·期末",
+  "merged_id": "unit7_114",
+  "original_id": "unit7_choice_114"
+},
+{
+  "id": "unit7_choice_115",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "护士记录病人的体温变化情况应选用（）统计图。",
+  "options": [
+    "A. 条形",
+    "B. 折线",
+    "C. 扇形"
+  ],
+  "answer": "B",
+  "solution": "【分析】扇形统计图能够清楚地看出各部分数量与总数量之间，部分与部分之间的关系；折线统计图不仅能看清数量的多少，还能通过折线的上升和下降表示数量的增减变化情况；条形统计图，用直条的长短表示数量的多少，从图中直观地看出数量的多少，便于比较；熟练掌握各种统计图的特点是解答本题的关键。\n【详解】护士记录病人的体温变化情况应选用折线统计图;故答案为: B",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "2021·江苏无锡·小升初真题",
+  "merged_id": "unit7_115",
+  "original_id": "unit7_choice_115"
+},
+{
+  "id": "unit7_choice_116",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "李阿姨计划从 2021 年 1 月开始运动减肥, 每月月末进行体重测量。如果她想清楚了解未来这一年的体重增减变化, 用 ( ) 最合适。",
+  "options": [
+    "A. 条形统计图",
+    "B. 折线统计图",
+    "C. 扇形统计图",
+    "D. 复式统计图"
+  ],
+  "answer": "B",
+  "solution": "【分析】一般来说，如果几个数量是并列的，只要求表示数量的多少时，选条形统计图。如果表示一个量或几个量增减变化和发展变化趋势，则选折线统计图。如果要求表示各部分数量与总数量之间的关系，则选扇形统计图。关键是熟悉各种统计图的特点，单纯比多少，反映变化趋势还是数量之间的关系，从而确定统计图。\n【详解】李阿姨计划从 2021 年 1 月开始运动减肥, 每月月末进行体重测量。如果她想清楚了解未来这一年的体重增减变化，用折线统计图最合适。故答案为: B",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "21-22 六年级上·浙江·期末",
+  "merged_id": "unit7_116",
+  "original_id": "unit7_choice_116"
+},
+{
+  "id": "unit7_choice_117",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "某商场要表示2012年到2019年销售额的增长变化情况，应选用（）。",
+  "options": [
+    "A. 条形统计图",
+    "B. 折线统计图",
+    "C. 扇形统计图"
+  ],
+  "answer": "B",
+  "solution": "【分析】条形统计图的特点：用直条的长短表示数量的多少，从图中直观地看出数量的多少，便于比较。折线统计图的特点：不仅能看清数量的多少，还能通过折线的上升和下降表示数量的增减变化情况。扇形统计图清楚地看出各部分数量与总数量之间，部分与部分之间的关系。一定要明确各种统计图的特点，进而选择出合适的统计方式。\n【详解】某商场要表示 2012 年到 2019 年销售额的增长变化情况，应选用折线统计图；故答案为：B",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "17-18六年级上·河南周口·期末",
+  "merged_id": "unit7_117",
+  "original_id": "unit7_choice_117"
+},
+{
+  "id": "unit7_choice_118",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "要表示某风景区内树木总量的变化情况，应选用________；若要表示各种树木所占的百分比，应选用________；若要表示各种树木数量的多少，选用________比较合适。下面答案全部正确的是（）。(1) 条形统计图②折线统计图③扇形统计图",
+  "options": [
+    "A. ①②③",
+    "B. ①③②",
+    "C. ②①③",
+    "D. ②③①"
+  ],
+  "answer": "D",
+  "solution": "【分析】一般来说，如果几个数量是并列的，只要求表示数量的多少时，选条形统计图。如果表示一个量或几个量增减变化和发展变化趋势，则选折线统计图。如果要求表示各部分数量与总数量之间的关系，则选扇形统计图。关键是熟悉各种统计图的特点。\n【详解】要表示某风景区内树木总量的变化情况，应选用折线统计图；若要表示各种树木所占的百分比，应选用扇形统计图；若要表示各种树木数量的多少，选用条形统计图比较合适。故答案为: D",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "20-21 六年级上·辽宁鞍山·期末",
+  "merged_id": "unit7_118",
+  "original_id": "unit7_choice_118"
+},
+{
+  "id": "unit7_choice_119",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "要想更清楚地了解各部分数量同总数之间的关系，应该选用（）。",
+  "options": [
+    "A. 条形统计图",
+    "B. 折线统计图",
+    "C. 扇形统计图"
+  ],
+  "answer": "C",
+  "solution": "【分析】一般来说，如果几个数量是并列的，只要求表示数量的多少时，选条形统计图；如果表示一个量或几个量增减变化和发展变化趋势，则选折线统计图；如果要求表示各部分数量与总数量之间的关系, 则选扇形统计图。关键是掌握各种统计图的特征，条形统计图便于比较，折线统计图能看出增减变化趋势，扇形统计图能看出部分数量与总数量之间的关系。\n【详解】要想更清楚地了解各部分数量同总数之间的关系，应该选用扇形统计图。故答案为: C",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "2017·全国·小升初真题",
+  "merged_id": "unit7_119",
+  "original_id": "unit7_choice_119"
+},
+{
+  "id": "unit7_choice_120",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "困难",
+  "question": "扇形统计图表示 ( ) 。",
+  "options": [
+    "A. 数量的多少",
+    "B. 数量的变化情况",
+    "C. 部分与整体的关系",
+    "D. 以上都不是"
+  ],
+  "answer": "C",
+  "solution": "【详解】条形统计图能很容易看出数量的多少；折现统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；故答案为：C。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "18-19六年级上·辽宁大连·期末",
+  "merged_id": "unit7_120",
+  "original_id": "unit7_choice_120"
+},
+{
+  "id": "unit7_choice_121",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "要统计永泰县各庄寨 2018 年国庆黄金周游客所占百分比的情况，应选用（）较合适。",
+  "options": [
+    "A. 条形统计图",
+    "B. 折线统计图",
+    "C. 扇形统计图",
+    "D. 以上都不是"
+  ],
+  "answer": "C",
+  "solution": "【分析】根据扇形统计图的特点判断即可。本题考查扇形统计图，解答本题的关键是掌握扇形统计图的特点。\n【详解】扇形统计图能反映部分占总体的百分之几。故答案为: C。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "20-21 六年级上·全国·单元测试",
+  "merged_id": "unit7_121",
+  "original_id": "unit7_choice_121"
+},
+{
+  "id": "unit7_choice_122",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "困难",
+  "question": "为了直观反映各科成绩与总成绩的关系，选择较为合理的统计图是（）。",
+  "options": [
+    "A. 条形统计图",
+    "B. 折线统计图",
+    "C. 扇形统计图"
+  ],
+  "answer": "C",
+  "solution": "【分析】扇形统计图是用整个圆表示总数（单位 1），用圆内过圆心 O 点的各个扇形的大小表示各部分量占总量的百分比。分之几，扇形统计图中各部分的百分比之和是单位 $1$ 。熟记扇形统计图的特点是解决本题的关键。\n【详解】为了直观反映各科成绩与总成绩的关系，选择较为合理的统计图是扇形统计图。故答案为: C",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "20-21六年级上·全国·单元测试",
+  "merged_id": "unit7_122",
+  "original_id": "unit7_choice_122"
+},
+{
+  "id": "unit7_choice_123",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "困难",
+  "question": "需要清楚地表示每个项目的具体数目应选择 ( )。",
+  "options": [
+    "A. 折线统计图",
+    "B. 扇形统计图",
+    "C. 条形统计图",
+    "D. 以上三者均可以"
+  ],
+  "answer": "C",
+  "solution": "【分析】条形统计图能直观地表示出每个项目的具体数目；折线统计图能直观地反映事物变化情况；扇形统计图能直观地表示出各部分在总体中所占的百分比。明确不同统计图的特点是解答本题的关键。\n【详解】需要清楚地表示每个项目的具体数目应选择条形统计图。故答案为: C。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "统计图的选择"
+  ],
+  "source": "2022·湖南衡阳·小升初真题",
+  "merged_id": "unit7_123",
+  "original_id": "unit7_choice_123"
+},
+{
+  "id": "unit7_choice_124",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "下列说法中，不正确的是（）。",
+  "options": [
+    "A. 可以很清楚地表示出各部分同总体之间关系的统计图是条形统计图",
+    "B. 要清楚地反映出数量增减变化的统计图是折线统计图",
+    "C. 为了清楚地知道你的各科成绩, 你可以选择制作条形统计图",
+    "D. 为了清楚地反映出全校人数同各年级人数之间的关系,应选择扇形统计图"
+  ],
+  "answer": "A",
+  "solution": "【分析】条形统计图能直观地表示出每个项目的具体数目；折线统计图能直观地反映事物变化情况；扇形统计图能直观地表示出各部分在总体中所占的百分比。明确各种统计图的特点是解答本题的关键。\n【详解】可以很清楚地表示出各部分同总体之间关系的统计图是扇形统计图，而不是条形统计图。故答案为: A。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "20-21 六年级上·全国·单元测试",
+  "merged_id": "unit7_124",
+  "original_id": "unit7_choice_124"
+},
+{
+  "id": "unit7_choice_125",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "下面的信息资料中，适合用复式条形统计图的是（）。",
+  "options": [
+    "A. 一位病人的体温变化情况",
+    "B. 四年级 6 名同学的身高情况",
+    "C. 四（1）班的和四（2）班语文、数学、英语的单科平均分"
+  ],
+  "answer": "C",
+  "solution": "【分析】根据条形统计图的特点：条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此对各题进行依次分析、进而得出结论。\n【详解】A．统计一位病人的体温变化情况，用折线统计图最合适；  B. 四年级 6 名同学的身高情况, 主要是表示出数量的多少,所以用单式条形统计图最合适;  C. 四（1）班的和四（2）班语文、数学、英语的单科平均分，主要是表示出两类数量的多少，所以用复式条形统计图最合适；故答案为: C。此题主要考查条形、折线和扇形统计图的特点，应熟练掌握其特点，并能灵活运用。",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "2020 六年级下·全国·专题练习",
+  "merged_id": "unit7_125",
+  "original_id": "unit7_choice_125"
+},
+{
+  "id": "unit7_choice_126",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "折线统计图可以（）。",
+  "options": [
+    "A. 表示出各种数量的多少",
+    "B. 表示出数量增减变化情况",
+    "C. 表示出部分数与总数、部分数与部分数之间的数量关系"
+  ],
+  "answer": "B",
+  "solution": "【分析】首先要清楚每一种统计图的特点：条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；据此判断即可。此题考查了各种统计图的特点, 要掌握每种统计图的作用, 根据实际情况选择合适的统计图。\n【详解】折线统计图可以表示出数量增减变化情况。故答案为: B",
+  "knowledge_points": [
+    "单式折线统计图",
+    "复式折线统计图"
+  ],
+  "source": "2020 六年级下·全国·专题练习",
+  "merged_id": "unit7_126",
+  "original_id": "unit7_choice_126"
+},
+{
+  "id": "unit7_choice_127",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "气象站观测6月份气温变化情况，选用（）比较恰当。",
+  "options": [
+    "A. 条形统计图",
+    "B. 折线统计图",
+    "C. 扇形统计图",
+    "D. 统计表"
+  ],
+  "answer": "B",
+  "solution": "【分析】略",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "19-20五年级下·全国·单元测试",
+  "merged_id": "unit7_127",
+  "original_id": "unit7_choice_127"
+},
+{
+  "id": "unit7_choice_128",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "如果想要表示长兴2018年12月份气温变化的情况，最适合的是选用（）统计图。",
+  "options": [
+    "A. 条形",
+    "B. 折线",
+    "C. 扇形"
+  ],
+  "answer": "B",
+  "solution": "【详解】条形统计图特点：可以清楚地看出数量的多少；折线统计图特点：不但可以表示数量的多少，还可以清楚的看出数量的增减变化情况；扇形统计图特点：可以看出各个部分数量与总数之间的关系，据此结合题意选择合适的统计图。",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "19-20 六年级下·山东·单元测试",
+  "merged_id": "unit7_128",
+  "original_id": "unit7_choice_128"
+},
+{
+  "id": "unit7_choice_129",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "下面四种情况，（ ）选用扇形统计图比较合适。",
+  "options": [
+    "A. 王姨 6 月份卖出女装的数量",
+    "B. 王姨 6 月份卖出女装占卖出服装总量的情况",
+    "C. 王姨 6 个月卖出女装增减变化情况",
+    "D. 王姨 6 个月卖出服装的总数量"
+  ],
+  "answer": "B",
+  "solution": "【分析】条形统计图特点：可以清楚地看出数量的多少；折线统计图特点：不但可以表示数量的多少，还可以清楚的看出数量的增减变化情况；扇形统计图特点：可以看出各个部分数量与总数之间的关系，据此结合题意选择合适的统计图。\n【详解】选项A，王姨6月份卖出女装的数量适合用条形统计图；选项B，王姨6月份卖出女装占卖出服装总量的情况适合用扇形统计图；选项C，王姨6个月卖出女装增减变化情况适合用折线统计图；选项 D, 王姨 6 个月卖出服装的总数量适合用条形统计图。故答案为: B。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "统计图的选择"
+  ],
+  "source": "19-20六年级下·江苏·单元测试",
+  "merged_id": "unit7_129",
+  "original_id": "unit7_choice_129"
+},
+{
+  "id": "unit7_choice_130",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "爸爸把家庭每月各种支出情况绘制成扇形统计图，是为了（）。",
+  "options": [
+    "A. 能直观地看出每项支出的多少",
+    "B. 能看出每项支出的变化趋势",
+    "C. 能直观地看出每项支出与月总支出的关系",
+    "D. 形象、美观"
+  ],
+  "answer": "C",
+  "solution": "【详解】扇形统计图特点：可以看出各个部分数量与总数之间的关系，据此解答。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "20-21 六年级上·北京·单元测试",
+  "merged_id": "unit7_130",
+  "original_id": "unit7_choice_130"
+},
+{
+  "id": "unit7_choice_131",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "利用收集的数据，画成扇形统计图，则所有的扇形的百分比之和为（）。",
+  "options": [
+    "A. $70\\%$",
+    "B. $90\\%$",
+    "C. $100\\%$"
+  ],
+  "answer": "C",
+  "solution": "【分析】根据扇形统计图的意义，把一个圆的面积看作单位'1'，各部分之和等于1，也就是利用收集的数据，画成扇形统计图，则所有的扇形的百分比之和为 $100\\%$ 。此题主要是考查扇形统计图的意义，扇形统计图中各部分之和等于圆面积，也就是所有扇形的百分率为 $100\\%$ 。\n【详解】根据分析可知, $1 = 100\\%$ 利用收集的数据，画成扇形统计图，则所有的扇形的百分比之和为 $100 \\%$ .故答案为: C",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "19-20六年级上·辽宁·单元测试",
+  "merged_id": "unit7_131",
+  "original_id": "unit7_choice_131"
+},
+{
+  "id": "unit7_choice_132",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "果农要清楚地表示果园内各种果树占果树总数量的百分比，需要绘制（）统计图.",
+  "options": [
+    "A. 条形",
+    "B. 折线",
+    "C. 扇形"
+  ],
+  "answer": "C",
+  "solution": "【分析】略",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "19-20六年级上·全国·单元测试",
+  "merged_id": "unit7_132",
+  "original_id": "unit7_choice_132"
+},
+{
+  "id": "unit7_choice_133",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "为了清楚地展示彩电全年销售的变化趋势，用( )统计图更合适。",
+  "options": [
+    "A. 条形",
+    "B. 折线",
+    "C. 扇形"
+  ],
+  "answer": "B",
+  "solution": "【分析】略",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "19-20六年级上·全国·单元测试",
+  "merged_id": "unit7_133",
+  "original_id": "unit7_choice_133"
+},{
+  "id": "unit7_choice_134",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "为反映空气的主要成分所占的体积与总体积的关系，最好用（ ）统计图。",
+  "options": [
+    "A. 条形",
+    "B. 折线",
+    "C. 扇形"
+  ],
+  "answer": "C",
+  "solution": "【详解】略",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "19-20六年级上·全国·单元测试",
+  "merged_id": "unit7_134",
+  "original_id": "unit7_choice_134"
+},
+{
+  "id": "unit7_choice_135",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "某地区 2004 年一2008年各年的降水量分别是800毫升、920毫升、850毫升、780 毫升、1200 毫升。为表示降水量的变化情况，采用（）统计图最合适.",
+  "options": [
+    "A. 条形",
+    "B. 折线",
+    "C. 扇形"
+  ],
+  "answer": "B",
+  "solution": "【分析】略",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "19-20 六年级上·全国·单元测试",
+  "merged_id": "unit7_135",
+  "original_id": "unit7_choice_135"
+},
+{
+  "id": "unit7_choice_136",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "要清楚反映牛肉中各种营养成分的含量，应选用( )统计图。",
+  "options": [
+    "A. 条形",
+    "B. 折线",
+    "C. 扇形"
+  ],
+  "answer": "C",
+  "solution": "【分析】略",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "19-20 六年级上·全国·单元测试",
+  "merged_id": "unit7_136",
+  "original_id": "unit7_choice_136"
+},
+{
+  "id": "unit7_choice_137",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "某村共种蔬菜40公顷，其中种西红柿8公顷若制成扇形统计图，则表示西红柿的扇形占整个圆的（）",
+  "options": [
+    "A. $8 \\%$",
+    "B. $20\\%$",
+    "C. $40\\%$"
+  ],
+  "answer": "B",
+  "solution": "【详解】 $8 \\div 40 = 20\\%$答：表示西红柿的扇形占整个圆的 $20\\%$ .故选 B.",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "19-20六年级上·全国·单元测试",
+  "merged_id": "unit7_137",
+  "original_id": "unit7_choice_137"
+},
+{
+  "id": "unit7_fill_138",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "小敏每 2 小时记录一次气温, 研究一天中气温变化情况, 用( )统计图更合适。",
+  "options": [],
+  "answer": "折线",
+  "solution": "【分析】根据题意，结合各个统计图的特点，直接选择合适的统计图即可。\n【详解】折线统计图可以清晰反映数据的变化情况，所以，要研究一天中气温变化情况，用折线统计图更合适。\n【点睛】本题考查了折线统计图, 明确折线统计图的特点是解题的关键。",
+  "knowledge_points": [
+    "单式折线统计图"
+  ],
+  "source": "21-22 六年级上·江西赣州·单元测试",
+  "merged_id": "unit7_138",
+  "original_id": "unit7_fill_138"
+},
+{
+  "id": "unit7_fill_139",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "扇形统计图用( )的面积表示总数, 用( )的面积表示各部分占总数的百分比。在某一扇形统计图中，所有扇形的百分比之和是( )。",
+  "options": [],
+  "answer": "圆 ；扇形 ；1",
+  "solution": "【分析】扇形统计图比较清楚地反映出部分与部分、部分与整体之间的数量关系，圆的面积表示整体数量，扇形的面积表示部分数量。根据扇形统计图的概念和意义可知圆代表整体，是把圆看作一个单位'1'，所以所有的百分比之和必须等于1。本题主要考查扇形统计图的意义及特征。\n【详解】扇形统计图中用圆的面积表示总数，用扇形的面积表示各部分占总数的百分比。在某一扇形统计图中，所有扇形的百分比之和是 $1$ 。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "21-22 六年级上·江西赣州·单元测试",
+  "merged_id": "unit7_139",
+  "original_id": "unit7_fill_139"
+},
+{
+  "id": "unit7_fill_140",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "某车间应出勤200人，有一天缺勤4人，这天的出勤率是( )%。",
+  "options": [],
+  "answer": "98",
+  "solution": "【分析】根据'出勤率 $=\\frac{\\text{出勤人数}}{\\text{总人数}} \\times 100\\%$ '，进行解答即可。明确出勤率的含义是解答本题的关键。\n【详解】 $\\frac{200 - 4}{200} \\times 100\\% = 98\\%$",
+  "knowledge_points": [
+    "求一个数是另一个数的百分之几（百分率问题）"
+  ],
+  "source": "21-22六年级上·全国·单元测试",
+  "merged_id": "unit7_140",
+  "original_id": "unit7_fill_140"
+},
+{
+  "id": "unit7_fill_141",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "( )统计图可以很清楚地表示各部分同总数之间的关系, 从( )统计图中可以清楚地看出数量增减的变化情况。",
+  "options": [],
+  "answer": "扇形 ；折线",
+  "solution": "【分析】折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。此题应根据折线统计图、扇形统计图各自的特点进行解答。\n【详解】扇形统计图可以很清楚地表示各部分同总数之间的关系；从折线统计图可以清楚地看出数量增减变化情况。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "单式折线统计图"
+  ],
+  "source": "19-20 六年级下·山东济南·期末",
+  "merged_id": "unit7_141",
+  "original_id": "unit7_fill_141"
+},
+{
+  "id": "unit7_fill_142",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "扇形统计图: 用整个圆的面积表示( ), 用扇形面积表示( )。扇形统计图可清楚地表示出( )。",
+  "options": [],
+  "answer": "总数 ；各部分数量 ；各部分数量同总数之间的关系",
+  "solution": "【分析】根据扇形统计图的概念和特征，直接填空即可。本题考查了扇形统计图, 明确扇形统计图的特征是解题的关键。\n【详解】用整个圆的面积表示总数，用扇形面积表示各部分数量。扇形统计图可清楚地表示出各部分数量同总数之间的关系。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "21-22 六年级上·全国·单元测试",
+  "merged_id": "unit7_142",
+  "original_id": "unit7_fill_142"
+},
+{
+  "id": "unit7_fill_143",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "从( )统计图中很容易看出数量的增减变化情况, 从( )统计图中可以清楚地表示出各部分数量同总量之间的关系。",
+  "options": [],
+  "answer": "折线 ；扇形",
+  "solution": "【分析】折线统计图：用不同位置的点表示数量的多少，并用折线的上升和下降表示数量的增减变化情况；扇形统计图：以一个圆的面积表示物体的总数量，以相应的扇形面积占整个圆面积的百分数表示各有关部分占总数量的百分数，据此解答。考查了折线统计图和扇形统计图的优点, 学生应掌握。\n【详解】从（折线）统计图中很容易看出数量的增减变化情况，从（扇形）统计图中可以清楚地表示出各部分数量同总量之间的关系。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "单式折线统计图",
+    "复式折线统计图"
+  ],
+  "source": "19-20 六年级下·河南南阳·期末",
+  "merged_id": "unit7_143",
+  "original_id": "unit7_fill_143"
+},
+{
+  "id": "unit7_fill_144",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "医生为了统计病人每天体温的变化情况应选用( )统计图。",
+  "options": [],
+  "answer": "折线",
+  "solution": "【分析】条形统计图特点：可以清楚地看出数量的多少；折线统计图特点：不但可以表示数量的多少，还可以清楚的看出数量的增减变化情况；扇形统计图特点：可以看出各个部分数量与总数之间的关系；据此结合题意选择合适的统计图。有 '变化' 一词就可以判断出应选用折线统计图,即可快速得出答案。\n【详解】医生为了统计病人每天体温的变化情况应选用折线统计图。",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "2020·湖南长沙·小升初真题",
+  "merged_id": "unit7_144",
+  "original_id": "unit7_fill_144"
+},
+{
+  "id": "unit7_fill_145",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "要表示某地年降水量的变化情况, 应选用_____统计图。",
+  "options": [],
+  "answer": "折线",
+  "solution": "【分析】条形统计图特点：可以清楚地看出数量的多少；折线统计图特点：不但可以表示数量的多少，还可以清楚的看出数量的增减变化情况，据此选择。\n【详解】根据折线统计图的特点可知：要表示某地年降水量的变化情况，应选用折线统计图。\n【点睛】本题主要考查统计图的选择，牢记折线、条形统计图的特点是解题的关键。",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "20-21 五年级下·江苏·单元测试",
+  "merged_id": "unit7_145",
+  "original_id": "unit7_fill_145"
+},
+{
+  "id": "unit7_fill_146",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "( )统计图、( )统计图、( )统计图是最常用的三种统计图。",
+  "options": [],
+  "answer": "条形 ；折线 ；扇形",
+  "solution": "【分析】常用的统计图有三种，条形统计图、折线统计图，扇形统计图，条形统计图可以清楚的看到各个数量的多少；折线统计图不但可以清楚的看到数量的多少，还能清楚地看出数量间的变化情况；扇形统计图能清楚地看出部分的量和总量之间的关系。此题考查的是对常用的统计图的掌握情况。\n【详解】常用的统计图有三种，条形统计图、折线统计图，扇形统计图。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "单式折线统计图",
+    "1格表示一个单位的单式条形统计图"
+  ],
+  "source": "20-21 六年级上·安徽宿州·期末",
+  "merged_id": "unit7_146",
+  "original_id": "unit7_fill_146"
+},
+{
+  "id": "unit7_fill_147",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "在扇形统计图中，所有扇形的百分比之和为( )，扇形统计图可以表示( )之间的关系。",
+  "options": [],
+  "answer": "1 ；各个部分和总体",
+  "solution": "【分析】扇形统计图：以一个圆的面积表示事物的总体，以扇形面积表示各部分占总体的百分数的统计图，叫做扇形统计图。结合扇形统计图的特点可知，它是用整个圆表示总数量，所以所有扇形的百分比之和就是 $100\\%$ ，即 1。\n【详解】由分析得：在扇形统计图中，所有扇形的百分比之和为（1），扇形统计图可以表示（各个部分和总体）之间的关系。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "19-20六年级上·河南南阳·期末",
+  "merged_id": "unit7_147",
+  "original_id": "unit7_fill_147"
+},
+{
+  "id": "unit7_fill_148",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "条形统计图能直观地表示出每个项目的( )；折线统计图能直观地反映事物( )；扇形统计图能直观地表示出各部分在总体中所占的( )。",
+  "options": [],
+  "answer": "具体数目 ；变化情况 ；百分比",
+  "solution": "【详解】条形统计图能直观地表示出每个项目的具体数目；折线统计图能直观地反映事物变化情况；扇形统计图能直观地表示出各部分在总体中所占的百分比。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "统计图的选择"
+  ],
+  "source": "20-21 六年级上·全国·单元测试",
+  "merged_id": "unit7_148",
+  "original_id": "unit7_fill_148"
+},
+{
+  "id": "unit7_fill_149",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "困难",
+  "question": "在第 28 届奥运会上，为了统计在奥运会上各国奖牌数所占的百分比，应选用( )统计图。",
+  "options": [],
+  "answer": "扇形",
+  "solution": "【分析】扇形统计图能直观地表示出各部分在总体中所占的百分比，由此解答即可。本题主要考查了扇形统计图的特点，一定要牢记。\n【详解】在第 28 届奥运会上, 为了统计在奥运会上各国奖牌数所占的百分比, 应选用扇形统计图。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "20-21 六年级上·全国·单元测试",
+  "merged_id": "unit7_149",
+  "original_id": "unit7_fill_149"
+},
+{
+  "id": "unit7_fill_150",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "我们学过（ ）统计图、（ ）统计图和（ ）统计图。（ ）统计图能清楚的表示出整体与部分之间的关系。",
+  "options": [],
+  "answer": "条形 ；折线 ；扇形 ；扇形",
+  "solution": "【分析】折线统计图能清楚的表示出数量的增减变化；扇形统计图能反映各部分同总数的关系；条形统计图能直接比较出数量的多少，据此解答即可。本题考查条形、折线、扇形统计图，解答本题的关键是掌握条形、折线、扇形统计图的特点。\n【详解】我们学过条形统计图、折线统计图和扇形统计图。扇形统计图能清楚的表示出整体与部分之间的关系。故答案为：条形；折线；扇形；扇形。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "单式折线统计图",
+    "1格表示一个单位的单式条形统计图"
+  ],
+  "source": "2020六年级下·内蒙古呼伦贝尔·专题练习",
+  "merged_id": "unit7_150",
+  "original_id": "unit7_fill_150"
+},
+{
+  "id": "unit7_fill_151",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "在学过的统计图中，要表示数量的增减变化情况用( )统计图，表示各部分占总体的百分比用( )统计图，描述全班同学最喜欢的体育运动情况用( )统计图。",
+  "options": [],
+  "answer": "折线 ；扇形 ；条形",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。此题应根据条形统计图、折线统计图、扇形统计图各自的特点进行解答。\n【详解】在学过的统计图中，要表示数量的增减变化情况用折线统计图，表示各部分占总体的百分比用扇形统计图，描述全班同学最喜欢的体育运动情况用条形统计图。",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "18-19六年级下·广东阳江·单元测试",
+  "merged_id": "unit7_151",
+  "original_id": "unit7_fill_151"
+},
+{
+  "id": "unit7_fill_152",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "根据所要描述的情况，填写合适的统计图。(1) 期末考试后, 要描述各分数段人数的分布情况, 用 ( )统计图。  (2) 护士要描述病人一天内体温的变化情况, 用( )统计图。  (3) 描述优秀学生人数占全校总人数的百分比情况, 用 ( )统计图。",
+  "options": [],
+  "answer": "条形； 折线 ；扇形",
+  "solution": "【详解】略",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "19-20六年级下·全国·单元测试",
+  "merged_id": "unit7_152",
+  "original_id": "unit7_fill_152"
+},
+{
+  "id": "unit7_fill_153",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "要表示各种数量的多少需要画( )统计图；要表示数量的增减变化的情况需要画( )统计图；要表示各部分和整体之间的关系需要画( )统计图。",
+  "options": [],
+  "answer": "条形 ；折线 ；扇形",
+  "solution": "【详解】要表示数量的多少需要画条形统计图，如：想具体了解一年中四个季度每个季度超市的营业额，就用条形统计图。要表示数量的增减变化情况需要画折线统计图，如：想了解一年四季中超市营业额的变化情况，就用折线统计图。要表示各部分数量与总数之间的关系需要画扇形统计图。如：想了解一年四季中各季度营业额与全年总营业额的关系，就用扇形统计图。",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "19-20六年级上·河北沧州·期末",
+  "merged_id": "unit7_153",
+  "original_id": "unit7_fill_153"
+},
+{
+  "id": "unit7_fill_154",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "要反映某食品各种营养成分的含量，最好选用( )统计图。",
+  "options": [],
+  "answer": "扇形",
+  "solution": "【分析】可以根据条形统计图、折线统计图和扇形统计图的特点进行选择用那种最合适。此题考查了三种常用统计图的优点。\n【详解】解：条形统计图优点：可以清楚的看清数量的多少；折线统计图优点：可以清楚的看清数量的变化趋势；扇形统计图优点: 可以清楚的表示出某一个量与总量的关系;根据上面的题意，所以要选择扇形统计图。故答案为：扇形。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "19-20 六年级上·湖南株洲·期末",
+  "merged_id": "unit7_154",
+  "original_id": "unit7_fill_154"
+},
+{
+  "id": "unit7_fill_155",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "小丽家上月的教育支出是全月总支出的 $25\\%$ ，绘制她家上月支出情况的扇形统计图时，圆的面积表示____。若爱心捐款支出是总支出的 $12.5\\%$ ，则教育支出是爱心捐款支出的____倍。",
+  "options": [],
+  "answer": "全月总支出 ；2",
+  "solution": "【详解】用一个圆形表示全月总支出，用扇形来表示各部分支出占总支出的百分率。用教育支出占的百分率除以爱心捐款支出的百分率即可求出二者的倍数关系。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "含百分数的运算"
+  ],
+  "source": "19-20六年级上·全国·单元测试",
+  "merged_id": "unit7_155",
+  "original_id": "unit7_fill_155"
+},
+{
+  "id": "unit7_fill_156",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "要反映六年级人数占全校人数的百分比，最好用_统计图。",
+  "options": [],
+  "answer": "扇形",
+  "solution": "【详解】略",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "19-20六年级上·全国·单元测试",
+  "merged_id": "unit7_156",
+  "original_id": "unit7_fill_156"
+},
+{
+  "id": "unit7_fill_157",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "要想清楚地反映出各车间工业产值占全厂工业总产值的百分比，应绘制( )统计图.",
+  "options": [],
+  "answer": "扇形",
+  "solution": "【分析】略",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "19-20 六年级上·全国·单元测试",
+  "merged_id": "unit7_157",
+  "original_id": "unit7_fill_157"
+},
+{
+  "id": "unit7_fill_158",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "为了能清楚地看出各部分数量同总数之间的关系，应选用( )统计图，为了能很容易地看出各种数量的多少，应选用( )统计图.",
+  "options": [],
+  "answer": "扇形 ；条形",
+  "solution": "【分析】略\n【详解】略",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "19-20四年级上·全国·单元测试",
+  "merged_id": "unit7_158",
+  "original_id": "unit7_fill_158"
+},
+{
+  "id": "unit7_fill_159",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "常见的统计图有( )要表示某学校各年度的毕业人数情况，可以选用( )统计图。",
+  "options": [],
+  "answer": "条形统计图 ；折线统计图 ；扇形统计图 ；条形",
+  "solution": "【分析】略\n【详解】略",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "统计图的选择",
+    "1 格表示多个单位的单式条形统计图"
+  ],
+  "source": "19-20 四年级上·全国·单元测试",
+  "merged_id": "unit7_159",
+  "original_id": "unit7_fill_159"
+},
+{
+  "id": "unit7_fill_160",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "要表示汽车四个车间六月份的产量与全厂六月份总量之间的关系，应绘制绘制( )统计图.",
+  "options": [],
+  "answer": "扇形",
+  "solution": "【分析】略\n【详解】略",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "19-20六年级上·全国·单元测试",
+  "merged_id": "unit7_160",
+  "original_id": "unit7_fill_160"
+},
+{
+  "id": "unit7_fill_161",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "为表示一个病人的体温变化情况, 应绘制( ).",
+  "options": [],
+  "answer": "折线统计图",
+  "solution": "【分析】略\n【详解】略",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "19-20 六年级上·全国·单元测试",
+  "merged_id": "unit7_161",
+  "original_id": "unit7_fill_161"
+},
+{
+  "id": "unit7_fill_162",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "可以直观、清楚地表示出各部分数量与总数之间的关系，用( )表示。",
+  "options": [],
+  "answer": "扇形",
+  "solution": "【分析】略\n【详解】略",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "19-20六年级上·全国·单元测试",
+  "merged_id": "unit7_162",
+  "original_id": "unit7_fill_162"
+},
+{
+  "id": "unit7_fill_163",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "瑞光小学开展课外小组活动，参加美术组的有180人，参加体育组的有130人，参加航模组的有190人。如果制成扇形统计图，那么表示体育组人数的那部分扇形所对应的百分比为( )%。",
+  "options": [],
+  "answer": "26",
+  "solution": "【分析】略\n【详解】略",
+  "knowledge_points": [
+    "百分数的四则运算"
+  ],
+  "source": "19-20六年级上·全国·单元测试",
+  "merged_id": "unit7_163",
+  "original_id": "unit7_fill_163"
+},
+{
+  "id": "unit7_fill_164",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "困难",
+  "question": "用统计图表示数量之间的关系比较形象，常用的统计图有( )、( )和( )三种。",
+  "options": [],
+  "answer": "条形统计图 ；折线统计图 ；扇形统计图",
+  "solution": "【分析】常见的统计图有条形统计图、折线统计图、扇形统计图，它们各有各的特点：条形统计图很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；据此填写即可。\n【详解】用统计图表示数量之间的关系比较形象，常用的统计图有条形统计图、折线统计图和扇形统计图三种。",
+  "knowledge_points": [
+    "复式条形统计图",
+    "扇形统计图的特点及绘制",
+    "单式折线统计图"
+  ],
+  "source": "19-20六年级上·全国·单元测试",
+  "merged_id": "unit7_164",
+  "original_id": "unit7_fill_164"
+},
+{
+  "id": "unit7_fill_165",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "含有( )的统计项目的统计表叫复式统计图.",
+  "options": [],
+  "answer": "两个或两个以上",
+  "solution": "【分析】略\n【详解】略",
+  "knowledge_points": [
+    "复式条形统计图",
+    "复式折线统计图"
+  ],
+  "source": "19-20 六年级上·全国·单元测试",
+  "merged_id": "unit7_165",
+  "original_id": "unit7_fill_165"
+},
+{
+  "id": "unit7_fill_166",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "( )统计图能清楚地表示数量的多少，( )统计图既能反映数量的多少，又能表示数量变化情况.",
+  "options": [],
+  "answer": "条形 ；折线",
+  "solution": "【分析】略\n【详解】略",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "19-20 六年级上·全国·单元测试",
+  "merged_id": "unit7_166",
+  "original_id": "unit7_fill_166"
+},
+{
+  "id": "unit7_fill_167",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "制作统计图一般分为四个步骤: ( )、( )、( )、( )。",
+  "options": [],
+  "answer": "收集数据 ；整理数据 ；设计算表 ；正式制表",
+  "solution": "【分析】略\n【详解】略",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "19-20 六年级上·全国·单元测试",
+  "merged_id": "unit7_167",
+  "original_id": "unit7_fill_167"
+},
+{
+  "id": "unit7_fill_168",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "条形统计图和折线统计图都有( )式和( )式两种。",
+  "options": [],
+  "answer": "单 ；复",
+  "solution": "【分析】略\n【详解】略",
+  "knowledge_points": [
+    "简单条形统计图"
+  ],
+  "source": "19-20 六年级上·全国·单元测试",
+  "merged_id": "unit7_168",
+  "original_id": "unit7_fill_168"
+},
+{
+  "id": "unit7_fill_169",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "要统计阳光小学六(1)班学生最喜欢的各类体育运动项目的人数，最好选用( )统计图.",
+  "options": [],
+  "answer": "条形",
+  "solution": "【分析】根据统计图的特点进行选择。本题考查了条形统计图的特点。\n【详解】条形统计图的特点: 从图中直观地看出数量的多少, 便于比较。故答案为: 条形",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "19-20 六年级上·全国·单元测试",
+  "merged_id": "unit7_169",
+  "original_id": "unit7_fill_169"
+},
+{
+  "id": "unit7_fill_170",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "可以清楚反映统计数据增减变化趋势的统计图是( )统计图，可以清楚地看出各种数据的统计图是( )统计图，如果要更清楚地了解各部分数量同总量之间的关系可以用( )统计图来表示..",
+  "options": [],
+  "answer": "折线 ；条形； 扇形",
+  "solution": "【分析】略\n【详解】略",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "统计图的选择"
+  ],
+  "source": "19-20六年级上·全国·单元测试",
+  "merged_id": "unit7_170",
+  "original_id": "unit7_fill_170"
+},
+{
+  "id": "unit7_fill_171",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "学校图书馆里有三种书, 其中故事书 200 本, 连环画 120 本, 文艺书 180 本,如果制成扇形统计图, 那么表示故事书, 连环画, 文艺书的扇形部分分别占圆面积( )%, ( )%, ( )%.",
+  "options": [],
+  "answer": "40 ；24 ；36",
+  "solution": "【分析】略\n【详解】略",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "19-20 六年级上·全国·课后作业",
+  "merged_id": "unit7_171",
+  "original_id": "unit7_fill_171"
+},
+{
+  "id": "unit7_fill_172",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "要比较两个城市某一段时间内的降水量情况，选用( )统计图更合适。",
+  "options": [],
+  "answer": "条形",
+  "solution": "【分析】略\n【详解】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "19-20六年级上·全国·课后作业",
+  "merged_id": "unit7_172",
+  "original_id": "unit7_fill_172"
+},
+{
+  "id": "unit7_fill_173",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "折线统计图表示( ), 扇形统计图表示( ), 条形统计图表示( ).",
+  "options": [],
+  "answer": "数量关系的多少和增减变化情况； 部分与总数的关系 ；数量的多少",
+  "solution": "【分析】略\n【详解】扇形统计图表示各部分数量同总数之间的关系,折线统计图能等直接反映数量的增减变化情况, 条形统计图能清楚地表示出每个项目的具体数目.",
+  "knowledge_points": [
+    "复式条形统计图",
+    "扇形统计图的特点及绘制",
+    "统计图的选择",
+    "单式折线统计图"
+  ],
+  "source": "19-20 六年级上·全国·课后作业",
+  "merged_id": "unit7_173",
+  "original_id": "unit7_fill_173"
+},
+{
+  "id": "unit7_judge_174",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "在制作统计图时, 大豆占总面积的 $40 \\%$ , 它所在扇形圆心角的度数是 $40^{\\circ}$( )",
+  "options": [],
+  "answer": "×",
+  "solution": "【分析】一周是 $360^{\\circ}$ , 用 $360^{\\circ} \\times 40\\%$ , 求出大豆所在扇形圆心角的度数, 从而判断题干的正误即可。本题考查了扇形统计图, 明确扇形统计图的绘制方法是判断的关键。\n【详解】 $360^{\\circ} \\times 40\\% = 144^{\\circ}$ , 所以大豆所在扇形圆心角的度数是 $144^{\\circ}$ 。所以判断错误。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "21-22 六年级上·江西赣州·单元测试",
+  "merged_id": "unit7_174",
+  "original_id": "unit7_judge_174"
+},
+{
+  "id": "unit7_judge_175",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "用多种统计图都可表示出某些信息, 但一定有一种统计图最适合。( )",
+  "options": [],
+  "answer": "√",
+  "solution": "【分析】因为每一种信息都有最适合它的统计图，据此解答。此题考查的是根据实际情况选择合适的统计图。\n【详解】由分析可得，用多种统计图都可表示出某些信息，但一定有一种统计图最适合。正确，故答案为: √ 。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "统计图的选择"
+  ],
+  "source": "21-22 六年级上·江西赣州·单元测试",
+  "merged_id": "unit7_175",
+  "original_id": "unit7_judge_175"
+},
+{
+  "id": "unit7_judge_176",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "容易",
+  "question": "在扇形统计图中, 扇形面积大的那部分占的百分比一定大。( )",
+  "options": [],
+  "answer": "√",
+  "solution": "【分析】扇形统计图比较清楚地反映出部分与部分、部分与整体之间的数量关系，圆的面积表示整体数量，扇形的面积表示部分数量。本题考查扇形统计图，明确扇形统计图表示的意义是解题的关键。\n【详解】由分析可知：圆的面积表示整体数量，扇形的面积越大则占圆的面积的百分比越大。故正确答案为: √",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "21-22 六年级上·江西赣州·单元测试",
+  "merged_id": "unit7_176",
+  "original_id": "unit7_judge_176"
+},
+{
+  "id": "unit7_judge_177",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "扇形统计图能直接看出各种量的多少。( )",
+  "options": [],
+  "answer": "×",
+  "solution": "【分析】扇形统计图可以直观地比较出各部分数量的相对大小，以及各部分数量与总数之间的关系，据此解答。掌握扇形统计图的特点是解答本题的关键。\n【详解】扇形统计图可以表示出各部分占整体的百分比,不能直接看出数量的多少。故答案为: ×",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "18-19 六年级下·四川·期末",
+  "merged_id": "unit7_177",
+  "original_id": "unit7_judge_177"
+},
+{
+  "id": "unit7_judge_178",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "表示某学生一学期的数学成绩变化, 选用扇形统计图比较合适。( )",
+  "options": [],
+  "answer": "×",
+  "solution": "【分析】一般来说，如果几个数量是并列的，只要求表示数量的多少时，选条形统计图。如果表示一个量或几个量增减变化和发展变化趋势，则选折线统计图。如果要求表示各部分数量与总数量之间的关系，则选扇形统计图。扇形统计图清楚地看出各部分数量与总数量之间，部分与部分之间的关系。\n【详解】表示某学生一学期的数学成绩变化，选用折线统计图比较合适，所以原题说法错误。",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "19-20 六年级下·四川巴中·期末",
+  "merged_id": "unit7_178",
+  "original_id": "unit7_judge_178"
+},
+{
+  "id": "unit7_judge_179",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "想要了解病人各时段的体温变化情况, 把收集的体温数据制成条形统计图比较好。( )",
+  "options": [],
+  "answer": "×",
+  "solution": "【分析】条形统计图可以清楚的看出数量的多少，折线统计图不仅能看出数量的多少，还能看出数量的变化情况，扇形统计图能清楚的看出部分与总量之间的关系，据此选择。此题考查了统计图的选择, 需牢记每种统计图的特点。\n【详解】想要了解病人各时段的体温变化情况，需要看出数据的变化，用折线统计图比较合适。原题说法错误。故答案为: ×",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "19-20 六年级下·河南南阳·期末",
+  "merged_id": "unit7_179",
+  "original_id": "unit7_judge_179"
+},
+{
+  "id": "unit7_judge_180",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "容易",
+  "question": "要反映某超市一周的营业额变化情况，制成折线统计图比较合适。（）",
+  "options": [],
+  "answer": "√",
+  "solution": "【分析】条形统计图特点：可以清楚地看出数量的多少；折线统计图特点：不但可以表示数量的多少，还可以清楚的看出数量的增减变化情况；扇形统计图特点：可以看出各个部分数量与总数之间的关系，据此结合题意选择合适的统计图。有 '变化' 一词就可以判断出应选用折线统计图,即可快速判断出答案。\n【详解】要反映某超市一周的营业额变化情况，制成折线统计图比较合适。故答案为: √。",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "18-19六年级下·天津·期末",
+  "merged_id": "unit7_180",
+  "original_id": "unit7_judge_180"
+},
+{
+  "id": "unit7_judge_181",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "要反映一周的气温变化情况，用扇形统计图比较好。（）",
+  "options": [],
+  "answer": "×",
+  "solution": "【分析】一般来说，如果几个数量是并列的，只要求表示数量的多少时，选条形统计图。如果表示一个量或几个量增减变化和发展变化趋势，则选折线统计图。如果要求表示各部分数量与总数量之间的关系，则选扇形统计图。用点、线、面等来表示相关联的量之间的数量关系的图形, 叫做统计图。用统计图表示有关数据量之间的关系更为直接、具体、形象。\n【详解】要反映一周的气温变化情况，用折线统计图比较好，所以原题说法错误。",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "2021六年级下·全国·专题练习",
+  "merged_id": "unit7_181",
+  "original_id": "unit7_judge_181"
+},
+{
+  "id": "unit7_judge_182",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "容易",
+  "question": "扇形统计图中，扇形的圆心角越大，就说明这一部分占总量的百分比就越大。（）",
+  "options": [],
+  "answer": "√",
+  "solution": "【分析】根据扇形统计图的意义可知，各部分占总体的百分比之和为 1 , 每部分占总体的百分比等于该部分所对应的扇形圆心角的度数与 360 度的比, 所以扇形的圆心角越大,说明这一部分占总量的百分比就越大, 由此判断即可。明确扇形统计图的意义是解答本题的关键。\n【详解】扇形统计图中，扇形的圆心角越大，就说明这一部分占总量的百分比就越大，原题说法正确；故答案为： √ 。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "2014 六年级·全国·课后作业",
+  "merged_id": "unit7_182",
+  "original_id": "unit7_judge_182"
+},
+{
+  "id": "unit7_judge_183",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "容易",
+  "question": "条形和折线统计图都有复式的, 扇形统计图也有复式的。( )",
+  "options": [],
+  "answer": "×",
+  "solution": "【详解】条形和折线统计图都有复式的，扇形统计图没有复式的；故答案为：×。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "19-20 六年级上·河南南阳·期末",
+  "merged_id": "unit7_183",
+  "original_id": "unit7_judge_183"
+},
+{
+  "id": "unit7_judge_184",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "容易",
+  "question": "要表现一组数据的变化趋势, 应选择条形统计图。( )",
+  "options": [],
+  "answer": "×",
+  "solution": "【分析】首先要清楚每一种统计图的特点：条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；据此判断即可。此题应根据条形统计图、折线统计图、扇形统计图各自的特点进行解答。\n【详解】要表现一组数据的变化趋势，应选择折线统计图。故答案为：×",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "19-20 四年级下·吉林长春·期末",
+  "merged_id": "unit7_184",
+  "original_id": "unit7_judge_184"
+},
+{
+  "id": "unit7_judge_185",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "描述新冠肺炎治愈人数占感染人数的百分比情况用折线统计图比较合适。（）",
+  "options": [],
+  "answer": "×",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择。此题应根据条形统计图、折线统计图、扇形统计图各自的特点进行解答。\n【详解】描述新冠肺炎治愈人数占感染人数的百分比情况用扇形统计图比较合适。故答案为：×",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "19-20六年级下·广东佛山·期末",
+  "merged_id": "unit7_185",
+  "original_id": "unit7_judge_185"
+},
+{
+  "id": "unit7_judge_186",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "容易",
+  "question": "在扇形统计图中,所有扇形的百分比之和为 1。( )",
+  "options": [],
+  "answer": "√",
+  "solution": "【分析】扇形统计图中, 将整个圆看作单位 '1', 则所有扇形的百分比之和为 1。本题主要考查扇形统计图的概念和意义, 圆代表整体, 即单位 '1', 各个扇形代表部分, 掌握扇形统计图的特点是关键。\n【详解】在扇形统计图中，所有扇形的百分比之和为 1 ，题目描述正确。故答案为: √",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "19-20 六年级下·江苏·单元测试",
+  "merged_id": "unit7_186",
+  "original_id": "unit7_judge_186"
+},
+{
+  "id": "unit7_judge_187",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "统计图的种类不同, 但反映的信息相同。( )",
+  "options": [],
+  "answer": "×",
+  "solution": "【分析】常见的三种统计图的特点是：条形统计图：能很容易看出各种数量的多少；折线统计图：不但能表示数量的多少，还能表示出数量增减变化；扇形统计图：能很清楚地表示出各部分数量同总数的关系，统计图的种类不同，反映的信息也是不同的。熟记三种统计图的特点，会根据题目的要求及统计图的特点选择合适的统计图。\n【详解】根据统计图的特点可知，统计图的种类不同，反映的信息也是不同的，故原题说法错误。故答案为: ×",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "19-20 四年级下·上海·期中",
+  "merged_id": "unit7_187",
+  "original_id": "unit7_judge_187"
+},
+{
+  "id": "unit7_judge_188",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "困难",
+  "question": "扇形统计图的特点是只能比较清楚地看出各部分量的多少。( )",
+  "options": [],
+  "answer": "×",
+  "solution": "【分析】扇形统计图描述的是各部分所占的百分比，条形统计图描述的是各部分具体的数量。本题考查扇形统计图的特征，熟记各类统计图的特征是解答此题的关键。\n【详解】扇形统计图的特点是只能清楚地看出各部分所占的百分比。故原说法错误。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "18-19 六年级下·浙江宁波·期末",
+  "merged_id": "unit7_188",
+  "original_id": "unit7_judge_188"
+},
+{
+  "id": "unit7_judge_189",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "容易",
+  "question": "要表示小丽同学 $7 \\sim 12$ 岁的身高变化情况, 应选用条形统计图。( )",
+  "options": [],
+  "answer": "×",
+  "solution": "【分析】反应某一数据的变化情况，应使用折线统计图，据此判断即可。\n【详解】根据分析可知, 要表示小丽同学 $7 \\sim 12$ 岁的身高变化情况, 应选用折线统计图。故答案为:×",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "19-20 六年级上·全国·期末",
+  "merged_id": "unit7_189",
+  "original_id": "unit7_judge_189"
+},
+{
+  "id": "unit7_judge_190",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "统计一个班某次数学考试成绩分布情况，最好选用扇形统计图。（）",
+  "options": [],
+  "answer": "√",
+  "solution": "【分析】略\n【详解】略",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "统计图的选择",
+    "单式折线统计图",
+    "1格表示一个单位的单式条形统计图"
+  ],
+  "source": "19-20六年级上·全国·期末",
+  "merged_id": "unit7_190",
+  "original_id": "unit7_judge_190"
+},
+{
+  "id": "unit7_judge_191",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "在一个扇形统计图中，经济作物的扇形圆心角是60度，则经济作物的面积占总面积的 $\\frac{1}{6}$ 。（）",
+  "options": [],
+  "answer": "√",
+  "solution": "【分析】因为圆周角是 $360^{\\circ}$ , 经济作物的扇形圆心角是 60 度, 说明经济作物占总面积的 $60^{\\circ} \\div 360^{\\circ} = \\frac{1}{6}$ 。据此解答即可。本题考查扇形统计图及相关计算。在扇形统计图中，每部分占总部分的分率等于该部分所对应的扇形圆心角的度数与 $360^{\\circ}$ 的比值。\n【详解】解: 由题意得: 经济作物种植面积占总面积的: $60^{\\circ} \\div 360^{\\circ} = \\frac{1}{6}$ 。答: 经济作物种植面积占总面积的 $\\frac{1}{6}$ 故正确答案为: √",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "2020·山西忻州·小升初真题",
+  "merged_id": "unit7_191",
+  "original_id": "unit7_judge_191"
+},
+{
+  "id": "unit7_judge_192",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "要反映全校这次视力检测的近视率、正常率要用折线统计图。（）",
+  "options": [],
+  "answer": "×",
+  "solution": "【分析】略\n【详解】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；据此判断即可。",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "19-20六年级上·全国·课后作业",
+  "merged_id": "unit7_192",
+  "original_id": "unit7_judge_192"
+},
+{
+  "id": "unit7_judge_193",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "扇形统计图可直观看出统计数据各部分占总体的比率（）",
+  "options": [],
+  "answer": "√",
+  "solution": "【分析】 略\n【详解】略",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "19-20六年级上·全国·单元测试",
+  "merged_id": "unit7_193",
+  "original_id": "unit7_judge_193"
+},
+{
+  "id": "unit7_judge_194",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "扇形统计图和其他统计图一样也要有标题和图例。（）",
+  "options": [],
+  "answer": "×",
+  "solution": "【分析】 略\n【详解】扇形统计图和其他统计图一样也要有标题, 但可以没有图例。故答案为: ×",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "2014 六年级·全国·课后作业",
+  "merged_id": "unit7_194",
+  "original_id": "unit7_judge_194"
+},
+{
+  "id": "unit7_judge_195",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "统计表与统计图是两个毫无关系的知识。（）",
+  "options": [],
+  "answer": "×",
+  "solution": "【分析】统计表和统计图都是统计数据的一种方法，用统计表的数据也可以用统计图来表示，并且用统计图表示比用统计表更直观。据此判断。此题考查的目的是理解掌握统计表、统计图的特点、作用及应用。\n【详解】由分析得: 统计表和统计图都是统计数据的一种方法, 它们既有联系也有区别。因此，统计表与统计图是两个毫无关系的知识。这种说法是错误的。故答案为: ×",
+  "knowledge_points": [
+    "复式统计表的特点及填补",
+    "统计图表的综合应用",
+    "扇形统计图的特点及绘制",
+    "复式折线统计图"
+  ],
+  "source": "19-20六年级上·全国·单元测试",
+  "merged_id": "unit7_195",
+  "original_id": "unit7_judge_195"
+},
+{
+  "id": "unit7_calc_196",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "计算题",
+  "difficulty": "困难",
+  "question": "解方程。(1) $90 - 3x = 12$ (2) $20\\% (x - 2.5) = 4$ (3) $\\frac{3}{x} + \\frac{1}{3x} = \\frac{5}{6}$",
+  "options": [],
+  "answer": "(1) $x = 26$ ; (2) $x = 22.5$ ; (3) $x = 4$",
+  "solution": "【分析】 (1) $90 - 3x = 12$ , 先写成 $3x = 90 - 12$ 的形式, 方程两边再同时除以 3 即可;(2) $20\\% (x - 2.5) = 4$ ，方程两边先同时除以 $20\\%$ ，再同时加 $2.5$ ; (3) $\\frac{3}{x} + \\frac{1}{3x} = \\frac{5}{6}$ , 方程两边同时乘 $3x$ , 再将方程两边同时 $\\times \\frac{2}{5}$ ; (注意 $x \\neq 0$ ),本题考查了解方程，解方程根据等式的基本性质。\n【详解】 (1) $90 - 3x = 12$\n解： $3x\\div 3 = 78\\div 3\\\\x = 2 6$\n(2) $20\\% (x - 2.5) = 4$\n解： $20\\% (x - 2.5) \\div 20\\% = 4 \\div 20\\%\\\\ x - 2. 5 + 2. 5 = 2 0 + 2. 5 \\\\ x = 2 2. 5 $\n(3) $\\frac{3}{x} + \\frac{1}{3x} = \\frac{5}{6}$\n解: $(\\frac{3}{x} + \\frac{1}{3 x}) \\times 3 x = \\frac{5}{6} \\times 3 x\\\\9 + 1 = \\frac {5}{2} x \\\\ \\frac {5}{2} x \\times \\frac {2}{5} = 1 0 \\times \\frac {2}{5} \\\\ x = 4 $",
+  "knowledge_points": [
+    "应用等式的性质1和2解方程"
+  ],
+  "source": "2019·浙江·小升初真题",
+  "merged_id": "unit7_196",
+  "original_id": "unit7_calc_196"
+},
+{
+  "id": "unit7_calc_197",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "计算题",
+  "difficulty": "中等",
+  "question": "解方程. $0.2 x- 7 \\times 1. 3 = 9. 9$ $\\frac {1}{2} \\times (x - 8) = 4$ $x - 0. 9 x = 3. 6$ $4 x - \\frac {1}{8} x = \\frac {1 3}{8}$",
+  "options": [],
+  "answer": "95 ；16 ；36 ； $\\frac{13}{31}$",
+  "solution": "【详解】 (1) $0.2 \\times 7 \\times 1.3 = 9.9\\\\0.2 x - 9. 1 = 9. 9 \\\\ 0. 2 x = 9. 9 + 9. 1 \\\\ 0. 2 x = 1 9\\\\x = 9 5 $\n(2) $\\frac{1}{2} \\times (x - 8) = 4\\\\x - 8 = 4 \\times 2 \\\\\\ x - 8 = 8 \\\\ x = 1 6 $\n(3) $x - 0.9x = 3.6 \\\\ 0. 1 x = 3. 6 \\\\ x = 3 6 $\n(4) $4x - \\frac{1}{8} x = \\frac{13}{8}\\\\3 2 x - x = 1 3 \\\\ 3 1 x = 1 3 \\\\ x = \\frac {1 3}{3 1}$",
+  "knowledge_points": [
+    "分数与分数的除法",
+    "应用等式的性质 1 和 2 解方程"
+  ],
+  "source": "19-20 六年级上·全国·单元测试",
+  "merged_id": "unit7_197",
+  "original_id": "unit7_calc_197"
+},
+{
+  "id": "unit7_draw_198",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "作图题",
+  "difficulty": "中等",
+  "question": "六年级有130人参加体育项目，各组人数如下。长跑组：30人；拔河组：20人；接力组：20人；短跑组：30人；掷铅球组：30人。根据数据制成扇形统计图。",
+  "options": [],
+  "answer": "见详解",
+  "solution": "【分析】利用除法先分别计算出各组人数占总人数的百分之几，再画出对应的扇形统计图即可。\n【详解】长跑组： $30 \\div 130 \\approx 23.08\\%$ ，拔河组：$20 \\div 130 \\approx 15.38\\%$ , 接力组: $20 \\div 130 \\approx 15.38\\%$ , 短跑组:$30 \\div 130 \\approx 23.08\\%$ , 掷铅球组: $30 \\div 130 \\approx 23.08 \\text{‰}$扇形统计图如下：![](images/3cd77e06c937274c650e29e2f62f7e94d06d4a551da8b861157f99d85c1447fe.jpg)\n【点睛】本题考查了扇形统计图, 会画扇形统计图是解题的关键。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "21-22六年级上·全国·单元测试",
+  "merged_id": "unit7_198",
+  "original_id": "unit7_draw_198"
+},
+{
+  "id": "unit7_solve_199",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "解答题",
+  "difficulty": "中等",
+  "question": "小明把2018年家里的所有支出分成'食品、还货款、教育、服装、水电、其他'这六类，他想制作一个一年支出的统计图，你觉得设计成什么统计图比较合适？并结合统计图的特点写出理由。",
+  "options": [],
+  "answer": "扇形统计图；原因见详解",
+  "solution": "【详解】设计成扇形统计图比较合适，因为扇形统计图可以看出各个部分数量与总数之间的百分比。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "统计图的选择"
+  ],
+  "source": "19-20六年级下·山东·单元测试",
+  "merged_id": "unit7_199",
+  "original_id": "unit7_solve_199"
+},
+{
+  "id": "unit7_solve_200",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "解答题",
+  "difficulty": "中等",
+  "question": "某食品店5月25日销售饮料情况如下：果茶24瓶，雪碧40瓶，可乐56瓶，橘子汁8瓶。请根据以上数据制作扇形统计图.某食品店5月25日销售饮料情况统计图",
+  "options": [],
+  "answer": "![](images/44ee8e6a3edb96911af53501b70f5196aa259be03d8b3f6d1e1188aba1b2b231.jpg)",
+  "solution": "【详解】 $24 + 40 + 56 + 8 = 128$ （瓶） 果茶: $24 \\div 128 = 0.1875 = 18.75 \\%$\n$360^{\\circ}\\times 18.75\\% = 67.5^{\\circ}$ 雪碧: $40 \\div 128 = 0.3125 = 31.25\\%$$360^{\\circ}\\times 31.25\\% = 112.5^{\\circ}$可乐: $56 \\div 128 = 0.4375 = 43.75\\%$$360^{\\circ} \\times 43.75 \\% = 157.5^{\\circ}$橘子汁: $8 \\div 128 = 0.0625 = 6.25 \\%$$360^{\\circ}\\times 6.25\\% = 22.5^{\\circ}$",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "19-20六年级上·全国·单元测试",
+  "merged_id": "unit7_200",
+  "original_id": "unit7_solve_200"
+},{
+  "id": "unit7_choice_201",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "不但能表示数量的多少，而且能清楚地表示出数量的增减变化情况是（）统计图。",
+  "options": [
+    "A. 条形",
+    "B. 折线",
+    "C. 扇形",
+    "D. 无法确定"
+  ],
+  "answer": "B",
+  "solution": "【详解】略",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "2015·全国·小升初真题",
+  "merged_id": "unit7_201",
+  "original_id": "unit7_choice_201"
+},
+{
+  "id": "unit7_choice_202",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "一个商场2006年上半年收入中，家电收入占 $55\\%$，针织品收入占 $20\\%$，其他收入占 $25\\%$。将此制成扇形统计图，其中表示家电收入的扇形的圆心角是（）.",
+  "options": [
+    "A. $198^{\\circ}$",
+    "B. $126^{\\circ}$",
+    "C. $36^{\\circ}$",
+    "D. $54^{\\circ}$"
+  ],
+  "answer": "A",
+  "solution": "【分析】略",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "19-20六年级上·全国·单元测试",
+  "merged_id": "unit7_202",
+  "original_id": "unit7_choice_202"
+},
+{
+  "id": "unit7_choice_203",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "要表示学校各年级学生的数量, 选用 ( ) 比较合适。",
+  "options": [
+    "A. 折线统计图",
+    "B. 条形统计图",
+    "C. 扇形统计图"
+  ],
+  "answer": "B",
+  "solution": "【详解】要表示学校各年级学生的数量，选用条形统计图比较合适。故答案为: B",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "18-19 六年级上·广东中山·期末",
+  "merged_id": "unit7_203",
+  "original_id": "unit7_choice_203"
+},
+{
+  "id": "unit7_choice_204",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "要统计几个发展中国家人口占世界总人口的百分比,需要用( )。",
+  "options": [
+    "A. 条形统计图",
+    "B. 折线统计图",
+    "C. 扇形统计图"
+  ],
+  "answer": "C",
+  "solution": "【分析】本题考查了统计图表的选择。\n【详解】略",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "19-20六年级上·全国·单元测试",
+  "merged_id": "unit7_204",
+  "original_id": "unit7_choice_204"
+},
+{
+  "id": "unit7_choice_205",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "在一个扇形统计图中，苹果树棵数的扇形圆心角是 $60^{\\circ}$，那么苹果树棵数占水果树总棵数的（）。",
+  "options": [
+    "A. $\\frac{1}{4}$",
+    "B. $\\frac{1}{6}$",
+    "C. $\\frac{1}{8}$"
+  ],
+  "answer": "B",
+  "solution": "【分析】周角是 $360^{\\circ}$, 先用 $60^{\\circ} \\div 360^{\\circ}$ 求出扇形面积占所在圆面积的几分之几, 即苹果树棵数占水果树总棵数的几分之几。扇形统计图是用整个圆表示总数，用圆内大小不同的扇形表示各部分数量占总数的百分比。\n【详解】$60^{\\circ} \\div 360^{\\circ}]\\\\ =\\frac {6 0}{3 6 0} \\\\ = \\frac {1}{6}$\n所以苹果树棵数占水果树总棵数的 $\\frac{1}{6}$。故答案为: B",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "求一个数占另一个数几分之几"
+  ],
+  "source": "19-20 六年级上·全国·单元测试",
+  "merged_id": "unit7_205",
+  "original_id": "unit7_choice_205"
+},
+{
+  "id": "unit7_choice_206",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "扇形统计图中, 所有扇形的百分比之和应该 ( )。",
+  "options": [
+    "A. 小于 1",
+    "B. 等于 1",
+    "C. 大于 1"
+  ],
+  "answer": "B",
+  "solution": "【分析】扇形统计图，它是用整个圆表示总数，用圆内各个扇形的大小表示各部分数量占总数的百分数。通过扇形统计图可以很清楚地表示出各部分数量同总数之间的关系。据此解答即可。本题考查学生对扇形统计图的理解和掌握。\n【详解】根据扇形统计图的含义，可以得知所有扇形的大小加起来就是整个圆的大小，也就是所有扇形的百分比之和应该等于 1。故答案为: B",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "19-20 六年级上·全国·单元测试",
+  "merged_id": "unit7_206",
+  "original_id": "unit7_choice_206"
+},
+{
+  "id": "unit7_choice_207",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "某校为表示学生体育锻炼达标情况，选用（）可以清楚地表示出'优''良''合格'各等次人数占全校总人数的百分比。",
+  "options": [
+    "A. 条形统计图",
+    "B. 折线统计图",
+    "C. 扇形统计图"
+  ],
+  "answer": "C",
+  "solution": "【分析】扇形统计图是用整个圆表示总数（单位'1'），用圆内过圆心 O 点的各个扇形的大小表示各部分量占总量的百分之几，通过扇形统计图可以很清楚地表示出各部分数量与总数之间的关系。\n【详解】选用扇形统计图可以清楚地表示出'优''良''合格' 各等次人数占全校总人数的百分比。故答案为: C",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "19-20六年级上·全国·单元测试",
+  "merged_id": "unit7_207",
+  "original_id": "unit7_choice_207"
+},
+{
+  "id": "unit7_choice_208",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "在一个扇形统计图中,油料作物所在扇形面积占整个圆的 $\\frac{1}{4}$, 则油料作物种植面积占总种植面积的( ).",
+  "options": [
+    "A. $50\\%$",
+    "B. $25\\%$",
+    "C. $17\\%$"
+  ],
+  "answer": "B",
+  "solution": "【详解】占整个圆的 $\\frac{1}{4}$, 则说明所占的比例为 $\\frac{1}{4}\\times 100 \\% = 25 \\%$",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "19-20 六年级上·全国·单元测试",
+  "merged_id": "unit7_208",
+  "original_id": "unit7_choice_208"
+},
+{
+  "id": "unit7_choice_209",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "光明小学为清楚地表示学校各年级的人数占全校学生总数的百分之几,应绘制( ).",
+  "options": [
+    "A. 条形统计图",
+    "B. 折线统计图",
+    "C. 扇形统计图"
+  ],
+  "answer": "C",
+  "solution": "【详解】扇形统计图能够反映各区域占整体的百分率",
+  "knowledge_points": [
+    "统计图的选择",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "19-20六年级上·全国·单元测试",
+  "merged_id": "unit7_209",
+  "original_id": "unit7_choice_209"
+},
+{
+  "id": "unit7_choice_210",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "关于选用统计图，下面说法合适的是（）。",
+  "options": [
+    "A. '二孩'政策后，为统计本区每个月新生儿出生变化情况，选用条形统计图。",
+    "B. 要统计一袋牛奶里的营养成分所占百分比情况, 选用扇形统计图。",
+    "C. 要了解超市每月销售额和利润额数据，选用折线统计图。",
+    "D. 以上都合适。"
+  ],
+  "answer": "B",
+  "solution": "【分析】折线统计图是以折线的上升或下降来表示统计数量的增减变化的统计图，折线统计图用折线的起伏表示数据的增减变化情况；扇形统计图是用整个圆表示总数，用圆内过圆心O点的各个扇形的大小表示各部分量占总量的百分之几，通过扇形统计图可以很清楚地表示出各部分数量与总数之间的关系；条形统计图是用一个单位长度表示一定的数量，根据数量的多少画成长短不同的直条，然后把这些直条按一定的顺序排列起来的统计图，从条形统计图中很容易看出各种数量的多少。\n【详解】A项中，'二孩'政策后，为统计本区每个月新生儿出生变化情况，选用折线统计图；B项中，要统计一袋牛奶里的营养成分所占百分比情况，选用扇形统计图；C项中，要了解超市每月销售额和利润额数据，选用条形统计图。所以合适的选B。故答案为：B。",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "19-20六年级上·全国·单元测试",
+  "merged_id": "unit7_210",
+  "original_id": "unit7_choice_210"
+},
+{
+  "id": "unit7_choice_211",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "气象站要制作一周的气温变化情况，应选用（）统计图。",
+  "options": [
+    "A. 条形",
+    "B. 折线",
+    "C. 扇形"
+  ],
+  "answer": "B",
+  "solution": "【详解】略",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "18-19六年级下·云南临沧·期末",
+  "merged_id": "unit7_211",
+  "original_id": "unit7_choice_211"
+},
+{
+  "id": "unit7_choice_212",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "只表示数量多少的统计图是 ( )。",
+  "options": [
+    "A. 条形统计图",
+    "B. 折线统计图",
+    "C. 扇形统计图"
+  ],
+  "answer": "A",
+  "solution": "【分析】条形统计图可以清楚的看出数量的多少；折线统计图不仅能看出数量的多少，还能看出数量的增减变化；扇形统计图可以清楚的看出整体和部分之间的关系，据此解答即可。\n【详解】由分析可知, 只表示数量多少的统计图是条形统计图。故答案为: A",
+  "knowledge_points": [
+    "1 格表示一个单位的单式条形统计图",
+    "单式折线统计图",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "19-20 四年级上·全国·课后作业",
+  "merged_id": "unit7_212",
+  "original_id": "unit7_choice_212"
+},
+{
+  "id": "unit7_choice_213",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "要清楚地反映各部分量和总数的关系，最好选用（）。",
+  "options": [
+    "A. 扇形统计图",
+    "B. 条形统计图",
+    "C. 折线统计图",
+    "D. 以上都可以"
+  ],
+  "answer": "A",
+  "solution": "【分析】扇形统计图是以一个圆的面积表示物体的总数量，以相应的扇形面积占整个圆面积的百分数表示各有关部分占总数量的百分数，能清楚地看出各部分数量与总数量之间的关系；条形统计图是用一个单位长度表示一定的数量，用直条的长短表示数量的多少，作用是从图中能清楚地看出各种数量的多少，便于相互比较；折线统计图是用不同位置的点表示数量的多少，根据数量的多少描出各点，然后把各点用线段顺次连接起来，不仅能看清数量的多少，还能通过折线的上升和下降表示数量的增减变化情况。据此解答。\n【详解】由分析得，要清楚地反映各部分量和总数的关系，最好选用扇形统计图。故答案为：A",
+  "knowledge_points": [
+    "统计图的选择（扇形统计图）"
+  ],
+  "source": "2018·江苏无锡·小升初真题",
+  "merged_id": "unit7_213",
+  "original_id": "unit7_choice_213"
+},
+{
+  "id": "unit7_choice_214",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "要反映某地区今年各类谷物的产量占粮食总产量的百分比，应绘制（）统计图。",
+  "options": [
+    "A. 条形",
+    "B. 扇形",
+    "C. 折线",
+    "D. 以上都可以"
+  ],
+  "answer": "B",
+  "solution": "【分析】条形统计图特点：可以清楚地看出数量的多少；折线统计图特点：不但可以表示数量的多少，还可以清楚的看出数量的增减变化情况；扇形统计图特点：可以看出各个部分数量与总数之间的关系，据此结合题意选择合适的统计图。\n【详解】略",
+  "knowledge_points": [
+    "1 格表示一个单位的单式条形统计图",
+    "单式折线统计图",
+    "统计图的选择",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "18-19六年级下·江苏连云港·期中",
+  "merged_id": "unit7_214",
+  "original_id": "unit7_choice_214"
+},
+{
+  "id": "unit7_choice_215",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "要把四个商场6月份甲、乙两种商品的销售量制成统计图，你觉得制成（）比较好。",
+  "options": [
+    "A. 复式条形统计图",
+    "B. 折线统计图",
+    "C. 复式折线统计图",
+    "D. 扇形统计图"
+  ],
+  "answer": "A",
+  "solution": "【分析】条形统计图的特点：用直条的长短表示数量的多少，从图中直观地看出数量的多少，便于比较；折线统计图的特点：不仅能看清数量的多少，还能通过折线的上升和下降表示数量的增减变化情况；复式折现统计可以表示两组不同量的变化情况。扇形统计图清楚地看出各部分数量与总数量之间，部分与部分之间的关系，由此进行选择即可。\n【详解】由分析可知: 要把四个商场 6 月份甲、乙两种商品的销售量制成统计图, 应选择复式折线统计图。故答案为：A",
+  "knowledge_points": [
+    "统计图的选择（扇形统计图）",
+    "复式条形统计图",
+    "统计图的选择（折线统计图）"
+  ],
+  "source": "18-19六年级下·全国·期末",
+  "merged_id": "unit7_215",
+  "original_id": "unit7_choice_215"
+},
+{
+  "id": "unit7_choice_216",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "甲班有学生 50 人, 上午有 1 人请了一天的病假, 下午又有 2 人请事假, 下午的出勤率是 ( )",
+  "options": [
+    "A. $98\\%$",
+    "B. $96\\%$",
+    "C. $94\\%$"
+  ],
+  "answer": "C",
+  "solution": "【详解】$\\frac{50 - 1 - 2}{50} \\times 100\\% = 94\\%$, 下午的出勤率是 $94 \\%$。故答案为: C",
+  "knowledge_points": [
+    "求一个数是另一个数的百分之几（百分率问题）"
+  ],
+  "source": "17-18 六年级上·湖南怀化·期末",
+  "merged_id": "unit7_216",
+  "original_id": "unit7_choice_216"
+},
+{
+  "id": "unit7_choice_217",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "鸡、鸭、鹅的只数比是 $3: 2: 1$, 画成扇形统计图后, 表示鸡的只数的扇形圆心角的度数是（）。",
+  "options": [
+    "A. $180^{\\circ}$",
+    "B. $90^{\\circ}$",
+    "C. $60^{\\circ}$",
+    "D. $30^{\\circ}$"
+  ],
+  "answer": "A",
+  "solution": "【分析】根据题意可知, 圆周角 $= 360^{\\circ}$, 根据鸡、鸭、鹅的只数比, 可以求出鸡占总只数的分率, 然后用圆周角的度数 $\\times$ 鸡占总数的分率 = 表示鸡的只数的扇形圆心角的度数,据此列式解答。\n【详解】$360^{\\circ} \\times \\frac{1}{3 + 2 + 1} = 180^{\\circ}$。故答案为: A",
+  "knowledge_points": [
+    "比的意义",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "18-19 六年级上·浙江·课后作业",
+  "merged_id": "unit7_217",
+  "original_id": "unit7_choice_217"
+},
+{
+  "id": "unit7_choice_218",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "要反映六年级学生在体育健康测试中各个项目合格人数所占的百分比情况，最好选用（）。",
+  "options": [
+    "A. 扇形统计图",
+    "B. 折线统计图",
+    "C. 条形统计图"
+  ],
+  "answer": "A",
+  "solution": "【详解】略",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "2019·全国·小升初模拟",
+  "merged_id": "unit7_218",
+  "original_id": "unit7_choice_218"
+},
+{
+  "id": "unit7_choice_219",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "绘制扇形统计图的优点是 ( )",
+  "options": [
+    "A. 表示数量的多少",
+    "B. 表示部分与整体的关系",
+    "C. 表示数量增减变化情况"
+  ],
+  "answer": "B",
+  "solution": "【详解】扇形统计图中把整体看成单位 '1', 较易表示出各部分占整体的百分之几或几分之几。故答案为: B",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "15-16 五年级·甘肃·期末",
+  "merged_id": "unit7_219",
+  "original_id": "unit7_choice_219"
+},
+{
+  "id": "unit7_choice_220",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "要反映某种儿童食品中各种营养成分的含量，最好选用（）统计图。",
+  "options": [
+    "A. 条形",
+    "B. 折线",
+    "C. 扇形"
+  ],
+  "answer": "C",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。掌握各统计图的特点是解答本题的关键。\n【详解】根据各统计图的特点可知, 要反映某种儿童食品中各种营养成分的含量, 就要选择能很好的体现部分和整体关系的统计图, 即扇形统计图。故答案为: C",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "16-17 六年级上·全国·单元测试",
+  "merged_id": "unit7_220",
+  "original_id": "unit7_choice_220"
+},
+{
+  "id": "unit7_choice_221",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "某地区 $2001 \\sim$2005年的年降水量分别是860毫米，950毫米，810毫米，1000毫米，980毫米．为表示降水量的情况，采用________统计图最合适.",
+  "options": [
+    "A. 条形",
+    "B. 折线",
+    "C. 扇形",
+    "D. 都可以"
+  ],
+  "answer": "A",
+  "solution": "【详解】对于统计图各自的特点我们要能够掌握，并能做出合适的选择。",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "15-16 四年级上·山东青岛·课后作业",
+  "merged_id": "unit7_221",
+  "original_id": "unit7_choice_221"
+},
+{
+  "id": "unit7_choice_222",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "要形象反映化肥厂2010年下半年每月的产量情况，最好选用（）。",
+  "options": [
+    "A. 条形统计图",
+    "B. 折线统计图",
+    "C. 扇形统计图",
+    "D. 以上答案都不对"
+  ],
+  "answer": "A",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。\n【详解】根据统计图的特点可知：要形象反映化肥厂2010年下半年每月的产量情况，最好选用条形统计图。故答案为: A",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "2015·四川绵阳·小升初真题",
+  "merged_id": "unit7_222",
+  "original_id": "unit7_choice_222"
+},
+{
+  "id": "unit7_choice_223",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "要表示出某地区2012年平均气温的变化情况，最好选（）",
+  "options": [
+    "A. 条形统计图",
+    "B. 折线统计图",
+    "C. 扇形统计图"
+  ],
+  "answer": "B",
+  "solution": "【分析】略\n【详解】略",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "18-19 五年级下·全国·课后作业",
+  "merged_id": "unit7_223",
+  "original_id": "unit7_choice_223"
+},
+{
+  "id": "unit7_choice_224",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "画统计图时, 要表示出数量的多少, 最好选 ( )",
+  "options": [
+    "A. 条形统计图",
+    "B. 折线统计图",
+    "C. 扇形统计图"
+  ],
+  "answer": "A",
+  "solution": "【分析】条形统计图可以清楚的看出数量的多少；折线统计图不仅能看出数量的多少，还能看出数量的增减变化；扇形统计图可以清楚的看出整体和部分之间的关系，据此选择。掌握各种统计图的特点是解题关键。\n【详解】由分析可知, 画统计图时, 要表示出数量的多少,最好选条形统计图。故答案为: A",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "18-19 五年级下·全国·课后作业",
+  "merged_id": "unit7_224",
+  "original_id": "unit7_choice_224"
+},
+{
+  "id": "unit7_choice_225",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "一个鸡蛋按质量计算, 蛋壳、蛋白、蛋黄约占整个鸡蛋的百分比分别为：15%、53%、32%，如果将数据画成统计图，选（）统计图较合适。",
+  "options": [
+    "A. 条形",
+    "B. 折线",
+    "C. 扇形",
+    "D. 复式条形"
+  ],
+  "answer": "C",
+  "solution": "【分析】扇形统计图表示的是部分在总体中所占的百分比,但一般不能直接从图中得到具体的数据;折线统计图表示的是事物的变化情况;条形统计图能清楚地表示出每个项目的具体数目。依此即可作出选择。\n【详解】要求表示的是蛋壳、蛋白、蛋黄在总体鸡蛋中所占的百分比，结合统计图各自的特点，应选择扇形统计图。故选 C.",
+  "knowledge_points": [
+    "1 格表示一个单位的单式条形统计图",
+    "单式折线统计图",
+    "统计图的选择",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "2014·全国·小升初真题",
+  "merged_id": "unit7_225",
+  "original_id": "unit7_choice_225"
+},
+{
+  "id": "unit7_choice_226",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "商家对某款手机销售的趋势变化情况进行分析，应该采用（）。",
+  "options": [
+    "A. 条形统计图",
+    "B. 扇形统计图",
+    "C. 折线统计图"
+  ],
+  "answer": "C",
+  "solution": "【详解】条形统计图可以直观地看出数量的多少；折线统计图不仅能看清数量的多少，还能反映数量的增减变化情况；扇形统计图能清楚地看出各部分数量与总数量之间的关系。据此，商家对某款手机销售的趋势变化情况进行分析，应该采用折线统计图。故答案为: C",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "2022·山东济宁·小升初真题",
+  "merged_id": "unit7_226",
+  "original_id": "unit7_choice_226"
+},
+{
+  "id": "unit7_choice_227",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "要统计东莞近五年降雨量的变化情况，选用（）统计图比较合适。",
+  "options": [
+    "A. 条形",
+    "B. 折线",
+    "C. 扇形",
+    "D. 不确定"
+  ],
+  "answer": "B",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。此题应根据条形统计图、折线统计图、扇形统计图各自的特点进行解答。\n【详解】由分析可得：要统计东莞近五年降雨量的变化情况，选折线统计图比较合适。故答案为: B",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "2019·全国·小升初模拟",
+  "merged_id": "unit7_227",
+  "original_id": "unit7_choice_227"
+},
+{
+  "id": "unit7_choice_228",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "折线统计图表示( ), 扇形统计图表示( ), 条形统计图表示( ).",
+  "options": [
+    "A. 数量的多少和增减变化情况",
+    "B. 数量的多少",
+    "C. 部分与总量的关系"
+  ],
+  "answer": "ACB",
+  "solution": "【详解】略",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "18-19 六年级上·全国·单元测试",
+  "merged_id": "unit7_228",
+  "original_id": "unit7_choice_228"
+},
+{
+  "id": "unit7_choice_229",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "要想很清楚的反映出各部分和整体之间的关系，绘制（）比较合适。",
+  "options": [
+    "A. 扇形统计图",
+    "B. 折线统计图",
+    "C. 条形统计图"
+  ],
+  "answer": "A",
+  "solution": "【分析】扇形统计图的特征：能很清楚地表示出各部分数量同总数的关系，据此解答。熟练掌握扇形统计图的特点是解决本题的关键。\n【详解】要想很清楚的反映出各部分和整体之间的关系,绘制扇形统计图比较合适。故答案为: A",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "2018 六年级上·全国·专题练习",
+  "merged_id": "unit7_229",
+  "original_id": "unit7_choice_229"
+},
+{
+  "id": "unit7_choice_230",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "要反映一个地区 12 个月平均气温的升降情况，选择（）。",
+  "options": [
+    "A. 条形统计图",
+    "B. 折线统计图",
+    "C. 扇形统计图"
+  ],
+  "answer": "B",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。此题应根据条形统计图、折线统计图、扇形统计图各自的特点进行解答\n【详解】由分析可得：反映一个地区 12 个月平均气温的升降情况，选择折线统计图。故答案为: B",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "2018 六年级上·全国·专题练习",
+  "merged_id": "unit7_230",
+  "original_id": "unit7_choice_230"
+},
+{
+  "id": "unit7_choice_231",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "要表示各班人数占全校总人数的百分比, 你应选择 ( ) 统计图。",
+  "options": [
+    "A. 条形",
+    "B. 折线",
+    "C. 扇形",
+    "D. 无法确定"
+  ],
+  "answer": "C",
+  "solution": "【分析】要根据统计图的特点来选择：条形统计图是用一个单位长度表示一定的数量，从条形统计图中很容易看出各种数量的多少；扇形统计图是用整个圆表示总数用圆内各个扇形的大小表示各部分数量占总数的百分数，可以很清楚的表示出各部分数量同总数之间的关系。折线统计图不但可以表示出数量的多少，而且还能够清楚的表示出数量增减变化的情况。本题的关键根据题目要求正确的选择合适的统计图。\n【详解】根据统计图的特点，要表示各班人数占全校总人数的百分比应选择扇形统计图。故选 C。",
+  "knowledge_points": [
+    "统计图的选择",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "17-18 六年级上·全国·期末",
+  "merged_id": "unit7_231",
+  "original_id": "unit7_choice_231"
+},
+{
+  "id": "unit7_choice_232",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "下面的信息资料中,不适合用统计图表示的是( ).",
+  "options": [
+    "A. 某县小学各年级的人数",
+    "B. 六年级各班做好事的件数",
+    "C. 病人体温变化情况",
+    "D. 全国小学教师的总人数"
+  ],
+  "answer": "D",
+  "solution": "【详解】根据统计图的特点选择.",
+  "knowledge_points": [
+    "统计图"
+  ],
+  "source": "18-19 六年级上·全国·课后作业",
+  "merged_id": "unit7_232",
+  "original_id": "unit7_choice_232"
+},
+{
+  "id": "unit7_choice_233",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "为了表示某地区一年内平均每月降雨量的变化情况, 可制成( )统计图.",
+  "options": [
+    "A. 条形",
+    "B. 折线",
+    "C. 扇形",
+    "D. 以上选择都错误"
+  ],
+  "answer": "B",
+  "solution": "【详解】折线统计图能清晰表示数量的增减变化趋势.",
+  "knowledge_points": [
+    "统计图"
+  ],
+  "source": "18-19 六年级上·全国·课后作业",
+  "merged_id": "unit7_233",
+  "original_id": "unit7_choice_233"
+}, {
+  "id": "unit7_choice_234",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "统计股票某日的走势情况，应绘制（ ）统计图。",
+  "options": [
+    "A. 条形",
+    "B. 折线",
+    "C. 扇形",
+    "D. 三种都可以"
+  ],
+  "answer": "B",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；据此解答。\n【详解】根据分析得，对比两只股票某日的走势情况，应绘制折线统计图。故答案为: B",
+  "knowledge_points": [
+    "统计图的选择（折线统计图）"
+  ],
+  "source": "18-19六年级上·全国·单元测试",
+  "merged_id": "unit7_234",
+  "original_id": "unit7_choice_234"
+},
+{
+  "id": "unit7_choice_235",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "要表示出陈老师家今年六月份各项生活支出占月总收入的百分比情况，用（ ）统计图比较合适。",
+  "options": [
+    "A. 扇形",
+    "B. 折线",
+    "C. 条形"
+  ],
+  "answer": "A",
+  "solution": "【详解】表示陈老师家六月份各项生活支出占月总收入的百分比情况，用扇形统计图比较合适。故答案为: A",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "16-17六年级上·广东广州·期末",
+  "merged_id": "unit7_235",
+  "original_id": "unit7_choice_235"
+},
+{
+  "id": "unit7_choice_236",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "（）统计图可以很容易地看出数量的多少。",
+  "options": [
+    "A. 条形",
+    "B. 折线",
+    "C. 扇形"
+  ],
+  "answer": "A",
+  "solution": "【分析】条形统计图能清楚地表示出数量的多少，折线统计图能表示数量的多少，还能表示数量的增减变化情况，扇形统计图能表示部分与整体之间的关系。\n【详解】条形统计图是用长短不同的长条来表示数量的多少，很容易看出数量的多少。故答案为: A\n【点睛】本题的关键是根据条形统计图，折线统计图以及扇形统计图的特点进行选择。",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "2018 六年级下·云南·专题练习",
+  "merged_id": "unit7_236",
+  "original_id": "unit7_choice_236"
+},
+{
+  "id": "unit7_fill_237",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "条形统计图和折线统计图都可用来表示( )的多少.",
+  "options": [],
+  "answer": "数量",
+  "solution": "【分析】条形统计图能清楚地看出数量的多少；折线统计图不仅能看出数量的多少，还能清楚地反映数量的增减变化情况，由此即可解答。\n【详解】条形统计图和折线统计图都可以用来表示数量的多少，折线统计图还可以反映数量的增减变化情况。故答案为：数量\n【点睛】此题主要考查学生对统计图的特点的掌握情况,解答此类习题要熟练掌握条形统计图和折线统计图各自的特点。",
+  "knowledge_points": [
+    "单式折线统计图",
+    "统计图的选择",
+    "复式折线统计图"
+  ],
+  "source": "19-20 六年级上·全国·单元测试",
+  "merged_id": "unit7_237",
+  "original_id": "unit7_fill_237"
+},
+{
+  "id": "unit7_fill_238",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "在制作统计表时，为了说明数量之间的关系，常常用到( )数。",
+  "options": [],
+  "answer": "百分",
+  "solution": "【分析】略",
+  "knowledge_points": [
+    "统计图表的综合应用",
+    "扇形统计图的特点及绘制",
+    "复式统计表的特点及填补"
+  ],
+  "source": "19-20六年级上·全国·单元测试",
+  "merged_id": "unit7_238",
+  "original_id": "unit7_fill_238"
+},
+{
+  "id": "unit7_fill_239",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "一本书 300 页, 红红第一天看了全书的 $20 \\%$, 第二天看了全书的 $30\\%$\n(1) 第一天看了  \n(2) 第二天看  \n(3) 还剩下____没有看?",
+  "options": [],
+  "answer": "60页 90页 150页",
+  "solution": "【分析】略",
+  "knowledge_points": [
+    "百分数简单应用题"
+  ],
+  "source": "19-20 六年级上·辽宁·期中",
+  "merged_id": "unit7_239",
+  "original_id": "unit7_fill_239"
+},
+{
+  "id": "unit7_fill_240",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "扇形统计图的特点是 , 折线统计图的特点是 .\nA: 能清楚地表示出各个项目的具体数目  \nB: 能清楚地反映事物的变化情况  \nC: 能清楚地表示出各部分在总体中所占的百分比  \nD: 以上都可以",
+  "options": [],
+  "answer": "C B",
+  "solution": "【分析】略",
+  "knowledge_points": [
+    "统计图的选择",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "19-20 六年级上·全国·单元测试",
+  "merged_id": "unit7_240",
+  "original_id": "unit7_fill_240"
+},
+{
+  "id": "unit7_fill_241",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "如果要统计上海世博会每天观众的变化情况，那么该选用( )统计图进行统计。",
+  "options": [],
+  "answer": "折线",
+  "solution": "【分析】折线统计图是用一个单位长度表示一定的数量，根据数量的多少描出各点，然后把各点用线段顺次连接起来。折线统计图不但可以表示出数量的多少，而且能够清楚地表示出数量增减变化的情况。\n【详解】要统计上海世博会每天观众的变化情况，重点突出的是数据的变化情况，所以应绘制折线统计图。故答案为：折线\n【点睛】本题考查根据条形统计图、折线统计图和扇形统计图各自的优点选择合适的统计图。",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "19-20 四年级上·全国·单元测试",
+  "merged_id": "unit7_241",
+  "original_id": "unit7_fill_241"
+},
+{
+  "id": "unit7_fill_242",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "六年级四班期中考试成绩情况如下：'优秀'（80分及以上）占 $70\\%$，'良好'(60~79 分) 占 $20 \\%$, '不及格'占 $10 \\%$。在绘制扇形统计图时, 表示 '优秀' 的扇形的圆心角是( )°, 表示 '良好' 的扇形的圆心角是( )°, 表示 '不及格' 的扇形的圆心角是( )°。",
+  "options": [],
+  "answer": "252  72 36",
+  "solution": "【分析】略",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "19-20六年级上·全国·单元测试",
+  "merged_id": "unit7_242",
+  "original_id": "unit7_fill_242"
+},
+{
+  "id": "unit7_fill_243",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "气象员记录一天的气温变化用( )统计图；某村表示各种农作物种植面积占总面积的百分比用( )统计图；学校表示各年级的学生人数用( )统计图.",
+  "options": [],
+  "answer": "折线 扇形 条形",
+  "solution": "【分析】略",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "19-20六年级上·全国·单元测试",
+  "merged_id": "unit7_243",
+  "original_id": "unit7_fill_243"
+},
+{
+  "id": "unit7_fill_244",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "小明的爸爸把一个月的工资做了一次消费计划,要想清楚地看到各部分消费与工资的关系,要选用( )统计图.",
+  "options": [],
+  "answer": "扇形",
+  "solution": "【分析】略",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "19-20六年级上·全国·单元测试",
+  "merged_id": "unit7_244",
+  "original_id": "unit7_fill_244"
+},
+{
+  "id": "unit7_fill_245",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "乐乐今年初中毕业考高中,妈妈想知道他各科成绩占总分的百分比,应选用( )统计图;杨校长想知道2014年上半年学校的支出情况,王会计应给杨校长绘制一张( )统计图;明明发烧住院,李医生想绘制一张统计图来了解他的体温变化情况,李医生应选用( )统计图.",
+  "options": [],
+  "answer": "扇形 条形 折线",
+  "solution": "【分析】略",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "19-20 六年级上·全国·单元测试",
+  "merged_id": "unit7_245",
+  "original_id": "unit7_fill_245"
+},
+{
+  "id": "unit7_fill_246",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "为了反映前进小学2008—2012年在校人数变化情况，最好用( )。为了反映某蔬菜基地 2012 年 7 月份蔬菜产量情况, 最好用( )。为了反映空气中的主要成分所占的体积与总体积的关系，最好用( )。\nA. 条形统计图\nB. 折线统计图\nC. 扇形统计图",
+  "options": [],
+  "answer": "B A C",
+  "solution": "【分析】略",
+  "knowledge_points": [
+    "统计图的选择",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "19-20 六年级上·全国·单元测试",
+  "merged_id": "unit7_246",
+  "original_id": "unit7_fill_246"
+},
+{
+  "id": "unit7_fill_247",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "对明星中学初二(5) 班的学生进行调查，发现有 21 人喜欢打篮球，有 13人喜欢打乒乓球, 有 15 人喜欢踢足球, 为了清楚地表示爱好各种球类活动的人数占全班人数的百分比，应该制作( )统计图.",
+  "options": [],
+  "answer": "扇形",
+  "solution": "【分析】略",
+  "knowledge_points": [
+    "统计图的选择",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "19-20 六年级上·全国·单元测试",
+  "merged_id": "unit7_247",
+  "original_id": "unit7_fill_247"
+},
+{
+  "id": "unit7_fill_248",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "既要表示数量的多少，又要清楚地表示数量增减变化的情况，应选用( )统计图.",
+  "options": [],
+  "answer": "折线",
+  "solution": "【分析】略",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "19-20六年级上·全国·单元测试",
+  "merged_id": "unit7_248",
+  "original_id": "unit7_fill_248"
+},
+{
+  "id": "unit7_fill_249",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "条形统计图可以清楚地表示出( )，如果要更清楚地了解各部分数量同总量之间的关系，可以用( )统计图来表示。扇形统计图用( )的面积表示总数，用扇形的面积表示( )。",
+  "options": [],
+  "answer": "数量的多少 扇形 整圆 各部分量占总量的百分比",
+  "solution": "【分析】略",
+  "knowledge_points": [
+    "统计图的选择",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "19-20六年级上·全国·单元测试",
+  "merged_id": "unit7_249",
+  "original_id": "unit7_fill_249"
+},
+{
+  "id": "unit7_fill_250",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "如果要更清楚地了解各部分数量同总数之间的关系，可以用( )统计图表示.( )统计图可以清楚地呈现出各种数量的多少.",
+  "options": [],
+  "answer": "扇形 条形",
+  "solution": "【分析】略",
+  "knowledge_points": [
+    "1 格表示一个单位的单式条形统计图",
+    "复式条形统计图",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "19-20 六年级上·全国·单元测试",
+  "merged_id": "unit7_250",
+  "original_id": "unit7_fill_250"
+},
+{
+  "id": "unit7_fill_251",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "只要求表示数量的多少，最好绘制成____统计图。要想清楚地表示出各部分与总数的关系选用_____统计图较好。",
+  "options": [],
+  "answer": "条形 扇形",
+  "solution": "【详解】条形统计图是用一个单位长度表示一定的数量，根据数量的多少画成长短不同的直条，然后把这些直条按一定的顺序排列起来。从条形统计图中很容易看出各种数量的多少。扇形统计图是用整个圆表示总数（单位'1'），用圆内过圆心O点的各个扇形的大小表示各部分量占总量的百分之几，通过扇形统计图可以很清楚地表示出各部分数量与总数之间的关系。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "19-20六年级上·全国·单元测试",
+  "merged_id": "unit7_251",
+  "original_id": "unit7_fill_251"
+},
+{
+  "id": "unit7_fill_252",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "人们经常要有目的地调查和分析一些问题,这就要搜集资料,整理数据绘制( )、( ).",
+  "options": [],
+  "answer": "统计表 统计图",
+  "solution": "【详解】略",
+  "knowledge_points": [
+    "统计图表的综合应用"
+  ],
+  "source": "19-20六年级上·全国·单元测试",
+  "merged_id": "unit7_252",
+  "original_id": "unit7_fill_252"
+},
+{
+  "id": "unit7_fill_253",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "统计九江镇各学校学生人数,应绘制( )统计图,统计各个学校人数占全镇学生的百分之几,应绘制( )统计图.",
+  "options": [],
+  "answer": "条形 扇形",
+  "solution": "【详解】略",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "19-20 六年级上·全国·单元测试",
+  "merged_id": "unit7_253",
+  "original_id": "unit7_fill_253"
+},
+{
+  "id": "unit7_fill_254",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "要反映某件衣服中各种纤维成分的含量，最好选用( )统计图。",
+  "options": [],
+  "answer": "扇形",
+  "solution": "【分析】条形统计图可以体现具体数据大小；折线统计图可以体现数据大小以及数据的增减变化；扇形统计图可以体现数据整体与部分之间的关系。此题考查统计图的选择，熟练掌握各种统计图表示数据的特点是解题的关键。\n【详解】反映某件衣服中各种纤维成分的含量，是各种纤维的含量与所有成分之间的关系，是整体与部分之间的关系，选择扇形统计图比较合适。",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "2016·全国·小升初真题",
+  "merged_id": "unit7_254",
+  "original_id": "unit7_fill_254"
+},
+{
+  "id": "unit7_fill_255",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "要反映张华家上个月各项支出占他家总支出的关系, 可选用_____统计图。如果要反映他家总收入的增减变化情况, 可选用_____统计图。",
+  "options": [],
+  "answer": "扇形 折线",
+  "solution": "【详解】扇形统计图是用整个圆表示总数，也就是单位'1'，用圆内过圆心O点的各个扇形的大小表示各部分量占总量的百分之几，扇形统计图中各部分的百分比之和是单位 '1'; 折线统计图是以折线的上升或下降来表示统计数量的增减变化的统计图; 折线统计图用折线的起伏表示数据的增减变化情况。",
+  "knowledge_points": [
+    "1 格表示一个单位的单式条形统计图",
+    "单式折线统计图",
+    "统计图的选择",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "18-19 六年级下·江苏淮安·期中",
+  "merged_id": "unit7_255",
+  "original_id": "unit7_fill_255"
+},
+{
+  "id": "unit7_fill_256",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "( )统计图可以清楚地表示各部分数量同总数量之间的关系。",
+  "options": [],
+  "answer": "扇形",
+  "solution": "【分析】扇形统计图是以一个圆的面积（看作单位'1'）表示物体的总数量，以相应的扇形面积占整个圆面积的百分数表示各有关部分占总数量的百分数的统计图。关键是熟悉扇形统计图的特点。\n【详解】由分析可得：扇形统计图可以清楚地表示各部分数量同总数量之间的关系。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "17-18 六年级上·广东广州·期末",
+  "merged_id": "unit7_256",
+  "original_id": "unit7_fill_256"
+},
+{
+  "id": "unit7_fill_257",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "医生每四个小时给病人测温度变化选用( )统计图，统计住院各科人数时，选用( )",
+  "options": [],
+  "answer": "折线 条形统计图",
+  "solution": "【详解】略",
+  "knowledge_points": [
+    "单式折线统计图",
+    "统计图的选择",
+    "1格表示多个单位的单式条形统计图",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "18-19 六年级下·全国·期末",
+  "merged_id": "unit7_257",
+  "original_id": "unit7_fill_257"
+},
+{
+  "id": "unit7_fill_258",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "实验小组的同学要对实验结果进行统计.(1) 观察实验组和对比组幼苗每星期高度的数据, 选择( )统计图较合适.  (2) 要统计甲、乙两种型号玉米的单棵产量，选择( )统计图较合适.\n        A. 单式条形 B. 复式条形 C. 复式折线",
+  "options": [],
+  "answer": "C A",
+  "solution": "【详解】略",
+  "knowledge_points": [
+    "统计图的选择",
+    "复式折线统计图",
+    "复式条形统计图",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "18-19五年级下·全国·单元测试",
+  "merged_id": "unit7_258",
+  "original_id": "unit7_fill_258"
+},
+{
+  "id": "unit7_fill_259",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "如果要反映2014年湖滨新城每个月的旅游收入的增减变化情况，可以用_______统计图表示．要反映每个月收入的多少，用______统计图．要反映每个月收入占全年收入的百分比用__统计图.",
+  "options": [],
+  "answer": "折线 条形 扇形",
+  "solution": "【详解】略",
+  "knowledge_points": [
+    "1 格表示一个单位的单式条形统计图",
+    "单式折线统计图",
+    "统计图的选择",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "16-17六年级下·江苏·阶段练习",
+  "merged_id": "unit7_259",
+  "original_id": "unit7_fill_259"
+},
+{
+  "id": "unit7_fill_260",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "在统计图中，____统计图能反映各部分占总数的百分比，____统计图能反映数据的变化趋势。",
+  "options": [],
+  "answer": "扇形 折线",
+  "solution": "【详解】根据扇形统计图和折线统计图的特点选择作答。在统计图中，扇形统计图能反映各部分占总数的百分比，折线统计图能反映数据的变化趋势。",
+  "knowledge_points": [
+    "统计图的选择",
+    "复式折线统计图",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "18-19六年级上·全国·单元测试",
+  "merged_id": "unit7_260",
+  "original_id": "unit7_fill_260"
+},
+{
+  "id": "unit7_fill_261",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "扇形统计图是利用圆和____表示____和部分的关系，圆代表的是总体，即 $100\\%$，扇形代表____.",
+  "options": [],
+  "answer": "扇形 总量 部分量",
+  "solution": "【详解】扇形统计图是用整个圆表示总数，用圆内各个扇形的大小表示各部分数量占总数的百分数，通过扇形统计图可以很清楚的表示出各部分数量同总数之间的关系，用整个圆的面积表示总数（单位1），用圆的扇形面积表示各部分占总数的百分数。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "18-19六年级下·四川·课后作业",
+  "merged_id": "unit7_261",
+  "original_id": "unit7_fill_261"
+},
+{
+  "id": "unit7_fill_262",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "调查学校五年级学生最喜欢的课外活动，要表示出喜欢每种课外活动的人数，应绘制____统计图。",
+  "options": [],
+  "answer": "条形",
+  "solution": "【详解】根据三种统计图的作用可知，条形统计图中很容易看出各种数量的多少；折线统计图不但可以表示出数量的多少，而且能够清楚地表示出数量增减变化的情况；扇形统计图可以很清楚地表示各部分数量同总数之间的关系，据此选择即可。故答案为条形.",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "2019 六年级下·全国·专题练习",
+  "merged_id": "unit7_262",
+  "original_id": "unit7_fill_262"
+},
+{
+  "id": "unit7_fill_263",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "( )统计图很清楚的,看出整体与部分的关系。",
+  "options": [],
+  "answer": "扇形",
+  "solution": "【详解】根据统计图的特点可知：要清楚地反映部分与整体的关系，用扇形统计图最适合；故答案为扇形。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "16-17 六年级上·全国·单元测试",
+  "merged_id": "unit7_263",
+  "original_id": "unit7_fill_263"
+},
+{
+  "id": "unit7_fill_264",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "扇形统计图是用整个圆表示( ), 用圆内各个扇形的大小表示( ).",
+  "options": [],
+  "answer": "总数 各部分数量占总数的百分数",
+  "solution": "【详解】扇形统计图是用整个圆表示总数，用圆内各个扇形的大小表示各部分数量占总数的百分数。由此可知答案。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "15-16 六年级上·全国·课后作业",
+  "merged_id": "unit7_264",
+  "original_id": "unit7_fill_264"
+},
+{
+  "id": "unit7_fill_265",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "我们学过的统计图有____.",
+  "options": [],
+  "answer": "条形统计图，折线统计图，扇形统计图.",
+  "solution": "【详解】略",
+  "knowledge_points": [
+    "1 格表示一个单位的单式条形统计图",
+    "单式折线统计图",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "12-13 六年级·全国·期末",
+  "merged_id": "unit7_265",
+  "original_id": "unit7_fill_265"
+},
+{
+  "id": "unit7_fill_266",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "要清楚地表示六年级各班人数的情况，可选用( )统计图；要表示某水站水位的变化情况应选用( )统计图.",
+  "options": [],
+  "answer": "条形 折线",
+  "solution": "【详解】略",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "18-19六年级上·全国·单元测试",
+  "merged_id": "unit7_266",
+  "original_id": "unit7_fill_266"
+},
+{
+  "id": "unit7_fill_267",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "我们学过的统计图有( )、( )和( )，其中( )统计图可以清楚地反映出各部分数量同总数之间的关系。",
+  "options": [],
+  "answer": "条形统计图 折线统计图 扇形统计图 扇形",
+  "solution": "【详解】略",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "18-19 六年级上·全国·单元测试",
+  "merged_id": "unit7_267",
+  "original_id": "unit7_fill_267"
+},
+{
+  "id": "unit7_fill_268",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "要反映某地一天气温的变化, 应选择( )统计图。",
+  "options": [],
+  "answer": "折线",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。\n【详解】根据折线统计图的特点可知: 要反映某地一天气温变化情况, 最好绘制折线统计图。",
+  "knowledge_points": [
+    "单式折线统计图",
+    "统计图的选择",
+    "复式条形统计图",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "17-18 六年级上·全国·期末",
+  "merged_id": "unit7_268",
+  "original_id": "unit7_fill_268"
+},
+{
+  "id": "unit7_fill_269",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "我们常用的统计图有()统计图、()统计图和()统计图,其中()统计图既能表示各种数量的多少,又能清楚地表示数量增减变化情况;( )统计图能清楚地表示各部分数量与总数之间的关系.",
+  "options": [],
+  "answer": "条形 扇形 折线 折线 扇形",
+  "solution": "【详解】略",
+  "knowledge_points": [
+    "统计图"
+  ],
+  "source": "18-19六年级上·辽宁·单元测试",
+  "merged_id": "unit7_269",
+  "original_id": "unit7_fill_269"
+},
+{
+  "id": "unit7_fill_270",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "要统计欣欣从6岁到12岁的身高变化情况，应选用( )统计图；要统计幸福村各种农作物种植面积占耕地总面积的百分比，应选用( )统计图；要统计某市各小学2017年秋在校学生人数，应选用( )统计图.",
+  "options": [],
+  "answer": "折线 扇形 条形",
+  "solution": "【详解】略",
+  "knowledge_points": [
+    "统计图"
+  ],
+  "source": "18-19六年级上·全国·单元测试",
+  "merged_id": "unit7_270",
+  "original_id": "unit7_fill_270"
+},
+{
+  "id": "unit7_fill_271",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "扇形统计图用一个圆表示( )，用圆内各个扇形的大小表示( )占( )的百分之几，扇形统计图可以表示出( )与( )之间的关系。",
+  "options": [],
+  "answer": "总数量 各部分 总数 部分 总数",
+  "solution": "【详解】略",
+  "knowledge_points": [
+    "统计图"
+  ],
+  "source": "18-19 六年级上·全国·单元测试",
+  "merged_id": "unit7_271",
+  "original_id": "unit7_fill_271"
+},
+{
+  "id": "unit7_fill_272",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "校园里有杨树、柳树和槐树三种树，其中杨树有60棵，占总数的 $30\\%$，柳树有 100 棵，占总数的( )%，如果画扇形统计图，槐树部分占圆面积的( )%。",
+  "options": [],
+  "answer": "50 20",
+  "solution": "【分析】根据百分法的意义，求出三种树的总棵树：$600\\div30\\%=200（棵）$，由总棵树和柳树的棵树可以求出柳树占总数的百分之几，$100\\div 200 \\times100\\%=50\\%$槐树部分在扇形统计图上占得百分比就是 $1-30\\%-50\\% $。已知一个数的百分之几是多少，求这个数用除法计算；已知一个数，求这个数的百分之几是多少用乘法计算。\n$60 \\div 30\\% = 200$（棵），由总棵树和柳树的棵树可以求出柳树占总数的百分之几，$100 \\div 200 \\times 100\\% = 50\\%$，槐树部分在扇形统计图上占得百分比就是 $1 - 30\\% - 50\\%$。\n【详解】三种树的总棵树：$60 \\div 30\\% = 200$（棵）柳树占总数的：\n$1 0 0 \\div 2 0 0 \\times 1 0 0 \\\\ = 0.5 \\times 100 \\% \\\\ = 50 \\% $\n槐树部分占圆面积的:$1 - 30\\% -50\\%$$70\\% -50\\% = 20 \\% $",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "已知一个数的百分之几是多少，求这个数"
+  ],
+  "source": "18-19六年级上·全国·单元测试",
+  "merged_id": "unit7_272",
+  "original_id": "unit7_fill_272"
+},
+{
+  "id": "unit7_fill_273",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "为了表示一年中每个月的平均气温变化情况，应绘制成( )统计图；为了表示某次测验优、良、及格、不及格的人数与班级总人数的关系，应绘制成( )统计图。",
+  "options": [],
+  "answer": "折线 扇形",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。本题考查了统计图的选择，掌握折线统计图、条形统计图以及扇形统计图的特征是解题的关键。\n【详解】为了表示一年中每个月的平均气温变化情况，应绘制成折线统计图；为了表示某次测验优、良、及格、不及格的人数与班级总人数的关系，应绘制成扇形统计图。",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "18-19六年级上·全国·单元测试",
+  "merged_id": "unit7_273",
+  "original_id": "unit7_fill_273"
+},
+{
+  "id": "unit7_fill_274",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "要清楚地表示各部分数量与总数之间的关系，选择( )统计图最合适。",
+  "options": [],
+  "answer": "扇形",
+  "solution": "【分析】根据扇形统计图的特征解答即可。\n【详解】要清楚地表示各部分数量与总数之间的关系，选择扇形统计图最合适。\n【点睛】本题考查扇形统计图，解答本题的关键掌握扇形统计图的特征。",
+  "knowledge_points": [
+    "统计图的选择",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "2022·广东梅州·小升初真题",
+  "merged_id": "unit7_274",
+  "original_id": "unit7_fill_274"
+},
+{
+  "id": "unit7_fill_275",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "在一所民族小学里,为了表示出各民族学生人数占总人数的百分比情况，可以选用统计图。",
+  "options": [],
+  "answer": "扇形",
+  "solution": "【分析】条形统计图能表示数量的多少；折线统计图不仅能表示数量的多少，还能表示数量的增减变化情况；扇形统计图表示部分与整体之间的关系。本题考查的是扇形统计图的特征，关键审清题意看需要做出什么样的效果。\n【详解】为了表示出各民族学生占的比例，因为要表示部分分与整体之间的关系, 所以可以选用扇形统计图。故答案为：扇形。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "2018 六年级下·云南·专题练习",
+  "merged_id": "unit7_275",
+  "original_id": "unit7_fill_275"
+},
+{
+  "id": "unit7_fill_276",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "扇形统计图的优点是可以很清楚地表示出( )。",
+  "options": [],
+  "answer": "部分与总数之间的关系",
+  "solution": "【分析】扇形统计图是一种用整个圆表示总数，通过圆内各个扇形的大小来表示各部分数量占总数的百分比的图形，其优点是可以很清楚地表示出部分与总数之间的关系。\n【详解】扇形统计图的优点是可以很清楚地表示出部分与总数之间的关系。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "17-18 六年级下·全国·课后作业",
+  "merged_id": "unit7_276",
+  "original_id": "unit7_fill_276"
+},
+{
+  "id": "unit7_judge_277",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "绘制统计图时, 为了很容易看出各种数量的多少, 应选用条形统计图. ( )",
+  "options": [],
+  "answer": "√",
+  "solution": "【分析】略",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "19-20 六年级上·全国·单元测试",
+  "merged_id": "unit7_277",
+  "original_id": "unit7_judge_277"
+},
+{
+  "id": "unit7_judge_278",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "容易",
+  "question": "条形、扇形和折线统计图都能表示统计量的数量多少. ( )",
+  "options": [],
+  "answer": "×",
+  "solution": "【分析】略",
+  "knowledge_points": [
+    "统计图的选择",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "19-20 六年级上·全国·单元测试",
+  "merged_id": "unit7_278",
+  "original_id": "unit7_judge_278"
+},
+{
+  "id": "unit7_judge_279",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "容易",
+  "question": "在一个扇形统计图中，所有扇形的百分比之和一定是 $100\\%$。（）",
+  "options": [],
+  "answer": "√",
+  "solution": "【分析】根据扇形统计图的意义，把一个圆的面积看作单位'1'，各部分之和等于1，也就是利用收集的数据，画成扇形统计图，则所有的扇形的百分比之和为 $100\\%$。此题主要是考查扇形统计图的意义，扇形统计图中各部分之和等于圆面积，也就是所有扇形的百分率为 $100\\%$。\n【详解】利用收集的数据，画成扇形统计图，则所有的扇形的百分比之和为 $100\\%$，原题说法正确。故答案为: √",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "19-20六年级上·全国·单元测试",
+  "merged_id": "unit7_279",
+  "original_id": "unit7_judge_279"
+},
+{
+  "id": "unit7_judge_280",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "容易",
+  "question": "要表示地球上山脉、河流、陆地的面积各是多少，应选用扇形统计图。（）",
+  "options": [],
+  "answer": "×",
+  "solution": "【分析】略",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "19-20六年级上·全国·单元测试",
+  "merged_id": "unit7_280",
+  "original_id": "unit7_judge_280"
+},
+{
+  "id": "unit7_judge_281",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "容易",
+  "question": "在扇形统计图中，萝卜用地占 $40\\%$，土豆用地占 $70\\%$。（）",
+  "options": [],
+  "answer": "×",
+  "solution": "【分析】在扇形统计图，每种种植的东西的占地面积总和不能超过 $100 \\%$, 据此判断。\n【详解】$40\\% + 70\\% = 110\\% > 100\\%$，原题说法错误。故答案为：×",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "19-20六年级上·全国·单元测试",
+  "merged_id": "unit7_281",
+  "original_id": "unit7_judge_281"
+},
+{
+  "id": "unit7_judge_282",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "容易",
+  "question": "某同学为了统计自己家一天消费的各类食物所占百分比情况，他应选用扇形统计图比较合适。（）",
+  "options": [],
+  "answer": "√",
+  "solution": "【详解】扇形统计图的优点：它可以清楚地表示出部分数量与总数、部分数量与部分数量之间的关系。重点考查扇形统计图的优点。\n【解答】解：根据扇形统计图的优点，某同学为了统计自己家一天消费的各类食物所占百分比情况，他应选用扇形统计图比较合适。故答案为：√",
+  "knowledge_points": [
+    "统计图的选择",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "19-20六年级上·全国·单元测试",
+  "merged_id": "unit7_282",
+  "original_id": "unit7_judge_282"
+},
+{
+  "id": "unit7_judge_283",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "要表示出果园里各种果树的棵数与总棵数之间的关系，应绘制扇形统计图。（）",
+  "options": [],
+  "answer": "√",
+  "solution": "【详解】扇形统计图可以反映各种果树的棵数占总数的百分比。故答案为:  √",
+  "knowledge_points": [
+    "统计图的选择",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "19-20六年级上·全国·单元测试",
+  "merged_id": "unit7_283",
+  "original_id": "unit7_judge_283"
+},
+{
+  "id": "unit7_judge_284",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "在所有统计图中扇形统计图最好，最适用。 （）",
+  "options": [],
+  "answer": "×",
+  "solution": "【详解】每种不同的统计图都有自己的优点，选择何种统计图要根据实际情况而定，所以原题说法错误。故答案为: ×",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "19-20 六年级上·全国·单元测试",
+  "merged_id": "unit7_284",
+  "original_id": "unit7_judge_284"
+},
+{
+  "id": "unit7_judge_285",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "容易",
+  "question": "要清楚反映全校人数与各年级人数之间的关系，应选择扇形统计图。（）",
+  "options": [],
+  "answer": "√",
+  "solution": "【分析】扇形统计图能清楚的反映出部分和总体之间的关系。本题考查统计图的选择，熟知各统计图的特点就能做出判断。\n【详解】全校人数是总体，各年级人数的部分，想反映它们之间的关系，用扇形统计图比较合适。故答案为:  √",
+  "knowledge_points": [
+    "统计图的选择",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "2017·重庆铜梁·小升初真题",
+  "merged_id": "unit7_285",
+  "original_id": "unit7_judge_285"
+},
+{
+  "id": "unit7_judge_286",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "为了清楚地表示电脑全年的销量变化趋势，用条形统计图比较合适。（）",
+  "options": [],
+  "answer": "×",
+  "solution": "【详解】略",
+  "knowledge_points": [
+    "单式折线统计图",
+    "统计图的选择",
+    "1格表示多个单位的单式条形统计图",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "18-19 五年级下·全国·单元测试",
+  "merged_id": "unit7_286",
+  "original_id": "unit7_judge_286"
+},
+{
+  "id": "unit7_judge_287",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "容易",
+  "question": "要反映全校患有龋齿的学生人数同全校人数之间的关系, 选用扇形统计图比较合适。( )",
+  "options": [],
+  "answer": "正确",
+  "solution": "【详解】扇形统计图是用整个圆面积表示总量，用每个扇形面积表示部分量，从题中条件可知，选用扇形统计图比较合适。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "2022·山西临汾·小升初真题",
+  "merged_id": "unit7_287",
+  "original_id": "unit7_judge_287"
+},
+{
+  "id": "unit7_judge_288",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "扇形统计图中的圆表示整体。( )",
+  "options": [],
+  "answer": "√",
+  "solution": "【详解】扇形统计图中的圆表示整体，扇形表示各部分占整体的百分率，由此判断即可。故答案为: √",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "17-18 六年级上·吉林·单元测试",
+  "merged_id": "unit7_288",
+  "original_id": "unit7_judge_288"
+},
+{
+  "id": "unit7_judge_289",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "容易",
+  "question": "扇形统计图、折线统计图都是用一个长度单位表示一定的数量。( )",
+  "options": [],
+  "answer": "×",
+  "solution": "【分析】根据条形、折线统计图的含义：条形统计图上的每个小格表示就是一个长度单位，每个长度单位表示的数量是相同的，由于数据不同，所以画出直条的长短也不同；折线统计图是用一个单位长度表示一定的数量，根据数量的多少描出各点，然后把各点用线段顺次连接起来；进行解答即可。\n【详解】由条形、折线统计图的含义可知: 都要用一个单位长度表示一定的数量的统计图是条形统计图和折线统计图, 所以本题说法错误。故答案为:×",
+  "knowledge_points": [
+    "单式折线统计图",
+    "复式折线统计图",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "14-15 六年级·山东青岛·期中",
+  "merged_id": "unit7_289",
+  "original_id": "unit7_judge_289"
+},
+{
+  "id": "unit7_judge_290",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "要反映部分与整体之间的关系，选用扇形统计图比较合适。（）",
+  "options": [],
+  "answer": "正确",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况解答即可。此题应根据条形统计图、折线统计图、扇形统计图各自的特点进行解答。\n【详解】由统计图的特点可知：要反映部分与整体之间的关系，选用扇形统计图比较合适；所以原题说法是正确的；故答案为正确。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "18-19六年级上·辽宁·单元测试",
+  "merged_id": "unit7_290",
+  "original_id": "unit7_judge_290"
+},
+{
+  "id": "unit7_judge_291",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "折线统计图不但可表示数量的多少，而且能够清楚地表示出数量增减变化的情况。",
+  "options": [],
+  "answer": "√",
+  "solution": "【分析】（1）条形统计图的特点：能清楚的表示出数量的多少；(2) 折线统计图的特点: 不但可以表示出数量的多少, 而且能看出各种数量的增减变化情况;  (3) 扇形统计图比较清楚地反映出部分与部分、部分与整体之间的数量关系; 据此进行解答即可。解答此题应根据条形、折线和扇形统计图的特点进行解答。\n【详解】由分析知：折线统计图不但可表示数量的多少，而且能够表示数量增减变化的情况。故答案为: √",
+  "knowledge_points": [
+    "单式折线统计图",
+    "复式折线统计图"
+  ],
+  "source": "2015·全国·小升初真题",
+  "merged_id": "unit7_291",
+  "original_id": "unit7_judge_291"
+},
+{
+  "id": "unit7_judge_292",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "表示一个星期的气温变化情况，选用扇形统计图比较合适。（）",
+  "options": [],
+  "answer": "×",
+  "solution": "【分析】折线统计图的特点：不仅能看清数量的多少，还能通过折线的上升和下降表示数量的增减变化情况；扇形统计图清楚地看出各部分数量与总数量之间，部分与部分之间的关系；据此解答。\n【详解】根据分析：表示一个星期的气温变化情况，选用折线统计图比较合适，而不是扇形统计图，原题说法错误。故答案为：×",
+  "knowledge_points": [
+    "统计图的选择（扇形统计图）",
+    "统计图的选择（折线统计图）"
+  ],
+  "source": "2019·全国·小升初模拟",
+  "merged_id": "unit7_292",
+  "original_id": "unit7_judge_292"
+},
+{
+  "id": "unit7_judge_293",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "容易",
+  "question": "为了清楚地看出核桃的营养成分，应制成折线统计图。（）",
+  "options": [],
+  "answer": "×",
+  "solution": "【详解】为了清楚地看出核桃的营养成分，应制成扇形统计图。故答案为:  ×",
+  "knowledge_points": [
+    "统计图的选择",
+    "复式折线统计图",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "16-17 六年级·全国·单元测试",
+  "merged_id": "unit7_293",
+  "original_id": "unit7_judge_293"
+},
+{
+  "id": "unit7_judge_294",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "容易",
+  "question": "扇形统计图能形象、直观地展示各部分数量与总数量间的关系，但不能明确各部分具体的数量。（）",
+  "options": [],
+  "answer": "√",
+  "solution": "【分析】扇形统计图的特点：可以直观地看出各个部分数量与总数之间的关系，据此解答此题。\n【详解】扇形统计图能形象、直观地展示各部分数量与总数量间的关系，但不能明确各部分具体的数量。故答案为: √\n【点睛】此题主要考查统计图的特征, 掌握扇形统计图的特点是解答本题的关键。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "2022·河北石家庄·小升初真题",
+  "merged_id": "unit7_294",
+  "original_id": "unit7_judge_294"
+},
+{
+  "id": "unit7_judge_295",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "容易",
+  "question": "扇形统计图中, 所有扇形的百分比之和必须小于 1 。( )",
+  "options": [],
+  "answer": "×",
+  "solution": "【分析】绘制扇形统计图时，是把圆看做一个单位'1'，所以所有的百分比之和必须等于1，据此作答。\n【详解】根据上述分析，在扇形统计图中，所有扇形的百分比之和必须等于1，所以原题说法错误。故答案为: ×",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "16-17 六年级·全国·单元测试",
+  "merged_id": "unit7_295",
+  "original_id": "unit7_judge_295"
+},
+{
+  "id": "unit7_judge_296",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "晶晶家 5 月份食品支出占生活总支出的 $30 \\%$, 在制作扇形统计图时, 表示食品支出的扇形的圆心角是 $30^{\\circ}$。( )",
+  "options": [],
+  "answer": "×",
+  "solution": "【分析】在扇形统计图中，表示部分的是占整个圆的百分之几，本题中支出的 $30\\%$ 并不是扇形的圆心角，可据此进行判断。深刻理解扇形统计图中的特点是解决本题的关键。\n【详解】根据扇形统计图的特点，食品支出占生活总支出的 $30 \\%$, 扇形的圆心角是:$360^{\\circ} \\times 30\\% = 108^{\\circ}$故答案为:×",
+  "knowledge_points": [
+    "弧、圆心角、扇形的认识",
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "16-17 六年级·全国·单元测试",
+  "merged_id": "unit7_296",
+  "original_id": "unit7_judge_296"
+},
+{
+  "id": "unit7_calc_297",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "计算题",
+  "difficulty": "中等",
+  "question": "直接写出下列各题的得数.\n$ \\frac {3}{4} + 0. 2 5 =  \\quad\\frac {5}{6} \\times 1. 2 = \\quad 3 \\div 8 \\times \\frac {1}{8} = \\quad 1 \\div \\frac {2}{5} = \\\\ \\frac {4}{9} \\times 1. 8 = \\quad \\frac {3}{2 0} \\div 0. 1 5 = \\quad \\frac {3}{8} \\times \\frac {4}{9} = 6\\quad \\frac {1}{3} \\div \\frac {1}{2} = \\\\ \\frac {3}{4} - \\frac {1}{2} =\\quad 4 \\div \\frac {1}{4} = \\quad\\frac {2 1}{4} \\times \\frac {4}{7} = \\quad \\frac {3}{8} \\times \\frac {2}{9} = $",
+  "options": [],
+  "answer": "1; 1;  $\\frac{3}{64}; \\frac{5}{2}$; 0.8; 1;  $\\frac{1}{6}; \\frac{2}{3}$  ;$ \\frac {1}{4}; 1 6; 3; \\frac {1}{1 2} $",
+  "solution": "【分析】本题属于口算综合题\n【详解】一个算式中既有分数又有小数化为同一种数更方便计算，分数相乘'对角线'位置可以约分",
+  "knowledge_points": [
+    "口算综合"
+  ],
+  "source": "19-20六年级上·全国·单元测试",
+  "merged_id": "unit7_297",
+  "original_id": "unit7_calc_297"
+},
+{
+  "id": "unit7_match_298",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "连线题",
+  "difficulty": "中等",
+  "question": "连连看\n(1) 同一时间里某医院同病房中 8 位患者的体温。 折线统计图\n(2) 玉林小学 2005—2009 年各年学生总人数。 条形统计图  \n(3) 玉林小学 1—6 年各年级人数占全校人数的百分比。扇形统计图",
+  "options": [],
+  "answer": "见详解",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。\n【详解】（1）是记录同一时间里8位患者的体温，主要是展示不同患者体温的具体数值，更适合用条形统计图来直观对比数量多少。(2) 是玉林小学 2005—2009 年各年学生总人数, 我们需要对比每年学生人数的多少, 用折线统计图能很清晰地展示出来。  (3) 是玉林小学 1—6 年级各年级人数占全校人数的百分比,正好符合扇形统计图的特点, 能直观地展示各部分占比情况。\n连线如下:\n![](images/121f4a2e8fab549d2eef475cf765a92c.png)",
+  "knowledge_points": [
+    "单式折线统计图",
+    "统计图的选择（扇形统计图）",
+    "扇形统计图的特点及绘制",
+    "统计图的选择（折线统计图）"
+  ],
+  "source": "19-20 六年级上·全国·单元测试",
+  "merged_id": "unit7_298",
+  "original_id": "unit7_match_298"
+},
+{
+  "id": "unit7_solve_299",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "解答题",
+  "difficulty": "中等",
+  "question": "一个养禽专业户去年养的鸡、鸭、鹅的只数分别是1300只、750只和450只。根据这些数据绘制扇形统计图。 某养禽专业户养鸡、鸭、鹅情况统计图",
+  "options": [],
+  "answer": "![](images/7723cf846c4c010ed4a2039ba07826f7e65045e0f19925e24cc8fa8ef85e9efa.jpg)",
+  "solution": "【分析】把这个养禽户养的鸡、鸭、鹅的总只数看作单位'1'，分别求出鸡、鸭、鹅分别点三种家禽总只数的百分比，再用 $360^{\\circ}$ 分别乘鸡、鸭、鹅所占的百分比，求出表示鸡、鸭、鹅扇形圆心角的度数, 即可画出表示这家养禽专业户去年养的鸡、鸭、鹅的只数的扇形统计图。绘制扇形统计图的关键是求出表示各部分的扇形圆心角的度数，另外还要标注出数据，写上标题，绘图时间等，所绘制图形要求美观、大方、简洁.\n【详解】三种家禽总只数：$1300 + 750 + 450 = 2500$（只）鸡所占的百分率： $1300 \\div 2500 = 0.52 = 52\\%$鸭所占的百分率： $750 \\div 2500 = 0.3 = 30\\%$鹅所占的百分率： $450 \\div 2500 = 0.18 = 18\\%$表示鸡只数的扇形圆心角： $360^{\\circ} \\times 52\\% = 187.2^{\\circ}$表示鸭只数的扇形圆心角： $360^{\\circ} \\times 30\\% = 108^{\\circ}$表示鹅只数的扇形圆心角:  $360^{\\circ} \\times 18\\% = 64.8^{\\circ}$根据以上数据绘制扇形统计图如下:![](images/fdcde23cf00e2260219f8bf454e2f44227c5c1038e41e6ca82d79f166cc2ab70.jpg)  某养禽专业户养鸡、鸭、鹅情况统计图",
+  "knowledge_points": [
+    "统计图的绘制"
+  ],
+  "source": "18-19六年级上·浙江·课后作业",
+  "merged_id": "unit7_299",
+  "original_id": "unit7_solve_299"
+},
+{
+  "id": "unit7_solve_300",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "解答题",
+  "difficulty": "中等",
+  "question": "东风小学六年级二班上学期期末的数学成绩情况为优秀 14 人, 良好 16 人, 及格8人,不及格2人.每种成绩的人数各占全班人数的百分之几?把这次的数学成绩用扇形统计图表示出来.",
+  "options": [],
+  "answer": "全班人数: $14 + 16 + 8 + 2 = 40$ (人) 优秀: $14 \\div 40 = 35 \\%$ $360^{\\circ} \\times 35 \\% = 126^{\\circ}$  良好: $16 \\div 40 = 40 \\%$ $360^{\\circ} \\times 40\\% = 144^{\\circ}$  及格: $8 \\div 40 = 20 \\%$ $360^{\\circ} \\times 20\\% = 72^{\\circ}$   不及格: $2 \\div 40 = 5 \\%$ $360^{\\circ} \\times 5 \\% = 18^{\\circ}$东风小学六年级二班上学期期末数学成绩统计图![](images/dc4a1c6bef6ea803d7f04eda52a9975b5bafd6d17784cfab945c33b3974ce2af.jpg)",
+  "solution": "【详解】略",
+  "knowledge_points": [
+    "统计图的绘制"
+  ],
+  "source": "18-19 六年级下·江苏·单元测试",
+  "merged_id": "unit7_300",
+  "original_id": "unit7_solve_300"
+},
+{
+  "id": "unit7_choice_301",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "要想很清楚的反映出各部分和整体之间的关系，绘制（）比较合适。",
+  "options": [
+    "A. 折线统计图",
+    "B. 扇形统计图",
+    "C. 条形统计图"
+  ],
+  "answer": "B",
+  "solution": "【分析】扇形统计图的特征：能很清楚地表示出各部分数量同总数的关系，据此解答。\n\n【详解】要想很清楚的反映出各部分和整体之间的关系,绘制扇形统计图比较合适。故答案为: B",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "统计图的选择"
+  ],
+  "source": "18-19六年级上·全国·单元测试",
+  "merged_id": "unit7_301",
+  "original_id": "unit7_choice_301"
+},
+{
+  "id": "unit7_choice_302",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "要表示各种蔬菜的种植面积各占总面积的百分比, 用 ( ) 统计图比较合适。",
+  "options": [
+    "A. 条形",
+    "B. 折线",
+    "C. 扇形"
+  ],
+  "answer": "C",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。\n\n【详解】根据统计图的特点可知, 要表示各种蔬菜的种植面积各占总面积的百分比, 用扇形统计图比较合适。故答案为: C",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "统计图的选择"
+  ],
+  "source": "16-17 六年级上·全国·期末",
+  "merged_id": "unit7_302",
+  "original_id": "unit7_choice_302"
+},
+{
+  "id": "unit7_choice_303",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "从（）统计图中，不但能清楚的看出数量增减变化情况，也能看出数量的多少。",
+  "options": [
+    "A. 条形",
+    "B. 折线",
+    "C. 扇形"
+  ],
+  "answer": "B",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。牢记各种统计图的特点是解答本题的关键。\n\n【详解】根据统计图的特点可知：从折线统计图中，不但能清楚的看出数量增减变化情况，也能看出数量的多少；故选B。",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "2017·云南楚雄·小升初真题",
+  "merged_id": "unit7_303",
+  "original_id": "unit7_choice_303"
+},
+{
+  "id": "unit7_choice_304",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "要反映中国 $1996 \\sim 2012$年在各届奥运会上获得的奖牌的变化情况，应选择（）。",
+  "options": [
+    "A. 条形统计图",
+    "B. 折线统计图",
+    "C. 扇形统计图",
+    "D. 复式统计表"
+  ],
+  "answer": "B",
+  "solution": "【详解】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "1格表示一个单位的单式条形统计图",
+    "单式折线统计图"
+  ],
+  "source": "2017·宁夏固原·小升初真题",
+  "merged_id": "unit7_304",
+  "original_id": "unit7_choice_304"
+},
+{
+  "id": "unit7_choice_305",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "淘气妈妈想统计今年家里各项支出的比例情况，选用（）统计图最合适。",
+  "options": [
+    "A. 条形",
+    "B. 折线",
+    "C. 扇形",
+    "D. 复式统计图"
+  ],
+  "answer": "C",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。本题主要考查统计图的特点。\n\n【详解】根据统计图的特点可知：统计今年家里各项支出的比例情况，用扇形统计图最合适。故答案为: C",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "16-17六年级上·广东深圳·期末",
+  "merged_id": "unit7_305",
+  "original_id": "unit7_choice_305"
+},
+{
+  "id": "unit7_choice_306",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "扇形统计图可以清楚地看出 ( )",
+  "options": [
+    "A. 数量的多少",
+    "B. 数量的增减变化情况",
+    "C. 各部分数量同总数之间的关系"
+  ],
+  "answer": "C",
+  "solution": "【分析】首先要清楚每一种统计图的特点：条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；据此选择即可。此题应根据条形统计图、折线统计图、扇形统计图各自的特点进行解答。\n\n【详解】根据扇形统计图的特点可知, 扇形统计图可以清楚地看出各部分数量同总数之间的关系。故正确答案为: C。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "统计图的选择"
+  ],
+  "source": "16-17 六年级·全国·期末",
+  "merged_id": "unit7_306",
+  "original_id": "unit7_choice_306"
+},
+{
+  "id": "unit7_choice_307",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "要统计某地去年月平均气温情况，最好选用（）。",
+  "options": [
+    "A. 折线统计图",
+    "B. 扇形统计图",
+    "C. 条形统计图"
+  ],
+  "answer": "A",
+  "solution": "【分析】选取的统计图要能够反映每个月气温的变化以及各组之间的差别，所以最好选用折线统计图。折线统计图能够清楚地反映出数据的变化情况；条形统计图能够清楚地反映出各部分数量的多少；扇形统计图能够清楚地反映出各部分与整体的关系。\n\n【详解】要统计某地去年月平均气温情况，最好选用折线统计图。故答案为: A",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "2014 六年级·全国·课后作业",
+  "merged_id": "unit7_307",
+  "original_id": "unit7_choice_307"
+},
+{
+  "id": "unit7_choice_308",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "扇形统计图甲中女生占 $56\\%$ ，扇形统计图乙中女生占 $45\\%$ ，甲乙两个统计图中所表示的女生人数（）。",
+  "options": [
+    "A. 甲比乙多",
+    "B. 甲比乙少",
+    "C. 不能确定"
+  ],
+  "answer": "C",
+  "solution": "【分析】因为这是两个统计图，且各自的总人数不确定，因此不能确定哪个统计图中的女生人数多少，根据此选择。本题考查的是扇形统计图的认识，需要明白，只有单位'1'相同的两个扇形统计图才能进行比较，单位'1'不确定，部分也不确定。\n\n【详解】两扇形统计图中的单位 '1'不一定相同, 因此, 两个表示的女生所占的百分率无法进行比较。故答案为: C",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "2024·山西长治·小升初真题",
+  "merged_id": "unit7_308",
+  "original_id": "unit7_choice_308"
+},
+{
+  "id": "unit7_choice_309",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "( ) 条形统计图能够清楚反映出各种数量的增减变化。",
+  "options": [
+    "A. 条形",
+    "B. 折线",
+    "C. 条形和折线",
+    "D. 扇形"
+  ],
+  "answer": "B",
+  "solution": "【详解】根据统计图的特点可知：折线条形统计图能够清楚反映出各种数量的增减变化。例如:条形统计图能反映数量的多少。![](images/4e01da87c11e5c7e6ced298e37b32fbebe34dc702a2fa39970661e849efdc2b7.jpg)\n\n折线统计图能反映数据的变化情况。![](images/ededb14f2e35fc89c883c9ff18e4eae8509e14ebd0a1053959e56b9b3b52d6b7.jpg)扇形统计图能表示出各部分占总数的百分比。\n\n![](images/91d9913bdc18c3c2a84a5751fdc68fb01bb4c4bdfcd3ade746d496a3be09cf16.jpg)故答案为: B",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "1格表示多个单位的单式条形统计图",
+    "单式折线统计图"
+  ],
+  "source": "15-16 五年级上·安徽滁州·单元测试",
+  "merged_id": "unit7_309",
+  "original_id": "unit7_choice_309"
+},
+{
+  "id": "unit7_choice_310",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "气象小组要绘制一幅统计图，清楚地表示出上周每天平均气温的高低和变化情况，宜选用（）统计图。",
+  "options": [
+    "A. 条形",
+    "B. 折线",
+    "C. 扇形"
+  ],
+  "answer": "B",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。此题应根据条形统计图、折线统计图、扇形统计图各自的特点进行解答。\n\n【详解】由分析可知：绘制一幅能清楚地表示出上周每天平均气温的高低和变化情况的统计图，选用折线统计图最合适；故答案为: B",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "13-14六年级·贵州贵阳·期末",
+  "merged_id": "unit7_310",
+  "original_id": "unit7_choice_310"
+},
+{
+  "id": "unit7_choice_311",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "( ) 统计图不仅可以表示数量的多少，而且可以表示数量增减变化的情况。",
+  "options": [
+    "A. 条形",
+    "B. 折线",
+    "C. 扇形"
+  ],
+  "answer": "B",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。此题应根据条形统计图、折线统计图、扇形统计图各自的特点进行解答。\n\n【详解】由折线统计图的特点可知: 折线统计图不仅容易看出数量的多少, 而且能反映数量的增减变化情况。\n\n故答案为: B",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "1格表示多个单位的单式条形统计图",
+    "单式折线统计图"
+  ],
+  "source": "2015·四川泸州·小升初真题",
+  "merged_id": "unit7_311",
+  "original_id": "unit7_choice_311"
+},
+{
+  "id": "unit7_choice_312",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "如两个扇形统计图，其中说法不正确的是（）\n\n![](images/9abf3cf49b91640ba9649f646b82a77e1da62fbbf5dad1ff0c4c9e455c5590f2.jpg)\n\n![](images/744e463a6e6874ca6479340e51ece2e1b2f05bdd64ceda83f72efd91dbcc6fb6.jpg)",
+  "options": [
+    "A. 甲班的女生占全班人数的 $\\frac{2}{5}$",
+    "B. 乙班的男生占全班人数的 $\\frac{7}{10}$",
+    "C. 乙班的男生一定比甲班的男生多"
+  ],
+  "answer": "C",
+  "solution": "【分析】根据扇形统计图表示出数据，然后结合选项进行分析解答即可。\n\n【详解】如图,![](images/8895cdc6b665ec6852673dcecf8427638cc1160d63f5320fcb5b9109d89a8b98.jpg)![](images/2570bccc52d478cabe342e0b6e490b31161ce1e939a0e720cb76fe9cb643d8a3.jpg)A.把甲班人数看作单位 '1', 其中男生人数占 $60 \\%$ , 化成分数是 $\\frac{3}{5}$ , 女生人数占 $40 \\%$ , 化成分数是 $\\frac{2}{5}$ 。因此, 甲班的女生占全班的 $\\frac{2}{5}$ 这种说法正确; B. 把乙班人数看作单位 '1', 其中男生人数占 $70 \\%$ , 化成分数是 $\\frac{7}{10}$ , 女生人数占 $30 \\%$ , 化成分数是 $\\frac{3}{10}$ 。因此, 乙班的男生占全班人数的 $\\frac{7}{10}$ 这种说法正确; C. 单从分数的大小比较来看, $\\frac{7}{10} > \\frac{3}{5}$ , 但甲、乙两个班的人数不一定相同, 也就是单位 '1' 不一定相同, 因此, 两个班的男生无法比较多少。\n\n故答案为: C",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "求一个数是另一个数的百分之几（百分率问题）"
+  ],
+  "source": "13-14六年级·全国·单元测试",
+  "merged_id": "unit7_312",
+  "original_id": "unit7_choice_312"
+},
+{
+  "id": "unit7_choice_313",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "如图是某农场种植三种蔬菜面积的扇形统计图，如果西红柿的种植面积为4.2公顷，黄瓜的种植面积是（ ）公顷。\n\n![](images/129eaeae09ee49975c31814ec4148eca625e6f856565918e1499804d5ccfcc74.jpg)",
+  "options": [
+    "A. 7.5",
+    "B. 1.05",
+    "C. 2.25"
+  ],
+  "answer": "C",
+  "solution": "【分析】用西红柿的种植面积除以它占总面积的百分率求出三种蔬菜总面积, 再根据一个数乘分数的意义即可求出黄瓜的面积。\n\n【详解】 $4.2 \\div 56\\% = 7.5$ （公顷）$7.5 \\times 30\\% = 2.25$ （公顷）答: 黄瓜的种植面积是 2.25 公顷。故答案为: C",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "15-16六年级·全国·期末",
+  "merged_id": "unit7_313",
+  "original_id": "unit7_choice_313"
+},
+{
+  "id": "unit7_choice_314",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "某公司有员工 700人参加元旦庆祝活动，参加各种活动的情况统计图如下图，规定每人只参加一项，不下围棋的共有（）人。![](images/7f9cbe84adba513791d26bafc7aa79d0e92c16718ba323fc70b836487c714684.jpg)",
+  "options": [
+    "A. 259",
+    "B. 441",
+    "C. 438",
+    "D. 700"
+  ],
+  "answer": "B",
+  "solution": "【详解】 $700 \\times (1 - 37\\%)\\\\=7 0 0 \\times 0. 6 3\\\\=441 (人)$故答案为: B",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "14-15 六年级下·全国·单元测试",
+  "merged_id": "unit7_314",
+  "original_id": "unit7_choice_314"
+},
+{
+  "id": "unit7_choice_315",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "下图是六 (3) 和六(4) 两个班级男、女生人数统计图, 下列说法正确的是 ( )。![](images/2f944269faa8e1d32613372405b9e4b929144a26633c56cc58bef232472c2865.jpg)![](images/7c8ca930f44dfacafe11b2388398ef1ccfe47c5888c1e362370ca243cf665e94.jpg)",
+  "options": [
+    "A. 六 (4) 班的男生比六 (3) 班的男生少",
+    "B. 六 (4) 班的女生比六 (3) 班的女生多",
+    "C. 六 (4) 班的学生比六 (3) 班的学生多",
+    "D. 无法比较"
+  ],
+  "answer": "D",
+  "solution": "【分析】由第一个图可知: 六 (3) 班的总人数是单位 '1'男生占 $54 \\%$ , 女生占 $46 \\%$ ;由第二个图可知: 六 (4) 班的总人数是单位 '1', 男生占 $52 \\%$ , 女生占 $48 \\%$ ; 由此求解;\n\n【详解】两个图的单位'1'不相同，一个是六（3）班的总人数，一个是六（4）班的总人数；而两个班的各有多少人未知，它们的百分之几是多少就无法求解，两个班之间的人数就无法比较。故答案为: D",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "统计图表的综合应用"
+  ],
+  "source": "2014 六年级·全国·课后作业",
+  "merged_id": "unit7_315",
+  "original_id": "unit7_choice_315"
+},
+{
+  "id": "unit7_choice_316",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "第 29 届北京奥运会上中国共获得金牌 51 枚，银牌 21 枚，铜牌 28 枚，表示中国获得奖牌情况的是（）幅图。",
+  "options": [
+    "A. A幅图 ![](images/57347b663c7372b5c395525971d8a5e63fc66c51875786a250dd310a755e137a.jpg)",
+    "B. B幅图 ![](images/2eee73e28d472eb5efde693ae39489090903a5c2d10870d90e0ea226692b7ef1.jpg)",
+    "C. C幅图 ![](images/bf863122a06a875fb13a0b3ebc908debdadc008ff044cf7ce5e27548204e1c80.jpg)",
+    "D. D幅图 ![](images/1fa25ba4c95a1355703306b72eae173da9970f78f46f80bbc6a7c8f90914da02.jpg)"
+  ],
+  "answer": "A",
+  "solution": "【分析】根据题意，可计算出中国在第29届奥运共得到了多少枚奖牌，然后再用得到的金牌数、银牌数、铜牌数分别除以得到的总的奖牌数就是得到的金牌、银牌、铜牌占总奖牌数的百分之几，然后再根据得到的百分数绘制扇形统计图即可得到答案。此题主要考查的是如何根据题干所提供的数据进行绘制扇形统计图。\n\n【详解】总的奖牌数为： $51 + 21 + 28 = 100$ （枚）金牌占总牌数的： $51 \\div 100 \\times 100\\% = 51\\%$银牌占总牌数的： $21 \\div 100 \\times 100\\% = 21\\%$铜牌占总牌数的： $28 \\div 100 \\times 100\\% = 28\\%$![](images/fed5620c5e7858f1bffb8e23197d11200a235b6268f75d9b0944f02443edda62.jpg)故答案为：A",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "求一个数是另一个数的百分之几（百分率问题）"
+  ],
+  "source": "2014 六年级·全国·课后作业",
+  "merged_id": "unit7_316",
+  "original_id": "unit7_choice_316"
+},
+{
+  "id": "unit7_choice_317",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "绿源小区种树情况如图,其中杨树有 18 棵, 那么松树有 ( ) 棵。![](images/15b66cd61d801abc4265e8c2ad2c03fcb4878932d4c25bcfa8577f8a8c2c22c9.jpg)",
+  "options": [
+    "A. 40",
+    "B. 16",
+    "C. 6"
+  ],
+  "answer": "B",
+  "solution": "【分析】根据题意，可用 18 除以 $45 \\%$ 计算出绿源小区共有树多少棵，然后再乘 $40 \\%$ 进行计算即可得到答案。解答此题的关键是确定绿源小区共有树多少棵,然后再用总数乘 $40 \\%$ 即可。\n\n【详解】 $18 \\div 45\\% \\times 40\\%\\\\= 40 \\times 40 \\% \\\\= 16 (棵)$\n\n那么松树有 16 棵。",
+  "knowledge_points": [
+    "求一个数的百分之几是多少"
+  ],
+  "source": "2014 六年级·全国·课后作业",
+  "merged_id": "unit7_317",
+  "original_id": "unit7_choice_317"
+},
+{
+  "id": "unit7_choice_318",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "某校对六年级 300 名学生的体育测试成绩进行了一次调查，统计结果如右图所示，则在不合格范围内的学生有（）名。![](images/16aea04733e2596558094746aad2857a165ac1e7e671ea92456a1651f3a0ebcf.jpg)",
+  "options": [
+    "A. 75",
+    "B. 60",
+    "C. 90"
+  ],
+  "answer": "A",
+  "solution": "【分析】把整个六年级的学生看作单位 '1', 用单位 '1'减去优秀、合格和良好占的百分数就是不合格占的百分数,然后再根据 '求一个数的百分之几是多少, 用乘法计算'求出不合格的人数即可.本题是考查如何从扇形统计图中获取信息，并对所获取的信息进行有关计算。扇形统计图的计算主要是百分数应用方面的计算。\n\n【详解】 $300 \\times (1 - 30\\% - 25\\% - 20\\%) = 300 \\times 25\\%=75 (名)$ ;答: 在不合格范围内的学生有 75 名.",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "2014 六年级·全国·课后作业",
+  "merged_id": "unit7_318",
+  "original_id": "unit7_choice_318"
+},
+{
+  "id": "unit7_choice_319",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "在一个 40 名学生的班级中选举班长，选举结果是：张强 刘莉 李浩 赵红\n\n20票 10票 4票 6票下面哪个圆圈图显示了这些结果？（ ）",
+  "options": [
+    "A. ![](images/ff08a00108eacf62ad8b99109cbb4018a0d69160e19b881ba09d2b41b4bc8db0.jpg)",
+    "B. ![](images/48aa49a8169e1b189c9e68398d5018d413a635e410a97430209c5da23a81d95a.jpg)",
+    "C. ![](images/39d7706db1aeba5a775bccd7b8ff33bb2cb4e6294228d5c2a53da476bf74470d.jpg)"
+  ],
+  "answer": "A",
+  "solution": "【分析】根据表中的数据知道, 张强获 20 票, 刘莉获 10 票,李浩获 4 票, 赵红获 6 票, 由此分别算出每人获得的票数占总人数的百分之几, 即可做出选择。\n\n【详解】 张强： $20 \\div 40 = 50\\%$ ；刘莉: $10 \\div 40 = 25\\%$ ;李浩: $4 \\div 40 = 10\\%$ ;赵红： $6 \\div 40 = 15 \\%$ ;A、完整的表示出来四人的得票情况; B、没有正确表示张强和刘莉的得票情况; C、没有正确表示才刘莉、李浩、赵红的得票情况;",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "2014 六年级·全国·课后作业",
+  "merged_id": "unit7_319",
+  "original_id": "unit7_choice_319"
+},
+{
+  "id": "unit7_choice_320",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "如图是某校六年级学生选择摄影、唱歌、书法、十字绣四个兴趣小组的扇形统计图，以下说法正确的是（）。![](images/a56ccad9b1682a176b547f66a4ab20656110de3d12dff6563c78948a0232c39c.jpg)",
+  "options": [
+    "A. 参加唱歌小组的学生比参加摄影小组的多 $15 \\%$",
+    "B. 参加摄影小组的学生占六年级学生的 $\\frac{1}{4}$",
+    "C. 参加摄影小组与书法小组的人数之和多于十字绣小组的学生人数",
+    "D. 参加唱歌小组与十字绣小组的人数之比为 $5: 8$"
+  ],
+  "answer": "D",
+  "solution": "【分析】根据统计图得出摄影占 $15 \\%$ , 书法展 $20 \\%$ , 唱歌的占 $90 \\div 360 = 25 \\%$ , 再根据选项中的问题分别进行分析做出判断。\n\n【详解】A、 $90 \\div 360 - 15\\% = 10\\%$ ,所以此题是说法是错误的;B、摄影占六年级学生是 $15 \\%$ , 即 $\\frac{3}{20}$ ,所以此题的说法是错误的;C、十字绣小组占： $1 - 15\\% - 20\\% - 90 \\div 360 = 40\\%$ ,参加摄影小组与书法小组的人数之和： $15\\% + 20\\% = 35\\%$ ,$35 \\% < 40\\%,$所以此题的说法是错误的;D、 $(90 \\div 360)$ : $(1 - 15\\% - 20\\% - 90 \\div 360)$ ,$ 25\\% : 40\\%, = 5: 8, $所以此题的说法是正确的;",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "2014 六年级·全国·课后作业",
+  "merged_id": "unit7_320",
+  "original_id": "unit7_choice_320"
+},
+{
+  "id": "unit7_choice_321",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "困难",
+  "question": "下面是两个学校男生和女生的统计图。甲校和乙校的女生人数相比，下面选项正确的是( ).![](images/1a1e8a378de8dad2e48fbc0276ced6ea790bce0dbd9de74479c255c442f7f20d.jpg)![](images/377c20e9b67afaba8056c0dca17be8b87b1a37033f142dc5866017140785aa16.jpg)",
+  "options": [
+    "A. 甲校多",
+    "B. 乙校多",
+    "C. 无法比较",
+    "D. 一样多"
+  ],
+  "answer": "C",
+  "solution": "【详解】解：当甲校学生=乙校学生时，甲校和乙校的女生人数比=50% ÷ 40%= 5:4; 当甲校学生≠乙校学生时，无法比较.故答案为 C. 因为甲、乙两校的学生数不明确, 也就是单位 '1'不统一, 分率标准不一致, 所以无法进行比较.",
+  "knowledge_points": [
+    "百分数应用题",
+    "统计表"
+  ],
+  "source": "17-18 六年级下·吉林·期末",
+  "merged_id": "unit7_321",
+  "original_id": "unit7_choice_321"
+},
+{
+  "id": "unit7_choice_322",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "如图所示，女生人数占学校人数的百分比大的学校是（）\n\n甲校![](images/2cbcee7fcc593fb7f19ba76a52bf331caf3f46a4a2c94f1852616e291de41cb3.jpg)  乙校![](images/38c85e2c14d7a4539f4b0227c73527ae82079a8cbe4f36cabd2ae481acb3120b.jpg)",
+  "options": [
+    "A. 甲校",
+    "B. 乙校",
+    "C. 一样多"
+  ],
+  "answer": "A",
+  "solution": "【详解】 $50\\% > 40 \\%$ , 所以甲校女生人数占学校人数的百分比大。故答案为: A根据百分数大小的比较方法判断两个百分率的大小, 这样就能判断出哪个学校女生人数占学校人数的百分比大。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "18-19六年级上·全国·单元测试",
+  "merged_id": "unit7_322",
+  "original_id": "unit7_choice_322"
+},
+{
+  "id": "unit7_choice_323",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "困难",
+  "question": "如下图所示，女生多的学校是（）\n\n甲校![](images/c64e4b13136d8afbc39df15e7c3a754f62705de53eae490358d6c2f02146f0eb.jpg)  乙校![](images/d0af390be3c6be9a96ae486c8351c5d312c238b5352b669dbb634ac8039cbe6e.jpg)",
+  "options": [
+    "A. 甲校",
+    "B. 乙校",
+    "C. 由学校的总人数而定"
+  ],
+  "answer": "C",
+  "solution": "【详解】不知道两个学校各自的总人数，就无法分别计算女生人数，也无法比较女生人数的多少.故答案为 C学校总人数 $\\times$ 女生占的百分率 $=$ 女生人数, 要比较女生人数的多少, 一定得知道学校总人数, 由此判断并选择即可.",
+  "knowledge_points": [
+    "统计图"
+  ],
+  "source": "18-19六年级上·全国·单元测试",
+  "merged_id": "unit7_323",
+  "original_id": "unit7_choice_323"
+},
+{
+  "id": "unit7_choice_324",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "中等",
+  "question": "李明把自己一周的支出情况用扇形统计图表示出来, 下面说法错误的是 ( )。\n\n![](images/8b6cbcd230c42e07c3f726e4d081e5b9b7cb6f8f241b182b85db006e9ed439ab.jpg)",
+  "options": [
+    "A. 从图中不可以看出具体消费数额",
+    "B. 从图中可以直接看出各项消费数额占总消费数额的百分比",
+    "C. 从图中不可以直接看出各项消费数额在一周中的具体变化情况",
+    "D. 从图中可以直接看出各项消费数额在一周中的具体变化情况"
+  ],
+  "answer": "D",
+  "solution": "【分析】扇形统计图的特点是: 可以很清楚地表示出各部分数量与总数之间的关系, 不能看出具体的数量和数量的变化情况, 据此解答。\n\n【详解】A．李明把自己一周的支出情况用扇形统计图表示出来，扇形统计图不能表示具体的数量，选项描述正确； B. 扇形统计图可以很清楚地表示出各部分数量与总数之间的关系, 即部分量占总量的百分之几, 选项描述正确; C. 折线统计图可以表示数量的变化情况, 扇形统计图不能表示数量的变化情况, 选项描述正确;D. 折线统计图可以表示数量的变化情况, 扇形统计图不能表示数量的变化情况, 选项描述错误; 故答案为: D",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "16-17 六年级·全国·单元测试",
+  "merged_id": "unit7_324",
+  "original_id": "unit7_choice_324"
+},
+{
+  "id": "unit7_choice_325",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "选择题",
+  "difficulty": "容易",
+  "question": "如图，六（1）班优秀的人数和六（2）班优秀的人数相比，（）。\n\n![](images/54abd1aadc568041a502348e66a058bf2e925ad01a4429290cc78d6edfa231d6.jpg)",
+  "options": [
+    "A. 同样多",
+    "B. 六 (1) 班多",
+    "C. 六 (2) 班多",
+    "D. 无法确定哪个班多"
+  ],
+  "answer": "D",
+  "solution": "【分析】从左图可知，六（1）班优秀的人数占六（1）班总人数的 $20 \\%$ , 即六 (1) 班总人数 $\\times 20 \\% =$ 六 (1) 班优秀的人数;从右图可知，六（2）班优秀的人数占六（2）班总人数的25%, 即六 (2) 班总人数 $\\times 25 \\% =$ 六 (2) 班优秀的人数;把两个班各自的总人数看作单位 '1', 题目没有说明两个班的总人数, 所以无法比较两个班的优秀人数。本题考查百分数乘法的意义，利用扇形统计图的特点及作用进行分析。明确单位 '1' 不同时, 不能进行比较。\n\n【详解】不知道六（1）班和六（2）班的总人数各是多少，即单位 '1'不确定，所以两个班的优秀人数无法比较，则无法确定哪个班多。故答案为: D",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "求一个数的百分之几是多少",
+    "单位'1'的认识与确定"
+  ],
+  "source": "2014 六年级·全国·课后作业",
+  "merged_id": "unit7_325",
+  "original_id": "unit7_choice_325"
+},
+{
+  "id": "unit7_fill_326",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "常用的统计图有()和( )。",
+  "options": [],
+  "answer": "条形统计图 折线统计图 扇形统计图",
+  "solution": "【分析】条形统计图重点突出每个信息的具体量；扇形统计图能够清楚地看到各类信息占总量的百分比；折线统计图能够清楚地看到某一量的变化情况，据此解答。\n\n【详解】常用的统计图有（条形统计图）、（折线统计图）和（扇形统计图）。",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "17-18 五年级下·全国·课后作业",
+  "merged_id": "unit7_326",
+  "original_id": "unit7_fill_326"
+},
+{
+  "id": "unit7_fill_327",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "一个养鸡专业户养了 120 只鸡, 其中有公鸡 24 只, 如果制成扇形统计图, 表示公鸡只数部分的扇形圆心角应为 ( )度.",
+  "options": [],
+  "answer": "72",
+  "solution": "【详解】 $24 \\div 120 = 20\\%$ ， $360^{\\circ} \\times 20\\% = 72^{\\circ}$\n\n【分析】用公鸡的只数除以鸡的总只数求出公鸡所占的百分率, 用 $360^{\\circ}$ 乘这个百分率就是表示公鸡只数部分的扇形圆心角的度数.",
+  "knowledge_points": [
+    "统计图"
+  ],
+  "source": "17-18 六年级下·全国·课后作业",
+  "merged_id": "unit7_327",
+  "original_id": "unit7_fill_327"
+},
+{
+  "id": "unit7_fill_328",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "扇形统计图用整个圆表示____，用圆内各个____的大小表示____数量占____的百分之几。",
+  "options": [],
+  "answer": "整体数量 扇形面积 部分 整体数量",
+  "solution": "【分析】扇形统计图比较清楚地反映出部分与部分、部分与整体之间的数量关系，圆的面积表示整体数量，扇形的面积表示部分数量。本题考查了扇形统计图的制作方法: 圆的面积表示整体数量, 扇形的面积表示部分数量。\n\n【详解】扇形统计图用整个圆表示整体数量，用圆内各个扇形面积的大小表示部分数量占整体数量的百分之几。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "2014 六年级·全国·课后作业",
+  "merged_id": "unit7_328",
+  "original_id": "unit7_fill_328"
+},
+{
+  "id": "unit7_fill_329",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "扇形统计图是用_表示总数，用圆内各个扇形表示__。",
+  "options": [],
+  "answer": "圆的面积 部分数量",
+  "solution": "【分析】扇形统计图比较清楚地反映出部分与部分、部分与整体之间的数量关系，圆的面积表示整体数量，扇形的面积表示部分数量。本题考查了扇形统计图的制作方法: 圆的面积表示整体数量, 扇形的面积表示部分数量。\n\n【详解】扇形统计图中圆的面积表示整体数量，扇形的面积表示部分数量。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "2014 六年级·全国·课后作业",
+  "merged_id": "unit7_329",
+  "original_id": "unit7_fill_329"
+},
+{
+  "id": "unit7_judge_330",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "六年级有 300 人，班委干部有 50 人，制成扇形统计图时占的圆心角是 $60^{\\circ}$ 。__。",
+  "options": [],
+  "answer": "√",
+  "solution": "【分析】先求出班委干部占总人数的几分之几，再求出它在扇形统计图中所对应的圆心角是多少，如果它们相等，则此题说法正确，反之，错误。\n\n抓住扇形统计图的绘制特点，即可解决此类问题。\n\n【详解】 $50 \\div 300 = \\frac{1}{6}$ $3 6 0 ^ {\\circ} \\times \\frac {1}{6} = 6 0 ^ {\\circ}$所以原题说法正确。故答案为: √",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "2014 六年级·全国·课后作业",
+  "merged_id": "unit7_330",
+  "original_id": "unit7_judge_330"
+},
+{
+  "id": "unit7_judge_331",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "扇形统计图用一个圆的面积表示部分数。____.",
+  "options": [],
+  "answer": "×",
+  "solution": "【分析】扇形统计图比较清楚地反映出部分与部分、部分与整体之间的数量关系, 圆的面积表示整体数量, 扇形的面积表示部分数量.\n\n【详解】扇形统计图中，一个圆的面积表示整体数量，扇形的面积才表示部分数量。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "2014 六年级·全国·课后作业",
+  "merged_id": "unit7_331",
+  "original_id": "unit7_judge_331"
+},
+{
+  "id": "unit7_fill_332",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "育英小学开展课外小组活动，参加美术组的有 180 人，体育组的有 130 人航模组的有 190 人，如果制成扇形统计图，那么体育组的人数占参加课外小组活动全部人数的 $\\_\\%$ ，美术组的人数占总人数的 $\\_\\%$ ，航模小组的人数占总人数的 $\\_\\%$ 。",
+  "options": [],
+  "answer": "26 36 38",
+  "solution": "【分析】先求出总人数，然后用体育组的人数除以总人数，求出体育组的人数占总人数的百分之几，用美术组的人数除以总人数，求出美术组的人数占总人数的百分之几，用航模组的人数除以总人数，求出航模小组的人数占总人数的百分之几。本题是求一个数是另一个数的百分之几, 关键是看把谁当成了单位 '1', 单位 '1'的量就为除数。\n\n【详解】 $180 + 130 + 190 = 500$ （人）；$130 \\div 500 = 26 \\%; \\quad 180 \\div 500 = 36 \\%; \\quad 190 \\div 500 = 38 \\% $\n\n答: 体育组的人数占参加课外小组活动全部人数的 $26 \\%$ ,美术组的人数占总人数的 $36 \\%$ , 航模小组的人数占总人数的 $38 \\%$",
+  "knowledge_points": [
+    "求一个数是另一个数的百分之几（百分率问题）"
+  ],
+  "source": "2014 六年级·全国·课后作业",
+  "merged_id": "unit7_332",
+  "original_id": "unit7_fill_332"
+},
+{
+  "id": "unit7_fill_333",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "为了反映小亮骑车行驶路程的变化情况，要制成（  ）统计图。",
+  "options": [],
+  "answer": "折线",
+  "solution": "【分析】此题应根据条形统计图、折线统计图、扇形统计图各自的特点进行解答.\n\n【详解】试题分析：条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。解：根据统计图的特点可知：为了反映小亮骑车行驶路程的变化情况，要制成折线统计图。故答案为折线.",
+  "knowledge_points": [
+    "统计图"
+  ],
+  "source": "16-17六年级上·全国·单元测试",
+  "merged_id": "unit7_333",
+  "original_id": "unit7_fill_333"
+}, {
+  "id": "unit7_fill_334",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "学校气象小组要绘制一幅统计图，公布上周每天平均气温的高低和变化情况，应选用_统计图。",
+  "options": [],
+  "answer": "折线",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。此题应根据条形统计图、折线统计图、扇形统计图各自的特点进行解答。\n\n【详解】根据统计图的特点可知：学校气象小组要绘制一幅统计图，公布上周每天平均气温的高低和变化情况，应选用折线统计图。",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "14-15 五年级·全国·期末",
+  "merged_id": "unit7_334",
+  "original_id": "unit7_fill_334"
+},
+{
+  "id": "unit7_fill_335",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "要统计某超市各月销售额增长情况，应绘制（  ）统计图。",
+  "options": [],
+  "answer": "折线",
+  "solution": "【分析】此题应根据条形统计图、折线统计图、扇形统计图各自的特点进行解答.\n\n【详解】试题分析：条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；由此根据情况选择即可。解：根据统计图的特点可知：要统计某超市各月销售额增长情况，应绘制折线统计图.故答案为折线.",
+  "knowledge_points": [
+    "统计图"
+  ],
+  "source": "12-13六年级上·贵州铜仁·期末",
+  "merged_id": "unit7_335",
+  "original_id": "unit7_fill_335"
+},
+{
+  "id": "unit7_fill_336",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "要反映某班学生在课外活动中参加各种小组的情况, 最好选用（  ）统计图.",
+  "options": [],
+  "answer": "扇形",
+  "solution": "【分析】此题应根据条形统计图、折线统计图、扇形统计图各自的特点进行解答.\n\n【详解】分析：首先要清楚每一种统计图的特点：条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；据此判断即可。解答：解：根据统计图的特点可知，要反映某班学生在课外活动中参加各种小组的情况，最好选用扇形统计图。故答案为扇形.",
+  "knowledge_points": [
+    "统计图"
+  ],
+  "source": "13-14 六年级·全国·单元测试",
+  "merged_id": "unit7_336",
+  "original_id": "unit7_fill_336"
+},
+{
+  "id": "unit7_fill_337",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "条形统计图能很容易地看出（  ），扇形统计图能清楚地表示（  ），折线统计图能清楚地表示（  ）.",
+  "options": [],
+  "answer": "数量的多少，部分与整体的关系，数量的增减变化情况",
+  "solution": "【分析】此题应根据条形统计图、折线统计图、扇形统计图各自的特点进行解答。\n\n【详解】分析：首先要清楚每一种统计图的特点：条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；据此判断即可。解答：解：根据统计图的特点可知，条形统计图能很容易地看出数量的多少，扇形统计图能清楚地表示部分与整体的关系，折线统计图能清楚地表示数量的增减变化情况。故答案为数量的多少, 部分与整体的关系, 数量的增减变化情况.",
+  "knowledge_points": [
+    "统计图"
+  ],
+  "source": "13-14 六年级·全国·单元测试",
+  "merged_id": "unit7_337",
+  "original_id": "unit7_fill_337"
+},
+{
+  "id": "unit7_fill_338",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "要想清楚地表示出各部分同总数之间的关系，应选用_统计图。",
+  "options": [],
+  "answer": "扇形",
+  "solution": "【详解】根据扇形统计图的特点和作用，要想清楚地表示出各部分同总数之间的关系，应选用扇形统计图。",
+  "knowledge_points": [
+    "统计图的选择"
+  ],
+  "source": "2014·全国·小升初真题",
+  "merged_id": "unit7_338",
+  "original_id": "unit7_fill_338"
+},
+{
+  "id": "unit7_fill_339",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "要统计某学校各年级的人数, 可以选用_____统计图。",
+  "options": [],
+  "answer": "条形",
+  "solution": "【详解】条形统计图能清楚地看出数量的多少, 所以要统计某学校各年级的人数, 可以选用条形统计图。",
+  "knowledge_points": [
+    "1 格表示一个单位的单式条形统计图",
+    "1 格表示多个单位的单式条形统计图",
+    "统计图的选择"
+  ],
+  "source": "15-16 六年级上·全国·课后作业",
+  "merged_id": "unit7_339",
+  "original_id": "unit7_fill_339"
+},
+{
+  "id": "unit7_fill_340",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "如图（1），如果用整个图表示总体,那么扇形________表示总体的 $\\frac{1}{3}$ , 扇形_表示总体的 $\\frac{1}{2}$ 图（1）.![](images/84d6a536525d8abf6cbb5ea10ca6e51dd1de48fd1d96e1a9839adab7f1f19ebf.jpg)",
+  "options": [],
+  "answer": "A   B",
+  "solution": "【详解】扇形统计图是用整个圆表示总数，用圆内各个扇形的大小表示各部分数量占总数的百分数。由此可知答案。总结：本题主要考查扇形统计图的特点的掌握情况。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "2014 三年级·全国·课后作业",
+  "merged_id": "unit7_340",
+  "original_id": "unit7_fill_340"
+},
+{
+  "id": "unit7_fill_341",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "红星村今年对农田秋季播种作物如图规划，且只种植这三种农作物，则该村种植的大麦占种植所有农作物的（  ）$\\%$.\n\n![](images/cab23b8468e8a865a722960231cc2439d47b31d54bee65478479e901a9214e68.jpg)",
+  "options": [],
+  "answer": "36",
+  "solution": "【详解】扇形统计图是用整个圆表示总数，用圆内各个扇形的大小表示各部分数量占总数的百分数，大麦占种植所有农作物的百分之几，就是 $450 \\div (450 + 500 + 300) \\times 100\\%$ ，由此可知答案.考点：扇形统计图的特点。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "15-16六年级上·全国·课后作业",
+  "merged_id": "unit7_341",
+  "original_id": "unit7_fill_341"
+},
+{
+  "id": "unit7_fill_342",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "光明中学对图书馆的书分成 3 类，A 表示科技类，B 表示科学类，C 表示艺术类，所占的百分比如图所示，如果该校共有图书 8500 册，则艺术书共有____册。\n\n![](images/d41f39672edd5df118844f43c1a59e4d0afa1c07531b2a56ef7f295b3a9785c0.jpg)",
+  "options": [],
+  "answer": "595",
+  "solution": "【分析】根据题意可知艺术类所占的百分比是 $1 - 57\\% - 36\\% = 7\\%$ ,所以艺术书共有 $7 \\% \\times 8500 = 595$ 册.\n\n【详解】艺术书共有 $(1 - 57\\%-36\\%)\\times8500 = 7 \\% \\times 8500 = 595$ (册)",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "14-15 六年级下·全国·单元测试",
+  "merged_id": "unit7_342",
+  "original_id": "unit7_fill_342"
+},
+{
+  "id": "unit7_fill_343",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "如图是某校六年级学生爱好各种球类人数分布情况统计图。\n\n![](images/9415faefd1b74d1e48f541d57103a56a1fa76c9bbe63e67ffaf6fffa8c5b856c.jpg)\n\n(1) 爱好_的人数最多。\n(2) 和 _的爱好者人数差不多。\n(3) 如果爱好足球的有 48 人, 六年级爱好羽毛球的有_人。",
+  "options": [],
+  "answer": "足球 乒乓球 羽毛球 27",
+  "solution": "【分析】（1）观图可知：把这个年级学生总数看作单位'1'，也就是 $100\\%$ ，爱乒乓球占 $23\\%$ 、足球占 $32\\%$ 、篮球占 $17\\%$ 、其它的占 $10\\%$ ，羽毛球的占 $18\\%$ ，所以爱好足球的人数最多；(2) 篮球占 $17 \\%$ , 羽毛球的占 $18 \\%$ , 所以羽毛球和乒乓球的爱好者人数差不多; (3) 根据百分数除法的意义, 用喜爱足球的人数除以喜爱足球人数所占的百分比就是这个年级的总人数, 再用总人数乘 $18 \\%$ , 即可求出六年级爱好羽毛球的有多少人, 解答即可。此题是考查如何从扇形统计图中获取信息，并根据所获取的信息进行有关计算等。\n\n【详解】由分析可知：(1) 爱好足球的人数最多;(2) 乒乓球和羽毛球的爱好者人数差不多;(3) $48 \\div 32\\% \\times 18\\%\\\\=150 \\times 18 \\% \\\\ = 2 7 (人) $\n\n六年级爱好羽毛球的有 27 人。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "2014·辽宁铁岭·小升初真题",
+  "merged_id": "unit7_343",
+  "original_id": "unit7_fill_343"
+},
+{
+  "id": "unit7_fill_344",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "一次数学测验后，张老师根据某班成绩绘制了如下图所示的扇形统计图（ $80\\sim 89$ 分的百分比因故模糊不清），若80分以上（含80分）为优秀，则本次测验这个班的优秀率为( )。\n\n![](images/a31e4eda444a8c39445af4a90d409cf9a00b039801ad9d548d7c88541ef9805a.jpg)",
+  "options": [],
+  "answer": "$68\\%$",
+  "solution": "【分析】扇形统计图中整个圆面表示的是总数量，每个小扇形表示的是部分量占总数量的百分比，即所有小扇形中表示的百分数的和应该是整体'1'，所以 $80\\sim 89$ 分的学生占总学生数的百分数是： $1 - 20\\% -12\\% -36\\% = 32\\%$ 。本题中要求的是优秀率，即80分 $\\sim 100$ 分的学生占学生总数的百分数，所以是 $32\\% +36\\% = 68\\%$ ，或者根据优秀率 $= 1-$ 非优秀率 $= 1-$ （ $20\\% +12\\%$ ） $= 68\\%$ 亦可得解。从扇形统计图中读出各部分所占百分数进行求解,是解决本题的关键。\n\n【详解】由分析可得：\n\n$1 - 20\\% - 12\\% - 36\\%\\\\ =80 \\% -12 \\% -36 \\% \\\\ = 68\\% -36\\% \\\\ = 32 \\%$\n\n$32 \\% + 36 \\% = 68 \\%$则本次测验这个班的优秀率为 $68 \\%$ 。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "含百分数的运算"
+  ],
+  "source": "2016·全国·小升初真题",
+  "merged_id": "unit7_344",
+  "original_id": "unit7_fill_344"
+},
+{
+  "id": "unit7_fill_345",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "根据统计图填空。![](images/de0adc0cbfef6246b49512f8686b33ebd36b952294fbcb6a0f0ed9ab62171d28.jpg)前进小学六年级学生喜欢的运动项目统计如图，其中喜欢足球的有40人。\n\n(1) 喜欢踢毽的占总人数的 $\\frac{1}{2}\\%$ .\n(2) 前进小学六年级一共有_人。\n(3) 喜欢乒乓球的人数比喜欢足球的人数多 $\\frac{1}{2}\\%$ 。",
+  "options": [],
+  "answer": "8 ;200 ;50",
+  "solution": "【详解】 (1) $1 - 20\\% - 19\\% - 30\\% - 23\\% = 8\\%$\n\n答: 喜欢踢毽的占总人数的 $8 \\%$ .![](images/adeb62fd60660c9dc4d2b28ab80339910866f779ac5741961284888717e3b27c.jpg)\n\n(2) $40 \\div 20\\% = 200$ （人）答: 前进小学六年级一共有 200 人。\n\n(3) $(30\\% - 20\\%) \\div 20\\%\\\\=10 \\% \\div 20 \\% \\\\ = 50 \\% $\n\n答: 喜欢乒乓球的人数比喜欢足球的人数多 $50 \\%$ .故答案为 8, 200, 50。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "2015·贵州遵义·小升初真题",
+  "merged_id": "unit7_345",
+  "original_id": "unit7_fill_345"
+},
+{
+  "id": "unit7_fill_346",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "某售票处门票情况如下图：![](images/402a042f5a55eacd1dff8782fe1a75240b227b1600f16fdfb0eeaa032f33b4c6.jpg)\n\n(1)甲级门票占总票数的 ______$\\%$\n(2)售出的______级门票最多，占总数的______%；\n(3)一共售出门票 400 张, 乙级门票比甲级门票多售出______张.",
+  "options": [],
+  "answer": "(1) 25   (2) 丙; 45    (3) 20",
+  "solution": "【分析】（1）整个圆是360度，代表售出的所有门票，甲级门票占了其中的90度；（2）从扇形图中可以直观的看出丙级门票售出最多，甲级+乙级+丙级售出的门票可以看作是单位'1'，则丙级门票占的量为 $1 - 25\\% - 30\\%$ ；（3）门票总数是400张，乙级门票比甲级门票多 $30\\% - 25\\%$ ，根据乘法的意义可以求出多出的具体数量。\n\n【详解】（1）甲级门票占票数的： $90^{\\circ} \\div 360^{\\circ} \\times 100\\% = 25\\%$ (2) 售出的丙门票最多，占总数的 $1 - 25\\% - 30\\% = 45\\%$ (3) 乙级门票比甲级门票多售出 $400 \\times (30\\% -25\\%) = 20$ （张）.故答案为（1）25；（2）丙；45；（3）20",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "统计图表的综合应用"
+  ],
+  "source": "16-17 六年级上·广东深圳·期末",
+  "merged_id": "unit7_346",
+  "original_id": "unit7_fill_346"
+},
+{
+  "id": "unit7_fill_347",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "幸福村今年农作物的种植面积如图所示，并且只种植这三种农作物。则该村种植的玉米占种植总面积的（  ）$\\%$ , 花生占种植总面积的（  ）$\\%$ .![](images/318184e9f42d3e89b0813c9c1a7d31ea119ee87bca0e59cc7f992d790500684b.jpg)",
+  "options": [],
+  "answer": "36 ;40",
+  "solution": "【详解】试题分析：根据题意，先求出三种农作物种植的总面积，再根据求一个数是另一个数的几分几，用除法计算.解答本题的关键是根据题意, 求出一共的种植面积, 再根据分数除法的意义解答.\n\n解: $300 + 450 + 500 = 1250$ (亩)\n\n$450 \\div 1250 \\times 100 \\% \\\\ = 0.36 \\times 100 \\% \\\\ = 36 \\%$\n\n$ \\\\ 500 \\div 1250 \\times 100 \\% \\\\ = 0.4 \\times 100 \\% \\\\ = 40 \\% $\n\n答：该村种植的玉米占种植总面积的 $36 \\%$ , 花生占种植总面积的 $40 \\%$ .故答案为 36 ; 40 .",
+  "knowledge_points": [
+    "统计图"
+  ],
+  "source": "16-17六年级·全国·单元测试",
+  "merged_id": "unit7_347",
+  "original_id": "unit7_fill_347"
+},
+{
+  "id": "unit7_fill_348",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "下面是世界各大洲面积占陆地总面积百分比的统计图.\n\n![](images/7c0760f38e542a3dceb3699c8dddfc455400571f165407078f8bb4000feee522.jpg)\n\n(1)全世界共有____大洲.\n(2)____洲面积最大.\n(3)____洲面积最小，占陆地总面积的____%。",
+  "options": [],
+  "answer": "6 ; 亚;  大洋 ; 6",
+  "solution": "【详解】略",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "统计图表的综合应用"
+  ],
+  "source": "17-18 六年级上·全国·课后作业",
+  "merged_id": "unit7_348",
+  "original_id": "unit7_fill_348"
+},
+{
+  "id": "unit7_fill_349",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "一个面积是 $120 \\mathrm{~m}^{2}$ 的花坛里种了三种花，各种花的种植面积如图。\n\n![](images/3f54a01eda9cb991cbe786fc5d8c85e47249c45436ec8b70415a73dfdd65c799.jpg)\n\n(1) 如图是一幅（  ）统计图。\n(2) 菊花的种植面积占花坛面积的（  ）$\\%$ 。\n(3) 玫瑰花的种植面积是（  ）$\\_ \\mathrm{m}^{2}$ 。\n(4) 菊花与玫瑰花种植面积的比是____:____。",
+  "options": [],
+  "answer": "扇形  25   60  1:2",
+  "solution": "【分析】（1）根据扇形统计图的意义可知，这是一幅扇形统计图；(2) 把这个花坛的总面积看作单位 '1', 即 $100 \\%$ , 用 1 减去其它花种植面积所占的百分率即可; (3) 用这个花坛的总面积乘玫瑰花种植面积所占的百分率就是玫瑰花的种植面积; (4) 用菊花种植面积所占的百分率比玫瑰花种植面积的所占的百分率化简即可。\n\n【详解】（1）如图是一幅扇形统计图。(2) $1 - 50\\% - 25\\% = 25\\%$ (3) $120 \\times 50\\%$ (4) $25\\%$ : $50\\% = 1:2$\n\n$ 1 2 0 \\times 0. 5 \\\\ = 6 0 (\\mathrm{m} ^ {2}) $",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "比的意义"
+  ],
+  "source": "2014 六年级·全国·课后作业",
+  "merged_id": "unit7_349",
+  "original_id": "unit7_fill_349"
+},
+{
+  "id": "unit7_fill_350",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "小军家 2011 年 1 月支出情况统计如下图。![](images/dab6cdf4f66c7c0a6e4942de4ea7bf210d63f87d8492fe3eeaa71222598738f3.jpg)\n\n(1) 购买衣物支出为 3600 元, 小军家这个月共支出 ( )元。\n(2) 小军家这个月水电支出( )元。\n(3) 小军家这个月购买衣物比文化教育少支出( )$\\%$。",
+  "options": [],
+  "answer": "18000 1440 20",
+  "solution": "【分析】把总支出看成单位'1'，其中购买衣物支出占20%，文化教育支出占25%，伙食支出占35%，其它支出占12%，水电支出占8%；(1) 购买衣物支出的 $20 \\%$ 对应的数量是 3600 元, 由此用除法求出总支出;(2) 然后用总支出乘水电支出占总支出的百分之几即可; (3) 先用文化教育支出占的百分数减去购买衣服的支出占的百分数, 然后除以文化教育支出的百分数即可。此题主要考查的是如何观察扇形统计图并且从统计图中获取信息，然后再进行计算、解答即可。【详解】 (1) $3600 \\div 20\\% = 18000$ (元）即小军家这个月共支出18000元。(2) $18000 \\times 8\\% = 1440$ （元）即小军家这个月水电支出1440元。(3) $(25\\% - 20\\%) \\div 25\\%$ $ 5 \\% \\div 25 \\% = 20 \\% $即小军家这个月购买衣物比文化教育少支出 $20 \\%$ 。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "2014 六年级·全国·课后作业",
+  "merged_id": "unit7_350",
+  "original_id": "unit7_fill_350"
+},
+{
+  "id": "unit7_fill_351",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "根据图形填空.如图是某校六年级同学最喜欢的运动项目统计图.\n\n(1) 从图中可看出, 喜欢____运动的人数最多.\n\n(2) 喜欢足球运动的同学比喜欢排球运动的多 _____ $\\%$ .\n\n(3) 喜欢篮球运动的同学有 38 人, 全年级喜欢运动的同学有 _____人.\n\n(4) 假如这是在你们学校统计的数据, 你想对体育老师提什么建议? ______________.\n\n![](images/b01e1ab016b79831424a068d55fac54b66fec61bd17b69d9e4c1a0c7af38264c.jpg)",
+  "options": [],
+  "answer": "乒乓球  7  200   增加一张乒乓球桌",
+  "solution": "【分析】（1）从统计图看出喜欢乒乓球运动的人数最多；(2) 喜欢足球运动的同学占 $25 \\%$ , 喜欢排球运动的占18%, 用喜欢足球运动的同学所占的百分率减去喜欢排球运动所占的百分率即可;(3) 喜欢篮球运动的同学占 $19 \\%$ , 用 38 除以喜欢篮球运动的同学所占的百分率就是全年级喜欢运动的同学的人数; (4) 根据图知道喜欢乒乓球运动的人数最多, 可以增加一张乒乓球桌.\n\n【详解】（1）从统计图看出喜欢乒乓球运动的人数最多；(2) $25\\% - 18\\% = 7\\%$ .答：喜欢足球运动的同学比喜欢排球运动的多 $7 \\%$ . (3) $38 \\div 19\\% = 200$ （人）. 答: 喜欢篮球运动的同学有 38 人, 全年级喜欢运动的同学有 200 人. (4) 可以增加一张乒乓球桌.",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "统计图表的综合应用"
+  ],
+  "source": "2014 六年级·全国·课后作业",
+  "merged_id": "unit7_351",
+  "original_id": "unit7_fill_351"
+},
+{
+  "id": "unit7_fill_352",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "根据图回答问题.\n\n(1) 鸡占总数的____.\n(2)这是一幅________统计图.\n(3) 已知鸡有 2000 只, 这个饲养场共养禽畜______只, 其中兔______只，鸭___只.![](images/34bca64546c56d9baa8a4a69d8aee6b9f5ab51bfff5ce2354d891a3278aad654.jpg)",
+  "options": [],
+  "answer": "$25\\%$  扇形  8000  4400  1600",
+  "solution": "【分析】（1）根据统计图知道，兔占总数的 $55\\%$ ，鸭占总数的 $20\\%$ ，由此得出鸡占总数的 $(1 - 55\\% - 20\\%)$ ；(2) 根据此统计图的特点, 判断这是一幅扇形统计图; (3) 用鸡的只数除以鸡占总数的百分数, 可以求出饲养场共养禽畜的只数, 再用饲养场共养禽畜的只数分别乘兔占总数的百分数、鸭占总数的百分数求出兔的只数、鸭的只数.关键是从统计图中获取与问题有关的信息，再根据基本的数量关系解答即可。\n\n【详解】 (1) $1 - 55\\% - 20\\% = 25\\%$ ;(2) 根据此统计图的特点，判断这是一幅扇形统计图；\n\n(3) $2000 \\div 25\\% = 8000$ （只），$8000 \\times 55 \\% = 4400 (只),$   $8000 \\times 20\\% = 1600$ （只）",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "统计图表的综合应用"
+  ],
+  "source": "2014 六年级·全国·课后作业",
+  "merged_id": "unit7_352",
+  "original_id": "unit7_fill_352"
+},
+{
+  "id": "unit7_fill_353",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "数学兴趣小组的同学在一次数学竞赛中的成绩统计如图。显然得优良和及格的同学都算达标，则数学兴趣小组的同学这次竞赛的达标率是 _______$\\%$ 。若全体同学的平均成绩是70分，达标同学的平均成绩是80分，则不及格同学的平均成绩是 ______分。\n\n![](images/c176693319cdf98697d5942443a5ad36752a0df951258fe5d5461fd436e15589.jpg)",
+  "options": [],
+  "answer": "75 40",
+  "solution": "【分析】（1）达标率就是优良和及格的同学占总人数百分比的和，由此求解；(2) 为了便于计算, 就令总人数是 100 人, 那么及格的有 $40 \\%$ , 优良的有 35 , 不及格的就是 25 人, 求出全班的总分。减去达标学生的总分, 就是不达标学生的总分, 然后再除以 25 人即可。此题主要考查的是如何观察扇形统计图并且从统计图中获取信息，然后再根据平均数的计算方法求解。\n\n【详解】 (1) $35\\% + 40\\% = 75\\%$ ;答：数学兴趣小组的同学这次竞赛的达标率是 $75 \\%$ 。(2) 令总人数是 100 人;\n\n那么及格的人数就是:\n\n$100 \\times 35 \\% + 100 \\times 40 \\%, \\\\ = 3 5 + 4 0, \\\\ = 7 5 (人) ; $\n\n不及格的就是：\n\n$100 \\times 25 \\% = 25 (人); \\quad 7 0 \\times 1 0 0 - 8 0 \\times 7 5 = 7 0 0 0 - 6 0 0 0= 1000 （分）$；\n\n$1000 \\div 25 = 40$ （分）；答: 不及格同学的平均成绩是 40 分。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "平均数的意义及求法"
+  ],
+  "source": "2020·浙江温州·小升初真题",
+  "merged_id": "unit7_353",
+  "original_id": "unit7_fill_353"
+},
+{
+  "id": "unit7_fill_354",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "观察扇形统计图, 并填写。\n\n(1)如果用这个圆代表总体，那么扇形_表示总体的 $45 \\%$ 。\n(2)如果用整个圆代表你们班级的总人数，那么扇形 B 大约代表____人。\n(3)如果用整个圆代表 9 公顷的稻田, 那扇形 A 大约代表____公顷。\n(4)如果用整个圆代表某校全体学生的人数，已知扇形 B 比扇形 A 多 $5 \\%$ , 且多 60 人, 全校_人。\n\n![](images/5e569ea9829cd595767992b4176492e1d5c6e8a078afccd9e280bc0801b13aae.jpg)",
+  "options": [],
+  "answer": "(1) C   (2)15   (3)2.25    (4)  $1200$",
+  "solution": "【分析】（1）B 占 $30\\%$ ， $45\\%$ 大于 $30\\%$ ，A 的区域小于 B 的区域，C 的区域大于 B 的区域；由此判断；(2) 我们班共有 50 人, 把 50 人看成单位 '1', 用 50 人乘 $30 \\%$ 就是 B 区域表示的人数; (3) 把总面积看成单位 '1', 先求出 A 区域大约占总数的百分之几, 然后用乘法求出 A 表示的面积;(4) 把全校的总人数看成单位 '1', 它的 $5 \\%$ 对应的数量是 60 人, 由此用除法求出总人数。本题考查了扇形统计图的特点，整个圆表示总量单位 '1', 每个扇形表示表示总量的百分之几 (或几分之几)。\n\n【详解】 (1) (1) $45\\% > 30\\%$用这个圆代表总体，扇形C表示总体的 $45\\%$ ； (2) 我们班有 50 人;\n\n$50 \\times 30 \\% = 15 (人)$答: 扇形 B 大约代表 15 人。 (3) $9 \\times (1 - 45\\% - 30\\%)=9 \\times 25 \\% = 2. 2 5 (公 顷) $\n\n答: 扇形 A 大约代表 2.25 公顷。\n\n(4) (4) $60 \\div (45\\% - 30\\%)\\\\= 60 \\div 5 \\% \\\\ = 1 2 0 0 (人) $\n\n答: 全校 1200 人。",
+  "knowledge_points": [
+    "求一个数的百分之几是多少",
+    "已知一个数的百分之几是多少，求这个数"
+  ],
+  "source": "2014 六年级·全国·课后作业",
+  "merged_id": "unit7_354",
+  "original_id": "unit7_fill_354"
+},
+{
+  "id": "unit7_fill_355",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "某学校对学生早上到学校的方式进行了调查，结果如图，其中 A 为骑自行车上学的学生，B 为乘公共汽车上学的学生，C 为走路上学的学生，D 为乘地铁上学的学生，所在扇形的圆心角为 $15^{\\circ}$ ，问其中走路上学的学生占受调查人数的（几分之几）。\n\n![](images/eebbcf9e19f5b38783b9fdca58e07332f1aac0aa27088ad2f74ef492ad7dc96a.jpg)",
+  "options": [],
+  "answer": "$\\frac{3}{8}$",
+  "solution": "【分析】根据图知道, A 对应的圆心角是 90 度, B 对应的圆心角是 120 度, D 对应的圆心角的 15 度, 而周角是 360 度, 由此求出 C 对应的圆心角, 再除以 360 求出走路上学的学生占受调查人数的几分之几。关键是明白扇形统计图的意义表示的是部分占整体的几分之几。\n\n【详解】 $(360 - 120 - 90 - 15) \\div 360\\\\= 1 3 5 \\div 3 6 0 \\\\ = \\frac {3}{8} $",
+  "knowledge_points": [
+    "可能性的大小"
+  ],
+  "source": "2014 六年级·全国·课后作业",
+  "merged_id": "unit7_355",
+  "original_id": "unit7_fill_355"
+},
+{
+  "id": "unit7_fill_356",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "下面是六（1）班学生喜欢的电视节目统计图。\n\n(1)喜欢《大风车》的同学人数占全班人数的_____ $\\%$ 。\n(2)喜欢《动物世界》、《走近科学》与《新闻联播》人数的比是____。\n(3)如果全班有 50 人, 那么, 喜欢《动物世界》的有_人。\n\n![](images/7cceba28ec850351a63c4e828b9b6dfca5b56bd0766e7095e9eed1dc46edc439.jpg)",
+  "options": [],
+  "answer": "(1)18 (2)  $20: 15: 6$   (3)20",
+  "solution": "【分析】由图可知：把六（1）班学生的总人数看成单位'1'，喜欢《动物世界》的占总人数的 $40 \\%$ ；喜欢《走近科学》的占总人数的 $30 \\%$ ，喜欢《新闻联播》的占总人数的 $12 \\%$(1) 先求出喜欢《动物世界》、《走近科学》与《新闻联播》的共占总人数的百分之几；然后用总人数 1 减去这部分就是喜欢《大风车》的占总人数的百分之几； (2) 用喜欢《动物世界》的分率比《走近科学》的分率比《新闻联播》的分率，然后再化简； (3) 用总人数乘喜《动物世界》的人数占的百分数即可。解决本题关键是读懂扇形统计图，再根据所求问 题找出需要的数量求解。\n\n【详解】（1） $1 - (40\\% + 12\\% + 30\\%)$ $= 1 - 82\\%$ $= 18\\%$ (2） $40\\% : 30\\% : 12\\%$ $= 40: 30: 12$ $= 20: 15: 6$ (3) $50 \\times 40\\% = 20$ （人）",
+  "knowledge_points": [
+    "比的应用",
+    "求一个数的百分之几是多少"
+  ],
+  "source": "2014 六年级·全国·课后作业",
+  "merged_id": "unit7_356",
+  "original_id": "unit7_fill_356"
+},
+{
+  "id": "unit7_fill_357",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "看图填空:(1) 茄子占总面积的_______ $\\%$ ，茄子的占地面积是 92 平方分米，黄瓜的占地面积是 _平方分米，青菜的占地面积是 _____________平方分米。(2) 在扇形统计图中表示黄瓜数量的圆心角是________度。\n\n![](images/5f2c28aeea06cdeb6616d370361e9a875b2fe012e02cc242deb75f7bcfbfe629.jpg)",
+  "options": [],
+  "answer": "5   644   1104    126",
+  "solution": "【分析】（1）根据题意，把菜地的总面积看作单位'1'，可用单位'1'分别减去 $35\\%$ 和 $60\\%$ 即可得到茄子占总面积的百分数；然后再用92除以茄子占总面积的百分数即可得到菜地的总面积，然后再用菜地的总面积乘黄瓜占总面积的百分数即可得到黄瓜的占地面积；然后再用青菜占总面积的百分数乘菜地的总面积即可得到青菜的占地面积，列式解答即可得到答案。(2) 因为黄瓜占 $35 \\%$ , 可用黄瓜占总面积的百分数乘 $360^{\\circ}$就是黄瓜数量的圆心角的度数。\n\n【详解】 (1) 茄子占总面积的: $1 - 35\\% - 60\\% = 5\\%$ .黄瓜占地面积为： $92\\div 5\\% \\times 35\\% = 644$ （平方分米）青菜占地面积为： $92 \\div 5\\% \\times 60\\% = 1104$ （平方分米）答: 茄子占总面积的 $5 \\%$ , 茄子的占地面积是 92 平方分米,黄瓜的占地面积是 644 平方分米, 青菜的占地面积是 1104 平方分米。(2) $360^{\\circ} \\times 35\\% = 126^{\\circ}$答: 在扇形统计图中表示黄瓜数量的圆心角是 126 度。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "求一个数是另一个数的百分之几（百分率问题）"
+  ],
+  "source": "2014 六年级·全国·课后作业",
+  "merged_id": "unit7_357",
+  "original_id": "unit7_fill_357"
+},
+{
+  "id": "unit7_fill_358",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "如图是笑笑对六年级（1）班40名同学进行的'你最喜欢的一项体育活动'的调查结果。从图中可以看出，喜欢______的同学最多。喜欢跳绳的比喜欢足球的多_____人.\n\n六(1)班同学最喜欢的体育活动统计图:![](images/4aa4d17c3908cb3934f19a0119b7e76b3231b9686c1b21baa4c560807132c791.jpg)",
+  "options": [],
+  "answer": "乒乓球 1",
+  "solution": "【详解】解：（1）从图中可以看出，因为 $35\\% > 25\\%$ 22.5% > 17.5%, 所以喜欢乒乓球的同学最多.\n\n(2) $40 \\times (25\\% - 22.5\\%\\\\= 40 \\times 2.5 \\% \\\\ = 1 (人) $\n\n答: 喜欢跳绳的比喜欢足球的多 1 人.故答案为乒乓球, 1.\n\n【分析】（1）从统计图看出喜欢乒乓球运动的人数最多； (2) 喜欢足球运动的同学占 $22.5 \\%$ , 喜欢跳绳运动的占25%, 用喜欢跳绳运动的同学所占的百分率减去喜欢足球运动所占的百分率, 然后用 40 去乘即可.",
+  "knowledge_points": [
+    "统计图"
+  ],
+  "source": "16-17 六年级上·北京海淀·期末",
+  "merged_id": "unit7_358",
+  "original_id": "unit7_fill_358"
+},
+{
+  "id": "unit7_fill_359",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "如图是一幅水果质量统计图, 水果的总质量是 1000 千克, 苹果占总质量的( )% , 重( )千克, 梨占水果总质量的( )%, 重( )千克, 香蕉和枣站水果总质量的( )%, 重( )千克. (提示: 求一个数的百分之几是多少, 用乘法计算, 用这个数乘百分数.)\n\n![](images/0c65478894cc20552b9a798a29103e612709315f65e69869a7a016b323d0e5a9.jpg)",
+  "options": [],
+  "answer": "$\\frac{1}{2}$  500   $\\frac{1}{4}$     250   $\\frac{1}{8}$ 125",
+  "solution": "【详解】略",
+  "knowledge_points": [
+    "统计图"
+  ],
+  "source": "17-18 五年级下·全国·单元测试",
+  "merged_id": "unit7_359",
+  "original_id": "unit7_fill_359"
+},
+{
+  "id": "unit7_fill_360",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "在一所民族小学里,为了表示出各民族学生人数占总人数的百分比情况，可以选用____统计图。",
+  "options": [],
+  "answer": "扇形",
+  "solution": "【分析】条形统计图能表示数量的多少；折线统计图不仅能表示数量的多少，还能表示数量的增减变化情况；扇形统计图表示部分与整体之间的关系。本题考查的是扇形统计图的特征，关键审清题意看需要做出什么样的效果。\n\n【详解】为了表示出各民族学生占的比例，因为要表示部分与整体之间的关系，所以可以选用扇形统计图。故答案为：扇形。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "2018 六年级下·云南·专题练习",
+  "merged_id": "unit7_360",
+  "original_id": "unit7_fill_360"
+},
+{
+  "id": "unit7_fill_361",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "容易",
+  "question": "扇形统计图的优点是可以很清楚地表示出( )。",
+  "options": [],
+  "answer": "部分与总数之间的关系",
+  "solution": "【分析】扇形统计图是一种用整个圆表示总数，通过圆内各个扇形的大小来表示各部分数量占总数的百分比的图形，其优点是可以很清楚地表示出部分与总数之间的关系。\n\n【详解】扇形统计图的优点是可以很清楚地表示出部分与总数之间的关系。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "17-18 六年级下·全国·课后作业",
+  "merged_id": "unit7_361",
+  "original_id": "unit7_fill_361"
+},
+{
+  "id": "unit7_fill_362",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "下图是六(1)班同学最喜欢喝的饮料情况统计图.\n\n![](images/cbf66365d45f7aaaf9b5a0e847d563a1269a7fcbfe61b9a37809a4557410ddd2.jpg)\n\n(1)最喜欢喝各种饮料的人数占全班总人数的百分比之和应该是( )%。\n(2)最喜欢喝( )的人数最多，占全班总人数的( )%。\n(3)如果全班有 40 人, 那么最喜欢喝奶茶的有( )人, 最喜欢喝可乐的有( )人.\n(4)如果最喜欢喝可乐的有 9 人, 那么全班有 ( )人.",
+  "options": [],
+  "answer": "100 绿茶 30 8 6 60",
+  "solution": "【详解】略",
+  "knowledge_points": [
+    "统计图"
+  ],
+  "source": "18-19六年级上·全国·单元测试",
+  "merged_id": "unit7_362",
+  "original_id": "unit7_fill_362"
+},
+{
+  "id": "unit7_fill_363",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "下面是六(1)班同学英语口语测试成绩的统计表和统计图, 请将它们补充完整.六（1）班英语口语测试成绩统计图:![](images/5b0778a78dfd43eadcf9663517efef9a5458f10ad3530930ed3dfcff2c319338.jpg)\n\n| 等级 | 优   | 良   | 及格 | 不及格 |\n| ---- | ---- | ---- | ---- | ------ |\n| 人数 | 18   | 16   |      |        |",
+  "options": [],
+  "answer": "及格：4 不及格：2 优：45 良：40",
+  "solution": "【详解】略",
+  "knowledge_points": [
+    "统计"
+  ],
+  "source": "18-19 六年级上·全国·单元测试",
+  "merged_id": "unit7_363",
+  "original_id": "unit7_fill_363"
+},
+{
+  "id": "unit7_fill_364",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "填空题",
+  "difficulty": "中等",
+  "question": "如图, 这是六(2)班图书角中各种图书所占百分比的不完整统计图.\n\n![](images/8deefecd918a69bc8d9c2bc984420345b9d1686458fa20b01164770f7b1fc87e.jpg)\n\n(1)这个圆代表的是____，科普书占图书总数的____%。\n(2)已知文学书有 150 本, 那么作文书有 \\_ 本, 科普书有 \\_ 本.",
+  "options": [],
+  "answer": "图书总数  35  125   175",
+  "solution": "【详解】略",
+  "knowledge_points": [
+    "统计图"
+  ],
+  "source": "18-19 六年级上·全国·单元测试",
+  "merged_id": "unit7_364",
+  "original_id": "unit7_fill_364"
+},
+{
+  "id": "unit7_judge_365",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "表示某地区一个星期的气温变化情况时应选用条形统计图比较合适。（）",
+  "options": [],
+  "answer": "×",
+  "solution": "【分析】条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系；据此判断即可。\n\n【详解】由分析可知：表示某地区一个星期的气温变化情况时应选用折线统计图比较合适。原题干说法错误。故答案为: ×",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "1格表示多个单位的单式条形统计图",
+    "统计图的选择",
+    "单式折线统计图"
+  ],
+  "source": "17-18六年级下·全国·期末",
+  "merged_id": "unit7_365",
+  "original_id": "unit7_judge_365"
+},
+{
+  "id": "unit7_judge_366",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "判断题",
+  "difficulty": "中等",
+  "question": "如果我想了解我们班同学的兴趣爱好占全班百分比情况，应选择条形统计图较好．（）",
+  "options": [],
+  "answer": "错误",
+  "solution": "【详解】如果想了解某班同学的兴趣爱好占全班百分比情况，应选择扇形统计图较好，原题说法错误。故答案为错误.\n\n【分析】根据扇形统计图的特点：扇形统计图能很清楚地表示出各部分数量同总数的关系，据此解答.",
+  "knowledge_points": [
+    "统计图"
+  ],
+  "source": "17-18六年级下·全国·课后作业",
+  "merged_id": "unit7_366",
+  "original_id": "unit7_judge_366"
+},
+{
+  "id": "unit7_correct_367",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "改错题",
+  "difficulty": "中等",
+  "question": "如图所示，王大妈批发的三种蔬菜中，茄子比黄瓜多 $20\\%$ 。这种说法对吗？若不对，请改正。\n\n![](images/01bbf20f4080d564cb17920be477c9c0e3ab4cf6eb62900c4681d4f3cd644995.jpg)",
+  "options": [],
+  "answer": "不对。改正：茄子比黄瓜多 $100\\%$ 。",
+  "solution": "【分析】从图中可以看出：茄子占三种蔬菜的 $40 \\%$ , 黄瓜占三种蔬菜的 $20 \\%$ , 要求茄子比黄瓜多百分之几, 用茄子占的百分比减去黄瓜占的百分比再除以黄瓜占的百分比即可。本题考查求一个数比另一个数多百分之几, 关键是把哪个量看作单位 '1', 单位 '1'作除数。\n\n【详解】 $(40\\% - 20\\%) \\div 20\\%\\\\= 20 \\% \\div 20 \\% \\\\ = 1 \\\\ = 100 \\% $\n\n答: 不对。茄子比黄瓜多 $100 \\%$ .",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "求一个数比另一个数多/少百分之几"
+  ],
+  "source": "16-17六年级·全国·单元测试",
+  "merged_id": "unit7_367",
+  "original_id": "unit7_correct_367"
+},
+{
+  "id": "unit7_solve_368",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "解答题",
+  "difficulty": "中等",
+  "question": "有 2 元和 5 元的人民币共 14 张，共计 43 元，问 2 元和 5 元的各有多少张？",
+  "options": [],
+  "answer": "5元的有5张，2元的有9张。",
+  "solution": "【分析】假设全是面值 5 元的人民币, 则应该是 $5 \\times 14 = 70$ 元, 这比已知的 43 元多出了 $70 - 43 = 27$ 元, 因为 1 张 5 元比 1 张 2 元的人民币多 $5 - 2 = 3$ 元, 由此即可得出面值是 2 元的人民币有 $27 \\div 3 = 9$ 张, 由此即可解答问题。此题属于鸡兔同笼问题，采用假设法即可解答。\n\n【详解】假设全是面值5元的人民币，则2元的人民币有：\n\n$(5 \\times 1 4 - 4 3) \\div (5 - 2) \\\\ = 2 7 \\div 3 \\\\ = 9 (张) $\n\n则 5 元的人民币有: $14 - 9 = 5$ (张)答: 5 元的有 5 张, 2 元的有 9 张。",
+  "knowledge_points": [
+    "鸡兔同笼"
+  ],
+  "source": "2014 六年级·全国·课后作业",
+  "merged_id": "unit7_368",
+  "original_id": "unit7_solve_368"
+},
+{
+  "id": "unit7_solve_369",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "解答题",
+  "difficulty": "中等",
+  "question": "如图是养兔专业户养的黑兔、灰兔、白兔的扇形统计图．如果这个养兔专业户共养兔2500只，算出三种兔各养了多少只？![](images/064af876a6d1a7d030fe4f25187d9a3f50a3f27c73de29d1df5ae068f8c6dfa6.jpg)",
+  "options": [],
+  "answer": "养兔场养白兔 450 只, 养灰兔 750 只, 养黑兔1300只.",
+  "solution": "【分析】根据题意, 把这个养兔专业户所养兔子的全部兔子看作单位 '1', 用养兔场的兔子的总只数分别乘白兔、灰兔、黑兔各占总兔的百分数即可得到养兔的白兔、灰兔、黑兔的各自的只数, 列式解答即可得到答案.\n\n【详解】白兔的只数为： $2500 \\times 18\\% = 450$ （只），灰兔的只数为： $2500 \\times 30\\% = 750$ （只），黑兔的只数为： $2500 \\times 52\\% = 1300$ （只），答：养兔场养白兔 450 只，养灰兔 750 只，养黑兔 1300 只。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "百分数的实际应用题"
+  ],
+  "source": "2015·河北·小升初真题",
+  "merged_id": "unit7_369",
+  "original_id": "unit7_solve_369"
+},
+{
+  "id": "unit7_solve_370",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "解答题",
+  "difficulty": "中等",
+  "question": "下图是养兔专业户养的黑兔、灰兔、白兔的扇形统计图。如果这个养兔专业户共养兔3000只，算出三种兔各养了多少只？![](images/ffd8ea552a6869c2cec6e90e5866d2a73c4d9a2e1eb5451e10ce9061bf7741e4.jpg)",
+  "options": [],
+  "answer": "白兔 540 只, 养灰兔 900 只, 养黑兔 1560 只。",
+  "solution": "【分析】整个圆的大小表示 3000 只兔子, 知道各类兔子所占的百分率, 直接计算即可。扇形统计图中，知道总体和百分率求部分，其实质上是分数乘法问题，总体即为单位一。\n\n【详解】白兔: $3000 \\times 18\\% = 540$ (只) 灰兔: $3000 \\times 30\\% = 900$ （只） 黑兔: $3000 \\times 52\\% = 1560$ （只)\n\n答: 白兔 540 只; 灰兔 900 只; 黑兔 1560 只。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "14-15六年级下·贵州贵阳·期中",
+  "merged_id": "unit7_370",
+  "original_id": "unit7_solve_370"
+},
+{
+  "id": "unit7_solve_371",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "解答题",
+  "difficulty": "中等",
+  "question": "学校图书室科技书、童话书、故事书三种图书本数的统计情况如图.![](images/55ba47f63ac361a2bdc719219ba73c6985b6f129c52ae26e2f0e2ebf95399451.jpg)\n\n(1)童话书和故事书各占总数的百分之几？\n(2)你还能提出什么数学问题？并解答.",
+  "options": [],
+  "answer": "(1) $(500 + 700) \\div (1 - 40\\%) = 2000$ (本) $500 \\div 2000 = 25 \\% 700 \\div 2000 = 35 \\%$答: 童话书占总数的 $25 \\%$ , 故事书占总数的 $35 \\%$ 。\n\n(2)故事书比童话书多百分之几？(所提问题不唯一) (700 - 500)÷500 = 40% 答: 故事书比童话书多 40%.",
+  "solution": "【详解】略",
+  "knowledge_points": [
+    "统计图"
+  ],
+  "source": "16-17 六年级·全国·单元测试",
+  "merged_id": "unit7_371",
+  "original_id": "unit7_solve_371"
+},
+{
+  "id": "unit7_solve_372",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "解答题",
+  "difficulty": "中等",
+  "question": "如图是朝阳蔬菜种植基地蔬菜种植面积的扇形统计图。已知西红柿的种植面积是4公顷，那么萝卜的种植面积是多少公顷？![](images/6b8e7352ecebf5f3fc686d5edc61e52a876d639832c7a9fbd651a9eb4630bcfd.jpg)",
+  "options": [],
+  "answer": "2.4 公顷",
+  "solution": "【分析】根据题意, 把蔬菜种植总面积看作单位 '1', 可用 4 除以 $50 \\%$ 计算出蔬菜种植的总面积, 然后再用总面积乘 $30 \\%$ 即可.\n\n【详解】 $4 \\div 50\\% \\times 30\\%\\\\= 8 \\times 30 \\% \\\\= 2.4 (公顷)$\n\n答: 萝卜的种植面积是 2.4 公顷.",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "统计图表的综合应用",
+    "百分数的实际应用题"
+  ],
+  "source": "17-18 六年级上·贵州黔南·期末",
+  "merged_id": "unit7_372",
+  "original_id": "unit7_solve_372"
+},
+{
+  "id": "unit7_solve_373",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "解答题",
+  "difficulty": "中等",
+  "question": "六年级学生一天的作息时间安排如图.![](images/87c8ba34d38240aafbddf2b7c22a79afcf52156a7a91180a261beb95ec801b32.jpg)\n\n(1) 六年级学生一天中上课时间是多少小时?\n(2) 据医学专家说, 小学生每天的睡眠时间应不少于 9 小时。六年级学生的睡眠时间符合要求吗?\n(3) 六年级学生每天校内外活动时间大约是多少小时?",
+  "options": [],
+  "answer": "6小时；符合；4小时",
+  "solution": "【详解】试题分析：（1）根据百分数乘法的意义，用一天的时间（24小时）乘上课时间所占的百分率就是上课的时间.(2) 同理, 用一天的时间乘睡眠时间所占的百分率就是睡眠时间，再与9小时进行比较，即可得出结论。(3) 同理, 用一天的时间乘学生在校内外活动时间所占的百分率就是在校内外活动时间.解：(1) $24 \\times 25\\% = 6$ （小时）。答: 六年级学生一天中上课时间是 6 小时.(2) $24 \\times 41.5\\% = 9.96$ （小时）9.96 小时 $> 9$ 小时答: 六年级学生的睡眠时间符合要求.(3) $24 \\times 16.7\\% \\approx 4$ （小时）答: 六年级学生每天校内外活动时间大约是 4 小时.\n\n【分析】此题是考查如何从扇形统计图中获取信息，并根据所获取的信息解决有关实际问题。此题主要是百分数乘法的应用方面的计算。",
+  "knowledge_points": [
+    "统计图"
+  ],
+  "source": "16-17 六年级·全国·单元测试",
+  "merged_id": "unit7_373",
+  "original_id": "unit7_solve_373"
+},
+{
+  "id": "unit7_solve_374",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "解答题",
+  "difficulty": "中等",
+  "question": "学校图书室科技书、童话书、故事书三种图书本数的统计情况如图。![](images/db47882ba75746445a4cc08ab1eb4ff2b05a0bc820c5b10a6405b7e641ed88fc.jpg)(1)童话书和故事书各占总数的百分之几？\n(2)你还能提出什么数学问题？并解答.",
+  "options": [],
+  "answer": "(1) $(500 + 700) \\div (1 - 40\\%) = 2000$ (本) $500 \\div 2000 = 25 \\% \\quad 700 \\div 2000 = 35 \\%$答: 童话书占总数的 $25 \\%$ , 故事书占总数的 $35 \\%$ 。\n\n(2)故事书比童话书多百分之几？(所提问题不唯一) (700 - 500)÷500 = 40% 答: 故事书比童话书多 40%.",
+  "solution": "【详解】略",
+  "knowledge_points": [
+    "统计图"
+  ],
+  "source": "17-18六年级上·全国·单元测试",
+  "merged_id": "unit7_374",
+  "original_id": "unit7_solve_374"
+},
+{
+  "id": "unit7_solve_375",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "解答题",
+  "difficulty": "中等",
+  "question": "如图是某蔬菜基地去年蔬菜种植情况示意图：![](images/be7c4ee68a660c08bcd1d90bf3cb187d4652da1cc9b29fbdc626e321f66aed73.jpg)\n\n(1) 已知白菜的种植面积是 30 公顷, 三种蔬菜去年的种植总面积是多少公顷?\n(2) 今年土豆的种植面积比去年减少了 $40 \\%$ , 今年土豆的种植面积是多少公顷?",
+  "options": [],
+  "answer": "（1）200公顷 (2）30公顷",
+  "solution": "【分析】（1）把蔬菜基地的面积看作单位'1'，已知白菜的种植面积是30公顷，占蔬菜基地面积的 $(1 - 60\\% - 25\\%)$ ，根据已知一个数的百分之几是多少求这个数，用除法解答。(2) 首先求出去年种土豆的面积: $200 \\times 25 \\%$ , 把去年种土豆的面积看作单位 '1', 今年土豆的种植面积是去年的 $1 - 40 \\%$ , 然后根据一个数乘百分数的意义, 用乘法解答。\n\n【详解】 (1) $30 \\div (1 - 60\\% - 25\\%)\\\\= 30 \\div 15 \\% \\\\ = 3 0 \\div 0. 1 5 \\\\ = 2 0 0 (公 顷) $\n\n答：三种蔬菜的种植总面积是 200 公顷\n\n(2) $200 \\times 25\\% \\times (1 - 40\\%)\\\\=2 0 0 \\times 0. 2 5 \\times 0. 6 \\\\ = 5 0 \\times 0. 6 \\\\ = 3 0 (公 顷) $\n\n答: 今年土豆的种植面积是 30 公顷。",
+  "knowledge_points": [
+    "比一个数多/少百分之几的数是多少",
+    "扇形统计图的特点及绘制",
+    "已知一个数的百分之几是多少，求这个数"
+  ],
+  "source": "16-17 六年级·全国·期末",
+  "merged_id": "unit7_375",
+  "original_id": "unit7_solve_375"
+},
+{
+  "id": "unit7_solve_376",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "解答题",
+  "difficulty": "中等",
+  "question": "聪聪家 2015 年 11月支出情况统计如图。聪聪家 2015 年 11 月的总支出是3600元。请你回答问题：![](images/a813ecd74a971c823009a126044a32d707e92530e08c7bcabd4d6f48576f5477.jpg)\n\n(1) 这个月哪项支出最多? 支出了多少元?\n(2) 购买衣物的支出比文化教育支出少百分之几? 少支出了多少元?",
+  "options": [],
+  "answer": "(1) 1260 元(2) 180 元",
+  "solution": "【分析】（1）由扇形统计图可以看出伙食支出最多，用总支出 3600 元乘伙食支出所占的百分率即可求出支出了多少元。(2) 用购买衣物比文化教育支出少的百分率（或钱数）除以文化教育所占的百分率（或钱数）；用总支出分别乘文化教育、购买衣物所占的百分率即可求出文化教育和购买衣物各支出了多少元，进而求出购买衣物的支出比文化教育支出少多少元。从扇形统计图中得出关键数据并进行计算是解决本题的关键所在。\n\n【详解】 (1) $3600 \\times 35\\% = 1260$ （元）答: 这个月伙食支出最多, 支出了 1260 元。\n\n(2) $(25\\% - 20\\%) \\div 25\\%\\\\=0.0 5 \\div 0. 2 5 \\\\ = 0. 2 \\\\ = 20 \\% $\n\n答：购买衣物的支出比文化教育支出少 $20 \\%$ 。\n\n$3600 \\times 25 \\% \\\\ = 3 6 0 0 \\times 0. 2 5 \\\\ = 9 0 0 (元) $\n\n$3600 \\times 20 \\% \\\\ = 3 6 0 0 \\times 0. 2 \\\\= 7 2 0 (元) $\n\n$ 9 0 0 - 7 2 0 = 1 8 0 (元) $\n\n答：少支出了180元。",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "17-18 六年级上·山东泰安·期中",
+  "merged_id": "unit7_376",
+  "original_id": "unit7_solve_376"
+},
+{
+  "id": "unit7_solve_377",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "解答题",
+  "difficulty": "中等",
+  "question": "![](images/f0503549c53e9e783a4266fdb4ebd9152a60af9c88d444b6439a54e742e837fa.jpg)上图是聪聪家十月份生活支出情况统计图.\n\n(1) 这是____统计图，从图中你知道了什么？\n(2) 如果聪聪家这个月的支出是 1600 元, 请你分别计算出各项支出的钱数.\n(3) 你还能提出什么问题?",
+  "options": [],
+  "answer": "(1) 扇形;从图中可知：食品支出占总支出的 $36 \\%$ , 赡养老人占总支出的 $16 \\%$ , 水电气占总支出的 $10 \\%$ ，文化占总支出的 $20 \\%$ ，服装占总支出的 $10 \\%$ ，其它占总支出的 $8 \\%$ 。(2) 食品支出 576 元，赡养老人支出 256 元，水电气支出的 160 元，文化支出 320 元，服装支出 160 元，其它支出 128 元。(3) 食品支出比文化支出多百分之几? $16 \\%$ （答案不唯一）",
+  "solution": "【分析】（1）根据统计图的形式知这是扇形统计图，从图中可以看出各种支出所占的百分比.(2) 总支出就是单位 '1', 用乘法可以算出各项的支出. (3) 根据图提出问题，并解答.\n\n【详解】（1）这是扇形统计图.故答案为扇形.从图中可知:食品支出占总支出的 $36 \\%$ , 赡养老人占总支出的 $16 \\%$ , 水电气占总支出的 $10 \\%$ , 文化占总支出的 $20 \\%$ , 服装占总支出的 $10 \\%$ , 其它占总支出的 $8 \\%$ 。(2) 食品: $1600 \\times 36\\% = 576$ （元）；赡养老人： $1600 \\times 16\\% = 256$ （元）；水电气: $1600 \\times 10\\% = 160$ （元）；\n\n文化： $1600 \\times 20\\% = 320$ （元）；服装： $1600 \\times 10\\% = 160$ （元）；其它: $1600 \\times 8 \\% = 128$ （元）.\n\n答: 食品支出 576 元, 赡养老人支出 256 元, 水电气支出的160元，文化支出320元，服装支出160元，其它支出128元。(3) 问题: 食品支出比文化支出多百分之几?$36 \\% - 20 \\% = 16 \\%$答: 食品支出比文化支出多 $16 \\%$ .",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "统计图表的综合应用"
+  ],
+  "source": "2014 六年级·全国·课后作业",
+  "merged_id": "unit7_377",
+  "original_id": "unit7_solve_377"
+},
+{
+  "id": "unit7_solve_378",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "解答题",
+  "difficulty": "中等",
+  "question": "下图是六(1)班同学最喜欢的运动项目统计图.\n\n![](images/a9c1ebc32e4e9c4515405a11dfacfc4063d09e5f8ed618bc9444cdc40b6ce10d.jpg)\n\n(1)从这个扇形统计图中，你知道了什么？\n(2)如果六(1)班有 60 人, 分别计算出最喜欢乒乓球、足球和跳绳的人数.",
+  "options": [],
+  "answer": "(1)最喜欢的运动项目中喜欢乒乓球的人数最多. (答案不唯一)   (2) 乒乓球: 18人 足球: 12人 跳绳: 9人",
+  "solution": "【详解】(2)乒乓球： $60 \\times 30\\% = 18$ （人） 足球: $60 \\times 20\\% = 12$ (人) 跳绳: $60 \\times 15\\% = 9$ (人)",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制",
+    "统计图表的综合应用"
+  ],
+  "source": "17-18 六年级上·吉林·单元测试",
+  "merged_id": "unit7_378",
+  "original_id": "unit7_solve_378"
+},
+{
+  "id": "unit7_solve_379",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "解答题",
+  "difficulty": "中等",
+  "question": "右图是某果园果树种植情况的统计图。请你算出一共有多少棵果树，其中桃树有多少棵？\n\n![](images/624f25c18bfd5ee760a339704a10da97c137303e689ae8a7af181efcf1bf6f30.jpg)",
+  "options": [],
+  "answer": "$450 \\div (1 - 45\\% - 30\\% - 10\\%) = 3000$ (棵)   $3000 \\times 30 \\% = 900 (棵)$  答: 一共有 3000 棵果树, 其中桃树有 900 棵.",
+  "solution": "【详解】略",
+  "knowledge_points": [
+    "统计图"
+  ],
+  "source": "17-18六年级上·全国·单元测试",
+  "merged_id": "unit7_379",
+  "original_id": "unit7_solve_379"
+},
+{
+  "id": "unit7_solve_380",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "解答题",
+  "difficulty": "中等",
+  "question": "新乡小学六年级有350人，视力情况如下图。(1)近视、假性近视和视力正常的各有多少人？![](images/c7b9a5da333e040bad3c2fcd0f371c162a1b1ecd3160c8150869bbfdbc8e9358.jpg)\n\n(2)视力不良(包括假性近视和近视)的人数比视力正常的人数多百分之几？",
+  "options": [],
+  "answer": "(1)近视： $350 \\times 28\\% = 98$ (人)   假性近视: $350 \\times (1 - 42\\% - 28\\%) = 105$ (人)  视力正常: $350 \\times 42\\% = 147$ (人)答：近视的有 98 人，假性近视的有 105 人，视力正常的有 147 人.\n\n(2)(98+105-147)÷147≈38.1%   答：视力不良(包括假性近视和近视)的人数比视力正常的人数大约多 $38.1\\%$ .",
+  "solution": "【详解】略",
+  "knowledge_points": [
+    "统计图"
+  ],
+  "source": "17-18六年级上·全国·单元测试",
+  "merged_id": "unit7_380",
+  "original_id": "unit7_solve_380"
+},
+{
+  "id": "unit7_solve_381",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "解答题",
+  "difficulty": "中等",
+  "question": "观察如图的扇形统计图，并填写。\n\n![](images/354bf5aa591dcce5266d29e70fae8fa207487ec6b22c4caa8652428ec7a2d2b6.jpg)\n\n(1) 如果用这个圆代表总体, 那么扇形_表示总体的 $45 \\%$ .\n(2) 如果用整个圆代表 9 公顷的稻田, 那扇形 A 大约代表（  ）公顷。\n(3) 如果用整个圆代有某校全体学生的人数, 已知扇形 B 比扇形 A 多 $5 \\%$ , 且多 60 人, 全校_人。",
+  "options": [],
+  "answer": "C; 2.25; 1200",
+  "solution": "【分析】（1）根据题意，把整个圆看作单位'1'，扇形A部分占单位'1'的 $\\frac{1}{4}$ ，可用 $100\\%$ 乘 $\\frac{1}{4}$ 即可得到扇形A占单位1的百分之几，再用单位'1'分别减去扇形A、B占单位'1'的百分数就是扇形C占单位'1'的百分数，列式解答即可得到答案；(2) 可直接用 9 成扇形 A 所占总数的百分数即可得到答案; (3) 可直接用多的 60 人除以扇形 B 比扇形 A 多的 $5 \\%$ 即可得到全校的人数, 列式解答即可得到答案。\n\n【详解】（1）扇形A占总数的： $100\\% \\times \\frac{1}{4} = 25\\%$ 扇形C占总数的： $1 - 25\\% -30\\% = 45\\%$ (2) $9 \\times 25\\% = 2.25$ （公顷）   (3) $60 \\div 5\\% = 1200$ （人），解答此题的关键是找准单位 '1', 然后再根据题意列式计算即可。",
+  "knowledge_points": [
+    "求一个数的百分之几是多少",
+    "已知一个数的百分之几是多少，求这个数"
+  ],
+  "source": "2014 六年级·全国·课后作业",
+  "merged_id": "unit7_381",
+  "original_id": "unit7_solve_381"
+},
+{
+  "id": "unit7_solve_382",
+  "unit": "第七单元：扇形统计图",
+  "chapter": "扇形统计图",
+  "question_type": "解答题",
+  "difficulty": "中等",
+  "question": "根据图回答下列问题:\n\n(1) 这个统计图叫做____统计图，可以看出它有一个明显的特点，能清楚地在图上表示出____和____之间的关系。\n(2) 本月饮食预算为 1200 元, 则总预算是______元, 用在购买衣服与文化教育的钱比用在饮食上的钱少______元.\n(3) 若本月的总预算增加 200 元, 那么饮食的经费增加____元.\n\n![](images/1923ee263ca4077b73b28ae5687528519605c98fecc8a449a41a2e87bcf7ed43.jpg)",
+  "options": [],
+  "answer": "扇形，部分，整体；2400，240；100.",
+  "solution": "【分析】（1）这是一个扇形统计图，扇形统计图可以清楚地表示出部分同整体之间的关系。(2) 把总钱数看成单位 '1', 饮食占 $50 \\%$ , 它对应的数量是 1200 元，由此用除法求出总钱数；先求出购买衣服与文化教育的钱比用在饮食上的钱少百分之几；然后用总钱数乘这个百分数就是少花的钱数。(3) 把增加的预算看成单位 '1', 饮食占其中的 $50 \\%$ , 由此用乘法求出饮食增加的钱数.本题先读图, 找出单位 '1', 以及各个数量, 然后根据题目要求和基本的数量关系找出合适的数据列式求解.\n\n【详解】（1）这是一个扇形统计图，它可以清楚地表示出部分同整体之间的关系。(2) $1200 \\div 50\\% = 2400$ （元）；$2400 \\times (50\\% - 25\\% - 15\\%) , \\\\ = 2400 \\times 10 \\% \\\\ = 2 4 0 (元) ; $\n\n答: 总预算是 2400 元, 用在购买衣服与文化教育的钱比用在饮食上的钱少 240 元.\n\n(3) $200 \\times 50\\% = 100$ （元）；答: 饮食的经费增加 100 元.",
+  "knowledge_points": [
+    "扇形统计图的特点及绘制"
+  ],
+  "source": "2014 六年级·全国·课后作业",
+  "merged_id": "unit7_382",
+  "original_id": "unit7_solve_382"
+}  ,
+
+{
+"id": "unit7_solve_383",
+"unit": "第七单元：扇形统计图",
+"chapter": "扇形统计图",
+"question_type": "解答题",
+"difficulty": "中等",
+"question": "今年'五一'黄金周李华一家三口去旅游，旅游各种费用如图。\n\n(1) 路费和食宿费占旅游总费用的几分之几?\n(2) 购物占旅游总费用的几分之几?\n\n![](images/8b76e608e3c6c503a3096ca0432870163a97cff59e8db09538601958c101f533.jpg)",
+"options": [],
+"answer": "(1) $\\frac{7}{10}$ ; (2) $\\frac{3}{10}$",
+"solution": "【分析】（1）由扇形统计图可以看出：路费占旅游总费用的 $\\frac{9}{20}$ ，食宿费占旅游总费用的 $\\frac{1}{4}$ ，二者相加就是路费和食宿费占旅游总费用的几分之几。本题是考查如何从扇形统计图中获取信息，再对这些信息进行整理、分析、计算等。\n\n(2) 把旅游各种费用看作单位 '1', 减去路费和食宿费占旅游总费用的几分之几就是购物占旅游总费用的几分之几。\n\n【详解】 (1) $\\frac{9}{20} + \\frac{1}{4}\\\\= \\frac {9}{2 0} + \\frac {5}{2 0} \\\\ = \\frac {7}{1 0} $\n\n答: 路费和食宿费占旅游总费用的 $\\frac{7}{10}$ ;\n\n(2) $1 - \\frac{7}{10} = \\frac{3}{10}$ 答: 购物占旅游总费用的 $\\frac{3}{10}$ 。",
+"knowledge_points": [
+  "扇形统计图的特点及绘制"
+],
+"source": "2014 六年级·全国·课后作业",
+"merged_id": "unit7_383",
+"original_id": "unit7_solve_383"
+},
+{
+"id": "unit7_solve_384",
+"unit": "第七单元：扇形统计图",
+"chapter": "扇形统计图",
+"question_type": "解答题",
+"difficulty": "中等",
+"question": "小敏家 1 月份生活支出情况如下面的统计图。\n\n(1) 从这个扇形统计图中, 你了解了什么?\n(2) 如果小敏家这个月的支出是 2000 元, 分别计算出各项支出的钱数。\n\n![](images/636069ba61ac5b49727091518dfa3dbd29c71d7c991d5fba9cbe4e9f69b91756.jpg)",
+"options": [],
+"answer": "（1）从统计图中，获得的信息为：小敏家 1 月份生活支出的情况为伙食占 $35 \\%$ , 文化教育占 $20 \\%$ , 赡养老人占 $15 \\%$ , 购买衣物占 $20 \\%$ , 水电费占 $10 \\%$ ;(2)伙食支出700元，文化教育支出400元，赡养老人支出300元，购买衣物支出400元，水电费支出200元。本题主要考查了从统计图中获取信息，提出问题，并能够根据基本的数量关系解决问题。",
+"solution": "【分析】（1）根据统计图知道：小敏家1月份生活支出的情况为，伙食占 $35\\%$ ，文化教育占 $20\\%$ ，赡养老人占 $15\\%$ ，购买衣物占 $20\\%$ ，水电费占 $10\\%$ ；(2) 把全部的支出看作单位 '1', 已知一个数, 求这个数的百分之几是多少, 根据分数乘法的意义列式解答即可。\n\n【详解】（1）从统计图中，获得的信息为：小敏家1月份生活支出的情况为伙食占 $35\\%$ ，文化教育占 $20\\%$ ，赡养老人占 $15\\%$ ，购买衣物占 $20\\%$ ，水电费占 $10\\%$ ；\n\n(2)伙食: $2000 \\times 35\\% = 700$ (元),文化教育： $2000 \\times 20\\% = 400$ （元），赡养老人： $2000 \\times 15\\% = 300$ （元），购买衣物: $2000 \\times 20\\% = 400$ （元），水电费: $2000 \\times 10\\% = 200$ （元），\n\n答: 伙食支出 700 元, 文化教育支出 400 元, 赡养老人支出 300 元, 购买衣物支出 400 元, 水电费支出 200 元。",
+"knowledge_points": [
+  "扇形统计图的特点及绘制",
+  "求一个数的百分之几是多少"
+],
+"source": "2014 六年级·全国·课后作业",
+"merged_id": "unit7_384",
+"original_id": "unit7_solve_384"
+},
+{
+"id": "unit7_solve_385",
+"unit": "第七单元：扇形统计图",
+"chapter": "扇形统计图",
+"question_type": "解答题",
+"difficulty": "中等",
+"question": "下面是某小学六年级学生视力情况统计图.① 近视人数占全年级人数的____%，视力不良（包括假性近视和近视人数）的人数占全年级人数的____%。② 视力正常的有76人，视力不良的有____人，六年级共有（  ）人. ③面对这个学校的六年级学生的视力状况，你有什么想法和好的建议吗？\n\n![](images/0f6ba045eb66a448bff019bf4955004c327bfac1776bb3d4f244aacfbd6f244e.jpg)",
+"options": [],
+"answer": "①30，62；②124，200。③同学们的视力太差，应注意保护眼睛，爱护眼睛。",
+"solution": "【分析】①把全年级的总人数看成单位'1'，先用总人数减去假性近视的百分率，再减去正常人数占的百分率就是近视人数占总人数的百分率；再用近视人数占的百分率加上假性近视人数占的百分率就是视力不良人数占的百分率。② 视力正常的人数占总人数的 $38 \\%$ , 它对应的数量是 76 人,由此用除法求出总人数; 再用总人数乘视力不良人数占的百分率, 就是视力不良的人数有多少人; ③从保护视力方面回答.\n\n【详解】① $1 - 32\\% - 38\\% = 30\\%$ ; $32 \\% + 30 \\% = 62 \\%;$答：近视人数占全年级人数的 $30 \\%$ , 视力不良 (包括假性近视和近视人数) 的人数占全年级人数的 $62 \\%$ 。② $76 \\div 38 \\% = 200$ （人）；$200 \\times 62\\% = 124$ （人）；答: 视力正常的有 76 人, 视力不良的有 124 人, 六年级共有 200 人.③ 同学们的视力太差，应注意保护眼睛，爱护眼睛。",
+"knowledge_points": [
+  "扇形统计图的特点及绘制",
+  "百分数的四则运算",
+  "统计图表的综合应用"
+],
+"source": "2014 六年级·全国·课后作业",
+"merged_id": "unit7_385",
+"original_id": "unit7_solve_385"
+},
+{
+"id": "unit7_solve_386",
+"unit": "第七单元：扇形统计图",
+"chapter": "扇形统计图",
+"question_type": "解答题",
+"difficulty": "中等",
+"question": "光明小学图书馆中各类图书情况如图所示。\n\n![](images/a76aaefe31751730cddbe7bdf4b1a29b052cfa7d314d232f019923af82d3bada.jpg)\n\n(1) 光明小学图书馆中哪类图书最多?\n(2) 若该图书馆共有图书 3500 本, 那么自然科学类的图书有多少本?\n(3) 其他图书有多少本?",
+"options": [],
+"answer": "(1) 文艺类图书;  (2) 875 本;   (3) 350本",
+"solution": "【分析】（1）扇形统计图中扇形面积最大的所表示的哪类图书最多。(2) 用该图书馆图书总数乘自然科学类图书所占的百分率就是自然科学类的图书的本数。(3) 根据各类图书所占的百分率求出其他图书所占的百分率, 用用该图书馆图书总数乘其他图书所占的百分率就是其他图书的本数。本题是考查如何根据计算需要，从扇形统计图中获取有用信息，然后进行有关计算。\n\n【详解】（1）从扇形统计图中可以看出，文艺类图书占的面积最大，此类图书最多。\n\n(2) $3500 \\times 25\\%\\\\= 3 5 0 0 \\times 0. 2 5 \\\\ = 8 7 5 (本) $\n\n答: 自然科学类的图书有 875 本。\n\n(3) $3500 \\times (1 - 20\\% - 25\\% - 45\\%)\\\\=3 5 0 0 \\times 0. 1 \\\\ = 3 5 0 (本) $\n\n答: 其他图书有 350 本。",
+"knowledge_points": [
+  "扇形统计图的特点及绘制",
+  "求一个数的百分之几是多少"
+],
+"source": "2014 六年级·全国·课后作业",
+"merged_id": "unit7_386",
+"original_id": "unit7_solve_386"
+},
+{
+"id": "unit7_solve_387",
+"unit": "第七单元：扇形统计图",
+"chapter": "扇形统计图",
+"question_type": "解答题",
+"difficulty": "中等",
+"question": "观察如图, 再解决问题.\n\n![](images/90d24a17362ebc28b8adf3a3b56e2b17ac1acff55e6184170a3a70238a571288.jpg)\n\n(1) 这是____统计图.\n(2) 图中 A、B、C 三部分的比是____.\n(3) 这是小明一家三口 '五一' 节旅游的各种费用统计图。其中 A 表示食宿费用，B 表示路费，C 表示购物费用。已知食宿费用是 2000 元，路费是多少元？",
+"options": [],
+"answer": "(1) 扇形  (2) $5: 9: 6$   (3) 3600元",
+"solution": "【分析】（1）根据统计图的特征，可知这是一幅扇形统计图；(2) 把总费用看作单位 '1', 根据食宿费用占的圆心角是 $90^{\\circ}$ , 可知食宿费用 (A) 占 $25 \\%$ , 购物费用 (C) 占 $30 \\%$ ,求出以上两种共占的百分率; 剩下的是路费 (B) 占的分率,用减法求出即可; 进而把 A、B、C 三部分占的分率相比, 再化简成最简比; (3) 根据食宿费用是 2000 元, 占总费用的 $25 \\%$ , 用 2000 除以对应分率 $25 \\%$ 即可求出总费用, 再用总费用乘路费占的分率即可求出路费.本题关键是根据扇形统计图找出单位 '1', 以及各部分所占的百分率, 再根据数量关系求解.\n\n【详解】（1）这是扇形统计图；(2) 食宿费用占的圆心角是 $90^{\\circ}$ , 可知食宿费用 (A) 占 $25 \\%$ ,\n\n路费（B）占的分率：1- $(25\\% + 30\\%) = 45\\%$A、B、C三部分的比： $25\\%:45\\%:30\\%=5:9:6;$\n\n(3) 总费用: $2000 \\div 25\\% = 8000$ (元),路费: $8000 \\times 45\\% = 3600$ （元）;答: 路费是 3600 元.",
+"knowledge_points": [
+  "扇形统计图的特点及绘制",
+  "统计图表的综合应用"
+],
+"source": "2014 六年级·全国·课后作业",
+"merged_id": "unit7_387",
+"original_id": "unit7_solve_387"
+},
+{
+"id": "unit7_solve_388",
+"unit": "第七单元：扇形统计图",
+"chapter": "扇形统计图",
+"question_type": "解答题",
+"difficulty": "中等",
+"question": "如图是某服装公司下属两个厂的产品销售情况统计图。小明看了两幅统计图后，说：'A厂的服装出口量一定比B厂的多'。你认为他说得对吗？说说理由。\n\n![](images/b88c5e4c7400ba314185c52d8270690484855fd2d522e247c6bacccd2169f540.jpg)\n\n![](images/8602e6fa132e2d5d7dd927b6f9d31af9c17deb1990d46adeddf9af3ecc945d8c.jpg)",
+"options": [],
+"answer": "不对；理由见详解",
+"solution": "【分析】判断哪个厂的服装出口量多，不能只看出口量所占产量的百分比，关键要找到正确的单位'1'，就是两个厂各自的总销售量，题中两个厂的总销售量不知道，也就是单位'1'可能不一样，出口量就不一定相同，据此解答即可。\n\n【详解】答: 'A 厂的服装出口量一定比 B 厂的多'。这种说法不对, 因为 A 厂和 B 厂的单位 '1' 不一定相同, 也就是销售量不一定相同, 因此, A 厂服装出口量的百分率与 B 厂服装出口量的百分率无法比较。",
+"knowledge_points": [
+  "扇形统计图的特点及绘制"
+],
+"source": "2014 六年级·全国·课后作业",
+"merged_id": "unit7_388",
+"original_id": "unit7_solve_388"
+},
+{
+"id": "unit7_solve_389",
+"unit": "第七单元：扇形统计图",
+"chapter": "扇形统计图",
+"question_type": "解答题",
+"difficulty": "中等",
+"question": "下面是明明一天的活动情况统计图。算出明明各种活动占用的时间。\n\n![](images/7117eff01aa096df73bfdf9c51fe5f4c0f4c70b5bf4637ed3b1fd34205a92714.jpg)",
+"options": [],
+"answer": "睡觉 9.84 时, 在校学习 7.92 时, 写作业 1.92 时, 玩 3.36 时,其他 0.96 时。",
+"solution": "【分析】一天共有 24 小时, 用一天的小时数分别乘各种活动占总时间的百分率即可求出各种时间所用的时间。\n\n【详解】睡觉： $24 \\times 41\\% = 9.84$ （时）在校学习: $24 \\times 33\\% = 7.92$ (时)写作业: $24 \\times 8\\% = 1.92$ (时)\n\n玩: $24 \\times 14\\% = 3.36$ (时)其他: $24 \\times 4\\% = 0.96$ (时)答: 睡觉 9.84 时, 在校学习 7.92 时, 写作业 1.92 时, 玩 3.36 时, 其他 0.96 时。",
+"knowledge_points": [
+  "统计图"
+],
+"source": "17-18 六年级下·全国·课后作业",
+"merged_id": "unit7_389",
+"original_id": "unit7_solve_389"
+},
+{
+"id": "unit7_solve_390",
+"unit": "第七单元：扇形统计图",
+"chapter": "扇形统计图",
+"question_type": "解答题",
+"difficulty": "中等",
+"question": "六年级同学参加兴趣小组情况统计图.\n\n![](images/8ac4a246924fd142d0862918d797a754c10254ef33de73e9147eccc170ec0a37.jpg)\n\n如果全年级共有 200 人，各种兴趣小组的同学分别有多少人？",
+"options": [],
+"answer": "解：科技组： $200 \\times 35\\% = 70$ （人）器乐组： $200 \\times 25\\% = 50$ (人) 绘画组： $200 \\times 30\\% = 60$ (人)\n\n其他小组： $200 \\times 10\\% = 20$ (人) 答: 科技小组有 70 人, 器乐小组有 50 人, 绘画小组有 60人，其他小组有 20 人。",
+"solution": "【详解】用总人数分别乘各种小组人数占总人数的百分率即可分别求出各种小组的人数.",
+"knowledge_points": [
+  "统计图"
+],
+"source": "17-18六年级下·全国·课后作业",
+"merged_id": "unit7_390",
+"original_id": "unit7_solve_390"
+},
+{
+"id": "unit7_solve_391",
+"unit": "第七单元：扇形统计图",
+"chapter": "扇形统计图",
+"question_type": "解答题",
+"difficulty": "中等",
+"question": "根据下图回答问题.\n\n![](images/7cc421facea2e10098d4877944afa781bd5bd6da36bdfb1f0791ff5c519b6e0a.jpg)\n\n(1) 如果用整个圆表示总体, 那么哪一个扇形表示总体的 $25\\%$ ?\n(2) 如果用整个圆代表我校的总人数(3000)，那么扇形B大约代表多少人？\n(3) 如果用整个圆代表 9 公顷的稻田, 那么扇形 B 大约代表多少公顷的稻田?",
+"options": [],
+"answer": "(1) A (2) 1000 人 (3) 扇形 B 大约代表 3 公顷的稻田",
+"solution": "【详解】（1）解：A 扇形表示总体的 $25\\%$(2) 解: $3000 \\times 33.3\\% \\approx 1000$ (人)\n\n答: 扇形 B 大约代表 1000 人.(3) 解: $9 \\times 33.3\\% = 3$ (公顷)答: 扇形 B 大约代表 3 公顷的稻田.\n\n扇形统计图中的数据列出百分数的乘法算式进行解答.",
+"knowledge_points": [
+  "统计图"
+],
+"source": "18-19六年级上·全国·单元测试",
+"merged_id": "unit7_391",
+"original_id": "unit7_solve_391"
+},
+{
+"id": "unit7_solve_392",
+"unit": "第七单元：扇形统计图",
+"chapter": "扇形统计图",
+"question_type": "解答题",
+"difficulty": "困难",
+"question": "某超市销售A、B、C三种饮料。十二月份三种饮料的销售量占总销售量的百分比如图。若 A 种饮料销售了 1200 瓶, 那么 C 种饮料比 A 种饮料少销售了多少瓶？\n\n![](images/c266fd9da9a4cfb96c83c57baba6f311320d9b154d16abf200c46ec43aa6ade2.jpg)",
+"options": [],
+"answer": "900 瓶",
+"solution": "【详解】解： $1200 - 1200 \\div 60\\% \\times 15\\% \\\\= 1200 - 2000 \\times 15\\%\\\\=1 2 0 0 - 3 0 0 \\\\ = 9 0 0 (瓶) $\n\n答: C 种饮料比 A 种饮料少销售了 900 瓶.把总销售量看作单位'1'，用A种饮料销售量除以它所占的百分比得出总销售量，用总销售量乘 C 种饮料所占的分率，即可求出 C 种饮料销售了多少瓶，再用 A 种饮料减去C 种饮料销售的瓶数即可.",
+"knowledge_points": [
+  "扇形统计图的特点及绘制"
+],
+"source": "18-19 六年级上·全国·单元测试",
+"merged_id": "unit7_392",
+"original_id": "unit7_solve_392"
+},
+{
+"id": "unit7_solve_393",
+"unit": "第七单元：扇形统计图",
+"chapter": "扇形统计图",
+"question_type": "解答题",
+"difficulty": "中等",
+"question": "下面是某学校小学学生人数统计图。已知低年级学生有230人，请解答下面问题：![](images/e400d0fc60e53cdde7801fc52aac95fc3c80a85bae7f92ac1679220b61745119.jpg)\n\n(1) 低年级人数占全校人数的百分之几？\n(2) 全校共有学生多少人？\n(3) 中年级比高年级少多少人？",
+"options": [],
+"answer": "(1) $25\\%$ ; (2) 920人 (3) 46人",
+"solution": "【分析】（1）把全校的学生看作单位'1'，用1减去高年级和中年级占全校的百分数求出低年级人数占全校人数的百分之几；（2）用低年级学生的人数除以低年级人数占全校人数的百分数求出全校的人数；（3）分别求出中年级与高年级的人数，再相减即可。此题主要考查的是如何观察扇形统计图并且从统计图中获取信息，然后再进行计算、解答即可。\n\n【详解】 (1) 解: $1 - 40\\% - 35\\% = 25\\%$ ,答：低年级人数占全校人数的 $25\\%$(2) 解: $230 \\div 25\\% = 920$ (人),\n\n答: 全校共有学生 920 人(3) 解: $920 \\times 40\\% - 920 \\times 35\\%$ ,$ 920 \\times 5 \\% = 4 6 (人), $答: 中年级比高年级少 46 人",
+"knowledge_points": [
+  "百分数应用题",
+  "统计表"
+],
+"source": "18-19 六年级上·全国·单元测试",
+"merged_id": "unit7_393",
+"original_id": "unit7_solve_393"
+},
+{
+"id": "unit7_solve_394",
+"unit": "第七单元：扇形统计图",
+"chapter": "扇形统计图",
+"question_type": "解答题",
+"difficulty": "中等",
+"question": "下面是幸福小学兴趣班人数的统计图，已知该小学总人数为800人，请问美术班的人数比音乐班的人数少多少人？![](images/cbe48d988ae89464cfbfe882e38aa7b52c3f3c0257cc6fce7c89256ead92b12a.jpg)",
+"options": [],
+"answer": "160人",
+"solution": "【详解】解： $800 \\times (35\\% - 15\\%) = 160$ （人）答: 美术班的人数比音乐班的人数少 160 人.已知总人数和各兴趣班占全部人数的百分比，用总人数 $\\times$ 各部分兴趣班占的百分比=各兴趣班的人数，然后用减法可以求出美术班的人数比音乐班的人数少几人，也可以用总人数×音乐班比美术班多的百分比=美术班的人数比音乐班的人数少的人数，据此解答。",
+"knowledge_points": [
+  "扇形统计图的特点及绘制"
+],
+"source": "18-19六年级上·全国·单元测试",
+"merged_id": "unit7_394",
+"original_id": "unit7_solve_394"
+},
+{
+"id": "unit7_solve_395",
+"unit": "第七单元：扇形统计图",
+"chapter": "扇形统计图",
+"question_type": "解答题",
+"difficulty": "困难",
+"question": "小军家 2012 年 11月支出情况统计如下图。小军家2012年11月的总支出是3600 元. 请你回答问题。\n\n![](images/85bb35c68a2ce3bd86655cad5d66adbe7f57a27837aafa729c7b4a96dd8c813d.jpg)\n\n(1) 这个月哪项支出最多? 支出了多少元?\n(2) 文化教育支出了多少元？购买衣物支出了多少元？\n(3) 购买衣物的支出比文化教育支出少百分之几?\n(4) 你还能提出什么问题? 并解决你所提出的问题?",
+"options": [],
+"answer": "（1）伙食支出最多，支出了 1260 元 (2) 900 元、720 元    (3) $20\\%$    (4) 文化教育支出比购买衣物支出多支出多少元? 180 元",
+"solution": "【详解】 (1) $3600 \\times 35\\% = 1260$ （元）\n\n答: 这个月伙食支出最多, 支出了 1260 元\n\n(2)\n\n$3600 \\times 25 \\% \\\\ = 3 6 0 0 \\times 0. 2 5 \\\\ = 9 0 0 (元) $\n\n$3600 \\times 20 \\%\\\\= 3 6 0 0 \\times 0. 2 \\\\ = 7 2 0 (元) $\n\n答：文化教育支出了900元，购买衣物支出了720元。\n\n(3) $(25\\% - 20\\%) \\div 25\\%\\\\=0.0 5 \\div 0. 2 5 \\\\ = 0. 2 \\\\ = 20 \\% $\n\n答: 购买衣物的支出比文化教育支出少 $20 \\%$ 。\n\n(4) 文化教育支出比购买衣物支出多支出多少元?\n\n$9 0 0 - 7 2 0 = 1 8 0 (元)$  答: 少支出了 180 元。",
+"knowledge_points": [
+  "扇形统计图的特点及绘制",
+  "统计图表的综合应用"
+],
+"source": "16-17六年级·山东青岛·课后作业",
+"merged_id": "unit7_395",
+"original_id": "unit7_solve_395"
+},
+{
+"id": "unit7_solve_396",
+"unit": "第七单元：扇形统计图",
+"chapter": "扇形统计图",
+"question_type": "解答题",
+"difficulty": "中等",
+"question": "下图是某中学一个班的学生平均每天上网时间情况统计图。\n\n![](images/a9a7e365c120d54e2eb68b1672f948735909648082489f14f580f58cf526f024.jpg)\n\n(1) 平均每天上网 1 小时以上的学生占本班学生人数的（）%；平均每天上网 1 小时以内（包括 1 小时）的学生占本班学生人数的（）%；不上网的学生占本班学生人数的（）%。\n(2) 若该班不上网的学生有 18 人, 则该班平均每天上网 1 小时以上的学生有多少人?",
+"options": [],
+"answer": "(1) $20; 50; 30$ ; (2) 12人",
+"solution": "【分析】（1）根据扇形统计图的信息，填空即可，最后一空，不上网的学生占本班学生人数的百分率 $= 1 -$ 上网 1 小时以上学说所占百分率一上网 1 小时以内学生所占百分率；\n(2) 不上网学生人数 $\\div$ 不上网学生所占百分率 $=$ 这个班的总人数, 总人数 $\\times$ 每天上网 1 小时以上学生所占百分率即可。\n\n【详解】 (1) $1 - 20\\% - 50\\% = 30\\%$平均每天上网 1 小时以上的学生占本班学生人数的（20）$\\%$ ；平均每天上网 1 小时以内（包括 1 小时）的学生占本班学生人数的（50）%；不上网的学生占本班学生人数的（30)。此题考查了扇形统计图的实际应用，根据问题在扇形统计图中寻找有效的数学信息解答即可。\n\n(2) $18 \\div 30\\% \\times 20\\%\\\\= 60 \\times 20 \\%\\\\= 1 2 (人)$\n\n答: 该班平均每天上网 1 小时以上的学生有 12 人。",
+"knowledge_points": [
+  "扇形统计图的特点及绘制"
+],
+"source": "16-17六年级·全国·单元测试",
+"merged_id": "unit7_396",
+"original_id": "unit7_solve_396"
+},
+{
+"id": "unit7_solve_397",
+"unit": "第七单元：扇形统计图",
+"chapter": "扇形统计图",
+"question_type": "解答题",
+"difficulty": "中等",
+"question": "学校图书室科技书、童话书、故事书三种图书本数的统计情况如图.\n\n![](images/870a436ef83bda47156540e05bf3c0030b36808aecb0b13576cca0cceaa4acf9.jpg)\n\n(1) 童话书和故事书各占总数的百分之几?\n(2) 你还能提出什么数学问题? 并解答.",
+"options": [],
+"answer": "(1) $35\\%$ (2) 答案不唯一",
+"solution": "【详解】(1)(500+700)÷(1-40%)=2000(本) 500÷2000=25% $700 \\div 2000 = 35\\%$(2)故事书比童话书多百分之几？(所提问题不唯一) (700 - 500)÷500 = 40% 答: 故事书比童话书多 40%.",
+"knowledge_points": [
+  "统计图"
+],
+"source": "18-19六年级上·全国·单元测试",
+"merged_id": "unit7_397",
+"original_id": "unit7_solve_397"
+},
+{
+"id": "unit7_solve_398",
+"unit": "第七单元：扇形统计图",
+"chapter": "扇形统计图",
+"question_type": "解答题",
+"difficulty": "困难",
+"question": "下图是某村今年植树的情况统计图，你知道杨树的棵数比柳树少百分之几吗？(百分号前保留一位小数)\n\n![](images/827ea3fb51a5a97303b944b00e755434.png)",
+"options": [],
+"answer": "$(30\\% - 20\\%) \\div 30\\% \\approx 33.3\\%$",
+"solution": "【详解】略",
+"knowledge_points": [
+  "扇形统计图的特点及绘制"
+],
+"source": "16-17六年级·全国·单元测试",
+"merged_id": "unit7_398",
+"original_id": "unit7_solve_398"
+},
+{
+"id": "unit7_solve_399",
+"unit": "第七单元：扇形统计图",
+"chapter": "扇形统计图",
+"question_type": "解答题",
+"difficulty": "中等",
+"question": "右图是某果园果树种植情况的统计图。请你算出一共有多少棵果树，其中桃树有多少棵？\n\n![](images/95fb0183632c9566ac4083f0567387b895a4d796b93fb650947a2221c9e00494.jpg)",
+"options": [],
+"answer": "3000棵 900棵",
+"solution": "【详解】 $450 \\div (1 - 45\\% - 30\\% - 10\\%) = 3000$ （棵） $3000 \\times 30\\% = 900$ （棵）答：一共有 3000 棵果树，其中桃树有 900 棵.",
+"knowledge_points": [
+  "统计图"
+],
+"source": "18-19六年级上·全国·单元测试",
+"merged_id": "unit7_399",
+"original_id": "unit7_solve_399"
+},
+{
+"id": "unit7_solve_400",
+"unit": "第七单元：扇形统计图",
+"chapter": "扇形统计图",
+"question_type": "解答题",
+"difficulty": "中等",
+"question": "新乡小学六年级有 350 人, 视力情况如下图.\n\n![](images/8b2619456da4f2d2c36947e3d35d2055f2fbe2c3ebc0f5a76a950aa6bd1dbb67.jpg)\n\n(1)近视、假性近视和视力正常的各有多少人？\n(2)视力不良(包括假性近视和近视)的人数比视力正常的人数多百分之几？",
+"options": [],
+"answer": "(1)98人 105人 147人 (2） $38.1\\%$",
+"solution": "【详解】(1)近视： $350 \\times 28\\% = 98$ （人）假性近视： $350 \\times (1 - 42\\% - 28\\%) = 105$ (人)视力正常： $350 \\times 42\\% = 147$ (人) (2)(98+105-147)÷147≈38.1%",
+"knowledge_points": [
+  "统计"
+],
+"source": "18-19 六年级上·全国·单元测试",
+"merged_id": "unit7_400",
+"original_id": "unit7_solve_400"
+}
+]
\ No newline at end of file
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new file mode 100644
index 0000000000000000000000000000000000000000..a6af65d0756bcfbfaa91125baa5415b6493cc658
--- /dev/null
+++ "b/\345\205\255\345\271\264\347\272\247\351\242\230\351\233\206/\345\205\255\345\271\264\347\272\247\344\270\212/\345\205\255\345\271\264\347\272\247\344\270\212\347\254\254\344\270\211\345\215\225\345\205\203\345\210\206\346\225\260\351\231\244\346\263\225.json"
@@ -0,0 +1,9673 @@
+[
+    {
+      "id": "unit3_choice_001",
+      "unit": "第三单元：分数除法",
+      "chapter": "倒数的认识",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "认识了倒数，六（1）班卓越组学生开展数学小讲堂活动。下面说法正确的是（）。",
+      "options": [
+        "A. $\\frac{5}{3}$ 的倒数是 $\\frac{5}{2}$",
+        "B. 一个假分数的倒数小于或等于1",
+        "C. $3$ 和 $\\frac{1}{3}$ 都是倒数",
+        "D. 一个分数的倒数都比这个分数大"
+      ],
+      "answer": "B",
+      "solution": "【分析】乘积是 1 的两个数互为倒数。求一个分数的倒数,把分子和分母调换位置即可。真分数的分子比分母小, 真分数小于 1 ; 它的倒数的分子就比分母大了, 所以真分数的倒数大于 1。所以真分数的倒数就大于本身了。假分数两种情况: ①这个假分数的分子和分母相等, 假分数等于 1 , 这个假分数的倒数也等于 1 , 所以倒数等于本身; ②这个假分数的分子大于分母, 假分数大于 1 , 这个假分数的倒数就小于 1 , 倒数小于本身。据此解答。\n\n【详解】A. $\\frac{5}{3}$ 的倒数是 $\\frac{3}{5}$, 原题干说法错误;B. 一个假分数的倒数小于或等于 1, 原题干说法正确; C. $3 \\text{和} \\frac{1}{3}$ 都是倒数这种说法不正确, 正确说法是 $3 \\text{和} \\frac{1}{3}$ 互为倒数, $3 \\text{是} \\frac{1}{3}$ 的倒数, $\\frac{1}{3}$ 是 3 的倒数; 原题干说法错误; D. 一个分数的倒数可能大于这个分数, 也可能小于这个分数, 也可能等于这个分数。原题干说法错误。故答案为: B",
+      "knowledge_points": [
+        "倒数的认识",
+        "真分数、假分数、带分数的认识"
+      ],
+      "source": "24-25 六年级上·全国·单元测试",
+      "merged_id": "unit3_001",
+      "original_id": "unit3_choice_001"
+    },
+    {
+      "id": "unit3_choice_002",
+      "unit": "第三单元：分数除法",
+      "chapter": "倒数的认识",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "下列商品中，两个单价互为倒数的是（）。",
+      "options": [
+        "A. 中性笔 3 元和橡皮 0.3 元",
+        "B. 彩纸 0.25 元和钢笔 4元",
+        "C. 尺子 0.5 元和铅笔 0.2 元",
+        "D. 彩铅 $\\frac{7}{6}$ 元和画笔 $\\frac{1}{6}$ 元"
+      ],
+      "answer": "B",
+      "solution": "【分析】要判断两个单价是否互为倒数，根据倒数的定义，若两个数的乘积为1，则这两个数互为倒数。所以我们需要分别计算每个选项中两个商品单价的乘积，看是否等于1。按照这个思路逐一分析选项，据此解答。\n\n【详解】A．中性笔单价3元，橡皮单价0.3元，它们的乘积为 $3 \\times 0.3 = 0.9 \\neq 1$，所以不互为倒数。B. 彩纸单价 0.25 元, 钢笔单价 4 元, 它们的乘积为 $0.25 \\times 4 = 1$, 所以互为倒数。C. 尺子单价 0.5 元, 铅笔单价 0.2 元, 它们的乘积为 $0.5 \\times 0.2 = 0.1 \\neq 1$, 所以不互为倒数。D. 彩铅单价 $\\frac{7}{6}$ 元, 画笔单价 ${ }^{1} \\frac{1}{6}$ 元 $= \\frac{7}{6}$ 元, 它们的乘积为 $\\frac{7}{6} \\times \\frac{7}{6} = \\frac{49}{36} \\neq 1$, 所以不互为倒数。故答案为: B",
+      "knowledge_points": [
+        "倒数的认识"
+      ],
+      "source": "24-25 六年级上·全国·单元测试",
+      "merged_id": "unit3_002",
+      "original_id": "unit3_choice_002"
+    },
+    {
+      "id": "unit3_choice_003",
+      "unit": "第三单元：分数除法",
+      "chapter": "倒数的认识",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "体育老师发现，小华和小丽完成跳绳训练的次数倒数和是 $\\frac{5}{6}$，请问小华和小丽完成跳绳训练的次数是（）。",
+      "options": [
+        "A. 1 次和 5 次",
+        "B. 2次和4次",
+        "C. 2次和3次",
+        "D. 5次和6次"
+      ],
+      "answer": "C",
+      "solution": "【分析】需要根据倒数的定义，分别计算每个选项中两个数的倒数之和，看是否等于 $\\frac{5}{6}$。倒数的定义是：若两个数的乘积为 1，则这两个数互为倒数。按照这个思路逐一分析选项，据此解答。\n\n【详解】A．1的倒数是1，5的倒数是 $\\frac{1}{5}$，它们的倒数和是 $1 + \\frac{1}{5} = \\frac{6}{5} \\neq \\frac{5}{6}$，所以该选项错误。B. 2 的倒数是 $\\frac{1}{2}$, 4 的倒数是 $\\frac{1}{4}$, 它们的倒数和是 $\\frac{1}{2} + \\frac{1}{4} = \\frac{3}{4} \\neq \\frac{5}{6}$, 所以该选项错误。C. 2 的倒数是 $\\frac{1}{2}$, 3 的倒数是 $\\frac{1}{3}$, 它们的倒数和是 $\\frac{1}{2} + \\frac{1}{3} = \\frac{5}{6}$,所以该选项正确。D. 5 的倒数是 $\\frac{1}{5}$, 6 的倒数是 $\\frac{1}{6}$, 它们的倒数和是 $\\frac{1}{5} + \\frac{1}{6} = \\frac{11}{30} \\neq \\frac{5}{6}$, 所以该选项错误。故答案为: C",
+      "knowledge_points": [
+        "倒数的认识"
+      ],
+      "source": "24-25 六年级上·全国·单元测试",
+      "merged_id": "unit3_003",
+      "original_id": "unit3_choice_003"
+    },
+    {
+      "id": "unit3_choice_004",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "如果数 A 与数 B 互为倒数，那么 $\\frac{5}{A} \\div \\frac{B}{2} =$ （）。",
+      "options": [
+        "A. $\\frac{2}{5}$",
+        "B. $\\frac{5}{2}$",
+        "C. 10"
+      ],
+      "answer": "C",
+      "solution": "【分析】已知数 A 与数 B 互为倒数，根据倒数的定义，可得 $A \\times B = 1$。要求 $\\frac{5}{A} \\div \\frac{B}{2}$ 的结果，需要将除法转化为乘法，再利用 $A \\times B = 1$ 进行计算，据此解答。\n\n【详解】因为 A 和 B 互为倒数，所以 $A \\times B = 1$。将除法 $\\frac{5}{A} \\div \\frac{B}{2}$ 转化为乘法: $\\frac{5}{A} \\times \\frac{2}{B}$。根据分数乘法法则，分子相乘作为积的分子，分母相乘作为积的分母，得到 $\\frac{5 \\times 2}{A \\times B} = \\frac{10}{A \\times B}$。把 $A \\times B = 1$ 代入上式，可得 $\\frac{10}{1} = 10$。故答案为: C",
+      "knowledge_points": [
+        "分数与分数的除法"
+      ],
+      "source": "25-26 六年级上·海南海口·单元测试",
+      "merged_id": "unit3_004",
+      "original_id": "unit3_choice_004"
+    },
+    {
+      "id": "unit3_choice_005",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "一瓶农药的 $\\frac{4}{5}$ 正好是 $\\frac{2}{5} \\mathrm{~L}$，这瓶农药有（） $\\mathrm{L}$。",
+      "options": [
+        "A. $\\frac{1}{2}$",
+        "B. 2",
+        "C. $\\frac{8}{25}$"
+      ],
+      "answer": "A",
+      "solution": "【分析】已知一瓶农药的 $\\frac{4}{5}$ 是 $\\frac{2}{5} \\mathrm{~L}$，要求这瓶农药的总量，根据'已知一个数的几分之几是多少，求这个数用除法'，用 $\\frac{2}{5}$ 除以 $\\frac{4}{5}$ 计算解答即可。\\n\\n【详解】$\\frac{2}{5} \\\\div \\\\frac{4}{5}$= $\\frac {2}{5} \\\\times \\\\frac {5}{4}$ $= \\frac {1}{2} \\\\quad (\\\\mathrm {L})$所以这瓶农药有 $\\frac{1}{2} \\mathrm{~L}$。故答案为：A",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "已知一个数的几分之几是多少，求这个数"
+      ],
+      "source": "25-26 六年级上·海南海口·单元测试",
+      "merged_id": "unit3_005",
+      "original_id": "unit3_choice_005"
+    },
+    {
+      "id": "unit3_choice_006",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "a 是一个非 0 自然数，下面各式中，（）的得数最大。",
+      "options": [
+        "A. $a \\div \\frac{1}{5}$",
+        "B. $a \\times \\frac{1}{5}$",
+        "C. $a \\div 1 \\frac{1}{5}$"
+      ],
+      "answer": "A",
+      "solution": "【分析】将选项中的除法算式转化为乘法算式，然后比较乘法算式中另一个因数的大小，因为 a 是一个非 0 自然数，根据'一个非 0 自然数乘一个大于 1 的数，积大于这个数；乘一个小于 1 的数，积小于这个数'来判断得数的大小，据此解答。\\n\\n【详解】A. $a \\\\div \\\\frac{1}{5} = a \\\\times 5, 5 > 1$ B. $a \\\\times \\\\frac{1}{5}, \\\\frac{1}{5} < 1$ C. $a \\\\div 1\\\\frac{1}{5} = a \\\\div \\\\frac{6}{5} = a \\\\times \\\\frac{5}{6}, \\\\frac{5}{6}< 1$因为 $5 > \\\\frac{5}{6} > \\\\frac{1}{5}$, 所以 $a \\\\times 5$ 的得数最大, 也就是 $a \\\\div \\\\frac{1}{5}$ 的得数最大。故答案为: A",
+      "knowledge_points": [
+        "因数和积的大小关系（分数乘法）",
+        "整数乘分数",
+        "分数与整数的除法"
+      ],
+      "source": "25-26 六年级上·海南海口·单元测试",
+      "merged_id": "unit3_006",
+      "original_id": "unit3_choice_006"
+    },
+    {
+      "id": "unit3_choice_007",
+      "unit": "第三单元：分数除法",
+      "chapter": "倒数的认识",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "真分数的倒数（）。",
+      "options": [
+        "A. 小于它本身",
+        "B. 大于它本身",
+        "C. 无法确定"
+      ],
+      "answer": "B",
+      "solution": "【分析】真分数是指分子小于分母的分数，其值小于 1。倒数是指乘积为 1 的两个数互为倒数。据此分析解答。\n\n【详解】真分数小于 1，那么它的倒数就是 1 除以这个真分数，结果会大于 1，而真分数本身小于 1，所以真分数的倒数大于它本身。故答案为: B",
+      "knowledge_points": [
+        "倒数的认识",
+        "真分数、假分数、带分数的认识"
+      ],
+      "source": "25-26 六年级上·海南海口·单元测试",
+      "merged_id": "unit3_007",
+      "original_id": "unit3_choice_007"
+    },
+    {
+      "id": "unit3_choice_008",
+      "unit": "第三单元：分数除法",
+      "chapter": "倒数的认识",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "下面说法正确的是（）。",
+      "options": [
+        "A. 1 的倒数是 1，0 的倒数是 0",
+        "B. 3 和 $\\frac{1}{3}$ 都是倒数",
+        "C. 6 和 $\\frac{1}{6}$ 互为倒数"
+      ],
+      "answer": "C",
+      "solution": "【分析】倒数的定义：乘积为 1 的两个数互为倒数。0 没有倒数，因为 0 乘任何数都为 0，不能得到 1。倒数是对两个数而言，是两个数相互依存的一个概念，不能单独说某个数是倒数。据此解答。\n\n【详解】A．1的倒数是1，但0没有倒数，所以选项错误。B. $3 \\times \\frac{1}{3} = 1$,3和 $\\frac{1}{3}$ 互为倒数，不能单独说都是倒数，所以选项错误。C. 因为 $6 \\times \\frac{1}{6} = 1$, 所以 6 和 $\\frac{1}{6}$ 互为倒数, 该选项正确。故答案为: C",
+      "knowledge_points": [
+        "倒数的认识",
+        "整数乘分数"
+      ],
+      "source": "25-26 六年级上·海南海口·单元测试",
+      "merged_id": "unit3_008",
+      "original_id": "unit3_choice_008"
+    },
+    {
+      "id": "unit3_choice_009",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "苗苗在公园里散步。她以 $2 \\mathrm{~km} / \\mathrm{h}$ 的速度走了总时长的一半，以 $3 \\mathrm{~km} / \\mathrm{h}$ 的速度走了总路程的一半，其余时间内的步行速度为 $4 \\mathrm{~km} / \\mathrm{h}$。苗苗以 $4 \\mathrm{~km} / \\mathrm{h}$ 的速度步行的时长占总时长的（）。",
+      "options": [
+        "A. $\\frac{1}{14}$",
+        "B. $\\frac{1}{12}$",
+        "C. $\\frac{1}{7}$",
+        "D. $\\frac{1}{5}$"
+      ],
+      "answer": "A",
+      "solution": "【分析】分析题目，可以假设总路程是 1，总时间是 t，根据路程 = 速度 × 时间，时间 = 路程 ÷ 速度分别求出速度是$2 \\mathrm{~km} / \\mathrm{h}$ 的路程和速度是 $3 \\mathrm{~km} / \\mathrm{h}$ 的时间，进而求出速度是 $4 \\mathrm{~km} / \\mathrm{h}$ 的路程和时间，再根据速度 × 时间 = 路程列出方程进一步求出 t 和速度是 $4 \\mathrm{~km} / \\mathrm{h}$ 的时长，最后用除法求出以 $4 \\mathrm{~km} / \\mathrm{h}$ 的速度步行的时长占总时长的几分之几即可。\n\n【详解】假设总路程是 1，总时长是 $t$。速度是 $2 \\mathrm{~km} / \\mathrm{h}$ 的路程: $2 \\times \\frac{1}{2} t = t$速度是 $3 \\mathrm{~km} / \\mathrm{h}$ 的时间: $1 \\times \\frac {1}{2} \\div 3 = \\frac {1}{2} \\times \\frac {1}{3} = \\frac {1}{6}$速度是 $4 \\mathrm{~km} / \\mathrm{h}$ 的路程: $1 - \\frac{1}{2} - t = \\frac{1}{2} - t$速度是 $4 \\mathrm{~km} / \\mathrm{h}$ 的时间: $t - \\frac{1}{2} t - \\frac{1}{6} = \\frac{1}{2} t - \\frac{1}{6}$ $4 \\times \\left(\\frac {1}{2} t - \\frac {1}{6}\\right) = \\frac {1}{2} - t$ $2 t - \\frac {2}{3} = \\frac {1}{2} - t$ $ 3 t = \\frac {1}{2} + \\frac {2}{3}$ $3 t = \\frac {7}{6}$ $ t = \\frac {7}{1 8}$ $\\frac {1}{2} t - \\frac {1}{6} = \\frac {1}{2} \\times \\frac {7}{1 8} - \\frac {1}{6} = \\frac {7}{3 6} - \\frac {6}{3 6} = \\frac {1}{3 6}$ $\\frac {1}{3 6} \\div \\frac {7}{1 8} = \\frac {1}{3 6} \\times \\frac {1 8}{7} = \\frac {1}{1 4}$故答案为: A",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "基础行程问题"
+      ],
+      "source": "24-25 六年级上·重庆渝中·期末",
+      "merged_id": "unit3_009",
+      "original_id": "unit3_choice_009"
+    },
+    {
+      "id": "unit3_choice_010",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数四则运算",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "下面算式中，计算结果最大的是（）。",
+      "options": [
+        "A. $\\frac{3}{5} + \\frac{1}{2}$",
+        "B. $\\frac{3}{5} -\\frac{1}{2}$",
+        "C. $\\frac{3}{5} \\times \\frac{1}{2}$",
+        "D. $\\frac{3}{5} \\div \\frac{1}{2}$"
+      ],
+      "answer": "D",
+      "solution": "【分析】一个不为 0 的数加上一个大于 0 的数，和大于这个数；减去一个大于 0 的数，差小于这个数；乘一个小于 1 的数，积就小于这个数；除以一个小于 1 的数，商就大于这个数。可知，$\\frac{3}{5} + \\frac{1}{2}$ 和 $\\frac{3}{5} \\div \\frac{1}{2}$ 结果都大于 $\\frac{3}{5}$，再根据分数加法和分数除法的方法进行计算，再比较结果的大小。\n\n【详解】A. 因为 $\\frac{1}{2} > 0$, 所以 $\\frac{3}{5} + \\frac{1}{2} > \\frac{3}{5}$;B. 因为 $\\frac{1}{2} > 0$, 所以 $\\frac{3}{5} - \\frac{1}{2} < \\frac{3}{5}$; C. 因为 $\\frac{1}{2} > 0$, 所以 $\\frac{3}{5} \\times \\frac{1}{2} < \\frac{3}{5}$; D. 因为 $\\frac{1}{2} > 0$, 所以 $\\frac{3}{5} \\div \\frac{1}{2} > \\frac{3}{5}$。$\\frac {3}{5} \\div \\frac {1}{2} = \\frac {3}{5} \\times 2 = \\frac {6}{5}$ $\\frac {6}{5} = \\frac {1 2}{1 0} > \\frac {1 1}{1 0}$故答案为: D",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "因数和积的大小关系（分数乘法）",
+        "异分母分数加、减法",
+        "被除数与商的大小关系（分数除法）"
+      ],
+      "source": "24-25 六年级上·河南三门峡·期末",
+      "merged_id": "unit3_010",
+      "original_id": "unit3_choice_010"
+    },
+    {
+      "id": "unit3_choice_011",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "我国有悠久的青铜器铸造历史，古籍《考工记》中就有关于冶炼青铜工具时铜锡配比的记载，在铸造青铜剑时，铜的含量是锡的 $\\frac{2}{5}$，如果铸造一把青铜剑，其中需要的锡、铜总量为 560 克，那么锡的含量是多少？如果把锡的含量设为 x 克，正确的方程是（）。",
+      "options": [
+        "A. $x + \\frac{2}{5} x = 560$",
+        "B. $x - \\frac{2}{5} x = 560$",
+        "C. $5 \\mathrm{x} + 2 \\mathrm{x} =560$",
+        "D. $5 \\mathrm{x} - 2 \\mathrm{x} = 560$"
+      ],
+      "answer": "A",
+      "solution": "【分析】将锡的含量看作单位'1'，锡的含量 $\\times$ 铜的对应分率 = 铜的含量，把锡的含量设为 $x$ 克，则铜的含量为 $\\frac{2}{5} x$，根据锡的含量 + 铜的含量 = 锡、铜总量，列出方程即可。\n\n【详解】解：设锡的含量是 $\\mathrm{x}$ 克。$x + \\frac {2}{5} x = 5 6 0$ $\\frac {7}{5} x = 5 6 0$ $ \\frac {7}{5} x \\div \\frac {7}{5} = 5 6 0 \\div \\frac {7}{5}$ x = $5 6 0 \\times \\frac {5}{7}$ x = 4 0 0 锡的含量是 400 克。正确的方程是 $x + \\frac{2}{5} x = 560$。故答案为：A",
+      "knowledge_points": [
+        "列方程解含两个未知数的问题"
+      ],
+      "source": "24-25 六年级上·河南郑州·期末",
+      "merged_id": "unit3_011",
+      "original_id": "unit3_choice_011"
+    },
+    {
+      "id": "unit3_choice_012",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数四则运算",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "下面三个算式中，计算结果最大的是（）。",
+      "options": [
+        "A. $\\frac{3}{4} + \\frac{2}{5}$",
+        "B. $\\frac{3}{4} \\times \\frac{2}{5}$",
+        "C. $\\frac{3}{4} \\div \\frac{2}{5}$"
+      ],
+      "answer": "C",
+      "solution": "【分析】异分母分数加减法的计算方法：先通分，再按照同分母分数加减法的计算法则进行计算。分数乘分数：用两个分数的分子相乘的积作分子，用两个分数的分母相乘的积作分母，在计算的过程中，如果有可以约分的数，可以先约分。分数除以分数：除以一个不为 0 数相当于乘这个数的倒数,据此先把除法化为乘法，再根据分数乘法的计算方法进行计算即可。先计算出每个选项，再比较。\n\n【详解】A. $\\frac{3}{4} + \\frac{2}{5}$ $ \\frac {1 5}{2 0} + \\frac {8}{2 0} = \\frac {2 3}{2 0}$ B. $\\frac{3}{4} \\times \\frac{2}{5} = \\frac{3}{10}$C. $\\frac{3}{4} \\div \\frac{2}{5}$ $\\frac {3}{4} \\times \\frac {5}{2} = \\frac {1 5}{8}$ $\\frac {2 3}{2 0} = 1. 1 5$ $\\frac {3}{1 0} = 0. 3$ $\\frac {1 5}{8} = 1. 8 7 5$ $1. 8 7 5 > 1. 1 5 > 0. 3$ $\\frac {1 5}{8} > \\frac {2 3}{2 0} > \\frac {3}{1 0}$ 计算结果最大的是 $\\frac{3}{4} \\div \\frac{2}{5}$。故答案为: C",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "分数乘分数",
+        "异分母分数加、减法"
+      ],
+      "source": "24-25 六年级上·重庆万州·期末",
+      "merged_id": "unit3_012",
+      "original_id": "unit3_choice_012"
+    },
+    {
+      "id": "unit3_choice_013",
+      "unit": "第三单元：分数除法",
+      "chapter": "倒数的认识",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "（）与 0.75 互为倒数。",
+      "options": [
+        "A. $\\frac{1}{75}$",
+        "B. 75",
+        "C. $\\frac{4}{3}$",
+        "D. $\\frac{3}{4}$"
+      ],
+      "answer": "C",
+      "solution": "【分析】乘积是 1 的两个数互为倒数，求一个非 0 数的倒数 $= 1 \\div$ 这个数。\n\n【详解】$1 \\div 0.75 = 1 \\div \\frac{3}{4} = 1 \\times \\frac{4}{3} = \\frac{4}{3}$ 所以 $\\frac{4}{3}$ 与 0.75 互为倒数。故答案为: C",
+      "knowledge_points": [
+        "倒数的认识",
+        "分数与整数的除法"
+      ],
+      "source": "24-25 六年级上·广东汕头·期末",
+      "merged_id": "unit3_013",
+      "original_id": "unit3_choice_013"
+    },
+    {
+      "id": "unit3_choice_014",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "已知 $x$、$y$ 互为倒数，下面计算结果等于 $\\frac{1}{5}$ 的是（）。",
+      "options": [
+        "A. $\\frac{x}{5} \\div \\frac{1}{y}$",
+        "B. $x \\times \\frac{5}{y}$",
+        "C. $\\frac{x}{5} \\div \\frac{y}{5}$",
+        "D. $\\frac{1}{x} \\div \\frac{y}{5}$"
+      ],
+      "answer": "A",
+      "solution": "【分析】若两个数的乘积是 1，我们就称这两个数互为倒数,则 $x y = 1$，逐项化简算式，再判断。\n\n【详解】A．已知 $x y = 1$，$\\frac{x}{5} \\div \\frac{1}{y} = \\frac{x}{5} \\times y = \\frac{x y}{5} = \\frac{1}{5}$，符合题意。B. $x \\times \\frac{5}{y} = \\frac{5 x}{y} \\neq \\frac{1}{5}$, 不符合题意。C. $\\frac{x}{5} \\div \\frac{y}{5} = \\frac{x}{5} \\times \\frac{5}{y} = \\frac{x}{y} \\neq \\frac{1}{5}$, 不符合题意。D. 已知 $x y = 1, \\frac{1}{x} \\div \\frac{y}{5} = \\frac{1}{x} \\times \\frac{5}{y} = \\frac{5}{x y} = 5 \\neq \\frac{1}{5}$, 不符合题意。故答案为: A",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "倒数的认识",
+        "用字母表示数",
+        "数量关系",
+        "含有字母式子的化简与求值"
+      ],
+      "source": "24-25 六年级上·河北沧州·期末",
+      "merged_id": "unit3_014",
+      "original_id": "unit3_choice_014"
+    },
+    {
+      "id": "unit3_choice_015",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "小明看一本科技书,每一天看了 $\\frac{3}{8}$，第二天看了剩下的 $\\frac{3}{5}$，第三天把剩下的 40 页看完了，这本书一共有多少页？正确的列式是（）。",
+      "options": [
+        "A. $40 \\div \\left(1 - \\frac{3}{8} - \\frac{3}{5}\\right)$",
+        "B. $40 \\times \\left(1 - \\frac{3}{8} - \\frac{3}{5}\\right)$",
+        "C. $40 \\div \\left(1 - \\frac{3}{8} - \\frac{3}{8} \\times \\frac{3}{5}\\right)$",
+        "D. $40 \\div \\left[1 - \\frac{3}{8} - \\left(1 - \\frac{3}{8}\\right) \\times \\frac{3}{5}\\right)$"
+      ],
+      "answer": "D",
+      "solution": "【分析】将这本书看作单位'1'，用单位'1'减去第一天看的分率，求出剩下的分率。将剩下的分率乘 $\\frac{3}{5}$，求出第二天看了全书的几分之几。将单位'1'减去第一天和第二天看的分率，求出剩下的 40 页是全书的几分之几。单位'1'未知,将 40 页除以对应的分率，即可求出这本书一共有多少页。据此列式，从而选出正确选项。\n\n【详解】求这本书一共有多少页，正确的列式是$4 0 \\div \\left[ 1 - \\frac {3}{8} - \\left(1 - \\frac {3}{8}\\right) \\times \\frac {3}{5} \\right]。$A. $40 \\div \\left(1 - \\frac{3}{8} - \\frac{3}{5}\\right)$, 列式错误; B. $40 \\times \\left(1 - \\frac{3}{8} - \\frac{3}{5}\\right)$, 列式错误; C. $40 \\div \\left(1 - \\frac{3}{8} - \\frac{3}{8} \\times \\frac{3}{5}\\right)$, 列式错误; D. $40 \\div \\left[1 - \\frac{3}{8} -\\left(1 - \\frac{3}{8}\\right) \\times \\frac{3}{5}\\right)$, 列式正确。故答案为: D",
+      "knowledge_points": [
+        "分数的四则混合运算",
+        "已知一部分量占总量的几分之几及另一部分量，求总量"
+      ],
+      "source": "24-25 六年级上·山东临沂·期中",
+      "merged_id": "unit3_015",
+      "original_id": "unit3_choice_015"
+    },
+    {
+      "id": "unit3_choice_016",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "$A \\times \\frac{12}{13} = B \\div \\frac{12}{13} = C \\times 1$(A、B、C 均大于 0)，把 $\\mathrm{A}$、$\\mathrm{~B}$、$\\mathrm{C}$ 按从小到大排列是（）。",
+      "options": [
+        "A. A>B>C",
+        "B. B>A>C",
+        "C. C>A>B",
+        "D. A>C>B"
+      ],
+      "answer": "D",
+      "solution": "【分析】假设 $A \\times \\frac{12}{13} = B \\div \\frac{12}{13} = C \\times 1 = 1$，分别求出 A、B、C 的结果,再比较大小即可。\n\n【详解】假设 $\\mathrm{A} \\times \\frac{12}{13} = \\mathrm{B} \\div \\frac{12}{13} = \\mathrm{C} \\times 1 = 1$，则：$ A = 1 \\div \\frac {1 2}{1 3} = 1 \\times \\frac {1 3}{1 2} = \\frac {1 3}{1 2}$ ${B} = 1 \\times \\frac {1 2}{1 3} = \\frac {1 2}{1 3}$ $A = 1 \\div \\frac {1 2}{1 3} = 1 \\times \\frac {1 3}{1 2} = \\frac {1 3}{1 2}$ $C = 1 \\div 1 = 1$因为 $\\frac{13}{12} > 1 > \\frac{12}{13}$所以 $\\mathrm{A} > \\mathrm{C} > \\mathrm{B}$故答案为: D",
+      "knowledge_points": [
+        "异分母异分子分数的大小比较",
+        "分数乘整数",
+        "整数乘分数",
+        "分数与整数的除法"
+      ],
+      "source": "24-25 六年级上·山东临沂·期中",
+      "merged_id": "unit3_016",
+      "original_id": "unit3_choice_016"
+    },
+    {
+      "id": "unit3_choice_017",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数四则运算",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "$14 \\times \\frac{5}{9} \\div 14 \\times \\frac{5}{9} =$ （）。",
+      "options": [
+        "A. 1",
+        "B. $1 \\frac{1}{9}$",
+        "C. $\\frac{25}{81}$",
+        "D. $\\frac{81}{25}$"
+      ],
+      "answer": "C",
+      "solution": "【分析】根据乘法交换律将原式变成 $14 \\div 14 \\times \\frac{5}{9} \\times \\frac{5}{9}$，再根据乘法结合律，将算式变成 $(14 \\div 14) \\times \\left(\\frac{5}{9} \\times \\frac{5}{9}\\right)$ 即可求出结果。\n\n【详解】$14 \\times \\frac{5}{9} \\div 14 \\times \\frac{5}{9}$ $1 4 \\div 1 4 \\times \\frac {5}{9} \\times \\frac {5}{9} = (1 4 \\div 1 4) \\times \\left(\\frac {5}{9} \\times \\frac {5}{9}\\right) = 1 \\times \\frac {2 5}{8 1} = \\frac {2 5}{8 1}$故答案为: C",
+      "knowledge_points": [
+        "整数乘法运算定律推广到分数乘法",
+        "分数的乘、除法的混合运算"
+      ],
+      "source": "23-24 六年级上·北京昌平·期中",
+      "merged_id": "unit3_017",
+      "original_id": "unit3_choice_017"
+    },
+    {
+      "id": "unit3_choice_018",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "读书是一种生活方式，它关乎人的心灵。为进一步打造'书香校园'，希望小学举办了校园第四届'读书节'活动。参加的女生人数是男生人数的 $\\frac{3}{4}$，已知女生共36人，男生有（）人。",
+      "options": [
+        "A. 24",
+        "B. 48",
+        "C. 27",
+        "D. 42"
+      ],
+      "answer": "B",
+      "solution": "【分析】参加的女生人数是男生人数的 $\\frac{3}{4}$，已知女生共 36 人，求男生人数，根据已知一个数的几分之几是多少，求这个数，用除法计算，即用 $36 \\div \\frac{3}{4}$ 即可解答。\n\n【详解】$36 \\div \\frac{3}{4}$ $= 3 6 \\times \\frac {4}{3} = 4 8 \\text {(人)}$所以，男生有 48 人。故答案为: B",
+      "knowledge_points": [
+        "已知一个数的几分之几是多少，求这个数"
+      ],
+      "source": "24-25六年级上·全国·单元测试",
+      "merged_id": "unit3_018",
+      "original_id": "unit3_choice_018"
+    },
+    {
+      "id": "unit3_choice_019",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "小马虎在计算一个数除以 $\\frac{3}{8}$ 时，看成了乘 $\\frac{3}{8}$，结果得到 $\\frac{9}{10}$，小马虎计算的那道题算式的正确结果应该是（）。",
+      "options": [
+        "A.$\\frac{9}{10} \\div \\frac{3}{8} \\times \\frac{3}{8}$",
+        "B.$\\frac{9}{10} \\div \\frac{3}{8} \\div \\frac{3}{8}$",
+        "C.$\\frac{9}{10} \\times \\frac{3}{8} \\times \\frac{3}{8}$",
+        "D.$\\frac{3}{8} \\times \\frac{3}{8}\\div \\frac{9}{10}$"
+      ],
+      "answer": "B",
+      "solution": "【分析】根据题意可知：将一个数除以 $\\frac{3}{8}$ 时，看成了乘 $\\frac{3}{8}$，结果得到 $\\frac{9}{10}$，即一个数 $\\times \\frac{3}{8} = \\frac{9}{10}$，根据一个因数 $=$ 积分另一个因数，用 $\\frac{9}{10} \\div \\frac{3}{8}$ 求出这个数；再用这个数除以 $\\frac{3}{8}$ 即可得正确结果。\n\n【详解】$\\frac{9}{10} \\div \\frac{3}{8} \\div \\frac{3}{8}$=$ \\frac {9}{1 0} \\times \\frac {8}{3} \\div \\frac {3}{8} = \\frac {1 2}{5} \\times \\frac {8}{3} = \\frac {3 2}{5}$故答案为: B",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "分数的连除运算",
+        "乘、除法的意义和各部分间的关系"
+      ],
+      "source": "24-25 六年级上·福建龙岩·期中",
+      "merged_id": "unit3_019",
+      "original_id": "unit3_choice_019"
+    },
+    {
+      "id": "unit3_choice_020",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "'甲数是 840，____，乙数是多少？'如果所列的算式为: $840 \\div \\left(1 + \\frac{2}{3}\\right)$，那么横线上应补充的条件是（）。",
+      "options": [
+        "A. 甲数比乙数多 $\\frac{2}{3}$",
+        "B. 甲数比乙数少 $\\frac{2}{3}$",
+        "C. 乙数比甲数多 $\\frac{2}{3}$",
+        "D. 乙数比甲数少 $\\frac{2}{3}$"
+      ],
+      "answer": "A",
+      "solution": "【分析】根据算式 $840 \\div \\left(1 + \\frac{2}{3}\\right)$ 可知，是把乙数看作单位'1'，甲数是乙数的 $\\left(1 + \\frac{2}{3}\\right)$，说明甲数比乙数多 $\\frac{2}{3}$，求乙数，据此解答。\n\n【详解】根据分析：'甲数是 840，_____，乙数是多少?'如果所列的算式为: $840 \\div \\left(1 + \\frac{2}{3}\\right)$，那么横线上应补充的条件是甲数比乙数多 $\\frac{2}{3}$。故答案为: A",
+      "knowledge_points": [
+        "已知比一个数多/少几分之几是多少，求这个数",
+        "提问题、填条件问题"
+      ],
+      "source": "24-25 六年级上·广东江门·期中",
+      "merged_id": "unit3_020",
+      "original_id": "unit3_choice_020"
+    },
+    {
+      "id": "unit3_choice_021",
+      "unit": "第三单元：分数除法",
+      "chapter": "倒数的认识",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "如果 $a$、$b$ 互为倒数,那么 $\\frac{3}{a} \\div \\frac{b}{7} = ()$。",
+      "options": [
+        "A. $\\frac{7}{3}$",
+        "B. $\\frac{3}{7}$",
+        "C. 21",
+        "D. $\\frac{1}{21}$"
+      ],
+      "answer": "C",
+      "solution": "【分析】乘积是 1 的两个数互为倒数。除以一个数 (0 除外)，就是乘这个数的倒数。据此解题。\n\n【详解】a、b互为倒数，那么 $a \\times b = 1$ $\\frac {3}{a} \\div \\frac {b}{7} = \\frac {3}{a} \\times \\frac {7}{b} = \\frac {2 1}{a b} = 2 1$故答案为: C",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "倒数的认识"
+      ],
+      "source": "23-24 六年级上·福建泉州·期中",
+      "merged_id": "unit3_021",
+      "original_id": "unit3_choice_021"
+    },
+    {
+      "id": "unit3_choice_022",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "《九章算术》中记载了一道有趣的数学题：今有凫（注：凫是一种水鸟）起南海，七日至北海；雁起北海，九日至南海。今凫、雁俱起（注释：同时起飞），（）日相遇。",
+      "options": [
+        "A. $\\frac{2}{63}$",
+        "B. $\\frac{63}{2}$",
+        "C. $\\frac{16}{63}$",
+        "D. $\\frac{63}{16}$"
+      ],
+      "answer": "D",
+      "solution": "【分析】把南海与北海之间的距离看作单位'1'，凫7天从南海飞到北海，雁9天从北海飞到南海，根据'速度=路程÷时间'，分别求出凫、雁各自的速度，相加即是它们的速度和，再根据'相遇时间=路程÷速度和'即可求解。\n\n【详解】$1 \\div \\left(\\frac{1}{7} + \\frac{1}{9}\\right)$ $ 1 \\div (\\frac {9}{6 3} + \\frac {7}{6 3}) = 1 \\div \\frac {1 6}{6 3} = 1 \\times \\frac {6 3}{1 6} = \\frac {6 3}{1 6} (\\text {天})$《九章算术》中记载了一道有趣的数学题：今有凫（注：凫是一种水鸟）起南海，七日至北海；雁起北海，九日至南海。今凫、雁俱起（注释：同时起飞），$\\frac{63}{16}$ 日相遇。故答案为: D",
+      "knowledge_points": [
+        "分数的四则混合运算",
+        "相遇问题"
+      ],
+      "source": "24-25 六年级上·广东东莞·期中",
+      "merged_id": "unit3_022",
+      "original_id": "unit3_choice_022"
+    },
+    {
+      "id": "unit3_choice_023",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "下面问题可以用算式 $\\frac{5}{12} \\div \\frac{5}{6}$ 解决的是（）。",
+      "options": [
+        "A. 一条绳子，第一次剪下 $\\frac{5}{12}$ 米，第二次剪下的是第一次的 $\\frac{5}{6}$。第二次剪下多少米?",
+        "B. 甲桶油重 $\\frac{5}{12}$ 千克，乙桶油的质量是甲桶油的 $\\frac{5}{6}$。乙桶油重多少千克?",
+        "C. 聪聪走 $\\frac{5}{6}$ 千米用了 $\\frac{5}{12}$ 时。照这样计算，聪聪每千米要走多少小时?",
+        "D. 一个梯形的面积为 $\\frac{5}{12}$ 平方分米，上底与下底的和是 $\\frac{5}{6}$ 分米。这个梯形的高是多少分米?"
+      ],
+      "answer": "C",
+      "solution": "【分析】（1）把第一次剪下的长度看作单位'1'，第二次剪下的是第一次的 $\\frac{5}{6}$，第二次剪下的长度 = 第一次剪下的长度 $\\times \\frac{5}{6}$;(2) 把甲桶油的质量看作单位'1'，乙桶油的质量是甲桶油的 $\\frac{5}{6}$，乙桶油的质量 = 甲桶油的质量 $\\times \\frac{5}{6}$; (3) 求聪聪每千米需要的小时数用除法计算，所求结果的单位与除法算式中被除数的单位保持一致，用题目中的小时数除以聪聪走的千米数; (4) 梯形的面积 = (上底 + 下底) × 高 ÷ 2，则梯形的高 = 梯形的面积 $\\times 2 \\div$ (上底 + 下底)，把选项中的数据代入计算即可。\n\n【详解】A. $\\frac{5}{12} \\times \\frac{5}{6} = \\frac{25}{72}$ （米）所以，第二次剪下 $\\frac{25}{72}$ 米。B. $\\frac{5}{12} \\times \\frac{5}{6} = \\frac{25}{72}$ (千克)所以，乙桶油重 $\\frac{25}{72}$ 千克。C. $\\frac{5}{12} \\div \\frac{5}{6}$ $= \\frac {5}{1 2} \\times \\frac {6}{5}$$= \\frac{1}{2}$ (小时)所以，聪聪每千米要走 $\\frac{1}{2}$ 小时。D. $\\frac{5}{12} \\times 2 \\div \\frac{5}{6}$$= 1$ (分米) $\\frac {5}{6} \\div \\frac {5}{6} = \\frac {5}{6} \\times \\frac {6}{5}$所以，这个梯形的高是 1 分米。故答案为: C",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "分数乘分数",
+        "求一个数的几分之几的问题",
+        "梯形面积的计算"
+      ],
+      "source": "24-25 六年级上·福建厦门·期中",
+      "merged_id": "unit3_023",
+      "original_id": "unit3_choice_023"
+    },
+    {
+      "id": "unit3_choice_024",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "一个数的 $\\frac{5}{6}$ 是 10，这个数的 $\\frac{3}{4}$ 是（）。",
+      "options": [
+        "A. 12",
+        "B. 10",
+        "C. 9"
+      ],
+      "answer": "C",
+      "solution": "【分析】根据已知一个数的几分之几是多少，求这个数用除法，先求出这个数，再根据求一个数的几分之几是多少用乘法，求出这个数的 $\\frac{3}{4}$。\n\n【详解】$10 \\div \\frac{5}{6} \\times \\frac{3}{4}$ $ = 1 0 \\times \\frac {6}{5} \\times \\frac {3}{4} = 1 2 \\times \\frac {3}{4} = 9$一个数的 $\\frac{5}{6}$ 是 10，这个数的 $\\frac{3}{4}$ 是 9。故答案为: C",
+      "knowledge_points": [
+        "分数的乘、除法的混合运算",
+        "求一个数的几分之几的问题",
+        "已知一个数的几分之几是多少，求这个数"
+      ],
+      "source": "24-25 六年级上·甘肃兰州·期中",
+      "merged_id": "unit3_024",
+      "original_id": "unit3_choice_024"
+    },
+    {
+      "id": "unit3_fill_025",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "30 是 35 的（）；（）是 48 的 $\\frac{1}{4}$；60 吨比（）吨少 $\\frac{1}{4}$。",
+      "options": [],
+      "answer": "$\\frac{6}{7}$；12；80",
+      "solution": "【分析】求 30 是 35 的几分之几，根据分数与除法的关系,用 30 除以 35 计算即可。将 48 看作单位'1'，求一个数的几分之几是多少，用乘法计算。用 48 乘 $\\frac{1}{4}$ 计算即可。把'所求吨数看作'1'，60吨比它少 $\\frac{1}{4}$，说明60吨对应整体的 $(1 - \\frac{1}{4})$；'已知一个数的几分之几是多少，求这个数'，用除法，即用60除以 $(1 - \\frac{1}{4})$ 即可。\n\n【详解】$30 \\div 35 = \\frac{6}{7}$ 所以 30 是 35 的 $\\frac{6}{7}$。$4 8 \\times \\frac {1}{4} = 1 2$所以 12 是 48 的 $\\frac{1}{4}$。把'所求吨数'看作'1'。$6 0 \\div (1 - \\frac {1}{4}) = 6 0 \\div \\frac {3}{4} = 6 0 \\times \\frac {4}{3}$=80 (吨) 60吨比80吨少 $\\frac{1}{4}$。",
+      "knowledge_points": [
+        "求一个数的几分之几的问题",
+        "已知比一个数多/少几分之几是多少，求这个数",
+        "求一个数占另一个数几分之几"
+      ],
+      "source": "25-26 六年级上·海南海口·单元测试",    
+      "merged_id": "unit3_025",
+      "original_id": "unit3_fill_025"
+    },
+    {
+      "id": "unit3_fill_026",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "把 $\\frac{3}{7}$ 米的绳子剪成每段长 $\\frac{3}{35}$ 米的小段，共剪成（）段，每段占全长的 $\\frac{（）}{（）}$。",
+      "options": [],
+      "answer": "5；$\\frac{1}{5}$",
+      "solution": "【分析】根据'段数 = 总长度 ÷ 每段长度'，已知绳子总长度是 $\\frac{3}{7}$ 米，每段长度是 $\\frac{3}{35}$ 米，把数据代入计算即可得出可以剪成多少段。把这根绳子的全长看作单位'1'，求每段占全长的几分之几,根据分数的意义：把单位'1'平均分成几份，每份占单位'1'的几分之一，解答即可。\n\n【详解】$\\frac{3}{7} \\div \\frac{3}{35}$ $= \\frac {3}{7} \\times \\frac {3 5}{3}$ $= 5$ (段)把这根绳子的全长看作单位'1'，平均剪成 5 段，每段占全长的 $\\frac{1}{5}$。把 $\\frac{3}{7}$ 米的绳子剪成每段长 $\\frac{3}{35}$ 米的小段，共剪成 5 段，每段占全长的 $\\frac{1}{5}$。",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "分数的意义"
+      ],
+      "source": "25-26 六年级上·海南海口·单元测试",
+      "merged_id": "unit3_026",
+      "original_id": "unit3_fill_026"
+    },
+    {
+      "id": "unit3_fill_027",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "不计算，在括号里填上＜、＞或＝。\\n$$\\n5 \\\\div \\\\frac {4}{5} (\\\\quad) 5 \\\\div 4 \\\\times 5 \\\\quad \\\\frac {4}{7} \\\\div \\\\frac {1}{5} (\\\\quad) \\\\frac {4}{7} \\\\quad \\\\frac {5}{9} \\\\times \\\\frac {4}{7} (\\\\quad) \\\\frac {4}{7}\\n$$",
+      "options": [],
+      "answer": "=;>;<",
+      "solution": "【分析】根据分数除法的计算法则，除以一个分数等于乘它的倒数。$5 \\div \\frac {4}{5} = 5 \\times \\frac {5}{4}, \\text {而} 5 \\div 4 \\times 5 = 5 \\times \\frac {1}{4} \\times 5 = 5 \\times \\left(\\frac {1}{4} \\times 5\\right) = 5 \\times \\frac {5}{4}, \\text {所 以} 5 \\div \\frac {4}{5} = 5 \\div 4 \\times 5。$一个数 (0 除外) 除以一个小于 1 的数 (0 除外)，商大于这个数。因为 $\\frac{1}{5}<1$，所以 $\\frac{4}{7} \\div \\frac{1}{5}>\\frac{4}{7}$。一个数 (0 除外) 乘一个小于 1 的数 (0 除外)，积小于这个数。因为 $\\frac{5}{9}<1$，所以 $\\frac{5}{9} \\times \\frac{4}{7}<\\frac{4}{7}$。\n\n【详解】$5 \\div \\frac{4}{5} = 5 \\times \\frac{5}{4},$ $5 \\div 4 \\times 5 = 5 \\times \\frac{1}{4} \\times 5 = 5 \\times \\frac{5}{4}$ 所以 $5 \\div \\frac{4}{5} = 5 \\div 4 \\times 5;$ $\\frac{1}{5} < 1$，所以 $\\frac{4}{7} \\div \\frac{1}{5} > \\frac{4}{7}$; $\\frac{5}{9} < 1$，所以 $\\frac{5}{9} \\times \\frac{4}{7} < \\frac{4}{7}$。",
+      "knowledge_points": [
+        "因数和积的大小关系（分数乘法）",
+        "被除数与商的大小关系（分数除法）",
+        "分数与整数的除法"
+      ],
+      "source": "25-26 六年级上·海南海口·单元测试",
+      "merged_id": "unit3_027",
+      "original_id": "unit3_fill_027"
+    },
+    {
+      "id": "unit3_fill_028",
+      "unit": "第三单元：分数除法",
+      "chapter": "倒数的认识",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "（） $\\times \\frac{3}{2} = 5 \\times$ （）$= 0.25 \\times (\\quad) = 1$。",
+      "options": [],
+      "answer": "$\\frac{2}{3}$;$\\frac{1}{5}$;4",
+      "solution": "【分析】根据'一个因数 $=$ 积 $\\div$ 另一个因数'，已知积都是1,分别用 $1 \\div$ 每个算式中的已知因数，即可求出括号里的数,据此解答。\n\n【详解】因为 $1 \\div \\frac{3}{2} = 1 \\times \\frac{2}{3} = \\frac{2}{3}$，所以第一个括号里填 $\\frac{2}{3}$；因为 $1 \\div 5 = \\frac{1}{5}$，所以第二个括号里填 $\\frac{1}{5}$；因为 $1 \\div 0.25 = 4$，所以第三个括号里填 4; $\\frac{2}{3} \\times \\frac{3}{2} = 5 \\times \\frac{1}{5} = 0.25 \\times 4 = 1$。",
+      "knowledge_points": [
+        "小数与整数的乘法",
+        "除数是小数的小数除法",
+        "分数与整数的除法"
+      ],
+      "source": "25-26 六年级上·海南海口·单元测试",
+      "merged_id": "unit3_028",
+      "original_id": "unit3_fill_028"
+    },
+    {
+      "id": "unit3_fill_029",
+      "unit": "第三单元：分数除法",
+      "chapter": "倒数的认识",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "$\\frac{2}{5}$ 的倒数是（）；0.2 与（）互为倒数。",
+      "options": [],
+      "answer": "$\\frac{5}{2}$;5",
+      "solution": "【分析】根据倒数的定义，乘积为 1 的两个数互为倒数。分数的倒数是将分子和分母交换位置。据此计算解答即可。\n\n【详解】$\\frac{2}{5}$ 的分子是 2，分母是 5，交换分子分母的位置，即 $\\frac{5}{2}$。$1 \\div 0.2 = 5$。$\\frac{2}{5}$ 的倒数是 $\\frac{5}{2}$；0.2 与 5 互为倒数。",
+      "knowledge_points": [
+        "倒数的认识"
+      ],
+      "source": "25-26 六年级上·海南海口·单元测试",
+      "merged_id": "unit3_029",
+      "original_id": "unit3_fill_029"
+    },
+    {
+      "id": "unit3_fill_030",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法综合",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "在括号里填上＞、＜或＝。$$15 \\\\times \\\\frac {3}{5} (\\\\quad) 15 \\\\quad 13 \\\\div \\\\frac {2}{3} (\\\\quad) 13 \\\\quad \\\\frac {7}{8} \\\\div \\\\frac {1}{4} (\\\\quad) \\\\frac {7}{8} \\\\times 4$$",
+      "options": [],
+      "answer": "<;>;=",
+      "solution": "【分析】根据积的变化规律，一个数（0 除外）乘一个小于 1 的数，积小于这个数本身；根据商的变化规律，一个数（0 除外）除以一个小于 1 的数（0 除外），商大于这个数本身；根据分数除法的运算法则，除以一个分数等于乘这个分数的倒数。\n\n【详解】$\\frac{3}{5}< 1$，所以 $15 \\times \\frac{3}{5}< 15$；$\\frac{2}{3}< 1$，所以 $13 \\div \\frac{2}{3}>13$；$\\frac{1}{4}$ 的倒数是 4，将 $\\frac{7}{8} \\div \\frac{1}{4}$ 转化为 $\\frac{7}{8} \\times 4$，所以 $\\frac{7}{8} \\div \\frac{1}{4} = \\frac{7}{8} \\times 4$。",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "因数和积的大小关系（分数乘法）",
+        "整数乘分数",
+        "被除数与商的大小关系（分数除法）"
+      ],
+      "source": "23-24 六年级下·河北邢台·期中",
+      "merged_id": "unit3_030",
+      "original_id": "unit3_fill_030"
+    },
+    {
+      "id": "unit3_fill_031",
+      "unit": "第三单元：分数除法",
+      "chapter": "倒数的认识",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "$\\frac{1}{8}$ 的倒数是____，1.5 与____互为倒数，____的倒数是 1。",
+      "options": [],
+      "answer": "8;$\\frac{2}{3}$;1",
+      "solution": "【分析】若两个数的乘积是 1，我们就称这两个数互为倒数,则求一个数的倒数可以用 1 除以这个数。\n\n【详解】$1 \\div \\frac{1}{8} = 1 \\times 8 = 8$；$1 \\div 1.5 = 1 \\div \\frac {3}{2} = 1 \\times \\frac {2}{3} = \\frac {2}{3}$；$1 \\div 1 = 1$。即 $\\frac{1}{8}$ 的倒数是 8，1.5 与 $\\frac{2}{3}$ 互为倒数，1 倒数是 1。",
+      "knowledge_points": [
+        "倒数的认识",
+        "分数与整数的除法"
+      ],
+      "source": "24-25 六年级下·海南海口·单元测试",
+      "merged_id": "unit3_031",
+      "original_id": "unit3_fill_031"
+    },
+    {
+      "id": "unit3_fill_032",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "一个最简分数，把它的分子缩小到原来的 $\\frac{1}{5}$，分母扩大到原来的 2 倍后是 $\\frac{5}{6}$，这个最简分数是_____。",
+      "options": [],
+      "answer": "$\\frac{25}{3}$或$8\\frac{1}{3}$",
+      "solution": "【分析】已知一个最简分数的的分子缩小到原来的 $\\frac{1}{5}$，即分子乘 $\\frac{1}{5}$，则分数值也乘 $\\frac{1}{5}$；分母扩大到原来的 2 倍，即分母乘 2，则分数值除以 2，最终得到分数；根据倒推法，用 $\\frac{5}{6}$ 乘 2，再除以 $\\frac{1}{5}$，即可得到原来的分数。\n\n【详解】$\\frac{5}{6} \\times 2 \\div \\frac{1}{5} = \\frac {5}{3} \\times 5 = \\frac {25}{3}$。一个最简分数，把它的分子缩小到原来的 $\\frac{1}{5}$，分母扩大到原来的 2 倍后是 $\\frac{5}{6}$，这个最简分数是 $\\frac{25}{3}$。",
+      "knowledge_points": [
+        "分数乘整数",
+        "分数与整数的除法"
+      ],
+      "source": "24-25 六年级下·海南海口·单元测试",
+      "merged_id": "unit3_032",
+      "original_id": "unit3_fill_032"
+    },
+    {
+      "id": "unit3_fill_033",
+      "unit": "第三单元：分数除法",
+      "chapter": "解方程",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "当 $A = 5$ 时，式子 $(25 - 2A) \\times \\frac{3}{5}$ 的值是_____。当 A = _________时，式子 $(24 - 2A) \\times \\frac{3}{5}$ 的值是 $0$。",
+      "options": [],
+      "answer": "9;12",
+      "solution": "【分析】A=5时，代入算式（25-2A）$\\times \\frac{3}{5}$，即可求出算式的值。算式的结果为0，把算式看作是方程，即（24-2A）$\\times \\frac{3}{5} = 0$，根据等式的性质2，方程两边同时除以 $\\frac{3}{5}$，再根据等式的性质1方程两边同时加上2A，再根据等式的性质 2，方程两边同时除以 2 即可解答。\n\n【详解】当 $A = 5$ 时：$(25 - 2A) \\times \\frac {3}{5} = (25 - 2 \\times 5) \\times \\frac {3}{5} = (25 - 10) \\times \\frac {3}{5} = 15 \\times \\frac {3}{5} = 9$\n\n$(24 - 2A) \\times \\frac {3}{5} = 0$\n\n解：$(24-2A) \\times \\frac{3}{5} \\div \\frac{3}{5} = 0 \\div \\frac{3}{5}$；$24 - 2A = 0$；$24 - 2A + 2A = 0 + 2A$；$2A = 24$；$2A \\div 2 = 24 \\div 2$；$A = 12$\n\n当 $A = 5$ 时，式子 $(25 - 2A) \\times \\frac{3}{5}$ 的值是 9。当 $A = 12$ 时，式子 $(24 - 2A) \\times \\frac{3}{5} = 0$ 的值是 0。",
+      "knowledge_points": [
+        "解分数方程",
+        "应用等式的性质 2 解方程",
+        "含有字母式子的化简与求值"
+      ],
+      "source": "24-25 六年级下·海南海口·单元测试",
+      "merged_id": "unit3_033",
+      "original_id": "unit3_fill_033"
+    },{
+      "id": "unit3_fill_034",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "从 $\\frac{11}{3} 、 \\frac{7}{15} 、 3 、 \\frac{7}{25}$ 中,任意选择两个数, 分别作为被除数和除数, 其中商最小的算式是( )。",
+      "options": [],
+      "answer": "$\\frac{7}{25}\\div\\frac{11}{3}$",
+      "solution": "【分析】被除数越大, 除数越小, 商越大; 被除数越小,除数越大, 商越小。据此比较 4 个数的大小, 用最小数 $\\div$ 最大数的商最小。\n【详解】 $\\frac{11}{3} = 3 \\frac{2}{3} > 3$ 、 $\\frac{7}{15} > \\frac{7}{25}$ 、 $\\frac{11}{3} > 3 > \\frac{7}{15} > \\frac{7}{25}$ ，商最小的算式是 $\\frac{7}{25} \\div \\frac{11}{3}$ 。",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "被除数与商的大小关系（分数除法）"
+      ],
+      "source": "24-25 六年级上·重庆渝中·期末",
+      "merged_id": "unit3_034",
+      "original_id": "unit3_fill_034"
+    },
+    {
+      "id": "unit3_fill_035",
+      "unit": "第三单元：分数除法",
+      "chapter": "倒数的认识",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "如果数 a 与数 b 互为倒数, 则 $\\frac{a}{4} \\times b = (\\quad)$ 。",
+      "options": [],
+      "answer": "$\\frac{1}{4}$",
+      "solution": "【分析】乘积是 1 的两个数互为倒数, 数 a 与数 b 互为倒数,则 $a \\times b = 1$ ; 化简 $\\frac{a}{4} \\times b$ , 即可解答。\n【详解】 $a \\times b = 1$ $\\frac {a}{4} \\times b = \\frac {\\mathrm {a b}}{4} = \\frac {1}{4} $如果数 $a$ 与数 $b$ 互为倒数, 则 $\\frac{a}{4} \\times b = \\frac{1}{4}$ 。",
+      "knowledge_points": [
+        "分数乘分数",
+        "倒数的认识"
+      ],
+      "source": "24-25 六年级上·重庆巫山·期末",
+      "merged_id": "unit3_035",
+      "original_id": "unit3_fill_035"
+    },
+    {
+      "id": "unit3_fill_036",
+      "unit": "第三单元：分数除法",
+      "chapter": "倒数的认识",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "$\\frac{1}{8}$ 的倒数是( )，( )和 $\\frac{3}{5}$ 互为倒数。",
+      "options": [],
+      "answer": "8;$\\frac{5}{3}$",
+      "solution": "【分析】倒数的意义, 乘积是 1 的两个数互为倒数。根据乘数等于积除以另一个乘数, 用 1 分别除以 $\\frac{1}{8} 、 \\frac{3}{5}$ 即可得解。\n【详解】 $1 \\div \\frac{1}{8} = 1 \\times 8 = 8$   $1 \\div \\frac {3}{5} = 1 \\times \\frac {5}{3} = \\frac {5}{3}$$\\frac{1}{8}$ 的倒数是 8, $\\frac{5}{3}$ 和 $\\frac{3}{5}$ 互为倒数。",
+      "knowledge_points": [
+        "倒数的认识",
+        "分数与整数的除法"
+      ],
+      "source": "24-25 六年级上·湖北襄阳·期末",
+      "merged_id": "unit3_036",
+      "original_id": "unit3_fill_036"
+    },
+    {
+      "id": "unit3_fill_037",
+      "unit": "第三单元：分数除法",
+      "chapter": "倒数的认识",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "$\\frac{8}{5} \\times( )$ = $0.75 \\times( )$=$1\\times()$=$1\\frac{1}{5}\\times()$=1",
+      "options": [],
+      "answer": "$\\frac{5}{8};\\frac{4}{3};1;\\frac{5}{6}$",
+      "solution": "【分析】两个数的乘积是 1 , 这两个数互为倒数。分数的倒数就是将分子和分数互换位置;小数的倒数，先将小数转化为分数，再互换分数分子分母的位置；1 的倒数是本身;带分数的倒数，先将带分数转化为假分数，再互换分数分子分母的位置；\n【详解】 $0.75 = \\frac{3}{4}$ $1 \\frac {1}{5} = \\frac {6}{5}$ 则 $\\frac{8}{5} \\times \\frac{5}{8}$ 0.75 × $\\frac{4}{3}$ = 1 × 1 =$1\\frac{1}{5}\\times\\frac{5}{6}=1$",
+      "knowledge_points": [
+        "倒数的认识",
+        "与倒数有关的综合计算",
+        "一位或多位小数化分数（约分）",
+        "假分数与带分数或整数的互化"
+      ],
+      "source": "24-25 六年级上·湖北黄石·期末",
+      "merged_id": "unit3_037",
+      "original_id": "unit3_fill_037"
+    },
+    {
+      "id": "unit3_fill_038",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "$\\frac{5}{6}$ 公顷阔叶林每天约释放氧气 $\\frac{7}{12}$ 吨, 那么 1 公顷阔叶林每天可以释放氧气约( )吨。",
+      "options": [],
+      "answer": "$\\frac{7}{10}或0.7$",
+      "solution": "【分析】根据把一个数平均分，求每份是多少，用除法计算，用释放的氧气量除以公顷数即可。\n【详解】 $\\frac{7}{12} \\div \\frac{5}{6} = \\frac{7}{12} \\times \\frac{6}{5} = \\frac{7}{10}$ （吨）$\\frac{5}{6}$ 公顷阔叶林每天约释放氧气 $\\frac{7}{12}$ 吨, 那么 1 公顷阔叶林每天可以释放氧气约 $\\frac{7}{10}$ 吨。",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "分数的平均分"
+      ],
+      "source": "24-25 六年级上·北京通州·期末",
+      "merged_id": "unit3_038",
+      "original_id": "unit3_fill_038"
+    },
+    {
+      "id": "unit3_fill_039",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "新能源汽车的驱动方式主要以电动为主，可节约燃油资源，减少废气排放，有效保护环境。一辆新能源汽车行驶8千米耗电 $\\frac{6}{5}$ 千瓦时，则平均每行驶1千米耗电( )千瓦时，耗电1千瓦时，可供行驶( )千米。",
+      "options": [],
+      "answer": "$\\frac{3}{20}$或0.15;$\\frac{20}{3}$或$6\\frac{2}{3}$",
+      "solution": "【分析】耗电量 $\\div$ 行驶路程 $= 1$ 千米耗电量; 行驶路程 $\\div$ 耗电量 $= 1$ 千瓦时可行驶路程, 据此列式计算。除以一个数等于乘这个数的倒数。\n【详解】 $\\frac{6}{5} \\div 8 = \\frac{6}{5} \\times \\frac{1}{8} = \\frac{3}{20}$ （千瓦时） $8 \\div \\frac {6}{5} = 8 \\times \\frac {5}{6} = \\frac {2 0}{3} (\\text {千 米})$平均每行驶 1 千米耗电 $\\frac{3}{20}$ 千瓦时, 耗电 1 千瓦时, 可供行驶 $\\frac{20}{3}$ 千米。",
+      "knowledge_points": [
+        "分数与整数的除法"
+      ],
+      "source": "24-25 六年级上·山东济南·期中",
+      "merged_id": "unit3_039",
+      "original_id": "unit3_fill_039"
+    },
+    {
+      "id": "unit3_fill_040",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "面塑作为一种传统的民间手工艺品，制作前需先配制好面团，捏面艺人向盆内倒入150克糯米粉，是倒入面粉质量的 $\\frac{1}{4}$ ，面粉有多重？设面粉重 $x$ 克，等量关系式为( )，可列方程( )。",
+      "options": [],
+      "answer": "倒入面粉质量$\\times\\frac{1}{4}=$倒入的糯米粉的质量$\\frac{1}{4}x=150$",
+      "solution": "【分析】从题意可知: 以面粉质量为单位 \"1\", 倒入的糯米粉是倒入面粉质量的 $\\frac{1}{4}$ , 根据分数乘法的意义, 可知: 倒入面粉质量 $\\times \\frac{1}{4}$ 就是倒入的糯米粉的质量。设面粉重 $x$ 克, 根据等量关系, 列方程即可。\n【详解】根据题意列出的等量关系式为：倒入面粉质量 $\\times \\frac{1}{4} =$ 倒入的糯米粉的质量，解：设面粉重 $x$ 克。$\\frac {1}{4} x = 1 5 0 $ $ \\frac {1}{4} x \\div \\frac {1}{4} = 1 5 0 \\div \\frac {1}{4} $ $ x = 1 5 0 \\times 4$   $ x = 6 0 0 $面粉重600克。所以设面粉重 $x$ 克, 等量关系式为倒入面粉质量 $\\times \\frac{1}{4} =$ 倒入的糯米粉的质量, 可列方程 $\\frac{1}{4} x = 150$ 。",
+      "knowledge_points": [
+        "等式的认识及列等量关系式",
+        "已知一个数的几分之几是多少，求这个数",
+        "列方程解含一个未知数的问题"
+      ],
+      "source": "24-25 六年级上·山西长治·期中",
+      "merged_id": "unit3_040",
+      "original_id": "unit3_fill_040"
+    },
+    {
+      "id": "unit3_fill_041",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "A 的$\\frac{1}{4}$ 与 B 的 $\\frac{1}{5}$相等,如果A是100,B是( )。",
+      "options": [],
+      "answer": "125",
+      "solution": "【分析】求一个数的几分之几是多少，用乘法，据此用 $100 \\times \\frac{1}{4}$ 列式求出 A 的 $\\frac{1}{4}$ 是多少，已知 B 的 $\\frac{1}{5}$ 是 $100 \\times \\frac{1}{4}$ ，根据已知一个数的几分之几是多少，求这个数用除法解答，据此用 $100 \\times \\frac{1}{4} \\div \\frac{1}{5}$ 求出 B 是多少。\n【详解】 $100 \\times \\frac{1}{4} \\div \\frac{1}{5}$ $= 2 5 \\times 5 = 1 2 5$ 所以B是 $125 \\mathrm{~g}$ 。",
+      "knowledge_points": [
+        "分数的乘、除法的混合运算",
+        "求一个数的几分之几的问题",
+        "已知一个数的几分之几是多少，求这个数"
+      ],
+      "source": "2024·四川乐山·小升初真题",
+      "merged_id": "unit3_041",
+      "original_id": "unit3_fill_041"
+    },
+    {
+      "id": "unit3_fill_042",
+      "unit": "第三单元：分数除法",
+      "chapter": "倒数的认识",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "3.2 的倒数是( ),$\\frac{1}{4}$ 的倒数与 8 的积是( )。",
+      "options": [],
+      "answer": "$\\frac{5}{16}$;32",
+      "solution": "【分析】根据倒数的意义, 即乘积是 1 的两个数, 互为倒数。求 3.2 的倒数可用 1 除以 3.2 即可得解, 用相同的方法求 $\\frac{1}{4}$ 的倒数再乘 8 , 计算可得解。\n【详解】 $1 \\div 3.2$ $=1 \\div \\frac {1 6}{5}=1 1 \\times \\frac {5}{1 6} =\\frac {5}{1 6} $    $1 \\div \\frac {1}{4} \\times 8= 1 \\times 4 \\times 8 =3 2 $3.2 的倒数是 $\\frac{5}{16}$ , $\\frac{1}{4}$ 的倒数与 8 的积是 $32$ 。",
+      "knowledge_points": [
+        "倒数的认识",
+        "与倒数有关的综合计算",
+        "分数与整数的除法"
+      ],
+      "source": "23-24 六年级上·湖南岳阳·期中",
+      "merged_id": "unit3_042",
+      "original_id": "unit3_fill_042"
+    },
+    {
+      "id": "unit3_fill_043",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "15 的 $\\frac{1}{5}$ 是( )； $\\frac{1}{8}$ 是 $\\frac{1}{4}$ 的( )；( )的 $\\frac{3}{4}$ 是 12。",
+      "options": [],
+      "answer": "3  $\\frac{1}{2}$;16",
+      "solution": "【分析】15 的 $\\frac{1}{5}$ 是多少, 求一个数的几分之几是多少, 用 15 乘 $\\frac{1}{5}$ 计算; $\\frac{1}{8}$ 是 $\\frac{1}{4}$ 的多少, 求一个数是另一个数的几分之几,用 $\\frac{1}{8}$ 除以 $\\frac{1}{4}$ 计算; 已知一个数的几分之几是多少, 求这个数,用 12 除以 $\\frac{3}{4}$ 计算。\n【详解】 (1) $15 \\times \\frac{1}{5} = 3$    (2) $\\frac{1}{8} \\div \\frac{1}{4}$$ =\\frac {1}{8} \\times 4=\\frac {1}{2}$ (3) $12 \\div \\frac{3}{4}=1 2 \\times \\frac {4}{3}=16$ 因此 15 的 $\\frac{1}{5}$ 是 3 ; $\\frac{1}{8}$ 是 $\\frac{1}{4}$ 的 $\\frac{1}{2}$ ; 16 的 $\\frac{3}{4}$ 是 12。",
+      "knowledge_points": [
+        "已知一个数的几分之几是多少，求这个数",
+        "求一个数占另一个数几分之几"
+      ],
+      "source": "23-24 六年级上·山东济宁·期中",
+      "merged_id": "unit3_043",
+      "original_id": "unit3_fill_043"
+    },
+    {
+      "id": "unit3_fill_044",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "填空题",
+      "difficulty": "困难",
+      "question": "如果 $a$ 是一个非 0 的自然数, 那么 $\\frac{1}{5} \\div a = (\\quad), a \\div \\frac{1}{5} = (\\quad)$ 。",
+      "options": [],
+      "answer": "$\\frac{1}{5a}$;$5a$",
+      "solution": "【分析】分数除法的计算法则：除以一个不为 0 的数等于乘这个数的倒数。据此解答。\n【详解】     $\\frac{1}{5} \\div a = \\frac{1}{5} \\times \\frac{1}{a} = \\frac{1}{5a}$ $a \\div \\frac {1}{5} = a \\times 5 = 5 a$如果 $a$ 是一个非 0 的自然数, 那么 $\\frac{1}{5} \\div a = \\frac{1}{5 a}, a \\div \\frac{1}{5} = 5 a$ 。",
+      "knowledge_points": [
+        "含有字母式子的化简与求值",
+        "分数与整数的除法"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_044",
+      "original_id": "unit3_fill_044"
+    },
+    {
+      "id": "unit3_judge_045",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "如果 $a \\times \\frac{4}{3} = b \\times \\frac{9}{4} = c \\div \\frac{5}{6}$ ,(且 a、b、c 都大于零), 最大的数是 $c$ 。 ( )",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】可以设最后的结果是 1 , 根据两个数的乘积是 1 ,这两个数互为倒数, a 就是 $\\frac{4}{3}$ 的倒数, b 就是 $\\frac{9}{4}$ 的倒数。两个相同的数相除结果是 1 得出 c。再比较大小解答。\n【详解】设 $a \\times \\frac{4}{3} = b \\times \\frac{9}{4} = c \\div \\frac{5}{6} = 1$即 $a = \\frac{3}{4}, b = \\frac{4}{9}, c = \\frac{5}{6}$则 $b < a < c$ , 最大的数是 $c$ 。故答案为:  √",
+      "knowledge_points": [
+        "倒数的认识",
+        "分数与整数的除法"
+      ],
+      "source": "24-25 六年级上·重庆永川·期末",
+      "merged_id": "unit3_045",
+      "original_id": "unit3_judge_045"
+    },
+    {
+      "id": "unit3_judge_046",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "同一项工程, 甲单独做 $\\frac{1}{5}$ 天完成, 乙单独做 $\\frac{1}{6}$ 天完成, 乙比甲做得快。( )",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】把整个工程量看作单位 \"1\", 工作效率 = 工作总量 ÷ 工作时间, 据此分别求出甲和乙的工作效率, 再进行比较即可。\n【详解】 $1 \\div \\frac{1}{5} = 1 \\times 5 = 5$   $1 \\div \\frac {1}{6} = 1 \\times 6 = 6$ 6 > 5 所以乙的工作效率比甲的工作效率高，即乙比甲做得快。故答案为:  √",
+      "knowledge_points": [
+        "分数与整数的除法"
+      ],
+      "source": "24-25 六年级上·贵州黔东南·期末",
+      "merged_id": "unit3_046",
+      "original_id": "unit3_judge_046"
+    },
+    {
+      "id": "unit3_judge_047",
+      "unit": "第三单元：分数除法",
+      "chapter": "倒数的认识",
+      "question_type": "判断题",
+      "difficulty": "困难",
+      "question": "运算得数为 1 的两个数（0除外）互为倒数。（ ）",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】根据倒数的意义: 乘积为 1 的两个数互为倒数。据此判断即可。\n【详解】乘积为 1 的两个数（0 除外）互为倒数。例如：$2 \\times 0.5 = 1, 2$ 和 $0.5$ 互为倒数。原题说法错误。故答案为:  ×",
+      "knowledge_points": [
+        "倒数的认识"
+      ],
+      "source": "24-25 六年级上·贵州黔东南·期末",
+      "merged_id": "unit3_047",
+      "original_id": "unit3_judge_047"
+    },
+    {
+      "id": "unit3_judge_048",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "判断题",
+      "difficulty": "困难",
+      "question": "一个数除以假分数,所得的商一定大于这个数。（）",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】分子比分母大或分子和分母相等的分数叫做假分数。一个数（0除外）除以大于1的数，商比原来的数小。  一个数（0除外）除以1，商等于原来的数。\n【详解】如： $\\frac{1}{2} \\div \\frac{3}{3} = \\frac{1}{2} \\div 1 = \\frac{1}{2}$ ，商与被除数相等； $\\frac{1}{2} \\div \\frac{3}{2} = \\frac{1}{2} \\times \\frac{2}{3} = \\frac{1}{3}, \\frac{1}{3} < \\frac{1}{2}$ ，商小于被除数；所以，一个数除以假分数，所得的商小于或等于这个数。原题说法错误。故答案为:  ×",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "真分数、假分数、带分数的认识",
+        "被除数与商的大小关系（分数除法）"
+      ],
+      "source": "24-25 六年级上·广东广州·期中",
+      "merged_id": "unit3_048",
+      "original_id": "unit3_judge_048"
+    },
+    {
+      "id": "unit3_judge_049",
+      "unit": "第三单元：分数除法",
+      "chapter": "倒数的认识",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "$\\frac{3}{4}$ 除以它的倒数, 商是 $1 。$ ( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】根据倒数的定义：乘积是 1 的两个数互为倒数，例如：如果 $a$ 、 $b$ 不为 0， $a \\times b = 1$ ，则 $a$ 是 $b$ 的倒数， $b$ 是 $a$ 的倒数。求一个分数的倒数，把分子和分母调换位置即可；除以一个不为 0 的数相当于乘这个数的倒数；据此解答。\n【详解】因为 $\\frac{3}{4} \\times \\frac{4}{3} = 1$所以 $\\frac{3}{4}$ 的倒数是 $\\frac{4}{3}$ 。$\\frac {3}{4} \\div \\frac {4}{3} = \\frac {3}{4} \\times \\frac {3}{4} = \\frac {9}{1 6}$所以 $\\frac{3}{4}$ 除以它的倒数, 商是 $\\frac{9}{16}$ 。原题说法错误。故答案为×",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "倒数的认识"
+      ],
+      "source": "24-25 六年级上·江西宜春·期中",
+      "merged_id": "unit3_049",
+      "original_id": "unit3_judge_049"
+    },
+    {
+      "id": "unit3_judge_050",
+      "unit": "第三单元：分数除法",
+      "chapter": "倒数的认识",
+      "question_type": "判断题",
+      "difficulty": "困难",
+      "question": "任意一个数都有倒数。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】乘积是 1 的两个数互为倒数，但 0 没有倒数，据此举例说明即可。\n【详解】0 乘任何数都是 0 , 因此 0 没有倒数, 所以原题说法错误。故答案为: ×",
+      "knowledge_points": [
+        "倒数的认识"
+      ],
+      "source": "2018·安徽·小升初真题",
+      "merged_id": "unit3_050",
+      "original_id": "unit3_judge_050"
+    },
+    {
+      "id": "unit3_judge_051",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "一个数除以 $\\frac{3}{9}$ , 商一定大于这个数。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】一个数（0 除外）除以一个小于 1 的数，商大于原数，据此判断。\n【详解】一个数除以 $\\frac{3}{9}$ , 商不一定大于这个数, 例如 $0 \\div \\frac{3}{9} = 0$ , 一个数 (0 除外) 除以 $\\frac{3}{9}$ , 商一定大于这个数, 以原题说法错误。故答案为:  ×",
+      "knowledge_points": [
+        "被除数与商的大小关系（分数除法）"
+      ],
+      "source": "23-24 六年级上·河北沧州·期中",
+      "merged_id": "unit3_051",
+      "original_id": "unit3_judge_051"
+    },
+    {
+      "id": "unit3_calc_052",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法计算",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "用你喜欢的方法计算。\n$3 2 \\div \\frac {4}{7}$          $(\\frac {3}{5} - \\frac {2}{7}) \\times 3 5$       $4 \\div \\frac {2}{3} \\div \\frac {6}{1 7}$        $\\frac {3}{8} \\times \\frac {4}{9} \\times \\frac {5}{1 2}   $    $\\frac {2}{7} \\times 8 + \\frac {5}{7} \\div \\frac {1}{8}$    $\\frac {1}{4} \\div (3 - \\frac {5}{1 3} - \\frac {8}{1 3})$",
+      "options": [],
+      "answer": "56;11;17;$\\frac{5}{7 2}$;8;$\\frac {1}{8}$",
+      "solution": "【分析】 $32 \\div \\frac{4}{7}$ , 根据除以一个分数等于乘它的倒数, 计算即可;$\\left(\\frac{3}{5} - \\frac{2}{7}\\right) \\times 35$  ，利用乘法分配律进行计算；乘法分配律：两个数的差与一个数相乘，可以用被减数和减数分别与这个数相乘，再相减，用字母表示： $(a - b) \\times c = a \\times c - b \\times c$  ，乘法分配律可以逆运用，用字母表示：  $a \\times c - b \\times c = (a - b) \\times c;$ $4 \\div \\frac{2}{3} \\div \\frac{6}{17}$  ，从左到右依次计算； $\\frac{3}{8} \\times \\frac{4}{9} \\times \\frac{5}{12}$  ，从左到右依次计算；$\\frac{2}{7} \\times 8 + \\frac{5}{7} \\div \\frac{1}{8}$ , 把除法转化为乘法, 然后利用乘法分配律逆运算进行计算; 乘法分配律: 两个数的和与一个数相乘, 可以先把它们与这个数分别相乘, 再相加, 这叫做乘法分配律,用字母表示:  $(a + b) \\times c = a \\times c + b \\times c$ ; 乘法分配律可以逆运用, 用字母表示:  $a \\times c + b \\times c = (a + b) \\times c$ ;$\\frac{1}{4} \\div \\left(3 - \\frac{5}{13} - \\frac{8}{13}\\right)$ , 利用减法的性质先计算括号内的式子, 然后再算括号外的除法; 减法的运算性质: 一个数连续减去两个数, 等于减去这两个数的和, 用字母表示为:  $a - b - c = a - (b + c)$ ; 据此计算。\n【详解】 $32 \\div \\frac{4}{7}= 3 2 \\times \\frac {7}{4}= 5 6 $ $(\\frac {3}{5} - \\frac {2}{7}) \\times 3 5 = \\frac {3}{5} \\times 3 5 - \\frac {2}{7} \\times 3 5 = 2 1 - 1 0 = 1 1 $ $4 \\div \\frac {2}{3} \\div \\frac {6}{1 7} = 4 \\times \\frac {3}{2} \\times \\frac {1 7}{6} = 6 \\times \\frac {1 7}{6} = 1 7$ $\\frac {3}{8} \\times \\frac {4}{9} \\times \\frac {5}{1 2} = \\frac {1}{6} \\times \\frac {5}{1 2} = \\frac {5}{7 2}$ $\\frac {2}{7} \\times 8 + \\frac {5}{7} \\div \\frac {1}{8} = \\frac {2}{7} \\times 8 + \\frac {5}{7} \\times 8 = \\left(\\frac {2}{7} + \\frac {5}{7}\\right) \\times 1 \\times 8 = 8 $ $\\frac {1}{4} \\div \\left(3 - \\frac {5}{1 3} - \\frac {8}{1 3}\\right) = \\frac {1}{4} \\div \\left[ 3 - \\left(\\frac {5}{1 3} + \\frac {8}{1 3}\\right) \\right] = \\frac {1}{4} \\div (3 - 1) = \\frac {1}{4} \\div 2 = \\frac {1}{4} \\times \\frac {1}{2} = \\frac {1}{8}$",
+      "knowledge_points": [
+        "整数乘法运算定律推广到分数乘法",
+        "分数的四则混合运算",
+        "分数加、减简便运算",
+        "分数与整数的除法"
+      ],
+      "source": "25-26 六年级上·海南海口·单元测试",
+      "merged_id": "unit3_052",
+      "original_id": "unit3_calc_052"
+    },
+    {
+      "id": "unit3_calc_053",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法计算",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "计算下面各题。(能简算的要用简便方法计算)\n$4 2 \\div \\frac {6}{7}$        $ \\left(\\frac {2}{3} - \\frac {1}{2} + \\frac {5}{6}\\right) \\div \\frac {1}{2 4}$         $ \\frac {3}{5} \\times \\frac {3}{7} + \\frac {2}{5} \\times \\frac {3}{7}$\n$\\frac {3}{1 4} \\div \\frac {6}{7} \\div 12$     $ (\\frac {1}{6} + \\frac {1}{2}) \\div \\frac {2}{3} \\times \\frac {3}{4} $     $\\frac {7}{9} \\div [ \\frac {2}{3} \\times (\\frac {5}{1 8} - \\frac {1}{9}) ]$",
+      "options": [],
+      "answer": "49;24;$\\frac{3}{7}$;$\\frac{1}{48}$;$\\frac{3}{4}$;7",
+      "solution": "【分析】 $42 \\div \\frac{6}{7}$ , 除以一个数等于乘这个数的倒数, 据此计算;  $\\left(\\frac{2}{3} - \\frac{1}{2} + \\frac{5}{6}\\right) \\div \\frac{1}{24}$ , 先将除法改写成乘法, 再根据乘法分配律 (a +b)  $\\times c = a \\times c + b \\times c$ , 把  $\\left(\\frac{2}{3} - \\frac{1}{2} + \\frac{5}{6}\\right) \\times 24$  变成  $\\frac{2}{3} \\times 24 - \\frac{1}{2} \\times 24 + \\frac{5}{6} \\times 24$  进行简算;$\\frac{3}{5} \\times \\frac{3}{7} + \\frac{2}{5} \\times \\frac{3}{7}$ , 根据乘法分配律  $a \\times c + b \\times c = (a + b) \\times c$ , 把算式变成  $\\left(\\frac{3}{5} + \\frac{2}{5}\\right) \\times \\frac{3}{7}$  进行简算;$\\frac{3}{14} \\div \\frac{6}{7} \\div 12$ , 从左往右依次计算;$\\left(\\frac{1}{6} + \\frac{1}{2}\\right) \\div \\frac{2}{3} \\times \\frac{3}{4}$ , 先算括号里的加法, 再算括号外的除法, 最后算乘法;$\\frac{7}{9} \\div \\left[\\frac{2}{3} \\times \\left(\\frac{5}{18} - \\frac{1}{9}\\right)\\right]$ , 先算小括号里的减法, 再算中括号里的乘法,最后算中括号外的除法。\n【详解】 $42 \\div \\frac{6}{7} = 4 2 \\times \\frac {7}{6} = 4 9 $ $(\\frac {2}{3} - \\frac {1}{2} + \\frac {5}{6}) \\div \\frac {1}{2 4}= \\left(\\frac {2}{3} - \\frac {1}{2} + \\frac {5}{6}\\right) \\times 2 4 = \\frac {2}{3} \\times 2 4 - \\frac {1}{2} \\times 2 4 + \\frac {5}{6} \\times 2 4 = 1 6 - 1 2 + 2 0 = 2 4$ $ \\frac {3}{5} \\times \\frac {3}{7} + \\frac {2}{5} \\times \\frac {3}{7} = \\left(\\frac {3}{5} + \\frac {2}{5}\\right) \\times \\frac {3}{7} = 1 \\times \\frac {3}{7} = \\frac {3}{7} $ $\\frac {3}{1 4} \\div \\frac {6}{7} \\div 1 2 = \\frac {3}{1 4} \\times \\frac {7}{6} \\times \\frac {1}{1 2} = \\frac {1}{4} \\times \\frac {1}{1 2}= \\frac {1}{4 8}$ $(\\frac {1}{6} + \\frac {1}{2}) \\div \\frac {2}{3} \\times \\frac {3}{4} = \\left(\\frac {1}{6} + \\frac {3}{6}\\right) \\times \\frac {3}{2} \\times \\frac {3}{4} = \\frac {2}{3} \\times \\frac {3}{2} \\times \\frac {3}{4} = 1 \\times \\frac {3}{4} = \\frac {3}{4} $ $\\frac {7}{9} \\div \\left[ \\frac {2}{3} \\times \\left(\\frac {5}{1 8} - \\frac {1}{9}\\right) \\right] = \\frac {7}{9} \\div \\left[ \\frac {2}{3} \\times \\left(\\frac {5}{1 8} - \\frac {2}{1 8}\\right) \\right] = \\frac {7}{9} \\div \\left[ \\frac {2}{3} \\times \\frac {1}{6}\\right) = \\frac {7}{9} \\div \\frac {1}{9} = \\frac {7}{9} \\times 9 = 7$",
+      "knowledge_points": [
+        "整数乘法运算定律推广到分数乘法",
+        "分数的四则混合运算",
+        "分数的连除运算",
+        "分数与整数的除法"
+      ],
+      "source": "25-26 六年级上·海南海口·单元测试",
+      "merged_id": "unit3_053",
+      "original_id": "unit3_calc_053"
+    },
+    {
+      "id": "unit3_calc_054",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法计算",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "解方程。\n$x + \\frac {5}{8} = \\frac {3}{4} \\quad x \\div \\frac {3}{7} = \\frac {1 4}{1 5} \\quad \\frac {3}{8} x \\div \\frac {1}{2} = 1 2$",
+      "options": [],
+      "answer": "$x =\\frac{1}{8}$;$x=\\frac{2}{5}$;x=16",
+      "solution": "【分析】 $x + \\frac{5}{8} = \\frac{3}{4}$ , 根据等式的性质 1 , 两边同时  $-\\frac{5}{8}$  即可;  $x \\div \\frac{3}{7} = \\frac{14}{15}$ , 根据等式的性质 2 , 两边同时  $\\times \\frac{3}{7}$  即可;$\\frac{3}{8} x \\div \\frac{1}{2} = 12$ , 根据等式的性质 2 , 两边同时  $\\times \\frac{1}{2}$ , 再同时  $\\div \\frac{3}{8}$  即可。\n【详解】 $x + \\frac{5}{8} = \\frac{3}{4}$解:  $x + \\frac{5}{8} - \\frac{5}{8} = \\frac{3}{4} - \\frac{5}{8}$   $x = \\frac {6}{8} - \\frac {5}{8}$  $x = \\frac {1}{8}$ $x \\div \\frac {3}{7} = \\frac {1 4}{1 5} $ 解:  $x \\div \\frac{3}{7} \\times \\frac{3}{7} = \\frac{14}{15} \\times \\frac{3}{7}$  $x = \\frac {2}{5}$ $\\frac {3}{8} x \\div \\frac {1}{2} = 1 2$  解:  $\\frac{3}{8} x \\div \\frac{1}{2} \\times \\frac{1}{2} = 12 \\times \\frac{1}{2}$  $\\frac {3}{8} x = 6$  $\\frac {3}{8} x \\div \\frac {3}{8} = 6 \\div \\frac {3}{8}$  $x = 6 \\times \\frac {8}{3}  x = 1 6$",
+      "knowledge_points": [
+        "解分数方程",
+        "应用等式的性质 2 解方程",
+        "应用等式的性质 1 解方程"
+      ],
+      "source": "25-26 六年级上·海南海口·单元测试",
+      "merged_id": "unit3_054",
+      "original_id": "unit3_calc_054"
+    },
+    {
+      "id": "unit3_calc_055",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法计算",
+      "question_type": "计算题",
+      "difficulty": "困难",
+      "question": "直接写出得数。\n$\\frac {1}{6} \\times 1 2 $  $   9 \\div \\frac {3}{1 3} =$      $\\frac {1 8}{1 9} \\div 6 =$    $ \\frac {4}{9} \\times \\frac {3}{8} =$   $\\\\ \\frac {1}{2} \\div \\frac {1}{3} = $    $\\frac {1}{3} \\times \\frac {3}{4} = $   $\\frac {7}{2 4} \\div 1 4 =$   $ \\frac {1 0}{1 1} \\times \\frac {2 2}{2 7} =$",
+      "options": [],
+      "answer": "2;39;$\\frac{3}{19};\\frac{1}{6}$;$\\frac {3}{2}$;$\\frac {1}{4}$;$\\frac{1}{48}$;$\\frac{2 0}{2 7}$",
+      "solution": "【解析】略",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "分数乘分数",
+        "分数乘整数",
+        "分数与整数的除法"
+      ],
+      "source": "25-26 六年级上·海南海口·单元测试",
+      "merged_id": "unit3_055",
+      "original_id": "unit3_calc_055"
+    },
+    {
+      "id": "unit3_calc_056",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法计算",
+      "question_type": "计算题",
+      "difficulty": "容易",
+      "question": "计算 $(1 \\frac{5}{11} + 3 \\frac{1}{99}+9 \\frac {1}{3 3}) \\div (1 \\frac {1}{1 1} + 3 \\frac {5}{9 9} + 9 \\frac {5}{3 3})$",
+      "options": [],
+      "answer": "1.04",
+      "solution": "【分析】观察原式, 可以先把两个括号中的数拆成自然数与分数的和, 由于分母 $99 = 33 \\times 3 = 11 \\times 9$ , 因此, 可以产生公因数 $(1 + 3 + 9)$ , 这样拆分后, 整数部分可以分别组合计算, 分数部分也便于寻找规律, 从而简化计算, 据此解答。能够把两个括号中的数拆成自然数与分数的和、熟练运用商不变的规律是解决本题关键。\n【详解】 $(1\\frac{5}{11} +3\\frac{1}{99} +9\\frac{1}{33})\\div (1\\frac{1}{11} +3\\frac{5}{99} +9\\frac{5}{33})=[ (1 + 3 + 9) + (\\frac {5}{1 1} + \\frac {1}{9 9} + \\frac {1}{3 3}) ] \\div [ (1 + 3 + 9) + (\\frac {1}{1 1} + \\left. \\frac {5}{9 9} + \\frac {5}{3 3}\\right) ] = [ 1 3 + \\frac {5}{1 1} \\times (1 + 3 + 9) ] \\div [ 1 3 + \\frac {1}{1 1} \\times (1 + 3 + 9) ] = [ 1 3 \\times (1 + \\frac {5}{1 1}) ] \\div [ 1 3 \\times (1 + \\frac {1}{1 1}) ] \\\\= (1 + \\frac {5}{1 1}) \\div (1 + \\frac {1}{1 1}) = \\frac {1 0 4}{9 9} \\div \\frac {1 0 0}{9 9} = \\frac {1 0 4}{9 9} \\times \\frac {9 9}{1 0 0} = 1. 0 4$",
+      "knowledge_points": [
+        "分数的四则混合运算"
+      ],
+      "source": "24-25 六年级上·全国·单元测试",
+      "merged_id": "unit3_056",
+      "original_id": "unit3_calc_056"
+    },
+    {
+      "id": "unit3_calc_057",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法计算",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "$2008 \\div 2008 \\frac{2008}{2009} + \\frac{1}{2010}$",
+      "options": [],
+      "answer": "1",
+      "solution": "【分析】把带分数化成假分数，同时把除法转化成乘法，然后根据乘法分配律的逆运算 $a \\times b + a \\times c = a \\times (b + c)$ ，把算式转化为 $2008 \\times \\frac{2009}{2008 \\times (2009 + 1)} + \\frac{1}{2010}$ ，先计算乘法，再计算加法即可。\n【详解】 $2008 \\div 2008 \\frac{2008}{2009} + \\frac{1}{2010}=2 0 0 8 \\div \\frac {2 0 0 8 \\times 2 0 0 9 + 2 0 0 8}{2 0 0 9} + \\frac {1}{2 0 1 0} = 2 0 0 8 \\times \\frac {2 0 0 9}{2 0 0 8 \\times 2 0 0 9 + 2 0 0 8} + \\frac {1}{2 0 1 0} = 2 0 0 8 \\times \\frac {2 0 0 9}{2 0 0 8 \\times (2 0 0 9 +1)} + \\frac {1}{2 0 1 0} = \\frac {2 0 0 9}{2 0 1 0} + \\frac {1}{2 0 1 0} = 1 $",
+      "knowledge_points": [
+        "分数的四则混合运算",
+        "假分数与带分数或整数的互化",
+        "整数乘法分配律",
+        "分数与整数的除法"
+      ],
+      "source": "24-25 六年级上·全国·单元测试",
+      "merged_id": "unit3_057",
+      "original_id": "unit3_calc_057"
+    },
+    {
+      "id": "unit3_calc_058",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法计算",
+      "question_type": "计算题",
+      "difficulty": "容易",
+      "question": "计算。\n$$\\frac {5}{1 \\times 2 \\times 3} + \\frac {7}{2 \\times 3 \\times 4} + \\frac {9}{3 \\times 4 \\times 5} + \\dots + \\frac {1 5}{6 \\times 7 \\times 8}$$",
+      "options": [],
+      "answer": "$\\frac{165}{112}$",
+      "solution": "【分析】根据通项$\\frac{a}{n(n + 1)(n + 2)} = \\frac{A}{n(n + 1)} - \\frac{B}{(n + 1)(n + 2)}$ 可以对原式中的分数进行拆分；将分数分成两组，第一组通过裂项可以凑出能够正负相消的分数，进行简算；第二组首先提取公因式，再进行裂项和正负抵消；最后按照四则运算顺序进行计算即可。\n【详解】解： $\\frac{5}{1 \\times 2 \\times 3} + \\frac{7}{2 \\times 3 \\times 4} + \\frac{9}{3 \\times 4 \\times 5} + \\dots + \\frac{15}{6 \\times 7 \\times 8}=\\frac {1}{1 \\times 2} + \\frac {1}{1 \\times 3} + \\frac {1}{2 \\times 3} + \\frac {1}{2 \\times 4} + \\frac {1}{3 \\times 4} + \\frac {1}{3 \\times 5} + \\dots + \\frac {1}{6 \\times 7} + \\frac {1}{6 \\times 8} \\\\= \\left(\\frac {1}{1 \\times 2} + \\frac {1}{2 \\times 3} + \\frac {1}{3 \\times 4} + \\dots + \\frac {1}{6 \\times 7}\\right) + \\left(\\frac {1}{1 \\times 3} + \\frac {1}{2 \\times 4} + \\frac {1}{3 \\times 5} + \\dots + \\frac {1}{6 \\times 8}\\right) = \\left(1 - \\frac {1}{2} + \\frac {1}{2} - \\frac {1}{3} + \\dots + \\frac {1}{6} - \\frac {1}{7}\\right) + \\left[ \\left(\\frac {1}{1} - \\frac {1}{3}\\right) \\times \\frac {1}{2} + \\left(\\frac {1}{2} - \\frac {1}{4}\\right) \\times \\frac {1}{2} + \\left(\\frac {1}{3} - \\frac {1}{5}\\right) \\times \\frac {1}{2} + \\dots + \\left(\\frac {1}{6} - \\frac {1}{8}\\right) \\times \\frac {1}{2} \\right]\\\\= \\left(1 - \\frac {1}{7}\\right)+ \\left(\\frac {1}{1} - \\frac {1}{3} + \\frac {1}{2} - \\frac {1}{4} + \\frac {1}{3} - \\frac {1}{5} + \\dots + \\frac {1}{6} - \\frac {1}{8}\\right) \\times \\frac {1}{2} \\\\= 1 - \\frac {1}{7} + \\left(1 +\\frac {1}{2} - \\frac {1}{7} - \\frac {1}{8}\\right) \\times \\frac {1}{2} \\\\= \\frac {6}{7} + \\frac {6 9}{5 6} \\times \\frac {1}{2} \\\\= \\frac {6}{7} +\\frac {6 9}{1 1 2} \\\\= \\frac {1 6 5}{1 1 2}$",
+      "knowledge_points": [
+        "分数的四则混合运算"
+      ],
+      "source": "24-25 六年级上·全国·单元测试",
+      "merged_id": "unit3_058",
+      "original_id": "unit3_calc_058"
+    },
+    {
+      "id": "unit3_calc_059",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法计算",
+      "question_type": "计算题",
+      "difficulty": "容易",
+      "question": "计算。\n$$(\\frac {1}{2} + \\frac {2}{3} + \\frac {3}{4} + \\dots + \\frac {9}{1 0}) ^ {2} + (\\frac {1}{2} + \\frac {2}{3} + \\frac {3}{4} + \\dots + \\frac {9}{1 0}) \\times \\frac {1}{2} - (1 + \\frac {1}{2} + \\frac {2}{3} + \\dots + \\frac {9}{1 0}) \\times (\\frac {2}{3}+\\frac {3}{4}+ \\dots+\\frac {9}{10})$$",
+      "options": [],
+      "answer": "$\\frac{1}{2}$",
+      "solution": "【分析】可以假设 $\\frac{1}{2} + \\frac{2}{3} + \\frac{3}{4} + \\dots + \\frac{9}{10} = t$ , 然后把 $t$ 代入原式后, 进行化简即可求解。\n【详解】解：设 $t = \\frac{1}{2} + \\frac{2}{3} + \\frac{3}{4} + \\dots + \\frac{9}{10}$ 。$(\\frac {1}{2} + \\frac {2}{3} + \\frac {3}{4} + \\dots + \\frac {9}{1 0}) ^ {2} + \\left(\\frac {1}{2} + \\frac {2}{3} + \\frac {3}{4} + \\dots + \\frac {9}{1 0}\\right) \\times \\frac {1}{2} - \\left(1 + \\frac {1}{2} + \\frac {2}{3} + \\dots + \\frac {9}{1 0}\\right) \\times (\\frac {2}{3}+\\frac {3}{4}+ \\dots+\\frac {9}{10}) \\\\= t ^ {2} + t \\times \\frac {1}{2} - (1 + t) (t - \\frac {1}{2})= t ^ {2} +\\frac {1}{2} t - (t ^ {2} + t - \\frac {t}{2} - \\frac {1}{2}) \\\\= t ^ {2} + \\frac {1}{2} t - t ^ {2} - t + \\frac {t}{2} + \\frac {1}{2} \\\\= t ^ {2} - t ^ {2} + \\frac {1}{2} t - t + \\frac {t}{2}+ \\frac {1}{2} \\\\= \\frac {1}{2}$",
+      "knowledge_points": [
+        "分数的四则混合运算"
+      ],
+      "source": "24-25 六年级上·全国·单元测试",
+      "merged_id": "unit3_059",
+      "original_id": "unit3_calc_059"
+    },
+    {
+      "id": "unit3_calc_060",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法计算",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "计算下面各题, 能简便计算的要简便计算。\n$3 9 \\times 1 0 1$    $4. 8 \\div \\frac {6}{7} \\times \\frac {5}{8} $     $ 3. 7 - 2. 4 1 + 6. 3 - 3. 5 9 $\n$(\\frac {1}{8} + \\frac {1 1}{1 2} - \\frac {5}{9}) \\times 7 2$     $ 3 2 \\times 1. 2 5 \\times 0. 2 5 $        $ \\frac {7}{1 2} \\div \\left[ \\left(\\frac {3}{4} + \\frac {3}{1 0}\\right) \\times \\frac {2}{3}\\right]$",
+      "options": [],
+      "answer": "3939;3.5;4;35;10;$\\frac{5}{6}$",
+      "solution": "【分析】\" $39 \\times 101$ \"将 101 写成 $100 + 1$ , 再根据乘法分配律$\\mathrm{a} \\times (\\mathrm{b} + \\mathrm{c}) = \\mathrm{a} \\times \\mathrm{b} + \\mathrm{a} \\times \\mathrm{c}$ 计算;\" $4.8 \\div \\frac{6}{7} \\times \\frac{5}{8}$ \"先计算除法，再计算乘法；\" $3.7 - 2.41 + 6.3 - 3.59$ \"带符号交换 2.41 和 6.3 的位置, 再根据减法的性质 $a - b - c = a - (b + c)$ 计算;$\\left(\\frac{1}{8} + \\frac{11}{12} - \\frac{5}{9}\\right) \\times 72$ 根据乘法分配律展开计算;\" $32 \\times 1.25 \\times 0.25$ \"将 32 写成 $4 \\times 8$ , 再根据乘法交换律、乘法结合律计算。乘法交换律: $a \\times b = b \\times a$ , 乘法结合律:$(a \\times b) \\times c = a \\times (b \\times c);$\" $\\frac{7}{12} \\div \\left[\\left(\\frac{3}{4} + \\frac{3}{10}\\right) \\times \\frac{2}{3}\\right]$ \"先计算小括号内的加法, 再计算中括号内的乘法, 最后计算括号外的除法。\n【详解】 $39 \\times 101 = 3 9 \\times (1 0 0 + 1) = 3 9 \\times 1 0 0 + 3 9 \\times 1 = 3 9 0 0 + 3 9 = 3 9 3 9 $ $4.8 \\div \\frac {6}{7} \\times \\frac {5}{8} =4. 8 \\times \\frac {7}{6} \\times \\frac {5}{8} = 5. 6 \\times \\frac {5}{8} = 3. 5 $ $3. 7 - 2. 4 1 + 6. 3 - 3. 5 9 = 3. 7 + 6. 3 - 2. 4 1 - 3. 5 9 = 1 0 - 6 = 4$ $(\\frac {1}{8} + \\frac {1 1}{1 2} - \\frac {5}{9}) \\times 7 2 = \\frac {1}{8} \\times 7 2 + \\frac {1 1}{1 2} \\times 7 2 - \\frac {5}{9} \\times 7 2 = 9 + 6 6 - 4 0 = 3 5$ $3 2 \\times 1. 2 5 \\times 0. 2 5 = 8 \\times 4 \\times 1. 2 5 \\times 0. 2 5 = (8 \\times 1. 2 5) \\times (4 \\times 0. 2 5) = 1 0 \\times 1 = 1 0 $ $\\frac {7}{1 2} \\div \\left[ \\left(\\frac {3}{4} + \\frac {3}{1 0}\\right) \\times \\frac {2}{3}\\right] =\\frac {7}{1 2} \\div \\left[ \\left(\\frac {1 5}{2 0} + \\frac {6}{2 0}\\right) \\times \\frac {2}{3} \\right]= \\frac {7}{1 2} \\div \\left[ \\frac {2 1}{2 0} \\times \\frac {2}{3}\\right] =\\frac {7}{1 2} \\div \\frac {7}{1 0} = \\frac {7}{1 2} \\times \\frac {1 0}{7} =\\frac {5}{6}$",
+      "knowledge_points": [
+        "整数乘法运算定律推广到分数乘法",
+        "分数的四则混合运算",
+        "与小数减法相关的简便运算",
+        "整数乘法运算定律推广到小数乘法"
+      ],
+      "source": "22-23 六年级下·青海西宁·期末",
+      "merged_id": "unit3_060",
+      "original_id": "unit3_calc_060"
+    },
+    {
+      "id": "unit3_calc_061",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法计算",
+      "question_type": "计算题",
+      "difficulty": "容易",
+      "question": "巧算。\n$\\frac {1 6}{2 3} \\times 2 7 + 1 6 \\times \\frac {1 9}{2 3} \\quad \\frac {4}{3 5} \\times \\frac {3}{3 2} + \\frac {3}{3 5} \\times \\frac {3 1}{3 2} \\quad 3 0 \\frac {1}{2 8} \\times \\frac {1}{2 9}$",
+      "options": [],
+      "answer": "$32;\\frac{3}{32};1;\\frac{1}{28}$",
+      "solution": "【分析】本题主要运用乘法分配律及其逆运算进行简便计算。观察式子中数字特点，通过变形构造出相同因数，再利用运算律简化计算。两个分数相乘，调换它们的分子或分母之间的位置，积不变。\n【详解】 $\\frac{16}{23} \\times 27 + 16 \\times \\frac{19}{23}$ $ \\frac {1 6}{2 3} \\times 2 7 + 1 9 \\times \\frac {1 6}{2 3} = \\frac {1 6}{2 3} \\times (2 7 + 1 9) = \\frac {1 6}{2 3} \\times 4 6 = 3 2$ $ \\frac {4}{3 5} \\times \\frac {3}{3 2} + \\frac {3}{3 5} \\times \\frac {3 1}{3 2} =\\frac {4}{3 5} \\times \\frac {3}{3 2} + \\frac {3 1}{3 5} \\times \\frac {3}{3 2} = \\frac {3}{3 2} \\times \\left(\\frac {4}{3 5} + \\frac {3 1}{3 5}\\right) = \\frac {3}{3 2} \\times 1 = \\frac {3}{3 2}$ $ 3 0 \\frac {1}{2 8} \\times \\frac {1}{2 9}=(2 9 + 1 + \\frac {1}{2 8}) \\times \\frac {1}{2 9} =(2 9 + \\frac {2 9}{2 8}) \\times \\frac {1}{2 9} =2 9 \\times \\frac {1}{2 9} + \\frac {2 9}{2 8} \\times \\frac {1}{2 9}= 1 + \\frac {1}{2 8} =1\\frac {1}{2 8} $",
+      "knowledge_points": [
+        "分数的四则混合运算"
+      ],
+      "source": "25-26 六年级上·全国·课后作业",
+      "merged_id": "unit3_061",
+      "original_id": "unit3_calc_061"
+    },
+    {
+      "id": "unit3_calc_062",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法计算",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "用简便方法计算。\n$9 3 0 0 \\div 5 \\div 6 \\qquad 4 0 4 \\times 2. 5 \\qquad \\left(3 \\frac {5}{1 2} - 1 \\frac {3}{7}\\right) - \\left(2 \\frac {4}{7} - 1 \\frac {7}{1 2}\\right)$\n$(\\frac {1}{6} + \\frac {7}{8} - \\frac {1}{4}) \\div \\frac {1}{2 4} \\quad \\quad(8 9 + 8 9 + 8 9 + 8 9) \\times 2 5 \\quad\\quad 7 \\frac {5}{7} \\times \\frac {3}{8} + \\frac {3}{7} \\times \\frac {2}{8}$",
+      "options": [],
+      "answer": "310;1010;1;19;8900;3",
+      "solution": "【分析】 (1) 根据除法的性质 $a \\div b \\div c = a \\div (b \\times c)$ 把 $9 3 0 0 \\div 5 \\div 6 $变成 $9 3 0 0 \\div (5 \\times 6), 再按顺序计算;$(2) 先把 404 拆成 $4 \\times 101$ , 然后根据乘法交换律 $a \\times b = b \\times a$把 $4 \\times 101 \\times 2.5$ 变成 $4 \\times 2.5 \\times 101$ , 再按顺序计算;(3) 先根据减法的性质 $a - (b - c) = a - b + c$ 去掉括号,算式变成 $3 \\frac{5}{12} - 1 \\frac{3}{7} - 2 \\frac{4}{7} + 1 \\frac{7}{12}$ , 然后根据加法交换律 $a + b = b + a$ . 减法的性质 $a - b - c = a - (b + c)$ 把算式变成 $\\left(3 \\frac{5}{12} + 1 \\frac{7}{12}\\right) - \\left(1 \\frac{3}{7} + 2 \\frac{4}{7}\\right)$ , 再按顺序计算;(4) 先把除法转化成乘法, 然后根据乘法分配律 $(a + b) \\times c = a \\times c + b \\times c$ 把 $\\left(\\frac{1}{6} + \\frac{7}{8} - \\frac{1}{4}\\right) \\times 24$ 变成 $\\frac{1}{6} \\times 24 + \\frac{7}{8} \\times 24 - \\frac{1}{4} \\times 24$ , 再按顺序计算;  (5) 先把 $(89 + 89 + 89 + 89) \\times 25$ 改写成 $89 \\times 4 \\times 25$ , 然后根据乘法结合律 $(a \\times b) \\times c = a \\times (b \\times c)$ 把算式变成 $89 \\times (4 \\times 25)$ , 再按顺序计算;  (6) 先根据积不变的规律把 $7 \\frac{5}{7} \\times \\frac{3}{8} + \\frac{3}{7} \\times \\frac{2}{8}$ 变成 $7 \\frac{5}{7} \\times \\frac{3}{8} + \\frac{2}{7} \\times \\frac{3}{8}$ , 然后根据乘法分配律 $a \\times c + b \\times c = (a + b) \\times c$ 把算式变成$\\left(7 \\frac{5}{7} + \\frac{2}{7}\\right) \\times \\frac{3}{8}$ , 再按顺序计算。\n【详解】 (1) $9300 \\div 5 \\div 6 =9 3 0 0 \\div (5 \\times 6) = 9 3 0 0 \\div 3 0 =310$ (2) $404 \\times 2.5= 4 \\times 1 0 1 \\times 2. 5=4 \\times 2. 5 \\times 1 0 1=1 0 \\times 1 0 1=1010$ (3)$ (3\\frac{5}{12} - 1 \\frac{3}{7}) - (2 \\frac{4}{7} - 1 \\frac{7}{12})= 3 \\frac {5}{1 2} - 1 \\frac {3}{7} - 2 \\frac {4}{7} + 1 \\frac {7}{1 2}=3 \\frac {5}{1 2} + 1 \\frac {7}{1 2} - 1 \\frac {3}{7} - 2 \\frac {4}{7}=(3 \\frac {5}{1 2} + 1 \\frac {7}{1 2}) - (1 \\frac {3}{7} + 2 \\frac {4}{7})=5-4=1$ (4) $(\\frac{1}{6} + \\frac{7}{8} - \\frac{1}{4}) \\div \\frac{1}{24}=(\\frac {1}{6} + \\frac {7}{8} - \\frac {1}{4}) \\times 2 4=\\frac {1}{6} \\times 2 4 + \\frac {7}{8} \\times 2 4 - \\frac {1}{4} \\times 2 4=4 + 2 1 - 6=19$ (5) $(89 + 89 + 89 + 89) \\times 25= 8 9 \\times 4 \\times 2 5=89\\times(4\\times25)=89\\times100=8900$ (6) $7 \\frac{5}{7} \\times \\frac{3}{8} + \\frac{3}{7} \\times \\frac{2}{8}= 7 \\frac {5}{7} \\times \\frac {3}{8} + \\frac {2}{7} \\times \\frac {3}{8}=(7 \\frac {5}{7} + \\frac {2}{7}) \\times \\frac {3}{8}=8 \\times \\frac{3}{8}=3$",
+      "knowledge_points": [
+        "整数乘法运算定律推广到分数乘法",
+        "运用转化法解决复杂的小数乘法简算问题",
+        "分数除法相关的简便计算",
+        "分数加、减简便运算"
+      ],
+      "source": "24-25 六年级下·海南海口·单元测试",
+      "merged_id": "unit3_062",
+      "original_id": "unit3_calc_062"
+    },
+    {
+      "id": "unit3_calc_063",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法计算",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "求未知数 $x$ 。\n$① 5 x - 4. 5 \\times 2 = 0. 5 \\quad ② \\frac {4}{5} x - 2. 4 \\times 5 = 3 6$",
+      "options": [],
+      "answer": "①x = 1.9;② x = 60",
+      "solution": "【分析】①先把方程化简成 $5 x - 9 = 0.5$ , 然后方程两边先同时加上 9 , 再同时除以 5 , 求出方程的解;② 先把方程化简成 $\\frac{4}{5} x - 12 = 36$ , 然后方程两边先同时加上 12, 再同时除以 $\\frac{4}{5}$ , 求出方程的解。\n【详解】① $5x - 4.5 \\times 2 = 0.5$解： $5x - 9 = 0.5 \\quad 5 x - 9 + 9 = 0. 5 + 9\\quad 5 x = 9. 5 \\quad 5 x \\div 5 = 9. 5 \\div 5 \\quad x = 1. 9 $ ② $\\frac {4}{5} x - 2. 4 \\times 5 = 3 6 $解: $\\frac{4}{5} x - 12 = 36 \\quad \\frac {4}{5} x - 1 2 + 1 2 = 3 6 + 1 2 \\quad \\frac {4}{5} x = 4 8 \\quad \\frac {4}{5} x \\div \\frac {4}{5} = 4 8 \\div \\frac {4}{5} \\quad x = 4 8 \\times \\frac {5}{4} \\quad x = 6 0$",
+      "knowledge_points": [
+        "解小数方程",
+        "应用等式的性质 1 和 2 解方程",
+        "解分数方程"
+      ],
+      "source": "24-25 六年级下·海南海口·单元测试",
+      "merged_id": "unit3_063",
+      "original_id": "unit3_calc_063"
+    },
+    {
+      "id": "unit3_calc_064",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法计算",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "脱式计算。\n$① 5 7 4 1 - 1 9 2 \\div 8 \\times 1 5 \\quad ② 7. 5 \\times 3. 6 \\div (3. 6 + 8. 4) \\quad③ \\frac {5}{6} \\div \\frac {2}{3} - \\frac {2}{3} \\div \\frac {5}{6} \\quad  ④ \\frac {5}{6} \\div[ \\frac {1}{4} + (\\frac {5}{6} - \\frac {2}{3})]$",
+      "options": [],
+      "answer": "① 5381;② 2.25;③$\\frac{9}{20}$;④ 2",
+      "solution": "【分析】①根据运算顺序，有加减法和乘除法，先算乘除，再算加减法，即这个算式先算除法，再算乘法，最后算减法即可；(2) 根据运算顺序, 先算小括号里的加法, 再按照从左到右的顺序计算即可;  (3) 根据运算顺序，先同时计算两边的除法，最后算减法；  (4) 根据运算顺序, 先算小括号里的减法, 再算中括号里的加法, 最后算括号外的除法。\n【详解】① $5741 - 192 \\div 8 \\times 15 = 5 7 4 1 - 2 4 \\times 1 5 = 5 7 4 1 - 3 6 0 = 5 3 8 1$ ② $7. 5 \\times 3. 6 \\div (3. 6 + 8. 4) = 7. 5 \\times 3. 6 \\div 1 2 = 2 7 \\div 1 2=2.25 $ ③$ \\frac {5}{6} \\div \\frac {2}{3} - \\frac {2}{3} \\div \\frac {5}{6} = \\frac {5}{6} \\times \\frac {3}{2} - \\frac {2}{3} \\times \\frac {6}{5} = \\frac {5}{4} - \\frac {4}{5} = \\frac {9}{2 0} $   $ ④ \\frac {5}{6} \\div \\left[ \\frac {1}{4} + \\left(\\frac {5}{6} - \\frac {2}{3}\\right) \\right] = \\frac {5}{6} \\div \\left[ \\frac {1}{4} + \\frac {1}{6}\\right] = \\frac {5}{6} \\div \\frac {5}{1 2} = \\frac {5}{6} \\times \\frac {1 2}{5} = 2$",
+      "knowledge_points": [
+        "分数的四则混合运算",
+        "乘加、乘减混合运算",
+        "小数的四则运算及法则"
+      ],
+      "source": "24-25 六年级下·海南海口·单元测试",
+      "merged_id": "unit3_064",
+      "original_id": "unit3_calc_064"
+    },
+    {
+      "id": "unit3_calc_065",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法计算",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "计算下面各题 (能简算的要用简便方法计算)。\n① $\\frac {2}{5} \\times \\frac {4}{7} \\times 1 5 \\times 2 1\\quad  ② (\\frac {5}{8} - \\frac {1}{3}) \\times 2 4$\n③$ \\frac {5}{1 8} \\times (\\frac {1}{5} + \\frac {1}{2 5})  \\quad④ \\frac {5}{2 2} \\times [ (\\frac {1}{3} + \\frac {2}{5}) \\div \\frac {5}{6}]$",
+      "options": [],
+      "answer": "①72;②7;③$\\frac{1}{15}$;④$\\frac{1}{5}$",
+      "solution": "【分析】①根据乘法交换律 $a \\times b = b \\times a$ , 乘法结合律 $a \\times b =b \\times a$ 把 $\\frac{2}{5} \\times \\frac{4}{7} \\times 15 \\times 21$ 变成 $\\left(\\frac{2}{5} \\times 15\\right) \\times \\left(\\frac{4}{7} \\times 21\\right)$ , 再按顺序计算;(2) 根据乘法分配律 $(a + b) \\times c = a \\times c + b \\times c$ 把 $\\left(\\frac{5}{8} - \\frac{1}{3}\\right) \\times 24$ 变成 $\\frac{5}{8} \\times 24 - \\frac{1}{3} \\times 24$ , 再按顺序计算;  (3) 先算括号里面的加法, 再算括号外面的乘法;  (4) 先算小括号里面的加法, 再算中括号里面的除法, 最后算中括号外面的乘法。\n【详解】① $\\frac{2}{5} \\times \\frac{4}{7} \\times 15 \\times 21=(\\frac {2}{5} \\times 1 5) \\times (\\frac {4}{7} \\times 2 1)=6\\times12=72$ ②   $\\left(\\frac{5}{8} - \\frac{1}{3}\\right) \\times 24=\\frac{5}{8} \\times 24- \\frac{1}{3} \\times 24=15-8=7$ ③ $\\frac{5}{18} \\times(\\frac{1}{5} + \\frac{1}{25})=\\frac{5}{18} \\times \\frac{6}{25} =\\frac{1}{15}$ ④ $\\frac{5}{22} \\times [(\\frac{1}{3} + \\frac{2}{5}) \\div \\frac{5}{6}] = \\frac {5}{2 2} \\times [(\\frac {5}{1 5} + \\frac {6}{1 5}) \\div \\frac {5}{6}] =  \\frac {5}{2 2} \\times[ \\frac {1 1}{1 5} \\div \\frac {5}{6}]= \\frac {5}{2 2} \\times [ \\frac {1 1}{1 5} \\times \\frac {6}{5}]= \\frac {1}{5}  $",
+      "knowledge_points": [
+        "整数乘法运算定律推广到分数乘法",
+        "分数的四则混合运算"
+      ],
+      "source": "24-25 六年级下·海南海口·单元测试",
+      "merged_id": "unit3_065",
+      "original_id": "unit3_calc_065"
+    },
+    {
+      "id": "unit3_calc_066",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法计算",
+      "question_type": "计算题",
+      "difficulty": "困难",
+      "question": "直接写出得数。\n$2 + 3. 5 7 = \\quad 0. 2 3 \\times 4 0 0= \\quad 7 \\div 1. 4 =  \\quad 4 0 0 \\div 2 5 \\div 8 =  \\\\ \\frac {5}{8} \\div \\frac {2}{3} =  \\quad 9 + \\frac {1}{3} =  \\quad \\frac {1}{2} - \\frac {1}{4} =  \\quad \\frac {3}{1 6} \\times \\frac {4}{9} =$",
+      "options": [],
+      "answer": "5.57;92;5;2;$\\frac {1 5}{1 6}$;$9\\frac {1}{3}$; $\\frac{1}{4}$;$\\frac{1}{1 2}$",
+      "solution": "【解析】略",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "分数乘分数",
+        "除数是小数的小数除法",
+        "异分母分数加、减法"
+      ],
+      "source": "24-25 六年级下·海南海口·单元测试",
+      "merged_id": "unit3_066",
+      "original_id": "unit3_calc_066"
+    },{
+      "id": "unit3_calc_067",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法计算",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "怎样简便就怎样计算。\n$1 2. 5 \\times 8 \\div 1 2. 5 \\times 8 \\quad\\frac {5}{6} \\times \\frac {4}{9} \\div \\frac {1 0}{3} + \\frac {5}{3}\\quad (\\frac {1}{4} + \\frac {5}{6} - \\frac {7}{8}) \\div \\frac {1}{1 2}\\\\ 3. 7 2 - \\frac {4}{9} + 2. 2 8 - 1 \\frac {5}{9} \\quad2. 5 \\div \\frac {5}{8} \\times \\frac {7}{4} \\quad3 2 \\times 0. 2 5 \\times 1 2. 5$",
+      "options": [],
+      "answer": "64;$\\frac{16}{9}$;$2\\frac{1}{2}$;4;7;100",
+      "solution": "【分析】（1）按照从左到右的顺序计算；(2) 按照从左到右的顺序计算;  (3) 先把除法变为乘法, 再根据乘法分配律:  $a \\times (b + c) = a \\times b + a \\times c$ , 把原式化为:  $\\left(\\frac{1}{4} + \\frac{5}{6} - \\frac{7}{8}\\right) \\times 12 = \\frac{1}{4} \\times 12 + \\frac{5}{6} \\times 12 - \\frac{7}{8} \\times 12$  进行简算;(4) 根据带符号搬家和减法的性质把原式化为:  $(3.72 + 2.28) - \\left(\\frac{4}{9} + 1\\frac{5}{9}\\right)$  进行简算;(5) 按照从左到右的顺序计算;  (6) 先把 32 拆成  $4 \\times 8$ , 再根据乘法交换律和结合律计算。\n【详解】 (1)  $12.5 \\times 8 \\div 12.5 \\times 8 = 1 0 0 \\div 1 2. 5 \\times 8  = 8 \\times 8  = 6 4 $(2)  $\\frac{5}{6} \\times \\frac{4}{9} \\div \\frac{10}{3} + \\frac{5}{3}  \\frac {1 0}{2 7} \\times \\frac {3}{1 0} + \\frac {5}{3}  = \\frac {1}{9} + \\frac {1 5}{9} = \\frac {1 6}{9}$(3)  $(\\frac{1}{4} + \\frac{5}{6} - \\frac{7}{8}) \\div \\frac{1}{12}(\\frac {1}{4} + \\frac {5}{6} - \\frac {7}{8}) \\times 1 2  = \\frac {1}{4} \\times 1 2 + \\frac {5}{6} \\times 1 2 - \\frac {7}{8} \\times 1 2  = 3 + 1 0 - 1 0 \\frac {1}{2}  = 2 \\frac {1}{2} $ (4)  $3.72 - \\frac{4}{9} + 2.28 - 1\\frac{5}{9}= (3. 7 2 + 2. 2 8) - (\\frac {4}{9} + 1 \\frac {5}{9}) = 6 - 2 = 4 $(5)  $2.5 \\div \\frac{5}{8} \\times \\frac{7}{4} = 2. 5 \\times \\frac {8}{5} \\times \\frac {7}{4} = 4 \\times \\frac {7}{4} = 7 $(6)  $32 \\times 0.25 \\times 12.5 = 8 \\times 4 \\times 0. 2 5 \\times 1 2. 5  = (0. 2 5 \\times 4) \\times (8 \\times 1 2. 5)  = 1 \\times 1 0 0= 1 0 0 $",
+      "knowledge_points": [
+        "整数乘法运算定律推广到分数乘法",
+        "分数的乘、除法的混合运算",
+        "分数的四则混合运算",
+        "整数乘法运算定律推广到小数乘法"
+      ],
+      "source": "23-24 六年级下·四川遂宁·期末",
+      "merged_id": "unit3_067",
+      "original_id": "unit3_calc_067"
+    },
+    {
+      "id": "unit3_calc_068",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法计算",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "我会解方程。\n$\\frac {5}{1 2} x = 1 0 \\quad \\frac {7}{1 5} x = 1 \\quad \\frac {2}{3} x \\div \\frac {1}{4} = 1 2 \\quad x - \\frac {4}{5} x - 7 = 2 3$",
+      "options": [],
+      "answer": "x = 24;$x=\\frac{15}{7}$; $x=\\frac{9}{2}$; x = 150",
+      "solution": "【分析】等式的基本性质 1: 等式的两边同时加或者减去一个相同的数，等式仍然成立。等式的基本性质 2: 等式的两边同时乘或者除以一个不为 0 的数, 等式仍然成立。利用等式的性质 2, 将等式的两边同时除以  $\\frac{5}{12}$ , 除以一个分数相当于乘这个分数的倒数;利用等式的性质 2, 将等式的两边同时除以  $\\frac{7}{15}$ ;利用等式的性质 2 , 先将等式的两边同时乘  $\\frac{1}{4}$ , 再同时除以  $\\frac{2}{3}$ ;利用乘法的分配律, 提出  $x$ , 得出  $\\frac{1}{5} x$ , 同时利用等式的性质 1, 将等式的两边同时加上 7 , 最后再利用等式的性质 2 , 将等式的两边同时除以  $\\frac{1}{5}$  。\n【详解】 $\\frac{5}{12} x = 10$解:  $x = 10 \\div \\frac{5}{12}\\quad = 1 0 \\times \\frac {1 2}{5}\\quad x = 2 4$$\\frac {7}{1 5} x = 1$解: $ x = 1 \\div \\frac{7}{15}  x = 1 \\times \\frac {1 5}{7} \\quad x = \\frac {1 5}{7}$$ \\frac {2}{3} x \\div \\frac {1}{4} = 1 2$解:  $\\frac{2}{3} x = 12 \\times \\frac{1}{4} \\frac {2}{3} x = 3 \\quad  x = 3 \\div \\frac {2}{3} \\quad x = 3 \\times \\frac {3}{2} x = \\frac {9}{2} $$\\\\ x - \\frac {4}{5} x - 7 = 2 3$解:  $\\frac{1}{5} x - 7 = 23 \\quad \\frac {1}{5} x = 2 3 + 7  \\frac {1}{5} x = 3 0  x = 3 0 \\div \\frac {1}{5}  x = 3 0 \\times 5  x = 1 5 0 $",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "应用等式的性质 1 和 2 解方程",
+        "分数乘整数",
+        "应用等式的性质 2 解方程"
+      ],
+      "source": "24-25 六年级上·全国·单元测试",
+      "merged_id": "unit3_068",
+      "original_id": "unit3_calc_068"
+    },
+    {
+      "id": "unit3_calc_069",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法计算",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "用你喜欢的方法计算。\n$\\frac {3}{8} \\times \\frac {4}{9} \\times \\frac {5}{1 2} \\quad \\frac {2}{7} \\times 8 + \\frac {5}{7} \\div \\frac {1}{8} \\quad \\frac {1}{4} \\div \\left(3 - \\frac {5}{1 3} - \\frac {8}{1 3}\\right)$",
+      "options": [],
+      "answer": "$\\frac{5}{72}$;8;$\\frac{1}{8}$",
+      "solution": "【分析】（1）能约分的先约分再从左往右计算。(2) 先把  $\\frac{5}{7} \\div \\frac{1}{8}$  转化为  $\\frac{5}{7} \\times 8$ , 再根据乘法分配律  $a \\times b + a \\times c = a \\times (b + c)$ , 进行简便计算。  (3) 根据减法的运算性质, 从一个数中依次减去两个数,等于这个数减去这两个数的和。把  $\\left(3 - \\frac{5}{13} - \\frac{8}{13}\\right)$  转化为 $\\left[3 - \\left(\\frac{5}{13} + \\frac{8}{13}\\right)\\right]$ , 先计算小括号里面的加法, 再计算中括号里面的减法, 最后计算括号外面的除法。\n【详解】 $\\frac{3}{8} \\times \\frac{4}{9} \\times \\frac{5}{12}$$=\\frac {1}{8} \\times \\frac {1}{9} \\times 5  \\frac {1}{7 2} \\times 5 =  \\frac {5}{7 2}$   $ \\frac {2}{7} \\times 8 + \\frac {5}{7} \\div \\frac {1}{8} =\\frac {2}{7} \\times 8 + \\frac {5}{7} \\times 8 =(\\frac {2}{7} + \\frac {5}{7}) \\times 8 =1 \\times 8=8 $   $ \\frac {1}{4} \\div(3 - \\frac {5}{1 3} - \\frac {8}{1 3}) =  \\frac {1}{4} \\div [ 3 - (\\frac {5}{1 3} + \\frac {8}{1 3})] = \\frac {1}{4} \\div [ 3 - 1] = \\frac {1}{4} \\div 2 =  \\frac {1}{4} \\times \\frac {1}{2} =\\frac {1}{8} $",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "整数乘法运算定律推广到分数乘法",
+        "分数的四则混合运算",
+        "分数加、减简便运算"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_069",
+      "original_id": "unit3_calc_069"
+    },
+    {
+      "id": "unit3_calc_070",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "看图列式计算。![](images/5cd4f26720a14ec9c968d2ddb410b94efdb462149599249bc22d8cbe8a3c4c03.jpg)",
+      "options": [],
+      "answer": "$45\\mathrm{~kg}$",
+      "solution": "【分析】从线段图中可知, 乙重  $30 \\mathrm{~kg}$ , 乙比甲少  $\\frac{1}{3}$ , 把甲的重量看作单位 \"1\", 则乙是甲的  $(1 - \\frac{1}{3})$ , 单位 \"1\"未知,用乙的重量除以  $(1 - \\frac{1}{3})$ , 即是甲的重量。\n【详解】 $30 \\div (1 - \\frac{1}{3})= 3 0 \\div \\frac {2}{3}  = 3 0 \\times \\frac {3}{2} = 4 5(k g)$甲重  $45 \\mathrm{~kg}$  。",
+      "knowledge_points": [
+        "分数的四则混合运算",
+        "已知比一个数多/少几分之几是多少，求这个数"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_070",
+      "original_id": "unit3_calc_070"
+    },
+    {
+      "id": "unit3_solve_071",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "在首届全国节能环保宣传周上，习主席强调：希望全社会都参与到节能减排中来，形成节约资源、保护环境的良好风尚。为响应习主席的号召，阳光小学开展了\"绿色校园低碳行动\"活动。小明家安装了太阳能节水系统，6天节约了  $\\frac{3}{8}$  吨水，照这样计算，多少天能节约  $\\frac{1}{2}$  吨水？",
+      "options": [],
+      "answer": "8天",
+      "solution": "【分析】先用分数除法求出 1 天节约水的质量, 即  $\\left(\\frac{3}{8} \\div 6\\right)$ 吨, 求多少天能节约  $\\frac{1}{2}$  吨水就是求  $\\frac{1}{2}$  吨里面有多少个  $\\left(\\frac{3}{8} \\div 6\\right)$ 吨, 用除法计算, 列式为  $\\frac{1}{2} \\div \\left(\\frac{3}{8} \\div 6\\right)$ , 据此解答。\n【详解】 $\\frac{1}{2} \\div (\\frac{3}{8} \\div 6 ) = \\frac {1}{2} \\div (\\frac {3}{8} \\times \\frac {1}{6})  = \\frac {1}{2} \\div \\frac {1}{1 6}= \\frac {1}{2} \\times 1 6  = 8 $答: 8 天能节约  $\\frac{1}{2}$  吨水。",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "分数与整数的除法"
+      ],
+      "source": "24-25六年级上·全国·单元测试",
+      "merged_id": "unit3_071",
+      "original_id": "unit3_solve_071"
+    },
+    {
+      "id": "unit3_solve_072",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "如图, 两个长方形A、B 重叠在一起，重叠部分的面积是 A 的  $\\frac{1}{3}$ ，是 B 的  $\\frac{1}{5}$ 。已知 B 的面积是 60 平方厘米，A 的面积是多少平方厘米？![](images/895660fa9d4dac6e0d262114d6c2b5e934875358a665bed478448739850998fe.jpg)",
+      "options": [],
+      "answer": "36 平方厘米",
+      "solution": "【分析】已知 B 的面积是 60 平方厘米, 重叠部分的面积是 B 的  $\\frac{1}{5}$ , 把 B 的面积看作单位 \"1\", 单位 \"1\" 已知, 用 B 的面积乘  $\\frac{1}{5}$ , 求出重叠部分的面积;已知重叠部分的面积是 A 的  $\\frac{1}{3}$ , 把 A 的面积看作单位 \"1\",单位 \"1\"未知, 用重叠部分的面积除以  $\\frac{1}{3}$ , 求出 A 的面积。\n【详解】重叠部分的面积：$60 \\times \\frac{1}{5} = 12$  （平方厘米）A 的面积:$1 2 \\div \\frac {1}{3} = 1 2 \\times 3 = 3 6 \\text {(平 方 厘 米)} $答: A 的面积是 36 平方厘米。",
+      "knowledge_points": [
+        "求一个数的几分之几的问题",
+        "已知一个数的几分之几是多少, 求这个数",
+        "整数乘分数",
+        "分数与整数的除法"
+      ],
+      "source": "24-25 六年级上·全国·单元测试",
+      "merged_id": "unit3_072",
+      "original_id": "unit3_solve_072"
+    },
+    {
+      "id": "unit3_solve_073",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "甲、乙、丙三人按约定分配劳务酬金。第一次：甲先取其中的  $\\frac{1}{4}$ ，乙取剩下的  $\\frac{1}{5}$ ，丙取乙取后剩下的  $\\frac{1}{6}$ 。接着，按原次序和比率再分配两次。第四次，只有甲、乙还按原次序和原比率分配，而丙分得剩下的 135 元。则甲 4 次共分得多少元？",
+      "options": [],
+      "answer": "843.75元",
+      "solution": "【分析】把总金额看成\"1\"，三个人第一次取后剩余：$1 \\times (1 - \\frac{1}{4}) \\times (1 - \\frac{1}{5}) \\times (1 - \\frac{1}{6}) = \\frac{1}{2}​$ , 也就是三人共取了  $\\frac{1}{2}​$ ,按原次序和比率再分配两次, 那么就是  $\\frac{1}{2} \\times \\frac{1}{2} \\times \\frac{1}{2}​$ , 第四次, 只有甲、乙还按原次序和原比率分配, 那就用前三次剩下的再乘  $\\left(1 - \\frac{1}{4}\\right)​$  再乘  $\\left(1 - \\frac{1}{5}\\right)​$  即可得到丙分得的金额对应的分率, 根据已知一个数的几分之几是多少, 求这个数用除法计算,可得总数, 由题意可知, 第一次甲取  $\\frac{1}{4}​$ ; 第一次剩余  $\\frac{1}{2}​$ , 第二次甲取  $\\frac{1}{2} \\times \\frac{1}{4} = \\frac{1}{8}​$ ; 第二次剩余  $\\frac{1}{2} \\times \\frac{1}{2} = \\frac{1}{4}​$ , 第三次甲取  $\\frac{1}{4} \\times \\frac{1}{4} = \\frac{1}{16}​$ ;第三次剩余  $\\frac{1}{2} \\times \\frac{1}{2} \\times \\frac{1}{2} = \\frac{1}{8}​$ , 第四次甲取  $\\frac{1}{8} \\times \\frac{1}{4} = \\frac{1}{32}​$  。把四次取的相加, 再根据求一个数的几分之几是多少, 用乘法计算即可得解。要先找出已知数 135 对应的分率, 即可得总数, 再根据求一个数的几分之几是多少的方法解答。\n【详解】三个人第一次取后剩余: $1 \\times (1 - \\frac {1}{4}) \\times (1 - \\frac {1}{5}) \\times (1 - \\frac {1}{6})= \\frac {3}{4} \\times \\frac {4}{5} \\times \\frac {5}{6}  = \\frac {1}{2}$每次分配率不变，则丙最后分得$\\frac {1}{2} \\times \\frac {1}{2} \\times \\frac {1}{2} \\times (1 - \\frac {1}{4}) \\times (1 - \\frac {1}{5})= \\frac {1}{8} \\times \\frac {3}{4} \\times \\frac {4}{5}  = \\frac {3}{4 0}$故总金额为  $135 \\div \\frac{3}{40} = 135 \\times \\frac{40}{3} = 1800$  （元）第一次甲取  $\\frac{1}{4}$ ; 第一次剩余  $\\frac{1}{2}$ , 第二次甲取  $\\frac{1}{2} \\times \\frac{1}{4} = \\frac{1}{8}$ ; 第二次剩余  $\\frac{1}{2} \\times \\frac{1}{2} = \\frac{1}{4}$ , 第三次甲取  $\\frac{1}{4} \\times \\frac{1}{4} = \\frac{1}{16}$ ; 第三次剩余  $\\frac{1}{2} \\times \\frac{1}{2} \\times \\frac{1}{2} = \\frac{1}{8}$ ,第四次甲取  $\\frac{1}{8} \\times \\frac{1}{4} = \\frac{1}{32}$所以甲四次共取了$1 8 0 0 \\times (\\frac {1}{4} + \\frac {1}{8} + \\frac {1}{1 6} + \\frac {1}{3 2}) = 1 8 0 0 \\times (\\frac {8}{3 2} + \\frac {4}{3 2} + \\frac {2}{3 2} + \\frac {1}{3 2}) = 1 8 0 0 \\times \\frac {1 5}{3 2} = 8 4 3. 7 5 \\text {(元)}$答: 甲 4 次共取了 843.75 元。",
+      "knowledge_points": [
+        "分数的四则混合运算",
+        "经济问题",
+        "求一个数的几分之几的问题",
+        "已知一个数的几分之几是多少，求这个数"
+      ],
+      "source": "24-25六年级上·全国·单元测试",
+      "merged_id": "unit3_073",
+      "original_id": "unit3_solve_073"
+    },
+    {
+      "id": "unit3_solve_074",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "工作犬可分为军警犬、搜救犬和服务犬等，它们是人类可靠的帮手。甲、乙两只警犬接到搜寻任务，警犬甲单独完成需要3小时，警犬乙单独完成需要 5 小时。现安排让甲、乙警犬一起搜寻, 合作搜寻 30 分钟后将甲警犬带走, 剩余任务由乙警犬单独完成，那么乙警犬完成此次任务一共用了多长时间？",
+      "options": [],
+      "answer": "$\\frac{25}{6}$小时",
+      "solution": "【分析】首先总的搜寻任务为单位 \"1\", 然后根据分数除法算出警犬甲和乙的效率, 然后通过单位的转化将分钟化为小时, 再根据等量关系 \" (甲的效率 + 乙的效率) ×时间\" 求出二者合作 30 分钟做出的工作, 之后求出剩余的工作量,最后再除以乙犬的工作效率, 得到乙所需要的时间, 然后将时间相加, 得到一共所需时间。本题主要考查工程问题中基础公式：工作效率  $\\times$  工作时间  $=$  工作总量，以及相应的两个除法公式的运用；按过程分类，分别去计算出，每个过程的工作时间和工作量，是解决这类问题的关键。\n【详解】将这个搜寻任务看作\"1\"警犬甲的效率:  $1 \\div 3 = \\frac{1}{3}$警犬乙的效率:  $1 \\div 5 = \\frac{1}{5}$30 分钟  $= \\frac{1}{2}$  小时合作了 30 分钟完成了:  $\\left(\\frac{1}{3} + \\frac{1}{5}\\right) \\times \\frac{1}{2} = \\frac{4}{15}$警犬乙单独完成剩下的任务用了:  $(1 - \\frac{4}{15}) \\div \\frac{1}{5} = \\frac{11}{3}$  (小时)$\\frac {1 1}{3} + \\frac {1}{2} = \\frac {2 5}{6}$ (小 时)答: 乙警犬完成此次任务一共用了  $\\frac{25}{6}$  小时。",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "分数乘分数",
+        "两人合作的工程问题"
+      ],
+      "source": "24-25 六年级上·全国·单元测试",
+      "merged_id": "unit3_074",
+      "original_id": "unit3_solve_074"
+    },
+    {
+      "id": "unit3_solve_075",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "搬运一个仓库的货物，单独做，甲需20小时，乙需24小时，丙需30小时，有同样的仓库A和B，甲在A仓库，乙在B仓库，同时开始搬运货物，丙开始帮助甲搬运，中途又转向帮助乙搬运，最后同时搬完两个仓库的货物，丙帮助甲、乙各搬运几小时？",
+      "options": [],
+      "answer": "丙帮助甲搬运了6小时;帮乙搬运10小时",
+      "solution": "【分析】根据题意，把一个仓库的货物看作单位\"1\"，先计算三人的工作效率，从整体上来看，整个工作的量为2，由甲乙丙三人共同完成，结合工作时间  $=$  工作总量  $\\div$  工作效率，则三人搬运的时间  $= 2 \\div \\left( \\frac{1}{20} + \\frac{1}{24} + \\frac{1}{30} \\right) = 16$  （小时），对于A仓库，甲搬了16小时，搬的量为  $\\frac{1}{20} \\times 16$  ，则剩余的是丙帮忙搬运的，则丙帮助甲搬运的时间  $=$  （1一甲的工作效率  $\\times 16$  ）÷丙的工作效率，同理，丙帮助乙搬运的时间  $=$  （1一乙的工作效率  $\\times 16$  ）÷丙的工作效率，分别代入数值计算即可。本题工作过程比较复杂，因此按人物分类，从整体入手，计算出整个工作时间，是解决此题的关键。一般我们解决工程问题有两条途径，一个按过程分类，一个按任务分类，当过程较复杂时，选择按人物分类会让题目变得简单。\n【详解】 $1 \\div 20 = \\frac{1}{20}\\quad1 \\div 2 4 = \\frac {1}{2 4} \\quad 1 \\div 3 0 = \\frac {1}{3 0}  \\quad 2 \\div (\\frac {1}{2 0} + \\frac {1}{2 4} + \\frac {1}{3 0})\\quad= 2 \\div (\\frac {1 2}{2 4 0} + \\frac {1 0}{2 4 0} + \\frac {8}{2 4 0})\\quad = 2 \\div \\frac {1}{8} =2 \\times 8 = 1 6 (\\text {小 时}) $丙帮助甲搬运的时间:  $(1 - \\frac{1}{20} \\times 16) \\div \\frac{1}{30}= (1 - \\frac {4}{5}) \\times 3 0  = \\frac {1}{5} \\times 3 0  = 6 \\text {(小 时)}$丙帮助乙搬运的时间:  $(1 - \\frac{1}{24} \\times 16) \\div \\frac{1}{30}= (1 - \\frac {2}{3}) \\times 3 0  = \\frac {1}{3} \\times 3 0  = 1 0 \\text {(小 时)} $答: 丙帮助甲搬运了 6 小时, 帮乙搬运 10 小时。",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "两人合作的工程问题",
+        "分数乘整数",
+        "分数与整数的除法"
+      ],
+      "source": "24-25六年级上·全国·单元测试",
+      "merged_id": "unit3_075",
+      "original_id": "unit3_solve_075"
+    },
+    {
+      "id": "unit3_solve_076",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "一只猴子摘了一堆桃子, 第一天它吃了这堆桃子的  $\\frac{1}{7}$ , 第二天它吃了余下桃子的  $\\frac{1}{6}$ , 第三天它吃了余下桃子的  $\\frac{1}{5}$ , 第四天它吃了余下桃子的  $\\frac{1}{4}$ , 第五天它吃了余下桃子的  $\\frac{1}{3}$ , 第六天它吃了余下桃子的  $\\frac{1}{2}$ , 这时还剩下 12 只桃子, 那么前三天猴子所吃桃子的总数是多少只?",
+      "options": [],
+      "answer": "36只",
+      "solution": "【分析】最后剩下的 12 只桃子是第六天吃剩的  $\\left(1 - \\frac{1}{2}\\right)$ ,于是可以求出第六天时有多少只桃子, 这个数又是第五天吃剩的  $\\left(1 - \\frac{1}{3}\\right)$ , 于是又可以求出第五天时有多少只桃子……, 倒着想, 可以求出这只猴子摘的桃子共有的只数。\n【详解】桃子的总数：$1 2 \\div (1 - \\frac {1}{2}) \\div (1 - \\frac {1}{3}) \\div (1 - \\frac {1}{4}) \\div (1 - \\frac {1}{5}) \\div (1 - \\frac {1}{6}) \\div (1 - \\frac {1}{7}) \\\\ = 1 2 \\div \\frac {1}{2} \\div \\frac {2}{3} \\div \\frac {3}{4} \\div \\frac {4}{5} \\div \\frac {5}{6} \\div \\frac {6}{7} \\\\ = 1 2 \\times 2 \\times \\frac {3}{2} \\times \\frac {4}{3} \\times \\frac {5}{4} \\times \\frac {6}{5} \\times \\frac {7}{6} \\\\ = 1 2 \\times 7 \\\\ = 8 4 \\text {(只)}$第一天吃:  $84 \\times \\frac{1}{7} = 12$  (只)  第二天吃：  $(84 - 12) \\times \\frac{1}{6} = 72 \\times \\frac{1}{6} = 12$  （只）第三天吃:  $(84 - 12 - 12) \\times \\frac{1}{5} = 60 \\times \\frac{1}{5} = 12$  (只)  三天共吃:  $12 + 12 + 12 = 36$  (只)答：前三天猴子所吃桃子的总数是36只。",
+      "knowledge_points": [
+        "运用转化法或倒推法解决稍复杂的分数应用题"
+      ],
+      "source": "24-25六年级上·全国·单元测试",
+      "merged_id": "unit3_076",
+      "original_id": "unit3_solve_076"
+    },
+    {
+      "id": "unit3_solve_077",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "一项工程, 甲、乙两队合作需要 12 天完成,乙、丙两队合作需要 15 天完成,甲、丙两队合作需要 20 天完成。(1) 甲、乙、丙三队合作需要多少天完成?  (2) 乙队单独完成需要多少天?",
+      "options": [],
+      "answer": "(1)10天;(2)20天",
+      "solution": "【分析】（1）要求出甲、乙、丙三队合作需要多少天完成，就要先求出甲、乙、丙三队的工作效率之和；若把这项工程的工作量看作\"1\"，则甲、乙两队的工作效率之和是  $\\frac{1}{12}$ ，乙、丙两队的工作效率之和是  $\\frac{1}{15}$ ，甲、丙两队的工作效率之和是  $\\frac{1}{20}$ ；因此  $\\left(\\frac{1}{12} + \\frac{1}{15} + \\frac{1}{20}\\right)$  就是甲、乙、丙三队的工作效率之和的2倍，由此可以求出三队的工作效率和，然后用工作总量÷工作效率和 = 合作的天数；(2) 把工作总量看作单位 \"1\", 工作总量  $\\div$  (三队的工作效率和一甲丙的工作效率和) =乙单独完成的天数, 据此列式解答。\n【详解】 (1)  $(\\frac{1}{12} + \\frac{1}{15} + \\frac{1}{20}) \\div 2=(\\frac {2 5}{3 0 0} + \\frac {2 0}{3 0 0} + \\frac {1 5}{3 0 0}) \\div 2 \\\\ = \\frac {1}{5} \\div 2 \\\\ = \\frac {1}{5} \\times \\frac {1}{2} \\\\ = \\frac {1}{1 0}$$1 \\div \\frac {1}{1 0} = 1 \\times 1 0 = 1 0 (\\text {天})$答: 甲、乙、丙三队合作需要 10 天。(2)  $1 \\div(\\frac{1}{10} - \\frac{1}{20})= 1 \\div (\\frac {2}{2 0} - \\frac {1}{2 0}) \\\\ = 1 \\div \\frac {1}{2 0} \\\\ = 1 \\times 2 0 \\\\ = 2 0 \\text {(天)}$答: 乙队单独完成需要 20 天。",
+      "knowledge_points": [
+        "分数的四则混合运算",
+        "两人合作的工程问题",
+        "异分母分数加、减法",
+        "分数与整数的除法"
+      ],
+      "source": "24-25 六年级上·全国·单元测试",
+      "merged_id": "unit3_077",
+      "original_id": "unit3_solve_077"
+    },
+    {
+      "id": "unit3_solve_078",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "泥人面塑是以泥巴陶土为主料的一门传统民间艺术，有一批泥人面塑订单，张师傅单独做需要  $8 \\frac{1}{2}$  天, 比李师傅单独做多用  $\\frac{1}{2}$  天。现两人合作 4 天后, 剩下的 24 个泥人面塑由张师傅单独去做, 自始至终张师傅共做了多少个?",
+      "options": [],
+      "answer": "408个",
+      "solution": "【分析】把工作总量看作单位\"1\"，首先根据工作总量÷工作时间 = 工作效率求出张师傅和李师傅的工作效率，再用他们的工作效率和乘 4 得到合作的工作量对应分率，用 1 减这个分率可得 24 对应的分率，根据已知一个数的几分之几是多少，求这个数用除法计算，可得工作总量，再根据求一个数的几分之几是多少，用乘法计算，可得张师傅 1 天的工作效率再乘 4 再加 24 即可得解。\n【详解】张师傅的工作效率:  $1 \\div 8 \\frac{1}{2} = 1 \\div \\frac{17}{2} = 1 \\times \\frac{2}{17} = \\frac{2}{17}$李师傅单独做需要用到的天数:  $8 \\frac{1}{2} - \\frac{1}{2} = 8$  (天)李师傅的工作效率:  $1 \\div 8 = \\frac{1}{8}$两人的工作效率和:  $\\frac{2}{17} + \\frac{1}{8} = \\frac{16}{136} + \\frac{17}{136} = \\frac{33}{136}$两人合作 4 天的工作量:  $\\frac{33}{136} \\times 4 = \\frac{132}{136}$剩下的工作量:  $1 - \\frac{132}{136} = \\frac{1}{34}$需要做的泥人面塑总数量:  $24 \\div \\frac{1}{34} = 24 \\times 34 = 816$  (个)张师傅1天制作的数量： $816 \\times \\frac{2}{17} = 96$ （个）张师傅4天制作的数量：  $96 \\times 4 = 384$  （个）张师傅一共制作的数量：$384 + 24 = 408$ （个）答：自始至终张师傅共做了408个。",
+      "knowledge_points": [
+        "两人合作的工程问题",
+        "异分母分数加、减法",
+        "整数乘分数",
+        "分数与整数的除法"
+      ],
+      "source": "24-25 六年级上·全国·单元测试",
+      "merged_id": "unit3_078",
+      "original_id": "unit3_solve_078"
+    },
+    {
+      "id": "unit3_solve_079",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "故宫的修复工作正在持续不断地进行。按原计划甲、乙两组合作30天可以完成某宫殿的修复工作，后因甲组人员增加，乙组人员减少，导致甲组的工作效率是原来的  $\\frac{3}{2}$ , 乙组的工作效率下降了50%，合作工期比原计划延后6天完成。若甲、乙两组人员未变动，则他们分别单独进行修复各需要多少天？",
+      "options": [],
+      "answer": "甲组 90 天, 乙组 45 天",
+      "solution": "【分析】把修复工作的工作总量看成单位\"1\"，原计划甲、乙合作的工作效率之和为  $\\frac{1}{30}$  。甲组的工作效率提高到原来的  $\\frac{3}{2}$ ，乙组的工作效率下降了  $50\\%$ ， $30 + 6 = 36$  天完成，工作效率之和为  $\\frac{1}{36}$ ，假设甲组、乙组的工作效率均提高到原来的  $\\frac{3}{2}$ ，那么工作效率之和就提高到原来的  $\\frac{3}{2}$ ，也就是  $\\frac{1}{30} \\times \\frac{3}{2} = \\frac{1}{20}$ ，假设的情况与实际情况相比，相差的是乙原来工作效率的  $\\frac{3}{2} - \\frac{1}{2}$ ，对应  $\\frac{1}{20} - \\frac{1}{36}$ ，再根据分数除法的意义求出乙的工作效率，进而求出甲的工作效率，再用1分别除以乙和甲的工作效率，即可求出乙组和甲组单独完成全部修复工作要用多少天。本题考查工程问题, 掌握工作效率、工作时间、工作总量之间的关系以及分数除法的应用是解题的关键。\n【详解】乙组的工作效率下降了  $50 \\%$ , 乙组的工作效率变为原来的  $\\frac{1}{2}$  。最终共用了  $30 + 6 = 36$  （天）所以此时甲、乙两组工作效率之和是  $1 \\div 36 = \\frac{1}{36}$如果甲组工作效率提高到原来的  $\\frac{3}{2}$ , 乙组的工作效率也提高到原来的  $\\frac{3}{2}$ , 此时甲、乙两组的工作效率之和是  $\\frac{1}{30} \\times \\frac{3}{2} = \\frac{1}{20}$  乙组原来的工作效率为:$(\\frac {1}{2 0} - \\frac {1}{3 6}) \\div (\\frac {3}{2} - \\frac {1}{2})=(\\frac {3 6}{7 2 0} - \\frac {2 0}{7 2 0}) \\div 1=  \\frac {1 6}{7 2 0} = \\frac {1}{4 5}$甲组原来的工作效率为:  $\\frac{1}{30} - \\frac{1}{45} = \\frac{3}{90} - \\frac{2}{90} = \\frac{1}{90}$乙组单独修复要用:  $1 \\div \\frac{1}{45} = 45$  (天)甲组单独修复要用  $1 \\div \\frac{1}{90} = 90$  (天)答: 若甲、乙两组人员未变动, 则他们分别单独进行修复。各需要 90 天和 45 天。",
+      "knowledge_points": [
+        "两人合作的工程问题",
+        "分数与整数的除法"
+      ],
+      "source": "24-25六年级上·全国·单元测试",
+      "merged_id": "unit3_079",
+      "original_id": "unit3_solve_079"
+    },
+    {
+      "id": "unit3_solve_080",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "小明的妈妈买来一篮鸡蛋，第一天吃了  $\\frac{1}{7}$  第二天吃了余下的  $\\frac{1}{4}$ ，第三、四天都吃了第二天余下的  $\\frac{1}{3}$ ，第五天吃了余下的  $\\frac{1}{2}$ ，还剩下3个鸡蛋。妈妈共买了多少个鸡蛋？",
+      "options": [],
+      "answer": "28个",
+      "solution": "【分析】把第五天的鸡蛋个数看成单位\"1\"，那么第五天剩下  $(1 - \\frac{1}{2})$ ，再把第三、四天的鸡蛋个数看成单位\"1\"，求出第三、四天剩下的鸡蛋是总鸡蛋的  $(1 - \\frac{1}{3} - \\frac{1}{3}) \\times (1 - \\frac{1}{2})$ ，再把第二天的鸡蛋个数看成单位\"1\"，求出第二天剩下的鸡蛋是总鸡蛋的  $(1 - \\frac{1}{4}) \\times (1 - \\frac{1}{3} - \\frac{1}{3}) \\times (1 - \\frac{1}{2})$ ，把第一天的鸡蛋个数看成单位\"1\"，可求出第一天剩下的鸡蛋是总鸡蛋的  $(1 - \\frac{1}{7}) \\times (1 - \\frac{1}{4}) \\times (1 - \\frac{1}{3} - \\frac{1}{3}) \\times (1 - \\frac{1}{2})$ ，正好是3个，再根据除法的意义解答即可。\n【详解】 $(1 - \\frac{1}{7}) \\times (1 - \\frac{1}{4}) \\times (1 - \\frac{1}{3} - \\frac{1}{3}) \\times (1 - \\frac{1}{2}) =\\frac {6}{7} \\times \\frac {3}{4} \\times \\frac {1}{3} \\times \\frac {1}{2}= \\frac {9}{1 4} \\times \\frac {1}{3} \\times \\frac {1}{2} = \\frac {3}{1 4} \\times \\frac {1}{2}= \\frac {3}{2 8} \\quad3 \\div \\frac {3}{2 8} = 3 \\times \\frac {2 8}{3} = 2 8 \\quad (\\text {个}) $答: 妈妈共买了 28 个鸡蛋。",
+      "knowledge_points": [
+        "分数的四则混合运算",
+        "单位 \"1\"的认识与确定",
+        "分数与整数的除法"
+      ],
+      "source": "24-25六年级上·全国·单元测试",
+      "merged_id": "unit3_080",
+      "original_id": "unit3_solve_080"
+    },
+    {
+      "id": "unit3_solve_081",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "北京颐和园是世界上保存最完整的古代皇家园林之一。据颐和园工作人员统计，某天游览时租用讲解器的人数比游览总人数的  $\\frac{1}{3}$  少 200人，未租用讲解器的人数比游览总人数的  $\\frac{4}{5}$  多 40 人，租用讲解器和未租用讲解器的人数分别是多少？",
+      "options": [],
+      "answer": "租用讲解器的有 200 人, 未租用讲解器的有 1000 人。",
+      "solution": "【分析】根据题目信息：租用讲解器的人数比总人数的  $\\frac{1}{3}$  少200人，可知未租用讲解器的人数是总人数的  $\\left(1 - \\frac{1}{3}\\right)$  多200人，又知未租用讲解器的人数比游览总人数的  $\\frac{4}{5}$  多40人，则总人数的  $\\frac{4}{5}$  比  $\\left(1 - \\frac{1}{3}\\right)$  多  $(200 - 40)$  人，根据已知一个数的几分之几是多少，求这个数用除法计算，可求出总人数，再根据求一个数的几分之几是多少，用乘法计算求出总人数的  $\\frac{1}{3}$  再减200可得租用讲解器的人数，再用总人数减租用讲解器的人数可得未租用讲解器的人数。要先找出一个具体数量占总人数的几分之几, 据此求出总人数, 再根据求一个数的几分之几是多少, 用乘法计算对应分率的人数即可。\n【详解】 $(200 - 40) \\div [\\frac{4}{5} - (1 - \\frac{1}{3})] = 1 6 0 \\div \\frac {2}{1 5}  = 1 6 0 \\times \\frac {1 5}{2}  = 1 2 0 0 (\\text {人}) $$ 1 2 0 0 \\times \\frac {1}{3} - 2 0 0  = 4 0 0 - 2 0 0 = 2 0 0 \\text {(人)} \\quad 1 2 0 0 - 2 0 0 = 1 0 0 0 \\text {(人)} $答: 租用讲解器的有 200 人, 未租用讲解器的有 1000 人。",
+      "knowledge_points": [
+        "分数的四则混合运算",
+        "求一个数的几分之几的问题",
+        "已知一个数的几分之几是多少，求这个数",
+        "分数与整数的除法"
+      ],
+      "source": "24-25六年级上·全国·单元测试",
+      "merged_id": "unit3_081",
+      "original_id": "unit3_solve_081"
+    },
+    {
+      "id": "unit3_solve_082",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "一只蜗牛从  $\\frac{5}{2}$  米深的井底往上爬, 白天爬  $\\frac{1}{2}$  米, 晚上下滑  $\\frac{1}{3}$  米, 它从某日早晨往上爬, 第几天爬到井口?",
+      "options": [],
+      "answer": "13天",
+      "solution": "【分析】蜗牛白天往上爬  $\\frac{1}{2}$  米, 晚上下滑  $\\frac{1}{3}$  米, 相当于每天只往上爬  $\\left(\\frac{1}{2} - \\frac{1}{3}\\right)$  米, 因为最后一天向上爬  $\\frac{1}{2}$  米后到达井口,不会再下滑, 所以最后一天之前蜗牛向上爬了  $\\left(\\frac{5}{2} - \\frac{1}{2}\\right)$  米, 用最后一天之前爬的距离除以每天爬的距离, 再加最后 1 天即可得解。\n【详解】 $\\frac{5}{2} -\\frac{1}{2} = 2$  （米）$\\frac {1}{2} - \\frac {1}{3} = \\frac {3}{6} - \\frac {2}{6} = \\frac {1}{6} (\\text {米}) \\quad 2 \\div \\frac {1}{6} = 2 \\times 6 = 1 2 (\\text {天}) \\quad 1 2 + 1 = 1 3 \\text {(天)}$ 答: 第 13 天爬到井口。",
+      "knowledge_points": [
+        "异分母分数加、减法",
+        "基础行程问题",
+        "分数与整数的除法"
+      ],
+      "source": "24-25 六年级上·全国·单元测试",
+      "merged_id": "unit3_082",
+      "original_id": "unit3_solve_082"
+    },
+    {
+      "id": "unit3_solve_083",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "一本故事书, 小华从第一页开始看起, 第一天看了总页数的  $\\frac{1}{4}$  多 3 页, 第二天看了余下的  $\\frac{1}{3}$  少 4 页，还剩 122 页没有看。这本故事书一共有多少页？",
+      "options": [],
+      "answer": "240页",
+      "solution": "【分析】本题可进行倒推, 从题意可知, 第二天看了余下的  $\\frac{1}{3}$  少 4 页, 说明剩下没有看的比余下的  $\\left(1 - \\frac{1}{3}\\right)$  多 4 页, 根据已知一个数的几分之几是多少, 求这个数用除法计算, 用除法求出第一天看后余下的页数, 同理第一天看了后余下的比总页数的  $\\left(1 - \\frac{1}{4}\\right)$  少 3 页, 用除法可求出这本故事书的总页数。\n【详解】 $(122 - 4) \\div \\left(1 - \\frac{1}{3}\\right)= 1 1 8 \\div \\frac {2}{3} = 1 1 8 \\times \\frac {3}{2}  = 1 7 7 \\text {(页)} \\quad(1 7 7 - 3) \\div \\left(1 - \\frac {1}{4}\\right)  = 1 8 0 \\div \\frac {3}{4}  = 1 8 0 \\times \\frac {4}{3}  = 2 4 0 \\text {(页)} $答: 这本故事书一共有 240 页。",
+      "knowledge_points": [
+        "分数的四则混合运算",
+        "已知一个数的几分之几是多少，求这个数",
+        "分数与整数的除法"
+      ],
+      "source": "24-25 六年级上·全国·单元测试",
+      "merged_id": "unit3_083",
+      "original_id": "unit3_solve_083"
+    },
+    {
+      "id": "unit3_solve_084",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "《弟子规》是依据孔子教诲编成的学童生活规范，形式为\"三字一句\"，核心思想是\"儒家的孝悌仁爱\"。学校举行\"传经典·学国学\"活动，小玲第一周背诵了总句数的  $\\frac{1}{5}$ ，第二周背诵了余下的  $\\frac{1}{3}$ ，还剩下192句没背，《弟子规》全文共有多少句？多少个字？",
+      "options": [],
+      "answer": "360 句;1080 个",
+      "solution": "【分析】把《弟子规》全文的总句数看作单位\"1\"，第一周背诵了总句数的  $\\frac{1}{5}$ ，则剩余  $1 - \\frac{1}{5} = \\frac{4}{5}$ ；第二周背诵了余下的  $\\frac{1}{3}$ ，则第二周背诵了总句数的  $\\frac{4}{5} \\times \\frac{1}{3} = \\frac{4}{15}$ ；用单位\"1\"分别减去第一周和第二周背诵的，计算出还剩下总句数的几分之几没有背诵；根据已知一个数的几分之几是多少，求这个数，用除法计算；最后根据\"三字一句\"，用总句数乘3，所得结果即为一共有多少个字。\n【详解】 $1 - \\frac{1}{5} -\\left(1 - \\frac{1}{5}\\right)\\times \\frac{1}{3}= \\frac {4}{5} - \\frac {4}{1 5}= \\frac {8}{1 5}$总句数:  $192 \\div \\frac{8}{15} =9 2 \\times \\frac {1 5}{8}=360 (句)$总字数:  $360 \\times 3 = 1080$  (个)   答: 《弟子规》全文共有 360 句, 1080 个字。",
+      "knowledge_points": [
+        "分数的四则混合运算",
+        "已知一部分量占总量的几分之几及另一部分量，求总量"
+      ],
+      "source": "23-24六年级下·重庆酉阳·期末",
+      "merged_id": "unit3_084",
+      "original_id": "unit3_solve_084"
+    },
+    {
+      "id": "unit3_solve_085",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "姐妹二人做幸运星,姐姐做的颗数是妹妹的  $\\frac{5}{3}$ , 姐妹两人一共做了 72 颗幸运星,姐姐做了多少颗幸运星? (用方程解决问题)",
+      "options": [],
+      "answer": "45颗",
+      "solution": "【分析】解: 设妹妹做了  $x$  颗幸运星, 姐姐做的颗数是妹妹的  $\\frac{5}{3}$ , 则姐姐做了  $\\frac{5}{3} x$  颗幸运星; 姐妹两人一共做了 72 颗幸运星, 列方程:  $x + \\frac{5}{3} x = 72$ , 解方程, 即可解答。\n【详解】解: 设妹妹做了  $x$  颗幸运星, 则姐姐做了  $\\frac{5}{3} x$  颗幸运星。 $ x + \\frac {5}{3} x = 7 2 \\quad \\frac {8}{3} x = 7 2 \\quad x = 7 2 \\div \\frac {8}{3} \\quad x = 7 2 \\times \\frac {3}{8}\\quad x = 2 7 $ 姐姐:  $27 \\times \\frac{5}{3} = 45$  (颗) 答: 姐姐做了 45 颗幸运星。",
+      "knowledge_points": [
+        "求一个数的几分之几的问题",
+        "解分数方程",
+        "列方程解含两个未知数的问题"
+      ],
+      "source": "23-24 六年级下·天津河东·期末",
+      "merged_id": "unit3_085",
+      "original_id": "unit3_solve_085"
+    },
+    {
+      "id": "unit3_solve_086",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "困难",
+      "question": "小萱在学校踢毽子比赛中踢了 60 个。她踢毽子的数量是小琳的  $\\frac{3}{4}$  。小琳踢了多少个?",
+      "options": [],
+      "answer": "80个",
+      "solution": "【分析】把小琳踢毽子的数量看作单位\"1\"，小萱踢毽子的数量是小琳的  $\\frac{3}{4}$ , 对应的是小萱踢毽子的数量 60 个, 求单位 \"1\", 用  $60 \\div \\frac{3}{4}$  解答。\n【详解】 $60 \\div \\frac{3}{4}= 6 0 \\times \\frac {4}{3}= 80  （个）$答: 小琳踢了 80 个。",
+      "knowledge_points": [
+        "已知一个数的几分之几是多少, 求这个数",
+        "分数与整数的除法"
+      ],
+      "source": "2024·辽宁盘锦·小升初真题",
+      "merged_id": "unit3_086",
+      "original_id": "unit3_solve_086"
+    },
+    {
+      "id": "unit3_solve_087",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "围棋起源于中国，春秋战国时期曾有记载，隋唐时经朝鲜传入日本，此后又流传到欧美各国，围棋蕴含着中华文化的丰富内涵，它是中国文化与文明的体现。盒子里现有黑白两种颜色的围棋子共361枚，乐乐拿出白棋子的  $\\frac{1}{5}$ ，明明拿出37枚黑棋子，剩下的白棋子数和黑棋子数正好相等。原有黑、白棋子各多少枚？",
+      "options": [],
+      "answer": "黑棋子 181 枚, 白棋子 180 枚",
+      "solution": "【分析】把原有白棋子数量看作单位\"1\"。拿出37枚黑棋子后，剩下棋子总数为  $361 - 37 = 324$  枚；此时剩下的白棋子是原来白棋子的  $1 - \\frac{1}{5} = \\frac{4}{5}$ ，且剩下白棋子数和黑棋子数相等，那么剩下棋子总数相当于原来白棋子数量的  $1 + \\frac{4}{5} = \\frac{9}{5}$  即324枚对应的分率是  $\\frac{9}{5}$ ，用  $324 \\div \\frac{9}{5}$  计算出单位\"1\"的量，也就是白棋子的数量，最后用总数量减去白棋子的数量就是原有黑棋子的数量。\n【详解】 $(361 - 37) \\div [1 + (1 - \\frac{1}{5})] = 3 2 4 \\div [ 1 + \\frac {4}{5} ]  = 3 2 4 \\div [ \\frac {5}{5} + \\frac {4}{5} ] = 3 2 4 \\div \\frac {9}{5}  = 3 2 4 \\times \\frac {5}{9} = 3 6 \\times 5= 180  （枚） $$361 - 180 = 181$  （枚）答：原有黑棋子 181 枚，白棋子 180 枚。",
+      "knowledge_points": [
+        "分数的四则混合运算",
+        "已知一个数的几分之几是多少，求这个数",
+        "分数的意义",
+        "单位\"1\"的认识与确定"
+      ],
+      "source": "2024·陕西安康·小升初真题",
+      "merged_id": "unit3_087",
+      "original_id": "unit3_solve_087"
+    },
+    {
+      "id": "unit3_solve_088",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "春节快到了、笑笑一家去采购年货，他们买了糖果、饮料和巧克力。买糖果用了120元，买饮料用的钱是糖果的  $\\frac{3}{4}$ ，是买巧克力所用钱的  $\\frac{9}{10}$ 。笑笑一家买糖果、饮料和巧克力共花了多少钱？",
+      "options": [],
+      "answer": "310元",
+      "solution": "【分析】将买糖果的钱数看作单位 \"1\", 买糖果的钱数  $\\times$  买饮料的对应分率  $=$  买饮料的钱数; 将买巧克力的钱数看作单位 \"1\", 买饮料的钱数  $\\div$  对应分率  $=$  买巧克力的钱数, 将买糖果、饮料和巧克力的钱数相加即可。\n【详解】 $120 \\times \\frac{3}{4} = 90（元）9 0 \\div \\frac {9}{1 0} = 9 0 \\times \\frac {1 0}{9} = 1 0 0 (\\text {元}) \\quad 1 2 0 + 9 0 + 1 0 0 = 3 1 0 \\text {(元)}$答: 笑笑一家买糖果、饮料和巧克力共花了 310 元钱。",
+      "knowledge_points": [
+        "求一个数的几分之几的问题",
+        "已知一个数的几分之几是多少，求这个数"
+      ],
+      "source": "24-25六年级上·山西晋中·期末",
+      "merged_id": "unit3_088",
+      "original_id": "unit3_solve_088"
+    },
+    {
+      "id": "unit3_solve_089",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "为了庆祝中国人民解放军建军 97 周年，学校组织了一次军体拳表演活动。参加军体拳表演的女生人数是男生的  $\\frac{4}{5}$ ，已知参加军体拳表演的男女生一共有 540 人, 参加军体拳表演的男生、女生各有多少人？(用方程解)",
+      "options": [],
+      "answer": "男生 300 人, 女生 240 人。",
+      "solution": "【分析】设男生为  $x$  人, 根据\"女生人数是男生的  $\\frac{4}{5}$ \", 可知女生有  $\\frac{4}{5} x$  人; 根据\"男女生一共有 540 人\"可列方程  $x + \\frac{4}{5} x = 540$ , 解方程，并求出女生人数即可。\n【详解】解: 设男生为  $x$  人, 则女生有  $\\frac{4}{5} x$  人$x + \\frac {4}{5} x = 5 4 0 \\quad \\frac {9}{5} x = 5 4 0   \\quad x = 5 4 0 \\div \\frac {9}{5} \\quad x = 5 4 0 \\times \\frac {5}{9} \\quad x = 3 0 0$女生:  $\\frac{4}{5} \\times 300 = 240$  （人）答: 参加军体拳表演的男生 300 人, 女生 240 人。",
+      "knowledge_points": [
+        "列方程解含两个未知数的问题",
+        "解分数方程",
+        "求一个数的几分之几的问题"
+      ],
+      "source": "24-25六年级上·全国·单元测试",
+      "merged_id": "unit3_089",
+      "original_id": "unit3_solve_089"
+    },
+    {
+      "id": "unit3_solve_090",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "'阅读理解、分析解答、回顾反思'是解决问题的三步骤，请你按照你的理解来看图解决问题（如下图所示）。![](images/0bb9d81160a5c343890f2607c5a408ad7c585db9cce7a74ce5de632718f19a70.jpg)(1) 阅读理解: 从图中知道全天参加活动的有 ( ) 人和 ( )。(2) 分析解答: 等量关系式是 ( )。列方程解答:(3) 回顾反思: (怎样知道你的解答是正确的?)",
+      "options": [],
+      "answer": "(1) 784;下午参加活动的人数是上午的  $(1 + \\frac{1}{3})$(2) 上午参加活动的人数 + 下午参加活动的人数 = 全天参加活动的人数列方程解答见详解(3) 上午参加活动的人数加上下午参加活动的人数是 784 人, 且它们的比是  $3: 4$ , 说明解答是正确的。",
+      "solution": "【分析】（1）把上午参加活动的人数看作单位'1'，下午参加活动的人数是上午的  $(1 + \\frac{1}{3})$ ，全天参加活动的有784人。据此解答。(2) 设上午参加活动的有  $x$  人, 则下午参加活动的人数是  $(1 + \\frac{1}{3}) x$ , 等量关系式是: 上午参加活动的人数 + 下午参加活动的人数 = 全天参加活动的人数, 据此列方程解答。(3) 上午参加活动的人数与下午参加活动的人数比是  $3: 4$ , 根据求出的  $x$  的值, 用上午参加活动的人数加上下午参加活动的人数，看是否等于 784 人，如果等于 784 人，且它们的等于 3:4，说明是正确的，否则不正确。\n【详解】（1）从图中知道全天参加活动的有 784 人和下午参加活动的人数是上午的  $(1 + \\frac{1}{3})$  。(2) 解: 设上午参加活动的有  $\\mathrm{x}$  人。等量关系式是：上午参加活动的人数 + 下午参加活动的人数 = 全天参加活动的人数。$ x + \\left(1 + \\frac {1}{3}\\right) x = 7 8 4 \\quad x + \\frac {4}{3} x = 7 8 4 \\quad \\frac {7}{3} x = 7 8 4 \\quad x = 7 8 4 \\times \\frac {3}{7}  x = 3 3 6 \\quad 3 3 6 \\times \\left(1 + \\frac {1}{3}\\right)  \\quad = 3 3 6 \\times \\frac {4}{3}  = 4 4 8 \\text {(人)} $答: 上午参加活动的有 336 人, 下午参加活动的有 448 人。(3)  $336 + 448 = 784 （人） \\quad3 3 6: 4 4 8  = (3 3 6 \\div 1 1 2): (4 4 8 \\div 1 1 2) = 3: 4$因为上午参加活动的人数加上下午参加活动的人数是 784 人,且它们的比是  $3: 4$ , 说明解答是正确的。",
+      "knowledge_points": [
+        "等式的认识及列等量关系式",
+        "解分数方程",
+        "列方程解含两个未知数的问题"
+      ],
+      "source": "24-25 六年级上·河南郑州·期末",
+      "merged_id": "unit3_090",
+      "original_id": "unit3_solve_090"
+    },
+    {
+      "id": "unit3_calc_091",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法计算",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "为了节约能源，昌盛工厂使用了节能灯，一盏节能灯1小时耗电  $\\frac{3}{250}$  千瓦时，这盏灯上个月共耗电  $\\frac{6}{5}$  千瓦时, 这盏灯上个月共使用了多少小时？",
+      "options": [],
+      "answer": "100小时",
+      "solution": "【分析】根据除法的包含意义，用上个月总的用电量除以每小时的耗电量，即可求出这节能灯的使用时间。\n【详解】  $\\frac{6}{5} \\div \\frac{3}{250}= \\frac {6}{5} \\times \\frac {2 5 0}{3} = 1 0 0 (\\text {小 时}) $答: 这盏灯上个月共使用了 100 小时。",
+      "knowledge_points": [
+        "分数与分数的除法"
+      ],
+      "source": "24-25 六年级上·吉林四平·期末",
+      "merged_id": "unit3_091",
+      "original_id": "unit3_calc_091"
+    },
+    {
+      "id": "unit3_solve_092",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "'智能亚运'是杭州第十九届亚运会的一大亮点，120件物品由甲、乙两只电子狗进行配送，甲电子狗单独送3小时完成，乙电子狗单独送4小时完成。如果两只电子狗合作完成，需要几小时送达？",
+      "options": [],
+      "answer": "$\\frac{12}{7}$小时",
+      "solution": "【分析】方法一：用需要配送的物品的总件数分别除以甲、乙用的时间，得出甲、乙电子狗每小时配送的件数，再用总件数除以甲、乙每小时配送的件数和，即可得两只电子狗合作，几小时可以送完。方法二：将这批物品看作工作总量\"1\"，甲电子狗单独送3小时完成，计算出甲的工作效率，乙电子狗单独送4小时完成，计算出乙的工作效率。合作完成，工作时间 = 工作总量 ÷ 工作效率之和，据此可以解答。选择喜欢的方法解决即可。\n【详解】方法一：$1 2 0 \\div (1 2 0 \\div 4 + 1 2 0 \\div 3) \\\\ = 1 2 0 \\div (3 0 + 4 0) \\\\ = 1 2 0 \\div 7 0 \\\\ = \\frac {1 2}{7} (\\text {小 时})$方法二：将这批物品看作工作总量\"1\"$1 \\div \\left(\\frac {1}{3} + \\frac {1}{4}\\right) \\\\ = 1 \\div \\left(\\frac {4}{1 2} + \\frac {3}{1 2}\\right) \\\\ = 1 \\div \\frac {7}{1 2} \\\\ = \\frac {1 2}{7} (\\text {小 时})$答: 如果两只电子狗合作完成, 需要  $\\frac{12}{7}$  小时送达。",
+      "knowledge_points": [
+        "分数的四则混合运算",
+        "两人合作的工程问题"
+      ],
+      "source": "24-25 六年级上·湖南岳阳·期中",
+      "merged_id": "unit3_092",
+      "original_id": "unit3_solve_092"
+    },
+    {
+      "id": "unit3_solve_093",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "困难",
+      "question": "2022年的某一天，光明学校的两位老师杨老师和刘老师进行体能训练。跑道为一个椭圆形状，他们同时从同地点出发，沿相反方向跑，每人跑完第一圈到达出发点后立即回头加速跑第二圈。跑第一圈时，刘老师的速度是杨老师的三分之二，杨老师跑第二圈时速度比第一圈提高了三分之一，刘老师跑第二圈时速度提高了五分之一，已知杨老师、刘老师第二次相遇点距第一次190米，那么你知道这条椭圆形跑道长多少米吗？",
+      "options": [],
+      "answer": "400米",
+      "solution": "【分析】设跑道长是  $x$  米, 杨老师跑第一圈的速度是  $a$ , 则刘老师跑第一圈的速度是  $\\frac{2}{3} a$  。根据题意, 把杨老师第一圈的速度看作单位\"1\"，则杨老师跑第二圈时的速度是（1+  $\\frac{1}{3}$ ） $a = \\frac{4}{3} a$ ；把刘老师第一圈的速度看作单位\"1\"，则刘老师跑第二圈的速度便是  $\\frac{2}{3} a \\times (1 + \\frac{1}{5}) = \\frac{4}{5} a$ 。那么杨老师跑第一圈所用时间是  $\\frac{x}{a}$ ，刘老师跑第一圈所用时间是  $x \\div \\frac{2}{3} a = \\frac{3x}{2a}$ ，比杨老师多用  $\\frac{3x}{2a} - \\frac{x}{a} = \\frac{x}{2a}$ 。杨老师、刘老师第一次相遇所用时间是  $x \\div (a + \\frac{2}{3} a)$ ，按杨老师前进方向距出发点为  $x \\div (a + \\frac{2}{3} a) \\times a = \\frac{3}{5} x$ （米）；当刘老师到达出发点时，杨老师自出发点回头又跑了  $\\frac{x}{2a} \\times \\frac{4}{3} a = \\frac{2}{3} x$ ；所以，杨老师、刘老师第二次相遇时，共同跑的路程是  $x - \\frac{2}{3} x = \\frac{1}{3} x$ ，按刘老师前进方向距出发点为  $\\frac{1}{3} x \\div (\\frac{4}{3} a + \\frac{4}{5} a) \\times \\frac{4}{5} a = \\frac{1}{8} x$ （米）。要注意刘老师跑第二圈的方向和杨老师跑第一圈的方向相同，已知杨老师、刘老师第二次相遇点距第一次190米，由此可得： $\\frac{3}{5} x - \\frac{1}{8} x = 190$ ，根据等式的性质解出方程即可。用含有字母的式子分别表示出两位老师第一圈和第二圈的速度、时间，从而用含有字母的式子表示出两次相遇点与出发点的距离是解题的关键。\n【详解】解：设跑道长是  $x$  米，杨老师跑第一圈的速度是  $a$  ，则刘老师跑第一圈的速度是  $\\frac{2}{3} a$  。杨老师跑第二圈时的速度:  $\\left(1 + \\frac{1}{3}\\right) a = \\frac{4}{3} a$刘老师跑第二圈的速度:  $\\frac{2}{3} a \\times (1 + \\frac{1}{5})=\\frac {2}{3} a \\times \\frac {6}{5} = \\frac {4}{5} a$杨老师跑第一圈所用时间:  $\\frac{x}{a}$刘老师跑第一圈所用时间:  $x \\div \\frac{2}{3} a  \\quad x \\times \\frac {3}{2 a} = \\frac {3 x}{2 a}$刘老师比杨老师多用时间:  $x \\div \\frac{2}{3} a - \\frac{x}{a}=x \\times \\frac {3}{2 a} - \\frac {x}{a} = \\frac {3 x}{2 a} - \\frac {x}{a} = \\frac {\\mathrm {x}}{2 \\mathrm {a}}$第一次相遇点按杨老师前进方向距出发点:  $x \\div (a + \\frac{2}{3} a) \\times a = x \\div \\frac {5}{3} a \\times a  = x \\times \\frac {3}{5 a} \\times a  = \\frac {3}{5} x (\\text {米}) $杨老师、刘老师第二次相遇时共同跑的路程:$ x - \\frac {x}{2 a} \\times \\frac {4}{3} a  = x - \\frac {2}{3} x  = \\frac {1}{3} x (\\text {米}) $第二次相遇按刘老师前进方向距出发点:$\\frac {1}{3} \\mathrm {x} \\div (\\frac {4}{3} \\mathrm {a} + \\frac {4}{5} \\mathrm {a}) \\times \\frac {4}{5} \\mathrm {a} = \\frac {1}{3} x \\div \\frac {3 2 a}{1 5} \\times \\frac {4}{5} a = \\frac {1}{3} x \\times \\frac {1 5}{3 2 a} \\times \\frac {4}{5} a  = \\frac {1}{8} x (\\text {米})$$ \\frac {3}{5} x - \\frac {1}{8} x = 1 9 0 \\quad \\frac {2 4}{4 0} x - \\frac {5}{4 0} x = 1 9 0  \\quad  \\frac {1 9}{4 0} x = 1 9 0  \\quad \\frac {1 9}{4 0} \\mathrm {x} \\times \\frac {4 0}{1 9} = 1 9 0 \\times \\frac {4 0}{1 9}\\quad  x = 4 0 0 $答: 这条椭圆形跑道长 400 米。",
+      "knowledge_points": [
+        "列方程解稍复杂的行程问题",
+        "含有字母式子的化简与求值",
+        "相遇问题",
+        "用字母表示稍复杂的数量关系"
+      ],
+      "source": "24-25六年级上·广东江门·期中",
+      "merged_id": "unit3_093",
+      "original_id": "unit3_solve_093"
+    },
+    {
+      "id": "unit3_solve_094",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "'国之重器' 东风 -17 高超音速导弹速度达 10 马赫（1 马赫 = 1 倍音速，音速为 340 米/秒），且发射的行动轨迹没有规律可循，进而无法对其进行拦截。但它的速度却比东风 -21 导弹的速度慢  $\\frac{3}{5}$ ，那么东风 -21 导弹的速度是多少米/秒？",
+      "options": [],
+      "answer": "8500米/秒",
+      "solution": "【分析】先将 10 马赫换算成  $340 \\times 10 = 3400$  米/秒。以东风 -21 的速度为单位 \"1\", 东风 -17 的速度是东风 -21 速度的  $(1 - \\frac{3}{5})$ , 已知东风 -17 的速度, 根据已知比一个数少几分之几是多少, 求这个数用除法。用东风 -17 的速度÷(1 -  $\\frac{3}{5}$  ) 即可。\n【详解】  $340 \\times 10 \\div (1 - \\frac{3}{5}) = 3 4 0 \\times 1 0 \\div \\frac {2}{5} = 3 4 0 0 \\times \\frac {5}{2} = 8 5 0 0 \\text {(米 / 秒)} $答: 东风一21导弹的速度是8500米/秒。",
+      "knowledge_points": [
+        "分数的四则混合运算",
+        "已知比一个数多/少几分之几是多少，求这个数"
+      ],
+      "source": "24-25 六年级上·河南信阳·期中",
+      "merged_id": "unit3_094",
+      "original_id": "unit3_solve_094"
+    },
+    {
+      "id": "unit3_solve_095",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "中国自主研发的'复兴号'正式运行，标志中国铁路技术装备达到'领跑世界'的先进水平。①'复兴号'的速度是350千米/时。②普通列车的速度是快速列车的  $\\frac{2}{3}$  。③'复兴号'速度比普通列车快  $\\frac{27}{8}$  。④快速列车的速度比'复兴号'慢 $\\frac{23}{35}$ 。(1) 要求普通列车的速度, 需要用到上面的信息有 ( )。(填序号)(2) 普通列车的速度是多少？",
+      "options": [],
+      "answer": "(1) ①③ (2) 80 千米/时",
+      "solution": "【分析】（1）求普通列车的速度，需要知道'复兴号'的速度是多少，以及'复兴号'速度与普通列车速度之间的关系，即需要用到①'复兴号'的速度是350千米/时；③'复兴号'速度比普通列车快  $\\frac{27}{8}$  （答案不唯一）。(2) 把普通列车的速度看作单位 \"1\", \"复兴号\" 速度是普通列车速度的  $(1 + \\frac{27}{8})$ , 对应的是 \"复兴号\" 的速度,求单位 \"1\", 用 \"复兴号\" 的速度  $\\div$ $(1 + \\frac{27}{8})$  解答。\n【详解】（1）要求普通列车的速度，需要用到上面的信息有①和③。（答案不唯一）(2)  $350 \\div (1 + \\frac{27}{8})= 3 5 0 \\div \\frac {3 5}{8} = 3 5 0 \\times \\frac {8}{3 5}  = 8 0 \\text {(千 米 / 时)}$答: 普通列车的速度是 80 千米/时。",
+      "knowledge_points": [
+        "已知比一个数多/少几分之几是多少，求这个数",
+        "\"提问题\"\"填条件\"问题"
+      ],
+      "source": "24-25六年级上·广东东莞·期中",
+      "merged_id": "unit3_095",
+      "original_id": "unit3_solve_095"
+    },
+    {
+      "id": "unit3_solve_096",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "国家设立科学技术奖的目的是营造尊重知识、尊重人才的良好社会氛围，鼓励广大科技工作者勤奋工作，为实施科教兴国战略做出更大的贡献。某年某省科学技术奖共授奖 300 项（人）。技术发明奖一等奖占科学技术奖总数的  $\\frac{1}{25}$ ，也是技术发明奖的  $\\frac{3}{10}$ 。科学技术成果推广奖数量是技术发明奖数量的  $\\frac{1}{4}$ ，科学技术成果推广奖比科技型中小企业创新奖少  $\\frac{1}{3}$ 。(1) 获评技术发明奖的有多少项?(2) 获评科技型中小企业创新奖的有多少项? (列方程解)",
+      "options": [],
+      "answer": "(1) 40 项 (2) 15项",
+      "solution": "【分析】（1）已知科学技术奖共授奖300项，技术发明奖一等奖占科学技术奖总数的  $\\frac{1}{25}$ ，把科学技术奖总数看作单位\"1\"，单位\"1\"已知，用科学技术奖总数乘  $\\frac{1}{25}$ ，求出技术发明奖一等奖数量；又已知技术发明奖一等奖也是技术发明奖的  $\\frac{3}{10}$ , 把技术发明奖数量看作单位 \"1\", 单位 \"1\" 未知, 用技术发明奖一等奖数量除以  $\\frac{3}{10}$ , 求出技术发明奖的数量。(2) 已知科学技术成果推广奖数量是技术发明奖数量的  $\\frac{1}{4}$ ,把技术发明奖数量看作单位\"1\"，单位\"1\"已知，用技术发明奖数量乘  $\\frac{1}{4}$ ，求出科学技术成果推广奖数量；已知科学技术成果推广奖比科技型中小企业创新奖少  $\\frac{1}{3}$ ，把科技型中小企业创新奖数量看作单位\"1\"，则科学技术成果推广奖数量是科技型中小企业创新奖的  $(1 - \\frac{1}{3})$ ，得出等量关系：科技型中小企业创新奖数量  $\\times (1 - \\frac{1}{3}) =$  科学技术成果推广奖数量，据此列出方程，并求解。\n【详解】 (1)  $300 \\times \\frac{1}{25} \\div \\frac{3}{10} = 1 2 \\div \\frac {3}{1 0}  = 1 2 \\times \\frac {1 0}{3}  = 4 0 \\text {(项)}$答: 获评技术发明奖的有 40 项。(2) 解: 设获评科技型中小企业创新奖的有  $x$  项。$(1 - \\frac {1}{3}) x = 4 0 \\times \\frac {1}{4} \\quad \\frac {2}{3} x = 1 0  \\quad x = 1 0 \\div \\frac {2}{3}   x = 1 0 \\times \\frac {3}{2}  x = 1 5$答: 获评科技型中小企业创新奖的有 15 项。",
+      "knowledge_points": [
+        "分数的乘、除法的混合运算",
+        "求一个数的几分之几的问题",
+        "已知一个数的几分之几是多少，求这个数",
+        "列方程解含一个未知数的问题"
+      ],
+      "source": "24-25六年级上·福建福州·期中",
+      "merged_id": "unit3_096",
+      "original_id": "unit3_solve_096"
+    },
+    {
+      "id": "unit3_solve_097",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "《诗经》是我国古代诗歌的开端, 共 305 篇, 在内容上分为《风》《雅》《颂》三部分，反映了周初至周晚期约五百年间的社会风貌。其中《雅》的篇数是《风》的  $\\frac{21}{32}$ ，《颂》的篇数是《风》的  $\\frac{1}{4}$ ，《诗经》中《风》《雅》《颂》各有多少篇？",
+      "options": [],
+      "answer": "《风》：160篇;《雅》：105篇;《颂》：40篇",
+      "solution": "【分析】（1）分析题目，把《风》的篇数看作单位\"1\"，则《雅》的篇数是  $\\frac{21}{32}$ ，《颂》的篇数是  $\\frac{1}{4}$ ，《风》的篇数的(1 +  $\\frac{21}{32} + \\frac{1}{4}$  ) 就是 305 , 据此根据已知一个数的几分之几是多少求这个数用除法列式求出《风》的篇数;(2) 根据求一个数的几分之几是多少用乘法, 用《风》的篇数分别乘  $\\frac{21}{32} 、 \\frac{1}{4}$  求出《雅》、《颂》的篇数。\n【详解】 《风》:  $305 \\div (1 + \\frac{21}{32} + \\frac{1}{4}) = 3 0 5 \\div (\\frac {5 3}{3 2} + \\frac {1}{4})= 3 0 5 \\div \\frac {6 1}{3 2} = 3 0 5 \\times \\frac {3 2}{6 1}= 1 6 0 \\text {(篇)} $《雅》:  $160 \\times \\frac{21}{32} = 105$  (篇)《颂》:  $160 \\times \\frac{1}{4} = 40$  (篇)答：《风》有160篇，《雅》有105篇，《颂》有40篇。",
+      "knowledge_points": [
+        "分数的四则混合运算",
+        "求一个数的几分之几的问题",
+        "已知一个数的几分之几是多少，求这个数"
+      ],
+      "source": "24-25 六年级上·山东济南·期中",
+      "merged_id": "unit3_097",
+      "original_id": "unit3_solve_097"
+    },
+    {
+      "id": "unit3_solve_098",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "飞跃家电城今天卖出两台电视机, 售价均为 3600 元, 一台赚  $\\frac{1}{5}$ , 一台亏  $\\frac{1}{5}$  。请你帮老板算一算, 他今天是赚了还是亏了? 若赚了, 赚了多少钱？若亏了，亏了多少钱？",
+      "options": [],
+      "answer": "亏了，亏了 300 元",
+      "solution": "【分析】根据题意，把赚  $\\frac{1}{5}$  的电视机进价看作单位\"1\"，则售价是进价的  $\\left(1 + \\frac{1}{5}\\right)$ ，用售价除以  $\\left(1 + \\frac{1}{5}\\right)$ ，即可求出赚  $\\frac{1}{5}$  的电视机进价；把亏  $\\frac{1}{5}$  的电视机进价看作单位\"1\"，则售价是进价的  $\\left(1 - \\frac{1}{5}\\right)$ ，用售价除以  $\\left(1 - \\frac{1}{5}\\right)$ ，即可求出亏  $\\frac{1}{5}$  的电视机进价；再把两台电视机的进价相加，与两台电视机的售价之和进行比较，即可求出老板今天是赚还是亏。本题考查分数除法的计算及应用，解答的关键是要找准单位\"1\"，注意正确计算进价和售价的差异。\n【详解】  $3600 \\div \\left(1 + \\frac{1}{5}\\right) = 3 6 0 0 \\div \\frac {6}{5}  = 3 6 0 0 \\times \\frac {5}{6}=3000（元）$$ 3 6 0 0 \\div \\left(1 - \\frac {1}{5}\\right) = 3 6 0 0 \\div \\frac {4}{5} = 3 6 0 0 \\times \\frac {5}{4}  = 4 5 0 0 \\text {(元)}$$3 0 0 0 + 4 5 0 0 = 7 5 0 0 \\text {(元)} \\\\ 3 6 0 0 + 3 6 0 0 = 7 2 0 0 \\text {(元)} \\\\ 7 5 0 0 > 7 2 0 0 \\\\ 7 5 0 0 - 7 2 0 0 = 3 0 0 (\\text {元})$答: 老板他今天是亏了, 亏了 300 元。",
+      "knowledge_points": [
+        "分数的四则混合运算",
+        "已知比一个数多/少几分之几是多少，求这个数"
+      ],
+      "source": "22-23六年级上·江西南昌·期中",
+      "merged_id": "unit3_098",
+      "original_id": "unit3_solve_098"
+    },
+    {
+      "id": "unit3_solve_099",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "甲、乙两筐苹果共重 135 千克, 如果从甲筐拿出  $\\frac{1}{5}$  后, 就和乙筐重量相等, 甲筐苹果重多少千克?",
+      "options": [],
+      "answer": "75千克",
+      "solution": "【分析】设甲筐苹果重  $\\mathrm{x}$  千克；则乙筐苹果重（  $135 - x$  ）千克；如果从甲筐拿出  $\\frac{1}{5}$  后，就和乙筐重量相等，即拿出  $\\frac{1}{5} x$  千克，甲筐还剩下  $(x - \\frac{1}{5} x)$  千克，等于乙筐苹果重量；列方程：  $x - \\frac{1}{5} x = 135 - x$  ，解方程，即可解答。\n【详解】解：设甲筐苹果重  $\\mathrm{x}$  千克，则乙筐苹果重（135- x）千克。$x - \\frac {1}{5} x = 1 3 5 - x   \\quad \\frac {4}{5} x + x = 1 3 5 \\quad \\frac {9}{5} x = 1 3 5 \\quad   x = 1 3 5 \\times \\frac {5}{9} \\quad x=75$答: 甲筐苹果重 75 千克。",
+      "knowledge_points": [
+        "列方程解含两个未知数的问题",
+        "分数与整数的除法"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_099",
+      "original_id": "unit3_solve_099"
+    },
+    {
+      "id": "unit3_solve_100",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "甲汽车每小时行 60 千米, 是乙汽车速度的  $\\frac{4}{5}$ , 乙汽车每小时行多少千米?",
+      "options": [],
+      "answer": "75千米",
+      "solution": "【分析】已知甲汽车每小时行 60 千米, 是乙汽车速度的  $\\frac{4}{5}$ ,把乙汽车的速度看作单位 \"1\", 单位 \"1\"未知, 用甲汽车的速度除以  $\\frac{4}{5}$ , 即是乙汽车的速度。\n【详解】  $60 \\div \\frac{4}{5} = 6 0 \\times \\frac {5}{4} = 7 5 \\text {(千 米 / 小 时)}$答: 乙汽车每小时行 75 千米。",
+      "knowledge_points": [
+        "已知一个数的几分之几是多少，求这个数",
+        "分数与整数的除法"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_100",
+      "original_id": "unit3_solve_100"
+    },{
+      "id": "unit3_choice_101",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "某体育器材室有 20 个篮球, 比足球的个数少 $\\frac{1}{5}$ , 该体育器材室有 ( ) 个足球。",
+      "options": [
+        "A. 15",
+        "B. 16",
+        "C. 25",
+        "D. 26"
+      ],
+      "answer": "C",
+      "solution": "【分析】把足球的个数看作单位 \"1\", 篮球的个数是足球的 $(1 - \\frac{1}{5})$ ; 篮球的个数已知, 根据已知一个数的几分之几是多少, 求这个数, 用除法计算; 用 20 除以 $(1 - \\frac{1}{5})$ , 所得结果即为足球的个数。\n【详解】 $20 \\div \\left(1 - \\frac{1}{5}\\right)$ $\\ 2 0 \\div \\frac {4}{5}$因此该体育器材室有25个足球。故答案为: C",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "已知比一个数多/少几分之几是多少，求这个数"
+      ],
+      "source": "23-24 六年级下·陕西榆林·期末",
+      "merged_id": "unit3_101",
+      "original_id": "unit3_choice_101"
+    },
+    {
+      "id": "unit3_choice_102",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "把 7 升水倒入容量为 $\\frac{2}{3}$ 升的杯子中, 需要 ( ) 个杯子。",
+      "options": [
+        "A. 10",
+        "B. 11",
+        "C. 12"
+      ],
+      "answer": "B",
+      "solution": "【分析】水的体积÷杯子容积 = 需要的杯子个数，除以一个数等于乘这个数的倒数，据此列式计算，最后无论剩下多少水都得需要一个杯子，因此结果用进一法保留整数。\n【详解】 $7 \\div \\frac{2}{3} = 7 \\times \\frac{3}{2} = 10.5 \\approx 11$ （个）需要 11 个杯子。故答案为: B",
+      "knowledge_points": [
+        "分数与整数的除法"
+      ],
+      "source": "22-23 五年级下·河北邢台·期末",
+      "merged_id": "unit3_102",
+      "original_id": "unit3_choice_102"
+    },
+    {
+      "id": "unit3_choice_103",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "把一根长 $\\frac{2}{5}$ 米的彩带对折三次, 沿折痕剪掉, 每段彩带长 ( ) 米。",
+      "options": [
+        "A. $\\frac{2}{15}$",
+        "B. $\\frac{1}{10}$",
+        "C. $\\frac{1}{20}$"
+      ],
+      "answer": "C",
+      "solution": "【分析】把一根彩带对折三次，平均分成 $(2 \\times 2 \\times 2)$ 段，彩带长度÷段数 = 每段长度，据此列式计算。\n【详解】 $\\frac{2}{5} \\div (2 \\times 2 \\times 2)$ $\\ {2}{5} \\div 8 = \\frac {2}{5} \\times \\frac {1}{8} = \\frac {1}{2 0} (\\text {米})$每段彩带长 $\\frac{1}{20}$ 米。故答案为: C",
+      "knowledge_points": [
+        "分数的平均分",
+        "分数与整数的除法"
+      ],
+      "source": "22-23 六年级上·江苏淮安·阶段练习",
+      "merged_id": "unit3_103",
+      "original_id": "unit3_choice_103"
+    },
+    {
+      "id": "unit3_choice_104",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "我国幅员辽阔，东西相距约5200千米，是南北距离的 $\\frac{52}{55}$ ，则南北距离是（）。",
+      "options": [
+        "A. 5100千米",
+        "B. 5300千米",
+        "C. 5500千米",
+        "D. 5700 千米"
+      ],
+      "answer": "C",
+      "solution": "【分析】把南北距离看作单位 \"1\", 东西距离是南北距离的 $\\frac{52}{55}$ , 对应的是东西距离 5200 千米, 求单位 \"1\", 用东西距离 $\\div \\frac{52}{55}$ , 即可解答。\n【详解】 $5200 \\div \\frac{52}{55} = 5 2 0 0 \\times \\frac {5 5}{5 2} = 5 5 0 0 \\text {(千 米)}$我国幅员辽阔，东西相距约5200千米，是南北距离的 $\\frac{52}{55}$ ，则南北距离是5500千米。故答案为: C",
+      "knowledge_points": [
+        "已知一个数的几分之几是多少，求这个数"
+      ],
+      "source": "23-24六年级上·湖南张家界·期末",
+      "merged_id": "unit3_104",
+      "original_id": "unit3_choice_104"
+    },
+    {
+      "id": "unit3_choice_105",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "一袋土豆，吃掉它的 $\\frac{2}{5}$ ，还剩下60千克，这袋土豆原有（）千克。",
+      "options": [
+        "A. 24",
+        "B. 36",
+        "C. 100"
+      ],
+      "answer": "C",
+      "solution": "【分析】把这袋土豆原有的质量看作单位 \"1\", 吃掉它的 $\\frac{2}{5}$ ,则还剩下的土豆质量占原有土豆的 $(1 - \\frac{2}{5})$ , 单位 \"1\"未知, 用还剩下的土豆质量除以 $(1 - \\frac{2}{5})$ , 即可求出这袋土豆原有的质量。\n【详解】 $60 \\div (1 - \\frac{2}{5})$ $\\begin{array}{l} = 6 0 \\div \\frac {3}{5} = 6 0 \\times \\frac {5}{3} = 1 0 0 \\text {(千 克)} \\\\ \\end{array}$这袋土豆原有 100 千克。故答案为: C",
+      "knowledge_points": [
+        "已知一部分量占总量的几分之几及另一部分量,求总量"
+      ],
+      "source": "23-24六年级上·新疆喀什·期末",
+      "merged_id": "unit3_105",
+      "original_id": "unit3_choice_105"
+    },
+    {
+      "id": "unit3_choice_106",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "一袋土豆, 吃了它的 $\\frac{2}{5}$ , 还剩 30 千克, 这袋土豆原有 ( ) 千克。",
+      "options": [
+        "A. 20",
+        "B. 50",
+        "C. 75"
+      ],
+      "answer": "B",
+      "solution": "【分析】将原有质量看作单位 \"1\", 吃了它的 $\\frac{2}{5}$ , 还剩 (1 - $\\frac{2}{5}$ ), 剩下的质量÷对应分率 = 原有质量, 据此列式计算。\n【详解】 $30 \\div (1 - \\frac{2}{5}) = 3 0 \\div \\frac {3}{5} = 3 0 \\times \\frac {5}{3} = 5 0 \\text {(千 克)}$这袋土豆原有 50 千克。故答案为: B",
+      "knowledge_points": [
+        "已知一个数的几分之几是多少，求这个数"
+      ],
+      "source": "19-20 六年级上·四川广元·期中",
+      "merged_id": "unit3_106",
+      "original_id": "unit3_choice_106"
+    },
+    {
+      "id": "unit3_choice_107",
+      "unit": "第三单元：分数除法",
+      "chapter": "倒数的认识",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "两个自然数的倒数和是 $\\frac{5}{6}$ ,这两个数是 ( )。",
+      "options": [
+        "A. 2和4",
+        "B. 5和6",
+        "C. 2和3"
+      ],
+      "answer": "C",
+      "solution": "【分析】假设这两个自然数为 a 和 b, 则这两个自然数的倒数为 $\\frac{1}{a}$ 和 $\\frac{1}{b}$ , 这两个自然数的倒数和等于 $\\left(\\frac{1}{a} + \\frac{1}{b}\\right)$ , 将各选项的数分别带入算式进行计算, 即可选出正确答案。\n【详解】根据分析, 将各选项数值带入 $\\left(\\frac{1}{a} + \\frac{1}{b}\\right)$\nA. $\\frac{1}{2} + \\frac{1}{4}$ $\\begin{array}{l} = \\frac {2}{4} + \\frac {1}{4} = \\frac {3}{4} \\end{array}$所以 A 不符合题意; B. $\\frac{1}{5} + \\frac{1}{6} = \\frac {6}{3 0} + \\frac {5}{3 0} = \\frac {1 1}{3 0} $所以 B 不符合题意; C. $\\frac{1}{2} + \\frac{1}{3} = \\frac {3}{6} + \\frac {2}{6} = \\frac {5}{6} $所以 C 符合题意; 故答案为: C",
+      "knowledge_points": [
+        "异分母分数加、减法",
+        "自然数与倒数的和或差的问题"
+      ],
+      "source": "2014 五年级·全国·课后作业",
+      "merged_id": "unit3_107",
+      "original_id": "unit3_choice_107"
+    },
+    {
+      "id": "unit3_choice_108",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "一个数的 $\\frac{4}{5}$ 是 $\\frac{2}{3}$ , 求这个数, 下列算式中正确的是 ( )。",
+      "options": [
+        "A. $\\frac{4}{5} +\\frac{2}{3}$",
+        "B. $\\frac{4}{5} \\times \\frac{2}{3}$",
+        "C. $\\frac{2}{3} \\div \\frac{4}{5}$",
+        "D. $\\frac{2}{3} \\times \\frac{4}{5}$"
+      ],
+      "answer": "C",
+      "solution": "【分析】根据已知一个数的几分之几是多少，求这个数，用除法计算, 即用 $\\frac{2}{3}$ 除以 $\\frac{4}{5}$ 即可求出这个数。\n【详解】由分析可知：一个数的 $\\frac{4}{5}$ 是 $\\frac{2}{3}$ , 求这个数, 列式为 $\\frac{2}{3} \\div \\frac{4}{5}$ 。 故答案为: C\n【点睛】本题考查已知一个数的几分之几是多少, 求这个数, 明确用除法是解题的关键。",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "分数除法的应用"
+      ],
+      "source": "22-23 六年级上·上海静安·期中",
+      "merged_id": "unit3_108",
+      "original_id": "unit3_choice_108"
+    },
+    {
+      "id": "unit3_choice_109",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "选择题",
+      "difficulty": "困难",
+      "question": "小明从学校步行回家，当走了全程的一半时下雨了，他继续走；当雨停了时，剩下路程是他在雨中步行路程的 $\\frac{5}{7}$ ，那么，小明在雨中步行的路程是全程的（）。",
+      "options": [
+        "A. $\\frac{5}{14}$",
+        "B. $\\frac{5}{24}$",
+        "C. $\\frac{7}{24}$",
+        "D. $\\frac{2}{7}$"
+      ],
+      "answer": "C",
+      "solution": "【分析】可设小明在雨中步行的路程为 7 份, 那么剩下的路程为 5 份; 这也就是说 $5 + 7 = 12$ 份是全程份数的一半, 即全程为 $12 \\times 2 = 24$ 份; 最后用步行的份数除以全程的份数即可。解答此题的关键是弄清 \"雨中步行路程、剩下路程与全程之间的关系\"。\n【详解】设小明在雨中步行的路程为 7 份, 剩下的路程为 5 份, 则得: $5 + 7 = 1 2 \\quad (\\text {份})$全程为 $12 \\times 2 = 24$ （份） $7 \\div 2 4 = \\frac {7}{2 4} $ 所以, 小明在雨中步行的路程是全程的 $\\frac{7}{24}$ 。 故答案为: C",
+      "knowledge_points": [
+        "求一个数占另一个数几分之几",
+        "分数的意义."
+      ],
+      "source": "2023 六年级上·全国·专题练习",
+      "merged_id": "unit3_109",
+      "original_id": "unit3_choice_109"
+    },
+    {
+      "id": "unit3_choice_110",
+      "unit": "第三单元：分数除法",
+      "chapter": "倒数的认识",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "以下各组中的两个数互为倒数的是 ( )。",
+      "options": [
+        "A. 4 和 $\\frac{1}{4}$",
+        "B. $1 \\frac{2}{3}$ 和 $\\frac{3}{2}$",
+        "C. 0.6 和 $\\frac{3}{5}$"
+      ],
+      "answer": "A",
+      "solution": "【分析】根据倒数的意义，乘积是 1 的两个数互为倒数；据此计算出各个选项中两个数的积，看看是不是 1 即可解题。正确理解倒数的意义，是解答此题的关键。\n【详解】A. $4 \\times \\frac{1}{4} = 1$ ，所以4和 $\\frac{1}{4}$ 互为倒数；B. $1 \\frac{2}{3} \\times \\frac{3}{2} = \\frac{5}{3} \\times \\frac{3}{2} = \\frac{5}{2}, \\frac{5}{2} \\neq 1$ , 所以, $1 \\frac{2}{3}$ 不是 $\\frac{3}{2}$ 的倒数;\nC. $0.6 \\times \\frac{3}{5} = \\frac{3}{5} \\times \\frac{3}{5} = \\frac{9}{25}, \\frac{9}{25} \\neq 1$ , 所以, 0.6 不是 $\\frac{3}{5}$ 的倒数。 故答案为: A",
+      "knowledge_points": [
+        "分数乘小数",
+        "倒数的认识及应用",
+        "分数乘分数"
+      ],
+      "source": "22-23六年级上·山西晋中·期末",
+      "merged_id": "unit3_110",
+      "original_id": "unit3_choice_110"
+    },
+    {
+      "id": "unit3_choice_111",
+      "unit": "第三单元：分数除法",
+      "chapter": "倒数的认识",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "如果 $x$ 、 $y$ 互为倒数,那么 \" $xy + 3$ \" 的计算结果是 ( )。",
+      "options": [
+        "A. 3",
+        "B. 4",
+        "C. 不能确定"
+      ],
+      "answer": "B",
+      "solution": "【分析】根据倒数的意义：乘积是 1 的两个数互为倒数，即xy=1, 由此即可求出 $x y + 3$ 的值。本题主要考查倒数的意义, 熟练掌握倒数的含义是解题的关键。\n【详解】由分析可知： $\\mathrm {x y} = 1$ $x y + 3 = 1 + 3 = 4$ 故答案为: B",
+      "knowledge_points": [
+        "倒数的认识及应用",
+        "含有字母式子的化简与求值。"
+      ],
+      "source": "22-23 五年级下·上海·单元测试",
+      "merged_id": "unit3_111",
+      "original_id": "unit3_choice_111"
+    },
+    {
+      "id": "unit3_choice_112",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "一项工作, 甲队 2 天完成这项工作的 $\\frac{1}{3}$ , 按照这个进度, ( )天可以完成这项工作。",
+      "options": [],
+      "answer": "6",
+      "solution": "【分析】将总天数看作单位\"1\"，甲队用的天数÷对应分率=总天数，据此列式计算。\n【详解】 $2 \\div \\frac{1}{3} = 2 \\times 3 = 6$ （天） 6天可以完成这项工作。",
+      "knowledge_points": [
+        "已知一个数的几分之几是多少，求这个数"
+      ],
+      "source": "23-24 六年级下·湖北武汉·期末",
+      "merged_id": "unit3_112",
+      "original_id": "unit3_choice_112"
+    },
+    {
+      "id": "unit3_choice_113",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "修一条路, 已经修了 $\\frac{1}{3}$ 正好是 3 千米, 这条路长( )千米。",
+      "options": [],
+      "answer": "9",
+      "solution": "【分析】把这条路的长度看作单位 \"1\", 依据题意可得: 这条路长度的 $\\frac{1}{3}$ 是 3 千米, 运用分数除法意义即可解答。\n【详解】 $3 \\div \\frac{1}{3} = 3 \\times 3 = 9$ （千米） 这条路长 9 千米。",
+      "knowledge_points": [
+        "已知一个数的几分之几是多少，求这个数"
+      ],
+      "source": "23-24 六年级下·江苏淮安·期中",
+      "merged_id": "unit3_113",
+      "original_id": "unit3_choice_113"
+    },
+    {
+      "id": "unit3_choice_114",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "一个县去年造林1200公顷，比原计划多造 $\\frac{1}{5}$ ，这个县去年计划造林( )公顷。",
+      "options": [],
+      "answer": "1000",
+      "solution": "【分析】已知去年造林面积比原计划多造 $\\frac{1}{5}$ , 把原计划造林面积看作单位 \"1\", 则去年造林面积是原计划的 $(1 + \\frac{1}{5})$ , 单位 \"1\" 未知, 用去年造林面积除以 $(1 + \\frac{1}{5})$ , 求出原计划造林面积。\n【详解】 $1200 \\div (1 + \\frac{1}{5})$ $\\begin{array}{l} = 1 2 0 0 \\div \\frac {6}{5} = 1 2 0 0 \\times \\frac {5}{6} = 1 0 0 0 \\text {(公 坝)} \\\\ \\end{array}$ 这个县去年计划造林 1000 公顷。",
+      "knowledge_points": [
+        "分数的四则混合运算",
+        "已知比一个数多/少几分之几是多少，求这个数"
+      ],
+      "source": "22-23六年级下·山东济南·开学考试",
+      "merged_id": "unit3_114",
+      "original_id": "unit3_choice_114"
+    },
+    {
+      "id": "unit3_choice_115",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "一堆货物, 运走了 $\\frac{5}{6}$ ,还剩 30 吨, 这堆货物原来有_吨。",
+      "options": [],
+      "answer": "180",
+      "solution": "【分析】将原来吨数看作单位 \"1\", 运走了 $\\frac{5}{6}$ , 还剩 $(1 - \\frac{5}{6})$ , 剩下的吨数 $\\div$ 对应分率 $=$ 原来的吨数, 据此列式计算。\n【详解】 $30 \\div (1 - \\frac{5}{6})$ $\\begin{array}{l} = 3 0 \\div \\frac {1}{6} = 3 0 \\times 6 = 1 8 0 (\\text {吨}) \\\\ \\end{array}$ 这堆货物原来有 180 吨。",
+      "knowledge_points": [
+        "已知一部分量占总量的几分之几及另一部分量,求总量"
+      ],
+      "source": "22-23 六年级上·山西忻州·期末",
+      "merged_id": "unit3_115",
+      "original_id": "unit3_choice_115"
+    },
+    {
+      "id": "unit3_choice_116",
+      "unit": "第三单元：分数除法",
+      "chapter": "倒数的认识",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "如果 $a$ 与 $b$ 互为倒数, 那么 $\\frac{a}{8} \\times \\frac{b}{7} + \\frac{1}{ab} = (\\quad)$ 。",
+      "options": [],
+      "answer": "$1 \\frac{1}{56}$",
+      "solution": "【分析】乘积是 1 的两个数互为倒数, 分数与分数相乘, 用分子相乘的积作为分子, 分母相乘的积作为分母。据此进行计算化简, 并将 $ab = 1$ 代入求值即可。\n【详解】a 与 b 互为倒数, 那么 $ab = 1$ , $\\frac{a}{8} \\times \\frac{b}{7} + \\frac{1}{ab} = \\frac{ab}{56} + \\frac{1}{ab} = \\frac{1}{56} + 1 = 1\\frac{1}{56}$ 。",
+      "knowledge_points": [
+        "分数乘分数",
+        "与倒数有关的综合计算"
+      ],
+      "source": "23-24 六年级上·湖北武汉·期末",
+      "merged_id": "unit3_116",
+      "original_id": "unit3_choice_116"
+    },
+    {
+      "id": "unit3_choice_117",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "一辆小轿车每行6千米耗油 $\\frac{3}{5}$ 升，平均每升汽油可以行驶( )千米，行1千米要耗油( )升。",
+      "options": [],
+      "answer": "$10\\frac{1}{10}$",
+      "solution": "【分析】由题意可知, 用小轿车行驶的距离除以耗油量即可求出平均每升汽油可以行驶多少千米; 用耗油量除以行驶的距离即可求出行 1 千米要耗油多少升。\n【详解】 $6 \\div \\frac{3}{5} = 6 \\times \\frac{5}{3} = 10$ （千米） $\\frac {3}{5} \\div 6 = \\frac {3}{5} \\times \\frac {1}{6} = \\frac {1}{1 0} (\\text {升})$ 则平均每升汽油可以行驶 10 千米, 行 1 千米要耗油 $\\frac{1}{10}$ 升。",
+      "knowledge_points": [
+        "分数与整数的除法"
+      ],
+      "source": "15-16六年级上·云南玉溪·期末",
+      "merged_id": "unit3_117",
+      "original_id": "unit3_choice_117"
+    },
+    {
+      "id": "unit3_choice_118",
+      "unit": "第三单元：分数除法",
+      "chapter": "倒数的认识",
+      "question_type": "填空题",
+      "difficulty": "困难",
+      "question": "两个连续的奇数的倒数差是 $\\frac{2}{143}$ , 这两个连续的奇数分别是( )和( )。",
+      "options": [],
+      "answer": "11 13",
+      "solution": "【分析】两个连续奇数一定是互质数，它们的倒数差的分母是这两个连续奇数的积。那么由 $143 = 11 \\times 13$ 可把 143 拆成两个奇数 11 和 13 的乘积即可得解。\n【详解】由分析可得: $143 = 11 \\times 13, \\frac{1}{11} - \\frac{1}{13} = \\frac{2}{143}$ , 那么这两个连续的奇数分别是 11 和 13。",
+      "knowledge_points": [
+        "奇数与偶数的认识",
+        "自然数与倒数的和或差的问题"
+      ],
+      "source": "23-24 六年级上·浙江杭州·期中",
+      "merged_id": "unit3_118",
+      "original_id": "unit3_choice_118"
+    },
+    {
+      "id": "unit3_choice_119",
+      "unit": "第三单元：分数除法",
+      "chapter": "倒数的认识",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "( )的倒数等于0.7; 比 20 克少 $\\frac{1}{4}$ 是( )克; 25 立方分米比( )立方分米多 $\\frac{2}{3}$ 。",
+      "options": [],
+      "answer": "$1 \\frac{3}{7}\\quad 15 \\quad15$",
+      "solution": "【分析】根据倒数的意义，乘积是 1 的两个数互为倒数；求一个小数的倒数，用 1 除以这个小数即可，可得： $1 \\div 0.7 = 1^{\\frac{3}{7}}$ ；要求比 20 克少 $\\frac{1}{4}$ 是多少克，用 20 乘 $(1 - \\frac{1}{4})$ ；要求 25 立方分米比多少立方分米多 $\\frac{2}{3}$ ，用 25 除以 $(1 + \\frac{2}{3})$ 即可。\n【详解】 $1 \\div 0.7$ $= 1 \\div \\frac {7}{1 0} = 1 \\times \\frac {1 0}{7} = \\frac {1 0}{7} = 1 \\frac {3}{7} $$2 0 \\times (1 - \\frac {1}{4}$ $\\begin{array}{l} = 2 0 \\times \\frac {3}{4} = 1 5 \\text {(克)} \\\\ \\end{array}$$2 5 \\div (1 + \\frac {2}{3}) = 2 5 \\div \\frac {5}{3} = 2 5 \\times \\frac {3}{5} = 1 5 \\text {(立 方 分 米)} $所以, $1 \\frac{3}{7}$ 的倒数等于 0.7 ; 比 20 克少 $\\frac{1}{4}$ 是 15 克; 25 立方分米比 15 立方分米多 $\\frac{2}{3}$ 。",
+      "knowledge_points": [
+        "倒数的认识及应用",
+        "分数的四则混合运算",
+        "求比一个数多/少几分之几的数是多少"
+      ],
+      "source": "2024 六年级下·全国·专题练习",
+      "merged_id": "unit3_119",
+      "original_id": "unit3_choice_119"
+    },
+    {
+      "id": "unit3_choice_120",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "近年来, 我国在治理雾霾方面取得了一定的成效，我国东部某市今年5月份有12 天雾霾天气，比去年减少了 $\\frac{1}{5}$ ，这个城市去年 5 月份有( )天雾霾天气。",
+      "options": [],
+      "answer": "15",
+      "solution": "【分析】把去年 5 月份的雾霾天数看作单位 \"1\", 今年 5 月份雾霾天数是去年的 $(1 - \\frac{1}{5})$ , 根据分数除法的意义, 用 $12 \\div (1 - \\frac{1}{5})$ 即可求出去年 5 月份雾霾天数。\n【详解】 $12 \\div (1 - \\frac{1}{5})$ $\\begin{array}{l} = 1 2 \\div \\frac {4}{5} = 1 2 \\times \\frac {5}{4} = 1 5 (\\text {天}) \\\\ \\end{array}$ 这个城市去年 5 月份有 15 天雾霾天气。",
+      "knowledge_points": [
+        "已知比一个数多/少几分之几是多少，求这个数"
+      ],
+      "source": "2024 六年级下·全国·专题练习",
+      "merged_id": "unit3_120",
+      "original_id": "unit3_choice_120"
+    },
+    {
+      "id": "unit3_choice_121",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "甲数比乙数多 $\\frac{1}{5}$ , 乙数比甲数少( )。",
+      "options": [],
+      "answer": "$\\frac{1}{6}$",
+      "solution": "【分析】甲数比乙数多 $\\frac{1}{5}$ , 把乙数看作单位 \"1\", 则甲数是乙数的 $(1 + \\frac{1}{5})$ , 用 1 乘 $(1 + \\frac{1}{5})$ 即可求出甲数。求一个数比另一个数多（或少）几分之几，先求出多（或少）的具体数量，再除以单位\"1\"数量即可解答。据此求乙数比甲数少几分之几，用两数的差除以甲数（单位\"1\"）即可解答。\n【详解】 $1 \\times (1 + \\frac{1}{5}) = 1 \\times 1 \\frac {1}{5} = 1 \\frac {1}{5} $$ \\left(1 \\frac {1}{5} - 1\\right) \\div 1 \\frac {1}{5} = \\frac {1}{5} \\div \\frac {6}{5} = \\frac {1}{5} \\times \\frac {5}{6}= \\frac {1}{6}$ 则乙数比甲数少 $\\frac{1}{6}$ 。",
+      "knowledge_points": [
+        "分数的四则混合运算"
+      ],
+      "source": "18-19 六年级下·江苏·期末",
+      "merged_id": "unit3_121",
+      "original_id": "unit3_choice_121"
+    },
+    {
+      "id": "unit3_choice_122",
+      "unit": "第三单元：分数除法",
+      "chapter": "倒数的认识",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "一个自然数与它倒数的和是 8.125, 则这个自然数与它倒数的差是( )。",
+      "options": [],
+      "answer": "7.875",
+      "solution": "【分析】把 8.125 转化成带分数 $8 \\frac{1}{8}$ , 发现自然数 8 和它的倒数 $\\frac{1}{8}$ 的和是 $8 \\frac{1}{8}$ , 则相减求出它们的差即可。\n【详解】 $8.125 = 8\\frac{1}{8}$ $8 - \\frac {1}{8} = 8 - 0. 1 2 5 \\dot {i} 7. 8 7 5$ 所以这个自然数与它倒数的差是 $7.875$ 。\n【点睛】本题考查倒数的认识, 解答本题的关键是掌握倒数的概念。",
+      "knowledge_points": [
+        "倒数的认识及应用",
+        "自然数与倒数的和或差的问题",
+        "分数化小数"
+      ],
+      "source": "23-24 六年级上·河南南阳·期中",
+      "merged_id": "unit3_122",
+      "original_id": "unit3_choice_122"
+    },
+    {
+      "id": "unit3_choice_123",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "有2吨货物，甲车每次运 $\\frac{1}{2}$ ，乙车每次运 $\\frac{1}{2}$ 吨。若单独运完这些货物，甲车需运( )次，乙车需运( )次。",
+      "options": [],
+      "answer": "2 4",
+      "solution": "【分析】把 2 吨货物看作单位 \"1\", 甲车每次运 $\\frac{1}{2}$ , 根据工作量 $\\div$ 工作效率 $=$ 工作时间, 用 1 除以 $\\frac{1}{2}$ 即可求出甲车需要运送的次数; 同样根据工作量 $\\div$ 工作效率 $=$ 工作时间, 用 2吨除以 $\\frac{1}{2}$ 吨, 即可求出乙车需要运送的次数。\n【详解】 $1 \\div \\frac{1}{2} = 2$ （次）$2 \\div \\frac{1}{2} = 2 \\times 2 = 4$ (次)即甲车需运2次，乙车需运4次。\n【点睛】本题考查知识点：依据工作时间，工作效率以及工作总量之间数量关系解决问题。",
+      "knowledge_points": [
+        "工程问题",
+        "分数与整数的除法"
+      ],
+      "source": "18-19六年级上·全国·单元测试",
+      "merged_id": "unit3_123",
+      "original_id": "unit3_choice_123"
+    },
+    {
+      "id": "unit3_choice_124",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "一种汽车行 $\\frac{3}{2}$ 千米耗油 $\\frac{3}{25}$ 升，那么，这种汽车行 1 千米耗油( )升，这种汽车耗 1 升油可行( )千米。",
+      "options": [],
+      "answer": "$\\frac{2}{25} / 0.08 \\quad \\frac{25}{2} / 12 \\frac{1}{2} / 12.5$",
+      "solution": "【分析】除以一个数等于乘这个数的倒数，耗油量÷路程=行1千米耗油量；路程÷耗油量=耗1升油可行驶路程，据此列式计算。\n【详解】 $\\frac{3}{25} \\div \\frac{3}{2} = \\frac{3}{25} \\times \\frac{2}{3} = \\frac{2}{25}$ （升）$\\frac{3}{2} \\div \\frac{3}{25} = \\frac{3}{2} \\times \\frac{25}{3} = \\frac{25}{2}$ （千米）这种汽车行 1 千米耗油 $\\frac{2}{25}$ 升, 这种汽车耗 1 升油可行 $\\frac{25}{2}$ 千米。\n【点睛】关键是理解数量关系，掌握分数除法的计算方法。",
+      "knowledge_points": [
+        "分数与分数的除法"
+      ],
+      "source": "2014 六年级·全国·课后作业",
+      "merged_id": "unit3_124",
+      "original_id": "unit3_choice_124"
+    },
+    {
+      "id": "unit3_choice_125",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "把 $\\frac{2}{5} \\mathrm{~m}$ 长的绳子平均成 4 段, 每段占全长的( ) , 每段长( ) $\\mathrm{m}$ 。",
+      "options": [],
+      "answer": "$\\frac{1}{4} \\quad \\frac{1}{10} / 0.1$",
+      "solution": "【分析】求每段占全长的几分之几，是把这根绳子的全长看作单位\"1\"，把\"1\"平均分成4段，用1除以4；\n求每段的长度, 是把 $\\frac{2}{5} \\mathrm{~m}$ 长的绳子平均分成 4 段, 用这根绳子的长度除以 4 。\n【详解】 $1 \\div 4 = \\frac{1}{4}$ $ \\frac {2}{5} \\div 4 = \\frac {2}{5} \\times \\frac {1}{4} = \\frac {1}{1 0} (m)$ 每段占全长的 $\\frac{1}{4}$ , 每段长 $\\frac{1}{10} \\mathrm{~m}$ 。\n【点睛】解决此题关键是弄请求的是\"分率\"还是\"具体的数量\"，求分率，平均分的是单位\"1\"；求具体的数量，平均分的是具体的数量。注意：分率不带单位名称，而具体的数量要带单位名称。",
+      "knowledge_points": [
+        "分数与除法的关系",
+        "分数的平均分",
+        "分数与整数的除法",
+        "单位 \"1\"的认识与确定"
+      ],
+      "source": "19-20 六年级上·全国·单元测试",
+      "merged_id": "unit3_125",
+      "original_id": "unit3_choice_125"
+    },
+    {
+      "id": "unit3_choice_126",
+      "unit": "第三单元：分数除法",
+      "chapter": "工程问题",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "小华看一本故事书, 每天看全书的 $\\frac{1}{7}$ ,( ) 天可以看完。",
+      "options": [],
+      "answer": "7",
+      "solution": "【分析】把这本故事书看作单位 \"1\", 每天看全书的 $\\frac{1}{7}$ , 根据 \"工作时间 = 工作总量 ÷ 工作效率\", 即可求出看的天数。\n【详解】 $1 \\div \\frac{1}{7}$ $\\begin{array}{l} = 1 \\times 7 = 7 \\text {(天)} \\end{array}$ 7天可以看完。\n【点睛】本题考查分数除法的应用，掌握工作效率、工作时间、工作总量之间的关系是解题的关键。",
+      "knowledge_points": [
+        "工程问题",
+        "分数与整数的除法"
+      ],
+      "source": "2014 三年级·全国·课后作业",
+      "merged_id": "unit3_126",
+      "original_id": "unit3_choice_126"
+    },
+    {
+      "id": "unit3_choice_127",
+      "unit": "第三单元：分数除法",
+      "chapter": "倒数的认识",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "一个数与它的倒数的差是 $12 \\frac{12}{13}$ , 这个数是( )。",
+      "options": [],
+      "answer": "13",
+      "solution": "【分析】乘积是 1 的两个数互为倒数。求一个真分数或假分数的倒数，只需要将分子、分母交换位置即可。\n已知一个数与它的倒数的差是 $12 \\frac{12}{13}$ , $12 \\frac{12}{13}$ 可以分解成 $13 - \\frac{1}{13}$ , 13 的倒数正好是 $\\frac{1}{13}$ , 据此解答。\n【详解】 $12\\frac{12}{13} = 13 - \\frac{1}{13}$ 13 的倒数是 $\\frac{1}{13}$ ; 所以，这个数是 $13$ 。\n【点睛】本题考查倒数的求法, 把带分数 $\\frac{12}{13}$ 分解成 $13 - \\frac{1}{13}$ 的形式是解题的关键。",
+      "knowledge_points": [
+        "真分数、假分数、带分数的认识",
+        "自然数与倒数的和或差的问题",
+        "同分母分数加、减法"
+      ],
+      "source": "21-22 六年级下·湖南永州·期中",
+      "merged_id": "unit3_127",
+      "original_id": "unit3_choice_127"
+    },
+    {
+      "id": "unit3_choice_128",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "把 $\\frac{5}{9} \\times 3 = \\frac{5}{3}$ 改写成两道除法算式是( )和( )。",
+      "options": [],
+      "answer": "$\\frac{5}{3} \\div 3 = \\frac{5}{9}$ $\\frac{5}{3} \\div \\frac{5}{9} = 3$",
+      "solution": "【分析】根据乘法算式各部分的名称和各部分之间的关系式：因数 $= \\text{积} \\div$ 另一个因数，据此解答即可。\n【详解】把 $\\frac{5}{9} \\times 3 = \\frac{5}{3}$ 改写成两道除法算式是 $\\frac{5}{3} \\div 3 = \\frac{5}{9}$ 和 $\\frac{5}{3} \\div \\frac{5}{9} = 3$ 。\n【点睛】本题考查的是乘法算式各部分之间关系的运用。",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "乘、除法的意义和各部分间的关系",
+        "分数与分数的除法"
+      ],
+      "source": "22-23 五年级上·山东·单元测试",
+      "merged_id": "unit3_128",
+      "original_id": "unit3_choice_128"
+    },
+    {
+      "id": "unit3_choice_129",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "小华从家到学校所用的时间由原来 30 分钟降低到 25 分钟, 速度提高了 $\\frac{1}{6}$ 。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】将从家到学校的路程看作单位 \"1\", 时间分之一可以看作速度, 速度差 $\\div$ 原来的速度 = 速度提高了几分之几,据此列式计算。\n【详解】 $\\left(\\frac{1}{25} -\\frac{1}{30}\\right)\\div \\frac{1}{30}$ $\\begin{array}{l} = \\frac {1}{1 5 0} \\div \\frac {1}{3 0} = \\frac {1}{1 5 0} \\times 3 0 = \\frac {1}{5} \\end{array}$小华从家到学校所用的时间由原来 30 分钟降低到 25 分钟,速度提高了 $\\frac{1}{5}$ , 所以原题说法错误。 故答案为: ×",
+      "knowledge_points": [
+        "求一个数占另一个数几分之几",
+        "基础行程问题",
+        "分数的四则混合运算"
+      ],
+      "source": "23-24六年级上·海南省直辖县级单位·期末",
+      "merged_id": "unit3_129",
+      "original_id": "unit3_choice_129"
+    },
+    {
+      "id": "unit3_choice_130",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "一个数 (0 除外) 除以假分数, 商一定小于被除数。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】分子比分母大或分子和分母相等的分数叫做假分数；假分数 $\\geqslant 1$ 。一个数（0除外）除以大于1的数，商比原来的数小。 一个数（0除外）除以1，商等于原来的数。\n【详解】因为假分数大于或等于 1 , 所以一个数 (0 除外)除以假分数, 商可能小于被除数, 也可能等于被除数。原题说法错误。故答案为: ×",
+      "knowledge_points": [
+        "被除数与商的大小关系（分数除法）",
+        "真分数、假分数、带分数的认识"
+      ],
+      "source": "2023·河北邯郸·小升初真题",
+      "merged_id": "unit3_130",
+      "original_id": "unit3_choice_130"
+    },
+    {
+      "id": "unit3_choice_131",
+      "unit": "第三单元：分数除法",
+      "chapter": "倒数的认识",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "如果 $a \\times b = 1$ , 那么 $a$ 与 $b$ 互为倒数。( )",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】根据倒数的意义: 乘积是 1 的两个数互为倒数, 据此解答。\n【详解】因为 $a \\times b = 1$ , 所以 $a$ 和 $b$ 互为倒数。原题干说法正确。 故答案为: √",
+      "knowledge_points": [
+        "倒数的认识",
+        "用字母表示数",
+        "数量关系"
+      ],
+      "source": "2014 六年级·全国·课后作业",
+      "merged_id": "unit3_131",
+      "original_id": "unit3_choice_131"
+    },
+    {
+      "id": "unit3_choice_132",
+      "unit": "第三单元：分数除法",
+      "chapter": "倒数的认识",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "因为 1 的倒数是 1，所以 0 的倒数是 0。（）",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】乘积是 1 的两个数互为倒数；据此判断。\n【详解】 $1 \\times 1 = 1$ ，1的倒数是1; 0 乘任何数都得 0 , 所以 0 没有倒数。 即 1 的倒数是 1, 0 没有倒数。 原题说法错误。 故答案为: ×",
+      "knowledge_points": [
+        "倒数的认识"
+      ],
+      "source": "2018·河南信阳·小升初真题",
+      "merged_id": "unit3_132",
+      "original_id": "unit3_choice_132"
+    },
+    {
+      "id": "unit3_choice_133",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "将 $\\frac{4}{5}$ 米长的丝带剪成同样长的 8 段, 每段占全长的 $\\frac{1}{10}$ 。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】把这条丝带的全长看作单位 \"1\", 把 \"1\" 平均分成 8 段, 用 1 除以 8 , 是求每段占全长的几分之几, 计算结果不带单位;把 $\\frac{4}{5}$ 米长的丝带平均分成 8 段, 用这条丝带的长度除以 8 ,是求每段的长度, 计算结果带单位。\n【详解】 $1 \\div 8 = \\frac{1}{8}$ $\\begin{array}{l} \\frac {4}{5} \\div 8 = \\frac {4}{5} \\times \\frac {1}{8} = \\frac {1}{1 0} (\\text {米}) \\end{array}$ 将 $\\frac{4}{5}$ 米长的丝带剪成同样长的 8 段, 每段占全长的 $\\frac{1}{8}$ , 每段长 $\\frac{1}{10}$ 米。 原题说法错误。故答案为: ×",
+      "knowledge_points": [
+        "分数与除法的关系",
+        "分数的平均分",
+        "分数与整数的除法",
+        "单位 \"1\" 的认识与确定"
+      ],
+      "source": "23-24 六年级上·河南南阳·期末",
+      "merged_id": "unit3_133",
+      "original_id": "unit3_choice_133"
+    },{
+      "id": "unit3_choice_134",
+      "unit": "第三单元：分数除法",
+      "chapter": "倒数的认识",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "面积是 1 的平行四边形相对应的底和高的长互为倒数。( )",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】乘积是 1 的两个数互为倒数。因为平行四边形的面积 $= \\text{底} \\times$ 高, 如果平行四边形底与高的积是 1 , 根据倒数的意义可知, 这个平行四边形相对应的底和高互为倒数, 据此判断。\n【详解】如：一个平行四边形的底是 2，高是 $\\frac{1}{2}$ ，则面积是 $2 \\times \\frac{1}{2} = 1$ ，其中 2 和 $\\frac{1}{2}$ 互为倒数。所以面积是 1 的平行四边形相对应的底和高的长互为倒数。 原题说法正确。 故答案为: √",
+      "knowledge_points": [
+        "平行四边形的面积",
+        "与倒数有关的综合计算"
+      ],
+      "source": "23-24 六年级上·内蒙古赤峰·期末",
+      "merged_id": "unit3_134",
+      "original_id": "unit3_choice_134"
+    },
+    {
+      "id": "unit3_choice_135",
+      "unit": "第三单元：分数除法",
+      "chapter": "倒数的认识",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "如果 a 和 b 互为倒数,那么 $\\frac{a}{5} \\div \\frac{6}{b} = 30$ 。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】根据倒数的意义: 乘积是 1 的两个数互为倒数; a 和 b 互为倒数, 则 $ab = 1$ ; 根据分数与分数除法的计算法则, 化简 $\\frac{a}{5} \\div \\frac{6}{b}$ , 进而求出 $\\frac{a}{5} \\div \\frac{6}{b}$ 的值, 再进行比较, 即可解答。\n【详解】ab=1 $\\begin{array}{l} \\frac {a}{5} \\div \\frac {6}{b} = \\frac {a}{5} \\times \\frac {b}{6} = \\frac {a b}{3 0} = \\frac {1}{3 0} \\\\ \\end{array}$ 如果 $a$ 和 $b$ 互为倒数, 那么 $\\frac{a}{5} \\div \\frac{6}{b} = \\frac{1}{30}$ 。 原题干说法错误。故答案为: ×",
+      "knowledge_points": [
+        "倒数的认识及应用",
+        "分数与分数的除法"
+      ],
+      "source": "23-24 五年级上·山东青岛·期末",
+      "merged_id": "unit3_135",
+      "original_id": "unit3_choice_135"
+    },
+    {
+      "id": "unit3_choice_136",
+      "unit": "第三单元：分数除法",
+      "chapter": "倒数的认识",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "任何真分数的倒数都是假分数。（ ）",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】真分数：分子小于分母的分数；假分数：分子大于或等于分母的分数；根据求一个倒数的方法，是把一个\n分数的分子分母互换位置就得到这个分数的倒数，真分数的分子分母互换位置后就变成假分数, 可以举例解答。\n【详解】如： $\\frac{2}{3}, \\frac{2}{3}$ 的倒数是 $\\frac{3}{2}, \\frac{3}{2}$ 是假分数。任何真分数的倒数都是假分数。原题干说法正确。故答案为: √",
+      "knowledge_points": [
+        "真分数、假分数、带分数的认识",
+        "倒数的认识"
+      ],
+      "source": "2021·广西南宁·小升初真题",
+      "merged_id": "unit3_136",
+      "original_id": "unit3_choice_136"
+    },
+    {
+      "id": "unit3_choice_137",
+      "unit": "第三单元：分数除法",
+      "chapter": "倒数的认识",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "$b$ 是自然数, $b$ 的倒数是$\\frac {1}{b}$()",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】根据题意，没有说明 $b \\neq 0, 0$ 是最小的自然数，0没有倒数。据此判断即可。\n【详解】 $b$ 是自然数, $b$ 的倒数是 $\\frac{1}{b}$ 。原题说法错误。故答案为: $\\times$",
+      "knowledge_points": [
+        "倒数的认识及应用",
+        "自然数的认识"
+      ],
+      "source": "2020·湖北黄冈·小升初真题",
+      "merged_id": "unit3_137",
+      "original_id": "unit3_choice_137"
+    },
+    {
+      "id": "unit3_choice_138",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "$\\frac{1}{6} \\div 2$ 与 $\\frac{1}{6} \\times \\frac{1}{2}$ 的意义相同, 计算结果也相同。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】 $\\frac{1}{6} \\div 2$ 表示把 $\\frac{1}{6}$ 平均分成 2 份; $\\frac{1}{6} \\div 2 = \\frac{1}{6} \\times \\frac{1}{2} = \\frac{1}{12}$ ; $\\frac{1}{6} \\times \\frac{1}{2}$ 表示 $\\frac{1}{6}$ 的 $\\frac{1}{2}$ 是多少, $\\frac{1}{6} \\times \\frac{1}{2} = \\frac{1}{12}$ ; 据此解答。本题考查分数除以整数和分数乘分数的意义, 要熟练掌握。\n【详解】根据分析可知, $\\frac{1}{6} \\div 2$ 与 $\\frac{1}{6} \\times \\frac{1}{2}$ 的意义不同, 计算结果相同。原题干说法错误。故答案为: ×",
+      "knowledge_points": [
+        "分数的平均分",
+        "分数与整数的除法",
+        "分数乘分数"
+      ],
+      "source": "22-23 六年级上·贵州黔西·期中",
+      "merged_id": "unit3_138",
+      "original_id": "unit3_choice_138"
+    },
+    {
+      "id": "unit3_choice_139",
+      "unit": "第三单元：分数除法",
+      "chapter": "倒数的认识",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "$\\frac{4}{5} \\times \\frac{1}{2} \\times \\frac{5}{2} = 1$ , 所以 $\\frac{4}{5}$ 、 $\\frac{1}{2}$ 、 $\\frac{5}{2}$ 互为倒数。（）",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】乘积是 1 的两个数互为倒数；据此判断。\n【详解】 $\\frac{4}{5} \\times \\frac{1}{2} \\times \\frac{5}{2} = 1$ ，三个数的乘积为 1，不符合倒数的定义。原题说法错误。故答案为: ×\n【点睛】本题考查倒数的意义及应用。",
+      "knowledge_points": [
+        "倒数的认识及应用"
+      ],
+      "source": "20-21 六年级上·山东济宁·期末",
+      "merged_id": "unit3_139",
+      "original_id": "unit3_choice_139"
+    },
+    {
+      "id": "unit3_choice_140",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "男生人数比女生人数多$\\frac{1}{5}$ , 那么女生人数比男生人数少 $\\frac{1}{5}$ 。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】把女生人数看作单位 \"1\", 则男生的人数为 $1 \\times (1 + \\frac{1}{5})$ ; 然后求出女生人数比男生人数少多少, 再除以男生人数即可, 据此判断。本题考查求一个数比另一个数少几分之几, 明确用除法是解题的关键。\n【详解】假设女生的人数为1，男生人数为： $ 1 \\times (1 + \\frac {1}{5}) = 1 \\times \\frac {6}{5} = \\frac {6}{5} $ $\\begin{array}{l} \\left(\\frac {6}{5} - 1\\right) \\div \\frac {6}{5} = \\frac {1}{5} \\div \\frac {6}{5} = \\frac {1}{5} \\times \\frac {5}{6} = \\frac {1}{6} \\\\ \\end{array}$则女生人数比男生人数少 $\\frac{1}{6}$ 。原题干说法错误。故答案为: ×",
+      "knowledge_points": [
+        "求一个数占另一个数几分之几",
+        "求比一个数多/少几分之几的数是多少",
+        "分数与分数的除法"
+      ],
+      "source": "2017·湖南邵阳·小升初真题",
+      "merged_id": "unit3_140",
+      "original_id": "unit3_choice_140"
+    },
+    {
+      "id": "unit3_choice_141",
+      "unit": "第三单元：分数除法",
+      "chapter": "倒数的认识",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "假分数的倒数一定小于 $1$ 。（）",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】分子等于或大于分母的分数就是假分数；把分数的分子和分母的位置颠倒就是该分数的倒数，据此判断即可。本题考查假分数和倒数，明确假分数和倒数的定义是解题的关键。\n【详解】如： $\\frac{5}{5}$ 是假分数， $\\frac{5}{5}$ 的倒数是 1，此时 $\\frac{5}{5}$ 的倒数等于 1。原题干说法错误。故答案为: ×",
+      "knowledge_points": [
+        "倒数的认识及应用",
+        "真分数、假分数、带分数的认识"
+      ],
+      "source": "2017·宁夏吴忠·小升初真题",
+      "merged_id": "unit3_141",
+      "original_id": "unit3_choice_141"
+    },
+    {
+      "id": "unit3_choice_142",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "一个数除以分数的商一定比原来的数大。（）",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】一个数（0 除外）除以一个小于 1 的分数，商比这个数大；一个数（0 除外）除以一个大于 1 的分数，商比这个数小；一个数（0 除外）除以一个等 于 1 的分数，商等于这个数。本题主要考查一个数除以一个大于 1 , 小于 1 或等于 1 的数, 这个数与商之间的关系。\n【详解】根据分析可知, 一个数除以分数的商不一定比原来的数大。故答案为: × 。",
+      "knowledge_points": [
+        "被除数和商的大小关系（小数除法）"
+      ],
+      "source": "21-22六年级上·天津蓟州·阶段练习",
+      "merged_id": "unit3_142",
+      "original_id": "unit3_choice_142"
+    },
+    {
+      "id": "unit3_choice_143",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "A 比 B 多 $\\frac{1}{4}$ , 也就是 B 比 A 少 $\\frac{1}{4}$ 。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】把B看作单位\"1\"，A比B多 $\\frac{1}{4}$ ，则A占B的 $(1 + \\frac{1}{4})$ ，B比A少的分率 $= (A - B) \\div A$ ，据此解答。本题主要考查分数除法的应用，掌握求一个数比另一个数少几分之几的计算方法是解答题目的关键\n【详解】 $(1 + \\frac{1}{4} - 1) \\div (1 + \\frac{1}{4})$ $\\begin{array}{l} = \\frac {1}{4} \\div \\frac {5}{4} = \\frac {1}{4} \\times \\frac {4}{5} = \\frac {1}{5} \\\\ \\end{array}$所以, A 比 B 多 $\\frac{1}{4}$ , 也就是 B 比 A 少 $\\frac{1}{5}$ 。故答案为:×。",
+      "knowledge_points": [
+        "分数的四则混合运算"
+      ],
+      "source": "21-22 六年级上·全国·期末",
+      "merged_id": "unit3_143",
+      "original_id": "unit3_choice_143"
+    },
+    {
+      "id": "unit3_choice_144",
+      "unit": "第三单元：分数除法",
+      "chapter": "倒数的认识",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "a 是一个整数，它的倒数是 $\\frac{1}{a}$ 。（ ）",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】根据倒数的意义，乘积是 1 的两个数互为倒数；1 的倒数是 1, 0 没有倒数；据此解答即可。\n【详解】根据分析可知:如果 $a = 0$ , 那么它就没有倒数。故答案为: ×\n【点睛】找到 a 的特殊情况是解题关键, 注意 0 没有倒数。",
+      "knowledge_points": [
+        "倒数的认识及应用"
+      ],
+      "source": "2014 六年级·全国·课后作业",
+      "merged_id": "unit3_144",
+      "original_id": "unit3_choice_144"
+    },
+    {
+      "id": "unit3_choice_145",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "甲比乙多 $\\frac{1}{3}$ , 也就是乙比甲少 $\\frac{1}{3}$ 。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】把乙看作单位 \"1\", 甲比乙多 $\\frac{1}{3}$ , 甲占乙的 $(1 + \\frac{1}{3})$ , 乙比甲少的分率 $=$ (甲一乙) ÷甲, 据此解答。确定题目中的单位 \"1\"并掌握求一个数比另一个数 多（少）几分之几的计算方法是解答题目的关键。\n【详解】假设乙为 1。 $1\\times (1 + \\frac {1}{3}) = 1 \\times \\frac {4}{3} = \\frac {4}{3} \\quad(\\frac {4}{3} - 1) \\div \\frac {4}{3} = \\frac {1}{3} \\div \\frac {4}{3} = \\frac {1}{3} \\times \\frac {3}{4} = \\frac {1}{4} $所以，甲比乙多 $\\frac{1}{3}$ , 也就是乙比甲少 $\\frac{1}{4}$ 。故答案为: ×。",
+      "knowledge_points": [
+        "求一个数占另一个数几分之几",
+        "分数的四则混合运算",
+        "单位 \"1\"的认识与确定",
+        "求比一个数多/少几分之几的数是多少。"
+      ],
+      "source": "15-16 六年级上·北京·课后作业",
+      "merged_id": "unit3_145",
+      "original_id": "unit3_choice_145"
+    },
+    {
+      "id": "unit3_choice_146",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "一个不为 0 的数除以 $\\frac{1}{3}$ ,相当于把这个数扩大到原来的 3 倍。( )",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】根据\"除以一个不为 0 的数，等于乘这个数的倒数\"，据此判断。掌握分数的计算方法是解题的关键。\n【详解】 $\\frac{1}{3}$ 的倒数是 3 ;一个不为 0 的数除以 $\\frac{1}{3}$ , 等于乘 3 , 相当于把这个数扩大到原来的 3 倍。原题说法正确。故答案为: √",
+      "knowledge_points": [
+        "倒数的认识及应用",
+        "分数与整数的除法",
+        "分数与分数的除法"
+      ],
+      "source": "2020·山西忻州·小升初真题",
+      "merged_id": "unit3_146",
+      "original_id": "unit3_choice_146"
+    },
+    {
+      "id": "unit3_choice_147",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法计算",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "计算。\n$2 0 1 6 \\div 2 0 1 7 \\frac {1}{2 0 1 5} + \\frac {1}{2 0 1 6}$",
+      "options": [],
+      "answer": "1",
+      "solution": "【分析】先把带分数化成假分数，用\"带分数的整数部分×分母+分子\"得到假分数的分子 $2017 \\times 2015 + 1$ ，然后把2017拆成 $2016 + 1$ ，根据乘法分配律 $(a + b) \\times c = a \\times c + b \\times c$ 进行简算；再把除法转化成乘法进行约分，最后算加法。\n【详解】 $2016 \\div 2017 \\frac{1}{2015} + \\frac{1}{2016}\\\\= 2 0 1 6 \\div \\frac {2 0 1 7 \\times 2 0 1 5 + 1}{2 0 1 5} + \\frac {1}{2 0 1 6}\\\\= 2 0 1 6 \\div \\frac {(2 0 1 6 + 1) \\times 2 0 1 5 + 1}{2 0 1 5} + \\frac {1}{2 0 1 6}\\\\= 2 0 1 6 \\div \\frac {2 0 1 6 \\times 2 0 1 5 + 2 0 1 5 + 1}{2 0 1 5} + \\frac {1}{2 0 1 6}\\\\= 2 0 1 6 \\div \\frac {2 0 1 6 \\times 2 0 1 5 + 2 0 1 6}{2 0 1 5} + \\frac {1}{2 0 1 6}\\\\=2 0 1 6 \\div \\frac {2 0 1 6 \\times (2 0 1 5 + 1)}{2 0 1 5} + \\frac {1}{2 0 1 6}\\\\= 2 0 1 6 \\div \\frac {2 0 1 6 \\times 2 0 1 6}{2 0 1 5} + \\frac {1}{2 0 1 6}\\\\=\\frac {2 0 1 5}{2 0 1 6}+\\frac {1}{2 0 1 6}$",
+      "knowledge_points": [
+        "分数除法相关的简便计算"
+      ],
+      "source": "23-24六年级上·全国·单元测试",
+      "merged_id": "unit3_147",
+      "original_id": "unit3_choice_147"
+    },
+    {
+      "id": "unit3_choice_148",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法计算",
+      "question_type": "计算题",
+      "difficulty": "困难",
+      "question": "计算。\n(1) $987 \\div 988 \\frac{1}{986}$\n(2) $286 \\div 286 \\frac{286}{287} + 288 \\div 289 \\frac{1}{287}$",
+      "options": [],
+      "answer": "(1) $\\frac{986}{987}$ ; (2) $\\frac{287}{144}$",
+      "solution": "【分析】（1）先把 $\\frac{988}{986}$ 改写成 $987 + 1 + \\frac{1}{986}$ ，合起来是 $\\frac{987}{986}$ ，这样算式变成 $987 \\div 987 \\times \\frac{987}{986}$ ；再根据商不变的规律，被除数和除数同时除以 987，商不变；然后把 \" $\\div 987$ \" 转化成 \" $\\times \\frac{1}{987}$ \"，被除数、除数同时约分后再计算即可。\n(2) 先算除法，再算加法。\n前部分 $286 \\div 286 \\frac{286}{287}$ , 根据商不变的规律, 被除数和除数同时除以 286 , 商不变; 然后把 \" $\\div 286$ \" 转化成 \" $\\times \\frac{1}{286}$ \", 被除数、除数同时约分即可;后部分 $288 \\div 289 \\frac{1}{287}$ , 先把 $289 \\frac{1}{287}$ 改写成 $288 + 1 + \\frac{1}{287}$ , 合起来是 $288 \\frac{288}{287}$ , 这样算式变成 $288 \\div 288 \\frac{288}{287}$ , 再根据商不变的规律, 被除数和除数同时除以 288 , 商不变; 然后把 \" $\\div 288$ \" 转化成 \" $\\times \\frac{1}{288}$ \", 被除数、除数同时约分即可;\n【详解】 (1) $987 \\div 988 \\frac{1}{986}$\n$\\ = 9 8 7 \\div 9 8 7 \\frac {9 8 7}{9 8 6} = (9 8 7 \\div 9 8 7) \\div \\left(9 8 7 \\frac {9 8 7}{9 8 6} \\div 9 8 7\\right) = \\left(9 8 7 \\times \\frac {1}{9 8 7}\\right) \\div \\left(9 8 7 \\frac {9 8 7}{9 8 6} \\times \\frac {1}{9 8 7}\\right) = 1 \\div 1 \\frac {1}{9 8 6} = 1 \\div \\frac {9 8 7}{9 8 6} = 1 \\times \\frac {9 8 6}{9 8 7} = \\frac {9 8 6}{9 8 7} $\n(2) $286 \\div 286 \\frac{286}{287} + 288 \\div 289 \\frac{1}{287}$\n$\\ = 2 8 6 \\div 2 8 6 \\frac {2 8 6}{2 8 7} + 2 8 8 \\div 2 8 8 \\frac {2 8 8}{2 8 7} = (2 8 6 \\div 2 8 6) \\div \\left(2 8 6 \\frac {2 8 6}{2 8 7} \\div 2 8 6\\right) + (2 8 8 \\div 2 8 8) \\div \\left(2 8 8 \\frac {2 8 8}{2 8 7} \\div 2 8 8\\right) = \\left(2 8 6 \\times \\frac {1}{2 8 6}\\right) \\div \\left(2 8 6 \\frac {2 8 6}{2 8 7} \\times \\frac {1}{2 8 6}\\right) + \\left(2 8 8 \\times \\frac {1}{2 8 8}\\right) \\div \\left(2 8 8 \\frac {2 8 8}{2 8 7} \\times \\frac {1}{2 8 8}\\right) = 1 \\div 1 \\frac {1}{2 8 7} + 1 \\div 1 \\frac {1}{2 8 7} = 1 \\div \\frac {2 8 8}{2 8 7} + 1 \\div \\frac {2 8 8}{2 8 7} = 1 \\times \\frac {2 8 7}{2 8 8} + 1 \\times \\frac {2 8 7}{2 8 8} = \\frac {2 8 7}{2 8 8} + \\frac {2 8 7}{2 8 8} = \\frac {2 8 7}{1 4 4} $",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "商不变的规律及应用",
+        "分数的四则混合运算",
+        "分数除法相关的简便计算"
+      ],
+      "source": "23-24六年级上·全国·单元测试",
+      "merged_id": "unit3_148",
+      "original_id": "unit3_choice_148"
+    },
+    {
+      "id": "unit3_choice_149",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法计算",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "计算。\n(1) $119 \\div 119 \\frac{119}{120}$\n(2) $2003 \\div 2003 \\frac{2003}{2004}$",
+      "options": [],
+      "answer": "(1) $\\frac{120}{121}$ ; (2) $\\frac{2004}{2005}$",
+      "solution": "【分析】 (1) (2) 除数都是带分数, 先把带分数化成假\n分数, 假分数的分子是 \"带分数的整数部分 $\\times$ 分母 + 分子\"得到, 分子部分根据乘法分配律 $a \\times c + b \\times c = (a + b) \\times c$ 进行简算, 然后把除法转化成乘法计算即可。\n【详解】 (1) $119 \\div 119 \\frac{119}{120}$\n$\\ 1 1 9 \\div \\frac {1 1 9 \\times 1 2 0 + 1 1 9}{1 2 0} == 1 1 9 \\div \\frac {1 1 9 \\times (1 2 0 + 1)}{1 2 0} = 1 1 9 \\div \\frac {1 1 9 \\times 1 2 1}{1 2 0}= 1 1 9 \\times \\frac {1 2 0}{ 1 2 1} = \\frac {1 2 0}{1 2 1}$ \\\n(2) $2003 \\div 2003 \\frac{2003}{2004}$\n$ 2 0 0 3 \\div \\frac {2 0 0 3 \\times 2 0 0 4 + 2 0 0 3}{2 0 0 4} = 2 0 0 3 \\div \\frac {2 0 0 3 \\times (2 0 0 4 + 1)}{2 0 0 4}= 2 0 0 3 \\div \\frac {2 0 0 3 \\times 2 0 0 5}{2 0 0 4}= 2 0 0 3 \\times \\frac {2 0 0 4}{2 0 0 3 \\times 2 0 0 5} = \\frac {2 0 0 4}{2 0 0 5}$",
+      "knowledge_points": [
+        "分数除法相关的简便计算"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_149",
+      "original_id": "unit3_choice_149"
+    },
+    {
+      "id": "unit3_choice_150",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法计算",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "用简便方法计算。\n(1) $155 \\div 155 \\frac{155}{156}$\n(2) $998 \\div 998 \\frac{998}{995}$",
+      "options": [],
+      "answer": "(1) $\\frac{156}{157}$ ; (2) $\\frac{995}{996}$",
+      "solution": "【分析】（1）先根据商不变的规律，被除数和除数同时除以 155，商不变；然后把 \" $\\div 155$ \" 转化成 \" $\\times \\frac{1}{155}$ \"，被除数、除数同时约分后再计算即可。 (2) 先根据商不变的规律, 被除数和除数同时除以 998,商不变; 然后把 \" $\\div$ 998\"转化成 \" $\\times \\frac{1}{998}$ \", 被除数、除数同时约分后再计算即可。\n【详解】 (1) $155 \\div 155 \\frac{155}{156}$\n$ (1 5 5 \\div 1 5 5) \\div \\left(1 5 5 \\frac {1 5 5}{1 5 6} \\div 1 5 5\\right) = \\left(1 5 5 \\times \\frac {1}{1 5 5}\\right) \\div \\left(1 5 5 \\frac {1 5 5}{1 5 6} \\times \\frac {1}{1 5 5}\\right) =1\\div 1 \\frac {1}{1 5 6}1 \\div \\frac {1 5 7}{1 5 6} = 1 \\times \\frac {1 5 6}{1 5 7} = \\frac {1 5 6}{1 5 7}$\n(2) $998 \\div 998 \\frac{998}{995}$\n$(9 9 8 \\div 9 9 8) \\div \\left(9 9 8 \\frac {9 9 8}{9 9 5} \\div 9 9 8\\right) =\\left(9 9 8 \\times \\frac {1}{9 9 8}\\right) \\div \\left(9 9 8 \\frac {9 9 8}{9 9 5} \\times \\frac {1}{9 9 8}\\right) = 1 \\div 1 \\frac {1}{9 9 5} =1 \\div \\frac {9 9 6}{9 9 5} =1 \\times \\frac {9 9 5}{9 9 6} = \\frac {9 9 5}{9 9 6} $",
+      "knowledge_points": [
+        "约分的认识及应用",
+        "分数与整数的除法",
+        "分数除法相关的简便计算",
+        "整数乘法运算定律推广到分数乘法"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_150",
+      "original_id": "unit3_choice_150"
+    },
+    {
+      "id": "unit3_choice_151",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数乘法计算",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "计算。\n(1) $72 \\frac{1}{15} \\times \\frac{1}{8}$\n(2) $54 \\frac{1}{7} \\times \\frac{7}{9}$",
+      "options": [],
+      "answer": "(1) $9 \\frac{1}{120}$ ; (2) $42 \\frac{1}{9}$",
+      "solution": "【分析】（1）（2）把带分数改写成\"整数+真分数\"的形式，再根据乘法分配律 $(a + b) \\times c = a \\times c + b \\times c$ 进行简算。\n(1) $72 \\frac{1}{15} \\times \\frac{1}{8}=7 2 \\times \\frac {1}{8} + \\frac {1}{1 5} \\times \\frac {1}{8}=9 + \\frac {1}{1 2 0}=9 \\frac{1}{120}$\n(2) $54\\frac{1}{7} \\times \\frac{7}{9}= \\left(5 4 + \\frac {1}{7}\\right) \\times \\frac {7}{9} =5 4 \\times \\frac {7}{9} + \\frac {1}{7} \\times \\frac {7}{9} = 42 + \\frac {1}{9} =4 2 \\frac {1}{9} $",
+      "knowledge_points": [
+        "整数乘法运算定律推广到分数乘法"
+      ],
+      "source": "23-24六年级上·全国·单元测试",
+      "merged_id": "unit3_151",
+      "original_id": "unit3_choice_151"
+    },
+    {
+      "id": "unit3_choice_152",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数乘法计算",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "能简算的要简算。\n(1) $298 \\frac{3}{199} \\times \\frac{199}{298}$\n(2) $2000 \\frac{1}{1998} \\times \\frac{1998}{1999}$",
+      "options": [],
+      "answer": "(1) $199 \\frac{3}{298}$ ; (2) 1999",
+      "solution": "【分析】（1）把假分数 $298 \\frac{3}{199}$ 改写成 $(298 + \\frac{3}{199})$ ，然后根据乘法分配律 $(a + b) \\times c = a \\times c + b \\times c$ 进行简算。\n(2) 把假分数 $2000 \\frac{1}{1998}$ 改写成 $\\left(2000 + \\frac{1}{1998}\\right)$ , 发现 2000 比 1999 多 1 , 再把 2000 拆成 $1999 + 1$ , 把 1 给分数部分, 然后根据乘法分配律 $(a + b) \\times c = a \\times c + b \\times c$ 进行简算。\n【详解】 (1) $298 \\frac{3}{199} \\times \\frac{199}{298}$\n(2) $2000 \\frac{1}{1998} \\times \\frac{1998}{1999}$\n$= (2 9 8 + \\frac {3}{1 9 9}) \\times \\frac {1 9 9}{2 9 8} = 2 9 8 \\times \\frac {1 9 9}{2 9 8} + \\frac {3}{1 9 9} \\times \\frac {1 9 9}{2 9 8} = 1 9 9 + \\frac {3}{2 9 8} = 1 9 9 \\frac {3}{2 9 8} = (2 0 0 0 + \\frac {1}{1 9 9 8}) \\times \\frac {1 9 9 8}{1 9 9 9} = (1 9 9 9 + 1 + \\frac {1}{1 9 9 8}) \\times \\frac {1 9 9 8}{1 9 9 9} = (1 9 9 9 + \\frac {1 9 9 9}{1 9 9 8}) \\times \\frac {1 9 9 8}{1 9 9 9} = 1 9 9 9 \\times \\frac {1 9 9 8}{1 9 9 9} + \\frac {1 9 9 9}{1 9 9 8} \\times \\frac {1 9 9 8}{1 9 9 9} = 1 9 9 8 + 1$",
+      "knowledge_points": [
+        "整数乘法运算定律推广到分数乘法"
+      ],
+      "source": "23-24六年级上·全国·单元测试",
+      "merged_id": "unit3_152",
+      "original_id": "unit3_choice_152"
+    },
+    {
+      "id": "unit3_choice_153",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数乘法计算",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "计算。\n(1) $\\frac{74}{75} \\times 8$\n(2) $\\frac{115}{114} \\times 113$",
+      "options": [],
+      "answer": "(1) $7 \\frac{67}{75}$ ; (2) $113 \\frac{113}{114}$",
+      "solution": "【分析】（1）观察分数 $\\frac{74}{75}$ ，分子比分母少 1，把 $\\frac{74}{75}$ 可以改写成 $\\left(1 - \\frac{1}{75}\\right)$ ，然后根据乘法分配律 $(a + b) \\times c = a \\times c + b \\times c$ 进行简算；(2) 观察分数 $\\frac{115}{114}$ , 分子比分母多 1 , 把 $\\frac{115}{114}$ 可以改写成 $\\left(1 + \\frac{1}{114}\\right)$ , 然后根据乘法分配律 $(a + b) \\times c = a \\times c + b \\times c$ 进行简算。\n【详解】 (1) $\\frac{74}{75} \\times 8=\\left(1 - \\frac {1}{7 5}\\right) \\times 8=1 \\times 8 - \\frac {1}{7 5} \\times 8 = 8 - \\frac {8}{7 5} = 7 \\frac {6 7}{7 5}$\n(2) ${\\frac{115}{114} \\times 113}= (1 + \\frac {1}{1 1 4}) \\times 1 1 3 = 1 \\times 1 1 3 + \\frac {1}{1 1 4} \\times 1 1 3 = 1 1 3 + \\frac {1 1 3}{1 1 4} = 1 1 3 \\frac {1 1 3}{1 1 4} $",
+      "knowledge_points": [
+        "整数乘法运算定律推广到分数乘法"
+      ],
+      "source": "23-24六年级上·全国·单元测试",
+      "merged_id": "unit3_153",
+      "original_id": "unit3_choice_153"
+    },
+    {
+      "id": "unit3_choice_154",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数乘法计算",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "计算。\n(1) $\\frac{29}{98} \\times 99$ (2) $67 \\times \\frac{15}{68}$",
+      "options": [],
+      "answer": "(1) $29 \\frac{29}{98}$ ; (2) $14 \\frac{53}{68}$",
+      "solution": "【分析】（1）先把99拆成 $98 + 1$ ，再根据乘法分配律 $(a + b) \\times c = a \\times c + b \\times c$ 进行简算；\n(2) 先把 67 拆成 $68 - 1$ , 再根据乘法分配律 $(a + b) \\times c =$ a×c+b×c进行简算。\n【详解】 (1) $\\frac{29}{98} \\times 99=\\frac {2 9}{9 8} \\times (9 8 + 1)= \\frac {2 9}{9 8} \\times 9 8 + \\frac {2 9}{9 8} \\times 1 = 2 9 + \\frac {2 9}{9 8} = 2 9 \\frac {2 9}{9 8} $\n(2) $67 \\times \\frac{15}{68}=( 6 8 - 1) \\times \\frac {1 5}{6 8} == 6 8 \\times \\frac {1 5}{6 8} - 1 \\times \\frac {1 5}{6 8}= 15 - \\frac {1 5}{6 8} =1 4 \\frac {5 3}{6 8} $",
+      "knowledge_points": [
+        "整数乘法运算定律推广到分数乘法"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_154",
+      "original_id": "unit3_choice_154"
+    },
+    {
+      "id": "unit3_choice_155",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数乘法计算",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "用简便方法计算下面各题。\n(1) $27 \\times \\frac{15}{26}$\n(2) $\\frac{44}{45} \\times 37$",
+      "options": [],
+      "answer": "(1) $15 \\frac{15}{26}$ ; (2) $36 \\frac{8}{45}$",
+      "solution": "【分析】（1）整数 27 比分母 26 多 1，可以把 27 拆成（26 + 1）,再根据乘法分配律 $(a + b) \\times c = a \\times c + b \\times c$ 进行简算； (2) 分数 $\\frac{44}{45}$ 的分子比分母少 1 , 可以把 $\\frac{44}{45}$ 改写成 $(1 - \\frac{1}{45})$ , 再根据乘法分配律 $(a + b) \\times c = a \\times c + b \\times c$ 进行简算。\n【详解】 (1) $27 \\times \\frac{15}{26}= (2 6 + 1) \\times \\frac {1 5}{2 6} = 2 6 \\times \\frac {1 5}{2 6} + 1 \\times \\frac {1 5}{2 6} = 1 5 + \\frac {1 5}{2 6} = 1 5 \\frac {1 5}{2 6}$\n(2) $\\frac{44}{45} \\times 37= (1 - \\frac {1}{4 5}) \\times 3 7 = 1 \\times 3 7 - \\frac {1}{4 5} \\times 3 7 = 3 7 - \\frac {3 7}{4 5} = 3 6 \\frac {8}{4 5}$",
+      "knowledge_points": [
+        "整数乘法运算定律推广到分数乘法"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_155",
+      "original_id": "unit3_choice_155"
+    },
+    {
+      "id": "unit3_choice_156",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数乘法计算",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "计算。\n(1) $\\frac{5}{17} \\times \\frac{9}{24} + \\frac{9}{17} \\times \\frac{7}{24}$\n(2) $\\frac{5}{9} \\times \\frac{9}{16} + \\frac{7}{9} \\times \\frac{5}{16}$",
+      "options": [],
+      "answer": "(1) $\\frac{9}{34}$ ; (2) $\\frac{5}{9}$",
+      "solution": "【分析】 (1) 先把 $\\frac{5}{17} \\times \\frac{9}{24}$ 改写成 $\\frac{9}{17} \\times \\frac{5}{24}$ , 然后根据乘法分配律 $a \\times c + b \\times c = (a + b) \\times c$ 进行简算;\n(2) 先把 $\\frac{7}{9} \\times \\frac{5}{16}$ 改写成 $\\frac{5}{9} \\times \\frac{7}{16}$ , 然后根据乘法分配律 $a \\times c + b \\times c = (a + b) \\times c$ 进行简算。\n【详解】 (1) $\\frac{5}{17} \\times \\frac{9}{24} + \\frac{9}{17} \\times \\frac{7}{24}= \\frac {9}{1 7} \\times \\frac {5}{2 4} + \\frac {9}{1 7} \\times \\frac {7}{2 4}=\\frac {9}{1 7} \\times \\frac {1}{2}=\\frac {9}{3 4}$\n(2) $\\frac{5}{9} \\times \\frac{9}{16} + \\frac{7}{9} \\times \\frac{5}{16}=\\frac {5}{9} \\times \\frac {9}{1 6} + \\frac {5}{9} \\times \\frac {7}{1 6}= \\frac {5}{9} \\times \\left(\\frac {9}{1 6} + \\frac {7}{1 6}\\right)= \\frac {5}{9} \\times 1 ={\\frac {5}{9}}$",
+      "knowledge_points": [
+        "整数乘法运算定律推广到分数乘法"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_156",
+      "original_id": "unit3_choice_156"
+    },
+    {
+      "id": "unit3_choice_157",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数乘法计算",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "计算。\n(1) $\\frac{11}{13} \\times \\frac{6}{19} + \\frac{6}{13} \\times \\frac{8}{19}$\n(2) $139 \\times \\frac{137}{138} - 137 \\times \\frac{1}{138}$",
+      "options": [],
+      "answer": "(1) $\\frac{6}{13}$ ; (2) 137",
+      "solution": "【分析】（1）计算分数乘法时，分子乘分子分母乘分母，即 $\\frac{11}{13} \\times \\frac{6}{19} = \\frac{11 \\times 6}{13 \\times 19}$ ，此时分子满足乘法交换律，也就是说可以交换分子： $\\frac{11}{13} \\times \\frac{6}{19} = \\frac{6}{13} \\times \\frac{11}{19}$ ，再根据乘法分配律 $a \\times c + b \\times c = (a + b) \\times c$ 进行简算。\n(2) $137 \\times \\frac{1}{138}$ 也可以交换因数 137 和分子的位置变为 $1 \\times \\frac{137}{138}$ ,再根据乘法分配律 $a \\times c + b \\times c = (a + b) \\times c$ 进行简算。【详解】 (1) $\\frac{11}{13} \\times \\frac{6}{19} + \\frac{6}{13} \\times \\frac{8}{19}= \\frac {6}{1 3} \\times \\frac {1 1}{1 9} + \\frac {6}{1 3} \\times \\frac {8}{1 9} = \\frac {6}{1 3} \\times (\\frac {1 1}{1 9} + \\frac {8}{1 9}) = \\frac {6}{1 3} \\times 1 = \\frac {6}{1 3} $\n(2) $139 \\times \\frac{137}{138} - 137 \\times \\frac{1}{138}= 1 3 9 \\times \\frac {1 3 7}{1 3 8} - 1 \\times \\frac {1 3 7}{1 3 8} = (1 3 9 - 1) \\times \\frac {1 3 7}{1 3 8} = 1 3 8 \\times \\frac {1 3 7}{1 3 8} = 1 3 7$",
+      "knowledge_points": [
+        "分数的四则混合运算",
+        "整数乘法运算定律推广到分数乘法"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_157",
+      "original_id": "unit3_choice_157"
+    },
+    {
+      "id": "unit3_choice_158",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法计算",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "计算。\n$4. 4 4 \\div 4 \\frac {5}{8} + \\frac {3 1}{3 7} \\div \\frac {2 5}{1 1 1} + \\frac {3 6}{3 7} \\times 4 \\frac {1 1}{2 5}$",
+      "options": [],
+      "answer": "9",
+      "solution": "【分析】先把 $4.44$ 化成 $\\frac{111}{25}$ , $4 \\frac{5}{8}$ 化成 $\\frac{37}{8}$ , $4 \\frac{11}{25}$ 化成 $\\frac{111}{25}$ , 然后把除法转化成乘法, 再根据乘法分配律 $a \\times c + b \\times c = (a + b) \\times c$ 进行简算。\n【详解】 $4.44 \\div 4\\frac{5}{8} + \\frac{31}{37} \\div \\frac{25}{111} + \\frac{36}{37} \\times 4\\frac{11}{25}=\\frac {1 1 1}{2 5} \\div \\frac {3 7}{8} + \\frac {3 1}{3 7} \\div \\frac {2 5}{1 1 1} + \\frac {3 6}{3 7} \\times \\frac {1 1 1}{2 5} =\\frac {1 1 1}{2 5} \\times \\frac {8}{3 7} + \\frac {3 1}{3 7} \\times \\frac {1 1 1}{2 5} + \\frac {3 6}{3 7} \\times \\frac {1 1 1}{2 5} = \\frac {1 1 1}{2 5} \\times \\left(\\frac {8}{3 7} + \\frac {3 1}{3 7} + \\frac {3 6}{3 7}\\right) =\\frac {1 1 1}{2 5} \\times \\frac {7 5}{3 7}=9$",
+      "knowledge_points": [
+        "分数除法相关的简便计算",
+        "整数乘法运算定律、推广到分数乘法"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_158",
+      "original_id": "unit3_choice_158"
+    },
+    {
+      "id": "unit3_choice_159",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法计算",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "能简算的要简算。\n(1) $4.6 \\times \\frac{11}{8} + 8.4 \\div \\frac{8}{11} - \\frac{11}{8} \\times 5$\n(2) $4.85 \\div \\frac{5}{18} - 3.6 + 6.15 \\times 3\\frac{3}{5}$",
+      "options": [],
+      "answer": "(1) 11; (2) 36",
+      "solution": "【分析】（1）先把除法转化成乘法，再根据乘法分配律 $a \\times c + b \\times c = (a + b) \\times c$ 进行简算；\n(2) 先把除法转化成乘法, 把 $3.6 、 3 \\frac{3}{5}$ 都化成 $\\frac{18}{5}$ , 再根据乘法分配律 $a \\times c + b \\times c = (a + b) \\times c$ 进行简算。\n【详解】 (1) $4.6 \\times \\frac{11}{8} + 8.4 \\div \\frac{8}{11} - \\frac{11}{8} \\times 5= 4. 6 \\times \\frac {1 1}{8} + 8. 4 \\times \\frac {1 1}{8} - \\frac {1 1}{8} \\times 5 = \\frac {1 1}{8} \\times (4. 6 + 8. 4 - 5) = \\frac {1 1}{8} \\times 8 = 1 1$\n(2) $4.85 \\div \\frac{5}{18} - 3.6 + 6.15 \\times 3\\frac{3}{5}= 4. 8 5 \\times \\frac {1 8}{5} - \\frac {1 8}{5} \\times 1 + 6. 1 5 \\times \\frac {1 8}{5} = (4. 8 5 - 1 + 6. 1 5) \\times \\frac {1 8}{5} = 1 0 \\times \\frac {1 8}{5} = 3 6$",
+      "knowledge_points": [
+        "分数除法相关的简便计算",
+        "整数乘法运算定律、推广到分数乘法"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_159",
+      "original_id": "unit3_choice_159"
+    },
+    {
+      "id": "unit3_choice_160",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数乘法计算",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "用简便方法计算下面各题。\n(1) $\\left(\\frac{8}{9} + \\frac{4}{27}\\right) \\times 27$\n(2) $\\frac{5}{6} \\times \\frac{5}{9} + \\frac{5}{9} \\times \\frac{1}{6}$",
+      "options": [],
+      "answer": "(1) 28; (2) $\\frac{5}{9}$",
+      "solution": "【分析】（1）根据乘法分配律 $(a + b) \\times c = a \\times c + b \\times c$ 进行简算； (2) 根据乘法分配律逆运算 $a \\times c + b \\times c = (a + b) \\times c$ 进行简算。\n【详解】 (1) $\\left(\\frac{8}{9} + \\frac{4}{27}\\right) \\times 27$ $= \\frac {8}{9} \\times 2 7 + \\frac {4}{2 7} \\times 2 7 = 2 4 + 4 = 2 8$\n(2) $\\frac{5}{6} \\times \\frac{5}{9} + \\frac{5}{9} \\times \\frac{1}{6}$ $= \\frac {5}{9} \\times (\\frac {5}{6} + \\frac {1}{6}) = \\frac {5}{9} \\times 1 = \\frac {5}{9}$",
+      "knowledge_points": [
+        "整数乘法运算定律推广到分数乘法"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_160",
+      "original_id": "unit3_choice_160"
+    },
+    {
+      "id": "unit3_choice_161",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数四则混合运算",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "下列各题怎样简便怎样算。\n$\\frac {5}{6} + \\frac {7}{2} \\div (\\frac {1}{2} + \\frac {2}{3}) \\quad 2 0 2 2 \\times \\frac {2 0 2 2}{2 0 2 3}\\quad \\frac {7}{1 3} \\times \\frac {1}{1 2} + \\frac {1 1}{1 3} \\times \\frac {7}{1 2}$",
+      "options": [],
+      "answer": "$3 \\frac{5}{6}; 2021 \\frac{1}{2023}; \\frac{7}{13}$",
+      "solution": "【分析】（1）先算括号里面的，再算除法，然后算加法。(2) 将 2022 转化成 (2023-1)，再利用乘法分配律计算。\n(3) 将其中一个分数变成 $\\frac{7}{12}$ , 再利用乘法分配律计算。\n【详解】 (1) $\\frac{5}{6} + \\frac{7}{2} \\div \\left(\\frac{1}{2} + \\frac{2}{3}\\right)$ $= \\frac {5}{6} + \\frac {7}{2} \\div \\frac {7}{6} = \\frac {5}{6} + \\frac {7}{2} \\times \\frac {6}{7} = \\frac {5}{6} + 3 = 3 \\frac {5}{6}$\n(2) $2022 \\times \\frac{2022}{2023}$ $= (2 0 2 3 - 1) \\times \\frac {2 0 2 2}{2 0 2 3} = 2 0 2 3 \\times \\frac {2 0 2 2}{2 0 2 3} - \\frac {2 0 2 2}{2 0 2 3} = 2 0 2 2 - \\frac {2 0 2 2}{2 0 2 3} = 2 0 2 1 \\frac {1}{2 0 2 3}$\n(3) $\\frac{7}{13} \\times \\frac{1}{12} + \\frac{11}{13} \\times \\frac{7}{12}$ $= \\frac {1}{1 3} \\times \\frac {7}{1 2}+ \\frac {1 1}{1 3} \\times \\frac {7}{1 2} = \\left(\\frac {1}{1 3} + \\frac {1 1}{1 3}\\right) \\times \\frac {7}{1 2} = \\frac {1 2}{1 3} \\times \\frac {7}{1 2} = \\frac {7}{1 3}$",
+      "knowledge_points": [
+        "分数的四则混合运算",
+        "整数乘法运算定律推广到分数乘法"
+      ],
+      "source": "22-23 六年级下·江西吉安·期末",
+      "merged_id": "unit3_161",
+      "original_id": "unit3_choice_161"
+    },
+    {
+      "id": "unit3_choice_162",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数四则混合运算",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "脱式计算。（能简算的要简算)\n$1. 2 5 \\times 0. 2 5 \\times 3. 2 \\quad 1 2 \\div \\left(\\frac {5}{8} \\times 1 8 + 0. 6 2 5 \\times 1 4\\right)$\n$\\left(\\frac {2}{3} - \\frac {1}{2} \\times \\frac {1}{3}\\right) \\div \\frac {1 5}{8} \\quad \\left(\\frac {1 7}{1 8} - \\frac {3}{4} + \\frac {1}{6}\\right) \\div \\frac {1}{3 6}$",
+      "options": [],
+      "answer": "1; $\\frac{3}{5}; \\frac {4}{1 5}; 1 3$",
+      "solution": "【分析】（1）先把 3.2 拆成 $0.8 \\times 4$ ，然后根据乘法交换律 $a \\times b = b \\times a$ ，乘法结合律 $(a \\times b) \\times c = a \\times (b \\times c)$ 进行简算；(2) 先算括号里面的, 根据乘法分配律逆运算 $a \\times c + b \\times c = (a + b) \\times c$ 进行简算, 再算括号外面的除法。 (3) 先算括号里面的乘法, 再算括号里面的减法, 最后算括号外面的除法; (4) 先把除法转化成乘法, 再根据乘法分配律 $(a + b) \\times c = a \\times c + b \\times c$ 进行简算。\n【详解】 (1) $1.25 \\times 0.25 \\times 3.2= 1. 2 5 \\times 0. 2 5 \\times (0. 8 \\times 4)= (1. 2 5 \\times 0. 8) \\times (0. 2 5 \\times 4) =1 \\times 1=1$\n(2) $12 \\div \\left(\\frac{5}{8} \\times 18 + 0.625 \\times 14\\right)=1 2 \\div \\left(\\frac {5}{8} \\times 1 8 + \\frac {5}{8} \\times 1 4\\right)=1 2 \\div \\left[ \\frac {5}{8} \\times (1 8 + 1 4) \\right]=1 2 \\div \\left[ \\frac {5}{8} \\times 3 2\\right)]= 1 2 \\div 2 0=\\frac {3}{5}$\n(3) $\\left(\\frac{2}{3} - \\frac{1}{2} \\times \\frac{1}{3}\\right) \\div \\frac{15}{8}= \\left(\\frac {2}{3} - \\frac {1}{6}\\right) \\div \\frac {1 5}{8}=\\left(\\frac {4}{6} - \\frac {1}{6}\\right) \\div \\frac {1 5}{8}= \\frac {1}{2} \\times \\frac {8}{15} =\\frac {4}{15}$\n(4) $\\left(\\frac{17}{18} -\\frac{3}{4} +\\frac{1}{6}\\right)\\div \\frac{1}{36}=\\left(\\frac {1 7}{1 8} - \\frac {3}{4} + \\frac {1}{6}\\right) \\times 3 6=\\frac {1 7}{1 8} \\times 3 6 - \\frac {3}{4} \\times 3 6 + \\frac {1}{6} \\times 3 6=3 4 - 2 7 + 6 = 113$",
+      "knowledge_points": [
+        "整数乘法运算定律推广到小数乘法",
+        "分数的四则混合运算",
+        "分数除法相关的简便计算",
+        "整数乘法运算定律推广到分数乘法"
+      ],
+      "source": "23-24 六年级下·湖南常德·期中",
+      "merged_id": "unit3_162",
+      "original_id": "unit3_choice_162"
+    },
+    {
+      "id": "unit3_choice_163",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数四则混合运算",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "计算下列各题, 能简便运算的要简便计算。\n(1) $\\frac{2}{3} + \\frac{5}{8} \\times \\frac{4}{15} - \\frac{7}{12}$\n(2) $25 \\times \\frac{1}{4} + 75 \\div 4$",
+      "options": [],
+      "answer": "(1) $\\frac{1}{4}$ ; (2) 25",
+      "solution": "【分析】（1）根据运算顺序，先计算乘法，再计算加法，最后计算减法；\n(2) 根据除以一个数等于乘它的倒数, 把式子转化为 $25 \\times \\frac{1}{4} + 75 \\times \\frac{1}{4}$ , 再根据乘法分配律进行简算。\n【详解】 (1) $\\frac{2}{3} + \\frac{5}{8} \\times \\frac{4}{15} - \\frac{7}{12}$ $= \\frac {2}{3} + \\frac {1}{6} - \\frac {7}{1 2} = \\frac {5}{6} - \\frac {7}{1 2} = \\frac {1}{4}$\n(2) $25 \\times \\frac{1}{4} + 75 \\div 4$ $= 2 5 \\times \\frac {1}{4} + 7 5 \\times \\frac {1}{4} = (2 5 + 7 5) \\times \\frac {1}{4} = 1 0 0 ^ {\\times} \\frac {1}{4} = 2 5 $",
+      "knowledge_points": [
+        "分数的四则混合运算",
+        "整数乘法运算定律推广到分数乘法"
+      ],
+      "source": "22-23 六年级下·四川绵阳·期中",
+      "merged_id": "unit3_163",
+      "original_id": "unit3_choice_163"
+    },
+    {
+      "id": "unit3_choice_164",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数四则混合运算",
+      "question_type": "计算题",
+      "difficulty": "容易",
+      "question": "脱式计算。\n$\\left(\\frac {4}{5} + \\frac {1}{4}\\right) \\div \\frac {7}{3} + \\frac {3}{1 0}\\quad \\frac {7}{1 2} \\times \\left(\\frac {1 0}{7} - \\frac {6}{7} \\div \\frac {3}{4}\\right) \\\\ \\frac {1}{6} \\times \\frac {8}{3} \\div \\left(0. 8 - \\frac {8}{1 5}\\right)\\quad5 \\div \\left[ \\left(\\frac {2}{3} + 0. 2\\right) \\times \\frac {1}{1 3}\\right) $",
+      "options": [],
+      "answer": "$\\frac{3}{4}; \\frac{1}{6}$ $\\frac{5}{3}$ ; 75",
+      "solution": "【分析】（1）先计算小括号里的加法，再计算括号外面的除法，最后计算加法；(2) 先计算除法, 再计算减法, 最后计算小括号外面的乘法; (3) 先通分, 计算小括号里的减法, 再计算小括号外面的除法, 最后计算乘法; (4) 先通分, 计算小括号里的加法, 再计算乘法, 最后计算小括号外面的除法。\n【详解】 (1) $\\left(\\frac{4}{5} + \\frac{1}{4}\\right) \\div \\frac{7}{3} + \\frac{3}{10}=\\left(\\frac {1 6}{2 0} + \\frac {5}{2 0}\\right) \\times \\frac {3}{7} + \\frac {3}{1 0}= \\frac {2 1}{2 0} \\times \\frac {3}{7} + \\frac {3}{1 0}= \\frac {9}{2 0} + \\frac {6}{2 0}=\\frac {3}{4}$\n(2) $\\frac{7}{12} \\times \\left(\\frac{10}{7} - \\frac{6}{7} \\div \\frac{3}{4}\\right)=\\frac {7}{1 2} \\times \\left(\\frac {1 0}{7} - \\frac {6}{7} \\times \\frac {4}{3}\\right)={\\frac {7}{1 2}} \\times \\left(\\frac {1 0}{7} - \\frac {8}{7}\\right)= \\frac {7}{1 2} \\times \\frac {2}{7}=\\frac {1}{6}$\n(3) $\\frac{1}{6} \\times \\frac{8}{3} \\div \\left(0.8 - \\frac{8}{15}\\right)=\\frac {1}{6} \\times \\frac {8}{3} \\div \\left(\\frac {4}{5} - \\frac {8}{1 5}\\right)= \\frac {1}{6} \\times \\frac {8}{3} \\div \\left(\\frac {1 2}{1 5} - \\frac {8}{1 5}\\right)=\\frac {1}{6} \\times \\frac {8}{3} \\times \\frac {1 5}{4}=\\frac {5}{3}$\n(4) $5 \\div \\left[\\left(\\frac{2}{3} + 0.2\\right) \\times \\frac{1}{13}\\right]= 5 \\div \\left[ \\left(\\frac {2}{3} + \\frac {1}{5}\\right) \\times \\frac {1}{1 3}\\right]=5 \\div \\left[ \\frac {1 3}{1 5} \\times \\frac {1}{1 3}\\right]= 5 \\times 1 5 = 7 5 $",
+      "knowledge_points": [
+        "异分母分数加、减法",
+        "分数乘分数",
+        "一位或多位小数化分数",
+        "分数与分数的除法"
+      ],
+      "source": "21-22 六年级上·天津南开·期末",
+      "merged_id": "unit3_164",
+      "original_id": "unit3_choice_164"
+    },
+    {
+      "id": "unit3_choice_165",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法计算",
+      "question_type": "计算题",
+      "difficulty": "容易",
+      "question": "计算。\n$\\frac {6}{5} \\div \\frac {3}{5} = \\quad \\frac {5}{8} \\div \\frac {3}{2} = \\\\\\frac {5}{8} \\div \\frac {2 5}{2 8} = \\quad\\frac {4}{5} \\div \\frac {3}{5} =$",
+      "options": [],
+      "answer": "$2; \\frac{5}{12}; \\frac{7}{10}; \\frac{4}{3}$",
+      "solution": "【分析】略",
+      "knowledge_points": [
+        "分数与分数的除法"
+      ],
+      "source": "22-23 五年级上·山东·单元测试",
+      "merged_id": "unit3_165",
+      "original_id": "unit3_choice_165"
+    },
+    {
+      "id": "unit3_choice_166",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "一堆糖果, 其中奶糖占 $\\frac{9}{20}$ , 再放入 16 块水果糖后, 奶糖就只占 $\\frac{1}{4}$ 。这一堆糖果原来一共有多少块?",
+      "options": [],
+      "answer": "20 块",
+      "solution": "【分析】放入水果糖后, 这堆糖果的总数量增加了, 但是奶糖的数量不变, 找出等量关系: 原来糖果数量 $\\times \\frac{9}{20} =$ (原有糖果数量 +16) $\\times \\frac{1}{4}$ ; 设原来这堆糖果有 x 块, 据此列出方程, 解方程即可解答。\n【详解】解：设这一堆糖果原来一共 $x$ 块。\n$\\frac {9}{2 0} x = (x + 1 6) \\times \\frac {1}{4}\\quad\\frac {9}{2 0} x = \\frac {1}{4} x + 1 6 \\times \\frac {1}{4}\\quad\\frac {9}{2 0} x = \\frac {1}{4} x + 4\\quad\\frac {9}{2 0} x - \\frac {1}{4} x = \\frac {1}{4} x + 4 - \\frac {1}{4} \\quad\\frac {1}{5} x = 4\\\\ \\frac {1}{5} x \\div \\frac {1}{5} = 4 \\div \\frac {1}{5}\\quad x = 4 \\times 5 \\quad x = 2 0$答: 这堆糖果原来一共有 20 块。",
+      "knowledge_points": [
+        "列方程解含两个未知数的问题",
+        "应用等式的性质 1 和 2 解方程",
+        "分数与分数的除法"
+      ],
+      "source": "23-24六年级上·全国·单元测试",
+      "merged_id": "unit3_166",
+      "original_id": "unit3_choice_166"
+    },
+    {
+      "id": "unit3_choice_167",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "困难",
+      "question": "乙队原有人数是甲队的 $\\frac{3}{7}$ ，现在甲队派10人到乙队，则乙队人数是甲队的 $\\frac{2}{3}$ 。甲、乙两队原来有多少人？",
+      "options": [],
+      "answer": "甲队原来有 70 人; 乙队原来有 30 人",
+      "solution": "【分析】设甲队原有 $x$ 人，则乙队原有 $\\frac{3}{7} x$ 人，现在甲队派 10 人到乙队，则现在甲队有 $(x - 10)$ 人，乙队有 $(\\frac{3}{7} x + 10)$ 人，根据等量关系：现在的甲队的人数 $\\times \\frac{2}{3} =$ 现在的乙队人数'列方程解答即可求出甲队原来的人数，再根据求一个数的几分之几是多少，用乘法求出乙队原来的人数。找出等量关系: '现在的甲队的人数 $\\times \\frac{2}{3} =$ 现在的乙队人数'是列方程解题的关键。\n【详解】解：设甲队原有 $x$ 人。$\\frac {2}{3} (x - 1 0) = \\frac {3}{7} x + 1 0 \\quad \\frac {2}{3} x - \\frac {3}{7} x = 1 0 + \\frac {2 0}{3}\\quad{5}{2 1} x = \\frac {5 0}{3} \\\\ x = 7 0 \\quad 7 0 \\times \\frac {3}{7} = 3 0 (\\text {人})$\n答: 甲队原来的 70 人, 乙队原来有 30 人。",
+      "knowledge_points": [
+        "列方程解含两个未知数的问题"
+      ],
+      "source": "23-24六年级上·全国·单元测试",
+      "merged_id": "unit3_167",
+      "original_id": "unit3_choice_167"
+    },
+    {
+      "id": "unit3_choice_168",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "乙队原有的人数是甲队人数的 $\\frac{3}{7}$ 。现在甲队派 30 人到乙队, 则乙队人数现在是甲队的 $\\frac{3}{5}$ 。原来两队一共有多少人?",
+      "options": [],
+      "answer": "400人",
+      "solution": "【分析】首先设甲队原来有 $x$ 人, 因为乙队原有的人数是甲队人数的 $\\frac{3}{7}$ , 所以乙队原来有人 $\\frac{3}{7} x$ 人。然后甲队派 30 人到乙队后, 甲队人数变为 $x - 30$ , 乙队人数变为 $\\frac{3}{7} x + 30$ 。而此 时乙队人数是甲队的 $\\frac{3}{5}$ , 这就可以列出方程, $\\frac{3}{5} (x - 30) = \\frac{3}{7} x + 30$ 。\n【详解】解：设甲队原有 $\\mathrm{x}$ 人，乙队原有 $\\frac{3}{7} \\mathrm{x}$ 人，\n$\\frac {3}{5} (x - 3 0) = \\frac {3}{7} x + 3 0$\n解: $\\frac{3}{5} x - 18 = \\frac{3}{7} x + 30$\n$\\frac {3}{5} x - 1 8 - \\frac {3}{7} x = \\frac {3}{7} x + 3 0 - \\frac {3}{7} x$\n$\\frac {3}{5} x - \\frac {3}{7} x - 1 8 = 3 0​$\n$\\frac {6}{3 5} x - 1 8 + 1 8 = 3 0 + 1 8$\n$\\frac {6}{3 5} x = 4 8$\n$\\frac {6}{3 5} x \\div \\frac {6}{3 5} = 4 8 \\div \\frac {6}{3 5}$\n$\\frac {6}{3 5} \\mathrm {x} \\times \\frac {3 5}{6} = 4 8 \\times \\frac {3 5}{6}$\n$x = 2 8 0$\n乙队: $280 \\times \\frac{3}{7} = 120$ （人）共: $280 + 120 = 400$ (人)\n答: 原来两队一共有 400 人。",
+      "knowledge_points": [
+        "列方程解含两个未知数的问题"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_168",
+      "original_id": "unit3_choice_168"
+    },
+    {
+      "id": "unit3_choice_169",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "一根电线, 用去全长的 $\\frac{1}{3}$ 还多 4 米, 这时剩下的比用去的多 10 米。这根电线原来长多少米?",
+      "options": [],
+      "answer": "54米",
+      "solution": "【分析】把全长看作单位 \"1\", 用去部分是全长的 $\\frac{1}{3}$ 多 4 米,剩下部分是全长的 $\\frac{2}{3}$ 少 4 米。找出等量关系: 剩下的部分一用去的部分 $= 10$ , 设原来这根电线长 $x$ 米, 据此列出方程,解方程即可解答。\n【详解】解：设这根电线原来长 $x$ 米。\n$\\begin{array}{l} \\left(\\frac {2}{3} x - 4\\right) - \\left(\\frac {1}{3} x + 4\\right) = 1 0 \\frac {2}{3} x - 4 - \\frac {1}{3} x - 4 = 1 0 \\frac {1}{3} x - 8 = 1 0 \\frac {1}{3} x - 8 + 8 = 1 0 + 8 \\\\ \\end{array}$\n${1}{3} x = 1 8 \\frac {1}{3} x \\div \\frac {1}{3} = 1 8 \\div \\frac {1}{3} x = 5 4 $\n答: 这根电线原来长 54 米。",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "列方程解含两个未知数的问题",
+        "应用等式的性质 1 和 2 解方程"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_169",
+      "original_id": "unit3_choice_169"
+    },
+    {
+      "id": "unit3_choice_170",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "某校六年级共有学生 195 人, 从中选出男生的 $15 \\%$ 和 10 名女生参加数学竞赛,剩下的男生人数与女生人数相等，那么，该校六年级参加数学竞赛的学生有多少人？",
+      "options": [],
+      "answer": "25人",
+      "solution": "【分析】设六年级男生有 x 人, 则女生有 (195-x) 人; 选出男生的 $15 \\%$ 和 10 名女生后, 男生还有 $(1 - 15 \\%) x$ 人, 女生还有 $(195 - x - 10)$ 人; 根据数量关系: 剩下的男女\n生人数正好相等，据此列方程，解方程求出六年级男生有多少人；最后用男生人数乘 $15 \\%$ 计算参加数学竞赛的男生人数，再加上 10 ，所得结果即为参加数学竞赛的学生人数。\n【详解】解：设六年级男生有 $x$ 人，则女生有 $(195 - x)$ 人。\n$(1 - 15 \\%) x = 195 - x - 10$\n$\\begin{array}{l} 85 \\% \\mathrm {x} = 185 - \\mathrm {x} 0. 8 5 x + x = 1 8 5 - x + x 1. 8 5 \\mathrm {x} = 1 8 5 1. 8 5 \\mathrm {x} \\div 1. 8 5 = 1 8 5 \\div 1. 8 5 x = 1 0 0 100 \\times 15 \\% + 10 = 1 5 + 1 0 = 2 5 \\text {(人)} \\\\ \\end{array}$\n答: 该校六年级参加数学竞赛的学生有 25 人。",
+      "knowledge_points": [
+        "列方程解含两个未知数的问题",
+        "应用等式的性质 1 和 2 解方程",
+        "求一个数的百分之几是多少"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_170",
+      "original_id": "unit3_choice_170"
+    },
+    {
+      "id": "unit3_choice_171",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "实验小学六年级有学生 152 人。现在要选出男生人数的 $\\frac{1}{11}$ 和女生 5 人, 到国际数学家大会与专家见面。学校按照上述要求选出若干名代表后，剩下的男、女生人数相等。实验小学六年级男生多少人？",
+      "options": [],
+      "answer": "77人",
+      "solution": "【分析】根据'选出男生人数的 $\\frac{1}{11}$ '，把男生的人数看作单位\"1\"，剩下的男生对应了 $\\frac{10}{11}$ ，剩下的女生人数等于从女生总人数中减去5人。根据题目信息写出等量关系：男生人数 $x\\left(1 - \\frac{1}{11}\\right) =$ 女生人数-5；设男生人数有 $\\mathbf{x}$ 人，则女生有 $(152 - x)$ 人，列方程： $x \\times (1 - \\frac{1}{11}) = 152 - x - 5$ ，解方程即可解答。\n【详解】解：设实验小学六年级有男生 $\\mathrm{x}$ 人，则女生有 $(152 - x)$ 人。\n$\\mathrm {x} \\times (1 - \\frac {1}{1 1}) = 1 5 2 - \\mathrm {x} - 5$\n${1 0}{1 1} \\mathrm {x} = 1 4 7 - \\mathrm {x} \\frac {1 0}{1 1} \\mathrm {x} + \\mathrm {x} = 1 4 7 \\frac {2 1}{1 1} x = 1 4 7 x = 1 4 7 \\div \\frac {2 1}{1 1} x = 1 4 7 \\times \\frac {1 1}{2 1} x = 7 7$\n答: 试验小学六年级有男生 77 人。",
+      "knowledge_points": [
+        "列方程解含两个未知数的问题"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_171",
+      "original_id": "unit3_choice_171"
+    },
+    {
+      "id": "unit3_choice_172",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "小光看一本书，第一天看了全书的 $20\\%$ ，第二天看了剩下的 $25\\%$ ，还剩下120页没有看。那么这本书共有多少页？",
+      "options": [],
+      "answer": "200页",
+      "solution": "【分析】设这本书共有 $x$ 页, 则第一天看了 $20 \\% x$ 页, 还剩下 $(x - 20 \\% x)$ 页, 再把剩下的页数看作单位 \"1\", 第二\n天看了剩下的 $25 \\%$ , 则第二天看了 $(x - 20 \\% x) \\times 25 \\%$ 页,根据等量关系: \"总页数 = 第一天看的页数 + 第二天看的页数 + 120 页\" 列方程解答即可。\n【详解】解：设这本书共有 $x$ 页。\n$x = 20 \\% x + (x - 20 \\% x) \\times 25 \\% + 120 x = 0. 2 x + 0. 2 5 x - 0. 0 5 x + 1 2 0 x - 0. 4 x = 1 2 0 0. 6 x = 1 2 0 x = 2 0 0$ 答: 这本书共有 200 页。",
+      "knowledge_points": [
+        "列方程解含一个未知数的问题"
+      ],
+      "source": "23-24六年级上·全国·单元测试",
+      "merged_id": "unit3_172",
+      "original_id": "unit3_choice_172"
+    },
+    {
+      "id": "unit3_choice_173",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "小明读一本小说,第一天读了全书的 $\\frac{4}{7}$ , 第二天又读了余下的 $\\frac{3}{5}$ , 还有 42 页没读完, 这本书有多少页?",
+      "options": [],
+      "answer": "245页",
+      "solution": "【分析】设这本书共有 $x$ 页, 则第一天读了 $\\frac{4}{7} x$ 页, 还余下 $(1 - \\frac{4}{7}) x$ 页, 第二天又读了余下的 $\\frac{3}{5}$ , 则第二天读了 $\\frac{3}{5} \\times (1 - \\frac{4}{7}) x$ 页, 根据等量关系: \"第一天读的页数 + 第二天读的页数 + 42 页 = 这本书的总页数\" 列方程解答。\n【详解】解：设这本书共有 $\\mathbf{X}$ 页。\n${4}{7} x + \\frac {3}{5} \\times (1 - \\frac {4}{7}) x + 4 2 = x \\frac {4}{7} x + \\frac {9}{3 5} x + 4 2 = x x - \\frac {2 9}{3 5} x = 4 2 \\frac {6}{3 5} x = 4 2 x = 2 4 5$ 答: 这本书共有 245 页。\n【点睛】找出等量关系：\"第一天读的页数 + 第二天读的页数 + 42 页 = 这本书的总页数\"是列方程解答的关键。",
+      "knowledge_points": [
+        "列方程解含一个未知数的问题"
+      ],
+      "source": "23-24六年级上·全国·单元测试",
+      "merged_id": "unit3_173",
+      "original_id": "unit3_choice_173"
+    },
+    {
+      "id": "unit3_choice_174",
+      "unit": "第三单元：分数除法",
+      "chapter": "工程问题",
+      "question_type": "解答题",
+      "difficulty": "困难",
+      "question": "有两个同样的仓库A 和 B，搬运一个仓库里的货物，甲需要 18 小时，乙需要 12 小时，丙需要 9 小时。甲、乙在 A 仓库，丙在 B 仓库，同时开始搬运。中途甲又转向帮助丙搬运。最后两个仓库同时搬完。甲帮助乙、丙各多少小时？",
+      "options": [],
+      "answer": "甲帮乙：6小时；甲帮丙：2小时",
+      "solution": "【分析】把每个仓库的货物看作单位\"1\"，根据工作效率 $=$ 工作总量 $\\div$ 工作时间，分别求出甲，乙、丙的工作效率。无论怎么样干活，最后一起干完，相当于三个人一块搬完两个仓库的货物，那么求出搬完一共用的时间。再根据工作总量 $=$ 工作效率 $\\times$ 工作时间，用乙的工作效率 $\\times$ 搬完一共用的时间，求出乙的工作总量，用1一乙的工作总量，求出甲在仓库A的工作总量，再除以甲的工作效率，即可求出甲在仓库A的工作时间，即甲帮乙的时间；再用一共用的时\n间减去甲帮乙时间，即可求出甲在仓库B的工作时间，也就是帮丙的时间，据此解答。\n【详解】把每个仓库的货物看作单位 \"1\", 则甲的工作效率是 $\\frac{1}{18}$ , 乙的工作效率是 $\\frac{1}{12}$ , 丙的工作效率是 $\\frac{1}{9}$ 。\n$ 2 \\div (\\frac {1}{1 8} + \\frac {1}{1 2} + \\frac {1}{9}) = 2 \\div (\\frac {2}{3 6} + \\frac {3}{3 6} + \\frac {4}{3 6}) = 2 \\div (\\frac {5}{3 6} + \\frac {4}{3 6}) = 2 \\div \\frac {1}{4} = 2 \\times 4 = 8 \\text {(小 时)}$\n$(1 - \\frac {1}{1 2} \\times 8) \\div \\frac {1}{1 8} = (1 - \\frac {2}{3}) \\div \\frac {1}{1 8} = \\frac {1}{3} \\times 1 8 = 6 \\text {(小 时)} $\n$8 - 6 = 2 \\text {(小 时)}$\n甲帮乙 6 小时, 甲帮丙 2 小时。\n【点睛】解答本题的关键是先求出三人同时搬运所需要的时间，再利用工作总量、工作效率和工作时间三者的关系，进行解答。",
+      "knowledge_points": [
+        "两人合作的工程问题",
+        "分数的四则混合运算"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_174",
+      "original_id": "unit3_choice_174"
+    },
+    {
+      "id": "unit3_choice_175",
+      "unit": "第三单元：分数除法",
+      "chapter": "工程问题",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "一项工程，甲乙合作6天可以完成。乙做7天、甲做4天可以完成这项工程的 $\\frac{13}{15}$ ，剩下的由乙单独做需要多少天？",
+      "options": [],
+      "answer": "2天",
+      "solution": "【分析】把这项工程总量看作单位 \"1\", 甲乙合作 6 天可以完成, 则甲乙的工作效率之和为 $\\frac{1}{6}$ ; 乙做 7 天, 甲做 4 天可以完成这项工程的 $\\frac{13}{15}$ , 则剩下的工程量为 $(1 - \\frac{13}{15})$ ; 设乙的工作效率为 $x$ , 则甲的工作效率为 $(\\frac{1}{6} - x)$ , 根据工作总量\n= 工作效率 $\\times$ 工作时间, 代入相应数值计算出乙的工作效率;最后用剩余的工程量除以乙的工作效率, 所得结果即为乙单独完成剩余的工程需要的天数。\n【详解】解：设乙的工作效率为 $x$ ，则甲的工作效率为($\\frac {1}{6} - x$）\n$ 7 x + 4 \\times \\left(\\frac {1}{6} - x\\right) = \\frac {1 3}{1 5} \\\\ 7 x + \\frac {2}{3} - 4 x = \\frac {1 3}{1 5} \\\\ 3 x + \\frac {2}{3} - \\frac {2}{3} = \\frac {1 3}{1 5} - \\frac {2}{3} \\\\ 3 x = \\frac {1}{5} \\\\ 3 x \\div 3 = \\frac {1}{5} \\div 3 \\\\ x = \\frac {1}{1 5} \\\\ $ $\\left(1 - \\frac {1 3}{1 5}\\right) \\div \\frac {1}{1 5} = \\frac {2}{1 5} \\div \\frac {1}{1 5} =2(天)$\n答: 剩下的由乙单独做需要 2 天。",
+      "knowledge_points": [
+        "两人合作的工程问题",
+        "分数的四则混合运算",
+        "列方程解含两个未知数的问题",
+        "应用等式的性质 1 和 2 解方程"
+      ],
+      "source": "23-24六年级上·全国·单元测试",
+      "merged_id": "unit3_175",
+      "original_id": "unit3_choice_175"
+    },
+    {
+      "id": "unit3_choice_176",
+      "unit": "第三单元：分数除法",
+      "chapter": "工程问题",
+      "question_type": "解答题",
+      "difficulty": "困难",
+      "question": "一件工作，甲工程队独做12小时可以完成，现在先甲、乙合做4小时，剩下的工作，全部交给乙完成，还需要2小时，乙单独完成这份工作需要多少天？",
+      "options": [],
+      "answer": "9天",
+      "solution": "【分析】把这件工作的总量看作单位\"1\"，甲工程队独做12小时可以完成，则甲的工作效率是 $\\frac{1}{12}$ ，完成这件工作，甲一共做了4小时，乙做了（ $4+2$ ）小时，根据\"工作量 $=$ 工作效率 $\\times$ 工作时间\"求出甲4小时完成的工作量，再用总工作量减去甲4小时完成的工作量就是乙（ $4+2$ ）小时完成的工作量，再根据\"工作量 $\\div$ 工作时间 $=$ 工作效率\"求出乙的工作效率，再根据\"工作量 $\\div$ 工作效率 $=$ 工作时间\"即可解答。本题考查了工作量、工作效率、工作时间的关系,求出乙的工作效率是解题的关键。\n【详解】 $(1 - \\frac{1}{12} \\times 4) \\div (4 + 2)={2}{3} \\div 6 = \\frac {1}{9}$$1 \\div \\frac {1}{9} = 9 \\quad (\\text {天})$答: 乙单独完成这份工作需要 9 天。",
+      "knowledge_points": [
+        "两人合作的工程问题",
+        "分数的四则混合运算"
+      ],
+      "source": "23-24六年级上·全国·单元测试",
+      "merged_id": "unit3_176",
+      "original_id": "unit3_choice_176"
+    },
+    {
+      "id": "unit3_choice_177",
+      "unit": "第三单元：分数除法",
+      "chapter": "工程问题",
+      "question_type": "解答题",
+      "difficulty": "困难",
+      "question": "师徒二人加工一批零件，师傅单独加工需12天完成，先师傅和徒弟一起加工7天，师傅因事离开，徒弟还需3天才能完成全部工程。那么，徒弟单独做完成全工程需多少天？",
+      "options": [],
+      "answer": "24天",
+      "solution": "【分析】根据工作效率 $=$ 工作总量 $\\div$ 工作时间；把这批零件的总量看作单位\"1\"，用 $1 \\div 12$ ，求出师傅的工作效率；再根据工作总量 $=$ 工作效率 $\\times$ 工作时间，用师傅的工作效率 $\\times 7$ ，求出师傅7天的工作量，再用1一师傅7天的工作量，求出徒弟的工作量，再用徒弟的工作量÷徒弟的工作时间，即可求出徒弟的工作效率，再用工作总量÷徒弟的工作效率，即可求出徒弟单独完成全工程需要的时间。求出徒弟的工作效率是解答本题的关键。\n【详解】 $(1 - \\frac{1}{12} \\times 7) \\div (7 + 3)$ $= (1 - \\frac {7}{1 2}) \\div 1 0 = \\frac {5}{1 2} \\div 1 0 = \\frac {5}{1 2} \\times \\frac {1}{1 0} = \\frac {1}{2 4}$$1 \\div \\frac {1}{2 4} = 1 \\times 2 4 = 2 4 \\text {(天)}$\n答: 徒弟单独完成全工程需要 24 天。",
+      "knowledge_points": [
+        "两人合作的工程问题",
+        "分数的四则混合运算"
+      ],
+      "source": "23-24六年级上·全国·单元测试",
+      "merged_id": "unit3_177",
+      "original_id": "unit3_choice_177"
+    },
+    {
+      "id": "unit3_choice_178",
+      "unit": "第三单元：分数除法",
+      "chapter": "工程问题",
+      "question_type": "解答题",
+      "difficulty": "困难",
+      "question": "加工一批零件, 甲单独做要15天，乙单独做要10天完成，甲乙同时工作，全做完时，甲做了2400个，这批零件共有多少个？",
+      "options": [],
+      "answer": "6000 个",
+      "solution": "【分析】甲单独做要 15 天, 乙单独做要 10 天完成, 则甲每天完成全部零件的 $1 \\div 15 = \\frac{1}{15}$ , 乙每天完成全部零件的 $1 \\div 10 = \\frac{1}{10}$ ,两人合作时每天完成全部零件的 $\\left(\\frac{1}{15} + \\frac{1}{10}\\right)$ , 据此求出两人合作完成的天数, 再求出两人合作时, 甲完成的占全部零件的分率, 用甲完成的数量除以他完成的占全部零件的分率,求出这批零件共有多少个即可。本题考查工程问题、分数乘除法，解答本题的关键是掌握题中的数量关系式。\n【详解】 合作完成天数: $1 \\div \\left(\\frac{1}{15} + \\frac{1}{10}\\right)= 1 \\div \\frac {1}{6}=6 (天)$这批零件数量: $2400 \\div \\left(\\frac{1}{15} \\times 6\\right)= 2 4 0 0 \\div \\frac {2}{5}=2 4 0 0 \\times \\frac {5}{2}=6000 (个)$答：这批零件共有 6000 个。",
+      "knowledge_points": [
+        "两人合作的工程问题",
+        "求一个数的几分之几的问题",
+        "已知一个数的几分之几是多少，求这个数"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_178",
+      "original_id": "unit3_choice_178"
+    },
+    {
+      "id": "unit3_choice_179",
+      "unit": "第三单元：分数除法",
+      "chapter": "工程问题",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "一批零件，师傅单独加工需要50天完成，徒弟单独加工需要75天完成，现师徒两人共同加工，完成任务时，徒弟比师傅少加工120个，这批零件共有多少个？",
+      "options": [],
+      "answer": "600 个",
+      "solution": "【分析】先算出两人单独的工作效率，用单位\"1\"除以工作效率之和可以算出合作完成所用的时间；再用两人的工作效率分别乘工作时间算出两人的工作量，120个对应的分率是工作量之差，用120除以工作量之差即可算出零件个数。\n【详解】师傅的工作效率: $1 \\div 50 = \\frac{1}{50}$徒弟的工作效率: $1 \\div 75 = \\frac{1}{75}$师徒合作时间: $1 \\div \\left(\\frac{1}{50} + \\frac{1}{75}\\right)= 1 \\div \\frac {1}{3 0}=30(天)$$1 2 0 \\div \\left(\\frac {1}{5 0} \\times 3 0 - \\frac {1}{7 5} \\times 3 0\\right)= 1 2 0 \\div \\left(\\frac {3}{5} - \\frac {2}{5}\\right)=1 2 0 \\div \\frac {1}{5}=600 (个)$\n答：这批零件共有600个。",
+      "knowledge_points": [
+        "两人合作的工程问题",
+        "分数的四则混合运算"
+      ],
+      "source": "23-24六年级上·全国·单元测试",
+      "merged_id": "unit3_179",
+      "original_id": "unit3_choice_179"
+    },
+    {
+      "id": "unit3_choice_180",
+      "unit": "第三单元：分数除法",
+      "chapter": "工程问题",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "加工一批零件甲单独做6小时完成，乙单独做要10小时完成。甲乙共同做了1小时后再由甲单独做2小时，此时还剩240个没有加工，这批零件共有多少个？",
+      "options": [],
+      "answer": "600 个",
+      "solution": "【分析】把这批零件的个数看作单位 \"1\", 根据工作效率 $=$工作总量÷工作时间，分别求出甲、乙的工作效率；根据工作量 = 工作效率 × 工作时间，求出甲乙共做了 1 小时的工作量；求出甲单独做2小时的工作量，用1减去甲乙共做了1小时的工作量，减去甲单独做2小时的工作量，求出还剩下几分之几没有完成，对应的是 240 个，再根据分数除法的意义，求出单位\"1\"，据此解答。\n【详解】 $240 \\div [1 - \\left(\\frac{1}{6} + \\frac{1}{10}\\right) \\times 1 - \\frac{1}{6} \\times 2]= 2 4 0 \\div [ 1 - (\\frac {5}{3 0} + \\frac {3}{3 0}) \\times 1 - \\frac {1}{3} ] = 2 4 0 \\div [ 1 - \\frac {4}{1 5} \\times 1 - \\frac {1}{3} ] = 2 4 0 \\div [ 1 - \\frac {4}{1 5} - \\frac {5}{1 5} ] = 2 4 0 \\div [ \\frac {1 1}{1 5} - \\frac {5}{1 5} ] = 2 4 0 \\div \\frac {2}{5} = 2 4 0 \\times \\frac {5}{2} = 6 0 0 (\\text {个})$\n答：这批零件共有 600 个。",
+      "knowledge_points": [
+        "两人合作的工程问题",
+        "分数的四则混合运算"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_180",
+      "original_id": "unit3_choice_180"
+    },
+    {
+      "id": "unit3_choice_181",
+      "unit": "第三单元：分数除法",
+      "chapter": "工程问题",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "加工一批零件, 甲独做8小时完成，乙独做10小时完成。甲乙合作2小时后，还剩下440个没有加工。这批零件共有多少个？",
+      "options": [],
+      "answer": "800 个",
+      "solution": "【分析】把这批零件的总数量看作单位\"1\"，根据工作效率 $=$ 工作总量÷工作时间，分别求出甲的工作效率和乙的工作效率；再根据工作总量 $=$ 工作效率 $\\times$ 工作时间，求出甲乙合作2小时的工作量；用1一甲乙合作2小时的工作量，求出还剩下几分之几没有完成，对应的是440个，再根据分数除法的意义，求出单位\"1\"，据此解答。\n【详解】 $440 \\div [1 - \\left(\\frac{1}{8} + \\frac{1}{10}\\right) \\times 2]= 4 4 0 \\div [ 1 - (\\frac {1}{4} + \\frac {1}{5}) ] = 4 4 0 \\div [ 1 - (\\frac {5}{2 0} + \\frac {4}{2 0}) ] = 4 4 0 \\div [ 1 - \\frac {9}{2 0} ] = 4 4 0 \\div \\frac {1 1}{2 0} = 4 4 0 \\times \\frac {2 0}{1 1} = 8 0 0 \\text {(个)}$\n答: 这批零件共 800 个。",
+      "knowledge_points": [
+        "两人合作的工程问题",
+        "分数的四则混合运算"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_181",
+      "original_id": "unit3_choice_181"
+    },
+    {
+      "id": "unit3_choice_182",
+      "unit": "第三单元：分数除法",
+      "chapter": "工程问题",
+      "question_type": "解答题",
+      "difficulty": "困难",
+      "question": "加工一批零件, 甲独做要12天，乙独做要15天，甲乙合作3天后，乙又做了2天后，还剩175个零件没有加工，这批零件共有多少个？",
+      "options": [],
+      "answer": "420 个",
+      "solution": "【分析】把工作总量看作单位 \"1\", 根据甲乙的独做时间分别求出他们的工作效率, 工作效率乘上他们对应的工作时间: 甲3天, 乙5天, 可以算出已经完成的工作量对应分率,用1减去已经完成的工作量对应分率, 就是剩余部分175个对应的分率, 用175除以剩余部分的分率即可。本题考查工程问题、分数乘除法，解答本题的关键是掌握题中的数量关系。\n【详解】甲的工作效率: $1 \\div 12 = \\frac{1}{12}$乙的工作效率: $1 \\div 15 = \\frac{1}{15}$已完成: $3 \\times \\frac{1}{12} + 5 \\times \\frac{1}{15}= \\frac {1}{4} + \\frac {1}{3}$零件总共有: $175 \\div (1 - \\frac{7}{12})=\\frac {7}{12} = 175 \\div \\frac {5}{1 2} = 1 7 5 \\times \\frac {1 2}{5} = 4 2 0 \\text {(个)}$答: 这批零件共有 420 个。",
+      "knowledge_points": [
+        "两人合作的工程问题",
+        "分数与整数的除法",
+        "分数乘整数"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_182",
+      "original_id": "unit3_choice_182"
+    },
+    {
+      "id": "unit3_choice_183",
+      "unit": "第三单元：分数除法",
+      "chapter": "工程问题",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "修一条长150米的铁路，甲队单独修完需要15天，乙队单独修完需要30天。现在甲乙一起合修了全长的 $\\frac{1}{5}$ ，问他们合修了多少天？",
+      "options": [],
+      "answer": "2天",
+      "solution": "【分析】根据甲乙单独完成工作的时间，可以用分率（把总量看作单位\"1\"）表示出甲乙的工作效率，再用 $\\frac{1}{5}$ 除以两人的工作效率之和即可。本题考查了工作量、工作时间、工作效率的关系,熟练掌握三者的关系是解题的关键。\n【详解】甲: $1 \\div 15 = \\frac{1}{15}$ 乙: $1 \\div 30 = \\frac{1}{30}$ ${1}{5} \\div (\\frac {1}{1 5} + \\frac {1}{3 0}) = \\frac {1}{5} \\div \\frac {1}{1 0} = 2 \\text {(天)}$答: 他们合修了 2 天。",
+      "knowledge_points": [
+        "两人合作的工程问题",
+        "分数的四则混合运算"
+      ],
+      "source": "23-24六年级上·全国·单元测试",
+      "merged_id": "unit3_183",
+      "original_id": "unit3_choice_183"
+    },
+    {
+      "id": "unit3_choice_184",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "某车间有 52 名工人,后来又调进 4 名女工人, 这时女工人数是男工人数的 $\\frac{3}{4}$ , 这车间原有女工多少人?",
+      "options": [],
+      "answer": "20人",
+      "solution": "【分析】由于女工人数增加所以总人数也跟着增加，不变的是男工人数，把男工人数看作单位\"1\"，现在女工人数是男工人数的 $\\frac{3}{4}$ ，总人数就是男工人数的 $1 + \\frac{3}{4} = \\frac{7}{4}$ ，对应的是 $52 + 4 = 56$ 人；已知一个数的几分之几是多少求这个数用除法计算，则用 56 除以 $\\frac{7}{4}$ 可以算出男工人数，之后再分别计算现在和原来的女工人数即可。\n【详解】男工：$(5 2 + 4) \\div (1 + \\frac {3}{4}) = 5 6 \\div \\frac {7}{4} = 5 6 \\times \\frac {4}{7} = 3 2 \\text {(人)}$女工原来:$3 2 \\times \\frac {3}{4} - 4 = 2 4 - 4 = 2 0 \\text {(人)}$答: 这车间原有女工 20 人。",
+      "knowledge_points": [
+        "求一个数的几分之几的问题",
+        "已知一个数的几分之几是多少，求这个数"
+      ],
+      "source": "23-24六年级上·全国·单元测试",
+      "merged_id": "unit3_184",
+      "original_id": "unit3_choice_184"
+    },
+    {
+      "id": "unit3_choice_185",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "一袋大米第一周吃了全部的 $\\frac{1}{3}$ 还多 6 千克, 后又加入 8 千克, 第二周又吃了剩下的 $\\frac{1}{3}$ , 这时袋里的大米恰好是 24 千克。这袋大米原来有多少千克?",
+      "options": [],
+      "answer": "51千克",
+      "solution": "【分析】本题从后往前推导，因此先把第二周吃前的大米数量看作单位\"1\"，吃掉了其中的 $\\frac{1}{3}$ ，那么24千克对应剩下的 $1 - \\frac{1}{3} = \\frac{2}{3}$ ，用24除以 $\\frac{2}{3}$ 可以求出第二周吃前大米有36千克。再把原来的大米看作单位 \"1\", 吃掉了 $\\frac{1}{3}$ 多 6 千克, 又加入 8 千克, 说明剩余部分比 $\\frac{2}{3}$ 多 2 千克。用 36 千克减去 2 千克剩下 34 千克, 刚好对应原来的 $\\frac{2}{3}$ , 用分数除法即可求出大米原有多少千克。\n【详解】 $24 \\div (1 - \\frac{1}{3})$ $= 2 4 \\div \\frac {2}{3}$ $= 2 4 \\times \\frac {3}{2}$ = 36 （千克) $(3 6 - 8 + 6) \\div (1 - \\frac {1}{3}) = 3 4 \\div \\frac {2}{3} = 3 4 \\times \\frac {3}{2}=51$答: 这袋大米原有 51 千克。",
+      "knowledge_points": [
+        "已知一部分量占总量的几分之几及另一部分量,求总量",
+        "分数的四则混合运算"
+      ],
+      "source": "23-24六年级上·全国·单元测试",
+      "merged_id": "unit3_185",
+      "original_id": "unit3_choice_185"
+    },
+    {
+      "id": "unit3_choice_186",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "修一条路, 第一天修了全长的 $\\frac{1}{4}$ , 第二天修了剩下的 $\\frac{1}{4}$ , 第一天比第二天多修 200 米。这条路长多少米?",
+      "options": [],
+      "answer": "3200 米",
+      "solution": "【分析】把一条路总长看作单位 \"1\", 第一天完成 $\\frac{1}{4}$ , 第二天完成 $\\left(1 - \\frac{1}{4}\\right) \\times \\frac{1}{4} = \\frac{3}{16}$ , 多出的 200 米对应 $\\left(\\frac{1}{4} - \\frac{3}{16}\\right)$ 。根据分数除法的意义, 用 200 除以 $\\frac{1}{4} - \\frac{3}{16}$ 的差可以求出单位 \"1\", 也就是路的总长度\n【详解】 $\\left(1 - \\frac{1}{4}\\right) \\times \\frac{1}{4}$ $ = \\frac {3}{4} \\times \\frac {1}{4} = \\frac {3}{1 6}$$2 0 0 \\div \\left(\\frac {1}{4} - \\frac {3}{1 6}\\right) = 2 0 0 \\div \\frac {1}{1 6} = 2 0 0 \\times 1 6= 3 2 0 0 (\\text {米})$答: 这条路长 3200 米。",
+      "knowledge_points": [
+        "已知一部分量占总量的几分之几及另一部分量,求总量",
+        "分数的四则混合运算"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_186",
+      "original_id": "unit3_choice_186"
+    },
+    {
+      "id": "unit3_choice_187",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "小明假期做作业,第一天完成了全部的 $\\frac{1}{3}$ , 第二天完成了剩余部分的 $\\frac{1}{3}$ , 第一天比第二天多做了 1 页, 小明还有多少页作业没做?",
+      "options": [],
+      "answer": "4页",
+      "solution": "【分析】把全部作业作为单位 \"1\", 第一天完成 $\\frac{1}{3}$ , 第二天完成 $\\frac{2}{3} \\times \\frac{1}{3} = \\frac{2}{9}$ , 多出的 1 页对应 $\\frac{1}{3} - \\frac{2}{9}$ , 用 $1$ 除以 $\\frac{1}{3} - \\frac{2}{9}$ 的差可以求出单位 \"1\", 再求有多少作业没做。\n【详解】第二天：$ \\left(1 - \\frac {1}{3}\\right) \\times \\frac {1}{3} = \\frac {2}{3} \\times \\frac {1}{3} = \\frac {2}{9} $总页数:$1 \\div \\left(\\frac {1}{3} - \\frac {2}{9}\\right) = 1 \\div \\frac {1}{9} \\\\ $$= 1 \\times 9 = 9 \\text {(页)} $剩下的页数:$9 \\times (1 - \\frac {1}{3} - \\frac {2}{9}) = 9 \\times \\left(\\frac {2}{3} - \\frac {2}{9}\\right) = 9 \\times \\frac {4}{9} = 4 \\text {(页)}$答: 小明还剩 4 页作业没做。",
+      "knowledge_points": [
+        "分数的四则混合运算"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_187",
+      "original_id": "unit3_choice_187"
+    },
+    {
+      "id": "unit3_choice_188",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "玲玲家十一月和十二月的电费共计 360 元, 十一月电费是十二月电费的 $\\frac{1}{3}$ , 玲玲家十一月和十二月的电费分别是多少?\n![](images/c3594d220ad83e7076798e8a3ef23b733f3cab37ade77376780868b43b19aaf6.jpg)",
+      "options": [],
+      "answer": "90元；270元",
+      "solution": "【分析】根据题意，设十二月电费为 $x$ 元，关系式为：十一月电费 + 十二月电费 = 360 元，列方程求解即可求出十二月的电费，然后再将十二月电费乘 $\\frac{1}{3}$ 即可。\n【详解】解：设十二月份电费为 $x$ 元，如图：\n$x + \\frac {1}{3} x = 3 6 0$ 解: $\\frac{4}{3} x = 360 \\quad {4}{3} x \\div \\frac {4}{3} = 3 6 0 \\div \\frac {4}{3} \\quad \\frac {4}{3} x \\times \\frac {3}{4} = 3 6 0 \\times \\frac {3}{4} x = 2 7 0 \\quad 2 7 0 \\times \\frac {1}{3} = 9 0 (\\text {元})$答: 玲玲家十一月电费是 90 元, 十二月的电费 270 元。",
+      "knowledge_points": [
+        "列方程解含两个未知数的问题",
+        "求一个数的几分之几的问题"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_188",
+      "original_id": "unit3_choice_188"
+    },
+    {
+      "id": "unit3_choice_189",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "修一条公路, 已经修了全部的 $\\frac{4}{5}$ , 没有修的还有 120 米, 这条路全长多少米?",
+      "options": [],
+      "answer": "600 米",
+      "solution": "【分析】把这条公路的全长看作单位 \"1\", 已经修了全部的 $\\frac{4}{5}$ , 那么没有修的 120 米占全长的 $(1 - \\frac{4}{5})$ , 单位 \"1\"未知,用没有修的长度除以 $(1 - \\frac{4}{5})$ , 即可求出这条路的全长。\n【详解】 $120 \\div \\left(1 - \\frac{4}{5}\\right)$ $= 1 2 0 \\div \\frac {1}{5} = 1 2 0 \\times 5 = 6 0 0 \\text {(米)}$答: 这条路全长 600 米。",
+      "knowledge_points": [
+        "分数的四则混合运算",
+        "已知一个数的几分之几是多少，求这个数"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_189",
+      "original_id": "unit3_choice_189"
+    },
+    {
+      "id": "unit3_choice_190",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "一个养兔场卖出兔子总只数的 $\\frac{5}{7}$ 后, 还剩 1200 只。这个养兔场原有兔子多少只?",
+      "options": [],
+      "answer": "4200只",
+      "solution": "【分析】养兔场原有的兔子数量看作单位 \"1\", 卖出部分占 $\\frac{5}{7}$ , 则剩余部分占 $1 - \\frac{5}{7} = \\frac{2}{7}$ , 也就是说 1200 只对应兔子总只数的 $\\frac{2}{7}$ , 要求单位 \"1\"就用 1200 除以 $\\frac{2}{7}$ 。\n【详解】 $1200 \\div (1 - \\frac{5}{7})$ $= 1 2 0 0 \\div \\frac {2}{7} = 1 2 0 0 \\times \\frac {7}{2} = 4 2 0 0 \\text {(只)}$答: 养兔场原有兔子 4200 只。",
+      "knowledge_points": [
+        "已知一个数的几分之几是多少，求这个数"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_190",
+      "original_id": "unit3_choice_190"
+    },
+    {
+      "id": "unit3_choice_191",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "为了丰富学生的课余生活，逸夫小学开展了学生课后社团活动。六年级原来参加围棋社的女生人数是男生人数的 $\\frac{2}{3}$ ，后来又增加了 10 名女生，这时男生人数是女生的 $\\frac{3}{4}$ 。原来参加围棋社的女生有多少名？",
+      "options": [],
+      "answer": "10名",
+      "solution": "【分析】男生人数没变，将男生人数看作单位\"1\"，后来又增加了10名女生，这时男生人数是女生的 $\\frac{3}{4}$ ，则女生人数是男生人数的 $\\frac{4}{3}$ ，增加的女生人数对应分率是 $(\\frac{4}{3} - \\frac{2}{3})$ ，增加的女生人数÷对应分率 = 男生人数，男生人数×原来女生对应分率 = 原来女生人数，据此列式解答。\n【详解】 $10 \\div \\left(\\frac{4}{3} - \\frac{2}{3}\\right)$ $= 1 0 \\div \\frac {2}{3} = 1 0 \\times \\frac {3}{2} = 1 5 \\text {(名)} \\quad 1 5 \\times \\frac {2}{3} = 1 0 (\\text {名})$答: 原来参加围棋社的女生有 10 名。",
+      "knowledge_points": [
+        "已知一部分量占总量的几分之几及另一部分量,求总量",
+        "求一个数的几分之几的问题"
+      ],
+      "source": "23-24六年级上·福建莆田·期末",
+      "merged_id": "unit3_191",
+      "original_id": "unit3_choice_191"
+    },
+    {
+      "id": "unit3_choice_192",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "一列火车的速度是 180 千米/时, 一辆小汽车的速度是这列火车的 $\\frac{5}{9}$ , 这辆小汽车的速度是一架喷气式飞机速度的 $\\frac{1}{9}$ 。喷气式飞机的速度是多少?",
+      "options": [],
+      "answer": "900千米/时",
+      "solution": "【分析】根据求一个数的几分之几是多少，用乘法计算，即用 $180$ 乘 $\\frac{5}{9}$ 即可求出这辆小汽车的速度；再根据已知一个数的几分之几，求这个数，用除法计算，即用这辆小汽车的速度除以 $\\frac{1}{9}$ 即可求出喷气式飞机的速度。\n【详解】 $180 \\times \\frac{5}{9} \\div \\frac{1}{9}$ $= 1 0 0 \\div \\frac {1}{9} = 1 0 0 \\times 9 = 9 0 0 \\text {(千 米 / 时)}$\n答: 喷气式飞机的速度是 900 千米/时。",
+      "knowledge_points": [
+        "分数的乘、除法的混合运算",
+        "求一个数的几分之几的问题"
+      ],
+      "source": "14-15 六年级下·黑龙江伊春·期末",
+      "merged_id": "unit3_192",
+      "original_id": "unit3_choice_192"
+    },
+    {
+      "id": "unit3_choice_193",
+      "unit": "第三单元：分数除法",
+      "chapter": "工程问题",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "'中欧班列'是往来于中国与欧洲及'一带一路'沿线各国的集装箱国际铁路联运班列。近日，'齐鲁号'中欧班列从哈萨克斯坦的阿拉木图出发，开往中国山东省济南市。列车到达济南市后，装载的货物由甲运输队单独搬运，需要8小时完成；由乙运输队单独搬运，需要10小时完成。现先由甲运输队搬运2小时，然后乙运输队加入，还需几个小时能搬运完？",
+      "options": [],
+      "answer": "$\\frac{10}{3}$ 小时",
+      "solution": "【分析】根据工作效率 $=$ 工作总量 $\\div$ 工作时间；把货物看作单位\"1\"，甲运输队单独搬运需要8小时，甲运输队的工作效率是 $1 \\div 8 = \\frac{1}{8}$ , 乙运输队单独搬运需要 10 小时, 乙运输队的工作效率是 $1 \\div 10 = \\frac{1}{10}$ ; 再根据工作总量 $=$ 工作效率 $\\times$ 工作时间，用 $\\frac{1}{8} \\times 2$ ，求出甲运输队 2 小时完成的这批货物数量；再用这批货物的总量一甲运输队 2 小时运输的货物的数量，求出还剩下货物的数量，再根据工作时间 $=$ 工作总量÷工作效率，用这批货物剩下的数量÷甲运输队与乙运输队的工作效率和，即可解答。【详解】 $(1 - \\frac{1}{8} \\times 2) \\div (\\frac{1}{8} + \\frac{1}{10})$ = $(1 - \\frac {1}{4}) \\div (\\frac {5}{4 0} + \\frac {4}{4 0}) = \\frac {3}{4} \\div \\frac {9}{4 0} = \\frac {3}{4} \\times \\frac {4 0}{9} = \\frac {1 0}{3} (\\text {小 时})$答: 还需要 $\\frac{10}{3}$ 小时能搬完。",
+      "knowledge_points": [
+        "工程问题",
+        "分数的四则混合运算"
+      ],
+      "source": "23-24 六年级上·河南三门峡·期中",
+      "merged_id": "unit3_193",
+      "original_id": "unit3_choice_193"
+    },
+    {
+      "id": "unit3_choice_194",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "乐山大佛是世界上最高的佛像，总高 71 米。大佛耳长是佛像总高的 $\\frac{7}{71}$，是头长的 $\\frac{10}{21}$ 。佛像的头长多少米？（用方程解答和直接列式解答）",
+      "options": [],
+      "answer": "14.7 米",
+      "solution": "【分析】大佛耳朵的长度是佛像总高的 $\\frac{7}{71}$ , 单位 \"1\" 是佛像的高度, 求一个数的几分之几是多少用乘法计算, 耳朵长度是头长的 $\\frac{10}{21}$ , 单位 \"1\" 是头长, 求单位 \"1\" 用除法计算。本题重点是能够找出等量关系, 其次是知道求一个数的几分之几是多少用乘法计算。\n【详解】方法一：解：设佛像的头长 $x$ 米。$\\frac {1 0}{2 1} x = 7 1 \\times \\frac {7}{7 1} \\quad \\frac {1 0}{2 1} x \\div \\frac {1 0}{2 1} = 7 \\div \\frac {1 0}{2 1} \\quad \\frac {1 0}{2 1} x \\times \\frac {2 1}{1 0} = 7 \\times \\frac {2 1}{1 0}\\quad x = 1 4. 7$ 方法二: $7 1 \\times \\frac {7}{7 1} \\div \\frac {1 0}{2 1} = 7 \\div \\frac {1 0}{2 1} = 7 \\times \\frac {2 1}{1 0}=$= 14.7 (米) 答: 佛像的头长 14.7 米。",
+      "knowledge_points": [
+        "列方程解含一个未知数的问题",
+        "分数除法的应用",
+        "求一个数的几分之几的问题"
+      ],
+      "source": "2024 六年级下·全国·专题练习",
+      "merged_id": "unit3_194",
+      "original_id": "unit3_choice_194"
+    },
+    {
+      "id": "unit3_choice_195",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "困难",
+      "question": "在武汉美食中，能将武汉人热情、爽快的性格特征与美食特性有机融合的当属热干面。某餐馆一天售出优质热干面和普通热干面共 60 碗，共收入 390 元。如果把售出的优质热干面和普通热干面的碗数交换一下，共收入 360 元。已知优质热干面的单价比普通热干面的单价高 $\\frac{1}{2}$ ，两种热干面的单价各是多少？",
+      "options": [],
+      "answer": "7.5元/碗；5元/碗",
+      "solution": "【分析】因为仅仅是把售出的优质热干面和普通热干面的碗数交换一下，所以用交换前后的收入和 $\\div 60 =$ 优质热干面和普通热干面的单价和。将普通热干面的单价看作单位 \"1\"，优质热干面的单价是普通热干面的 $(1 + \\frac{1}{2})$ ，优质热干面和普通热干面的单价和是普通热干面的 $\\left(1 + \\frac{1}{2} + 1\\right)$ ，优质热干面和普通热干面的单价和 $\\div$ 对应分率 $=$ 普通热干面的单价，优质热干面和普通热干面的单价和一普通热干面的单价 $=$ 优质热干面的单价，据此列式解答。\n【详解】 单价和： $(390 + 360) \\div 60=75 0 \\div 6 0=12.5 (元)$\n普通热干面单价: $12.5 \\div \\left(1 + \\frac{1}{2} + 1\\right)=12.5\\div \\frac{5}{2}=1 2. 5 \\times \\frac {2}{5}5 (元/碗)$优质热干面单价： $12.5 - 5 = 7.5$ （元/碗）答: 优质热干面的单价是 7.5 元/碗, 普通热干面的单价是 5 元/碗。\n【点睛】关键是先求出两种热干面的单价和，确定单位\"1\"，理解分数除法的意义。",
+      "knowledge_points": [
+        "已知比一个数多/少几分之几是多少, 求这个数",
+        "和差倍问题"
+      ],
+      "source": "2024 六年级下·全国·专题练习",
+      "merged_id": "unit3_195",
+      "original_id": "unit3_choice_195"
+    },
+    {
+      "id": "unit3_choice_196",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "学校要举行校运会,小思要买一套运动服价钱共 300 元, 其中裤子的价格是上衣的 $\\frac{2}{3}$ 。上衣和裤子的价格分别是多少元? (用方程解)",
+      "options": [],
+      "answer": "180元，120元",
+      "solution": "【分析】由题意可知, 设上衣的价格是 $x$ 元, 则裤子的价格是 $\\frac{2}{3} x$ 元, 再根据等量关系: 上衣的价格 + 裤子的价格 = 300 , 据此列方程解答即可。\n【详解】解：设上衣的价格是 $x$ 元，则裤子的价格是 $\\frac{2}{3} x$ 元。\n$x + \\frac {2}{3} x = 3 0 0 \\\\ \\left(1 + \\frac {2}{3}\\right) x = 3 0 0 \\\\ \\frac {5}{3} x = 3 0 0 \\\\ \\frac {5}{3} x \\div \\frac {5}{3} = 3 0 0 \\div \\frac {5}{3} \\\\ x = 3 0 0 \\times \\frac {3}{5} \\\\ x = 1 8 0 \\\\ 3 0 0 - 1 8 0 = 1 2 0 \\quad (\\text {元}) $\n答: 上衣和裤子的价格分别是 180 元、120 元。\n【点睛】本题考查用方程解决实际问题, 明确等量关系是解题的关键。",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "列方程解含两个未知数的问题",
+        "分数与分数的除法"
+      ],
+      "source": "14-15 六年级上·宁夏中卫·期中",
+      "merged_id": "unit3_196",
+      "original_id": "unit3_choice_196"
+    },
+    {
+      "id": "unit3_choice_197",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "李师傅加工一批零件，已经加工了全部零件的 $\\frac{1}{3}$ 还多 18 个，余下没有加工的零件比已加工的还多 48 个，则这批零件共有多少个？",
+      "options": [],
+      "answer": "252 个",
+      "solution": "【分析】本题可列方程解答，设共有零件 $x$ 个，已经加工了全部零件的 $\\frac{1}{3}$ 还多 18 个，则已加工 $\\left(\\frac{1}{3} x + 18\\right)$ 个零件，此时还剩下 $x - \\left(\\frac{1}{3} x + 18\\right)$ 个零件，余下没有加工的零件比已加工的还多 48 个，等量关系式：余下没有加工的零件数量 = 已加工的零件数量 + 48 个，据此列方程解答。本题主要考查列方程解决实际问题, 明确题目中存在的等量关系是解答本题的关键\n【详解】解：设这批零件共有 $x$ 个。\n$x - \\left(\\frac {1}{3} x + 1 8\\right) = \\frac {1}{3} x + 1 8 + 4 8$\n$x - \\frac {1}{3} x - 1 8 = \\frac {1}{3} x + 6 6$\n$\\frac {2}{3} x - 1 8 = \\frac {1}{3} x + 6 6$\n$\\frac {2}{3} x - \\frac {1}{3} x = 6 6 + 1 8$\n$\\frac {1}{3} x = 8 4$\n$x = 8 4 \\div \\frac {1}{3}$\n$x = 8 4 \\times 3$\n$x = 252$\n答: 这批零件共有 252 个。",
+      "knowledge_points": [
+        "列方程解含一个未知数的问题",
+        "分数与整数的除法"
+      ],
+      "source": "2023 六年级上·全国·专题练习",
+      "merged_id": "unit3_197",
+      "original_id": "unit3_choice_197"
+    },
+    {
+      "id": "unit3_choice_198",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "学校图书室有故事书和科技书共 3000 本, 其中科技书的本数是故事书的 $\\frac{2}{3}$ , 故事书和科技书各有多少本?",
+      "options": [],
+      "answer": "故事书 1800 本; 科技书 1200 本",
+      "solution": "【分析】根据\"科技书的本数是故事书的 $\\frac{2}{3}$ \"，设故事书有 $x$ 本，则科技书有 $\\frac{2}{3} x$ 本。等量关系：故事书的本数十科技书的本数 = 故事书和科技书的总本数，据此列出方程，并求解。\n【详解】解：设故事书有 $x$ 本，则科技书有 $\\frac{2}{3} x$ 本。\n$x + \\frac {2}{3} x = 3 0 0 0 \\\\ \\frac {5}{3} x = 3 0 0 0 \\\\ \\frac {5}{3} x \\div \\frac {5}{3} = 3 0 0 0 \\div \\frac {5}{3} \\\\ x = 3 0 0 0 \\times \\frac {3}{5} \\\\ x = 1 8 0 0 $\n科技书: $3000 - 1800 = 1200$ (本)答: 故事书有 1800 本, 科技书有 1200 本。\n【点睛】本题考查列方程解决问题, 从题目中找到等量关系, 按等量关系列出方程。",
+      "knowledge_points": [
+        "应用等式的性质 2 解方程",
+        "列方程解含两个未知数的问题"
+      ],
+      "source": "19-20 六年级下·四川广安·期末",
+      "merged_id": "unit3_198",
+      "original_id": "unit3_choice_198"
+    },
+    {
+      "id": "unit3_choice_199",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "一杯 250 毫升的鲜牛奶含有 $\\frac{3}{10}$ 克的钙质, 占一个成年人一天所需钙质的 $\\frac{3}{8}$ 。一个成年人一天需要多少钙质?",
+      "options": [],
+      "answer": "$\\frac{4}{5}$ 克",
+      "solution": "【分析】将成年人一天所需的钙质看成单位 \"1\", 则用 $\\frac{3}{10}$ 克对应单位 \"1\"的 $\\frac{3}{8}$ , 根据除法的意义用 $\\frac{3}{10} \\div \\frac{3}{8}$ 计算即可解答。本题的关键是找出单位 \"1\", 并找出单位 \"1\"的分率对应的数量, 用除法就可以求出单位 \"1\"的量。\n【详解】 $\\frac{3}{10} \\div \\frac{3}{8}= \\frac {3}{1 0} \\times \\frac {8}{3} = \\frac {4}{5} (\\text {克})$答: 一个成年人一天需要 $\\frac{4}{5}$ 克钙。",
+      "knowledge_points": [
+        "分数除法的应用"
+      ],
+      "source": "18-19 五年级下·辽宁·单元测试",
+      "merged_id": "unit3_199",
+      "original_id": "unit3_choice_199"
+    },
+    {
+      "id": "unit3_choice_200",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法应用",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "'奋斗者号'载人潜水器是我国研制的万米载人潜水器。为了向学生展示我国载人潜水器的发展历史, 实验小学开展了一场相关讲座,六年级有 320 人参加, 四年级参加的人数是六年级的 $\\frac{3}{8}$ , 是五年级的 $\\frac{15}{16}$ , 五年级有多少人参加了这场讲座?",
+      "options": [],
+      "answer": "128人",
+      "solution": "【分析】已知六年级有 320 人参加, 四年级参加的人数是六年级的 $\\frac{3}{8}$ , 把六年级参加的人数看作单位 \"1\", 根据分数乘法的意义, 用 $320 \\times \\frac{3}{8}$ 即可求出四年级参加人数; 又已知四年级参加人数是五年级的 $\\frac{15}{16}$ , 则把五年级参加人数看作单位 \"1\", 根据分数除法的意义, 用 $320 \\times \\frac{3}{8} \\div \\frac{15}{16}$ 即可求出五年级参加人数。本题考查了分数乘除法的混合应用，明确求一个数的几分之几是多少，用乘法计算以及已知一个数的几分之几是多少，求这个数用除法计算。\n【详解】 $320 \\times \\frac{3}{8} \\div \\frac{15}{16}= 1 2 0 \\div \\frac {1 5}{1 6} = 1 2 0 \\times \\frac {1 6}{1 5} = 1 2 8 (\\text {人}) $答: 五年级有 128 人参加了这场讲座。",
+      "knowledge_points": [
+        "分数的乘、除法的混合运算",
+        "分数除法的应用",
+        "求一个数的几分之几的问题"
+      ],
+      "source": "22-23六年级上·河南南阳·期末",
+      "merged_id": "unit3_200",
+      "original_id": "unit3_choice_200"
+    }, {
+      "id": "unit3_choice_201",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "甲数的 $\\frac{3}{5}$ 等于乙数的 $\\frac{3}{7}$（甲数与乙数均不为 0），这两个数中比较大的是（）。",
+      "options": [
+        "A. 甲数",
+        "B. 乙数",
+        "C. 一样大",
+        "D. 无法比较"
+      ],
+      "answer": "B",
+      "solution": "【分析】由题意可知，甲数 $\\times \\frac{3}{5} =$ 乙数 $\\times \\frac{3}{7}$，然后令甲数 $\\times \\frac{3}{5} =$ 乙数 $\\times \\frac{3}{7} = 1$，根据分数乘法各部分之间的关系，积$\\div$一个因数 $=$ 另一个因数，据此分别求出甲数和乙数，最后再进行比较即可。\n\n【详解】由分析可知：甲数 $\\times \\frac{3}{5} =$ 乙数 $\\times \\frac{3}{7}$令甲数 $\\times \\frac{3}{5} =$ 乙数 $\\times \\frac{3}{7} = 1$则甲数 $= 1 \\div \\frac{3}{5} = \\frac{5}{3}$，乙数 $= 1 \\div \\frac{3}{7} = \\frac{7}{3}$因为 $\\frac{5}{3}< \\frac{7}{3}$，所以甲数 $<$ 乙数，也就是乙数比较大。故答案为：B",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "求一个数的几分之几的问题",
+        "分数比较大小的方法"
+      ],
+      "source": "2014 三年级·全国·课后作业",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_201",
+      "original_id": "unit3_choice_201"
+    },
+    {
+      "id": "unit3_choice_202",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "一根绳子，剪去 $\\frac{1}{3}$ 后还剩 $\\frac{1}{2}$ 米，这根绳子原来有多长，下面列式正确的是（）。",
+      "options": [
+        "A. $\\frac{1}{2} + \\frac{1}{3}$",
+        "B. $\\frac{1}{2} \\div (1 - \\frac{1}{3})$",
+        "C. $\\frac{1}{2} + \\frac{1}{2} \\times \\frac{1}{3}$",
+        "D. $\\frac{1}{2} \\div \\frac{1}{3}$"
+      ],
+      "answer": "B",
+      "solution": "【分析】把这个绳子原来的长度看作单位'1'，已知剪去 $\\frac{1}{3}$，则剩余的占总长的 $(1 - \\frac{1}{3})$，又已知还剩 $\\frac{1}{2}$ 米，根据分数除法的意义，用 $\\frac{1}{2} \\div (1 - \\frac{1}{3})$ 即可求出绳子原来的长度。本题考查了分数除法的应用，明确已知一个数的几分之几是多少，求这个数用除法计算。\n\n【详解】$\\frac{1}{2} \\div (1 - \\frac{1}{3})$$=\\frac {1}{2} \\div \\frac {2}{3} = \\frac {1}{2} \\times \\frac {3}{2} = \\frac {3}{4} (\\text {米})$这根绳子原来有 $\\frac{3}{4}$ 米，列式正确的是 $\\frac{1}{2} \\div (1 - \\frac{1}{3})$。故答案为：B",
+      "knowledge_points": [
+        "分数除法的应用"
+      ],
+      "source": "19-20 六年级上·全国·单元测试",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_202",
+      "original_id": "unit3_choice_202"
+    },
+    {
+      "id": "unit3_choice_203",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "（）的倒数大于它本身。",
+      "options": [
+        "A. 整数",
+        "B. 假分数",
+        "C. 真分数"
+      ],
+      "answer": "C",
+      "solution": "【分析】倒数的定义为：两个数相乘得一，称这两个数互为倒数，根据选项举例即可解题。解决此题的关键是可以适当举例用排除法即可求解。\n\n【详解】A．假设整数为1，1的倒数为1，因为 $1 = 1$，即等于它本身，错误；B. 假设假分数为 $\\frac{3}{2}, \\frac{3}{2}$ 的倒数为 $\\frac{2}{3}$，因为 $\\frac{2}{3}<\\frac{3}{2}$，即小于它本身，错误； C. 假设真分数为 $\\frac{1}{2}, \\frac{1}{2}$ 的倒数为 2，因为 $2 > \\frac{1}{2}$，即大于它本身，正确。故答案为：C",
+      "knowledge_points": [
+        "倒数的认识及应用",
+        "真分数、假分数、带分数的认识"
+      ],
+      "source": "22-23 五年级下·全国·单元测试",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_203",
+      "original_id": "unit3_choice_203"
+    },
+    {
+      "id": "unit3_choice_204",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "若 a 是非零自然数，下列算式中的计算结果最大的是（）。",
+      "options": [
+        "A. $a \\times \\frac{5}{8}$",
+        "B. $a \\div \\frac{5}{8}$",
+        "C. $a \\div \\frac{3}{2}$",
+        "D. $\\frac{3}{2} \\div a$"
+      ],
+      "answer": "B",
+      "solution": "【分析】若 a 是非零自然数，假设 $a = 120$，代入到 4 个选项中，利用分数乘法和分数除法的计算法则，求出算式的结果，再比较大小即可得解。此题的解题关键是采用赋值法，利用分数乘法和分数除法的计算法则，再通过比较大小得解。\n\n【详解】假设 $a = 120$，代入到各个算式中，A. $a \\times \\frac{5}{8} = 120 \\times \\frac{5}{8} = 75$ B. $a \\div \\frac{5}{8} = 120 \\div \\frac{5}{8} = 120 \\times \\frac{8}{5} = 192$ C. $a \\div \\frac{3}{2} = 120 \\div \\frac{3}{2} = 120 \\times \\frac{2}{3} = 80$ D. $\\frac{3}{2} \\div a = \\frac{3}{2} \\div 120 = \\frac{3}{2} \\times \\frac{1}{120} = \\frac{1}{80}$$\\frac {1}{8 0} < 7 5 < 8 0 < 1 9 2$所以算式中的计算结果最大的是 $a \\div \\frac{5}{8}$。故答案为：B",
+      "knowledge_points": [
+        "分数与整数的乘法",
+        "分数与整数的除法"
+      ],
+      "source": "2016·全国·小升初真题",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_204",
+      "original_id": "unit3_choice_204"
+    },
+    {
+      "id": "unit3_choice_205",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "计算结果小于 $\\frac{9}{23}$ 的算式是（）。",
+      "options": [
+        "A. $0 \\div \\frac{9}{23}$",
+        "B. $\\frac{9}{23} \\times 1$",
+        "C. $\\frac{9}{23} \\div \\frac{2}{3}$",
+        "D. $\\frac{9}{23} \\times \\frac{9}{5}$"
+      ],
+      "answer": "A",
+      "solution": "【分析】0 除以任何数都得 0；一个大于零的数，乘 1 还得原数；乘大于 1 的数，积大于这个数；除以小于 1 的数，商大于这个数；据此解答。\n\n【详解】A. $0 \\div \\frac{9}{23} = 0, 0 < \\frac{9}{23}$，符合题意；B. $\\frac{9}{23} \\times 1 = \\frac{9}{23}, \\frac{9}{23} = \\frac{9}{23}$，不符合题意； C. $\\frac{2}{3}< 1$，所以 $\\frac{9}{23} \\div \\frac{2}{3}>\\frac{9}{23}$，不符合题意； D. $\\frac{9}{5} > 1$，所以 $\\frac{9}{23} \\times \\frac{9}{5} > \\frac{9}{23}$，不符合题意；故答案为：A",
+      "knowledge_points": [
+        "分数乘分数",
+        "分数与分数的除法"
+      ],
+      "source": "21-22 六年级上·广东肇庆·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_205",
+      "original_id": "unit3_choice_205"
+    },
+    {
+      "id": "unit3_choice_206",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "在下面的选项中，互为倒数的是（）。",
+      "options": [
+        "A. $\\frac{1}{5}$ 与 0.5",
+        "B. $\\frac{1}{7}$ 和 7",
+        "C. $1 \\frac{4}{5}$ 和 $1 \\frac{5}{4}$"
+      ],
+      "answer": "B",
+      "solution": "【分析】乘积是 1 的两个数互为倒数。根据倒数的意义，分别计算 $\\frac{1}{5} \\times 0.5$、$\\frac{1}{7} \\times 7$、$1 \\frac{4}{5} \\times 1 \\frac{5}{4}$，根据乘积是否等于 1 进行选择。明确倒数的意义是解决此题的关键。\n\n【详解】A．因为 $\\frac{1}{5} \\times 0.5 = \\frac{1}{5} \\times \\frac{1}{2} = \\frac{1}{10}$， $\\frac{1}{10} \\neq 1$，所以 $\\frac{1}{5}$ 和 0.5 不是互为倒数。B. 因为 $\\frac{1}{7} \\times 7 = 1$，所以 $\\frac{1}{7}$ 和 7 互为倒数。 C. 因为 $1 \\frac{4}{5} \\times 1 \\frac{5}{4} = \\frac{9}{5} \\times \\frac{9}{4} = \\frac{81}{20}, \\frac{81}{20} \\neq 1$，所以 $1 \\frac{4}{5}$ 和 $1 \\frac{5}{4}$ 不是互为倒数。故答案为：B",
+      "knowledge_points": [
+        "倒数的认识及应用"
+      ],
+      "source": "22-23 五年级上·山东济南·期中",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_206",
+      "original_id": "unit3_choice_206"
+    },
+    {
+      "id": "unit3_choice_207",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "豆豆把一个数除以 $\\frac{2}{3}$，错算成乘 $\\frac{2}{3}$，得出结果是 6，正确的计算结果应该是（）。",
+      "options": [
+        "A. 13.5",
+        "B. 4",
+        "C. 1.3",
+        "D. 6"
+      ],
+      "answer": "A",
+      "solution": "【分析】先根据'因数 $=$ 积 $\\div$ 另一个因数'求出被除数，再根据'商 $=$ 被除数 $\\div$ 除数'求出正确的结果，据此解答。掌握乘除法运算中各部分之间的关系是解答题目的关键。\n\n【详解】$6 \\div \\frac{2}{3} \\div \\frac{2}{3}$$=9 \\div \\frac {2}{3} = 1 3. 5$故答案为：A",
+      "knowledge_points": [
+        "分数的连除运算",
+        "乘、除法的意义和各部分间的关系"
+      ],
+      "source": "18-19 六年级上·全国·课后作业",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_207",
+      "original_id": "unit3_choice_207"
+    },
+    {
+      "id": "unit3_choice_208",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "水结成冰，体积约增加 $\\frac{1}{10}$；那么冰化成水，体积约减少（）。",
+      "options": [
+        "A. $\\frac{1}{9}$",
+        "B. $\\frac{1}{10}$",
+        "C. $\\frac{1}{11}$",
+        "D. $\\frac{1}{12}$"
+      ],
+      "answer": "C",
+      "solution": "【分析】根据题意，水结成冰，体积约增加 $\\frac{1}{10}$，是把水的体积看作单位'1'，冰的体积是水的 $(1 + \\frac{1}{10})$；求冰化成水，体积约减少几分之几，就是求水的体积比冰的体积减少几分之几，用水和冰的体积之差除以冰的体积即可。找准单位'1'，明确求一个数比另一个数多或少几分之几，用两者的差值除以另一个数。\n\n【详解】$\\frac{1}{10} \\div (1 + \\frac{1}{10})$$=\\frac {1}{1 0} \\div \\frac {1 1}{1 0}= \\frac {1}{1 0} \\times \\frac {1 0}{1 1} = \\frac {1}{1 1}$故答案为：C",
+      "knowledge_points": [
+        "分数除法的应用"
+      ],
+      "source": "19-20 六年级上·全国·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_208",
+      "original_id": "unit3_choice_208"
+    },
+    {
+      "id": "unit3_choice_209",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "打印一份稿件，4分钟完成了全部任务的 $\\frac{1}{3}$，照这样计算，余下的任务还要（）分钟。",
+      "options": [
+        "A. 4",
+        "B. 8",
+        "C. 12"
+      ],
+      "answer": "B",
+      "solution": "【分析】根据'工作效率 = 工作总量 $\\div$ 工作时间'，代入数值，把全部任务看作单位'1'，用1减去 $\\frac{1}{3}$ 求出余下任务占全部任务的几分之几，再根据'工作时间 $=$ 工作总量 $\\div$ 工作效率'，求出余下的任务需要的时间，据此解答。本题主要考查了工作总量、工作效率和工作时间三者之间的关系。\n\n【详解】$\\frac{1}{3} \\div 4 = \\frac{1}{12}\\quad1 - \\frac {1}{3} = \\frac {2}{3} \\quad \\frac {2}{3} \\div \\frac {1}{1 2} = 8 (\\text {分钟})$故答案为：B",
+      "knowledge_points": [
+        "工程问题",
+        "分数与分数的除法",
+        "单位'1'的认识与确定"
+      ],
+      "source": "17-18 五年级下·全国·课后作业",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_209",
+      "original_id": "unit3_choice_209"
+    },
+    {
+      "id": "unit3_choice_210",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "$\\frac{3}{5}$ 和它倒数的和是（）。",
+      "options": [
+        "A. 1",
+        "B. $\\frac{34}{15}$",
+        "C. $\\frac{9}{15}$"
+      ],
+      "answer": "B",
+      "solution": "【分析】将 $\\frac{3}{5}$ 分子分母调换位置，先写出它的倒数，再求和即可。本题考查了倒数，乘积是 1 的两个数互为倒数。\n\n【详解】$\\frac{3}{5} + \\frac{5}{3} = \\frac{34}{15}$所以，$\\frac{3}{5}$ 和它倒数的和是 $\\frac{34}{15}$。故答案为：B",
+      "knowledge_points": [
+        "倒数的认识及应用",
+        "异分母分数加、减法"
+      ],
+      "source": "22-23 六年级上·全国·单元测试",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_210",
+      "original_id": "unit3_choice_210"
+    },
+    {
+      "id": "unit3_choice_211",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "中国书法是中国文化的一部分，每一个方方正正的汉字都是艺术的载体。在汉字硬笔书写比赛中，王杨 $\\frac{5}{6}$ 分钟写了15个汉字。照这样计算，他12分钟能写多少个？下面四种解决问题的算式中，不正确的是（）。",
+      "options": [
+        "A. $15 \\div \\frac{5}{6} \\times 12$",
+        "B. $12 \\div (\\frac{5}{6} \\div 15)$",
+        "C. $12 \\div \\frac{5}{6}\\times15$",
+        "D. $15 \\times \\frac{5}{6} \\times 12$"
+      ],
+      "answer": "D",
+      "solution": "【分析】王杨 $\\frac{5}{6}$ 分钟写了 15 个汉字，求照这样计算，他 12 分钟能写多少个。可以先求出 15 分钟里面包含多少个 $\\frac{5}{6}$ 分钟，就是能写多少个 15 个汉字，用 15 分钟除以 $\\frac{5}{6}$ 分钟，再乘 12；或先求他写每个汉字需要多少分钟，再求 12 分钟里面包含多少个写每个汉字所需要的分钟数，用 12 分钟除以括号内的 $\\frac{5}{6}$ 除以 15 的商；或先用 15 除以 $\\frac{5}{6}$，求出每分钟能写汉字的个数，再用每分钟写的个数乘 12。此题是考查分数除法、乘法的意义。此题属于归一问题。\n\n【详解】A. $15 \\div \\frac{5}{6} \\times 12$$1 8 \\times 1 2 = 2 1 6 (\\text {个})$ B. $12 \\div (\\frac{5}{6} \\div 15)$$ 1 2 \\div \\frac {1}{1 8} = 2 1 6 \\text {(个)}$ C. $12 \\div \\frac{5}{6} \\times 15$$\\frac {7 2}{5} \\times 1 5= 2 1 6 \\text {(个)}$ D. $15 \\times \\frac{5}{6} \\times 12$ $\\frac {7 5}{6} \\times 1 2 = 1 5 0 (\\text {个})$ 故答案为：D",
+      "knowledge_points": [
+        "分数的乘、除法的混合运算",
+        "归一问题"
+      ],
+      "source": "2022·河南洛阳·小升初真题",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_211",
+      "original_id": "unit3_choice_211"
+    },
+    {
+      "id": "unit3_choice_212",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "某校六年级（1）班有 30 名女生，已知女生人数是全班人数的 $\\frac{3}{5}$，则该班有（）名学生。",
+      "options": [
+        "A. 16",
+        "B. 40",
+        "C. 50",
+        "D. 60"
+      ],
+      "answer": "C",
+      "solution": "【分析】已知女生人数是全班人数的 $\\frac{3}{5}$，把全班人数看作单位'1'，单位'1'未知，用女生人数除以 $\\frac{3}{5}$，求出全班人数。已知一个数的几分之几是多少，求这个数，用除法计算。\n\n【详解】$30 \\div \\frac{3}{5}$$ 3 0 \\times \\frac {5}{3} = 5 0 \\text {(名)}$故答案为：C",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "单位'1'的认识与确定"
+      ],
+      "source": "20-21 六年级上·贵州安顺·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_212",
+      "original_id": "unit3_choice_212"
+    },
+    {
+      "id": "unit3_choice_213",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "假分数的倒数（）1。",
+      "options": [
+        "A. 大于",
+        "B. 小于",
+        "C. 小于或等于"
+      ],
+      "answer": "C",
+      "solution": "【分析】根据倒数的意义，乘积是 1 的两个数互为倒数。因为假分数的分子等于或大于分母，把分子和分母调换位置后，则成了分子小于或等于分母，所以假分数的倒数小于或等于 1。据此解答。此题考查的目的是理解倒数、假分数的意义，掌握求一个数的倒数的方法。\n\n【详解】根据分析得，假分数的倒数小于或等于 1。故答案为：C",
+      "knowledge_points": [
+        "倒数的认识及应用",
+        "假分数与带分数或整数的互化"
+      ],
+      "source": "2014 六年级·全国·课后作业",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_213",
+      "original_id": "unit3_choice_213"
+    },
+    {
+      "id": "unit3_choice_214",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "已知 $a \\div \\frac{5}{6} = b \\div 1 = c \\div \\frac{1}{13}$，并且 a、b、c 都大于 0，那么 a、b、c 这三个数中最大的数是（）。",
+      "options": [
+        "A. a",
+        "B. b",
+        "C. c"
+      ],
+      "answer": "B",
+      "solution": "【分析】先将除法改写成乘法形式，再结合乘数和积的关系，判断出 a、b、c 之间的大小关系。本题考查了分数除法，除以一个数就等于乘它的倒数。\n\n【详解】因为 $a \\div \\frac{5}{6} = b \\div 1 = c \\div \\frac{1}{13}$，所以 $a \\times \\frac{6}{5} = b = c \\times 13$，因为 $\\frac{6}{5} > 1, 13 > 1$，所以 $a < b, c < b$，所以 $a、b、c$ 这三个数中最大的数是 $b$。故答案为：B。",
+      "knowledge_points": [
+        "因数和积的大小关系（小数乘法）",
+        "分数与整数的除法",
+        "分数与分数的除法"
+      ],
+      "source": "19-20 六年级上·全国·期中",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_214",
+      "original_id": "unit3_choice_214"
+    },
+    {
+      "id": "unit3_choice_215",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "下面除法算式中，得数小于被除数的是（）",
+      "options": [
+        "A. $\\frac{9}{10} \\div \\frac{1}{12}$",
+        "B. $\\frac{3}{5} \\div 8$",
+        "C. $\\frac{7}{2} \\div 1$"
+      ],
+      "answer": "B",
+      "solution": "【分析】一个数（0 除外）除以一个大于 1 的数，结果比原来的数小；一个数（0 除外）除以一个小于 1 的数，结果比原来的数大；一个数除以 1，还得原数。本题考查分数除法，明确分数除法的计算方法是解题的关键。\n\n【详解】A. 因为 $\\frac{1}{12}< 1$，所以 $\\frac{9}{10} \\div \\frac{1}{12}>\\frac{9}{10}$。B. 因为 $8 > 1$，所以 $\\frac{3}{5} \\div 8 < \\frac{3}{5}$。C. $\\frac{7}{2} \\div 1 = \\frac{7}{2}$，得数等于被除数。故答案为：B",
+      "knowledge_points": [
+        "被除数和商的大小关系（小数除法）",
+        "分数与整数的除法",
+        "分数与分数的除法"
+      ],
+      "source": "20-21 六年级上·湖北黄冈·期中",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_215",
+      "original_id": "unit3_choice_215"
+    },
+    {
+      "id": "unit3_choice_216",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "某冰箱厂去年全年生产冰箱 220 万台，其中上半年产量是下半年的 $\\frac{5}{6}$。这个冰箱厂去年下半年的产量是（）万台。",
+      "options": [
+        "A. 120",
+        "B. 100",
+        "C. $\\frac{550}{3}$"
+      ],
+      "answer": "A",
+      "solution": "【分析】上半年产量是下半年的 $\\frac{5}{6}$，把下半年的产量看作单位'1'，则全年产量是下半年的 $(1 + \\frac{5}{6})$，根据分数除法的意义，下半年产量是 $220 \\div (1 + \\frac{5}{6})$ 万台，由此进行列式解答即可。根据分数加法的意义求出全年产量占下半年产量的分率是完成本题的关键。\n\n【详解】$220 \\div (1 + \\frac{5}{6})= 2 2 0 \\div \\frac {11}{6}= 120$ （万台）故答案为：A",
+      "knowledge_points": [
+        "分数除法的应用"
+      ],
+      "source": "19-20六年级上·吉林白山·期中",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_216",
+      "original_id": "unit3_choice_216"
+    },
+    {
+      "id": "unit3_choice_217",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "狮子奔跑时的最高时速可以达到 60 千米/时，比猎豹慢 $\\frac{5}{11}$。猎豹奔跑时的最高时速是多少？列式正确的是（）。",
+      "options": [
+        "A. $60 \\times \\frac{5}{11}$",
+        "B. $60 \\times (1 - \\frac{5}{11})$",
+        "C. $60 \\div (1 - \\frac{5}{11})$"
+      ],
+      "answer": "C",
+      "solution": "【分析】把猎豹的速度看成单位'1'，它的 $(1 - \\frac{5}{11})$ 就是狮子的速度 60 千米/时，由此用除法求出猎豹的速度。本题先找出单位'1'，已知一个数的几分之几是多少，求这个数用除法。\n\n【详解】$60 \\div (1 - \\frac{5}{11})= 6 0 \\div \\frac {6}{1 1}= 110$ （千米/时）故答案为：C",
+      "knowledge_points": [
+        "分数与整数的除法"
+      ],
+      "source": "19-20 六年级上·吉林白山·期中",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_217",
+      "original_id": "unit3_choice_217"
+    },
+    {
+      "id": "unit3_choice_218",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "判断下面算式的结果，最大的是（）。",
+      "options": [
+        "A. $233 \\times \\frac{2}{5}$",
+        "B. $233 + \\frac{2}{5}$",
+        "C. $233 \\div \\frac{2}{5}$",
+        "D. $233 - \\frac{2}{5}$"
+      ],
+      "answer": "C",
+      "solution": "【分析】一个数（0 除外），乘小于 1 的数，积比原数小；加上一个非零数，和比原数大；除以小于 1 的数，商比原数大；减去一个非零数，差比原数小。先将结果小于原数的排除，比较加法和除法的结果即可。关键是掌握分数四则运算的计算方法，除以一个数等于乘这个数的倒数。\n\n【详解】A. $233 \\times \\frac{2}{5} < 233$ B. $233 + \\frac{2}{5} > 233$ C. $233 \\div \\frac{2}{5} > 233$ D. $233 - \\frac{2}{5} < 233$ $233 + \\frac{2}{5} = 233\\frac{2}{5}$ $233 \\div \\frac{2}{5} = 233 \\times \\frac{5}{2} = 582.5$ $233 \\frac{2}{5}<582.5$故答案为：C",
+      "knowledge_points": [
+        "分数与整数的乘法",
+        "分数与整数的除法",
+        "异分母分数加、减法"
+      ],
+      "source": "20-21 六年级下·江西上饶·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_218",
+      "original_id": "unit3_choice_218"
+    },
+    {
+      "id": "unit3_choice_219",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "苹果树有 100 棵，_______，梨树有多少棵？横线上要补充下面（），列式才是 $100 \\div (1 - \\frac{1}{5})$。",
+      "options": [
+        "A. 苹果树是梨树的 $\\frac{1}{5}$",
+        "B. 梨树比苹果树多 $\\frac{1}{5}$",
+        "C. 苹果树比梨树少 $\\frac{1}{5}$",
+        "D. 梨树比苹果树少 $\\frac{1}{5}$"
+      ],
+      "answer": "C",
+      "solution": "【分析】根据列式可知，是求单位'1'的量，即梨树是单位'1'，并且是 $1 - \\frac{1}{5}$。所以，把梨树棵数看作单位'1'，苹果树的棵数比梨树少 $\\frac{1}{5}$，求梨树有多少棵就用除法计算。据此解答。此题考查的是分数应用题的列式，要先找准单位'1'，再据题中的数量关系解题。\n\n【详解】横线上补充：苹果树比梨树少 $\\frac{1}{5}$，列式就是：$100 \\div (1 - \\frac{1}{5})$。故答案为：C",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "已知比一个数多/少几分之几是多少，求这个数",
+        "\"提问题\"\"填条件\"问题"
+      ],
+      "source": "20-21 六年级上·福建厦门·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_219",
+      "original_id": "unit3_choice_219"
+    },
+    {
+      "id": "unit3_choice_220",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "下面算式符合如图所示图形图意的是（）。\n\n![](images/dd96720fc7d1d14b96a56053ccccb304dc33e619b7ff7e69986a1698ec1132e6.jpg)",
+      "options": [
+        "A. $\\frac{2}{3} \\times \\frac{2}{5}$",
+        "B. $\\frac{2}{5} \\div \\frac{2}{3}$",
+        "C. $\\frac{2}{3} \\times \\frac{3}{5}$",
+        "D. $\\frac{1}{3} \\times \\frac{1}{5}$"
+      ],
+      "answer": "A",
+      "solution": "【分析】把整个长方形看作单位'1'，第一次把长方形平均分成三行，将上 2 行涂色，涂色部分为 $\\frac{2}{3}$，第二次将长方形平均分 5 列，将第一次涂色部分的左边 2 列涂色，第二次涂色部分为第一次涂色部分的 $\\frac{2}{5}$。此题重点考查分数乘分数的意义，利用图形涂色的方法，更直观地看出分数乘分数的积是多少。\n\n【详解】双重阴影部分为 $\\frac{2}{3}$ 的 $\\frac{2}{5}$，列式为：$\\frac{2}{3} \\times \\frac{2}{5}$。故答案为：A",
+      "knowledge_points": [
+        "分数乘分数",
+        "分数与分数的除法"
+      ],
+      "source": "20-21 六年级上·福建厦门·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_220",
+      "original_id": "unit3_choice_220"
+    },
+    {
+      "id": "unit3_choice_221",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "花店有 40 盆兰花，兰花的盆数比菊花多 $\\frac{3}{5}$，则（）。",
+      "options": [
+        "A. 兰花的盆数比菊花多 15 盆",
+        "B. 兰花的盆数比菊花多 24 盆",
+        "C. 兰花的盆数比菊花少 $\\frac{3}{5}$",
+        "D. 菊花有 24 盆"
+      ],
+      "answer": "A",
+      "solution": "【分析】根据题意，兰花的盆数比菊花多 $\\frac{3}{5}$，把菊花的盆数看作单位'1'，兰花的盆数是菊花的 $(1 + \\frac{3}{5})$，单位'1'未知，用兰花的盆数除以 $(1 + \\frac{3}{5})$，即可求出菊花的盆数；再逐项分析计算，选择正确的说法。掌握分数除法的意义及应用，明确已知比一个数多或少几分之几的数是多少，求这个数，用除法计算；求一个数比另一个数多或少几分之几，用多或少的量除以另一个数。\n\n【详解】菊花的盆数：$4 0 \\div (1 + \\frac {3}{5})$$ 4 0 \\div \\frac {8}{5} = 4 0 \\times \\frac {5}{8} = 2 5 \\text {(盆)} $A. 兰花的盆数比菊花多： $40 - 25 = 15$ (盆)，原题说法正确； B. 兰花的盆数比菊花多： $40 - 25 = 15$ (盆)，原题说法错误； C. 兰花的盆数比菊花多：$ (4 0 - 2 5) \\div 25 = 1 5 \\div 2 5 = \\frac {3}{5} $原题说法错误； D. 菊花盆数有 25 盆，原题说法错误。故答案为：A",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "已知比一个数多/少几分之几是多少，求这个数"
+      ],
+      "source": "20-21 六年级上·河北邯郸·期中",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_221",
+      "original_id": "unit3_choice_221"
+    },
+    {
+      "id": "unit3_choice_222",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "图书室有名著 240 本，名著是科技书的 $\\frac{5}{6}$，科技书有几本？请你根据条件选择合适的算式（）。",
+      "options": [
+        "A. $240 \\times \\frac{5}{6}$",
+        "B. $240 \\div \\frac{5}{6}$",
+        "C. $240 \\div (1 + \\frac{1}{5})$",
+        "D. $240 \\times (1 + \\frac{1}{5})$"
+      ],
+      "answer": "B",
+      "solution": "【分析】把科技书的本数看作单位'1'，已知科技书的 $\\frac{5}{6}$ 是 240 本，求单位'1'，用 $240 \\div \\frac{5}{6}$，即可解答。根据已知一个数的几分之几是多少，求这个数的知识进行解答。\n\n【详解】$240 \\div \\frac{5}{6} =2 4 0 \\times \\frac {6}{5}= 2 8 8 \\text {(本)}$答：科技书有 288 本。故答案为：B",
+      "knowledge_points": [
+        "分数与整数的除法"
+      ],
+      "source": "20-21 六年级上·浙江宁波·期中",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_222",
+      "original_id": "unit3_choice_222"
+    },
+    {
+      "id": "unit3_choice_223",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "$\\frac{5}{13} \\div 3 + \\frac{1}{13} \\times \\frac{1}{3} =$ （）。",
+      "options": [
+        "A. 16",
+        "B. $\\frac{2}{13}$",
+        "C. 7",
+        "D. 55"
+      ],
+      "answer": "B",
+      "solution": "【分析】先把除以3化为乘 $\\frac{1}{3}$，再利用乘法分配律简便计算。\n\n【详解】$\\frac{5}{13} \\div 3 + \\frac{1}{13} \\times \\frac{1}{3}=\\frac {5}{1 3} \\times \\frac {1}{3} + \\frac {1}{1 3} \\times \\frac {1}{3} = (\\frac {5}{1 3} + \\frac {1}{1 3}) \\times \\frac {1}{3} = \\frac {6}{1 3} \\times \\frac {1}{3} = \\frac {2}{1 3}$故答案为：B",
+      "knowledge_points": [
+        "整数乘法运算定律推广到分数乘法",
+        "分数除法相关的简便计算"
+      ],
+      "source": "2022 六年级下·全国·专题练习",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_223",
+      "original_id": "unit3_choice_223"
+    },
+    {
+      "id": "unit3_choice_224",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "小强 $\\frac{1}{5}$ 小时走 $\\frac{5}{12}$ 千米，他走 1 千米要多少小时？正确列式是（）。",
+      "options": [
+        "A. $\\frac{5}{12} \\div \\frac{1}{5}$",
+        "B. $\\frac{5}{12} \\times \\frac{1}{5}$",
+        "C. $\\frac{1}{5} \\div \\frac{5}{12}$",
+        "D. $\\frac{5}{12} \\times \\frac{12}{5}$"
+      ],
+      "answer": "C",
+      "solution": "【分析】走 1 千米需要的时间 = 总共用去的时间 ÷ 总共走的千米数，代入数值即可得解答。此题属于分数除法应用题，列式容易出错，一定要注意题干所要求的问题。\n\n【详解】走 1 千米需要的时间 $= \\frac{1}{5} \\div \\frac{5}{12}$故答案为：C",
+      "knowledge_points": [
+        "分数与分数的除法"
+      ],
+      "source": "2020·陕西宝鸡·小升初真题",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_224",
+      "original_id": "unit3_choice_224"
+    },
+    {
+      "id": "unit3_choice_225",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "下面计算结果在 $\\frac{1}{4}$ 和 $\\frac{8}{9}$ 之间的是（）。",
+      "options": [
+        "A. $\\frac{1}{4} \\times \\frac{9}{10}$",
+        "B. $\\frac{1}{4} \\div \\frac{9}{10}$",
+        "C. $\\frac{8}{9} \\times \\frac{3}{2}$",
+        "D. $\\frac{1}{4} \\times 4$"
+      ],
+      "answer": "B",
+      "solution": "【分析】首先根据分数乘除法的运算方法，求出每个算式的结果是多少；然后把所得的结果和 $\\frac{1}{4}$ 和 $\\frac{8}{9}$ 比较大小，判断出哪题计算的结果在 $\\frac{1}{4}$ 和 $\\frac{8}{9}$ 之间即可。此题主要考查了分数乘除法的运算方法，以及分数大小比较的方法，要熟练掌握，解答此题的关键是求出每个算式的结果是多少。\n\n【详解】A. $\\frac{1}{4} \\times \\frac{9}{10} = \\frac{9}{40}, \\frac{1}{4} = \\frac{10}{40}, \\frac{9}{40} < \\frac{10}{40}$，所以 $\\frac{1}{4} \\times \\frac{9}{10}$ 的积不在 $\\frac{1}{4}$ 和 $\\frac{8}{9}$ 之间；B. $\\frac{1}{4} \\div \\frac{9}{10} = \\frac{10}{36}, \\frac{8}{9} > \\frac{10}{36} > \\frac{1}{4}$，所以 $\\frac{1}{4} \\div \\frac{9}{10}$ 的积在 $\\frac{1}{4}$ 和 $\\frac{8}{9}$ 之间；C. $\\frac{8}{9} \\times \\frac{3}{2} = \\frac{24}{18}, \\frac{24}{18} > 1,$所以 $\\frac{8}{9} \\times \\frac{3}{2}$ 的积不在 $\\frac{1}{4}$ 和 $\\frac{8}{9}$ 之间；D. $\\frac{1}{4} \\times 4 = 1, 1 > \\frac{8}{9}$，所以 $\\frac{1}{4} \\times 4$ 的积不在 $\\frac{1}{4}$ 和 $\\frac{8}{9}$ 之间；故答案为：B",
+      "knowledge_points": [
+        "异分母异分子分数的大小比较",
+        "分数乘分数",
+        "分数与分数的除法"
+      ],
+      "source": "20-21 五年级下·浙江金华·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_225",
+      "original_id": "unit3_choice_225"
+    },
+    {
+      "id": "unit3_choice_226",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "小熊遮住了甲、乙的一部分（如图），原来的甲、乙相比，（）。\n\n![](images/69d53749ea0c3d9dce711c2f6d4b95944cba645c37063f526886ba208f6eee6b.jpg)",
+      "options": [
+        "A. 乙比甲长",
+        "B. 甲比乙长",
+        "C. 甲和乙一样长",
+        "D. 无法比较"
+      ],
+      "answer": "A",
+      "solution": "【分析】观察线段图可知，甲 $\\times \\frac{1}{2} =$ 乙 $\\times \\frac{1}{3}$，假设甲 $\\times \\frac{1}{2} =$ 乙 $\\times \\frac{1}{3}$ $= 1$，分别求出甲和乙，然后进行对比即可。本题考查分数除法，明确甲和乙的关系是解题的关键。\n\n【详解】假设甲 $\\times \\frac{1}{2} =$ 乙 $\\times \\frac{1}{3} = 1$则甲 $= 1 \\div \\frac{1}{2} = 2$$乙 = 1 \\div \\frac {1}{3} = 3$所以甲<乙。故答案为：A",
+      "knowledge_points": [
+        "分数与整数的除法"
+      ],
+      "source": "2017·四川雅安·小升初真题",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_226",
+      "original_id": "unit3_choice_226"
+    },
+    {
+      "id": "unit3_choice_227",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "下面能用简便方法计算的是（）。",
+      "options": [
+        "A. $\\frac{3}{10} \\times 98$",
+        "B. $\\frac{3}{10} \\div 98$",
+        "C. $98 \\div \\frac{3}{10}$"
+      ],
+      "answer": "A",
+      "solution": "【分析】计算 $\\frac{3}{10} \\times 98$ 时，把 98 化为 $(100 - 2)$，再利用乘法分配律简便计算；计算分数除法时，先把分数除法化为分数乘法，再约分计算。掌握分数除法的计算方法以及乘法分配律的应用是解答题目的关键。\n\n【详解】A. $\\frac{3}{10} \\times 98 \\frac {3}{1 0} \\times (1 0 0 - 2) = \\frac {3}{1 0} \\times 1 0 0 - \\frac {3}{1 0} \\times 2 = 3 0 - \\frac {3}{5}= 2 9 \\frac {2}{5}$ B. $\\frac{3}{10} \\div 98= \\frac {3}{1 0} \\times \\frac {1}{9 8} = \\frac {3}{9 8 0}$ C. $98 \\div \\frac{3}{10}=9 8 \\times \\frac {1 0}{3} = \\frac {9 8 0}{3}$故答案为：A",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "分数乘法运算律"
+      ],
+      "source": "20-21 五年级下·广西来宾·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_227",
+      "original_id": "unit3_choice_227"
+    },
+    {
+      "id": "unit3_choice_228",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "因为 $\\frac{3}{5} \\times \\frac{5}{3} = 1$，所以（）。",
+      "options": [
+        "A. $\\frac{3}{5}$ 是倒数",
+        "B. $\\frac{5}{3}$ 是倒数",
+        "C. $\\frac{3}{5}$ 和 $\\frac{5}{3}$ 都是倒数",
+        "D. $\\frac{3}{5}$ 和 $\\frac{5}{3}$ 互为倒数"
+      ],
+      "answer": "D",
+      "solution": "【分析】由倒数的定义可知，两个数乘积是 1 的数互为倒数。据此判断。此题的解题关键是掌握倒数的概念并作出正确的判断。\n\n【详解】因为 $\\frac{3}{5} \\times \\frac{5}{3} = 1$，所以 $\\frac{3}{5}$ 和 $\\frac{5}{3}$ 互为倒数。故答案为：D",
+      "knowledge_points": [
+        "倒数的认识及应用"
+      ],
+      "source": "21-22 六年级上·辽宁盘锦·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_228",
+      "original_id": "unit3_choice_228"
+    },
+    {
+      "id": "unit3_choice_229",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "希望文具店 12 月的营业额为 30 万元，比 11 月多 $\\frac{1}{5}$。设 11 月营业额为 $x$ 万元，则下列方程正确的是（）。",
+      "options": [
+        "A. $x + \\frac{1}{5} = 30$",
+        "B. $x - \\frac{1}{5} = 30$",
+        "C. $x - \\frac{1}{5} x = 30$",
+        "D. $x + \\frac{1}{5} x = 30$"
+      ],
+      "answer": "D",
+      "solution": "【分析】把 11 月营业额看作单位'1'设为 $x$ 万元，则 12 月份就是 $x + \\frac{1}{5} x$，根据题意得数量关系式 11 月营业额+ 比 11 月份多的部分 = 12 月份营业额，据此选择。此题考查的是列方程解决问题，理解题意明确等量关系是解答此题关键。\n\n【详解】由分析得，解：设11月营业额为 $x$ 万元，根据题意得$x + \\frac {1}{5} x = 3 0$ $\\frac {6}{5} x = 3 0$ $x = 2 5$故选：D",
+      "knowledge_points": [
+        "应用等式的性质 2 解方程",
+        "分数与整数的除法",
+        "列方程解含一个未知数的问题"
+      ],
+      "source": "21-22 六年级上·山西阳泉·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_229",
+      "original_id": "unit3_choice_229"
+    },
+    {
+      "id": "unit3_choice_230",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "要修一条 14 千米的路。如果甲队单独修，需要 8 天；如果乙队单独修，需要 6 天。现在两队合作，需要（）天完成。",
+      "options": [
+        "A. $\\frac{24}{7}$",
+        "B. 48",
+        "C. 1"
+      ],
+      "answer": "A",
+      "solution": "【分析】把这项工程的量看作单位'1'，先依据工作总量 $=$ 工作时间 $\\times$ 工作效率，求出甲队和乙队的工作效率，两队合作后，把两队工作效率相加，最后根据工作时间 $=$ 工作总量 $\\div$ 工作效率和即可解答。此题的解题关键是依据工作时间、工作效率以及工作总量之间数量关系来解决实际的工程问题。\n\n【详解】$1 \\div (\\frac{1}{8} + \\frac{1}{6})=1 \\div (\\frac {3}{2 4} + \\frac {4}{2 4})= \\frac {24}{7} (\\text {天})$故答案为：A",
+      "knowledge_points": [
+        "工程问题",
+        "分数与整数的除法"
+      ],
+      "source": "21-22 六年级上·河南郑州·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_230",
+      "original_id": "unit3_choice_230"
+    },
+    {
+      "id": "unit3_choice_231",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "若 $a \\times \\frac{3}{4} = b \\div \\frac{2}{5} = c \\times 2$（a、b、c均大于0），那么a、b、c按从小到大的顺序排列是（）。",
+      "options": [
+        "A. b $<$ a $<$ c",
+        "B. b<c<a",
+        "C. c $< a< b$"
+      ],
+      "answer": "B",
+      "solution": "【分析】假设等式的值为 1，求出 a、b、c 的值，最后比较三个数的大小，据此解答。掌握用求倒数比较大小的方法是解答题目的关键。\n\n【详解】假设 $a \\times \\frac{3}{4} = b \\div \\frac{2}{5} = c \\times 2 = 1$，则 $a = \\frac{4}{3}, b = \\frac{2}{5}, c = \\frac{1}{2}$ 因为 $\\frac{2}{5} < \\frac{1}{2} < \\frac{4}{3}$，所以 $b < c < a$。故答案为：B",
+      "knowledge_points": [
+        "异分母异分子分数的大小比较",
+        "倒数的认识及应用"
+      ],
+      "source": "21-22 六年级上·河南郑州·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_231",
+      "original_id": "unit3_choice_231"
+    },
+    {
+      "id": "unit3_fill_232",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "用 $\\frac{9}{10}$ 千克小麦可以磨出 $\\frac{3}{4}$ 千克面粉，每千克小麦可以磨面粉（）千克，要磨 1 千克面粉需要小麦（）千克。",
+      "options": [],
+      "answer": "$\\frac{5}{6}$；$1\\frac{1}{5}$",
+      "solution": "【分析】求 1 千克小麦可以磨面粉质量，用面粉除以小麦质量即可；求 1 千克面粉需要小麦质量，用小麦质量除以面粉质量即可。关键是理解数量关系，掌握分数除法的计算。\n\n【详解】1 千克小麦可以磨面粉质量：$\\frac{3}{4} \\div \\frac{9}{10}$$\\frac {3}{4} \\times \\frac {1 0}{9} = \\frac {5}{6} (\\text {千克})$ 1 千克面粉需要小麦质量：$\\frac{9}{10} \\div \\frac{3}{4}$$ \\frac {9}{1 0} \\times \\frac {4}{3} = \\frac {6}{5} = 1 \\frac {1}{5} (\\text {千克})$",
+      "knowledge_points": [
+        "分数除法的应用"
+      ],
+      "source": "18-19六年级上·全国·单元测试",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_232",
+      "original_id": "unit3_fill_232"
+    },
+    {
+      "id": "unit3_fill_233",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "把 $\\frac{5}{6}$ 米长的绳子平均分成 5 段，每段长是这根绳子的（），每段长（）米。",
+      "options": [],
+      "answer": "$\\frac{1}{5}$；$\\frac{1}{6}$",
+      "solution": "【分析】求每段长是这根绳子的几分之几，平均分的是单位'1'，表示把单位'1'平均分成 5 份，求的是每一份占的分率，用除法计算。把 $\\frac{5}{6}$ 米长的绳子平均分成 5 段，可用除法算出每段的长度。解决此题关键是弄清求得是分率还是具体的数量，求分率平均分的是单位'1'，求具体的数量平均分的是具体的数量，要注意分率不能带单位名称，而具体的数量要带单位名称。\n\n【详解】$1 \\div 5 = \\frac{1}{5}$$\\frac {5}{6} \\div 5 = \\frac {5}{6} \\times \\frac {1}{5} = \\frac {1}{6} (\\text {米})$即每段长是这根绳子的 $\\frac{1}{5}$，每段长 $\\frac{1}{6}$ 米。",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "分数的意义",
+        "单位'1'的认识与确定",
+        "分数与除法的关系"
+      ],
+      "source": "19-20 六年级上·全国·单元测试",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_233",
+      "original_id": "unit3_fill_233"
+    }, {
+      "id": "unit3_fill_234",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "把一根 $\\frac{4}{5}$ 米长的绳子，平均截成 5 段，每段占全长的（），每段长（）米。",
+      "options": [],
+      "answer": "$\\frac{1}{5}$；$\\frac{4}{25}$",
+      "solution": "【分析】求每段长占全长的几分之几，平均分的是单位'1'，表示把单位'1'平均分成5份，求的是每一份占的分率，用除法计算。把一根 $\\frac{4}{5}$ 米长的绳子，平均截成 5 段，可用除法算出每段的长度。解决此题关键是弄清求得是分率还是具体的数量，求分率平均分的是单位'1'，求具体的数量平均分的是具体的数量，要注意分率不能带单位名称，而具体的数量要带单位名称。\n\n【详解】$1 \\div 5 = \\frac{1}{5}$所以每段占全长的 $\\frac{1}{5}$；$\\frac {4}{5} \\div 5 = \\frac {4}{5} \\times \\frac {1}{5} = \\frac {4}{2 5} (\\text {米})$所以每段长 $\\frac{4}{25}$ 米。",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "分数的意义",
+        "分数与除法的关系"
+      ],
+      "source": "2022·湖南郴州·小升初真题",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_234",
+      "original_id": "unit3_fill_234"
+    },
+    {
+      "id": "unit3_fill_235",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "在括号里填上'>' '<'或'='。\n\n$\\frac {3}{7} \\times \\frac {2}{3} (\\quad) \\frac {3}{7}\\quad \\frac {5}{9} \\div \\frac {3}{4} (\\quad) \\frac {5}{9} \\times \\frac {3}{4}\\quad  \\frac {1}{3} \\times 3 (\\quad) \\frac {2}{5} \\times \\frac {5}{2}$",
+      "options": [],
+      "answer": "<；>；=",
+      "solution": "【分析】（1）一个数（0除外）乘小于1的数，积比原来的数小；(2) 一个数 (0 除外) 除以小于 1 的数，商比原来的数大；一个数 (0 除外) 乘小于 1 的数，积比原来的数小； (3) 根据分数乘法的计算法则，计算出两个乘法算式的结果，再比较。本题考查判断积与因数之间大小关系、商与被除数之间大小关系的方法、分数乘法的计算。\n\n【详解】(1) 因为 $\\frac{2}{3}<1$，所以 $\\frac{3}{7} \\times \\frac{2}{3}<\\frac{3}{7}$；(2) $\\frac{3}{4}<1$，则 $\\frac{5}{9} \\div \\frac{3}{4}> \\frac{5}{9}, \\frac{5}{9} \\times \\frac{3}{4}< \\frac{5}{9}$；所以 $\\frac{5}{9} \\div \\frac{3}{4} > \\frac{5}{9} \\times \\frac{3}{4}$；(3) $\\frac{1}{3} \\times 3 = 1$$\\frac {2}{5} \\times \\frac {5}{2} = 1$所以 $\\frac{1}{3} \\times 3 = \\frac{2}{5} \\times \\frac{5}{2}$。",
+      "knowledge_points": [
+        "被除数和商的大小关系（小数除法）",
+        "因数和积的大小关系（小数乘法）",
+        "分数乘分数",
+        "分数与分数的除法"
+      ],
+      "source": "22-23 六年级上·全国·单元测试",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_235",
+      "original_id": "unit3_fill_235"
+    },
+    {
+      "id": "unit3_fill_236",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "把 $\\frac{8}{9}$ 米长的绳子平均分成 4 段，每段占全长的（），每段长（）米。",
+      "options": [],
+      "answer": "$\\frac{1}{4}$；$\\frac{2}{9}$",
+      "solution": "【分析】求每段长是全长的几分之几，平均分的是单位'1'，表示把单位'1'平均分成 4 份，求的是每一份占的分率，用除法计算。把 $\\frac{8}{9}$ 米长的绳子平均分成 4 段，可用除法算出一段的长度。解决此题关键是弄清求得是分率还是具体的数量，求分率平均分的是单位'1'，求具体的数量平均分的是具体的数量，要注意分率不能带单位名称，而具体的数量要带单位名称。\n\n【详解】$1 \\div 4 = \\frac{1}{4} \\quad\\frac {8}{9} \\div 4 = \\frac {2}{9} (\\text {米})$即每段占全长的 $\\frac{1}{4}$，每段长 $\\frac{2}{9}$ 米。",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "分数的意义",
+        "单位'1'的认识与确定",
+        "分数与除法的关系"
+      ],
+      "source": "2021·山东济宁·小升初真题",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_236",
+      "original_id": "unit3_fill_236"
+    },
+    {
+      "id": "unit3_fill_237",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "（）$\\times \\frac{11}{5} =$ （）$\\times 1 7 = \\frac {1 1}{1 5} \\times$ （）$= 0. 5 \\times$ （）$= 1$",
+      "options": [],
+      "answer": "$\\frac{5}{11}$；$\\frac{1}{17}$；$\\frac{15}{11}$；2",
+      "solution": "【分析】由题意四个算式得结果都为 1，在乘法计算中，互为倒数的两个数乘积为 1；据此解答即可。掌握倒数的意义是解答题目的关键。\n\n【详解】$1 \\div \\frac{11}{5} = \\frac{5}{11}$ $1 \\div 1 7 = \\frac {1}{1 7}$ $1 \\div \\frac {1 1}{1 5} = \\frac {1 5}{1 1}$ $1 \\div 0. 5 = 2$",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "除数是小数的小数除法",
+        "倒数的认识及应用"
+      ],
+      "source": "22-23 六年级上·辽宁·单元测试",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_237",
+      "original_id": "unit3_fill_237"
+    },
+    {
+      "id": "unit3_fill_238",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "$\\frac{27}{5} \\times$ （）$= 0.8 \\times$ （）$= ($（）$) \\times 29 = 1$",
+      "options": [],
+      "answer": "$\\frac{5}{27}$；$1.25 / \\frac{5}{4}$；$\\frac{1}{29}$",
+      "solution": "【分析】观察式子，三个算式的乘积都是 1，所以本题相当于填 $\\frac{27}{5}$、0.8 和 29 的倒数。据此解题。本题考查了倒数的认识，乘积是 1 的两个数互为倒数，求分数的倒数，将分子分母调换位置即可。\n\n【详解】$\\frac{27}{5} \\times \\frac{5}{27} = 0.8 \\times 1.25 = \\frac{1}{29} \\times 29 = 1$或 $\\frac{27}{5} \\times \\frac{5}{27} = 0.8 \\times \\frac{5}{4} = \\frac{1}{29} \\times 29 = 1$。",
+      "knowledge_points": [
+        "倒数的认识及应用"
+      ],
+      "source": "22-23 六年级上·全国·单元测试",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_238",
+      "original_id": "unit3_fill_238"
+    },
+    {
+      "id": "unit3_fill_239",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "分数单位为 $\\frac{1}{6}$ 的最大真分数与最小假分数的和是（），再加上（）个这样的分数单位就可得到最小的质数。",
+      "options": [],
+      "answer": "$\\frac{11}{6}$；1",
+      "solution": "【分析】分母是几分数单位就是几分之一，真分数的分子小于分母，假分数的分子等于或大于分母。分数单位为 $\\frac{1}{6}$ 的最大真分数是 $\\frac{5}{6}$，最小假分数是 $\\frac{6}{6}$，求它们和是多少，用加法计算；最小的质数是 2，减去和，再除以 $\\frac{1}{6}$，即可解答。关键是理解真分数和假分数的含义，理解分数单位的意义，除以一个数等于乘这个数的倒数。\n\n【详解】$\\frac{5}{6} +\\frac{6}{6} = \\frac{11}{6}$ $ (2 - \\frac {1 1}{6}) \\div \\frac {1}{6} = \\frac {1}{6} \\div \\frac {1}{6} = 1$",
+      "knowledge_points": [
+        "分数单位的认识与确定",
+        "分数与分数的除法",
+        "同分母分数加、减法",
+        "真分数、假分数、带分数的认识"
+      ],
+      "source": "2021·湖南益阳·小升初真题",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_239",
+      "original_id": "unit3_fill_239"
+    },
+    {
+      "id": "unit3_fill_240",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "$\\frac{8}{9}$ 米长的铁丝重 $\\frac{4}{9}$ 千克，1千克重的铁丝长（）米。",
+      "options": [],
+      "answer": "2",
+      "solution": "【分析】求 1 千克重的铁丝的长度，用铁丝的长度除以铁丝的质量即可。本题考查分数除法的意义及应用，掌握分数除法的计算法则是解题的关键。\n\n【详解】$\\frac{8}{9} \\div \\frac{4}{9}=\\frac {8}{9} \\times \\frac {9}{4} = 2 \\text {(米)}$",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "分数除法的应用"
+      ],
+      "source": "21-22 五年级下·河南商丘·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_240",
+      "original_id": "unit3_fill_240"
+    },
+    {
+      "id": "unit3_fill_241",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "在括号里填上'>''<'或'='。\n\n$\\frac {8}{9} \\times 4 (\\quad) \\frac {8}{9}\\quad \\frac {7}{6} (\\quad) 4. 2 \\times \\frac {7}{6}\\quad \\frac {5}{9} \\div \\frac {7}{4} (\\quad) \\frac {5}{9} \\times \\frac {4}{7}$",
+      "options": [],
+      "answer": ">;$<$;=",
+      "solution": "【分析】(1) (2) 一个数 (0 除外) 乘一个比 1 大的数，积比这个数大；(3) 分数除法转化成分数乘法计算：除以一个不为 0 的数等于乘这个数的倒数。本题考查不用计算判断积与因数之间大小关系的方法以及分数除法的计算法则。\n\n【详解】(1) 因为 $4 > 1$，所以 $\\frac{8}{9} \\times 4 > \\frac{8}{9}$；(2) 因为 $4.2 > 1$，所以 $\\frac{7}{6} < 4.2 \\times \\frac{7}{6}$； (3) $\\frac{5}{9} \\div \\frac{7}{4} = \\frac{5}{9} \\times \\frac{4}{7}$。",
+      "knowledge_points": [
+        "分数乘小数",
+        "分数与整数的乘法",
+        "分数乘分数",
+        "分数与分数的除法"
+      ],
+      "source": "20-21 六年级上·广西玉林·期中",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_241",
+      "original_id": "unit3_fill_241"
+    },
+    {
+      "id": "unit3_fill_242",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "$\\frac{1}{6}$ 与它的倒数的积是（），商是（）。",
+      "options": [],
+      "answer": "1；$\\frac{1}{36}$",
+      "solution": "【分析】根据倒数的定义，乘积为 1 的两个数互为倒数，所以 $\\frac{1}{6}$ 与它的倒数的积是 1； $\\frac{1}{6}$ 的倒数是 6，用 $\\frac{1}{6}$ 除以 6 即可求出商的值。此题的解题关键是灵活运用倒数的定义来求解。\n\n【详解】根据分析得， $\\frac{1}{6}$ 与它的倒数的积是 $1; \\frac{1}{6} \\div 6 = \\frac{1}{36}$，所以商是 $\\frac{1}{36}$。",
+      "knowledge_points": [
+        "倒数的认识及应用"
+      ],
+      "source": "20-21 六年级上·贵州安顺·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_242",
+      "original_id": "unit3_fill_242"
+    },
+    {
+      "id": "unit3_fill_243",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "周末，乐乐和爸爸到黄石市父子山休闲绿道骑车，他们 $\\frac{5}{4}$ 分钟骑行了 $\\frac{2}{5}$ 千米，平均每分钟骑行（）千米，骑行 1 千米需要（）分钟。",
+      "options": [],
+      "answer": "$\\frac{8}{25}$；$\\frac{25}{8}$",
+      "solution": "【分析】$\\frac{2}{5}$ 千米是骑行的路程， $\\frac{5}{4}$ 分钟是所用时间，用路程除以时间得速度，再用路程 1 千米除以速度得骑行 1 千米所用的时间。此题主要考查了行程问题中速度、时间和路程的关系：速度 $\\times$ 时间 = 路程，路程 $\\div$ 时间 = 速度，路程 $\\div$ 速度 = 时间，要熟练掌握。\n\n【详解】$\\frac{2}{5} \\div \\frac{5}{4} = \\frac{2}{5} \\times \\frac{4}{5} = \\frac{8}{25}$ （千米/分钟）$1 \\div \\frac{8}{25} = \\frac{25}{8}$ (分)",
+      "knowledge_points": [
+        "分数与分数的除法"
+      ],
+      "source": "21-22 六年级下·河南南阳·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_243",
+      "original_id": "unit3_fill_243"
+    },
+    {
+      "id": "unit3_fill_244",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "在括号里填上'>''<'或'='。\n\n$\\frac {9}{1 0} \\div \\frac {1}{6} (\\quad) \\frac {9}{1 0} \\quad \\frac {3}{4} \\div \\frac {1}{2} (\\quad) \\frac {3}{4} \\times 3$",
+      "options": [],
+      "answer": ">;$<$",
+      "solution": "【分析】一个数（0 除外）除以小于 1 的数，结果比原来的数大；根据除以一个不为 0 的数，等于乘这个数的倒数，其中一个因数相同，另一个因数越大，则积越大。据此解答即可。本题考查分数乘除法，明确分数乘除法的计算方法是解题的关键。\n\n【详解】由分析可知：$\\frac {9}{1 0} \\div \\frac {1}{6} > \\frac {9}{1 0} \\quad \\frac {3}{4} \\div \\frac {1}{2} < \\frac {3}{4} \\times 3$",
+      "knowledge_points": [
+        "被除数和商的大小关系（小数除法）",
+        "分数与整数的乘法",
+        "分数与分数的除法"
+      ],
+      "source": "20-21 六年级上·辽宁盘锦·期中",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_244",
+      "original_id": "unit3_fill_244"
+    },
+    {
+      "id": "unit3_fill_245",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "学校合唱小组的女生人数占全组人数的 $\\frac{12}{25}$。如果男生人数比女生人数多 2 人，那么学校合唱小组一共有（）人。",
+      "options": [],
+      "answer": "50",
+      "solution": "【分析】将全组人数看成单位'1'，女生占 $\\frac{12}{25}$，则男生占 $1-\\frac{12}{25}=\\frac{13}{25}$。所以 2 人对应的分率为 $\\frac{13}{25}-\\frac{12}{25}=\\frac{1}{25}$，根据分数除法的意义，用 $2 \\div \\frac{1}{25}$ 求出全组人数；据此解答。找准单位'1'，并找出与已知量对应的分率是解题的关键。\n\n【详解】$2 \\div (1 - \\frac{12}{25} - \\frac{12}{25})$ $ 2 \\div \\frac {1}{2 5} = 5 0 \\text {(人)}$",
+      "knowledge_points": [
+        "已知比一个数多/少几分之几是多少，求这个数"
+      ],
+      "source": "19-20六年级上·云南楚雄·期中",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_245",
+      "original_id": "unit3_fill_245"
+    },
+    {
+      "id": "unit3_fill_246",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "5 的 $\\frac{2}{3}$ 是（），把 $\\frac{3}{4}$ 米平均分成 3 份，每份是（）米。",
+      "options": [],
+      "answer": "$\\frac{10}{3}$；$\\frac{1}{4}$",
+      "solution": "【分析】求 5 的 $\\frac{2}{3}$ 是多少，用 $5 \\times \\frac{2}{3}$ 即可；用 $\\frac{3}{4}$ 米除以平均分的份数，即可求出每一份是多少米。根据求一个数的百分之几是多少，以及分数与整数的除法的应用。\n\n【详解】$5 \\times \\frac{2}{3} = \\frac{10}{3}$ $\\frac {3}{4} \\div 3 = \\frac {3}{4} \\times \\frac {1}{3}= \\frac{1}{4}(米)$",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "求一个数的几分之几的问题"
+      ],
+      "source": "21-22 六年级上·四川广安·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_246",
+      "original_id": "unit3_fill_246"
+    },
+    {
+      "id": "unit3_fill_247",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "等候公共汽车的人整齐的排成一列，六年级的小飞同学也在其中。他数了一下人数，发现排在他前面的人数占总人数的 $\\frac{2}{3}$，排在他后面的人数占总人数的 $\\frac{1}{4}$。从前往后数，小飞同学排在第（）个。",
+      "options": [],
+      "answer": "9",
+      "solution": "【分析】把等候坐公交车的人数看作单位'1'，从前往后数，排在他前面的人数占总人数的 $\\frac{2}{3}$，也就是小飞和他后面的人数占总人数的 $1 - \\frac{2}{3} = \\frac{1}{3}$。由此可知：小飞 (1 人) 占总人数的 $(1 - \\frac{1}{3} - \\frac{1}{4})$，根据已知一个数的几分之几是多少，求这个数，用除法求出总人数，再根据一个数乘分数的意义，用乘法求出他前面的人数，然后用他前面的人数加上 1 就是小飞从前面数排在第几个。此题解答关键是求出小飞（1人）占总人数的几分之几，据此求出总人数，进而求出小飞排在第几个。\n\n【详解】$1 \\div (1 - \\frac{2}{3} - \\frac{1}{4})= 1 \\div (\\frac {1}{3} - \\frac {1}{4})=1 \\div \\frac {1}{1 2}=12 (人)$ $12 \\times \\frac {2}{3} + 1=8 + 1=9 (个)$答：小飞同学排在第 9 个。",
+      "knowledge_points": [
+        "分数的四则混合运算",
+        "分数与整数的除法",
+        "异分母分数加、减法",
+        "认识一个整体的几分之几及应用"
+      ],
+      "source": "14-15六年级·河南三门峡·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_247",
+      "original_id": "unit3_fill_247"
+    },
+    {
+      "id": "unit3_fill_248",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "$\\frac{3}{7}$ 与（）互为倒数；0.25 的倒数是（）。",
+      "options": [],
+      "answer": "$\\frac{7}{3}$；4",
+      "solution": "【分析】乘积为 1 的两个数互为倒数；求一个分数的倒数，只需要将分子、分母交换位置即可；求一个小数的倒数，先将小数化成分数，再根据分数求倒数的方法求解。掌握倒数的定义以及求一个数的倒数的方法是解题的关键。\n\n【详解】$\\frac{3}{7}$ 与 $\\frac{7}{3}$ 互为倒数；$0.2 5 = \\frac {1}{4}$， $\\frac{1}{4}$ 的倒数是 $4$。",
+      "knowledge_points": [
+        "倒数的认识及应用"
+      ],
+      "source": "19-20 六年级上·河北衡水·期中",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_248",
+      "original_id": "unit3_fill_248"
+    },
+    {
+      "id": "unit3_fill_249",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "在（）里填上'>''<'或'='。\n\n$\\frac {9}{1 0} \\div \\frac {1}{6} (\\quad) \\frac {9}{1 0} \\quad \\frac {3}{4} \\div \\frac {1}{2} (\\quad) \\frac {3}{4} \\times 2 \\quad \\frac {5}{6} \\div \\frac {1}{3} (\\quad) \\frac {5}{6} \\times \\frac {1}{3}$",
+      "options": [],
+      "answer": ">;=;>",
+      "solution": "【分析】（1）根据：一个非零数除以一个小于 1 (0 除外)的数，商大于这个数，可得： $\\frac{9}{10} \\div \\frac{1}{6} > \\frac{9}{10}$。(2) 根据一个数除以一个分数，等于这个数乘这个分数的倒数，可得： $\\frac{3}{4} \\div \\frac{1}{2} = \\frac{3}{4} \\times 2$ (3) 根据： $\\frac{5}{6} \\div \\frac{1}{3} > \\frac{5}{6}, \\frac{5}{6} \\times \\frac{1}{3} < \\frac{5}{6}$，可得： $\\frac{5}{6} \\div \\frac{1}{3} > \\frac{5}{6} \\times \\frac{1}{3}$。解答此题的关键是要明确：（1）一个非零数乘一个小于 1 的数，积小于这个数；一个非零数乘一个等于 1 的数，积等于这个数；一个非零数乘一个大于 1 的数，积大于这个数。（2）一个非零数除以一个小于 1（0 除外）的数，商大于这个数；一个非零数除以一个等于 1 的数，商等于这个数；一个非零数除以一个大于 1 的数，商小于这个数。\n\n【详解】$\\frac{9}{10} \\div \\frac{1}{6} > \\frac{9}{10}; \\frac{3}{4} \\div \\frac{1}{2} = \\frac{3}{4} \\times 2; \\frac{5}{6} \\div \\frac{1}{3} > \\frac{5}{6} \\times \\frac{1}{3}$",
+      "knowledge_points": [
+        "被除数和商的大小关系（小数除法）",
+        "因数和积的大小关系（小数乘法）",
+        "分数乘分数",
+        "分数与分数的除法"
+      ],
+      "source": "20-21 六年级上·浙江绍兴·期中",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_249",
+      "original_id": "unit3_fill_249"
+    },
+    {
+      "id": "unit3_fill_250",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "$1 \\mathrm{~m}^{3}$ 的水结成冰，体积比水增加 $\\frac{1}{10}$，这块冰再融化成水，体积减小（）。(填分数)",
+      "options": [],
+      "answer": "$\\frac{1}{11}$",
+      "solution": "【分析】把水的体积看作单位'1'，冰的体积比水增加 $\\frac{1}{10}$，则冰的体积占水的 $(1 + \\frac{1}{10})$，用分数乘法求出冰的体积；求水比冰体积减少的分率时，把冰的体积看作单位'1'，水的体积比冰的体积减少的分率 = (冰的体积 - 水的体积) ÷冰的体积。本题主要考查分数除法的应用，找准题目中的单位'1'是解答题目的关键。\n\n【详解】冰的体积： $1 \\times (1 + \\frac{1}{10}) =1 \\times \\frac {1 1}{1 0} = \\frac {1 1}{1 0} ( {m} ^ {3})$水的体积比冰的体积减少的分率： $(\\frac{11}{10} - 1) \\div \\frac{11}{10}$$ \\frac {1}{1 0} \\div \\frac {1 1}{1 0} = \\frac {1}{1 1}$所以，这块冰再融化成水，体积减小 $\\frac{1}{11}$。",
+      "knowledge_points": [
+        "分数的四则混合运算",
+        "已知比一个数多/少几分之几是多少，求这个数",
+        "单位'1'的认识与确定",
+        "求比一个数多/少几分之几的数是多少"
+      ],
+      "source": "21-22 六年级上·山东济南·期中",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_250",
+      "original_id": "unit3_fill_250"
+    },
+    {
+      "id": "unit3_fill_251",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "一个数的 $\\frac{4}{5}$ 正好是 20，这个数是（），它的倒数是（）。",
+      "options": [],
+      "answer": "25；$\\frac{1}{25}$",
+      "solution": "【分析】根据题意，把要求的数看作单位'1'，它的 $\\frac{4}{5}$ 是 20，求单位'1'，用除法计算；明确已知一个数的几分之几是多少，求这个数，用除法计算。\n\n根据'乘积是1的两个数互为倒数'，用1除以这个数，即可求出这个数的倒数。\n\n【详解】$20 \\div \\frac{4}{5}= 2 0 \\times \\frac {5}{4} = 2 5 \\quad 1 \\div 2 5 = \\frac {1}{2 5}$25 的倒数是 $\\frac{1}{25}$。",
+      "knowledge_points": [
+        "分数除法的应用",
+        "倒数的认识及应用"
+      ],
+      "source": "21-22 五年级下·河南商丘·阶段练习",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_251",
+      "original_id": "unit3_fill_251"
+    },
+    {
+      "id": "unit3_fill_252",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "一部分书稿，甲打字员打完全书要 20 天，乙打字员用同样的时间只能完成书稿的 $\\frac{4}{5}$，甲、乙两人合打这部书稿要（）天完成。",
+      "options": [],
+      "answer": "$11 \\frac{1}{9}$",
+      "solution": "【分析】把这份书稿的工作总量看作单位'1'，根据工作效率 $=$ 工作总量 $\\div$ 工作时间，分别求出甲、乙单独完成时的工作效率，再根据工作时间 $=$ 工作总量 $\\div$ 工作效率，用 1 除以甲、乙的工作效率之和，就是两人合作完成的天数。掌握工程问题中工作总量、工作效率、工作时间之间的关系是解题的关键。\n\n【详解】甲的工作效率： $1 \\div 20 = \\frac{1}{20}$。\n\n乙的工作效率：$ \\frac {4}{5} \\div 2 0 = \\frac {4}{5} \\times \\frac {1}{2 0} = \\frac {1}{2 5}$甲、乙两人合作完成的天数：$\\div (\\frac {1}{2 0} + \\frac {1}{2 5}) = 1 \\div (\\frac {5}{1 0 0} + \\frac {4}{1 0 0}) = 1 \\div \\frac {9}{1 0 0} = 1 1 \\frac {1}{9} (\\text {天})$",
+      "knowledge_points": [
+        "分数的四则混合运算",
+        "工程问题",
+        "单位'1'的认识与确定"
+      ],
+      "source": "19-20 六年级下·甘肃嘉峪关·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_252",
+      "original_id": "unit3_fill_252"
+    },
+    {
+      "id": "unit3_fill_253",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "池塘有 4 只鹅，正好是鸭的只数的 $\\frac{1}{3}$，池塘里有（）只鸭。",
+      "options": [],
+      "answer": "12",
+      "solution": "【分析】将鸭的只数看作单位'1'，鹅的只数 $\\div$ 对应分率 $=$ 鸭的只数，据此列式计算。关键是确定单位'1'，理解分数除法的意义。\n\n【详解】$4 \\div \\frac{1}{3} = 12$ （只）",
+      "knowledge_points": [
+        "分数除法的应用"
+      ],
+      "source": "20-21 五年级下·广东韶关·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_253",
+      "original_id": "unit3_fill_253"
+    },
+    {
+      "id": "unit3_fill_254",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "从 1 里面每次减去 $\\frac{1}{7}$，减去（）次后得 0。",
+      "options": [],
+      "answer": "7",
+      "solution": "【分析】本题可以理解为：求 1 里面有几个 $\\frac{1}{7}$，用 1 除以 $\\frac{1}{7}$ 即可。本题考查了除法的包含意义：求一个数里面有几个另一个数，用除法求解。\n\n【详解】$1 \\div \\frac{1}{7} = 1 \\times 7 = 7$",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "分数的初步认识"
+      ],
+      "source": "20-21 五年级下·内蒙古阿拉善盟·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_254",
+      "original_id": "unit3_fill_254"
+    },
+    {
+      "id": "unit3_judge_255",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "哥哥比弟弟高 $\\frac{1}{4}$，也就是弟弟比哥哥矮 $\\frac{1}{5}$。（）",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】哥哥比弟弟高 $\\frac{1}{4}$，将弟弟身高看作单位'1'，则哥哥身高是弟弟的 $(1 + \\frac{1}{4})$，弟弟与哥哥的身高差 $\\div$ 哥哥身高 =弟弟比哥哥矮几分之几，据此分析。关键是理解分数的意义，确定单位'1'，差÷较大数=少几分之几。\n\n【详解】$(1 + \\frac{1}{4} - 1) \\div (1 + \\frac{1}{4})=\\frac {1}{4} \\div \\frac {5}{4} = \\frac {1}{4} \\times \\frac {4}{5} = \\frac {1}{5}$哥哥比弟弟高 $\\frac{1}{4}$，也就是弟弟比哥哥矮 $\\frac{1}{5}$，说法正确。故答案为：√",
+      "knowledge_points": [
+        "求一个数占另一个数几分之几",
+        "分数的四则混合运算"
+      ],
+      "source": "22-23 六年级上·湖北十堰·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_255",
+      "original_id": "unit3_judge_255"
+    },
+    {
+      "id": "unit3_judge_256",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "得数是 1 的两个数互为倒数。（）",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】如果两个数的乘积为 1，我们就说这两个数互为倒数，或者说一个数是另一个数的倒数，据此解答。本题主要考查倒数的认识，掌握倒数的意义是解答题目的关键。\n\n【详解】分析可知，乘积为 1 的两个数互为倒数，如： $\\frac{1}{2} \\times 2 = 1$， $\\frac{1}{2}$ 和 2 互为倒数，所以题目说法不正确。\n\n故答案为：×",
+      "knowledge_points": [
+        "倒数的认识及应用"
+      ],
+      "source": "2014 六年级·全国·课后作业",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_256",
+      "original_id": "unit3_judge_256"
+    },
+    {
+      "id": "unit3_judge_257",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "珊珊的爸爸今年 48 岁，正好是爷爷年龄的 $\\frac{4}{7}$，爷爷今年 84 岁。（）",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】将爷爷年龄看作单位'1'，爸爸年龄 $\\div$ 对应分率 $=$ 爷爷年龄，据此分析。关键是确定单位'1'，理解分数除法的意义。\n\n【详解】$48 \\div \\frac{4}{7} = 48 \\times \\frac{7}{4} = 84$ （岁）珊珊的爸爸今年 48 岁，正好是爷爷年龄的 $\\frac{4}{7}$，爷爷今年 84 岁，原题说法正确。故答案为：√",
+      "knowledge_points": [
+        "已知一个数的几分之几是多少，求这个数"
+      ],
+      "source": "22-23 六年级上·广东广州·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_257",
+      "original_id": "unit3_judge_257"
+    },
+    {
+      "id": "unit3_judge_258",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "一个不为 0 的数除以假分数，商一定小于被除数。（）",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】假分数是只大于或等于 1 的分数；在除法中，当除数小于 1（0 除外），商大于被除数；当除数等于 1，商等于被除数；当除数大于 1，商小于被除数。由此即可做出判断。此题考查了根据除数与 1 的大小关系来判断商与被除数的大小关系方法的运用。\n\n【详解】一个不为 0 的数除以假分数时，所得的商小于或等于被除数，原题说法错误；故答案为：×",
+      "knowledge_points": [
+        "被除数和商的大小关系（小数除法）",
+        "分数与分数的除法",
+        "真分数、假分数、带分数的认识"
+      ],
+      "source": "2014 六年级·全国·课后作业",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_258",
+      "original_id": "unit3_judge_258"
+    },
+    {
+      "id": "unit3_judge_259",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "$\\frac{5}{6} \\div 3$ 表示把 $\\frac{5}{6}$ 平均分成3份，求每份是多少。（）",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】根据平均分的意义，总数÷数量 = 每份数，据此解答。本题考查分数除法的意义，根据除法中平均分的意义解答。\n\n【详解】$\\frac{5}{6} \\div 3$ 表示把 $\\frac{5}{6}$ 平均分成 3 份，求每份是多少。原题说法正确。故答案为：√",
+      "knowledge_points": [
+        "分数的平均分"
+      ],
+      "source": "21-22 六年级上·云南昭通·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_259",
+      "original_id": "unit3_judge_259"
+    },
+    {
+      "id": "unit3_judge_260",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "假分数的倒数都小于 1。（）",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】分子比分母大或分子和分母相等的分数叫做假分数；假分数 $\\geqslant 1$；乘积是 1 的两个数互为倒数；求一个假分数的倒数，只需要将分子、分母交换位置即可；据此判断。本题考查倒数、假分数的意义以及假分数的倒数的求法，明确 1 的倒数是 1，0 没有倒数。\n\n【详解】例如：设假分数为 $\\frac{5}{2}$， $\\frac{5}{2}$ 的倒数是 $\\frac{2}{5}$， $\\frac{2}{5}<1$；设假分数为 $\\frac{3}{3}$， $\\frac{3}{3}$ 的倒数是 $\\frac{3}{3}$， $\\frac{3}{3}=1$；所以假分数的倒数都小于或等于 $1$。原题说法错误。故答案为：×",
+      "knowledge_points": [
+        "倒数的认识及应用",
+        "真分数、假分数、带分数的认识"
+      ],
+      "source": "21-22 六年级下·吉林通化·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_260",
+      "original_id": "unit3_judge_260"
+    },
+    {
+      "id": "unit3_judge_261",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "假分数的倒数一定小于它本身。（）",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】假分数的分子大于等于分母，假分数大于等于 1。据此，再结合倒数的认识，分析解题即可。本题考查了倒数的认识，乘积是 1 的两个数互为倒数，1 的倒数是本身。\n\n【详解】$\\frac{5}{2}$ 的倒数是 $\\frac{2}{5}, \\frac{2}{5}<\\frac{5}{2}$，此时，假分数的倒数小于它本身；当假分数分子和分母相等，即假分数等于 1 时，它的倒数仍然是 1。所以，假分数的倒数不一定小于它本身，也有可能和本身相等。原题说法错误。故答案为：×",
+      "knowledge_points": [
+        "倒数的认识及应用",
+        "真分数、假分数、带分数的认识"
+      ],
+      "source": "2022·广东汕头·小升初真题",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_261",
+      "original_id": "unit3_judge_261"
+    },
+    {
+      "id": "unit3_judge_262",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "甲数比乙数多 $\\frac{1}{5}$，那么乙数比甲数少 $\\frac{1}{6}$。（）",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】由'甲数比乙数多 $\\frac{1}{5}$'，把乙数看作单位'1'，甲数相当于乙数的 $(1 + \\frac{1}{5})$；要求乙数比甲数少几分之几，也就是求乙数比甲数少的部分占甲数的几分之几。据此解答。此题的解决关键是找出单位'1'的量，先求得比较量对应的分率，进而解决问题。\n\n【详解】$(1 + \\frac{1}{5} - 1) \\div (1 + \\frac{1}{5})=\\frac {1}{5} \\div \\frac {6}{5} = \\frac {1}{6}$即乙数比甲数少 $\\frac{1}{6}$。\n\n故答案为：√",
+      "knowledge_points": [
+        "已知比一个数多/少几分之几是多少，求这个数",
+        "单位'1'的认识与确定",
+        "求比一个数多/少几分之几的数是多少"
+      ],
+      "source": "18-19 六年级上·全国·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_262",
+      "original_id": "unit3_judge_262"
+    },
+    {
+      "id": "unit3_judge_263",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "把 $\\frac{4}{5}$ 米长的绳子平均分成 4 份，每份占全长的 $\\frac{1}{4}$，也就是 $\\frac{1}{5}$ 米。（）",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】求每段长是这根绳子的几分之几，平均分的是单位'1'，求的是分率；求每段长的米数，平均分的是具体的数量 $\\frac{4}{5}$ 米，求的是具体的数量；都用除法计算。解决此题关键是弄清求得是具体的数量还是分率，求具体的数量平均分的是具体的数量；求分率平均分的是单位'1'。\n\n【详解】$1 \\div 4 = \\frac{1}{4}$ $ \\frac {4}{5} \\div 4 = \\frac {1}{5} (\\text {米})$原题说法正确；故答案为：√",
+      "knowledge_points": [
+        "分数的意义",
+        "单位'1'的认识与确定",
+        "分数与除法的关系"
+      ],
+      "source": "21-22 五年级下·河南南阳·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_263",
+      "original_id": "unit3_judge_263"
+    },
+    {
+      "id": "unit3_judge_264",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "$\\frac{1}{4} \\times \\frac{1}{4} \\div \\frac{1}{4} \\times \\frac{1}{4} = 1$。（）",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】根据四则混合运算的顺序计算出结果即可。解答本题时要注意除法性质的正确运用。\n\n【详解】$\\frac{1}{4} \\times \\frac{1}{4} \\div \\frac{1}{4} \\times \\frac{1}{4}= \\frac{1}{16}$，原题说法错误；$ \\frac {1}{1 6} \\div \\frac {1}{4} \\times \\frac {1}{4} = \\frac {1}{4} \\times \\frac {1}{4}=\\frac{1}{16}$原题说法错误；故答案为：×。",
+      "knowledge_points": [
+        "分数的乘、除法的混合运算"
+      ],
+      "source": "21-22 六年级下·黑龙江佳木斯·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_264",
+      "original_id": "unit3_judge_264"
+    },
+    {
+      "id": "unit3_judge_265",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "$\\frac{2}{3}$ 吨水泥，用去 $\\frac{1}{3}$ 吨，还剩 $\\frac{1}{2}$。（）",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】根据减法的意义，用水泥的总吨数减去用去的吨数，就是还剩的水泥的吨数；再用还剩的水泥吨数除以总吨数，即可求出还剩的水泥占总吨数的几分之几；据此判断。本题考查分数除法的意义及应用；求一个数是另一个数的几分之几，用除法计算。\n\n【详解】$\\frac{2}{3} -\\frac{1}{3} = \\frac{1}{3}$ （吨）$ \\frac {1}{3} \\div \\frac {2}{3} = \\frac {1}{3} \\times \\frac {3}{2} = \\frac {1}{2}$原题说法正确。故答案为：√",
+      "knowledge_points": [
+        "求一个数占另一个数几分之几",
+        "分数与分数的除法"
+      ],
+      "source": "20-21 六年级上·广西玉林·期中",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_265",
+      "original_id": "unit3_judge_265"
+    },
+    {
+      "id": "unit3_judge_266",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "一个自然数（0 除外）除以一个真分数，商一定大于这个自然数。（）",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】根据真分数的定义可知，真分数小于 1，一个非 0 数除以一个小于 1 的数，商大于被除数；除以一个大于 1 的数，商小于被除数；据此解答。本题主要考查分数除法中商与除数的关系。\n\n【详解】根据分析得，一个自然数（0 除外）除以一个真分数，商一定大于这个自然数。故答案为：√",
+      "knowledge_points": [
+        "被除数和商的大小关系（小数除法）",
+        "分数与分数的除法",
+        "真分数、假分数、带分数的认识"
+      ],
+      "source": "19-20 六年级上·全国·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_266",
+      "original_id": "unit3_judge_266"
+    },{
+      "id": "unit3_judge_267",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "如果男生比女生多 $\\frac{1}{4}$，那么女生就比男生少 $\\frac{1}{4}$。（）",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】如果男生比女生多 $\\frac{1}{4}$，就是把女生人数看作单位'1'，男生人数是女生人数的 $1 + \\frac{1}{4}$，求女生比男生少几分之几，是以男生人数为标准，用 $\\frac{1}{4} \\div (1 + \\frac{1}{4})$ 解答。此题考查的是分数除法的应用，解答本题关键是找准单位'1'，弄清以谁为标准。\n\n【详解】女生比男生少：$\\frac {1}{4} \\div (1 + \\frac {1}{4}) = \\frac {1}{4} \\div \\frac {5}{4} = \\frac {1}{5}$故答案为：×",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "求比一个数多/少几分之几的数是多少"
+      ],
+      "source": "2018·河北秦皇岛·小升初真题",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_267",
+      "original_id": "unit3_judge_267"
+    },
+    {
+      "id": "unit3_judge_268",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "$20 \\div \\frac{4}{5}$ 是表示把 20 平均分成 5 份，取其中的 4 份。（）",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】已知一个数的几分之几是多少，求这个数，用除法计算，据此解答。本题主要考查了分数除法的意义，已知单位'1'的几分之几是多少，求单位'1'用除法计算。\n\n【详解】$20 \\div \\frac{4}{5}$：表示一个数的 $\\frac{4}{5}$ 是 20，求这个数是多少。 $20 \\times \\frac{4}{5}$：表示求 20 的 $\\frac{4}{5}$ 是多少，即把 20 平均分成 5 份，取其中的 4 份。故答案为：×",
+      "knowledge_points": [
+        "分数与整数的除法"
+      ],
+      "source": "20-21 六年级上·湖南永州·期中",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_268",
+      "original_id": "unit3_judge_268"
+    },
+    {
+      "id": "unit3_judge_269",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "a 是一个整数，它的倒数一定是 $\\frac{1}{a}$。（）",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】根据乘积是1的两个数互为倒数，举例说明即可。找非0自然数的倒数，可以用1除以这个数。\n\n【详解】0 是整数，但是 0 没有倒数。故答案为：×",
+      "knowledge_points": [
+        "倒数的认识及应用"
+      ],
+      "source": "2021·河北保定·小升初真题",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_269",
+      "original_id": "unit3_judge_269"
+    },
+    {
+      "id": "unit3_judge_270",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "两个数相除，商一定大于被除数。（）",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】一个数（0 除外），除以大于 1 的数，商比原数小；除以小于 1 的数，商比原数大，举例说明即可。从小数除法开始，商和被除数的大小就有这样的关系。\n\n【详解】$1 \\div 2 = 0.5$ $1 \\div \\frac {1}{2} = 2$两个数相除，商可能大于被除数，也可能小于被除数，原题说法错误。故答案为：×",
+      "knowledge_points": [
+        "被除数和商的大小关系（小数除法）"
+      ],
+      "source": "2014 三年级·全国·课后作业",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_270",
+      "original_id": "unit3_judge_270"
+    },
+    {
+      "id": "unit3_judge_271",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "$\\frac{5}{21} \\times \\frac{7}{10} \\div \\frac{7}{10} \\times \\frac{5}{21} = 1$。（）",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】根据分数乘除法的计算方法计算出结果，即可判断。本题主要考查分数乘除法的计算方法。\n\n【详解】$\\frac{5}{21} \\times \\frac{7}{10} \\div \\frac{7}{10} \\times \\frac{5}{21}=\\frac {1}{6} \\div \\frac {7}{1 0} \\times \\frac {5}{2 1} = \\frac {1}{6} \\times \\frac {1 0}{7} \\times \\frac {5}{2 1} = \\frac {5}{2 1} \\times \\frac {5}{2 1} = \\frac {2 5}{4 4 1}$故答案为：×",
+      "knowledge_points": [
+        "分数的乘、除法的混合运算"
+      ],
+      "source": "21-22 六年级上·河北邯郸·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_271",
+      "original_id": "unit3_judge_271"
+    },
+    {
+      "id": "unit3_judge_272",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "$a \\times \\frac{2}{3} = b \\div \\frac{2}{3}$（a、b 都为不 0），那么 $a > b$。（）",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】将除法改成乘法，根据积一定，一个数乘的数越大其本身越小，进行分析。关键是掌握分数乘除法的计算方法，除以一个数等于乘这个数的倒数。\n\n【详解】$a \\times \\frac{2}{3} = b \\div \\frac{2}{3} = b \\times \\frac{3}{2}, \\frac{2}{3} < \\frac{3}{2}$，所以 $a > b$。故答案为：√",
+      "knowledge_points": [
+        "分数乘分数",
+        "分数与分数的除法"
+      ],
+      "source": "21-22 六年级上·湖北随州·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_272",
+      "original_id": "unit3_judge_272"
+    },
+    {
+      "id": "unit3_judge_273",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "$\\frac{5}{8}$ 除以一个假分数，商一定比 $\\frac{5}{8}$ 小。（）",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】一个数（0 除外），除以大于 1 的数，商比原数小，举例说明即可。关键是掌握分数除法的计算方法，假分数大于或等于 1。\n\n【详解】$\\frac{5}{8} \\div \\frac{8}{8} = \\frac{5}{8}$，所以原题说法错误。",
+      "knowledge_points": [
+        "分数与分数的除法"
+      ],
+      "source": "21-22 六年级上·广东佛山·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_273",
+      "original_id": "unit3_judge_273"
+    },
+    {
+      "id": "unit3_judge_274",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "分数的倒数大于它本身。（）",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】乘积是 1 的两个数互为倒数，举例说明即可。关键是理解比的意义，明白分数可以等于、小于或大于 1，其倒数可能等于、大于或小于 1。\n\n【详解】$\\frac{3}{2}$ 的倒数是 $\\frac{2}{3}, \\frac{2}{3} < \\frac{3}{2}$，所以原题说法错误。",
+      "knowledge_points": [
+        "倒数的认识及应用"
+      ],
+      "source": "21-22 六年级上·湖北鄂州·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_274",
+      "original_id": "unit3_judge_274"
+    },
+    {
+      "id": "unit3_calc_275",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "用简便方法计算。\n\n$\\frac {3}{7} \\times 1 2. 5 \\times \\frac {6}{7} \\times 8 \\quad 4.6 \\mathrm {-} 1. 2 5 \\mathrm {-} 1. 7 5 + 0. 4 \\quad5 4 \\frac {2}{5} \\div 1 7$",
+      "options": [],
+      "answer": "$\\frac{1800}{49}$；2；$3\\frac{1}{5}$",
+      "solution": "【分析】（1）根据乘法交换律 $a \\times b = b \\times a$，乘法结合律$(a \\times b) \\times c = a \\times (b \\times c)$ 进行简算；(2) 根据加法交换律 $a + b = b + a$，加法结合律 $(a + b) + c = a + (b + c)$，减法的性质 $a - b - c = a - (b + c)$ 进行简算；(3) 把 $54 \\frac{2}{5}$ 分解成 $51 + \\frac{17}{5}$，除法转化成乘法，然后根据乘法分配律 $(a + b) \\times c = a \\times c + b \\times c$ 进行简算。\n\n【详解】(1) $\\frac{3}{7} \\times 12.5 \\times \\frac{6}{7} \\times 8$\n\n$=(\\frac {3}{7} \\times \\frac {6}{7}) \\times (1 2. 5 \\times 8) = \\frac {1 8}{4 9} \\times 1 0 0= \\frac {1 8 0 0}{4 9}$\n\n(2) $4.6 - 1.25 - 1.75 + 0.4$\n\n$= (4. 6 + 0. 4) - (1. 2 5 + 1. 7 5) = 5 - 3 = 2$\n\n(3) $54 \\frac{2}{5} \\div 17$\n\n$=(5 1 + \\frac {1 7}{5}) \\times \\frac {1}{1 7} = 5 1 \\times \\frac {1}{1 7} + \\frac {1 7}{5} \\times \\frac {1}{1 7} = 3 + \\frac {1}{5} = 3 \\frac {1}{5}$",
+      "knowledge_points": [
+        "小数加、减法简便运算",
+        "小数的加、减法混合运算",
+        "分数乘法运算律"
+      ],
+      "source": "18-19六年级下·全国·单元测试",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_275",
+      "original_id": "unit3_calc_275"
+    },
+    {
+      "id": "unit3_calc_276",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "计算题",
+      "difficulty": "困难",
+      "question": "解方程。\n\n$\\frac {2}{3} x + (1 3 5 - x ) \\times \\frac {2}{5} = 1 9 8$\n\n$\\frac {x}{5} - \\frac {1 7 - x}{3} = 1$",
+      "options": [],
+      "answer": "$x = 540$；$x = 12.5$",
+      "solution": "【分析】$\\frac{2}{3} x + (135 - x) \\times \\frac{2}{5} = 198$，先用乘法分配律去掉括号，然后根据等式的性质1和2，在方程左右两边同时减去54，再将方程左边合并为 $\\frac{4}{15} x$，最后在方程左右两边同时除以 $\\frac{4}{15}$ 即可；$\\frac{x}{5} - \\frac{17 - x}{3} = 1$，先在方程左右两边同时乘 15，方程左边化为$3x - (17 - x) \\times 5$，然后根据乘法分配律去掉括号，方程变为$8 x - 85 = 15$，最后根据等式的性质 1 和 2，在方程左右两边同时加上 85，再同时除以 8 即可。\n\n【详解】$\\frac{2}{3} x + (135 - x) \\times \\frac{2}{5} = 198$\n\n解： $\\frac{2}{3} x + 135 \\times \\frac{2}{5} - x \\times \\frac{2}{5} = 198$\n\n$=\\frac {2}{3} x + 5 4 - x \\times \\frac {2}{5} = 1 9 8$\n\n$=\\frac {2}{3} x + 5 4 - x \\times \\frac {2}{5} - 5 4 = 1 9 8 - 5 4$\n\n$\\frac {2}{3} x - \\frac {2}{5} x = 1 4 4 \\\\ \\frac {4}{1 5} x = 1 4 4 \\\\ \\frac {4}{1 5} x \\div \\frac {4}{1 5} = 1 4 4 \\div \\frac {4}{1 5} \\\\ x = 5 4 0 \\\\ \\frac {x}{5} - \\frac {1 7 - x}{3} = 1$\n\n解： $3x - (17 - x)\\times 5 = 1\\times 15$\n\n$3 x - 8 5 + 5 x = 1 5 \\\\ 8 x - 8 5 = 1 5 \\\\ 8 x - 8 5 + 8 5 = 1 5 + 8 5 \\\\ 8 x = 1 0 0 \\\\ 8 x \\div 8 = 1 0 0 \\div 8 \\\\ x = 1 2. 5$",
+      "knowledge_points": [
+        "应用等式的性质1和2解方程"
+      ],
+      "source": "2021·新疆乌鲁木齐·小升初真题",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_276",
+      "original_id": "unit3_calc_276"
+    },
+    {
+      "id": "unit3_calc_277",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "直接写出得数。\n\n$\\frac {2}{3} + \\frac {1}{4} = \\quad \\frac {2}{5} \\times \\frac {3}{4} \\div \\frac {2}{5} \\times \\frac {3}{4}$ $1 8 0 0 - 7 9 9 =$\n\n$5 6 0 \\div 8 \\times 7 = $ $(1 - \\frac {7}{8}) \\times 1 6 = $ $\\frac {1}{3} \\times 2 4 \\div \\frac {1}{3} =$",
+      "options": [],
+      "answer": "$\\frac{11}{12}$；$\\frac{9}{16}$；1001；490；2；24",
+      "solution": "【分析】略",
+      "knowledge_points": [
+        "分数与整数的乘法",
+        "分数乘分数",
+        "分数与分数的除法",
+        "异分母分数加、减法"
+      ],
+      "source": "2021·上海·小升初真题",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_277",
+      "original_id": "unit3_calc_277"
+    },
+    {
+      "id": "unit3_calc_278",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "下列各题，怎样简便就怎样计算。\n\n$2 4 3 \\times 8 \\frac {1}{4} + 7 5. 7 \\times 8 2. 5 \\quad\\frac {3}{5} \\div [ (\\frac {7}{9} + \\frac {1}{3}) \\times \\frac {3}{2}]$\n\n$(\\frac {1}{1 0} + \\frac {1}{1 7}) \\times 1 3 + 4 \\div 1 7\\quad \\frac {3}{1 4} \\times [ 7 - 4 \\div (\\frac {2}{1 5} + \\frac {2}{3})]$",
+      "options": [],
+      "answer": "8250；$\\frac{9}{25}$；$\\frac {2 3}{1 0}$；$\\frac {3}{7}$",
+      "solution": "【分析】根据积不变的性质，把 $243 \\times 8 \\frac{1}{4} + 75.7 \\times 82.5$ 改写成$243 \\times 8.25 + 757 \\times 8.25$，再根据乘法分配律进行简算；$\\frac{3}{5} \\div [(\\frac{7}{9} + \\frac{1}{3}) \\times \\frac{3}{2}]$，先算小括号中的加法，再算中括号中的分数乘法，最后算除法；$(\\frac{1}{10} + \\frac{1}{17}) \\times 13 + 4 \\div 17$，先改写成 $\\frac{13}{10} + \\frac{13}{17} + \\frac{4}{17}$，根据加法结合律进行简算；$\\frac{3}{14} \\times [7 - 4 \\div (\\frac{2}{15} + \\frac{2}{3})]$，先算小括号中的加法，再算中括号中的除法，接着算中括号中的减法，最后算乘法。\n\n【详解】$243 \\times 8\\frac{1}{4} + 75.7 \\times 82.5$\n\n$ 2 4 3 \\times 8. 2 5 + 7 5 7 \\times 8. 2 5 \\\\ = (2 4 3 + 7 5 7) \\times 8. 2 5 \\\\ = 1 0 0 0 \\times 8. 2 5 \\\\ = 8 2 5 0 \\\\ \\frac {3}{5} \\div [ (\\frac {7}{9} + \\frac {1}{3}) \\times \\frac {3}{2}] \\\\ = \\frac {3}{5} \\div [ \\frac {1 0}{9} \\times \\frac {3}{2}]$\n\n$ \\frac {3}{5} \\div \\frac {5}{3} \\\\ = \\frac {9}{2 5} \\\\ (\\frac {1}{1 0} + \\frac {1}{1 7}) \\times 1 3 + 4 \\div 1 7 \\\\ = \\frac {1 3}{1 0} + \\frac {1 3}{1 7} + \\frac {4}{1 7} \\\\ = \\frac {1 3}{1 0} + (\\frac {1 3}{1 7} + \\frac {4}{1 7}) \\\\ = \\frac {1 3}{1 0} + 1 \\\\ = \\frac {2 3}{1 0} \\\\ \\frac {3}{1 4} \\times [ 7 - 4 \\div (\\frac {2}{1 5} + \\frac {2}{3}) ] \\\\ = \\frac {3}{1 4} \\times [ 7 - 4 \\div \\frac {1 2}{1 5}] \\\\ = \\frac {3}{1 4} \\times [ 7 - 4 \\times \\frac {1 5}{1 2}] \\\\ = \\frac {3}{1 4} \\times [ 7 - 5] \\\\ = \\frac {3}{1 4} \\times 2 \\\\ = \\frac {3}{7}$",
+      "knowledge_points": [
+        "分数加、减简便运算",
+        "分数的四则混合运算",
+        "小数的四则运算及法则",
+        "分数乘法运算律"
+      ],
+      "source": "21-22 六年级下·湖南永州·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_278",
+      "original_id": "unit3_calc_278"
+    },
+    {
+      "id": "unit3_calc_279",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "解方程。\n\n(1) $\\frac{11}{15} x = \\frac{22}{45}$\n\n(2) $x \\div \\frac{2}{3} = \\frac{5}{8}$\n\n(3) $\\frac{3}{4} x \\div 6 = \\frac{4}{5}$",
+      "options": [],
+      "answer": "(1) $x = \\frac{2}{3}$；(2) $x = \\frac{5}{12}$；(3) $x = \\frac{32}{5}$",
+      "solution": "【分析】（1）根据等式的性质，算式两边同时除以 $\\frac{11}{15}$ 即可；(2) 根据等式的性质，算式两边同时乘 $\\frac{2}{3}$ 即可；\n\n(3) 根据等式的性质，算式两边同时乘 6，再同时除以 $\\frac{3}{4}$ 即可。\n\n【详解】(1) $\\frac{11}{15} x = \\frac{22}{45}$\n\n解： $\\frac{11}{15} x \\div \\frac{11}{15} = \\frac{22}{45} \\div \\frac{11}{15}$\n\n$x = \\frac {2 2}{4 5} \\times \\frac {1 5}{1 1}$\n\n$x = \\frac {2}{3}$\n\n(2) $x \\div \\frac{2}{3} = \\frac{5}{8}$\n\n解： $x \\div \\frac{2}{3} \\times \\frac{2}{3} = \\frac{5}{8} \\times \\frac{2}{3}$\n\n$x = \\frac {5}{1 2}$\n\n(3) $\\frac{3}{4} x \\div 6 = \\frac{4}{5}$\n\n解： $\\frac{3}{4} x \\div 6 \\times 6 = \\frac{4}{5} \\times 6$\n\n$ \\frac {3}{4} x = \\frac {2 4}{5} \\\\ \\frac {3}{4} x \\div \\frac {3}{4} = \\frac {2 4}{5} \\div \\frac {3}{4} \\\\ x = \\frac {2 4}{5} \\times \\frac {4}{3} \\\\ x = \\frac {3 2}{5}$",
+      "knowledge_points": [
+        "应用等式的性质 1 和 2 解方程",
+        "解分数方程"
+      ],
+      "source": "21-22 五年级下·河南商丘·阶段练习",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_279",
+      "original_id": "unit3_calc_279"
+    },
+    {
+      "id": "unit3_calc_280",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "$\\frac{1}{3} \\div \\frac{6}{5} \\times \\frac{9}{10}$\n\n$\\frac {7}{1 5} \\div 1 0 \\times \\frac {2 5}{2 1}$\n\n$\\frac {5}{1 5} \\times 2 4 \\div \\frac {4}{2 1}$",
+      "options": [],
+      "answer": "$\\frac{1}{4}$；$\\frac{1}{18}$；42",
+      "solution": "【分析】（1）先把分数除法化为分数乘法，再利用乘法结合律简便计算；\n\n(2) 先把分数除法化为分数乘法，再利用乘法交换律简便计算； \n(3) 先把 $\\frac{5}{15}$ 化为最简分数，再按照从左往右的顺序计算。\n\n【详解】(1) $\\frac{1}{3} \\div \\frac{6}{5} \\times \\frac{9}{10}$\n\n$ \\frac {1}{3} \\times \\frac {5}{6} \\times \\frac {9}{1 0} \\\\ = \\frac {1}{3} \\times (\\frac {5}{6} \\times \\frac {9}{1 0}) \\\\ = \\frac {1}{3} \\times \\frac {3}{4} \\\\ = \\frac {1}{4}$\n\n(2) $\\frac{7}{15} \\div 10 \\times \\frac{25}{21}$\n\n$\\frac {7}{1 5} \\times \\frac {1}{1 0} \\times \\frac {2 5}{2 1} \\\\ = \\frac {7}{1 5} \\times \\frac {2 5}{2 1} \\times \\frac {1}{1 0} \\\\ = \\frac {5}{9} \\times \\frac {1}{1 0} \\\\ = \\frac {1}{1 8}$\n\n(3) $\\frac{5}{15} \\times 24 \\div \\frac{4}{21}$\n\n$ =\\frac {1}{3} \\times 2 4 \\times \\frac {2 1}{4} \\\\ = 8 \\times \\frac {2 1}{4} \\\\ = 4 2$",
+      "knowledge_points": [
+        "分数的乘、除法的混合运算",
+        "分数乘法运算律"
+      ],
+      "source": "2022 六年级下·全国·专题练习",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_280",
+      "original_id": "unit3_calc_280"
+    },
+    {
+      "id": "unit3_solve_281",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "古人云：'下大事必作于细，古今兴盛皆在于实。'崔校长以科技节为契机，用创新引领未来，重视培养科技能力。繁峙县实验小学在11月18日举办的科技节上，由实验人自己研发的红色纸飞机每秒飞行150米，太原科技馆的灰色飞机的飞行速度是红色纸飞机的 $\\frac{4}{5}$，又是科技馆蓝色飞机的 $\\frac{2}{3}$，蓝色飞机的飞行速度是多少？",
+      "options": [],
+      "answer": "180 米/秒",
+      "solution": "【分析】先把红色纸飞机的飞行速度看作单位'1'，灰色飞机的飞行速度是红色纸飞机的 $\\frac{4}{5}$，根据求一个数的几分之几是多少，用红色纸飞机的飞行速度乘 $\\frac{4}{5}$，求出灰色飞机的飞行速度；再把蓝色飞机的飞行速度看作单位'1'，灰色飞机的飞行速度又是蓝色飞机的 $\\frac{2}{3}$，根据已知一个数的几分之几是多少，求这个数，用灰色飞机的飞行速度除以 $\\frac{2}{3}$，即可求出蓝色飞机的飞行速度。找出单位'1'，区分两个单位'1'的不同；单位'1'已知，根据分数乘法的意义列式计算；单位'1'未知，根据分数除法的意义列式计算。\n\n【详解】$150 \\times \\frac{4}{5} \\div \\frac{2}{3}$\n\n$ 1 2 0 \\div \\frac {2}{3} \\\\ = 1 2 0 \\times \\frac {3}{2} \\\\ = 1 8 0 \\text {(米 / 秒)}$答：蓝色飞机的飞行速度是 180 米/秒。",
+      "knowledge_points": [
+        "分数的乘、除法的混合运算",
+        "分数除法的应用",
+        "求一个数的几分之几的问题"
+      ],
+      "source": "21-22 六年级上·山西忻州·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_281",
+      "original_id": "unit3_solve_281"
+    },
+    {
+      "id": "unit3_solve_282",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "学校举行朗诵比赛，获三等奖的有 120 人，获一等奖的人数是获三等奖 $\\frac{1}{4}$，是获二等奖 $\\frac{2}{3}$。获二等奖的有多少人？",
+      "options": [],
+      "answer": "45人",
+      "solution": "【分析】根据求一个数的几分之几是多少，用乘法计算：用 $120$ 乘 $\\frac{1}{4}$ 即可求出获一等奖的人数；再根据已知一个数的几分之几是多少，求这个数，用除法计算：用获一等奖的人数除以 $\\frac{2}{3}$ 即可求出获二等奖的人数。本题考查已知一个数的几分之几是多少，求这个数，明确用除法是解题的关键\n\n【详解】$120 \\times \\frac{1}{4} \\div \\frac{2}{3}$\n\n$= 3 0 \\div \\frac {2}{3} \\\\ = 4 5 \\text {(人)}$答：获二等奖的有 45 人。",
+      "knowledge_points": [
+        "分数的乘、除法的混合运算",
+        "求一个数的几分之几的问题"
+      ],
+      "source": "15-16 六年级·全国·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_282",
+      "original_id": "unit3_solve_282"
+    },
+    {
+      "id": "unit3_solve_283",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "一个粮食仓库存有一批粮食，运走 $\\frac{2}{3}$ 后，又运来 5.6 吨，这时现有的粮食是原来存粮的 $\\frac{4}{5}$。粮库原有存粮多少吨？",
+      "options": [],
+      "answer": "12吨",
+      "solution": "【分析】把粮库原有存粮 $\\mathrm{x}$ 吨，把粮库原有存粮看作单位'1'，运走 $\\frac{2}{3}$，还剩 $(1 - \\frac{2}{3})$，还剩下 $(1 - \\frac{2}{3}) \\mathrm{x}$ 吨，又运来 5.6 吨，这时现有的粮食是原来存粮的 $\\frac{4}{5}$，即剩下的存粮 +5.6 吨 = 原来存粮的吨数 $\\times \\frac{4}{5}$，列方程： $(1 - \\frac{2}{3}) x + 5.6 = \\frac{4}{5} x$，解方程，即可解答。根据方程的实际应用，利用原来存粮吨数与运走和运来吨数之间的关系，设出未知数，找出相关的量，列方程，解方程。\n\n【详解】解：设粮库原有存粮 $\\mathbf{X}$ 吨。\n\n$(1 - \\frac {2}{3}) x + 5. 6 = \\frac {4}{5} x$\n\n$\\frac {1}{3} x + 5. 6 = \\frac {4}{5} x$\n\n$\\frac {4}{5} x - \\frac {1}{3} x = 5. 6$\n\n$\\frac {1 2}{1 5} x - \\frac {5}{1 5} x = 5. 6$\n\n$\\frac {7}{1 5} x = 5. 6$\n\n$x = 5. 6 \\div \\frac {7}{1 5}$\n\n$x = 5. 6 \\times \\frac {1 5}{7} \\\\ x = 1 2$答：粮库原有存粮 12 吨。",
+      "knowledge_points": [
+        "列方程解含一个未知数的问题"
+      ],
+      "source": "21-22 六年级上·广东揭阳·期中",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_283",
+      "original_id": "unit3_solve_283"
+    },
+    {
+      "id": "unit3_solve_284",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "困难",
+      "question": "有两个同样的仓库 A 和 B，搬运一个仓库里的货物，甲需要 10 小时，乙需要 12 小时，丙需要 15 小时。甲和丙在 A 仓库，乙在 B 仓库，同时开始搬运。中途丙转向帮助乙搬运。最后，两个仓库同时搬完，丙帮助甲、乙各多少时间？",
+      "options": [],
+      "answer": "丙帮助甲3小时，帮助乙5小时",
+      "solution": "【分析】把一个仓库的工作量看作单位'1'，根据已知条件'搬运一个仓库里的货物，甲需要 10 小时，乙需要 12 小时，丙需要 15 小时'可知，甲的工作效率是 $\\frac{1}{10}$，乙的工作效率是 $\\frac{1}{12}$，丙的工作效率是 $\\frac{1}{15}$；三人同时搬运，合作工作效率是 $(\\frac{1}{10} + \\frac{1}{12} + \\frac{1}{15})$；又因两个仓库是同样的仓库，两个仓库的工作总量是'2'；先看成两个仓库的货物三人合作完成，根据'合作工作时间 = 工作总量 ÷ 合作工作效率'，求出三人同时搬运 2 个仓库需要的时间；根据'工作量 = 工作效率 × 工作时间'，求出合作工作时间内甲在 A 仓库的工作量，再用'1'减去甲在 A 仓库的工作量，剩下的就是丙在 A 仓库的工作量，除以丙的工作效率，即可求出丙在 A 仓库的搬运时间，用合作工作时间减去丙在 A 仓库的搬运时间，就是丙在 B 仓库的搬运时间。本题考查复杂的工程问题，掌握工作效率、工作时间，工作量之间的关系是解题的关键。\n\n【详解】2个仓库三人合作工作时间：\n\n$2 \\div (\\frac {1}{1 0} + \\frac {1}{1 2} + \\frac {1}{1 5})$\n\n$ =2 \\div (\\frac {6}{6 0} + \\frac {5}{6 0} + \\frac {4}{6 0}) \\\\ = 2 \\div \\frac {1}{4} \\\\ = 2 \\times 4 \\\\ = 8 \\text {(小时)}$\n\n丙在 A 仓库的工作时间：\n\n$ (1 - \\frac {1}{1 0} \\times 8) \\div \\frac {1}{1 5} \\\\ = (1 - \\frac {4}{5}) \\div \\frac {1}{1 5} \\\\ = \\frac {1}{5} \\times 1 5 \\\\ = 3 \\text {(小时)}$\n\n丙在B仓库的搬运时间：\n\n$8 - 3 = 5 \\text {(小时)}$\n\n答：丙帮助甲3小时，帮助乙5小时。",
+      "knowledge_points": [
+        "分数的四则混合运算",
+        "工程问题"
+      ],
+      "source": "2022·重庆·小升初真题",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_284",
+      "original_id": "unit3_solve_284"
+    },
+    {
+      "id": "unit3_solve_285",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "一根电线，第一次用去它的 $\\frac{2}{5}$，第二次用去余下的 $\\frac{2}{3}$，还剩 $60 \\mathrm{~m}$，这根电线原来长多少米？",
+      "options": [],
+      "answer": "300 米",
+      "solution": "【分析】一根电线，第一次用去它的 $\\frac{2}{5}$，还剩 $1 - \\frac{2}{5} = \\frac{3}{5}$，第二次用去余下的 $\\frac{2}{3}$，也就是 $\\frac{3}{5}$ 的 $\\frac{2}{3}$， $\\frac{3}{5} \\times \\frac{2}{3} = \\frac{2}{5}$，第二次用去了整体的 $\\frac{2}{5}$，第一次和第二次共用去 $\\frac{2}{5} + \\frac{2}{5} = \\frac{4}{5}$，还乘整体的 $1 - \\frac{4}{5} = \\frac{1}{5}$，由题意知：60 米对应着分率 $\\frac{1}{5}$，用 60 除以 $\\frac{1}{5}$ 即是电线的总长。据此解答。求得 60 米对应的分率，用分数除法计算是解答本题的关键。\n\n【详解】$(1 - \\frac{2}{5}) \\times \\frac{2}{3} + \\frac{2}{5}$\n\n$ \\frac {3}{5} \\times \\frac {2}{3} + \\frac {2}{5} \\\\ = \\frac {2}{5} + \\frac {2}{5} \\\\ = \\frac {4}{5}$\n\n$ 6 0 \\div (1 - \\frac {4}{5}) = 6 0 \\div \\frac {1}{5} = 3 0 0 \\text {(米)}$答：这根电线的原来长 300 米。",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "分数除法的应用"
+      ],
+      "source": "21-22 六年级下·湖南永州·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_285",
+      "original_id": "unit3_solve_285"
+    },
+    {
+      "id": "unit3_solve_286",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "困难",
+      "question": "一件工作，若由甲单独做72天完成，现在甲做1天后，乙加入一起工作，合作2天后，丙也一起工作，三人再一起工作4天，完成全部工作的 $\\frac{1}{3}$，又过了8天，完成了全部工作的 $\\frac{5}{6}$，若余下的工作由丙单独完成，还需要几天？",
+      "options": [],
+      "answer": "6天",
+      "solution": "【分析】要求余下的工作由丙单独完成，还需要几天，就要求出丙的工作效率，在此需要一步步推算，甲乙丙3人8天完成： $\\frac{5}{6} - \\frac{1}{3} = \\frac{1}{2}$，则甲乙丙3人每天完成（即3人的工作效率）： $\\frac{1}{2} \\div 8 = \\frac{1}{16}$，甲乙丙3人4天完成： $\\frac{1}{16} \\times 4 = \\frac{1}{4}$，则甲做一天后乙做 2 天要做： $\\frac{1}{3} - \\frac{1}{4} = \\frac{1}{12}$，那么乙一天做： $[\\frac{1}{12} - \\frac{1}{72} \\times 3] \\div 2 = \\frac{1}{48}$，则丙一天做： $\\frac{1}{16} - \\frac{1}{72} - \\frac{1}{48} = \\frac{1}{36}$，那么余下的由丙做要： $(1 - \\frac{5}{6}) \\div \\frac{1}{36} = 6$ (天)。此题属于工程问题，工作效率、工作时间、工作量三者之间的数量关系不明显，所以就要寻求一些特殊的思路，如综合转化、整体思考等方法来解决。\n\n【详解】$1 \\div 72 = \\frac{1}{72}$\n\n$ (\\frac {5}{6} - \\frac {1}{3}) \\div 8 \\\\ = \\frac {1}{2} \\div 8 \\\\ = \\frac {1}{1 6} \\\\ (\\frac {1}{3} - \\frac {1}{1 6} \\times 4 - \\frac {1}{7 2} \\times 3) \\div 2 \\\\ = (\\frac {1}{3} - \\frac {1}{4} - \\frac {1}{3 6}) \\div 2 \\\\ = \\frac {1}{2 4} \\div 2 \\\\ = \\frac {1}{4 8}$\n\n$\\frac {1}{1 6} - \\frac {1}{7 2} - \\frac {1}{4 8} \\\\ = \\frac {7}{1 4 4} - \\frac {1}{4 8} \\\\ = \\frac {1}{3 6}$\n\n$ (1 - \\frac {5}{6}) \\div \\frac {1}{3 6} \\\\ = \\frac {1}{6} \\div \\frac {1}{3 6} \\\\=6(天)$\n\n答：还需要6天。",
+      "knowledge_points": [
+        "分数的四则混合运算",
+        "工程问题"
+      ],
+      "source": "2015·山东·小升初真题",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_286",
+      "original_id": "unit3_solve_286"
+    },
+    {
+      "id": "unit3_solve_287",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "师徒二人加工一批零件，师傅加工了 150 个，恰好是这批零件的 $\\frac{5}{8}$，而此时徒弟加工的个数占这批零件的 $\\frac{1}{6}$，徒弟加工了多少个零件？",
+      "options": [],
+      "answer": "40 个",
+      "solution": "【分析】把这批零件的总数看成单位'1'，师傅加工了 150 个，恰好是这批零件的 $\\frac{5}{8}$，用 150 除以 $\\frac{5}{8}$，求出零件的总数，徒弟加工的个数占这批零件的 $\\frac{1}{6}$，用零件总数乘 $\\frac{1}{6}$，即可求出徒弟加工了多少个零件。解答此题的关键是找出单位'1'，求单位'1'的几分之几用乘法求解；已知单位'1'的几分之几是多少，求单位'1'用除法求解。\n\n【详解】$150 \\div \\frac{5}{8} \\times \\frac{1}{6}$\n\n$=2 4 0 \\times \\frac {1}{6} \\\\ = 4 0 (\\text {个})$答：徒弟加工了 40 个零件。",
+      "knowledge_points": [
+        "分数的四则混合运算",
+        "工程问题"
+      ],
+      "source": "21-22 六年级上·黑龙江黑河·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_287",
+      "original_id": "unit3_solve_287"
+    },
+    {
+      "id": "unit3_solve_288",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "直播带货受到越来越多人的喜爱。某网店本月通过直播带货达成的交易额有25560元，比该网店本月销售总额的 $\\frac{4}{5}$ 多236元，该网店本月的销售总额是多少元？",
+      "options": [],
+      "answer": "31655元",
+      "solution": "【分析】把该网店本月销售总额看作单位'1'，用本月直播带货达成的交易额减去 236 即是该网店本月销售总额的 $\\frac{4}{5}$，根据除法的意义，用除法解答即可。本题考查分数除法，明确单位'1'是解题的关键。\n\n【详解】$(25560 - 236) \\div \\frac{4}{5}$\n\n$= 2 5 3 2 4 \\div \\frac {4}{5} \\\\ = 3 1 6 5 5 \\text {(元)}$答：该网店本月的销售总额是 31655 元。",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "已知比一个数多/少几分之几是多少，求这个数"
+      ],
+      "source": "21-22六年级下·河南南阳·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_288",
+      "original_id": "unit3_solve_288"
+    },
+    {
+      "id": "unit3_solve_289",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "一个饲养场，养鸭 300 只，养的鸭比鸡多 $\\frac{1}{5}$，养鸡多少只？",
+      "options": [],
+      "answer": "250只",
+      "solution": "【分析】把鸡的只数看作单位'1'，养的鸭比鸡多 $\\frac{1}{5}$，也就是养鸭的只数是鸡的 $1 + \\frac{1}{5}$，依据分数除法意义即可解答。本题主要考查学生依据分数除法意义解决问题的能力\n\n【详解】$300 \\div (1 + \\frac{1}{5})= 3 0 0 \\div \\frac {6}{5} = 2 5 0 \\text {(只)}$",
+      "knowledge_points": [
+        "分数的四则混合运算",
+        "已知比一个数多/少几分之几是多少，求这个数"
+      ],
+      "source": "21-22 六年级上·黑龙江佳木斯·期中",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_289",
+      "original_id": "unit3_solve_289"
+    },
+    {
+      "id": "unit3_solve_290",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "冰化成水后，水的体积是冰的体积的 $\\frac{9}{10}$。现有一块冰，融化成水后的体积是 $36\\mathrm{dm}^3$，这块冰的体积是多少 $\\mathrm{dm}^3$？",
+      "options": [],
+      "answer": "$40 \\mathrm{dm}^{3}$",
+      "solution": "【分析】已知冰融化成水后的体积是冰的 $\\frac{9}{10}$，则把冰的体积看做单位'1'，又知冰融化成水后的体积是 $36 \\mathrm{dm}^{3}$，也就是冰体积的 $\\frac{9}{10}$ 是 $36 \\mathrm{dm}^{3}$，根据已知一个数的几分之几是多少，求这个数，用除法列式解答即可。解答此题的关键是确定单位'1'，然后根据分数除法的意义列式解答即可。\n\n【详解】$36 \\div \\frac{9}{10} = 40 \\, (\\mathrm{dm}^3)$答：这块冰的体积是 $40 \\mathrm{dm}^{3}$。",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "分数除法的应用"
+      ],
+      "source": "19-20六年级上·广西来宾·期中",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_290",
+      "original_id": "unit3_solve_290"
+    },
+    {
+      "id": "unit3_solve_291",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "一批零件，每小时加工60个，5小时后正好完成零件总数的 $\\frac{2}{5}$。这批零件共有多少个？",
+      "options": [],
+      "answer": "750 个",
+      "solution": "【分析】根据工作效率 $\\times$ 工作时间 = 工作总量，计算 5 小时共加工的零件数，加工完成的零件数是零件总数的 $\\frac{2}{5}$，已知一个数的 $\\frac{2}{5}$ 是多少，求这个数，根据分数除法的意义，用除法计算出这批零件的总数量。此题的解题关键是根据工作效率、工作时间、工作总量三者之间关系以及分数除法的意义，解决实际问题。\n\n【详解】$60 \\times 5 \\div \\frac{2}{5}= 3 0 0 \\div \\frac {2}{5} = 7 5 0 \\text {(个)}$答：这批零件共有 750 个。",
+      "knowledge_points": [
+        "工程问题",
+        "分数除法的应用"
+      ],
+      "source": "20-21六年级上·黑龙江哈尔滨·期中",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_291",
+      "original_id": "unit3_solve_291"
+    },
+    {
+      "id": "unit3_solve_292",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "某商场八月份的营业额是 480 万元，比七月份增加了 $\\frac{1}{4}$。七月份的营业额是多少？",
+      "options": [],
+      "answer": "384万元",
+      "solution": "【分析】把八月份的营业额看成单位'1'，七月份的营业额就是八月份的 $1 + \\frac{1}{4}$，求七月份的营业额用除法。本题的关键是找出单位'1'，并找出单位'1'的几分之几对应的数量，用除法就可以求出单位'1'的量。\n\n【详解】$480 \\div (1 + \\frac{1}{4})=4 8 0 \\div \\frac {5}{4} = 3 8 4 (\\text {万元})$答：七月份营业额是 384 万元。",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "求比一个数多/少几分之几的数是多少"
+      ],
+      "source": "21-22 五年级下·湖北武汉·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_292",
+      "original_id": "unit3_solve_292"
+    },
+    {
+      "id": "unit3_solve_293",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "某小区积极开展'垃圾分类进社区'活动，东、西区一个月共收集到废电池288节，西区收集到废电池的节数是东区的 $\\frac{7}{9}$，东、西区各收集了多少节废电池？",
+      "options": [],
+      "answer": "东区收集了 162 节，西区收集了 126 节",
+      "solution": "【分析】由题意可知，设东区收集到的废电池有 $\\mathrm{x}$ 节，则西区收集到废电池的节数有 $\\frac{7}{9} \\mathrm{x}$ 节，根据东、西区一个月共收集到废电池 288 节，据此列方程即可。本题考查用方程解决实际问题，明确数量关系是解题的关键。\n\n【详解】解：设东区收集到的废电池有 $x$ 节，则西区收集到废电池的节数有 $\\frac{7}{9} x$ 节。\n\n$x + \\frac {7}{9} x = 2 8 8$\n\n$\\frac {1 6}{9} x = 2 8 8$\n\n$x = 1 6 2$\n\n$1 6 2 \\times \\frac {7}{9} = 1 2 6 (\\text {节})$答：东区收集了 162 节，西区收集了 126 节。",
+      "knowledge_points": [
+        "分数与整数的乘法",
+        "分数与整数的除法",
+        "求一个数的几分之几的问题",
+        "列方程解含两个未知数的问题"
+      ],
+      "source": "20-21 六年级下·山东济南·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_293",
+      "original_id": "unit3_solve_293"
+    },
+    {
+      "id": "unit3_solve_294",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "乘坐公交车如果刷IC卡每人每次需支付1.6元，比投币便宜了 $\\frac{1}{5}$，投币支付每人每次需多少元？",
+      "options": [],
+      "answer": "2元",
+      "solution": "【分析】把投币支付每人每次需支付的钱数看作单位'1'，刷IC卡每人每次需支付的钱数相当于投币支付 $(1 - \\frac{1}{5})$，根据分数除法的意义，用刷IC卡每人每次需支付的钱数除以 $(1 - \\frac{1}{5})$，就是投币支付每人每次需要支付的钱数。此题是考查分数除法的意义及应用。已知一个数的几分之几是多少，求这个数，用已知数除以它所对应的分率。\n\n【详解】$1.6 \\div (1 - \\frac{1}{5})=1.6 \\div \\frac {4}{5}=2 (元)$答：投币支付每人每次需 2 元。",
+      "knowledge_points": [
+        "已知比一个数多/少几分之几是多少，求这个数"
+      ],
+      "source": "21-22六年级上·浙江台州·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_294",
+      "original_id": "unit3_solve_294"
+    },
+    {
+      "id": "unit3_solve_295",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "一架喷气式飞机的速度是 900 千米/时。一列火车的速度是这架飞机的 $\\frac{2}{15}$，是一辆小汽车的 $\\frac{3}{2}$，这辆小汽车的速度是多少？",
+      "options": [],
+      "answer": "80千米/时",
+      "solution": "【分析】以喷气式飞机的速度为单位 1，可以求出火车的速度，再根据量率对应，喷气式飞机的速度是一辆小汽车的 $\\frac{3}{2}$，即可求出小汽车的速度。本题考查分数乘法和除法的应用问题，找准单位 1 是关键。\n\n【详解】$900 \\times \\frac{2}{15} \\div \\frac{3}{2}=1 2 0 \\times \\frac {2}{3} = 8 0 \\text {(千米 / 时)}$答：这辆小汽车的速度是每小时 80 千米。",
+      "knowledge_points": [
+        "分数除法的应用",
+        "求一个数的几分之几的问题"
+      ],
+      "source": "19-20 六年级上·四川攀枝花·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_295",
+      "original_id": "unit3_solve_295"
+    },
+    {
+      "id": "unit3_solve_296",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "困难",
+      "question": "加工一批零件，甲、乙合作24天可以完成。现在由甲先做16天，然后由乙再做12天，正好完成这批零件的 $\\frac{3}{5}$。已知甲每天比乙多加工5个零件。这批零件一共有多少个？",
+      "options": [],
+      "answer": "600 个",
+      "solution": "【分析】把这批零件看作单位'1'，甲、乙合作24天可以完成，那么甲、乙两人的合作工作效率是 $\\frac{1}{24}$；现在由甲先做16天，然后由乙再做12天，正好完成这批零件的 $\\frac{3}{5}$，相当于甲、乙合作12天，甲单独做（16-12）天，正好完成这批零件的 $\\frac{3}{5}$；由此可以求出甲、乙的工作效率；又知甲每天比乙多加工5个零件，对应的分率是甲的工作效率减去乙的工作效率，即得出5个零件占这批零件的几分之几，然后根据已知一个数的几分之几是多少，求这个数，用除法解答。本题是较复杂的工程问题，熟练掌握工作效率、工作时间、工作量之间的关系；关键是把'由甲先做16天，然后由乙再做12天，正好完成这批零件的 $\\frac{3}{5}$'，转化为'甲、乙合作12天，甲单独做（16- 12）天，正好完成这批零件的 $\\frac{3}{5}$'，进而求出甲、乙的工作效率。\n\n【详解】$\\frac{1}{24} \\times 12 = \\frac{1}{2}$甲的工作效率：$(\\frac {3}{5} - \\frac {1}{2}) \\div (1 6 - 1 2)= (\\frac {6}{1 0} - \\frac {5}{1 0}) \\div 4 =\\frac {1}{1 0} \\div 4 = \\frac {1}{1 0} \\times \\frac {1}{4} = \\frac {1}{4 0}$乙的工作效率：$\\frac {1}{2 4} - \\frac {1}{4 0} = \\frac {5}{1 2 0} - \\frac {3}{1 2 0} = \\frac {1}{6 0}$这批零件共有：$ 5 \\div (\\frac {1}{4 0} - \\frac {1}{6 0}) = 5 \\div (\\frac {3}{1 2 0} - \\frac {2}{1 2 0}) = 5 \\div \\frac {1}{1 2 0} = 5 \\times 1 2 0 = 6 0 0 \\quad (\\text {个})$答：这批零件共有 600 个。",
+      "knowledge_points": [
+        "分数的四则混合运算",
+        "工程问题",
+        "已知比一个数多/少几分之几是多少，求这个数"
+      ],
+      "source": "20-21六年级上·湖南永州·期中",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_296",
+      "original_id": "unit3_solve_296"
+    },
+    {
+      "id": "unit3_solve_297",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "用长 27 米的篱笆围一块长方形地，已知宽是长的 $\\frac{4}{5}$，请问这个长方形的面积是多少平方米？(列方程解决)",
+      "options": [],
+      "answer": "45平方米",
+      "solution": "【分析】设长方形的长为 $x$ 米，则宽为 $\\frac{4}{5} x$ 米，根据长方形周长 $=$ （长 + 宽） $\\times 2$，列出方程求出长方形的长和宽，再根据长方形面积 $=$ 长 $\\times$ 宽，代入数据求解即可。解决本题的关键是先根据长方形周长 $=$ （长 + 宽） $\\times 2$，列出方程求出长方形的长和宽，再根据长方形面积 $=$ 长 $\\times$ 宽求出长方形的面积。\n\n【详解】解：设长方形的长为 $x$ 米，则宽为 $\\frac{4}{5} x$。\n\n$(x + \\frac {4}{5} x) \\times 2 = 2 7$\n\n$\\frac {9}{5} \\mathrm {x} \\times 2 = 2 7 \\\\ \\frac {9}{5} x \\times 2 \\div 2 = 2 7 \\div 2 \\\\ \\frac {9}{5} x = \\frac {2 7}{2} \\\\ \\frac {9}{5} x \\div \\frac {9}{5} = \\frac {2 7}{2} \\div \\frac {9}{5} \\\\ x = \\frac {1 5}{2}$则宽为： $\\frac{4}{5} \\times \\frac{15}{2} = 6$ (米)则长方形的面积是： $\\frac{15}{2} \\times 6 = 45$ (平方米)答：这个长方形的面积是 45 平方米。",
+      "knowledge_points": [
+        "长方形的周长",
+        "长方形的面积",
+        "分数与分数的除法",
+        "列方程解含两个未知数的问题"
+      ],
+      "source": "20-21 六年级上·浙江宁波·期中",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_297",
+      "original_id": "unit3_solve_297"
+    },
+    {
+      "id": "unit3_solve_298",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "一段路，第一周修了全长的 $\\frac{1}{5}$，第二周修了余下的 $\\frac{1}{4}$，这时还剩下300米没有修，这段路全长多少米？",
+      "options": [],
+      "answer": "500 米",
+      "solution": "【分析】这段路的全长 $=$ 还剩下没修的米数 $\\div [1 -$ 第一天修了全长的几分之几一第二天修了全长的几分之几]，第二天修了全长的几分之几 $=$ （1一第一天修了全长的几分之几）×第二天修了余下的几分之几，代入数值计算即可。找到单位'1'，根据分数除法的意义进行解答即可。\n\n【详解】$300 \\div [1 - \\frac{1}{5} - (1 - \\frac{1}{5}) \\times \\frac{1}{4}]=3 0 0 \\div [ 1 - \\frac {1}{5} - \\frac {4}{5} \\times \\frac {1}{4} ]= 3 0 0 \\div [ 1 - \\frac {1}{5} - \\frac {1}{5} ] = 3 0 0 \\div \\frac {3}{5} = 5 0 0 \\text {(米)}$答：这段路全长 500 米。",
+      "knowledge_points": [
+        "分数除法的应用"
+      ],
+      "source": "2022六年级下·全国·专题练习",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_298",
+      "original_id": "unit3_solve_298"
+    },
+    {
+      "id": "unit3_solve_299",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "困难",
+      "question": "学校组织六年级的同学参加美化城市劳动，上午派出60名学生参加，下午派出的学生比余下的学生的 $\\frac{2}{5}$ 多12人，这样学校先后派出150名学生参加劳动，六年级共有多少名学生？",
+      "options": [],
+      "answer": "255名",
+      "solution": "【分析】上午派出 60 名学生，余下的学生就是六年级学生数 -60 人，下午派出的学生比余下的学生的 $\\frac{2}{5}$ 多 12 人，下午派出的学生就是 $\\frac{2}{5} \\times$ (六年级学生数 -60 人) +12 人，这样学校先后派出 150 名学生参加劳动，则上午派出的学生 +下午派出的学生 = 150 人，此题的等量关系为： $60 \\text{人} +\\frac{2}{5} \\times$ （六年级学生数 -60 人）+12 人=150 人，设六年级有 x 名学生，根据等量关系列方程求解。此题考查列方程解决实际问题，关键是根据题意找出等量关系。\n\n【详解】解：设六年级共有 $x$ 名学生\n\n$6 0 + \\frac {2}{5} (x - 6 0) + 1 2 = 1 5 0$\n\n$ \\frac {2}{5} (x - 6 0) = 7 8 \\\\ x - 6 0 = 7 8 \\div \\frac {2}{5} \\\\ x - 6 0 = 1 9 5 \\\\ x = 1 9 5 + 6 0 \\\\ x = 2 5 5$答：六年级共有 255 名学生。",
+      "knowledge_points": [
+        "列方程解含一个未知数的问题"
+      ],
+      "source": "2022 六年级下·全国·专题练习",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_299",
+      "original_id": "unit3_solve_299"
+    },
+    {
+      "id": "unit3_solve_300",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "中国农历中的冬至是一年中白昼最短，黑夜最长的一天。这一天某地白昼时间只有黑夜时间的 $\\frac{5}{7}$。该地冬至这天的白昼和黑夜各有多少小时？(用方程解答)",
+      "options": [],
+      "answer": "白昼有 10 小时，黑夜有 14 小时。",
+      "solution": "【分析】设黑夜的时间有 $x$ 小时，则白昼的时间有 $\\frac{5}{7} x$ 小时，根据一天有 24 小时，据此列方程，解方程即可。本题考查用方程解决问题，明确数量关系是解题的关键。\n\n【详解】解：设黑夜时间有 $x$ 小时，则白昼的时间有 $\\frac{5}{7} x$ 小时。\n\n$\\frac {5}{7} x + x = 2 4 \\\\ \\frac {1 2}{7} x = 2 4 \\\\ x = 1 4$\n\n白昼时间： $24 - 14 = 10$ (小时)答：该地冬至这天的白昼有 10 小时，黑夜有 14 小时。",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "列方程解含两个未知数的问题"
+      ],
+      "source": "21-22 六年级上·云南玉溪·期末",
+      "create_time": "2024-10-11 00:00:00",
+      "merged_id": "unit3_300",
+      "original_id": "unit3_solve_300"
+    },{
+      "id": "unit3_choice_301",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "一款空调降价 $\\frac{1}{4}$ 后是 5200 元, 求这款空调的原价列式正确的是 ( )。",
+      "options": [
+        "A. $5200 \\div \\frac{1}{4}$",
+        "B. $5200 \\div (1 + \\frac{1}{4})$",
+        "C. $5200 \\div (1 - \\frac{1}{4})$"
+      ],
+      "answer": "C",
+      "solution": "【分析】根据题意，一款空调降价 $\\frac{1}{4}$ 后是 5200 元，意思是空调的现价 5200 元比原价降低 $\\frac{1}{4}$，把原价看作单位'1'，则现价是原价的 $(1 - \\frac{1}{4})$，单位'1'未知，用现价除以 $(1 - \\frac{1}{4})$，即是原价。\n\n【详解】$5200 \\div (1 - \\frac{1}{4})=5200 \\div \\frac{3}{4} = 5200 \\times \\frac{4}{3}= \\frac{20800}{3} (\\text{元})$ 这款空调的原价是 $\\frac{20800}{3}$ 元。求这款空调的原价列式正确的是 $5200 \\div (1 - \\frac{1}{4})$。故答案为: C",
+      "knowledge_points": [
+        "已知比一个数多/少几分之几是多少，求这个数"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_301",
+      "original_id": "unit3_choice_301"
+    },
+    {
+      "id": "unit3_choice_302",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "下列算式中, 计算结果大于 8 的是 ( )。",
+      "options": [
+        "A. $\\frac{2}{3} \\times 8$",
+        "B. $\\frac{2}{3} \\div 8$",
+        "C. $8 \\div \\frac{2}{3}$"
+      ],
+      "answer": "C",
+      "solution": "【分析】A. 一个数 (0 除外) 乘小于 1 的数, 积比原来的数小。B. 一个数 (0 除外) 除以大于 1 的数, 商比原来的数小。C. 一个数 (0 除外) 除以小于 1 的数, 商比原来的数大。\n\n【详解】A. $\\frac{2}{3}<1$, 所以 $\\frac{2}{3} \\times 8<8$, 不符合题意; B. $8 > 1$, 则 $\\frac{2}{3} \\div 8 < \\frac{2}{3}$, 则 $\\frac{2}{3} \\div 8 < 8$, 不符合题意; C. $\\frac{2}{3}< 1$, 所以 $8 \\div \\frac{2}{3}>8$, 符合题意。故答案为: C",
+      "knowledge_points": [
+        "因数和积的大小关系（分数乘法）",
+        "被除数与商的大小关系（分数除法）"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_302",
+      "original_id": "unit3_choice_302"
+    },
+    {
+      "id": "unit3_choice_303",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "今年比去年多 $\\frac{1}{3}$, 则今年的产量相当于去年的 ( )。",
+      "options": [
+        "A. $\\frac{2}{3}$",
+        "B. $\\frac{1}{4}$",
+        "C. $\\frac{4}{3}$"
+      ],
+      "answer": "C",
+      "solution": "【分析】把去年的产量看作单位 '1', 则今年的产量是 $1 + \\frac{1}{3} = \\frac{4}{3}$, 求今年的产量相当于去年的几分之几, 用今年的产量除以去年的产量即可。\n\n【详解】$(1 + \\frac{1}{3}) \\div 1= \\frac{4}{3} \\div 1= \\frac{4}{3}$ 所以今年的产量相当于去年的 $\\frac{4}{3}$。故答案为: C",
+      "knowledge_points": [
+        "求一个数占另一个数几分之几",
+        "单位'1'的认识与确定"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_303",
+      "original_id": "unit3_choice_303"
+    },
+    {
+      "id": "unit3_choice_304",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "下面说法错误的是 ( )。",
+      "options": [
+        "A. 1 的倒数是 1",
+        "B. 3 和 $\\frac{1}{3}$ 都是倒数",
+        "C. 6 和 $\\frac{1}{6}$ 互为倒数"
+      ],
+      "answer": "B",
+      "solution": "【分析】乘积是 1 的两个数互为倒数。注意倒数是对两个数而言, 是两个数相互依存的一个概念, 不能单独说某个数是倒数。0 没有倒数, 1 的倒数是 1。\n\n【详解】A．1的倒数是1，原选项说法正确 B. $3 \\text{和} \\frac{1}{3}$ 互为倒数, 不能说 $3 \\text{和} \\frac{1}{3}$ 都是倒数, 原选项说法错误; C. 6 和 $\\frac{1}{6}$ 互为倒数, 原选项说法正确。故答案为: B",
+      "knowledge_points": [
+        "倒数的认识"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_304",
+      "original_id": "unit3_choice_304"
+    },
+    {
+      "id": "unit3_choice_305",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "梨树占果树总棵数的 $\\frac{5}{9}$, 这句话的数量关系可以表示为 ( )。",
+      "options": [
+        "A. 果树总棵数 $\\times \\frac{5}{9} =$ 梨树的棵数",
+        "B. 梨树的棵数 $\\times \\frac{5}{9} =$ 果树总棵数",
+        "C. 果树总棵数 $\\div \\frac{5}{9} =$ 梨树的棵数",
+        "D. 梨树的棵数 $+\\frac{5}{9} =$ 果树总棵数"
+      ],
+      "answer": "A",
+      "solution": "【分析】梨树占果树总棵数的 $\\frac{5}{9}$, 将果树总棵数看作单位 '1', 根据整体数量 $\\times$ 部分对应分率 $=$ 部分数量, 部分数量 $\\div$ 对应分率 $=$ 整体数量, 确定数量关系即可。\n\n【详解】梨树占果树总棵数的 $\\frac{5}{9}$, 根据分析, 这句话的数量关系可以表示为果树总棵数 $\\times \\frac{5}{9} =$ 梨树的棵数, 或梨树的棵数 $\\div \\frac{5}{9} =$ 果树总棵数。故答案为: A",
+      "knowledge_points": [
+        "等式的认识及列等量关系式",
+        "求一个数的几分之几的问题",
+        "已知一个数的几分之几是多少，求这个数"
+      ],
+      "source": "23-24 六年级上·湖南张家界·期中",
+      "merged_id": "unit3_305",
+      "original_id": "unit3_choice_305"
+    },
+    {
+      "id": "unit3_choice_306",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "若 $a \\div \\frac{1}{2} = b \\times \\frac{1}{2} (a \\neq 0)$, 则 ( )。",
+      "options": [
+        "A. a>b",
+        "B. $a = b$",
+        "C. a $<$ b",
+        "D. 无法确定"
+      ],
+      "answer": "C",
+      "solution": "【分析】求一个分数的倒数，只要把这个分数的分子、分母调换位置。设 $a \\div \\frac{1}{2} = b \\times \\frac{1}{2} = 1$, 则 $a = 1 \\times \\frac{1}{2} = \\frac{1}{2}$; $b$ 和 $\\frac{1}{2}$ 互为倒数, $b = 2$。据此解答。\n\n【详解】通过分析可得：设 $a \\div \\frac{1}{2} = b \\times \\frac{1}{2} = 1$, 则 $a = \\frac{1}{2}, b = 2$。$\\frac{1}{2} < 2$, 则 $a < b$。故答案为: C",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "整数乘分数",
+        "倒数的认识"
+      ],
+      "source": "24-25 六年级上·全国·单元测试",
+      "merged_id": "unit3_306",
+      "original_id": "unit3_choice_306"
+    },
+    {
+      "id": "unit3_choice_307",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "下列各组数中, ( ) 互为倒数。",
+      "options": [
+        "A. 0.8和 $\\frac{4}{5}$",
+        "B. $\\frac{1}{3}$ 和 0",
+        "C. $\\frac{1}{2}$ 和 2",
+        "D. $\\frac{7}{10}$ 和 $\\frac{1}{7}$"
+      ],
+      "answer": "C",
+      "solution": "【分析】根据倒数的定义：乘积是 1 的两个数互为倒数，例如：如果 $a$、$b$ 不为 0，$a \\times b = 1$，则 $a$ 是 $b$ 的倒数，$b$ 是 $a$ 的倒数。注意 0 没有倒数，据此判断每个选项即可。\n\n【详解】A. $0.8 \\times \\frac{4}{5}=\\frac{4}{5} \\times \\frac{4}{5} = \\frac{16}{25}$ 0.8 和 $\\frac{4}{5}$ 并不互为倒数; B. 0 没有倒数, 所以 $\\frac{1}{3}$ 和 0 不符合; C. $\\frac{1}{2} \\times 2 = 1$ $\\frac{1}{2}$ 和 2 互为倒数; D. $\\frac{7}{10} \\times 1\\frac{2}{7}$ $\\frac{7}{10} \\times \\frac{9}{7} = \\frac{9}{10}$ $\\frac{7}{10}$ 和 $1\\frac{2}{7}$ 并不互为倒数。故答案为: C",
+      "knowledge_points": [
+        "分数乘分数",
+        "分数乘小数",
+        "倒数的认识"
+      ],
+      "source": "24-25 六年级上·全国·单元测试",
+      "merged_id": "unit3_307",
+      "original_id": "unit3_choice_307"
+    },
+    {
+      "id": "unit3_choice_308",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "一个蓄水池，有甲、乙两个进水管，单开甲管2小时将水注满，单开乙管3小时将水注满。两管同时开放，（）小时能将水注满。",
+      "options": [
+        "A. 1",
+        "B. 1.2",
+        "C. 无法确定"
+      ],
+      "answer": "B",
+      "solution": "【分析】首先确定将蓄水池容积看作单位'1'。由于单开甲管2小时注满，所以甲管的注水效率为 $1 \\div 2 = \\frac{1}{2}$；单开乙管3小时注满，乙管的注水效率为 $1 \\div 3 = \\frac{1}{3}$。两管同时开放时，注水效率为甲、乙两管效率之和，即 $\\frac{1}{2} + \\frac{1}{3} = \\frac{3}{6} + \\frac{2}{6} = \\frac{5}{6}$，然后根据时间 $=$ 工作量 $\\div$ 工作效率，用 $1 \\div \\frac{5}{6}$，求出几小时能将水注满。\n\n【详解】$1 \\div 2 = \\frac{1}{2}$ $1 \\div 3 = \\frac{1}{3}$ $\\frac{1}{2} + \\frac{1}{3} = \\frac{3}{6} + \\frac{2}{6} = \\frac{5}{6}$ $1 \\div \\frac{5}{6} = 1 \\times \\frac{6}{5} = 1.2 (\\text{小时})$ 两管同时开放，1.2 小时能将水注满。故答案为: B",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "两人合作的工程问题",
+        "单位'1'的认识与确定"
+      ],
+      "source": "24-25 六年级上·全国·课后作业",
+      "merged_id": "unit3_308",
+      "original_id": "unit3_choice_308"
+    },
+    {
+      "id": "unit3_choice_309",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "妈妈买回苹果和橘子共 $6kg$, 其中苹果是橘子的 $\\frac{2}{3}$。橘子和苹果各买回多少千克?如果设买回苹果 $x \\mathrm{kg}$, 列式为 ( ) ; 如果设买回橘子 $x \\mathrm{kg}$, 列式为 ( )。",
+      "options": [
+        "A. $x + \\frac{2}{3} x = 6, \\frac{2}{3} x + \\frac{3}{2} x = 6$",
+        "B. $x + \\frac{3}{2} x = 6, x +\\frac{2}{3} x = 6$",
+        "C. $x + \\frac{3}{2} x = 6, \\frac{2}{3} x + \\frac{3}{2} x = 6$"
+      ],
+      "answer": "B",
+      "solution": "【分析】根据题意, 如果设买回苹果 $xkg$, 把买回的橘子看作单位 '1', 买回苹果是橘子的 $\\frac{2}{3}$, 对应的是买回苹果 $\\mathrm{xkg}$, 求单位 '1', 用买回的苹果 $\\div \\frac{2}{3}$, 求买回的橘子, 即 $x \\div \\frac{2}{3} = \\frac{3}{2} \\mathrm{xkg}$; 买回苹果和橘子共 $6kg$, 列方程: $x + \\frac{3}{2} x = 6$; 如果设买回桔子 $xkg$, 把买回桔子看作单位 '1', 买回苹果是橘子 $\\frac{2}{3}$, 用买回橘子质量 $\\times \\frac{2}{3}$, 求出买回苹果的质量, 即 $\\frac{2}{3} xkg$, 买回苹果和橘子共 $6kg$, 列方程: $x + \\frac{2}{3} x = 6$, 据此解答。\n\n【详解】根据分析可知，妈妈买回苹果和橘子共 $6kg$，其中苹果是橘子的 $\\frac{2}{3}$。橘子和苹果各买回多少千克？如果设买回苹果 $\\mathrm{xkg}$, 列式为 $x + \\frac{3}{2} x = 6$; 如果设买回橘子 $\\mathrm{xkg}$, 列式为 $x + \\frac{2}{3} x = 6$。故答案为: B",
+      "knowledge_points": [
+        "求一个数的几分之几的问题",
+        "已知一个数的几分之几是多少，求这个数",
+        "列方程解含两个未知数的问题"
+      ],
+      "source": "24-25 六年级上·全国·课后作业",
+      "merged_id": "unit3_309",
+      "original_id": "unit3_choice_309"
+    },
+    {
+      "id": "unit3_choice_310",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "下面算式中, 商大于被除数的是 ( )。",
+      "options": [
+        "A. $\\frac{6}{7} \\div 3$",
+        "B. $6 \\div \\frac{5}{4}$",
+        "C. $\\frac{4}{5} \\div \\frac{4}{5}$"
+      ],
+      "answer": "C",
+      "solution": "【分析】一个数（0 除外）除以大于 1 的数，结果比原来的数小；一个数（0 除外）除以小于 1 的数，结果比原来的数大。一个数（0 除外）除以这个数的本身等于 1。据此选择即可。本题考查分数除法，明确商与被除数之间的关系是解题的关键。\n\n【详解】A. 因为 $3 > 1$, 所以 $\\frac{6}{7} \\div 3 < \\frac{6}{7}$, 商小于被除数; B. 因为 $\\frac{5}{4} > 1$, 所以 $6 \\div \\frac{5}{4}< 6$, 商小于被除数; C. $\\frac{4}{5} \\div \\frac{4}{5} = 1 > \\frac{4}{5}$, 商大于被除数。故答案为: C",
+      "knowledge_points": [
+        "被除数和商的大小关系（小数除法）",
+        "分数与整数的除法",
+        "分数与分数的除法"
+      ],
+      "source": "19-20 六年级下·湖南株洲·期末",
+      "merged_id": "unit3_310",
+      "original_id": "unit3_choice_310"
+    },
+    {
+      "id": "unit3_choice_311",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "在暑期夏令营义卖活动中，'扬帆班'筹得义卖款860元，比'启航班'少 $\\frac{1}{3}$。'启航班'筹得义卖款多少元？正确解答方式是（）。",
+      "options": [
+        "A. $860 \\times (1 - \\frac{1}{3})$",
+        "B. 解：设'启航班'筹得义卖款 $x$ 元, $(1 + \\frac{1}{3}) x = 860$",
+        "C. $860 \\div (1 + \\frac{1}{3})$",
+        "D. 解: 设 '启航班' 筹得义卖款 $x$ 元, $(1 - \\frac{1}{3}) x = 860$"
+      ],
+      "answer": "D",
+      "solution": "【分析】方法一：把'启航班'筹得义卖款的钱数看作单位'1'，'扬帆班'筹得义卖款是'启航班'的（$1 - \\frac{1}{3}$），对应的是'杨航班'筹得义卖款860元，求单位'1'，用 $860 \\div (1 - \\frac{1}{3})$ 解答。方法二：设'启航班'筹得义卖款 $x$ 元，'杨航班'筹得义卖款比'启航班'少 $\\frac{1}{3}$，即'杨航班'筹得义卖款是'启航班'的 $(1 - \\frac{1}{3})$，用'启航班'筹得义卖款 $\\times (1 - \\frac{1}{3}) =$ '扬帆班'筹得义卖款，列方程：$x \\times (1 - \\frac{1}{3}) = 860$，据此解答。\n\n【详解】方法一：$860 \\div (1 - \\frac{1}{3}) = 860 \\div \\frac{2}{3}= 860 \\times \\frac{3}{2} = 1290 (\\text{元})$ 方法二:解：设'启航班'筹得义卖款 $x$ 元。$(1 - \\frac{1}{3}) x = 860$ $\\frac{2}{3} x = 860$ $x = 860 \\div \\frac{2}{3}$ $x = 860 \\times \\frac{3}{2}$ $x =1290$ 在暑期夏令营义卖活动中，'扬帆班'筹得义卖款860元，比'启航班'少 $\\frac{1}{3}$。'启航班'筹得义卖款多少元？正确解答方式是 $860 \\div (1 - \\frac{1}{3})$ 或解：设'启航班'筹得义卖款 $x$ 元。$x \\times (1 - \\frac{1}{3}) = 860$。故答案为: D",
+      "knowledge_points": [
+        "已知比一个数多/少几分之几是多少，求这个数",
+        "列方程解含一个未知数的问题"
+      ],
+      "source": "23-24 六年级上·福建莆田·期中",
+      "merged_id": "unit3_311",
+      "original_id": "unit3_choice_311"
+    },
+    {
+      "id": "unit3_choice_312",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "加工同一种零件,甲 $\\frac{4}{5}$ 小时加工 80 个, 乙加工 20 个要用 15 分钟, 甲、乙加工零件的速度相比, ( )。",
+      "options": [
+        "A. 甲快",
+        "B. 乙快",
+        "C. 一样快",
+        "D. 无法比较"
+      ],
+      "answer": "A",
+      "solution": "【分析】根据工作效率 $=$ 工作总量 $\\div$ 工作时间, 求出甲、乙平均 1 小时加工零件的个数, 再进行大小比较, 即可解答。\n\n【详解】15 分钟 $= \\frac{1}{4}$ 小时 甲: $80 \\div \\frac{4}{5} = 80 \\times \\frac{5}{4} = 100$ (个) 乙: $20 \\div \\frac{1}{4} = 20 \\times 4 = 80$ (个)​, 因此甲、乙加工零件的速度相比, 甲快。$100 > 80$, 因此甲、乙加工零件的速度相比, 甲快。故答案为: A",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "有具体量的工程问题",
+        "时、分的认识及换算"
+      ],
+      "source": "23-24 六年级上·全国·课后作业",
+      "merged_id": "unit3_312",
+      "original_id": "unit3_choice_312"
+    },
+    {
+      "id": "unit3_choice_313",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "王叔叔和李叔叔合开了一家公司，王叔叔投资240万元，____，李叔叔投资多少万元？如果设李叔叔投资 $x$ 万，列式为 $(1 + \\frac{1}{7}) x = 240$，那么横线上应该填的条件是( )。",
+      "options": [
+        "A. 李叔叔的投资比王叔叔多 $\\frac{1}{7}$",
+        "B. 比李叔叔的投资多 $\\frac{1}{7}$",
+        "C. 比李叔叔的投资少 $\\frac{1}{7}$"
+      ],
+      "answer": "B",
+      "solution": "【分析】从'如果设李叔叔投资 $x$ 万元，列式为 $(1 + \\frac{1}{7}) x = 240$'可得等量关系：李叔叔的投资 $\\times (1 + \\frac{1}{7}) =$ 王叔叔的投资，根据等量关系，逐项判断即可。\n\n【详解】A. 从 '李叔叔的投资比王叔叔多 $\\frac{1}{7}$' 可知, 以王叔叔的投资为单位 '1', 李叔叔的投资相当于王叔叔的投资的 $(1 + \\frac{1}{7})$, 用王叔叔的投资 $\\times (1 + \\frac{1}{7}) =$ 李叔叔的投资。该选项不符合题意。B. 从 '比李叔叔的投资多 $\\frac{1}{7}$' 可知, 以李叔叔的投资为单位 '1', 王叔叔的投资相当于李叔叔的投资的 $(1 + \\frac{1}{7})$, 用李叔叔的投资 $\\times (1 + \\frac{1}{7}) =$ 王叔叔的投资。该选项符合题意。C. 从 '比李叔叔的投资少 $\\frac{1}{7}$' 可知, 以李叔叔的投资为单位 '1', 王叔叔的投资相当于李叔叔的投资的 $(1 - \\frac{1}{7})$, 用李叔叔的投资 $\\times (1 - \\frac{1}{7}) =$ 王叔叔的投资。该选项不符合题意。故答案为: B",
+      "knowledge_points": [
+        "单位'1'的认识与确定",
+        "求比一个数多/少几分之几的数是多少",
+        "已知比一个数多/少几分之几是多少, 求这个数",
+        "'提问题' '填条件' 问题"
+      ],
+      "source": "23-24 六年级上·辽宁·单元测试",
+      "merged_id": "unit3_313",
+      "original_id": "unit3_choice_313"
+    },
+    {
+      "id": "unit3_choice_314",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "已知 $a > 0$, 下面各式中, 结果最大的是 ( )。",
+      "options": [
+        "A. $a \\times \\frac{5}{4}$",
+        "B. $a \\times \\frac{1}{4}$",
+        "C. $a \\times 0$",
+        "D. $a \\div \\frac{4}{7}$"
+      ],
+      "answer": "D",
+      "solution": "【分析】计算出每个算式的结果，再根据与 a 相乘的数的大小，确定结果的大小即可。\n\n【详解】$a \\div \\frac{4}{7} = a \\times \\frac{7}{4}$ $a \\times 0 = 0$ $\\frac{7}{4} > \\frac{5}{4} > \\frac{1}{4}$ 所以 $a \\div \\frac{4}{7} > a \\times \\frac{5}{4} > a \\times \\frac{1}{4} > a \\times 0$, 结果最大的是 $a \\div \\frac{4}{7}$。故答案为: D",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "分数乘分数",
+        "含有字母式子的化简与求值"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_314",
+      "original_id": "unit3_choice_314"
+    },
+    {
+      "id": "unit3_choice_315",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "一台榨油机 $\\frac{9}{5}$ 小时榨油 $\\frac{3}{4} t$, 求平均每小时榨油多少吨, 正确的列式是 ( )。",
+      "options": [
+        "A. $\\frac{9}{5} \\div \\frac{3}{4}$",
+        "B. $\\frac{3}{4} \\div \\frac{9}{5}$",
+        "C. $\\frac{9}{5} \\times \\frac{3}{4}$",
+        "D. $\\frac{4}{3} \\div \\frac{9}{5}$"
+      ],
+      "answer": "B",
+      "solution": "【分析】用榨油总量除以榨油时间，求出平均每小时榨油多少吨，据此列式计算即可。\n\n【详解】根据分析可知, 平均每小时榨油: $\\frac{3}{4} \\div \\frac{9}{5}$。故答案为: B",
+      "knowledge_points": [
+        "分数与分数的除法"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_315",
+      "original_id": "unit3_choice_315"
+    },
+    {
+      "id": "unit3_choice_316",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "水结冰后, 体积增加 $\\frac{1}{10}$, 4 立方分米的冰融化后的体积是 ( ) 升。",
+      "options": [
+        "A. $\\frac{11}{10}$",
+        "B. $\\frac{10}{11}$",
+        "C. $\\frac{20}{11}$",
+        "D. $\\frac{40}{11}$"
+      ],
+      "answer": "D",
+      "solution": "【分析】把水的体积看作单位 '1', 水结冰后, 体积增加 $\\frac{1}{10}$, 即冰的体积比水的体积增加 $\\frac{1}{10}$, 冰的体积是水的体积的 (1 + $\\frac{1}{10}$), 根据已知一个数比另一个数多几分之几的数是多少,求这个数, 用除法计算, 据此解答。\n\n【详解】$4 \\div (1 + \\frac{1}{10})= 4 \\div \\frac{11}{10} = 4 \\times \\frac{10}{11} = \\frac{40}{11} (\\text{立方分米})$ $\\frac{40}{11}$ 立方分米 $= \\frac{40}{11}$ 升 即 4 立方分米的冰融化后的体积是 $\\frac{40}{11}$ 升。故答案为: D",
+      "knowledge_points": [
+        "已知比一个数多/少几分之几是多少, 求这个数"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_316",
+      "original_id": "unit3_choice_316"
+    },
+    {
+      "id": "unit3_choice_317",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "一个衣柜降价 $\\frac{1}{6}$ 后比原价少 520 元, 这个衣柜原价是多少钱? 正确的列式是 ()。",
+      "options": [
+        "A. $520 \\div \\frac{1}{6}$",
+        "B. $520 \\times \\frac{1}{6}$",
+        "C. $520 \\div (1 - \\frac{1}{6})$",
+        "D. $520 \\times (1 + \\frac{1}{6})$"
+      ],
+      "answer": "A",
+      "solution": "【分析】现在的售价比原来降低了 $\\frac{1}{6}$, 现在比原来少了 520 元，用对应量除以对应份数就是原价。\n\n【详解】$520 \\div \\frac{1}{6}=520 \\times 6 = 3120 (\\text{元})$ 这个衣柜原价是 3120 元, 正确的列式是 $520 \\div \\frac{1}{6}$。故答案为: A",
+      "knowledge_points": [
+        "分数与整数的除法"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_317",
+      "original_id": "unit3_choice_317"
+    },
+    {
+      "id": "unit3_choice_318",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "某畜牧场山羊的数量是 480 只的 $\\frac{5}{6}$, 绵羊数量的 $\\frac{8}{7}$ 是 400 只, 山羊和绵羊的数量相比, ( )。",
+      "options": [
+        "A. 山羊多",
+        "B. 绵羊多",
+        "C. 一样多",
+        "D. 无法判断"
+      ],
+      "answer": "A",
+      "solution": "【分析】把山羊的总数量看作单位 '1', 求它的 $\\frac{5}{6}$ 是多少只,用 $480 \\times \\frac{5}{6}$, 求出山羊的总数量; 把绵羊的总数量看作单位 '1', 它的 $\\frac{8}{7}$ 对应的是 400 只, 求单位 '1', 用 $400 \\div \\frac{8}{7}$, 求出绵羊的只数, 再进行比较, 即可解答。\n\n【详解】山羊: $480 \\times \\frac{5}{6} = 400$ (只) 绵羊: $400 \\div \\frac{8}{7}= 400 \\times \\frac{7}{8} = 350 (\\text{只})$ $400 > 350$，山羊多。某畜牧场山羊的数量是 480 只的 $\\frac{5}{6}$, 绵羊数量的 $\\frac{8}{7}$ 是 400 只,山羊和绵羊的数量相比, 山羊多。故答案为: A",
+      "knowledge_points": [
+        "求一个数的几分之几的问题",
+        "已知一个数的几分之几是多少，求这个数"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_318",
+      "original_id": "unit3_choice_318"
+    },
+    {
+      "id": "unit3_choice_319",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "与 $\\frac{16}{45} \\div \\frac{2}{3} \\times \\frac{3}{5}$ 的结果不相等的是 ( )。",
+      "options": [
+        "A. $\\frac{16}{45} \\times \\frac{3}{5} \\div \\frac{2}{3}$",
+        "B. $\\frac{16}{45} \\times \\frac{3}{2} \\times \\frac{3}{5}$",
+        "C. $\\frac{16}{45} \\div (\\frac{2}{3} \\times \\frac{3}{5})$",
+        "D. $\\frac{16}{45} \\times \\frac{3}{5} \\times \\frac{3}{2}$"
+      ],
+      "answer": "C",
+      "solution": "【分析】根据分数乘除的计算方法，分别求出题干与各项算式的值，再进行对比即可。\n\n【详解】$\\frac{16}{45} \\div \\frac{2}{3} \\times \\frac{3}{5}= \\frac{16}{45} \\times \\frac{3}{2} \\times \\frac{3}{5} = \\frac{8}{15} \\times \\frac{3}{5} = \\frac{8}{25}$ A. $\\frac{16}{45} \\times \\frac{3}{5} \\div \\frac{2}{3}=\\frac{16}{45} \\times \\frac{3}{5} \\times \\frac{3}{2} = \\frac{16}{75} \\times \\frac{3}{2} = \\frac{8}{25}$ B. $\\frac{16}{45} \\times \\frac{3}{2} \\times \\frac{3}{5}=\\frac{8}{15} \\times \\frac{3}{5} = \\frac{8}{25}$ C. $\\frac{16}{45} \\div (\\frac{2}{3} \\times \\frac{3}{5})=\\frac{16}{45} \\div \\frac{2}{5} = \\frac{16}{45} \\times \\frac{5}{2} = \\frac{8}{9}$ D. $\\frac{16}{45} \\times \\frac{3}{5} \\times \\frac{3}{2}=\\frac{8}{15} \\times \\frac{3}{5} = \\frac{8}{25}$ 故答案为: C",
+      "knowledge_points": [
+        "分数的乘、除法的混合运算"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_319",
+      "original_id": "unit3_choice_319"
+    },
+    {
+      "id": "unit3_choice_320",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "a，b，c均不为0，且已知 $a \\times \\frac{6}{7} = b \\div \\frac{6}{7} = c \\div \\frac{1}{7}$。将a，b，c按从大到小的顺序排列为（）。",
+      "options": [
+        "A. a>b>c",
+        "B. b>a>c",
+        "C. c>a>b",
+        "D. c>b>a"
+      ],
+      "answer": "A",
+      "solution": "【分析】设 $a \\times \\frac{6}{7} = b \\div \\frac{6}{7} = c \\div \\frac{1}{7} = 1$, 分别求出 a、b、c 的值,再进行比较, 即可解答。\n\n【详解】设 $a \\times \\frac{6}{7} = b \\div \\frac{6}{7} = c \\div \\frac{1}{7} = 1$ $a \\times \\frac{6}{7} = 1$ $a = 1 \\div \\frac{6}{7}$ $a = 1 \\times \\frac{7}{6}$ $a=\\frac{7}{6}$ $b \\div \\frac{6}{7} = 1$ $b = 1 \\times \\frac{6}{7}$ $b = \\frac{6}{7}$ $c \\div \\frac{1}{7} = 1$ $c = 1 \\times \\frac{1}{7}$ $c = \\frac{1}{7}$ $\\frac{7}{6} > \\frac{6}{7} > \\frac{1}{7}$, 即 $a > b > c$。a, b, c 均不为 0, 且已知 $a \\times \\frac{6}{7} = b \\div \\frac{6}{7} = c \\div \\frac{1}{7}$。将 a, b, c 按从大到小的顺序排列为 $a > b > c$。故答案为: A",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "整数乘分数",
+        "用字母表示数",
+        "数量关系"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_320",
+      "original_id": "unit3_choice_320"
+    },
+    {
+      "id": "unit3_choice_321",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "一个数 (0 除外) 除以 $\\frac{1}{3}$, 那么这个数就 ( )。",
+      "options": [
+        "A. 缩小到原来的 $\\frac{1}{3}$",
+        "B. 扩大到原来的 3 倍",
+        "C. 不变",
+        "D. 可能变大也可能变小"
+      ],
+      "answer": "B",
+      "solution": "【分析】除以一个数相当于乘这个数的倒数，一个数（0 除外）除以 $\\frac{1}{3}$，相当于是给这个数乘 3，那么将扩大到原来的 3 倍。据此解答。\n\n【详解】如：$5 \\div \\frac{1}{3} = 5 \\times 3 = 15$ 一个数 (0 除外) 除以 $\\frac{1}{3}$, 那么这个数就扩大到原来的 3 倍。故答案为: B",
+      "knowledge_points": [
+        "分数与整数的除法"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_321",
+      "original_id": "unit3_choice_321"
+    },
+    {
+      "id": "unit3_choice_322",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "下面各组数中, 互为倒数的一组是 ( )。",
+      "options": [
+        "A. $\\frac{1}{8}$ 和 $\\frac{7}{8}$",
+        "B. $\\frac{13}{10}$ 和 $\\frac{3}{10}$",
+        "C. $\\frac{1}{4}$ 和 4",
+        "D. 0 和 1"
+      ],
+      "answer": "C",
+      "solution": "【分析】乘积为 1 的两个数互为倒数，据此选项验证即可。\n\n【详解】A. $\\frac{1}{8} \\times \\frac{7}{8} = \\frac{7}{64}, \\frac{7}{64} \\neq 1$，所以 $\\frac{1}{8}$ 和 $\\frac{7}{8}$ 不互为倒数；B. $\\frac{13}{10} \\times \\frac{3}{10} = \\frac{39}{100}, \\frac{39}{100} \\neq 1$, 所以 $\\frac{13}{10}$ 和 $\\frac{3}{10}$ 不互为倒数; C. $\\frac{1}{4} \\times 4 = 1, 1 = 1$, 所以 $\\frac{1}{4}$ 和 4 互为倒数; D. $0 \\times 1 = 0, 0 \\neq 1$, 所以 0 和 1 不互为倒数; 故答案为: C。",
+      "knowledge_points": [
+        "分数乘分数",
+        "倒数的认识"
+      ],
+      "source": "23-24 五年级下·陕西渭南·期末",
+      "merged_id": "unit3_322",
+      "original_id": "unit3_choice_322"
+    },
+    {
+      "id": "unit3_fill_323",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "( )吨比 120 吨多它的 $\\frac{1}{4}$; 60 吨比( )吨少它的 $\\frac{1}{4}$。",
+      "options": [],
+      "answer": "150; 80",
+      "solution": "【分析】求一个数的几分之几是多少，用乘法，据此用 120 乘 $\\frac{1}{4}$，求出比 120 吨多 $\\frac{1}{4}$ 是多少吨，再加上 120 吨就是多少吨比 120 吨多它的 $\\frac{1}{4}$；求 60 吨比多少吨少它的 $\\frac{1}{4}$，即已知一个数的 $(1 - \\frac{1}{4})$ 是 60 吨，求这个数用除法解答，列式为 $60 \\div (1 - \\frac{1}{4})$。\n\n【详解】$120 + 120 \\times \\frac{1}{4}= 120 + 30 = 150 (\\text{吨})$ $60 \\div (1 - \\frac{1}{4})=60 \\div \\frac{3}{4} = 60 \\times \\frac{4}{3} = 80 \\text{(吨)}$ 所以 150 吨比 120 吨多它的 $\\frac{1}{4}$, 60 吨比 80 吨少它的 $\\frac{1}{4}$。",
+      "knowledge_points": [
+        "求比一个数多/少几分之几的数是多少",
+        "已知比一个数多/少几分之几是多少，求这个数"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_323",
+      "original_id": "unit3_fill_323"
+    },
+    {
+      "id": "unit3_fill_324",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "36 是 40 的( )；( )是 48 的 $\\frac{1}{4}$; 45 是( )的 $\\frac{1}{5}$。",
+      "options": [],
+      "answer": "$\\frac{9}{10}$; 12; 225",
+      "solution": "【分析】求 36 是 40 的几分之几, 用 36 除以 40 即可; 求 48 的 $\\frac{1}{4}$ 是多少, 用 48 乘 $\\frac{1}{4}$ 即可; 已知一个数的 $\\frac{1}{5}$, 求这个数,用 45 除以 $\\frac{1}{5}$ 即可。\n\n【详解】$36 \\div 40 = \\frac{9}{10}$ $48 \\times \\frac{1}{4} = 12$ $45 \\div \\frac{1}{5} = 45 \\times 5 = 225$ 36 是 40 的 $\\frac{9}{10}$; 12 是 48 的 $\\frac{1}{4}$; 45 是 225 的 $\\frac{1}{5}$。",
+      "knowledge_points": [
+        "求一个数的几分之几的问题",
+        "已知一个数的几分之几是多少，求这个数",
+        "求一个数占另一个数几分之几"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_324",
+      "original_id": "unit3_fill_324"
+    },
+    {
+      "id": "unit3_fill_325",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "把 $\\frac{5}{6}$ 米的绳子剪成每段长 $\\frac{5}{24}$ 米的小段, 共剪成 ( ) 段, 每段占全长的 $\\frac{()}{()}$。",
+      "options": [],
+      "answer": "4; $\\frac{1}{4}$",
+      "solution": "【分析】根据题意, 用绳子的全长除以每段的长度, 即可求出共剪成的段数; 把绳子的全长看作单位 '1', 用 1 除以剪成的段数, 即是每一段占全长的几分之几。\n\n【详解】$\\frac{5}{6} \\div \\frac{5}{24}= \\frac{5}{6} \\times \\frac{24}{5} = 4 \\text{(段)}$ $1 \\div 4 = \\frac{1}{4}$ 共剪成 4 段, 每段占全长的 $\\frac{1}{4}$。",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "分数的意义"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_325",
+      "original_id": "unit3_fill_325"
+    },
+    {
+      "id": "unit3_fill_326",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "不计算, 在括号里填上 \"<\" \">\" 或 \"=\"。\n\n① $\\frac{7}{15} \\div \\frac{2}{5}$ () $\\frac{7}{15}$\n\n② $\\frac{3}{7} \\div 3$ () $\\frac{3}{7}$\n\n③ $18 \\div \\frac{2}{3}$ () $18 \\div 2 \\times 3$",
+      "options": [],
+      "answer": ">; <; =",
+      "solution": "【分析】①一个数（0 除外）除以小于 1 的数，商比原来的数大。②一个数 (0 除外) 除以大于 1 的数, 商比原来的数小。③先根据分数除法的计算法则算出 $18 \\div \\frac{2}{3}$ 的结果, 再与 $18 \\div 2 \\times 3$ 的计算结果进行比较, 得出结论。\n\n【详解】① $\\frac{2}{5} < 1$, 所以 $\\frac{7}{15} \\div \\frac{2}{5} > \\frac{7}{15}$; ② $3 > 1$, 所以 $\\frac{3}{7} \\div 3 < \\frac{3}{7}$; ③ $18 \\div \\frac{2}{3} = 18 \\times \\frac{3}{2} = 27$ $18 \\div 2 \\times 3 = 9 \\times 3 = 27$ $27 = 27$, 所以 $18 \\div \\frac{2}{3} = 18 \\div 2 \\times 3$。",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "分数与分数的除法",
+        "被除数与商的大小关系（分数除法）"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_326",
+      "original_id": "unit3_fill_326"
+    },
+    {
+      "id": "unit3_fill_327",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "( ) $\\times \\frac{7}{4} = 8 \\times$ ( ) $= 0.25 \\times$ ( ) $= 1$。",
+      "options": [],
+      "answer": "$\\frac{4}{7}$; $\\frac{1}{8}$; 4",
+      "solution": "【分析】乘积是 1 的两个数互为倒数。求一个真分数或假分数的倒数，只需要将分子、分母交换位置即可。求整数 (0 除外) 的倒数时, 先把整数看作分母是 1 的假分数再交换分子、分母的位置。小数求倒数时，先把小数化成最简真分数或假分数，再按分数求倒数的方法求解。\n\n【详解】$\\frac{7}{4}$ 的倒数是 $\\frac{4}{7}$, 所以 $\\frac{4}{7} \\times \\frac{7}{4} = 1$; 8的倒数是$\\frac{1}{8}$,所以$8 \\times \\frac{1}{8}=1$; 0.25=$\\frac{1}{4}$,$\\frac{1}{4}$的倒数是4，所以0.25$\\times$4=1 即 $\\frac{4}{7} \\times \\frac{7}{4} = 8 \\times \\frac{1}{8} = 0.25 \\times 4 = 1$。",
+      "knowledge_points": [
+        "一位或多位小数化分数（约分）",
+        "倒数的认识"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_327",
+      "original_id": "unit3_fill_327"
+    },
+    {
+      "id": "unit3_fill_328",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "10 的倒数是( )；0.4 与( )互为倒数。",
+      "options": [],
+      "answer": "$\\frac{1}{10}$; $\\frac{5}{2}$",
+      "solution": "【分析】乘积是 1 的两个数互为倒数；10 化为 $\\frac{10}{1}$；把小数化为分数 $0.4 = \\frac{2}{5}$，把分子分母调换位置，即可求出倒数，据此解答。\n\n【详解】$10 = \\frac{10}{1}$, $\\frac{10}{1}$ 的倒数是 $\\frac{1}{10}$。$0.4 = \\frac{2}{5}$; $\\frac{2}{5}$ 的倒数是 $\\frac{5}{2}$。10 的倒数是 $\\frac{1}{10}$; 0.4 与 $\\frac{5}{2}$ 互为倒数。",
+      "knowledge_points": [
+        "一位或多位小数化分数（约分）",
+        "倒数的认识"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_328",
+      "original_id": "unit3_fill_328"
+    },
+    {
+      "id": "unit3_fill_329",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "甲数的 $\\frac{1}{4}$ 等于乙数的 $\\frac{1}{5}$, 如甲数是60，那么乙数是( )。",
+      "options": [],
+      "answer": "75",
+      "solution": "【分析】先把甲数看作单位'1'，求甲数的 $\\frac{1}{4}$ 是多少，单位'1'已知，用甲数乘 $\\frac{1}{4}$ 求解；再把乙数看作单位 '1', 甲数的 $\\frac{1}{4}$ 等于乙数的 $\\frac{1}{5}$, 单位 '1'未知, 用甲数的 $\\frac{1}{4}$ 除以 $\\frac{1}{5}$, 即可求出乙数。\n\n【详解】$60 \\times \\frac{1}{4} \\div \\frac{1}{5}=15 \\div \\frac{1}{5} = 15 \\times 5 = 75$ 那么乙数是 $75$。",
+      "knowledge_points": [
+        "分数的乘、除法的混合运算",
+        "求一个数的几分之几的问题",
+        "已知一个数的几分之几是多少，求这个数"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_329",
+      "original_id": "unit3_fill_329"
+    },
+    {
+      "id": "unit3_fill_330",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "一辆小汽车行驶 $\\frac{5}{2}$ 千米用了 $\\frac{5}{24}$ 升汽油。照这样计算, 1 升汽油能行驶 ( ) 千米,行驶 1 千米要用汽油 ( ) 升。",
+      "options": [],
+      "answer": "12; $\\frac{1}{12}$",
+      "solution": "【分析】用行驶的千米数除以耗油量即可求出平均每升汽油可以行驶多少千米；用耗油量除以千米数求出1千米耗油多少升。\n\n【详解】$\\frac{5}{2} \\div \\frac{5}{24}= \\frac{5}{2} \\times \\frac{24}{5}= 12$ (千米) $\\frac{5}{24} \\div \\frac{5}{2}= \\frac{5}{24} \\times \\frac{2}{5}= \\frac{1}{12}$(升) 一辆小汽车行驶 $\\frac{5}{2}$ 千米用了 $\\frac{5}{24}$ 升汽油。照这样计算, 1 升汽油能行驶 12 千米, 行驶 1 千米要用汽油 $\\frac{1}{12}$ 升。",
+      "knowledge_points": [
+        "分数与分数的除法"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_330",
+      "original_id": "unit3_fill_330"
+    },
+    {
+      "id": "unit3_fill_331",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "在括号里填上\">\" \"<\"或\"=\"。\n\n$\\frac{2}{3} \\div \\frac{5}{7}$ ( ) $\\frac{2}{3}$ $\\frac{3}{4} \\times \\frac{1}{9}$ ( ) $\\frac{3}{4} \\div 9$ $\\frac{3}{8} \\div \\frac{6}{5}$ ( ) $\\frac{3}{8}$ $\\frac{1}{4} \\div \\frac{5}{8}$ ( ) $\\frac{5}{8} \\div \\frac{1}{4}$",
+      "options": [],
+      "answer": ">; =; <; <",
+      "solution": "【分析】第一、三小题：一个非 0 数，除以大于 1 的数，商小于被除数；一个非 0 数，除以小于 1 的数，商大于被除数；据此解答；第二小题：把除法换算成乘法，再进行比较，即可解答；第四小题：计算出两边算式的结果，再进行比较，据此解答。\n\n【详解】$\\frac{2}{3} \\div \\frac{5}{7}$ 和 $\\frac{2}{3}$ 因为 $\\frac{5}{7}< 1$, 所以 $\\frac{2}{3} \\div \\frac{5}{7}> \\frac{2}{3}$ $\\frac{3}{4} \\times \\frac{1}{9}$ 和 $\\frac{3}{4} \\div 9$ $\\frac{3}{4} \\div 9 = \\frac{3}{4} \\times \\frac{1}{9}$ 因为 $\\frac{3}{4} \\times \\frac{1}{9} = \\frac{3}{4} \\times \\frac{1}{9}$, 所以 $\\frac{3}{4} \\times \\frac{1}{9} = \\frac{3}{4} \\div 9$ $\\frac{3}{8} \\div \\frac{6}{5}$ 和 $\\frac{3}{8}$ 因为 $\\frac{6}{5} > 1$, 所以 $\\frac{3}{8} \\div \\frac{6}{5} < \\frac{3}{8}$ $\\frac{1}{4} \\div \\frac{5}{8}$ 和 $\\frac{5}{8} \\div \\frac{1}{4}$ $\\frac{1}{4} \\div \\frac{5}{8} = \\frac{1}{4} \\times \\frac{8}{5} = \\frac{2}{5}$ $\\frac{5}{8} \\div \\frac{1}{4} = \\frac{5}{8} \\times 4 = \\frac{5}{2}$ 因为 $\\frac{2}{5}<\\frac{5}{2}$, 所以 $\\frac{1}{4} \\div \\frac{5}{8}<\\frac{5}{8} \\div \\frac{1}{4}$",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "分数乘分数",
+        "因数和积的大小关系（分数乘法）",
+        "被除数与商的大小关系（分数除法）"
+      ],
+      "source": "24-25 六年级上·全国·单元测试",
+      "merged_id": "unit3_331",
+      "original_id": "unit3_fill_331"
+    },
+    {
+      "id": "unit3_fill_332",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "两个因数的积是 1 ,其中一个因数是 $\\frac{4}{7}$, 另一个因数是( )。",
+      "options": [],
+      "answer": "$\\frac{7}{4}$",
+      "solution": "【分析】根据除法的意义，已知两个因数的积是 1, 其中一个因数是 $\\frac{4}{7}$, 求另一个因数, 用除法解答。\n\n【详解】$1 \\div \\frac{4}{7}=1 \\times \\frac{7}{4} = \\frac{7}{4}$ 另一个因数是 $\\frac{7}{4}$。",
+      "knowledge_points": [
+        "分数与整数的除法"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_332",
+      "original_id": "unit3_fill_332"
+    },
+    {
+      "id": "unit3_fill_333",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "'一束花的 $\\frac{3}{4}$ 是玫瑰花', 这句话中把( )看作单位 '1', 由此可得: $( ) = ( ) \\div \\frac{3}{4}$。",
+      "options": [],
+      "answer": "一束花的数量; 一束花的数量; 玫瑰花的数量",
+      "solution": "【分析】根据判断单位 '1' 的方法, 一般是把分率 '的'字前面的量看作单位 '1', 或把 '是、占、比' 后面的量看作单位 '1'。已知 '一束花的 $\\frac{3}{4}$ 是玫瑰花', 根据已知一个数的几分之几是多少, 求这个数, 用除法计算, 据此写出等量关系。\n\n【详解】由分析可得: '一束花的 $\\frac{3}{4}$ 是玫瑰花', 这句话中把一束花的数量看作单位 '1', 由此可得: 一束花的数量 = 玫瑰花的数量 $\\div \\frac{3}{4}$。",
+      "knowledge_points": [
+        "已知一个数的几分之几是多少，求这个数",
+        "单位'1'的认识与确定"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_333",
+      "original_id": "unit3_fill_333"
+    }, {
+      "id": "unit3_fill_334",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "( )是 50 的 $\\frac{2}{5}$, 30 是( )的 $\\frac{3}{4}$。",
+      "options": [],
+      "answer": "20; 40",
+      "solution": "【分析】求 50 的 $\\frac{2}{5}$ 是多少, 用 50 乘 $\\frac{2}{5}$ 即可; 已知一个数的 $\\frac{3}{4}$ 是 30, 求这个数, 用 30 除以 $\\frac{3}{4}$ 即可。\n\n【详解】$50 \\times \\frac{2}{5} = 20$ $30 \\div \\frac{3}{4} = 30 \\times \\frac{4}{3} = 40$ 20是50的 $\\frac{2}{5}$, 30是40的 $\\frac{3}{4}$。",
+      "knowledge_points": [
+        "求一个数的几分之几的问题",
+        "已知一个数的几分之几是多少，求这个数"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_334",
+      "original_id": "unit3_fill_334"
+    },
+    {
+      "id": "unit3_fill_335",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "1 的倒数是( )，5 的倒数是( ), $\\frac{7}{5}$ 的倒数是( )。",
+      "options": [],
+      "answer": "1; $\\frac{1}{5}$; $\\frac{5}{7}$",
+      "solution": "【分析】乘积是 1 的两个数互为倒数；1 的倒数是 1，求一个数的倒数，用 1 除以这个数，据此解答。\n\n【详解】1 的倒数是 1; $1 \\div 5 = \\frac{1}{5}$ $1 \\div \\frac{7}{5} = 1 \\times \\frac{5}{7} = \\frac{5}{7}$ 1 的倒数是 1, 5 的倒数是 $\\frac{1}{5}, \\frac{7}{5}$ 的倒数是 $\\frac{5}{7}$。",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "倒数的认识"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_335",
+      "original_id": "unit3_fill_335"
+    },
+    {
+      "id": "unit3_fill_336",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "小亮 $\\frac{5}{2}$ 小时走 $\\frac{15}{14}$ 千米,他平均每小时走( )千米, 走 1 千米需( )小时。",
+      "options": [],
+      "answer": "$\\frac{3}{7}$; $\\frac{7}{3}$",
+      "solution": "【分析】先根据速度 $=$ 路程 $\\div$ 时间, 算出小亮平均每小时走多少千米, 再根据时间 $=$ 路程 $\\div$ 速度, 求出走 1 千米需多少小时, 据此解答。\n\n【详解】$\\frac{15}{14} \\div \\frac{5}{2} = \\frac{15}{14} \\times \\frac{2}{5} = \\frac{3}{7}$ （千米/小时）$1 \\div \\frac{3}{7} = 1 \\times \\frac{7}{3} = \\frac{7}{3} (\\text{小时})$ 即他平均每小时走 $\\frac{3}{7}$ 千米, 走 1 千米需 $\\frac{7}{3}$ 小时。",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "分数与分数的除法",
+        "基础行程问题"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_336",
+      "original_id": "unit3_fill_336"
+    },
+    {
+      "id": "unit3_fill_337",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "钢琴班的人数比绘画班的人数少 $\\frac{1}{8}$, 钢琴班有 35 人, 绘画班有( )人。",
+      "options": [],
+      "answer": "40",
+      "solution": "【分析】把绘画班的人数看作单位'1'，钢琴班的人数比绘画班的人数少 $\\frac{1}{8}$, 则钢琴班的人数是绘画班的人数的 $(1 - \\frac{1}{8})$, 根据已知比一个数少几分之几的数是多少，求这个数，用除法计算，据此解答。\n\n【详解】$35 \\div (1 - \\frac{1}{8})=35 \\div \\frac{7}{8}=35 \\times \\frac{8}{7}=\\text{40 (人)}$ 即绘画班有 40 人。",
+      "knowledge_points": [
+        "已知比一个数多/少几分之几是多少，求这个数"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_337",
+      "original_id": "unit3_fill_337"
+    },
+    {
+      "id": "unit3_fill_338",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "$\\frac{3}{8} \\div \\frac{5}{4} = \\frac{3}{8} \\times \\frac{()}{()} = \\frac{()}{()}$\n\n$\\frac{2}{5} \\div 4 = \\frac{2}{5} \\times \\frac{()}{()} = \\frac{()}{()}$",
+      "options": [],
+      "answer": "$\\frac{4}{5}$; $\\frac{3}{10}$; $\\frac{1}{4}$; $\\frac{1}{10}$",
+      "solution": "【分析】一个数除以另一个数，等于这个数乘另一个数的倒数，据此解答。\n\n【详解】$\\frac{3}{8} \\div \\frac{5}{4}$, 因为 $\\frac{5}{4}$ 的倒数是 $\\frac{4}{5}$, 所以 $\\frac{3}{8} \\div \\frac{5}{4} = \\frac{3}{8} \\times \\frac{4}{5}$, 然后分子乘分子, 分母乘分母, 得到 $\\frac{12}{40}$, 约分后为 $\\frac{3}{10}$。对于 $\\frac{2}{5} \\div 4$, 4 可以写成 $\\frac{4}{1}$, 其倒数是 $\\frac{1}{4}$, 所以 $\\frac{2}{5} \\div 4 = \\frac{2}{5} \\times \\frac{1}{4}$, 计算可得 $\\frac{2}{20}$, 约分后为 $\\frac{1}{10}$。\n\n$\\frac{3}{8} \\div \\frac{5}{4} = \\frac{3}{8} \\times \\frac{4}{5} = \\frac{3}{10}; \\quad \\frac{2}{5} \\div 4 = \\frac{2}{5} \\times \\frac{1}{4} = \\frac{1}{10}$",
+      "knowledge_points": [
+        "分数与分数的除法"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_338",
+      "original_id": "unit3_fill_338"
+    },
+    {
+      "id": "unit3_fill_339",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "困难",
+      "question": "一桶油连桶共重 50 千克, 第一次倒出油的一半少 4 千克, 第二次倒出余下的油的 $\\frac{3}{4}$ 多 $\\frac{2}{3}$ 千克, 这时剩下的油和桶共重 $\\frac{1}{3}$ 千克。原来桶中有油( )千克。",
+      "options": [],
+      "answer": "48",
+      "solution": "【分析】根据题意, 可以设原来桶中有油 x 千克, 桶重 $(50 - x)$ 千克。第一次倒出油的一半少 4 千克, 那么第一次倒出的油量为 $(\\frac{x}{2} - 4)$ 千克, 此时剩下的油量为 $x - (\\frac{x}{2} - 4) = \\frac{x}{2} + 4$; 第二次倒出余下的油的 $\\frac{3}{4}$ 多 $\\frac{2}{3}$ 千克, 第一次倒出后剩下 $(\\frac{x}{2} + 4)$ 千克, 所以第二次倒出的油量为 $[(\\frac{x}{2} + 4) \\times \\frac{3}{4} + 2\\frac{2}{3}]$ 千克。可列方程为: 油的总重量一第一次倒出油一第二次倒出的油 + 桶重 $= 6\\frac{1}{3}$, 据此解答。本题通过复杂的倒油情境，综合考查了分数的运算和方程的应用。解题关键在于准确理解题目中每次倒出的油量与剩余油量之间的关系，巧妙设未知数并根据已知条件列出方程。同时，要注意在计算过程中对分数的正确处理，以及运用整体与部分的思维方式来分析问题。通过本题的练习，可以提升同学们分析问题、建立数学模型并求解的能力\n\n【详解】解：设原来桶中有油 $x$ 千克，桶重（50-x）千克。$x - (\\frac{x}{2} - 4) - (\\frac{x}{2} + 4) \\times \\frac{3}{4} - 2\\frac{2}{3} + (50 - x) = 6\\frac{1}{3}$ $\\frac{x}{2} + 4 - \\frac{3}{8}x - 5\\frac{2}{3} + 50 - x = 6\\frac{1}{3}$ $\\frac{x}{2} - \\frac{3}{8} x - x = 6\\frac{1}{3} - 50 + 5\\frac{2}{3} - 4 \\\\ \\frac{4}{8} x - \\frac{3}{8} x - \\frac{8}{8} x = \\frac{19}{3} - \\frac{150}{3} + \\frac{17}{3} - \\frac{12}{3} \\\\ \\frac{7}{8} x = \\frac{126}{3} \\\\ x = 48$ 原来桶中有油 48 千克。",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "分数的四则混合运算",
+        "列方程解含两个未知数的问题"
+      ],
+      "source": "24-25 六年级上·全国·课后作业",
+      "merged_id": "unit3_339",
+      "original_id": "unit3_fill_339"
+    },
+    {
+      "id": "unit3_fill_340",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "困难",
+      "question": "金星小学体育队原来有队员 120 人。今年女队员增加 $\\frac{1}{8}$, 男队员减少 $\\frac{1}{6}$ 后, 现在有队员 114 人。现在男队员有( )人, 女队员有( )人。",
+      "options": [],
+      "answer": "60; 54",
+      "solution": "【分析】首先明确问题中有两个未知量，即原来男队员人数和女队员人数。考虑设原来男队员人数为未知数 $x$，那么女队员人数就可以用总人数 120 减去 $x$ 来表示。接着分析变化情况，女队员增加 $\\frac{1}{8}$，男队员减少 $\\frac{1}{6}$ 后总人数变为 114 人。把金星小学体育队原来队员人数看作单位'1'。根据变化后的情况列出方程，方程左边是变化后的女队员人数加上变化后的男队员人数，右边是 114。可列出方程（$120 - x$) × $(1 + \\frac{1}{8}) + (1 - \\frac{1}{6}) x = 114$，计算出结果后，再分别计算出现在男、女队员的人数。本题主要考查了分数运算和二元一次方程组的应用。解题的关键在于根据已知条件设出合适的未知数，建立方程组来求解男、女队员原来的人数，进而得出现在男、女队员的人数。通过本题的练习, 可以锻炼同学们分析复杂数量关系和运用数学方法解决实际问题的能力。\n\n【详解】解：设男队员有 $x$ 人，则女队员有（$120 - x$）人 $(120 - x) \\times (1 + \\frac{1}{8}) + (1 - \\frac{1}{6}) x = 114 \\\\ 120 \\times \\frac{9}{8} - \\frac{9}{8} x + \\frac{5}{6} x = 114 \\\\ 135 - \\frac{27}{24} x + \\frac{20}{24} x = 114 \\\\ 135 - \\frac{7}{24} x = 114 \\\\ \\frac{7}{24} x = 135 - 114 \\\\ \\frac{7}{24} x = 21 \\\\ \\frac{7}{24} x \\div \\frac{7}{24} = 21 \\div \\frac{7}{24} \\\\ x = 21 \\times \\frac{24}{7} \\\\ x = 72$ 原来男队员有 72 人, 现在男队员人数为 $(1 - \\frac{1}{6}) \\times 72 =$ 60 (人)现在女队员人数为 $114 - 60 = 54$ （人）现在男队员有60人，女队员有54人。",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "求比一个数多/少几分之几的数是多少",
+        "列方程解含两个未知数的问题"
+      ],
+      "source": "24-25 六年级上·全国·课后作业",
+      "merged_id": "unit3_340",
+      "original_id": "unit3_fill_340"
+    },
+    {
+      "id": "unit3_fill_341",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "困难",
+      "question": "兴仁放马坪高山草原景区，地貌奇特，具有'高原塞外'之称。放马坪景区风光旖旎，分为天然草场和天然林，天然草场面积比景区总面积的 $\\frac{15}{19}$ 少 100 公顷，天然林面积是天然草场面积的 $\\frac{5}{14}$。放马坪景区总面积( )公顷。",
+      "options": [],
+      "answer": "1900",
+      "solution": "【分析】根据一个数乘分数的意义，求一个数的几分之几是多少，用乘法计算。把马坪景区总面积为 $x$ 公顷，所以，马坪景区总面积 $\\times \\frac{15}{19} - 100$ 公顷 = 天然草场面积，天然林面积 = $\\frac{5}{14} \\times$ 天然草场面积，天然草场面积 + 天然林面积 = 景区总面积。根据等量关系，列出方程，再利用等式的基本性质解出未知数。本题考查分数乘法的应用, 解答此题的关键是找到数量间的等量关系。\n\n【详解】解：设放马坪景区总面积为 $x$ 公顷。$\\frac{15}{19} x - 100 + \\frac{5}{14} \\times (\\frac{15}{19} x - 100) = x$ $(1 + \\frac{5}{14}) \\times (\\frac{15}{19} x - 100) = x$ $\\frac{19}{14} \\times (\\frac{15}{19} x - 100) = x$ $\\frac{19}{14} \\times (\\frac{15}{19} x - 100) \\times \\frac{14}{19} = x \\times \\frac{14}{19}$ $\\frac{15}{19} x - 100 = \\frac{14}{19} x$ $\\frac{15}{19} x - \\frac{14}{19} x = 100$ $\\frac{1}{19} x = 100$ $x = 100 \\times 19$ $x = 1900$ 放马坪景区总面积1900公顷。",
+      "knowledge_points": [
+        "列方程解决稍复杂的实际问题",
+        "求一个数的几分之几的问题",
+        "解分数方程"
+      ],
+      "source": "23-24 六年级下·贵州黔西·期末",
+      "merged_id": "unit3_341",
+      "original_id": "unit3_fill_341"
+    },
+    {
+      "id": "unit3_fill_342",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "甲、乙、丙三数之和是 180, 甲是乙的 $\\frac{1}{2}$, 乙是丙的 $\\frac{1}{3}$, 则甲是( ), 乙是( ), 丙是( )。",
+      "options": [],
+      "answer": "20; 40; 120",
+      "solution": "【分析】设丙是 $x$, 则乙是 $\\frac{1}{3} x$, 甲是 $\\frac{1}{3} x \\times \\frac{1}{2} = \\frac{1}{6} x$。根据题意,甲数十乙数十丙数 $= 180$, 据此列方程即可解答。\n\n【详解】解：设丙是 $x$，则乙是 $\\frac{1}{3} x$，甲是 $\\frac{1}{6} x$。$\\frac{1}{6} x + \\frac{1}{3} x + x = 180$ $\\frac{3}{2} x = 180$ $\\frac{3}{2} x \\times \\frac{2}{3} = 180 \\times \\frac{2}{3}$ $x = 120$ 乙: $120 \\times \\frac{1}{3} = 40$ 甲: $40 \\times \\frac{1}{2} = 20$ 则甲是 20, 乙是 40, 丙是 $120 \\mathrm{~g}$。",
+      "knowledge_points": [
+        "列方程解决稍复杂的实际问题"
+      ],
+      "source": "23-24 六年级下·新疆乌鲁木齐·期末",
+      "merged_id": "unit3_342",
+      "original_id": "unit3_fill_342"
+    },
+    {
+      "id": "unit3_fill_343",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "一件工作, 甲单独要做20天完成，乙单独要做12天完成，这件工作由甲做了若干天后，由乙继续做完，从开始到完工共用了14天，这件工作由甲先做了( )天。",
+      "options": [],
+      "answer": "5",
+      "solution": "【分析】从题意可知，以这件工作为单位'1'，甲每天可完成这项工作的 $\\frac{1}{20}$，乙每天可完成这项工作的 $\\frac{1}{12}$，假设这项工程是乙单独做了 14 天，那么乙完成了这项工作的 $\\frac{14}{12}$，就比单位'1'多了 $\\frac{14}{12} - 1 = \\frac{1}{6}$，这是因为乙每天比甲每天多做 $\\frac{1}{12} - \\frac{1}{20} = \\frac{1}{30}$，最后用 $\\frac{1}{6} \\div \\frac{1}{30}$ 即可求出这件工作由甲先做的天数。据此解答。\n\n【详解】$1 \\div 20 = \\frac{1}{20}$ $1 \\div 12 = \\frac{1}{12}$ $(\\frac{1}{12} \\times 14 - 1) \\div (\\frac{1}{12} - \\frac{1}{20}) \\\\ = (\\frac{14}{12} - 1) \\div \\frac{1}{30} \\\\ = \\frac{1}{6} \\times 30 \\\\ = 5 \\text{(天)}$ 这件工作由甲先做了 5 天。",
+      "knowledge_points": [
+        "两人合作的工程问题",
+        "分数的四则混合运算",
+        "假设法解鸡兔同笼"
+      ],
+      "source": "23-24 六年级下·天津南开·期末",
+      "merged_id": "unit3_343",
+      "original_id": "unit3_fill_343"
+    },
+    {
+      "id": "unit3_fill_344",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "a、b 互为倒数且都不等于 0, 则 $\\frac{8}{a} \\times \\frac{10}{b} = (\\quad)$, $\\frac{a}{8} \\div \\frac{10}{b} = (\\quad)$。",
+      "options": [],
+      "answer": "80; $\\frac{1}{80}$",
+      "solution": "【分析】乘积是 1 的两个数互为倒数, 由题可知, $a \\times b = 1$。分数乘分数: 分子乘分子, 分母乘分母, 能约分的可先约分。分数除法: 除以一个数就是乘这个数的倒数。据此解题。\n\n【详解】a、b 互为倒数且都不等于 0, 那么 $\\mathrm{a} \\times \\mathrm{b} = 1$。$\\frac{8}{a} \\times \\frac{10}{b} = \\frac{8 \\times 10}{a \\times b} = 80 \\\\ \\frac{a}{8} \\div \\frac{10}{b} = \\frac{a}{8} \\times \\frac{b}{10} = \\frac{1}{80}$",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "分数乘分数",
+        "倒数的认识"
+      ],
+      "source": "23-24 五年级下·湖北省直辖县级单位·期末",
+      "merged_id": "unit3_344",
+      "original_id": "unit3_fill_344"
+    },
+    {
+      "id": "unit3_fill_345",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "小明步行 $\\frac{4}{5}$ 千米用 $\\frac{1}{4}$ 小时。照这样计算, 他每小时行( )千米, 行 1 千米需要( )小时。",
+      "options": [],
+      "answer": "$\\frac{16}{5}$; $\\frac{5}{16}$",
+      "solution": "【分析】求每小时行多少千米，用总千米数除以总时间；求行 1 千米需要多少小时，用总时间除以总千米数。\n\n【详解】$\\frac{4}{5} \\div \\frac{1}{4}= \\frac{16}{5}$ (千米) $\\frac{1}{4} \\div \\frac{4}{5} \\\\ = \\frac{1}{4} \\times \\frac{5}{4}\\\\= \\frac{5}{16}$(小时) 他每小时行 $\\frac{16}{5}$ 千米, 行 1 千米需要 $\\frac{5}{16}$ 小时。",
+      "knowledge_points": [
+        "分数与分数的除法"
+      ],
+      "source": "22-23 六年级上·江苏盐城·期中",
+      "merged_id": "unit3_345",
+      "original_id": "unit3_fill_345"
+    },
+    {
+      "id": "unit3_fill_346",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "$\\frac{3}{7}$ 的倒数是( )；6的倒数是( )。",
+      "options": [],
+      "answer": "$\\frac{7}{3}$; $\\frac{1}{6}$",
+      "solution": "【分析】倒数的意义: 乘积为 1 的两个数互为倒数。求一个数的倒数用 1 除以这个数, 据此解答。\n\n【详解】$1 \\div \\frac{3}{7} = 1 \\times \\frac{7}{3} = \\frac{7}{3}$ $1 \\div 6 = \\frac{1}{6}$ 即 $\\frac{3}{7}$ 的倒数是 $\\frac{7}{3}$; 6 的倒数是 $\\frac{1}{6}$。",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "倒数的认识"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_346",
+      "original_id": "unit3_fill_346"
+    },
+    {
+      "id": "unit3_fill_347",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "在括号里填上\">\" \"<\"或\"=\"。\n\n$\\frac{5}{8} \\div 5$ ( ) $\\frac{5}{8}$ $\\frac{2}{9} \\div 1$ ( ) $\\frac{2}{9}$ $\\frac{3}{11} \\div \\frac{3}{4}$ ( ) $\\frac{3}{11}$",
+      "options": [],
+      "answer": "<; =; >",
+      "solution": "【分析】一个非 0 数除以小于 1 的数，商大于被除数；一个非 0 数除以大于 1 的数，商小于被除数。一个非 0 数除以 1，商等于被除数。\n\n【详解】因为 $5 > 1$，所以 $\\frac{5}{8} \\div 5 < \\frac{5}{8}$ 因为 $1 = 1$，所以 $\\frac{2}{9} \\div 1 = \\frac{2}{9}$ 因为 $\\frac{3}{4} < 1$，所以 $\\frac{3}{11} \\div \\frac{3}{4} > \\frac{3}{11}$。",
+      "knowledge_points": [
+        "被除数与商的大小关系（分数除法）"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_347",
+      "original_id": "unit3_fill_347"
+    },
+    {
+      "id": "unit3_fill_348",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "《九章算术》是中国古代的一部数学著作，书中解决分数除法问题的具体方法是'经分术'。根据'经分术'，如果被除数和除数都是分数，要先将两个分数通分，再使分子相除，如下所示：$\\frac{a}{b} \\div \\frac{c}{d} = \\frac{ad}{bd} \\div \\frac{bc}{bd} = \\frac{ad}{bc}$ (b，c，d均不为0)，按照以上方法，可以这样计算 $\\frac{2}{7} \\div \\frac{3}{5} = (\\quad)$。",
+      "options": [],
+      "answer": "$\\frac{10}{21}$",
+      "solution": "【分析】观察可知，《九章算术》计算分数除法，先通分，第一个分数的分子作分子，第二个分数的分子作分母，据此计算。\n\n【详解】$\\frac{2}{7} \\div \\frac{3}{5} = \\frac{2 \\times 5}{7 \\times 5} \\div \\frac{3 \\times 7}{5 \\times 7} = \\frac{2 \\times 5}{3 \\times 7} = \\frac{10}{21}$ 根据'经分术', 如果被除数和除数都是分数, 要先将两个分数通分, 再使分子相除, 如下所示: $\\frac{a}{b} \\div \\frac{c}{d} = \\frac{ad}{bd} \\div \\frac{bc}{bd} = \\frac{ad}{bc}$ (b, c, d 均不为 0), 按照以上方法, 计算出 $\\frac{2}{7} \\div \\frac{3}{5} = \\frac{10}{21}$。",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "通分的认识及应用"
+      ],
+      "source": "23-24 六年级下·北京海淀·期末",
+      "merged_id": "unit3_348",
+      "original_id": "unit3_fill_348"
+    },
+    {
+      "id": "unit3_sort_349",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "排序题",
+      "difficulty": "中等",
+      "question": "已知: $A \\times \\frac{7}{8} = B \\div \\frac{5}{4} = C \\times \\frac{5}{6} = \\frac{6}{7} \\div D$, 把 A, B, C, D 按从小到大的顺序排列。",
+      "options": [],
+      "answer": "D $< \\mathrm{A} < \\mathrm{C} < \\mathrm{B}$",
+      "solution": "【分析】假设 $A \\times \\frac{7}{8} = B \\div \\frac{5}{4} = C \\times \\frac{5}{6} = \\frac{6}{7} \\div D = 1$, 分别求出 A, B, C, D 的值, 再比较大小即可。\n\n【详解】假设 $A \\times \\frac{7}{8} = B \\div \\frac{5}{4} = C \\times \\frac{5}{6} = \\frac{6}{7} \\div D = 1$。则 $A = 1 \\div \\frac{7}{8} = \\frac{8}{7}$, $B = 1 \\times \\frac{5}{4} = \\frac{5}{4}$, $C = 1 \\div \\frac{5}{6} = \\frac{6}{5}$, 因为 $\\frac{6}{7} < \\frac{8}{7} < \\frac{6}{5} < \\frac{5}{4}$ 所以 $D < A < C < B$",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "异分母异分子分数的大小比较"
+      ],
+      "source": "23-24 六年级上·全国·课后作业",
+      "merged_id": "unit3_349",
+      "original_id": "unit3_sort_349"
+    },
+    {
+      "id": "unit3_sort_350",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "排序题",
+      "difficulty": "中等",
+      "question": "已知:甲 $\\times \\frac{4}{5} =$ 乙 $\\times \\frac{2}{3} =$ 丙 $\\times \\frac{6}{7} =$ 丁 $\\times \\frac{3}{4}$, 把甲, 乙, 丙、丁按从小到大的照序排列。",
+      "options": [],
+      "answer": "丙<甲<丁<乙",
+      "solution": "【分析】假设 $甲 \\times \\frac{4}{5} = 乙 \\times \\frac{2}{3} = 丙 \\times \\frac{6}{7} = 丁 \\times \\frac{3}{4} = 1$, 分别求出甲, 乙,丙、丁的结果, 再比较大小即可。\n\n【详解】设甲 $\\times \\frac{4}{5} =$ 乙 $\\times \\frac{2}{3} =$ 丙 $\\times \\frac{6}{7} =$ T $\\times \\frac{3}{4} = 1$ 则: 甲 $= \\frac{5}{4}$, 乙 $= \\frac{3}{2}$, 丙 $= \\frac{7}{6}$, 丁 $= \\frac{4}{3}$ 因为 $\\frac{5}{4} = \\frac{15}{12}, \\frac{3}{2} = \\frac{18}{12}, \\frac{7}{6} = \\frac{14}{12}, \\frac{4}{3} = \\frac{16}{12}$ 因为 $\\frac{14}{12} < \\frac{15}{12} < \\frac{16}{12} < \\frac{18}{12}$ 所以 $\\frac{7}{6}<\\frac{5}{4}<\\frac{4}{3}<\\frac{3}{2}$ 所以丙 $<$ 甲 $<$ 丁 $<$ 乙。",
+      "knowledge_points": [
+        "异分母异分子分数的大小比较",
+        "倒数的认识"
+      ],
+      "source": "23-24 六年级上·全国·课后作业",
+      "merged_id": "unit3_350",
+      "original_id": "unit3_sort_350"
+    },
+    {
+      "id": "unit3_judge_351",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "$\\frac{1}{2} \\times \\frac{1}{3} \\div \\frac{1}{2} \\times \\frac{1}{3} = \\frac{1}{6} \\div \\frac{1}{6} = 1$。（）",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】整数的四则混合运算在分数中同样适用，先算乘除再算加减，所以 $\\frac{1}{2} \\times \\frac{1}{3} \\div \\frac{1}{2} \\times \\frac{1}{3}$ 从左往右依次计算即可。\n\n【详解】$\\frac{1}{2} \\times \\frac{1}{3} \\div \\frac{1}{2} \\times \\frac{1}{3}= \\frac{1}{2} \\times \\frac{1}{3} \\times 2 \\times \\frac{1}{3}= \\frac{1}{6} \\times 2 \\times \\frac{1}{3} = \\frac{1}{3} \\times \\frac{1}{3} = \\frac{1}{9}$ 原题干说法错误。故答案为: ×",
+      "knowledge_points": [
+        "分数的乘、除法的混合运算"
+      ],
+      "source": "24-25 六年级上·全国·单元测试",
+      "merged_id": "unit3_351",
+      "original_id": "unit3_judge_351"
+    },
+    {
+      "id": "unit3_judge_352",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "两个数的积一定大于这两个数的商。（）",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】一个非 0 数, 乘小于 1 的数, 积小于原数; 一个非 0 数, 除以小于 1 的数, 商大于被除数, 据此举例解答。\n\n【详解】如: $3 \\times \\frac{1}{3}$ 和 $3 \\div \\frac{1}{3}$ $3 \\times \\frac{1}{3} = 1$ $3 \\div \\frac{1}{3} = 3 \\times 3 = 9$ 因为 $1 < 9$, 所以 $3 \\times \\frac{1}{3} < 3 \\div \\frac{1}{3}$。由此可知，两个数的积不一定大于这两个数的商。原题干说法错误。故答案为: ×",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "分数乘整数",
+        "因数和积的大小关系（分数乘法）",
+        "被除数与商的大小关系（分数除法）"
+      ],
+      "source": "24-25 六年级上·全国·单元测试",
+      "merged_id": "unit3_352",
+      "original_id": "unit3_judge_352"
+    },
+    {
+      "id": "unit3_judge_353",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "$6 \\div \\frac{3}{2} = 6 \\div 3 \\div 2 = 1$。（）",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】在数学运算中, 除法的运算规则是不能随意将除数进行拆分的。当我们计算 $6 \\div \\frac{3}{2}$ 时, 应该将除法转化为乘法,即 $6 \\times \\frac{2}{3}$, 这样计算得到的结果是 4, 据此解答。\n\n【详解】$6 \\div \\frac{3}{2} = 6 \\times \\frac{2}{3} = 4$, 而 $6 \\div 3 \\div 2 = 2 \\div 2 = 1$, 两者结果不同。故答案为: ×",
+      "knowledge_points": [
+        "分数与整数的除法"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_353",
+      "original_id": "unit3_judge_353"
+    },
+    {
+      "id": "unit3_judge_354",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "一个非零数除以假分数, 商一定小于这个非零数。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】假分数是大于或等于 1 的分数, 可以分为等于 1 和大于 1 两种来分类说明即可。\n\n【详解】当这个假分数等于 1 时, 比如 $3 \\div \\frac{2}{2} = 3 \\div 1 = 3$, 此时商等于这个非零数。当这个假分数大于 1 时, 比如 $3 \\div \\frac{3}{2} = 3 \\times \\frac{2}{3} = 2$, 此时商小于这个非零数。所以一个非零数除以假分数，商不一定小于这个数，故原题该说法错误。故答案为: ×",
+      "knowledge_points": [
+        "被除数与商的大小关系（分数除法）",
+        "真分数、假分数、带分数的认识"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_354",
+      "original_id": "unit3_judge_354"
+    },
+    {
+      "id": "unit3_judge_355",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "山羊有 80 只, 比绵羊少 $\\frac{3}{5}$, 绵羊有多少只? 正确列式为 $80 \\times (1 - \\frac{3}{5})$。 ()",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】把绵羊只数看作单位 '1', 山羊是绵羊的 $(1 - \\frac{3}{5})$, 已知山羊有 80 只, 根据已知一个数的几分之几是多少,求这个数用除法计算, 用山羊只数除以其对应的分率, 即可求出绵羊的只数。\n\n【详解】由分析可知, 绵羊的只数为: $80 \\div (1 - \\frac{3}{5})$。原说法错误。故答案为: ×",
+      "knowledge_points": [
+        "已知比一个数多/少几分之几是多少，求这个数"
+      ],
+      "source": "24-25 六年级上·全国·课后作业",
+      "merged_id": "unit3_355",
+      "original_id": "unit3_judge_355"
+    },
+    {
+      "id": "unit3_judge_356",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "求一个数 (0 除外)的倒数，就是把这个数的分子、分母交换位置。（）",
+      "options": [],
+      "answer": "✔",
+      "solution": "【分析】乘积是 1 的两个数互为倒数；整数和小数都可以化成分数，求一个数（0 除外）的倒数，可以交换分子和分母的位置，据此判断即可。\n\n【详解】求一个数（0 除外）的倒数，就是把这个数的分子、分母交换位置，说法正确。故答案为: ✔",
+      "knowledge_points": [
+        "倒数的认识"
+      ],
+      "source": "24-25 六年级上·全国·课后作业",
+      "merged_id": "unit3_356",
+      "original_id": "unit3_judge_356"
+    },
+    {
+      "id": "unit3_judge_357",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "黑麦的质量比小米多 $\\frac{1}{5}$, 也就是说小米的质量比黑麦少 $\\frac{1}{5}$。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】把小米的质量看作单位 '1', 则黑麦的质量为 $1 \\times (1 + \\frac{1}{5})$, 然后求出小米的质量比黑麦少多少, 再除以黑麦的质量即可。\n\n【详解】$1 \\times (1 + \\frac{1}{5}) = 1 \\times \\frac{6}{5} = \\frac{6}{5}$ $(\\frac{6}{5} - 1) \\div \\frac{6}{5} = \\frac{1}{5} \\div \\frac{6}{5} = \\frac{1}{5} \\times \\frac{5}{6} = \\frac{1}{6}$ 则小米的质量比黑麦少 $\\frac{1}{6}$, 原题干说法错误。故答案为: ×",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "求比一个数多/少几分之几的数是多少"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_357",
+      "original_id": "unit3_judge_357"
+    },
+    {
+      "id": "unit3_judge_358",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "小华每分钟跑 $260 \\mathrm{~m}$, 比小雪快 $\\frac{1}{5}$, 这里是把小雪跑的速度看作单位 '1'。（）",
+      "options": [],
+      "answer": "✓",
+      "solution": "【分析】'小华每分钟跑 $260 \\mathrm{~m}$, 比小雪快', 这表明小华的速度是小雪速度的 $1 + \\frac{1}{5} = \\frac{6}{5}$ 倍。所以是以小雪的速度为标准, 把小雪跑的速度看作单位 '1'。\n\n【详解】因为小华速度是小雪速度的 $\\frac{6}{5}$ 倍，所以小雪的速度为 $260 \\div \\frac{6}{5} = 260 \\times \\frac{5}{6} = \\frac{650}{3}$ （米/分钟）。小华比小雪快的速度为 $\\frac{650}{3} \\times \\frac{1}{5} = \\frac{130}{3}$ （米/分钟），小雪的速度 $\\frac{650}{3}$ 加上快的速度 $\\frac{130}{3}$ 等于小华的速度 $260 \\mathrm{~m}$ /分钟，这说明是以小雪的速度为单位'1'来进行比较的。故答案为: ✓",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "基础行程问题",
+        "单位'1'的认识与确定",
+        "求比一个数多/少几分之几的数是多少"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_358",
+      "original_id": "unit3_judge_358"
+    },
+    {
+      "id": "unit3_judge_359",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "正方形的周长是 $\\frac{4}{5}$ 米,则它的面积是 $\\frac{1}{25}$ 平方米。（）",
+      "options": [],
+      "answer": "✓",
+      "solution": "【分析】正方形的边长 $\\times 4 =$ 周长, 正方形的边长 $=$ 周长 $\\div 4$, 据此求出正方形的边长, 再根据正方形的面积 $=$ 边长 $\\times$ 边长解答。\n\n【详解】$\\frac{4}{5} \\div 4=\\frac{4}{5} \\times \\frac{1}{4} = \\frac{1}{5} (\\text{米})$ $\\frac{1}{5} \\times \\frac{1}{5} = \\frac{1}{25}$ （平方米）所以正方形的面积是 $\\frac{1}{25}$ 平方米。所以原题说法正确。故答案为: ✓",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "分数乘分数",
+        "正方形的周长",
+        "正方形的面积"
+      ],
+      "source": "23-24 五年级下·陕西渭南·期末",
+      "merged_id": "unit3_359",
+      "original_id": "unit3_judge_359"
+    },
+    {
+      "id": "unit3_calc_360",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "解方程。\n\n(1) $\\frac{3}{8} x = 24$ (2) $x \\div \\frac{3}{7} = \\frac{14}{15}$ (3) $2x \\div \\frac{11}{20} = \\frac{4}{15}$",
+      "options": [],
+      "answer": "(1) $x = 64$; (2) $x = \\frac{2}{5}$; (3) $x = \\frac{11}{150}$",
+      "solution": "【分析】(1) 根据等式的性质, 方程两边同时除以 $\\frac{3}{8}$ 即可; (2) 根据等式的性质, 方程两边同时乘 $\\frac{3}{7}$ 即可; (3) 根据等式的性质, 方程两边同时乘 $\\frac{11}{20}$, 再同时除以 2 即可。\n\n【详解】(1) $\\frac{3}{8} x = 24$ 解: $\\frac{3}{8} x \\div \\frac{3}{8} = 24 \\div \\frac{3}{8}$ $x = 24 \\times \\frac{8}{3}$ $x = 64$ (2) $x \\div \\frac{3}{7} = \\frac{14}{15}$ 解: $x \\div \\frac{3}{7} \\times \\frac{3}{7} = \\frac{14}{15} \\times \\frac{3}{7}$ $x = \\frac{2}{5}$ (3) $2x \\div \\frac{11}{20} = \\frac{4}{15}$ 解: $2x \\div \\frac{11}{20} \\times \\frac{11}{20} = \\frac{4}{15} \\times \\frac{11}{20}$ $2x = \\frac{11}{75} \\\\ 2x \\div 2 = \\frac{11}{75} \\div 2 \\\\ x = \\frac{11}{75} \\times \\frac{1}{2} \\\\ x = \\frac{11}{150}$",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "分数与分数的除法",
+        "应用等式的性质 2 解方程",
+        "解分数方程"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_360",
+      "original_id": "unit3_calc_360"
+    },
+    {
+      "id": "unit3_calc_361",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "计算下面各题。(能简算的要用简便方法计算)\n\n(1) $\\frac{3}{5} \\times \\frac{3}{7} + \\frac{2}{5} \\times \\frac{3}{7}$ ② $(\\frac{2}{3} - \\frac{1}{2} + \\frac{5}{6}) \\div \\frac{1}{24}$ ③ $\\frac{3}{14} \\div \\frac{6}{7} \\div 12$ ④ $\\frac{1}{8} \\div (\\frac{3}{4} \\div 3)$ ⑤ $(\\frac{1}{6} + \\frac{1}{2}) \\div \\frac{2}{3} \\times \\frac{3}{4}$ ⑥ $\\frac{7}{9} \\div [\\frac{2}{3} \\times (\\frac{5}{18} - \\frac{1}{9})]$",
+      "options": [],
+      "answer": "① $\\frac{3}{7}$; ② $24$; ③ $\\frac{1}{48}$; ④ $\\frac{1}{2}$; ⑤ $\\frac{3}{4}$; ⑥ $7$",
+      "solution": "【分析】①根据乘法分配律逆运算 $a \\times c + b \\times c = (a + b) \\times c$ 进行简算；②先把除法转化成乘法，再根据乘法分配律 $(a + b) \\times c = a \\times c + b \\times c$ 进行简算；(3) 先把除法转化成乘法, 再从左往右依次计算; (4) 先算括号里面的除法, 再算括号外面的除法; (5) 先算括号里面的加法, 再算括号外面的除法、乘法; (6) 先算小括号里面的减法, 再算中括号里面的乘法, 最后算中括号外面的除法。\n\n【详解】① $\\frac{3}{5} \\times \\frac{3}{7} + \\frac{2}{5} \\times \\frac{3}{7}$ $(\\frac{3}{5} + \\frac{2}{5}) \\times \\frac{3}{7} \\\\ =1 \\times \\frac{3}{7} \\\\ =\\frac{3}{7}$ ② $(\\frac{2}{3} - \\frac{1}{2} + \\frac{5}{6}) \\div \\frac{1}{24}$ $=(\\frac{2}{3} - \\frac{1}{2} + \\frac{5}{6}) \\times 24$ $=\\frac{2}{3} \\times 24 - \\frac{1}{2} \\times 24 + \\frac{5}{6} \\times 24$ $=16 - 12 + 20$ =24 ③ $\\frac{3}{14} \\div \\frac{6}{7} \\div 12$ $=\\frac{3}{14} \\times \\frac{7}{6} \\times \\frac{1}{12} \\\\ = \\frac{1}{4} \\times \\frac{1}{12} \\\\ = \\frac{1}{48}$ ④ $\\frac{1}{8} \\div (\\frac{3}{4} \\div 3)$ $= \\frac{1}{8} \\div (\\frac{3}{4} \\times \\frac{1}{3}) \\\\ = \\frac{1}{8} \\div \\frac{1}{4} \\\\ = \\frac{1}{8} \\times 4 \\\\ = \\frac{1}{2}$ ⑤ $(\\frac{1}{6} + \\frac{1}{2}) \\div \\frac{2}{3} \\times \\frac{3}{4}$ $= (\\frac{1}{6} + \\frac{3}{6}) \\div \\frac{2}{3} \\times \\frac{3}{4} \\\\ = \\frac{2}{3} \\div \\frac{2}{3} \\times \\frac{3}{4} \\\\ = 1 \\times \\frac{3}{4} \\\\ = \\frac{3}{4}$ ⑥ $\\frac{7}{9} \\div [\\frac{2}{3} \\times (\\frac{5}{18} - \\frac{1}{9})]$ $= \\frac{7}{9} \\div [\\frac{2}{3} \\times (\\frac{5}{18} - \\frac{2}{18})] \\\\ = \\frac{7}{9} \\div [\\frac{2}{3} \\times \\frac{1}{6}] \\\\ = \\frac{7}{9} \\div \\frac{1}{9} \\\\ = \\frac{7}{9} \\times 9 \\\\ = 7$",
+      "knowledge_points": [
+        "整数乘法运算定律推广到分数乘法",
+        "分数的四则混合运算",
+        "分数的连除运算",
+        "分数除法相关的简便计算"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_361",
+      "original_id": "unit3_calc_361"
+    },
+    {
+      "id": "unit3_calc_362",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "计算题",
+      "difficulty": "容易",
+      "question": "直接写出得数。\n\n$3 \\div \\frac{1}{6} =$ $9 \\times \\frac{2}{3} =$ $\\frac{3}{5} - \\frac{1}{2} =$ $\\frac{4}{9} \\times \\frac{3}{8} =$ $\\frac{3}{5} \\div \\frac{4}{5} =$ $\\frac{4}{9} \\times 1.8 =$ $14 \\div \\frac{7}{15} =$ $(\\frac{1}{6} - \\frac{1}{9}) \\times 18 =$",
+      "options": [],
+      "answer": "18; 6; $\\frac{1}{10}$; $\\frac{1}{6}$; $\\frac{3}{4}$; 0.8; 30; 1",
+      "solution": "【详解】略",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "整数乘法运算定律推广到分数乘法",
+        "分数乘分数",
+        "异分母分数加、减法"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_362",
+      "original_id": "unit3_calc_362"
+    },
+    {
+      "id": "unit3_calc_363",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "计算。能用简便方法计算的要用简便方法计算。\n\n① $\\frac{4}{7} \\times 76 + \\frac{4}{7}$ ② $(1 - \\frac{3}{10} \\div \\frac{6}{7}) \\times \\frac{10}{13}$ ③ $\\frac{2}{5} \\div [\\frac{4}{7} \\times (\\frac{1}{2} + \\frac{3}{8})]$",
+      "options": [],
+      "answer": "①44；② $\\frac{1}{2}$；③ $\\frac{4}{5}$",
+      "solution": "【分析】①利用乘法分配律可知 $\\frac{4}{7} \\times 76 + \\frac{4}{7} = \\frac{4}{7} \\times (76 + 1)$, 再根据加法法则及分数的乘法法则即可解答;②先算小括号里面的除法, 再算括号里面的减法, 最后算括号外面的乘法即可;③ 先算小括号里面的加法, 再算中括号里面的乘法, 最后算除法。本题考查了分数的除法运算法则，分数的乘法运算法则，乘法的分配律，熟练运算乘法的运算法则是解题的关键。\n\n【详解】$1 \\frac{4}{7} \\times 76 + \\frac{4}{7}$ $\\frac{4}{7} \\times (76 + 1) \\\\ = \\frac{4}{7} \\times 77 \\\\ = 44$ (2) $(1 - \\frac{3}{10} \\div \\frac{6}{7}) \\times \\frac{10}{13}$ $(1 - \\frac{3}{10} \\times \\frac{7}{6}) \\times \\frac{10}{13} \\\\ = (1 - \\frac{7}{20}) \\times \\frac{10}{13} \\\\ = \\frac{13}{20} \\times \\frac{10}{13} \\\\ = \\frac{1}{2}$ (3) $\\frac{2}{5} \\div [\\frac{4}{7} \\times (\\frac{1}{2} + \\frac{3}{8})]$ $\\frac{2}{5} \\div [\\frac{4}{7} \\times \\frac{7}{8}] \\\\ = \\frac{2}{5} \\div \\frac{1}{2} \\\\ = \\frac{2}{5} \\times 2 \\\\= \\frac{4}{5}$",
+      "knowledge_points": [
+        "整数乘法运算定律推广到分数乘法",
+        "分数的四则混合运算"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_363",
+      "original_id": "unit3_calc_363"
+    },
+    {
+      "id": "unit3_calc_364",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "计算题",
+      "difficulty": "容易",
+      "question": "口算。\n\n$12 \\times \\frac{3}{11} =$ ② $0.75 \\div \\frac{3}{4} =$ ③ $\\frac{1}{5} + 1.8 =$ ④ $(\\frac{1}{6} - \\frac{1}{8}) \\times 24 =$ ⑤ $\\frac{2}{3} + \\frac{3}{4} =$ ⑥ $88 \\div 16 =$ ⑦ $3 - \\frac{5}{7} + \\frac{1}{2} =$ ⑧ $\\frac{3}{5} \\times \\frac{5}{12} - 0.25 =$",
+      "options": [],
+      "answer": "①6; ②1; ③2; ④1; ⑤ $\\frac{17}{12}$; ⑥ $\\frac{11}{2}$; ⑦ $\\frac{39}{14}$; ⑧ 0",
+      "solution": "【详解】略",
+      "knowledge_points": [
+        "分数的加、减法混合运算",
+        "分数的四则混合运算",
+        "异分母分数加、减法",
+        "整数乘分数"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_364",
+      "original_id": "unit3_calc_364"
+    },
+    {
+      "id": "unit3_calc_365",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "解下列方程。\n\n(1) $\\frac{3}{20} x = \\frac{6}{5}$ (2) $\\frac{5}{12} \\div x = \\frac{15}{8}$ (3) $\\frac{1}{3} x + \\frac{3}{5} = \\frac{13}{10}$",
+      "options": [],
+      "answer": "(1) $x = 8$; (2) $x = \\frac{2}{9}$; (3) $x = \\frac{21}{10}$",
+      "solution": "【分析】（1）根据等式的性质，方程两边同时除以 $\\frac{3}{20}$ 即可；(2) 根据等式的性质, 方程两边同时乘 $x$, 再同时除以 $\\frac{15}{8}$ 即可; (3) 根据等式的性质, 方程两边同时减去 $\\frac{3}{5}$, 再同时除以 $\\frac{1}{3}$ 即可。\n\n【详解】(1) $\\frac{3}{20} x = \\frac{6}{5}$ 解: $\\frac{3}{20} x \\div \\frac{3}{20} = \\frac{6}{5} \\div \\frac{3}{20}$ $x = \\frac{6}{5} \\times \\frac{20}{3}$ $x = 8$ (2) $\\frac{5}{12} \\div x = \\frac{15}{8}$ 解: $\\frac{5}{12} \\div x \\times x = \\frac{15}{8} \\times x$ $\\frac{15}{8} x = \\frac{5}{12}$ $\\frac{15}{8} x \\div \\frac{15}{8} = \\frac{5}{12} \\div \\frac{15}{8} \\\\ x = \\frac{5}{12} \\times \\frac{8}{15} \\\\ x = \\frac{2}{9}$ (3) $\\frac{1}{3} x + \\frac{3}{5} = \\frac{13}{10}$ 解: $\\frac{1}{3} x + \\frac{3}{5} - \\frac{3}{5} = \\frac{13}{10} - \\frac{3}{5}$ $\\frac{1}{3} x = \\frac{7}{10} \\\\ \\frac{1}{3} x \\div \\frac{1}{3} = \\frac{7}{10} \\div \\frac{1}{3} \\\\ x = \\frac{7}{10} \\times 3 \\\\ x = \\frac{21}{10}$",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "分数乘分数",
+        "异分母分数加、减法",
+        "应用等式的性质 1 和 2 解方程"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_365",
+      "original_id": "unit3_calc_365"
+    },
+    {
+      "id": "unit3_calc_366",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "计算下列各题, 能简算的要简算。\n\n$\\frac{2}{13} \\times \\frac{5}{8} \\div \\frac{15}{26}$ $\\frac{3}{5} \\div \\frac{7}{9} + \\frac{2}{5} \\div \\frac{7}{9}$ $(\\frac{7}{3} - \\frac{5}{4}) \\div \\frac{5}{12}$ $\\frac{5}{14} \\div \\frac{1}{4} \\div \\frac{15}{7}$ $\\frac{13}{25} \\div (\\frac{2}{5} + \\frac{9}{10})$ $\\frac{7}{10} \\times \\frac{3}{4} + \\frac{1}{10} \\div \\frac{4}{7}$",
+      "options": [],
+      "answer": "$\\frac{1}{6}$; $\\frac{9}{7}$; $\\frac{13}{5}$; $\\frac{2}{3}$; $\\frac{2}{5}$; $\\frac{7}{10}$",
+      "solution": "【分析】（1）化除法为乘法，把原式化为 $\\frac{2}{13} \\times \\frac{5}{8} \\times \\frac{26}{15}$，再按照从左到右的运算顺序进行计算即可；(2) 化除法为乘法, 把原式化为 $\\frac{3}{5} \\times \\frac{9}{7} + \\frac{2}{5} \\times \\frac{9}{7}$, 再运用乘法分配律化为 $(\\frac{3}{5} + \\frac{2}{5}) \\times \\frac{9}{7}$, 依此进行计算即可; (3) 先算小括号里面的减法, 再算小括号外面的除法即可; (4) 化除法为乘法, 把原式化为 $\\frac{5}{14} \\times 4 \\times \\frac{7}{15}$, 再运用乘法交换律化为 $\\frac{5}{14} \\times \\frac{7}{15} \\times 4$, 依此进行计算即可; (5) 先算小括号里面的加法，再算括号外面的除法即可；(6) 化除法为乘法, 把原式化为 $\\frac{7}{10} \\times \\frac{3}{4} + \\frac{1}{10} \\times \\frac{7}{4}$, 再根据分数乘法的计算方法化为 $\\frac{7}{10} \\times \\frac{3}{4} + \\frac{7}{10} \\times \\frac{1}{4}$, 最后运用乘法分配律化为 $\\frac{7}{10} \\times (\\frac{3}{4} + \\frac{1}{4})$, 依此进行计算即可。\n\n【详解】$\\frac{2}{13} \\times \\frac{5}{8} \\div \\frac{15}{26}$ $= \\frac{2}{13} \\times \\frac{5}{8} \\times \\frac{26}{15}$ $=\\frac{5}{52} \\times \\frac{26}{15} \\\\ = \\frac{1}{6}$ $\\frac{3}{5} \\div \\frac{7}{9} + \\frac{2}{5} \\div \\frac{7}{9}$ $= \\frac{3}{5} \\times \\frac{9}{7} + \\frac{2}{5} \\times \\frac{9}{7} \\\\ = (\\frac{3}{5} + \\frac{2}{5}) \\times \\frac{9}{7} \\\\ = 1 \\times \\frac{9}{7} \\\\ = \\frac{9}{7}$ $(\\frac{7}{3} - \\frac{5}{4}) \\div \\frac{5}{12}$ $= \\frac{13}{12} \\div \\frac{5}{12} \\\\ = \\frac{13}{12} \\times \\frac{12}{5} \\\\ = \\frac{13}{5}$ $\\frac{5}{14} \\div \\frac{1}{4} \\div \\frac{15}{7}$ $= \\frac{5}{14} \\times 4 \\times \\frac{7}{15} \\\\ = \\frac{5}{14} \\times \\frac{7}{15} \\times 4 \\\\ = \\frac{1}{6} \\times 4 \\\\ = \\frac{2}{3}$ $\\frac{13}{25} \\div (\\frac{2}{5} + \\frac{9}{10})$ $= \\frac{13}{25} \\div \\frac{13}{10} \\\\ = \\frac{13}{25} \\times \\frac{10}{13} \\\\ = \\frac{2}{5}$ $\\frac{7}{10} \\times \\frac{3}{4} + \\frac{1}{10} \\div \\frac{4}{7}$ $= \\frac{7}{10} \\times \\frac{3}{4} + \\frac{1}{10} \\times \\frac{7}{4} \\\\ = \\frac{7}{10} \\times \\frac{3}{4} + \\frac{7}{10} \\times \\frac{1}{4} \\\\ = \\frac{7}{10} \\times (\\frac{3}{4} + \\frac{1}{4}) \\\\ = \\frac{7}{10} \\times 1 \\\\ = \\frac{7}{10}$",
+      "knowledge_points": [
+        "整数乘法运算定律推广到分数乘法",
+        "分数的加、减法混合运算",
+        "分数的四则混合运算"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_366",
+      "original_id": "unit3_calc_366"
+    },{
+      "id": "unit3_calc_367",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "先写出等量关系,再列方程解答。一个数的 $\\frac{3}{5}$ 比它的 $\\frac{1}{7}$ 多 $\\frac{8}{21}$, 这个数是多少?",
+      "options": [],
+      "answer": "一个数 $\\times \\frac{3}{5}$ 一这个数 $\\times \\frac{1}{7} = \\frac{8}{21}$; $\\frac{5}{6}$",
+      "solution": "【分析】从题意可知：以这个数为单位'1'，根据求一个数的几分之几是多少，用乘法计算，就有等量关系：一个数 $\\times \\frac{3}{5}$ 一这个数 $\\times \\frac{1}{7} = \\frac{8}{21}$；设这个数为 $x$，根据等量关系，列出方程求出 $x$ 的值即可。\n\n【详解】根据分析, 可得等量关系: 一个数 $\\times \\frac{3}{5}$ 一这个数 $\\times \\frac{1}{7}$ $= \\frac{8}{21}$ 解: 设这个数为 $x$。$\\frac{3}{5} x - \\frac{1}{7} x = \\frac{8}{21}$ $\\frac{21}{35} x - \\frac{5}{35} x = \\frac{8}{21}$ $\\frac{16}{35} x = \\frac{8}{21}$ $x = \\frac{8}{21} \\div \\frac{16}{35}$ $x = \\frac{8}{21} \\times \\frac{35}{16}$ $x = \\frac{5}{6}$ 这个数是 $\\frac{5}{6}$。",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "分数的四则混合运算",
+        "列方程解含一个未知数的问题"
+      ],
+      "source": "23-24 六年级上·辽宁·课后作业",
+      "merged_id": "unit3_367",
+      "original_id": "unit3_calc_367"
+    },
+    {
+      "id": "unit3_calc_368",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "脱式计算。（能简算的要简算)\n\n$(\\frac{1}{2} - \\frac{1}{3}) \\div (\\frac{3}{4} - \\frac{5}{8})$ $\\frac{25}{7} \\times \\frac{3}{11} - \\frac{3}{7} \\div \\frac{11}{3}$ $(\\frac{7}{24} + \\frac{5}{12} - \\frac{1}{4}) \\times 48$",
+      "options": [],
+      "answer": "$\\frac{4}{3}$; $\\frac{6}{7}$; $22$",
+      "solution": "【分析】（1）先算括号里的减法，再算括号外的除法即可；(2) 先把除法改成乘法, 再利用乘法分配律进行简便计算即可; (3) 利用乘法分配律进行简便计算即可。\n\n【详解】(1) $(\\frac{1}{2} - \\frac{1}{3}) \\div (\\frac{3}{4} - \\frac{5}{8})$ $=\\frac{1}{6} \\div \\frac{1}{8}$ $= \\frac{1}{6} \\times 8$ $=\\frac{4}{3}$ (2) $\\frac{25}{7} \\times \\frac{3}{11} - \\frac{3}{7} \\div \\frac{11}{3}$ $=\\frac{25}{7} \\times \\frac{3}{11} - \\frac{3}{7} \\times \\frac{3}{11}$ $=(\\frac{25}{7} - \\frac{3}{7}) \\times \\frac{3}{11}$ $=\\frac{22}{7} \\times \\frac{3}{11}$ $= \\frac{6}{7}$ (3) $(\\frac{7}{24} + \\frac{5}{12} - \\frac{1}{4}) \\times 48$ $\\frac{7}{24} \\times 48 + \\frac{5}{12} \\times 48 - \\frac{1}{4} \\times 48 \\\\ = 14 + 20 - 12 \\\\ =22$",
+      "knowledge_points": [
+        "整数乘法运算定律推广到分数乘法",
+        "分数的四则混合运算"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_368",
+      "original_id": "unit3_calc_368"
+    },
+    {
+      "id": "unit3_solve_369",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "甲、乙两筐苹果共重 135 千克, 如果从甲筐拿出 $\\frac{1}{5}$ 后, 就和乙筐重量相等, 甲筐苹果重多少千克?",
+      "options": [],
+      "answer": "75千克",
+      "solution": "【分析】设甲筐苹果重 $x$ 千克; 则乙筐苹果重 $(135 - x)$ 千克; 如果从甲筐拿出 $\\frac{1}{5}$ 后, 就和乙筐重量相等, 即拿出 $\\frac{1}{5} x$ 千克, 甲筐还剩下 $(x - \\frac{1}{5} x)$ 千克, 等于乙筐苹果重量; 列方程: $x - \\frac{1}{5} x = 135 - x$, 解方程, 即可解答。\n\n【详解】解：设甲筐苹果重 $x$ 千克，则乙筐苹果重（135- $x$）千克。$x - \\frac{1}{5} x = 135 - x \\\\ \\frac{4}{5} x + x = 135 \\\\ \\frac{9}{5} x = 135 \\\\ x = 135 \\div \\frac{9}{5} \\\\ x = 135 \\times \\frac{5}{9} \\\\ x = 75$ 答: 甲筐苹果重 75 千克。",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "列方程解含两个未知数的问题"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_369",
+      "original_id": "unit3_solve_369"
+    },
+    {
+      "id": "unit3_solve_370",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "甲汽车每小时行 60 千米, 是乙汽车速度的 $\\frac{4}{5}$, 乙汽车每小时行多少千米?",
+      "options": [],
+      "answer": "75 千米",
+      "solution": "【分析】已知甲汽车每小时行 60 千米, 是乙汽车速度的 $\\frac{4}{5}$,把乙汽车的速度看作单位 '1', 单位 '1'未知, 用甲汽车的速度除以 $\\frac{4}{5}$, 即是乙汽车的速度。\n\n【详解】$60 \\div \\frac{4}{5}$ $= 60 \\times \\frac{5}{4}$ =75 (千米/小时) 答: 乙汽车每小时行 75 千米。",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "已知一个数的几分之几是多少，求这个数"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_370",
+      "original_id": "unit3_solve_370"
+    },
+    {
+      "id": "unit3_solve_371",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "某工程队挖一条水渠，第一天挖了全长的 $\\frac{1}{5}$，第二天比第一天少挖了 50 米，这时还剩下 200 米没有挖。这条水渠全长多少米？",
+      "options": [],
+      "answer": "250 米",
+      "solution": "【分析】把这条水渠的全长看作单位 '1', 第一天挖了全长的 $\\frac{1}{5}$, 第二天挖的比全长的 $\\frac{1}{5}$ 少 50 米, 还剩下 200 米没有挖,用剩下的 200 米减去 50 米, 对应的分率是 $1 - \\frac{1}{5} - \\frac{1}{5}$, 单位 '1'位置, 用对应的数量除以对应的分率即可解答。\n\n【详解】$(200 - 50) \\div (1 - \\frac{1}{5} - \\frac{1}{5})$ $150 \\div (\\frac{4}{5} - \\frac{1}{5}) \\\\ = 150 \\div \\frac{3}{5} \\\\ = 150 \\times \\frac{5}{3} \\\\ = 250 \\text{(米)}$ 答: 这条水渠全长 250 米。",
+      "knowledge_points": [
+        "已知一部分量占总量的几分之几及另一部分量,求总量"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_371",
+      "original_id": "unit3_solve_371"
+    },
+    {
+      "id": "unit3_solve_372",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "一堆煤，第一天运走了60吨，第二天运走了120吨，两天共运了这堆煤的 $\\frac{3}{5}$，这堆煤共有多少吨？",
+      "options": [],
+      "answer": "300吨",
+      "solution": "【分析】把这堆煤的总吨数看作单位 '1', 两天一共运走 $(60 + 120)$ 吨, 占总吨数的 $\\frac{3}{5}$, 单位 '1'未知, 用两天一共运走的吨数除以 $\\frac{3}{5}$, 即可求出总吨数。\n\n【详解】$(60 + 120) \\div \\frac{3}{5}$ $=180 \\div \\frac{3}{5} \\\\ = 180 \\times \\frac{5}{3} \\\\ = 300 (\\text{吨})$ 答: 这堆煤共有 300 吨。",
+      "knowledge_points": [
+        "分数的四则混合运算",
+        "已知一个数的几分之几是多少，求这个数"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_372",
+      "original_id": "unit3_solve_372"
+    },
+    {
+      "id": "unit3_solve_373",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "买一套演出服一共要花240元。裤子的单价是上衣单价的 $\\frac{3}{5}$，裤子和上衣的单价分别是多少元？",
+      "options": [],
+      "answer": "裤子 90 元; 上衣 150 元",
+      "solution": "【分析】将上衣的单价看作单位 '1', 根据 '裤子的单价是上衣单价的 $\\frac{3}{5}$', 可以设上衣的单价是 $x$ 元, 则裤子的单价是 $\\frac{3}{5} x$ 元; 根据'买一套演出服一共要花240元'可得出等量关系：上衣的单价+裤子的单价=一套演出服的价钱，据此列出方程，并求解。\n\n【详解】解：设上衣的单价是 $x$ 元，则裤子的单价是 $\\frac{3}{5} x$ 元。$x + \\frac{3}{5} x = 240 \\\\ \\frac{8}{5} x = 240 \\\\ x = 240 \\div \\frac{8}{5} \\\\ x = 240 \\times \\frac{5}{8} \\\\ x = 150$ 裤子: $240 - 150 = 90$ (元) 答: 裤子的单价是 90 元, 上衣的单价是 150 元。",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "应用等式的性质 2 解方程",
+        "解分数方程",
+        "列方程解含两个未知数的问题"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_373",
+      "original_id": "unit3_solve_373"
+    },
+    {
+      "id": "unit3_solve_374",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "加工一批零件, 如由李师傅单独加工, 需要 8 天完成, 如由林师傅单独加工,需要 12 天完成。如由李师傅和林师傅两人合作, 多少天能完成这批零件的 $\\frac{5}{6}$?",
+      "options": [],
+      "answer": "4天",
+      "solution": "【分析】把这批零件的总量看作单位'1'，根据工作效率 =工作总量÷工作时间，分别用 $1 \\div 8$ 和 $1 \\div 12$ 求得李师傅和林师傅各自的工作效率，然后根据工作时间 = 工作总量 ÷ 工作效率和。用 $\\frac{5}{6}$ 除以两人的工作效率和，即可求出多少天能完成这批零件的 $\\frac{5}{6}$。\n\n【详解】$1 \\div 8 = \\frac{1}{8}$ $1 \\div 12 = \\frac{1}{12} \\\\ \\frac{5}{6} \\div (\\frac{1}{8} + \\frac{1}{12}) \\\\ = \\frac{5}{6} \\div \\frac{5}{24} \\\\ = \\frac{5}{6} \\times \\frac{24}{5} \\\\ = 4 \\text{(天)}$ 答: 4 天能完成这批零件的 $\\frac{5}{6}$。",
+      "knowledge_points": [
+        "两人合作的工程问题",
+        "分数的四则混合运算"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_374",
+      "original_id": "unit3_solve_374"
+    },
+    {
+      "id": "unit3_solve_375",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "一桶油的 $\\frac{2}{5}$ 重 80 千克, 这桶油的 $\\frac{3}{4}$ 重多少千克?",
+      "options": [],
+      "answer": "150千克",
+      "solution": "【分析】将这桶油的重量看作单位 '1', 已知一个数的几分之几是多少, 求这个数用除法, 据此用 80 除以 $\\frac{2}{5}$ 求出这桶油的质量, 求一个数的几分之几是多少, 用乘法, 据此用这桶油的质量乘 $\\frac{3}{4}$ 即可解答。\n\n【详解】$80 \\div \\frac{2}{5} \\times \\frac{3}{4}$ $80 \\times \\frac{5}{2} \\times \\frac{3}{4} \\\\ = 200 \\times \\frac{3}{4} \\\\ = 150 \\text{(千克)}$ 答: 这桶油的 $\\frac{3}{4}$ 重 150 千克。",
+      "knowledge_points": [
+        "分数的乘、除法的混合运算",
+        "求一个数的几分之几的问题",
+        "已知一个数的几分之几是多少，求这个数"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_375",
+      "original_id": "unit3_solve_375"
+    },
+    {
+      "id": "unit3_solve_376",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "某水果店今天上午购进苹果和梨共 210 千克。当天卖出苹果总数的 $\\frac{2}{5}$, 同时也卖出了梨总数的 $\\frac{2}{5}$, 这时苹果还剩下 56 千克。请问: 这时梨还剩下多少千克?",
+      "options": [],
+      "answer": "70千克",
+      "solution": "【分析】把苹果总数看作单位'1'，已知当天卖出苹果总数的 $\\frac{2}{5}$，苹果还剩下56千克，剩余的苹果数量占苹果总数的 $(1 - \\frac{2}{5})$，根据分数除法的意义，用剩余的苹果数量÷（$1 - \\frac{2}{5}$）即可求出苹果总数；然后用苹果和梨的总数减去苹果总数，即可求出梨的总数；已知卖出了梨总数的 $\\frac{2}{5}$，把梨的总数看作单位'1'，剩余的梨的数量占梨总数的 $(1 - \\frac{2}{5})$，根据分数乘法的意义，用梨总数乘 $(1 - \\frac{2}{5})$ 即可求出剩余的梨的数量。\n\n【详解】苹果总数：$56 \\div (1 - \\frac{2}{5}) \\\\ = 56 \\div \\frac{3}{5}\\\\=56 \\times \\frac{5}{3} \\\\ = \\frac{280}{3} (\\text{千克})$ 梨的总数: $210 - \\frac{280}{3} = \\frac{350}{3}$ (千克) 剩余的梨的数量:$\\frac{350}{3} \\times (1 - \\frac{2}{5}) \\\\ = \\frac{350}{3} \\times \\frac{3}{5} \\\\ = 70 \\text{(千克)}$ 答: 这时梨还剩下 70 千克。",
+      "knowledge_points": [
+        "已知一部分量占总量的几分之几及另一部分量,求总量",
+        "已知总量及一部分分率, 求另一部分量"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_376",
+      "original_id": "unit3_solve_376"
+    },
+    {
+      "id": "unit3_solve_377",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "我会列式。（只列式，不计算）一堆苹果，第一辆车运走了 $\\frac{1}{4}$，第二辆车运走了 $\\frac{1}{5}$。第一辆车比第二辆车多运 $100kg$，这堆苹果共多少千克？",
+      "options": [],
+      "answer": "$100 \\div (\\frac{1}{4} - \\frac{1}{5})$",
+      "solution": "【分析】将这堆苹果的总质量的看成单位 '1', 第一次比第二次多运走的分率 = 第一次运走的分率 - 第二次运走的分率。也就是多运走的苹果的质量是总质量的 $(\\frac{1}{4} - \\frac{1}{5})$, 多运走的数量是 $100kg$, 已知一个数的几分之几是多少, 求这个数用除法。\n\n【详解】$100 \\div (\\frac{1}{4} - \\frac{1}{5})$ $=100 \\div \\frac{1}{20} \\\\ = 100 \\times 20 \\\\ = 2000 (\\mathrm{kg})$ 则列式为: $100 \\div (\\frac{1}{4} - \\frac{1}{5})$",
+      "knowledge_points": [
+        "分数的四则混合运算",
+        "已知一个数的几分之几是多少，求这个数"
+      ],
+      "source": "24-25 六年级上·全国·单元测试",
+      "merged_id": "unit3_377",
+      "original_id": "unit3_solve_377"
+    },
+    {
+      "id": "unit3_solve_378",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "我会列式。（只列式，不计算）将一段长为 $\\frac{9}{10}$ 米的木料锯 9 次, 平均每段长多少米?",
+      "options": [],
+      "answer": "$\\frac{9}{10} \\div (9 + 1)$",
+      "solution": "【分析】锯的段数 = 锯的次数 + 1，据此确定段数，木料长度 ÷ 段数 = 每段长度，据此列式。\n\n【详解】$\\frac{9}{10} \\div (9 + 1)$ $=\\frac{9}{10} \\div 10 \\\\ = \\frac{9}{10} \\times \\frac{1}{10} \\\\ = \\frac{9}{100} (\\text{米})$ 答: 平均每段长 $\\frac{9}{100}$ 米。",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "植树问题（两端都不栽）"
+      ],
+      "source": "24-25 六年级上·全国·单元测试",
+      "merged_id": "unit3_378",
+      "original_id": "unit3_solve_378"
+    },
+    {
+      "id": "unit3_solve_379",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "我会列式。（只列式，不计算）某书店第一季度售出图书 1600 本, 第一季度比第二季度少售出 $\\frac{3}{5}$。第二季度售出图书多少本?",
+      "options": [],
+      "answer": "$1600 \\div (1 - \\frac{3}{5})$",
+      "solution": "【分析】将第二季度售出本数看作单位 '1', 第一季度售出本数是第二季度的 $(1 - \\frac{3}{5})$, 第一季度售出本数 $\\div$ 对应分率 $=$ 第二季度售出本数, 据此列式。\n\n【详解】$1600 \\div (1 - \\frac{3}{5})$ $=1600 \\div \\frac{2}{5} \\\\ = 1600 \\times \\frac{5}{2} \\\\ = 4000 (\\text{本})$ 答: 第二季度售出图书 4000 本。",
+      "knowledge_points": [
+        "已知比一个数多/少几分之几是多少，求这个数"
+      ],
+      "source": "24-25 六年级上·全国·单元测试",
+      "merged_id": "unit3_379",
+      "original_id": "unit3_solve_379"
+    },
+    {
+      "id": "unit3_solve_380",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "某超市一天卖出纯奶 25 箱，是卖出酸奶箱数的 $\\frac{5}{8}$，卖出酸奶多少箱？",
+      "options": [],
+      "answer": "40箱",
+      "solution": "【分析】已知一个数的几分之几是多少, 求这个数, 用除法计算, 据此用卖出纯奶的箱数除以 $\\frac{5}{8}$, 即可求出卖出酸奶多少箱。\n\n【详解】$25 \\div \\frac{5}{8}$ $=25 \\times \\frac{8}{5} \\\\ = 40 \\text{(箱)}$ 答: 卖出酸奶 40 箱。",
+      "knowledge_points": [
+        "已知一个数的几分之几是多少，求这个数"
+      ],
+      "source": "24-25 六年级上·全国·单元测试",
+      "merged_id": "unit3_380",
+      "original_id": "unit3_solve_380"
+    },
+    {
+      "id": "unit3_solve_381",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "明明和爷爷一起去散步。明明绕花园走一圈需要 $\\frac{1}{12}$ 小时, 爷爷需要 $\\frac{1}{10}$ 小时。\n\n(1) 如果两人同时同地出发，相背而行，几小时后相遇？\n(2) 如果两人同时同地出发, 同方向而行, 几小时后明明都超出爷爷一整圈?",
+      "options": [],
+      "answer": "(1) $\\frac{1}{22}$ 小时; (2) $\\frac{1}{2}$ 小时",
+      "solution": "【分析】（1）假设绕花园走一圈的路程为 1，先根据'速度 = 路程 ÷ 时间'表示出明明的速度和爷爷的速度，再根据'相遇时间 = 总路程 ÷ 速度和'求出两人的相遇时间；(2) 假设绕花园走一圈的路程为 1, 明明超出爷爷一整圈,说明两人的路程差为 1, 根据 '路程差÷速度差 = 追及时问' 求出明明超出爷爷一整圈需要的时间, 据此解答。\n\n【详解】(1) $1 \\div (1 \\div \\frac{1}{10} + 1 \\div \\frac{1}{12})$ $=1 \\div (1 \\times 10 + 1 \\times 12) \\\\ = 1 \\div (10 + 12) \\\\ = 1 \\div 22$ $\\frac{1}{22}$ (时) 答: 如果两人同时同地出发, 相背而行, $\\frac{1}{22}$ 小时后相遇。(2) $1 \\div (1 \\div \\frac{1}{12} - 1 \\div \\frac{1}{10})$ $1 \\div (1 \\times 12 - 1 \\times 10) \\\\ = 1 \\div (12 - 10)\\\\= 1 \\div 2\\\\=\\frac{1}{2}$ (时) 答: 如果两人同时同地出发, 同方向而行, $\\frac{1}{2}$ 小时后明明超出爷爷一整圈。",
+      "knowledge_points": [
+        "追及问题",
+        "分数的四则混合运算",
+        "相遇问题"
+      ],
+      "source": "24-25 六年级上·全国·单元测试",
+      "merged_id": "unit3_381",
+      "original_id": "unit3_solve_381"
+    },
+    {
+      "id": "unit3_solve_382",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "菜场运来番茄 300 千克, 是黄瓜的 $\\frac{4}{5}$, 运来黄瓜多少千克? (列方程解答)",
+      "options": [],
+      "answer": "375千克",
+      "solution": "【分析】据题意可知, 把黄瓜的质量看作单位 '1', 假设运来黄瓜 $x$ 千克, 则运来的番茄有 $\\frac{4}{5} x$。根据番茄的质量等于 300, 列方程解答, 即可得解。\n\n【详解】解：设运来黄瓜 $x$ 千克。$\\frac{4}{5} x = 300\\\\x = 300 \\div \\frac{4}{5} \\\\ x = 300 \\times \\frac{5}{4} \\\\ x = 375$ 答: 运来黄瓜 375 千克。",
+      "knowledge_points": [
+        "已知一个数的几分之几是多少，求这个数",
+        "列方程解含一个未知数的问题"
+      ],
+      "source": "24-25 六年级上·全国·课后作业",
+      "merged_id": "unit3_382",
+      "original_id": "unit3_solve_382"
+    },
+    {
+      "id": "unit3_solve_383",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "某工程队挖一条 8千米长的隧道, 已经挖了 3 天, 平均每天挖 $\\frac{6}{5}$ 千米, 剩下的要 4 天挖完, 这 4 天平均每天挖多少千米?",
+      "options": [],
+      "answer": "1.1 千米",
+      "solution": "【分析】平均数 $\\times$ 份数 = 总数量，设平均每天要挖 $x$ 千米，根据剩下平均每天挖的距离 $\\times$ 剩下天数 + 平均每天挖的距离 $\\times$ 已经挖的天数 = 隧道总长度，列出方程解答即可。\n\n【详解】解：设这4天平均每天挖 $x$ 千米。$4x + \\frac{6}{5} \\times 3 = 8$ $4x + \\frac{18}{5} = 8 \\\\ 4x + \\frac{18}{5} - \\frac{18}{5} = 8 - \\frac{18}{5} \\\\ 4x = 4.4 \\\\ 4x \\div 4 = 4.4 \\div 4 \\\\ x = 1.1$ 答: 这 4 天平均每天挖 1.1 千米。",
+      "knowledge_points": [
+        "有具体量的工程问题",
+        "分数乘整数",
+        "列方程解含一个未知数的问题"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_383",
+      "original_id": "unit3_solve_383"
+    },
+    {
+      "id": "unit3_solve_384",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "小明的爸爸买了一筐苹果和一筐梨，一筐苹果重50千克，苹果质量的 $\\frac{3}{5}$ 与梨的 $\\frac{2}{3}$ 正好相等，这筐梨重多少千克？",
+      "options": [],
+      "answer": "45千克",
+      "solution": "【分析】已知一筐苹果重 50 千克, 苹果质量的 $\\frac{3}{5}$ 与梨的 $\\frac{2}{3}$ 正好相等, 则用苹果的质量乘 $\\frac{3}{5}$ 求出梨的 $\\frac{2}{3}$ 是多少, 再用梨的 $\\frac{2}{3}$除以 $\\frac{2}{3}$ 即可求出梨的质量。\n\n【详解】$50 \\times \\frac{3}{5} \\div \\frac{2}{3}$ $=30 \\div \\frac{2}{3} \\\\ = 30 \\times \\frac{3}{2} \\\\ = 45 \\text{(千克)}$ 答: 这筐梨重 45 千克。",
+      "knowledge_points": [
+        "分数的乘、除法的混合运算",
+        "已知一个数的几分之几是多少，求这个数"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_384",
+      "original_id": "unit3_solve_384"
+    },
+    {
+      "id": "unit3_solve_385",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "困难",
+      "question": "《九章算术》中有一道题: '今有人持米出三关, 外关三而取一, 中关五而取一，内关七而取一，余米五斗。问持米几何？'题意：有人背米过关卡，经过外关时，用全部米的 $\\frac{1}{3}$ 纳税，过中时用所余米的 $\\frac{1}{5}$ 纳税, 经过内关时再用余米的 $\\frac{1}{7}$ 纳税, 最后还剩下5斗米。这个人原来背多少斗米出关？",
+      "options": [],
+      "answer": "$10\\frac{15}{16}$ 斗",
+      "solution": "【分析】将过内关时剩余米的斗数看作单位'1'，最后剩的米的斗数是过内关时剩余米的 $(1 - \\frac{1}{7})$，最后剩的米的斗数÷对应分率 $=$ 过内关时剩余米的斗数；再将过中关时剩余米的斗数看作单位'1'，过内关时剩余米的斗数是过中关时剩余米的 $(1 - \\frac{1}{5})$，过中关时剩余米的斗数÷对应分率 $=$ 过中关时剩余米的斗数；最后将背的米的总斗数看作单位'1'，过中关时剩余米的斗数是背的米的总斗数的 $(1 - \\frac{1}{3})$，过中关时剩余米的斗数÷对应分率 $=$ 背的米的总斗数，据此列式解答。关键是确定单位 '1', 理解分数除法的意义, 根据部分数量 $\\div$ 对应分率 $=$ 整体数量, 列式解答。\n\n【详解】$5 \\div (1 - \\frac{1}{7}) \\div (1 - \\frac{1}{5}) \\div (1 - \\frac{1}{3})$ $= 5 \\div \\frac{6}{7} \\div \\frac{4}{5} \\div \\frac{2}{3} \\\\ = 5 \\times \\frac{7}{6} \\times \\frac{5}{4} \\times \\frac{3}{2} \\\\ = \\frac{35}{6} \\times \\frac{5}{4} \\times \\frac{3}{2} \\\\ =\\frac{175}{24} \\times \\frac{3}{2} \\\\ = 10\\frac{15}{16} (\\text{斗})$ 答: 这个人原来背 $10\\frac{15}{16}$ 斗米出关。",
+      "knowledge_points": [
+        "运用转化法或倒推法解决稍复杂的分数应用题",
+        "分数的四则混合运算",
+        "已知一个数的几分之几是多少，求这个数"
+      ],
+      "source": "23-24 六年级上·全国·课后作业",
+      "merged_id": "unit3_385",
+      "original_id": "unit3_solve_385"
+    },
+    {
+      "id": "unit3_solve_386",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "丢番图的墓志铭中写道: '童年占六分之一, 又过十二分之一, 两颊长胡,再过七分之一，点燃起结婚的蜡烛，五年之后天赐贵子，可怜迟到的宁馨儿，享年仅及其父之半，便进入冰冷的墓。悲伤只能用数论研究去弥补，又过四年，他也走完了人生的旅途。'你能计算出丢番图的寿命吗？",
+      "options": [],
+      "answer": "84岁",
+      "solution": "【分析】根据题意, 把丢番图的寿命看作单位 '1', 题中的数量关系可用下图表示:\n\n![](images/6a5ea4a79cd6081b46bbb20acd62538b4295a4c3e4b59ecdae15e63a8aa42311.jpg)\n\n从图中可以找到已知具体数 $(5 + 4)$ 对应的分率 $(1 - \\frac{1}{6} - \\frac{1}{12} - \\frac{1}{7} - \\frac{1}{2})$, 根据已知一个数的几分之几是多少, 求这个数用除法计算,用已知具体数除以其对应的分率。据此解答。\n\n【详解】$(5 + 4) \\div (1 - \\frac{1}{6} - \\frac{1}{12} - \\frac{1}{7} - \\frac{1}{2})$ $9 \\div (\\frac{5}{6} - \\frac{1}{12} - \\frac{1}{7} - \\frac{1}{2}) \\\\=9 \\div (\\frac{9}{12} - \\frac{1}{7} - \\frac{1}{2}) \\\\ = 9 \\div (\\frac{17}{28} - \\frac{1}{2}) \\\\ =9 \\div \\frac{3}{28} \\\\ =9 \\times \\frac{28}{3} \\\\ =\\text{84 (岁)}$ 答: 我能计算出丢番图的寿命, 是 84 岁。",
+      "knowledge_points": [
+        "已知一部分量占总量的几分之几及另一部分量,求总量",
+        "分数的四则混合运算"
+      ],
+      "source": "23-24 六年级上·全国·课后作业",
+      "merged_id": "unit3_386",
+      "original_id": "unit3_solve_386"
+    },
+    {
+      "id": "unit3_solve_387",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "困难",
+      "question": "一组割草人要把两片草地的草割掉，大片草地面积是小片草地面积的2倍。上午大家在大片草地上割草，午后分成两组，一半人继续在大片草地上割草，到傍晚收工时恰好割完；另一半人到小片草地上割草，到傍晚还剩一小块，这一小块第二天由一个人去割，恰好需要一天时间。请你算一下，这组割草人共有多少人？",
+      "options": [],
+      "answer": "8人",
+      "solution": "【分析】上午所有人都在大草地割草，下午一半人继续在大草地割草到傍晚割完，所以大草地的割草量相当于半组人割了3个半天，把大片草地面积看作单位'1'，那么一半割草人半天时间可割大片草地面积的 $\\frac{1}{3}$；大片草地面积是小片草地面积的 2 倍, 即小片草地面积是大片草地面积的 $\\frac{1}{2}$,因此小片草地上留下的没割的一小块是大片草地的 $\\frac{1}{2} - \\frac{1}{3} = \\frac{1}{6}$,这样, 一个割草人每天能割大片草地的 $\\frac{1}{6}$, 而第一天割草一共是割了大片草地的 $1 + \\frac{1}{3} = \\frac{4}{3}$, 用 $\\frac{4}{3}$ 除以 $\\frac{1}{6}$, 即可求出这组割草人共有多少人, 据此解答。这道题关键是根据题意得出：大草地的割草量相当于半组人割了 3 个半天, 求出一半割草人半天时间能割草的面积, 再求出算出小草地割草的剩余量, 从而得出单人割草的效率, 最后再求出全组人数。\n\n【详解】$1 \\div (2 + 1)$ $=1 \\div 3 \\\\ = \\frac{1}{3}$ $\\frac{1}{2} - \\frac{1}{3} = \\frac{1}{6} \\\\ (1 + \\frac{1}{3}) \\div \\frac{1}{6} \\\\ = \\frac{4}{3} \\div \\frac{1}{6} \\\\ = \\frac{4}{3} \\times 6 \\\\ = 8 \\text{(人)}$ 答: 这组割草人共有 8 人。",
+      "knowledge_points": [
+        "两人合作的工程问题",
+        "分数的四则混合运算"
+      ],
+      "source": "23-24 六年级上·全国·课后作业",
+      "merged_id": "unit3_387",
+      "original_id": "unit3_solve_387"
+    },
+    {
+      "id": "unit3_solve_388",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "星光服装厂一月份计划生产一批防护服，实际上半月完成计划的 $\\frac{5}{8}$，下半月完成计划的 $\\frac{2}{3}$，结果超额完成 2100 件防护服。计划生产多少件防护服？",
+      "options": [],
+      "answer": "7200件",
+      "solution": "【分析】把一月份计划生产一批防护服的件数看作单位'1'，实际上半月完成计划的分率加上下半月完成计划的分率，再减去1，求出实际一个月超额完成了计划的几分之几，对应的是 2100 件, 再根据已知一个数的几分之几是多少,求这个数用除法解答, 列式为: $2100 \\div (\\frac{5}{8} + \\frac{2}{3} - 1)$。\n\n【详解】$2100 \\div (\\frac{5}{8} + \\frac{2}{3} - 1)$ $=2100 \\div (\\frac{15}{24} +\\frac{16}{24} - 1) \\\\ = 2100 \\div \\frac{7}{24} \\\\ = 2100 \\times \\frac{24}{7} \\\\ = 7200 \\text{(件)}$ 答：计划生产 7200 件防护服。",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "分数的四则混合运算",
+        "已知一个数的几分之几是多少，求这个数",
+        "单位'1'的认识与确定"
+      ],
+      "source": "23-24 六年级上·辽宁·单元测试",
+      "merged_id": "unit3_388",
+      "original_id": "unit3_solve_388"
+    },
+    {
+      "id": "unit3_solve_389",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "六年级有男生 330人, 女生 280 人, 六年级学生正好是全校学生人数的 $\\frac{1}{10}$, 全校有学生多少人?",
+      "options": [],
+      "answer": "6100人",
+      "solution": "【分析】先求出六年级学生的总数, 把全校学生总数看作单位 '1', 用六年级学生的总数除以所占的份数即可求出全校共有的学生数。\n\n【详解】$(330 + 280) \\div \\frac{1}{10}$ $=610 \\div \\frac{1}{10} \\\\ = 610 \\times 10 \\\\ = 6100 \\text{(人)}$ 答: 全校有学生 6100 人。",
+      "knowledge_points": [
+        "已知一个数的几分之几是多少，求这个数"
+      ],
+      "source": "22-23 六年级上·江苏盐城·期中",
+      "merged_id": "unit3_389",
+      "original_id": "unit3_solve_389"
+    },
+    {
+      "id": "unit3_solve_390",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "小明步行 $\\frac{4}{3}$ 千米用 $\\frac{1}{4}$ 小时。照这样计算, 他每小时行多少千米, 行 1 千米需要多少小时?",
+      "options": [],
+      "answer": "$\\frac{16}{3}$ 千米; $\\frac{3}{16}$ 小时",
+      "solution": "【分析】已知小明步行 $\\frac{4}{3}$ 千米用 $\\frac{1}{4}$ 小时, 根据: 速度 = 路程 ÷ 时间, 用 $\\frac{4}{3} \\div \\frac{1}{4}$, 即可求出每小时行驶多少千米; 再用 1 千米除以速度, 即可求出需要多少小时。\n\n【详解】$\\frac{4}{3} \\div \\frac{1}{4} = \\frac{4}{3} \\times 4 = \\frac{16}{3}$ （千米）$1 \\div \\frac{16}{3} = 1 \\times \\frac{3}{16} = \\frac{3}{16}$ (小时)答: 小明每小时行 $\\frac{16}{3}$ 千米, 行 1 千米需要 $\\frac{3}{16}$ 小时。",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "分数与分数的除法",
+        "基础行程问题"
+      ],
+      "source": "22-23 六年级上·江苏盐城·期中",
+      "merged_id": "unit3_390",
+      "original_id": "unit3_solve_390"
+    },
+    {
+      "id": "unit3_solve_391",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "光明小学六年级共有学生 193 人, 选出女同学的 $\\frac{1}{10}$ 和 3 名男同学参加歌唱比赛,剩下的男、女同学人数刚好相等。六年级男、女同学各有多少人?",
+      "options": [],
+      "answer": "女生：100人；男生：93人",
+      "solution": "【分析】设女同学有 $x$ 人, 则男同学有 $(193 - x)$, 选出女同学的 $\\frac{1}{10}$, 即选出女同学 $\\frac{1}{10} x$ 人, 还剩下 $(x - \\frac{1}{10} x)$ 人; 男同学选出 3 名, 男同学还剩下 (193-x-3) 人, 剩下的男、女同学人数刚好相等, 剩下的女同学人数 = 剩下的男同学人数, 列方程: $x - \\frac{1}{10} x = 193 - x - 3$, 解方程, 即可解答。\n\n【详解】解：设女同学有 $\\mathbf{x}$ 人，则男同学有（193-x）人。$x - \\frac{1}{10} x = 193 - x - 3 \\\\ \\frac{9}{10} x = 190 - x \\\\ \\frac{9}{10} x + x = 190 \\\\ \\frac{19}{10} x = 190 \\\\ x = 190 \\div \\frac{19}{10} \\\\ x = 190 \\times \\frac{10}{19} \\\\ x = 100$ 男同学: $193 - 100 = 93$ (人)答: 女同学有 100 人, 男同学有 93 人。",
+      "knowledge_points": [
+        "列方程解含两个未知数的问题"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_391",
+      "original_id": "unit3_solve_391"
+    },
+    {
+      "id": "unit3_solve_392",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "小芳同学在计算一道除法算式时，不小心把被除数写成了原来的5倍，再除以6, 结果是 $\\frac{1}{30}$。请算出这道除法算式正确的被除数是多少?",
+      "options": [],
+      "answer": "$\\frac{1}{25}$",
+      "solution": "【分析】先根据错误的计算求出错误的被除数，因为错误的被除数是原来的 5 倍，所以再除以 5 就能得到正确的被除数。\n\n【详解】错误的被除数为: $\\frac{1}{30} \\times 6 = \\frac{1}{5}$, 因为这个错误的被除数是原来被除数的 5 倍, 所以正确的被除数为: $\\frac{1}{5} \\div 5 = \\frac{1}{5} \\times \\frac{1}{5} = \\frac{1}{25}$答: 这道除法算式正确的被除数是 $\\frac{1}{25}$。",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "商的变化规律及应用",
+        "乘、除法的意义和各部分间的关系"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_392",
+      "original_id": "unit3_solve_392"
+    },
+    {
+      "id": "unit3_solve_393",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "一辆汽车从甲地开往乙地，已经行驶了全程的 $\\frac{3}{7}$，还剩 160 千米，甲、乙两地相距多少千米？",
+      "options": [],
+      "answer": "280千米",
+      "solution": "【分析】由题意可知, 将甲乙两地间的全程看作单位 '1',平均分成 7 份, 已经行驶了全程的 $\\frac{3}{7}$, 则还剩 $(1 - \\frac{3}{7})$。已知比较量求单位 '1', 用比较量除以对应分率即可。据此解答。\n\n【详解】$160 \\div (1 - \\frac{3}{7})$ $=160 \\div \\frac{4}{7} \\\\ = 160 \\times \\frac{7}{4} \\\\ = 280 \\text{(千米)}$ 答: 甲、乙两地相距 280 千米。",
+      "knowledge_points": [
+        "已知一部分量占总量的几分之几及另一部分量,求总量"
+      ],
+      "source": "23-24 五年级下·陕西宝鸡·期末",
+      "merged_id": "unit3_393",
+      "original_id": "unit3_solve_393"
+    },
+    {
+      "id": "unit3_solve_394",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "只列综合算式, 不计算。一项工作，甲单独做需12天，乙每天完成这项工作的 $\\frac{1}{20}$,甲、乙合作这项工作需几天完成？\n\n列式:_____________________________________",
+      "options": [],
+      "answer": "$1 \\div (\\frac{1}{12} + \\frac{1}{20})$",
+      "solution": "【分析】把这项工作看作单位'1'，甲单独做12天完成，每天完成这项工作的 $\\frac{1}{12}$; 甲乙合作一天就可以完成这项工作的 $(\\frac{1}{12} + \\frac{1}{20})$, 根据工作时间 $=$ 工作总量 $\\div$ 工作效率, 即可列式解答。\n\n【详解】根据分析可得：$1 \\div (\\frac{1}{12} + \\frac{1}{20}) \\\\ = 1 \\div \\frac{2}{15} \\\\ = 1 \\times \\frac{15}{2} \\\\ = 7.5 \\text{(天)}$ 答: 甲、乙合作这项工作需 7.5 天完成。",
+      "knowledge_points": [
+        "两人合作的工程问题",
+        "分数的四则混合运算"
+      ],
+      "source": "2024·浙江宁波·小升初真题",
+      "merged_id": "unit3_394",
+      "original_id": "unit3_solve_394"
+    },
+    {
+      "id": "unit3_solve_395",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "某品牌羽毛球今年六月份的价格是每筒90元，比一月份上涨了 $\\frac{1}{9}$。该品牌羽毛球一月份的价格是每筒多少元？",
+      "options": [],
+      "answer": "81元",
+      "solution": "【分析】把一月份的羽毛球价格看作单位 '1', 六月份价格比一月份上涨了 $\\frac{1}{9}$, 六月份的羽毛球价格是 1 月份价格的 (1 + $\\frac{1}{9}$), 为 90 元。因此一月份的价格是每筒 $90 \\div (1 + \\frac{1}{9})$ 元。\n\n【详解】$90 \\div (1 + \\frac{1}{9})$ $=90 \\div \\frac{10}{9} \\\\ = 90 \\times \\frac{9}{10} \\\\ = 81 \\text{(元)}$ 答: 该品牌羽毛球一月份的价格是每筒 81 元。",
+      "knowledge_points": [
+        "已知比一个数多/少几分之几是多少，求这个数"
+      ],
+      "source": "23-24 六年级下·陕西渭南·期末",
+      "merged_id": "unit3_395",
+      "original_id": "unit3_solve_395"
+    },
+    {
+      "id": "unit3_solve_396",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "中国天文科技发展日新月异，'中国天眼'是全球最大且最灵敏的射电望远镜，超越了美国著名的天文望远镜阿雷西博。已知阿雷西博的口径长350米，是'中国天眼'口径的 $\\frac{7}{10}$，'中国天眼'的口径是多少米？（先画图分析，再列方程解答）",
+      "options": [],
+      "answer": "500米；图见解析",
+      "solution": "【分析】根据题意, 可设 '中国天眼' 的口径是 $x$ 米, '中国天眼'的口径 $\\times \\frac{7}{10} =$ 阿雷西博的口径, 据此解答即可。\n\n【详解】\n\n![](images/dc41bb825cdbe595c4282ee2eceac5553bf67c27d00b2ca6adcab7d040172671.jpg)\n\n解：设'中国天眼'的口径是 $x$ 米，根据题意列方程：$\\frac{7}{10} x = 350$ $\\frac{7}{10} x \\div \\frac{7}{10} = 350 \\div \\frac{7}{10} \\\\ x = 500$ 答: '中国天眼' 的口径是 500 米。",
+      "knowledge_points": [
+        "已知一个数的几分之几是多少，求这个数",
+        "列方程解含一个未知数的问题"
+      ],
+      "source": "23-24 五年级下·广西贺州·期末",
+      "merged_id": "unit3_396",
+      "original_id": "unit3_solve_396"
+    },
+    {
+      "id": "unit3_solve_397",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "困难",
+      "question": "有一位数学家, 他生命的 $\\frac{1}{6}$ 是幸福的童年, 又过了童年的 $\\frac{1}{2}$ 后, 脸上长了细细的胡须, 他结婚后度过了他人生的 $\\frac{1}{7}$, 又过 5 年得到了一个可爱的儿子, 但他孩子的寿命只有这位数学家寿命的 $\\frac{1}{2}$, 这位数学家在孩子死后悲痛地度过了 4 年后离开了人间, 这位数学家是古希腊人, 请你推算一下他活了多少岁?",
+      "options": [],
+      "answer": "84岁",
+      "solution": "【分析】把数学家一生的年龄看作单位 '1', 已知他生命的 $\\frac{1}{6}$ 是幸福的童年, 则童年的 $\\frac{1}{2}$ 相当于他生命的 $\\frac{1}{6} \\times \\frac{1}{2}$, 根据题意可知, 数学家的 (5+4) 年占生命长度的 $(1 - \\frac{1}{6} - \\frac{1}{6} \\times \\frac{1}{2} - \\frac{1}{7} - \\frac{1}{2})$, 根据分数除法的意义, 用 (5+4) $\\div$ $(1 - \\frac{1}{6} - \\frac{1}{6} \\times \\frac{1}{2} - \\frac{1}{7} - \\frac{1}{2})$ 即可求出数学家一生的年龄。本题主要考查了分数除法的应用, 找到对应量以及对应的分率是解答本题的关键。\n\n【详解】$(5 + 4) \\div (1 - \\frac{1}{6} - \\frac{1}{6} \\times \\frac{1}{2} - \\frac{1}{7} - \\frac{1}{2})$ $(5 + 4) \\div (1 - \\frac{1}{6} - \\frac{1}{12} - \\frac{1}{7} - \\frac{1}{2}) \\\\ = 9 \\div \\frac{9}{84} \\\\ = 9 \\times \\frac{84}{9} \\\\ = 84 \\text{(岁)}$ 答: 他活了 84 岁。",
+      "knowledge_points": [
+        "分数的四则混合运算",
+        "已知一个数的几分之几是多少，求这个数"
+      ],
+      "source": "2020·广西柳州·小升初模拟",
+      "merged_id": "unit3_397",
+      "original_id": "unit3_solve_397"
+    },
+    {
+      "id": "unit3_solve_398",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "甲、乙两人共有邮票若干张, 其中甲占 $\\frac{9}{20}$, 若乙给甲 12 张, 则乙余下的张数占总数的 $\\frac{2}{5}$。两人共有邮票多少张?",
+      "options": [],
+      "answer": "80 张",
+      "solution": "【分析】把两人共有的邮票总数看作单位 '1'。设两人共有邮票 x 张, 那么甲原来有 $\\frac{9}{20} x$ 张, 乙原来有 $x-\\frac{9}{20} x = \\frac{11}{20} x$ 张,乙给甲 12 张后, 乙余下的张数为总数的 $\\frac{2}{5}$, 可列出方程 $\\frac{11}{20} x-12=\\frac{2}{5} x$, 算出结果即可。\n\n【详解】解：设两人共有邮票 $x$ 张，乙原来有 $x - \\frac{9}{20} = \\frac{11}{20} x$, 乙给甲 12 张后, 乙余下的张数为 $\\frac{11}{20} x - 12$, 此时乙余下的张数占总数的 $\\frac{2}{5}$, 可列方程: $x - \\frac{9}{20} x - 12 = \\frac{2}{5} x$ 解: $\\frac{11}{20} x - 12 + 12 = \\frac{2}{5} x + 12$ $\\frac{11}{20} x = \\frac{2}{5} x + 12$ $\\frac{11}{20} x - \\frac{2}{5} x = \\frac{2}{5} x + 12 - \\frac{2}{5} x$ $\\frac{3}{20} x = 12$ $\\frac{3}{20} x \\div \\frac{3}{20} = 12 \\div \\frac{3}{20}$ $\\frac{3}{20} x \\times \\frac{20}{3} = 12 \\times \\frac{20}{3} \\\\ x = 80$ 答: 两人共有邮票 80 张。",
+      "knowledge_points": [
+        "列方程解含两个未知数的问题"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_398",
+      "original_id": "unit3_solve_398"
+    },
+    {
+      "id": "unit3_solve_399",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "困难",
+      "question": "某科技发明兴趣小组中女生占 $\\frac{7}{12}$, 后来又转来了 15 名女生, 这样女生占总人数的 $\\frac{3}{5}$。这个兴趣小组的男生有多少人?",
+      "options": [],
+      "answer": "150人",
+      "solution": "【分析】根据题意可知：设原来兴趣小组有 $x$ 人，则女生有 $\\frac{7}{12} x$ 人；后来又转来了 15 名女生，后来兴趣小组就有 $(x + 15)$ 人，后来的女生人数是 $(x + 15) \\times \\frac{3}{5}$，根据原来的女生人数 $+15 =$ 后来的女生人数，列出方程，求出原来兴趣小组的人数，再乘 $(1 - \\frac{7}{12})$，即求出男生人数。理解题意, 找出合适的等量关系, 进而列出方程是解此题的关键。\n\n【详解】解：设这个兴趣小组原有 $x$ 人。$\\frac{7}{12} x + 15 = (x + 15) \\times \\frac{3}{5} \\\\ \\frac{7}{12} x + 15 = \\frac{3}{5} x + 15 \\times \\frac{3}{5} \\\\ \\frac{7}{12} x + 15 = \\frac{3}{5} x + 9 \\\\ \\frac{3}{5} x - \\frac{7}{12} x = 15 - 9 \\\\ \\frac{1}{60} x = 6 \\\\ x = 6 \\div \\frac{1}{60} \\\\ x = 6 \\times 60 \\\\ x = 360 \\\\ 360 \\times (1 - \\frac{7}{12}) \\\\ = 360 \\times \\frac{5}{12} \\\\= \\text{150} (\\text{人})$ 答: 这个兴趣小组的男生有 150 人。",
+      "knowledge_points": [
+        "求一个数的几分之几的问题",
+        "列方程解含一个未知数的问题"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_399",
+      "original_id": "unit3_solve_399"
+    },
+    {
+      "id": "unit3_solve_400",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "一堆糖果, 其中奶糖占 $\\frac{9}{20}$, 再放入 16 块水果糖后, 奶糖就只占 $\\frac{1}{4}$。这一堆糖果原来一共有多少块?",
+      "options": [],
+      "answer": "20块",
+      "solution": "【分析】放入水果糖后, 这堆糖果的总数量增加了, 但是奶糖的数量不变, 找出等量关系: 原来糖果数量 $\\times \\frac{9}{20} =$ (原有糖果数量 +16) $\\times \\frac{1}{4}$; 设原来这堆糖果有 x 块, 据此列出方程, 解方程即可解答。\n\n【详解】解：设这一堆糖果原来一共 $x$ 块。$\\frac{9}{20} x = (x + 16) \\times \\frac{1}{4}$ $\\frac{9}{20} x = \\frac{1}{4} x + 16 \\times \\frac{1}{4}$ $\\frac{9}{20} x = \\frac{1}{4} x + 4$ $\\frac{9}{20} x - \\frac{1}{4} x = \\frac{1}{4} x + 4 - \\frac{1}{4} x$ $\\frac{1}{5} x = 4$ $\\frac{1}{5} x \\div \\frac{1}{5} = 4 \\div \\frac{1}{5} \\\\ x = 4 \\times 5 \\\\ x = 20$ 答: 这堆糖果原来一共有 20 块。",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "列方程解含两个未知数的问题",
+        "应用等式的性质 1 和 2 解方程"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit3_400",
+      "original_id": "unit3_solve_400"
+    },
+    {
+      "id": "unit3_choice_401",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "东东用 $\\frac{5}{12}$ 小时完成了一幅拼图的 $\\frac{3}{4}$，照这样计算，完成这幅拼图要用多长时间？下面列式正确的是（）。",
+      "options": [
+        "A. $\\frac{3}{4} \\times \\frac{5}{12}$",
+        "B. $\\frac{3}{4} \\div \\frac{5}{12}$",
+        "C. $\\frac{5}{12} \\div \\frac{3}{4}$",
+        "D. 以上三项均不正确"
+      ],
+      "answer": "C",
+      "solution": "【分析】把完成整幅拼图需要的时间看作单位'1'，完成整幅拼图的 $\\frac{3}{4}$ 需要 $\\frac{5}{12}$ 小时，根据'量÷对应的分率'即可求得完成这幅拼图要用的时间。本题考查分数除法的应用，找出量和对应的分率是解答题目的关键。\n\n【详解】分析可知，$\\frac{5}{12} \\div \\frac{3}{4} = \\frac{5}{9}$（小时）故答案为：C",
+      "knowledge_points": [
+        "分数除法的应用"
+      ],
+      "source": "21-22六年级上·贵州黔南·期末",
+      "merged_id": "unit3_401",
+      "original_id": "unit3_choice_401"
+    },
+    {
+      "id": "unit3_choice_402",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "最小的合数的倒数比最小的质数的倒数小（）。",
+      "options": [
+        "A. $\\frac{1}{2}$",
+        "B. $\\frac{1}{4}$",
+        "C. $\\frac{1}{6}$",
+        "D. $\\frac{1}{8}$"
+      ],
+      "answer": "B",
+      "solution": "【分析】根据倒数的意义：乘积是1的两个数互为倒数。最小的质数是2，最小的合数是4，求一个大于1的自然数的倒数，用1除以这个数即可，用最小质数的倒数减最小合数的倒数，据此解答。此题考查的目的是理解倒数的意义，掌握求一个数的倒数的方法，明确：最小的质数是2，最小的合数是4。\n\n【详解】最小合数的倒数是 $\\frac{1}{4}$，最小质数的倒数是 $\\frac{1}{2}$，$\\frac{1}{2} - \\frac{1}{4} = \\frac{1}{4}$ 最小的合数的倒数比最小的质数的倒数小 $\\frac{1}{4}$。故选：B",
+      "knowledge_points": [
+        "质数与合数",
+        "倒数的认识及应用"
+      ],
+      "source": "2022·甘肃天水·小升初真题",
+      "merged_id": "unit3_402",
+      "original_id": "unit3_choice_402"
+    },
+    {
+      "id": "unit3_choice_403",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "$1\\frac{1}{5}$ 千克菜籽能榨油0.5千克，那么1千克菜籽能榨油多少千克？算式（）。",
+      "options": [
+        "A. $1\\frac{1}{5} \\div 0.5$",
+        "B. $1\\frac{1}{5} \\times 0.5$",
+        "C. $0.5 \\div 1\\frac{1}{5}$",
+        "D. $0.5 \\times 1^{\\frac{1}{5}}$"
+      ],
+      "answer": "C",
+      "solution": "【分析】1千克菜籽能榨油的质量 = 油的质量 ÷ 菜籽的质量，据此解答。所求结果的单位和除法算式中被除数的单位保持一致。\n\n【详解】$0.5 \\div 1\\frac{1}{5} = \\frac{1}{2} \\div \\frac{6}{5} = \\frac{5}{12}$（千克）故答案为：C",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "分数除法的应用"
+      ],
+      "source": "21-22六年级上·湖南长沙·期末",
+      "merged_id": "unit3_403",
+      "original_id": "unit3_choice_403"
+    },
+    {
+      "id": "unit3_choice_404",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "如果a是一个大于零的自然数，那么下列各式中得数最大的是（）。",
+      "options": [
+        "A.$a \\times \\frac{7}{8}$",
+        "B.$a \\div \\frac{7}{8}$",
+        "C.$\\frac{7}{8} \\div a$"
+      ],
+      "answer": "B",
+      "solution": "【分析】假设出a的值并计算出选项中各式的结果，最后比较大小即可求得。掌握分数乘除法的计算方法是解答题目的关键。\n\n【详解】假设 $a = 2$ A. $a \\times \\frac{7}{8} = 2 \\times \\frac{7}{8} = \\frac{7}{4}$；B. $a \\div \\frac{7}{8} = 2 \\div \\frac{7}{8} = \\frac{16}{7}$；C. $\\frac{7}{8} \\div a = \\frac{7}{8} \\div 2 = \\frac{7}{16}$。因为 $\\frac{16}{7} > \\frac{7}{4} > \\frac{7}{16}$，所以 $a \\div \\frac{7}{8} > a \\times \\frac{7}{8} > \\frac{7}{8} \\div a$。故答案为：B",
+      "knowledge_points": [
+        "异分母异分子分数的大小比较",
+        "分数与整数的乘法",
+        "分数与整数的除法"
+      ],
+      "source": "17-18六年级上·河北保定·期中",
+      "merged_id": "unit3_404",
+      "original_id": "unit3_choice_404"
+    },
+    {
+      "id": "unit3_choice_405",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "已知 $a \\times \\frac{10}{3} = \\frac{11}{12} \\times b = \\frac{11}{15} \\times c$（a, b, c均不为0），则（）。",
+      "options": [
+        "A.a>b>c",
+        "B.b>a>c",
+        "C.c>b>a"
+      ],
+      "answer": "C",
+      "solution": "【分析】假设它们的结果为1，分别求出a, b, c的值，再比较得结论。此题考查的是分数大小比较，解答此题关键是假设它们的结果为1，分别求出a, b, c的值，再比较得结论。\n\n【详解】假设 $a \\times \\frac{10}{3} = \\frac{11}{12} \\times b = \\frac{11}{15} \\times c = 1$，则 $a = \\frac{3}{10}; b = \\frac{12}{11}; c = \\frac{15}{11};$ 因为 $\\frac{15}{11} > \\frac{12}{11} > \\frac{3}{10}$，所以，c>b>a 故选：C",
+      "knowledge_points": [
+        "异分母异分子分数的大小比较",
+        "分数与整数的除法"
+      ],
+      "source": "21-22六年级上·湖南怀化·单元测试",
+      "merged_id": "unit3_405",
+      "original_id": "unit3_choice_405"
+    },
+    {
+      "id": "unit3_choice_406",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "一条公路已修800米，占全长的 $\\frac{4}{5}$，公路全长多少米？设全长为 $x$ 米，列方程是（）。",
+      "options": [
+        "A.$800 \\times \\frac{4}{5} = x$",
+        "B.$\\frac{4}{5}x = 800$",
+        "C.$800^{x} = \\frac{4}{5}$"
+      ],
+      "answer": "B",
+      "solution": "【分析】分析题意找出等量关系式：全长 $\\times \\frac{4}{5} =$ 已修800米，据此解答。找出等量关系式是列方程的关键。\n\n【详解】解：设全长为 $x$ 米。列方程为：$\\frac{4}{5}x = 800$；故选：B。",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "列方程解含一个未知数的问题"
+      ],
+      "source": "20-21六年级上·湖南怀化·单元测试",
+      "merged_id": "unit3_406",
+      "original_id": "unit3_choice_406"
+    },
+    {
+      "id": "unit3_choice_407",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "若 $a \\times \\frac{3}{7} = b = c \\div \\frac{1}{3}$（a、b、c不为零），则a、b、c的大小关系是（）。",
+      "options": [
+        "A.a>b>c",
+        "B.c>b>a",
+        "C.b>c>a"
+      ],
+      "answer": "A",
+      "solution": "【分析】由题意可知，令 $a \\times \\frac{3}{7} = b = c \\div \\frac{1}{3} = 1$，分别求出a、b、c的值，然后进行对比即可。本题考查分数乘除法，明确分数乘除法的计算方法法是解题的关键。\n\n【详解】令 $a \\times \\frac{3}{7} = b = c \\div \\frac{1}{3} = 1$ 则 $a = 1 \\div \\frac{3}{7} = \\frac{7}{3}$，$b = 1$，$c = 1 \\times \\frac{1}{3} = \\frac{1}{3}$，$\\frac{7}{3} > 1 > \\frac{1}{3}$，所以 $a > b > c$。故答案为：A",
+      "knowledge_points": [
+        "分数与整数的乘法",
+        "分数与整数的除法"
+      ],
+      "source": "20-21六年级上·湖南怀化·单元测试",
+      "merged_id": "unit3_407",
+      "original_id": "unit3_choice_407"
+    },
+    {
+      "id": "unit3_choice_408",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "甲数比乙数少 $\\frac{1}{5}$，乙数比甲数多（）。",
+      "options": [
+        "A.$\\frac{1}{5}$",
+        "B.$\\frac{1}{4}$",
+        "C.$\\frac{4}{5}$"
+      ],
+      "answer": "B",
+      "solution": "【分析】甲数比乙数少 $\\frac{1}{5}$，即以乙数为单位'1'，甲数就是 $1 - \\frac{1}{5} = \\frac{4}{5}$；要求乙数比甲数多几分之几，就是用乙数比甲数多的部分除以甲数，即 $\\frac{1}{5} \\div \\frac{4}{5}$，计算即可。本题考查学生对单位'1'的确定，甲数比乙数少 $\\frac{1}{5}$，是把乙数看作单位'1'，乙数比甲数多百分之几，是把甲数看作单位'1'，这样才能列出正确的算式，不致出错。\n\n【详解】$\\frac{1}{5} \\div (1 - \\frac{1}{5}) = \\frac{1}{5} \\div \\frac{4}{5} = \\frac{1}{5} \\times \\frac{5}{4} = \\frac{1}{4}$ 则乙数比甲数多 $\\frac{1}{4}$。故选：B",
+      "knowledge_points": [
+        "求比一个数多/少几分之几的数是多少",
+        "分数与分数的除法"
+      ],
+      "source": "20-21六年级上·湖南怀化·单元测试",
+      "merged_id": "unit3_408",
+      "original_id": "unit3_choice_408"
+    },
+    {
+      "id": "unit3_choice_409",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "算式 $\\frac{3}{4} \\times \\frac{5}{6}$ 与 $\\frac{3}{4} \\div \\frac{6}{5}$ 相比较，下面结论中正确的是（）。",
+      "options": [
+        "意义相同",
+        "结果相同",
+        "意义与结果都相同",
+        "意义与结果都不同"
+      ],
+      "answer": "B",
+      "solution": "【分析】根据分数乘除法的计算方法计算出两式的值，再根据乘除法的意义分析即可。除法运算是已知两个数的积和其中一个因数，求另一个因数的运算，乘法和除法意义不相同。\n\n【详解】$\\frac{3}{4} \\times \\frac{5}{6} = \\frac{5}{8}$，表示求 $\\frac{3}{4}$ 和 $\\frac{5}{6}$ 的积是多少；$\\frac{3}{4} \\div \\frac{6}{5} = \\frac{3}{4} \\times \\frac{5}{6} = \\frac{5}{8}$，表示求 $\\frac{3}{4}$ 和 $\\frac{6}{5}$ 的商是多少；所以 $\\frac{3}{4} \\times \\frac{5}{6}$ 与 $\\frac{3}{4} \\div \\frac{6}{5}$ 相比较，结果相同，意义不同。故答案为：B",
+      "knowledge_points": [
+        "分数乘分数",
+        "分数与分数的除法"
+      ],
+      "source": "21-22六年级上·江西赣州·单元测试",
+      "merged_id": "unit3_409",
+      "original_id": "unit3_choice_409"
+    },
+    {
+      "id": "unit3_choice_410",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "一条绳子两次用完，第一次用去 $90\\mathrm{~cm}$，__________。这条绳子长多少厘米？横线上要补充哪个条件，才能用算式：$90 \\div (1 - \\frac{4}{7})$ 来解决（）。",
+      "options": [
+        "A.第一次用去了全长的 $\\frac{4}{7}$",
+        "B.第二次比第一次少 $\\frac{4}{7}$",
+        "C.第二次用去全长的 $\\frac{4}{7}$",
+        "D.第一次比第二次少 $\\frac{4}{7}$"
+      ],
+      "answer": "C",
+      "solution": "【分析】已知这条绳子分两次用完，且第一次用去了 $90\\mathrm{~cm}$，要补充适当的条件，可先看所列算式，$90 \\div (1 - \\frac{4}{7})$，这是把这条绳子的长度看作单位'1'，且 $90\\mathrm{~cm}$ 占 $(1 - \\frac{4}{7})$，这就是说第二次用了全长的 $\\frac{4}{7}$，据此解答。结合分数除法的意义，以及题目里看作单位'1'的具体量可知：要补充的条件为：第二次用去全长的 $\\frac{4}{7}$。故答案为：C。用除法计算，意味着单位'1'是未知的，再依据量率对应的关系，可知题目里具体量所对应的分率是几分之几，再结合题意可确定答案。\n\n【详解】结合分数除法的意义，以及题目里看作单位'1'的具体量可知：要补充的条件为：第二次用去全长的 $\\frac{4}{7}$。故答案为：C。",
+      "knowledge_points": [
+        "分数的四则混合运算",
+        "'提问题''填条件'问题"
+      ],
+      "source": "20-21六年级上·福建厦门·单元测试",
+      "merged_id": "unit3_410",
+      "original_id": "unit3_choice_410"
+    },
+    {
+      "id": "unit3_choice_411",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "一项工程，甲队单独做需要10天完成，乙队单独做需要12天完成。下面说法有错误的是（）。",
+      "options": [
+        "A.甲每天可以完成这项工程的 $\\frac{1}{10}$",
+        "B.两队合作每天可以完成这项工程的 $\\frac{1}{10} + \\frac{1}{12}$",
+        "C.甲的工作效率比乙的工作效率低",
+        "D.甲乙两队合作一共需要 $\\frac{60}{11}$ 天"
+      ],
+      "answer": "C",
+      "solution": "【分析】把总工作量看作'1'，根据'工作量÷工作效率=工作时间'判断即可。此题属于工程问题，掌握'工作量 $\\div$ 工作效率 = 工作时间'是解题关键。\n\n【详解】A．甲的工作效率：$1 \\div 10 = \\frac{1}{10}$，正确；B. 乙工作效率：$1 \\div 12 = \\frac{1}{12}$，两队合作工作效率为：$\\frac{1}{10} + \\frac{1}{12}$，所以正确；C. 因为 $\\frac{1}{10} > \\frac{1}{12}$，所以甲的工作效率高于乙的工作效率，所以错误；D. 甲乙两队合作一共需要时间：$1 \\div (\\frac{1}{10} + \\frac{1}{12}) = \\frac{60}{11}$（天），所以正确。故答案为：C",
+      "knowledge_points": [
+        "工程问题",
+        "分数与整数的除法"
+      ],
+      "source": "20-21六年级上·福建厦门·单元测试",
+      "merged_id": "unit3_411",
+      "original_id": "unit3_choice_411"
+    },
+    {
+      "id": "unit3_choice_412",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "修路队修一条2千米的路，9天修了全长的 $\\frac{3}{5}$，照这样算，还要（）天才能修完这条路。",
+      "options": [
+        "A.5天",
+        "B.6天",
+        "C.10天",
+        "D.15天"
+      ],
+      "answer": "B",
+      "solution": "【分析】把这条路的全长看成'1'，根据'工作量÷工作时间=工作效率'求出修路队的工作效率，再根据'工作量÷工作效率=工作时间'求出修路队修完这条路共用的时间，再减去9天即可求出还需要的天数。\n\n【详解】$1 \\div (\\frac{3}{5} \\div 9) - 9 = 1 \\div \\frac{1}{15} - 9 = 15 - 9 = 6$（天）故答案为：B\n\n【点睛】此题属于工程问题，解答此题应注意用这种方法解，全长2千米是多余条件。",
+      "knowledge_points": [
+        "工程问题",
+        "分数与整数的除法"
+      ],
+      "source": "20-21六年级上·福建厦门·单元测试",
+      "merged_id": "unit3_412",
+      "original_id": "unit3_choice_412"
+    },
+    {
+      "id": "unit3_choice_413",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "两个真分数相乘的积与它们相除的商相比较，（）。",
+      "options": [
+        "A.积大",
+        "B.商大",
+        "C.相等",
+        "D.无法确定"
+      ],
+      "answer": "B",
+      "solution": "【分析】根据题意可知，真分数都小于1，所以两个真分数相乘，积一定小于其中任何一个数，在分数除法中，如果除数是真分数，那么商一定大于被除数，即可解答。\n\n【详解】根据分析可知，两个真分数相乘的积与它们的商相比，商大于积。故答案选：B\n\n【点睛】本题考查乘法中因数与积的关系；以及商与被除数的关系。",
+      "knowledge_points": [
+        "分数乘分数",
+        "真分数、假分数、带分数的认识",
+        "分数与分数的除法"
+      ],
+      "source": "2023·山东济南·小升初真题",
+      "merged_id": "unit3_413",
+      "original_id": "unit3_choice_413"
+    },
+    {
+      "id": "unit3_choice_414",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "G是一个非零的自然数，下列算式中得数最大的是（）。",
+      "options": [
+        "A.$G \\div \\frac{2}{5}$",
+        "B.$G \\times \\frac{2}{5}$",
+        "C.$G \\div \\frac{7}{5}$"
+      ],
+      "answer": "A",
+      "solution": "【分析】把选项中的算式根据分数除法的计算方法，都变成G乘一个数的形式，因为因数G相同，比较另一个因数，另一个因数大的算式的运算结果就大。此题考查的是分数乘除的应用，解答本题关键根据两个乘法算式中，一个因数相同，另一个因数大的积就大，进行比较。\n\n【详解】A. $G \\div \\frac{2}{5} = G \\times \\frac{5}{2}$ B. $G \\times \\frac{2}{5}$ C. $G \\div \\frac{7}{5} = G \\times \\frac{5}{7}$ 因为 $\\frac{5}{2} > \\frac{5}{7} > \\frac{2}{5}$，所以 $G \\div \\frac{2}{5} > G \\div \\frac{7}{5} > G \\times \\frac{2}{5}$ 故选：A",
+      "knowledge_points": [
+        "分数与整数的乘法",
+        "分数与整数的除法",
+        "分数与分数的除法"
+      ],
+      "source": "20-21六年级上·江西赣州·单元测试",
+      "merged_id": "unit3_414",
+      "original_id": "unit3_choice_414"
+    },
+    {
+      "id": "unit3_choice_415",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "如果 $\\frac{6}{7} \\div \\mathrm{B} > \\frac{6}{7}$，则B一定是（）。",
+      "options": [
+        "A.真分数",
+        "B.假分数",
+        "C.带分数",
+        "D.任何小于1的数"
+      ],
+      "answer": "A",
+      "solution": "【分析】根据商的变化规律解答：两个非零的数相除，当除数大于1，商小于被除数；当除数小于1，商大于被除数；当除数等于1，商等于被除数。本题考查了商的变化规律的灵活应用，掌握规律是解答本题的关键。\n\n【详解】由分析得，如果 $\\frac{6}{7} \\div \\mathrm{B} > \\frac{6}{7}$，那么 $\\mathrm{B}$ 一定是小于1的分数但不为0，即真分数。故选：A",
+      "knowledge_points": [
+        "分数与分数的除法"
+      ],
+      "source": "20-21六年级上·江西赣州·单元测试",
+      "merged_id": "unit3_415",
+      "original_id": "unit3_choice_415"
+    },
+    {
+      "id": "unit3_choice_416",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "一个大于0的数除以 $\\frac{1}{4}$，就是把这个数（）。",
+      "options": [
+        "A.缩小到原来的 $\\frac{1}{4}$",
+        "B.扩大到原来的4倍",
+        "C.比原来减少 $\\frac{1}{4}$",
+        "D.不变"
+      ],
+      "answer": "B",
+      "solution": "【分析】根据分数除法的计算方法，除以一个数等于乘这个数的倒数，一个大于0的数除以 $\\frac{1}{4}$，相当于这个数乘4，据此解答。掌握分数除法的计算方法是解答题目的关键。\n\n【详解】假设这个数为 $\\frac{1}{4}$ $\\frac{1}{4} \\div \\frac{1}{4} = \\frac{1}{4} \\times 4 = 1$ 一个大于0的数除以 $\\frac{1}{4}$，就是把这个数扩大到原来的4倍。故答案为：B",
+      "knowledge_points": [
+        "分数与分数的除法"
+      ],
+      "source": "17-18五年级下·全国·课后作业",
+      "merged_id": "unit3_416",
+      "original_id": "unit3_choice_416"
+    },
+    {
+      "id": "unit3_choice_417",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "习总书记提出'一带一路'伟大战略构想，给沿线国家带来福祉，'一带一路'的新运力'中欧班列'运送一批货物，其中茶叶已经运了 $\\frac{4}{5}$，还剩 $\\frac{1}{5}$ 吨，这批茶叶有（）吨。",
+      "options": [
+        "A.$\\frac{1}{5}$",
+        "B.$\\frac{4}{5}$",
+        "C.1"
+      ],
+      "answer": "C",
+      "solution": "【分析】把这批茶叶质量看作单位'1'，已经运了 $\\frac{4}{5}$，还剩 $\\frac{1}{5}$ 吨，数量 $\\frac{1}{5}$ 吨所对应的分率是 $(1 - \\frac{4}{5}) = \\frac{1}{5}$，单位'1'未知用除法。\n\n【详解】$\\frac{1}{5} \\div (1 - \\frac{4}{5}) = \\frac{1}{5} \\div \\frac{1}{5} = 1$（吨）故选：C\n\n【点睛】此题考查的是分数除法的应用，掌握数量÷对应分率 = 单位'1'的量是解题关键。",
+      "knowledge_points": [
+        "分数除法的应用"
+      ],
+      "source": "19-20六年级上·湖北十堰·阶段练习",
+      "merged_id": "unit3_417",
+      "original_id": "unit3_choice_417"
+    },
+    {
+      "id": "unit3_choice_418",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "在一卷公元前1600年左右遗留下来的埃及草卷中，记载着一些数学问题，其中一个问题翻译过来是：'啊哈，它的全部，它的 $\\frac{1}{7}$，其和等于19'。如果把'它'看作 $x$，下列符合题意的方程是（）。",
+      "options": [
+        "A.$\\frac{1}{7}x = 19$",
+        "B.$1 + \\frac{1}{7}x = 19$",
+        "C.$x + \\frac{1}{7}x = 19$",
+        "D.$x - \\frac{1}{7}x = 19$"
+      ],
+      "answer": "C",
+      "solution": "【分析】由题意可知，$x$ 和它的 $\\frac{1}{7}$ 的和是19，其中 $x$ 的 $\\frac{1}{7}$ 表示为 $\\frac{1}{7}x$，据此列方程。\n\n【详解】由分析可知：$x + \\frac{1}{7}x = 19$\n\n解：$\\frac{8}{7}x = 19$\n\n$x = 19 \\div \\frac{8}{7}$\n\n$x = 19 \\times \\frac{7}{8}$\n\n$x = \\frac{133}{8}$\n\n所以，它是 $\\frac{133}{8}$。故答案为：C\n\n【点睛】根据题意找出等量关系式是解答题目的关键。",
+      "knowledge_points": [
+        "应用等式的性质2解方程",
+        "分数与整数的除法",
+        "列方程解含一个未知数的问题"
+      ],
+      "source": "2022·贵州铜仁·小升初真题",
+      "merged_id": "unit3_418",
+      "original_id": "unit3_choice_418"
+    },
+    {
+      "id": "unit3_choice_419",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "若a是非零自然数，下列算式中的计算结果最大的是（）。",
+      "options": [
+        "A.$a \\times \\frac{5}{8}$",
+        "B.$a \\div \\frac{5}{8}$",
+        "C.$a \\times \\frac{2}{3}$",
+        "D.$a \\div \\frac{2}{3}$"
+      ],
+      "answer": "B",
+      "solution": "【分析】a是非零自然数，所以可以将其假设为1，据此分别求出各个式子的结果，从而选出计算结果最大的式子。本题考查了分数乘除法，有一定的运算能力是解题的关键。\n\n【详解】令 $a$ 为1，那么 $a \\times \\frac{5}{8} = \\frac{5}{8}, a \\div \\frac{5}{8} = \\frac{8}{5}, a \\times \\frac{2}{3} = \\frac{2}{3}, a \\div \\frac{2}{3} = \\frac{3}{2}$，所以计算结果最大的式子是 $a \\div \\frac{5}{8}$。故答案为：B",
+      "knowledge_points": [
+        "分数与整数的乘法",
+        "分数与整数的除法"
+      ],
+      "source": "19-20六年级上·江苏盐城·期中",
+      "merged_id": "unit3_419",
+      "original_id": "unit3_choice_419"
+    },
+    {
+      "id": "unit3_choice_420",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "一根铁丝长 $\\frac{4}{5}$ 米，用它围成一个最大的正方形，这个正方形的面积是（）平方米。",
+      "options": [
+        "A.$\\frac{16}{25}$",
+        "B.$\\frac{1}{25}$",
+        "C.$\\frac{4}{25}$",
+        "D.$\\frac{1}{5}$"
+      ],
+      "answer": "B",
+      "solution": "【分析】铁丝的长度相当于正方形的周长，根据正方形的周长计算出正方形的边长，再利用正方形的面积公式即可求得。掌握分数乘除法的计算方法是解答题目的关键。\n\n【详解】$(\\frac{4}{5} \\div 4) \\times (\\frac{4}{5} \\div 4) = (\\frac{4}{5} \\times \\frac{1}{4}) \\times (\\frac{4}{5} \\times \\frac{1}{4}) = \\frac{1}{5} \\times \\frac{1}{5} = \\frac{1}{25}$（平方米）故答案为：B",
+      "knowledge_points": [
+        "分数的乘、除法的混合运算",
+        "长、正方形周长的应用",
+        "正方形的面积"
+      ],
+      "source": "18-19五年级下·辽宁·单元测试",
+      "merged_id": "unit3_420",
+      "original_id": "unit3_choice_420"
+    },
+    {
+      "id": "unit3_fill_421",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "一箱橘子 $\\frac{3}{4}$ 卖出，正好是12千克，这箱橘子重（）千克。",
+      "options": [],
+      "answer": "16",
+      "solution": "【分析】将这箱橘子的质量看作单位'1'，卖出的质量÷对应分率 = 这箱橘子的质量。关键是确定单位'1'，理解分数除法的意义。\n\n【详解】$12 \\div \\frac{3}{4} = 16$（千克）",
+      "knowledge_points": [
+        "分数除法的应用"
+      ],
+      "source": "18-19六年级上·湖北武汉·期末",
+      "merged_id": "unit3_421",
+      "original_id": "unit3_fill_421"
+    },
+    {
+      "id": "unit3_fill_422",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "六(1)共有学生55人，其中女生人数是男生的 $\\frac{5}{6}$，女生有（）人。",
+      "options": [],
+      "answer": "25",
+      "solution": "【分析】六（1）班共有学生55人，其中女生人数是男生的 $\\frac{5}{6}$，把男生人数看作单位'1'，则总人数是男生的 $1 + \\frac{5}{6}$，根据分数除法的意义，男生有 $55 \\div (1 + \\frac{5}{6})$ 人，则用总人数减男生人数，即得女生多少人。此题考查的是和倍问题，解答此题首先根据分数加法的意义求出总人数是男生人数的几分之几是完成本题的关键。\n\n【详解】$55 \\div (1 + \\frac{5}{6}) = 55 \\div \\frac{11}{6} = 30$（人）女生：$55 - 30 = 25$（人）",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "分数除法的应用"
+      ],
+      "source": "21-22六年级上·湖南怀化·单元测试",
+      "merged_id": "unit3_422",
+      "original_id": "unit3_fill_422"
+    },
+    {
+      "id": "unit3_fill_423",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "一批水果重 $\\frac{3}{5}$ 吨，5天卖完，平均每天卖（）吨，平均每天卖这批水果的（）。",
+      "options": [],
+      "answer": "$\\frac{3}{25}$; $\\frac{1}{5}$",
+      "solution": "【分析】水果质量÷卖的天数 = 平均每天卖的质量；将水果质量看作单位'1'，$1 \\div$ 卖的天数 = 平均每天卖这批水果的几分之几。除以一个数等于乘这个数的倒数。\n\n【详解】$\\frac{3}{5} \\div 5 = \\frac{3}{25}$（吨）$1 \\div 5 = \\frac{1}{5}$",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "求一个数占另一个数几分之几"
+      ],
+      "source": "20-21六年级上·湖南怀化·单元测试",
+      "merged_id": "unit3_423",
+      "original_id": "unit3_fill_423"
+    },
+    {
+      "id": "unit3_fill_424",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "（）米的 $\\frac{3}{7}$ 是6米；12千克的 $\\frac{1}{4}$ 是（）千克。",
+      "options": [],
+      "answer": "14; 3",
+      "solution": "【分析】第一个空，用已知长度 $\\div$ 对应分率 $=$ 所求长度；第二个空，用已知质量 $\\times$ 所求部分对应分率 $=$ 所求质量。整体数量 $\\times$ 部分对应分率 $=$ 部分数量，部分数量÷对应分率 $=$ 整体数量。\n\n【详解】$6 \\div \\frac{3}{7} = 14$（米）$12 \\times \\frac{1}{4} = 3$（千克）",
+      "knowledge_points": [
+        "求一个数的几分之几的问题",
+        "分数除法的应用"
+      ],
+      "source": "20-21六年级上·湖南怀化·单元测试",
+      "merged_id": "unit3_424",
+      "original_id": "unit3_fill_424"
+    },
+    {
+      "id": "unit3_fill_425",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "一本故事书有120页，小明第一天读了全书的 $\\frac{1}{6}$，第二天读了余下的 $\\frac{7}{10}$，第三天应从第（）页读起。",
+      "options": [],
+      "answer": "91",
+      "solution": "【分析】根据求一个数的几分之几是多少，用乘法；用 $120 \\times \\frac{1}{6}$ 是第一天读书的页数，用全书页数减去第一天看的页数，再乘 $\\frac{7}{10}$，即是第二天看的页数，最后将第一天和第二天读的页数再加1，即是第三天开始看的页数。此题主要考查分数乘法的实际应用，需要掌握求一个数的几分之几是多少，用乘法。\n\n【详解】第一天看的页数：$120 \\times \\frac{1}{6} = 20$（页）第二天看的页数：$(120 - 20) \\times \\frac{7}{10} = 100 \\times \\frac{7}{10} = 70$（页）第三天开始看的页数：$20 + 70 + 1 = 91$（页）",
+      "knowledge_points": [
+        "求一个数的几分之几的问题"
+      ],
+      "source": "20-21六年级上·湖南怀化·单元测试",
+      "merged_id": "unit3_425",
+      "original_id": "unit3_fill_425"
+    },
+    {
+      "id": "unit3_fill_426",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "现在国际市场黄金涨价了 $\\frac{1}{10}$，现价是原价的（）。",
+      "options": [],
+      "answer": "$\\frac{11}{10}$",
+      "solution": "【分析】假设黄金的原价是1，现在黄金的价格为 $1 \\times (1 + \\frac{1}{10})$，然后用现价除以原价即可。本题考查一个数是另一个数的几分之几，明确用除法是解题的关键。\n\n【详解】$1 \\times (1 + \\frac{1}{10}) \\div 1 = \\frac{11}{10} \\div 1 = \\frac{11}{10}$",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "求一个数占另一个数几分之几",
+        "求比一个数多/少几分之几的数是多少"
+      ],
+      "source": "20-21六年级上·湖南怀化·单元测试",
+      "merged_id": "unit3_426",
+      "original_id": "unit3_fill_426"
+    },
+    {
+      "id": "unit3_fill_427",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "一本故事书9天看完，每天看完这本故事书的（）。",
+      "options": [],
+      "answer": "$\\frac{1}{9}$",
+      "solution": "【分析】把这本故事书的总页数看作单位'1'，把它平均分成9份，每份是它的 $\\frac{1}{9}$，即平均每天看这本故事书的 $\\frac{1}{9}$。此题是考查分数的意义。把单位'1'平均分成若干份，用分数表示，分母是分成的份数，分子是要表示的份数。注意一定是把单位'1'平均分。\n\n【详解】由分析可知：一本故事书9天看完，平均每天看这本故事书的九分之一，所以每天看这本故事书的 $\\frac{1}{9}$。",
+      "knowledge_points": [
+        "分数与除法的关系"
+      ],
+      "source": "20-21六年级上·湖南怀化·单元测试",
+      "merged_id": "unit3_427",
+      "original_id": "unit3_fill_427"
+    },
+    {
+      "id": "unit3_fill_428",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "甲数是4，乙数是5，甲数是乙数的（），甲数比乙数少（）；乙数比甲数多（）。",
+      "options": [],
+      "answer": "$\\frac{4}{5}$; $\\frac{1}{5}$; $\\frac{1}{4}$",
+      "solution": "【分析】求甲数是乙数的几分之几，用甲数除以乙数即可；求甲数比乙数少几分之几，先求出甲数比乙数少多少，然后再除以乙数即可；求乙数比甲数多几分之几，先求出乙数比甲数多多少，然后再除以甲数即可。本题考查求一个数比另一个数多（少）几分之几，明确单位'1'是解题的关键。\n\n【详解】$4 \\div 5 = \\frac{4}{5}$ $(5 - 4) \\div 5 = 1 \\div 5 = \\frac{1}{5}$ $(5 - 4) \\div 4 = 1 \\div 4 = \\frac{1}{4}$",
+      "knowledge_points": [
+        "分数与除法的关系",
+        "求一个数占另一个数几分之几"
+      ],
+      "source": "20-21六年级上·湖南怀化·单元测试",
+      "merged_id": "unit3_428",
+      "original_id": "unit3_fill_428"
+    },
+    {
+      "id": "unit3_fill_429",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "乘积是1的两个数互为（），$\\frac{6}{7}$ 与（）互为倒数；（）的倒数是1；$1\\frac{1}{6}$ 的倒数是（）；（）的倒数是0.25。",
+      "options": [],
+      "answer": "倒数; $\\frac{7}{6}$; 1; $\\frac{6}{7}$; 4",
+      "solution": "【分析】互为倒数的两个数的乘积是1，1的倒数还是1，据此解答即可。本题考查倒数的定义，明确倒数的定义是解题的关键。\n\n【详解】$1 \\div \\frac{6}{7} = \\frac{7}{6}$，$1 \\div 1\\frac{1}{6} = \\frac{6}{7}$，$1 \\div 0.25 = 4$，所以乘积是1的两个数互为倒数，$\\frac{6}{7}$ 与 $\\frac{7}{6}$ 互为倒数；1的倒数是1；$1\\frac{1}{6}$ 的倒数是 $\\frac{6}{7}$；4的倒数是0.25。",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "倒数的认识及应用"
+      ],
+      "source": "20-21六年级上·湖南怀化·单元测试",
+      "merged_id": "unit3_429",
+      "original_id": "unit3_fill_429"
+    },
+    {
+      "id": "unit3_fill_430",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "把一根长为 $\\frac{4}{7}$ 米长铁丝平均分成5段，每段占全长的（），每段长（）米。",
+      "options": [],
+      "answer": "$\\frac{1}{5}$; $\\frac{4}{35}$",
+      "solution": "【分析】（1）求每段长是全长的几分之几，就是把这根铁丝的全长看作单位'1'，平均分为5份，求一份是这根铁丝全长的几分之几，用 $1 \\div 5$ 解答；(2)求每段长多少米，用这根铁丝的全长除以段数即可。本题主要考查分数的意义，注意求每段长是这根铁丝的几分之几，用1除以段数；求每段长多少米，用这根铁丝的全长除以段数。\n\n【详解】$1 \\div 5 = \\frac{1}{5}$ $\\frac{4}{7} \\div 5 = \\frac{4}{7} \\times \\frac{1}{5} = \\frac{4}{35}$（米）",
+      "knowledge_points": [
+        "分数与除法的关系",
+        "分数的平均分"
+      ],
+      "source": "20-21六年级上·湖南怀化·单元测试",
+      "merged_id": "unit3_430",
+      "original_id": "unit3_fill_430"
+    },
+    {
+      "id": "unit3_fill_431",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "3千克的 $\\frac{2}{5}$ 是（）；（）千米的 $\\frac{1}{24}$ 是36千米。",
+      "options": [],
+      "answer": "$\\frac{6}{5}$; 864",
+      "solution": "【分析】根据求一个数的几分之几是多少，用乘法进行计算即可；已知一个数的几分之几是多少，用除法进行计算即可。本题考查分数乘除法，明确分数乘除法的计算方法是解题的关键。\n\n【详解】$3 \\times \\frac{2}{5} = \\frac{6}{5}$（千克）$36 \\div \\frac{1}{24} = 864$（千米）",
+      "knowledge_points": [
+        "求一个数的几分之几的问题",
+        "分数与整数的乘法",
+        "分数与整数的除法"
+      ],
+      "source": "20-21六年级上·湖南怀化·单元测试",
+      "merged_id": "unit3_431",
+      "original_id": "unit3_fill_431"
+    },
+    {
+      "id": "unit3_fill_432",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "'松树棵数相当于柏树棵数的 $\\frac{2}{5}$'是把（）看作单位'1'；'九月份用水量比八月份节约了 $\\frac{5}{2}$'中，单位'1'是（）。",
+      "options": [],
+      "answer": "柏树棵数; 八月份用水量",
+      "solution": "【分析】根据判断单位'1'的方法：一般是把'比、占、是、相当于'后面的量看作单位'1'，即分数'的'字前面的量看作单位'1'，进行解答即可。掌握判断单位'1'的方法是解题的关键。\n\n【详解】'松树棵数相当于柏树棵数的 $\\frac{2}{5}$'是把（柏树棵数）看作单位'1'；'九月份用水量比八月份节约了 $\\frac{5}{2}$'中，单位'1'是（八月份用水量）。",
+      "knowledge_points": [
+        "单位'1'的认识与确定"
+      ],
+      "source": "20-21六年级上·湖南怀化·单元测试",
+      "merged_id": "unit3_432",
+      "original_id": "unit3_fill_432"
+    },
+    {
+      "id": "unit3_fill_433",
+      "unit": "第三单元：分数除法",
+      "chapter": "分数除法",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "一辆汽车 $\\frac{2}{5}$ 小时行24千米，照这样的速度1小时行（）千米，行1千米需要（）小时。",
+      "options": [],
+      "answer": "60; $\\frac{1}{60}$",
+      "solution": "【分析】求1小时行驶距离，用路程 $\\div$ 时间；求行1千米需要的时间，用时间 $\\div$ 路程，据此列式计算。关键是掌握分数乘除法的计算方法。\n\n【详解】$24 \\div \\frac{2}{5} = 60$（千米）$\\frac{2}{5} \\div 24 = \\frac{1}{60}$（小时）照这样的速度1小时行60千米，行1千米需要 $\\frac{1}{60}$ 小时。",
+      "knowledge_points": [
+        "分数与整数的除法",
+        "基础行程问题"
+      ],
+      "source": "14-15六年级上·贵州遵义·阶段练习",
+      "merged_id": "unit3_433",
+      "original_id": "unit3_fill_433"
+    },
+    {
+        "id": "unit3_fill_434",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "人的脚印长大约是身高的 $\\frac{1}{7}$，公安机关在某案发现场发现了一个长25厘米的脚印，那么可以推测犯罪嫌疑人的身高是（）厘米。",
+        "options": [],
+        "answer": "175",
+        "solution": "【分析】根据人的脚印长大约是身高的 $\\frac{1}{7}$，求出犯罪嫌疑人的身高即可。本题考查分数除法，解答本题的关键是掌握已知一个数的几分之几是多少，求这个数的解决方法。\n\n【详解】$25 \\div \\frac{1}{7} = 175$（厘米）",
+        "knowledge_points": [
+          "分数除法的应用"
+        ],
+        "source": "20-21六年级上·湖南怀化·单元测试",
+        "merged_id": "unit3_434",
+        "original_id": "unit3_fill_434"
+      },
+      {
+        "id": "unit3_fill_435",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "一项工作，甲单独做要8小时，乙单独做要10小时，现在甲、乙合做要（）小时。",
+        "options": [],
+        "answer": "$\\frac{40}{9}$",
+        "solution": "【分析】将工作总量看作单位'1'，时间分之一可以看作效率，用工作总量 $\\div$ 效率和 = 工作时间，据此列式计算。关键是理解工作效率、工作时间、工作总量之间的关系。\n\n【详解】$1 \\div (\\frac{1}{8} + \\frac{1}{10}) = 1 \\div \\frac{9}{40} = \\frac{40}{9}$（小时）",
+        "knowledge_points": [
+          "工程问题"
+        ],
+        "source": "20-21六年级上·湖南怀化·单元测试",
+        "merged_id": "unit3_435",
+        "original_id": "unit3_fill_435"
+      },
+      {
+        "id": "unit3_fill_436",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "把 $\\frac{3}{7}$ 米铁丝平均分成3份，每份长（）米，每份占全长的（）。",
+        "options": [],
+        "answer": "$\\frac{1}{7}$; $\\frac{1}{3}$",
+        "solution": "【分析】求每段长的米数，平均分的是具体的数量 $\\frac{3}{7}$ 米，求的是具体的数量；求每段长是这根铁丝的几分之几，平均分的是单位'1'，求的是分率；都用除法计算。此题考查的是分数的意义，解决此题关键是弄清求得是具体的数量还是分率，求具体的数量平均分的是具体的数量；求分率平均分的是单位'1'。\n\n【详解】每段长的米数：$\\frac{3}{7} \\div 3 = \\frac{1}{7}$（米）每段占全长的分率：$1 \\div 3 = \\frac{1}{3}$",
+        "knowledge_points": [
+          "分数与整数的除法"
+        ],
+        "source": "17-18六年级上·贵州黔南·期末",
+        "merged_id": "unit3_436",
+        "original_id": "unit3_fill_436"
+      },
+      {
+        "id": "unit3_fill_437",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "90米比（）米多 $\\frac{1}{8}$；比（）千克少 $\\frac{1}{7}$ 是42千克。",
+        "options": [],
+        "answer": "80; 49",
+        "solution": "【分析】90米比几米多 $\\frac{1}{8}$，可将求的数看作单位'1'，则90米为 $(1 + \\frac{1}{8})$ 份，运用分数除法得出答案；42千克比几千克少 $\\frac{1}{7}$，可将几千克看作单位'1'，则所求数位 $(1 - \\frac{1}{7})$ 份，运用分数除法得出答案。本题主要考查的是分数除法的运用，解题的关键是合理运用单位'1'，在运用分数除法运算法则，进而得出答案。\n\n【详解】$90 \\div (1 + \\frac{1}{8}) = 90 \\div \\frac{9}{8} = 80$（米）$42 \\div (1 - \\frac{1}{7}) = 42 \\div \\frac{6}{7} = 49$（千克）",
+        "knowledge_points": [
+          "分数与整数的除法",
+          "单位'1'的认识与确定",
+          "求比一个数多/少几分之几的数是多少"
+        ],
+        "source": "21-22六年级上·江西赣州·单元测试",
+        "merged_id": "unit3_437",
+        "original_id": "unit3_fill_437"
+      },
+      {
+        "id": "unit3_fill_438",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "把 $\\frac{4}{5}$ 米长的丝带剪成同样长的8段，每段丝带有多长就是求（），列出算式是（）。",
+        "options": [],
+        "answer": "$\\frac{4}{5}$ 除以8的商是多少？; $\\frac{4}{5} \\div 8$",
+        "solution": "【分析】每段丝带的长度 = 丝带的总长度 ÷ 平均分成的段数，据此解答。掌握分数除法的意义是解答题目的关键。\n\n【详解】$\\frac{4}{5} \\div 8 = \\frac{1}{10}$（米）所以，每段丝带长 $\\frac{1}{10}$ 米。",
+        "knowledge_points": [
+          "分数的平均分",
+          "分数与整数的除法"
+        ],
+        "source": "21-22六年级上·江西赣州·单元测试",
+        "merged_id": "unit3_438",
+        "original_id": "unit3_fill_438"
+      },
+      {
+        "id": "unit3_fill_439",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "把 $\\frac{2}{9} \\times 4 = \\frac{8}{9}$ 改写成两道除法算式是（）和（）。",
+        "options": [],
+        "answer": "$\\frac{8}{9} \\div 4 = \\frac{2}{9}$; $\\frac{8}{9} \\div \\frac{2}{9} = 4$",
+        "solution": "【分析】$\\frac{2}{9} \\times 4 = \\frac{8}{9}$，分数乘法中，乘数 $=$ 积分被乘数，被乘数=积÷乘数，据此可得出答案。本题主要考查的是分数乘法与除法的关系，解题的关键是合理利用分数乘法、除法之间的关系，进而得出答案。\n\n【详解】$\\frac{2}{9} \\times 4 = \\frac{8}{9}$，则可得到除法算式：$\\frac{8}{9} \\div 4 = \\frac{2}{9}, \\frac{8}{9} \\div \\frac{2}{9} = 4$。",
+        "knowledge_points": [
+          "分数乘分数",
+          "分数与分数的除法"
+        ],
+        "source": "21-22六年级上·江西赣州·单元测试",
+        "merged_id": "unit3_439",
+        "original_id": "unit3_fill_439"
+      },
+      {
+        "id": "unit3_fill_440",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "B与C互为倒数，则 $1 + \\mathrm{B} \\div \\frac{1}{C} = ($   $)$。",
+        "options": [],
+        "answer": "2",
+        "solution": "【分析】根据互为倒数的两个数的乘积是1，除以一个不为0的数等于乘这个数的倒数，据此解答即可。本题考查倒数的定义，明确倒数的定义是解题的关键。\n\n【详解】由分析可知：$B \\times C = 1 + B \\div \\frac{1}{C} = 1 + B \\times C = 1 + 1 = 2$",
+        "knowledge_points": [
+          "分数与整数的除法",
+          "倒数的认识及应用",
+          "分数与分数的除法"
+        ],
+        "source": "20-21六年级上·福建厦门·单元测试",
+        "merged_id": "unit3_440",
+        "original_id": "unit3_fill_440"
+      },
+      {
+        "id": "unit3_fill_441",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "将 $\\frac{6}{11}$ 米的丝带平均截成6段，每段是这根丝带的（），每段长（）米。",
+        "options": [],
+        "answer": "$\\frac{1}{6}$; $\\frac{1}{11}$",
+        "solution": "【分析】丝带的总长度为单位'1'，将其平均分成6段，根据分数的意义可知，每段是这根丝带的 $\\frac{1}{6}$；用丝带的总长度除以平均分成的段数即可求出每段长多少米。解答本题时要注意区分求关系还是具体的数，求关系根据分数的意义解答，求具体的数根据除法的意义解答。\n\n【详解】每段是这根丝带的 $\\frac{1}{6}$；$\\frac{6}{11} \\div 6 = \\frac{1}{11}$（米）",
+        "knowledge_points": [
+          "分数与整数的除法",
+          "分数的意义"
+        ],
+        "source": "20-21六年级上·福建厦门·单元测试",
+        "merged_id": "unit3_441",
+        "original_id": "unit3_fill_441"
+      },
+      {
+        "id": "unit3_fill_442",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "0.15的倒数是（），最小质数的倒数是（）。",
+        "options": [],
+        "answer": "$\\frac{20}{3}$; $\\frac{1}{2}$",
+        "solution": "【分析】求一个数的倒数，可颠倒这个数的分子分母，可先把题目中的小数、整数化为分数，再求它们的倒数。求一个数（0除外）的倒数并不难，关键是明白分子分母相倒并且两个乘积是1的数互为倒数。\n\n【详解】$0.15 = \\frac{15}{100} = \\frac{3}{20}$ $\\frac{3}{20}$ 的倒数是 $\\frac{20}{3}$；最小的质数是2，$\\frac{2}{1}$ 的倒数是 $\\frac{1}{2}$。",
+        "knowledge_points": [
+          "倒数的认识及应用"
+        ],
+        "source": "20-21六年级上·福建厦门·单元测试",
+        "merged_id": "unit3_442",
+        "original_id": "unit3_fill_442"
+      },
+      {
+        "id": "unit3_fill_443",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "（）米的 $\\frac{3}{5}$ 是27米；3.6吨的 $\\frac{7}{12}$ 是（）吨。",
+        "options": [],
+        "answer": "45; 2.1",
+        "solution": "【分析】已知一个数的几分之几是多少，求这个数，用除法解答；求一个数的几分之几是多少，用乘法解答。熟练掌握分数乘、除法的意义是解答本题的关键。\n\n【详解】$27 \\div \\frac{3}{5} = 45$（米）$3.6 \\times \\frac{7}{12} = 2.1$（吨）",
+        "knowledge_points": [
+          "求一个数的几分之几的问题",
+          "分数除法的应用"
+        ],
+        "source": "20-21六年级上·福建厦门·单元测试",
+        "merged_id": "unit3_443",
+        "original_id": "unit3_fill_443"
+      },
+      {
+        "id": "unit3_fill_444",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "把 $\\frac{4}{5}\\mathrm{~L}$ 的果汁分装在 $0.2\\mathrm{~L}$ 的小瓶里，可以装（）瓶。",
+        "options": [],
+        "answer": "4",
+        "solution": "【分析】求的是 $\\frac{4}{5}$ 里有几个0.2，根据除法意义列式解答即可。此题考查的是分数除法的应用，解答此题应注意分数除以小数，把分数化成小数再相除。\n\n【详解】由分析得，$\\frac{4}{5} \\div 0.2 = 0.8 \\div 0.2 = 4$（瓶）",
+        "knowledge_points": [
+          "分数与分数的除法"
+        ],
+        "source": "20-21六年级上·福建厦门·单元测试",
+        "merged_id": "unit3_444",
+        "original_id": "unit3_fill_444"
+      },
+      {
+        "id": "unit3_fill_445",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如果甲数是7，乙数是5，那么甲数是乙数的（），乙数是甲数的（）倍。",
+        "options": [],
+        "answer": "$\\frac{7}{5}$; $\\frac{5}{7}$",
+        "solution": "【分析】根据一个数是另一个数的（几倍）几分之几用除法，据此解答。此题考查的是分数的意义，明确谁是平均分得的总份数是解题关键。\n\n【详解】由分析得，$7 \\div 5 = \\frac{7}{5}$ $5 \\div 7 = \\frac{5}{7}$",
+        "knowledge_points": [
+          "求一个数是另一个数的几倍",
+          "求一个数占另一个数几分之几"
+        ],
+        "source": "20-21六年级上·江西赣州·单元测试",
+        "merged_id": "unit3_445",
+        "original_id": "unit3_fill_445"
+      },
+      {
+        "id": "unit3_fill_446",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "一根铁丝用去 $\\frac{1}{6}$，还剩下 $\\frac{1}{6}$ 米，这根铁丝原来长（）米。",
+        "options": [],
+        "answer": "$\\frac{1}{5}$",
+        "solution": "【分析】根据题意可知，还剩下这根铁丝的 $1 - \\frac{1}{6} = \\frac{5}{6}$，正好是 $\\frac{1}{6}$ 米，再根据分数除法的意义解答即可。明确 $\\frac{1}{6}$ 米正好对应全长的 $\\frac{5}{6}$ 是解答本题的关键，再根据分数除法的意义解答。\n\n【详解】$\\frac{1}{6} \\div (1 - \\frac{1}{6}) = \\frac{1}{6} \\div \\frac{5}{6} = \\frac{1}{5}$（米）",
+        "knowledge_points": [
+          "分数除法的应用"
+        ],
+        "source": "20-21六年级上·江西赣州·单元测试",
+        "merged_id": "unit3_446",
+        "original_id": "unit3_fill_446"
+      },
+      {
+        "id": "unit3_fill_447",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "$750\\mathrm{~cm}^{3} =$ （）dm³；80克=（）千克；12分=（）时。",
+        "options": [],
+        "answer": "0.75; 0.08; 0.2",
+        "solution": "【分析】1立方分米 = 1000立方厘米，1千克 = 1000克，1时 = 60分；大单位变小单位乘进率，小单位变大单位除以进率，由此解答即可。\n\n【详解】$750\\mathrm{~cm}^{3} = 0.75\\mathrm{dm}^{3}$；80克 $= 0.08$ 千克；12分=0.2时\n\n【点睛】熟练掌握体积单位、质量单位、时间单位之间的进率是解答本题的关键。",
+        "knowledge_points": [
+          "克、千克之间的换算与比较",
+          "时、分的认识及换算",
+          "体积单位间的进率与换算（立方厘米、立方分米和立方米）"
+        ],
+        "source": "20-21六年级上·江西赣州·单元测试",
+        "merged_id": "unit3_447",
+        "original_id": "unit3_fill_447"
+      },
+      {
+        "id": "unit3_fill_448",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "把 $\\frac{3}{5}$ 米彩带剪成相同的6段，每段长（）米，是全长的（）。",
+        "options": [],
+        "answer": "$\\frac{1}{10}$; $\\frac{1}{6}$",
+        "solution": "【分析】用彩带的总长度除以平均剪成的段数即可求出每段长多少米；彩带的总长度为单位'1'，将其剪成相同的6段，根据分数的意义可知，每段是全长的 $\\frac{1}{6}$。解答本题时要注意区分求关系还是具体的数，求关系根据分数的意义解答，求具体的数根据除法的意义解答。\n\n【详解】$\\frac{3}{5} \\div 6 = \\frac{1}{10}$（米）；每段是全长的 $\\frac{1}{6}$。",
+        "knowledge_points": [
+          "分数与整数的除法",
+          "分数的意义"
+        ],
+        "source": "20-21六年级上·江西赣州·单元测试",
+        "merged_id": "unit3_448",
+        "original_id": "unit3_fill_448"
+      },
+      {
+        "id": "unit3_fill_449",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（）是18的 $\\frac{5}{9}$；72吨是（）吨的 $\\frac{9}{8}$；（）比64千米多 $\\frac{1}{8}$；54dm比（）dm少 $\\frac{1}{7}$。",
+        "options": [],
+        "answer": "10; 64; 72; 63",
+        "solution": "【分析】根据求一个数的几分之几是多少，用乘法计算即可；已知一个数的几分之几是多少，求这个数，用除法即可；求比一个数多几分之几，用乘法即可；求一个数比另一个数少几分之几，用除法解答即可。本题考查求一个数的几分之几是多少，明确用乘法是解题的关键。\n\n【详解】由分析可知：$18 \\times \\frac{5}{9} = 10, 72 \\div \\frac{9}{8} = 64, 64 \\times (1 + \\frac{1}{8}) = 64 \\times \\frac{9}{8} = 72, 54 \\div (1 - \\frac{1}{7}) = 54 \\div \\frac{6}{7} = 63$",
+        "knowledge_points": [
+          "求一个数的几分之几的问题",
+          "分数与整数的乘法",
+          "分数与整数的除法",
+          "求比一个数多/少几分之几的数是多少"
+        ],
+        "source": "20-21六年级上·江西赣州·单元测试",
+        "merged_id": "unit3_449",
+        "original_id": "unit3_fill_449"
+      },
+      {
+        "id": "unit3_fill_450",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "今年比去年增产 $\\frac{1}{8}$，表示今年产量是去年的（）。",
+        "options": [],
+        "answer": "$\\frac{9}{8}$",
+        "solution": "【分析】根据题意可知，去年产量为单位'1'，今年产量是去年的 $1 + \\frac{1}{8} = \\frac{9}{8}$，据此解答即可。\n\n【详解】今年比去年增产 $\\frac{1}{8}$，今年产量是去年的 $1 + \\frac{1}{8} = \\frac{9}{8}$。\n\n【点睛】本题较易，找准单位'1'是解答本题的关键。",
+        "knowledge_points": [
+          "单位'1'的认识与确定",
+          "分数的意义"
+        ],
+        "source": "20-21六年级上·江西赣州·单元测试",
+        "merged_id": "unit3_450",
+        "original_id": "unit3_fill_450"
+      },
+      {
+        "id": "unit3_fill_451",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "$\\frac{6}{5} - ($ $) = ($ $) + \\frac{4}{9} = \\frac{7}{8} \\times ($ $) = \\frac{6}{11} \\div ($ $)$",
+        "options": [],
+        "answer": "$\\frac{1}{5}$; $\\frac{5}{9}$; $\\frac{8}{7}$; $\\frac{6}{11}$",
+        "solution": "【分析】根据题干可知，需把他们的结果假设为1，进而求出各数即可。此题考查的是分数的计算，本题关键是用假设法，假设结果为1，分别求出各数即可。\n\n【详解】假设 $\\frac{6}{5} - ( ) = ( ) + \\frac{4}{9} = \\frac{7}{8} \\times ( ) = \\frac{6}{11} \\div ( ) = 1$，得 $\\frac{6}{5} - \\frac{1}{5} = \\frac{5}{9} + \\frac{4}{9} = \\frac{7}{8} \\times \\frac{8}{7} = \\frac{6}{11} \\div \\frac{6}{11}$。",
+        "knowledge_points": [
+          "分数乘分数",
+          "分数与整数的除法",
+          "分数与分数的除法"
+        ],
+        "source": "20-21六年级上·江西赣州·单元测试",
+        "merged_id": "unit3_451",
+        "original_id": "unit3_fill_451"
+      },
+      {
+        "id": "unit3_fill_452",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "把 $\\frac{5}{8} \\times \\frac{2}{3} = \\frac{5}{6}$ 这个算式改写成两个除法算式是（）和（）。",
+        "options": [],
+        "answer": "$\\frac{5}{6} \\div \\frac{2}{3} = \\frac{5}{8}$; $\\frac{5}{6} \\div \\frac{5}{8} = \\frac{2}{3}$",
+        "solution": "【分析】根据'积÷一个因数=另一个因数'，写出两个除法算式即可。\n\n【详解】把 $\\frac{5}{8} \\times \\frac{2}{3} = \\frac{5}{6}$ 这个算式改写成两个除法算式是 $\\frac{5}{6} \\div \\frac{2}{3} = \\frac{5}{8}$ 和 $\\frac{5}{6} \\div \\frac{5}{8} = \\frac{2}{3}$。\n\n【点睛】本题较易，明确乘、除法属于互逆运算是解答本题的关键。",
+        "knowledge_points": [
+          "分数乘分数",
+          "分数与分数的除法"
+        ],
+        "source": "20-21六年级上·江西赣州·单元测试",
+        "merged_id": "unit3_452",
+        "original_id": "unit3_fill_452"
+      },
+      {
+        "id": "unit3_fill_453",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "在括号里填上'>'、'<'、'='。$\\frac{2}{5} \\div 4$ (  ) $\\frac{2}{5}$ $\\frac{6}{7} \\div \\frac{1}{4}$ (  ) $\\frac{6}{7} \\times \\frac{1}{4}$ $\\frac{4}{5} \\div \\frac{2}{3}$ (  ) $\\frac{4}{5} \\div 2 \\times 3$",
+        "options": [],
+        "answer": "<; >; =",
+        "solution": "【分析】（1）根据一个数除以大于1的数，商小于被除数；(2)根据一个数除以小于1的数，商大于被除数；根据一个数乘小于1的数，积小于原数；(3)算出结果再比较。此题考查的是分数乘除法的计算规律的应用，掌握规律是解答本题的关键。\n\n【详解】(1)因为 $4 > 1$，所以 $\\frac{2}{5} \\div 4 < \\frac{2}{5}$；(2)因为 $\\frac{1}{4}<1, \\frac{6}{7} \\div \\frac{1}{4}>\\frac{6}{7}, \\frac{6}{7} \\times \\frac{1}{4}<\\frac{6}{7}$，所以 $\\frac{6}{7} \\div \\frac{1}{4}>\\frac{6}{7} \\times \\frac{1}{4}$；(3)$\\frac{4}{5} \\div \\frac{2}{3} = \\frac{6}{5}, \\frac{4}{5} \\div 2 \\times 3 = \\frac{6}{5}, \\frac{6}{5} = \\frac{6}{5}$，所以 $\\frac{4}{5} \\div \\frac{2}{3} = \\frac{4}{5} \\div 2 \\times 3$。",
+        "knowledge_points": [
+          "分数乘分数",
+          "分数与整数的除法",
+          "分数与分数的除法"
+        ],
+        "source": "20-21六年级上·江西赣州·单元测试",
+        "merged_id": "unit3_453",
+        "original_id": "unit3_fill_453"
+      },
+      {
+        "id": "unit3_fill_454",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "$1\\frac{1}{8}$ 的倒数是（），2.5的倒数是（）。",
+        "options": [],
+        "answer": "$\\frac{8}{9}$; $\\frac{2}{5}$",
+        "solution": "【分析】求小数倒数的方法：先将小数化成分数，再分子和分母互换位置即可；求分数的倒数的方法：分子和分母互换位置即可。熟练掌握求一个数的倒数的方法是解答本题的关键。\n\n【详解】$1\\frac{1}{8}$ 的倒数是 $\\frac{8}{9}$，2.5的倒数是 $\\frac{2}{5}$",
+        "knowledge_points": [
+          "倒数的认识及应用"
+        ],
+        "source": "20-21六年级上·江西赣州·单元测试",
+        "merged_id": "unit3_454",
+        "original_id": "unit3_fill_454"
+      },
+      {
+        "id": "unit3_fill_455",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "把 $\\frac{6}{7}\\mathrm{~m}$ 长的铁丝平均分成2段，每段长（）$\\mathrm{m}$，每段占全长的（）。",
+        "options": [],
+        "answer": "$\\frac{3}{7}$; $\\frac{1}{2}$",
+        "solution": "【分析】（1）每段铁丝的长度 $=$ 铁丝的总长度 $\\div$ 铁丝平均分成的段数；(2)把这根铁丝看作单位'1'，每段占的分率 $=$ 单位'1'÷铁丝平均分成的段数。掌握分数除法的意义是解答题目的关键。\n\n【详解】(1)$\\frac{6}{7} \\div 2 = \\frac{6}{7} \\times \\frac{1}{2} = \\frac{3}{7}$（米）(2)$1 \\div 2 = \\frac{1}{2}$",
+        "knowledge_points": [
+          "分数与除法的关系",
+          "分数的平均分",
+          "分数与整数的除法"
+        ],
+        "source": "2014六年级·全国·课后作业",
+        "merged_id": "unit3_455",
+        "original_id": "unit3_fill_455"
+      },
+      {
+        "id": "unit3_sort_456",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "排序题",
+        "difficulty": "中等",
+        "question": "已知 $a \\times \\frac{6}{7} = b \\times \\frac{6}{5} = c$，$a、b、c$ 都是不为0的自然数，把 $a、b、c$ 这三个数按从小到大的顺序排列起来是（）。",
+        "options": [],
+        "answer": "$b < c < a$",
+        "solution": "【分析】假设它们的结果为1，分别求出 $a、b、c$ 的值再比较。此题考查的是分数大小比较，解答此题关键是假设它们的结果为1，分别求出 $a, b, c$ 的值再比较。\n\n【详解】假设 $a \\times \\frac{6}{7} = b \\times \\frac{6}{5} = c = 1$，则 $a = \\frac{7}{6}, b = \\frac{5}{6}, c = 1$，因为 $\\frac{5}{6}< 1< \\frac{7}{6}$，所以 $b<c<a$。",
+        "knowledge_points": [
+          "分数与整数的乘法",
+          "分数乘分数",
+          "分数与整数的除法"
+        ],
+        "source": "19-20六年级上·全国·单元测试",
+        "merged_id": "unit3_456",
+        "original_id": "unit3_sort_456"
+      },
+      {
+        "id": "unit3_sort_457",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "排序题",
+        "difficulty": "中等",
+        "question": "已知 $a \\times \\frac{3}{4} = \\frac{4}{5} \\times b = c \\div \\frac{5}{6}$，并且a、b、c都不等于0。那么，a、b、c按从小到大的顺序排列是（）。",
+        "options": [],
+        "answer": "c<b<a",
+        "solution": "【分析】假设 $a \\times \\frac{3}{4} = \\frac{4}{5} \\times b = c \\div \\frac{5}{6} = 1$，分别计算出a、b、c，再比较即可。本题采用了假设法，使题目变得具体化，简单化，分别计算出a、b、c，再比较。\n\n【详解】假设 $a \\times \\frac{3}{4} = \\frac{4}{5} \\times b = c \\div \\frac{5}{6} = 1$；则 $a = \\frac{4}{3}, b = \\frac{5}{4}, c = \\frac{5}{6}$；因为 $\\frac{5}{6}<\\frac{5}{4}<\\frac{4}{3}$，所以 $c<b<a$。",
+        "knowledge_points": [
+          "分数乘分数",
+          "倒数的认识及应用",
+          "分数与分数的除法"
+        ],
+        "source": "19-20六年级上·全国·单元测试",
+        "merged_id": "unit3_457",
+        "original_id": "unit3_sort_457"
+      },
+      {
+        "id": "unit3_judge_458",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "如果a与b互为倒数，则 $\\frac{1}{a} = b$。（）",
+        "options": [],
+        "answer": "✓",
+        "solution": "【分析】互为倒数的两个数乘积为1，用积1除以其中一个数等于另一个数，据此判断即可。本题考查倒数的认识，解答本题的关键是掌握倒数的概念。\n\n【详解】如果 $a$ 与 $b$ 互为倒数，$ab = 1$，所以 $b = \\frac{1}{a}$，说法正确。故答案为：✓",
+        "knowledge_points": [
+          "倒数的认识及应用"
+        ],
+        "source": "18-19六年级下·广东汕头·单元测试",
+        "merged_id": "unit3_458",
+        "original_id": "unit3_judge_458"
+      },
+      {
+        "id": "unit3_judge_459",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "因为 $\\frac{5}{6} \\times 6 \\times \\frac{1}{5} = 1$，所以 $\\frac{5}{6}$、6和 $\\frac{1}{5}$ 互为倒数。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】根据倒数的意义，乘积是1的两个数互为倒数，由此判断即可。此题主要考查倒数的意义，要充分理解倒数的概念中几个关键的字眼，比如：'乘积''两个''互为'，并结合题意展开分析。\n\n【详解】倒数是就两个因数而言的，题干中3个数不能称作互为倒数，所以原题说法错误。故答案为：×",
+        "knowledge_points": [
+          "倒数的认识及应用"
+        ],
+        "source": "22-23六年级上·全国·假期作业",
+        "merged_id": "unit3_459",
+        "original_id": "unit3_judge_459"
+      },
+      {
+        "id": "unit3_judge_460",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "如果 $a \\times \\frac{5}{7} = b \\div \\frac{5}{7}$（a、b不为零），则 $a > b$。（）",
+        "options": [],
+        "answer": "✓",
+        "solution": "【分析】假设式子的值为1，求出a和b的值，最后比较大小即可。掌握用求倒数比较大小的方法是解答题目的关键。\n\n【详解】假设 $a \\times \\frac{5}{7} = b \\div \\frac{5}{7} = 1$ 则 $a = \\frac{7}{5}, b = \\frac{5}{7}$ 因为 $\\frac{7}{5} > \\frac{5}{7}$，所以 $a > b$。故答案为：✓",
+        "knowledge_points": [
+          "倒数的认识及应用"
+        ],
+        "source": "19-20六年级上·四川绵阳·期末",
+        "merged_id": "unit3_460",
+        "original_id": "unit3_judge_460"
+      },
+      {
+        "id": "unit3_judge_461",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "一个自然数除以真分数，商一定大于这个自然数。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】一个数（0除外），除以小于1的数，商比原数大，举例说明即可。关键是注意特殊情况，0除以任何不是0的数都得0。\n\n【详解】0也是自然数，$0 \\div \\frac{1}{3} = 0$，所以原题说法错误。",
+        "knowledge_points": [
+          "分数与整数的除法"
+        ],
+        "source": "20-21六年级上·湖南怀化·单元测试",
+        "merged_id": "unit3_461",
+        "original_id": "unit3_judge_461"
+      },
+      {
+        "id": "unit3_judge_462",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "因为 $\\frac{5}{7} \\times \\frac{2}{5} \\times \\frac{7}{2} = 1$，所以 $\\frac{5}{7} \\cdot \\frac{2}{5}$ 和 $\\frac{7}{2}$ 互为倒数。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】根据乘积是1的两个数互为倒数，进行分析。关键是理解倒数的含义。\n\n【详解】倒数是两个数之间的关系，所以原题说法错误。故答案为：×",
+        "knowledge_points": [
+          "倒数的认识及应用"
+        ],
+        "source": "20-21六年级上·湖南怀化·单元测试",
+        "merged_id": "unit3_462",
+        "original_id": "unit3_judge_462"
+      },
+      {
+        "id": "unit3_judge_463",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "甲数比乙数多 $\\frac{1}{4}$，则乙数比甲数少 $\\frac{1}{4}$。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】可先假设乙数是单位'1'，则甲数就是 $1 + \\frac{1}{4} = \\frac{5}{4}$，要计算乙数比甲数少几分之几可列式：$(\\frac{5}{4} - 1) \\div \\frac{5}{4} = \\frac{1}{5}$，据此判断。\n\n【详解】假设乙数是单位'1'，甲数可以用 $1 + \\frac{1}{4} = \\frac{5}{4}$ 来表示；最后计算乙数比甲数少几分之几：$(\\frac{5}{4} - 1) \\div \\frac{5}{4} = \\frac{1}{4} \\div \\frac{5}{4} = \\frac{1}{4} \\times \\frac{4}{5} = \\frac{1}{5}$ 即乙数比甲数少 $\\frac{1}{5}$，原题说法错误。故答案为：×\n\n【点睛】甲数比乙数多几分之几，是把乙数看作单位'1'；乙数比甲数少几分之几，是把甲数看作单位'1'，因为单位'1'不同，导致两个占比有所差异。",
+        "knowledge_points": [
+          "分数的四则混合运算"
+        ],
+        "source": "14-15六年级上·湖北孝感·期中",
+        "merged_id": "unit3_463",
+        "original_id": "unit3_judge_463"
+      },
+      {
+        "id": "unit3_judge_464",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "甲数比乙数多 $\\frac{3}{5}$，就是乙数比甲数少 $\\frac{3}{5}$。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】由'甲数比乙数多 $\\frac{3}{5}$'可知，乙数是1，甲数是 $(1 + \\frac{3}{5})$。根据'(甲数一乙数)÷甲数'求出乙数比甲数少几分之几即可。明确前后单位'1'不一致是解答本题的关键。\n\n【详解】$\\frac{3}{5} \\div (1 + \\frac{3}{5}) = \\frac{3}{5} \\div \\frac{8}{5} = \\frac{3}{8}$ 故答案为：×",
+        "knowledge_points": [
+          "求比一个数多/少几分之几的数是多少"
+        ],
+        "source": "21-22六年级上·新疆克孜勒苏·期中",
+        "merged_id": "unit3_464",
+        "original_id": "unit3_judge_464"
+      },
+      {
+        "id": "unit3_judge_465",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "$\\frac{3}{11} \\times \\frac{3}{11} \\div \\frac{3}{11} \\times \\frac{3}{11} = 1$。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】利用交换结合律进行简算，再与结果对照即可。关键是注意运算顺序，整数四则混合运算的运算顺序和简便计算方法同样适用于分数。\n\n【详解】$\\frac{3}{11} \\times \\frac{3}{11} \\div \\frac{3}{11} \\times \\frac{3}{11}$ $(\\frac{3}{11} \\div \\frac{3}{11}) \\times (\\frac{3}{11} \\times \\frac{3}{11}) = 1 \\times \\frac{9}{121} = \\frac{9}{121}$ 故答案为：×",
+        "knowledge_points": [
+          "分数乘法运算律",
+          "分数的乘、除法的混合运算"
+        ],
+        "source": "2018六年级上·全国·专题练习",
+        "merged_id": "unit3_465",
+        "original_id": "unit3_judge_465"
+      },
+      {
+        "id": "unit3_judge_466",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "桃子的重量比梨多 $\\frac{2}{9}$，则梨的重量就比桃子少 $\\frac{2}{9}$。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】桃子的重量比梨多 $\\frac{2}{9}$，是把梨的重量看作单位'1'，桃子的重量 = 梨的重量 $\\times (1 + \\frac{2}{9})$，计算梨的重量比桃子少几分之几时，是把桃子的重量看作单位'1'，据此解答。找准单位'1'是解答题目的关键。\n\n【详解】![](images/b10de7e0f970bf11c7321192fd3c06c945d3f8d62f9da68571cc25b6985cca98.jpg)\n\n由图可知，梨有9份，桃子有 $(9 + 2)$ 份。$2 \\div (9 + 2) = 2 \\div 11 = \\frac{2}{11}$ 所以，梨的重量比桃子少 $\\frac{2}{11}$。故答案为：×",
+        "knowledge_points": [
+          "单位'1'的认识与确定",
+          "已知比一个数多/少几分之几是多少，求这个数"
+        ],
+        "source": "21-22六年级上·江西赣州·单元测试",
+        "merged_id": "unit3_466",
+        "original_id": "unit3_judge_466"
+      },{
+        "id": "unit3_judge_467",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "$\\frac{2}{3} \\times \\frac{2}{5} \\times \\frac{15}{4} = 1$，所以 $\\frac{2}{3}, \\frac{2}{5}, \\frac{15}{4}$ 互为倒数。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】乘积为1的两个数互为倒数，据此可知，倒数是两个数之间的关系，由此解答即可。理解倒数的含义是解答本题的关键。\n\n【详解】$\\frac{2}{3} \\times \\frac{2}{5} \\times \\frac{15}{4} = 1$，但不能说 $\\frac{2}{3}$、$\\frac{2}{5}$、$\\frac{15}{4}$ 互为倒数，原题说法错误；故答案为：×\n\n【点睛】",
+        "knowledge_points": [
+          "倒数的认识及应用"
+        ],
+        "source": "20-21六年级上·江西赣州·单元测试",
+        "merged_id": "unit3_467",
+        "original_id": "unit3_judge_467"
+      },
+      {
+        "id": "unit3_judge_468",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "一个数的倒数一定小于它本身。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】根据乘积是1的两个数互为倒数，举例说明即可。关键是理解倒数的含义，真分数的倒数大于它本身。\n\n【详解】$\\frac{3}{4}$ 的倒数是 $\\frac{4}{3}, \\frac{4}{3} > \\frac{3}{4}$，所以原题说法错误。故答案为：×",
+        "knowledge_points": [
+          "倒数的认识及应用"
+        ],
+        "source": "2014六年级·全国·课后作业",
+        "merged_id": "unit3_468",
+        "original_id": "unit3_judge_468"
+      },
+      {
+        "id": "unit3_judge_469",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "如果 $a \\div b = \\frac{1}{5}$，那么 $b$ 就是 $a$ 的5倍。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】把a看作被除数，b看作除数，$\\frac{1}{5}$ 看作商，利用'除数 = 被除数 ÷ 商'表示出b即可求得。掌握除法计算中各部分之间的关系是解答题目的关键。\n\n【详解】$a \\div b = \\frac{1}{5}$ $b = a \\div \\frac{1}{5}$ $b=5a$ 所以b是a的5倍。故答案为：√",
+        "knowledge_points": [
+          "乘、除法的意义和各部分间的关系",
+          "分数与分数的除法"
+        ],
+        "source": "19-20六年级上·全国·单元测试",
+        "merged_id": "unit3_469",
+        "original_id": "unit3_judge_469"
+      },
+      {
+        "id": "unit3_calc_470",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "4.6减去 $\\frac{8}{5}$ 的差被 $\\frac{3}{5}$ 除，结果是多少？",
+        "options": [],
+        "answer": "$\\frac{1}{5}$",
+        "solution": "【分析】第一步计算 $4.6$ 减去 $\\frac{8}{5}$，第二步用 $\\frac{3}{5}$ 除以它们的差，据此计算出结果即可。\n\n【详解】$\\frac{3}{5} \\div (4.6 - \\frac{8}{5}) = \\frac{3}{5} \\div 3 = \\frac{1}{5}$",
+        "knowledge_points": [
+          "分数的四则混合运算"
+        ],
+        "source": "18-19六年级下·广东汕头·单元测试",
+        "merged_id": "unit3_470",
+        "original_id": "unit3_calc_470"
+      },
+      {
+        "id": "unit3_calc_471",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "计算下面各题，能简算的要写出必要的简算过程。$68 \\times (\\frac{3}{4} - \\frac{11}{17})$ $(1 - \\frac{3}{10} \\div \\frac{6}{7}) \\times \\frac{10}{13}$ $3.7 \\times \\frac{6}{5} + 1.3 \\div \\frac{5}{6}$ $\\frac{8}{15} \\times [\\frac{5}{6} \\div (\\frac{7}{9} - \\frac{1}{3})]$",
+        "options": [],
+        "answer": "$7; \\frac{1}{2}; 6; 1$",
+        "solution": "【分析】(1) (3) 利用乘法分配律简便计算；(2) (4) 按照四则混合运算顺序计算。\n\n【详解】(1) $68 \\times (\\frac{3}{4} - \\frac{11}{17}) = 68 \\times \\frac{3}{4} - 68 \\times \\frac{11}{17} = 51 - 44 = 7$\n\n(2) $(1 - \\frac{3}{10} \\div \\frac{6}{7}) \\times \\frac{10}{13} = (1 - \\frac{3}{10} \\times \\frac{7}{6}) \\times \\frac{10}{13} = (1 - \\frac{7}{20}) \\times \\frac{10}{13} = \\frac{13}{20} \\times \\frac{10}{13} = \\frac{1}{2}$\n\n(3) $3.7 \\times \\frac{6}{5} + 1.3 \\div \\frac{5}{6} = 3.7 \\times \\frac{6}{5} + 1.3 \\times \\frac{6}{5} = (3.7 + 1.3) \\times \\frac{6}{5} = 5 \\times \\frac{6}{5} = 6$\n\n(4) $\\frac{8}{15} \\times [\\frac{5}{6} \\div (\\frac{7}{9} - \\frac{1}{3})] = \\frac{8}{15} \\times [\\frac{5}{6} \\div \\frac{4}{9}] = \\frac{8}{15} \\times \\frac{5}{6} \\times \\frac{9}{4} = \\frac{4}{9} \\times \\frac{9}{4} = 1$",
+        "knowledge_points": [
+          "分数乘法运算律",
+          "分数的四则混合运算"
+        ],
+        "source": "21-22六年级上·湖南怀化·单元测试",
+        "merged_id": "unit3_471",
+        "original_id": "unit3_calc_471"
+      },
+      {
+        "id": "unit3_calc_472",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "解方程。$x \\div \\frac{6}{25} = \\frac{5}{12}$ $\\frac{6}{7} - 3x = \\frac{3}{14}$ $\\frac{7}{18} \\div x = \\frac{14}{27}$",
+        "options": [],
+        "answer": "$x = \\frac{1}{10}; x = \\frac{3}{14}; x = \\frac{3}{4}$",
+        "solution": "【分析】$x \\div \\frac{6}{25} = \\frac{5}{12}$，根据等式的性质2，两边同时 $\\times \\frac{6}{25}$ 即可；$\\frac{6}{7} - 3x = \\frac{3}{14}$，先写成 $3x + \\frac{3}{14} = \\frac{6}{7}$ 的形式，根据等式的性质1和2，两边先同时 $-\\frac{3}{14}$，再同时 $\\div 3$ 即可；$\\frac{7}{18} \\div x = \\frac{14}{27}$，先写成 $\\frac{14}{27}x = \\frac{7}{18}$ 的形式，根据等式的性质2，两边同时 $\\times \\frac{27}{14}$ 即可。\n\n【详解】$x \\div \\frac{6}{25} = \\frac{5}{12}$\n\n解：$x \\div \\frac{6}{25} \\times \\frac{6}{25} = \\frac{5}{12} \\times \\frac{6}{25}$\n\n$x = \\frac{1}{10}$\n\n$\\frac{6}{7} - 3x = \\frac{3}{14}$\n\n解：$3x + \\frac{3}{14} - \\frac{3}{14} = \\frac{6}{7} - \\frac{3}{14}$\n\n$3x \\div 3 = \\frac{9}{14} \\div 3$ $x = \\frac{3}{14}$ $\\frac{7}{18} \\div x = \\frac{14}{27}$\n\n解：$\\frac{14}{27}x \\times \\frac{27}{14} = \\frac{7}{18} \\times \\frac{27}{14}$\n\n$x = \\frac{3}{4}$",
+        "knowledge_points": [
+          "应用等式的性质1和2解方程"
+        ],
+        "source": "20-21六年级上·湖南怀化·单元测试",
+        "merged_id": "unit3_472",
+        "original_id": "unit3_calc_472"
+      },
+      {
+        "id": "unit3_calc_473",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "解下列方程。$\\frac{2}{9}x \\div \\frac{1}{3} = 3$ $\\frac{2}{3}x - \\frac{1}{5}x = 2.8$",
+        "options": [],
+        "answer": "$x = \\frac{9}{2}; x = 6$",
+        "solution": "【分析】$\\frac{2}{9}x \\div \\frac{1}{3} = 3$，根据等式的性质2，两边先同时 $\\times \\frac{1}{3}$，再同时 $\\times \\frac{9}{2}$ 即可；$\\frac{2}{3}x - \\frac{1}{5}x = 2.8$，先将左边进行合并，再根据等式的性质2解方程。\n\n【详解】$\\frac{2}{9}x \\div \\frac{1}{3} = 3$\n\n解：$\\frac{2}{9}x \\div \\frac{1}{3} \\times \\frac{1}{3} = 3 \\times \\frac{1}{3}$\n\n$\\frac{2}{9}x \\times \\frac{9}{2} = 1 \\times \\frac{9}{2}$ $x = \\frac{9}{2}$ $\\frac{2}{3}x - \\frac{1}{5}x = 2.8$\n\n解：$\\frac{7}{15}x \\times \\frac{15}{7} = 2.8 \\times \\frac{15}{7}$\n\n$x = 6$",
+        "knowledge_points": [
+          "应用等式的性质2解方程"
+        ],
+        "source": "20-21六年级上·湖南怀化·单元测试",
+        "merged_id": "unit3_473",
+        "original_id": "unit3_calc_473"
+      },
+      {
+        "id": "unit3_calc_474",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "直接写结果。$\\frac{5}{4} \\times \\frac{6}{5} =$ $\\frac{3}{7} \\div 3 =$ $5 \\div \\frac{5}{8} =$ $3.2 \\times \\frac{5}{8} =$ $\\frac{2}{7} \\div \\frac{2}{21} =$",
+        "options": [],
+        "answer": "$\\frac{3}{2}; \\frac{1}{7}; 8; 2; 3$",
+        "solution": "【解析】略",
+        "knowledge_points": [
+          "分数乘分数",
+          "分数与分数的除法"
+        ],
+        "source": "19-20六年级上·内蒙古赤峰·期末",
+        "merged_id": "unit3_474",
+        "original_id": "unit3_calc_474"
+      },
+      {
+        "id": "unit3_calc_475",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "解下列方程。$x \\div \\frac{15}{16} = \\frac{4}{5}$ $\\frac{4}{7}x \\div 2 = 40$",
+        "options": [],
+        "answer": "$x = \\frac{3}{4}; x = 140$",
+        "solution": "【分析】分数除法中，除以一个数相当于乘它的倒数；解方程时根据等式的基本性质求解，据此可得出本题答案。\n\n【详解】$x \\div \\frac{15}{16} = \\frac{4}{5}$\n\n解：$x = \\frac{4}{5} \\times \\frac{15}{16}$\n\n$x = \\frac{3}{4}$\n\n$\\frac{4}{7}x \\div 2 = 40$\n\n解：$\\frac{4}{7}x = 80$\n\n$x = 80 \\div \\frac{4}{7}$\n\n$x = 140$",
+        "knowledge_points": [
+          "应用等式的性质1和2解方程",
+          "分数乘分数",
+          "分数与整数的除法",
+          "分数与分数的除法"
+        ],
+        "source": "21-22六年级上·江西赣州·单元测试",
+        "merged_id": "unit3_475",
+        "original_id": "unit3_calc_475"
+      },
+      {
+        "id": "unit3_calc_476",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "计算题",
+        "difficulty": "容易",
+        "question": "解方程。$\\frac{1}{8}x + 1 = 1.5$ $x - \\frac{1}{6}x = \\frac{5}{8}$ $x \\div \\frac{5}{7} = 4.2$",
+        "options": [],
+        "answer": "$x = 4; x = \\frac{3}{4}; x = 3$",
+        "solution": "【分析】①先应用等式的性质1，将方程左右两边同时减去1；再应用等式性质2，将方程左右两边同时除以 $\\frac{1}{8}$，得到方程的解；②逆用乘法分配律，将方程整理成 $ax = b$ 的形式，再应用等式性质2，将方程左右两边同时除以 $\\frac{5}{6}$，得到方程的解；③应用等式的性质2，将方程左右两边同时乘 $\\frac{5}{7}$，得到方程的解。\n\n【详解】$\\frac{1}{8}x + 1 = 1.5$\n\n解：$\\frac{1}{8}x = 1.5 - 1$\n\n$\\frac{1}{8}x = \\frac{1}{2}$ $x = \\frac{1}{2} \\div \\frac{1}{8}$ $x = \\frac{1}{2} \\times 8$ $x = 4$\n\n解：$(1 - \\frac{1}{6})x = \\frac{5}{8}$\n\n$x - \\frac{1}{6}x = \\frac{5}{8}$ $\\frac{5}{6}x = \\frac{5}{8}$ $x = \\frac{5}{8} \\div \\frac{5}{6}$ $x = \\frac{5}{8} \\times \\frac{6}{5}$ $x = \\frac{3}{4}$\n\n$x \\div \\frac{5}{7} = 4.2$\n\n解：$x = 4.2 \\times \\frac{5}{7}$\n\n$x = 3$",
+        "knowledge_points": [
+          "应用等式的性质1和2解方程",
+          "分数的四则混合运算"
+        ],
+        "source": "20-21六年级上·福建厦门·单元测试",
+        "merged_id": "unit3_476",
+        "original_id": "unit3_calc_476"
+      },
+      {
+        "id": "unit3_calc_477",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "脱式计算。（能简算的要简算）$\\frac{7}{8} \\times 7 + \\frac{7}{8}$ $\\frac{8}{15} \\times [\\frac{5}{6} \\div (\\frac{7}{9} - \\frac{1}{3})]$ $(\\frac{1}{4} + \\frac{2}{9}) \\times 0.36$ $0.75 \\times 26 + 75 \\times \\frac{3}{4} - \\frac{3}{4}$",
+        "options": [],
+        "answer": "7; 1; 0.17; 75",
+        "solution": "【分析】利用乘法分配律计算；先算减法，再算除法，最后算乘法；利用乘法分配律计算；把小数化成分数，利用乘法分配律计算。\n\n【详解】$\\frac{7}{8} \\times 7 + \\frac{7}{8} = \\frac{7}{8} \\times (7 + 1) = 7$ $\\frac{8}{15} \\times [\\frac{5}{6} \\div (\\frac{7}{9} - \\frac{1}{3})] = \\frac{8}{15} \\times [\\frac{5}{6} \\div \\frac{4}{9}] = \\frac{8}{15} \\times \\frac{15}{8} = 1$ $(\\frac{1}{4} + \\frac{2}{9}) \\times 0.36 = \\frac{1}{4} \\times 0.36 + \\frac{2}{9} \\times 0.36 = 0.17$ $0.75 \\times 26 + 75 \\times \\frac{3}{4} - \\frac{3}{4} = \\frac{3}{4} \\times (26 + 75 - 1) = 75$",
+        "knowledge_points": [
+          "分数乘法运算律",
+          "分数的四则混合运算"
+        ],
+        "source": "21-22六年级上·河南南阳·期中",
+        "merged_id": "unit3_477",
+        "original_id": "unit3_calc_477"
+      },
+      {
+        "id": "unit3_calc_478",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "能简算的要简算。$\\frac{7}{15} \\times 23 - 8 \\div \\frac{15}{7}$ $(\\frac{4}{5} + \\frac{1}{4} - \\frac{1}{3}) \\div \\frac{1}{60}$",
+        "options": [],
+        "answer": "7; 43",
+        "solution": "【分析】（1）把除法化为乘法，然后运用乘法分配律进行计算即可。(2)运用乘法分配律进行计算即可。\n\n【详解】$\\frac{7}{15} \\times 23 - 8 \\div \\frac{15}{7} = \\frac{7}{15} \\times 23 - 8 \\times \\frac{7}{15} = \\frac{7}{15} \\times (23 - 8) = \\frac{7}{15} \\times 15 = 7$ $(\\frac{4}{5} + \\frac{1}{4} - \\frac{1}{3}) \\div \\frac{1}{60} = (\\frac{4}{5} + \\frac{1}{4} - \\frac{1}{3}) \\times 60 = \\frac{4}{5} \\times 60 + \\frac{1}{4} \\times 60 - \\frac{1}{3} \\times 60 = 48 + 15 - 20 = 63 - 20 = 43$",
+        "knowledge_points": [
+          "分数乘法运算律",
+          "分数与整数的除法"
+        ],
+        "source": "20-21六年级上·江西赣州·单元测试",
+        "merged_id": "unit3_478",
+        "original_id": "unit3_calc_478"
+      },
+      {
+        "id": "unit3_calc_479",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "解下列方程。$\\frac{5}{9}x = \\frac{5}{6}$ $x + \\frac{5}{8}x = 39$ $\\frac{5}{9}x - \\frac{1}{12} = \\frac{7}{12}$",
+        "options": [],
+        "answer": "$x = \\frac{3}{2}; x = 24; x = \\frac{6}{5}$",
+        "solution": "【分析】根据等式的性质：等式的性质1：等式两边加上或减去同一个数，左右两边仍然相等；等式的性质2：等式两边乘同一个数，或除以同一个不为0的数，左右两边仍然相等；据此计算。\n\n【详解】(1) $\\frac{5}{9}x = \\frac{5}{6}$\n\n解：$x = \\frac{5}{6} \\div \\frac{5}{9}$\n\n$x = \\frac{5}{6} \\times \\frac{9}{5}$\n\n$x = \\frac{3}{2}$\n\n(2) $x + \\frac{5}{8}x = 39$\n\n解：$\\frac{13}{8}x = 39$\n\n$x = 39 \\div \\frac{13}{8}$\n\n$x = 39 \\times \\frac{8}{13}$\n\n$x = 24$\n\n(3) $\\frac{5}{9}x - \\frac{1}{12} = \\frac{7}{12}$\n\n解：$\\frac{5}{9}x = \\frac{7}{12} + \\frac{1}{12}$\n\n$\\frac{5}{9}x = \\frac{2}{3}$\n\n$x = \\frac{2}{3} \\div \\frac{5}{9}$\n\n$x = \\frac{2}{3} \\times \\frac{9}{5}$ $x = \\frac{6}{5}$",
+        "knowledge_points": [
+          "应用等式的性质1和2解方程",
+          "分数与分数的除法"
+        ],
+        "source": "21-22六年级上·内蒙古·期中",
+        "merged_id": "unit3_479",
+        "original_id": "unit3_calc_479"
+      },
+      {
+        "id": "unit3_calc_480",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "解方程。$\\frac{6}{7}x + \\frac{1}{4} = \\frac{3}{2}$ $x \\div \\frac{7}{8} = \\frac{3}{5}$",
+        "options": [],
+        "answer": "$x = \\frac{35}{24}; x = \\frac{21}{40}$",
+        "solution": "【分析】根据等式的性质：1.在等式两边同时加或减去一个相同的数，等式仍然成立。2.在等式两边同时乘或除以一个相同的数（0除外），等式仍然成立。据此进行解方程即可。\n\n【详解】$\\frac{6}{7}x + \\frac{1}{4} = \\frac{3}{2}$\n\n解：$\\frac{6}{7}x = \\frac{3}{2} - \\frac{1}{4}$\n\n$\\frac{6}{7}x = \\frac{5}{4}$ $x = \\frac{5}{4} \\div \\frac{6}{7}$ $x = \\frac{35}{24}$ $x \\div \\frac{7}{8} = \\frac{3}{5}$\n\n解：$x = \\frac{3}{5} \\times \\frac{7}{8}$\n\n$x = \\frac{21}{40}$",
+        "knowledge_points": [
+          "应用等式的性质1和2解方程",
+          "分数乘分数",
+          "分数与分数的除法"
+        ],
+        "source": "20-21六年级上·江西赣州·单元测试",
+        "merged_id": "unit3_480",
+        "original_id": "unit3_calc_480"
+      },
+      {
+        "id": "unit3_calc_481",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "用你喜欢的方法计算。$24 \\times (\\frac{3}{4} + \\frac{2}{3} - \\frac{1}{8})$ $\\frac{8}{15} \\times [\\frac{5}{6} \\div (\\frac{7}{9} - \\frac{1}{3})]$ $2 - \\frac{1}{4} + \\frac{3}{4}$ $0.4 \\times 4 \\div 4 + 0.4$",
+        "options": [],
+        "answer": "31; 1; $2\\frac{1}{2}$; 0.8",
+        "solution": "【分析】（1）运用乘法分配律进行计算即可。(2)有中括号和小括号，先算小括号里面的，再算中括号里面的，最后算括号外面的。(3)运用加法交换律，带符号交换位置，然后运用加法结合律进行计算即可。(4)根据四则运算法则，先算乘除法，后算加减法即可。\n\n【详解】$24 \\times (\\frac{3}{4} + \\frac{2}{3} - \\frac{1}{8}) = 24 \\times \\frac{3}{4} + 24 \\times \\frac{2}{3} - 24 \\times \\frac{1}{8} = 18 + 16 - 3 = 34 - 3 = 31$\n\n$\\frac{8}{15} \\times [\\frac{5}{6} \\div (\\frac{7}{9} - \\frac{1}{3})] = \\frac{8}{15} \\times [\\frac{5}{6} \\div \\frac{4}{9}] = \\frac{8}{15} \\times \\frac{15}{8} = 1$\n\n$2 - \\frac{1}{4} + \\frac{3}{4} = 2 + (\\frac{3}{4} - \\frac{1}{4}) = 2 + \\frac{1}{2} = 2\\frac{1}{2}$\n\n$0.4 \\times 4 \\div 4 + 0.4 = 1.6 \\div 4 + 0.4 = 0.4 + 0.4 = 0.8$",
+        "knowledge_points": [
+          "分数乘法运算律",
+          "分数加、减简便运算",
+          "有括号的运算顺序",
+          "分数与分数的除法"
+        ],
+        "source": "20-21六年级上·江西赣州·单元测试",
+        "merged_id": "unit3_481",
+        "original_id": "unit3_calc_481"
+      },
+      {
+        "id": "unit3_solve_482",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "一项工程，甲独做要20小时完成，现乙先独做8小时，剩下的甲乙再合做4小时完成。乙单独完成全工程一共需多少小时？",
+        "options": [],
+        "answer": "15小时",
+        "solution": "【分析】先求出甲的工作效率，再求出4小时甲的工作量，剩下的由乙完成，再用乙的工作量除以乙用的12小时，求出乙的工作效率，据此求出乙单独完成全工程的时间。\n\n【详解】$1 \\div 20 \\times 4 = \\frac{1}{20} \\times 4 = \\frac{1}{5}$\n\n$(1 - \\frac{1}{5}) \\div (8 + 4)$\n\n$\\frac{4}{5} \\div 12$\n\n$= \\frac{1}{15}$\n\n$1 \\div \\frac{1}{15} = 15$（小时）答：乙单独完成全工程一共需15小时。\n\n【点睛】本题考查分数除法、工程问题，解答本题的关键是掌握工程问题中的数量关系式。",
+        "knowledge_points": [
+          "工程问题",
+          "分数除法的应用"
+        ],
+        "source": "18-19六年级下·广东汕头·单元测试",
+        "merged_id": "unit3_482",
+        "original_id": "unit3_solve_482"
+      },
+      {
+        "id": "unit3_solve_483",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "学校的合唱团有男、女生共40人，如果选出5名女生和男生的 $\\frac{1}{4}$ 站在前，则剩下的男、女生人数正好相等。合唱团原有男生多少人？",
+        "options": [],
+        "answer": "20人",
+        "solution": "【分析】根据题意可得，剩下男生的 $\\frac{3}{4}$ 与剩下的女生人数相等，设男生人数为 $x$ 人，则女生人数为 $(\\frac{3}{4}x + 5)$ 人，再根据男、女生共40人列出方程解答即可。本题考查列方程解决问题、分数除法，解答本题的关键是掌握题中的等量关系式。\n\n【详解】解：设男生人数为 $x$ 人。\n\n$x + (\\frac{3}{4}x + 5) = 40$ $\\frac{7}{4}x + 5 = 40$ $\\frac{7}{4}x = 35$ $x = 20$\n\n答：唱团原有男生20人。",
+        "knowledge_points": [
+          "列方程解含一个未知数的问题",
+          "分数除法的应用"
+        ],
+        "source": "18-19六年级下·广东汕头·单元测试",
+        "merged_id": "unit3_483",
+        "original_id": "unit3_solve_483"
+      },
+      {
+        "id": "unit3_solve_484",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "一批大米，第一次运走 $\\frac{2}{9}$，第二次运走 $\\frac{1}{4}$ 多2吨，还剩36吨，这批大米共有多少吨？",
+        "options": [],
+        "answer": "72吨",
+        "solution": "【分析】把这批大米的总质量看作单位'1'，第一次运走 $\\frac{2}{9}$，第二次运走 $\\frac{1}{4}$，这批大米还剩 (36 + 2) 吨，根据'量÷对应的分率'求出这批大米的总质量，据此解答。本题考查分数除法的应用，找出量和对应的分率是解答题目的关键。\n\n【详解】$(36 + 2) \\div (1 - \\frac{2}{9} - \\frac{1}{4}) = 38 \\div (\\frac{7}{9} - \\frac{1}{4}) = 38 \\div \\frac{19}{36} = 72$（吨）答：这批大米共有72吨。\n\n【点睛】",
+        "knowledge_points": [
+          "分数的四则混合运算",
+          "分数除法的应用"
+        ],
+        "source": "18-19六年级下·广东汕头·单元测试",
+        "merged_id": "unit3_484",
+        "original_id": "unit3_solve_484"
+      },
+      {
+        "id": "unit3_solve_485",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "修一条道路，甲施工队单独修需要12天能完成，乙施工队单独修需要18天能完成。如果两队合修8天能修完这条路吗？",
+        "options": [],
+        "answer": "能修完",
+        "solution": "【分析】把这项工程看做单位'1'，那么甲的工作效率是 $\\frac{1}{12}$，乙的工作效率是 $\\frac{1}{18}$，根据工作时间 = 工作总量 ÷ 工作效率求出甲乙两队合修需要的天数即可解答。此题主要考查工作时间、工作效率、工作总量三者之间的数量关系，解答时往往把工作总量看做1，再利用它们的数量关系解答。\n\n【详解】$1 \\div (\\frac{1}{12} + \\frac{1}{18}) = 1 \\div (\\frac{3}{36} + \\frac{2}{36}) = 1 \\div \\frac{5}{36} = 7.2$（天）\n\n答：如果两队合修8天能修完这条路。",
+        "knowledge_points": [
+          "工程问题",
+          "分数与整数的除法",
+          "异分母分数加、减法"
+        ],
+        "source": "21-22六年级上·湖南怀化·单元测试",
+        "merged_id": "unit3_485",
+        "original_id": "unit3_solve_485"
+      },
+      {
+        "id": "unit3_solve_486",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "学校把一批图书分给五年级和六年级，五年级分得80本，是六年级分得图书本数的 $\\frac{5}{8}$，六年级分得图书多少本？",
+        "options": [],
+        "answer": "128本",
+        "solution": "【分析】将六年级分得图书本数看作单位'1'，五年级分得本数 $\\div$ 对应分率 $=$ 六年级分得本数，据此列式解答。关键是确定单位'1'，理解分数除法的意义。\n\n【详解】$80 \\div \\frac{5}{8} = 128$（本）答：六年级分得图书128本。",
+        "knowledge_points": [
+          "分数除法的应用"
+        ],
+        "source": "20-21六年级上·湖南怀化·单元测试",
+        "merged_id": "unit3_486",
+        "original_id": "unit3_solve_486"
+      },
+      {
+        "id": "unit3_solve_487",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "哥哥每天吃的食物总量是1200克，其中主食相当于副食的 $\\frac{2}{3}$，哥哥每天吃主食和副食各多少克？",
+        "options": [],
+        "answer": "480克、720克",
+        "solution": "【分析】将副食质量看作单位'1'，吃的食物总量 $\\div$ 对应分率 $=$ 副食质量，吃的食物总量一副食质量 $=$ 主食质量，据此列式解答。关键是确定单位'1'，理解分数除法的意义。\n\n【详解】$1200 \\div (1 + \\frac{2}{3}) = 1200 \\div \\frac{5}{3} = 720$（克）\n\n$1200 - 720 = 480$（克）\n\n答：哥哥每天吃主食和副食各480克、720克。",
+        "knowledge_points": [
+          "分数除法的应用"
+        ],
+        "source": "20-21六年级上·湖南怀化·单元测试",
+        "merged_id": "unit3_487",
+        "original_id": "unit3_solve_487"
+      },
+      {
+        "id": "unit3_solve_488",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "某工程队工人的人数是技术员的 $\\frac{5}{2}$，已知工人的人数比技术员多210人，那么工人和技术员各有多少人？",
+        "options": [],
+        "answer": "工人：350人；技术员：140人",
+        "solution": "【分析】已知这个工程队里，工人的人数是技术员的 $\\frac{5}{2}$，且又知工人比技术员多210人，则可假设技术员为 $x$ 人，那么工人就有 $\\frac{5}{2}x$ 人，结合具体题意可列方程：$\\frac{5}{2}x - x = 210$。像这样的题目，通常要结合题目里某一个条件，假设一倍量为标准量 $x$，则比较量就可以用含有未知数的式子表示；再根据另一个具体条件，找到等量关系列出方程。\n\n【详解】解：设技术员有 $x$ 名。\n\n$\\frac{5}{2}x - x = 210$\n\n$(\\frac{5}{2} - 1)x = 210$\n\n$\\frac{3}{2}x = 210$\n\n$x = 210 \\div \\frac{3}{2}$\n\n$x = 210 \\times \\frac{2}{3}$\n\n$x = 140$\n\n$140 \\times \\frac{5}{2} = 350$（人）\n\n答：工人有350人；技术员有140人。",
+        "knowledge_points": [
+          "分数的四则混合运算",
+          "列方程解含两个未知数的问题"
+        ],
+        "source": "20-21六年级上·福建厦门·单元测试",
+        "merged_id": "unit3_488",
+        "original_id": "unit3_solve_488"
+      },
+      {
+        "id": "unit3_solve_489",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "一批零件，师傅单独做要10小时，徒弟单独做需15小时，师傅每小时比徒弟多做180个，这批零件共几个？",
+        "options": [],
+        "answer": "5400个",
+        "solution": "【分析】根据题意可知，这批零件总量为单位'1'，师傅的工作效率为 $\\frac{1}{10}$，徒弟的工作效率为 $\\frac{1}{15}$，师傅每小时比徒弟多做的180个对应的分率为 $(\\frac{1}{10} - \\frac{1}{15})$，再根据分数除法的意义解答即可。明确师傅每小时比徒弟多做的180个零件对应的分辨率是解答本题的关键。\n\n【详解】$180 \\div (\\frac{1}{10} - \\frac{1}{15}) = 180 \\div \\frac{1}{30} = 5400$（个）；\n\n答：这批零件共5400个。",
+        "knowledge_points": [
+          "已知比一个数多/少几分之几是多少，求这个数"
+        ],
+        "source": "20-21六年级上·福建厦门·单元测试",
+        "merged_id": "unit3_489",
+        "original_id": "unit3_solve_489"
+      },
+      {
+        "id": "unit3_solve_490",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "工厂要包装369千克的水果糖，每袋装 $\\frac{3}{5}$ 千克，已知包装了 $\\frac{1}{4}$，已经包装了多少袋？",
+        "options": [],
+        "answer": "153袋",
+        "solution": "【分析】根据求一个数的几分之几是多少，用乘法求出已经包装的重量是多少，然后用已经包装的重量除以每袋装的重量，其结果要根据实际情况运用去尾法保留整数即可。本题考查求一个数的几分之几是多少，明确用乘法先求出水果糖的 $\\frac{1}{4}$ 是多少是解题的关键。\n\n【详解】$369 \\times \\frac{1}{4} \\div \\frac{3}{5} = 92.25 \\div \\frac{3}{5} \\approx 153$（袋）\n\n答：已经包装了153袋。",
+        "knowledge_points": [
+          "分数的乘、除法的混合运算"
+        ],
+        "source": "20-21六年级上·福建厦门·单元测试",
+        "merged_id": "unit3_490",
+        "original_id": "unit3_solve_490"
+      },
+      {
+        "id": "unit3_solve_491",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "青草晒干后，约失去原来重量的 $\\frac{7}{25}$，如果晒干后要得到36千克，需要青草多少千克？",
+        "options": [],
+        "answer": "50千克",
+        "solution": "【分析】根据题意可知，'青草的质量 $\\times (1 - \\frac{7}{25}) = 36$'，据此解答即可。\n\n【详解】$36 \\div (1 - \\frac{7}{25}) = 36 \\div \\frac{18}{25} = 50$（千克）\n\n答：需要青草50千克。\n\n【点睛】明确题目中存在的等量关系式是解答本题的关键。",
+        "knowledge_points": [
+          "分数的四则混合运算",
+          "分数除法的应用"
+        ],
+        "source": "20-21六年级上·福建厦门·单元测试",
+        "merged_id": "unit3_491",
+        "original_id": "unit3_solve_491"
+      },
+      {
+        "id": "unit3_solve_492",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "只列式不计算。挖一条水渠，王大爷要10天，李叔叔每天可以挖这条水渠的 $\\frac{1}{15}$。两人合作，几天能挖完？",
+        "options": [],
+        "answer": "$1 \\div (\\frac{1}{10} + \\frac{1}{15})$",
+        "solution": "【分析】已知挖这条水渠，王大爷要10天，即王大爷每天挖水渠全长的 $\\frac{1}{10}$，再结合李叔叔的工效每天挖这条水渠的 $\\frac{1}{15}$ 可得，两人合作一天可挖 $(\\frac{1}{10} + \\frac{1}{15})$，最后根据把挖这条水渠的工作总量看成单位'1'，以及'工作时间 = 工作总量 ÷ 工效'，可得：要计算几天能挖完可列式 $1 \\div (\\frac{1}{10} + \\frac{1}{15})$。本题需要我们把挖这条水渠的工作量看成单位'1'，再将两人的工作时间转化为工效，最后用单位'1'除以工效之和，结果就是工作天数。\n\n【详解】$1 \\div (\\frac{1}{10} + \\frac{1}{15}) = 1 \\div (\\frac{6}{60} + \\frac{4}{60}) = 1 \\div \\frac{10}{60} = 6$（天）\n\n答：两人合作，6天能挖完。",
+        "knowledge_points": [
+          "分数的四则混合运算"
+        ],
+        "source": "20-21六年级上·福建厦门·单元测试",
+        "merged_id": "unit3_492",
+        "original_id": "unit3_solve_492"
+      },
+      {
+        "id": "unit3_solve_493",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "某工程队修一条路，第一天修了全长的 $\\frac{1}{3}$，第二天修了全长的 $\\frac{1}{4}$，第一天比第二天多修 $200\\mathrm{~m}$，这条路全长是多少米？",
+        "options": [],
+        "answer": "2400米",
+        "solution": "【分析】根据题意可知，第一天比第二天多修 $200\\mathrm{~m}$ 占全长的 $(\\frac{1}{3} - \\frac{1}{4})$，再根据分数除法的意义解答即可。熟练掌握分数除法的意义是解答本题的关键。\n\n【详解】$200 \\div (\\frac{1}{3} - \\frac{1}{4}) = 200 \\div \\frac{1}{12} = 2400$（米）\n\n答：这条路全长是2400米。",
+        "knowledge_points": [
+          "已知比一个数多/少几分之几是多少，求这个数"
+        ],
+        "source": "19-20五年级下·河北·单元测试",
+        "merged_id": "unit3_493",
+        "original_id": "unit3_solve_493"
+      },
+      {
+        "id": "unit3_solve_494",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "一瓶油连瓶重3.4千克，用去一半后，连瓶还重1.9千克。原来有油多少千克？瓶子重多少千克？",
+        "options": [],
+        "answer": "油3千克；瓶子0.4千克",
+        "solution": "【分析】先计算出这瓶油的一半的质量，再根据'量 $\\div$ 对应的分率'求出这瓶油的总质量；瓶子的质量 = 油和瓶子的总质量 - 这瓶油的质量。本题考查分数除法的应用，找准量和对应的分率是解答题目的关键。\n\n【详解】油：$(3.4 - 1.9) \\div \\frac{1}{2} = 1.5 \\div \\frac{1}{2} = 3$（千克）\n\n瓶：$3.4 - 3 = 0.4$（千克）\n\n答：原来有油3千克，瓶子重0.4千克。",
+        "knowledge_points": [
+          "分数除法的应用"
+        ],
+        "source": "17-18三年级下·北京·单元测试",
+        "merged_id": "unit3_494",
+        "original_id": "unit3_solve_494"
+      },
+      {
+        "id": "unit3_solve_495",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "一个长方体盒子，宽是长的 $\\frac{2}{3}$，长是高的 $\\frac{5}{6}$，它的宽是 $20$ 厘米。这个长方体盒子的体积是多少立方厘米？",
+        "options": [],
+        "answer": "21600立方厘米",
+        "solution": "【分析】根据已知一个数的几分之几是多少，求这个数，用除法即可求出这个长方体的长和高，然后根据长方体的体积 $=$ 长 $\\times$ 宽 $\\times$ 高，据此代入数值进行计算即可。本题考查长方体的体积，熟记公式是解题的关键。\n\n【详解】$20 \\div \\frac{2}{3} = 30$（厘米）\n\n$30 \\div \\frac{5}{6} = 36$（厘米）\n\n$30 \\times 20 \\times 36 = 600 \\times 36 = 21600$（立方厘米）\n\n答：这个长方体盒子的体积是21600立方厘米。",
+        "knowledge_points": [
+          "长方体的体积",
+          "分数与整数的除法"
+        ],
+        "source": "2021六年级上·全国·专题练习",
+        "merged_id": "unit3_495",
+        "original_id": "unit3_solve_495"
+      },
+      {
+        "id": "unit3_solve_496",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "学校图书室里，女同学人数占 $\\frac{4}{9}$，后来又来了2名女同学，这时女同学占 $\\frac{9}{19}$，图书室原来有多少同学？",
+        "options": [],
+        "answer": "36人",
+        "solution": "【分析】假设图书馆原来有 $x$ 人，原来女同学数为 $\\frac{4}{9}x$ 人，根据题意列出等式计算出结果即可。\n\n【详解】解：设图书馆原来有 $\\mathrm{x}$ 人，原来女同学数为 $\\frac{4}{9}\\mathrm{x}$ 人。本题关键在于找出题中的等量关系式：原来的女生人数 $+2 =$ 现在的女生人数。\n\n$\\frac{4}{9}x + 2 = (x + 2) \\times \\frac{9}{19}$ $\\frac{4}{9}x + 2 = \\frac{9}{19}x + \\frac{18}{19}$ $2 - \\frac{18}{19} = \\frac{9}{19}x - \\frac{4}{9}x$ $\\frac{20}{19} = \\frac{81}{171}x - \\frac{76}{171}x$ $\\frac{5}{171}x = \\frac{20}{19}$ $x = \\frac{20}{19} \\div \\frac{5}{171}$ $x = \\frac{20}{19} \\times \\frac{171}{5}$ $x = 36$\n\n答：原来有36人。",
+        "knowledge_points": [
+          "分数除法的应用"
+        ],
+        "source": "20-21六年级上·江西赣州·单元测试",
+        "merged_id": "unit3_496",
+        "original_id": "unit3_solve_496"
+      },
+      {
+        "id": "unit3_solve_497",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "打一份文稿，小张单独完成需要10小时，小李单独完成需要8小时，两人一起打5小时能完成任务吗？",
+        "options": [],
+        "answer": "能",
+        "solution": "【分析】把这份稿件的总数看作单位'1'，先求出两人合打的工作效率和，再依据工作时间 = 工作总量 ÷ 工作效率和，求出两人合作完成这项工作的时间，再和5小时作比较即可。本题属于工程问题，能灵活运用等量关系式：工作时间 = 工作总量 ÷ 工作效率是解答此题的关键。\n\n【详解】$1 \\div (\\frac{1}{10} + \\frac{1}{8}) = 1 \\div \\frac{9}{40} = 4\\frac{4}{9}$（小时）\n\n$4\\frac{4}{9}<5$，所以能完成。\n\n答：两人一起打5小时能完成任务。",
+        "knowledge_points": [
+          "工程问题",
+          "分数与整数的除法"
+        ],
+        "source": "20-21六年级上·江西赣州·单元测试",
+        "merged_id": "unit3_497",
+        "original_id": "unit3_solve_497"
+      },
+      {
+        "id": "unit3_solve_498",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "公平商店里中性笔的单价是钢笔的 $\\frac{1}{3}$，聪聪买了一支钢笔和一支中性笔共用去24元，钢笔和中性笔的单价各是多少元？",
+        "options": [],
+        "answer": "18元，6元",
+        "solution": "【分析】假设钢笔的价钱为 $x$ 元，则中性笔的价格是 $\\frac{1}{3}x$ 元，根据题意列出等式求解即可。本题关键在于根据题意列出方程并进行求解，关键是分数除法的计算。\n\n【详解】设钢笔的价钱为 $x$ 元，\n\n$x + \\frac{1}{3}x = 24$ $\\frac{4}{3}x = 24$ $x = 24 \\div \\frac{4}{3}$ $x = 24 \\times \\frac{3}{4}$ $x = 18$\n\n则中性笔的价格为：$\\frac{1}{3} \\times 18 = 6$（元）\n\n答：钢笔的单价为18元，中性笔的单价为6元。",
+        "knowledge_points": [
+          "列方程解含一个未知数的问题",
+          "分数除法的应用"
+        ],
+        "source": "20-21六年级上·江西赣州·单元测试",
+        "merged_id": "unit3_498",
+        "original_id": "unit3_solve_498"
+      },
+      {
+        "id": "unit3_solve_499",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "一种钢材，$\\frac{1}{4}\\mathrm{~m}$ 重 $\\frac{5}{8}$，现有这样的钢材 $6\\mathrm{~m}$，重多少 $\\mathrm{t}$？",
+        "options": [],
+        "answer": "15t",
+        "solution": "【分析】用 $\\frac{5}{8} \\div \\frac{1}{4}$ 求出1米重多少吨，再乘 $6\\mathrm{~m}$ 即可。先求出1米重多少吨是解答本题的关键。\n\n【详解】$\\frac{5}{8} \\div \\frac{1}{4} \\times 6 = \\frac{5}{2} \\times 6 = 15$（吨）\n\n答：现有这样的钢材 $6\\mathrm{~m}$，重 $15\\mathrm{t}$。",
+        "knowledge_points": [
+          "分数与整数的乘法",
+          "分数与分数的除法"
+        ],
+        "source": "20-21六年级上·江西赣州·单元测试",
+        "merged_id": "unit3_499",
+        "original_id": "unit3_solve_499"
+      },
+      {
+        "id": "unit3_solve_500",
+        "unit": "第三单元：分数除法",
+        "chapter": "分数除法",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "有60吨的货物，由甲车单独运需要10次完成，由乙车单独运需要15次。用两车同时运，多少次可以运完？（用两种方法解答）",
+        "options": [],
+        "answer": "6次",
+        "solution": "【分析】方法一：可运用货物重除以运的次数，即可得出甲、乙两车每次运货物的数量，再解出答案；方法二：可将这些货物看作单位'1'，用1除以次数即可得到甲车、乙车的效率，再用1除以甲乙效率之和，即可得出答案。本题主要考查的是单位'1'及分数的四则运算，解题的关键是找出甲车、乙车的效率，进而得出答案。\n\n【详解】方法一：\n\n$60 \\div (60 \\div 10 + 60 \\div 15) = 60 \\div (6 + 4) = 60 \\div 10 = 6$（次）\n\n答：两车同时运，6次可以运完。\n\n方法二：\n\n将这些货物看作单位'1'，则甲车的效率为 $\\frac{1}{10}$，乙车效率为 $\\frac{1}{15}$，\n\n$1 \\div (\\frac{1}{10} + \\frac{1}{15}) = 1 \\div \\frac{1}{6} = 6$\n\n答：两车同时运，6次可以运完。",
+        "knowledge_points": [
+          "分数的四则混合运算",
+          "工程问题"
+        ],
+        "source": "20-21六年级上·福建厦门·单元测试",
+        "merged_id": "unit3_500",
+        "original_id": "unit3_solve_500"
+      }
+  ]
\ No newline at end of file
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new file mode 100644
index 0000000000000000000000000000000000000000..b595de4b36737976837e3ec94a4f41bbc21d4195
--- /dev/null
+++ "b/\345\205\255\345\271\264\347\272\247\351\242\230\351\233\206/\345\205\255\345\271\264\347\272\247\344\270\212/\345\205\255\345\271\264\347\272\247\344\270\212\347\254\254\344\272\214\345\215\225\345\205\203\344\275\215\347\275\256\344\270\216\346\226\271\345\220\221\346\225\260\346\215\256\351\233\206.json"
@@ -0,0 +1,10009 @@
+[
+    {
+        "id": "unit2_choice_1",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "明明家位于学校的东偏南 $60^{\\circ}$ 方向 1000 米处, 则学校位于明明家的 ( ) 方向 1000 米处。",
+        "options": [
+          "A. 北偏西 $60^{\\circ}$",
+          "B. 西偏北 $60^{\\circ}$",
+          "C. 南偏东 $60^{\\circ}$",
+          "D. 南偏东 $30^{\\circ}$"
+        ],
+        "answer": "B",
+        "solution": "根据方向的相对性：方向相反，角度不变，距离不变；据此选择即可。根据方向的相对性可得：明明家位于学校的东偏南 $60^{\\circ}$ 方向 1000 米处，则学校位于明明家的西偏北 $60^{\\circ}$（或北偏西 $30^{\\circ}$）方向 1000 米处。故答案为: B",
+        "knowledge_points": [
+          "东、南、西、北方向",
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级下·天津津南·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_001",
+        "original_id": "unit2_choice_1"
+      },
+      {
+        "id": "unit2_choice_2",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "下列条件中能确定学校位置的是 ( )。",
+        "options": [
+          "A. 学校在图书馆的南偏东 $30^{\\circ}$ 方向处",
+          "B. 学校在离体育馆 600 米处",
+          "C. 学校在书店正北方向 500 米处",
+          "D. 学校在医院的正东方向处"
+        ],
+        "answer": "C",
+        "solution": "确定物体的位置需要方向和距离两个要素。据此分析。A. 仅给出方向（南偏东 $30^{\\circ}$），缺少距离，无法确定位置，此选项错误。B. 仅给出距离 (600 米), 缺少方向, 无法确定位置, 此选项错误。C. 同时给出方向（正北）和距离（500 米），可以确定位置，此选项正确。D. 仅给出方向 (正东), 缺少距离, 无法确定位置, 此选项错误。故答案为: C",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级下·江西南昌·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_002",
+        "original_id": "unit2_choice_2"
+      },
+      {
+        "id": "unit2_choice_3",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "小华家在外婆家的北偏西 $30^{\\circ}$ 方向 2 千米处, 外婆家在小华家的 ( )。",
+        "options": [
+          "A. 北偏西 $30^{\\circ}$ 方向 2 千米处",
+          "B. 南偏东 $30^{\\circ}$ 方向 2 千米处",
+          "C. 南偏西 $30^{\\circ}$ 方向 2 千米处"
+        ],
+        "answer": "B",
+        "solution": "根据方向的相对性：方向相对，角度相同。即与北相对的是南，与西相对的是东，据此解答。小华家在外婆家的北偏西 $30^{\\circ}$ 方向 2 千米处, 外婆家在小华家的南偏东 $30^{\\circ}$ (或东偏南 $60^{\\circ}$) 方向 2 千米处。故答案为: B",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级下·湖南邵阳·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_003",
+        "original_id": "unit2_choice_3"
+      },
+      {
+        "id": "unit2_choice_4",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "如果 $A$ 地在 $B$ 地的南偏东 $50^{\\circ}$ 方向上, 那么 $B$ 地在 $A$ 地的 ( ) 方向上。",
+        "options": [
+          "A. 北偏西 $50^{\\circ}$",
+          "B. 南偏西 $40^{\\circ}$",
+          "C. 北偏西 $40^{\\circ}$",
+          "D. 南偏东 $50^{\\circ}$"
+        ],
+        "answer": "A",
+        "solution": "根据方向的相对性，它们的方向相反，角度相等，距离相等；据此解答（答案不唯一）。如果 $A$ 地在 $B$ 地的南偏东 $50^{\\circ}$ 方向上，那么 $B$ 地在 $A$ 地的北偏西 $50^{\\circ}$ 方向上。故答案为：A",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级下·河北邢台·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_004",
+        "original_id": "unit2_choice_4"
+      },
+      {
+        "id": "unit2_choice_5",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "小明家在学校的东偏南 $75^{\\circ}$ 方向, 那么学校在小明家的 ( ) 方向。",
+        "options": [
+          "A. 南偏东 $75^{\\circ}$",
+          "B. 北偏西 $25^{\\circ}$",
+          "C. 西偏北 $75^{\\circ}$",
+          "D. 东偏南 $25^{\\circ}$"
+        ],
+        "answer": "C",
+        "solution": "根据方向的相对性：方向相反，角度相同。即与东相对的是西，与南相对的是北，据此选择即可。小明家在学校的东偏南 $75^{\\circ}$ 方向, 那么学校在小明家的西偏北 $75^{\\circ}$ (或北偏西 $15^{\\circ}$) 方向。故答案为: C",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级下·湖南张家界·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_005",
+        "original_id": "unit2_choice_5"
+      },
+      {
+        "id": "unit2_choice_6",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "下列说法, 不正确的是（）。",
+        "options": [
+          "A. 3 厘米, 8 厘米, 4 厘米的三根小棒不能拼成三角形",
+          "B. 在同一平面内, 两条直线如果不相交, 就一定互相平行",
+          "C. 图中一共有 9 条线段 ![](images/9b9bfebc-19c4-47d1-b8bf-4fa58e35c9ae.png)",
+          "D. 甲在乙的东偏北 $40^{\\circ}$ 方向, 乙在甲的西偏南 $40^{\\circ}$ 方向"
+        ],
+        "answer": "C",
+        "solution": "A. 根据三角形任意两边之和大于第三边, 任意两边之差小于第三边; B. 根据平行线的定义判断: 在同一个平面内, 不相交的两条直线叫做平行线; C. 线段: 2 个端点, 可以度量长度, 是直线的一部分; D. 根据方向的相对性, 它的方向相反, 角度相等, 距离相等; 据此解答。A. 因为 $3 + 4 < 8$, 所以 3 厘米, 8 厘米, 4 厘米的三根小棒不能拼成三角形, 选项说法正确; B. 在同一平面内, 两条直线如果不相交就一定平行, 选项说法正确; C. 基本的线段有 4 条, 由 2 条基本的线段组成的线段有 3 条, 由 3 条基本的线段组成的线段有 2 条, 由 4 条基本的线段组成的线段有 1 条, 图中一共有 10 条线段, 选项说法错误; D. 甲在乙的东偏北 $40^{\\circ}$ 方向, 乙在甲的西偏南 $40^{\\circ}$ 方向, 选项说法正确。故答案为: C",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置",
+          "三角形三边关系",
+          "平行的特征及性质",
+          "线段、直线、射线的认识及特征"
+        ],
+        "source": "23-24 六年级下·湖北随州·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_006",
+        "original_id": "unit2_choice_6"
+      },
+      {
+        "id": "unit2_choice_7",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "小华从家出发向北偏东 $40^{\\circ}$ 方向走 300 米到图书馆, 原路返回时应向 ( ) 方向走 300 米。",
+        "options": [
+          "A. 南偏西 $40^{\\circ}$",
+          "B. 北偏西 $40^{\\circ}$",
+          "C. 南偏西 $50^{\\circ}$"
+        ],
+        "answer": "A",
+        "solution": "小华从家出发向北偏东 $40^{\\circ}$ 方向走 300 米到图书馆,原路返回时应向相反方向, 相等距离返回, 根据方向相对性, 北偏东 $40^{\\circ}$ 方向和南偏西 $40^{\\circ}$ 相对, 据此解答。![](images/cf97ca65d457305a81514c5b53f873cb9a59a6d93dd4dcedda5b2a312618db54.jpg) 小华从家出发向北偏东 $40^{\\circ}$ 方向走 300 米到图书馆, 原路返回时应向南偏西 $40^{\\circ}$ (或西偏南 $50^{\\circ}$) 方向走 300 米。故答案为: A",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 四年级下·湖南长沙·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_007",
+        "original_id": "unit2_choice_7"
+      },
+      {
+        "id": "unit2_choice_8",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "东偏北 $60^{\\circ}$, 与 ( ) 描述的是同一方向。",
+        "options": [
+          "A. 北偏东 $60^{\\circ}$",
+          "B. 北偏东 $30^{\\circ}$",
+          "C. 西偏南 $60^{\\circ}$"
+        ],
+        "answer": "B",
+        "solution": "东和北之间的夹角是 $90^{\\circ}$, 东偏北也可以说成北偏东, 角度 $= 90^{\\circ} -$ 东偏北的角度, 据此分析。$90^{\\circ} - 60^{\\circ} = 30^{\\circ}$ 东偏北 $60^{\\circ}$, 与北偏东 $30^{\\circ}$ 描述的是同一方向。故答案为: B",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 五年级下·江西赣州·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_008",
+        "original_id": "unit2_choice_8"
+      },
+      {
+        "id": "unit2_choice_9",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "一架飞机从某飞机场向南偏东 $40^{\\circ}$ 方向飞行 1000 千米，返航时，飞机要向（）。",
+        "options": [
+          "A. 北偏西 $50^{\\circ}$ 方向飞行 1000 千米",
+          "B. 东偏南 $40^{\\circ}$ 方向飞行1000千米",
+          "C. 北偏西 $40^{\\circ}$ 方向飞行1000千米"
+        ],
+        "answer": "C",
+        "solution": "一架飞机从某飞机场向南偏东 $40^{\\circ}$ 方向飞行 1000 千米, 此时以机场为坐标原点, 向上为正北方向, 左为正西方向, 右为正东方向, 下为正南方向。这架飞机是向下偏右方向, 与向下夹角为 $40^{\\circ}$。飞机要返航, 则此时以飞机为坐标原点, 机场位置在飞机的左偏上位置, 与正上方夹角为 $40^{\\circ}$, 即北偏西 $40^{\\circ}$, 距离应该是 1000 千米。据分析可得: 飞机返航时, 飞机要向北偏西 $40^{\\circ}$ 方向飞行 1000 千米。故答案为: C",
+        "knowledge_points": [
+          "根据方向、角度和距离描述路线图"
+        ],
+        "source": "24-25 六年级下·河南三门峡·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_009",
+        "original_id": "unit2_choice_9"
+      },
+      {
+        "id": "unit2_choice_10",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "小明家在超市北偏东 $40^{\\circ}$ 方向，那么超市在小明家（）方向。",
+        "options": [
+          "A. 南偏西 $40^{\\circ}$",
+          "B. 南偏西 $50^{\\circ}$",
+          "C. 南偏东 $50^{\\circ}$",
+          "D. 南偏东 $40^{\\circ}$"
+        ],
+        "answer": "A",
+        "solution": "小明家在超市北偏东 $40^{\\circ}$ 方向, 坐标中向上为北方,向右为东方, 向左为西方, 向下为南方, , 即小明家在超市的右上角 $40^{\\circ}$, 与正上方的角度为 $40^{\\circ}$。超市位于小明家的方向, 即左下方, 与正下方角度为 $40^{\\circ}$, 即南偏西 $40^{\\circ}$,据此可得出答案。小明家在超市北偏东 $40^{\\circ}$ 方向, 那么超市在小明家南偏西 $40^{\\circ}$。故答案为: A",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "2024·河南郑州·小升初真题",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_010",
+        "original_id": "unit2_choice_10"
+      },
+      {
+        "id": "unit2_choice_11",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "小明从家出发向东偏北 $25^{\\circ}$ 方向走 500 米到学校，放学回家时应向（）方向走 500 米。",
+        "options": [
+          "A. 西偏南 $25^{\\circ}$",
+          "B. 东偏南 $25^{\\circ}$",
+          "C. 西偏南 $65^{\\circ}$"
+        ],
+        "answer": "A",
+        "solution": "![](images/7002faa5687917c10a050aa2853e310f6aab2b42058cf5af8f45b4e1fbb4cc2f.jpg) 如图: 学校和家的位置是相对的,此题方向相反，距离角度不变，由此求解。小明从家出发向东偏北 $25^{\\circ}$ 方向走 500 米到学校,放学回家时应向西偏南 $25^{\\circ}$ 方向走 500 米。故答案为：A",
+        "knowledge_points": [
+          "根据方向、角度和距离描述路线图"
+        ],
+        "source": "24-25 四年级下·湖南岳阳·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_011",
+        "original_id": "unit2_choice_11"
+      },
+      {
+        "id": "unit2_choice_12",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "一架飞机从正定机场向北偏东 $60^{\\circ}$ 方向飞行了 1500 千米，返回时飞机（）。",
+        "options": [
+          "A. 向北偏西 $60^{\\circ}$ 方向飞行 1500 千米。",
+          "B. 向南偏西 $60^{\\circ}$ 方向飞行 1500 千米。",
+          "C. 向北偏西 $30^{\\circ}$ 方向飞行 1500 千米。",
+          "D. 向南偏西 $30^{\\circ}$ 方向飞行 1500 千米。"
+        ],
+        "answer": "B",
+        "solution": "两个地点的相对位置，方向相反，角度和距离不变。一架飞机从正定机场向北偏东 $60^{\\circ}$ 方向飞行，那么返回时的方向是南偏西 $60^{\\circ}$，距离不变。$90^{\\circ} - 60^{\\circ} = 30^{\\circ}$, 一架飞机从正定机场向北偏东 $60^{\\circ}$ 方向飞行了 1500 千米, 返回时飞机应向南偏西 $60^{\\circ}$ 或西偏南 $30^{\\circ}$ 方向飞行 1500 千米。故答案为: B",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "2024·河北邯郸·小升初真题",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_012",
+        "original_id": "unit2_choice_12"
+      },
+      {
+        "id": "unit2_choice_13",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "小明家在学校南偏西 $60^{\\circ}$ 的方向上，那么学校在小明家（）的方向上。",
+        "options": [
+          "A. 东偏北 $60^{\\circ}$",
+          "B. 南偏西 $30^{\\circ}$",
+          "C. 北偏东 $60^{\\circ}$"
+        ],
+        "answer": "C",
+        "solution": "根据位置的相对性可知, 它们的方向相反, 角度相等, 距离相等, 据此解答。小明家在学校南偏西 $60^{\\circ}$ 的方向上, 那么学校在小明家北偏东 $60^{\\circ}$ 的方向上。故答案为: C",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "2024·贵州安顺·小升初真题",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_013",
+        "original_id": "unit2_choice_13"
+      },
+      {
+        "id": "unit2_choice_14",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "超市在学校东偏南 $30^{\\circ}$ 方向，那么学校在超市的（）方向上。",
+        "options": [
+          "A. 西偏北 $30^{\\circ}$",
+          "B. 西偏北 $60^{\\circ}$",
+          "C. 南偏东 $30^{\\circ}$",
+          "D. 南偏东 $60^{\\circ}$"
+        ],
+        "answer": "A",
+        "solution": "根据位置的相对性（方向相反，度数不变）来解答本题；如：我在你的西偏南 $40^{\\circ}$ 方向上，你就在我的东偏北 $40^{\\circ}$ 方向上。因为：东与西相对，南与北相对，超市在学校东偏南 $30^{\\circ}$ 方向，所以学校在超市的西偏北 $30^{\\circ}$ 方向上。故答案为：A",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25六年级下·海南海口·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_014",
+        "original_id": "unit2_choice_14"
+      },
+      {
+        "id": "unit2_choice_15",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "学校在红红家的东偏北 $60^{\\circ}$ 方向上，则红红家在学校的（）方向上。",
+        "options": [
+          "A. 南偏东 $60^{\\circ}$",
+          "B. 南偏西 $60^{\\circ}$",
+          "C. 西偏南 $30^{\\circ}$",
+          "D. 西偏南 $60^{\\circ}$"
+        ],
+        "answer": "D",
+        "solution": "根据方向的相对性，它们的方向相反，角度相等，距离相等，据此解答。$90^{\\circ} - 60^{\\circ} = 30^{\\circ}$ 学校在红红家的东偏北 $60^{\\circ}$ 方向上, 则红红家在学校的西偏南 $60^{\\circ}$ (或南偏西 $30^{\\circ}$) 方向上。故答案为: D",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "2024·河北承德·小升初真题",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_015",
+        "original_id": "unit2_choice_15"
+      },
+      {
+        "id": "unit2_choice_16",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "周末，小宁从家出发沿东偏北 $25^{\\circ}$ 方向走 580 米到宁波图书馆借书，借完后他原路返回（）方向走 580 米就可以到家。",
+        "options": [
+          "A. 北偏东 $25^{\\circ}$",
+          "B. 东偏南 $25^{\\circ}$",
+          "C. 西偏南 $25^{\\circ}$",
+          "D. 西偏北 $25^{\\circ}$"
+        ],
+        "answer": "C",
+        "solution": "根据方向的相对性，它们的方向相反，角度相等，距离相等；据此解答。$90^{\\circ} - 25^{\\circ} = 65^{\\circ}$ 周末，小宁从家出发沿东偏北 $25^{\\circ}$ 方向走 580 米到宁波图书馆借书，借完后他原路返回西偏南 $25^{\\circ}$ （或南偏西 $65^{\\circ}$）方向走 580 米就可以到家。故答案为: C",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "2024·浙江宁波·小升初真题",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_016",
+        "original_id": "unit2_choice_16"
+      },
+      {
+        "id": "unit2_choice_17",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "北斗卫星导航系统是中国自主研发的卫星导航系统之一，广泛应用于军事、民用、商业等领域。根据我国北斗导航定位显示：台风中心位于A市北偏西 $25^{\\circ}$ 方向，还可以说成台风中心位于A市（）方向。",
+        "options": [
+          "A. 南偏东 $65^{\\circ}$",
+          "B. 西偏北 $65^{\\circ}$",
+          "C. 东偏南 $65^{\\circ}$",
+          "D. 西偏北 $25^{\\circ}$"
+        ],
+        "answer": "B",
+        "solution": "’北偏西 $25^{\\circ}$’是以正北方向为起始边，逆时针旋转 $25^{\\circ}$ 到目标方向，’北偏西 $25^{\\circ}$’在正北和正西方向之间，我们也可以以正西方向为起始边，到目标方向则需要顺时针旋转，且旋转的角度就是 $90^{\\circ} - 25^{\\circ} = 65^{\\circ}$。因为西和北之间是 $90^{\\circ}$, $90^{\\circ} - 25^{\\circ} = 65^{\\circ}$ 所以在描述方向时, 台风中心位于 A 市北偏西 $25^{\\circ}$ 方向, 还可以说成台风中心位于 A 市西偏北 $65^{\\circ}$ 方向。故答案为: B",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "2024·陕西安康·小升初真题",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_017",
+        "original_id": "unit2_choice_17"
+      },
+      {
+        "id": "unit2_choice_18",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "小红家在小强家北偏东 $42^{\\circ}$ 的方向上，小强家在小红家（）的方向上。",
+        "options": [
+          "A. 南偏西 $42^{\\circ}$",
+          "B. 北偏东 $48^{\\circ}$",
+          "C. 南偏西 $48^{\\circ}$",
+          "D. 北偏东 $42^{\\circ}$"
+        ],
+        "answer": "A",
+        "solution": "小红家在小强家北偏东 $42^{\\circ}$ 的方向上, 是以小强家为观测点; 小强家在小红家的方向是以小红家为观测点;根据位置的相对性可知, 观测点不同, 方向相反, 夹角的度数相同, 距离相同;由此可知, 北偏东 $42^{\\circ}$ 相对的是南偏西 $42^{\\circ}$, 南和西之间的夹角是 $90^{\\circ}, 90^{\\circ} - 42^{\\circ} = 48^{\\circ}$, 所以南偏西 $42^{\\circ}$ 方向, 还可以说成西偏南 $48^{\\circ}$ 方向。$90^{\\circ} - 42^{\\circ} = 48^{\\circ}$ 小红家在小强家北偏东 $42^{\\circ}$ 的方向上, 小强家在小红家南偏西 $42^{\\circ}$ 或西偏南 $48^{\\circ}$ 的方向上。故答案为: A",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "2024·重庆大渡口·小升初真题",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_018",
+        "original_id": "unit2_choice_18"
+      },
+      {
+        "id": "unit2_choice_19",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "超市在学校北偏东 $30^{\\circ}$ 方向上，学校在超市（）方向上。",
+        "options": [
+          "A. 东偏北 $30^{\\circ}$",
+          "B. 南偏西 $30^{\\circ}$",
+          "C. 西偏南 $30^{\\circ}$"
+        ],
+        "answer": "B",
+        "solution": "根据位置的相对性可知, 它们的方向相反, 角度相等, 距离相等, 据此解答。超市在学校北偏东 $30^{\\circ}$ 方向上，学校在超市南偏西 $30^{\\circ}$ 方向上。故答案为: B",
+        "knowledge_points": [
+          "东、南、西、北方向",
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "2024·河南周口·小升初真题",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_019",
+        "original_id": "unit2_choice_19"
+      },
+      {
+        "id": "unit2_choice_20",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "如图，李明的座位用数对表示为（1，3），小英的座位在李明东偏南 $45^{\\circ}$ 方向上，小英的座位用数对表示可能是（）。![](images/4fab77e92aa3cfa4d39cbf73bf45ca2164ea3ccf8f2cdc5b1cb1d76e7f3d3387.jpg)",
+        "options": [
+          "A. (0, 2)",
+          "B. (1, 2)",
+          "C. (2, 2)",
+          "D. (2, 4)"
+        ],
+        "answer": "C",
+        "solution": "以李明为观测点，以图上的’上北下南，左西右东’确定方向，在图中画出李明的东偏南 $45^{\\circ}$ 方向的线段,这条线段与方格的交点都可能是小英的座位, 用数对表示出小英的座位即可。用数对表示位置的方法：数对的第一个数字表示列，第二个数字表示行。如图：![](images/d4fe7190696eab5694d2786a9e23b287904b79cb36da4b2945736bfb023ef36f.jpg) 小英的座位用数对表示可能是（2，2），或（3，1），或（4，0）。故答案为: C",
+        "knowledge_points": [
+          "用数对表示位置",
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "2025 六年级下·云南·专题练习",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_020",
+        "original_id": "unit2_choice_20"
+      },
+      {
+        "id": "unit2_choice_21",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "学校在聪聪家北偏东 $30^{\\circ}$ 的方向上，则聪聪家在学校的（）方向上。",
+        "options": [
+          "A. 南偏西 $60^{\\circ}$",
+          "B. 南偏东 $30^{\\circ}$",
+          "C. 西偏南 $30^{\\circ}$",
+          "D. 西偏南 $60^{\\circ}$"
+        ],
+        "answer": "D",
+        "solution": "学校在聪聪家北偏东 $30^{\\circ}$ 的方向上，是以聪聪家为观测点；聪聪家在学校的方向是以学校为观测点；根据位置的相对性可知，观测点不同，方向相反，夹角的度数相同，距离相同。根据位置的相对性可知: 北偏东 $30^{\\circ}$ 相对的是南偏西 $30^{\\circ}$, 南和西之间的夹角是 $90^{\\circ}, 90^{\\circ} - 30^{\\circ} = 60^{\\circ}$, 所以南偏西 $30^{\\circ}$ 方向, 还可以说成西偏南 $60^{\\circ}$ 方向, 所以聪聪家在学校的南偏西 $30^{\\circ}$ (或西偏南 $60^{\\circ}$) 方向上。故答案为: D",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级下·广西南宁·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_021",
+        "original_id": "unit2_choice_21"
+      },
+      {
+        "id": "unit2_choice_22",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "北斗卫星导航系统是中国自主研发的卫星导航系统之一。根据我国北斗卫星定位显示：台风中心位于M市南偏西 $35^{\\circ}$ 方向，还可以说成台风中心位于M市（）方向。",
+        "options": [
+          "A. 北偏东 $35^{\\circ}$",
+          "B. 东偏北 $55^{\\circ}$",
+          "C. 东偏南 $35^{\\circ}$",
+          "D. 西偏南 $55^{\\circ}$"
+        ],
+        "answer": "D",
+        "solution": "由于南偏西 $35^{\\circ}$ 方向, 是从南往西偏 $30^{\\circ}$, 也可以把这两个方向反着说, 即西偏南的方向, 正南和正西之间夹角是 $90^{\\circ}$, 由于从西往南偏, 那么角度是 $90^{\\circ} - 35^{\\circ} = 55^{\\circ}$, 据此即可选择。由分析可知：北斗卫星导航系统是中国自主研发的卫星导航系统之一。根据我国北斗卫星定位显示：台风中心位于M市南偏西 $35^{\\circ}$ 方向，还可以说成台风中心位于M市西偏南 $55^{\\circ}$ 方向。故答案为: D",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25六年级上·重庆九龙坡·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_022",
+        "original_id": "unit2_choice_22"
+      },
+      {
+        "id": "unit2_choice_23",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "一艘轮船由 A 地向南偏西$45^{\\circ}$ 的方向行驶 40 海里到达 B 地, 再由 B 地向北偏西 $15^{\\circ}$ 的方向行驶 40 海里到达 C 地，则 A、C 相距 ( ) 海里。",
+        "options": [
+          "A. 30",
+          "B. 40",
+          "C. 60",
+          "D. 50"
+        ],
+        "answer": "B",
+        "solution": "根据’轮船由A地向南偏西 $45^{\\circ}$ 的方向行驶40海里到达B地，再由B地向北偏西 $15^{\\circ}$ 的方向行驶40海里到达C地’作图，如图所示：![](images/3276a3b3b3057aae76b389f4ced0cdafce8ef9e5fc7328f41925a5fc529d35cd.jpg) 连接 AC，由图可以判定三角形 ABC 是等边三角形，从而可知 AC 的距离。根据题意作图，如图所示：![](images/d7662b744c9454e18e03c6bd434982f9c1b6db1b9dcd1e4b3522b55da2b16a9a.jpg) 由图可知: $\\angle ABC = 45^{\\circ} + 15^{\\circ} = 60^{\\circ}, \\quad AB = BC = 40$ 海里,所以三角形 $ABC$ 为等边三角形, 即 $AC = AB = 40$ 海里。故答案为: $B$。",
+        "knowledge_points": [
+          "东南、西南、东北、西北方向",
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "2025 六年级·全国·竞赛",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_023",
+        "original_id": "unit2_choice_23"
+      },
+      {
+        "id": "unit2_choice_24",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "在地图上万达广场在虎英公园的北偏西 $50^{\\circ}$ 方向上，那么虎英公园在万达广场的（ ）方向上。",
+        "options": [
+          "A. 南偏东 $40^{\\circ}$",
+          "B. 北偏东 $40^{\\circ}$",
+          "C. 南偏东 $50^{\\circ}$",
+          "D. 北偏东 $50^{\\circ}$"
+        ],
+        "answer": "C",
+        "solution": "根据方向的相对性：方向相反，角度相同。即北相反是南，西相反是东，据此选择即可。据分析可知，在地图上万达广场在虎英公园的北偏西 $50^{\\circ}$ 方向上，那么虎英公园在万达广场的南偏东 $50^{\\circ}$ 方向上。故答案为: C",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "2024·广东东莞·小升初真题",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_024",
+        "original_id": "unit2_choice_24"
+      },
+      {
+        "id": "unit2_choice_25",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "老师从办公室向西偏北 $40^{\\circ}$ 走 $70 \\mathrm{~m}$ 到教室, 下课后他从原路返回办公室, 应向 ( ) 走 $70 \\mathrm{~m}$。",
+        "options": [
+          "A. 北偏西 $40^{\\circ}$",
+          "B. 东偏南 $40^{\\circ}$",
+          "C. 南偏东 $40^{\\circ}$",
+          "D. 西偏北 $40^{\\circ}$"
+        ],
+        "answer": "B",
+        "solution": "老师从办公室向西偏北 $40^{\\circ}$ 走 $70 \\mathrm{~m}$ 到教室, 是以从办公室为观测点; 下课后他从原路返回办公室, 是以教室为观测点; 根据位置的相对性可知, 观测点不同, 方向相反, 夹角的度数相同, 距离相同, 据此解答。老师从办公室向西偏北 $40^{\\circ}$ 走 $70 \\mathrm{~m}$ 到教室, 下课后他从原路返回办公室，应向东偏南 $40^{\\circ}$（或南偏东 $50^{\\circ}$）走 $70 \\mathrm{~m}$。故答案为: B",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "2024·河北张家口·小升初真题",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_025",
+        "original_id": "unit2_choice_25"
+      },
+      {
+        "id": "unit2_choice_26",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "下图是一个等边三角形, 点 A 在点 C 的 ( ) 方向。![](images/ad58138739710fbbeaf095566ee01e42a1a2f08ddd83cbb0e6ebe15fca771e99.jpg)",
+        "options": [
+          "A. 东偏南 $60^{\\circ}$",
+          "B. 南偏东 $60^{\\circ}$",
+          "C. 西偏北 $60^{\\circ}$",
+          "D. 北偏西 $60^{\\circ}$"
+        ],
+        "answer": "C",
+        "solution": "等边三角形的每个角都为 $60^{\\circ}$, 根据平面图上方向的规定: 上北下南, 左西右东, 以点 C 为观测点即可确定点 A 的方向。根据分析可知, 点 A 在点 C 的西偏北 $60^{\\circ}$, 或北偏西 $30^{\\circ}$ 的方向上。故答案为: C",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置",
+          "等腰三角形和等边三角形的认识及特征"
+        ],
+        "source": "24-25 六年级上·广东云浮·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_026",
+        "original_id": "unit2_choice_26"
+      },
+      {
+        "id": "unit2_choice_27",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "丁丁先向东偏北 $45^{\\circ}$方向走了 $120 \\mathrm{~m}$, 又向南偏东 $45^{\\circ}$ 方向走了 $120 \\mathrm{~m}$, 他现在的位置在起点的 ( ) 方向。",
+        "options": [
+          "A. 正北",
+          "B. 正东",
+          "C. 东南"
+        ],
+        "answer": "B",
+        "solution": "地图是上北下南，左西右东，根据丁丁描述的方向、距离、角度，画出路线图，再判断即可。![](images/b8efd54bb0ae291326b936de01e41398fabe331884b38141d7bb021fd8806f80.jpg) 从图中可知：丁丁现在的位置在起点的正东方向。故答案为: B",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·河南焦作·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_027",
+        "original_id": "unit2_choice_27"
+      },
+      {
+        "id": "unit2_choice_28",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "小明家在学校的北偏西 40 度, 则学校在小明家 ( )。",
+        "options": [
+          "A. 东偏南 40 度",
+          "B. 南偏东 50 度",
+          "C. 南偏东40度",
+          "D. 东偏西 50 度"
+        ],
+        "answer": "C",
+        "solution": "根据方向的相对性，方向相反，角度相等，距离相等。以小明家为观测点，描述学校与小明家的位置关系。根据分析可得：小明家在学校的北偏西 40 度（或西偏北 $50^{\\circ}$），则学校在小明家南偏东 40 度（或东偏南 $50^{\\circ}$）。故答案为: C",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·山东菏泽·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_028",
+        "original_id": "unit2_choice_28"
+      },
+      {
+        "id": "unit2_choice_29",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "在表示方向时,灯塔在舰艇的南偏东 $40^{\\circ}$ 也可以说成灯塔在舰艇的 ( )。",
+        "options": [
+          "A. 北偏西 $40^{\\circ}$",
+          "B. 北偏西 $50^{\\circ}$",
+          "C. 东偏南 $50^{\\circ}$",
+          "D. 北偏东 $40^{\\circ}$"
+        ],
+        "answer": "C",
+        "solution": "用方位角描述方向时，通常以正北或正南方向为角的始边，以对象所处的射线为终边，故描述方位角时，一般先叙述北或南，再叙述偏东或偏西，偏多少度；也可以先叙述东或西，再叙述偏南或偏北，偏多少度。$90^{\\circ} - 40^{\\circ} = 50^{\\circ}$ 南偏东 $40^{\\circ}$ 也可以说成东偏南 $50^{\\circ}$。故答案为: C",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·贵州黔东南·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_029",
+        "original_id": "unit2_choice_29"
+      },
+      {
+        "id": "unit2_choice_30",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "小刚放学回家时往北偏西 $35^{\\circ}$ 方向走，那么他上学时应该往（）方向走。",
+        "options": [
+          "A. 西偏北 $35^{\\circ}$",
+          "B. 东偏北 $55^{\\circ}$",
+          "C. 南偏东 $35^{\\circ}$",
+          "D. 南偏东 $55^{\\circ}$"
+        ],
+        "answer": "C",
+        "solution": "小刚放学回家时往北偏西 $35^{\\circ}$ 方向走, 是以学校为观测点; 那么他上学时是以家为观测点; 根据位置的相对性可知, 观测点不同, 方向相反, 夹角的度数相同, 距离相同; 由此可知, 北偏西 $35^{\\circ}$ 相对的是南偏东 $35^{\\circ}$。小刚放学回家时往北偏西 $35^{\\circ}$ 方向走, 那么他上学时应该往南偏东 $35^{\\circ}$ (或东偏南 $55^{\\circ}$) 方向走。故答案为: C",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·湖北十堰·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_030",
+        "original_id": "unit2_choice_30"
+      },
+      {
+        "id": "unit2_choice_31",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "下列条件中, 能确定位置的是 ( )。",
+        "options": [
+          "A. 学校的东偏南 $45^{\\circ}$ 方向处",
+          "B. 学校的南偏北方向 200米处",
+          "C. 离学校 500 米处",
+          "D. 学校的正东方向 100 米处"
+        ],
+        "answer": "D",
+        "solution": "确定位置需要两个条件：一是方向，二是距离。将方向和距离结合起来描述位置时，要注意三个要素：一是观测点，二是方向，三是距离。据此分析解答。A. 学校的东偏南 $45^{\\circ}$ 方向处, 有方向, 但无距离,不符合题意; B. 学校的南偏北方向 200 米处, 有距离, 但方向没讲清角度，而且南偏北的说法有误，不符合题意。C. 离学校 500 米处, 有距离, 但无方向, 不符合题意。D. 学校的正东方向 100 米处, 有方向, 有距离, 符合题意。故答案为: D",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·湖北襄阳·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_031",
+        "original_id": "unit2_choice_31"
+      },
+      {
+        "id": "unit2_choice_32",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "确定观测点后,’北偏东 $30^{\\circ}$’与（）表示的方向是一样的。",
+        "options": [
+          "A. 东偏北 $60^{\\circ}$",
+          "B. 南偏西 $60^{\\circ}$",
+          "C. 东偏南 $30^{\\circ}$",
+          "D. 西偏南 $30^{\\circ}$"
+        ],
+        "answer": "A",
+        "solution": "相邻两个方向之间的夹角是 $90^{\\circ}$, 一个方向可以有两种说法, 如东偏南也可以看作南偏东, 注意角度即可。据此解答。$90^{\\circ} - 30^{\\circ} = 60^{\\circ}$ 北偏东 $30^{\\circ}$ 也可以说东偏北 $60^{\\circ}$。确定观测点后，’北偏东 $30^{\\circ}$’与东偏北 $60^{\\circ}$ 表示的方向是一样的。故答案为: A",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·广东东莞·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_032",
+        "original_id": "unit2_choice_32"
+      },
+      {
+        "id": "unit2_choice_33",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "小明家在体育馆的东偏南 $30^{\\circ}$ 方向上，则体育馆在小明家（）方向上。",
+        "options": [
+          "A. 南偏东 $30^{\\circ}$",
+          "B. 西偏北 $30^{\\circ}$",
+          "C. 北偏西 $30^{\\circ}$",
+          "D. 西偏北 $60^{\\circ}$"
+        ],
+        "answer": "B",
+        "solution": "根据方向的相对性：方向相反，角度相同。即东相反是西，南相反是北，据此选择即可。据分析可知, 小明家在体育馆的东偏南 $30^{\\circ}$ 方向上,则体育馆在小明家西偏北 $30^{\\circ}$ 方向上。故答案为: B",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25六年级上·西藏拉萨·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_033",
+        "original_id": "unit2_choice_33"
+      },
+    {
+        "id": "unit2_choice_34",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "游泳馆在萌萌家西偏南 20 度方向 800 米处, 萌萌家在游泳馆 ( ) 20 度方向 800 米处。",
+        "options": [
+          "A. 南偏西",
+          "B. 西偏北",
+          "C. 东偏北"
+        ],
+        "answer": "C",
+        "solution": "分析题目，根据位置的相对性可知：游泳馆和萌萌家方向相反，角度相等，距离相等，据此结合西对东，南对北可知：西偏南20度的反方向是东偏北20度方向；距离都是800米，据此解答。根据位置的相对性可知：游泳馆在萌萌家西偏南 20 度方向 800 米处，萌萌家在游泳馆东偏北 20 度方向 800 米处。故答案为: C",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·湖南永州·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_034",
+        "original_id": "unit2_choice_34"
+      },
+      {
+        "id": "unit2_choice_35",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "如果小明在小军的北偏西 $45^{\\circ}$ 方向100米处，那么小军在小明的（）方向100米处。",
+        "options": [
+          "A. 西偏南 $45^{\\circ}$",
+          "B. 东偏南 $45^{\\circ}$",
+          "C. 东偏北 $45^{\\circ}$"
+        ],
+        "answer": "B",
+        "solution": "根据方向的相对性：方向相反，角度不变，距离不变；据此判断即可。由分析可得: 如果小明在小军的北偏西 $45^{\\circ}$ 方向 100 米处, 那么小军在小明的南偏东 $45^{\\circ}$ (或东偏南 $45^{\\circ}$) 方向 100 米处。故答案为: B",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25六年级上·吉林松原·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_035",
+        "original_id": "unit2_choice_35"
+      },
+      {
+        "id": "unit2_choice_36",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "小明家在小强家南偏西 $35^{\\circ}$ 方向上，那么小强家在小明家（）方向上。",
+        "options": [
+          "A. 北偏东 $35^{\\circ}$",
+          "B. 北偏东 $55^{\\circ}$",
+          "C. 东偏南 $55^{\\circ}$",
+          "D. 西偏南 $55^{\\circ}$"
+        ],
+        "answer": "A",
+        "solution": "根据位置的相对性：方向相反，角度相同，距离相等；南相反就是北，西相反就是东，进行解答即可。小明家在小强家南偏西 $35^{\\circ}$ 方向上, 那么小强家在小明家北偏东 $35^{\\circ}$ 方向上。故答案为：A",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级下·吉林白城·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_036",
+        "original_id": "unit2_choice_36"
+      },
+      {
+        "id": "unit2_choice_37",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "国庆期间, 依依一家去湄洲岛游玩。在游玩过程中，爸爸去超市买水，让依依和妈妈在原地等候。已知爸爸向东偏南 $25^{\\circ}$ 方向走了 $50 \\mathrm{~m}$到达超市，爸爸若按原路返回，应该向（）方向走 $50 \\mathrm{~m}$。",
+        "options": [
+          "A. 东偏南 $25^{\\circ}$",
+          "B. 西偏北 $25^{\\circ}$",
+          "C. 北偏东 $25^{\\circ}$",
+          "D. 东偏南 $65^{\\circ}$"
+        ],
+        "answer": "B",
+        "solution": "东偏南对西偏北，角度不变，据此选择；西和北之间的夹角是 $90^{\\circ}$，西偏北也可以说成北偏西，角度 $= 90^{\\circ}$ -西偏北的角度；据此解答。$90^{\\circ} - 25^{\\circ} = 65^{\\circ}$ 已知爸爸向东偏南 $25^{\\circ}$ 方向走了 $50 \\mathrm{~m}$ 到达超市, 爸爸若按原路返回, 应该向西偏北 $25^{\\circ}$ 方向走 $50 \\mathrm{~m}$ 或北偏西 $65^{\\circ}$ 方向走 $50 \\mathrm{~m}$。故答案为: B",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·湖南衡阳·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_037",
+        "original_id": "unit2_choice_37"
+      },
+      {
+        "id": "unit2_choice_38",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "学校在小明家北偏东 $30^{\\circ}$ 方向上, 则小明家在学校的 ( ) 方向上。",
+        "options": [
+          "A. 东偏北 $30^{\\circ}$",
+          "B. 南偏西 $30^{\\circ}$",
+          "C. 西偏南 $30^{\\circ}$",
+          "D. 南偏西 $60^{\\circ}$"
+        ],
+        "answer": "B",
+        "solution": "根据方向的相对性：方向相反，角度相同。即北相反是南，东相反是西，据此选择即可。据分析可知, 学校在小明家北偏东 $30^{\\circ}$ 方向上, 则小明家在学校的南偏西 $30^{\\circ}$ 方向上（或西偏南 $60^{\\circ}$ 方向上）。故答案为: B",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置",
+          "根据方向、角度和距离描述路线图"
+        ],
+        "source": "23-24 六年级下·四川内江·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_038",
+        "original_id": "unit2_choice_38"
+      },
+      {
+        "id": "unit2_choice_39",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "北偏西 $25^{\\circ}$ 可以说成是 ( )。",
+        "options": [
+          "A. 西偏北 $25^{\\circ}$",
+          "B. 西偏北 $65^{\\circ}$",
+          "C. 西偏南 $65^{\\circ}$",
+          "D. 北偏东 $25^{\\circ}$"
+        ],
+        "answer": "B",
+        "solution": "根据地图上方向的规定和辨别方法‘上北下南，左西右东‘，以及相邻两个方向间的夹角是 $90^{\\circ}$，据此判断解答。因为北和西之间是 $90^{\\circ}$。$90^{\\circ} - 25^{\\circ} = 65^{\\circ}$北偏西 $25^{\\circ}$ 可以说成西偏北 $65^{\\circ}$。故答案为: B",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·海南海口·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_039",
+        "original_id": "unit2_choice_39"
+      },
+      {
+        "id": "unit2_choice_40",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "南偏东 $50^{\\circ}$ 也可以说 ( )。",
+        "options": [
+          "A. 北偏西 $50^{\\circ}$",
+          "B. 东偏南 $40^{\\circ}$",
+          "C. 东偏北 $50^{\\circ}$",
+          "D. 北偏西 $40^{\\circ}$"
+        ],
+        "answer": "B",
+        "solution": "根据东、西、南、北四个基本方向中，相邻的两个方向的夹角是 90 度，即可解答本题。由分析可知：$90 - 50 = 40$ (度),南偏东 $50^{\\circ}$ 也可以说东偏南 $40^{\\circ}$。故答案为: B",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·江西上饶·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_040",
+        "original_id": "unit2_choice_40"
+      },
+      {
+        "id": "unit2_choice_41",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "若实验楼在教学楼的东偏南 $40^{\\circ}$ 方向上, 则教学楼在实验楼的 ( ) 方向上。",
+        "options": [
+          "A. 西偏北 $50^{\\circ}$",
+          "B. 北偏西 $50^{\\circ}$",
+          "C. 西偏南 $40^{\\circ}$",
+          "D. 南偏东 $50^{\\circ}$"
+        ],
+        "answer": "B",
+        "solution": "实验楼在教学楼的东偏南 $40^{\\circ}$ 方向上, 是以教学楼为观测点; 则教学楼在实验楼的方向是以实验楼为观测点;根据位置的相对性可知, 观测点不同, 方向相反, 夹角的度数相同, 距离不变;由此可知, 东偏南 $40^{\\circ}$ 相对的是西偏北 $40^{\\circ}$, 西和北之间的夹角是 $90^{\\circ}, 90^{\\circ} - 40^{\\circ} = 50^{\\circ}$, 所以西偏北 $40^{\\circ}$ 方向, 还可以说成北偏西 $50^{\\circ}$ 方向。若实验楼在教学楼的东偏南 $40^{\\circ}$ 方向上, 则教学楼在实验楼的西偏北 $40^{\\circ}$ 或北偏西 $50^{\\circ}$ 方向上。故答案为: B",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·江西吉安·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_041",
+        "original_id": "unit2_choice_41"
+      },
+      {
+        "id": "unit2_choice_42",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "北偏西 $20^{\\circ}$, 还可以说成 ( )。",
+        "options": [
+          "A. 南偏西 $20^{\\circ}$",
+          "B. 西偏北 $20^{\\circ}$",
+          "C. 西偏北 $70^{\\circ}$"
+        ],
+        "answer": "C",
+        "solution": "北和西之间的夹角是 $90^{\\circ}, 90^{\\circ} - 20^{\\circ} = 70^{\\circ}$, 所以北偏西 $20^{\\circ}$ 方向, 还可以说成西偏北 $70^{\\circ}$ 方向。$90^{\\circ} - 20^{\\circ} = 70^{\\circ}$ 北偏西 $20^{\\circ}$, 还可以说成西偏北 $70^{\\circ}$。故答案为:B",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·海南三亚·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_042",
+        "original_id": "unit2_choice_42"
+      },
+      {
+        "id": "unit2_choice_43",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "如图, 以小红家为观测点, 商场在小红家的 ( ) 方向上。![](images/08447eb132abcb8afa6a4ed3248bbe575e12acf67a3392c8d8109d0432e242ac.jpg)",
+        "options": [
+          "A. 南偏东 $30^{\\circ}$",
+          "B. 北偏西 $30^{\\circ}$",
+          "C. 西偏北 $30^{\\circ}$"
+        ],
+        "answer": "C",
+        "solution": "将方向和距离结合起来描述位置时，要注意三个要素：一是观测点，二是方向，三是距离。地图上按上北下南左西右东确定方向，本题不需要描述距离，说准确方向和角度即可。以小红家为观测点，商场在小红家的西偏北 $30^{\\circ}$ 或北偏西 $60^{\\circ}$ 方向上。故答案为: C",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·河南信阳·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_043",
+        "original_id": "unit2_choice_43"
+      },
+      {
+        "id": "unit2_choice_44",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "在学校的平面图上，图书馆在体育馆的南偏西 $30^{\\circ}$ 方向，那么体育馆在图书馆的（）方向。",
+        "options": [
+          "A. 北偏东 $30^{\\circ}$",
+          "B. 东偏北 $30^{\\circ}$",
+          "C. 南偏西 $30^{\\circ}$",
+          "D. 北偏西 $30^{\\circ}$"
+        ],
+        "answer": "A",
+        "solution": "根据方向的相对性，它们的方向相反，角度相等，距离相等；据此解答。$90^{\\circ} - 30^{\\circ} = 60^{\\circ}$ 在学校的平面图上, 图书馆在体育馆的南偏西 $30^{\\circ}$ 方向, 那么体育馆在图书馆的北偏东 $30^{\\circ}$ (或东偏北 $60^{\\circ}$) 方向。故答案为: A",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25六年级上·河南南阳·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_044",
+        "original_id": "unit2_choice_44"
+      },
+      {
+        "id": "unit2_choice_45",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "飞机朝东偏南 $30^{\\circ}$ 方向飞行的, 因有紧急任务, 需要立即返航。返航时它应朝 ( ) 方向飞行。",
+        "options": [
+          "A. 西偏北 $30^{\\circ}$",
+          "B. 北偏西 $30^{\\circ}$",
+          "C. 南偏东 $60^{\\circ}$",
+          "D. 北偏东 $60^{\\circ}$"
+        ],
+        "answer": "A",
+        "solution": "根据方向的相对性，他们的方向，角度相等，距离相等，所以东偏南 $30^{\\circ}$ 和西偏北 $30^{\\circ}$ 相对。据此解答即可。飞机朝东偏南 $30^{\\circ}$ 方向飞行的，因有紧急任务，需要立即返航。返航时它应朝西偏北 $30^{\\circ}$ （或北偏西 $60^{\\circ}$）方向飞行。故答案为: A",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·福建莆田·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_045",
+        "original_id": "unit2_choice_45"
+      },
+      {
+        "id": "unit2_choice_46",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "学校在公园的东偏北 $40^{\\circ}$ 方向, 那么公园在学校的 ( ) 方向。",
+        "options": [
+          "A. 北偏东 $40^{\\circ}$",
+          "B. 西偏南 $40^{\\circ}$",
+          "C. 北偏东 $50^{\\circ}$",
+          "D. 西偏南 $50^{\\circ}$"
+        ],
+        "answer": "B",
+        "solution": "根据位置的相对性可知, 它们的方向相反, 角度相等, 距离相等, 据此解答。学校在公园的东偏北 $40^{\\circ}$ 方向, 那么公园在学校的西偏南 $40^{\\circ}$ 或南偏西 $50^{\\circ}$ 方向。故答案为: B",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·广东广州·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_046",
+        "original_id": "unit2_choice_46"
+      },
+      {
+        "id": "unit2_fill_47",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "小华家在小明家的南偏东 $35^{\\circ}$ 方向 1.8 千米处。小华周末从家出发去小明家玩耍, 应该向( )偏( )( )°方向走( )千米。",
+        "options": [],
+        "answer": "北;西;35;1.8",
+        "solution": "根据方向的相对性，南偏东对北偏西，角度和距离不变，进行填空。西和北之间的夹角是 $90^{\\circ}$，北偏西也可以说成西偏北，角度 $= 90^{\\circ}$ 一北偏西的角度。$90^{\\circ} - 35^{\\circ} = 55^{\\circ}$ 小华周末从家出发去小明家玩耍，应该向北偏西 $35^{\\circ}$ 或西偏北 $55^{\\circ}$ 方向走 1.8 千米。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级下·浙江绍兴·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_047",
+        "original_id": "unit2_fill_47"
+      },
+      {
+        "id": "unit2_fill_48",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "在一次实验中，智能小汽车从 A 点出发，第一次向正东行驶了 5 米。第二次又向正南行驶了 5 米。第三次再向正西行驶 5 米，这时小汽车在 A 点的( )方向，距离 A 点( )米。",
+        "options": [],
+        "answer": "正南;5",
+        "solution": "根据上北下南左西右东进行绘制，首先智能小汽车从A点出发，第一次向正东行驶了5米，然后向正南行驶5米，再向正西行驶 5 米, 如图所示, A 5 米 向正西行驶 5 米, 如图所示, ![](images/55396b23-5f3a-4dda-8486-0de59951117c.png)此时在 A 点的正南方向, 连接 A 点与目前所在的位置, 此时构成一个正方形, 四条边相等, 所以距离 A 点 5 米。在一次实验中，智能小汽车从A点出发，第一次向正东行驶了 5 米。第二次又向正南行驶了 5 米。第三次再向正西行驶 5 米, 这时小汽车在 A 点的正南方向, 距离 A 点 5 米。",
+        "knowledge_points": [
+          "东、南、西、北方向",
+          "根据方向、角度和距离确定物体的位置",
+          "正方形的概念及特点"
+        ],
+        "source": "24-25 四年级下·北京房山·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_048",
+        "original_id": "unit2_fill_48"
+      },
+      {
+        "id": "unit2_fill_49",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "飞机从某机场向南偏东 $30^{\\circ}$ 方向飞行了 $1200 \\mathrm{~km}$, 返回时飞机要向( )方向飞行( ) $\\mathrm{km}$。",
+        "options": [],
+        "answer": "北偏西$30^{\\circ}$;1200",
+        "solution": "地图的方位是上北下南左西右东。位置相对方向相反。南对北，东对西，因此南偏东对北偏西，度数不变。飞机从某机场向南偏东 $30^{\\circ}$ 方向飞行了 $1200 \\mathrm{~km}$ ,返回时飞机要向北偏西 $30^{\\circ}$ 方向（答案不唯一）飞行 $1200 \\mathrm{~km}$。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "2024·福建福州·小升初真题",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_049",
+        "original_id": "unit2_fill_49"
+      },
+      {
+        "id": "unit2_fill_50",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "学校在少年宫的南偏西 $45^{\\circ}$ 方向800米处，少年宫在学校的( )偏( )()°方向。",
+        "options": [],
+        "answer": "北;东;45",
+        "solution": "西南方向正对的是东北方向，根据已知条件，可知站在学校的位置看少年宫在东北方向，因为学校在少年宫的南偏西 $45^{\\circ}$ 方向，所以不论是北偏东还是东偏北，角度都是 $45^{\\circ}$。学校在少年宫的南偏西 $45^{\\circ}$ 方向 800 米处, 少年宫在学校的北偏东 $45^{\\circ}$ 方向。",
+        "knowledge_points": [
+          "东南、西南、东北、西北方向",
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 四年级下·湖南岳阳·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_050",
+        "original_id": "unit2_fill_50"
+      },
+      {
+        "id": "unit2_fill_51",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "雯雯家在慧慧家西边 150 米,聪聪家在雯雯家东边 230 米, 那么聪聪家离慧慧家( )米。",
+        "options": [],
+        "answer": "80",
+        "solution": "根据题意，用聪聪家和雯雯家之间的距离减去雯雯家和慧慧家之间的距离，求出聪聪家离慧慧家多少米即可。$230 - 150 = 80$ （米）所以聪聪家离慧慧家 80 米。",
+        "knowledge_points": [
+          "东、南、西、北方向"
+        ],
+        "source": "2023 四年级·全国·竞赛",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_051",
+        "original_id": "unit2_fill_51"
+      },
+      {
+        "id": "unit2_fill_52",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "书店在电影院的东偏北 $35^{\\circ}$ 方向 $5 \\mathrm{~km}$ 处，邮局在电影院的东偏北 $35^{\\circ}$ 方向 $7 \\mathrm{~km}$ 处，那么书店在邮局的( )方向( )km处。",
+        "options": [],
+        "answer": "南偏西$55^{\\circ}$;2",
+        "solution": "将方向和距离结合起来描述位置时，要注意三个要素：一是观测点，二是方向，三是距离。读题可知，以电影院为观测点，书店和邮局在同一个方向，邮局远，书店近，以邮局为观测点，根据方向的相对性，东偏北对西偏南，角度不变，确定方向和角度，邮局距电影院的距离一书店距电影院的距离 = 书店和邮局之间的距离。$7 - 5 = 2(\\mathrm{km})$ 书店在邮局的西偏南 $35^{\\circ}$ 或南偏西 $55^{\\circ}$ 方向 $2 \\mathrm{~km}$ 处。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25六年级下·安徽宣城·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_052",
+        "original_id": "unit2_fill_52"
+      },
+      {
+        "id": "unit2_fill_53",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "小明家在学校北偏东45°方向上，那么学校在小明家的( )偏( )（ ）度方向上。",
+        "options": [],
+        "answer": "南;西;45",
+        "solution": "根据方向的相对性：方向相反，角度相同，距离相同。即北相反是南，东相反是西，据此解答即可。学校在小明家的南偏西 45 度方向上。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "2024·河南周口·小升初真题",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_053",
+        "original_id": "unit2_fill_53"
+      },
+      {
+        "id": "unit2_fill_54",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "甲地在乙地东偏北 $30^{\\circ}$ 的方向上, 距离乙地 3 千米, 那么, 乙地在甲地( )的方向上, 距离甲地( )千米。",
+        "options": [],
+        "answer": "西偏南$30^{\\circ}$;3",
+        "solution": "明确方向相对性原理：在描述位置关系时，观测点互换，方向是相对的，角度不变，距离也不变。确定乙地相对甲地的方向：已知甲地在乙地东偏北 $30^{\\circ}$ 方向，以甲地为观测点时，乙地方向应是西偏南 $30^{\\circ}$。确定乙地与甲地的距离：因为两者之间的距离是固定的，甲地距离乙地 3 千米，所以乙地距离甲地也是 3 千米。如图：![](images/df49d58714f86d449a342bb92feb039ce935a83e5721db79a32e3177f8bd107a.jpg) $90^{\\circ} - 30^{\\circ} = 60^{\\circ}$,甲地在乙地东偏北 $30^{\\circ}$ 的方向上, 距离乙地 $3 \\mathrm{~km}$, 那么乙地在甲地西偏南 $30^{\\circ}$ 或南偏西 $60^{\\circ}$ 的方向上, 距离甲地 $3 \\mathrm{~km}$。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "2024·黑龙江齐齐哈尔·小升初真题",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_054",
+        "original_id": "unit2_fill_54"
+      },
+      {
+        "id": "unit2_fill_55",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "李老师家在学校的东偏南 $45^{\\circ}$ 方向 2500 米处，那么学校就在李老师家( )偏( )°方向 2500 米处。",
+        "options": [],
+        "answer": "西;北45",
+        "solution": "方向是上北下南，左西右东；方向的相对性是东北与西南相对，西北与东南相对。根据方向的相对性，东偏南对西偏北，角度和距离不变，进行填空。东偏南 $45^{\\circ}$ 的相反方向是西偏北 $45^{\\circ}$, 距离不变。李老师家在学校的东偏南 $45^{\\circ}$ 方向 2500 米处, 那么学校就在李老师家西偏北 $45^{\\circ}$ 方向 2500 米处。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "2024·黑龙江佳木斯·小升初真题",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_055",
+        "original_id": "unit2_fill_55"
+      },
+      {
+        "id": "unit2_fill_56",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "学校教学楼在足球场的东偏北 $30^{\\circ}$ 方向约 400 米处，那么足球场在教学楼的( )方向约 400 米处。",
+        "options": [],
+        "answer": "西偏南;$30^{\\circ}$",
+        "solution": "根据两个物体的位置相对性，分别以它们为观测点，看到对方的方向相反，角度和距离相等，据此解答。学校教学楼在足球场的东偏北 $30^{\\circ}$ 方向约 400 米处,那么足球场在教学楼的西偏南 $30^{\\circ}$ (或南偏西 $60^{\\circ}$) 方向约 400 米处。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "2025 六年级下·云南·专题练习",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_056",
+        "original_id": "unit2_fill_56"
+      },
+      {
+        "id": "unit2_fill_57",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "小明家在文具店南偏东 $20^{\\circ}$ 方向, 距离 350 米的位置, 则文具店在小明家( )偏( ) $70^{\\circ}$ 方向( )米的位置。",
+        "options": [],
+        "answer": "西;北;350",
+        "solution": "方向和距离两个条件才能确定物体的位置，根据位置的相对性，可知两处位置观测点不同，它们的方向相反，角度相等，据此解答。由分析可得: 小明家在文具店南偏东 $20^{\\circ}$ 方向, 距离 350 米的位置, 则文具店在小明家北偏西 $20^{\\circ}$ 方向也就是西偏北 $70^{\\circ}$ 方向 350 米的位置。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·湖南长沙·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_057",
+        "original_id": "unit2_fill_57"
+      },
+      {
+        "id": "unit2_fill_58",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "汕头南澳岛在小公园东偏北 $30^{\\circ}$ 的方向上, 距离小公园 $44 \\mathrm{~km}$, 那么小公园在南澳岛( )偏( ) $30^{\\circ}$ 方向上, 距离南澳岛( ) $\\mathrm{km}$。",
+        "options": [],
+        "answer": "西;南;44",
+        "solution": "分析题目，根据位置的相对性可知：东对西，北对南，且角度相同，距离不变，据此可知东偏北 $30^{\\circ}$ 相对的是西偏南 $30^{\\circ}$，距离还是 $44 \\mathrm{~km}$，据此解答。以南澳岛为观测点，小公园在南澳岛的西偏南 $30^{\\circ}$ 方向，且小公园距离南澳岛 $44 \\mathrm{~km}$。汕头南澳岛在小公园东偏北 $30^{\\circ}$ 的方向上, 距离小公园 $44 \\mathrm{~km}$, 那么小公园在南澳岛西偏南 $30^{\\circ}$ 方向上, 距离南澳岛 $44 \\mathrm{~km}$",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "2024·广东汕头·小升初真题",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_058",
+        "original_id": "unit2_fill_58"
+      },
+      {
+        "id": "unit2_fill_59",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "红红在丫丫的北偏东 $40^{\\circ}$ 方向, 那么丫丫在红红的( )方向。",
+        "options": [],
+        "answer": "南偏西$40^{\\circ}$",
+        "solution": "方向和距离两个条件才能确定物体的位置，根据位置的相对性，可知两处位置观测点不同，它们的方向相反，角度相等，据此解答。红红在丫丫的北偏东 $40^{\\circ}$ 方向, 那么丫丫在红红的南偏西 $40^{\\circ}$ 方向。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·河北石家庄·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_059",
+        "original_id": "unit2_fill_59"
+      },
+      {
+        "id": "unit2_fill_60",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "鄂州在武汉的东偏北约 $30^{\\circ}$ 的方向上，那么武汉在鄂州的( )偏( )约 $30^{\\circ}$ 方向上。",
+        "options": [],
+        "answer": "西;南",
+        "solution": "相对的两个位置距离和角度不变，方向相反，据此解答即可。鄂州在武汉的东偏北约 $30^{\\circ}$ 的方向上, 那么武汉在鄂州的西偏南约 $30^{\\circ}$ 方向上。",
+        "knowledge_points": [
+          "东、南、西、北方向",
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25六年级上·湖北鄂州·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_060",
+        "original_id": "unit2_fill_60"
+      },
+      {
+        "id": "unit2_fill_61",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "妈妈从单位出发去菜市场，沿西偏北 $30^{\\circ}$ 方向前行800米到达菜市场，买菜结束后妈妈沿东偏南 $30^{\\circ}$ 方向1500米到达家里。家在单位的____偏 ____ $30^{\\circ}$ 方向距离________米处。",
+        "options": [],
+        "answer": "东;南;700",
+        "solution": "两地的相对位置，方向相反，角度和距离不变。妈妈从单位出发到菜市场，需要沿着西偏北 $30^{\\circ}$ 方向行 800 米，那么妈妈买完菜后，向东偏南 $30^{\\circ}$ 方向行走 800 米可回到单位。根据题意可知，妈妈继续向东偏南 $30^{\\circ}$ 方向行 $(1500 - 800)$ 米可回到家中。这说明，家在单位的东偏南 $30^{\\circ}$ 方向，距离是 $(1500 - 800)$ 米。$1500 - 800 = 700$ （米）所以家在单位的东偏南 $30^{\\circ}$ 方向距离 700 米处。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25六年级上·重庆·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_061",
+        "original_id": "unit2_fill_61"
+      },
+      {
+        "id": "unit2_fill_62",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "亮亮看丫丫在北偏东 $50^{\\circ}$, 则丫丫看亮亮在( )偏( )°。",
+        "options": [],
+        "answer": "南;西50",
+        "solution": "根据两个物体的位置相对性，分别以它们为观测点，看到对方的方向相反，角度和距离相等，据此解答。由分析可知：亮亮看丫丫在北偏东 $50^{\\circ}$，则丫丫看亮亮在南偏西 $50^{\\circ}$ （或西偏南 $40^{\\circ}$）。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 五年级上·河北沧州·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_062",
+        "original_id": "unit2_fill_62"
+      },
+      {
+        "id": "unit2_fill_63",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "小美从家到学校要沿西偏南 $30^{\\circ}$ 方向走 $1.2 \\mathrm{~km}$, 她放学回家时, 应沿( )方向走( )km。",
+        "options": [],
+        "answer": "东偏北;$30^{\\circ}$;1.2",
+        "solution": "根据方向的相对性：方向相反，角度相同，距离相同。即西相反是东，南相反是北，据此解答即可。据分析可知, 小美从家到学校要沿西偏南 $30^{\\circ}$ 方向走 $1.2 \\mathrm{~km}$, 她放学回家时, 应沿东偏北 $30^{\\circ}$ 方向走 $1.2 \\mathrm{~km}$。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置",
+          "根据方向、角度和距离描述路线图"
+        ],
+        "source": "24-25 六年级上·河南南阳·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_063",
+        "original_id": "unit2_fill_63"
+      },
+      {
+        "id": "unit2_fill_64",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "丁丁家在学校的北偏东 $30^{\\circ}$ 方向 300 米处, 学校在丁丁家( )方向, 距离( )米处。",
+        "options": [],
+        "answer": "南偏西$30^{\\circ}$;300",
+        "solution": "以学校为方位，丁丁家在学校的北偏东 $30^{\\circ}$ 方向，距离学校 300 米的表示方法，则以丁丁家为方位，根据上北下南，左西右东，学校的位置是南偏西 $30^{\\circ}$ 方向，距离丁丁家 300 米处。由分析可知, 学校在丁丁家南偏西 $30^{\\circ}$ 方向, 距离 300 米处。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·江西抚州·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_064",
+        "original_id": "unit2_fill_64"
+      },
+      {
+        "id": "unit2_fill_65",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "李琳上学从家往东偏北35度方向走500米到学校，那么她放学应该往( )()度方向走( )米到家。",
+        "options": [],
+        "answer": "西偏南;35;500",
+        "solution": "根据两个物体的位置相对性，分别以它们为观测点，看到对方的方向相反，角度和距离相等，据此解答。李琳上学从家往东偏北 35 度方向走 500 米到学校,那么她放学应该往西偏南 35 度方向走 500 米到家。",
+        "knowledge_points": [
+          "根据方向、角度和距离描述路线图"
+        ],
+        "source": "24-25 六年级上·江西宜春·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_065",
+        "original_id": "unit2_fill_65"
+      },
+      {
+        "id": "unit2_fill_66",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "明明家在儿童剧院的东偏南 $35^{\\circ}$ 方向上，距离 600 米，也可以说明明家在儿童剧院的( )方向上，距离 600 米；那儿童剧院在明明家的( )方向上，距离 600 米。",
+        "options": [],
+        "answer": "南偏东$55^{\\circ}$;西偏北 $35^{\\circ}$或北偏西$55^{\\circ}$",
+        "solution": "明明家在儿童剧院的东偏南 $35^{\\circ}$ 方向上, 距离 600 米, 因为东和南之间的夹角是 $90^{\\circ}, 90^{\\circ} - 35^{\\circ} = 55^{\\circ}$, 所以东偏南 $35^{\\circ}$ 方向还可以说成南偏东 $55^{\\circ}$;明明家在儿童剧院的东偏南 $35^{\\circ}$ 方向上, 距离 600 米, 是以儿童剧院为观测点; 而儿童剧院在明明家的方向, 是以明明家为观测点, 根据位置的相对性可知, 观测点不同, 方向相反, 夹角的度数相同, 距离相同; 据此解答。$90^{\\circ} - 35^{\\circ} = 55^{\\circ}$ 明明家在儿童剧院的东偏南 $35^{\\circ}$ 方向上，距离 600 米，也可以说明明家在儿童剧院的（南偏东 $55^{\\circ}$）方向上，距离 600 米；那儿童剧院在明明家的西偏北 $35^{\\circ}$ （或北偏西 $55^{\\circ}$）方向上，距离 600 米。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·江西吉安·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_066",
+        "original_id": "unit2_fill_66"
+      },
+      {
+        "id": "unit2_fill_67",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如果小雪看小明在北偏东 $30^{\\circ}$ 的方向上，距离是 $3 \\mathrm{~m}$，那么小明看小雪在（ ）偏（ ）（ ）的方向上，距离是（ ） $\\mathrm{m}$。",
+        "options": [],
+        "answer": "南;西;$30^{\\circ}$ 3",
+        "solution": "两人的相对位置，方向相反，角度和距离不变。北偏东 $30^{\\circ}$ 的相反方向是南偏西 $30^{\\circ}$, 南偏西 $30^{\\circ}$ 也可写成西偏南 $60^{\\circ}$。据此填空。如果小雪看小明在北偏东 $30^{\\circ}$ 的方向上, 距离是 $3 \\mathrm{~m}$, 那么小明看小雪在南偏西 $30^{\\circ}$ (或西偏南 $60^{\\circ}$) 的方向上, 距离是 $3 \\mathrm{~m}$。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25六年级上·河南开封·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_067",
+        "original_id": "unit2_fill_67"
+      },
+      {
+        "id": "unit2_fill_68",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "冷水滩城区在普利桥镇南偏西 40 度方向 29 千米处, 普利桥镇在冷水滩城区( )方向( )千米处。",
+        "options": [],
+        "answer": "北偏东40度;29",
+        "solution": "冷水滩城区在普利桥镇南偏西 40 度方向 29 千米处,是以普利桥镇为观测点; 普利桥镇在冷水滩城区的方向是以冷水滩城区为观测点;根据位置的相对性可知，观测点不同，方向相反，夹角的度数相同，距离相同；由此可知，南偏西 40 度相对的是北偏东 40 度，北和东之间的夹角是 90 度， $90 - 40 = 50$ 度，所以北偏东 40 度方向，还可以说成东偏北 50 度方向。冷水滩城区在普利桥镇南偏西40度方向29千米处，普利桥镇在冷水滩城区北偏东 40 度（或东偏北 50 度）方向 29 千米处。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·湖南永州·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_068",
+        "original_id": "unit2_fill_68"
+      },
+      {
+        "id": "unit2_fill_69",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "如果图书馆在学校南偏西 $15^{\\circ}$ 方向上, 距离 400 米, 那么还可以说图书馆在学校的( )方向上, 距离( )米。",
+        "options": [],
+        "answer": "西偏南;$75^{\\circ}$; 400",
+        "solution": "已知图书馆在学校南偏西 $15^{\\circ}$ 方向上, 南和西之间的夹角是 $90^{\\circ}, 90^{\\circ} - 15^{\\circ} = 75$, 所以南偏西 $15^{\\circ}$ 方向上, 还可以说成西偏南 $75^{\\circ}$ 方向。$90^{\\circ} - 15^{\\circ} = 75^{\\circ}$ 如果图书馆在学校南偏西 $15^{\\circ}$ 方向上, 距离 400 米, 那么还可以说图书馆在学校的西偏南 $75^{\\circ}$ 方向上, 距离 400 米。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·四川凉山·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_069",
+        "original_id": "unit2_fill_69"
+      },
+      {
+        "id": "unit2_fill_70",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "丽丽面向北站立,向右转 $40^{\\circ}$ 后, 所面对的方向是( ), 再向左转 $90^{\\circ}$ 后, 背面的方向是( )。",
+        "options": [],
+        "answer": "北偏东;$40^{\\circ}$;南偏东;$50^{\\circ}$",
+        "solution": "面向北，北的右边就是东，所以向右转 $40^{\\circ}$ 后，面对的方向是北偏东 $40^{\\circ}$;再向左转，转动角度是 $90^{\\circ}$，即向北边转后再向西转，那么面对的方向是北偏西（ $90^{\\circ} - 40^{\\circ}$）。背面方向正相反，方向相反，角度不变。$90^{\\circ} - 40^{\\circ} = 50^{\\circ}$ 丽丽面向北站立，向右转 $40^{\\circ}$ 后，所面对的方向是北偏东 $40^{\\circ}$，再向左转 $90^{\\circ}$ 后，背面的方向是南偏东 $50^{\\circ}$。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·山东临沂·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_070",
+        "original_id": "unit2_fill_70"
+      },
+      {
+        "id": "unit2_fill_71",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "学校在广场的东偏北 $35^{\\circ}$ 方向 $1.5 \\mathrm{~km}$ 处, 这是以( )为观测点, 如果改变观测点，广场在学校的( )方向( )km处。",
+        "options": [],
+        "answer": "广场;西偏南$35^{\\circ}$;1.5",
+        "solution": "在后面的地点就是观测点，根据方向的相对性：方向相反，角度相同，距离相同。即东相反是西，北相反是南，据此解答即可。学校在广场的东偏北 $35^{\\circ}$ 方向 $1.5 \\mathrm{~km}$ 处, 这是以广场为观测点, 如果改变观测点, 广场在学校的西偏南 $35^{\\circ}$ (或南偏西 $55^{\\circ}$) 方向 $1.5 \\mathrm{~km}$ 处。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置",
+          "根据方向、角度和距离描述路线图"
+        ],
+        "source": "24-25 六年级上·河南三门峡·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_071",
+        "original_id": "unit2_fill_71"
+      },
+      {
+        "id": "unit2_fill_72",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "小明家位于学校东偏南30度方向，距离学校300米，也可以说小明家位于学校( )偏( )（ ）方向上300米处。",
+        "options": [],
+        "answer": "南;东;$60$度/$60^{\\circ}$",
+        "solution": "小明家位于学校东偏南 30 度方向，距离学校 300 米，是以学校为观测点，小明家的位置如图所示，也可以说小明家位于学校南偏东方向上，角度是（90 度 - 30 度），距离不变还是 300 米。![](images/25b17731608a177443a161690853bbb0186fcbfc76cceff203810aeb4a845881.jpg) 90度-30度=60度 因此小明家位于学校东偏南 30 度方向，距离学校 300 米，也可以说小明家位于学校南偏东 60 度方向上 300 米处。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25六年级上·内蒙古通辽·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_072",
+        "original_id": "unit2_fill_72"
+      },
+      {
+        "id": "unit2_fill_73",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "李兰家在吴辉家东偏南 $40^{\\circ}$ 方向 200 米处, 还可以说李兰家在吴辉家( )方向()米处。",
+        "options": [],
+        "answer": "南偏东$50^{\\circ}$或南偏东50度; 200",
+        "solution": "东和南之间的夹角是 $90^{\\circ}$, 东偏南也可以说成南偏东, 角度 $= 90^{\\circ} -$ 东偏南的角度, 距离不变, 据此分析。$90^{\\circ} - 40^{\\circ} = 50^{\\circ}$ 李兰家在吴辉家东偏南 $40^{\\circ}$ 方向 200 米处, 还可以说李兰家在吴辉家南偏东 $50^{\\circ}$ 方向 200 米处。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·河南信阳·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_073",
+        "original_id": "unit2_fill_73"
+      },
+      {
+        "id": "unit2_fill_74",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "操场上, 萌萌面向北站立, 他向右转 $40^{\\circ}$, 此时, 她面向( )方向站立。",
+        "options": [],
+        "answer": "北偏东$40^{\\circ}$",
+        "solution": "面向北站立，左边是西，右边是东，萌萌面向北站立，他向右转，即从北偏向东，向右转的角度就是北偏东的角度，据此分析。操场上，萌萌面向北站立，他向右转 $40^{\\circ}$，此时，她面向北偏东 $40^{\\circ}$ 或东偏北 $50^{\\circ}$ 方向站立。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·河南信阳·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_074",
+        "original_id": "unit2_fill_74"
+      },
+      {
+        "id": "unit2_fill_75",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "书店在科技馆的东偏北 $65^{\\circ}$ 方向 $5 \\mathrm{~km}$ 处, 科技馆在书店的( )方向( )km 处。",
+        "options": [],
+        "answer": "西偏南$65^{\\circ}$;5",
+        "solution": "根据方向的相对性，东偏北对西偏南，角度和距离不变，进行填空。书店在科技馆的东偏北 $65^{\\circ}$ 方向 $5 \\mathrm{~km}$ 处, 根据分析,科技馆在书店的西偏南 $65^{\\circ}$ (或南偏西 $25^{\\circ}$) 方向 $5 \\mathrm{~km}$ 处。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·河南三门峡·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_075",
+        "original_id": "unit2_fill_75"
+      },
+      {
+        "id": "unit2_fill_76",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "以学校为观测点，小明家在学校的北偏东 $30^{\\circ}$ 方向，距离学校 $500 \\mathrm{~m}$ 处，那么以小明家为观测点，学校在小明家( )偏( )（ ）°方向。",
+        "options": [],
+        "answer": "南;西;30",
+        "solution": "根据位置的相对性，方向相反，角度相同，距离相等；相邻的两个方向之间的度数为 $90^{\\circ}$，据此填空即可。以学校为观测点，小明家在学校的北偏东 $30^{\\circ}$ 方向，距离学校 $500 \\mathrm{~m}$ 处，那么以小明家为观测点，学校在小明家的南偏西 $30^{\\circ}$ （或西偏南 $60^{\\circ}$）方向，距离小明家 $500 \\mathrm{~m}$ 处。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25六年级上·湖南娄底·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_076",
+        "original_id": "unit2_fill_76"
+      },
+      {
+        "id": "unit2_fill_77",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "小智和小东是一对好朋友，他们在同一所学校上学。小智家在学校东偏南 $20^{\\circ}$ 方向150米处，小东家在学校北偏西 $70^{\\circ}$ 方向250米处。小智家在小东家( )偏( )（ ）方向，距离是( )米。",
+        "options": [],
+        "answer": "东;南;$20^{\\circ}$;400",
+        "solution": "根据上北下南，左西右东，图上1厘米表示100米，画图如下，由此判断小智家在小东家的哪个位置上，据此解答。![](images/8be09c112e3fc40c8814f312f0722c5c892c838b3501baf9ccd72058653e3ad0.jpg) 由分析得:$250 + 150 = 400 \\text{(米)}$则小智家在小东家东偏南 $20^{\\circ}$ 方向, 距离是 400 米。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25六年级上·福建三明·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_077",
+        "original_id": "unit2_fill_77"
+      },
+      {
+        "id": "unit2_judge_78",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "如果甲在乙的南偏西 $30^{\\circ}$ 方向上, 那么乙在甲的北偏东 $60^{\\circ}$ 方向上。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "根据方向的相对性，南偏西对北偏东，角度不变，进行分析。北和东之间的夹角是 $90^{\\circ}$，北偏东也可以说成东偏北，角度 $= 90^{\\circ}$ 一北偏东的角度。如果甲在乙的南偏西 $30^{\\circ}$ 方向上, 那么乙在甲的北偏东 $30^{\\circ}$ 或东偏北 $60^{\\circ}$ 方向上, 原题说法错误。故答案为: $×$",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级下·新疆阿克苏·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_078",
+        "original_id": "unit2_judge_78"
+      },
+      {
+        "id": "unit2_judge_79",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "甲在乙的北偏东 $30^{\\circ}$ 方向上, 则乙在甲的东偏北 $60^{\\circ}$ 方向上。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "根据位置的相对性，方向相反，角度相同，距离相等。据此解答即可。甲在乙的北偏东 $30^{\\circ}$ 方向上，则乙在甲的南偏西 $30^{\\circ}$ 方向上。原题说法错误。故答案为： $×$",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级下·云南楚雄·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_079",
+        "original_id": "unit2_judge_79"
+      },
+      {
+        "id": "unit2_judge_80",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "判断题",
+        "difficulty": "困难",
+        "question": "丁丁在笑笑的北偏东 $50^{\\circ}$ 方向 100 米处，笑笑就在丁丁的南偏西 $50^{\\circ}$ 方向 100 米处。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "根据位置的相对性，当两个物体的位置相对时，它们的方向相反，角度相等，距离保持不变。北偏东 $50^{\\circ}$ 的相反方向是南偏西 $50^{\\circ}$，距离不变。丁丁在笑笑的北偏东 $50^{\\circ}$ 方向 100 米处, 说明以笑笑为观测点, 丁丁位于北偏东 $50^{\\circ}$ 方向。根据位置相对性,以丁丁为观测点时, 笑笑的方向应为北偏东 $50^{\\circ}$ 的相反方向,即南偏西 $50^{\\circ}$, 距离仍为 100 米。因此, 原题说法正确。故答案为: $√$",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "2024·甘肃兰州·小升初真题",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_080",
+        "original_id": "unit2_judge_80"
+      },
+      {
+        "id": "unit2_judge_81",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "判断题",
+        "difficulty": "困难",
+        "question": "若轮船在灯塔的北偏东 $30^{\\circ}$ 方向上, 则灯塔就在轮船的南偏西 $30^{\\circ}$ 方向上。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "当以灯塔为观测点时，轮船在灯塔的北偏东 $30^{\\circ}$ 方向，这是基于灯塔这个固定点来确定轮船的位置；当观测点变为轮船时，此时灯塔相对于轮船的位置与轮船相对于灯塔的位置正好是相反的方向，北的相反方向是南，东的相反方向是西，角度保持不变，所以灯塔就在轮船的南偏西 $30^{\\circ}$ 方向。观测点改变时，灯塔相对于轮船的位置与轮船相对于灯塔的位置正好是相反的方向，北偏东方向变为南偏西方向，角度保持不变，所以灯塔就在轮船的南偏西 $30^{\\circ}$ 方向。故答案为: $√$",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "2024·河南新乡·小升初真题",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_081",
+        "original_id": "unit2_judge_81"
+      },
+      {
+        "id": "unit2_judge_82",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "如果淘气在笑笑东偏北 $40^{\\circ}$ 方向上，那么笑笑在淘气西偏南 $40^{\\circ}$ 方向上。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "根据方向的相对性：方向相反，角度不变。东和西相对，南和北相对。据此判断即可。根据方向的相对性可得: 如果淘气在笑笑东偏北 $40^{\\circ}$ 方向上, 那么笑笑在淘气西偏南 $40^{\\circ}$ 方向上。所以原题说法正确。故答案为: $√$",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "2024·广东梅州·小升初真题",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_082",
+        "original_id": "unit2_judge_82"
+      },
+      {
+        "id": "unit2_judge_83",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "判断题",
+        "difficulty": "困难",
+        "question": "电影院在奶奶家南偏西 $60^{\\circ}$ 方向 $800 \\mathrm{~m}$ 处，则奶奶家在电影院北偏东 $30^{\\circ}$ 方向 $800 \\mathrm{~m}$ 处。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "根据方向的相对性，东对西，南对北，所以南偏西对北偏东，角度和距离不变，进行解答。电影院在奶奶家南偏西 $60^{\\circ}$ 方向 $800 \\mathrm{~m}$ 处, 则奶奶家在电影院北偏东 $60^{\\circ}$ (东偏北 $30^{\\circ}$) 方向 $800 \\mathrm{~m}$ 处, 原题说法错误。故答案为: $×$",
+        "knowledge_points": [
+          "东南、西南、东北、西北方向",
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "2024·河南信阳·小升初真题",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_083",
+        "original_id": "unit2_judge_83"
+      },
+      {
+        "id": "unit2_judge_84",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "判断题",
+        "difficulty": "困难",
+        "question": "南偏东 $40^{\\circ}$ 也可以说是东偏南 $50^{\\circ}$。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "东和南之间的夹角是 $90^{\\circ}$, 南偏东 $40^{\\circ}$ 是以正南为基准, 向东偏转 $40^{\\circ}$, 此时与正东方向之间的夹角为 $90^{\\circ} - 40^{\\circ} = 50^{\\circ}$。$90^{\\circ} - 40^{\\circ} = 50^{\\circ}$ 南偏东 $40^{\\circ}$ 也可以说是东偏南 $50^{\\circ}$。原题说法正确。故答案为: $√$",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24 五年级下·河南驻马店·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_084",
+        "original_id": "unit2_judge_84"
+      },
+      {
+        "id": "unit2_judge_85",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "学校在小明家南偏东 $30^{\\circ}$ 方向上, 则小明家在学校东偏南 $30^{\\circ}$ 方向上。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "描述物体的位置时，先确定观测点，再根据‘上北下南，左西右东‘结合题中角度描述方向，最后根据两地之间的距离确定物体的位置，据此解答。![](images/17b095ae8befe292576b57abd43358ad7d676debb432434d0cf910321a335ccb.jpg) 由题意可知，以小明家为观测点时，学校在小明家南偏东$30^{\\circ}$ 方向上，那么以学校为观测点时，小明家在学校北偏西$30^{\\circ}$ 或者西偏北 $60^{\\circ}$ 方向上，两地之间的距离不变。故答案为: $×$",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级下·河南漯河·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_085",
+        "original_id": "unit2_judge_85"
+      },
+      {
+        "id": "unit2_judge_86",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "判断题",
+        "difficulty": "困难",
+        "question": "欢欢从家里出发向南偏东 $56^{\\circ}$ 方向上走 $2 \\mathrm{~km}$ 到学校, 那么, 她放学时可以从学校往北偏西 $56^{\\circ}$ 方向上走 $2 \\mathrm{~km}$ 回家。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "欢欢从家里出发向南偏东 $56^{\\circ}$ 方向上走 $2 \\mathrm{~km}$ 到学校,是以欢欢家为观测点; 她放学时可以从学校出发是以学校为观测点;根据位置的相对性可知，观测点不同，方向相反，夹角的度数相同，距离相同；由此可知，向南偏东 $56^{\\circ}$ 相对的是北偏西 $56^{\\circ}$。欢欢从家里出发向南偏东 $56^{\\circ}$ 方向上走 $2 \\mathrm{~km}$ 到学校,那么, 她放学时可以从学校往北偏西 $56^{\\circ}$ (或西偏北 $34^{\\circ}$)方向上走 $2 \\mathrm{~km}$ 回家。原题说法正确。故答案为: $√$",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·湖北荆州·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_086",
+        "original_id": "unit2_judge_86"
+      },
+      {
+        "id": "unit2_judge_87",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "判断题",
+        "difficulty": "困难",
+        "question": "聪聪家在学校的南偏东 $25^{\\circ}$ 方向, 学校则在聪聪家的北偏西 $25^{\\circ}$ 方向。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "根据方向的相对性，位置相反，角度相同，距离相等，据此即可判断。由分析可知：聪聪家在学校的南偏东 $25^{\\circ}$ 方向, 学校则在聪聪家的北偏西 $25^{\\circ}$ 方向。原题说法正确。故答案为: $√$",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·江西赣州·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_087",
+        "original_id": "unit2_judge_87"
+      },
+      {
+        "id": "unit2_judge_88",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "小新家在学校南偏东 $40^{\\circ}$ 方向1000米处，那么学校在小新家北偏东 $40^{\\circ}$ 方向1000米处。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "根据方向的相对性，它们的方向相反，角度相等，距离相等；据此解答。$90^{\\circ} - 40^{\\circ} = 50^{\\circ}$ 小新家在学校南偏东 $40^{\\circ}$ 方向 1000 米处, 那么学校在小新家北偏西 $40^{\\circ}$ （或西偏北 $50^{\\circ}$）方向 1000 米处。原题干说法错误。故答案为: $×$",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25六年级上·四川广元·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_088",
+        "original_id": "unit2_judge_88"
+      },
+      {
+        "id": "unit2_judge_89",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "乐乐家在欢欢家北偏西 $30^{\\circ}$ 方向上，则欢欢家在乐乐家南偏东 $30^{\\circ}$ 方向上。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "根据位置的相对性可知, 它们的方向相反, 角度相等，距离相等，据此解答。乐乐家在欢欢家北偏西 $30^{\\circ}$ 方向上，则欢欢家在乐乐家南偏东 $30^{\\circ}$ 方向上。所以原题说法正确。故答案为: $√$",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·四川凉山·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_089",
+        "original_id": "unit2_judge_89"
+      },
+      {
+        "id": "unit2_judge_90",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "A 店在 B 店的西偏北 $30^{\\circ}$ 方向 200 米处, 则 B 店在 A 店的东偏南 $30^{\\circ}$ 方向 200 米处。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "描述物体的位置时，先确定观测点，再根据‘上北下南，左西右东‘结合题中角度描述方向，最后根据两地之间的距离确定物体的位置，据此解答。![](images/46c3bdc74b7504539889c72280c795b027e6517e757c6f7c02933bfc58054255.jpg) 如图所示, 以 B 店为观测点时, A 店在 B 店的西偏北 $30^{\\circ}$ 方向, 那么以 A 店为观测点时, B 店在 A 店的东偏南 $30^{\\circ}$ 方向,两地之间的距离不变都是 200 米, 所以题目说法正确。故答案为: $√$",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·宁夏石嘴山·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_090",
+        "original_id": "unit2_judge_90"
+      },
+      {
+        "id": "unit2_judge_91",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "西偏北 30 度和北偏西 60 度是同一个方向。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "描述物体的位置时，先确定观测点，再根据‘上北下南，左西右东‘结合题中角度描述方向，最后根据两地之间的距离确定物体的位置，据此解答。![](images/e6dc477d07180c904eeedf32d10a54c1264b7f4543032016eb3990c18bd8097a.jpg) 如图所示, 西偏北 30 度和北偏西 60 度是同一个方向。故答案为: $√$",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·湖南永州·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_091",
+        "original_id": "unit2_judge_91"
+      },
+      {
+        "id": "unit2_judge_92",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "A 点在 B 点的北偏东 $30^{\\circ}$ 方向上，那么 B 点在 A 点的西偏南 $30^{\\circ}$ 的方向上。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "根据 ‘A 点在 B 点的北偏东 $30^{\\circ}$ 方向上‘, 作图如下:![](images/554c2eb802a5bf833a49ea7a26320ed558bf6458636acaaee42feb4b60d67a78.jpg) A 点在 B 点的北偏东 $30^{\\circ}$ 方向上, 是以 B 点为观测点; B 点在 A 点的方向是以 A 点为观测点; 根据位置的相对性, 观测点不同, 方向相反, 夹角的度数相同, 距离相同。由此判断。A点在B点的北偏东 $30^{\\circ}$ 方向上，那么B点在A点的南偏西 $30^{\\circ}$ （或西偏南 $60^{\\circ}$）方向上，原题说法错误。故答案为: $×$",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "2024 六年级上·全国·专题练习",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_092",
+        "original_id": "unit2_judge_92"
+      },
+      {
+        "id": "unit2_judge_93",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "判断题",
+        "difficulty": "困难",
+        "question": "邮局在公园的南偏西 $20^{\\circ}$ 方向, 也可以说公园在邮局的西偏南 $20^{\\circ}$ 方向。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "已知邮局在公园的南偏西 $20^{\\circ}$ 方向, 是以公园为观测点; 公园在邮局的方向是以邮局为观测点; 观测点不同,方向相反, 夹角的度数相同, 据此判断。如图所示：邮局在公园的南偏西 $20^{\\circ}$ 方向，也可以说公园在邮局的北偏东 $20^{\\circ}$ 方向，原题干的说法是错误的。![](images/c38a1cdca29f75a895b96ea6cc11ef0a57c3b9267741d9b331818ff291dd76e0.jpg) 故答案为: $×$",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 五年级上·河北保定·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_093",
+        "original_id": "unit2_judge_93"
+      },
+      {
+        "id": "unit2_drawing_94",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "在一次军事演习中，一艘军舰从起点先向东偏北 $60^{\\circ}$ 方向行驶72千米后，再向正东方向行驶36千米，最后向东偏南 $30^{\\circ}$ 方向行驶24千米到达终点。(1) 根据上面的描述, 把军舰行驶的路线图画完整。(2) 如果从终点按原路匀速返回起点用了 4 小时, 那么这艘军舰返回时的速度是 ( ) 千米/小时。",
+        "options": [],
+        "answer": "（1）见详解 (2) 33",
+        "solution": "（1）根据上北下南，左西右东，结合角度和距离画图即可；(2) 返回时的路程等于 $(72 + 36 + 24)$ 千米, 根据 ‘速度 = 路程 ÷ 时间‘ 解答。(1) 如图:![](images/86e0a3a6d05da6e0c93155f62b885c4c68a4bd59bb840282cf033b938bfd7590.jpg) (2) $(72 + 36 + 24) \\div 4 = (108 + 24) \\div 4 = 132 \\div 4 =33$ (千米/小时) 所以这艘军舰返回时的速度是 33 千米/小时。",
+        "knowledge_points": [
+          "根据方向、角度和距离画线路图",
+          "基础行程问题"
+        ],
+        "source": "24-25六年级下·云南楚雄·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_094",
+        "original_id": "unit2_drawing_94"
+      },
+      {
+        "id": "unit2_drawing_95",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "根据下面的描述,在平面图上标出各场所的位置。![](images/b6ecf9d8d389c408e081256042cdd1a0464edb54c1fe3fe532bb4651fb36eeaa.jpg) (1) 图书馆在学校的北偏西 $30^{\\circ}$ 方向 300 米处。(2) 派出所在学校的东偏北 $45^{\\circ}$ 方向 500 米处。(3) 银行在学校的南偏西 $45^{\\circ}$ 方向 400 米处。(4) 邮局在学校的东偏南 $20^{\\circ}$ 方向 600 米处。",
+        "options": [],
+        "answer": "见详解",
+        "solution": "图上通常是按照‘上北下南、左西右东‘的方向绘制的，图上用一段线段代表 200 米，计算出各个地点需要用几段线段表示与学校的距离。以学校为观测点，确定各个地点对于学校的方向，以及方向之间的角度和计算出的线段长度来作图。(1) $300 \\div 200 = 1.5$ （段）(2) $500 \\div 200 = 2.5$ (段) (3) $400 \\div 200 = 2$ (段) (4) $600 \\div 200 = 3$ (段) 如图所示:![](images/6e5c9f5a0d2ce91df2d5029a5081d05cd51829dc42ec6427722da9c176bb7623.jpg)",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·湖南娄底·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_095",
+        "original_id": "unit2_drawing_95"
+      },
+      {
+        "id": "unit2_drawing_96",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "明明从家里出发,向东偏南 $40^{\\circ}$ 方向上走了 3 千米到超市, 然后向东偏北 $25^{\\circ}$ 方向走了2千米到了广场，最后再向正东方向走1.5千米到达游泳馆（用1厘米表示1千米）。请在图上表示超市、广场和游泳馆的位置。![](images/c26966c965b586ccd6b654f382f8cfe59ba3ccf931c8ebf2488f455a7949f590.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "solution": "由图可知：以明明家为观测点，超市在东偏南 $40^{\\circ}$ 的方向上，又因图上距离 1 厘米表示实际距离 1 千米，而明明家与超市实际距离 3 千米，于是就可以求出明明家与超市的图上距离，据此画出超市的具体位置，同理画出广场和游泳馆的位置。$3 \\div 1 = 3$ （厘米），即明明家到超市的图上距离是 3 厘米；$2 \\div 1 = 2$ （厘米），即超市到广场的图上距离是 $2$ 厘米；1.5÷1=1.5（厘米），即广场到游泳馆的图上距离是1.5厘米。作图如下:![](images/2a22d71fe51a6fbb122aa12e1e90caecf24cc1e922d493356979bc5802d68170.jpg)",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·福建漳州·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_096",
+        "original_id": "unit2_drawing_96"
+      },
+      {
+        "id": "unit2_drawing_97",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "操作。![](images/94d57c39d8cb20744fa13c2194126793f0bb9078f1e75e3a5408a9e7ba56f666.jpg) (1) 体育场在东东的北偏东 $45^{\\circ}$ 方向 1000 米处。(2) 火车站在东东的南偏西 $70^{\\circ}$ 方向 1500 米处。(3) 东东家在广场的西偏北 $50^{\\circ}$ 方向 1000 米处。",
+        "options": [],
+        "answer": "(1) (2) (3) 见详解",
+        "solution": "（1）根据题意可知，1 厘米表示 500 米，计算出体育场到东东家的图上距离，再根据地图上方向的规定‘上北下南，左西右东‘，以东东家为观测点，画出体育场的位置；(2) 计算出火车站到东东家的图上距离, 再以东东家为观测点, 画出火车站的位置; (3) 计算出东东家到广场的图上距离; 再根据方向的相对性: 根据方向的相对性, 它的方向相反, 角度相等, 距离相等; 以东东家为观测点, 确定出广场的位置, 再画出广场的位置，据此解答。(1) $1000 \\div 500 = 2$ （厘米）如下图：(2) $1500 \\div 500 = 3$ （厘米）如下图:(3) $90^{\\circ} - 50^{\\circ} = 40^{\\circ}$ 广场在东东家东偏南 $50^{\\circ}$ （或南偏东 $40^{\\circ}$）方向 1000 米处。$1000 \\div 500 = 2$ （厘米）如下图：![](images/6e43470f897caf57808995f861e0be298b573481cb41b39257a9e1664689f524.jpg)",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·湖南衡阳·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_097",
+        "original_id": "unit2_drawing_97"
+      },
+      {
+        "id": "unit2_drawing_98",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "请根据下面的描述画出平面图。超市在学校西偏北 $25^{\\circ}$ 方向 200 米处，电影院在学校东偏南$40^{\\circ}$ 方向 300 米处。",
+        "options": [],
+        "answer": "见详解",
+        "solution": "根据题意，用 1 厘米表示 100 米，先计算出超市到学校的图上距离；学校到电影院的图上距离；再根据地图上方向的规定‘上北下南，左西右东‘，以学校为观测点，画出超市、电影院的位置，据此解答。1 厘米表示 100 米。$200 \\div 100 = 2 \\text{(厘米)}$ $300 \\div 100 = 3 (\\text{厘米})$ 如图:![](images/524fde472521d553f71f0fbbe87fbfdd494973a5b891bce72b0af56b5b7285fc.jpg)",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "2024六年级上·全国·专题练习",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_098",
+        "original_id": "unit2_drawing_98"
+      },
+      {
+        "id": "unit2_solve_99",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "东东在小区草坪上玩定向运动游戏，他从黄桷树下出发，向南偏西 $45^{\\circ}$ 方向走了3米到达墙角，再向北偏东 $45^{\\circ}$ 方向走了7米到达花坛。这时东东离黄桷树有多远？请先画出线路图再分析作答。(1 厘米代表 1 米)![](images/f8a564c15076e7e31e92c4cea684876b23dd498d6e4858eb6d2d01e0b88fafe3.jpg) 黄桷树",
+        "options": [],
+        "answer": "4 米；作图见详解",
+        "solution": "先确定观测点是黄桷树，根据上北下南，左西右东确定方向和角度，再根据1厘米代表1米，确定距离，据此画图；观察所画的线路图可知，用(7-3)可得这时东东离黄桷树的距离。$3 \\div 1 = 3$ （厘米） $7 \\div 1 = 7$ （厘米）作图如下:![](images/af89b5ea86fe5af6225a83a317f82b0d6cc2ae33e590a14ab8d3a703590d543a.jpg) $7 - 3 = 4$ (米) 答: 这时东东离黄桷树有 4 米远。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置",
+          "根据方向、角度和距离画线路图"
+        ],
+        "source": "24-25 六年级上·重庆南岸·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_099",
+        "original_id": "unit2_solve_99"
+      },
+      {
+        "id": "unit2_solve_100",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "下图是丁菲周末去体育场的路线图。![](images/e1b31a87e5c199511f2d0619e67051d5fe742ff8bd839c465d403e960b09efd0.jpg) (1) 丁菲去体育场，应该先向（）偏（）（）方向走（）米到达超市，再向（）方向走（）米到达体育场。(2) 亮亮家到体育场的路程比丁菲家到体育场的路程少 $\\frac{3}{4}$。亮亮家到体育场的路程是多少米？",
+        "options": [],
+        "answer": "（1）北；东； $30^{\\circ}$ ；800；正东；1600 (2) 600 米",
+        "solution": "（1）图上的方向是上北下南、左西右东，图上距离 $1 \\mathrm{~cm}$ 表示实际距离 400 米，据此确定方向、距离和角度。(2) 先计算出丁菲家到体育场的路程, 再以丁菲家到体育场的路程为单位 ’1‘, 亮亮家到体育场的路程相当于丁菲家到体育场路程的 $(1 - \\frac{3}{4})$, 根据求比一个数少几分之几的数是多少, 用乘法计算, 用丁菲家到体育场路程 $\\times \\left(1 - \\frac{3}{4}\\right)$ 即可求出亮亮家到体育场的路程。(1) $400 \\times 2 = 800$ (米) $400 \\times 4 = 1600$ (米)丁菲去体育场，应该先向北偏东 $30^{\\circ}$ （或东偏北 $60^{\\circ}$）方向走 800 米到达超市，再向正东方向走 1600 米到达体育场。(2) $(800 + 1600) \\times (1 - \\frac{3}{4}) = 2400 \\times \\frac{1}{4} = 600 \\text{(米)}$ 答: 亮亮家到体育场的路程是 600 米。",
+        "knowledge_points": [
+          "分数的四则混合运算",
+          "根据方向、角度和距离描述路线图",
+          "求比一个数多/少几分之几的数是多少"
+        ],
+        "source": "24-25 六年级上·山西长治·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_100",
+        "original_id": "unit2_solve_100"
+      }, {
+        "id": "unit2_choice_101",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "观察下面示意图, 描述正确的是 ( )。![](images/3dd8f4ca4a4f51a749748ea56145c5e03f9e467eacfff151d198bec7a92e8b48.jpg)",
+        "options": [
+          "A. 火车站在电视塔东偏北  $50^{\\circ}$  方向  $4000 \\mathrm{~m}$  处",
+          "B. 电影院在电视塔西偏北  $34^{\\circ}$  方向  $7000 \\mathrm{~m}$  处",
+          "C. 东方宾馆在电视塔南偏西  $71^{\\circ}$  方向  $3000 \\mathrm{~m}$  处",
+          "D. 超市在电视塔西偏南  $63^{\\circ}$  方向  $6000 \\mathrm{~m}$  处"
+        ],
+        "answer": "B",
+        "solution": "根据地图上的方向“上北下南，左西右东”及其他信息（角度、距离）来确定位置即可。\n\nA.  $5 \\times 1000 = 5000$  (m)\n火车站在电视塔东偏北  $50^{\\circ}$  （北偏东  $40^{\\circ}$  ）方向  $5000\\mathrm{m}$  处，选项说法错误；\nB.  $7 \\times 1000 = 7000 \\, (\\mathrm{m})$\n电影院在电视塔西偏北  $34^{\\circ}$  （北偏西  $56^{\\circ}$  ）方向  $7000\\mathrm{m}$  处，选项说法正确；\nC.  $3 \\times 1000 = 3000 \\, (\\mathrm{m})$\n东方宾馆在电视塔南偏西  $19^{\\circ}$  （西偏南  $71^{\\circ}$  ）方向  $3000\\mathrm{m}$  处，选项说法错误；\nD.  $6 \\times 1000 = 6000 \\, (\\mathrm{m})$\n超市在电视塔东偏南  $63^{\\circ}$  （南偏东  $27^{\\circ}$  ）方向  $6000\\mathrm{m}$  处，选项说法错误；\n故答案为：B",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·全国·单元测试",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_101",
+        "original_id": "unit2_choice_101"
+      },
+      {
+        "id": "unit2_choice_102",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "如图, 若喷泉在苗圃南偏西  $60^{\\circ}$  的方向上, 则苗圃在喷泉 ( ) 的方向上。![](images/ffaa0445ba612bdb9837a6b773d817e5b0c9fa4761ae8e1bdc3757e5d89eebe5.jpg)",
+        "options": [
+          "A. 东偏北  $60^{\\circ}$",
+          "B. 东偏北  $30^{\\circ}$",
+          "C. 西偏南  $60^{\\circ}$"
+        ],
+        "answer": "B",
+        "solution": "根据位置的相对性，当一个事物在另一个事物某个方向和距离处时，另一个事物就在这个事物相反方向，相同距离处。据此解答。\n\n喷泉在苗圃南偏西  $60^{\\circ}$  的方向上, 则苗圃在喷泉北偏东  $60^{\\circ}$  的方向或东偏北  $30^{\\circ}$  方向上。\n故答案为：B",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "2022·甘肃陇南·小升初真题",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_102",
+        "original_id": "unit2_choice_102"
+      },
+      {
+        "id": "unit2_choice_103",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "学校在电影院的南偏西  $50^{\\circ}$  方向上，电影院在学校的（）方向上。",
+        "options": [
+          "A. 北偏西  $40^{\\circ}$",
+          "B. 北偏东  $50^{\\circ}$",
+          "C. 北偏东  $40^{\\circ}$",
+          "D. 西偏南  $40^{\\circ}$"
+        ],
+        "answer": "B",
+        "solution": "学校在电影院的南偏西  $50^{\\circ}$  方向上, 是以电影院为观测点; 电影院在学校的方向是以学校为观测点; 根据位置的相对性可知, 观测点不同, 方向相反, 夹角的度数相同, 距离相同。由此可知, 南偏西  $50^{\\circ}$  相对的是北偏东  $50^{\\circ}$ , 东和北之间的夹角是  $90^{\\circ}, 90^{\\circ} - 50^{\\circ} = 40^{\\circ}$ , 所以北偏东  $50^{\\circ}$  方向, 还可以说成东偏北  $40^{\\circ}$  方向。\n\n学校在电影院的南偏西  $50^{\\circ}$  方向上，电影院在学校的北偏东  $50^{\\circ}$  （或东偏北  $40^{\\circ}$ ）方向上。\n故答案为：B",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级下·江苏盐城·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_103",
+        "original_id": "unit2_choice_103"
+      },
+      {
+        "id": "unit2_choice_104",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "如图, 有一个等边三角形花圃, 边长 50 米。小明想要从点 A 走到点 B, 可以 ()。![](images/af3bb93237d9994cf8560e2e82740227fd87058bf9f91e30c51e66d6d8152b12.jpg)",
+        "options": [
+          "A. 向北偏东  $30^{\\circ}$  方向走50米",
+          "B. 向北偏东  $60^{\\circ}$  方向走 50 米",
+          "C. 向南偏西  $30^{\\circ}$  方向走 50 米",
+          "D. 向南偏西  $60^{\\circ}$  方向走 50 米"
+        ],
+        "answer": "A",
+        "solution": "将方向和距离结合起来描述位置时，要注意三个要素：一是观测点，二是方向，三是距离。等边三角形的三个内角都是  $60^{\\circ}$ ，东和北之间的夹角是  $90^{\\circ}$ ，东偏北也可以说成北偏东，角度  $= 90^{\\circ}$  一东偏北的角度。\n\n$90^{\\circ} - 60^{\\circ} = 30^{\\circ}$ 小明想要从点 A 走到点 B, 可以向东偏北  $60^{\\circ}$  或北偏东  $30^{\\circ}$  方向走 50 米。\n故答案为：A",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级下·江苏盐城·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_104",
+        "original_id": "unit2_choice_104"
+      },
+      {
+        "id": "unit2_choice_105",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "下列说法错误的是（）。",
+        "options": [
+          "A. 在一个盒子里, 装有 9 个红色球和一个白色球, 任意摸一个球, 一定能摸出红球",
+          "B. 5个点最多可以连10条线段",
+          "C. 同一钟面上, 当时针旋转了 30 度, 分针就旋转了 360 度",
+          "D. 若 A 点在 B 点的北偏西 30 度方向, 则 B 点在 A 点的南偏东 30 度方向"
+        ],
+        "answer": "A",
+        "solution": "A. 盒子里只要有的球，任意摸一个球，都有可能摸到，据此分析；B. 线段有 2 个端点, 任意 1 个点都可与其余 (5-1) 个点连成线段, 共  $5 \\times (5 - 1)$  条线段, 这样重复计算了一遍, 再除以 2 即可;C. 钟面 1 个大格是 30 度, 1 圈是 360 度, 时针转 1 个大格分针转 1 圈, 据此分析;D. 根据方向的相对性, 北偏西对南偏东, 角度不变, 据此分析。\n\nA. 在一个盒子里, 装有 9 个红色球和一个白色球, 任意摸一个球, 可能摸出红球, 也可能摸出白球, 摸出红球的可能性比较大, 选项说法错误;B.  $5 \\times (5 - 1) \\div 2$  $= 5\\times 4\\div 2$  $= 10$  （条）5个点最多可以连10条线段，选项说法正确；C. 同一钟面上, 当时针旋转了 30 度是 1 小时, 分针就旋转了 360 度也是 1 小时, 选项说法正确;  D. 若 A 点在 B 点的北偏西 30 度方向, 则 B 点在 A 点的南偏东 30 度方向, 选项说法正确。说法错误的是在一个盒子里, 装有 9 个红色球和一个白色球, 任意摸一个球, 一定能摸出红球。故答案为：A",
+        "knowledge_points": [
+          "旋转三要素及旋转图形",
+          "线段的初步认识",
+          "事件的确定性与不确定性",
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级下·四川凉山·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_105",
+        "original_id": "unit2_choice_105"
+      },
+      {
+        "id": "unit2_choice_106",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "邮局在书店北偏东  $60^{\\circ}$  方向  $400 \\mathrm{~m}$  处, 书店在邮局 ( )。",
+        "options": [
+          "A. 北偏东  $60^{\\circ}$  方向  $400 \\mathrm{~m}$  处",
+          "B. 南偏西  $60^{\\circ}$  方向  $400 \\mathrm{~m}$  处",
+          "C. 东偏北  $60^{\\circ}$  方向  $400 \\mathrm{~m}$  处",
+          "D. 西偏南  $60^{\\circ}$  方向  $400 \\mathrm{~m}$  处"
+        ],
+        "answer": "B",
+        "solution": "已知邮局在书店北偏东  $60^{\\circ}$  方向  $400 \\mathrm{~m}$  处, 是以书店为观测点; 书店在邮局的方向是以邮局为观测点; 观测点不同, 方向相反, 夹角的度数相同, 距离相同; 据此解答。\n\n邮局在书店北偏东  $60^{\\circ}$  方向  $400\\mathrm{m}$  处，书店在邮局南偏西  $60^{\\circ}$  方向  $400\\mathrm{m}$  处。\n故答案为：B",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级下·河南周口·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_106",
+        "original_id": "unit2_choice_106"
+      },
+      {
+        "id": "unit2_choice_107",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "芳芳从家出发, 先向北偏东  $45^{\\circ}$  方向走 100 米, 再向东偏南  $45^{\\circ}$  方向走 100 米, 家在芳芳现在位置的 ( ) 方向。",
+        "options": [
+          "A. 正西",
+          "B. 正东",
+          "C. 正南"
+        ],
+        "answer": "A",
+        "solution": "根据平面图上方向的规定：上北下南，左西右东；确定芳芳位置的方向，据此画出芳芳的线路图，然后再判断芳芳家的位置即可。\n\n如图：![](images/5050b3fdfb84c502a4b80d99253e2a5cbdbf02aae48814e5d8a4a54a9c581e56.jpg)\n根据图可知，芳芳家的位置在正西方向。芳芳从家出发，先向北偏东  $45^{\\circ}$  方向走 100 米，再向东偏南  $45^{\\circ}$  方向走 100 米，家在芳芳现在位置的正西方向。故答案为：A",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离描述路线图",
+          "根据方向",
+          "角度和距离确定物体的位置【分析】根据平面图上方向的规定：上北下南",
+          "左西右东；确定芳芳位置的方向",
+          "据此画出芳芳的线路图",
+          "然后再判断芳芳家的位置即可"
+        ],
+        "source": "23-24 六年级下·重庆巫山·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_107",
+        "original_id": "unit2_choice_107"
+      },
+      {
+        "id": "unit2_choice_108",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "文具店、书店和玩具店依次坐落在一条东西走向的大街上, 文具店在书店西边  $20 \\mathrm{~m}$  处, 玩具店位于书店东边  $100 \\mathrm{~m}$  处, 小明从书店沿街向东走了  $40 \\mathrm{~m}$ , 接着又向西走了  $60 \\mathrm{~m}$ , 此时小明的位置在 ( )。",
+        "options": [
+          "A. 文具店",
+          "B. 玩具店",
+          "C. 文具店西  $40 \\mathrm{~m}$  处",
+          "D. 玩具店西  $60 \\mathrm{~m}$  处"
+        ],
+        "answer": "A",
+        "solution": "小明从书店先向东走  $40 \\mathrm{~m}$ , 此时没有到达玩具店, 然后又向西走  $60 \\mathrm{~m}$ , 这  $60 \\mathrm{~m}$  包括  $40 \\mathrm{~m}$  和  $20 \\mathrm{~m}$  。最初的向西走  $40 \\mathrm{~m}$ , 回到书店, 再向西走  $20 \\mathrm{~m}$  到达文具店。\n\n小明从书店沿街向东走了  $40 \\mathrm{~m}$ , 接着又向西走了  $60 \\mathrm{~m}$ , 此时小明的位置在文具店。故答案为: A",
+        "knowledge_points": [
+          "根据方向和距离描述简单的路线"
+        ],
+        "source": "2014 六年级·全国·课后作业",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_108",
+        "original_id": "unit2_choice_108"
+      },
+      {
+        "id": "unit2_choice_109",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "小英家位于学校的东偏北  $30^{\\circ}$ , 那么学校位于小英家的 ( )。",
+        "options": [
+          "A. 西偏南  $60^{\\circ}$",
+          "B. 北偏东  $30^{\\circ}$",
+          "C. 西偏南  $30^{\\circ}$",
+          "D. 东偏北  $30^{\\circ}$"
+        ],
+        "answer": "C",
+        "solution": "本题考查位置的相对性, 即当一个地点 A 位于地点 B 的某个方向某个角度时, 那么地点 B 就位于地点 A 的相对的方向上、角度不变。\n\n根据方向的相对性原理可知：小英家在学校的东偏北  $30^{\\circ}$  方向上, 那么学校就应该位于小英家的相对的方向上、角度不变, 即西偏南  $30^{\\circ}$  。\n故答案为：C",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级下·湖南株洲·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_109",
+        "original_id": "unit2_choice_109"
+      },
+      {
+        "id": "unit2_choice_110",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "北偏西  $30^{\\circ}$ , 还可以说成 ( )。",
+        "options": [
+          "A. 南偏西  $30^{\\circ}$",
+          "B. 西偏北  $30^{\\circ}$",
+          "C. 西偏北  $60^{\\circ}$",
+          "D. 西偏南  $60^{\\circ}$"
+        ],
+        "answer": "C",
+        "solution": "因为西和北的夹角是  $90^{\\circ}, 90^{\\circ} - 30^{\\circ} = 60^{\\circ}$ , 所以北偏西  $30^{\\circ}$  方向, 还可以说成西偏北  $60^{\\circ}$  方向。\n\n北偏西  $30^{\\circ}$  ，还可以说成西偏北  $60^{\\circ}$  。\n故答案为：C",
+        "knowledge_points": [
+          "东",
+          "南",
+          "西",
+          "北方向",
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "22-23 三年级下·湖北黄冈·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_110",
+        "original_id": "unit2_choice_110"
+      },
+      {
+        "id": "unit2_choice_111",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "如下图所示, 医院在东环公园的 ( )。![](images/80fffdc48595d863f0e2a8d55bee17c80f4c62aa5065a8b3d9ce9b6e399de809.jpg)",
+        "options": [
+          "A. 东偏南  $77^{\\circ}$  的方向上",
+          "B. 西偏北  $77^{\\circ}$  的方向上",
+          "C. 北偏东  $77^{\\circ}$  的方向上",
+          "D. 南偏西  $77^{\\circ}$  的方向上"
+        ],
+        "answer": "B",
+        "solution": "以东环公园为观测点，以图上的“上北下南，左西右东”为准，结合方向和角度，得出医院和东环公园的位置关系。\n\n如上图，医院在东环公园的西偏北  $77^{\\circ}$  （或北偏西  $13^{\\circ}$  ）的方向上。\n故答案为：B",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "22-23 六年级下·河北保定·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_111",
+        "original_id": "unit2_choice_111"
+      },
+      {
+        "id": "unit2_choice_112",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "如图, 一个平行四边形,  $\\angle D = 65^{\\circ}$ . 那么点  $D$  在点 A 的 ( )。![](images/b4fe806d86002e1b22186a69749d33b3758201ce2722d40aa74f1f64f2e2905a.jpg)",
+        "options": [
+          "A. 南偏西  $25^{\\circ}$  方向",
+          "B. 西偏南  $25^{\\circ}$  方向",
+          "C. 北偏东  $25^{\\circ}$  方向",
+          "D. 东偏北  $25^{\\circ}$  方向"
+        ],
+        "answer": "A",
+        "solution": "上北下南，左西右东。所以，点 A 在点 D 的东偏北  $65^{\\circ}$  方向上。两点的相对位置，方向相反，角度和距离不变，那么点 D 就在点 A 的西偏南  $65^{\\circ}$  方向上。西偏南  $65^{\\circ}$ ，也就是南偏西  $25^{\\circ}$ 。\n\n$90^{\\circ} - 65^{\\circ} = 25^{\\circ}$ 点 A 在点 D 的东偏北  $65^{\\circ}$  方向上, 那么点 D 就在点 A 的西偏南  $65^{\\circ}$  (南偏西  $25^{\\circ}$ ) 方向上。故答案为: A",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级下·云南昆明·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_112",
+        "original_id": "unit2_choice_112"
+      },
+      {
+        "id": "unit2_choice_113",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "如果明明看丽丽的方向是南偏西  $35^{\\circ}$ , 那么丽丽看明明的方向是 ( )。",
+        "options": [
+          "A. 北偏西  $55^{\\circ}$",
+          "B. 北偏西  $35^{\\circ}$",
+          "C. 北偏东  $35^{\\circ}$",
+          "D. 北偏东  $55^{\\circ}$"
+        ],
+        "answer": "C",
+        "solution": "根据方向的相对性，它们的方向相反，角度相等，距离相等；据此解答。\n\n$90^{\\circ} - 35^{\\circ} = 55^{\\circ}$如果明明看丽丽的方向是南偏西  $35^{\\circ}$ , 那么丽丽看明明的方向是北偏东  $35^{\\circ}$  (或东偏北  $55^{\\circ}$  )。故答案为：C",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 五年级下·河南郑州·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_113",
+        "original_id": "unit2_choice_113"
+      },
+      {
+        "id": "unit2_choice_114",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "下列描述中指的是同一方向的是（）。",
+        "options": [
+          "A. 南偏西 20 度与北偏东 20 度",
+          "B. 东偏北20度与北偏东70度",
+          "C. 西偏南 20 度与南偏东 70 度",
+          "D. 东偏北 35 度与北偏东 35 度"
+        ],
+        "answer": "B",
+        "solution": "根据东南西北四个基本方位中，相邻两个方位之间的夹角是  $90^{\\circ}$ ，一个方向可以有两种说法，如东偏南  $50^{\\circ}$  也可以看作南偏东  $40^{\\circ}$ ，注意角度即可。\n\nA．南偏西20度与西偏南70度是同一个方向，选项说法错误；B. 东偏北 20 度与北偏东 70 度是同一个方向, 选项说法正确;  C. 西偏南 20 度与南偏西 70 度是同一个方向, 选项说法错误;  D. 东偏北 35 度与北偏东 55 度是同一个方向, 选项说法错误。故答案为：B",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "22-23六年级上·河南驻马店·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_114",
+        "original_id": "unit2_choice_114"
+      },
+      {
+        "id": "unit2_choice_115",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "动物园在植物园的北偏西  $35^{\\circ}$ , 那么植物园在动物园的 ( )。",
+        "options": [
+          "A. 北偏东  $35^{\\circ}$",
+          "B. 南偏西  $35^{\\circ}$",
+          "C. 南偏东  $35^{\\circ}$"
+        ],
+        "answer": "C",
+        "solution": "以植物园为观测点, 动物园在北偏西  $35^{\\circ}$  方向上, 如果以动物园为观测点, 植物园在东偏南  $55^{\\circ}$  (或南偏东  $35^{\\circ}$ ) 方向上, 据此解答。\n\n动物园在植物园的北偏西  $35^{\\circ}$ , 那么植物园在动物园的东偏南  $55^{\\circ}$  (或南偏东  $35^{\\circ}$ ) 方向上。\n故答案为：C",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "22-23 五年级上·广西贵港·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_115",
+        "original_id": "unit2_choice_115"
+      },
+      {
+        "id": "unit2_choice_116",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "北京距离昆明北偏东  $42^{\\circ}$  方向 2000 千米, 那么昆明距离北京 ( )。",
+        "options": [
+          "A. 北偏东  $42^{\\circ}$  方向 2000 千米",
+          "B. 东偏北  $42^{\\circ}$  方向 2000 千米",
+          "C. 西偏南  $42^{\\circ}$  方向 2000 千米",
+          "D. 南偏西  $42^{\\circ}$  方向 2000 千米"
+        ],
+        "answer": "D",
+        "solution": "一个事物在另一个事物的某个方向一定度数的位置，那么另一个事物在这个事物相对的方向相同度数的位置，据此解答即可。\n\n由分析可得: 北京距离昆明北偏东  $42^{\\circ}$  方向 2000 千米, 那么昆明距离北京南偏西  $42^{\\circ}$  方向 2000 千米。\n故答案为：D",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "2024 六年级下·四川·专题练习",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_116",
+        "original_id": "unit2_choice_116"
+      },
+      {
+        "id": "unit2_choice_117",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "书店在学校西偏北  $25^{\\circ}$  的方向上, 学校在书店 ( ) 方向上。",
+        "options": [
+          "A. 西偏北  $25^{\\circ}$",
+          "B. 北偏西  $25^{\\circ}$",
+          "C. 东偏南  $25^{\\circ}$",
+          "D. 南偏东  $25^{\\circ}$"
+        ],
+        "answer": "C",
+        "solution": "用角度表示方向利用平面图上方向规定：上北下南左西右东，当位置相对时，将原方向的相反的方向列出来，角度不变，就是与原方向相对的位置。\n\n书店在学校西偏北  $25^{\\circ}$  的方向上，那么学校在书店东偏南  $25^{\\circ}$  方向上。\n故答案为：C",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "2024 六年级下·山东·专题练习",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_117",
+        "original_id": "unit2_choice_117"
+      },
+      {
+        "id": "unit2_choice_118",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "我国军舰正在亚丁湾向正北方向航行，突然发现两点钟方向有一艘海盗船。两点钟方向是指海盗船在我国军舰（）。",
+        "options": [
+          "A. 北偏东  $20^{\\circ}$  方向",
+          "B. 北偏东  $60^{\\circ}$  方向",
+          "C. 南偏东  $60^{\\circ}$  方向"
+        ],
+        "answer": "B",
+        "solution": "时钟面上有 12 个大格, 表针转一周是  $360^{\\circ}$ , 那么每经过一个大格是  $360^{\\circ} \\div 12 = 30^{\\circ}$  。根据题意, 以我国军舰为观测点, 根据 “上北下南, 左西右东” 可知, 钟面上的 12 时作为正北方向, 钟面上的 6 时为正南方向, 3 时为正东方向, 9 时为正西方向; 那么钟面上的数字 “2”与 “12” 的夹角是  $30^{\\circ} \\times 2 = 60^{\\circ}$ , 钟面上的数字 “2”与 “3”的夹角是  $30^{\\circ}$ , 据此解答。\n\n我国军舰正在亚丁湾向正北方向航行，突然发现两点钟方向有一艘海盗船。两点钟方向是指海盗船在我国军舰北偏东  $60^{\\circ}$  方向或东偏北  $30^{\\circ}$  方向。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "2024 六年级下·江苏·专题练习",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_118",
+        "original_id": "unit2_choice_118"
+      },
+      {
+        "id": "unit2_choice_119",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "对于下图，下列说法正确的是（）。![](images/bf2a2cc8b7d925498b9ddf0f42cb8856ef3977f4283c855abc70ae12f87fee9a.jpg)",
+        "options": [
+          "A. A 在 B 的东偏南  $30^{\\circ}$  方向上。",
+          "B. B 在 A 的西偏北  $30^{\\circ}$  方向上。",
+          "C. B 在 A 的南偏东  $60^{\\circ}$  方向上。"
+        ],
+        "answer": "C",
+        "solution": "将方向和距离结合起来描述位置时，要注意三个要素：一是观测点，二是方向，三是距离。南和东之间的夹角是  $90^{\\circ}$ ，南偏东也可以说成东偏南，角度  $= 90^{\\circ}$  一南偏东的角度。\n\nA．A在B的西偏北  $30^{\\circ}$  方向上，选项说法错误。  B.  $90^{\\circ} - 60^{\\circ} = 30^{\\circ}$ , B 在 A 的东偏南  $30^{\\circ}$  方向上, 选项说法错误。  C. B 在 A 的南偏东  $60^{\\circ}$  方向上, 说法正确。故答案为：C",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "22-23六年级上·河北张家口·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_119",
+        "original_id": "unit2_choice_119"
+      },
+      {
+        "id": "unit2_choice_120",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "小明从家出发向北偏西  $30^{\\circ}$  方向步行 600 米到书店, 它从书店回家时的路线是 ( )。",
+        "options": [
+          "A. 北偏东  $30^{\\circ} 600$  米",
+          "B. 南偏东  $30^{\\circ} 600$  米",
+          "C. 南偏东  $60^{\\circ} 600$  米",
+          "D. 南偏西  $30^{\\circ} 600$  米"
+        ],
+        "answer": "B",
+        "solution": "已知方向和距离两个条件才能确定物体的位置，根据位置的相对性，可知两处位置观测点不同，它们的方向相反，角度相等，据此解答。\n\n小明从家出发向北偏西  $30^{\\circ}$  方向步行 600 米到书店, 他从书店回家时的路线是南偏\n东  $30^{\\circ}600$  米。\n故答案为：B",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "22-23 三年级下·山东济南·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_120",
+        "original_id": "unit2_choice_120"
+      },
+      {
+        "id": "unit2_choice_121",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "奇思在笑笑的西偏南  $60^{\\circ}$  方向 150 米处, 那么笑笑在奇思的 ( ) 方向 150 米处。![](images/511e52e05406ef519de340943141aa9be8e2384adac7637239f6619526ee603e.jpg)",
+        "options": [
+          "A. 南偏西  $30^{\\circ}$",
+          "B. 东偏北  $30^{\\circ}$",
+          "C. 东偏北  $60^{\\circ}$"
+        ],
+        "answer": "C",
+        "solution": "物体的位置是相对的，对于两个物体来说，分别以自身为观测点，则它们的方向相反，距离和角度是不变的。据此可知，笑笑在奇思的东偏北  $60^{\\circ}$  方向 150 米处。还可以这样表示，如下图所示，笑笑在奇思的北偏东  $30^{\\circ}$  方向 150 米处。![](images/59f8870132aa3cc8675ef9b67dc91c8f649b6f97ae35c68faa936a0052220d52.jpg)\n\n通过分析可得：奇思在笑笑的西偏南  $60^{\\circ}$  方向 150 米处，则笑笑在奇思的东偏北  $60^{\\circ}$  方向 150 米处，或笑笑在奇思的北偏东  $30^{\\circ}$  方向 150 米处。故答案为：C",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "22-23 六年级下·广东茂名·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_121",
+        "original_id": "unit2_choice_121"
+      },
+      {
+        "id": "unit2_choice_122",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "公园在英英家北偏西  $40^{\\circ}$  方向 500 米处。英英从公园回家，应向（）方向走 500 米。",
+        "options": [
+          "A. 北偏东  $40^{\\circ}$",
+          "B. 南偏西  $40^{\\circ}$",
+          "C. 南偏东  $40^{\\circ}$"
+        ],
+        "answer": "C",
+        "solution": "根据位置的相对性可知, 英英从公园回家与从公园到英英家的方向相反, 角度相等, 距离相等, 据此解答。\n\n公园在英英家北偏西  $40^{\\circ}$  方向 500 米处。英英从公园回家, 应向南偏东  $40^{\\circ}$  方向走 500 米。\n故答案为：C",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "22-23 五年级下·山东济宁·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_122",
+        "original_id": "unit2_choice_122"
+      },
+      {
+        "id": "unit2_choice_123",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "医院位于公园的西偏南  $30^{\\circ}$  方向 800 米处。从医院去公园, 要往 ( ) 方向走 800 米。",
+        "options": [
+          "A. 东偏北  $30^{\\circ}$",
+          "B. 西偏南  $30^{\\circ}$",
+          "C. 西偏南  $60^{\\circ}$",
+          "D. 东偏北  $60^{\\circ}$"
+        ],
+        "answer": "A",
+        "solution": "以公园为观测点，医院位于公园的西偏南  $30^{\\circ}$  方向，由位置的相对性可知，以医院为观测点，公园在医院东偏北  $30^{\\circ}$  或北偏东  $60^{\\circ}$  方向，两地之间的距离不变，据此解答。\n\n![](images/117a385f22b249f212e56d90a7539e00120df3ed6f374253084f8c0c850f9a6b.jpg)\n由图可知, 医院位于公园的西偏南  $30^{\\circ}$  方向 800 米处。从医院去公园, 要往东偏北  $30^{\\circ}$  或北偏东  $60^{\\circ}$  方向走 800 米。故答案为：A",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·广东东莞·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_123",
+        "original_id": "unit2_choice_123"
+      },
+      {
+        "id": "unit2_choice_124",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "王青家在学校的东偏北30度方向上，学校在王青家（）。",
+        "options": [
+          "A. 北偏东 30 度",
+          "B. 西偏南 30 度",
+          "C. 北偏东 60 度",
+          "D. 南偏西 30 度"
+        ],
+        "answer": "B",
+        "solution": "王青家在学校的东偏北 30 度方向上, 是以学校为观测点; 学校在王青家的方向是以王青家为观测点; 根据位置的相对性可知, 观测点不同, 方向相反, 夹角的度数相同,所以东偏北 30 度相对的是西偏南 30 度, 由此解答。\n\n王青家在学校的东偏北30度方向上，学校在王青家西偏南30度。\n故答案为：B",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·福建厦门·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_124",
+        "original_id": "unit2_choice_124"
+      },
+      {
+        "id": "unit2_choice_125",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "张爷爷做晨运, 他先面向正北方向站立, 然后全身向左转了  $30^{\\circ}$ , 这时他面向 ( ) 的方向。",
+        "options": [
+          "A. 北偏西  $30^{\\circ}$",
+          "B. 北偏东  $30^{\\circ}$",
+          "C. 西偏北  $30^{\\circ}$",
+          "D. 东偏北  $30^{\\circ}$"
+        ],
+        "answer": "A",
+        "solution": "根据方向的规定“上北下南，左西右东”，张爷爷先面向正北方向站立，然后全身向左转了  $30^{\\circ}$ ，是从北往西旋转了  $30^{\\circ}$ ，结合角度，得出这时他面对的方向。\n\n张爷爷做晨运, 他先面向正北方向站立, 然后全身向左转了  $30^{\\circ}$ , 这时他面向北偏西  $30^{\\circ}$  (或西偏北  $60^{\\circ}$ ) 的方向。故答案为：A",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·广东广州·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_125",
+        "original_id": "unit2_choice_125"
+      },
+      {
+        "id": "unit2_choice_126",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "与“南偏西  $30^{\\circ}$  表示同一方向的是( )。",
+        "options": [
+          "A. 南偏西  $60^{\\circ}$",
+          "B. 北偏西  $60^{\\circ}$",
+          "C. 西偏南  $60^{\\circ}$"
+        ],
+        "answer": "C",
+        "solution": "南和西之间的夹角是  $90^{\\circ}$ , 南偏西也可以说成西偏南, 角度  $= 90^{\\circ} -$  南偏西的角度,据此分析。\n\n$90^{\\circ} - 30^{\\circ} = 60^{\\circ}$ 与“南偏西  $30^{\\circ}$  表示同一方向的是西偏南  $60^{\\circ}$ 。故答案为：C",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·广西玉林·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_126",
+        "original_id": "unit2_choice_126"
+      },
+      {
+        "id": "unit2_choice_127",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "北京在哈尔滨的西偏南  $10^{\\circ}$  方向上, 哈尔滨在北京的 ( ) 方向上。",
+        "options": [
+          "A. 西偏南  $80^{\\circ}$",
+          "B. 东偏北  $10^{\\circ}$",
+          "C. 南偏西  $80^{\\circ}$",
+          "D. 北偏东  $10^{\\circ}$"
+        ],
+        "answer": "B",
+        "solution": "根据位置的相对性可知, 描述两个物体之间的相对位置时, 可以方向相反, 角度相等; 也可以用不同的方向和角度进行描述。如下图所示, 以北京为观测点, 哈尔滨在北京的东偏北  $10^{\\circ}$  方向上;  $90^{\\circ} - 10^{\\circ} = 80^{\\circ}$ , 也可以说哈尔滨在北京的北偏东  $80^{\\circ}$  方向上。\n![](images/1481170a0e314a7e42885c62b68a21f4429a0c26561c78086b703561e5fc0e7f.jpg)\n\n通过分析可得: 北京在哈尔滨的西偏南  $10^{\\circ}$  方向上, 哈尔滨在北京的北偏东  $80^{\\circ}$  或东偏北  $10^{\\circ}$  方向上。只有选项 B 符合题意。故答案为：B",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·福建莆田·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_127",
+        "original_id": "unit2_choice_127"
+      },
+      {
+        "id": "unit2_choice_128",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "如图，图中三角形ABC是一个边长为  $5\\mathrm{cm}$  的等边三角形，那么点A在点B的（）。 ![](images/283df1c78750b57479343b283b99ff210659d6a5874ff5350310440971183ac1.jpg)",
+        "options": [
+          "A. 西偏北  $60^{\\circ}$  方向  $5 \\mathrm{~cm}$  处",
+          "B. 西偏北  $30^{\\circ}$  方向  $5 \\mathrm{~cm}$  处",
+          "C. 东偏北  $60^{\\circ}$  方向  $5 \\mathrm{~cm}$  处"
+        ],
+        "answer": "C",
+        "solution": "根据等边三角形的特征可知，等边三角形的3条边相等，3个内角都是  $60^{\\circ}$  。以点 B 为观测点, 以图上的 “上北下南, 左西右东” 为准, 根据图上的方向、角度和距离,得出点 A 与点 B 的位置关系。\n\n![](images/faa018f5f44aac2d6243032a79e708034b612a905fdbaa4feb7b9c2c0f2b7f63.jpg)\n图中三角形 ABC 是一个边长为  $5 \\mathrm{~cm}$  的等边三角形, 那么点 A 在点 B 的东偏北  $60^{\\circ}$  (或北偏东  $30^{\\circ}$ ) 方向  $5 \\mathrm{~cm}$  处。故答案为：C",
+        "knowledge_points": [
+          "等腰三角形和等边三角形的认识及特征",
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·河南周口·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_128",
+        "original_id": "unit2_choice_128"
+      },
+      {
+        "id": "unit2_choice_129",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "下面描述中指的是同一方向的是（）。",
+        "options": [
+          "A. 东偏南  $20^{\\circ}$  与南偏西  $70^{\\circ}$",
+          "B. 东偏南  $20^{\\circ}$  与南偏东  $70^{\\circ}$",
+          "C. 东偏南  $20^{\\circ}$  与东偏北  $70^{\\circ}$",
+          "D. 东偏南  $20^{\\circ}$  与西偏南  $70^{\\circ}$"
+        ],
+        "answer": "B",
+        "solution": "东和南之间的夹角是  $90^{\\circ}$ , 东偏南也可以说成南偏东, 角度  $= 90^{\\circ}$  一东偏南的角度,据此分析。\n\n$90^{\\circ} - 20^{\\circ} = 70^{\\circ}$ 东偏南  $20^{\\circ}$  也可以说成南偏东  $70^{\\circ}$ , 描述中指的是同一方向的是东偏南  $20^{\\circ}$  与南偏东  $70^{\\circ}$  。故答案为：B\n# 二、填空题",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·湖北黄冈·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_129",
+        "original_id": "unit2_choice_129"
+      },
+      {
+        "id": "unit2_choice_130",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "学校在体育场的西偏南  $45^{\\circ}$  方向，体育场在学校的（ ）方向。",
+        "options": [],
+        "answer": "东偏北  $45^{\\circ}$",
+        "solution": "根据方向的相对性，西偏南对东偏北，角度不变，进行填空。\n\n学校在体育场的西偏南  $45^{\\circ}$  方向，体育场在学校的东偏北  $45^{\\circ}$  或北偏东  $45^{\\circ}$  方向。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级下·河南三门峡·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_130",
+        "original_id": "unit2_choice_130"
+      },
+      {
+        "id": "unit2_choice_131",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "商店在学校南偏东  $35^{\\circ}$  方向 800 米处, 则学校在商店( )方向( )米处。",
+        "options": [],
+        "answer": "北偏西  $35^{\\circ}$  800",
+        "solution": "根据位置的相对性可知，描述两个物体之间的相对位置时，方向相反，角度相等，据此解答。\n\n商店在学校南偏东  $35^{\\circ}$  方向 800 米处, 则学校在商店北偏西  $35^{\\circ}$  方向 800 米处。所以学校在商店北偏    西  $35^{\\circ}$  方向 800 米处。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "2024·河南南阳·小升初真题",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_131",
+        "original_id": "unit2_choice_131"
+      },
+      {
+        "id": "unit2_choice_132",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如图是小王家到学校的路线图, 根据路线图填空。 ![](images/2f3011f23c85653c1b95fea4ed7a9e4ad9869bf6fa7c0b88a0ede67d9cfa36b3.jpg) (1)小王从家出发向东走到达星星超市后，需向( )偏( ) ( )°行( )米到达百果园。 (2)小王每天早晨7：40从家出发，要想7：55前到校，他每分钟最少走( )米。",
+        "options": [],
+        "answer": "(1) 东   北    50    300 (2)60",
+        "solution": "（1）根据图示，我们看路线图，小王从家往东走到星星超市后，再往东北方向走，图中已给出角度为  $50^{\\circ}$ ，即东偏北  $50^{\\circ}$ ，通过数格子可以知道走了 2 格，因为图上 1 格代表 150 米，所以走的距离是  $150 \\times 2 = 300$  米，据此解答。(2) 先算出从家到学校一共走了 6 格, 那么路程就是  $6 \\times 150 = 900$  米, 然后计算出所用时间是 7 时 55 分减 7 时 40 分等于 15 分钟, 最后根据速度  $=$  路程 ÷ 时间, 得出  $900 \\div 15 = 60$  米/分钟。\n\n(1) 小王从家出发向东走到达星星超市后, 是向东偏北的方向走到百果园, 通过图中比例尺可知图上 1 格代表 150 米, 从星星超市到百果园是 2 格, 所以距离是  $150 \\times 2 =$300米。小王从家出发向东走到达星星超市后，需向东偏北  $50^{\\circ}$  （北偏东  $40^{\\circ}$  ）行300米到达百果园。(2) 小王从家到学校一共走了 6 格, 每格 150 米, 所以路程是  $6 \\times 150 = 900$  米, 7 时 55 分减去 7 时 40 分等于 15 分钟, 根据速度  $=$  路程 ÷ 时间, 可得  $900 \\div 15 = 60$  (米/分钟)。小王每天早晨7：40从家出发，要想7：55前到校，他每分钟最少走60米。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离描述路线图",
+          "东",
+          "南",
+          "西",
+          "北方向",
+          "基础行程问题",
+          "时",
+          "分",
+          "秒时间的推算"
+        ],
+        "source": "23-24 六年级下·吉林·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_132",
+        "original_id": "unit2_choice_132"
+      },
+      {
+        "id": "unit2_choice_133",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "早上, 冬冬在小区门口坐上 20 路公交车, 公交车先向正北行驶  $2 \\mathrm{~km}$  到十字路口后右转, 行  $3 \\mathrm{~km}$  后到下一个十字路口, 右转后又行  $2 \\mathrm{~km}$  后到达学校门口, 学校在冬冬家( )方约( )km 处。",
+        "options": [],
+        "answer": "东   3",
+        "solution": "根据上北下南、左西右东，结合方向画图分析即可，如下图，可知学校在冬冬家东方约  $3 \\mathrm{~km}$  处。![](images/64e2a1144a032f353b1bddf4dd2febe37b54f7673dcd3b4f4f6f2274d35b07d8.jpg)\n\n由分析可知：早上，冬冬在小区门口坐上 20 路公交车，公交车先向正北行驶  $2 \\mathrm{~km}$  到十字路口后右转，行  $3 \\mathrm{~km}$  后到下一个十字路口，右转后又行  $2 \\mathrm{~km}$  后到达学校门口，学校在冬冬家东方约  $3 \\mathrm{~km}$  处。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级下·河南南阳·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_133",
+        "original_id": "unit2_choice_133"
+      }, {
+        "id": "unit2_choice_134",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "已知学校在小红家西偏北 40 度的方向上, 那么小红放学应往( )偏( )（）度的方向走回家。",
+        "options": [],
+        "answer": "东     南     40",
+        "solution": "【分析】根据题意作图如下：\n\n![](images/8dd5007e97a11bb5855172fb07d3c3f0468917243b910bdcfac64adc08475a01.jpg)\n\n学校在小红家西偏北 40 度的方向上, 那么小红家应该在学校的东南方向。由图可知, 在直角三角形中, $\\angle 1 + 40^{\\circ} = 90^{\\circ}$ , 可以先用减法算出  $\\angle 1$  的度数是  $50^{\\circ}$  。  $\\angle 1 + \\angle 2 = 90^{\\circ}$ , 可以再用减法算出  $\\angle 2$  的度数是  $40^{\\circ}$  。那么小红家在学校的东偏南  $40^{\\circ}$  方向上, 或南偏东  $50^{\\circ}$  方向上。\n\n【详解】  $90^{\\circ} - 40^{\\circ} = 50^{\\circ}$ $9 0 ^ {\\circ} - 5 0 ^ {\\circ} = 4 0 ^ {\\circ}$故学校在小红家西偏北40度的方向上，那么小红放学应往东偏南40度的方向走回家。或小红放学应往南偏东50度的方向走回家。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离描述路线图"
+        ],
+        "source": "22-23 四年级下·湖南湘西·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_134",
+        "original_id": "unit2_choice_134"
+      },
+      {
+        "id": "unit2_choice_135",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "下图中, 只用方向和角度描述, 学校在广场的( )。\n![](images/b15125fcde9035f62138f58dc29680683dce8eadc4b0667b7620740bf9ab27e3.jpg)",
+        "options": [],
+        "answer": "北偏东  $30^{\\circ}$  方向/东偏北  $60^{\\circ}$  方向",
+        "solution": "【分析】以广场为观测点，以图上的上北下南，左西右东为准，结合方向和角度，得出学校和广场的位置关系。\n\n【详解】从图中可知，$9 0 ^ {\\circ} - 3 0 ^ {\\circ} = 6 0 ^ {\\circ}$学校在广场的北偏东  $30^{\\circ}$  方向，或学校在广场的东偏北  $60^{\\circ}$  方向。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级下·江西抚州·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_135",
+        "original_id": "unit2_choice_135"
+      },
+      {
+        "id": "unit2_choice_136",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "星期日奇思从家经过学校去书店买书，路线如下：\n\n![](images/2d45f02f359ba889b511083b8f05e9ec1dede4b17de61a3e98f1b0070348878c.jpg)\n\n(1)奇思从家出发先向( )偏( )  $20^{\\circ}$  的方向走  $900 \\mathrm{~m}$  到学校, 再向( )偏( )  $50^{\\circ}$  的方向走  $600 \\mathrm{~m}$  到达书店。  (2)奇思买完书返回学校后，应当向( )偏西( )度的方向走  $900 \\mathrm{~m}$  到家。",
+        "options": [],
+        "answer": "(1) 东  南  东  北 (2) 北  70",
+        "solution": "【分析】（1）根据地图上的方向：上北下南，左西右东，以家为观测点，找出学校的方向角度与距离；再以学校为观测点，找出书店的方向角度与距离；  (2) 根据地图上的方向: 上北下南, 左西右东, 描述出奇思先从书店到学校的方向和距离,再从学校到家的方向和距离即可。\n\n【详解】（1）奇思从家出发先向东偏南  $20^{\\circ}$  的方向走  $900 \\mathrm{~m}$  到学校，再向东偏北  $50^{\\circ}$  的方向走  $600 \\mathrm{~m}$  到达书店。  (2) 奇思买完书返回学校后，应当向北偏西 70 度的方向走  $900 \\mathrm{~m}$  到家。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离描述路线图"
+        ],
+        "source": "23-24五年级下·陕西西安·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_136",
+        "original_id": "unit2_choice_136"
+      },
+      {
+        "id": "unit2_choice_137",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "已知果果家在潜潜家的正东方, 学校在潜潜家的北偏东  $35^{\\circ}$ 方向上, 果果家到学校的距离与潜潜家到学校的距离相等。那么学校在果果家的____方向。",
+        "options": [],
+        "answer": "北偏西  $35^{\\circ}$",
+        "solution": "【分析】根据题目的意思画出示意图, 学校到潜潜家和学校到果果家的距离是相等的, 从图中可以得出学校、潜潜家、果果家的连接在一起是一个等腰三角形。则以果果家为观测点, 可以得出学校在果果家的北偏西的方向, 度数恰好就是  $35^{\\circ}$  。![](images/cd2eddcaf443b5f4c8dccda26f3fbbbc150520fa335301fdcc41b5c18f9580f1.jpg)\n\n【详解】根据图形分析，则学校在果果家的北偏西  $35^{\\circ}$  方向。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "2012 六年级·全国·竞赛",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_137",
+        "original_id": "unit2_choice_137"
+      },
+      {
+        "id": "unit2_choice_138",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如图所示, 从车站出发向东方向走( )米到达书店, 再向( )偏( )  $30^{\\circ}$  方向走 2000 米到达邮局, 最后向( )偏( ) ()的方向走 2000 米就到达学校。\n![](images/33c670dbc87a64345f24437e93253c4eb02a1d00db06fbbc7b02e1777cbb0b7e.jpg)",
+        "options": [],
+        "answer": "3000  东  北  东  南  $35^{\\circ}$",
+        "solution": "【分析】以图上的 “上北下南, 左西右东” 为准, 图例表示图上 1 厘米相当于实际距离 1000 米;确定路线时，注意起始点与目的地，起始点是观测点，根据方向、角度和距离描述路线。\n\n【详解】  $3 \\times 1000 = 3000$  （米）从车站出发向东方向走 3000 米到达书店，再向东偏北  $30^{\\circ}$  方向走 2000 米到达邮局，最后向东偏南  $35^{\\circ}$  （或南偏东  $55^{\\circ}$ ）方向走 2000 米就到达学校。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离描述路线图"
+        ],
+        "source": "23-24 六年级下·湖南邵阳·期中",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_138",
+        "original_id": "unit2_choice_138"
+      },
+      {
+        "id": "unit2_choice_139",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "小东面向南站立, 他向左转  $55^{\\circ}$  后, 所面对的方向是( )偏( )°。",
+        "options": [],
+        "answer": "南 东55",
+        "solution": "【分析】根据“上北下南，左西右东”可得，小东面向南站立，向左转  $55^{\\circ}$ ，朝东转了  $55^{\\circ}$ ，据此解答。\n\n【详解】  $90^{\\circ} - 55^{\\circ} = 35^{\\circ}$ 小东面向南站立，他向左转  $55^{\\circ}$  后，所面对的方向是南偏东  $55^{\\circ}$  （或东偏南  $35^{\\circ}$  ）。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 五年级下·黑龙江大庆·期中",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_139",
+        "original_id": "unit2_choice_139"
+      },
+      {
+        "id": "unit2_choice_140",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "图图面向北偏西  $20^{\\circ}$  方向站立, 在他向左转  $90^{\\circ}$  后,他面向的是( )偏( )（ ）°方向。",
+        "options": [],
+        "answer": "西  南  20",
+        "solution": "【分析】根据“上北下南，左西右东”可得，将北偏西  $20^{\\circ}$  向左转  $90^{\\circ}$  后，变成了西偏南  $20^{\\circ}$  方向。据此填空。\n\n【详解】图图面向北偏西  $20^{\\circ}$  方向站立，在他向左转  $90^{\\circ}$  后，他面向的是西偏南  $20^{\\circ}$  （或南偏西  $70^{\\circ}$ ）方向。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级下·江苏南京·期中",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_140",
+        "original_id": "unit2_choice_140"
+      },
+      {
+        "id": "unit2_choice_141",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "小明从游乐场到动物园, 他向( )偏( )()度步行了( )米, 到达动物园 (图上 1 厘米表示 500 米)。\n![](images/d1c6e87de2c8bcf68de0d9943718b5ce145b85059e3e680853ff595f3c38444a.jpg)",
+        "options": [],
+        "answer": "东 北   45   1000",
+        "solution": "【分析】图中 1 小段表示 500 米, 从游乐场到动物园有 2 小段, 表示是 (2×500) 米; 以游乐场为观测点, 动物园在北偏东 45 度（或东偏北 45 度）方向上, 距离游乐场有 (2×500) 米, 据此解答。\n\n【详解】  $2 \\times 500 = 1000$  （米）小明从游乐场到动物园，他向北偏东45度（或东偏北45度）步行了1000米，到达动物园。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离描述路线图",
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "20-21 六年级上·河南洛阳·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_141",
+        "original_id": "unit2_choice_141"
+      },
+      {
+        "id": "unit2_choice_142",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "以广场为观测点, 动物园在东偏( )  $30^{\\circ}$  方向上,距离是( )米; 小明从广场到学校, 要向( )偏( ) ( )°方向走 500 米。\n![](images/2a7ec2b0b52c68d5c9b229c8518427e59dfc04ef9af97d3c85de927823ac8edd.jpg)",
+        "options": [],
+        "answer": "北  300   南   东   40",
+        "solution": "【分析】将方向和距离结合起来描述位置时，要注意三个要素：一是观测点，二是方向，三是距离。看图可知，图上1段表示100米，图上有几段就是几百米；南和东之间的夹角是  $90^{\\circ}$ ，南偏东也可以说成东偏南，角度  $= 90^{\\circ}$  —南偏东的角度。\n\n【详解】  $90^{\\circ} - 40^{\\circ} = 50^{\\circ}$ 以广场为观测点, 动物园在东偏北  $30^{\\circ}$  方向上, 距离是 300 米; 小明从广场到学校, 要向南偏东  $40^{\\circ}$  或东偏南  $50^{\\circ}$  方向走 500 米。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "21-22 六年级上·河南洛阳·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_142",
+        "original_id": "unit2_choice_142"
+      },
+      {
+        "id": "unit2_choice_143",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "小明家在学校南偏东  $30^{\\circ}$  方向  $600 \\mathrm{~m}$  处, 则学校在小明家( )°方向( )m 处。",
+        "options": [],
+        "answer": "北偏西30 600",
+        "solution": "【分析】小明家在学校南偏东  $30^{\\circ}$  方向  $600 \\mathrm{~m}$  处, 是以学校为观测点; 学校在小明家的方向是以小明家为观测点; 根据位置的相对性可知, 观测点不同, 方向相反, 夹角的度数相同,所以南偏东  $30^{\\circ}$  相对的是北偏西  $30^{\\circ}$ , 由此答。\n\n【详解】小明家在学校南偏东  $30^{\\circ}$  方向  $600 \\mathrm{~m}$  处, 则学校在小明家北偏西  $30^{\\circ}$  (或西偏北  $60^{\\circ}$ ) 方向  $600 \\mathrm{~m}$  处。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "2024 六年级下·山东·专题练习",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_143",
+        "original_id": "unit2_choice_143"
+      },
+      {
+        "id": "unit2_choice_144",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "淘气家在学校的北偏东  $60^{\\circ}$  方向上  $700 \\mathrm{~m}$  处。则学校\n在淘气家( )  $60^{\\circ}$  方向上( )  $\\mathrm{m}$  处。",
+        "options": [],
+        "answer": "南偏西 700",
+        "solution": "【分析】以学校为观测点，淘气家在北偏东  $60^{\\circ}$  方向上  $700 \\mathrm{~m}$  处，根据位置的相对性可知，以淘气家为观测点，学校在南偏西  $60^{\\circ}$  方向上，距离不变，据此解答。\n\n【详解】以学校为观测点, 淘气家在北偏东  $60^{\\circ}$  方向上  $700 \\mathrm{~m}$  处; 以淘气家为观测点, 学校在南偏西  $60^{\\circ}$  方向上, 距离不变是  $700 \\mathrm{~m}$  处。因此淘气家在学校的北偏东  $60^{\\circ}$  方向上  $700 \\mathrm{~m}$  处。则学校在淘气家南偏西  $60^{\\circ}$  方向上  $700 \\mathrm{~m}$  处。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "2024 六年级下·四川·专题练习",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_144",
+        "original_id": "unit2_choice_144"
+      },
+      {
+        "id": "unit2_choice_145",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "确定物体的位置时, 需要知道方向、( )和( )三个条件。",
+        "options": [],
+        "answer": "观测点 距离",
+        "solution": "【详解】将方向和距离结合起来描述位置时，要注意三个要素：一是观测点，二是方向，三是距离。所以，确定物体的位置时，需要知道方向（包括角度）、观测点和距离三个条件。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "22-23 六年级上·湖南永州·期中",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_145",
+        "original_id": "unit2_choice_145"
+      },
+      {
+        "id": "unit2_choice_146",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "描述线路图时, 要先按行走路线确定每一个____,然后以每一个观测点建立____，描述到下一个目标行走的____和____。",
+        "options": [],
+        "answer": "观测点  方向标  方向  路程",
+        "solution": "【分析】行走过程中的观测点在不断变化，描述路线图时，要先按行驶的路线确定每一个观测点，然后以每一个观测点为中心建立方向标，根据方向和距离描述到下一个目标的过程。\n\n【详解】根据分析可知，描述线路图时，要先按行走路线确定每一个观测点，然后以每一个观测点建立方向标，描述到下一个目标行走的方向和路程。",
+        "knowledge_points": [
+          "根据方向和距离描述简单的路线"
+        ],
+        "source": "22-23 六年级上·山西忻州·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_146",
+        "original_id": "unit2_choice_146"
+      },
+      {
+        "id": "unit2_choice_147",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如图, 三角形  $A B C$  是一个等腰直角三角形, 点  $B$  在点  $C$  的( )偏( )()度方向上。![](images/b04d12823243a78e923d9de1a973835443964fa47ae6d5bb408f5cbdff1ed808.jpg)",
+        "options": [],
+        "answer": "西   北   45",
+        "solution": "【分析】三角形 ABC 是等腰直角三角形， $\\angle B = 90$  度，所以  $\\angle A = \\angle B = 45$  度，再根据地图上方向的规定：上北下南，左西右东；以点 C 为观测点，确定出点 B 的位置，据此解答。\n\n【详解】因为  $\\angle B = 90$  度所以  $\\angle A = \\angle B = 45^{\\circ}$点B在西偏北（北偏西）45度方向上。如图，三角形ABC是一个等腰直角三角形，点B在点C的西偏北（北偏西）45度方向上。![](images/9b53af4222511d19bd9a2f72f965356c824c20bdaf55dd492b7f530a79abf952.jpg)",
+        "knowledge_points": [
+          "等腰三角形和等边三角形的认识及特征",
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·福建莆田·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_147",
+        "original_id": "unit2_choice_147"
+      },
+      {
+        "id": "unit2_choice_148",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "鲤中永辉超市在鲤北方圆荟南偏西  $45^{\\circ}$  方向上, 那么鲤北方圆荟在鲤中永辉超市( )方向上。",
+        "options": [],
+        "answer": "西偏南  $45^{\\circ}$ / 南偏西  $45^{\\circ}$",
+        "solution": "【分析】根据方向的相对性, 北偏东对南偏西, 角度不变, 进行选择。南和西之间的夹角是  $90^{\\circ}$ , 南偏西也可以说成西偏南, 角度  $= 90^{\\circ}$  一南偏西的角度, 据此解答。\n\n【详解】  $90^{\\circ} - 45^{\\circ} = 45^{\\circ}$ 那么鲤北方圆荟在鲤中永辉超市西偏南  $45^{\\circ}$  方向上。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·福建莆田·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_148",
+        "original_id": "unit2_choice_148"
+      },
+      {
+        "id": "unit2_choice_149",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "如图, 学校在奇奇家偏 方向上, 距离是 米。![](images/99d6522816f7f64ad565b1552c2033c0bd109b0213879ca7fd4c17f60abd487e.jpg)",
+        "options": [],
+        "answer": "东  北  45 400",
+        "solution": "【分析】以奇奇家为观测点，学校在东偏北  $45^{\\circ}$  方向上；图中一小格表示实际距离是 200 米, 奇奇家距离学校有 2 小格, 也就是  $(2 \\times 200)$  米, 据此解答。\n\n【详解】  $2 \\times 200 = 400$  （米）以奇奇家为观测点，学校在东偏北  $45^{\\circ}$  （或北偏东  $45^{\\circ}$  ）方向上，距离有 400 米。因此学校在奇奇家东偏北  $45^{\\circ}$  （或北偏东  $45^{\\circ}$  ）方向上，距离是 400 米。",
+        "knowledge_points": [
+          "东",
+          "南",
+          "西",
+          "北方向",
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "22-23 六年级下·福建莆田·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_149",
+        "original_id": "unit2_choice_149"
+      },
+      {
+        "id": "unit2_choice_150",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "偏( )（ ）方向上，距离是( )米。![](images/ff91e47dd0e5b6214d673e17665a7201a4a09c479850cc76e4480d5b4671d35b.jpg)",
+        "options": [],
+        "answer": "南    西    $45^{\\circ} / 45$  度    400",
+        "solution": "【分析】根据“上北下南，左西右东”的图上方向，结合比例尺和图上距离求出实际距离，分析解答即可。\n\n【详解】  $200 \\times 2 = 400$  （米）奇奇家在学校南偏西  $45^{\\circ}$  （或西偏南  $45^{\\circ}$  ）方向上，距离是 400 米。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离描述路线图",
+          "图上距离与实际距离的换算"
+        ],
+        "source": "22-23 六年级下·福建莆田·期末）如图，奇奇家在学校( ",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_150",
+        "original_id": "unit2_choice_150"
+      },
+      {
+        "id": "unit2_choice_151",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "量一量、填一填。![](images/3c9e5aee8317abc941778d49004602c1472ed116c62033ba85959dc567368552.jpg)\n\n从学校出发，沿( )偏( )（ )°方向走( )m到达书店；然后改变方向，沿( )偏( )（ )°方向走( )m到达小华家。",
+        "options": [],
+        "answer": "北 东 45 600  南  东 60  400",
+        "solution": "【分析】根据图上的  $1 \\mathrm{~cm}$  表示  $200 \\mathrm{~m}$ , 计算出学校到书店的实际距离; 和书店到小华家的实际距离; 根据地图上方向的规定 \"上北下南, 左西右东\", 以学校为观测点, 确定出书店的位置; 再以书店为观测点, 确定出小华家的位置即可。\n\n【详解】  $3 \\times 200 = 600$  (m)      $2 0 0 \\times 2 = 4 0 0 (m)$ 从学校出发，沿北偏东  $45^{\\circ}$  方向走  $600 \\mathrm{~m}$  到达书店；然后改变方向，沿南偏东  $60^{\\circ}$  方向走  $400 \\mathrm{~m}$  到达小华家或从学校出发，沿东偏北  $45^{\\circ}$  方向走  $600 \\mathrm{~m}$  到达书店；然后改变方向，沿东偏南  $30^{\\circ}$  方向走  $400 \\mathrm{~m}$  到达小华家",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·广东广州·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_151",
+        "original_id": "unit2_choice_151"
+      },
+      {
+        "id": "unit2_choice_152",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "（ ）度方向走( ）米到家。",
+        "options": [],
+        "answer": "西偏南 60 800",
+        "solution": "【分析】根据方向的相对性，它们的方向相反，角度相等，距离相等；据此解答。\n\n【详解】  $90 - 60 = 30$  （度）李琳上学从家往东偏北 60 度方向走 800 米到学校, 那么她放学应该往西偏南 60 度（或南偏西 30 度）方向走 800 米到家。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·江西宜春·期末）李琳上学从家往东偏北60度方向走800米到学校，那么她放学应该往( ",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_152",
+        "original_id": "unit2_choice_152"
+      },
+      {
+        "id": "unit2_choice_153",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "小兰在小华的南偏西  $35^{\\circ}$  方向  $100 \\mathrm{~m}$  处, 那么小华在小兰的( )偏( )（ ）°方向( ）m 处。",
+        "options": [],
+        "answer": "北东 35 100",
+        "solution": "【分析】小兰在小华的南偏西  $35^{\\circ}$  方向  $100 \\mathrm{~m}$  处, 是以小华为观测点; 小华在小兰的方向是以小兰为观测点; 观测点不同, 方向相反, 夹角的度数相同, 距离相同; 所以南偏西  $35^{\\circ}$ 相对的是北偏东  $35^{\\circ}$ , 由此解答。\n\n【详解】小兰在小华的南偏西  $35^{\\circ}$  方向  $100 \\mathrm{~m}$  处, 那么小华在小兰的北偏东  $35^{\\circ}$  (或东偏北  $55^{\\circ}$ ) 方向  $100 \\mathrm{~m}$  处。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·河南周口·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_153",
+        "original_id": "unit2_choice_153"
+      },
+      {
+        "id": "unit2_choice_154",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "已知外婆家在小明家的正东方, 学校在外婆家的北偏西  $40^{\\circ}$ ,外婆家到学校与小明家到学校的距离相等, 则学校在小明家的____![](images/0f0821929bd281a3f1b70b256e8bbc39acc13f15df248a5d2226a9fe5bcca2f6.jpg)",
+        "options": [],
+        "answer": "北偏东  $40^{\\circ}$",
+        "solution": "【分析】如图：![](images/3e79757a5bd6cc7006b33c3472ba41e73d05bc16ac1a87114116c09094ee09e7.jpg)\n\n外婆家到学校与小明家到学校的距离相等, 则学校、小明家、外婆家组成了一个等腰三角形。由于学校在外婆家的北偏西  $40^{\\circ}$ , 那么学校在外婆家的西偏北  $50^{\\circ}$ , 则组成的等腰三角形的底角是  $50^{\\circ}$ , 另外一个底角也是  $50^{\\circ}$ , 以小明家为观测点, 学校在小明家的北偏东  $40^{\\circ}$  。【详解】如图![](images/0564e6dbfd070d98774e2382cd956fadd2d83174aecc4b8d27d89da8893e33be.jpg)则学校在小明家的北偏东  $40^{\\circ}$  。",
+        "knowledge_points": [
+          "等腰三角形和等边三角形的认识及特征",
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "2020 六年级·全国·竞赛",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_154",
+        "original_id": "unit2_choice_154"
+      },
+      {
+        "id": "unit2_choice_155",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "在一幅地图上, A 村在 B 村的西偏北  $25^{\\circ}$  的方向上相距 1000 米处, 也可以说 B 村在 A 村的( )方向上相距( )米处。",
+        "options": [],
+        "answer": "东偏南  $25^{\\circ}$  1000",
+        "solution": "【分析】根据位置的相对性可知，描述两个物体之间的相对位置时，方向相反，角度相等，距离不变；据此解答。\n\n【详解】由分析可得：在一幅地图上，A 村在 B 村的西偏北  $25^{\\circ}$  的方向上相距 1000 米处，也可以说 B 村在 A 村的东偏南  $25^{\\circ}$  方向上相距 1000 米处。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·江西赣州·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_155",
+        "original_id": "unit2_choice_155"
+      },
+      {
+        "id": "unit2_choice_156",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "若点 A 在点 B 的正北方向上, 则点 B 在点 A 的( )方向上。",
+        "options": [],
+        "answer": "正南",
+        "solution": "【分析】根据位置的相对性可知，它们的方向相反，角度相等，距离相等；据此解答。\n\n【详解】若点A在点B的正北方向上，则点B在点A的正南方向上。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·湖南张家界·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_156",
+        "original_id": "unit2_choice_156"
+      },
+      {
+        "id": "unit2_choice_157",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "超市在学校北偏东  $30^{\\circ}$ 方向上, 这是以( )为观测点, 从学校的正北方向开始, 向东偏转  $(\\quad)$  。",
+        "options": [],
+        "answer": "学校   30",
+        "solution": "【分析】找准观测点，以观测点的“上北下南，左西右东”为准，根据题目给出的方向解答。\n\n【详解】超市在学校北偏东  $30^{\\circ}$  方向上，这是以学校为观测点，从学校的正北方向开始，向东偏转  $30^{\\circ}$  。\n\n# 三、判断题",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·湖南株洲·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_157",
+        "original_id": "unit2_choice_157"
+      },
+      {
+        "id": "unit2_choice_158",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题填空题",
+        "difficulty": "中等",
+        "question": "学校在公园北偏东  $30^{\\circ}$  约 500 米处, 公园在学校的南偏西  $30^{\\circ}$  约 500 米处。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】根据位置的相对性，方向相反，角度相同，距离相等，据此判断即可。\n\n【详解】由分析可知：学校在公园北偏东  $30^{\\circ}$  约 500 米处, 公园在学校的南偏西  $30^{\\circ}$  约 500 米处。说法正确故答案为:  $\\sqrt{}$",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级下·湖南娄底·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_158",
+        "original_id": "unit2_choice_158"
+      },
+      {
+        "id": "unit2_choice_159",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "一架飞机从某机场向北偏东  $50^{\\circ}$  方向飞行了1200千米，原路返回时要向南偏西  $50^{\\circ}$  方向飞行1200千米。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】根据位置的相对性可知：位置相对的两个物体所在的方向相反、角度相同、距离不变；据此解答。\n\n【详解】由分析可得: 一架飞机从某机场向北偏东  $50^{\\circ}$  方向飞行了 1200 千米, 原路返回时要向南偏西  $50^{\\circ}$  方向飞行 1200 千米, 原题说法正确。故答案为:  √",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级下·河南信阳·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_159",
+        "original_id": "unit2_choice_159"
+      },
+      {
+        "id": "unit2_choice_160",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "“小东家在南偏西  $60^{\\circ}$ , 距离  $2 \\mathrm{~km}$  的方向上”通过这句话可以确定小东家的位置。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】确定一个物体的位置时，先确定观测点，再根据方向、角度、距离确定物体位置，据此解答。\n\n【详解】“小东家在南偏西  $60^{\\circ}$ , 距离  $2 \\mathrm{~km}$  的方向上”没有确定观测点, 所以不能根据这句话确定小东家的位置。原题说法错误。故答案为： $\\times$",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 五年级下·广东韶关·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_160",
+        "original_id": "unit2_choice_160"
+      },
+      {
+        "id": "unit2_choice_161",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "学校在广场的北偏西  $30^{\\circ}$  方向 50 米处, 那么广场在学校的南偏东  $30^{\\circ}$  方向 50 米处。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】根据上北下南左西右东，结合方向具有相对性可知，学校在广场的北偏西处，相反即为广场在学校的南偏东处，角度和距离不变。\n\n【详解】学校在广场的北偏西  $30^{\\circ}$  方向 50 米处, 那么广场在学校的南偏东  $30^{\\circ}$  方向 50 米处。原题说法正确故答案为：√",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级下·四川凉山·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_161",
+        "original_id": "unit2_choice_161"
+      },
+      {
+        "id": "unit2_choice_162",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "确定物体的位置, 必须知道方向和距离。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】确定物体的位置要有三个步骤：（1）定观察点，（2）量方向，（3）算距离，据此即可进行解答。\n\n【详解】由分析可得：要确定一个物体的位置首先要知道这个物体在另一个物体的什么方向上，还要知道两个物体之间的距离，这样才能确定一个物体在图上的位置，所以原题说法正确。故答案为：√",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 五年级下·河南郑州·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_162",
+        "original_id": "unit2_choice_162"
+      },
+      {
+        "id": "unit2_choice_163",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "甜甜早晨上学要向北偏东  $30^{\\circ}$  方向走 200 米, 那么她下午放学回家应向南偏西  $30^{\\circ}$  方向走 200 米。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】此题考查了方向的相对性，应明确北偏东和南偏西相对,方向和距离两个条件才能确定物体的位置，根据位置的相对性，可知两处位置观测点不同，它们的方向相反，角度相等，据此解答。\n\n【详解】根据分析可知, 甜甜早晨上学要向北偏东  $30^{\\circ}$  方向走 200 米, 那么她下午放学回家应向南偏西  $30^{\\circ}$  方向走 200 米。原题干说法正确。故答案为：√",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级下·山东青岛·期中",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_163",
+        "original_id": "unit2_choice_163"
+      },
+      {
+        "id": "unit2_choice_164",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "东偏南  $40^{\\circ}$  和南偏东  $50^{\\circ}$  的方向是相同的。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】东和南之间的夹角是  $90^{\\circ}, 90^{\\circ} - 40^{\\circ} = 50^{\\circ}$ , 所以东偏南  $40^{\\circ}$  方向, 还可以说成南偏东  $50^{\\circ}$  方向, 据此判断。\n\n【详解】东偏南  $40^{\\circ}$  和南偏东  $50^{\\circ}$  的方向是相同的。原题说法正确。故答案为： $\\sqrt{}$",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "22-23六年级上·河南驻马店·期中",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_164",
+        "original_id": "unit2_choice_164"
+      },
+      {
+        "id": "unit2_choice_165",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "南偏西  $30^{\\circ}$  与西偏南  $60^{\\circ}$  表示的方向是一样的。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】南和西之间的夹角是  $90^{\\circ}$ , 南偏西也可以说成西偏南, 西偏南的角度  $= 90^{\\circ} -$  南偏西的角度, 据此分析。\n\n【详解】  $90^{\\circ} - 30^{\\circ} = 60^{\\circ}$ 即南偏西  $30^{\\circ}$  也可以说是西偏南  $60^{\\circ}$ , 原题说法正确;故答案为： $\\sqrt{}$",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "22-23 六年级上·山西长治·期中",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_165",
+        "original_id": "unit2_choice_165"
+      },
+      {
+        "id": "unit2_choice_166",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "超市在广场北偏东  $40^{\\circ}$  方向上, 那么广场在超市南偏西  $50^{\\circ}$  方向上。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】根据“上北下南左西右东”及角度，找准观测点来确定物体的位置，观测点不同，方向表示不同。据此解答。\n\n【详解】超市在广场北偏东  $40^{\\circ}$  方向上，是以广场为观察点；广场在超市南偏西  $40^{\\circ}$  （或西偏南  $50^{\\circ}$  ）方向上。原题说法错误。\n\n故答案为：×",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "22-23 六年级下·河南洛阳·期中",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_166",
+        "original_id": "unit2_choice_166"
+      },
+      {
+        "id": "unit2_choice_167",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "如果学校在公园的西偏北  $45^{\\circ}$  方向, 那公园在学校的南偏东  $45^{\\circ}$  方向。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】学校在公园的西偏北  $45^{\\circ}$  方向, 是以公园为观测点; 公园在学校的方向是以学校为观测点; 根据位置的相对性可知, 观测点不同, 方向相反, 夹角的度数相同, 所以西偏北  $45^{\\circ}$  相对的是东偏南  $45^{\\circ}$ , 东偏南  $45^{\\circ}$  也可以说成南偏东  $45^{\\circ}$ , 据此判断。\n\n【详解】如果学校在公园的西偏北  $45^{\\circ}$  方向, 那公园在学校的东偏南  $45^{\\circ}$  或者南偏东  $45^{\\circ}$  方向。原题说法正确。故答案为：√",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·山东临沂·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_167",
+        "original_id": "unit2_choice_167"
+      },
+      {
+        "id": "unit2_choice_168",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "小红在小南的西偏南 30 度方向上, 那小南就在小红的东偏北 30 度方向上。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】根据位置的相对性可知，描述两个物体之间的相对位置时，方向相反，角度相等，据此解答。\n\n【详解】通过分析可得: 小红在小南的西偏南 30 度方向上, 那小南就在小红的东偏北 30 度方向上。原题说法正确。故答案为: √",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "2024·湖南衡阳·小升初真题",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_168",
+        "original_id": "unit2_choice_168"
+      },
+      {
+        "id": "unit2_choice_169",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "北偏西  $60^{\\circ}$  方向也可以说成西偏北  $30^{\\circ}$  方向。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】根据上北下南，左西右东，四个方向，先画出北偏西  $60^{\\circ}$ ，因为北和西之间的夹角是  $90^{\\circ}, 90^{\\circ} - 60^{\\circ} = 30^{\\circ}$ ，所以北偏西  $60^{\\circ}$  方向可以看成是西偏北  $30^{\\circ}$ ，发现是同一个方向。\n\n【详解】据分析，画图：![](images/8f7090547c97f1ab5db8c1f86abb0634a4b8c3a0dccc0b9cd0fe650b57c1edb2.jpg)\n\n北偏西  $60^{\\circ}$  方向也可以说成西偏北  $30^{\\circ}$  方向。\n\n故答案为:  √\n\n# 四、作图题",
+        "knowledge_points": [
+          "东南",
+          "西南",
+          "东北",
+          "西北方向",
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 五年级上·河北沧州·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_169",
+        "original_id": "unit2_choice_169"
+      },
+      {
+        "id": "unit2_choice_170",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "六（1）班同学在春游时做了一个找宝藏的游戏,他们需要从驻地出发, 先依次到 A、B、C 三地猜对谜语, 才能到 D 处找到宝藏。已知 A、B、C、D 四地的位置关系如下:\n(1) A 地在驻地西偏北  $30^{\\circ}$  方向  $600 \\mathrm{~m}$  处。\n(2) B地在A地东偏北  $45^{\\circ}$  方向  $250\\mathrm{m}$  处。\n(3) C地在B地正南方向  $400\\mathrm{m}$  处。\n(4) D地在C地南偏东  $60^{\\circ}$  方向  $1000\\mathrm{m}$  处。\n请根据题意画出同学们找到宝藏的路线图。",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】用  $1 \\mathrm{~cm}$  表示  $100 \\mathrm{~m}$ , 根据一个地点的位置是由方向和距离决定的, 结合上北下南、左西右东画图即可。\n\n【详解】 (1) (2) (3) (4) 如下图：![](images/1cd65471655cacb0a31ab7e554d31d3491ac4d090a44ae68d69333057f960829.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_170",
+        "original_id": "unit2_choice_170"
+      },
+      {
+        "id": "unit2_choice_171",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如图, 小明每天从家去学校, 应向 ( ) 偏 ( )  $50^{\\circ}$  方向走 1.5 千米, 如果小明从学校到图书馆, 可以向 ( ) 偏 ( ) ( )°走 ( ) 千米。车站在图书馆东偏北  $30^{\\circ}$  方向 1 千米处, 请将车站的位置用  $\\triangle$  出来, 注意留下作图痕迹。![](images/eb22858ca1be155c8a762367b862d876684be1224bc99db9bd779a6fe84c3d8e.jpg)",
+        "options": [],
+        "answer": "北；东；南；东；40；1；见详解",
+        "solution": "【分析】根据图示，结合上北下南左西右东以及角度可知，小明从家去学校，应向北偏东  $50^{\\circ}$  方向走 1.5 千米；根据题意可知，小明家到学校的路线和学校到图书馆的路线成直角，在学校的位置作一条辅助线，平行于正北方向线，可以看出图书馆位于学校  $40^{\\circ}$  方向，3 格表示 1.5 千米，1 格则表示 0.5 千米，所以小明从学校到图书馆，可以向南偏东  $40^{\\circ}$  走了$(0.5 \\times 2)$  千米；根据给出的方向、角度、距离画出车站位置即可。\n\n【详解】  $1.5 \\div 3 = 0.5$  （千米） $0. 5 \\times 2 = 1 (\\text {千 米})$所以小明每天从家去学校，应向北偏东  $50^{\\circ}$  方向走 1.5 千米；如果小明从学校到图书馆，可以向南偏东  $40^{\\circ}$  走 1 千米。$1 \\div 0. 5 = 2 \\quad (\\text {格})$\n\n如图：![](images/66fa891ec99177b414bea3874746ab2b20cbb6f44b891064fa6ebac87296285d.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 五年级下·湖北省直辖县级单位·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_171",
+        "original_id": "unit2_choice_171"
+      },
+      {
+        "id": "unit2_choice_172",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "以公园为观测点：![](images/5c34167ba92afc6427e50160a08f808c0590aeeaa04b73f46296bfd69232ee10.jpg)\n\n(1) 超市在公园东偏（ ）（ ）°方向上，它们的距离是（ ）米。  \n(2) 学校在公园西偏北  $45^{\\circ}$  的方向上, 距离公园 300 米, 请画出学校的位置。",
+        "options": [],
+        "answer": "(1) 北；40；450 (2）见详解",
+        "solution": "【分析】（1）图上方位是上北下南，左西右东，以公园为观测点，超市在邮局的东往北偏点，偏了  $40^{\\circ}$ ，每小段是 150 米，3 小段是  $150 \\times 3 = 450$  米，据此即可填空；(2) 以公园为观测点, 300 米相当于 2 小段, 再根据地图上的方向, 找出学校的位置即可。\n\n【详解】（1）由分析可知：超市在公园的东偏北  $40^{\\circ}$  方向上，它们的距离是 450 米。![](images/36964121536f3c641bec784a067b568461a55d6b78d0c6225229a35fa8aa259b.jpg)  \n\n(2) 如下图所示:",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24五年级下·广东韶关·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_172",
+        "original_id": "unit2_choice_172"
+      },
+      {
+        "id": "unit2_choice_173",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "五年级的同学在进行徒步活动。他们从学校出发,先向正东方向走了 3 千米, 然后向正北方向走了 2 千米, 接着向正西方向走了 1 千米, 最后沿西偏南  $45^{\\circ}$  方向返回学校。请根据以上信息, 画出同学们徒步路线图。![](images/cee823b3d7645c6e38f3a8b27e23a74d6747f00fd048f21a79b443dbdd57dc63.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】由图可知, 图上 1 厘米代表实际距离 1 千米, 由此计算出行走的各个路段的图上距离, 利用平面图上方向规定: 上北下南, 左西右东, 依据题意结合图示去解答。\n\n【详解】  $3 \\div 1 = 3$  （厘米）$2 \\div 1 = 2 (\\text {厘 米})$如图：![](images/161c950130dac130fef75e17d312ff4b0805119c4754e7d3cd557e2b82476305.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离画线路图"
+        ],
+        "source": "23-24 五年级下·福建泉州·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_173",
+        "original_id": "unit2_choice_173"
+      },
+      {
+        "id": "unit2_choice_174",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "开展全民阅读活动是我国构建公共文化服务体系的一项重要部署，对培育和践行社会主义核心价值观、提高国民思想道德素质和科学文化素质、建设社会主义文化强国、增强国家文化软实力、实现中华民族伟大复兴中国梦具有重要意义。陈莹同学本周末准备去图书馆读书，请按要求完成下面的操作。陈莹从家出发，先向正东方向走200米到超市，又向东偏北  $30^{\\circ}$  方向走150米到达图书馆。请在下图中，把陈莹行走的路线图画完整，并标出超市和图书馆的位置。（相关角度和实际距离也要标注清楚![](images/24778b338f74f6274804ea1f6aca6ea4e0494c8da51ef8013055319bb3f0e45b.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】根据“上北下南、左西右东”先找出正东方向，再在此方向上的  $200 \\div 100 = 2$  （厘米）处找出超市的位置；以超市为观测点，找出超市的东偏北  $30^{\\circ}$  方向，再在此方向上的  $150 \\div 100 = 1.5$  （厘米）处找到图书馆的位置。据此，画出陈莹的路线图。\n\n【详解】如图：![](images/8bfaeef09c21045c68e8df0978d4af2f3693497e47fc4f7e82d7eb1485af115d.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置",
+          "根据方向",
+          "角度和距离画线路图"
+        ],
+        "source": "23-24六年级下·河南新乡·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_174",
+        "original_id": "unit2_choice_174"
+      },
+      {
+        "id": "unit2_choice_175",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "玲玲家正南方向  $800 \\mathrm{~m}$  是医院, 医院正西方向  $600 \\mathrm{~m}$ 是图书馆, 在下图中画出医院和图书馆的位置平面图。![](images/4c69dfe08d23466e5df66b29331c650ebb50eee6aa4e11c37cbb5a5b255b6e50.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】由图可知, 图上  $1 \\mathrm{~cm}$  代表实际距离  $400 \\mathrm{~m}$ , 由此计算出玲玲家与医院, 医院与图书馆的图上距离, 利用平面图上的方向规定: 上北下南左西右东, 依据题意结合图示去解答。\n\n【详解】  $800 \\div 400 = 2$  (cm)    $ 6 0 0 \\div 4 0 0 = 1. 5 (\\mathrm {c m})$\n\n如图:![](images/cd8d363bf2e9b784852111e732f8a9cb2f3c1e09fb95a76131de64a10fe83ddf.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级下·吉林松原·期中",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_175",
+        "original_id": "unit2_choice_175"
+      },
+      {
+        "id": "unit2_choice_176",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "为了加强医院疫情防控工作, 避免院内交叉感染,玉溪市中医医院在医院大门南偏西  $30^{\\circ}$  方向上 200 米处上新增设了核酸检测采样点, 请你在下面的平面图上确定核酸采样点的位置, 并用字母 A 表示。![](images/e6831ee6bb1860be06857045d77c1f6879566c1e53edd1d214e2825383aea2ac.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】1 厘米表示 100 米, 用  $200 \\div 100 = 2$  厘米, 求出医院大门与核酸检测采样点的图上距离, 再根据地图上方向的规定 “上北下南, 左西右东”, 以医院大门为观测点, 画出核酸检测采样点的位置。\n\n【详解】  $200 \\div 100 = 2$  （厘米）如图：![](images/c82bb3d6d1970d9fabc8c82ddf4c52e99c4ec7201d3b00163470d0b8f9488110.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "22-23 六年级上·云南玉溪·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_176",
+        "original_id": "unit2_choice_176"
+      },
+      {
+        "id": "unit2_choice_177",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "小明从百草园出发, 先向东走  $300 \\mathrm{~m}$  到达饮冰室,再向南走  $200 \\mathrm{~m}$  达到陋室, 然后向北偏东  $60^{\\circ}$  走  $400 \\mathrm{~m}$  到达三味书屋。在下图中画出小明的行走路线, 并标出各地点的位置。![](images/845b7840416f0458cd5173c7e703c110b51d968daa24fbc7ad4cc9bbbdd2c193.jpg)1:10000",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】用方向和距离结合来描述路线时，要注意三个要素：一是观测点（即参照物），二是方向，三是距离。根据图上距离  $=$  实际距离  $\\times$  比例尺，换算出图上距离，注意观测点，确定好方向和角度，作图即可。\n\n【详解】  $300\\mathrm{m} = 30000\\mathrm{cm}$  、  $30000 \\times \\frac{1}{10000} = 3$  (cm)   $2 0 0 \\mathrm {m} = 2 0 0 0 0 \\mathrm {c m}, 2 0 0 0 0 \\times \\frac {1}{1 0 0 0 0} = 2 (\\mathrm {c m})$\n\n$4 0 0 \\mathrm {m} = 4 0 0 0 0 \\mathrm {c m}, 4 0 0 0 0 \\times \\frac {1}{1 0 0 0 0} = 4 (\\mathrm {c m})$![](images/03b9a178081448611b72d411cef42b04da5c05f2eea48894c6cb8ffef06b110e.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离画线路图"
+        ],
+        "source": "22-23 六年级下·山东济宁·期中",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_177",
+        "original_id": "unit2_choice_177"
+      },
+      {
+        "id": "unit2_choice_178",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "根据下面的描述，在平面图上标出各场所的位置。![](images/2cba09d4bef6fe179b350d680970ff8a3f4aa60a537e15529ea91132ecadb834.jpg)\n\n① 小丽家在广场北偏西  $20^{\\circ}$  方向 600 米处。\n② 小彬家在广场西偏南  $45^{\\circ}$  方向 1200 米处。\n(3) 柳柳家在广场南偏东  $30^{\\circ}$  方向 900 米处。\n④ 军军在广场东偏北  $50^{\\circ}$  方向 1500 米处。",
+        "options": [],
+        "answer": "$①②③④$  见详解",
+        "solution": "【分析】①根据题意可知, 1 厘米表示 300 米, 计算出小丽家到广场的图上距离; 再根据地图上方向的规定 “上北下南, 左西右东”, 以广场为观测点, 画出小丽家的位置;  \n② 计算出小彬家到广场的图上距离，再以广场为观测点，画出小彬家的位置；  \n③ 计算出柳柳家到广场的图上距离，以广场为观测点，画出柳柳家的位置；  \n④ 计算出军军家到广场的图上距离，以广场为观测点，画出军军家的位置。\n\n【详解】  $① 6 0 0 \\div 3 0 0 = 2$  （厘米）如下图：$2 1 2 0 0 \\div 3 0 0 = 4 (\\text {厘 米})$如下图：$③ 9 0 0 \\div 3 0 0 = 3 (\\text {厘 米})$如下图：$④ 1 5 0 0 \\div 3 0 0 = 5 (\\text {厘 米}$\n\n如图：![](images/ef38fdb030fa7c1d275925393a7e0a33d3813eb1b4857bf5cf23e434a6232de4.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "22-23六年级下·河南洛阳·期中",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_178",
+        "original_id": "unit2_choice_178"
+      },
+      {
+        "id": "unit2_choice_179",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "在图上标出大剧院和会展中心的位置。\n\n![](images/59d8749baf4b72f62c1e8544761100b4caa3ee4f3f8d52150719c4b215ef5e48.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】弄清要标示的物体在哪个方位上, 有多少度, 按要求的方位和度数准确画图; 注意各场所离中心点的距离, 根据要求画出相应的长度。图上 1 厘米表示实际 200 米, 实际距离  $\\div 200 =$  要画的厘米数。\n\n【详解】  $400 \\div 200 = 2$  （厘米）$2 0 0 \\div 2 0 0 = 1 \\text {(厘 米)}$![](images/8c37ec21eb7541b83d9f5233e2daa7a55c01cace46d06954e635e75414f3326f.jpg)",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "22-23五年级下·山东济宁·期中",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_179",
+        "original_id": "unit2_choice_179"
+      },
+      {
+        "id": "unit2_choice_180",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "外卖方便、快捷深受大家喜爱。松鼠妈妈也在家里开了一家 “森林外卖”, 小松鼠成为送外卖的一名骑手, 刚刚接了山羊和小猴各下的一单。 妈妈告诉小松鼠说: “山羊家在我们家北偏西  $50^{\\circ}$  方向 400 米处, 小猴家在我们家东偏南  $30^{\\circ}$  方向 500 米处。” 你能帮小松鼠找到山羊家和小猴家吗? 在下图中表示出来。(用 1 厘米表示 200 米)![](images/e2757203ad9cf45922d799970191cdb94011698619469ca06b19842988677eba.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】1 厘米表示 200 米, 则松鼠家到山羊家应画  $400 \\div 200 = 2$  厘米, 松鼠家到小猴家应画  $500 \\div 200 = 2.5$  厘米; 再根据上北下南左西右东及方向角确定山羊家及小猴家的位置即可。\n\n【详解】  $400 \\div 200 = 2$  （厘米）$5 0 0 \\div 2 0 0 = 2. 5 \\text {(厘 米)}$画图如下：![](images/ecadc4518855af83a7783e2eddeefbd9772dbaa0e523b0494a27f84db94fa256.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 五年级下·辽宁·课后作业",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_180",
+        "original_id": "unit2_choice_180"
+      },
+      {
+        "id": "unit2_choice_181",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "根据下面的描述，在图中标出学校、游泳馆和汽车站的位置。![](images/5d5418d34735dfffb7c20b0c0775044848d7c5feeec2116d80b5194e8a80b603.jpg)\n\n(1) 学校在市政府的南偏东  $30^{\\circ}$  方向上，距离是 400 米。  \n(2) 游泳馆在市政府的北偏西  $20^{\\circ}$  方向上, 距离是 200 米。  \n(3) 市政府在汽车站的南偏西  $35^{\\circ}$  方向上，距离是 400 米。",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】以图上的 “上北下南, 左西右东” 为准, 图例表示图上 1 厘米相当于实际距离 200 米。(1) 以市政府为观测点，在市政府的南偏东  $30^{\\circ}$  方向上画  $400 \\div 200 = 2$  厘米长的线段，即是学校。  (2) 以市政府为观测点，在市政府的北偏西  $20^{\\circ}$  方向上画  $200 \\div 200 = 1$  厘米长的线段，即是游泳馆。(3) 市政府在汽车站的南偏西  $35^{\\circ}$  方向上, 距离是 400 米, 根据位置的相对性可知, 汽车站在市政府的北偏东  $35^{\\circ}$  方向上, 距离是 400 米;在市政府的北偏东  $35^{\\circ}$  方向上画  $400 \\div 200 = 2$  厘米长的线段，即是汽车站。\n\n【详解】如图：![](images/e5bad0338b18ab7eda06497fc45e82e9eeb88b51df325808eabcdcdc4f10c43c.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级下·江苏·课后作业",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_181",
+        "original_id": "unit2_choice_181"
+      },
+      {
+        "id": "unit2_choice_182",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "早上, 乐乐从家出发, 沿东偏北  $40^{\\circ}$  方向走 300 米到达书店; 再沿南偏东  $40^{\\circ}$  方向走 200 米到达学校。请你根据上面的描述, 画出乐乐上学的路线, 标出书店和学校的位置。(1 厘米代表 100 米)\n![](images/b8ccf0c72b9a388050e615597c25613f33202770af6513b4999d9e4954d81de2.jpg)",
+        "options": [],
+        "answer": "画图见详解",
+        "solution": "【分析】确定物体位置的三大要素是：方向、角度、距离，同时根据图上的方向：上北下南、左西右东，以及角度和距离就可以确定方位；\n\n由图可知，图上1厘米代表实际距离的100米，用实际距离除以100可求出图上距离，据此画图即可。\n\n【详解】  $300 \\div 100 = 3$  （厘米）$2 0 0 \\div 1 0 0 = 2 \\text {(厘 米)}$据此画图：![](images/764a6a1ff8a29c3b4cd8d98f051bb5bfa99c03e78412aae218bb114c0de7a6c1.jpg)",
+        "knowledge_points": [
+          "图上距离与实际距离的换算",
+          "根据方向",
+          "角度和距离确定物体的位置",
+          "根据方向",
+          "角度和距离画线路图"
+        ],
+        "source": "23-24六年级上·福建莆田·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_182",
+        "original_id": "unit2_choice_182"
+      },
+      {
+        "id": "unit2_choice_183",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "根据下面的描述，画出小亮一家的行车路线图。![](images/303dcc7ee56a1a56a3f94452728d3ba138702e4b48810e98fb014031dcce638a.jpg)\n\n小亮家在十字路口的中心位置。周末，小亮一家人开车去郊游，先向北偏东  $30^{\\circ}$  方向行驶 8 千米到动物园, 又向东偏南  $20^{\\circ}$  方向行驶 9 千米到达果园。",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】用方向和距离结合来画路线时, 要注意三个要素: 一是观测点 (即参照物), 二是方向, 三是距离。看图可知, 图上  $1 \\mathrm{~cm}$  表示实际  $500 \\mathrm{~m}$ , 实际米数  $\\div 500 =$  图上厘米数,据此作图。\n\n【详解】  $1\\mathrm{km} = 1000\\mathrm{m}$  $1 0 0 0 \\div 5 0 0 = 2 \\text {(厘 米)}$ $5 0 0 \\div 5 0 0 = 1 \\text {(厘 米)}$ $1 5 0 0 \\div 5 0 0 = 3 (\\text {厘 米})$![](images/85c52745e89389fa6408585f5bfb0670643f34a49e24f2592e4a531e7f355207.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离画线路图"
+        ],
+        "source": "23-24六年级上·江西宜春·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_183",
+        "original_id": "unit2_choice_183"
+      },
+      {
+        "id": "unit2_choice_184",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "李叔叔开车从 A 地到 C 地办事, 从 A 地出发向东偏南  $30^{\\circ}$  方向行驶 40 千米到达 B 地, 再向东偏北  $20^{\\circ}$  方向行驶 30 千米到达 C 地, 请画出李叔叔驾车行驶路线图。![](images/a067969834e98ccfbab1ccc3adc161dedde4795cc9f2534e8d3c25ec6c8f4663.jpg)\n![](images/2fd4469d4bf808fc0723aa2b250e7520813fbbb13f244cc329047a028390f38e.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】由图可知：图上1厘米表示实际10千米，先算出A到B，B到C的图上距离即  $40 \\div 10 = 4$  （厘米）、 $30 \\div 10 = 3$  （厘米），再根据方向、角度、距离确定B地、C地的位置，即可画出李叔叔驾车行驶路线图。\n\n【详解】如图所示：![](images/cdbc73e26461e9e60d1a290f5117f858fbefe30bfb761ff5859f50e2614d219e.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离画线路图"
+        ],
+        "source": "23-24 六年级上·福建漳州·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_184",
+        "original_id": "unit2_choice_184"
+      },
+      {
+        "id": "unit2_choice_185",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "以游泳池为观测点。请画出足球场、停车场的位置。![](images/32b207cea0e4940f14077df150f40af0ddfa3f9327a3fdf44f0c0641b85a7d3e.jpg)\n\n(1) 足球场在游泳池南偏东  $55^{\\circ}$  方向 800 米处。  \n(2) 停车场在游泳池北偏西  $30^{\\circ}$  方向 600 米处。",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】以游泳池为观测点，以图上的“上北下南，左西右东”为准，图例表示图上1厘米相当于实际距离200米。在游泳池的东偏北  $45^{\\circ}$  的方向上画  $600 \\div 200 = 3$  厘米长的线段, 即是足球场;在游泳池的南偏西  $30^{\\circ}$  的方向上画  $800 \\div 200 = 4$  厘米长的线段, 即是停车场。\n\n【详解】如图：![](images/0bc579f8bd45c7de9874b770732744922e5f806b08d0531d1e7545e379fbf3c8.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·山东临沂·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_185",
+        "original_id": "unit2_choice_185"
+      },
+      {
+        "id": "unit2_choice_186",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "容易",
+        "question": "根据下面的描述, 在平面图上标出各场所的位置。![](images/98eff70c3b43b3785da9265e89573bf00ecb4fee7643c67abcddcb24c6c9f16b.jpg)\n(1) 图书馆在办公楼的北偏西偏  $30^{\\circ}$  方向 500 米处。\n(2) 餐厅在办公楼的西偏南  $45^{\\circ}$  方向 600 米处。",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】（1）根据平面图上方向的辨别“上北下南，左西右东”，以办公楼的位置为观测点，图书馆在办公楼的北偏西偏  $30^{\\circ}$  方向，图中 1 厘米表示实际距离 200 米，则图书馆与办公楼的图上距离为  $500 \\div 200 = 2.5$  厘米，进而即可画出图书馆的位置。  \n(2) 同理, 以办公楼的位置为观测点, 餐厅在办公楼的西偏南  $45^{\\circ}$  方向, 图上距离为  $600 \\div 200 = 3$  厘米处, 画出餐厅的位置即可。\n\n【详解】  $500 \\div 200 = 2.5$  （厘米）$6 0 0 \\div 2 0 0 = 3 \\quad (\\text {厘 米})$\n\n根据分析作图如下：![](images/509c3e66529a06bab57da462bac126699f29c2cda93fee67deaae116f6bbb46a.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·山东济南·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_186",
+        "original_id": "unit2_choice_186"
+      },
+      {
+        "id": "unit2_choice_187",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "以大门为观测点, 根据给出的条件在平面图上标出正确的位置。![](images/017255cfa94d38c4cf18522b6cfc8fa05f91e6a4afee91352edea638b631468d.jpg)\n(1) 喷泉在正北方向 600 米处。  (2) 在北偏东  $35^{\\circ}$  方向 300 米处有一幢高楼。  (3) 休闲室位于西偏南  $45^{\\circ}$  方向400米处。",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】以大门为观测点，以图上的“上北下南，左西右东”为准，图例表示图上1厘米相当于实际距离200米。\n\n(1) 在大门的正北方向上画  $600 \\div 200 = 3$  厘米长的线段, 即是喷泉;  (2) 在大门的北偏东  $35^{\\circ}$  方向上画  $300 \\div 200 = 1.5$  厘米长的线段, 即是一幢高楼;  (3) 在大门的西偏南  $45^{\\circ}$  方向上画  $400 \\div 200 = 2$  厘米长的线段, 即是休闲室。\n\n【详解】如图：![](images/779345e1d01e886a60fb45b0f8bd9c6d8a5defd89c69e7808563755dc7cfb9e9.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·河北保定·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_187",
+        "original_id": "unit2_choice_187"
+      },
+      {
+        "id": "unit2_choice_188",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "我国自主研发的北斗导航系统日渐成熟。现通过卫星定位到草原上一只猎豹和一头狮子正在 O 点争食。突然它们受到了惊吓, 猎豹以每分钟 2 千米的最快速度向北偏东  $30^{\\circ}$  方向奔跑, 狮子以每分钟 1 千米的最快速度向南偏西  $60^{\\circ}$  方向奔跑, 请你在图中标出 2 分钟后它们各自的位置。![](images/aaf9f502607c90d2a0ac4c26015c13ee46dede07c2ce7b5cb549c30de80fea91.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】以 O 点为观测点, 以图上的 “上北下南, 左西右东” 为准, 图例表示图上 1 厘米相当于实际距离 1 千米。\n\n已知猎豹速度是 2 千米/分, 狮子的速度是 1 千米/分, 先根据 “速度  $\\times$  时间 = 路程”, 分别求出猎豹 2 分钟跑了 4 千米, 狮子 2 分钟跑了 2 千米;在 O 点的北偏东  $30^{\\circ}$  方向上画  $4 \\div 1 = 4$  厘米长的线段, 即是猎豹 2 分钟后的位置;在 O 点的南偏西  $60^{\\circ}$  方向上画  $2 \\div 1 = 2$  厘米长的线段, 即是狮子 2 分钟后的位置。\n\n​【详解】猎豹：  $2\\times 2 = 4$  （千米）$4 \\div 1 = 4$  （厘米）狮子：  $1\\times 2 = 2$  （千米）${2}\\div{1}=2$ (厘米)\n\n$2 \\div 1 = 2$  （厘米）\n\n如图：![](images/63f5b94b79369f722e9e40687796695a785486eb8dcecc1875440599f84ea052.jpg)\n\n# 五、解答题",
+        "knowledge_points": [
+          "基础行程问题",
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·重庆黔江·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_188",
+        "original_id": "unit2_choice_188"
+      },
+      {
+        "id": "unit2_choice_189",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "某文化宫广场周围环境如下图所示。\n![](images/390cc36b2500ee45946bed8a0919f8939e0d2c9b397e8bf723cef4e071e1a8fe.jpg)\n(1)体育馆在文化宫( )方向上。  (2)张刚以60米/分的速度从学校沿着人民路向东走，3分钟后他在文化宫( )面( )米处。",
+        "options": [],
+        "answer": "(1)北偏东  $45^{\\circ}$ (2) 西 20",
+        "solution": "【分析】（1）根据图中确定方向的方法：上北下南、左西右东，结合观测点，找准方向和角度即可；(2) 根据速度  $x$  时间 = 路程可知, 代入数据求出张刚 3 分钟走了  $60 \\times 3 = 180$  米, 而量得图中学校和文化宫的图上距离是 2 厘米, 据此求出学校和文化宫的实际距离是  $2 \\times 100 = 200$  (米), 说明李小明离文化宫还有 (200 - 180) 米, 据此解答。\n\n【详解】（1）体育馆在文化宫北偏东  $45^{\\circ}$  方向上或体育馆在文化宫东偏北  $45^{\\circ}$  方向上。(2)  $60 \\times 3 = 180$ (米)$ 2 \\times 1 0 0 = 2 0 0 (\\text {米}) $    $2 0 0 - 1 8 0 = 2 0 \\text {(米)}$ 即3分钟后他在文化宫西面20米处。",
+        "knowledge_points": [
+          "基础行程问题",
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级下·四川甘孜·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_189",
+        "original_id": "unit2_choice_189"
+      },
+      {
+        "id": "unit2_choice_190",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "填一填，画一画。![](images/18de1690fb88554791ff4adb8e803d839e772390817b1e797333e3cef6d8a435.jpg)\n\n(1) 公园在学校（）偏（）（）方向上，距离是（）米；  \n(2) 医院在学校（）偏（）（）方向上，距离是（）米；  \n(3) 学校在商场的（）偏（）（）方向上，距离是（）米；  \n(4) 少年宫在学校的东偏北  $45^{\\circ}$  方向上，距离是 1000 米，请在图中标出少年宫的位置。",
+        "options": [],
+        "answer": "(1) 西；北；  $30^{\\circ}$  ；800 (2）北；东；  $15^{\\circ}$  ；400 (3）见详解",
+        "solution": "【分析】（1）图中已知四个方位，书店的位置与正西方夹角为  $30^{\\circ}$ ，偏向北方；图中的一段线表示 200 米，到书店距离有 4 段线段，则有 800 米，据此得出答案；(2) 以图书馆为原点, 则学校的位置与与正北方夹角为  $15^{\\circ}$ , 偏向东方; 图中的一段线表示 200 米, 从图书馆到学校距离有 2 段线段, 则有 400 米, 据此得出答案;  (3) 少年宫的位置与正东方夹角  $20^{\\circ}$ , 偏向北方, 距离 1000 米, 即有 5 段这样的线段, 找出少年宫位置; 公园的位置与正东方夹角  $45^{\\circ}$ , 偏向南方, 距离 600 米, 即有 3 段这样的线段, 找出公园的位置。\n\n【详解】（1）书店在学校的西偏北  $30^{\\circ}$  方向上，距离是  $200 \\times 4 = 800$  （米）。(2) 学校在图书馆的北偏东  $15^{\\circ}$  方向上，距离是  $200 \\times 2 = 400$  （米）。  (3) 少年宫和公园的位置如图：![](images/6e8665841154c167a9063d54f44144b35e921b7178eac556817ec8696db8e12e.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离描述路线图",
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "21-22六年级下·湖南长沙·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_190",
+        "original_id": "unit2_choice_190"
+      },
+      {
+        "id": "unit2_choice_191",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "12 路公共汽车从起点站出发, 先向西偏北  $40^{\\circ}$  行驶 3 千米走到 A 站, 然后再向正西行驶 5 千米到达 B 站, 最后向西偏南  $50^{\\circ}$  行驶 4 千米到达终点站。(1) 根据上面的描述，把 12 路公共汽车行驶的路线图画完整。  (2) 根据路线图，描述 12 路公共汽车原路返回时所行驶的方向和路程。![](images/b8322600fd2ca4eec8488da63f69239ea956c9a730335c8ec0da536e4e7d5583.jpg)",
+        "options": [],
+        "answer": "(1) (2) 见详解",
+        "solution": "【分析】（1）根据地图上的方位是“上北下南、左西右东”，根据题意可知，图上1个单位长度表示1千米，据此画出12路公共汽车行驶的线路图即可；(2) 根据方向的相对性，写出公共汽车原路返回的行驶路线。\n\n【详解】 (1) 作图如下:![](images/7546901ae67ebf88ba64c9e3cfc18c6ae7b608f91f9b8bc4eea8ea539e4f6955.jpg)\n\n(2) 12 路公交车从终点站返回起点: 先向东偏北  $50^{\\circ}$  行驶 4 千米到达 B 点, 再向正东行驶 5 千米达 A 点, 然后向东偏南  $40^{\\circ}$  行驶 3 千米到达起点。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离描述路线图",
+          "根据方向",
+          "角度和距离画线路图"
+        ],
+        "source": "22-23 六年级下·陕西咸阳·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_191",
+        "original_id": "unit2_choice_191"
+      },
+      {
+        "id": "unit2_choice_192",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "六年级5班到营地开展军训，分为男、女小分队。![](images/f122c8cf242714032f00c3f2f1318fed34f917935cdb13030d98efd0169edddb.jpg)\n(1) 男生小分队在营地中心（）偏（）°方向距离约是（）千米处。  (2) 女生小分队在营地中心南偏西  $40^{\\circ}$  方向上，距离约是 3 千米处。请在图中画出女生小分队的准确位置。",
+        "options": [],
+        "answer": "（1）东；北30；2；",
+        "solution": "【分析】以营地中心为观测点，以图上的“上北下南，左西右东”为准，图例表示图上1厘米相当于实际的1千米。\n\n(1) 从图中可知, 男生小分队与营地中心相距 2 厘米, 那么实际相距  $1 \\times 2 = 2$  千米, 根据图上的方向、角度和距离, 得出男生小分队与营地中心的位置关系。  (2）在营地中心的南偏西  $40^{\\circ}$  方向上画  $3\\div 1 = 3$  厘米长的线段，即是女生小分队的位置。\n\n【详解】 (1)  $1 \\times 2 = 2$  （千米）男生小分队在营地中心东偏北  $30^{\\circ}$  （或北偏东  $60^{\\circ}$  ）方向距离约是2千米处。\n\n(2)  $3 \\div 1 = 3$  （厘米）女生小分队的位置如下图：![](images/1630019bc25d95700e4257998cb0c0a99fc2edc2349bc4041f51637b515f1973.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "2023·广西柳州·小升初真题",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_192",
+        "original_id": "unit2_choice_192"
+      },
+      {
+        "id": "unit2_choice_193",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "按要求填空或画图。![](images/9054d79055dec3390b61b11f8979dc8a40cc984793d7edd86b914c2bf2d42062.jpg)\n\n(1）电影院在少年宫的（）偏（）（）的方向上，距离是（）米。  \n(2) 体育馆在少年宫的（）偏（）（）的方向上，距离是（）米。  \n(3) 市场在少年宫南偏东  $30^{\\circ}$  的方向上，距离是 1200 米，在图中画出市场的位置。",
+        "options": [],
+        "answer": "（1）北；东；30；1000(2）见详解  (3) 南；东；60；1500；西；南；40；500",
+        "solution": "【分析】（1）根据题意可知，1 厘米表示 500 米；计算出公园到小明家的实际距离；根据平面图上方向的规定：上北下南，左西右东，以公园为观测点即可确定小明家的方向；(2) 1.5 千米  $= 1500$  米, 先计算出书店到公园的图上距离, 再以公园为观测点, 画出书店的位置;  (3) 根据图上 1 厘米表示实际 500 米, 求出公园到超市的实际距离。根据平面图上方向的规定: 上北下南, 左西右东, 以书店为观测点即可确定公园的方向; 再以公园为观测点即可确定超市的方向, 据此解答。\n\n【详解】 (1)  $500 \\times 2 = 1000$  （米）$9 0 ^ {\\circ} - 3 0 ^ {\\circ} = 6 0 ^ {\\circ}$小明家在公园北偏东  $30^{\\circ}$  （或东偏北  $60^{\\circ}$  ）方向，1000 米处；(2) 1.5千米  $= 1500$  米$1500 \\div 500 = 3$  （厘米）\n\n如图：![](images/b0946fc8b8e2e906074b1df3054bd919645102fa867905db9f77f4af624eff88.jpg)\n\n(3)  $500 \\times 1 = 500$  （米）$ 9 0 ^ {\\circ} - 6 0 ^ {\\circ} = 3 0 ^ {\\circ} $       $ 9 0 ^ {\\circ} - 4 0 ^ {\\circ} = 5 0 ^ {\\circ}$小明从书店经公园去超市。他先沿南偏东  $60^{\\circ}$  （或东偏南  $30^{\\circ}$  ）方向走1500米到公园，再沿西偏南  $40^{\\circ}$  （南偏西  $50^{\\circ}$  ）方向走500米就到了。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离描述路线图",
+          "根据方向",
+          "角度和距离确定物体的位置",
+          "根据方向",
+          "角度和距离画线路图"
+        ],
+        "source": "22-23六年级上·河南南阳·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_193",
+        "original_id": "unit2_choice_193"
+      },
+      {
+        "id": "unit2_choice_194",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "李叔叔驾车从家向东偏北  $35^{\\circ}$  方向行驶 2 千米后向西行驶 5 千米, 最后向西偏南  $30^{\\circ}$  行驶 3 千米到达森林公园。(1) 根据上面的描述，把李叔叔驾车行驶的路线图画完整。  (2) 根据路线图，说说李叔叔驾车原路返回时所行驶的方向和路程。![](images/2f9b16fce807125f67a81fd3317ce3ef0aa0a9257fe8f3f6ab640a125d51bfe3.jpg)",
+        "options": [],
+        "answer": "(1)(2)见详解",
+        "solution": "【分析】（1）用方向和距离结合来描述和画路线时，要注意三个要素：一是观测点（即参照物），二是方向，三是距离。看图可知，图上 1 厘米表示实际 1 千米，实际几千米就在图上画几厘米即可。  (2) 根据方向的相对性, 西偏南对东偏北, 西对东, 东偏北对西偏南, 角度和距离不变,描述原路返回时所行驶的方向和路程即可。\n\n【详解】(1)做图如下：![](images/502fecc183fc752ee8348443da96030795cd80e296d68635ab559ff044f0c438.jpg)\n\n(2) 李叔叔从森林公园向东偏北  $30^{\\circ}$  方向行驶 3 千米后向东行驶 5 千米, 最后向西偏南  $35^{\\circ}$ 方向行驶 2 千米到家。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离描述路线图",
+          "根据方向",
+          "角度和距离画线路图"
+        ],
+        "source": "22-23 六年级上·河南南阳·阶段练习",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_194",
+        "original_id": "unit2_choice_194"
+      },
+      {
+        "id": "unit2_choice_195",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "操作。![](images/950c5c31b8a85c962b2f94d1c5ca86c0682a625cf107977827ab3e2195b25489.jpg)\n\n(1) 医院在书店东偏南  $35^{\\circ}$  的方向上，距离为 400 米，请在图上标出医院的位置。  \n(2) 在书店、医院、超市和公园四个建筑物中，距离书店最近的建筑是（）。",
+        "options": [],
+        "answer": "(1) 北；东40；400  (2）图见解析",
+        "solution": "【分析】（1）根据地图上的方向的规定：上北下南，左西右东，以大转盘的位置为观测点，即可确定电影院位置的方向，然后再量出电影院与大转盘的图上距离，根据电影院与大转盘的图上距离及图中所标注的线段比例尺即可求出影院与大转盘的实际距离。  \n(2) 同理, 以大转盘的位置为观测点, 即可确定超市位置的方向, 然后超市与大转盘的图上距离及图中所标注的线段比例尺即可求出超市与大转盘的图上距离, 从而画出超市的位置。\n\n【详解】（1）量得影院到大转盘图上距离是2厘米 $2 0 0 \\times 2 = 4 0 0 (\\text {米})$电影院在大转盘北偏东40度  $^\\circ 400$  米处或电影院在大转盘东偏北50度  $^\\circ 400$  米处。(2) 0.5千米  $= 500$  米 $5 0 0 \\div 2 0 0 = 2. 5 \\text {(厘 米)}$即超市在大转盘南偏西40度方向图上距离2.5厘米处。在如图中标出超市的位置如下：![](images/c8b0586d0bea1893ba2ca170d7fa5d4453d334980be08d147b47c2b035bb53a8.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "22-23六年级下·河南平顶山·期中",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_195",
+        "original_id": "unit2_choice_195"
+      },
+      {
+        "id": "unit2_choice_196",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "某餐厅用机器人天天给顾客送餐，下面是送餐路线图。![](images/972eab49087ed1dfd07954688a51cc05ee9f570b8ca7977e5c7ec07cb574467f.jpg)\n\n(1) 天天先沿（）偏（）（）方向走了（）米送到①号桌。  \n(2) 然后天天再沿东偏北  $30^{\\circ}$  方向走 200 米送到②号桌，请在图中标出②号桌的位置（用●表示）。  \n(3) 天天完成送餐任务后，再按原路返回，用\"→\"画出返回的路线。",
+        "options": [],
+        "answer": "(1) 西，北，42，2；(2）见详解",
+        "solution": "【分析】由图可知, 图上 1 厘米代表实际距离 2 米, 由此计算出发点到 A 点的实际距离, C 点到 B 点的图上距离, 利用平面图上方向规定: 上北下南左西右东, 依据题意结合图示去解答。\n\n【详解】由图可知, 图上 1 厘米代表实际距离 2 米, 则出发点到 A 点的实际距离:  $2 \\times 1 =$  2 (米), C 点到 B 点的图上距离:  $4 \\div 2 = 2$  (厘米)\n\n(1) 天天从出发点出发, 向西偏北 42 度（北偏西 58 度）方向, 行走 2 米可以到达 A 点。  \n(2) 作图如下:![](images/69632f9beb619d29819d5dcb066c402118d2a4f2975567e80e194fad9ff11898.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离描述路线图",
+          "图上距离与实际距离的换算"
+        ],
+        "source": "23-24六年级上·甘肃庆阳·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_196",
+        "original_id": "unit2_choice_196"
+      },
+      {
+        "id": "unit2_choice_197",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如图是城市定向阅读打卡赛的地图。\n\n![](images/835a9541df99baa160731c15ffb0531415edaa7154ebe6006f0c1afb23a8d749.jpg)\n\n(1) 生活书店在新知书店的（）偏（）（）方向上，距离是（）米。  \n(2) 乐乐要从生活书店去城市书店, 城市书店在生活书店南偏东  $40^{\\circ}$  方向, 距离 600 米的位置, 请在图上标出城市书店的位置。",
+        "options": [],
+        "answer": "（1）南；西；  $70^{\\circ}$  ；1000(2）见详解",
+        "solution": "【分析】以图上的 \"上北下南, 左西右东\" 为准, 图例表示图上 1 厘米相当于实际距离 200 米。(1) 以新知书店为观测点, 从图中可知, 生活书店与新知书店相距 5 厘米, 那么实际相距  $(200 \\times 5)$  米, 根据图上的方向、角度和距离, 得出生活书店与新知书店的位置关系。  (2) 以生活书店为观测点，在生活书店南偏东  $40^{\\circ}$  方向上画  $600 \\div 200 = 3$  厘米长的线段，即是城市书店。\n\n【详解】 (1)  $200 \\times 5 = 1000$  （米）生活书店在新知书店的南偏西  $70^{\\circ}$  （或西偏南  $20^{\\circ}$  ）方向上，距离是 1000 米。\n\n(2）如图：![](images/6487dd372ce7c836e0adf4cd5ef32f3c41ccb5d875ee0b59ee071074ad798279.jpg)",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·云南玉溪·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_197",
+        "original_id": "unit2_choice_197"
+      },
+      {
+        "id": "unit2_choice_198",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "某餐馆里机器人按照桌号顺序依次送餐, 从工作站出发, 向北偏西  $60^{\\circ}$  方向, 行走 4 米, 到达①号桌。再向西偏北  $45^{\\circ}$  方向走 6 米到②号桌, 再向南偏西  $20^{\\circ}$  方向走 8 米到③号桌。然后原路返回工作站。![](images/d0da297f1eb5333911401b876850e636642fcc1bf266065b2f312100c2e235fd.jpg)\n(1) 根据上面的描述将路线图补充完整。\n(2) 根据第（1）小题完整的线路图，用自己的语言叙述机器人原路返回时走的方向和路程。",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】（1）从①号桌向西偏北  $45^{\\circ}$  方向走 6 米到②号桌，就是以①号桌为观测点，用量角器量出西偏北  $45^{\\circ}$ ，从图中可知，一段表示 2 米，则 6 米是这样的 3 段。同理以②号桌为观测点，用量角器测出度数，则 8 米就是这样的 4 段。  (2) 将方向和距离结合起来描述位置时, 要注意三个要素: 一是观测点, 二是方向, 三是距离。\n\n【详解】 (1)\n\n![](images/f3666da05b275837521a39d22057d97142c7863833d5cccffa508be9dd18d7ae.jpg)\n\n(2) 从③号桌向北偏东  $20^{\\circ}$  方向走 8 米到②号桌, 再向东偏南  $45^{\\circ}$  方向走 6 米到①号桌, 最后向南偏东  $60^{\\circ}$  方向走 4 米就到了工作站。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·四川绵阳·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_198",
+        "original_id": "unit2_choice_198"
+      },
+      {
+        "id": "unit2_choice_199",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "![](images/726a9d1c882decd08910869d2f14fdafc848a1b9c8602def31b7d0d1ac03417e.jpg)\n(1) 根据描述画图: 兰兰周末去图书馆看书, 她从家出发向北偏东  $30^{\\circ}$  方向走 400 米, 再向正西方向走 600 米到图书馆, 看完书后向南偏东方向走回家。  (2) 兰兰行走的路线连起来是什么图形？",
+        "options": [],
+        "answer": "(1) 图见详解 (2）三角形",
+        "solution": "【分析】（1）以每一次移动的出发点为观测点，根据上北下南左西右东确定方向，最后根据比例尺算出图上距离，从兰兰家开始，一步一步的画出路线图；  (2) 把路线连接起来，再判断是什么图形。\n\n【详解】 作图如下：![](images/80f92d4da024169e1a97e3469c1829b08c473627a0fefd359df1632728a05d8a.jpg)(2) 兰兰行走的路线连起来是三角形。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置",
+          "根据方向",
+          "角度和距离画线路图",
+          "长方形",
+          "正方形",
+          "三角形和圆的初步认识"
+        ],
+        "source": "23-24六年级上·河南驻马店·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_199",
+        "original_id": "unit2_choice_199"
+      },
+      {
+        "id": "unit2_choice_200",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "![](images/258d3f70bf31d6c6383d98a053b5ec7d55528accc314bdbf4119cfb2d2165954.jpg)\n(1）银行在丽丽家（）偏（）（）方向（）米处。\n(2) 少年宫在丽丽家西偏北  $45^{\\circ}$  方向 300 米处, 请在图上标出少年宫的位置。",
+        "options": [],
+        "answer": "(1) 东；北； $30^{\\circ}$ ；200",
+        "solution": "【分析】(1) 1 厘米表示 100 米, 用  $100 \\times 2 = 200$  米, 求出丽丽家到银行的路程, 根据地图上方向的规定 “上北下南, 左西右东”, 以丽丽家为观测点, 确定银行的位置;  (2) 用  $300 \\div 100 = 3$  厘米; 先求出丽丽家到少年宫的图上距离, 再以丽丽家为观测点, 画出少年宫的位置。\n\n【详解】 (1)  $100 \\times 2 = 200$  （米）$9 0 ^ {\\circ} - 3 0 ^ {\\circ} = 6 0 ^ {\\circ}$银行在丽丽家东偏北  $30^{\\circ}$  （或北偏东  $60^{\\circ}$  ）方向200米处。(2)  $300 \\div 100 = 3$  （厘米）\n\n如图：![](images/f0e08e94549f1dbcf6ab7d7e9a5185222cdb687adf668eb61689ba31fcddce5f.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·福建三明·期末",
+        "create_time": "2024-10-08 10:00:00",
+        "merged_id": "unit2_200",
+        "original_id": "unit2_choice_200"
+      },
+      {
+        "id": "unit2_choice_201",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "如图所示,成都在北京的()方向。![](images/2720ce1f9e37bfc3c836509a222465951e3de33dcea142433f6915e9d86ec7ff.jpg)",
+        "options": [
+          "A. 西南",
+          "B. 东北",
+          "C. 西北"
+        ],
+        "answer": "A",
+        "solution": "![](images/d61eff7ce5cde0baf499882bac36ef0e9237e3bd5f9264f9c15d1e5d3f10b1fc.jpg) 如图,地图上按上北下南左西右东确定方向,据此以北京为观测点,确定成都的方向即可。根据分析,成都在北京的西南方向。故答案为: A",
+        "knowledge_points": [
+          "东南、西南、东北、西北方向"
+        ],
+        "source": "25-26六年级上·海南海口·单元测试",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_201",
+        "original_id": "unit2_choice_201"
+      },
+      {
+        "id": "unit2_choice_202",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "如下图所示。![](images/5a13efecaaccb4534696487f1891adda749950c7942de2969d7027ae99d08884.jpg) 电影院在小亮家()方向,体育馆在游泳池()方向。① 南偏西 $25^{\\circ}$ ② 东偏北 $25^{\\circ}$ ③ 南偏西 $60^{\\circ}$ ④ 东偏北 $60^{\\circ}$",
+        "options": [
+          "A.④; ①",
+          "B. ③; ②",
+          "C. ③; ①",
+          "D. ④; ②"
+        ],
+        "answer": "B",
+        "solution": "以图上的'上北下南,左西右东'为准,电影院在小亮家的方向,是以小亮家为观测点;体育馆在游泳池的方向,是以游泳池为观测点;结合图上的方向、角度得出电影院与小亮家的位置关系,体育馆与游泳池的位置关系。电影院在小亮家南偏西 $60^{\\circ}$ 方向,体育馆在游泳池东偏北 $25^{\\circ}$ 方向。故答案为:B",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "25-26六年级上·海南海口·单元测试",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_202",
+        "original_id": "unit2_choice_202"
+      },
+      {
+        "id": "unit2_choice_203",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "第三届'一带一路'田径邀请赛暨2025年全国田径大奖赛(第3站)于5月14—16日在长寿区体育中心举办。假如在比赛期间碧桂园酒店、体育中心和菩提古镇成为国际参赛选手主要的活动区域,下列描述正确的是()。![](images/d4fe1c1dd71085a6dc17b7168789530afe11c76e868761756daa787c2d005712.jpg)",
+        "options": [
+          "A. 从碧桂园酒店向西偏南 $39^{\\circ}$ 方向大约走 2 千米到体育中心。",
+          "B. 从体育中心向西偏北 $21^{\\circ}$ 方向走大约 3 千米到菩提古镇。",
+          "C. 从体育中心向东偏北 $39^{\\circ}$ 方向走大约 2 千米到碧桂园酒店。",
+          "D. 从碧桂园酒店向南偏西 $8^{\\circ}$ 方向大约走 3.5 千米到菩提古镇。"
+        ],
+        "answer": "B",
+        "solution": "判断选项时,要依据'方向和距离确定位置'的核心知识:先明确观测点(观测点不同,方向描述可能相反);方向需以'东、南、西、北'为基准,说明偏向方向和夹角(锐角);按这个思路,逐一核对每个选项的观测点、方向(含角度)和距离是否与图中信息一致,就能判断对错。A.从碧桂园酒店到体育中心,观测点是碧桂园酒店。正确方向应为南偏西 $39^{\\circ}$,而非西偏南 $39^{\\circ}$,距离约2千米。因此A选项方向错误。B. 从体育中心到菩提古镇,观测点是体育中心。方向以正西为基准,向北偏 $21^{\\circ}$,即西偏北 $21^{\\circ}$,距离约3千米。角度和距离均合理,因此B选项描述正确。C. 从体育中心到碧桂园酒店, 观测点是体育中心。以正北为基准, 向东偏 $39^{\\circ}$, 正确方向是北偏东 $39^{\\circ}$, 而非东偏北 $39^{\\circ}$, 因此 C 选项方向错误。D. 从碧桂园酒店到菩提古镇,观测点是碧桂园酒店。以正西为基准,向南偏 $8^{\\circ}$,正确方向是西偏南 $8^{\\circ}$,而南偏西 $8^{\\circ}$,因此 D 选项方向错误。故答案为: B",
+        "knowledge_points": [
+          "根据方向、角度和距离描述路线图"
+        ],
+        "source": "24-25六年级下·重庆长寿·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_203",
+        "original_id": "unit2_choice_203"
+      },
+      {
+        "id": "unit2_choice_204",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "一架飞机从某飞机场向南偏东 $40^{\\circ}$ 方向飞行 1000 千米, 返航时, 飞机要向 ( )。",
+        "options": [
+          "A. 北偏西 $50^{\\circ}$ 方向飞行 1000 千米",
+          "B. 东偏南 $40^{\\circ}$ 方向飞行1000千米",
+          "C. 北偏西 $40^{\\circ}$ 方向飞行 1000 千米"
+        ],
+        "answer": "C",
+        "solution": "一架飞机从某飞机场向南偏东 $40^{\\circ}$ 方向飞行1000千米, 此时以机场为坐标原点, 向上为正北方向, 左为正西方向, 右为正东方向, 下为正南方向。这架飞机是向下偏右方向, 与向下夹角为 $40^{\\circ}$。飞机要返航, 则此时以飞机为坐标原点, 机场位置在飞机的左偏上位置, 与正上方夹角为 $40^{\\circ}$, 即北偏西 $40^{\\circ}$, 距离应该是1000千米。据分析可得: 飞机返航时, 飞机要向北偏西 $40^{\\circ}$ 方向飞行 1000 千米。故答案为: C",
+        "knowledge_points": [
+          "根据方向、角度和距离描述路线图"
+        ],
+        "source": "24-25 六年级下·河南三门峡·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_204",
+        "original_id": "unit2_choice_204"
+      },
+      {
+        "id": "unit2_choice_205",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "车站在学校北偏东 $30^{\\circ}$ 方向 $40 \\mathrm{~km}$ 处, 用字母表示车站是 ( )。![](images/95d2f9f2d8705e7f5615302eb18a5f41fc4ee3017527bdc8a1b583c9c604e7c0.jpg)",
+        "options": [
+          "A. A",
+          "B. B",
+          "C. C",
+          "D. E"
+        ],
+        "answer": "C",
+        "solution": "以学校为观测点,以图上的'上北下南,左西右东'为准,图例表示图上1厘米相当于实际距离 $20\\mathrm{km}$,在学校的北偏东 $30^{\\circ}$ 方向上画 $40 \\div 20 = 2\\mathrm{cm}$ 长的线段,即是车站的位置。$40 \\div 20 = 2(\\mathrm{cm})$ 即车站与学校图上相距 $2 \\mathrm{~cm}$。车站在学校北偏东 $30^{\\circ}$ 方向 $40\\mathrm{km}$ 处,用字母表示车站是C。故答案为:C",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "25-26 六年级上·全国·课前预习",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_205",
+        "original_id": "unit2_choice_205"
+      },
+      {
+        "id": "unit2_choice_206",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "小明从家出发向东偏北 $25^{\\circ}$ 方向走500米到学校,放学回家时应向()方向走500米。",
+        "options": [
+          "A. 西偏南 $25^{\\circ}$",
+          "B. 东偏南 $25^{\\circ}$",
+          "C. 西偏南 $65^{\\circ}$"
+        ],
+        "answer": "A",
+        "solution": "![](images/24a5cda737fed39b3b8734c2ab6df511d03b42861cfd1787bab0196cba1ad7b3.jpg) 如图:学校和家的位置是相对的,此题方向相反,距离角度不变,由此求解。小明从家出发向东偏北 $25^{\\circ}$ 方向走 500 米到学校,放学回家时应向西偏南 $25^{\\circ}$ 方向走 500 米。故答案为:A",
+        "knowledge_points": [
+          "根据方向、角度和距离描述路线图"
+        ],
+        "source": "24-25四年级下·湖南岳阳·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_206",
+        "original_id": "unit2_choice_206"
+      },
+      {
+        "id": "unit2_choice_207",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "黄鹤楼是我国古代四大名楼之一, 位于湖北省武汉市。唐代诗人李白曾在黄鹤楼留下千古佳句'故人西辞黄鹤楼, 烟花三月下扬州'。李叔叔从家出发向北偏东 $60^{\\circ}$ 方向走了 1200 米到达黄鹤楼, 那么他从黄鹤楼回家应该往 ( ) 方向走 1200 米。",
+        "options": [
+          "A. 北偏东 $30^{\\circ}$",
+          "B. 北偏东 $60^{\\circ}$",
+          "C. 南偏西 $30^{\\circ}$",
+          "D. 南偏西 $60^{\\circ}$"
+        ],
+        "answer": "D",
+        "solution": "根据方向的相对性:方向相反,角度不变,距离不变;据此选择即可。$90^{\\circ} - 60^{\\circ} = 30^{\\circ}$,李叔叔从家出发向北偏东 $60^{\\circ}$ 方向走了1200米到达黄鹤楼,那么他从黄鹤楼回家应该往南偏西 $60^{\\circ}$ 方向或西偏南 $30^{\\circ}$ 方向走1200米。故答案为:D",
+        "knowledge_points": [
+          "根据方向、角度和距离描述路线图"
+        ],
+        "source": "2024·湖北襄阳·小升初真题",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_207",
+        "original_id": "unit2_choice_207"
+      },
+      {
+        "id": "unit2_choice_208",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "如图, 已知 $\\angle 1 = 36^{\\circ}$, 那么点 P 在点 O 的 ( ) 处。![](images/5640e877e20bd023beadc39086f9b75b5e57b9671b691883d800210d09153c9a.jpg)",
+        "options": [
+          "A. 南偏东 $54^{\\circ}$, $3 \\mathrm{~km}$",
+          "B. 南偏东 $36^{\\circ}$, $3 \\mathrm{~km}$",
+          "C. 北偏西 $54^{\\circ}$, $3 \\mathrm{~km}$",
+          "D. 北偏西 $36^{\\circ}$, 3 \\mathrm{~km}$"
+        ],
+        "answer": "B",
+        "solution": "以点 O 为观测点, 以图上的 '上北下南,左西右东' 为准, 图上的线段比例尺表示图上 1 个单位长度相当实际距离 $1 \\mathrm{~km}$; 从图中可知, 点 P 与点 O 相距 3 个单位长度, 则实际相距 $(1 \\times 3) \\mathrm{~km}$, 结合方向、角度和距离得出点 P 与点 O 的位置关系。$90^{\\circ} - 36^{\\circ} = 54^{\\circ}$ $1 \\times 3 = 3 (\\mathrm {k m})$ 图中,点P在点O的南偏东 $36^{\\circ}$ 或东偏南 $54^{\\circ}$, $3\\mathrm{km}$ 处。故答案为: B",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "2024·福建厦门·小升初真题",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_208",
+        "original_id": "unit2_choice_208"
+      },
+      {
+        "id": "unit2_choice_209",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "超市在学校北偏东 $30^{\\circ}$ 方向上, 学校在超市 ( ) 方向上。",
+        "options": [
+          "A. 东偏北 $30^{\\circ}$",
+          "B. 南偏西 $30^{\\circ}$",
+          "C. 西偏南 $30^{\\circ}$"
+        ],
+        "answer": "B",
+        "solution": "根据位置的相对性可知,它们的方向相反,角度相等,距离相等,据此解答。超市在学校北偏东 $30^{\\circ}$ 方向上,学校在超市南偏西 $30^{\\circ}$ 方向上。故答案为: B",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置、东、南、西、北方向"
+        ],
+        "source": "2024·河南周口·小升初真题",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_209",
+        "original_id": "unit2_choice_209"
+      },
+      {
+        "id": "unit2_choice_210",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "5月23日至24日, 中国人民解放军东部战区组织兵力, 在台湾岛周边开展'联合利剑 - 2024A'军事演习。如图中点A、B、C为主要演习区域, 三点构成一个边长约400千米的等边三角形,则B点在A点的()处。![](images/36085f151ad27824dde29209c51631c4493ae4d99e0df1befff0f4ed916ba050.jpg)",
+        "options": [
+          "A. 东偏北 $60^{\\circ}400$ 千米",
+          "B. 东偏北 $30^{\\circ} 400$ 千米",
+          "C. 西偏南 $30^{\\circ} 400$ 千米",
+          "D. 南偏西 $30^{\\circ} 400$ 千米"
+        ],
+        "answer": "D",
+        "solution": "依据题意结合图示可知, 利用等边三角形的特点可得 $\\angle B A C = 60^{\\circ}$, 利用平面图上方向规定: 上北下南左西右东, 先确定观测点、方向角度、距离, 依据题意结合图示去解答。由分析可知, $90^{\\circ} - 30^{\\circ} = 60^{\\circ}$,B点在A点的南偏西 $30^{\\circ}$ 或西偏南 $60^{\\circ}$ 方向400千米处。故答案为:D",
+        "knowledge_points": [
+          "等腰三角形和等边三角形的认识及特征、根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "2024·浙江宁波·小升初真题",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_210",
+        "original_id": "unit2_choice_210"
+      },
+      {
+        "id": "unit2_choice_211",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "如图, 奶奶家在超市的西偏北 $40^{\\circ}$ 方向 $300 \\mathrm{~m}$ 处, 奶奶从家去超市, 要向 ( )。![](images/1118eb56f8308e943758bf8cba6f73b43493ceb055fcc8d490f777a0d869721d.jpg)",
+        "options": [
+          "A.东偏南 $50^{\\circ}$ 方向走 $500\\mathrm{m}$",
+          "B.北偏西 $50^{\\circ}$ 方向走 $300\\mathrm{m}$",
+          "C. 南偏东 $40^{\\circ}$ 方向走 $500 \\mathrm{~m} \\mathrm{D}$",
+          "D. 东偏南 $40^{\\circ}$ 方向走 $300 \\mathrm{~m}$"
+        ],
+        "answer": "D",
+        "solution": "图中是按照'上北下南,左西右东'的方向绘制的,奶奶家在超市的西偏北 $40^{\\circ}$ 方向 $300 \\mathrm{~m}$ 处,根据方向的相对性,以奶奶家为观测点,超市在奶奶家东偏南 $40^{\\circ}$(或南偏东 $50^{\\circ}$)方向,距离不变。据此解答。由分析可知:奶奶家在超市的西偏北 $40^{\\circ}$ 方向 $300\\mathrm{m}$ 处,奶奶从家去超市,要向东偏南 $40^{\\circ}$ (或南偏东 $50^{\\circ}$)方向走 $300\\mathrm{m}$。故答案为: D",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24 五年级下·河南驻马店·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_211",
+        "original_id": "unit2_choice_211"
+      },
+      {
+        "id": "unit2_choice_212",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "北斗卫星导航系统是中国自主研发的卫星导航系统之一。根据我国北斗卫星定位显示: 台风中心位于M市南偏西 $35^{\\circ}$ 方向, 还可以说成台风中心位于M市 ( ) 方向。",
+        "options": [
+          "A. 北偏东 $35^{\\circ}$",
+          "B. 东偏北 $55^{\\circ}$",
+          "C. 东偏南 $35^{\\circ}$",
+          "D. 西偏南 $55^{\\circ}$"
+        ],
+        "answer": "D",
+        "solution": "由于南偏西 $35^{\\circ}$ 方向, 是从南往西偏 $30^{\\circ}$, 也可以把这两个方向反着说, 即西偏南的方向, 正南和正西之间夹角是 $90^{\\circ}$, 由于从西往南偏, 那么角度是 $90^{\\circ} - 35^{\\circ} = 55^{\\circ}$, 据此即可选择。由分析可知:北斗卫星导航系统是中国自主研发的卫星导航系统之一。根据我国北斗卫星定位显示:台风中心位于M市南偏西 $35^{\\circ}$ 方向,还可以说成台风中心位于M市西偏南 $55^{\\circ}$ 方向。故答案为: D",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25六年级上·重庆九龙坡·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_212",
+        "original_id": "unit2_choice_212"
+      },
+      {
+        "id": "unit2_choice_213",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "如图是学校和小芳家的位置方向图。下面的描述中正确的是()。![](images/6e3a46f62083bb6e7a8da4156ce52836af148dfb02e939567c8396193b0ab8ed.jpg)",
+        "options": [
+          "A. 学校在小芳家南偏东 $30^{\\circ}$ 方向 $200 \\mathrm{~m}$ 处",
+          "B. 小芳家在学校西偏北 $30^{\\circ}$ 方向 $200 \\mathrm{~m}$ 处",
+          "C. 学校在小芳家北偏西 $30^{\\circ}$ 方向 $200 \\mathrm{~m}$ 处"
+        ],
+        "answer": "A",
+        "solution": "将方向和距离结合起来描述位置时,要注意三个要素:一是观测点,二是方向,三是距离。看图可知, 学校在小芳家南偏东 $30^{\\circ}$ 方向 $200 \\mathrm{~m}$ 处。小芳家在学校北偏西 $30^{\\circ}$ 方向 $200 \\mathrm{~m}$ 处。故答案为:A",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·河南南阳·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_213",
+        "original_id": "unit2_choice_213"
+      },
+      {
+        "id": "unit2_choice_214",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "一艘轮船由 A 地向南偏西$45^{\\circ}$ 的方向行驶 40 海里到达 B 地, 再由 B 地向北偏西 $15^{\\circ}$ 的方向行驶 40 海里到达 C 地, 则 A、C 相距 () 海里。",
+        "options": [
+          "A. 30",
+          "B. 40",
+          "C. 60",
+          "D. 50"
+        ],
+        "answer": "B",
+        "solution": "根据'轮船由A地向南偏西 $45^{\\circ}$ 的方向行驶40海里到达B地,再由B地向北偏西 $15^{\\circ}$ 的方向行驶40海里到达C地'作图, 如图所示:![](images/3f7a44a7dc07dc8d9497ce16102b0e58e403cc43e6abe180bc7f96c103850cbd.jpg) 连接 AC, 由图可以判定三角形 ABC 是等边三角形, 从而可知 AC 的距离。根据题意作图,如图所示:![](images/152c4723c4cedfd870506b977f6b093903d842ba7925458cfc57288db0b7681a.jpg) 由图可知: $\\angle ABC = 45^{\\circ} + 15^{\\circ} = 60^{\\circ}$, $AB = BC = 40$ 海里, 所以三角形 $ABC$ 为等边三角形, 即 $AC = AB = 40$ 海里。故答案为: $B$。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置、东南、西南、东北、西北方向"
+        ],
+        "source": "2025 六年级·全国·竞赛",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_214",
+        "original_id": "unit2_choice_214"
+      },
+      {
+        "id": "unit2_choice_215",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "在地图上万达广场在虎英公园的北偏西 $50^{\\circ}$ 方向上, 那么虎英公园在万达广场的()方向上。",
+        "options": [
+          "A. 南偏东 $40^{\\circ}$",
+          "B. 北偏东 $40^{\\circ}$",
+          "C. 南偏东 $50^{\\circ}$",
+          "D. 北偏东 $50^{\\circ}$"
+        ],
+        "answer": "C",
+        "solution": "根据方向的相对性:方向相反,角度相同。即北相反是南,西相反是东,据此选择即可。据分析可知,在地图上万达广场在虎英公园的北偏西 $50^{\\circ}$ 方向上,那么虎英公园在万达广场的南偏东 $50^{\\circ}$ 方向上。故答案为:C",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "2024·广东东莞·小升初真题",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_215",
+        "original_id": "unit2_choice_215"
+      },
+      {
+        "id": "unit2_choice_216",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "以广场为观察点,学校在北偏西 $30^{\\circ}$ 的方向上,下图中正确的是()。",
+        "options": [
+          "A.![](images/c593e3a47cefbeba29dd8e9415d77799c0fbaa4b170c2293627bd1a9c71ac68f.jpg)",
+          "B.![](images/a1ead8735ca61f6f65ba5d9edec7a88981c1d0772c2ce7cb7469cad7b2496ebd.jpg)",
+          "C.![](images/0e92d0ad5876c2d2a19f8364107e5162b6bf75a4d9b498549909ec6c5f45e313.jpg)",
+          "D.![](images/1ada725115f47c89aed34372c8b06674cc7880c0b776545b467a2813fb612552.jpg)"
+        ],
+        "answer": "C",
+        "solution": "根据地图上方向的规定'上北下南,左西右东',以广场为观测点,确定出学校的位置,据此逐项分析,进行解答。A.![](images/1c5910f9ea8b7c8c651a82a03840a340b5752dd3539f8a8bcfb254540c004bb3.jpg),以广场为观测点,学校在东偏北 $30^{\\circ}$ 方向上,不符合题意。B.![](images/443d5e5e17bd0bd5f6512cbe56da20ab5171a4ba406d7f1e89f0c39a1e580abf.jpg),以广场为观测点,学校在北偏东 $30^{\\circ}$ 方向上,不符合题意。C.![](images/7f975506448e4a24df4bb5fb27a1c1ecf9a213724f05cd36024374ca4b250a2c.jpg),以广场为观测点,学校在北偏西 $30^{\\circ}$ 方向上,符合题意。D.![](images/17f411f1b8fcaff3689a09691330bc2ffcb0fb866fc37706b4e9f442ab42df34.jpg),以广场为观测点,学校在西偏南 $30^{\\circ}$ 方向上,不符合题意。以广场为观察点,学校在北偏西 $30^{\\circ}$ 的方向上,正确的是![](images/1e6531b196918efe887c70bd921e614e6aac1a3f065ffbd70aa4cb6224cf8f75.jpg)故答案为:C",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级下·广东韶关期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_216",
+        "original_id": "unit2_choice_216"
+      },
+      {
+        "id": "unit2_choice_217",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "聪聪从起点出发,先沿北偏东 $45^{\\circ}$ 方向走了 70 米,又沿东偏南 $45^{\\circ}$ 方向走了 70 米,他现在的位置在起点的 ( ) 方向。",
+        "options": [
+          "A. 正南",
+          "B. 正北",
+          "C. 正东",
+          "D. 正西"
+        ],
+        "answer": "C",
+        "solution": "按照'上北下南,左西右东'确定方向,结合用方向、角度和距离确定聪聪现在的位置。如图:![](images/746948c04822ed402c857d1042cad0ae81214ca1026ef9ef81f6f176c7772e93.jpg) 他现在的位置在起点的正东方向。故答案为:C",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·重庆江北·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_217",
+        "original_id": "unit2_choice_217"
+      },
+      {
+        "id": "unit2_choice_218",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "学校在小明家北偏东 $30^{\\circ}$ 方向上, 则小明家在学校的 ( ) 方向上。",
+        "options": [
+          "A. 东偏北 $30^{\\circ}$",
+          "B. 南偏西 $30^{\\circ}$",
+          "C. 西偏南 $30^{\\circ}$",
+          "D. 南偏西 $60^{\\circ}$"
+        ],
+        "answer": "B",
+        "solution": "根据方向的相对性:方向相反,角度相同。即北相反是南,东相反是西,据此选择即可。据分析可知, 学校在小明家北偏东 $30^{\\circ}$ 方向上, 则小明家在学校的南偏西 $30^{\\circ}$ 方向上(或西偏南 $60^{\\circ}$ 方向上)。故答案为: B",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置、根据方向、角度和距离描述路线图"
+        ],
+        "source": "23-24 六年级下·四川内江·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_218",
+        "original_id": "unit2_choice_218"
+      },
+      {
+        "id": "unit2_choice_219",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "小红家在学校东偏北 $60^{\\circ}$ 方向上, 距离 800 米。那么学校在小红家的 ( ) 方向上, 距离是 800 米。",
+        "options": [
+          "A. 东偏南 $60^{\\circ}$",
+          "B. 南偏东 $60^{\\circ}$",
+          "C. 西偏南 $30^{\\circ}$",
+          "D. 南偏西 $30^{\\circ}$"
+        ],
+        "answer": "D",
+        "solution": "小红家在学校东偏北 $60^{\\circ}$ 方向上, 距离 800 米, 是以学校为观测点; 那么学校在小红家的方向, 是以小红家为观测点;根据位置的相对性可知,观测点不同,方向相反,夹角的度数相同,距离相同;由此可知,东偏北 $60^{\\circ}$ 相对的是西偏南 $60^{\\circ}$, 西和南之间的夹角是 $90^{\\circ}, 90^{\\circ} - 60^{\\circ} = 30^{\\circ}$, 所以西偏南 $60^{\\circ}$ 方向, 还可以说成南偏西 $30^{\\circ}$ 方向。小红家在学校东偏北 $60^{\\circ}$ 方向上,距离800米。那么学校在小红家的西偏南 $60^{\\circ}$ (或南偏西 $30^{\\circ}$) 方向上,距离是800米。故答案为:D",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25六年级上·海南省直辖县级单位·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_219",
+        "original_id": "unit2_choice_219"
+      },
+      {
+        "id": "unit2_choice_220",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "李老师先向北步行到达超市, 再向南偏东方向步行到达邮局, 最后向东偏北方向步行到达图书馆。图 ( ) 可以表示李老师的步行路线。",
+        "options": [
+          "A.![](images/f447bf61fd9fde53933247c81f1761d23fa3e414a64d07a6b96abed492bee945.jpg)",
+          "B.![](images/856e42b52596dd1c90f12feb3776645ec97789c6237d7ed2bee558601318a5a2.jpg)",
+          "C.![](images/99cf2023537d7e13c2ca38211ba01ced3d1a22f1ae132fbb1ab45498b81fdaad.jpg)"
+        ],
+        "answer": "C",
+        "solution": "以李老师出发点为观测点,超市在正北方向;再以超市为观测点,邮局在南偏东(或东偏南)方向;最后以邮局为观测点,图书馆在东偏北(或北偏东)方向,据此可以确定李老师的步行路线图。A. 李老师先向北走到超市,再向南偏东(或东偏南)方向走到邮局,最后向南偏西(或西偏南)方向走到图书馆,不符合题意;B. 李老师先向东步行到超市, 不符合题意; C. 李老师先向北步行到超市, 再向南偏东(或东偏南)方向走到邮局,最后向东偏北(或北偏东)方向走到图书馆,符合题意; D. 李老师先向西偏北(或北偏西)方向走到超市,不符合题意。故答案为:C",
+        "knowledge_points": [
+          "根据方向、角度和距离画线路图"
+        ],
+        "source": "24-25六年级上·贵州六盘水·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_220",
+        "original_id": "unit2_choice_220"
+      },
+      {
+        "id": "unit2_choice_221",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "某市气象局发布消息:台风中心已到达A市东偏南 $30^{\\circ}$ 方向上,距离是300千米,正确表示台风中心位置的是()",
+        "options": [
+          "A.![](images/0e6c7982694ab1ae5168af9a7d8b059bcfe50b1cb71a1a9f102c695663ec9891.jpg)",
+          "B.![](images/7a5999f293519523b25b83d72eef42a2254920f2adb6c6146711df7531cf6714.jpg)",
+          "C.![](images/286ac6bc0e9d243211a8348c1a9ffa05ff2ccb62486652375c0a5c81197b3b5b.jpg)",
+          "D.![](images/6ac6a254ccfe4f25c784619e7abfedc9748893028c4c9800674ae3407b069ad1.jpg)"
+        ],
+        "answer": "A",
+        "solution": "地图方向是上北下南,左西右东,图上单位距离是 $100 \\mathrm{~km}$,再根据地图描述物体的位置逐项分析即可。A.由图可知,台风中心在A市的东偏南 $30^{\\circ}$ 方向上,距离是300千米,符合题意;B. 由图可知, 台风中心在 A 市的西偏南 $30^{\\circ}$ 方向上, 距离是 300 千米, 不符合题意; C. 由图可知, 台风中心在 A 市的东偏北 $30^{\\circ}$ 方向上, 距离是 300 千米, 不符合题意; D. 由图可知, 台风中心在 A 市的北偏西 $30^{\\circ}$ 方向上, 距离是 300 千米, 不符合题意。故答案为:A",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "22-23六年级上·陕西安康·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_221",
+        "original_id": "unit2_choice_221"
+      },
+      {
+        "id": "unit2_choice_222",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "观察如图的位置关系, 下面说法错误的是 ( )。![](images/0ce60c3a1c1055b505ba480595b2dc01ae4a0cbd65af89efdbdc2bf9976e83db.jpg)",
+        "options": [
+          "A.书店在体育馆东偏北 $35^{\\circ}$ 方向 $600 \\mathrm{~m}$ 处",
+          "B.状状家在体育馆西偏北 $30^{\\circ}$ 方向 $800 \\mathrm{~m}$ 处",
+          "C. 体育馆在状状家东偏南 $30^{\\circ}$ 方向 $800 \\mathrm{~m}$ 处",
+          "D. 体育馆在书店南偏西 $35^{\\circ}$ 方向 $600 \\mathrm{~m}$ 处"
+        ],
+        "answer": "D",
+        "solution": "图上 1 厘米表示实际距离 200 米, 用 200 分别乘两个地点之间的图上距离即可求出实际距离。以 '在' 字后面的场所为观测点, 根据地图 '上北下南,左西右东'的规定, 结合方向、角度和距离描述各场所的位置。根据位置的相对性可知, 描述两个物体之间的相对位置时,方向相反,角度相等,距离相等。据此解答。A. $200 \\times 3 = 600(\\mathrm{~m})$,以体育馆为观测点,书店在体育馆东偏北 $35^{\\circ}$ 方向 $600\\mathrm{m}$ 处,此选项说法正确; B. $200 \\times 4 = 800(\\mathrm{~m})$, 图中体育馆在状状家东偏南 $30^{\\circ}$ 方向 $800 \\mathrm{~m}$ 处, 根据位置的相对性, 以体育馆为观测点, 状状家在体育馆西偏北 $30^{\\circ}$ 方向 $800 \\mathrm{~m}$ 处, 此选项说法正确; C. 以状状家为观测点, 体育馆在状状家东偏南 $30^{\\circ}$ 方向 $800 \\mathrm{~m}$ 处, 此选项说法正确; D. 书店在体育馆东偏北 $35^{\\circ}$ 方向 $600 \\mathrm{~m}$ 处, 根据位置的相对性, 以书店为观测点, 体育馆在书店西偏南 $35^{\\circ}$ 方向 $600 \\mathrm{~m}$ 处, 此选项说法错误。故答案为:D",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·吉林四平·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_222",
+        "original_id": "unit2_choice_222"
+      },
+      {
+        "id": "unit2_choice_223",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "陈嘉庚纪念馆是厦门市首批研学实践教育基地,小明从家出发去纪念馆,他向西偏北 $36^{\\circ}$ 方向走了 $1200\\mathrm{m}$。他原路返回时应该向 ()方向走 $1200\\mathrm{m}$。",
+        "options": [
+          "A. 西偏北 $36^{\\circ}$",
+          "B. 北偏西 $36^{\\circ}$",
+          "C. 南偏东 $36^{\\circ}$",
+          "D. 东偏南 $36^{\\circ}$"
+        ],
+        "answer": "D",
+        "solution": "小明从家出发去纪念馆,是以小明家为观测点;原路返回时是以纪念馆为观测点;观测点不同,方向相反,夹角的度数相同,距离相同;西偏北 $36^{\\circ}$ 相对的是东偏南 $36^{\\circ}$, 东和南之间的夹角是 $90^{\\circ}$, $90^{\\circ} - 36^{\\circ} = 54^{\\circ}$, 所以东偏南 $36^{\\circ}$ 方向, 还可以说成南偏东 $54^{\\circ}$ 方向。小明从家出发去纪念馆,他向西偏北 $36^{\\circ}$ 方向走了 $1200 \\mathrm{~m}$。他原路返回时应该向东偏南 $36^{\\circ}$(或南偏东 $54^{\\circ}$)方向走 $1200 \\mathrm{~m}$。故答案为:D",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·福建厦门·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_223",
+        "original_id": "unit2_choice_223"
+      },
+      {
+        "id": "unit2_choice_224",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "某公园摆渡车的行驶路线是从正门向正东行驶 $2 \\mathrm{~km}$ 后, 再向西偏南 $60^{\\circ}$ 方向行驶 $1 \\mathrm{~km}$, 然后向正西方向行驶 $2 \\mathrm{~km}$, 最后向北偏东 $30^{\\circ}$ 方向行驶, 驿回正门, 正确的路线图是 ( )。",
+        "options": [
+          "A.![](images/fd15ea41c0d7efecd8a9c3a0c15a69a1f0433656c31032fa9588a55719002b8d.jpg)",
+          "B.![](images/a9a6c24aa190df0563c3fd2453fe858d8847661c5f46143149801e9e53b30e70.jpg)",
+          "C.![](images/c22e76f7cb5c9f4ebd2d2270d2fbbe442cb6ea2837fc21771df8691bdb9c6858.jpg)",
+          "D.![](images/92b4dabdc39669de0ea105d7ab8d8ea19904eb6899ff3062e97bf286c2102bfc.jpg)"
+        ],
+        "answer": "A",
+        "solution": "地图方向是上北下南,左西右东,一格代表 $1 \\mathrm{~km}$,根据方向、距离、角度画出正确的线路图,找到正确答案即可。A.从正门向正东行驶 $2\\mathrm{km}$ 后,再向西偏南 $60^{\\circ}$ 方向行驶 $1 \\mathrm{~km}$, 然后向正西方向行驶 $2 \\mathrm{~km}$, 最后向北偏东 $30^{\\circ}$ 方向行驶, 驿回正门; 符合题意;B. 从正门向正东行驶 $1 \\mathrm{~km}$ 后, 再向东偏南 $60^{\\circ}$ 方向行驶 $1 \\mathrm{~km}$, 然后向正西方向行驶 $2 \\mathrm{~km}$, 最后向北偏东 $30^{\\circ}$ 方向行驶, 驳回正门; 不符合题意; C. 从正门向正东行驶 $2 \\mathrm{~km}$ 后, 再向东偏南 $60^{\\circ}$ 方向行驶 $1 \\mathrm{~km}$, 然后向正西方向行驶 $2 \\mathrm{~km}$, 最后向西偏北 $60^{\\circ}$ 方向行驶, 驳回正门; 不符合题意; D. 从正门向正东行驶 $2 \\mathrm{~km}$ 后, 再向西偏南 $60^{\\circ}$ 方向行驶 $1 \\mathrm{~km}$, 然后向正西方向行驶 $1 \\mathrm{~km}$, 最后向西偏北 $60^{\\circ}$ 方向行驶, 驳回正门; 不符合题意;故答案为:A",
+        "knowledge_points": [
+          "根据方向、角度和距离描述路线图"
+        ],
+        "source": "23-24六年级上·贵州六盘水·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_224",
+        "original_id": "unit2_choice_224"
+      },
+      {
+        "id": "unit2_choice_225",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "海警船正在巡逻时,接收到一艘渔船发出的求助信息: 雷达显示,你们(海警船)在我们渔船的北偏东 $40^{\\circ}$ 方向5海里处。因我船故障无法航行,请求帮助。海警船立即开展救援行动,应该向( )方向航行5海里。",
+        "options": [
+          "A. 南偏西 $40^{\\circ}$",
+          "B. 南偏东 $40^{\\circ}$",
+          "C. 北偏东 $40^{\\circ}$",
+          "D. 北偏西 $40^{\\circ}$"
+        ],
+        "answer": "A",
+        "solution": "根据方向的相对性,北偏东对南偏西,角度和距离不变,进行分析。海警船立即开展救援行动,应该向南偏西 $40^{\\circ}$ 方向或西偏南 $50^{\\circ}$ 航行 5 海里。故答案为:A",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级下·浙江杭州·期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_225",
+        "original_id": "unit2_choice_225"
+      },
+      {
+        "id": "unit2_choice_226",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "下图中, 小明家在学校()方向。![](images/5f447781154e14c4c2199c40f98aa94f345d9a85b8847437248ac8d935fb3396.jpg)",
+        "options": [
+          "A. 南偏东 $60^{\\circ}$",
+          "B. 东偏南 $60^{\\circ}$",
+          "C. 西偏北 $60^{\\circ}$"
+        ],
+        "answer": "A",
+        "solution": "根据上北下南,左西右东,确定观测点是学校,先读正边方向,再读斜边偏向的方向,最后读角度,据此解答。据分析可知,小明家在学校南偏东 $60^{\\circ}$ 方向。故答案为:A",
+        "knowledge_points": [
+          "根据方向、角度和距离描述路线图"
+        ],
+        "source": "23-24 六年级上·全国·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_226",
+        "original_id": "unit2_choice_226"
+      },
+      {
+        "id": "unit2_choice_227",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "![](images/8272f942a1163da88cde6fead5612e39e41b92b824fea320d2bea7f738f27894.jpg) (1)以超市为观测点,邮局在()。A. 正南方 B. 正东方 C. 正西方 (2)以超市为观测点,学校在()。A. 东偏南 $30^{\\circ}$ B. 南偏东 $30^{\\circ}$ C. 西偏北 $30^{\\circ}$ (3)从福源小区出发,走()个站就到达学校了。A. 5 B. 6 C. 7",
+        "options": [],
+        "answer": "C A B ",
+        "solution": "(1)以超市为观测点,画出正北、正南方向,根据画出的图形即可判断邮局的位置;(2) 以超市为观测点,画出正北、正南方向,根据画出的图形即可判断学校的位置; (3) 根据图示可知从福源小区出发, 经过银行、邮局、电信局、粮油局、超市、学校6站点就到学校了。(1)![](images/02099135db6c23c6962a895c91a5e58a7a492c7c25aa1226867be68edd1252ab.jpg) 如图所示:以超市为观测点,邮局在正西方向上,故答案为:C (2)![](images/611d00db726328701683a5991fa29b5fd9d7df130de4a00cc8803a982d079f20.jpg) 如图所示:以超市为观测点,学校在东偏南 $30^{\\circ}$ 方向上,故答案为:A (3) 如图所示: 从从福源小区出发, 经过银行、邮局、电信局、粮油局、超市、学校6站点,故答案为: B",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_227",
+        "original_id": "unit2_choice_227"
+      },
+      {
+        "id": "unit2_choice_228",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "早上, 小明面向太阳站着,当他逆时针方向转 $90^{\\circ}$ 时就面向()方。",
+        "options": [
+          "A. 北",
+          "B. 西",
+          "C. 南"
+        ],
+        "answer": "A",
+        "solution": "早上,太阳从东方升起,因此小明早上面向东方,面向东方时,左边是北方,右边是南方,后面是西方,当他逆时针方向旋转 $90^{\\circ}$ 时,它面向北,依此选择。根据分析可知, 早上, 小明面向太阳站着, 当他逆时针方向转 $90^{\\circ}$ 时就面向北方。故答案为:A",
+        "knowledge_points": [
+          "旋转三要素及旋转图形、东、南、西、北方向"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_228",
+        "original_id": "unit2_choice_228"
+      },
+      {
+        "id": "unit2_choice_229",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "如图, 小明看小冬在 (), 小冬看小明在 ()。()![](images/7c49e9540e07e83972b8c09c6d8b7cfcbb36df10eace23c85eb9e4c7ea287f6e.jpg)",
+        "options": [
+          "A. 北偏东 $50^{\\circ}$ 的方向上; 西偏南 $40^{\\circ}$ 的方向上",
+          "B. 东偏北 $50^{\\circ}$ 的方向上; 西偏南 $40^{\\circ}$ 的方向上",
+          "C. 北偏东 $50^{\\circ}$ 的方向上; 东偏北 $50^{\\circ}$ 的方向上"
+        ],
+        "answer": "A",
+        "solution": "根据地图上方向的规定'上北下南,左西右东',以小明为观测点,确定出小冬的位置;再根据方向的相对性:它们的方向相反,角度相等,距离相等;据此解答。$90^{\\circ} - 50^{\\circ} = 40^{\\circ}$ $9 0 ^ {\\circ} - 4 0 ^ {\\circ} = 5 0 ^ {\\circ}$ 小明看小冬在北偏东 $50^{\\circ}$(东偏北 $40^{\\circ}$)方向上。小冬看小明在西偏南 $40^{\\circ}$(或南偏西 $50^{\\circ}$)方向上。小明看小冬在北偏东 $50^{\\circ}$ (东偏北 $40^{\\circ}$) 方向上, 小冬看小明在西偏南 $40^{\\circ}$ (或南偏西 $50^{\\circ}$) 方向上。故答案为: A",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_229",
+        "original_id": "unit2_choice_229"
+      },
+      {
+        "id": "unit2_choice_230",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "北偏西 $35^{\\circ}$, 还可以说成 ( )。",
+        "options": [
+          "A. 南偏西 $35^{\\circ}$",
+          "B. 西偏北 $35^{\\circ}$",
+          "C. 西偏北 $55^{\\circ}$"
+        ],
+        "answer": "C",
+        "solution": "![](images/5664d2735e78f2888b70495c4945492977b7d8f99ecac36dd5b5b4a02b943009.jpg) 如图,根据方位角的定义画图以及位置的相对性可知,北偏西 $35^{\\circ}$ 也可以说成西偏北 $55^{\\circ}$。因为图中表示北偏西 $35^{\\circ}$,所以图中也表示西偏北 $90^{\\circ} - 35^{\\circ} = 55^{\\circ}$ 故答案为: C",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_230",
+        "original_id": "unit2_choice_230"
+      },
+      {
+        "id": "unit2_choice_231",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "确定物体的位置,需要 ( ) 条件。",
+        "options": [
+          "A. 方向一个",
+          "B. 距离一个",
+          "C. 方向和距离两个"
+        ],
+        "answer": "C",
+        "solution": "要确定物体的位置必须用角度描述出物体在观测点的什么方向,还有这个物体与观测点之间的距离,据此解答。根据分析可知,确定物体位置,需要方向和距离两个条件。故答案为:C",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_231",
+        "original_id": "unit2_choice_231"
+      },
+      {
+        "id": "unit2_choice_232",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "小明从家出发向北偏东 $30^{\\circ}$ 方向走了 $200 \\mathrm{~m}$ 到达超市, 购物后原路返回, 他需要向 ( ) 方向走 $200 \\mathrm{~m}$ 才能回到家。",
+        "options": [
+          "A. 南偏西 $60^{\\circ}$",
+          "B. 南偏西 $30^{\\circ}$",
+          "C. 西偏南 $30^{\\circ}$",
+          "D. 东偏北 $60^{\\circ}$"
+        ],
+        "answer": "B",
+        "solution": "返回时,方向相反、角度和距离不变。东和西相对,南和北相对,去时向北偏东 $30^{\\circ}$ 方向走了 $200\\mathrm{m}$,返回时就向南偏西 $30^{\\circ}$ 方向走 $200\\mathrm{m}$。据此解答。由分析可得: 小明从家出发向北偏东 $30^{\\circ}$ 方向走了 $200 \\mathrm{~m}$ 到达超市, 购物后原路返回, 他需要向南偏西 $30^{\\circ}$ 方向走 $200 \\mathrm{~m}$。故答案为:B",
+        "knowledge_points": [
+          "根据方向、角度和距离描述路线图"
+        ],
+        "source": "23-24六年级下·湖南株洲期末",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_232",
+        "original_id": "unit2_choice_232"
+      },
+      {
+        "id": "unit2_choice_233",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "学校在电影院的南偏西 $50^{\\circ}$ 方向上,电影院在学校的()方向上。",
+        "options": [
+          "A. 北偏西 $40^{\\circ}$",
+          "B. 北偏东 $50^{\\circ}$",
+          "C. 北偏东 $40^{\\circ}$",
+          "D. 西偏南 $40^{\\circ}$"
+        ],
+        "answer": "B",
+        "solution": "学校在电影院的南偏西 $50^{\\circ}$ 方向上,是以电影院为观测点;电影院在学校的方向是以学校为观测点;根据位置的相对性可知,观测点不同,方向相反,夹角的度数相同,距离相同。由此可知,南偏西 $50^{\\circ}$ 相对的是北偏东 $50^{\\circ}$, 东和北之间的夹角是 $90^{\\circ}, 90^{\\circ} - 50^{\\circ} = 40^{\\circ}$, 所以北偏东 $50^{\\circ}$ 方向, 还可以说成东偏北 $40^{\\circ}$ 方向。学校在电影院的南偏西 $50^{\\circ}$ 方向上,电影院在学校的北偏东 $50^{\\circ}$ (或东偏北 $40^{\\circ}$) 方向上。故答案为:B",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级下·江苏盐城·期中",
+        "create_time": "2024-10-07 15:30:00",
+        "merged_id": "unit2_233",
+        "original_id": "unit2_choice_233"
+      },
+      {
+        "id": "unit2_choice_234",
+        "question": "小莹的妈妈步行健身, 这天她从家A出发, 先向东偏北  $45^{\\circ}$  走,再转向北偏西  $30^{\\circ}$  走, 然后转向西偏南  $30^{\\circ}$  走, 最后转向南一直走到家。那么下面四条线路中, 小莹妈妈走的路线是 ( )。\n\n![](images/0ae7e0150fb782d7ae63671dea86e54e2b73dc3a527227c873f24eddd63f8631.jpg)",
+        "options": [
+          "A. A  $\\rightarrow$  B  $\\rightarrow$  C  $\\rightarrow$  E  $\\rightarrow$  A",
+          "B. A  $\\rightarrow$  F  $\\rightarrow$  C  $\\rightarrow$  E  $\\rightarrow$  A",
+          "C. A  $\\rightarrow$  B  $\\rightarrow$  C  $\\rightarrow$  D  $\\rightarrow$  A",
+          "D. A  $\\rightarrow$  B  $\\rightarrow$  C  $\\rightarrow$  G  $\\rightarrow$  A"
+        ],
+        "answer": "A",
+        "source": "23-24 五年级下·湖北省直辖县级单位·期末",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离描述路线图"
+        ],
+        "solution": "由图可知，标出了北方的位置，指向上面的是北方，直接根据'上北下南,左西右东'的方法确定小莹妈妈所走的路线。\n\n由分析可得：下面四条线路中，小莹妈妈走的路线是  $\\mathrm{A}\\to \\mathrm{B}\\to \\mathrm{C}\\to \\mathrm{E}\\to \\mathrm{A}$  。\n\n故答案为：A",
+        "merged_id": "unit2_234",
+        "original_id": "unit2_choice_234",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "选择题"
+      },
+      {
+        "id": "unit2_choice_235",
+        "question": "如图, 有一个等边三角形花圃, 边长 50 米。小明想要从点 A 走到点 B, 可以 ()。\n\n![](images/7a1a6335de012cc636f61a231b70b18503c4a72701736b71b61b8045b9959d94.jpg)",
+        "options": [
+          "A. 向北偏东  $30^{\\circ}$  方向走50米",
+          "B. 向北偏东  $60^{\\circ}$  方向走50米",
+          "C. 向南偏西  $30^{\\circ}$  方向走 50 米",
+          "D. 向南偏西  $60^{\\circ}$  方向走 50 米"
+        ],
+        "answer": "A",
+        "source": "23-24 六年级下·江苏盐城·期末",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "solution": "将方向和距离结合起来描述位置时，要注意三个要素：一是观测点，二是方向，三是距离。等边三角形的三个内角都是  $60^{\\circ}$ ，东和北之间的夹角是  $90^{\\circ}$ ，东偏北也可以说成北偏东，角度  $= 90^{\\circ}$  一东偏北的角度。\n\n$90^{\\circ} - 60^{\\circ} = 30^{\\circ}$\n\n小明想要从点A走到点B, 可以向东偏北  $60^{\\circ}$  或北偏东  $30^{\\circ}$  方向走50米。\n\n故答案为：A",
+        "merged_id": "unit2_235",
+        "original_id": "unit2_choice_235",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "选择题"
+      },
+      {
+        "id": "unit2_choice_236",
+        "question": "如图中, 小红的位置在小明的 ( )。\n\n![](images/1eff8c6b2e630ced8989e623cdbd1357e4a1a88a1b349da467dbba043760df05.jpg)",
+        "options": [
+          "A. 北偏东  $40^{\\circ} 500$  米处",
+          "B. 南偏西  $50^{\\circ} 500$  米处",
+          "C. 北偏东  $50^{\\circ} 500$  米处"
+        ],
+        "answer": "B",
+        "source": "23-24 六年级下·山东滨州·期末",
+        "difficulty": "容易",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "solution": "因为位置具有相对性, 所以当小明在小红的某个方向和距离处时, 小红就在小明相反的方向, 角度相同, 距离不变。\n\n由图可知, 小明的位置在小红的东偏北  $40^{\\circ} 500$  米处, 根据位置的相对性, 小红的位置在小明的西偏南  $40^{\\circ} 500$  米处。小红的位置在小明的西偏南  $40^{\\circ} 500$  米处还可以描述为小红的位置在小明的南偏西  $50^{\\circ} 500$  米处。\n\n故答案为：B",
+        "merged_id": "unit2_236",
+        "original_id": "unit2_choice_236",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "选择题"
+      },
+      {
+        "id": "unit2_choice_237",
+        "question": "文具店、书店和玩具店依次坐落在一条东西走向的大街上, 文具店在书店西边  $20 \\mathrm{~m}$  处, 玩具店位于书店东边  $100 \\mathrm{~m}$  处, 小明从书店沿街向东走了  $40 \\mathrm{~m}$ , 接着又向西走了  $60 \\mathrm{~m}$ ,此时小明的位置在 ( )。",
+        "options": [
+          "A. 文具店",
+          "B. 玩具店",
+          "C. 文具店西  $40 \\mathrm{~m}$  处",
+          "D. 玩具店西  $60 \\mathrm{~m}$  处"
+        ],
+        "answer": "A",
+        "source": "2014 六年级·全国·课后作业",
+        "difficulty": "容易",
+        "knowledge_points": [
+          "根据方向和距离描述简单的路线"
+        ],
+        "solution": "小明从书店先向东走  $40\\mathrm{m}$  ，此时没有到达玩具店，然后又向西走  $60\\mathrm{m}$  ，这  $60\\mathrm{m}$  包括  $40\\mathrm{m}$  和  $20\\mathrm{m}$  。最初的向西走  $40\\mathrm{m}$  ，回到书店，再向西走  $20\\mathrm{m}$  到达文具店。\n\n小明从书店沿街向东走了  $40 \\mathrm{~m}$ , 接着又向西走了  $60 \\mathrm{~m}$ , 此时小明的位置在文具店。\n\n故答案为：A",
+        "merged_id": "unit2_237",
+        "original_id": "unit2_choice_237",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "选择题"
+      },
+      {
+        "id": "unit2_choice_238",
+        "question": "看图, 填一填。\n\n![](images/c7fb061cb50049f676831fe8f8cae68cbc8db1e155bffdd9179d7a55b3a88af8.jpg)\n\n小军去书店买书，他从家里出发，先沿着西偏北  $30^{\\circ}$  方向走了（）米，到超市买了一瓶矿泉水，然后沿着（）方向走了 300 米，到达书店。",
+        "options": [],
+        "answer": "200 正西/西",
+        "source": "24-25 六年级下·宁夏吴忠·期末",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离描述路线图"
+        ],
+        "solution": "（1）以小军家为观测点，根据'上北下南,左西右东'的方位辨别方法确定位置，再根据图上1厘米表示实际的100米确定出实际距离；\n\n(2) 以超市为观测点, 根据 '上北下南,左西右东'的方位辨别方法确定位置, 再根据图上 1 厘米表示实际的 100 米确定出距离。\n\n$2 \\times 100 = 200$  （米）\n\n小军去书店买书，他从家里出发，先沿着西偏北  $30^{\\circ}$  方向走了 200 米，到超市买了一瓶矿泉水，然后沿着正西方向走了 300 米，到达书店。",
+        "merged_id": "unit2_238",
+        "original_id": "unit2_choice_238",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "填空题"
+      },
+      {
+        "id": "unit2_choice_239",
+        "question": "看图填空。\n\n![](images/2392fc2b30f161cd3c2059ba6d52788712835eeca84d6394e0dfcdcb4c59d9e3.jpg)\n\n放学后，小红从学校出发，先向()（）°方向走() 米到达少年宫，再向()（）°方向走()米到达书店，她在书店买了一本故事书后，最后向()（）方向走()米回到家。",
+        "options": [],
+        "answer": "北偏西 40 300 南偏西 50 300 北偏西 30 200",
+        "source": "25-26 六年级上·海南海口·单元测试",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离描述路线图"
+        ],
+        "solution": "用方向和距离结合来描述路线时，要注意三个要素：一是观测点（即参照物），二是方向，三是距离。\n\n放学后，小红从学校出发，先向北偏西  $40^{\\circ}$  或西偏北  $50^{\\circ}$  方向走 300 米到达少年宫，再向南偏西  $50^{\\circ}$  或西偏南  $40^{\\circ}$  方向走 300 米到达书店，她在书店买了一本故事书后，最后向北偏西  $30^{\\circ}$  或西偏北  $60^{\\circ}$  方向走 200 米回到家。",
+        "merged_id": "unit2_239",
+        "original_id": "unit2_choice_239",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "填空题"
+      },
+      {
+        "id": "unit2_choice_240",
+        "question": "看图填空。\n\n![](images/6f1e3d72a42157fd8ad0f17d49e4aed46c7c832c0eb34c572f7f8d78fc776be6.jpg)\n\n(1)猴山在正门()偏()()°方向上，距离是()m。  \n(2)狮虎山在正门()偏()（）°方向上，距离是()m。  \n(3)大象馆在正门()偏()()°方向上，距离是()m。",
+        "options": [],
+        "answer": "(1)北 东  $20^{\\circ}$ 400\n\n(2)北 西 30 400  \n(3) 西 北 15 300",
+        "source": "25-26 六年级上·海南海口·单元测试",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "solution": "将方向和距离结合起来描述位置时，要注意三个要素：一是观测点，二是方向，三是距离。看图可知，图上几厘米就是几百米。\n\n（1）猴山在正门北偏东  $20^{\\circ}$  或东偏北  $70^{\\circ}$  方向上，距离是  $400\\mathrm{m}$  。\n\n(2) 狮虎山在正门北偏西  $30^{\\circ}$  或西偏北  $60^{\\circ}$  方向上, 距离是  $400 \\mathrm{~m}$  。  \n(3) 大象馆在正门西偏北  $15^{\\circ}$  或北偏西  $75^{\\circ}$  方向上, 距离是  $300 \\mathrm{~m}$  。",
+        "merged_id": "unit2_240",
+        "original_id": "unit2_choice_240",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "填空题"
+      },
+      {
+        "id": "unit2_choice_241",
+        "question": "沧浪亭、狮子林分别是苏州宋代和元代的园林代表, 下面是奇妙制作的沧浪亭一狮子林一金鸡湖一日游玩攻略图。\n\n![](images/1fdc68ca647a1e3f4ec893821c3e486fcc58ae2d8ca9b53b4de3699356b84722.jpg)\n\n(1)根据攻略图描述出具体路线：妙妙一家驾车从酒店出发向()方向行驶()km到达超市，继续向()方向行驶()km到达沧浪亭，再向()方向行驶()km到达网师园，接着向()方向行驶()km到狮子林，最后向()方向行驶()km到达金鸡湖景区。\n\n(2) (1) 中的游玩路线全程长()km。  \n(3) 游玩结束后, 妙妙一家打算驾车以  $40 \\mathrm{~km} / \\mathrm{h}$  的速度按最短路线从金鸡湖景区返回酒店, 需要 ( ) 分钟才能回到酒店。",
+        "options": [],
+        "answer": "(1)南偏东  $20^{\\circ}$ 1  正东/东 3 北偏东  $10^{\\circ}$ 1.2 北偏东  $25^{\\circ}$ 4.8  南偏东  $70^{\\circ}$ 10\n\n(2)20  \n(3)24",
+        "source": "24-25六年级上·全国·单元测试",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "基础行程问题、时、分、秒之间的换算与比较、根据方向、角度和距离描述路线图、东、南、西、北方向"
+        ],
+        "solution": "（1）先确定观测点，再根据上北下南，左西右东确定方向和角度，观察线段确定距离。其中观察可知狮子林在网师园的北偏东  $25^{\\circ}$  方向上，根据方向的相对性可知，网师园在狮子林的南偏西  $25^{\\circ}$  的方向上，再用  $(95^{\\circ} - 25^{\\circ})$  可求出金鸡湖景区在狮子林的什么方向上；\n\n(2) 用加法计算各路线的距离即可得解;  \n(3) 先确定最短路线为金鸡湖景区一沧浪亭一超市一酒店,算出最短距离,再根据'时间 = 路程 ÷ 速度'求得结果单位转化为分即可。\n\n(1)  $95^{\\circ} - 25^{\\circ} = 70^{\\circ}$\n\n妙妙一家驾车从酒店出发向南偏东  $20^{\\circ}$  （或东偏南  $70^{\\circ}$  ）方向行驶  $1\\mathrm{km}$  到达超市，继续向正东方向行驶  $3\\mathrm{km}$  到达沧浪亭，再向北偏东  $10^{\\circ}$  （或东偏北  $80^{\\circ}$  ）方向行驶  $1.2\\mathrm{km}$  到达网师园，接着向北偏东  $25^{\\circ}$  （或东偏北  $65^{\\circ}$  ）方向行驶  $4.8\\mathrm{km}$  到狮子林，最后向南偏东  $70^{\\circ}$  （或东偏南  $20^{\\circ}$  ）方向行驶 $10\\mathrm{km}$  到达金鸡湖景区;\n\n(2)  $1 + 3 + 1.2 + 4.8 + 10 = 20(\\mathrm{km})$  \n(1) 中的游玩路线全程长  $20 \\mathrm{~km}$ ;  \n(3) 最短路线为金鸡湖景区一沧浪亭一超市一酒店，全程长  $12 + 3 + 1 = 16(\\mathrm{km})$\n\n$16 \\div 40 = 0.4 (\\text {时}) = 24 (\\text {分})$\n\n游玩结束后，妙妙一家打算驾车以  $40 \\mathrm{~km} / \\mathrm{h}$  的速度按最短路线从金鸡湖景区返回酒店，需要24分钟才能回到酒店。",
+        "merged_id": "unit2_241",
+        "original_id": "unit2_choice_241",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "填空题"
+      },
+      {
+        "id": "unit2_choice_242",
+        "question": "某特种部队在执行敌后救援任务时, 使用军用无人侦察机（机器狗）探测友军位置。已知侦察机与救援队的具体位置如图。\n\n![](images/9abcdff2aff3c7f23189f574925a88c0fb4dc5c32ee5349c4cfd78b974b570d1.jpg)\n\n(1)无人侦察机位于救援队的()偏()（）方向上，距离是()米。  \n(2)侦察机探测到友军信号在其南偏西  $30^{\\circ}$  方向 300 米处。图中标记点 A、B、C、D 中, 正确位置是 ( )。",
+        "options": [],
+        "answer": "(1)北 西  $70^{\\circ}$ 600\n\n(2)D",
+        "source": "24-25六年级上·全国·单元测试",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置、根据方向、角度和距离描述路线图"
+        ],
+        "solution": "（1）根据图中可知，上北下南，左西右东，图上1厘米，相当于实际150米；无人侦察机位于救援队左上角，也就是北偏西方向，结合图中度数  $90^{\\circ} - 20^{\\circ}$  得到方向，距离是4个150米，据此解答；  \n(2) 侦察机探测到友军信号在其南偏西  $30^{\\circ}$  方向, 也就是左下角, 从南向西偏向  $30^{\\circ}$  的方向, 符合条件的是 C 和 D, 距离是  $300 \\div 150 = 2$  厘米, 符合条件的只有 D; 据此解答。\n\n（1）根据分析，  $90^{\\circ} - 20^{\\circ} = 70^{\\circ}$  ，  $4\\times 150 = 600$  （米）\n\n无人侦察机位于救援队的北偏西  $70^{\\circ}$  方向上, 距离是 600 米。(答案不唯一)\n\n(2) 根据分析,  $300 \\div 150 = 2$  (厘米)\n\n侦察机探测到友军信号在其南偏西  $30^{\\circ}$  方向 300 米处如下图:\n\n![](images/1744780639d82df23f244af5002f444ddef24fe61538bccae97b8113e40a8a97.jpg)\n\n所以，图中标记点A、B、C、D中，正确位置是D。",
+        "merged_id": "unit2_242",
+        "original_id": "unit2_choice_242",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "填空题"
+      },
+      {
+        "id": "unit2_choice_243",
+        "question": "飞机从某机场向南偏东  $30^{\\circ}$  方向飞行了  $1200 \\mathrm{~km}$ , 返回时飞机要向 ( )方向飞行 ( ) km。",
+        "options": [],
+        "answer": "北偏西  $30^{\\circ}$ 1200",
+        "source": "2024·福建福州·小升初真题",
+        "difficulty": "容易",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "solution": "地图的方位是上北下南左西右东。位置相对方向相反。南对北，东对西，因此南偏东对北偏西，度数不变。\n\n飞机从某机场向南偏东  $30^{\\circ}$  方向飞行了  $1200 \\mathrm{~km}$ , 返回时飞机要向北偏西  $30^{\\circ}$  方向（答案不唯一）飞行  $1200 \\mathrm{~km}$  。",
+        "merged_id": "unit2_243",
+        "original_id": "unit2_choice_243",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "填空题"
+      },
+      {
+        "id": "unit2_choice_244",
+        "question": "小松鼠先沿()方向走()m 可以吃到松果, 再沿()偏()()°方向走()m 还可以吃到松果。\n\n![](images/0e10ec1cd0d24a735e1a1fa86e70a69dd53669e0f277e773f047d4a28d4efa1b.jpg)",
+        "options": [],
+        "answer": "正东 80 北 东 60 120",
+        "source": "25-26 六年级上·全国·课后作业",
+        "difficulty": "容易",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置、根据方向、角度和距离描述路线图、东、南、西、北方向、东南、西南、东北、西北方向"
+        ],
+        "solution": "通常地图是按照'上北下南,左西右东'来绘制的，从图中可以看到，小松鼠到第一个松果是沿着水平方向走的，图上标了  $80 \\mathrm{~m}$  。水平方向在地图上对应的是东西方向，这里小松鼠是向右走，也就是正东方向。所以，小松鼠先沿正东方向走  $80 \\mathrm{~m}$  可以吃到第一个松果。从第一个松果到第二个松果，图上标了  $120 \\mathrm{~m}$  。图中还有一个  $60^{\\circ}$  的角，这个角是相对于正北方向向东测量的，所以方向是北偏东  $60^{\\circ}$  。因此，小松鼠再沿北偏东  $60^{\\circ}$  方向走  $120 \\mathrm{~m}$  还可以吃到第二个松果。\n\n小松鼠先沿着正东方向走  $80\\mathrm{m}$  可以吃到松果，再沿着北偏东  $60^{\\circ}$  方向走  $120\\mathrm{m}$  还可以吃到松果。",
+        "merged_id": "unit2_244",
+        "original_id": "unit2_choice_244",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "填空题"
+      },
+      {
+        "id": "unit2_choice_245",
+        "question": "汕头南澳岛在小公园东偏北  $30^{\\circ}$  的方向上, 距离小公园  $44 \\mathrm{~km}$ , 那么小公园在南澳岛()偏()  $30^{\\circ}$  方向上, 距离南澳岛() km。",
+        "options": [],
+        "answer": "西 南 44",
+        "source": "2024·广东汕头·小升初真题",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "solution": "分析题目，根据位置的相对性可知：东对西，北对南，且角度相同，距离不变，据此可知东偏北  $30^{\\circ}$  相对的是西偏南  $30^{\\circ}$ ，距离还是  $44 \\mathrm{~km}$ ，据此解答。\n\n以南澳岛为观测点，小公园在南澳岛的西偏南  $30^{\\circ}$  方向，且小公园距离南澳岛  $44\\mathrm{km}$  。\n\n汕头南澳岛在小公园东偏北  $30^{\\circ}$  的方向上，距离小公园  $44 \\mathrm{~km}$ ，那么小公园在南澳岛西偏南  $30^{\\circ}$  方向上，距离南澳岛  $44 \\mathrm{~km}$ 。",
+        "merged_id": "unit2_245",
+        "original_id": "unit2_choice_245",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "填空题"
+      },
+      {
+        "id": "unit2_choice_246",
+        "question": "红红在丫丫的北偏东  $40^{\\circ}$  方向, 那么丫丫在红红的()方向。",
+        "options": [],
+        "answer": "南偏西  $40^{\\circ}$",
+        "source": "24-25 六年级上·河北石家庄·期末",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "solution": "方向和距离两个条件才能确定物体的位置，根据位置的相对性，可知两处位置观测点不同，它们的方向相反，角度相等，据此解答。\n\n红红在丫丫的北偏东  $40^{\\circ}$  方向，那么丫丫在红红的南偏西  $40^{\\circ}$  方向。",
+        "merged_id": "unit2_246",
+        "original_id": "unit2_choice_246",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "填空题"
+      },
+      {
+        "id": "unit2_choice_247",
+        "question": "妈妈从单位出发去菜市场，沿西偏北  $30^{\\circ}$  方向前行800米到达菜市场，买菜结束后妈妈沿东偏南  $30^{\\circ}$  方向1500米到达家里。家在单位的 偏  $30^{\\circ}$  方向距离 米处。",
+        "options": [],
+        "answer": "东 南 700",
+        "source": "24-25 六年级上·重庆·期末",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "solution": "两地的相对位置，方向相反，角度和距离不变。妈妈从单位出发到菜市场，需要沿着西偏北  $30^{\\circ}$  方向行800米，那么妈妈买完菜后，向东偏南  $30^{\\circ}$  方向行走800米可回到单位。根据题意可知，妈妈继续向东偏南  $30^{\\circ}$  方向行(1500-800)米可回到家中。这说明，家在单位的东偏南  $30^{\\circ}$  方向，距离是（1500-800）米。\n\n$1500 - 800 = 700$  （米）所以家在单位的东偏南  $30^{\\circ}$  方向距离700米处。",
+        "merged_id": "unit2_247",
+        "original_id": "unit2_choice_247",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "填空题"
+      },
+      {
+        "id": "unit2_choice_248",
+        "question": "小美从家到学校要沿西偏南  $30^{\\circ}$  方向走  $1.2 \\mathrm{~km}$ , 她放学回家时, 应沿 ( )方向走 ( ) km。",
+        "options": [],
+        "answer": "东偏北  $30^{\\circ}$ 1.2",
+        "source": "24-25 六年级上·河南南阳·期末",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置、根据方向、角度和距离描述路线图"
+        ],
+        "solution": "根据方向的相对性：方向相反，角度相同，距离相同。即西相反是东，南相反是北，据此解答即可。\n\n据分析可知, 小美从家到学校要沿西偏南  $30^{\\circ}$  方向走  $1.2 \\mathrm{~km}$ , 她放学回家时, 应沿东偏北  $30^{\\circ}$  方向走  $1.2 \\mathrm{~km}$  。",
+        "merged_id": "unit2_248",
+        "original_id": "unit2_choice_248",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "填空题"
+      },
+      {
+        "id": "unit2_choice_249",
+        "question": "李琳上学从家往东偏北 35 度方向走 500 米到学校, 那么她放学应该往()（）度方向走()米到家。",
+        "options": [],
+        "answer": "西 偏南 35 500",
+        "source": "24-25 六年级上·江西宜春·期末",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离描述路线图"
+        ],
+        "solution": "根据两个物体的位置相对性，分别以它们为观测点，看到对方的方向相反，角度和距离相等，据此解答。\n\n李琳上学从家往东偏北35度方向走500米到学校,那么她放学应该往西偏南35度方向走500米到家。",
+        "merged_id": "unit2_249",
+        "original_id": "unit2_choice_249",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "填空题"
+      },
+      {
+        "id": "unit2_choice_250",
+        "question": "红红家在学校北偏西  $50^{\\circ}$  方向 600 米处。这天上学, 她忘记了带数学课本,需要原路返回拿书, 她该往()偏()°方向走()米回家。",
+        "options": [],
+        "answer": "南 东 50 600",
+        "source": "24-25 六年级上·湖南永州·期末",
+        "difficulty": "容易",
+        "knowledge_points": [
+          "根据方向、角度和距离描述路线图"
+        ],
+        "solution": "根据两个物体的位置相对性，分别以它们为观测点，看到对方的方向相反，角度和距离相等，据此解答。\n\n根据两个物体的位置相对性可知：红红家在学校北偏西  $50^{\\circ}$  方向 600 米处。这天上学，她忘记了带数学课本，需要原路返回拿书，她该往南偏东  $50^{\\circ}$  （或者东偏南  $40^{\\circ}$ ）方向走 600 米回家。",
+        "merged_id": "unit2_250",
+        "original_id": "unit2_choice_250",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "填空题"
+      },
+      {
+        "id": "unit2_choice_251",
+        "question": "小玲从家去书店买书后, 回家的路线是: 先向()偏()走()米到达商场, 然后向()偏()走()米返回家中。\n\n![](images/83d1c0a16e9a3d96e89afe04f02133c7a92aee18a494e8ef71e348ef6fe25028.jpg)",
+        "options": [],
+        "answer": "东 北  $45^{\\circ}$ 400  东 南  $30^{\\circ}$ 1000",
+        "source": "24-25 六年级上·内蒙古呼伦贝尔·期末",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离描述路线图"
+        ],
+        "solution": "图中一小段代表200米，以书店为观测点，商场在东偏北方向，角度是  $(90^{\\circ} - 45^{\\circ})$  ，或北偏东  $45^{\\circ}$  方向，距离书店有2小段长度；以商场为观测点，小玲家在东偏南  $30^{\\circ}$  方向，或南偏东方向，角度是  $(90^{\\circ} - 30^{\\circ})$  ，距离小玲家有5小段长度；据此解答。\n\n$90^{\\circ} - 45^{\\circ} = 45^{\\circ}$ $2\\times 200 = 400(米)$ $90^{\\circ} - 30^{\\circ} = 60^{\\circ}$ $5\\times 200 = 1000(米)$\n\n因此小玲从家去书店买书后，回家的路线是：先向东偏北  $45^{\\circ}$  （或北偏东  $45^{\\circ}$  ）方向走 400 米到达商场，然后向东偏南  $30^{\\circ}$  （或南偏东  $60^{\\circ}$  ）方向走 1000 米返回家中。",
+        "merged_id": "unit2_251",
+        "original_id": "unit2_choice_251",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "填空题"
+      },
+      {
+        "id": "unit2_choice_252",
+        "question": "![](images/d86927305c6d2acf3933f1c2f732e06c90e4df8689ed19e3ccc49b3ecfd69667.jpg)\n\n(1)小刚家在动物园()方向上，距离是()m。  \n(2)小红家在动物园()偏()()°方向上，距离是()m。  \n(3)小雪家在动物园()偏()()°方向上，距离是()m。",
+        "options": [],
+        "answer": "(1) 正西 800\n\n(2)北 东 45 1200  \n(3) 南 东 30 400",
+        "source": "23-24 六年级上·全国·单元测试",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "solution": "用东、西、南、北、东北、东南、西南、西北等方向来表述位置或用方向和距离相结合的方法来描述位置,既可以用来确定现实空间中物体的位置, 也可以用来确定平面上物体的位置。将方向和距离结合起来描述位置时,要注意三个要素: 一是观测点, 二是方向, 三是距离。1、弄清要标示的物体在哪个方位上, 有多少度。2、注意各场所离中心点的距离。在确定夹角时, 要根据方向来确定, 比如北偏东, 就是把正北方向对应量角器上的  $0^{\\circ}$  刻度线。南偏东就是把正南方向对应量角器商的  $0^{\\circ}$  刻度线。最后量出对应的度数。(1) 根据上北下南, 左西右东, 确定小刚在动物园的正西方向, 从图中可知一段代表  $400 \\mathrm{~m}$ , 则 2 段就是  $800 \\mathrm{~m}$ ; (2) 小红在动物园的北偏东方向, 或者东偏北方向, 再用量角器量出对应的度数, 则 3 段就是  $1200 \\mathrm{~m}$ ; (3) 小雪在动物园的南偏东方向, 东偏南方向上, 再用量角器量出对应的度数, 则 1 段就是  $400 \\mathrm{~m}$  。\n\n注意：不同方向上的度数是不一样的。确定方向后再确定度数。\n\n(1)  $400 \\times 2 = 800(\\mathrm{m})$\n\n小刚家在动物园正西方向上，距离  $800 \\mathrm{~m}$  。\n\n(2)  $400 \\times 3 = 1200(\\mathrm{m})$  \n小红家在动物园北偏东  $45^{\\circ}$  方向上, 距离是  $1200 \\mathrm{~m}$  。  \n(3) 小雪家在动物园南偏东  $30^{\\circ}$  方向上, 距离是  $400 \\mathrm{~m}$  。",
+        "merged_id": "unit2_252",
+        "original_id": "unit2_choice_252",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "填空题"
+      },
+      {
+        "id": "unit2_choice_253",
+        "question": "下面是某市旅游专线的行驶路线图, 看图后回答。\n\n![](images/429e0830148824c98add59b47e96c374da5c79abebbddba67881043a3e0dd795.jpg)\n\n从汽车站出发向_方向走约_米到达植物园，再向_方向走约_米到达动物园，然后向_方向走约_米到达科技馆，最后向_偏_方向走约_米就到达海滨浴场了。",
+        "options": [],
+        "answer": "东 1000 东北 3000 南 1000 东 南  $10^{\\circ}$  4000",
+        "source": "23-24 六年级上·全国·单元测试",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离描述路线图"
+        ],
+        "solution": "根据方向：上北下南、左西右东；用方向和距离结合来描述路线时，要注意三个要素：一是观测点（即参照物），二是方向，三是距离。图上1厘米表示实际1000米，先以汽车站为观测点即可确定植物园的方向，汽车站和植物园在图上相距1厘米；然后以植物园为观测点即可确定动物园的方向，植物园和动物园在图上相距3厘米；再以动物园为观测点即可确定科技馆的方向，动物园和科技馆在图上相距1厘米；最后以科技馆为观测点即可确定海滨浴场的方向，科技馆和海滨浴场在图上相距4厘米。据此解答。\n\n$1 \\times 1000 = 1000$  （米）\n\n$3 \\times 1000 = 3000 (\\text {米})$\n\n$4 \\times 1000 = 4000 (\\text {米})$\n\n从汽车站出发向东方向走约 1000 米到达植物园，再向东北方向走约 3000 米到达动物园，然后向南方向走约 1000 米到达科技馆，最后向东偏南  $10^{\\circ}$  （南偏东  $80^{\\circ}$  ）方向走约 4000 米就到达海滨浴场了。",
+        "merged_id": "unit2_253",
+        "original_id": "unit2_choice_253",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "填空题"
+      },
+      {
+        "id": "unit2_choice_254",
+        "question": "![](images/22c7e3155dc7ffafa088c35c97936bd1602f7028fc671ed146059b2ceb77a246.jpg)\n\n以A点为观测点。\n\n(1)小兔家在A点（_偏_）的方向上，距离_米。  \n(2)小猴家在 A 点（_偏_）的方向上，距离_米。  \n(3)小象家在A点（_偏_）的方向上，距离_米。  \n(4)小鹿家在A点（_偏_）的方向上，距离_米。",
+        "options": [],
+        "answer": "(1)西 北  $50^{\\circ}$  1000\n\n(2)东 北  $35^{\\circ}$  600  \n(3)东 南  $20^{\\circ}$ 800  \n(4)南 西  $30^{\\circ}$ 800",
+        "source": "23-24 六年级上·全国·单元测试",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "solution": "根据题意可知, 1 厘米表示 200 米, 分别计算出 A 点到小兔、小猴、小象、小鹿的实际距离, 再根据地图上方向的规定 '上北下南,左西右东', 以 A 点为观测点, 确定出小兔、小猴、小象、小鹿的位置。\n\n(1)  $200 \\times 5 = 1000$  （米）\n\n$90^{\\circ} - 50^{\\circ} = 40^{\\circ}$\n\n小兔家在 A 点西偏北  $50^{\\circ}$  （或北偏西  $40^{\\circ}$  ）的方向上，距离 1000 米。\n\n(2)  $200 \\times 3 = 600$  (米)\n\n$90^{\\circ} - 35^{\\circ} = 55^{\\circ}$\n\n小猴家在A点东偏北  $35^{\\circ}$  （或北偏东  $55^{\\circ}$  ）的方向上，距离600米。\n\n(3)  $200 \\times 4 = 800$  (米)\n\n$90^{\\circ} - 20^{\\circ} = 70^{\\circ}$\n\n小象家在A点东偏南  $20^{\\circ}$  （或南偏东  $70^{\\circ}$  ）的方向上，距离800米。\n\n(4)  $200 \\times 4 = 800$  (米)\n\n$90^{\\circ} - 30^{\\circ} = 60^{\\circ}$\n\n小鹿家在A点南偏西  $30^{\\circ}$  （或西偏南  $60^{\\circ}$  ）的方向上，距离800米。",
+        "merged_id": "unit2_254",
+        "original_id": "unit2_choice_254",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "填空题"
+      },
+      {
+        "id": "unit2_choice_255",
+        "question": "观察图, 完成下列各题。\n\n![](images/7d01b55f39c5498b287194c2e17bed599484393baa7f5837792e6f1a70bb2a82.jpg)\n\n(1)以小红为观测点，小亮在小红的()偏()方向()m处。  \n(2)以小亮为观测点, 小红在小亮的()偏()方向()m处。",
+        "options": [],
+        "answer": "(1)北 东  $60^{\\circ}$ 850\n\n(2) 西 南  $30^{\\circ}$ 850",
+        "source": "23-24 六年级上·全国·单元测试",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "solution": "（1）根据地图上的方向'上北下南,左西右东'及角度、距离来确定位置即可。\n\n(2) 一个事物在另一个事物的某个方向一定度数的位置,那么另一个事物在这个事物相对的方向相同度数的位置,据此解答即可。\n\n（1）以小红为观测点，小亮在小红的北偏东  $60^{\\circ}$  方向  $850\\mathrm{m}$  处或小亮在小红的东偏北  $30^{\\circ}$  方向  $850\\mathrm{m}$  处。  \n(2) 以小亮为观测点, 小红在小亮的西偏南  $30^{\\circ}$  方向  $850 \\mathrm{~m}$  处或小红在小亮的南偏西  $60^{\\circ}$  方向  $850 \\mathrm{~m}$  处。",
+        "merged_id": "unit2_255",
+        "original_id": "unit2_choice_255",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "填空题"
+      },
+      {
+        "id": "unit2_choice_256",
+        "question": "如图, 植物园在博物馆的()偏()  $60^{\\circ}$  方向, 广场在博物馆的()偏() $30^{\\circ}$  方向, 距离博物馆() km。() 在广场的南偏东  $35^{\\circ}$  方向, 距离广场  $3\\mathrm{km}$  处。\n\n![](images/2dbc1c51a10602f5ad0c16e490f44aec7906dc63cd8c6a45952f331979fcfad3.jpg)",
+        "options": [],
+        "answer": "西 北 东 北 5 图书馆",
+        "source": "23-24 五年级下·陕西渭南·期末",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "solution": "以博物馆为观察点，根据地图'上北下南,左西右东'的规定，结合图上给出的角度确定植物园在博物馆的什么方位；同理确定广场在博物馆的什么方位。以广场为观察点，找出它的南偏东  $35^{\\circ}$  方向即可确定。\n\n通过图示可知：植物园在博物馆的西偏北  $60^{\\circ}$  方向，广场在博物馆的东偏北  $30^{\\circ}$  方向，距离博物馆  $5\\mathrm{km}$  。图书馆在广场的南偏东  $35^{\\circ}$  方向，距离广场  $3\\mathrm{km}$  处。",
+        "merged_id": "unit2_256",
+        "original_id": "unit2_choice_256",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "填空题"
+      },
+      {
+        "id": "unit2_choice_257",
+        "question": "市政府在电视塔东偏北  $50^{\\circ}$  方向上。()\n\n![](images/fdb9dbc4f42ddfe7012379b9522470146ff78bb9223081d893ef65790ac9f29c.jpg)",
+        "options": [],
+        "answer": "×",
+        "source": "2024六年级上·全国·专题练习",
+        "difficulty": "容易",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "solution": "以电视塔为观测点，以图上的'上北下南,左西右东'为准，根据方向和角度确定市政府和电视塔的位置关系。\n\n从图中可知，市政府在电视塔北偏东  $50^{\\circ}$  或东偏北  $40^{\\circ}$  方向上。\n\n原题说法错误。\n\n故答案主为：×",
+        "merged_id": "unit2_257",
+        "original_id": "unit2_choice_257",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "判断题"
+      },
+      {
+        "id": "unit2_choice_258",
+        "question": "确定一个物体的位置, 只要知道它的方向就可以了。( )",
+        "options": [],
+        "answer": "×",
+        "source": "22-23 六年级上·江西南昌·期中",
+        "difficulty": "容易",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "solution": "物体位置对于某一观察点来说, 是由一定的方向和距离确定的, 只知道方向或距离不能确定物体的位置。据此解答即可。\n\n由分析可知：\n\n确定一个物体的位置，需要知道它的方向，角度距离才可以确定。原说法错误。\n\n故答案为： $\\times$",
+        "merged_id": "unit2_258",
+        "original_id": "unit2_choice_258",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "判断题"
+      },
+      {
+        "id": "unit2_choice_259",
+        "question": "看图连一连。\n\n![](images/658867f102c59a157f7f81a92ad90c07941872ea2765c7db174fabbe0da6678b.jpg)\n\n公园在学校东偏北  $30^{\\circ}$  方向  $200 \\mathrm{~m}$  处\n\n商场在学校西偏北  $45^{\\circ}$  方向  $800 \\mathrm{~m}$  处\n\n邮局在学校东偏南  $50^{\\circ}$  方向  $400\\mathrm{m}$  处\n\n广场在学校西偏南  $30^{\\circ}$  方向  $600\\mathrm{m}$  处",
+        "options": [],
+        "answer": "见详解",
+        "source": "25-26 六年级上·海南海口·单元测试",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "solution": "将方向和距离结合起来描述位置时，要注意三个要素：一是观测点，二是方向，三是距离。看图可知，图上  $1\\mathrm{cm}$  表示实际  $200\\mathrm{m}$ ，图上  $\\mathrm{cm}$  数  $\\times 1\\mathrm{cm}$  表示的  $\\mathrm{m}$  数 = 实际  $\\mathrm{m}$  数。\n\n公园:  $4 \\times 200 = 800(\\mathrm{m})$\n\n商场:  $1 \\times 200 = 200(\\mathrm{m})$\n\n邮局:  $2 \\times 200 = 400(\\mathrm{m})$\n\n广场:  $3 \\times 200 = 600(\\mathrm{m})$\n\n![](images/6710073d54bef327bd32a8f5a043c37d931d497d68535e814199331f3c5e4065.jpg)",
+        "merged_id": "unit2_259",
+        "original_id": "unit2_choice_259",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "连线题"
+      },
+      {
+        "id": "unit2_choice_260",
+        "question": "看图连一连。\n\n![](images/3b27ff5340152037ebad27b4d065aca344557699097cf55d832763f9cabfec5c.jpg)\n\n公园在学校东偏北  $30^{\\circ}$  方向  $200\\mathrm{m}$  处\n\n商场在学校西偏北  $45^{\\circ}$  方向800m处\n\n邮局在学校东偏南  $50^{\\circ}$  方向  $400\\mathrm{m}$  处\n\n广场在学校西偏南  $30^{\\circ}$  方向  $600\\mathrm{m}$  处",
+        "options": [],
+        "answer": "见详解",
+        "source": "23-24 六年级上·全国·单元测试",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "solution": "根据题意可知, 1 厘米表示  $200 \\mathrm{~m}$ ; 计算出学校到公园、商场、邮局、广场的实际距离; 再根据地图上方向的规定: 上北下南, 左西右东, 以学校为观测点, 确定出公园、商场、邮局、广场的位置, 再连一连。\n\n学校到公园的距离:  $200 \\times 4 = 800(\\mathrm{~m})$\n\n学校到商场的距离:  $200 \\times 1 = 200(\\mathrm{~m})$\n\n学校到邮局的距离:  $200 \\times 2 = 200(\\mathrm{~m})$\n\n学校到广场的距离:  $200 \\times 3 = 600(\\mathrm{~m})$\n\n如图：\n\n![](images/97a0795f154202526c225d55feeae8f932c2e7d49e7b77cdf6e3dd7793094636.jpg)",
+        "merged_id": "unit2_260",
+        "original_id": "unit2_choice_260",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "连线题"
+      },
+      {
+        "id": "unit2_choice_261",
+        "question": "近年来, 新能源汽车产业蓬勃兴起, 近期小米集团最新新能源车型SU7进行了试驾, 它从公司沿东偏北  $30^{\\circ}$  方向行驶了 4 千米, 又往正东方向行驶了 2 千米, 最后往南偏东  $40^{\\circ}$  方向行驶了 3 千米到达终点站。请根据描述画出车子行驶的路线图。\n\n![](images/92c4a40d8420b493d061f0dce6d9560f16580dbd182ad4abaa33939896f0d75f.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "source": "24-25六年级下·江西赣州·期末",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离画线路图"
+        ],
+        "solution": "先根据图上1厘米表示实际1千米求出对应的图上距离，再根据图上确定方向的方法：上北下南、左西右东，确定方向，最后根据给出的角度确定具体的位置并作图即可。\n\n$4 \\div 1 = 4$  （厘米）\n\n$2 \\div 1 = 2 (\\text {厘 米})$\n\n$3 \\div 1 = 3 (\\text {厘 米})$\n\n作图如下：\n\n![](images/972dfda5f9981679d0f7e4f50af5784347a0b9a888034d2febb7d8be9cee6532.jpg)",
+        "merged_id": "unit2_261",
+        "original_id": "unit2_choice_261",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "作图题"
+      },
+      {
+        "id": "unit2_choice_262",
+        "question": "观察下图, 然后完成各题。\n\n![](images/a2263bd142dbf3bb743ac526d312794bd6dfcaa3e96b92a3bf800c694eb992d0.jpg)\n\n(1)聪聪家在学校（）偏（）方向上，距离学校（）米处。  \n(2) 小丽家在学校正西方向 100 米处, 请你在图上标出小丽家的位置。  \n(3) 小亮家在学校南偏东  $45^{\\circ}$  方向 200 米处, 请你在图上标出小亮家的位置。  \n(4) 小丽沿着图中路线从家到学校走了 2 分钟, 接着又从学校到小亮家走了 3 分钟。小丽走完这段路程的平均速度是多少?",
+        "options": [],
+        "answer": "(1)北；西；  $55^{\\circ}$  ；300\n\n(2)(3)图见详解  \n(4)60米/分",
+        "source": "24-25 六年级上·全国·单元测试",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "基础行程问题、根据方向、角度和距离确定物体的位置"
+        ],
+        "solution": "（1）根据上北下南，左西右东，以及图中给出数据, 可得以学校为观测点,  $90^{\\circ} - 35^{\\circ} = 55^{\\circ}$ , 即聪聪家在学校北偏西  $55^{\\circ}$  方向上, 距离学校 300 米处。\n\n(2) 由图可知图上1厘米表示实际100米, 根据小丽家在学校正西方向100米处, 在图上学校正西方向1厘米处标注小丽家。  \n(3) 按照小亮家在学校南偏东  $45^{\\circ}$  方向 200 米处,  $200 \\div 100 = 2$  (厘米), 在图上学校南偏东  $45^{\\circ}$  方向 2 厘米处标注小亮家。  \n(4) 小丽从家到学校走了 100 米, 从学校到小亮家走了 200 米, 总路程为:  $100 + 200 = 300$  (米), 总时间为: $2 + 3 = 5$  (分钟), 根据平均速度  $=$  路程  $\\div$  时间, 即可求出平均速度。\n\n（1）聪聪家在学校北偏西  $55^{\\circ}$  方向上，距离学校300米处。（答案不唯一）\n\n(2) (3) 如图：\n\n![](images/08b8a2384108ce1c4b518780c48fa53287a03aadfabe8bb66e092e0ddbf36b48.jpg)\n\n(4)  $(100 + 200)\\div (2 + 3)$\n\n$= 300 \\div 5$\n\n$= 60 (\\text {米 / 分})$\n\n答: 小丽走完这段路程的平均速度是 60 米/分。",
+        "merged_id": "unit2_262",
+        "original_id": "unit2_choice_262",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "作图题"
+      },
+      {
+        "id": "unit2_choice_263",
+        "question": "在越野定位赛中,选手们习惯用几点钟方向表示方位,例如 3 点钟方向 10 千米处,即  $O$  点正东方向 10 千米处(如下图)。\n\n![](images/514f64537f23652696019480f5b534c36eb947a3c30e0d7998bc4563ecf68cbc.jpg)\n\n(1)  $A$  点在  $O$  点（）偏（）°方向，即（）点钟方向，距离（）千米处。  \n(2)  $B$  点在  $O$  点西偏南  $30^{\\circ}$  的方向, 距离  $O$  点 20 千米处, 请标出  $B$  点的位置。  \n(3) 下午将有一场暴雨, 以点  $O$  为中心, 覆盖半径为 20 千米, 请在图中画出暴雨覆盖的区域。",
+        "options": [],
+        "answer": "（1）东（或南）；南（或东）；30（或60）；4；40\n\n(2)(3)见解答",
+        "source": "24-25 六年级上·全国·单元测试",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "圆的概念及特点、几点钟方向、根据方向、角度和距离确定物体的位置、画圆"
+        ],
+        "solution": "（1） $90^{\\circ}$  所对的弧被平均分成3份，1份的弧所对的角是  $30^{\\circ}$ ，据此以点  $O$  为观察点，根据上北下南，左西右东确定方向和角度，根据钟面的刻度可知几点钟方向，根据圆的特征，同一圆上的半径都相等，观察可知半径有4个10千米可确定距离；\n\n(2) 找一个地方在另一个地方什么位置，就以另一个地方为观测点，根据方向、角度、距离确定物体的位置；  \n(3) 圆的画法: 确定圆心  $O$ , 圆规针尖固定在这一点, 拉开圆规的两脚, 使两脚之间的距离为 2 段, 然后旋转一周, 即可画出半径为 20 千米的圆;  \n(1) 东; 南; 30; 4; 40\n\n(1)  $10 \\times 4 = 40$  （千米）\n\n$90^{\\circ} \\div 3 = 30^{\\circ}$\n\nA点在O点东（或南）偏南（或东）  $30^{\\circ}$  （或  $60^{\\circ}$  ）方向，即4点钟方向，距离40千米处。\n\n(2)  $20 \\div 10 = 2$  （段）\n\n作图如下：\n\n![](images/7f22c03dbd242d03e916506bcfc01031ab545a407c43a0d552774bb00525f748.jpg)\n\n(3) 覆盖半径为 20 千米, 就是圆的半径是 2 段, 作图如下:\n\n![](images/ff4921670b2e59116bada2fe1d0af9e60575c481513a3b87b96376ddea369f48.jpg)",
+        "merged_id": "unit2_263",
+        "original_id": "unit2_choice_263",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "作图题"
+      },
+      {
+        "id": "unit2_choice_264",
+        "question": "张叔叔从  $A$  市驾车到  $C$  市, 途经  $B$  市。\n\n(1)  $B$  市在  $A$  市的正东方向,  $C$  市在  $B$  市的东偏北  $30^{\\circ}$  方向上。  \n(2) 从  $A$  市到  $C$  市的路程是400千米, 从  $A$  市到  $B$  市的路程是  $B$  市到  $C$  市的路程的  $\\frac{1}{3}$  。  \n(3) 从  $A$  市出发时加满了油, 到达  $B$  市时, 油还剩下  $\\frac{2}{3}$  箱。  \n(1) 根据以上信息，在下图中补充绘制张叔叔的路线图。\n\n(提示：需要先计算路程)。\n\n![](images/a2e398f5ff4624f5fe0dc9b6ca5e71e83811cc8f983def7c252c27ddb36073e3.jpg)\n\n(2) 张叔叔能否用剩下的油开到终点？请你尝试说明理由 (假设每千米的耗油量不变)。",
+        "options": [],
+        "answer": "(1)(2) 见解答",
+        "source": "24-25 六年级上·全国·单元测试",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "比的应用、根据方向、角度和距离确定物体的位置、分数乘整数"
+        ],
+        "solution": "（1）观察图可知，此图是按'上北下南,左西右东'来规定方向的，图上距离1厘米表示实际距离100千米，找准参照物，观察图可知，从A市到B市的路程相当于B市到C市的  $\\frac{1}{3}$ ，根据分数的意义可知，B市到C市的路程可平均分成3份，A市到B市的路程相当于其中的1份，即A市到C市的路程有  $1 + 3 = 4$  （份），用400除以4可得1份是100千米，100乘3可得B市到C市的路程是300千米。根据方向和距离找出B市和C市的位置，完成路线图；\n\n(2) 假设每千米的耗油量不变，剩下的路程需要的油是从A市到B市耗油量的3倍，把一箱油看作单位'1'，用1减  $\\frac{2}{3}$  据此解答。\n\n(1)  $400 \\div (1 + 3)$\n\n$= 400 \\div 4$\n\n$= 100 (\\text {千 米})$\n\n作图如下：\n\n![](images/8e3335ab576a6eea3542682cf9fcc018d02dff49797e7f0db7b64790e95c97de.jpg)  \n(2)  $\\left(1 - \\frac{2}{3}\\right)\\times 3 = \\frac{1}{3}\\times 3 = 1 < \\frac{2}{3}$\n\n答: 张叔叔不能用剩下的油开到终点, 从  $A$  市到  $B$  市的路程相当于  $B$  市到  $C$  市的  $\\frac{1}{3}$ , 从  $A$  市出发时加满了油, 到达  $B$  市时, 油还剩下  $\\frac{2}{3}$  箱, 说明从  $A$  市到  $B$  市用掉了  $\\frac{1}{3}$  箱油, 假设每千米的耗油量不变, 剩下的路程还需要 1 箱油才能到达。",
+        "merged_id": "unit2_264",
+        "original_id": "unit2_choice_264",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "作图题"
+      },
+      {
+        "id": "unit2_choice_265",
+        "question": "五年级的同学们计划从学校出发, 前往新建的校园科技馆进行研学活动。他们从学校(起点)出发, 向正东方向步行3千米, 到达'生态植物园'参观, 接着向正北方向步行2千米, 到达'天文观测台'进行学习, 然后向正西方向步行1千米, 到达'机器人创新实验室'体验科技项目, 最后沿西偏南  $45^{\\circ}$  方向返回学校, 完成整个研学活动。\n\n请根据以上信息，在图中画出同学们的参观路线图。\n\n![](images/953d8a748e046f8164fa6044e06ecc73012e70e6a2fdd7b83a0c2df7ab941619.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "source": "24-25六年级上·全国·单元测试",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置、根据方向、角度和距离画线路图"
+        ],
+        "solution": "由图可知，上北下南，左西右东，图上1厘米相当于实际1千米，据此从学校向右（东）画出3厘米的线段，标注生态植物园；再向上（北）画出2厘米的线段，标注天文观测台；然后向左（西）画1厘米标注机器人创新实验室；最后向左下角（西南）连线，画出返回学校即可。\n\n根据分析，作图如下：\n\n![](images/22f1de5fdf9923f79f86fc5a2dd6fe59a5e59bff086298ba6c4be67e779d641d.jpg)",
+        "merged_id": "unit2_265",
+        "original_id": "unit2_choice_265",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "作图题"
+      },
+      {
+        "id": "unit2_choice_266",
+        "question": "在一次军事演习中,一艘军舰从起点先向东偏北  $60^{\\circ}$  方向行驶 72 千米后, 再向正东方向行驶 36 千米, 最后向东偏南  $30^{\\circ}$  方向行驶 24 千米到达终点。\n\n(1) 根据上面的描述, 把军舰行驶的路线图画完整。  \n(2) 如果从终点按原路匀速返回起点用了 4 小时, 那么这艘军舰返回时的速度是 ( ) 千米/小时。",
+        "options": [],
+        "answer": "(1) 见详解\n\n(2) 33",
+        "source": "24-25 六年级下·云南楚雄·期末",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "基础行程问题、根据方向、角度和距离画线路图"
+        ],
+        "solution": "（1）根据上北下南，左西右东，结合角度和距离画图即可；\n\n(2)返回时的路程等于  $(72 + 36 + 24)$  千米，根据'速度=程÷时间'解答。\n\n(1)如图：\n\n![](images/833c81a90e99fb9c86786947eafee865d144f5237c83f7f0d1048fbfe108f84a.jpg)  \n(2)  $(72 + 36 + 24) \\div 4 = (108 + 24) \\div 4 = 132 \\div 4 = 33$  （千米/小时）\n\n所以这艘军舰返回时的速度是33千米/小时。",
+        "merged_id": "unit2_266",
+        "original_id": "unit2_choice_266",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "作图题"
+      },
+      {
+        "id": "unit2_choice_267",
+        "question": "某老年社团举行了'弘扬红色精神，展现老年风采'的活动，其中有一项活动是骑自行车接力赛，赛程安排如下：1号骑手从起点沿东偏北  $40^{\\circ}$  方向骑行  $15 \\mathrm{~km}$ ，到达第一接力点；2号骑手从第一接力点沿北偏西  $60^{\\circ}$  方向骑行  $15 \\mathrm{~km}$ ，到达第二接力点；3号骑手从第二接力点沿南偏西  $45^{\\circ}$  方向骑行  $10 \\mathrm{~km}$ ，到达终点。请你按照描述画出接力赛的路线图。\n\n![](images/054166d38d30777b09e05943a7359dde389741580845fd44148fa668c20c69f5.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "source": "25-26 六年级上·全国·课后作业",
+        "difficulty": "容易",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置、根据方向、角度和距离画线路图"
+        ],
+        "solution": "【分析】在平面图上通常按照'上北下南,左西右东'来确定方向，图中给出了线段比例尺，代表图上  $1\\mathrm{cm}$  距离对应实际距离  $5\\mathrm{km}$ ，通过计算实际距离与  $5\\mathrm{km}$  的比值，可得到图上距离。\n\n【详解】1号骑手从起点沿东偏北  $40^{\\circ}$  方向骑行  $15 \\mathrm{~km}$ , 则图上距离为  $15 \\div 5 = 3(\\mathrm{~cm})$ ; 所以在东偏北  $40^{\\circ}$  方向上量取 3 厘米长度, 确定为第一接力点的位置。\n\n2号骑手从第一接力点沿北偏西  $60^{\\circ}$  方向骑行  $15\\mathrm{km}$ ，则图上距离为  $3\\mathrm{cm}$ ；所以在北偏西  $60^{\\circ}$  方向上量取3厘米的长度，确定为第二接力点的位置。\n\n3号骑手从第二接力点沿南偏西  $45^{\\circ}$  方向骑行  $10\\mathrm{km}$ ，则图上距离为  $10 \\div 5 = 2(\\mathrm{cm})$ ；所以在南偏西  $45^{\\circ}$  方向上量取2厘米的长度，确定终点的位置。\n\n![](images/7fb3c1c46d3870d175691e3575e2988d962bd41ad047ea6f881d21465dafb892.jpg)\n\n答：如下图所示",
+        "merged_id": "unit2_267",
+        "original_id": "unit2_choice_267",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "作图题"
+      },
+      {
+        "id": "unit2_choice_268",
+        "question": "按照要求填空和作图。\n\n![](images/6728a8cdfaf557cd24f49f74af239c60f575f1615b785077165f7a5de09e256a.jpg)\n\n(1) 丽丽从家出发去图书馆, 她应先向 ( ) 偏 ( ) ( ) 方向走 200 米来到游乐园, 然后再向正东方向走 ( ) 米到图书馆。\n\n(2)丽丽看完书要去图书馆东偏南  $60^{\\circ}$ , 距离 200 米的欢欢家。请你在上图中补充丽丽从图书馆到欢欢家的路线图。",
+        "options": [],
+        "answer": "(1)北; 东; 60; 200\n\n(2) 见详解",
+        "source": "23-24 六年级下·福建漳州·期末",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置、根据方向、角度和距离画线路图"
+        ],
+        "solution": "【分析】（1）从图中可知：图上 1 厘米表示实际 100 米，游乐园到图书馆的实际距离是  $2 \\times 100 = 200$  米。根据地图是上北下南，左西右东，分别以丽丽家和游乐园为观测点，结合角度方向距离即可确定游乐园、图书馆的位置。（答案不唯一）\n\n(2) 图上 1 厘米表示实际 100 米, 图书馆与欢欢家的图上距离是  $200 \\div 100 = 2$  厘米。以图书馆为观测点, 根据上北下南, 左西右东, 结合角度方向距离即可确定欢欢家的位置, 并画图即可。\n\n【详解】（1）丽丽从家出发去图书馆，她应先向北偏东  $60^{\\circ}$  方向走200米来到游乐园，然后再向正东方向走200米到图书馆。\n\n(2) 如图:\n\n![](images/b0b5394036de7bcf4c37183f7666bd2d52a0aef0452d82c7b7c97c1628d92a45.jpg)",
+        "merged_id": "unit2_268",
+        "original_id": "unit2_choice_268",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "作图题"
+      },
+      {
+        "id": "unit2_choice_269",
+        "question": "填一填，量一量，画一画。\n\n![](images/ada52329fcea1a7236c21f2ee02747e88871f379ef01769bd113583e3a8fd231.jpg)\n\n(1) 小红家在学校()偏()()°方向()米处。\n\n(2)小方家在学校()偏()()°方向()米处。\n\n(3) 小海家在小红家正西方向 800 米处, 请你在图上标出小海家的位置。\n\n(4) 邮局在小红家东偏南  $40^{\\circ}$  方向 600 米处，请你在图上标出邮局的位置。",
+        "options": [],
+        "answer": "(1)东；北；  $40^{\\circ}$  ；800;\n\n(2)东;南;  $25^{\\circ}$ ; 500;\n\n(3) 见详解;\n\n(4) 见详解",
+        "source": "24-25六年级下·海南海口·单元测试",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "solution": "【分析】（1）以图上的'上北下南,左西右东'为准，图例表示图上1厘米相当于实际距离200米；以学校为观测点，先判断方向，再用量角器量出对应的角度，再测量出学校到小红家的图上距离，用图上距离乘200即可得到实际距离；同理完成（2）；\n\n(3) 以小红家为观测点, 先用实际距离除以 200 求出图上的距离, 再根据方向和图上距离确定小海家的位置即可;\n\n(4) 以小红家为观测点, 先用实际距离除以 200 求出图上的距离, 再根据方向、角度和图上距离确定出邮局的位置即可。\n\n【详解】(1)  $4 \\times 200 = 800$  （米）\n\n小红家在学校东偏北  $40^{\\circ}$  方向800米处。\n\n(2)  $2.5 \\times 200 = 500$  (米)\n\n小方家在学校东偏南  $25^{\\circ}$  方向 500 米处。\n\n(3)  $800 \\div 200 = 4$  （厘米）\n\n(4)  $600 \\div 200 = 3$  (厘米)\n\n(3) (4) 作图如下:\n\n![](images/f2b7e5f0cd457ac62e53f3a699e9d4459cc7fe07c9c43c23728aa8cf8583cc00.jpg)",
+        "merged_id": "unit2_269",
+        "original_id": "unit2_choice_269",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "作图题"
+      },
+      {
+        "id": "unit2_choice_270",
+        "question": "A楼在B楼正北方向  $600\\mathrm{m}$  处，C楼在B楼东偏北  $40^{\\circ}$  方向  $900\\mathrm{m}$  处。请你在下图中分别画出A楼和C楼的位置。\n\n![](images/ab6ff032960224ed80e9e240966cf7442a0f0397d81c399013cc88f73f225dbf.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "source": "24-25六年级下·海南海口·期中",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离画线路图"
+        ],
+        "solution": "【分析】根据地图方向的规定'上北下南,左西右东'，以B楼为观测点即可确定A楼、C楼的方向，再根据图上1cm表示实际300m，用A楼、C楼与B楼的实际距离分别除以300m，即可分别求出A楼、C楼与B楼的图上距离，从而画出A楼、C楼的位置。\n\n【详解】  $600 \\div 300 = 2 \\, (\\mathrm{cm})$ $900 \\div 300 = 3 \\, (\\mathrm{cm})$\n\n![](images/bc9123da6c4bfd75f340d86ab152ce175dbd9ba1406ae8418c65238398bb04b1.jpg)",
+        "merged_id": "unit2_270",
+        "original_id": "unit2_choice_270",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "作图题"
+      },
+      {
+        "id": "unit2_choice_271",
+        "question": "![](images/62ba1437654f7fc88a34a9dd98ee4db1b6bd9502cb383f4ef3adac488d16c56b.jpg)\n\n(1) 公园在小军家（）偏（）  $30^{\\circ}$  方向（）米处。\n\n(2) 小军的妈妈每天从家出发, 先沿南偏西  $20^{\\circ}$  方向步行 $400 \\mathrm{~m}$  到小区 3 号门, 再从小区 3 号门向正东方向步行  $1000 \\mathrm{~m}$  到地铁站, 乘坐地铁上班。根据描述在上图中画出小军妈妈从家到地铁站的路线。",
+        "options": [],
+        "answer": "①东；北；400\n\n(2) 见详解",
+        "source": "24-25五年级上·北京东城·期末",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置、根据方向、角度和距离画线路图、根据方向、角度和距离描述路线图"
+        ],
+        "solution": "【分析】①先确定观测点，再根据上北下南，左西右东确定方向，观察可知图上1厘米表示实际距离200米，用200乘2得到实际距离。据此解答。\n\n(2) 先确定观测点, 再根据上北下角, 左西右东确定方向和角度, 再用实际距离除以 200 确定图上距离, 据此画出路线图。\n\n【详解】  $200 \\times 2 = 400$  （米）\n\n据分析可知，公园在小军家东偏北  $30^{\\circ}$  方向 400 米处。\n\n(2)  $400 \\div 200 = 2$  （厘米）\n\n$1000 \\div 200 = 5$  （厘米）\n\n据分析画图如下：\n\n![](images/3d26d3e4195f772356458cfa134ed57d89fa3c09c19a96ceec79c8b8e6c0a558.jpg)",
+        "merged_id": "unit2_271",
+        "original_id": "unit2_choice_271",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "作图题"
+      },
+      {
+        "id": "unit2_choice_272",
+        "question": "下面是李叔叔家和通信公司的位置示意图。\n\n![](images/4fb5036d6261bc859c04ecbadeb2d48bee8be1d39f71f085562e828eaae35430.jpg)\n\n(1)通信公司在李叔叔家（）偏（）（）方向，距离（）千米处。\n\n(2) 由于通信信号塔出现故障，通信公司派李叔叔去维修。信号塔在通信公司西偏北  $45^{\\circ}$  方向 3 千米处，请在图中标出信号塔的位置。",
+        "options": [],
+        "answer": "(1)东；北；  $30^{\\circ}$  ；2\n\n(2）见详解",
+        "source": "24-25六年级上·福建三明·期末",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "solution": "【分析】（1）从图中可知：图上1厘米表示实际1千米，通信公司到李叔叔家是2千米。以李叔叔家为观测点，地图是上北下南，左西右东，结合角度方向和距离，即可描述通信公司与李叔叔家的位置关系。\n\n(2) 从图中可知: 图上 1 厘米表示实际 1 千米, 信号塔到通信公司的图上距离是 3 厘米, 以通信公司为观测点, 结合角度方向和距离, 即可确定信号塔的位置。\n\n【详解】(1)  $1 \\times 2 = 2$  （千米）\n\n通信公司在李叔叔家东偏北  $30^{\\circ}$  （或北偏东  $60^{\\circ}$  ）方向，距离2千米处。\n\n(2)  $3 \\div 1 = 3$  (厘米)\n\n信号塔的位置如图：\n\n![](images/3af902dc1d3018204276310e0bf4f52d1a218aa48577309f19e371c81427827e.jpg)",
+        "merged_id": "unit2_272",
+        "original_id": "unit2_choice_272",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "作图题"
+      },
+      {
+        "id": "unit2_choice_273",
+        "question": "如图是妙想家的扫地机器人在客厅进行清理工作时的路线示意图, 请你认真观察并回答问题。\n\n![](images/1504731362438f42ef6d313a634763ec3efeafc895d564ae11821e58a071bccc.jpg)\n\n(1) 电视柜在起点的 ( ) 方向, 起点在电视柜的 ( ) 方向。\n\n(2) 在绿植的西偏北  $30^{\\circ}$  方向 3 米处有一个书报架, 请你在图中标出书报架的位置。\n\n(3) 在一次清理工作中，扫地机器人从起点出发，先行进到茶几处，再行进到绿植处，最后行进到书报架处。请你描述扫地机器人的行进路线。",
+        "options": [],
+        "answer": "（1）西偏北  $45^{\\circ}$  ；东偏南  $45^{\\circ}$\n\n(2) 图见详解\n\n(3)行进路线见详解",
+        "source": "24-25 六年级上·北京房山·期末",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置、根据方向、角度和距离描述路线图"
+        ],
+        "solution": "【分析】以图上的'上北下南,左西右东'为准，图例表示图上1厘米相当于实际距离1.5米。(1)电视柜在起点的方向，是以起点为观测点，结合方向、角度确定电视柜与起点的位置关系。起点在电视柜的方向，是以电视柜为观测点，观测点不同，方向相反，夹角的度数相同。\n\n(2) 以绿植为观测点，在绿植的西偏北  $30^{\\circ}$  方向上画  $3 \\div 1.5 = 2$  厘米长的线段，即是书报架的位置。\n\n(3) 起点到茶几处的图上距离是 3 厘米, 相当于实际距离  $1.5 \\times 3 = 4.5$  米; 茶几处到绿植的图上距离是 2 厘米, 相当于实际距离  $1.5 \\times 2 = 3$  米; 绿植到书报架的图上距离是 2 厘米, 相当于实际距离  $1.5 \\times 2 = 3$  米;\n\n根据方向、角度和距离描述扫地机器人的行进路线。\n\n【详解】（1）电视柜在起点的西偏北  $45^{\\circ}$  （或北偏西  $45^{\\circ}$  ）方向，起点在电视柜的东偏南  $45^{\\circ}$  （或南偏东  $45^{\\circ}$  ）方向。\n\n(2)  $3 \\div 1.5 = 2$  （厘米）\n\n如图：\n\n![](images/fbdecb4ee6d002c24018a83c0db9475171962efda9556358271b78d032e2b203.jpg)\n\n(3)扫地机器人从起点出发，先向东偏北  $30^{\\circ}$  （或北偏东  $60^{\\circ}$  ）方向行进4.5米到达茶几处，再向正北方向行进3米到达绿植处，最后向西偏北  $30^{\\circ}$  （或北偏西  $60^{\\circ}$  ）方向行进3米到达书报架处。\n\n![](images/cb7d80c191140485313f74f9f07c85ec53ebccf0c8e8ccbb195aeee489aff924.jpg)",
+        "merged_id": "unit2_273",
+        "original_id": "unit2_choice_273",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "作图题"
+      },
+      {
+        "id": "unit2_choice_274",
+        "question": "操作。\n\n(1) 明明从学校出发, 向北偏西  $35^{\\circ}$  方向, 走  $200 \\mathrm{~m}$  到华侨商店; ②再从华侨商店向正西方向走  $300 \\mathrm{~m}$  到国光超市; ③然后从国光超市向西偏南  $45^{\\circ}$  方向, 走  $150 \\mathrm{~m}$  到邮局; ④最后从邮局向南偏东  $40^{\\circ}$  方向, 走  $250 \\mathrm{~m}$  到家。请根据上面的描述, 把路线图画完整。",
+        "options": [],
+        "answer": "见详解",
+        "source": "24-25 六年级上·江西吉安·期末",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离画线路图"
+        ],
+        "solution": "【分析】①图上  $1\\mathrm{cm}$  表示实际距离  $100\\mathrm{m},200\\div 100 = 2$  （cm），以学校为观测点，根据地图'上北下南,左西右东'的规定，向北偏西  $35^{\\circ}$  方向画一条  $2\\mathrm{cm}$  的线段，另一个端点就是华侨商店的位置；②  $300\\div 100 = 3$  （cm），以华侨商店为观测点，向正西方向画一条  $3\\mathrm{cm}$  的线段，端点是国光超市的位置；③  $150\\div 100 = 1.5$  （cm），以国光超市为观测点，向西偏南  $45^{\\circ}$  方向画出  $1.5\\mathrm{cm}$  的线段到邮局；④  $250\\div 100 = 2.5$  （cm），以邮局为观测点，向南偏东  $40^{\\circ}$  方向画出  $2.5\\mathrm{cm}$  的线段到明明家。据此作图。\n\n![](images/7c52d2ff7af349c9e769c47a322ce30d30cab7c8df7b5b9f2e623acf22d01ceb.jpg)\n\n【详解】",
+        "merged_id": "unit2_274",
+        "original_id": "unit2_choice_274",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "作图题"
+      },
+      {
+        "id": "unit2_choice_275",
+        "question": "动手动脑, 我能行。'1路公共汽车从起点站向西偏北  $50^{\\circ}$  方向行驶 2 千米, 然后向西行驶 3 千米, 最后向南偏西  $30^{\\circ}$  方向行驶 1 千米到达终点站。'请根据描述, 把 1 路公共汽车行驶的路线图画完整。\n\n![](images/68a61cb8df39d6687227f8d163ec208344285c7d9e1ddc73fa8e63460b931715.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "source": "24-25 六年级上·青海果洛·期末",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离画线路图"
+        ],
+        "solution": "【分析】以图上的'上北下南,左西右东'为准，图例表示图上1厘米相当于实际距离1千米；1路公共汽车从起点站向西偏北  $50^{\\circ}$  方向行驶2千米，即在起点站的西偏北  $50^{\\circ}$  方向上画  $2 \\div 1 = 2$  厘米长的线段；\n\n然后向西行驶3千米，即继续向正西方向上画  $3 \\div 1 = 3$  厘米长的线段；最后向南偏西  $30^{\\circ}$  方向行驶1千米到达终点站，即继续向南偏西  $30^{\\circ}$  方向上画  $1 \\div 1 = 1$  厘米长的线段，即是终点站。\n\n![](images/4f97ea0c61ef15fd0077f901d18f26dc3f0c1824d1d4e1392355aa8be9308ba7.jpg)\n\n【详解】如图：",
+        "merged_id": "unit2_275",
+        "original_id": "unit2_choice_275",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "作图题"
+      },
+      {
+        "id": "unit2_choice_276",
+        "question": "某机场停机坪采用无人配送车给旅客们配送行李，下面是配送车1号某次配送路线。\n\n![](images/dfd37533165fb27663672ef603ea264aec6bba091f8169ab69e8e0f82818c720.jpg)\n\n(1) 配送车 1 号从行李存放处出发, 到达 A 号客机后, 继续向东偏北  $30^{\\circ}$  方向, 距离  $150 \\mathrm{~m}$  处配送行李, 请标出此刻配送车 1 号的位置。\n\n(2) A 号客机的正北方向  $100 \\mathrm{~m}$  处停放着配送车 2 号, 请标出配送车 2 号的位置。",
+        "options": [],
+        "answer": "见详解",
+        "source": "24-25 六年级上·广东东莞·期末",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "solution": "【分析】（1）根据图上 1 厘米表示实际 50 米，那么实际 150 米就是图上  $150 \\div 50 = 3$  厘米。以 A 号客机为观测点，地图是上北下南，左西右东，结合角度方向和距离，即可标出此刻配送车 1 号的位置。\n\n(2) 根据图上 1 厘米表示实际 50 米, 那么实际 100 米就是图上  $100 \\div 50 = 2$  厘米。以 A 号客机为观测点, 地图是上北下南, 左西右东, 结合角度方向和距离, 即可标出配送车 2 号的位置。\n\n【详解】(1)  $150 \\div 50 = 3$  （厘米）\n\n(2)  $100 \\div 50 = 2$  （厘米）\n\n如图：\n\n![](images/8c479a6f4fadc69702238ce2be3af1ce6989dec407510a68f7cb5ee86e4bcb9a.jpg)",
+        "merged_id": "unit2_276",
+        "original_id": "unit2_choice_276",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "作图题"
+      },
+      {
+        "id": "unit2_choice_277",
+        "question": "中国是一个航运大国，我国自主研发的北斗导航系统对船舶实行全球、全天候、实时定位导航，为船舶营运安全和高效提供了可靠保证。在某海域一艘客轮和一艘货轮均需物资救助，船长发出SOS信号。指挥中心通过北斗定位发现：营救船'海巡一号'在客轮的东偏北  $45^{\\circ}$  方向60千米处，客轮在货轮北偏东  $30^{\\circ}$  方向40千米处。\n\n![](images/ce795ad7890b19f463502058ba987797d3c46873627498846e03e6f7f1356c63.jpg)\n\n(1) 请在图中标出'海巡一号'和货轮的位置。\n\n(2)'海巡一号'的平均航速是35千米/时，接到信号后最快大约()小时可到达客轮所在地。\n\n(3) 按计划，'海巡一号'应先将物资送到客轮所在地，然后再前往货轮所在地，'海巡一号'应该怎样航行？请描述它的航行路线。",
+        "options": [],
+        "answer": "(1) 见详解\n\n(2)  $\\frac{12}{7}$\n\n(3) 见详解",
+        "source": "23-24六年级上·重庆·期末",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置、根据方向、角度和距离描述路线图"
+        ],
+        "solution": "【分析】（1）图例表示图上1厘米相当于实际的20千米， $60 \\div 20 = 3$  （厘米），可知'海巡一号'与客轮的图上距离是3厘米，结合角度和图上距离标出'海巡一号'的位置；客轮在货轮北偏东  $30^{\\circ}$  方向40千米处，即货轮在客轮的南偏西  $30^{\\circ}$  方向40千米处， $40 \\div 20 = 2$  （厘米），即图上距离是2厘米；\n\n(2) 根据路程  $\\div$  速度  $=$  时间, 用  $60 \\div 35$  列式解答;\n\n(3) 以'海巡一号'为观测点, 结合角度和距离解答即可。\n\n【详解】(1)如图：\n\n![](images/62eb512ecd5609b990b4dead0cd2b3bf35bba815eacb910c89647195c3e457d9.jpg)\n\n(2)  $60 \\div 35 = \\frac{12}{7}$  (小时) 所以接到信号后最快大约  $\\frac{12}{7}$  小时可到达客轮所在地。\n\n(3)'海巡一号'先向西偏南  $45^{\\circ}$  方向行驶60千米到达客轮所在地，再向南偏西  $30^{\\circ}$  方向行驶40千米到达货轮所在地。",
+        "merged_id": "unit2_277",
+        "original_id": "unit2_choice_277",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "作图题"
+      },
+      {
+        "id": "unit2_choice_278",
+        "question": "![](images/09e1e91eca323ad86510d5a58a42ea8c8e63227172112ce8a08def4f283f0d40.jpg)\n\n如图所示，以学校为观测点：\n\n(1) 幼儿园在学校()偏()的方向上, 距离是()m。\n\n(2)书店在学校（）偏（）的方向上，距离是（）m。\n\n(3) 商店在学校（）偏（）的方向上，距离是（）m。\n\n(4) 银行在学校西偏南  $35^{\\circ}$  的方向上, 距离  $300 \\mathrm{~m}$ , 请在图中标出来。",
+        "options": [],
+        "answer": "(1)北; 西  $45^{\\circ}$ ; 300\n\n(2)东;北  $25^{\\circ}$ ; 200\n\n(3) 南; 东  $33^{\\circ}$ ; 300\n\n(4) 见详解",
+        "source": "24-25六年级上·内蒙古通辽·期中",
+        "difficulty": "容易",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "solution": "【分析】上北下南，左西右东。\n\n(1) 看图, 幼儿园在学校的北偏西  $45^{\\circ}$  方向上, 距离是  $3 \\times 100 = 300(\\mathrm{~m})$ ;\n\n(2) 看图, 书店在学校的东偏北  $25^{\\circ}$  方向上, 距离是  $2 \\times 100 = 200(\\mathrm{~m})$ ;\n\n(3) 看图, 商店在学校的南偏东  $33^{\\circ}$  方向上, 距离是  $3 \\times 100 = 300(\\mathrm{~m})$ ;\n\n(4) 先找出学校的西偏南  $35^{\\circ}$  方向, 银行在此方向上的  $300 \\div 100 = 3(\\mathrm{~cm})$  处。\n\n【详解】（1）幼儿园在学校北偏西  $45^{\\circ}$  的方向上，距离是  $300 \\mathrm{~m}$  。\n\n(2)书店在学校东偏北  $25^{\\circ}$  的方向上, 距离是  $200 \\mathrm{~m}$  。\n\n(3) 商店在学校南偏东  $33^{\\circ}$  的方向上, 距离是  $300 \\mathrm{~m}$  。\n\n![](images/93bbb57b825b0b49d518afc60df7f000bacc4c90f703254d2497c2a8c7537ff4.jpg)\n\n(4) 如图:",
+        "merged_id": "unit2_278",
+        "original_id": "unit2_choice_278",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "作图题"
+      },
+      {
+        "id": "unit2_choice_279",
+        "question": "李叔叔是一名智能快递机器人分拣系统的程序员, 他为机器人'小橙人'设计了一张行走路线图。\n\n![](images/3f49c37df52d8b084ebd95ac3c85be2e22afd34a140bb612b97505a6347945e2.jpg)\n\n(1)'小橙人'在出发点'驼上'包裹，先向（）偏（）（）°方向行走（）米到达A地扫码、称重，再向（）偏（）（）°方向行走（）米把包裹送到B地，然后自动掀起托盘，将包裹投入分拣口。\n\n(2) 完成上述工作后，'小橙人'从 B 地回到自己的'家'C地。C地位于B地西偏北  $40^{\\circ}$  方向9米处，请在图中标出'小橙人'的家'C'地。",
+        "options": [],
+        "answer": "（1）西；北；25；12；东；北；10；16\n\n(2) 见详解",
+        "source": "24-25六年级上·河南三门峡·期中",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置、根据方向、角度和距离描述路线图"
+        ],
+        "solution": "【分析】（1）根据题意可知，1厘米表示4米，先计算出各个路段的实际距离；确定线路时，注意起始点与目的地，起始点是观测点，再按照地图上'上北下南,左西右东'确定方向，用方向、角度和距离描述'小橙人'的行驶路线。\n\n(2) 先计算出 B 地到 C 地的图上距离, 再以 B 地为观测点, 画出 C 地的位置, 即可解答。\n\n【详解】(1)  $4 \\times 3 = 12$  （米）；  $4 \\times 4 = 16$  （米）  $90^{\\circ} - 25^{\\circ} = 65^{\\circ}$  ；  $90^{\\circ} - 10^{\\circ} = 80^{\\circ}$\n\n'小橙人'在出发点'驼上'包裹，先向西偏北  $25^{\\circ}$  （或北偏西  $65^{\\circ}$  ）方向行走12米到达A地扫码、称重，再向东偏北  $10^{\\circ}$  （或北偏东  $80^{\\circ}$  ）方向走16米把包裹送到B地，然后自动掀起托盘，将包裹投入分拣口。\n\n(2)  $9 \\div 4 = 2.25$  （厘米）\n\n如图：\n\n![](images/a41cda26b038d25f9a05d994d51521c715968244e4e6fb3f4a83e5da0e799a97.jpg)",
+        "merged_id": "unit2_279",
+        "original_id": "unit2_choice_279",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "作图题"
+      },
+      {
+        "id": "unit2_choice_280",
+        "question": "星光电器商行接到一份外卖订单。外卖员从星光电器商行出发, 先向东偏南  $45^{\\circ}$  方向行驶1000米到万达广场, 再向西偏南  $30^{\\circ}$  方向行驶1500米到达体育中心, 再向北偏西  $70^{\\circ}$  方向行驶1000米到达电影院, 最后正西方向行驶500米到达绿苑小区。根据上面的描述画出外卖员的路线图。\n\n![](images/96ca72ba459d3296417239f01f25039e0239500c87464001cffb12a75af78e5f.jpg)",
+        "options": [],
+        "answer": "图见详解",
+        "source": "24-25六年级上·福建福州·期中",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离画线路图"
+        ],
+        "solution": "【分析】由题意知: 以星光电器商行为参照点建立方向标,  $1000 \\div 500 = 2$  (个), 从星光电器商行向东偏南  $45^{\\circ}$  方向画 2 个单位长度, 标出万达广场; 再以万达广场为参照点建立方向标,  $1500 \\div 500 = 3$  (个), 从万达广场向西偏南  $30^{\\circ}$  方向画 3 个单位长度, 标出体育中心; 再以体育中心为参照点建立方向标, 从体育中心向北偏西70°方向画出2个单位长度，标出电影院；最后以电影院为参照点建立方向标，从电影院向正西方向画1个单位长度，标出绿苑小区。据此作图即可。\n\n【详解】据分析作图如下：\n\n![](images/3ee7f5ef25990345bef9cd9124eaaa473586a848e1d35e78da6c74a8491b8cfd.jpg)",
+        "merged_id": "unit2_280",
+        "original_id": "unit2_choice_280",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "作图题"
+      },
+      {
+        "id": "unit2_choice_281",
+        "question": "下面是奇奇写的一篇关于参观消防站的日记。\n\n我从大门出发，向东行走100米到达执勤楼，再向正北方向行走100米到达模拟训练场。同行的奇妙则从大门出发，朝东偏北  $45^{\\circ}$  方向行走，与我在模拟训练场相遇。\n\n请你根据奇异的描述画出他们两人行走的路线。\n\n![](images/a7bfb95e34c8323dba6cca71c58e119bf4cedcfe1a2bb8ad8dd398ca0a1a4768.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "source": "24-25 六年级上·广东东莞·期中",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置、根据方向、角度和距离画线路图、东、南、西、北方向"
+        ],
+        "solution": "【分析】根据题意可知，1厘米表示实际距离是50米，先用除法计算出图上的长度；再根据地图上方法的规定'上北下南,左西右东'，找准观测点，确定下一个要到的地点的方向和距离，据此解答。\n\n【详解】  $100 \\div 50 = 2$  （厘米）\n\n如图：\n\n![](images/c9496a676a3c30997b6bff3b09923ce16f7b3e4a7ca07621a3fd3d7b46360ba1.jpg)",
+        "merged_id": "unit2_281",
+        "original_id": "unit2_choice_281",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "作图题"
+      },
+      {
+        "id": "unit2_choice_282",
+        "question": "每年的农历九月初九是重阳节，也称敬老节，是中国传统节日，明明一家带着爷爷奶奶到某景区游玩，以下是路线图，请根据要求完成下面各题。（一个小正方形的对角线长10米)\n\n![](images/6fc9cca27a523070ea64b02a290847f9627aeca13e9ba558f9246fae2e342e5f.jpg)\n\n(1) 从景区 A 出发, 向北偏东  $45^{\\circ}$  方向走 30 米后到达景区 B, 再向东偏南  $45^{\\circ}$  方向走 30 米到达景区 C, 请标出景区 B 和 C 的位置。\n\n(2) 连接景区 A、B、C，观察这是一个（）。\n\n(3) 计算这个图形的面积。",
+        "options": [],
+        "answer": "(1) 见详解;\n\n(2) 等腰直角三角形;\n\n(3) 450 平方米",
+        "source": "23-24六年级上·福建厦门·期中",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "等腰三角形和等边三角形的认识及特征、三角形面积的计算、三角形的分类、根据方向、角度和距离画线路图"
+        ],
+        "solution": "【分析】（1）以图上的'上北下南,左西右东'为准，图中一个小正方形的对角线相当于实际距离10米；\n\n在景区A的北偏东  $45^{\\circ}$  方向上画  $30 \\div 10 = 3$  条对角线的长度, 即是景区B;\n\n在景区B的东偏南  $45^{\\circ}$  方向上画  $30 \\div 10 = 3$  条对角线的长度,即是景区C。\n\n(2) 连接 A、B、C 三点, 发现这是一个三角形, 两条腰都是30米，且有一个内角是直角，由此可知这是一个等腰直角三角形。\n\n(3) 这个等腰直角三角形的两条直角边是 30 米, 即这个三角形的底和高都是 30 米; 根据三角形的面积  $= \\text{底} \\times \\text{高} \\div 2$ , 代入数据计算即可求出它的面积。\n\n【详解】（1）景区B和C的位置如下图：\n\n![](images/3edb3a87d56856cb30ef08268ade52c819363cbc5ad6e7f2fc758aa2d86ed841.jpg)\n\n(2) 连接景区 A、B、C，观察这是一个等腰直角三角形。\n\n(3)  $30 \\times 30 \\div 2 = 900 \\div 2 = 450$  （平方米）\n\n答: 这个图形的面积是 450 平方米。",
+        "merged_id": "unit2_282",
+        "original_id": "unit2_choice_282",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "作图题"
+      },
+      {
+        "id": "unit2_choice_283",
+        "question": "淞江实验小学的同学们在参加暑假夏令营活动时, 从基地出发, 先向东偏北  $30^{\\circ}$  方向行进  $400 \\mathrm{~m}$  后到达A点, 再向正西方向行进  $400 \\mathrm{~m}$  到达B点, 再向南偏西  $45^{\\circ}$  方向行进  $300 \\mathrm{~m}$  到达C点, 最后再向东偏南  $40^{\\circ}$  方向行进  $200 \\mathrm{~m}$  到达目的地。\n\n![](images/552d2af750d7213fbcb5581b89148dfbea7f15beeff8f49d2dea51737eb912a8.jpg)\n\n![](images/9e0de965f2121d5db414dd9feacee249b67639a74f246cbcd92288b0e206886e.jpg)\n\n(1) 根据描述, 画出行进路线图。\n\n(2) 活动结束后，请描述出返程路线。\n\n(3) 已知从基地到目的地用时 10 分钟, 返程平均速度是去时的  $\\frac{10}{13}$ , 返回程用时 ( ) 分钟。",
+        "options": [],
+        "answer": "(1) 见详解\n\n(2) 见详解\n\n(3) 13",
+        "source": "23-24六年级上·贵州遵义·期中",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "基础行程问题、根据方向、角度和距离画线路图、根据方向、角度和距离描述路线图"
+        ],
+        "solution": "【分析】（1）（2）确定观测点，根据图中方向上北下南，左西右东，确定方向、角度和距离。\n\n(3) 先用加法计算总路程, 根据路程  $\\div$  时间 = 速度, 算出去时的速度, 再根据求一个数的几分之几是多少, 用乘法计算,求出返程平均速度, 最后根据路程  $\\div$  速度 = 时间, 计算时间即可。\n\n【详解】(1)如下图：\n\n![](images/ab9355fd17f4598a56e8d4faa85f3082ae2ef3dd2b1bc7cba78eb6f567718a6d.jpg)\n\n(2)返程路线:从目的地出发,先向西偏北  $40^{\\circ}$  (或北偏西  $50^{\\circ}$  ）方向行进  $200 \\mathrm{~m}$  到达 C 点，再向北偏东  $45^{\\circ}$  （或东偏北  $45^{\\circ}$  ）方向行进  $300 \\mathrm{~m}$  到达 B 点，再向正东方向行进  $400 \\mathrm{~m}$  到达 A 点，最后再向西偏南  $30^{\\circ}$  （或南偏西  $60^{\\circ}$  ）方向行进  $400 \\mathrm{~m}$  到达基地。\n\n(3)  $(400 + 400 + 300 + 200) \\div 10 \\times \\frac{10}{13} = 1300 \\div 10 \\times \\frac{10}{13} = 130 \\times \\frac{10}{13} = 100$  (米/分钟)\n\n$(400 + 400 + 300 + 200) \\div 100 = 1300 \\div 100 = 13$  (分钟)返程用时13分钟。",
+        "merged_id": "unit2_283",
+        "original_id": "unit2_choice_283",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "作图题"
+      },
+      {
+        "id": "unit2_choice_284",
+        "question": "1路公共汽车从起点站先向正西方向行驶2千米，再向西偏北  $40^{\\circ}$  方向行驶4千米，最后向南偏西  $30^{\\circ}$  方向行驶6千米到达终点站。\n\n(1) 根据上面的描述，画出 1 路公共汽车行驶的路线图。单位长度为  $2 \\mathrm{~km}$\n\n![](images/96f3eb8507acc6d966d8bd0f1a76caf6715d5af62e39a19c195532bbd9c343d5.jpg)\n\n(2) 1 路公共汽车从起点站到终点站共用了 15 分钟, 平均每分钟行驶多少米?",
+        "options": [],
+        "answer": "（1）见详解\n\n(2)800米",
+        "source": "23-24六年级上·湖北黄冈·期中",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "基础行程问题、根据方向、角度和距离画线路图"
+        ],
+        "solution": "【分析】（1）地图方向是上北下南，左西右东，图上1厘米代表实际距离2千米，根据1路公共汽车从起点站出发的线路图，画出即可；\n\n(2) 用总路程除以时间, 求出 1 路公共汽车的速度, 据此解答即可。\n\n【详解】(1)如图：\n\n![](images/7e96eec60c87f8813535d56f6cb156a961fd463be9a815e355b2d851b66c2bb2.jpg)\n\n(2) 路程:  $2 + 4 + 6 = 6 + 6 = 12$  (千米)\n\n12千米=12000米\n\n速度:  $12000 \\div 15 = 800$  (米/分)\n\n答: 平均每分钟行驶 800 米。",
+        "merged_id": "unit2_284",
+        "original_id": "unit2_choice_284",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "作图题"
+      },
+      {
+        "id": "unit2_choice_285",
+        "question": "根据描述, 在下图中标出各场所的位置。\n\n(1)书店在广场北偏东  $40^{\\circ}$  方向，600米处。\n\n(2) 超市在广场北偏西  $55^{\\circ}$  方向, 400 米。\n\n(3) 公园在广场南偏西  $60^{\\circ}$  方向, 200 米处。\n\n(4) 体育馆在广场南偏东  $30^{\\circ}$  方向, 600 米处。\n\n![](images/3f400c24133c8964547ae67d04b4412322964bd9c190e6f95b7f3d8870ac342b.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "source": "23-24六年级上·全国·单元测试",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "solution": "【分析】根据题意可知，1厘米表示实际距离200米，分别计算出广场到书店、超市、公园、体育场的图上距离，再根据地图上方向的规定：上北下南，左西右东；以广场为观测点，分别画出书店、超市、公园、体育场的位置，据此解答。\n\n【详解】  $600 \\div 200 = 3$  （厘米）\n\n$$\n4 0 0 \\div 2 0 0 = 2 (\\text {厘 米})\n$$\n\n$$\n2 0 0 \\div 2 0 0 = 1 (\\text {厘 米})\n$$\n\n$$\n6 0 0 \\div 2 0 0 = 3 (\\text {厘 米})\n$$\n\n如图：\n\n![](images/cd1c517c5141339e5e55594f43e9921c18bc01c168ef9658d5a95c28e701f53c.jpg)",
+        "merged_id": "unit2_285",
+        "original_id": "unit2_choice_285",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "作图题"
+      },
+      {
+        "id": "unit2_choice_286",
+        "question": "根据下面的描述, 在平面示意图上标出各位同学的位置。\n\n![](images/affe08b6cf4c545972a9ac965a1c149cdb0e7502148d4ad90947ce351d70dd42.jpg)\n\n(1) 李云在驻地东偏北  $45^{\\circ}$  的方向上, 距离是 200 米。\n\n(2) 张红在驻地的正西方向上, 距离是 100 米。\n\n(3) 杜江在驻地北偏西  $60^{\\circ}$  的方向上, 距离是 50 米。\n\n(4) 刘伟在驻地东偏南  $50^{\\circ}$  的方向上, 距离是 300 米。",
+        "options": [],
+        "answer": "(1)(2)(3)(4)见详解",
+        "source": "23-24 六年级上·全国·单元测试",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "solution": "【分析】（1）根据题意可知，1 厘米表示实际距离 50 米，计算出驻地与李云之间的图上距离，再根据地图上方向的规定'上北下南,左西右东'，以驻地为观测点，东偏北  $45^{\\circ}$  方向上 200 米处，画出李云的位置；\n\n(2) 计算出驻地与张红的图上距离, 再以驻地为观测点, 正西方向上 100 米处, 画出张红的位置;\n\n(3) 计算出驻地与杜江的图上距离, 再以驻地为观测点, 北偏西  $60^{\\circ}$  方向上 50 米处, 画出杜江的位置;\n\n(4) 计算出驻地与刘伟的图上距离, 再以驻地为观测点, 东偏南  $50^{\\circ}$  方向上 300 米处, 画出刘伟的位置。\n\n【详解】(1)  $200 \\div 50 = 4$  （厘米）\n\n(2)  $100 \\div 50 = 2$  （厘米）\n\n(3)  $50 \\div 50 = 1$  (厘米)\n\n(4)  $300 \\div 50 = 6$  (厘米)\n\n![](images/a1846b06e5e95561ad0ae947332d47946407e001e40d1c9743f5e3eba1cb75c7.jpg)\n\n图如下：",
+        "merged_id": "unit2_286",
+        "original_id": "unit2_choice_286",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "作图题"
+      },
+      {
+        "id": "unit2_choice_287",
+        "question": "野生动物观察员观测到一条蛇在甲处，1小时后这条蛇沿直线爬到了乙处，又经过半小时爬到了丙处。\n\n![](images/25d00a49c8508bd2bddb7450c7712d9633b3af490b53f30115de1d006eb07a27.jpg)\n\n(1) 请描述这条蛇从甲处经乙处到丙处的行走路线。\n\n(2) 最后, 这条蛇从丙处向西偏北  $45^{\\circ}$  方向行走  $200 \\mathrm{~m}$  到达丁处, 请在图中画出这条蛇从丙处到丁处的行走路线。",
+        "options": [],
+        "answer": "见详解",
+        "source": "23-24 六年级上·全国·单元测试",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离画线路图、根据方向、角度和距离描述路线图"
+        ],
+        "solution": "【分析】（1）以甲为观测点，乙在甲的东偏南  $35^{\\circ}$  （或南偏东  $55^{\\circ}$  ）方向  $100 \\times 4 = 400$  米处；以乙为观测点，丙在乙的东偏北  $25^{\\circ}$  （或北偏东  $65^{\\circ}$  ）方向  $3 \\times 100 = 300$  米处；\n\n(2) 以丙为观测点, 在丙的西偏北  $45^{\\circ}$  方向截取  $200 \\div 100 = 2$  个单位长度, 终点处标注丁即可。\n\n【详解】（1）蛇从甲处出发向东偏南  $35^{\\circ}$  （或南偏东  $55^{\\circ}$  ）方向飞行400米到达乙处，再从乙处出发向东偏北  $25^{\\circ}$  （或北偏东  $65^{\\circ}$  ）方向飞行300米到达丙处。\n\n(2)如图：\n\n![](images/4312f3d403f8a630cb9d86d9447326f57371dee99f57e35699d71312c537076c.jpg)",
+        "merged_id": "unit2_287",
+        "original_id": "unit2_choice_287",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "作图题"
+      },
+      {
+        "id": "unit2_choice_288",
+        "question": "六 (1) 班同学在春游时做了一个找宝藏的游戏, 他们需要从驻地出发, 先依次到 A、B、C 三地猜对谜语, 才能到 D 处找到宝藏。已知 A、B、C、D 四地的位置关系如下:\n\n(1) A 地在驻地西偏北  $30^{\\circ}$  方向  $600 \\mathrm{~m}$  处。\n\n(2) B 地在 A 地东偏北  $45^{\\circ}$  方向  $250 \\mathrm{~m}$  处。\n\n(3) C 地在 B 地正南方向  $400 \\mathrm{~m}$  处。\n\n(4) D 地在 C 地南偏东  $60^{\\circ}$  方向  $1000 \\mathrm{~m}$  处。\n\n请根据题意画出同学们找到宝藏的路线图。",
+        "options": [],
+        "answer": "见详解",
+        "source": "23-24 六年级上·全国·单元测试",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "solution": "【分析】用  $1 \\mathrm{~cm}$  表示  $100 \\mathrm{~m}$ , 根据一个地点的位置是由方向和距离决定的, 结合上北下南、左西右东画图即可。\n\n【详解】(1)(2)(3)(4)如下图：\n\n![](images/57d64baf9478ea46d1f3acb865e8ebf5e76bc2d183375b6e8b2252e97fd4e49a.jpg)",
+        "merged_id": "unit2_288",
+        "original_id": "unit2_choice_288",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "作图题"
+      },
+      {
+        "id": "unit2_choice_289",
+        "question": "五年级的同学在进行徒步活动。他们从学校出发, 先向正东方向走了 3 千米, 然后向正北方向走了 2 千米, 接着向正西方向走了 1 千米, 最后沿西偏南  $45^{\\circ}$  方向返回学校。请根据以上信息, 画出同学们徒步路线图。\n\n![](images/c11c2bf3e3e439a61b4530b8303bbb6035c3db26b8f2951fa99323ee13cb2b43.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "source": "23-24 五年级下·福建泉州·期末",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离画线路图"
+        ],
+        "solution": "【分析】由图可知, 图上1厘米代表实际距离1千米, 由此计算出行走的各个路段的图上距离, 利用平面图上方向规定: 上北下南, 左西右东, 依据题意结合图示去解答。\n\n【详解】  $3 \\div 1 = 3$  （厘米）\n\n$$\n2 \\div 1 = 2 (\\text {厘 米})\n$$\n\n如图：\n\n![](images/dc184c76046e3c69cbcb345facf4f2ac0d23d9c62761ccf91bfc644258844f2e.jpg)",
+        "merged_id": "unit2_289",
+        "original_id": "unit2_choice_289",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "作图题"
+      },
+      {
+        "id": "unit2_choice_290",
+        "question": "根据动物园示意图完成下面各题。\n\n![](images/52351168b767482e3b950102e38cf7a950abd0f2f69d0d0edd21a511d744efc9.jpg)\n\n(1) 河马馆在入口往北 200 米, 再往东 200 米处, 河马馆的位置用数对表示是 ( ) ; 天鹅湖在海豚馆往西 50 米, 再往南 200 米处, 天鹅湖的位置用数对表示是 ( ) 。请在图中标出河马馆和天鹅湖的位置。\n\n(2) 海豚馆的位置用数对表示是 (1, 8), 从海豚馆向 ( ) 走 ( ) 米到达狒狒馆, 狐狒馆的位置用数对表示是 ( )。\n\n(3) '五一'小长假, 乐乐一家在动物园的游览路线是这样的: 他们从入口往北走 100 米后向西走 200 米, 然后向北走 200 米, 再向北走 100 米, 最后往东走 450 米后接着向南走 250 米。乐乐一家游览了哪些动物馆? 请你按先后顺序写下来。",
+        "options": [],
+        "answer": "（1）河马馆（9，4）；天鹅湖（0，4）；图见详解(2)南;100;(1,6)(3)熊猫馆  $\\rightarrow$  狒狒馆  $\\rightarrow$  海豚馆  $\\rightarrow$  大象馆",
+        "source": "25-26 五年级上·全国·单元测试",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "用数对表示位置、根据方向、角度和距离确定物体的位置、除数是整十数的口算除法、根据数对找位置"
+        ],
+        "solution": "【分析】（1）用数对表示位置时，前面一个数表示第几列，后面一个数表示第几行；图中每个正方形边长是50米，河马馆在入口往北200米，再往东走200米，也就是入口处先往上数  $(200 \\div 50 = 4)$  格, 再往右数  $(200 \\div 50 = 4)$  格。天鹅湖在海豚馆往西 50 米, 再往南 200 米, 也就是海豚馆先往西数  $(50 \\div 50 = 1)$  格, 再往南数  $(200 \\div 50 = 4)$  格。\n\n(2) 海豚馆向下走 2 格到达狒狒馆, 狗狒馆在第 1 列, 第 6 行, 据此可用数对表示出狒狒馆的位置。\n\n(3) 走的格数 = 路程 ÷ 平均每格的长度, 再根据上北下南, 左西右东的方位特征, 据此描述游览路线。\n\n【详解】(1)  $200 \\div 50 = 4$  （格）\n\n$$\n\\begin{array}{l} 0 + 4 = 4, 5 + 4 = 9 \\\\ 1 - 1 = 0, 8 - 4 = 4 \\\\ \\end{array}\n$$\n\n因此河马馆的位置在9列，第4行，用数对表示是(9, 4)；天鹅湖的位置在0列，第4行，用数对表示是(0,4)。在图中的位置如下图:\n\n![](images/d0b69436b588b38f07ab88be71856f4cbf777fab6fc17bd9a3dec3e362c7f8d0.jpg)\n\n(2) 海豚馆和狒狒馆所在的列数相同, 海豚馆在第 8 行, 狍狒馆在第 6 行, 两者之间相差 2 格, 即  $50 \\times 2 = 100$  (米)。因此海豚馆的位置用数对表示是 (1, 8), 从海豚馆向南走 100 米到达狒狒馆, 狗狒馆的位置在 1 列, 第 6 行, 用数对表示是 (1, 6)。\n\n(3)  $100 \\div 50 = 2$  (格)\n\n$200 \\div 50 = 4$  (格)\n\n$450 \\div 50 = 9$  (格)\n\n$250 \\div 50 = 5$  (格)\n\n从入口往北走100米到熊猫馆，后向西走200米，然后向北走200米到狒狒馆，再向北走100米到海豚馆，最后往东走450米后接着向南走250米到大象馆。\n\n答：乐乐一家从入口出发，依次游览了熊猫馆、狒狒馆、海豚馆、最后到大象馆。",
+        "merged_id": "unit2_290",
+        "original_id": "unit2_choice_290",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "解答题"
+      },
+      {
+        "id": "unit2_choice_291",
+        "question": "适逢节假日,聪聪一家打算去洛阳游玩,下图是他们规划的自驾游路线。\n\n![](images/7c848c52b50835ab5a421ddb87e74c8342c4a9ceab6bc83ee8f8c2830b983617.jpg)\n\n(1) 聪聪爸爸计划以 30 千米/时的速度驾车, 根据聪聪一家的自驾游路线图填写下表。\n\n<table><tr><td>路段</td><td>方向</td><td>路程/千米</td><td>用时/分</td></tr><tr><td>王城公园→应天门</td><td></td><td></td><td></td></tr><tr><td>应天门→十字街</td><td></td><td></td><td></td></tr><tr><td>十字街→洛邑古城</td><td></td><td></td><td></td></tr><tr><td>洛邑古城→河洛古城</td><td></td><td></td><td></td></tr></table>\n\n(2) 根据 (1) 中的信息, 若他们上午 10:00 从十字街出发, 在洛邑古城停留 2 小时, 在 12:30 能到达河洛古城吗?",
+        "options": [],
+        "answer": "(1) 填表见详解\n\n(2) 能",
+        "source": "24-25六年级上·全国·单元测试",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "基础行程问题、根据方向、角度和距离确定物体的位置、时、分、秒时间的推算、除数是小数的小数除法"
+        ],
+        "solution": "【分析】（1）根据距离和方向描述路线图，并通过'时间 = 路程 ÷ 速度'求出每一段路所需要的时间。\n\n(2) 本题是简单的路程与时间问题, 通过 (1) 中得到的各段路线所需要的时间加起来得到最后所需的时间, 以10:00 为起始时间，加上中间经过的时间，得到最后到达的时间，和 12:30 比较，即可判断是否能在 12:30 到达河洛古城。\n\n【详解】(1)  $① 90^{\\circ} - 80^{\\circ} = 10^{\\circ}$\n\n应天门位于王城公园北偏东  $80^{\\circ}$  （或东偏北  $10^{\\circ}$  ）方向，距离是4.2千米处。\n\n30千米/时=0.5千米/分\n\n$4.2 \\div 0.5 = 8.4$  (分钟)\n\n从王城公园到应天门需要花费8.4分钟。\n\n(2)  $90^{\\circ} - 15^{\\circ} = 75^{\\circ}$\n\n十字街位于应天门东偏北  $15^{\\circ}$  （或北偏东  $75^{\\circ}$  ）方向，距离是 3.2 千米处。\n\n$3.2 \\div 0.5 = 6.4$  (分钟)\n\n因此从应天门到十字街需要花费6.4分钟。\n\n(3)  $180^{\\circ} - 110^{\\circ} - 15^{\\circ} = 55^{\\circ}, 90^{\\circ} - 55^{\\circ} = 35^{\\circ}$\n\n洛邑古城位于十字街南偏东  $35^{\\circ}$  （或东偏南  $55^{\\circ}$  ）方向，距离是 1.5 千米处。\n\n$1.5 \\div 0.5 = 3$  (分钟)\n\n因此从十字街到洛邑古城需要花费3分钟。\n\n(4)  $90^{\\circ} - 74^{\\circ} = 16^{\\circ}$\n\n河洛古城位于洛邑古城北偏东  $74^{\\circ}$  （或东偏北  $16^{\\circ}$  ）方向，距离是 7.5 千米处。\n\n$7.5 \\div 0.5 = 15$  (分钟)\n\n因此从洛邑古城到河洛古城需花费15分钟。\n\n填表如下：\n\n<table><tr><td>路段</td><td>方向</td><td>路程/千米</td><td>用时/分</td></tr><tr><td>王城公园→应天门</td><td>北偏东 80°</td><td>4.2</td><td>8.4</td></tr><tr><td>应天门→十字街</td><td>东偏北 15°</td><td>3.2</td><td>6.4</td></tr><tr><td>十字街→洛邑古城</td><td>南偏东 35°</td><td>1.5</td><td>3</td></tr><tr><td>洛邑古城→河洛古城</td><td>北偏东 74°</td><td>7.5</td><td>15</td></tr></table>\n\n(2) 从十字街到河洛古城需要:  $3 + 15 = 18$  (分钟)\n\n10时+2小时+18分钟=12时18分\n\n12时18分  $< {12}$  时30分\n\n答: 他们12:30能到达河洛古城。",
+        "merged_id": "unit2_291",
+        "original_id": "unit2_choice_291",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "解答题"
+      },
+      {
+        "id": "unit2_choice_292",
+        "question": "龙门石窟是世界上造像最多、规模最大的石刻艺术宝库, 被称为'中国石刻艺术的最高峰', 如图是聪聪查找的龙门石窟游玩路线示意图。\n\n![](images/5b9e4a43461599fb6cc44ed460e5e4b9eac71654db3e2489d878281c9572fdfc.jpg)\n\n(1) 从这份游玩路线图可看出, 卢舍那大像龛在宾阳三洞的 ( ) 方向, 距离宾阳三洞 ( ) 米。\n\n(2)聪聪一家现在的位置在东山石窟,他们需要向()偏()的方向走()米才能到达香山寺。\n\n(3) 从东山石窟返回西山石窟入口, 以下两条路线哪条路线更近?\n\n路线一\n\n东山石窟  $\\rightarrow$  香山寺  $\\rightarrow$  龙门桥  $\\rightarrow$  禹王池  $\\rightarrow$  西山石窟入口\n\n路线二\n\n东山石窟  $\\rightarrow$  伊阙桥  $\\rightarrow$  卢舍那大像龛  $\\rightarrow$  宾阳三洞  $\\rightarrow$  禹王池  $\\rightarrow$  西山石窟入口",
+        "options": [],
+        "answer": "(1)南偏西  $12^{\\circ}$ ; 490\n\n(2)北;东;10;779\n\n(3) 路线二",
+        "source": "24-25 六年级上·全国·单元测试",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置、角度的计算、东、南、西、北方向、两、三位数的连加运算"
+        ],
+        "solution": "【分析】（1）以宾阳三洞为观测点，卢舍那大像龛在南偏西  $12^{\\circ}$  方向，或西偏南方向，角度是  $(90^{\\circ} - 12^{\\circ})$ ；距离宾阳三洞490米。\n\n(2) 以东山石窟为观测点, 香山寺在北偏东方向, 角度是  $\\left[68^{\\circ} - \\left(90^{\\circ} - 32^{\\circ}\\right)\\right]$ , 或东偏北方向, 角度是  $\\left(180^{\\circ} - 68^{\\circ} - 32^{\\circ}\\right)$ , 距离东山石窟 779 米。\n\n(3) 根据线路图, 分别计算出两条路线的长度, 再比较大小, 选择路线较短的即可。\n\n【详解】(1)  $90^{\\circ} - 12^{\\circ} = 78^{\\circ}$\n\n以宾阳三洞为观测点，卢舍那大像龛在南偏西  $12^{\\circ}$  方向，或西偏南  $78^{\\circ}$  方向，距离宾阳三洞 490 米。\n\n因此从这份游玩路线图可看出, 卢舍那大像龛在宾阳三洞的南偏西  $12^{\\circ}$  (或西偏南  $78^{\\circ}$ ) 方向, 距离宾阳三洞 490 米。\n\n(2)  $68^{\\circ} - (90^{\\circ} - 32^{\\circ})$\n\n$$\n\\begin{array}{l} = 6 8 ^ {\\circ} - 5 8 ^ {\\circ} \\\\ = 1 0 ^ {\\circ} \\\\ \\end{array}\n$$\n\n以东山石窟为观测点，香山寺在北偏东  $10^{\\circ}$  方向，距离东山石窟 779 米。\n\n$$\n1 8 0 ^ {\\circ} - 6 8 ^ {\\circ} - 3 2 ^ {\\circ} = 8 0 ^ {\\circ}\n$$\n\n以东山石窟为观测点，香山寺在东偏北  $80^{\\circ}$  方向，距离东山石窟 779 米。\n\n因此聪聪一家现在的位置在东山石窟，他们需要向北偏东  $10^{\\circ}$  （或东偏北  $80^{\\circ}$  ）的方向走 779 米才能到达香山寺。\n\n(3) 路线一全长:  $779 + 300 + 525 = 1604$  (米)\n\n路线二全长:  $270 + 490 + 125 + 525 = 1410$  (米)\n\n因为  $1604 > 1410$  ，所以路线二全长较短，选择路线二更近。\n\n答：从东山石窟返回西山石窟入口，路线二更近。",
+        "merged_id": "unit2_292",
+        "original_id": "unit2_choice_292",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "解答题"
+      },
+      {
+        "id": "unit2_choice_293",
+        "question": "请根据下面的描述画图。\n\n![](images/9a0efa0ce1420209ebe8f497a3f90c0049ecfc131ff8bd1756f0041fc1b7740b.jpg)\n\n(1) 点 B 在点 A 东偏北  $60^{\\circ}$  方向, 距离点 A2 厘米; 点 C 在点 B 西偏北  $30^{\\circ}$  方向, 距离点 B3 厘米; 点 D 在点 C 西偏南  $60^{\\circ}$  方向, 距离点 C2 厘米。在图中画出点 B、C、D。\n\n(2) 顺次连接点 A、B、C、D、A，得到一个什么图形？在其中截取一个最大的正方形，余下部分的面积是多少平方厘米？",
+        "options": [],
+        "answer": "（1）见详解；（2）长方形；2平方厘米",
+        "source": "24-25 六年级上·全国·单元测试",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置、正方形的面积、长方形的面积"
+        ],
+        "solution": "【分析】（1）先根据描述确定方向，再利用距离确定点B、C、D的位置，最后连接各点。\n\n(2) 连接点 A、B、C、D、A，得到一个长方形。长方形的长为 3 厘米，宽为 2 厘米，截取的最大正方形的边长为 2 厘米，余下部分的面积 = 长方形面积 - 正方形面积。根据长方形面积公式：面积  $=$  长  $\\times$  宽；正方形面积公式：面积  $=$  边长  $\\times$  边长，把数据代入计算即可。\n\n【详解】\n\n![](images/fa1c76327206cf1d97c4b30ba4de66d988d9dcb83fd701be25aa9b68e47a04bc.jpg)\n\n(1) 如图:\n\n(2) 连接点 A、B、C、D、A，如图\n\n![](images/42633ea755a426ead56e8e74d098c327716581e7e87b2c951d57529ddf49449d.jpg)\n\n，得到一个长方形。\n\n$$\n\\begin{array}{l} 3 \\times 2 - 2 \\times 2 \\\\ = 6 - 4 \\\\ \\end{array}\n$$\n\n# $= 2$  （平方厘米）\n\n答: 得到一个长方形, 截取后余下部分的面积是 2 平方厘米。",
+        "merged_id": "unit2_293",
+        "original_id": "unit2_choice_293",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "解答题"
+      },
+      {
+        "id": "unit2_choice_294",
+        "question": "下面是从李彤家到赵晶家的路线图。\n\n![](images/673a9653158f8fd33c6b68b4e52fb78aab75df4004eaa6f7be8ac899471ba7a4.jpg)\n\n(1) 填表。\n\n<table><tr><td>路段</td><td>方向</td><td>路程</td><td>时间</td></tr><tr><td>李彤家→曹燕家</td><td></td><td></td><td>12分</td></tr><tr><td>曹燕家→赵晶家</td><td></td><td></td><td>8分</td></tr><tr><td>赵晶家→曹燕家</td><td></td><td></td><td>10分</td></tr><tr><td>曹燕家→李彤家</td><td></td><td></td><td>20分</td></tr><tr><td>全程</td><td>——</td><td></td><td></td></tr></table>\n\n(2) 李彤走完全程的平均速度是多少？",
+        "options": [],
+        "answer": "(1) 见详解;\n\n(2) 40米/分",
+        "source": "24-25六年级上·全国·单元测试",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "基础行程问题、根据方向、角度和距离描述路线图"
+        ],
+        "solution": "【分析】（1）描述路线图时，先确定观测点，再根据'上北下南,左西右东'结合题中角度描述方向，最后根据两地之间的距离确定目的地的位置；\n\n(2) 由表可知, 全程是 2000 米, 行完全程需要 50 分, 根据 '速度 = 路程 ÷ 时间' 即可求得李彤走完全程的平均速度, 据此解答。\n\n【详解】\n\n![](images/2d2501b5e0529bc598114ab3385e239dd2883071da7cf73b4281da8a26118546.jpg)\n\n填表如下：\n\n<table><tr><td>路段</td><td>方向</td><td>路程</td><td>时间</td></tr><tr><td>李彤家→曹燕家</td><td>北偏西45°或西偏北45°</td><td>600米</td><td>12分</td></tr><tr><td>曹燕家→赵晶家</td><td>南偏西60°或西偏南30°</td><td>400米</td><td>8分</td></tr><tr><td>赵晶家→曹燕家</td><td>北偏东60°或东偏北30°</td><td>400米</td><td>10分</td></tr><tr><td>曹燕家→李彤家</td><td>南偏东45°或东偏南45°</td><td>600米</td><td>20分</td></tr><tr><td>全程</td><td>——</td><td>2000米</td><td>50分</td></tr></table>\n\n$$\n\\begin{array}{l} 6 0 0 + 4 0 0 + 4 0 0 + 6 0 0 \\\\ = (6 0 0 + 4 0 0) + (4 0 0 + 6 0 0) \\\\ = 1 0 0 0 + 1 0 0 0 \\\\ = 2 0 0 0 (\\text {米}) \\\\ \\end{array}\n$$\n\n$$\n\\begin{array}{l} 1 2 + 8 + 1 0 + 2 0 \\\\ = 2 0 + 1 0 + 2 0 \\\\ = 5 0 (\\text {分}) \\\\ \\end{array}\n$$\n\n(2)  $2000 \\div 50 = 40$  （米/分）\n\n答: 李彤走完全程的平均速度是 40 米/分。",
+        "merged_id": "unit2_294",
+        "original_id": "unit2_choice_294",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "解答题"
+      },
+      {
+        "id": "unit2_choice_295",
+        "question": "郑州市二七纪念馆是河南省第一批红色教育基地之一。其中二七纪念塔修建于 1971 年 7 月 1 日, 是为纪念京汉铁路工人大罢工中牺牲的烈士, 发扬'二七'革命传统而修建的纪念性建筑物。下图是从二七纪念塔到河南博物院的一条交通线路。\n\n![](images/7e9b2d4de7ca2e3ff536a2f67481e54baf8613abf2f1bcee88c446ee73a3a809.jpg)\n\n(1) 看图填空。\n\n乘坐地铁1号线从二七广场站出发，向（）偏（）(°行驶（)站到达紫荆山站。\n\n(2) 请你写出从紫荆山站到河南博物院站的交通行驶路线。",
+        "options": [],
+        "answer": "(1)北; 东; 50; 2\n\n(2) 见详解",
+        "source": "23-24六年级下·重庆酉阳·期末",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置、根据方向、角度和距离描述路线图"
+        ],
+        "solution": "【分析】（1）首先看方向，图中有角度标识  $40^{\\circ}$ ，以二七广场站为观测点，根据'上北下南,左西右东'的原则，地铁1号线从二七广场站出发是向东北方向行驶，且  $90^{\\circ} - 40^{\\circ} = 50^{\\circ}$ ，所以是北偏东  $50^{\\circ}$ ；然后数站点，从二七广场站到紫荆山站，中间经过的站点数是2站（人民路站、紫荆山站，起始站二七广场站不算，到紫荆山站是2站）。\n\n(2) 以紫荆山站为观测点, 看地铁 2 号线的行驶方向是正北方向, 数站点: 从紫荆山站到黄河南路站是 1 站, 黄河南路站到关虎屯站是 1 站, 然后在关虎屯站换乘 B18 路公交车, 关虎屯站到河南省博物馆站是 1 站。\n\n【详解】（1）乘坐地铁1号线从二七广场站出发是向东北方向行驶，经过人民路站到达紫荆山站，且  $90^{\\circ} - 40^{\\circ} = 50^{\\circ}$ ，所以向北偏东  $50^{\\circ}$  行驶2站到达紫荆山站。\n\n(2) 从紫荆山站乘坐地铁2号线行驶2站，经过黄河路站到达关虎屯站，然后换乘B18路公交车行驶1站到达河南省博物院站。",
+        "merged_id": "unit2_295",
+        "original_id": "unit2_choice_295",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "解答题"
+      },
+      {
+        "id": "unit2_choice_296",
+        "question": "在图中标出各个景点的位置。\n\n![](images/e3ceca6856bccc8a9195929e5f7b6dac1b8caf3f860dff75689110aa811bab10.jpg)\n\n(1) 儿童乐园在人工湖的东偏北  $30^{\\circ}$  方向  $600 \\mathrm{~m}$  处。\n\n(2) 西游记宫在人工湖的北偏西  $45^{\\circ}$  方向  $400 \\mathrm{~m}$  处。\n\n(3) 水上世界在人工湖的西偏南  $20^{\\circ}$  方向  $400 \\mathrm{~m}$  处。\n\n(4) 迷宫在人工湖的南偏东  $35^{\\circ}$  方向  $200\\mathrm{m}$  处。",
+        "options": [],
+        "answer": "如图所示：\n\n![](images/36f721983647e5b6633fc6c93dbc11dcb8a3e7290ce6f7746107cfe104561e34.jpg)",
+        "source": "25-26六年级上·全国·课后作业",
+        "difficulty": "容易",
+        "knowledge_points": [
+          "图上距离与实际距离的换算、根据方向、角度和距离画线路图"
+        ],
+        "solution": "【分析】先依据'实际距离÷比例尺 = 图上距离'算出各景点与人工湖的图上距离，再结合方向（角度），以人工湖为观测点来确定各景点位置。\n\n【详解】计算儿童乐园图上距离:  $600 \\div 200 = 3$  (段), 以人工湖为中心, 沿东偏北  $30^{\\circ}$  方向画3段长度, 确定儿童乐园位置。\n\n计算西游记宫图上距离:  $400 \\div 200 = 2$  (段), 以人工湖为中心, 沿北偏西  $45^{\\circ}$  方向画 2 段长度, 确定西游记宫位置。\n\n计算水上世界图上距离:  $400 \\div 200 = 2$  (段), 以人工湖为中心, 沿西偏南  $20^{\\circ}$  方向画 2 段长度, 确定水上世界位置。\n\n计算迷宫图上距离:  $200 \\div 200 = 1$  (段), 以人工湖为中心, 沿南偏东  $35^{\\circ}$  方向画 1 段长度, 确定迷宫位置。\n\n如图所示：\n\n![](images/d39da1bbb6e81dcc9cd15919121760628dee58a6a36024c9d3800cf80ac4e0e4.jpg)",
+        "merged_id": "unit2_296",
+        "original_id": "unit2_choice_296",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "解答题"
+      },
+      {
+        "id": "unit2_choice_297",
+        "question": "看线路回答问题。\n\n![](images/f7e019393a7301ba34fc17588771cdde26b0422a767e24d5f4a459a3dec20f19.jpg)\n\n(1) 琪琪和妈妈从图书馆坐 3 路车向 ( ) 偏 ( )  $80^{\\circ}$  方向行驶 ( ) 站到动物园。\n\n(2) 爸爸从广播大厦坐 ( ) 路车向 ( ) 偏 ( )  $75^{\\circ}$  方向行驶 ( ) 站到海洋馆, 从海洋馆向南偏东  $45^{\\circ}$  方向行驶 2 站到 ( )。\n\n(3) 如果琪琪和妈妈从图书馆乘车出发到儿童乐园与爸爸会合, 他们应该怎样乘车? 请写出相关的乘车路线和方向。",
+        "options": [],
+        "answer": "（1）北；东；3\n\n(2) 2; 北; 东; 1; 儿童乐园\n\n(3) 见详解",
+        "source": "24-25 五年级上·河北保定·期中",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离描述路线图"
+        ],
+        "solution": "【分析】地图方向是上北下南，左西右东，到达一个地点，参照物也不同。确定路线时，注意起始点与目的地，起始点是观测点，以图上的'上北下南，左西右东确定方向，根据方向、角度和距离描述路线。\n\n(1) 从图书馆出发, 即以图书馆为观测点, 图上夹角为  $80^{\\circ}$ , 即北偏东  $80^{\\circ}$ , 终点站是动物园, 要行驶 3 站。\n\n(2) 从广播大厦出发, 即以广播大厦为观测点, 图上夹角为  $75^{\\circ}$ , 即北偏东  $75^{\\circ}$ , 要行驶 1 站到海洋馆换乘, 再以海洋馆为观测点, 图上夹角为  $45^{\\circ}$ , 南偏东  $45^{\\circ}$ , 再行驶 2 站, 到儿童乐园。\n\n(3) 先以图书馆为观测点，向北偏东  $80^{\\circ}$  方向行驶，再以动物园为观测点，向南偏东  $45^{\\circ}$  方向行驶即可。\n\n【详解】（1）琪琪和妈妈从图书馆坐3路车向北偏东  $80^{\\circ}$  方向行驶3站到动物园。\n\n(2) 爸爸从广播大厦坐 2 路车向北偏东  $75^{\\circ}$  方向行驶 1 站到海洋馆, 从海洋馆向南偏东  $45^{\\circ}$  方向行驶 2 站到儿童乐园。\n\n(3) 答: 琪琪和妈妈从图书馆坐 3 路车向北偏东  $80^{\\circ}$  方向行驶 3 站到动物园, 再从海洋馆向南偏东  $45^{\\circ}$  方向行驶 1 站到儿童乐园。",
+        "merged_id": "unit2_297",
+        "original_id": "unit2_choice_297",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "解答题"
+      },
+      {
+        "id": "unit2_choice_298",
+        "question": "根据路线图, 回答问题。\n\n![](images/c5bf0130c0c6a54a74caefb2611fd1f2f60ad4a2852a91aec3d3724d0553b811.jpg)\n\n(1) 在一次研学活动中, 同学们从军营大门出发, 向 ( ) 偏 ( )  $30^{\\circ}$  方向跑 ( ) 米到达训练场, 接着向 ( ) 偏 ( )  $25^{\\circ}$  方向跑 ( ) 米到达靶场, 再向北偏东  $30^{\\circ}$  方向跑 500 米到达终点。\n\n(2) 请在图中标出终点的位置。\n\n(3) 花花各段路程用的时间分别是 1.4 分钟、2.6 分钟和 2 分钟, 花花跑完全程, 平均每分钟跑 ( ) 米。",
+        "options": [],
+        "answer": "（1）东；北；400；东；南；600\n\n(2) 画图见详解\n\n(3) 250",
+        "source": "23-24 六年级上·河北唐山·期中",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置、根据方向、角度和距离画线路图"
+        ],
+        "solution": "【分析】（1）通过题干中的图片信息，判断从军营大门出发，向东偏北  $30^{\\circ}$  方向跑400米到达训练场，接着向东偏南  $25^{\\circ}$  方向跑600米到达靶场。\n\n(2) 确定好观测点，找出距离是 500 米，1 厘米对应的是200米，500米对应2.5厘米，再绘制路线图。\n\n(3) 平均每分钟的路程 = 总路程 ÷ 总时间; 总路程 = 400 + 600 + 500; 总时间 = 1.4 + 2.6 + 2。\n\n【详解】（1）从军营大门出发，向东偏北  $30^{\\circ}$  方向跑400米到达训练场，接着向东偏南  $25^{\\circ}$  方向跑600米到达靶场，再向北偏东  $30^{\\circ}$  方向跑500米到达终点。\n\n(2)\n\n![](images/3b374fbec6d5f52ac3447d186273a09649d3b38d3cef2728d02abd1cd266eb19.jpg)\n\n(3)  $400 + 600 + 500 = 1500$  (米)\n\n$1.4 + 2.6 + 2 = 6$  (分钟)\n\n$1500 \\div 6 = 250$  (米/分钟)\n\n答: 平均每分钟跑 250 米。",
+        "merged_id": "unit2_298",
+        "original_id": "unit2_choice_298",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "解答题"
+      },
+      {
+        "id": "unit2_choice_299",
+        "question": "下图是小丁、小丽、小亮、小白四位同学家的平面示意图, 请回答下列问题。\n\n![](images/ab27136889c96297d5f1412bcf444e3ed1d8e3194e9c843fb1eb5475bb12f013.jpg)\n\n(1) 小丁家在学校的什么位置？\n\n(2) 哪位同学家离学校最远？在学校的什么位置？\n\n(3) 你还能确定谁家的位置？请写出来。",
+        "options": [],
+        "answer": "（1）西偏北  $60^{\\circ}$  （或北偏西  $30^{\\circ}$  ）方向  $500\\mathrm{m}$  处。\n\n(2) 小白家; 南偏东  $50^{\\circ}$  (或东偏南  $40^{\\circ}$ ) 方向  $1000 \\mathrm{~m}$  处。\n\n(3) 小亮家; 西偏南  $45^{\\circ}$  (或南偏西  $45^{\\circ}$ ) 方向  $700 \\mathrm{~m}$  处。",
+        "source": "23-24 六年级上·全国·单元测试",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "solution": "【分析】（1）根据地图上的方向'上北下南,左西右东'及角度、距离来确定位置即可。\n\n(2) (3) 根据图上距离越短，距离越近，图上距离越远，距离越远，再根据根据地图上的方向'上北下南,左西右东'及角度、距离来确定位置即可。\n\n【详解】(1)  $5 \\times 100 = 500(\\mathrm{m})$\n\n小丁家在学校的西偏北  $60^{\\circ}$  （或北偏西  $30^{\\circ}$  ）方向  $500\\mathrm{m}$  处。\n\n(2)  $10 \\times 100 = 1000(\\mathrm{m})$\n\n小白家离学校最远，小白家在学校的南偏东  $50^{\\circ}$  （或东偏南  $40^{\\circ}$  ）方向  $1000 \\mathrm{~m}$  处。\n\n(3)  $7 \\times 100 = 700(\\mathrm{m})$\n\n小亮家在学校的西偏南  $45^{\\circ}$  （或南偏西  $45^{\\circ}$  ）方向  $700 \\mathrm{~m}$  处。",
+        "merged_id": "unit2_299",
+        "original_id": "unit2_choice_299",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "解答题"
+      },
+      {
+        "id": "unit2_choice_300",
+        "question": "学校举行科技节活动, 下图是智能机器人的行走路线图。\n\n(1) 智能机器人从起点出发，先向（）偏（）（）。\n\n方向行走了15米，到达A点。\n\n(2) 智能机器人最后从 B 点向南行走了 13 米, 到达目的地C点。请在图上标出C点的位置。\n\n![](images/fc2db678c9fc5accfa68730ce2f7ff1b68e6be85f5f23f874f90e020f55c2031.jpg)",
+        "options": [],
+        "answer": "(1)北;东;30\n\n(2）见详解",
+        "source": "23-24 六年级下·湖南邵阳·期末",
+        "difficulty": "中等",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "solution": "【分析】（1）图中1小段代表实际5米，根据地图上方向的规定'上北下南,左西右东'，以起点为观测点，确定出到A位置；\n\n(2) 以点 B 为观测点, 确定出 C 点位置, 画出 C 点的位置, 据此解答。\n\n【详解】(1)  $90^{\\circ} - 30^{\\circ} = 60^{\\circ}$\n\n智能机器人从起点出发，先向北偏东  $30^{\\circ}$  （或东偏北  $60^{\\circ}$  ）方向行走了15米，到达A点。\n\n(2)  $13 \\div 5 = 2.6$  (厘米)\n\n如图：\n\n![](images/62e5d5c4cb54197d14bfedec2c60e87d5cb9a02905160e71394f154468272591.jpg)",
+        "merged_id": "unit2_300",
+        "original_id": "unit2_choice_300",
+        "unit": "第二单元:位置与方向",
+        "chapter": "位置与方向综合练习",
+        "create_time": "2024-10-07 15:30:00",
+        "question_type": "解答题"
+      },{
+        "id": "unit2_choice_201",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "如图，下面说法正确的是（）。![](images/8bbcaef9107918e8ef1fe6e37ccdddef26c7c564ce7df04afcef9c16c2fbb8f5.jpg)",
+        "options": [
+          "A. 妙妙在甜甜北偏东  $30^{\\circ}$  方向",
+          "B. 甜甜在妙妙西偏南  $60^{\\circ}$  方向",
+          "C. 甜甜在妙妙的南偏西  $60^{\\circ}$  方向",
+          "D. 妙妙在甜甜的东偏北  $60^{\\circ}$  方向"
+        ],
+        "answer": "C",
+        "solution": "【分析】以图上的 “上北下南, 左西右东” 为准, 妙妙在甜甜的方向, 是以甜甜为观测点;甜甜在奇妙的方向, 是以奇妙为观测点; 根据方向、角度和距离确定奇妙与甜甜的位置关系。\n\n【详解】A．妙妙在甜甜东偏北  $30^{\\circ}$  （或北偏东  $60^{\\circ}$  ）方向，原选项说法错误；B. 甜甜在妙妙南偏西  $60^{\\circ}$  (或西偏南  $30^{\\circ}$ ) 方向, 原选项说法错误;  C. 甜甜在奇妙的南偏西  $60^{\\circ}$  方向, 原选项说法正确;  D. 妙妙在甜甜的东偏北  $30^{\\circ}$  (或北偏东  $60^{\\circ}$ ) 方向, 原选项说法错误。故答案为：C",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·浙江绍兴·期末",
+        "create_time": "2025-10-09 15:45:34",
+        "merged_id": "unit2_201",
+        "original_id": "unit2_choice_201"
+      },
+      {
+        "id": "unit2_choice_202",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "图书馆在剧院的东偏南 $30^{\\circ}$ 方向上，那么剧院在图书馆的（）。",
+        "options": [
+          "A. 东偏南  $30^{\\circ}$  方向",
+          "B. 南偏东  $60^{\\circ}$  方向",
+          "C. 西偏北  $30^{\\circ}$  方向"
+        ],
+        "answer": "C",
+        "solution": "【分析】图书馆在剧院的东偏南  $30^{\\circ}$  方向上，是以剧院为坐标原点，向上为正北方向，向右为正东方向，向左为正西方向，向下为正南方向。剧院在图书馆的相对方位，是以图书馆为坐标原点，在西偏北  $30^{\\circ}$  。据此可得出答案。\n\n【详解】图书馆在剧院的东偏南  $30^{\\circ}$  方向上，那么剧院在图书馆的西偏北  $30^{\\circ}$  方向。故答案为：C",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·山东临沂·期中",
+        "create_time": "2025-10-09 15:45:34",
+        "merged_id": "unit2_202",
+        "original_id": "unit2_choice_202"
+      },
+      {
+        "id": "unit2_choice_203",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "陈嘉庚纪念馆是厦门市首批研学实践教育基地, 小明从家出发去纪念馆, 他向西偏北 $36^{\\circ}$ 方向走了 $1200 \\mathrm{~m}$ 。他原路返回时应该向 ( ) 方向走 $1200 \\mathrm{~m}$ 。",
+        "options": [
+          "A. 西偏北  $36^{\\circ}$",
+          "B. 北偏西  $36^{\\circ}$",
+          "C. 南偏东  $36^{\\circ}$",
+          "D. 东偏南  $36^{\\circ}$"
+        ],
+        "answer": "D",
+        "solution": "【分析】小明从家出发去纪念馆，是以小明家为观测点；原路返回时是以纪念馆为观测点；观测点不同，方向相反，夹角的度数相同，距离相同；西偏北  $36^{\\circ}$  相对的是东偏南  $36^{\\circ}$ ，东和南之间的夹角是  $90^{\\circ}$ ，$90^{\\circ}-$ $36^{\\circ}$=$54^{\\circ}$，所以东偏南  $36^{\\circ}$  方向，还可以说成南偏东  $54^{\\circ}$  方向。\n\n【详解】小明从家出发去纪念馆，他向西偏北  $36^{\\circ}$  方向走了  $1200 \\mathrm{~m}$  。他原路返回时应该向东偏南  $36^{\\circ}$  （或南偏东  $54^{\\circ}$  ）方向走  $1200 \\mathrm{~m}$  。故答案为：D",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·福建厦门·期中",
+        "create_time": "2025-10-09 15:45:34",
+        "merged_id": "unit2_203",
+        "original_id": "unit2_choice_203"
+      },
+      {
+        "id": "unit2_choice_204",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "书店在学校北偏西 $30^{\\circ}$ 方向上, 那么学校就在书店 ( ) 的方向上。",
+        "options": [
+          "A. 西偏北  $30^{\\circ}$",
+          "B. 北偏东  $60^{\\circ}$",
+          "C. 南偏东  $30^{\\circ}$",
+          "D. 南偏东  $60^{\\circ}$"
+        ],
+        "answer": "C",
+        "solution": "【分析】根据位置的相对性可知, 描述两个物体之间的相对位置时, 方向相反, 角度相等。书店在学校北偏西  $30^{\\circ}$  方向上, 那么学校就在书店南偏东  $30^{\\circ}$  的方向上。而正南与正东方向的夹角是  $90^{\\circ}$ , 则南偏东  $30^{\\circ}$  也可以说是东偏南  $90 - 30^{\\circ} = 60^{\\circ}$  。\n\n【详解】通过分析可得：书店在学校北偏西  $30^{\\circ}$  方向上，那么学校就在书店南偏东  $30^{\\circ}$  的方向或东偏南  $60^{\\circ}$  方向。故答案为：C",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·山东临沂·期中",
+        "create_time": "2025-10-09 15:45:34",
+        "merged_id": "unit2_204",
+        "original_id": "unit2_choice_204"
+      },
+      {
+        "id": "unit2_choice_205",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "一架飞机向西偏南 $45^{\\circ}$ 方向飞行, 接到指令后, 改向相反的方向飞行, 那么这架飞机应向 ( ) 方向飞行。",
+        "options": [
+          "A. 东偏北  $45^{\\circ}$",
+          "B. 西偏北  $45^{\\circ}$",
+          "C. 东偏南  $45^{\\circ}$"
+        ],
+        "answer": "A",
+        "solution": "【分析】这道题运用了方向与位置的数学概念。首先，西偏南  $45^{\\circ}$  的相反方向，是以原来的位置为观测点，角度不变，方向完全相反。西的相反方向是东，南的相反方向是北，所以西偏南  $45^{\\circ}$  的相反方向是东偏北  $45^{\\circ}$\n\n【详解】飞机最初的飞行方向是西偏南  $45^{\\circ}$ ，相反的方向就是东偏北  $45^{\\circ}$ 。故答案为：A",
+        "knowledge_points": [
+          "东南",
+          "西南",
+          "东北",
+          "西北方向",
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·山东菏泽·期中",
+        "create_time": "2025-10-09 15:45:34",
+        "merged_id": "unit2_205",
+        "original_id": "unit2_choice_205"
+      },
+      {
+        "id": "unit2_choice_206",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "8 路公交车的行驶路线如图所示, 下列说法错误的是 ( )。![](images/e2cac56c7768398ea7062664752f57e03f82dd670dd30a7a1ce30341fc6dc9af.jpg)",
+        "options": [
+          "A. 文兴苑在万象城的正北方向。",
+          "B. 春光超市在图书馆的西偏南方向。",
+          "C. 从人民公园到体育馆要向东偏北方向行驶。"
+        ],
+        "answer": "C",
+        "solution": "【分析】依据地图上的方向辨别方法，即“上北下南，左西右东”以及图上标注的其他信息，即可解答。\n\n【详解】A．文兴苑在万象城的正北方向。原题干说法正确；  B. 春光超市在图书馆的西偏南方向。原题干说法正确;  C. 从人民公园到体育馆要向西偏南方向行驶。原题干说法错误。故答案为：C",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·山东菏泽·期中",
+        "create_time": "2025-10-09 15:45:34",
+        "merged_id": "unit2_206",
+        "original_id": "unit2_choice_206"
+      },
+      {
+        "id": "unit2_choice_207",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "如图中，果园在竹林的（）处。![](images/6f0832119f937cb52904407a265aa78a56ccaeb38f77b77c0390e3411a65c3c1.jpg)",
+        "options": [
+          "A. 东偏北  $30^{\\circ}$  方向2.4千米",
+          "B. 西偏北  $30^{\\circ}$  方向 2.4 千米",
+          "C. 西偏南  $30^{\\circ}$  方向 2.4 千米",
+          "D. 东偏南  $30^{\\circ}$  方向2.4千米"
+        ],
+        "answer": "A",
+        "solution": "【分析】地图的方位是上北下南左西右东，比例尺是图上1厘米表示实际800米；果园在竹林的东偏北  $30^{\\circ}$  方向  $800 \\times 3 = 2400$  米 = 2.4 千米处。\n\n【详解】$800 \\times 3 = 2400$  （米）  $= 2.4$  （千米）所以果园在竹林的东偏北  $30^{\\circ}$  （北偏东  $60^{\\circ}$  ）方向2.4千米处。故答案为：A",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·河北邢台·期中",
+        "create_time": "2025-10-09 15:45:34",
+        "merged_id": "unit2_207",
+        "original_id": "unit2_choice_207"
+      },
+      {
+        "id": "unit2_choice_208",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "乐乐家在学校的东偏南 $30^{\\circ}$ 方向上, 那么学校就在乐乐家的 ( ) 方向上。",
+        "options": [
+          "A. 东偏南  $30^{\\circ}$",
+          "B. 南偏东  $60^{\\circ}$",
+          "C. 西偏北  $30^{\\circ}$"
+        ],
+        "answer": "C",
+        "solution": "【分析】根据位置的相对性，以乐乐家、学校为观察中心时，看到对方的方向相反，角度和距离相等，据此解答。\n\n【详解】由题可得：乐乐家在学校的东偏南  $30^{\\circ}$  方向上，那么学校就在乐乐家的西偏北  $30^{\\circ}$  方向上。故答案为：C",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·广东汕尾·期中",
+        "create_time": "2025-10-09 15:45:34",
+        "merged_id": "unit2_208",
+        "original_id": "unit2_choice_208"
+      },
+      {
+        "id": "unit2_choice_209",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "在中国地图上, 测量得到北京在上海的北偏西 $40^{\\circ}$ 方向，距离4厘米处，那么在这幅地图上，上海在北京的（）方向，距离4厘米处。",
+        "options": [
+          "A. 东偏南  $40^{\\circ}$",
+          "B. 西偏北  $40^{\\circ}$",
+          "C. 南偏东  $40^{\\circ}$",
+          "D. 南偏东  $50^{\\circ}$"
+        ],
+        "answer": "C",
+        "solution": "【分析】根据位置的相对性, 可知北京和上海的方向相反, 角度相同相等, 距离相等, 据此解答。\n\n【详解】由分析得：北京在上海的北偏西  $40^{\\circ}$  方向, 距离 4 厘米处, 那么在这幅地图上, 上海在北京的南偏东  $40^{\\circ}$  方向, 距离 4 厘米处。故答案为：C",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·贵州遵义·期中",
+        "create_time": "2025-10-09 15:45:34",
+        "merged_id": "unit2_209",
+        "original_id": "unit2_choice_209"
+      },
+      {
+        "id": "unit2_choice_210",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "早上壮壮面向太阳站立, 然后向左转了 $30^{\\circ}$ , 这时壮壮面向的是 ( ) 方向。",
+        "options": [
+          "A. 东偏南  $30^{\\circ}$",
+          "B. 东偏北  $30^{\\circ}$",
+          "C. 北偏东  $30^{\\circ}$"
+        ],
+        "answer": "B",
+        "solution": "【分析】根据方向的规定：上北下南，左西右东；早上壮壮面向太阳，太阳在东方，所以壮壮面向东方，左边是北方，然后向左转了  $30^{\\circ}$ ，是从东往北转了  $30^{\\circ}$ ，结合角度，得出这时壮壮的方向。\n\n【详解】$90^{\\circ} - 30^{\\circ} = 60^{\\circ}$早上壮壮面向太阳站立，然后向左转了  $30^{\\circ}$ ，这时壮壮面向的是东偏北  $30^{\\circ}$ （或北偏东  $60^{\\circ}$ ）方向。故答案为：B",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 五年级上·广西贵港·期中",
+        "create_time": "2025-10-09 15:45:34",
+        "merged_id": "unit2_210",
+        "original_id": "unit2_choice_210"
+      },
+      {
+        "id": "unit2_choice_211",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "小明的家在公园的东偏南 $60^{\\circ}$ 方向上, 则公园在小明家 ( ) 方向上。",
+        "options": [
+          "A. 南偏东  $60^{\\circ}$",
+          "B. 西偏北  $30^{\\circ}$",
+          "C. 北偏西  $30^{\\circ}$",
+          "D. 西偏南  $60^{\\circ}$"
+        ],
+        "answer": "C",
+        "solution": "【分析】根据方向的相对性，它们的方向相反，角度相等，距离相等；据此解答。\n\n【详解】$90^{\\circ} - 60^{\\circ} = 30^{\\circ}$小明的家在公园的东偏南  $60^{\\circ}$  方向上, 则公园在小明家西偏北  $60^{\\circ}$  (或北偏西  $30^{\\circ}$ ) 方向上。故答案为：C",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·福建厦门·期中",
+        "create_time": "2025-10-09 15:45:34",
+        "merged_id": "unit2_211",
+        "original_id": "unit2_choice_211"
+      },
+      {
+        "id": "unit2_choice_212",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "彩旗向东南方向飘动, 说明风是从 ( ) 方向吹来的。",
+        "options": [
+          "A. 东北",
+          "B. 西北",
+          "C. 西南"
+        ],
+        "answer": "B",
+        "solution": "【分析】彩旗飘向的方向与风吹来的方向相反；风从西北方向吹来，吹往东南方向，则彩旗就飘向东南方，依此填空。\n\n【详解】由分析可知，彩旗向东南方向飘动，说明风是从西北方向吹来的。故答案为：B",
+        "knowledge_points": [
+          "东南",
+          "西南",
+          "东北",
+          "西北方向"
+        ],
+        "source": "23-24六年级上·广西玉林·期中",
+        "create_time": "2025-10-09 15:45:34",
+        "merged_id": "unit2_212",
+        "original_id": "unit2_choice_212"
+      },
+      {
+        "id": "unit2_choice_213",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "2022年11月，第14届中国国际航空航天博览会在珠海举办。展演期间，我国最新型隐形战斗机歼一20向东偏北 $35^{\\circ}$ 方向飞行，接到指令后改向相反方向飞行，现在的飞行方向是（）。",
+        "options": [
+          "A. 东偏南  $35^{\\circ}$",
+          "B. 西偏南  $35^{\\circ}$",
+          "C. 西偏南  $55^{\\circ}$"
+        ],
+        "answer": "B",
+        "solution": "【分析】一个事物在另一个事物的某个方向一定度数的位置，那么另一个事物在这个事物相对的方向相同度数的位置，东与西相对，北与南相对。据此解答即可。\n\n【详解】我国最新型隐形战斗机歼一20向东偏北  $35^{\\circ}$  方向飞行，接到指令后改向相反方向飞行，与东偏北相反的方向是西偏南，所以现在的飞行方向是西偏南  $35^{\\circ}$  。故答案为：B",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·广西玉林·期中",
+        "create_time": "2025-10-09 15:45:34",
+        "merged_id": "unit2_213",
+        "original_id": "unit2_choice_213"
+      },
+      {
+        "id": "unit2_choice_214",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "学校在小明家南偏西 $35^{\\circ}$ 的方向上, 距离 $800 \\mathrm{~m}$ , 小明家在学校 ( ) , 距离是 $800 \\mathrm{~m}$ 。",
+        "options": [
+          "A. 西偏南  $55^{\\circ}$",
+          "B. 南偏西  $55^{\\circ}$",
+          "C. 东偏北  $35^{\\circ}$",
+          "D. 北偏东  $35^{\\circ}$"
+        ],
+        "answer": "D",
+        "solution": "【分析】物体的位置是相对的，对于两个物体来说，分别以自身为观测点，则它们的方向相反，距离和角度是不变的，据此解答即可。南对北，东对西。\n\n【详解】学校在小明家南偏西  $35^{\\circ}$  的方向上, 距离  $800 \\mathrm{~m}$ , 小明家在学校北偏东  $35^{\\circ}$  方向,距离是  $800 \\mathrm{~m}$  。故答案为：D\n\n15．(23-24六年级上·江西南昌·期中）南偏西  $35^{\\circ}$  ，也可以说成（）。\n\nA. 西偏南  $35^{\\circ}$\n\nB. 西偏南  $55^{\\circ}$\n\nC. 西偏南  $65^{\\circ}$\n\nD. 东偏南  $35^{\\circ}$\n\n【答案】B\n\n【难度】0.65\n\n【知识点】根据方向、角度和距离确定物体的位置\n\n【分析】南偏西  $35^{\\circ}$  说明在正南方向左向西偏转了  $35^{\\circ}$ ; 由于正南方向和正西方向夹角为  $90^{\\circ}$ , 所以这个方向也可以表示为在正西方向下向南偏转了  $(90^{\\circ} - 35^{\\circ})$ , 计算即可得出答案。\n\n【详解】  $90^{\\circ} - 35^{\\circ} = 55^{\\circ}$据分析可知，南偏西  $35^{\\circ}$ ，也可以说成西偏南  $55^{\\circ}$ 。故答案为：B\n\n16．（23-24六年级下·浙江杭州·期末）海警船正在巡逻时，接收到一艘渔船发出的求助信息：雷达显示，你们（海警船）在我们渔船的北偏东  $40^{\\circ}$  方向5海里处。因我船故障无法航行，请求帮助。海警船立即开展救援行动，应该向（）方向航行5海里。\n\nA. 南偏西  $40^{\\circ}$\n\nB. 南偏东  $40^{\\circ}$\n\nC. 北偏东  $40^{\\circ}$\n\nD. 北偏西  $40^{\\circ}$\n\n【答案】A\n\n【难度】0.65\n\n【知识点】根据方向、角度和距离确定物体的位置\n\n【分析】根据方向的相对性，北偏东对南偏西，角度和距离不变，进行分析。\n\n【详解】海警船立即开展救援行动，应该向南偏西  $40^{\\circ}$  方向或西偏南  $50^{\\circ}$  航行5海里。故答案为：A\n\n17．(23-24六年级上·全国·期中）下图中，小明家在学校（）方向。![](images/00a399df5d61c0f628895eaae53e94aefbedfbd52601631a325033cd892ad86e.jpg)\n\nA. 南偏东  $60^{\\circ}$\n\nB. 东偏南  $60^{\\circ}$\n\nC. 西偏北  $60^{\\circ}$\n\n【答案】A\n\n【难度】0.65\n\n【知识点】根据方向、角度和距离描述路线图\n\n【分析】根据上北下南，左西右东，确定观测点是学校，先读正边方向，再读斜边偏向的方向，最后读角度，据此解答。\n\n【详解】据分析可知，小明家在学校南偏东  $60^{\\circ}$  方向。故答案为：A",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·贵州遵义·期中",
+        "create_time": "2025-10-09 15:45:34",
+        "merged_id": "unit2_214",
+        "original_id": "unit2_choice_214"
+      },
+      {
+        "id": "unit2_choice_215",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "南偏西 $35^{\\circ}$ ，也可以说成（）。",
+        "options": [
+          "A. 西偏南 $35^{\\circ}$",
+          "B. 西偏南 $55^{\\circ}$",
+          "C. 西偏南 $65^{\\circ}$",
+          "D. 东偏南 $35^{\\circ}$"
+        ],
+        "answer": "B",
+        "solution": "【分析】南偏西 $35^{\\circ}$ 说明在正南方向左向西偏转了 $35^{\\circ}$; 由于正南方向和正西方向夹角为 $90^{\\circ}$, 所以这个方向也可以表示为在正西方向下向南偏转了 $(90^{\\circ} - 35^{\\circ})$, 计算即可得出答案。\n\n【详解】$90^{\\circ} - 35^{\\circ} = 55^{\\circ}$据分析可知，南偏西 $35^{\\circ}$，也可以说成西偏南 $55^{\\circ}$。故答案为：B",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·江西南昌·期中",
+        "create_time": "2025-10-09 17:28:58",
+        "merged_id": "unit2_215",
+        "original_id": "unit2_choice_215"
+      },
+      {
+        "id": "unit2_choice_216",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "海警船正在巡逻时，接收到一艘渔船发出的求助信息：雷达显示，你们（海警船）在我们渔船的北偏东 $40^{\\circ}$ 方向5海里处。因我船故障无法航行，请求帮助。海警船立即开展救援行动，应该向（）方向航行5海里。",
+        "options": [
+          "A. 南偏西 $40^{\\circ}$",
+          "B. 南偏东 $40^{\\circ}$",
+          "C. 北偏东 $40^{\\circ}$",
+          "D. 北偏西 $40^{\\circ}$"
+        ],
+        "answer": "A",
+        "solution": "【分析】根据方向的相对性，北偏东对南偏西，角度和距离不变，进行分析。\n\n【详解】海警船立即开展救援行动，应该向南偏西 $40^{\\circ}$ 方向或西偏南 $50^{\\circ}$ 航行5海里。故答案为：A",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级下·浙江杭州·期末",
+        "create_time": "2025-10-09 17:28:58",
+        "merged_id": "unit2_216",
+        "original_id": "unit2_choice_216"
+      },
+      {
+        "id": "unit2_choice_217",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "下图中，小明家在学校（）方向。![](images/00a399df5d61c0f628895eaae53e94aefbedfbd52601631a325033cd892ad86e.jpg)",
+        "options": [
+          "A. 南偏东 $60^{\\circ}$",
+          "B. 东偏南 $60^{\\circ}$",
+          "C. 西偏北 $60^{\\circ}$"
+        ],
+        "answer": "A",
+        "solution": "【分析】根据上北下南，左西右东，确定观测点是学校，先读正边方向，再读斜边偏向的方向，最后读角度，据此解答。\n\n【详解】据分析可知，小明家在学校南偏东 $60^{\\circ}$ 方向。故答案为：A",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离描述路线图"
+        ],
+        "source": "23-24六年级上·全国·期中",
+        "create_time": "2025-10-09 17:28:58",
+        "merged_id": "unit2_217",
+        "original_id": "unit2_choice_217"
+      },
+      {
+        "id": "unit2_choice_218",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "早上, 小明面向太阳站着, 当他逆时针方向转 $90^{\\circ}$ 时就面向 ( ) 方。",
+        "options": [
+          "A. 北",
+          "B. 西",
+          "C. 南"
+        ],
+        "answer": "A",
+        "solution": "【分析】早上，太阳从东方升起，因此小明早上面向东方，面向东方时，左边是北方，右边是南方，后面是西方，当他逆时针方向旋转  $90^{\\circ}$  时，它面向北，依此选择。\n\n【详解】根据分析可知，早上，小明面向太阳站着，当他逆时针方向转  $90^{\\circ}$  时就面向北方。故答案为：A",
+        "knowledge_points": [
+          "东",
+          "南",
+          "西",
+          "北方向",
+          "旋转三要素及旋转图形"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "create_time": "2025-10-09 15:45:34",
+        "merged_id": "unit2_218",
+        "original_id": "unit2_choice_218"
+      },
+      {
+        "id": "unit2_choice_219",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "如图, 小明看小冬在 ( ) , 小冬看小明在 ( )。 ( )![](images/6d2dbfd71a5349b91d1f34944799c294532a796b3e09d442898b6f42fedee903.jpg)",
+        "options": [
+          "A. 北偏东  $50^{\\circ}$  的方向上; 西偏南  $40^{\\circ}$  的方向上",
+          "B. 东偏北  $50^{\\circ}$  的方向上; 西偏南  $40^{\\circ}$  的方向上",
+          "C. 北偏东  $50^{\\circ}$  的方向上; 东偏北  $50^{\\circ}$  的方向上"
+        ],
+        "answer": "A",
+        "solution": "【分析】根据地图上方向的规定“上北下南，左西右东”，以小明为观测点，确定出小冬的位置；再根据方向的相对性：它们的方向相反，角度相等，距离相等；据此解答。\n\n【详解】$90^{\\circ} - 50^{\\circ} = 40^{\\circ}$$9 0 ^ {\\circ} - 4 0 ^ {\\circ} = 5 0 ^ {\\circ}$小明看小冬在北偏东  $50^{\\circ}$  （东偏北  $40^{\\circ}$  ）方向上。小冬看小明在西偏南  $40^{\\circ}$  （或南偏西  $50^{\\circ}$  ）方向上。小明看小冬在北偏东  $50^{\\circ}$  (东偏北  $40^{\\circ}$ ) 方向上, 小冬看小明在西偏南  $40^{\\circ}$  (或南偏西  $50^{\\circ}$ )方向上。故答案为：A",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·全国·单元测试",
+        "create_time": "2025-10-09 15:45:34",
+        "merged_id": "unit2_219",
+        "original_id": "unit2_choice_219"
+      },
+      {
+        "id": "unit2_choice_220",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "北偏西 $35^{\\circ}$ , 还可以说成 ( )。",
+        "options": [
+          "A. 南偏西  $35^{\\circ}$",
+          "B. 西偏北  $35^{\\circ}$",
+          "C. 西偏北  $55^{\\circ}$"
+        ],
+        "answer": "C",
+        "solution": "【分析】![](images/4879e9729cd67321f7d6c0f8528c9a3599ad9a91450b3724465749008a4f26b5.jpg)如图，根据方位角的定义画图以及位置的相对性可知，北偏西\n\n$35^{\\circ}$  也可以说成西偏北  $55^{\\circ}$  。\n\n【详解】因为图中表示北偏西  $35^{\\circ}$ ,所以图中也表示西偏北  $90^{\\circ} - 35^{\\circ} = 55^{\\circ}$故答案为：C 。本题考查了方位角的定义及方位角的表示，根据题意画出图形是解题的关。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "create_time": "2025-10-09 15:45:34",
+        "merged_id": "unit2_220",
+        "original_id": "unit2_choice_220"
+      },
+      {
+        "id": "unit2_choice_221",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "确定物体的位置，需要（）条件。",
+        "options": [
+          "A. 方向一个",
+          "B. 距离一个",
+          "C. 方向和距离两个"
+        ],
+        "answer": "C",
+        "solution": "【分析】要确定物体的位置必须用角度描述出物体在观测点的什么方向，还有这个物体与观测点之间的距离，据此解答。\n\n【详解】根据分析可知，确定物体位置，需要方向和距离两个条件。故答案为：C",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·全国·单元测试",
+        "create_time": "2025-10-09 15:45:34",
+        "merged_id": "unit2_221",
+        "original_id": "unit2_choice_221"
+      },
+      {
+        "id": "unit2_choice_222",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "兰兰家在超市的东偏北 $30^{\\circ}$ 方向 $500 \\mathrm{~m}$ 处, 超市在兰兰家的 ( )。",
+        "options": [
+          "A. 西偏南  $60^{\\circ}$  方向  $500 \\mathrm{~m}$  处",
+          "B. 西偏南  $30^{\\circ}$  方向  $500 \\mathrm{~m}$  处",
+          "C. 东偏北  $60^{\\circ}$  方向  $500 \\mathrm{~m}$  处"
+        ],
+        "answer": "B",
+        "solution": "【分析】兰兰家在超市的东偏北  $30^{\\circ}$  方向  $500 \\mathrm{~m}$  处, 是以超市为观测点; 超市在兰兰家的方向是以兰兰家为观测点; 根据位置的相对性可知, 观测点不同, 方向相反, 夹角的度数相同, 距离相同。\n\n【详解】兰兰家在超市的东偏北  $30^{\\circ}$  方向  $500 \\mathrm{~m}$  处, 超市在兰兰家的西偏南  $30^{\\circ}$  (或南偏西  $60^{\\circ}$ ) 方向  $500 \\mathrm{~m}$  处。故答案为：B",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·湖南岳阳·期中",
+        "create_time": "2025-10-09 15:45:34",
+        "merged_id": "unit2_222",
+        "original_id": "unit2_choice_222"
+      },
+      {
+        "id": "unit2_choice_223",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "如图，学校在垟仔洲公园的（）。![](images/f8ed5ec7e65e04800604613cd55a3b7b6d838f74b798f17ed8b8898293866754.jpg)",
+        "options": [
+          "A. 北偏东  $40^{\\circ}$  方向距离 800 米",
+          "B. 东偏北  $40^{\\circ}$  方向距离 800 米",
+          "C. 西偏南  $40^{\\circ}$  方向距离 800 米",
+          "D. 南偏西  $40^{\\circ}$  方向距离 800 米"
+        ],
+        "answer": "D",
+        "solution": "【分析】根据上北下南，左西右东，以榉仔洲公园为观测点，即可确定学校的方向，已知学校和榉仔洲公园在图上相距 4 厘米，根据图上 1 厘米表示实际 200 米，求出两地的实际距离。\n\n【详解】$4 \\times 200 = 800$  （米）学校在垟仔洲公园的南偏西  $40^{\\circ}$  （西偏南  $50^{\\circ}$  ）方向距离800米。故答案为：D",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级下·福建龙岩·期末",
+        "create_time": "2025-10-09 15:45:34",
+        "merged_id": "unit2_223",
+        "original_id": "unit2_choice_223"
+      },
+      {
+        "id": "unit2_choice_224",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "食堂在操场的南偏东 $50^{\\circ}$ 方向上, 则操场在食堂的 ( ) 方向上。",
+        "options": [
+          "A. 北偏东  $50^{\\circ}$",
+          "B. 北偏西  $50^{\\circ}$",
+          "C. 北偏西  $40^{\\circ}$"
+        ],
+        "answer": "B",
+        "solution": "【分析】根据位置的相对性可知, 它们的方向相反, 角度相等, 距离相等, 食堂在操场的南偏东  $50^{\\circ}$  方向上, 是以操场为观测点, 而操场在食堂的什么方向上, 是以食堂为观测点,再结合西和北之间的角度是  $90^{\\circ}$  解答即可。\n\n【详解】食堂在操场的南偏东  $50^{\\circ}$  方向上，也可以说操场在食堂的北偏西  $50^{\\circ}$  的方向上或西偏北  $90^{\\circ} - 50^{\\circ} = 40^{\\circ}$  。故答案为：B",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 五年级下·广东河源·期末",
+        "create_time": "2025-10-09 15:45:34",
+        "merged_id": "unit2_224",
+        "original_id": "unit2_choice_224"
+      },
+      {
+        "id": "unit2_choice_225",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "丽丽看红红在（）方向上，红红看丽丽在（）方向上。![](images/5ae1ea0f89255302b83585d21e99416083e52471b6ccc3323c6287b493173b60.jpg)",
+        "options": [
+          "A. 北偏东  $50^{\\circ}$ ; 西偏南  $40^{\\circ}$",
+          "B. 西偏南  $40^{\\circ}$ ; 东偏北  $50^{\\circ}$",
+          "D. 南偏西  $40^{\\circ}$ ; 西偏南  $40^{\\circ}$"
+        ],
+        "answer": "A",
+        "solution": "【分析】图上方向按上北下南，左西右东来绘制的，东和北之间是东北方向，东和南之间是东南方向，西和北之间是西北方向，西和南之间是西南方向。描述物体的方向时，先确定观测点，再描述方向和角度。方向具有相对性，看物体的方位不同，其角度也就不同，按照先确定准基本点再判定方向的方法。\n\n【详解】由分析可知：丽丽看红红所在的位置，是以丽丽为观测点，丽丽看红红在北偏东  $50^{\\circ}$  （东偏北  $40^{\\circ}$  ）方向上。红红看丽丽所在的位置，是以红红为观测点，红红看丽丽在西偏南  $40^{\\circ}$  （南偏西  $50^{\\circ}$  ）方向上。故答案为：A",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "21-22六年级上·福建莆田·期末",
+        "create_time": "2025-10-09 15:45:34",
+        "merged_id": "unit2_225",
+        "original_id": "unit2_choice_225"
+      },
+      {
+        "id": "unit2_choice_226",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "明明家在亮亮家东偏南 $35^{\\circ}$ 约 1000 米处, 则亮亮在明明家的 ( ) 约 1000 米处。",
+        "options": [
+          "A. 西偏北  $55^{\\circ}$",
+          "B. 西偏北  $35^{\\circ}$",
+          "C. 南偏东  $55^{\\circ}$",
+          "D. 南偏东  $35^{\\circ}$"
+        ],
+        "answer": "B",
+        "solution": "【分析】根据位置的相对性可知，他们的方向相反，角度和距离相等，据此解答即可。\n\n【详解】明明家在亮亮家东偏南  $35^{\\circ}$  约 1000 米处, 则亮亮在明明家的西偏北  $35^{\\circ}$  (北偏西  $55^{\\circ}$ ) 约 1000 米处。故答案为：B",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级下·湖北黄冈·期末",
+        "create_time": "2025-10-09 15:45:34",
+        "merged_id": "unit2_226",
+        "original_id": "unit2_choice_226"
+      },
+      {
+        "id": "unit2_choice_227",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "一只猎狗追踪野兔, 先沿正东方向跑到 1 号地, 再沿南偏东方向跑到 2 号地, 最后沿西偏北方向跑到 3 号地。下面可以表示该猎狗追踪路线的是 ( )。",
+        "options": [
+          "A. ![](images/f26e95e8b82dcfb4078df4be97b4f4accdaeffb1f83b26c528d4a6c13f37b28e.jpg)",
+          "B. ![](images/4f8089028cbf4465a9b694bc9e14c6bacd1cabe578bffd364b008ec0a8bb75a3.jpg)",
+          "C. ![](images/5ee31108530de42aadfe9a3f649f491b9d7496e94b780b91d89b6b0fe19074de.jpg)",
+          "D. ![](images/ce6ef3dfa13275f540c1980857dca71cac2777cfb5ba04cf9db6f8df9d3edc8e.jpg)"
+        ],
+        "answer": "C",
+        "solution": "【分析】按照“上北下南，左西右东”的方向逐项判断各选项表示的含义，再作选择。\n\n【详解】A．表示野兔先沿正东方向跑到1号地，再沿北偏西方向跑到2号地，不符合题意；  B. 表示野兔先沿正东方向跑到 1 号地, 再沿东偏北跑到 2 号地, 不符合题意;  C. 表示野兔先沿正东方向跑到 1 号地, 再沿南偏东方向跑到 2 号地, 最后沿西偏北方向跑到 3 号地, 符合题意;  D. 表示野兔先沿正东方向跑到 1 号地, 再沿南偏东方向跑到 2 号地, 最后沿东偏北方向跑到 3 号地, 与题意不符。故答案为：C",
+        "knowledge_points": [
+          "根据方向和距离描述简单的路线"
+        ],
+        "source": "23-24六年级上·福建莆田·期中",
+        "create_time": "2025-10-09 15:45:34",
+        "merged_id": "unit2_227",
+        "original_id": "unit2_choice_227"
+      },
+      {
+        "id": "unit2_choice_228",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "如下图所示, 三角形 ABC 是一个等边三角形, 则小明家在学校的 ( ) 方向。(上北下南, 左西右东)![](images/de1e28355ad5a9894d36f3d1b21f74007909c097c4c984128ece33d21b47e9d6.jpg)",
+        "options": [
+          "A. 东偏北  $30^{\\circ}$",
+          "B. 北偏西  $30^{\\circ}$",
+          "C. 南偏东  $30^{\\circ}$",
+          "D. 南偏东  $60^{\\circ}$"
+        ],
+        "answer": "C",
+        "solution": "【分析】要看小明家在学校的什么方向, 就是以学校为观测点, 看小明家, 确定上北下南左西右东, 结合等边三角形是轴对称图形, 可以确定小明家在学校的南偏东  $30^{\\circ}$  方向。如图:![](images/67c7fd920091a79a8754ecd53d59f43907e425c80e4a06007ac202227c05d783.jpg)\n\n【详解】由分析可知，小明家在学校的南偏东  $30^{\\circ}$  方向。故答案为：C",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "22-23 六年级上·福建莆田·期中",
+        "create_time": "2025-10-09 15:45:34",
+        "merged_id": "unit2_228",
+        "original_id": "unit2_choice_228"
+      },
+      {
+        "id": "unit2_choice_229",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "学校在小芳家的南偏东 $30^{\\circ}$ 方向 $3 \\mathrm{~km}$ 处, 也可以说小芳家在学校的 ( )。",
+        "options": [
+          "A. 北偏西  $60^{\\circ}$  方向  $3 \\mathrm{~km}$  处",
+          "B. 西偏北  $30^{\\circ}$  方向  $3 \\mathrm{~km}$  处",
+          "C. 北偏西  $30^{\\circ}$  方向  $3 \\mathrm{~km}$  处",
+          "D. 东偏南  $30^{\\circ}$  方向  $3 \\mathrm{~km}$  处"
+        ],
+        "answer": "C",
+        "solution": "【分析】根据位置的相对性可知, 描述两个物体之间的相对位置时, 方向相反, 角度相等,距离相等, 则可以说小芳家在学校的北偏西  $30^{\\circ}$  方向  $3 \\mathrm{~km}$  处; 北和西两个方向的夹角是  $90^{\\circ}$ , 北偏西  $30^{\\circ}$  方向也就是西偏北  $90^{\\circ} - 30^{\\circ} = 60^{\\circ}$ , 所以也可以说小芳家在学校的西偏北  $60^{\\circ}$  方向  $3 \\mathrm{~km}$  处。\n\n【详解】通过分析可得: 学校在小芳家的南偏东  $30^{\\circ}$  方向  $3 \\mathrm{~km}$  处, 也可以说小芳家在学校的北偏西  $30^{\\circ}$  方向  $3 \\mathrm{~km}$  处, 或小芳家在学校的西偏北  $60^{\\circ}$  方向  $3 \\mathrm{~km}$  处。故答案为：C",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级下·湖南常德·期末",
+        "create_time": "2025-10-09 15:45:34",
+        "merged_id": "unit2_229",
+        "original_id": "unit2_choice_229"
+      },
+      {
+        "id": "unit2_choice_230",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "小明从家出发向北偏东 $30^{\\circ}$ 方向走了 $200\\mathrm{m}$ 到达超市， 购物后原路返回，他需要向（）方向走 $200\\mathrm{m}$ 才能回到家。",
+        "options": [
+          "A. 南偏西  $60^{\\circ}$",
+          "B. 南偏西  $30^{\\circ}$",
+          "C. 西偏南  $30^{\\circ}$",
+          "D. 东偏北  $60^{\\circ}$"
+        ],
+        "answer": "B",
+        "solution": "【分析】返回时，方向相反、角度和距离不变。东和西相对，南和北相对，去时向北偏东  $30^{\\circ}$  方向走了  $200 \\mathrm{~m}$ ，返回时就向南偏西  $30^{\\circ}$  方向走  $200 \\mathrm{~m}$ 。据此解答。\n\n【详解】由分析可得: 小明从家出发向北偏东  $30^{\\circ}$  方向走了  $200 \\mathrm{~m}$  到达超市, 购物后原路返回, 他需要向南偏西  $30^{\\circ}$  方向走  $200 \\mathrm{~m}$  。\n\n故答案为：B",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离描述路线图"
+        ],
+        "source": "23-24六年级下·湖南株洲·期末",
+        "create_time": "2025-10-09 15:45:34",
+        "merged_id": "unit2_230",
+        "original_id": "unit2_choice_230"
+      },
+      {
+        "id": "unit2_choice_231",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "巡洋舰和潜水艇与雷达站之间的位置关系如下图所示。以雷达站为观测点, 下面选项中位置描述正确的是 ( )。![](images/153654a16e80874841501133552fb83281ed8f287832ebfac20c2a0ac3bbc7bd.jpg)",
+        "options": [
+          "A. 潜水艇的位置是东偏北  $60^{\\circ}$ , 距离雷达站  $80 \\mathrm{~km}$",
+          "B. 潜水艇的位置是北偏东  $60^{\\circ}$ , 距离雷达站  $80 \\mathrm{~km}$",
+          "C. 巡洋舰的位置是北偏西  $15^{\\circ}$ , 距离雷达站  $100 \\mathrm{~km}$",
+          "D. 巡洋舰的位置是西偏北  $75^{\\circ}$ , 距离雷达站  $100 \\mathrm{~km}$"
+        ],
+        "answer": "B",
+        "solution": "【分析】将方向和距离结合起来描述位置时，要注意三个要素：一是观测点，二是方向，三是距离。北和东之间的夹角是  $90^{\\circ}$ ，北偏东也可以说成东偏北，角度  $= 90^{\\circ}$  一北偏东的角度；西和北之间的夹角是  $90^{\\circ}$ ，西偏北也可以说成北偏西，角度  $= 90^{\\circ}$  一西偏北的角度。\n\n【详解】A.  $90^{\\circ} - 60^{\\circ} = 30^{\\circ}$ , 潜水艇的位置是东偏北  $30^{\\circ}$ , 距离雷达站  $80 \\mathrm{~km}$ , 选项说法错误;  B. 潜水艇的位置是北偏东  $60^{\\circ}$ , 距离雷达站  $80 \\mathrm{~km}$ , 说法正确;  C.  $90^{\\circ} - 15^{\\circ} = 75^{\\circ}$ , 巡洋舰的位置是北偏西  $75^{\\circ}$ , 距离雷达站  $100 \\mathrm{~km}$ , 选项说法错误;  D. 巡洋舰的位置是西偏北  $15^{\\circ}$ , 距离雷达站  $100 \\mathrm{~km}$ , 选项说法错误。描述正确的是潜水艇的位置是北偏东  $60^{\\circ}$ , 距离雷达站  $80\\mathrm{km}$ 。故答案为：B",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 五年级下·北京海淀·期末",
+        "create_time": "2025-10-09 15:45:34",
+        "merged_id": "unit2_231",
+        "original_id": "unit2_choice_231"
+      },
+      {
+        "id": "unit2_choice_232",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "玲玲一家去八面山露营, 玲玲站在帐篷的位置, 爸爸从帐篷位置出发, 先向正东方向走了 100 米, 接着向正南方向走了 100 米。这时, 爸爸在玲玲的 ( ) 方向。",
+        "options": [
+          "A. 西偏南  $45^{\\circ}$",
+          "B. 北偏东  $30^{\\circ}$",
+          "C. 东偏南  $60^{\\circ}$",
+          "D. 南偏东  $45^{\\circ}$"
+        ],
+        "answer": "D",
+        "solution": "【分析】根据题意, 利用图上确定方向的方法: “上北下南, 左西右东” 作图, 先画出此时爸爸的位置, 再根据位置关系以玲玲为观测点, 观察爸爸的方向即可。\n\n【详解】如图：![](images/ba070ce239a681c396c92ded255a1f72371aed97b960dac96f379ca6f338f0f6.jpg)玲玲一家去八面山露营，玲玲站在帐篷的位置，爸爸从帐篷位置出发，先向正东方向走了 100 米，接着向正南方向走了 100 米。这时，爸爸在玲玲的南偏东  $45^{\\circ}$  （或东偏南  $45^{\\circ}$  ）方向。故答案为：D\n\n# 二、填空题",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "22-23 六年级下·湖南湘西·期末",
+        "create_time": "2025-10-09 15:45:34",
+        "merged_id": "unit2_232",
+        "original_id": "unit2_choice_232"
+      },
+      {
+        "id": "unit2_choice_233",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "小明从家里出发往南走 100 米, 再往西走 100 米, 这时，他正好在家的( )偏( )（）°方向上。",
+        "options": [],
+        "answer": "西  南  45",
+        "solution": "【分析】可画出相关图帮助理解：以小明家为观测点，先向南走 100 米，再向西走 100 米，此时小明的位置正好在他家的西偏南  $45^{\\circ}$ （或南偏西  $45^{\\circ}$ ）方向上。\n\n【详解】如图所示：![](images/5ed09648538e12cf703c2833beae1fe1ce0395265399ad3e1fdc95b48d5ddf7e.jpg)小明从家里出发往南走 100 米, 再往西走 100 米, 这时, 他正好在家的西偏南  $45^{\\circ}$  (或南偏西  $45^{\\circ}$ ) 方向上。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·湖北武汉·期中",
+        "create_time": "2025-10-09 15:45:34",
+        "merged_id": "unit2_233",
+        "original_id": "unit2_choice_233"
+      },
+      {
+        "id": "unit2_choice_234",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "公园在小红家的东偏南 $60^{\\circ}$ 方向, 距离 300 米。那么小红家在公园的( )偏 (_____, _____)°方向, 距离 (_____)米。",
+        "options": [],
+        "answer": "西  北   60  300",
+        "solution": "【分析】公园在小红家的东偏南  $60^{\\circ}$  方向上，是以小红家为观测点；小红家在公园的方向是以公园为观测点；根据位置的相对性可知，观测点不同，方向相反，夹角的度数相同，所以东偏南  $60^{\\circ}$  相对的是西偏北  $60^{\\circ}$ ，由此解答。\n\n【详解】公园在小红家的东偏南  $60^{\\circ}$  方向，距离300米。那么小红家在公园的西偏北  $60^{\\circ}$ （或北偏西  $30^{\\circ}$ ）方向，距离300米。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·山东临沂·期中",
+        "create_time": "2025-10-09 17:00:18",
+        "merged_id": "unit2_234",
+        "original_id": "unit2_choice_234"
+      },
+      {
+        "id": "unit2_choice_235",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "小红家在小明家的西偏北 35 度方向上, 距离是 15 千米, 那么小明家在小红家的( )偏( )( )方向上, 距离是( )千米。",
+        "options": [],
+        "answer": "东   南  35度  15",
+        "solution": "【分析】根据位置的相对性，方向相反，角度相同，距离相等，据此解答即可。\n\n【详解】小红家在小明家的西偏北 35 度方向上，距离是 15 千米，那么小明家在小红家的东偏南 35 度（南偏东 55 度）方向上，距离是 15 千米。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·河南郑州·期中",
+        "create_time": "2025-10-09 17:00:18",
+        "merged_id": "unit2_235",
+        "original_id": "unit2_choice_235"
+      },
+      {
+        "id": "unit2_choice_236",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "下面是芳芳从家去学校的路线图。![](images/505029af7b3263bdce5eeb139a3382d76d0bff265ef0f8a11572a0b2fae12b28.jpg)芳芳上学的路线：芳芳先向( )方向走( )m到达商场；再向( )方向走( )m到达公园，最后向( )方向走( )m到达学校。",
+        "options": [],
+        "answer": "正东  600  东偏南    $45^{\\circ}$  300  东偏北   $60^{\\circ}$  400",
+        "solution": "【分析】确定路线时，注意起始点与目的地，起始点是观测点，在地图上按照“上北下南，左西右东”确定方向，用方向和距离描述芳芳上学的路线。\n\n【详解】芳芳上学的路线：芳芳先向正东方向走  $600 \\mathrm{~m}$  到达商场；再向东偏南  $45^{\\circ}$  （或南偏东  $45^{\\circ}$  ）方向走  $300 \\mathrm{~m}$  到达公园，最后向东偏北  $60^{\\circ}$  （或北偏东  $30^{\\circ}$  ）方向走  $400 \\mathrm{~m}$  到达学校。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离描述路线图"
+        ],
+        "source": "23-24 六年级上·湖北黄冈·期中",
+        "create_time": "2025-10-09 17:00:18",
+        "merged_id": "unit2_236",
+        "original_id": "unit2_choice_236"
+      },
+      {
+        "id": "unit2_choice_237",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "东偏北 $50^{\\circ}$ , 也可以说成( ), 小明在小东的西偏南 $36^{\\circ}$ 的方向上, 那小东在小明的( )方向上。",
+        "options": [],
+        "answer": "北偏东 $40^{\\circ}$  南偏西  $36^{\\circ}$",
+        "solution": "【分析】根据地图方向的特征，一个位置可以有两种说法，方向相反，角度相加和为  $90^{\\circ}$ ;根据方向的相对性，方向相反，角度相同，距离相同，据此解答即可。\n\n【详解】东偏北  $50^{\\circ}$ , 也可以说成北偏东  $40^{\\circ}$ , 小明在小东的西偏南  $36^{\\circ}$  的方向上, 那小东在小明的南偏西  $36^{\\circ}$  方向上。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·广东阳江·期中",
+        "create_time": "2025-10-09 17:00:18",
+        "merged_id": "unit2_237",
+        "original_id": "unit2_choice_237"
+      },
+      {
+        "id": "unit2_choice_238",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "小明在小强北偏东 $30^{\\circ}$ 的方向上距离 50 米, 那么小强在小明( )偏( )（） $\\circ$ 方向上, 距离( )米处。",
+        "options": [],
+        "answer": "南西  30  50",
+        "solution": "【分析】根据方向的相对性，它们的方向相反，角度相等，距离相等；据此解答。\n\n【详解】$90^{\\circ} - 30^{\\circ} = 60^{\\circ}$\n\n小明在小强北偏东  $30^{\\circ}$  的方向上距离 50 米, 那么小强在小明南偏西  $30^{\\circ}$  (西偏南  $60^{\\circ}$ ) 方向上, 距离 50 米。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·广东江门·期中",
+        "create_time": "2025-10-09 17:00:18",
+        "merged_id": "unit2_238",
+        "original_id": "unit2_choice_238"
+      },
+      {
+        "id": "unit2_choice_239",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "以芙蓉学校为观测点, 贝贝家在学校的西偏北 $40^{\\circ}$ 方向上, 距离 800 米, 那么以贝贝家为观测点, 芙蓉学校在贝贝家的( ) ( )°方向上, 距离是( )米。",
+        "options": [],
+        "answer": "东偏南  40   800",
+        "solution": "【分析】根据方向的相对性，它们的方向相反，角度相等，距离相等；据此解答。\n\n【详解】$90^{\\circ} - 40^{\\circ} = 50^{\\circ}$以芙蓉学校为观测点，贝贝家在学校的西偏北  $40^{\\circ}$  方向上，距离 800 米，那么以贝贝家为观测点，芙蓉学校在贝贝家的东偏南  $40^{\\circ}$  （南偏东  $50^{\\circ}$ ）方向上，距离是 800 米。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·湖南张家界·期中",
+        "create_time": "2025-10-09 17:00:18",
+        "merged_id": "unit2_239",
+        "original_id": "unit2_choice_239"
+      },
+      {
+        "id": "unit2_choice_240",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如果一只小蚂蚁从 $C$ 点到 D 点要向北偏西 $30^{\\circ}$ 爬 1.2 米, 那么它从 D 点沿原路返回要向( )偏( )（）°方向爬( )米。",
+        "options": [],
+        "answer": "南 东 30  1.2",
+        "solution": "【分析】小蚂蚁从 C 点到 D 点的方向和距离确定, 返回时是沿原路返回, 方向相反, 角度不变, 距离不变。去的时候向北偏西  $30^{\\circ}$ , 返回时就应该向南偏东  $30^{\\circ}$ , 距离仍为 1.2 米;或者东偏南  $60^{\\circ}$ , 距离仍为 1.2 米; 据此解答。\n\n【详解】根据分析：它从 D 点沿原路返回要向南偏东  $30^{\\circ}$  方向爬 1.2 米，或者东偏南  $60^{\\circ}$  方向爬 1.2 米。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离描述路线图",
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·湖南常德·期中",
+        "create_time": "2025-10-09 17:00:18",
+        "merged_id": "unit2_240",
+        "original_id": "unit2_choice_240"
+      },
+      {
+        "id": "unit2_choice_241",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "下面是某市旅游专线的行驶路线图, 看图后回答。![](images/545a61bbd1489a2dab30b66a0ccd1a88cd0127a0aba876feee66cd1173071884.jpg) 从汽车站出发向____方向走约____米到达植物园，再向____方向走约____米到达动物园，然后向____方向走约____米到达科技馆，最后向____偏____ ____ 方向走约____米就到达海滨浴场了。",
+        "options": [],
+        "answer": "东  1000   东北   3000    南   1000   东  南  $10^{\\circ}$  4000",
+        "solution": "【分析】根据方向：上北下南、左西右东；用方向和距离结合来描述路线时，要注意三个要素：一是观测点（即参照物），二是方向，三是距离。图上1厘米表示实际1000米，先以汽车站为观测点即可确定植物园的方向，汽车站和植物园在图上相距1厘米；然后以植物园为观测点即可确定动物园的方向，植物园和动物园在图上相距3厘米；再以动物园为观测点即可确定科技馆的方向，动物园和科技馆在图上相距1厘米；最后以科技馆为观测点即可确定海滨浴场的方向，科技馆和海滨浴场在图上相距4厘米。据此解答。\n\n【详解】$1 \\times 1000 = 1000$  （米）   $3 \\times 1 0 0 0 = 3 0 0 0 (\\text {米})$      $4 \\times 1 0 0 0 = 4 0 0 0 (\\text {米})$从汽车站出发向东方向走约 1000 米到达植物园, 再向东北方向走约 3000 米到达动物园,然后向南方向走约 1000 米到达科技馆, 最后向东偏南  $10^{\\circ}$  (南偏东  $80^{\\circ}$ ) 方向走约 4000 米就到达海滨浴场了。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离描述路线图"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "create_time": "2025-10-09 17:00:18",
+        "merged_id": "unit2_241",
+        "original_id": "unit2_choice_241"
+      },
+      {
+        "id": "unit2_choice_242",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "东偏南 $25^{\\circ}$ 也可以说是( ) $65^{\\circ}$ 。",
+        "options": [],
+        "answer": "南偏东",
+        "solution": "【分析】如下图所示, 正东方向和正南方向的夹角是  $90^{\\circ}, 90^{\\circ} - 65^{\\circ} = 25^{\\circ}$ , 则东偏南  $25^{\\circ}$  也可以说是南偏东  $65^{\\circ}$  。![](images/43babd3b37e395f50a76b13e7a0b4d48e05c4eac2a444dd731ef2db379f92276.jpg)\n\n【详解】通过分析可得：东偏南  $25^{\\circ}$  也可以说是南偏东  $65^{\\circ}$  。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·湖南株洲·期中",
+        "create_time": "2025-10-09 17:00:18",
+        "merged_id": "unit2_242",
+        "original_id": "unit2_choice_242"
+      },
+      {
+        "id": "unit2_choice_243",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "小红放学回家时, 从学校出发, 先向( )方向走( )m 到达商店, 再向( )方向走( )m 到达养鸡场, 又向( )走( )m 到达打谷场, 最后向( )方向走( )m 到家。![](images/e75fb85da9c9dea646db769f416d1a389a799dd5e257403b4000e583a07bdaca.jpg)",
+        "options": [],
+        "answer": "北偏西  $45^{\\circ}$  40  南偏西  $60^{\\circ}$  50    西   75   北偏西  $55^{\\circ}$  60",
+        "solution": "【分析】根据上北下南，左西右东的方向，结合角度写出路线即可解答。\n\n【详解】小红放学回家时, 从学校出发, 先向北偏西  $45^{\\circ}$  方向走  $40 \\mathrm{~m}$  到达商店, 再向南偏西  $60^{\\circ}$  方向走  $50 \\mathrm{~m}$  达到养殖场, 又向西走  $75 \\mathrm{~m}$  达到打谷场, 最后向北偏西  $55^{\\circ}$  方向走  $60 \\mathrm{~m}$  到家。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离描述路线图"
+        ],
+        "source": "24-25 六年级上·全国·课后作业",
+        "create_time": "2025-10-09 17:00:18",
+        "merged_id": "unit2_243",
+        "original_id": "unit2_choice_243"
+      },
+      {
+        "id": "unit2_choice_244",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "观察下图，完成下列各题。![](images/9fff37bcfeb2c420b875dcc56668a609438fc7e361ce0e747fecce780dcb5aa8.jpg) (1)体育场在学校( )偏( )()方向( )m处。 (2)电影院在学校( )偏( )(()方向( )m处。 (3)图书馆在学校( )偏( )()方向( )m处。 (4)音乐厅在学校( )偏( )()方向( )m处。",
+        "options": [],
+        "answer": "(1) 西 北  $45^{\\circ}$  500 (2) 东 北  $60^{\\circ}$  900  (3) 南 西  $45^{\\circ}$  1200  (4) 东 南  $15^{\\circ}$  600",
+        "solution": "【分析】地图方向是上北下南，左西右东，根据各个地点在观测点学校的方向、角度、距离去描述即可。\n\n【详解】（1）体育场在学校西偏北  $45^{\\circ}$  方向  $500\\mathrm{m}$  处。(2）电影院在学校东偏北  $60^{\\circ}$  方向  $900\\mathrm{m}$  处。  (3) 图书馆在学校南偏西  $45^{\\circ}$  方向  $1200 \\mathrm{~m}$  处。  (4) 音乐厅在学校东偏南  $15^{\\circ}$  方向  $600 \\mathrm{~m}$  处。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·全国·单元测试",
+        "create_time": "2025-10-09 17:00:18",
+        "merged_id": "unit2_244",
+        "original_id": "unit2_choice_244"
+      },
+      {
+        "id": "unit2_choice_245",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "观察图, 完成下列各题。![](images/861887ad71bccc54159bfe8293e5c7b8a678e64c30d64ad2b53c3a58fe8b1129.jpg) (1)以小红为观测点，小亮在小红的( )偏( )()方向( )m处。 (2)以小亮为观测点，小红在小亮的( )偏( )()方向( )m处。",
+        "options": [],
+        "answer": "(1) 北 东  $60^{\\circ}$  850 (2) 西  南  $30^{\\circ}$  850",
+        "solution": "【分析】（1）根据地图上的方向“上北下南，左西右东”及角度、距离来确定位置即可。(2) 一个事物在另一个事物的某个方向一定度数的位置，那么另一个事物在这个事物相对的方向相同度数的位置，据此解答即可。\n\n【详解】（1）以小红为观测点，小亮在小红的北偏东  $60^{\\circ}$  方向  $850 \\mathrm{~m}$  处或小亮在小红的东偏北  $30^{\\circ}$  方向  $850 \\mathrm{~m}$  处。  (2) 以小亮为观测点, 小红在小亮的西偏南  $30^{\\circ}$  方向  $850 \\mathrm{~m}$  处或小红在小亮的南偏西  $60^{\\circ}$  方向  $850 \\mathrm{~m}$  处。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "create_time": "2025-10-09 17:00:18",
+        "merged_id": "unit2_245",
+        "original_id": "unit2_choice_245"
+      },
+      {
+        "id": "unit2_choice_246",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "以钟面的正中心为中心点, 钟面上的数字 4 在中心点的( )偏( )（ ）°的方向上。",
+        "options": [],
+        "answer": "东 南 30",
+        "solution": "【分析】时钟面上有 12 个大格, 指针转一周是  $360^{\\circ}$ , 那么两个相邻数字之间的夹角是  $360^{\\circ} \\div 12 = 30^{\\circ}$  。\n\n根据“上北下南，左西右东”可知，钟面上的3时为正东方向，6时为正南方向；根据方向和角度得出数字4的位置。\n\n【详解】如图：![](images/7bd3761c967734812051404fc2feb297aa13ddc60ec945fe2f655e21adc1eaae.jpg)以钟面的正中心为中心点，钟面上的数字 4 在中心点的东偏南  $30^{\\circ}$  （或南偏东  $60^{\\circ}$  ）的方向上。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "22-23 六年级上·福建莆田·期中",
+        "create_time": "2025-10-09 17:00:18",
+        "merged_id": "unit2_246",
+        "original_id": "unit2_choice_246"
+      },
+      {
+        "id": "unit2_choice_247",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "下面是以南湖动物园为观测点画成的示意图。![](images/4ebc6f517ab06a4a3c76d23589b185485fb79ff343a3ade5fbcd8bd82ffaa7fa.jpg)旱冰场在南湖动物园的( )偏( )（ ）方向( )m处。",
+        "options": [],
+        "answer": "东  南  $35^{\\circ}$  500",
+        "solution": "【分析】以南湖动物园为观测点，以图上的“上北下南，左西右东”为准，图例表示图上1个单位长度相当于实际距离  $100 \\mathrm{~m}$  。从图中可知, 旱冰场与南湖动物园相距 5 个单位长度, 那么实际相距  $(100 \\times 5) \\mathrm{~m}$ , 根据图上的方向、角度和距离, 得出旱冰场与南湖动物园的位置关系。\n\n【详解】$100 \\times 5 = 500$  (m)旱冰场在南湖动物园的东偏南  $35^{\\circ}$  （或南偏东  $55^{\\circ}$  ）方向  $500\\mathrm{m}$  处。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "2024 六年级上·全国·专题练习",
+        "create_time": "2025-10-09 17:00:18",
+        "merged_id": "unit2_247",
+        "original_id": "unit2_choice_247"
+      },
+      {
+        "id": "unit2_choice_248",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "交通大厦在火车站的____偏____ ____ 方向走约____米![](images/717c88ec33d5ff496945a9207219ee7649550b285e7be9ef9d62f25c3c0d84ae.jpg)",
+        "options": [],
+        "answer": "南   西   $45^{\\circ} / 45$  度    250",
+        "solution": "【分析】将方向和距离结合起来描述位置时，要注意三个要素：一是观测点，二是方向，三是距离。西和南之间的夹角是  $90^{\\circ}$ ，南偏西也可以说成西偏南，角度  $= 90^{\\circ}$  一南偏西的角度。\n\n【详解】$90^{\\circ} - 45^{\\circ} = 45^{\\circ}$交通大厦在火车站的南偏西  $45^{\\circ}$  或西偏南  $45^{\\circ}$  方向上，距离约是  $250\\mathrm{m}$ 。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·全国·课后作业",
+        "create_time": "2025-10-09 17:00:18",
+        "merged_id": "unit2_248",
+        "original_id": "unit2_choice_248"
+      },
+      {
+        "id": "unit2_choice_249",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "小红家在学校东偏北 $30^{\\circ}$ 方向上, 也可以说小红家在学校北偏())( )度的方向上。",
+        "options": [],
+        "answer": "东   60",
+        "solution": "【分析】在地图上的方位是上北，下南，左西，右东；东和西相对，南和北相对；西南和东北相对，西北和东南相对。\n\n【详解】![](images/2f74ca2fa58041d41fff74df721afa1eb7c52da718dacf10fcae2fcd8d0cd733.jpg)根据图示可知, 东和北之间的角度为  $90^{\\circ}$ , 小红家在学校东偏北  $30^{\\circ}$  方向上,  $90 - 30 = 60$  (度), 也可以说小红家在学校北偏东 60 度的方向上。",
+        "knowledge_points": [
+          "东",
+          "南",
+          "西",
+          "北方向",
+          "东南",
+          "西南",
+          "东北",
+          "西北方向",
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 三年级下·江西宜春·期末",
+        "create_time": "2025-10-09 17:00:18",
+        "merged_id": "unit2_249",
+        "original_id": "unit2_choice_249"
+      },
+      {
+        "id": "unit2_choice_250",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "如图, 在甲、乙两地之间修一条笔直的公路, 从甲地测得公路的走向是北偏东 $60^{\\circ}$ 。甲、乙两地同时开工。若干天后, 公路准确接通, 则乙地所修公路的走向是南偏西( )°。![](images/4291a16f4d997d12365edf506132e18a87f3983176fe6fec81cc3a3a4175e593.jpg)",
+        "options": [],
+        "answer": "60",
+        "solution": "【分析】根据方向的相对性，它们的方向相反，角度相等，距离相等。据此解答。\n\n【详解】在甲、乙两地之间修一条笔直的公路，从甲地测得公路的走向是北偏东  $60^{\\circ}$  。甲、乙两地同时开工。若干天后，公路准确接通，则乙地所修公路的走向是南偏西  $60^{\\circ}$  。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·全国·课后作业",
+        "create_time": "2025-10-09 17:00:18",
+        "merged_id": "unit2_250",
+        "original_id": "unit2_choice_250"
+      },
+      {
+        "id": "unit2_choice_251",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "明明面向西站立, 向右转 $60^{\\circ}$ 后所面对的方向是( )偏( )（ ）°或( )偏( )（ ）。",
+        "options": [],
+        "answer": "西   北  60  北  西  30",
+        "solution": "【分析】地图上按上北下南左西右东确定方向，面向西站立，向右转是转向北，向右旋转的角度就是西偏北的角度。西和北之间的夹角是  $90^{\\circ}$ ，西偏北也可以说成北偏西，角度  $= 90^{\\circ} -$  西偏北的角度。\n\n【详解】$90^{\\circ} - 60^{\\circ} = 30^{\\circ}$明明面向西站立，向右转  $60^{\\circ}$  后所面对的方向是西偏北  $60^{\\circ}$  或北偏西  $30^{\\circ}$  。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·全国·课后作业",
+        "create_time": "2025-10-09 17:00:18",
+        "merged_id": "unit2_251",
+        "original_id": "unit2_choice_251"
+      },
+      {
+        "id": "unit2_choice_252",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "小丽在东东西偏北 $30^{\\circ}$ 方向 $50 \\mathrm{~m}$ 处, 则东东在小丽的( )方向( )m 处。",
+        "options": [],
+        "answer": "东偏南 $30^{\\circ}$  50",
+        "solution": "【分析】根据方向的相对性，西偏北对东偏南，角度和距离不变，进行填空。东和南之间的夹角是  $90^{\\circ}$ ，东偏南也可以说成南偏东，角度  $= 90^{\\circ}$  一东偏南的角度。\n\n【详解】$90^{\\circ} - 30^{\\circ} = 60^{\\circ}$小丽在东东西偏北  $30^{\\circ}$  方向  $50 \\mathrm{~m}$  处，则东东在小丽的东偏南  $30^{\\circ}$  或南偏东  $60^{\\circ}$  方向  $50 \\mathrm{~m}$  处。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·全国·课后作业",
+        "create_time": "2025-10-09 17:00:18",
+        "merged_id": "unit2_252",
+        "original_id": "unit2_choice_252"
+      },
+      {
+        "id": "unit2_choice_253",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "下图表示篮球场与图书馆的位置关系。篮球场在图书馆的( )偏( ) $30^{\\circ}$ 方向( )m处。 ![](images/d8b92d4db0b76af080caa5eeaf8ea3836e50d22b7d664c343ba92b241061e57f.jpg)",
+        "options": [],
+        "answer": "西   北  600",
+        "solution": "【分析】用东、西、南、北、东北、东南、西南、西北等方向来表述位置或用方向和距离相结合的方法来描述位置，既可以用来确定现实空间中物体的位置，也可以用来确定平面上物体的位置。将方向和距离结合起来描述位置时，要注意三个要素：一是观测点，二是方向，三是距离。以图书馆为观测点，根据上北下南，左西右东，从图片中看出篮球场的位置在西偏北  $30^{\\circ}$  方向。图中一段图上的距离代表实际距离  $200 \\mathrm{~m}$ ，图上一共 3 段，则代表实际距离  $600 \\mathrm{~m}$ 。\n\n【详解】篮球场在图书馆的西偏北  $30^{\\circ}$  方向  $600\\mathrm{m}$  处。\n\n# 三、判断题",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级下·北京朝阳·期末",
+        "create_time": "2025-10-09 17:00:18",
+        "merged_id": "unit2_253",
+        "original_id": "unit2_choice_253"
+      },
+      {
+        "id": "unit2_choice_254",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "抗震纪念碑的南偏东 $40^{\\circ}$ 的方向, 与抗震纪念碑的东偏南 $40^{\\circ}$ 的方向相同。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】根据正东和正南的夹角是  $90^{\\circ}, 90^{\\circ} - 40^{\\circ} = 50^{\\circ}$ , 所以南偏东  $40^{\\circ}$  也可以说成东偏南  $50^{\\circ}$  。\n\n【详解】$90^{\\circ} - 40^{\\circ} = 50^{\\circ}$所以，抗震纪念碑的南偏东  $40^{\\circ}$  的方向，与抗震纪念碑的东偏南  $50^{\\circ}$  的方向相同。原题说法错误。故答案为： $\\times$",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·河北唐山·期中",
+        "create_time": "2025-10-09 17:00:18",
+        "merged_id": "unit2_254",
+        "original_id": "unit2_choice_254"
+      },
+      {
+        "id": "unit2_choice_255",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "东偏南 $40^{\\circ}$ 方向是西偏南 $40^{\\circ}$ 方向的反方向。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】根据位置的相对性可知，它们的方向相反，角度相等，据此解答。\n\n【详解】东偏南  $40^{\\circ}$  方向是西偏北  $40^{\\circ}$  方向的反方向。所以原题说法错误。故答案为：×。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·河北唐山·期中",
+        "create_time": "2025-10-09 17:00:18",
+        "merged_id": "unit2_255",
+        "original_id": "unit2_choice_255"
+      },
+      {
+        "id": "unit2_choice_256",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "小丽从家向东偏北 $40^{\\circ}$ 方向走到学校, 她放学从学校回家要向西偏南 $50^{\\circ}$ 方向走。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】根据位置的相对性：方向相反，角度和距离不变，原路返回时，与去时的方向相反，角度不变，据此解答。\n\n【详解】小丽从家向东偏北  $40^{\\circ}$  方向走到学校, 她放学从学校回家要向西偏南  $40^{\\circ}$  方向走,所以原题说法错误。\n\n故答案为： $\\times$",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·河北沧州·期中",
+        "create_time": "2025-10-09 17:00:18",
+        "merged_id": "unit2_256",
+        "original_id": "unit2_choice_256"
+      },
+      {
+        "id": "unit2_choice_257",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "欢欢在乐乐的西偏南 $30^{\\circ}$ 方向 $50 \\mathrm{~m}$ 处, 那么乐乐在欢欢的北偏东 $60^{\\circ}$ 方向 $50 \\mathrm{~m}$ 处。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】根据方向的相对性，它们的方向相反，角度相等，距离相等；据此解答。\n\n【详解】$90^{\\circ} - 30^{\\circ} = 60^{\\circ}$ 欢欢在乐乐的西偏南  $30^{\\circ}$  方向  $50 \\mathrm{~m}$  处, 那么乐乐在欢欢的东偏北  $30^{\\circ}$  (或北偏东  $60^{\\circ}$ ) 方向  $50 \\mathrm{~m}$  处。原题干说法正确。故答案为：√",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·河南南阳·期中",
+        "create_time": "2025-10-09 17:00:18",
+        "merged_id": "unit2_257",
+        "original_id": "unit2_choice_257"
+      },
+      {
+        "id": "unit2_choice_258",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "西偏北 $30^{\\circ}$ 与北偏西 $30^{\\circ}$ 表示同一方向。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】根据东南西北四个基本方位可知，西和北之间的夹角是  $90^{\\circ}$ ,  $90^{\\circ} - 30^{\\circ} = 60^{\\circ}$ , 所以西偏北  $30^{\\circ}$  方向, 还可以说成北偏西  $60^{\\circ}$  方向, 据此判断。\n\n【详解】$90^{\\circ} - 30^{\\circ} = 60^{\\circ}$西偏北  $30^{\\circ}$  与北偏西  $60^{\\circ}$  表示同一方向。原题说法错误。故答案为： $\\times$",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·贵州铜仁·期中",
+        "create_time": "2025-10-09 17:00:18",
+        "merged_id": "unit2_258",
+        "original_id": "unit2_choice_258"
+      },
+      {
+        "id": "unit2_choice_259",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "文化馆在游泳馆的南偏东 $30^{\\circ}$ 方向上, 也可以说文化馆在游泳馆的东偏南 $60^{\\circ}$ 方向上。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】以游泳馆为观测点, 文化馆在游泳馆的南偏东  $30^{\\circ}$  方向上, 观测点不变, 因为东和南之间的夹角是  $90^{\\circ}, 90^{\\circ} - 30^{\\circ} = 60^{\\circ}$ , 所以南偏东  $30^{\\circ}$  方向, 还可以说成东偏南  $60^{\\circ}$  方向。\n\n【详解】$90^{\\circ} - 30^{\\circ} = 60^{\\circ}$文化馆在游泳馆的南偏东  $30^{\\circ}$  方向上，也可以说文化馆在游泳馆的东偏南  $60^{\\circ}$  方向上。原题说法正确。故答案为：√",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·湖北黄冈·期中",
+        "create_time": "2025-10-09 17:00:18",
+        "merged_id": "unit2_259",
+        "original_id": "unit2_choice_259"
+      },
+      {
+        "id": "unit2_choice_260",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "早上, 小明面向太阳站着, 他先向左转。再向后转, 这时他就面向南方。( )理由：______________________________________________________",
+        "options": [],
+        "answer": "$\\sqrt{}$  早上太阳从东方升起，小明面向太阳，就是面向东方；先向左转，他面对的是北方，再向后转，面对的是南方。",
+        "solution": "【分析】根据地图上方向的规定：上北下南，左西右东；先说出小明面对太阳的方向，再说出向左转面对的方向，最后再向后转面对的方向，进而解答。\n\n【详解】早上，小明面向太阳站着，他先向左转。再向后转，这时他就面向南方。（√）理由：早上太阳从东方升起，小明面向太阳，就是面向东方；先向左转，他面对的是北方，再向后转，面对的是南方。",
+        "knowledge_points": [
+          "东",
+          "南",
+          "西",
+          "北方向",
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·全国·期中",
+        "create_time": "2025-10-09 17:00:18",
+        "merged_id": "unit2_260",
+        "original_id": "unit2_choice_260"
+      },
+      {
+        "id": "unit2_choice_261",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "小红从家出发, 向南偏西 $30^{\\circ}$ 走 800 米到学校, 放学时他向东偏北 $60^{\\circ}$ 走 800 米回家。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】根据方向的相对性，南偏西对北偏东，角度和距离不变进行分析，而北和东之间的夹角是  $90^{\\circ}$ ，因此北偏东也可以说成东偏北，角度  $= 90^{\\circ}$  一北偏东的角度。\n\n【详解】$90^{\\circ} - 30^{\\circ} = 60^{\\circ}$小红从家出发，向南偏西  $30^{\\circ}$  走 800 米到学校，放学时他向北偏东  $30^{\\circ}$  或东偏北  $60^{\\circ}$  走 800 米回家，原题说法正确。故答案为:  $\\sqrt{}$",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级下·湖南郴州·期末",
+        "create_time": "2025-10-09 17:00:18",
+        "merged_id": "unit2_261",
+        "original_id": "unit2_choice_261"
+      },
+      {
+        "id": "unit2_choice_262",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "学校在公园北偏东 $30^{\\circ}$ 约 500 米处, 公园在学校的南偏西 $30^{\\circ}$ 约 500 米处。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】根据位置的相对性，方向相反，角度相同，距离相等，据此判断即可。\n\n【详解】由分析可知：学校在公园北偏东  $30^{\\circ}$  约500米处，公园在学校的南偏西  $30^{\\circ}$  约500米处。说法正确。\n\n故答案为： $\\sqrt{}$\n\n# 四、作图题",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级下·湖南娄底·期末",
+        "create_time": "2025-10-09 17:00:18",
+        "merged_id": "unit2_262",
+        "original_id": "unit2_choice_262"
+      },
+      {
+        "id": "unit2_choice_263",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "根据下面的描述, 在平面图上标出各场所的位置。![](images\\65cbd9659b229a8f56b1eb96daea34683fbca895ed9188605521a7d55f0c151d.jpg) (1) 聪聪家在公园北偏西 $30^{\\circ}$ 方向 450 米处。 (2) 思思家在公园南偏西 $45^{\\circ}$ 方向 900 米处。 (3) 涵涵家在公园南偏东 $60^{\\circ}$ 方向 600 米处。 (4) 晶晶家在公园北偏东 $40^{\\circ}$ 方向 300 米处。",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】将方向和距离结合起来描述位置时，要注意三个要素：一是观测点，二是方向，三是距离。弄清要标示的物体在哪个方位上，有多少度，按要求的方位和度数准确画图；注意各场所离中心点的距离，根据要求画出相应的长度。\n\n【详解】$450 \\div 300 = 1.5$  （厘米）$900 \\div 300 = 3$  （厘米）$600 \\div 300 = 2$  （厘米）$300 \\div 300 = 1$  （厘米）![](images/0220fafe4e28c82b71af0073622f6e2503e9157aa8d4ddb42e5d4db3d29f2285.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "21-22 六年级上·河北保定·期末",
+        "create_time": "2025-10-09 17:00:18",
+        "merged_id": "unit2_263",
+        "original_id": "unit2_choice_263"
+      },
+      {
+        "id": "unit2_choice_264",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "作图。![](images/473078ecdecbfdb60680a6df2f2b78b30be1625f5c2f94ff26dc9bd3583cee1e.jpg)某勘探队在 A 城南偏西 $40^{\\circ}$ 约 60 千米处发现稀有金属矿, 请在图中标出该金属矿的位置。",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】根据纸上方向：上北下南，左西右东。由图可知，图上1厘米表示实际距离20千米，用60除以20得可用图上3厘米表示，再根据方向角度和距离在图中标出金属矿的位置。\n\n【详解】$60 \\div 20 = 3$  （厘米）作图如下：![](images/c29488c394fb58bca7951a3bafa86017f6c6bdff83ade6f33a189c22ec671737.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·河北邢台·期中",
+        "create_time": "2025-10-09 17:00:18",
+        "merged_id": "unit2_264",
+        "original_id": "unit2_choice_264"
+      },
+      {
+        "id": "unit2_choice_265",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "周六下午，宏宏约强强一起去书店。 (1) 宏宏从家里出发，沿（）偏（）（）°方向走（）米到达公园，再向（）方向走400米到达书店。 (2) 强强家在书店的东偏南 $30^{\\circ}$ 方向 800 米处, 请在图中画出强强家的位置。 ![](images/3292594d7f67d854c266dbbe571f2f797b86cafed5c5f9c393f0ed6138775988.jpg)",
+        "options": [],
+        "answer": "（1）北；东   65；600；正东",
+        "solution": "【分析】以图上的“上北下南，左西右东”为准，图例表示图上1厘米相当于实际距离200米。\n\n(1) 从图中可知, 宏宏家与公园相距 3 厘米, 则实际相距  $(200 \\times 3)$  米, 结合方向、角度和距离描述宏宏从家到公园的路线图。  (2) 在书店的东偏南  $30^{\\circ}$  方向上画  $800 \\div 200 = 4$  厘米长的线段, 即是强强家。\n\n【详解】(1)  $200 \\times 3 = 600$  （米）宏宏从家里出发，沿北偏东  $65^{\\circ}$  （或东偏北  $25^{\\circ}$  ）方向走 600 米到达公园，再向正东方向走 400 米到达书店。(2)  $800 \\div 200 = 4$  （厘米）\n\n作图如下：![](images/8725b0ad7a487681ea2adfe73f0790618b7347ba969ad793e5316d3730bd60bb.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离描述路线图",
+          "根据方向",
+          "角度和距离确定物体的位置。"
+        ],
+        "source": "23-24六年级上·浙江温州·期中",
+        "create_time": "2025-10-09 17:00:18",
+        "merged_id": "unit2_265",
+        "original_id": "unit2_choice_265"
+      },
+      {
+        "id": "unit2_choice_266",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "在平面图上标出各个球的位置。![](images/95893271ed46d9bf14c82e3db7e8c53dc1db1fcd7b29e8bc70643dcbaf299cc5.jpg) (1) 5号球在黑球东偏北 $20^{\\circ}$ 的方向40厘米处。 (2) 3号球在黑球西偏北 $40^{\\circ}$ 的方向60厘米处。 (3) 6号球在黑球南偏西 $45^{\\circ}$ 的方向60厘米处。 (4) 10号球在黑球东偏南 $30^{\\circ}$ 的方向30厘米处。",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】将方向和距离结合起来描述位置时，要注意三个要素：一是观测点，二是方向，三是距离。\n\n1、弄清要标示的物体在哪个方位上，有多少度，按要求的方位和度数准确画图  2、注意各场所离中心点的距离，根据要求的比例画出相应的长度。在确定夹角时，要根据方向来确定，比如北偏东，就是把正北方向对应量角器上的  $0^{\\circ}$  刻度线。南偏东就是把正南方向对应量角器商的  $0^{\\circ}$  刻度线。根据图中的线段图，得出一段也就是1厘米表示20厘米。(1) 40 厘米里面有 2 个 20 厘米, 则需要画两段;  (2) 60 厘米里面有 3 个 20 厘米, 则需要画三段;  \n\n(3) 60 厘米里面有 3 个 20 厘米, 则需要画三段;  (4) 30 厘米里面有 1.5 个 20 厘米, 则需要画 1.5 段;\n\n【详解】$40 \\div 20 = 2$ (段) $6 0 \\div 3 0 = 3 {(段)}$     $3 0 \\div 2 0 = 1. 5 (段)$\n\n![](images/4fe5fe2b30e5a52c22f15ce336c3798535174daa71b24c93a10bc6cadf15d962.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·河北沧州·期中",
+        "create_time": "2025-10-09 17:00:18",
+        "merged_id": "unit2_266",
+        "original_id": "unit2_choice_266"
+      },
+       {
+        "id": "unit2_choice_267",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "根据下面的描述，在平面图上标出各场所的位置。 ![](images/e669a36c2eea75494d3045274d11435c2aa54f8782d74bfe764f3e191ea3f320.jpg) (1) 体育馆在公园的东偏南 $30^{\\circ}$ 方向 1 千米处。(2) 图书馆在公园西偏南 $45^{\\circ}$ 方向 1500 米处。 (3) 海洋馆在公园的西偏北 $50^{\\circ}$ 方向 2 千米处。 (4) 超市在公园的北偏东 $60^{\\circ}$ 方向 1500 米处。",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】以公园为观测点，以图上的“上北下南，左西右东”为准，图例表示图上1厘米相当于实际距离500米。\n\n(1) 1 千米 = 1000 米, 在公园的东偏南  $30^{\\circ}$  方向上画  $1000 \\div 500 = 2$  厘米长的线段, 即是体育馆;  (2) 在公园西偏南  $45^{\\circ}$  方向上画  $1500 \\div 500 = 3$  厘米长的线段, 即是图书馆;  (3) 2千米 = 2000 米, 在公园的西偏北  $50^{\\circ}$  方向上画  $2000 \\div 500 = 4$  厘米长的线段, 即是海洋馆;  (4) 在公园的北偏东  $60^{\\circ}$  方向上画  $1500 \\div 500 = 3$  厘米长的线段, 即是超市。\n\n【详解】(1) 1千米  $= 1000$  米  $1 0 0 0 \\div 5 0 0 = 2 \\text {(厘 米)}$     (2)  $1500 \\div 500 = 3$  （厘米）  (3) 2千米  $= 2000$  米  $2 0 0 0 \\div 5 0 0 = 4 (\\text {厘 米})$(4)  $1500 \\div 500 = 3$  （厘米）\n\n如图:![](images/1bee4c262fb8f736fd4f3db0cea1a78e2f23e8e9ea9472b1493fe17be4548dd5.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·贵州铜仁·期中",
+        "create_time": "2025-10-09 17:06:53",
+        "merged_id": "unit2_267",
+        "original_id": "unit2_choice_267"
+      },
+      {
+        "id": "unit2_choice_268",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "101 路公共汽车从胡家湾向东偏北 $35^{\\circ}$ 方向行驶 4 千米后到达兴庆公园, 再向东行驶 3 千米到达省博物馆, 然后向北偏东 $60^{\\circ}$ 方向行驶 5 千米到达市第二医院, 最后向南行驶 2 千米到达人民广场。(根据上面的描述, 把公共汽车行驶的路线图画完整。)![](images/db21142794545edc2dda3d485fe8023aaee67c183544ca95a57d8dad659c95c3.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】确定路线时，注意起始点与目的地，起始点是观测点，以图上的“上北下南，左西右东”确定方向，图例表示图上1厘米相当于实际距离1千米，根据方向、角度和距离画出路线图。\n\n【详解】$4 \\div 1 = 4$  （厘米）  $3 \\div 1 = 3 (\\text {厘 米})$$5 \\div 1 = 5$  （厘米） $2 \\div 1 = 2$  （厘米）\n\n作图如下：![](images/843bda71aeec66c74d5d75642c319794c4bd1d0bcf979c5c633d2439e6dadfc4.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离画线路图"
+        ],
+        "source": "23-24 六年级上·甘肃武威·期中",
+        "create_time": "2025-10-09 17:06:53",
+        "merged_id": "unit2_268",
+        "original_id": "unit2_choice_268"
+      },
+      {
+        "id": "unit2_choice_269",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "“1 路公共汽车从起点站向西偏北 $40^{\\circ}$ 方向行驶 $3 \\mathrm{~km}$ 后向西行驶 $2 \\mathrm{~km}$ , 再向南偏西 $30^{\\circ}$ 方向行驶 $3.5 \\mathrm{~km}$ 到达终点站。”根据上面的描述, 把公共汽车行驶的路线图画完整。![](images\\2b174781-04ee-4d65-b7ca-b0c3575b9acd.png)",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】图上的方向是上北下南、左西右东，据题意可用图上1厘米表示实际1千米，先确定方向，然后根据实际距离确定图上距离，根据图上的方向、夹角的度数和图上距离画出路线图即可。\n\n【详解】据分析作图如下：![](images/f67a74e2635d9ef10702a1ac0f0414a174d9aec6f7be0638e289c51b388aad26.jpg)",
+        "knowledge_points": [
+          "东",
+          "南",
+          "西",
+          "北方向",
+          "根据方向",
+          "角度和距离确定物体的位置",
+          "根据方向",
+          "角度和距离画线路图"
+        ],
+        "source": "21-22 六年级上·河南信阳·期末",
+        "create_time": "2025-10-09 17:06:53",
+        "merged_id": "unit2_269",
+        "original_id": "unit2_choice_269"
+      },
+      {
+        "id": "unit2_choice_270",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "根据下面的描述, 在平面图上画出各动物场馆的位置。 ![](images/948d2f065895cf219bb79a1e6fb7c583c439a854f4fcdb68bcc55ff113509506.jpg) (1) 海豚馆在动物园大门的西偏南 $60^{\\circ}$ 方向 300 米处。 (2) 大象馆在动物园大门的北偏东 $30^{\\circ}$ 方向 400 米处。 (3) 熊猫馆在动物园大门的正西方向100米处。 (4) 河马馆在动物园大门的东偏北 $45^{\\circ}$ 方向 250 米处。",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】以图上的 “上北下南, 左西右东” 为准, 图例表示图上 1 厘米相当于实际距离 100 米。(1) 在动物园大门的西偏南  $60^{\\circ}$  方向上画  $300 \\div 100 = 3$  厘米长的线段, 即是海豚馆;(2) 在动物园大门的北偏东  $30^{\\circ}$  方向上画  $400 \\div 100 = 4$  厘米长的线段, 即是大象馆;  (3) 在动物园大门的正西方向上画  $100 \\div 100 = 1$  厘米长的线段, 即是熊猫馆;  (4) 在动物园大门的东偏北  $45^{\\circ}$  方向上画  $250 \\div 100 = 2.5$  厘米长的线段, 即是河马馆。\n\n【详解】(1)  $300 \\div 100 = 3$  （厘米）(2)  $400 \\div 100 = 4$  （厘米）  (3)  $100 \\div 100 = 1$  （厘米）  (4)  $250 \\div 100 = 2.5$  （厘米）\n\n如图：![](images/394cb60f0379a5edcb45ad6e9a495fd7cc9b6a1e40db65d95ebbb71b8eef13f7.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·河南焦作·期中",
+        "create_time": "2025-10-09 17:06:53",
+        "merged_id": "unit2_270",
+        "original_id": "unit2_choice_270"
+      },
+      {
+        "id": "unit2_choice_271",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "1 路公共汽车从起点站出发, 向西偏南 $30^{\\circ}$ 方向行驶3千米，然后向正西方向行驶4千米，最后向北偏西 $40^{\\circ}$ 方向行驶2千米到达终点站。(1) 根据上面的描述, 把 1 路公共汽车的行驶路线补充完整。![](images/a97443322b97b561439f6653b968e413be3edc9778bcbfc0172993d39155e020.jpg) (2) 根据路线图，填写公共汽车沿原路返回时行驶的方向和路程。首先，从终点站出发，向（）方向走（）千米；然后向（）方向走（）千米；最后向（）方向走（）千米到起点站。",
+        "options": [],
+        "answer": "(1) 见详解(2) 南偏东 $40^{\\circ}$ ; 2; 正东; 4; 东偏北  $30^{\\circ}$ ; 3",
+        "solution": "【分析】（1）根据地图上的方位是“上北下南、左西右东”，根据题意可知，图上1个单位长度表示1千米，据此画出1路公共汽车行驶的线路图即可；(2) 根据方向的相对性, 方向相反, 距离和角度不变, 写出公共汽车原路返回的行驶路线。\n\n【详解】(1)  $4 \\div 1 = 4$  （个）$2 \\div 1 = 2 (\\text {个})$\n\n如图：![](images/660eea86551b9fca5a3bd2022fcddc137437d384364d90d5a68e6bcec0f80fd5.jpg)\n\n(2) 首先, 从终点站出发, 向南偏东  $40^{\\circ}$  (东偏南  $60^{\\circ}$ ) 方向走 2 千米; 然后向正东方向走4千米；最后向东偏北  $30^{\\circ}$  （北偏东  $60^{\\circ}$  ）方向走3千米到起点站。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离描述路线图"
+        ],
+        "source": "23-24 六年级上·福建漳州·期中",
+        "create_time": "2025-10-09 17:06:53",
+        "merged_id": "unit2_271",
+        "original_id": "unit2_choice_271"
+      },
+      {
+        "id": "unit2_choice_272",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "淞江实验小学的同学们在参加暑假夏令营活动时，从基地出发，先向东偏北 $30^{\\circ}$ 方向行进 $400\\mathrm{m}$ 后到达A点，再向正西方向行进 $400\\mathrm{m}$ 到达B点，再向南偏西 $45^{\\circ}$ 方向行进 $300\\mathrm{m}$ 到达C点，最后再向东偏南 $40^{\\circ}$ 方向行进 $200\\mathrm{m}$ 到达目的地。![](images/0e04c837fb9c3b1549c8b1a06b728ab20c1720481b9d9907a6c5141f74ed0fdb.jpg)![](images/daf0d160acea9acf39864481439bab214cacd1c31c9dbe645fa5e3254b2dde6d.jpg)\n\n(1) 根据描述，画出行进路线图。\n(2) 活动结束后，请描述出返程路线。\n(3) 已知从基地到目的地用时 10 分钟, 返程平均速度是去时的 $\\frac{10}{13}$ , 返回程用时 ( ) 分钟。",
+        "options": [],
+        "answer": "(1) 见详解(2）见详解  (3) 13",
+        "solution": "【分析】（1）（2）确定观测点，根据图中方向上北下南，左西右东，确定方向、角度和距离。\n\n(3) 先用加法计算总路程, 根据路程 $\\div$ 时间 = 速度, 算出去时的速度, 再根据求一个数的几分之几是多少, 用乘法计算, 求出返程平均速度, 最后根据路程 $\\div$ 速度 = 时间, 计算时间即可。\n\n【详解】（1）如下图：\n\n![](images/888da6855c0eb08c94d429846e6aeda64b686225c54d1228930ce6b885b4e5a4.jpg)\n\n(2) 返程路线: 从目的地出发, 先向西偏北 $40^{\\circ}$ (或北偏西 $50^{\\circ}$) 方向行进 $200 \\mathrm{~m}$ 到达 C 点,再向北偏东 $45^{\\circ}$ (或东偏北 $45^{\\circ}$) 方向行进 $300 \\mathrm{~m}$ 到达 B 点, 再向正东方向行进 $400 \\mathrm{~m}$ 到达 A 点, 最后再向西偏南 $30^{\\circ}$ (或南偏西 $60^{\\circ}$) 方向行进 $400 \\mathrm{~m}$ 到达基地。\n\n(3) $(400 + 400 + 300 + 200)\\div 10\\times \\frac{10}{13} = 1 3 0 0 \\div 1 0 \\times \\frac {1 0}{1 3} = 1 3 0 \\times \\frac {1 0}{1 3}$ = 100 （米/分钟）$(4 0 0 + 4 0 0 + 3 0 0 + 2 0 0) \\div 1 0 0 = 1 3 0 0 \\div 1 0 0 = 1 3 \\quad (\\text {分 钟})$返程用时13分钟。",
+        "knowledge_points": [
+          "根据方向、角度和距离描述路线图",
+          "根据方向、角度和距离画线路图",
+          "基础行程问题"
+        ],
+        "source": "23-24六年级上·贵州遵义·期中",
+        "create_time": "2025-10-09 17:32:50",
+        "merged_id": "unit2_272",
+        "original_id": "unit2_choice_272"
+      },
+      {
+        "id": "unit2_choice_273",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "1路公共汽车从起点站先向正西方向行驶2千米，再向西偏北 $40^{\\circ}$ 方向行驶4千米，最后向南偏西 $30^{\\circ}$ 方向行驶6千米到达终点站。(1) 根据上面的描述，画出 1 路公共汽车行驶的路线图。![](images/70d95c8eb1ec2f6ecdb383217e8b0eff1818623225056d1ae7130b7bdd01aefc.jpg)(2) 1 路公共汽车从起点站到终点站共用了 15 分钟, 平均每分钟行驶多少米?",
+        "options": [],
+        "answer": "(1) 见详解 (2) 800米",
+        "solution": "【分析】（1）地图方向是上北下南，左西右东，图上1厘米代表实际距离2千米，根据1路公共汽车从起点站出发的线路图，画出即可；(2) 用总路程除以时间, 求出 1 路公共汽车的速度, 据此解答即可。\n\n【详解】(1) 如图:![](images/f9935e80974e394a146709c01fec932a9eff5482a60bfc03b29db3fe20e0f830.jpg)\n\n(2）路程：  $2 + 4 + 6 = 6 + 6 = 12$  （千米）12千米  $= 12000$  米 速度：  $12000\\div 15 = 800$  （米/分）\n\n答：平均每分钟行驶800米。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离画线路图",
+          "基础行程问题"
+        ],
+        "source": "23-24六年级上·湖北黄冈·期中",
+        "create_time": "2025-10-09 17:06:53",
+        "merged_id": "unit2_273",
+        "original_id": "unit2_choice_273"
+      },
+      {
+        "id": "unit2_choice_274",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "在平面图上画出校园内各建筑物的位置。 ![](images/99e09b90f5753ee51c58c4a4ab615e93ca1f19db787caddf9ffe430b294be285.jpg) (1) 教学楼在校门的正北方向 $200 \\mathrm{~m}$ 处。 (2) 图书馆在校门的北偏东 $45^{\\circ}$ 方向 $200 \\mathrm{~m}$ 处。 (3) 体育馆在校门的西偏北 $60^{\\circ}$ 方向 $300 \\mathrm{~m}$ 处。",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】弄清要标示的物体在哪个方位上，有多少度，按要求的方位和度数准确画图；注意各场所离中心点的距离，根据要求画出相应的长度。\n\n【详解】![](images/0a24a12e049b374cf156c16d0565a060091809db24315dc64cfadd71f3925922.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·广东佛山·期中",
+        "create_time": "2025-10-09 17:06:53",
+        "merged_id": "unit2_274",
+        "original_id": "unit2_choice_274"
+      },
+      {
+        "id": "unit2_choice_275",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "一艘渔船在途中遇特大风浪即将沉没，船长发出了SOS信号，如图是距离这艘渔船较近的几艘营救船所在位置的平面图。 ![](images/4d1f2ecc82f3388f1dd08a4fd07738432b688ae2f888cb08690cd3eedf8c540e.jpg) (1) 此时离渔船最近的海上搜救船在渔船北偏东 $20^{\\circ}$ 方向 250 海里处, 请在平面图中画图标记出这艘海上搜救船的位置。 (2) “海神”一号在这艘渔船（）方向（）海里处。“海神”二号在这艘渔船（）方向（）海里处。",
+        "options": [],
+        "answer": "（1）图见解析；",
+        "solution": "【分析】（1）由图示可知向上的方向为北，向右的方向为东，一段表示100海里，250表示两段另加半段，如图所示即为所求；(2) 由图示可知向上的方向为北, 向右的方向为东, “海神”一号在这艘渔船东偏北  $15$  方向, 一段表示 100 海里那么三段就表示 300 海里; “海神”二号在这艘渔船西偏北  $30^{\\circ}$  方向, 一段表示 100 海里那么三段就表示 300 海里。\n\n【详解】（1）如图搜救船在渔船北偏东  $20^{\\circ}$  方向250海里处，![](images/bf6106965e1f83c5eceeacd1de5f94da99c50c3891fb5d865c1a7cef192a7b87.jpg)\n\n(2) 因为  $2 \\times 100 = 200$  （海里），所以 “海神” 一号在这艘渔船东偏北  $15^{\\circ}$  方向  $200$  海里处, 或北偏东  $75^{\\circ}$  方向  $200$  海里处。因为  $3 \\times 100 = 300$  (海里),所以“海神”二号在这艘渔船西偏北  $30^{\\circ}$  方向300海里处，或北偏西  $60^{\\circ}$  方向300海里处。本题考查了方位角的定义，理解方位角的定义是解题的关键。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·湖南张家界·期中",
+        "create_time": "2025-10-09 17:06:53",
+        "merged_id": "unit2_275",
+        "original_id": "unit2_choice_275"
+      },
+      {
+        "id": "unit2_choice_276",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "一辆汽车从起点站出发，先向东偏南 $45^{\\circ}$ 方向行驶 $15\\mathrm{km}$ 后，然后向正南方面行驶 $5\\mathrm{km}$ ，再向正东方向行驶 $10\\mathrm{km}$ ，最后向东偏北 $30^{\\circ}$ 方向行驶 $10\\mathrm{km}$ 到达终点站。请把汽车行驶的路线图画完整。 ![](images\\f6fb8691-01fc-4fa0-ad31-333ae2079ac1.png)",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】由图可知, 图上  $1 \\mathrm{~cm}$  表示实际距离  $5 \\mathrm{~km}$ , 用实际距离除以 5 , 得到图上距离, 再根据图上表示的方向: 上北下南, 左西右东, 结合距离和角度求解, 据此解答。\n\n【详解】$15 \\div 5 = 3$  (cm)  $1 0 \\div 5 = 2 \\quad (\\mathrm {c m})$\n\n作图如下：![](images/aabd551ae1424f9f3e7906a095862d88025df077a3335919031a6a884dcffbf3.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离画线路图"
+        ],
+        "source": "23-24六年级上·湖南娄底·期中",
+        "create_time": "2025-10-09 17:06:53",
+        "merged_id": "unit2_276",
+        "original_id": "unit2_choice_276"
+      },
+      {
+        "id": "unit2_choice_277",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "根据描述, 在下图中标出各场所的位置。 (1) 书店在广场北偏东 $40^{\\circ}$ 方向，600 米处。 (2) 超市在广场北偏西 $55^{\\circ}$ 方向, 400 米。 (3) 公园在广场南偏西 $60^{\\circ}$ 方向, 200 米处。 (4) 体育馆在广场南偏东 $30^{\\circ}$ 方向, 600 米处。 ![](images/3ab0c4ac02615c635dc0c773f38c8ecc033041336b018939dd2fa258e446ff77.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】根据题意可知, 1 厘米表示实际距离 200 米, 分别计算出广场到书店、超市、公园、体育场的图上距离, 再根据地图上方向的规定: 上北下南, 左西右东; 以广场为观测点, 分别画出书店、超市、公园、体育场的位置, 据此解答。\n\n【详解】$600 \\div 200 = 3$  （厘米）$4 0 0 \\div 2 0 0 = 2 \\text {(厘 米)}$  $2 0 0 \\div 2 0 0 = 1 \\text {(厘 米)}$ $6 0 0 \\div 2 0 0 = 3 (\\text {厘 米})$如图：![](images/80b825cade7b2fcc65bf16889c59611dddd392797456ec8f37244d30469e43c6.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "create_time": "2025-10-09 17:06:53",
+        "merged_id": "unit2_277",
+        "original_id": "unit2_choice_277"
+      },
+      {
+        "id": "unit2_choice_278",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "操作。![](images/f089d3db3fead61af101029a802f402ba24be6ec006f9582e76182a2d3720fbd.jpg)图书馆在教学楼的南偏西 $45^{\\circ}$ 的 80 米处，科技馆在教学楼的东偏北 $60^{\\circ}$ 的 60 米处，体育馆在教学楼的正北方向 40 米处。请你在平面图中标出它们的位置。",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】根据右下角可知一格（一厘米）表示 20 米，先计算出教学楼到图书馆、科技馆、体育馆的图上距离，再根据地图上方向的规定：上北下南，左西右东，以教学楼为观测点，画出图书馆、科技馆、体育馆的位置。\n\n【详解】图书馆：  $80\\div 20 = 4$  （格）科技馆：  $60\\div 20 = 3$  （格）体育馆：  $40\\div 20 = 2$  （格）\n\n如图：![](images/ac31b2499daba8a13c229ec526cefb62657a0d5d579ca443719a4026b3876031.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·湖南衡阳·期中",
+        "create_time": "2025-10-09 17:06:53",
+        "merged_id": "unit2_278",
+        "original_id": "unit2_choice_278"
+      },
+      {
+        "id": "unit2_choice_279",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "同学们参加军训, 沿军营向东偏北 $45^{\\circ}$ 行进约 300 米到龙山后, 向东行进约 200 米到分路口, 最后再向东偏南 $30^{\\circ}$ 行进约 400 米到达宿营地。根据教官的描述, 你能画出同学们行军的路线图吗? 试试看。 ![](images/f0bede6394857bd75f308d3498a36b271d8d21a78d1c76ecd2f06d865cec1d53.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】画路线图时，要按行走路线确定每一个观测点，然后以每一个观测点为参照物，确定到下一个目标所行走的方向、角度和距离。从图上可以看出，图上1厘米的距离表示实际距离100米。 $300 \\div 100 = 3$ （厘米）， $200 \\div 100 = 2$ （厘米）， $400 \\div 100 = 4$ （厘米），据此根据地图“上北下南，左西右东”的规定，结合方向和角度即可化成行军路线图。\n\n【详解】$300 \\div 100 = 3$  （厘米）$ 2 0 0 \\div 1 0 0 = 2 \\text {(厘 米)}$   $4 0 0 \\div 1 0 0 = 4 \\text {(厘 米)}$\n\n\n\n![](images/0a1f2df087e45c0ccea12cfc5c11f6d7b2e1d1271dba8e1add840a2442837840.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离画线路图"
+        ],
+        "source": "23-24六年级上·湖南株洲·期中",
+        "create_time": "2025-10-09 17:06:53",
+        "merged_id": "unit2_279",
+        "original_id": "unit2_choice_279"
+      },
+      {
+        "id": "unit2_choice_280",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "操作。 ![](images/b5a270360f58b6e51487612a2d2b96f2cff5223b53917be3f6a8852f8e693aa7.jpg) （1）银行在丽丽家的（）偏（），（）°方向上，距离是（）米。说一说丽丽从银行回家的路线：（）。 (2) 图书大厦在丽丽家南偏西 $45^{\\circ}$ 方向上, 距离是 200 米, 在图上标出图书大厦的位置。",
+        "options": [],
+        "answer": "（1）东；北；30；200；沿着西偏南  $30^{\\circ}$  方向走200米  (2）见详解",
+        "solution": "【分析】（1）图中一小段表示实际距离是 100 米，以丽丽家为观测点，银行在东偏北  $30^{\\circ}$  （或北偏东  $60^{\\circ}$  ）方向上，有两小段距离；以银行为观测点，丽丽家在西偏南  $30^{\\circ}$  （或南偏西  $60^{\\circ}$  ）方向上，有两小段距离。  \n(2) 以丽丽家为观测点, 图书大厦在南偏西  $45^{\\circ}$  方向上, 距离是 200 米, 也就是图中两小段长度, 据此作图。\n\n【详解】(1)  $2 \\times 100 = 200$  （米）银行在丽丽家的东偏北  $30^{\\circ}$  (或北偏东  $60^{\\circ}$ ) 方向上, 距离是 200 米; 丽丽从银行回家的路线: 沿着西偏南  $30^{\\circ}$  (或南偏西  $60^{\\circ}$  ) 方向走 200 米。(2) 如图所示:![](images/9fd13dcc85660fe4f920db3d2b48d67fc500b57b591ea8e597e72c749f804dfd.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离描述路线图",
+          "根据方向",
+          "角度和距离确定物体的位置",
+          "三位数与一位数的乘法口算"
+        ],
+        "source": "23-24六年级上·湖南永州·期中",
+        "create_time": "2025-10-09 17:06:53",
+        "merged_id": "unit2_280",
+        "original_id": "unit2_choice_280"
+      },
+      {
+        "id": "unit2_choice_281",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "根据要求填一填, 画一画。 ![](images/323568583f533943511058c7f3b42d1a2121a8240fe180cde6f43f7eac0e9561.jpg) (1) 超市位于学校（）偏（）（）°方向上，距学校（）米。 (2) 小华家位于学校东偏北 $45^{\\circ}$ 方向 300 米处, 请在平面图上标出小华家的位置。 (3) 这天小华从家步行到学校, 平均每分钟走 50 米, 请你算一算, 从家到学校一共用了多少分钟?",
+        "options": [],
+        "answer": "（1）西；北；30；400(2）图见详解  (3) 6分钟",
+        "solution": "【分析】（1）根据题意可知，1 厘米表示 100 米；计算出学校到超市的实际距离，再根据地图上方向的规定“上北下南，左西右东”，以学校为观测点，说出超市的位置；(2) 计算出小华到学校的图上距离，以学校为观测点，画出小华家的位置；  (3) 根据时间  $=$  路程  $\\div$  速度, 用学校到小华家的路程  $\\div$  小华的速度, 即可解答。\n\n【详解】(1)  $100 \\times 4 = 400$  （米）$9 0 ^ {\\circ} - 3 0 ^ {\\circ} = 6 0 ^ {\\circ}$超市位于学校西偏北  $30^{\\circ}$  （或北偏西  $60^{\\circ}$  ）方向上，400米。(2)  $300 \\div 100 = 3$  （厘米）\n\n如图：![](images/acb5077284311ebd046f4b1b9f577068a3633038c79762dcbe08414033f45a35.jpg)\n\n(3)  $300 \\div 50 = 6$  （分钟）答：从家到学校一共用了6分钟。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置",
+          "基础行程问题"
+        ],
+        "source": "21-22 六年级上·福建莆田·期末",
+        "create_time": "2025-10-09 17:06:53",
+        "merged_id": "unit2_281",
+        "original_id": "unit2_choice_281"
+      },
+      {
+        "id": "unit2_choice_282",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "请在图中标出各停靠站的名称以及相应的距离和方向。 ![](images/68275df95d1f04d5ca6d6f1a0cb9d477f9c4270df1bf15030dd8dc42a55044a7.jpg) 3路公交汽车从起点站出发，先向东行1千米到医院；再向北偏东 $30^{\\circ}$ 的方向行1千米到百货大楼；又向南行1.5千米到市民广场，再向北偏东 $60^{\\circ}$ 方向行2千米到公园，最后向南偏东 $70^{\\circ}$ 的方向行1千米到达终点站。",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】以图上的“上北下南，左西右东”为准，图上1个单位长度相当于实际的1千米；根据3路公交汽车的行驶路线，在图中标出各停靠站的名称以及相应的距离和方向。\n\n【详解】如图：![](images/b9bc65a43d6d8c38ef8b05a8ca5edbcf76050e964d6892f800f644e3faba5236.jpg)",
+        "knowledge_points": [],
+        "source": "24-25 六年级上·全国·课后作业",
+        "create_time": "2025-10-09 17:06:53",
+        "merged_id": "unit2_282",
+        "original_id": "unit2_choice_282"
+      },
+      {
+        "id": "unit2_choice_283",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "按要求做。![](images/12d4fd40339d1a7ee55437c309cce32b0fd085cbe20af82c2ae0ef431cc09089.jpg) 1cm表示100m 淘气和笑笑计划暑假时一起游览北京中轴线的古建筑，第一站他们想去故宫，请你先帮助 他们找一找故宫里的这些著名景点吧。 (1) 武英殿在太和殿偏，的方向米处。 (2) 九龙壁在太和殿东偏北 $20^{\\circ}$ 的方向, 距离太和殿 320 米, 请在图中用 “ $\\bigcirc$ ” 标出九龙壁的位置。 (3) 慈宁宫在太和殿北偏西 $50^{\\circ}$ 的方向, 距离太和殿 300 米, 请你在图中用 “ $\\triangle$ ” 标出慈宁宫的位置。",
+        "options": [],
+        "answer": "（1）西；南；40；300(2）见详解  (3）见详解",
+        "solution": "【分析】（1）根据上北下南，左西右东，以及图中给出数据，可知武英殿在太和殿西偏南  $40^{\\circ}$  （或南偏西  $50^{\\circ}$  ）的方向300米处。  (2) 以太和殿为观测点, 以图中的 “上北下南, 左西右东” 为准, 图中东北方向内, 已画有两条 320 米长的线段, 靠近东的线段是在太和殿东偏北  $20^{\\circ}$  方向, 靠近北的线段是在太和殿北偏东  $20^{\\circ}$  的方向, 据此选择九龙壁正确的位置, 并用 “  $\\bigcirc$  ” 标出它的位置。  (3) 图上 1 厘米表示 100 米,  $300 \\div 100 = 3$  (厘米), 以太和殿为中心, 在太和殿北偏西  $50^{\\circ}$  的方向画出一条射线, 并量出 3 厘米的长度, 这个位置就是慈宁宫, 在图中用 “ $\\triangle$ ” 标出慈宁宫的位置即可。\n\n【详解】（1）武英殿在太和殿西偏南  $40^{\\circ}$  （或南偏西  $50^{\\circ}$  ）的方向300米处。\n\n(2) (3) 如图：![](images/c33bc08ca0f2a551f27cf304892a88ba3be7c9374a2c680db81036614139d6cb.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24五年级下·北京海淀·期末",
+        "create_time": "2025-10-09 17:06:53",
+        "merged_id": "unit2_283",
+        "original_id": "unit2_choice_283"
+      },
+      {
+        "id": "unit2_choice_284",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "操作。 ![](images/5c56030ad0c94b7761fee427ae3f011a65247b560546b802ca842623c22da9f0.jpg) (1) 实验楼在教学楼的 ( ) 偏 ( ) ( )°方向上, 距教学楼 ( ) 米处。科技楼在教学楼的 ( ) 偏 ( ) ( )°方向上, 距教学楼 ( ) 米处。 (2) 大门在教学楼的北偏西 $30^{\\circ}$ 方向 200 米处, 请在图上标出大门的位置。",
+        "options": [],
+        "answer": "（1）南；东；50；100；南；西；45；300",
+        "solution": "【分析】（1）由图可知，实验楼在教学楼的南边偏东的方向，已知一小段  $= 100$  米，所以实验楼距教学楼为 100 米。  (2) 根据题意的已知条件, 大门在教学楼北偏西  $30^{\\circ}$  方向 200 米处, 所以, 以教学楼为原点, 以纵向轴为角的其中一条边, 向北偏西画一条线段, 画一个  $30^{\\circ}$  的角即可, 据此解答。\n\n【详解】(1)  $1 \\times 100 = 100$  (米),  $3 \\times 100 = 300$  (米), 实验楼在教学楼的南偏东  $50^{\\circ}$  方向上, 距教学楼 100 米处。科技楼在教学楼的南偏西  $45^{\\circ}$  方向上, 距教学楼 300 米处。  实验楼在教学楼的南偏东  $50^{\\circ}$  方向上，距教学楼 100 米处。  (2)  $200 \\div 100 = 2$  （厘米），如图：![](images/0f06a0043ccd3ddade72f6d1c44d13fe3d980382be692c31804e19905529758b.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "2024·福建莆田·小升初真题",
+        "create_time": "2025-10-09 17:06:53",
+        "merged_id": "unit2_284",
+        "original_id": "unit2_choice_284"
+      },
+      {
+        "id": "unit2_choice_285",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "根据下面的描述，在平面图上标出各场所的位置。 ![](images/878923ffa6dd152a033baada2c181d85bb54e8f55329c2f4b7d6bffc46fd1d20.jpg) (1) 超市在广场的东偏北 $30^{\\circ}$ 方向 200 米处。 (2) 学校在广场的南偏西 $45^{\\circ}$ 方向 150 米处。",
+        "options": [],
+        "answer": "(1) (2) 见详解",
+        "solution": "【分析】（1）东偏北  $30^{\\circ}$  方向 200 米，也就是正东往北偏  $30^{\\circ}$ ，将量角器 0 刻度线对着东，往北面量出  $30^{\\circ}$  即可，观察发现图中左上角一小段表示 100 米， $200 \\div 100 = 2$ （小段），那么距离画 2 小段这样的长度即可，然后标上超市；(2) 南偏西  $45^{\\circ}$  方向 150 米, 也就是正南往西偏  $45^{\\circ}$ , 将量角器 0 刻度线对着南, 往西面量出  $45^{\\circ}$  即可, 观察发现图中左上角一小段表示 100 米,  $150 \\div 100 = 1.5$  (小段), 那么距离画 1.5 小段这样的长度即可, 然后标上学校; 据此作图。\n\n【详解】(1) (2) 如图：![](images/a7f931e5beac765a5fd5c695b91f3be31fb659810e397bddcee66de27f9b7a69.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24三年级下·河北邯郸·期末",
+        "create_time": "2025-10-09 17:06:53",
+        "merged_id": "unit2_285",
+        "original_id": "unit2_choice_285"
+      },
+      {
+        "id": "unit2_choice_286",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "根据下面的描述，在平面示意图上标出各位同学的位置。 ![](images/7bb05d0c628a53da1dc12e24b0f7f4a8d71f04d38c8358bdc87faeda293f2968.jpg) (1) 李云在驻地东偏北 $45^{\\circ}$ 的方向上, 距离是 200 米。 (2) 张红在驻地的正西方向上，距离是 100 米。 (3) 杜江在驻地北偏西 $60^{\\circ}$ 的方向上，距离是 50 米。 (4) 刘伟在驻地东偏南 $50^{\\circ}$ 的方向上, 距离是 300 米。",
+        "options": [],
+        "answer": "(1) (2) (3) (4) 见详解",
+        "solution": "【分析】（1）根据题意可知，1 厘米表示实际距离 50 米，计算出驻地与李云之间的图上距离，再根据地图上方向的规定“上北下南，左西右东”，以驻地为观测点，东偏北  $45^{\\circ}$  方向上 200 米处，画出李云的位置；\n\n(2) 计算出驻地与张红的图上距离, 再以驻地为观测点, 正西方向上 100 米处, 画出张红的位置;  \n(3) 计算出驻地与杜江的图上距离, 再以驻地为观测点, 北偏西  $60^{\\circ}$  方向上 50 米处, 画出杜江的位置;  \n(4) 计算出驻地与刘伟的图上距离, 再以驻地为观测点, 东偏南  $50^{\\circ}$  方向上 300 米处, 画出刘伟的位置。\n\n【详解】(1)  $200 \\div 50 = 4$  （厘米）(2)  $100 \\div 50 = 2$  （厘米）  (3)  $50 \\div 50 = 1$  （厘米）  (4)  $300 \\div 50 = 6$  （厘米）图如下：![](images/cfd0a39f503355248fd4880d99e48af451df24bf8855241e226d59f961071c87.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·全国·单元测试",
+        "create_time": "2025-10-09 17:06:53",
+        "merged_id": "unit2_286",
+        "original_id": "unit2_choice_286"
+      },
+      {
+        "id": "unit2_choice_287",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "根据下面的描述标出各场所的位置。 ![](images/eeebdbc82ecd6ea3dbb7b478092657809b4a25e994789011313bec17c8c08507.jpg) (1) 体育馆在学校的东偏南 $45^{\\circ}$ 方向 $1500 \\mathrm{~m}$ 处。 (2) 公园在学校的正西方向 $1000 \\mathrm{~m}$ 处。 (3) 医院在学校的西偏南 $30^{\\circ}$ 方向 $3000 \\mathrm{~m}$ 处。",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】将方向和距离结合起来描述位置时，要注意三个要素：一是观测点，二是方向，三是距离。地图上按上北下南左西右东确定方向，图上的 1 段表示  $500 \\mathrm{~m}$ ，实际距离  $\\div 1$  段表示的距离 = 要画的段数。\n\n【详解】$1500 \\div 500 = 3$  （段）$1 0 0 0 \\div 5 0 0 = 2 \\text {(段)}$  $3 0 0 0 \\div 5 0 0 = 6 \\text {(段)}$\n\n![](images/07aae4f46e470d3334506c68cb3ef1811f82f55f194be4d093a3f497b26dc31e.jpg)\n\n88．（24-25六年级上·全国·课后作业）星期日，小月全家开车去游玩。从家出发向正西方向行驶  $15\\mathrm{km}$  ，再向西偏南  $20^{\\circ}$  方向行驶  $20\\mathrm{km}$  ，最后向南偏东  $22^{\\circ}$  方向行驶  $5\\mathrm{km}$  到达假日旅游度假村。根据上面的描述，画出小月全家出行的路线图。\n\n![](images/7fdd64ed0f5730892607f12890b1e3f3adce4d53c0105dab9d022c081485aced.jpg)\n\n【答案】见详解\n\n【难度】0.65\n\n【知识点】根据方向、角度和距离画线路图\n\n【分析】根据题意得：坐标向上为正北方向，左为正西方，右为正东方，下为正南方。以小月家为坐标原点，题中的线段表示  $5 \\mathrm{~km}$ ，从家出发向正西方向行驶  $15 \\mathrm{~km}$ ，即向左方走3个线段距离，再向左下方  $20^{\\circ}$  方向上走4个线段距离；最后向右下方  $22^{\\circ}$  方向走1个线段的距离到达旅游度假村。\n\n【详解】画出出行的路线图如下：![](images/80ba268136fcf7318101c7c0729dbf7de93f5f64db82b3ce0b41ca8ab3aa3be3.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "2024 六年级上·全国·专题练习",
+        "create_time": "2025-10-09 17:06:53",
+        "merged_id": "unit2_287",
+        "original_id": "unit2_choice_287"
+      },
+      {
+        "id": "unit2_choice_288",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "星期日，小月全家开车去游玩。从家出发向正西方向行驶 $15\\mathrm{km}$ ，再向西偏南 $20^{\\circ}$ 方向行驶 $20\\mathrm{km}$ ，最后向南偏东 $22^{\\circ}$ 方向行驶 $5\\mathrm{km}$ 到达假日旅游度假村。根据上面的描述，画出小月全家出行的路线图。\n\n![](images/7fdd64ed0f5730892607f12890b1e3f3adce4d53c0105dab9d022c081485aced.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】根据题意得：坐标向上为正北方向，左为正西方，右为正东方，下为正南方。以小月家为坐标原点，题中的线段表示 $5 \\mathrm{~km}$ ，从家出发向正西方向行驶 $15 \\mathrm{~km}$ ，即向左方走3个线段距离，再向左下方 $20^{\\circ}$ 方向上走4个线段距离；最后向右下方 $22^{\\circ}$ 方向走1个线段的距离到达旅游度假村。\n\n【详解】画出出行的路线图如下：![](images/80ba268136fcf7318101c7c0729dbf7de93f5f64db82b3ce0b41ca8ab3aa3be3.jpg)",
+        "knowledge_points": [
+          "根据方向、角度和距离画线路图"
+        ],
+        "source": "24-25六年级上·全国·课后作业",
+        "create_time": "2025-10-09 17:32:50",
+        "merged_id": "unit2_288",
+        "original_id": "unit2_choice_288"
+      },
+      {
+        "id": "unit2_choice_289",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "一辆汽车从起点站向西偏南 $35^{\\circ}$ 方向行驶6千米，再向东偏南 $30^{\\circ}$ 方向行驶16千米，最后向正东方向行驶6千米到达终点站。(1) 根据上面的描述，把汽车行驶的路线图画完整。![](images/006bd8e5d1ff0a1066a807ad3f663c06a9e8811f64bc0a783bf17b9e0be5de03.jpg)(2) 根据路线图，说一说汽车沿原路返回时所行驶的方向和路程。",
+        "options": [],
+        "answer": "（1）见详解；（2）见详解",
+        "solution": "【分析】（1）根据平面图上方向的规定：上北下南，左西右东，以每次转换方向的位置，确定每次转变的方向，再根据图上1厘米表示实际2千米，求出6千米和16千米的图上距离，据此作图。  \n(2) 按原路返回时, 以终点站为观测点, 先向正西方向行驶 6 千米, 再向西偏北  $30^{\\circ}$  (或北偏西  $60^{\\circ}$ ) 方向行驶 16 千米, 最后向东偏北  $35^{\\circ}$  (或北偏东  $55^{\\circ}$ ) 方向行驶 6 千米到达起点站。\n\n【详解】(1)  $6 \\div 2 = 3$  （厘米）$1 6 \\div 2 = 8 (\\text {厘 米})$\n\n如图：![](images/0d36344f7c34a584ae9d536035225368aea7e055f91d31a9e971564bc5cf8ac0.jpg)(2) 汽车从终点站出发先向正西方向行驶 6 千米, 再向西偏北  $30^{\\circ}$  (或北偏西  $60^{\\circ}$ ) 方向行驶 16 千米, 最后向东偏北  $35^{\\circ}$  (或北偏东  $55^{\\circ}$ ) 方向行驶 6 千米到达起点站。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离描述路线图",
+          "根据方向",
+          "角度和距离画线路图"
+        ],
+        "source": "24-25六年级上·全国·课后作业",
+        "create_time": "2025-10-09 17:06:53",
+        "merged_id": "unit2_289",
+        "original_id": "unit2_choice_289"
+      },
+      {
+        "id": "unit2_choice_290",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "春游时, 同学们做一个寻找宝箱的游戏, 四个宝箱的位置分别是: (1) A宝箱在驻地北偏东 $40^{\\circ}$ 方向上，距驻地 $120\\mathrm{m}$ 的树上。 (2) B宝箱在驻地正南方向上，距驻地 $80 \\mathrm{~m}$ 的坑里。 (3) C宝箱在驻地西偏北 $25^{\\circ}$ 方向上，距驻地 $80 \\mathrm{~m}$ 的草丛里。 (4) D宝箱在驻地南偏东 $45^{\\circ}$ 方向上, 距驻地 $120 \\mathrm{~m}$ 的花丛里。 请你在平面图上标出它们的位置。![](images/2b3062903ce705d0bcc0df44958da32beea30d881aedfe457b6ad0b8117fda36.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】根据“上北下南，左西右东”的原则，按照给出的线段比例尺，图上距离1厘米代表实际距离40米，以及给出的方向即可确定各个宝箱的位置。要在地图上利用方向和距离确定一个位置，需要知道这个位置在什么方向什么角度，还需要知道与中心的实际距离是多少，实际距离需要利用比例尺换算成图上距离。\n\n【详解】(1) 图上距离为  $120 \\div 40 = 3$  (厘米), A 宝箱在驻地北偏东  $40^{\\circ}$  方向, 位置如图所示;(2) 图上距离为  $80 \\div 40 = 2$  (厘米), B宝箱在驻地正南方向, 位置如图所示;  (3) 图上距离为  $80 \\div 40 = 2$  (厘米), C宝箱在驻地西偏北  $25^{\\circ}$  方向, 位置如图所示;  (4) 图上距离为  $120 \\div 40 = 3$  (厘米), D宝箱在驻地南偏东  $45^{\\circ}$  方向, 位置如图所示。![](images/f53eea8ce50577159ea9bab812ca0bfd1729030c5b837ea035a341375c505a5b.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·全国·课后作业",
+        "create_time": "2025-10-09 17:06:53",
+        "merged_id": "unit2_290",
+        "original_id": "unit2_choice_290"
+      },
+      {
+        "id": "unit2_choice_291",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "石油勘探队在A城南偏西 $45^{\\circ}$ 方向600千米处打一口油井。你能在图上确定油井的位置吗? ![](images/d4a49546edd49cab5d9d09ced762833088f1bee42847dfb66a8e36530c20ceb8.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】根据“上北下南，左西右东”认识方向的方法，在南偏西  $45^{\\circ}$  方向先画出一条线，由图可知图上 1 厘米表示实际距离 300 千米，通过计算 600 千米里有几个 300 千米，就应画几个单位长度，最后在图中标出油井的位置即可。\n\n【详解】$600 \\div 300 = 2$  （厘米）油井的位置如下图：![](images/65bc0e448177bc49bd1ad0152eff268a2001e9cad0b94409a557caa6ef6f0f30.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·全国·课后作业",
+        "create_time": "2025-10-09 17:06:53",
+        "merged_id": "unit2_291",
+        "original_id": "unit2_choice_291"
+      },
+      {
+        "id": "unit2_choice_292",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "如图，超市位于学校北偏东 $40^{\\circ}$ 方向，同时又位于公园北偏西 $40^{\\circ}$ 方向。请画出超市的位置。![](images\\1d036852-e58f-4422-a85d-d6aee2ec3e38.png)",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】根据地图上方向的规定“上北下南，左西右东”，分别以学校和公园为观测点，画出两条线，相交的点就是超市的位置，据此画图。\n\n【详解】如图：![](images/477e58ebaa7a6401a589bf14cd3d69e37bdf784247382ed06a61260f978d63a4.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·全国·课后作业",
+        "create_time": "2025-10-09 17:06:53",
+        "merged_id": "unit2_292",
+        "original_id": "unit2_choice_292"
+      },
+      {
+        "id": "unit2_choice_293",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "张华从家往正东方向走 600 米到红绿灯处, 再往西北方向走 300 米到书店, 最后往东偏北 $30^{\\circ}$ 方向走 450 米到学校。画出张华从家到学校的路线示意图。",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】以图上的 “上北下南, 左西右东” 为准, 根据题目中所给的数据, 先确定图上 1 厘米相当于实际距离 150 米。张华从家往正东方向走 600 米, 即在家的正东方向上画  $600 \\div 150 = 4$  厘米长的线段, 即是红绿灯处;再往西北方向走 300 米到书店, 即在红绿灯处往西偏北  $45^{\\circ}$  方向上画  $300 \\div 150 = 2$  厘米长的线段, 即是书店;最后往东偏北  $30^{\\circ}$  方向走 450 米到学校, 即在书店往东偏北  $30^{\\circ}$  方向上画  $450 \\div 150 = 3$  厘米长的线段, 即是学校。\n\n【详解】如图：![](images/afb3296fc9b0b81995a65b03d4de39ba3dd63d817627875944d316a306a40d7c.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离画线路图"
+        ],
+        "source": "23-24 六年级上·全国·课后作业",
+        "create_time": "2025-10-09 17:06:53",
+        "merged_id": "unit2_293",
+        "original_id": "unit2_choice_293"
+      },
+      {
+        "id": "unit2_choice_294",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "某文化宫周围环境如图所示。 ![](images/b4c0b5f330978a78790dba182abee25e945e68bbd53d4f5c85241bd50cbdb63a.jpg) (1) 文化宫东面 350 米处, 有一条商业街与人民路互相垂直。在图中画直线表示这条街,并标上 “商业街”。 (2）体育馆在文化宫（）偏（） $45^{\\circ}$ 方向（）米处。 (3) 李小明以 60 米/分的速度从学校沿着人民路向东走, 3 分钟后他在文化宫 ( ) 面 ( ) 米处。",
+        "options": [],
+        "answer": "(1) 见详解",
+        "solution": "【分析】（1）1 厘米表示 100 米，计算出文化宫与商业街的图上距离，再根据地图上的方向“上北下南，左西右东”来确定商业街的位置即可；(2) 先计算出文化馆到体育馆的实际距离, 再以文化宫为观测点, 确定出体育馆的位置;  (3) 3 分钟李小明总共走了 180 米, 根据题图可知, 学校到文化宫的图上距离为 2 厘米,即 200 米, 说明李小明还未到文化宫, 所以 3 分钟后他在文化宫西面 20 米处。\n\n【详解】(1)  $350 \\div 100 = 3.5$  （厘米）如图：![](images/ac410ae5d54510724f1e0406e64f3d2d9714aef7dc42323ad7a856ea38326cba.jpg)\n\n(2)  $100 \\times 2.5 = 250$  （米）体育馆在文化宫北（东）偏东（北） $45^{\\circ}$ 方向250米处。(3)  $100 \\times 2 = 100$  （米）  $6 0 \\times 3 = 1 8 0 (\\text {米})$   $2 0 0 - 1 8 0 = 2 0 \\text {(米)} $李小明以60米/分的速度从学校沿着人民路向东走，3分钟后他在文化宫西面20米处。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置",
+          "基础行程问题"
+        ],
+        "source": "23-24六年级下·湖南邵阳·期末",
+        "create_time": "2025-10-09 17:06:53",
+        "merged_id": "unit2_294",
+        "original_id": "unit2_choice_294"
+      },
+      {
+        "id": "unit2_choice_295",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "根据下面的描述。在下边平面图上画出各场所的位置和名称。![](images/6944e9e6c21c3e809c82b76d7c04016d8417ac5f25bef6f7eacf238e19439474.jpg) (1) 公园在中心广场的东偏北 $60^{\\circ}$ 方向 $250 \\mathrm{~m}$ 处。 (2) 游乐场在中心广场的西偏南 $30^{\\circ}$ 方向 $350 \\mathrm{~m}$ 处。 (3) 图书馆在中心广场的北偏西 $45^{\\circ}$ 方向 $100 \\mathrm{~m}$ 处。",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】（1）以中心广场为参照点建立方向标，1个单位长度表示  $100 \\mathrm{~m}$ ，从中心广场向东偏北  $60^{\\circ}$  方向画  $250 \\div 100 = 2.5$  个单位长度，标出公园；(2) 以中心广场为参照点建立方向标, 1 个单位长度表示 100 米, 从中心广场向西偏南  $30^{\\circ}$  方向画  $350 \\div 100 = 3.5$  个单位长度, 标出游乐场;  (3) 以中心广场为参照点建立方向标, 1 个单位长度表示 100 米, 从中心广场向北偏西  $45^{\\circ}$  方向画 1 个单位长度, 标出图书馆。\n\n【详解】根据分析，画图如下：![](images/8ed7097bce6db7a6105e6a0bcf8a5256b440965b1742d95998309cce2d674565.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级下·湖南岳阳·期末",
+        "create_time": "2025-10-09 17:06:53",
+        "merged_id": "unit2_295",
+        "original_id": "unit2_choice_295"
+      },
+      {
+        "id": "unit2_choice_296",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "王叔叔开车去段叔叔家, 先向东偏南 $60^{\\circ}$ 方向行驶 15 千米, 然后向正东方向行驶 10 千米, 最后向北偏东 $30^{\\circ}$ 方向行驶 15 千米达到段叔叔家。![](images/7d722fef2efad4f7dd59ec7889649c5e641ca6f18aa535409ea60df4a37bffbf.jpg) (1) 根据描述，画出汽车行驶的路线图。 (2) 根据所画的图，判断段叔叔家在王叔叔家的什么方向上？",
+        "options": [],
+        "answer": "(1) 见详解 (2) 段叔叔家王叔叔家的正东方向上。",
+        "solution": "【分析】（1）图上可以看出向北为正方向，每个单位长度表示 10 千米；先根据上北下南左西右东四个基本的十字方向，以王叔叔家为中心确定出王叔叔家的四个基本方向，找到东偏南  $60^{\\circ}$  方向，并画出  $(15 \\div 10)$  格的长度，接着以此点为中心画出四个基本方向，向正东方向画出 1 格的长度，接着再以此时的点为中心，画出四个基本方向，向北偏东  $30^{\\circ}$  方向画出  $(15 \\div 10)$  格的长，即到段叔叔家。(2) 先确定以王叔叔家为中心, 根据上北下南左西右东四个基本的十字方向, 即可确定段叔叔家在王叔叔家的什么方向。\n\n【详解】(1)  $15 \\div 10 = 1.5$  （格）$1 0 \\div 1 0 = 1 \\quad (\\text {格})$路线图如下：![](images/d2bfaa0877a6a9aef7d557e6fa92ede6bfe69e6d554cd57cf0bd870676fe4df9.jpg)(2) 段叔叔家王叔叔家的正东方向上。\n\n# 五、解答题",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置",
+          "根据方向",
+          "角度和距离画线路图"
+        ],
+        "source": "22-23六年级下·湖南娄底·期末",
+        "create_time": "2025-10-09 17:06:53",
+        "merged_id": "unit2_296",
+        "original_id": "unit2_choice_296"
+      },
+      {
+        "id": "unit2_choice_297",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "下图是两个城市的位置图, 先观察, 再回答下列问题。 ![](images/ddf8413e3c96b7d1a5ebd9744370c3b83c7b61d9a0e2e45d7552bf15e59df73c.jpg) (1) 从青城看，蓝城位于（）偏（）（）°方向；从蓝城看，青城位于（）偏（）（）°方向。 (2) 有两辆汽车分别同时从青城和蓝城相对开出, 客车速度为 80 千米/时, 货车速度为 50 千米/时。请你在图中用 $\\triangle$ 标出两车相遇的大致位置。",
+        "options": [],
+        "answer": "（1）东；北；40；西；南；40  (2）见详解",
+        "solution": "【分析】（1）根据“上北下南，左西右东”及角度信息和距离填空即可；  (2) 根据相遇问题中, 相遇路程  $\\div$  速度和 = 相遇时间, 据此求出客车和货车行驶的时间,再根据速度  $\\times$  时间 = 路程, 求出客车行驶的路程, 进而标出两车相遇的大致位置。\n\n【详解】（1）从青城看，蓝城位于东偏北  $40^{\\circ}$  （北偏东  $50^{\\circ}$  ）方向；从蓝城看，青城位于西偏南  $40^{\\circ}$  （南偏西  $50^{\\circ}$  ）方向。\n\n(2)   $650 \\div (80 + 50)$  $ = 6 5 0 \\div 1 3 0$ $= 5$  （小时）$8 0 \\times 5 = 4 0 0 (\\text {千 米})$如图所示：![](images/e2acde84c4a4c9a473bd18d5e75e6578ba869d7a6e26e597cb60e7933ddd1d22.jpg)",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置",
+          "相遇问题"
+        ],
+        "source": "23-24 六年级上·浙江温州·期中",
+        "create_time": "2025-10-09 17:06:53",
+        "merged_id": "unit2_297",
+        "original_id": "unit2_choice_297"
+      },
+      {
+        "id": "unit2_choice_298",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "下面是小伟家附近部分街道平面图。请你根据图示说一 说小伟从家到大港小学行走的方向和路程。![](images/688495baaad885f7b53f4408872ada39df82cbc8a879b4df7965b25f4789157d.jpg)",
+        "options": [],
+        "answer": "小伟从家出发先向东走 180 米到达超市, 再从超市向北偏东  $60^{\\circ}$  方向走 240 米到达医院, 最后从医院再向正北方向走 150 米到达大港小学。",
+        "solution": "【分析】图上的方向为上北下南左西右东，每到一个十字路口，用方向、角度表示继续行进的方向，用图上标注的距离表示每段路的长度，据此应用方向、角度和距离描述整个路线图。\n\n【详解】小伟从家出发先向东走 180 米到达超市, 再从超市向北偏东  $60^{\\circ}$  方向走 240 米到达医院, 最后从医院再向正北方向走 150 米到达大港小学。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离描述路线图"
+        ],
+        "source": "24-25 六年级上·全国·期末",
+        "create_time": "2025-10-09 17:06:53",
+        "merged_id": "unit2_298",
+        "original_id": "unit2_choice_298"
+      },
+      {
+        "id": "unit2_choice_299",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "如图，上午8时，一艘船从海港A出发，以每小时15海里的速度航向在北偏东 $60^{\\circ}$ 的小岛B，10时整到达B岛，这时船在海港A的什么位置？从B看A在什么位置？\n\n![](images/73c1d53224bb3149a28c9b54f97d71dbe5160e056ff0035b7f2ae0ec40240f65.jpg)",
+        "options": [],
+        "answer": "北偏东 $60^{\\circ}$ 方向30海里处；南偏西 $60^{\\circ}$ 方向30海里处",
+        "solution": "【分析】从上午8时到上午10时，经过了2小时，因为这艘船每小时航行15海里，所以从 A 点到 B 点的距离是 30 海里。因此, 这时船在海港 A 的北偏东 $60^{\\circ}$ 方向 30 海里处。以 B 为观测点, A 在 B 的南偏西 $60^{\\circ}$ 方向 30 海里处, 据此解答。\n\n【详解】由分析可得: 船在海港 A 的北偏东 $60^{\\circ}$ 方向 30 海里处; 从 B 看 A 在南偏西 $60^{\\circ}$ 方向 30 海里处。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·全国·课后作业",
+        "create_time": "2025-10-09 17:32:50",
+        "merged_id": "unit2_299",
+        "original_id": "unit2_choice_299"
+      },
+      {
+        "id": "unit2_choice_300",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "学校举行冬季越野赛，比赛路线如图。 ![](images/698436301f10d40e450e058037221134698bc299839f2cb1f18322bfbd9b8b46.jpg) (1) 根据路线图，说说小明参加比赛所经过的方向和路程（角度自己用量角器量一量），完成下表。 | 路段 | 方向 | 路程 | 时间 | |------------|------|------|--------| | 学校→公园 | | | 5分钟 | | 公园→新村 | | | 3分钟 | | 新村→学校 | | | 7分钟 | (2) 小明的平均速度是多少？",
+        "options": [],
+        "answer": "（1）见详解",
+        "solution": "【分析】（1）根据地图上方向的规定“上北下南，左西右东”，根据示意图，用量角器量出从学校到公园，方向是北偏东  $55^{\\circ}$ ，路程是  $500 \\times 3 = 1500$  （米）；从公园到新村，方向是南偏东  $60^{\\circ}$ ，路程是  $500 \\times 2 = 1000$  （米）；从新村到学校，方向是南偏西  $75^{\\circ}$ ，路程是  $500 \\times 4 = 2000$  （米）；把求得的数据填入表格即可；  \n(2) 根据速度  $=$  路程  $\\div$  时间, 用总路程  $\\div$  总时间, 即可解答。\n\n【详解】(1)  $90^{\\circ} - 55^{\\circ} = 35^{\\circ}$ ;  $500 \\times 3 = 1500$  （米） $90^{\\circ} - 60^{\\circ} = 30^{\\circ}$ ;  $500 \\times 2 = 1000$  （米）$9 0 ^ {\\circ} - 7 5 ^ {\\circ} = 1 5 ^ {\\circ}; 5 0 0 \\times 4 = 2 0 0 0 (\\text {米})$\n\n| 路段       | 方向                         | 路程 | 时间   |\n|------------|------------------------------|------|--------|\n| 学校→公园 | 北偏东55°（或东偏北35°）     | 1500 | 5分钟  |\n| 公园→新村 | 南偏东60°（或东偏南30°）     | 1000 | 3分钟  |\n| 新村→学校 | 南偏西75°（或西偏南15°）     | 2000 | 7分钟  |\n(2)  $(1500 + 1000 + 2000)\\div (5 + 3 + 7)$ $= (2 5 0 0 + 2 0 0 0) \\div (8 + 7) = 4 5 0 0 \\div 1 5  = 3 0 0 \\text {(米 / 分)}$\n\n答：小明的平均速度是300米/分。",
+        "knowledge_points": [
+          "根据方向",
+          "角度和距离描述路线图",
+          "基础行程问题"
+        ],
+        "source": "23-24六年级上·全国·课后作业",
+        "create_time": "2025-10-09 17:06:53",
+        "merged_id": "unit2_300",
+        "original_id": "unit2_choice_300"
+      },{
+        "id": "unit2_choice_401",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "如图中，广场在学校的（）方向上。![](images/104a51cf0914a0fe2b1371a1366c64e53464e9a96f7200a6dfc67241e2277538.jpg)",
+        "options": [
+          "A. 北偏西 $30^{\\circ}$",
+          "B. 西偏北 $30^{\\circ}$",
+          "C. 东偏南 $60^{\\circ}$",
+          "D. 南偏东 $60^{\\circ}$"
+        ],
+        "answer": "C",
+        "solution": "根据\"上北下南，左西右东\"的图上方向，结合题意分析解答即可。如图：![](images/e5d5804553c8f6f93e5711c57bef6eeb0240fd5cbc6c9f2d37aa96f3df26bade.jpg)图中广场在学校的东偏南 $60^{\\circ}$ 方向上。故答案为：C",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25六年级上·福建漳州·期中",
+        "merged_id": "unit2_401",
+        "original_id": "unit2_choice_401"
+      },
+      {
+        "id": "unit2_choice_402",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "小猫的家在小狗家西偏南 $55^{\\circ}$ 方向上, 则小狗家在小猫家 ( ) 方向上。",
+        "options": [
+          "A. 西偏南 $55^{\\circ}$",
+          "B. 南偏西 $35^{\\circ}$",
+          "C. 东偏北 $55^{\\circ}$"
+        ],
+        "answer": "C",
+        "solution": "根据方向的相对性：方向相反，角度相同。即西相反是东，南相反是北，据此选择即可。小猫的家在小狗家西偏南 $55^{\\circ}$ 方向上，则小狗家在小猫家东偏北 $55^{\\circ}$ （或北偏东 $35^{\\circ}$ ）方向上。故答案为：C",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·内蒙古通辽·期中",
+        "merged_id": "unit2_402",
+        "original_id": "unit2_choice_402"
+      },
+      {
+        "id": "unit2_choice_403",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "图书馆位于体育场的西偏南 $30^{\\circ}$ 方向 1000 米处, 那么体育场位于图书馆的 ( ) 方向 1000 米。",
+        "options": [
+          "A. 西偏南 $30^{\\circ}$",
+          "B. 南偏西 $30^{\\circ}$",
+          "C. 东偏北 $30^{\\circ}$",
+          "D. 北偏东 $30^{\\circ}$"
+        ],
+        "answer": "C",
+        "solution": "前后观测点发生了变化，根据方向的相对性，西偏南对东偏北，角度和距离不变，据此进行分析即可解答。根据分析可知, 图书馆位于体育场的西偏南 $30^{\\circ}$ 方向 1000 米处, 那么体育场位于图书馆的东偏北 $30^{\\circ}$ 或北偏东 $60^{\\circ}$ 方向 1000 米。故答案为：C",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·广东东莞·期中",
+        "merged_id": "unit2_403",
+        "original_id": "unit2_choice_403"
+      },
+      {
+        "id": "unit2_choice_404",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "较难",
+        "question": "球球队在学校的北偏东 $30^{\\circ}$ 方向, 那么学校在球球队的 ( ) 方向。",
+        "options": [
+          "A. 南偏西 $30^{\\circ}$",
+          "B. 北偏西 $30^{\\circ}$",
+          "C. 南偏东 $60^{\\circ}$"
+        ],
+        "answer": "A",
+        "solution": "根据位置的相对性可知：位置相对的两个物体所在的方向相反、角度相同、距离不变；据此解答。根据位置的相对性可知：球队家在学校的北偏东 $30^{\\circ}$ 方向，那么学校在球队家的南偏西 $30^{\\circ}$ 方向。故答案为：A",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·山西长治·期中",
+        "merged_id": "unit2_404",
+        "original_id": "unit2_choice_404"
+      },
+      {
+        "id": "unit2_choice_405",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "西偏北 $35^{\\circ}$ 也可以说成（）。",
+        "options": [
+          "A. 南偏东 $55^{\\circ}$",
+          "B. 北偏西 $55^{\\circ}$",
+          "C. 东偏北 $35^{\\circ}$",
+          "D. 北偏东 $35^{\\circ}$"
+        ],
+        "answer": "B",
+        "solution": "西和北之间的夹角是 $90^{\\circ}$ , 根据方向的相对性, 西偏北也可以说成北偏西, 同时角度要变为: 北偏西的角度 $= 90^{\\circ}$ 一西偏北的角度, 据此分析。$90^{\\circ} - 35^{\\circ} = 55^{\\circ}$西偏北 $35^{\\circ}$ 也可以说成北偏西 $55^{\\circ}$ 。故答案为：B",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25六年级上·广西河池·期中",
+        "merged_id": "unit2_405",
+        "original_id": "unit2_choice_405"
+      },
+      {
+        "id": "unit2_choice_406",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "较难",
+        "question": "小红家在学校东偏北 $60^{\\circ}$ 方向上, 距离 800 米。那么学校在小红家的 ( ) 方向上, 距离是 800 米。",
+        "options": [
+          "A. 东偏南 $60^{\\circ}$",
+          "B. 南偏东 $60^{\\circ}$",
+          "C. 西偏南 $30^{\\circ}$",
+          "D. 南偏西 $30^{\\circ}$"
+        ],
+        "answer": "D",
+        "solution": "小红家在学校东偏北 $60^{\\circ}$ 方向上, 距离 800 米, 是以学校为观测点; 那么学校在小红家的方向, 是以小红家为观测点;根据位置的相对性可知，观测点不同，方向相反，夹角的度数相同，距离相同；由此可知, 东偏北 $60^{\\circ}$ 相对的是西偏南 $60^{\\circ}$ , 西和南之间的夹角是 $90^{\\circ}, 90^{\\circ} - 60^{\\circ} = 30^{\\circ}$ , 所以西偏南 $60^{\\circ}$ 方向, 还可以说成南偏西 $30^{\\circ}$ 方向。小红家在学校东偏北 $60^{\\circ}$ 方向上, 距离 800 米。那么学校在小红家的西偏南 $60^{\\circ}$（或南偏西 $30^{\\circ}$ ）方向上，距离是800米。故答案为：D",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·海南省直辖县级单位·期中",
+        "merged_id": "unit2_406",
+        "original_id": "unit2_choice_406"
+      },
+      {
+        "id": "unit2_choice_407",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "如图, 以公园为观测点, 学校在 ( ) 方向。![](images/fe238de1fed32cfd208e5f27382ad1d742ee7147697a3bc923236cb3489fd723.jpg)",
+        "options": [
+          "A. 东偏南 $45^{\\circ}$",
+          "B. 西偏南 $45^{\\circ}$",
+          "C. 北偏西 $45^{\\circ}$",
+          "D. 南偏西 $45^{\\circ}$"
+        ],
+        "answer": "A",
+        "solution": "先确定观测点，再根据上北下南，左西右东，确定方向和角度。观察可知，以公园为观测点，学校在东偏南 $45^{\\circ}$ 方向。故答案为：A",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·新疆克孜勒苏·期中",
+        "merged_id": "unit2_407",
+        "original_id": "unit2_choice_407"
+      },
+      {
+        "id": "unit2_choice_408",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "孙奶奶做晨练, 她先面向正北方向站立, 然后全身向左转了 $30^{\\circ}$ , 这时她面向 ( ) 的方向。",
+        "options": [
+          "A. 北偏西 $30^{\\circ}$",
+          "B. 北偏东 $30^{\\circ}$",
+          "C. 西偏北 $30^{\\circ}$"
+        ],
+        "answer": "A",
+        "solution": "根据方向的规定\"上北下南，左西右东\"，先面向正北方向站立，然后全身向左转了 $30^{\\circ}$ ，是从北往西旋转了 $30^{\\circ}$ ，结合角度，得出这时面对的方向。孙奶奶做晨练, 她先面向正北方向站立, 然后全身向左转了 $30^{\\circ}$ , 这时她面向北偏西 $30^{\\circ}$ (或西偏北方向 $60^{\\circ}$ ) 的方向。故答案为：A",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·湖南岳阳·期中",
+        "merged_id": "unit2_408",
+        "original_id": "unit2_choice_408"
+      },
+      {
+        "id": "unit2_choice_409",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "较难",
+        "question": "北京在鄂尔多斯市东偏北 $10^{\\circ}$ 方向, 直线距离约 450 千米。则下列说法正确的是 ( )。",
+        "options": [
+          "A. 北京在鄂尔多斯市南偏西 $10^{\\circ}$ 方向",
+          "B. 鄂尔多斯在北京市南偏西 $10^{\\circ}$ 方向",
+          "C. 鄂尔多斯在北京市西偏南 $10^{\\circ}$ 方向",
+          "D. 鄂尔多斯在北京市北偏东 $10^{\\circ}$ 方向"
+        ],
+        "answer": "C",
+        "solution": "根据位置的相对性可知, 它们的方向相反, 角度相等, 距离相等。北京在鄂尔多斯市东偏北 $10^{\\circ}$ 方向, 即鄂尔多斯在北京市西偏南 $10^{\\circ}$ 方向。据此解答即可。由分析可知: 北京在鄂尔多斯市东偏北 $10^{\\circ}$ 方向, 则鄂尔多斯在北京市西偏南 $10^{\\circ}$ 方向。故答案为：C",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "22-23 六年级上·内蒙古鄂尔多斯·期中",
+        "merged_id": "unit2_409",
+        "original_id": "unit2_choice_409"
+      },
+      {
+        "id": "unit2_choice_410",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "较难",
+        "question": "北偏西 $60^{\\circ}$ 还可以说成（）。",
+        "options": [
+          "A. 南偏西 $60^{\\circ}$",
+          "B. 西偏北 $60^{\\circ}$",
+          "C. 西偏北 $30^{\\circ}$"
+        ],
+        "answer": "C",
+        "solution": "北和西之间的夹角是 $90^{\\circ}, 90^{\\circ} - 30^{\\circ} = 60^{\\circ}$ , 所以北偏西 $60^{\\circ}$ 方向上, 还可以说成西偏北 $30^{\\circ}$ 方向。$90^{\\circ} - 60^{\\circ} = 30^{\\circ}$北偏西 $60^{\\circ}$ 还可以说成西偏北 $30^{\\circ}$ （或南偏东 $60^{\\circ}$ ）。故答案为：C",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25六年级上·河南南阳·期中",
+        "merged_id": "unit2_410",
+        "original_id": "unit2_choice_410"
+      },
+      {
+        "id": "unit2_choice_411",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "较难",
+        "question": "广场在小红家的东偏南 $40^{\\circ}$ 的方向上，那么小红在广场的（）方向上。",
+        "options": [
+          "A. 东偏南 $40^{\\circ}$",
+          "B. 西偏南 40",
+          "C. 东偏北 $40^{\\circ}$",
+          "D. 西偏北 $40^{\\circ}$"
+        ],
+        "answer": "D",
+        "solution": "广场在小红家的东偏南 $40^{\\circ}$ 的方向上，是以小红家为观测点；小红在广场的方向是以广场为观测点；根据位置的相对性可知，观测点不同，方向相反，夹角的度数相同，距离相同；由此可知, 东偏南 $40^{\\circ}$ 相对的是西偏北 $40^{\\circ}$ , 西和北之间的夹角是 $90^{\\circ}, 90^{\\circ} - 40^{\\circ} = 50^{\\circ}$ , 所以西偏北 $40^{\\circ}$ 方向, 还可以说成北偏西 $50^{\\circ}$ 方向。广场在小红家的东偏南 $40^{\\circ}$ 的方向上, 那么小红在广场的西偏北 $40^{\\circ}$ (或北偏西 $50^{\\circ}$ ) 方向上。故答案为：D",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25六年级上·湖北武汉·期中",
+        "merged_id": "unit2_411",
+        "original_id": "unit2_choice_411"
+      },
+      {
+        "id": "unit2_choice_412",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "新华书店在人民体育场北偏东 $40^{\\circ}$ 方向 600 米处。乐乐从新华书店出发, 应该向 ( ) 方向走 600 米, 才能到达人民体育场。",
+        "options": [
+          "A. 南偏东 $40^{\\circ}$",
+          "B. 南偏西 $40^{\\circ}$",
+          "C. 北偏东 $40^{\\circ}$",
+          "D. 北偏西 $40^{\\circ}$"
+        ],
+        "answer": "B",
+        "solution": "两个地点的相对位置，方向相反，角度和距离不变。北的相反方向是南，东的相反方向是西，北偏东的相反方向是南偏西。据此解题。新华书店在人民体育场北偏东 $40^{\\circ}$ 方向 600 米处, 那么人民体育场在新华书店的南偏西 $40^{\\circ}$ 方向 600 米处, 所以乐乐从新华书店出发, 应该向南偏西 $40^{\\circ}$ 方向走 600 米, 才能到达人民体育场。故答案为：B",
+        "knowledge_points": [
+          "根据方向、角度和距离描述路线图",
+          "东、南、西、北方向",
+          "东南、西南、东北、西北方向"
+        ],
+        "source": "24-25 六年级上·福建厦门·期中",
+        "merged_id": "unit2_412",
+        "original_id": "unit2_choice_412"
+      },
+      {
+        "id": "unit2_choice_413",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "和南偏东 $20^{\\circ}$ 表示同一个方向的是（）。",
+        "options": [
+          "A. 东偏南 $20^{\\circ}$",
+          "B. 北偏西 $20^{\\circ}$",
+          "C. 东偏南 $70^{\\circ}$"
+        ],
+        "answer": "C",
+        "solution": "南偏东 $20^{\\circ}$ 表示从正南方向往东偏 $20^{\\circ}$ , 正南和正东方向的夹角是 $90^{\\circ}, 90^{\\circ} - 20^{\\circ} = 70^{\\circ}$ , 则同一个观测点的南偏东 $20^{\\circ}$ 和东偏南 $70^{\\circ}$ 是一个方向。在地理和方向表示中, 南偏东 $20^{\\circ}$ 和东偏南 $70^{\\circ}$ 实际上是指同一个方向。这是因为方向是相对的, 南偏东 $20^{\\circ}$ 意味着从正南方向向东偏移 $20^{\\circ}$ , 而东偏南 $70^{\\circ}$ 则意味着从正东方向向南偏移 $70^{\\circ}$ , 两者在空间上是等价的。和南偏东 $20^{\\circ}$ 表示同一个方向的是东偏南 $70^{\\circ}$ 。故答案为：C",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25五年级上·河北石家庄·期中",
+        "merged_id": "unit2_413",
+        "original_id": "unit2_choice_413"
+      },
+      {
+        "id": "unit2_choice_414",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "较难",
+        "question": "海岛在军舰的东偏北 $40^{\\circ}$ 方向上，那么军舰在海岛的（）方向上。",
+        "options": [
+          "A. 西偏北 $40^{\\circ}$",
+          "B. 东偏南 $40^{\\circ}$",
+          "C. 西偏南 $40^{\\circ}$"
+        ],
+        "answer": "C",
+        "solution": "根据方向的相对性，东偏北对西偏南，角度不变，进行分析。海岛在军舰的东偏北 $40^{\\circ}$ 方向上，那么军舰在海岛的西偏南 $40^{\\circ}$ 或南偏西 $50^{\\circ}$ 方向上。故答案为：C",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·山东菏泽·期中",
+        "merged_id": "unit2_414",
+        "original_id": "unit2_choice_414"
+      },
+      {
+        "id": "unit2_choice_415",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "根据图中信息, 下列说法正确的是 ( )。![](images/0e9255cd436056a016b43e57e2d62c3e70a8ea49db28ae861d5429b730c117df.jpg)",
+        "options": [
+          "A. 从小玲家向西偏北 $30^{\\circ}$ 方向走 1000 米到商场",
+          "B. 从商场向东偏北 $45^{\\circ}$ 方向走 400 米到书店",
+          "C. 从书店向西偏北 $45^{\\circ}$ 方向走 400 米到商场",
+          "D. 从商场向南偏东 $30^{\\circ}$ 方向走 1000 米到小玲家"
+        ],
+        "answer": "A",
+        "solution": "确定线路时，注意起始点与目的地，起始点是观测点，再按照地图上\"上北下南，左西右东\"确定方向，用方向、角度和距离进行解答。A. $90^{\\circ} - 30^{\\circ} = 60^{\\circ}$ ； $200\\times 5 = 1000$ （米）从小玲家向西偏北 $30^{\\circ}$ （或北偏西 $60^{\\circ}$ ）方向走1000米到商场；原题干说法正确；B. $90^{\\circ} - 45^{\\circ} = 45^{\\circ}$ ; $200 \\times 2 = 400$ （米）从商场向南偏西 $45^{\\circ}$ （或西偏南 $45^{\\circ}$ ）方向走400米到书店；原题干说法错误；C. $90^{\\circ} - 45^{\\circ} = 45^{\\circ}; 200 \\times 2 = 400$ （米）从书店向东偏北 $45^{\\circ}$ （或北偏东 $45^{\\circ}$ ）方向走400米到商场；原题干说法错误；D. $90^{\\circ} - 30^{\\circ} = 60^{\\circ}$ ; $200 \\times 5 = 1000$ (米)从商场向南偏东 $60^{\\circ}$ （或东偏南 $30^{\\circ}$ ）方向走1000米到小玲家；原题干说法错误。说法正确的是从小玲家向西偏北 $30^{\\circ}$ 方向走1000米到商场。故答案为：A",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·陕西安康·期中",
+        "merged_id": "unit2_415",
+        "original_id": "unit2_choice_415"
+      },
+      {
+        "id": "unit2_choice_416",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "较难",
+        "question": "一架侦察机从机场向南偏西 $30^{\\circ}$ 方向飞行了 $1200 \\mathrm{~km}$ , 返回时这架侦察机要向 ( )。",
+        "options": [
+          "A. 南偏东 $30^{\\circ}$ 方向飞行 $1200 \\mathrm{~km}$",
+          "B. 北偏东 $60^{\\circ}$ 方向飞行 $1200 \\mathrm{~km}$",
+          "C. 北偏东 $30^{\\circ}$ 方向飞行 $1200 \\mathrm{~km}$",
+          "D. 南偏东 $60^{\\circ}$ 方向飞行 $1200 \\mathrm{~km}$"
+        ],
+        "answer": "C",
+        "solution": "根据题意，侦察机原路返回，根据位置的相对性可知，它飞行的方向与原来相反，夹角的度数相同，距离相同，据此解答。一架侦察机从机场向南偏西 $30^{\\circ}$ 方向飞行了 $1200 \\mathrm{~km}$ , 返回时这架侦察机要向北偏东 $30^{\\circ}$ (或东偏北 $60^{\\circ}$ ) 方向飞行 $1200 \\mathrm{~km}$ 。故答案为：C",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·山东菏泽·期中",
+        "merged_id": "unit2_416",
+        "original_id": "unit2_choice_416"
+      },
+      {
+        "id": "unit2_choice_417",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "较难",
+        "question": "要在甲、乙两地之间修一条笔直的公路, 从甲地测得公路的走向是东偏北 $47^{\\circ}$ , 从甲、乙两地同时开工, 要使若干天后公路准确接通, 乙地所修公路的走向是 ( )。",
+        "options": [
+          "A. 东偏北 $47^{\\circ}$",
+          "B. 北偏东 43",
+          "C. 西偏南 47",
+          "D. 南偏西 47"
+        ],
+        "answer": "C",
+        "solution": "根据方向的相对性，方向相反，角度相同；即东相反是西，北相反是南，据此解答即可。要在甲、乙两地之间修一条笔直的公路，从甲地测得公路的走向是东偏北 $47^{\\circ}$ ，从甲、乙两地同时开工，要使若干天后公路准确接通，乙地所修公路的走向是西偏南 $47^{\\circ}$ 。故答案为：C",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·福建厦门·期中",
+        "merged_id": "unit2_417",
+        "original_id": "unit2_choice_417"
+      },
+      {
+        "id": "unit2_choice_418",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "小明家在小华家的南偏西 $35^{\\circ}$ 方向上，则小华家在小明家的（）方向上。",
+        "options": [
+          "A. 北偏东 $35^{\\circ}$",
+          "B. 北偏西 $35^{\\circ}$",
+          "C. 北偏东 $55^{\\circ}$"
+        ],
+        "answer": "A",
+        "solution": "根据方向、角度和距离确定物体的位置，观测点不同，方向相反，夹角的度数相同，据此分析解答。小明家在小华家的南偏西 $35^{\\circ}$ 方向上，是以小华家为观测点；当以小明家为观测点观察小华家的位置时，小华家在小明家的北偏东 $35^{\\circ}$ 方向上。故答案为：A",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·陕西安康·期中",
+        "merged_id": "unit2_418",
+        "original_id": "unit2_choice_418"
+      },
+      {
+        "id": "unit2_choice_419",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "东偏南 $30^{\\circ}$ , 还可以说成 ( )。",
+        "options": [
+          "A. 南偏东 $30^{\\circ}$",
+          "B. 南偏东 60",
+          "C. 西偏北 $30^{\\circ}$"
+        ],
+        "answer": "B",
+        "solution": "正东和正南方向之间的夹角是 $90^{\\circ}$ , $90^{\\circ} - 30^{\\circ} = 60^{\\circ}$ , 则东偏南 $30^{\\circ}$ 还可以说成南偏东 $60^{\\circ}$ 。通过分析可得：东偏南 $30^{\\circ}$ 还可以说成南偏东 $60^{\\circ}$ 。故答案为：B",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "22-23 六年级上·陕西安康·期中",
+        "merged_id": "unit2_419",
+        "original_id": "unit2_choice_419"
+      },
+      {
+        "id": "unit2_choice_420",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "小红在小强的东偏南 $40^{\\circ}$ 的方向上, 那么小强在小红 ( )。",
+        "options": [
+          "A. 北偏西 $40^{\\circ}$ 的方向上",
+          "B. 西偏北 $40^{\\circ}$ 的方向上",
+          "C. 东偏北 $40^{\\circ}$ 的方向上",
+          "D. 北偏东 $40^{\\circ}$ 的方向上"
+        ],
+        "answer": "B",
+        "solution": "根据方向的相对性：方向相反，角度相同。即东相反是西，南相反是北，据此选择即可。由分析可知, 小红在小强的东偏南 $40^{\\circ}$ 的方向上, 那么小强在小红西偏北 $40^{\\circ}$ 的方向上。故答案为：B",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "22-23 六年级上·福建福州·期中",
+        "merged_id": "unit2_420",
+        "original_id": "unit2_choice_420"
+      },
+      {
+        "id": "unit2_choice_421",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "较难",
+        "question": "超市在书店的东偏北 $30^{\\circ}$ 方向上, 那么书店在超市的 ( ) 方向。",
+        "options": [
+          "A. 东偏北 $60^{\\circ}$",
+          "B. 西偏南 $60^{\\circ}$",
+          "C. 西偏南 $30^{\\circ}$",
+          "D. 北偏东 $60^{\\circ}$"
+        ],
+        "answer": "C",
+        "solution": "超市在书店的东偏北 $30^{\\circ}$ 方向上，是以书店为观测点；书店在超市的方向是以超市为观测点；根据位置的相对性可知，观测点不同，方向相反，夹角的度数相同，距离相同。由此可知, 东偏北 $30^{\\circ}$ 相对的是西偏南 $30^{\\circ}$ , 南和西之间的夹角是 $90^{\\circ}, 90^{\\circ} - 30^{\\circ} = 60^{\\circ}$ , 所以西偏南 $30^{\\circ}$ 方向, 还可以说成南偏西 $60^{\\circ}$ 方向。超市在书店的东偏北 $30^{\\circ}$ 方向上, 那么书店在超市的西偏南 $30^{\\circ}$ (或南偏西 $60^{\\circ}$ )方向。故答案为：C",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "22-23 六年级上·江西南昌·期中",
+        "merged_id": "unit2_421",
+        "original_id": "unit2_choice_421"
+      },
+      {
+        "id": "unit2_choice_422",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "学校在小李家的北偏西 $35^{\\circ}$ 的方向, 则小李的在学校的 ( ) 方向。",
+        "options": [
+          "A. 北偏西 $35^{\\circ}$",
+          "B. 南偏西 $35^{\\circ}$",
+          "C. 北偏东 $35^{\\circ}$",
+          "D. 南偏东 $35^{\\circ}$"
+        ],
+        "answer": "D",
+        "solution": "两个地方，按照不同的观测点，度数不变，方向是相对的，北对应南，西对应东。学校在小李家的北偏西 $35^{\\circ}$ 的方向，以学校观测点，则小李在学校的南偏东 $35^{\\circ}$ 方向。故答案为：D",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "2024·四川绵阳·小升初真题",
+        "merged_id": "unit2_422",
+        "original_id": "unit2_choice_422"
+      },
+      {
+        "id": "unit2_choice_423",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "较难",
+        "question": "和南偏东 $75^{\\circ}$ 表示同一方向的是（）。",
+        "options": [
+          "A. 东偏南 $75^{\\circ}$",
+          "B. 东偏南 $15^{\\circ}$",
+          "C. 北偏西 $75^{\\circ}$"
+        ],
+        "answer": "B",
+        "solution": "根据东南西北四个基本方位中，相邻两个方位之间的夹角是 $90^{\\circ}$ ，找到南偏东 $75^{\\circ}$ 方向表示的同一方位。![](images/11787c9c021f61a5bc2045396011fcb4a11517c86b430adeecc91270a4f4df7e.jpg)和南偏东 $75^{\\circ}$ 表示同一方向的是东偏南 $15^{\\circ}$ 。故答案为：B",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·广西玉林·期末",
+        "merged_id": "unit2_423",
+        "original_id": "unit2_choice_423"
+      },
+      {
+        "id": "unit2_choice_424",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "较难",
+        "question": "如图，下面说法正确的是（）。![](images/fad52cb6ddf03901f8d5cb72ed25db86ebc1d1683cde53a6fbccda8ebc795f0b.jpg)",
+        "options": [
+          "A. 妙妙在甜甜北偏东 $30^{\\circ}$ 方向",
+          "B. 甜甜在妙妙西偏南 $60^{\\circ}$ 方向",
+          "C. 甜甜在妙妙的南偏西 $60^{\\circ}$ 方向",
+          "D. 妙妙在甜甜的东偏北 $60^{\\circ}$ 方向"
+        ],
+        "answer": "C",
+        "solution": "以图上的 \"上北下南, 左西右东\" 为准, 妙妙在甜甜的方向, 是以甜甜为观测点;甜甜在奇妙的方向, 是以奇妙为观测点; 根据方向、角度和距离确定奇妙与甜甜的位置关系。A．妙妙在甜甜东偏北 $30^{\\circ}$ （或北偏东 $60^{\\circ}$ ）方向，原选项说法错误；B. 甜甜在妙妙南偏西 $60^{\\circ}$ (或西偏南 $30^{\\circ}$ ) 方向, 原选项说法错误;C. 甜甜在奇妙的南偏西 $60^{\\circ}$ 方向, 原选项说法正确; D. 妙妙在甜甜的东偏北 $30^{\\circ}$ (或北偏东 $60^{\\circ}$ ) 方向, 原选项说法错误。故答案为：C",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·浙江绍兴·期末",
+        "merged_id": "unit2_424",
+        "original_id": "unit2_choice_424"
+      },
+      {
+        "id": "unit2_choice_425",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "图书馆在剧院的东偏南 $30^{\\circ}$ 方向上, 那么剧院在图书馆的 ( )。",
+        "options": [
+          "A. 东偏南 $30^{\\circ}$ 方向",
+          "B. 南偏东 $60^{\\circ}$ 方向",
+          "C. 西偏北 $30^{\\circ}$ 方向"
+        ],
+        "answer": "C",
+        "solution": "图书馆在剧院的东偏南 $30^{\\circ}$ 方向上，是以剧院为坐标原点，向上为正北方向，向右为正东方向，向左为正西方向，向下为正南方向。剧院在图书馆的相对方位，是以图书馆为坐标原点，在西偏北 $30^{\\circ}$ 。据此可得出答案。图书馆在剧院的东偏南 $30^{\\circ}$ 方向上，那么剧院在图书馆的西偏北 $30^{\\circ}$ 方向。故答案为：C",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·山东临沂·期中",
+        "merged_id": "unit2_425",
+        "original_id": "unit2_choice_425"
+      },
+      {
+        "id": "unit2_choice_426",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "陈嘉庚纪念馆是厦门市首批研学实践教育基地, 小明从家出发去纪念馆, 他向西偏北 $36^{\\circ}$ 方向走了 $1200 \\mathrm{~m}$ 。他原路返回时应该向 ( ) 方向走 $1200 \\mathrm{~m}$ 。",
+        "options": [
+          "A. 西偏北 $36^{\\circ}$",
+          "B. 北偏西 $36^{\\circ}$",
+          "C. 南偏东 $36^{\\circ}$",
+          "D. 东偏南 $36^{\\circ}$"
+        ],
+        "answer": "D",
+        "solution": "小明从家出发去纪念馆，是以小明家为观测点；原路返回时是以纪念馆为观测点；观测点不同，方向相反，夹角的度数相同，距离相同；西偏北 $36^{\\circ}$ 相对的是东偏南 $36^{\\circ}$ ，东和南之间的夹角是 $90^{\\circ}$ ， $90^{\\circ} -36^{\\circ} = 54^{\\circ}$ ，所以东偏南 $36^{\\circ}$ 方向，还可以说成南偏东 $54^{\\circ}$ 方向。小明从家出发去纪念馆，他向西偏北 $36^{\\circ}$ 方向走了 $1200 \\mathrm{~m}$ 。他原路返回时应该向东偏南 $36^{\\circ}$ （或南偏东 $54^{\\circ}$ ）方向走 $1200 \\mathrm{~m}$ 。故答案为：D",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·福建厦门·期中",
+        "merged_id": "unit2_426",
+        "original_id": "unit2_choice_426"
+      },
+      {
+        "id": "unit2_fill_427",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "兰兰早上从家沿南偏西 $30^{\\circ}$ 方向走 300 米到达学校,放学后他要向( )方向走( )米才能到家。",
+        "options": [],
+        "answer": "北偏东 $30^{\\circ}$ 东偏北 $60^{\\circ}$ 300",
+        "solution": "根据方向的相对性, 东对西, 南对北。以兰兰家为观测点, 学校在兰兰家南偏西 $30^{\\circ}$ 方向上, 以学校为观测点, 兰兰家在学校北偏东 $30^{\\circ}$ (或东偏北 $60^{\\circ}$ ) 方向上, 两地之间距离不变。据此解答。由分析可知：兰兰早上从家沿南偏西 $30^{\\circ}$ 方向走 300 米到达学校, 放学后他要向北偏东 $30^{\\circ}$ (或东偏北 $60^{\\circ}$ ) 方向走 300 米才能到家。",
+        "knowledge_points": [
+          "根据方向、角度和距离描述路线图"
+        ],
+        "source": "24-25 六年级上·湖北鄂州·期中",
+        "merged_id": "unit2_427",
+        "original_id": "unit2_fill_427"
+      },
+      {
+        "id": "unit2_fill_428",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "小东家在小阳家的西偏北 35 度方向上, 距离是 15 千米, 那么小阳家在小东家的____偏____方向上, 距离是____千米。",
+        "options": [],
+        "answer": "东南35度 15",
+        "solution": "在描述两个物体之间的相对位置时，如果知道一个物体在另一个物体的某个方向上，那么另一个物体就在第一个物体的相反方向上，角度相同，距离相等。据此进行填空。小阳家在小东家的东偏南35度或南偏东55度方向上，距离是15千米。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·广东阳江·期中",
+        "merged_id": "unit2_428",
+        "original_id": "unit2_fill_428"
+      },
+      {
+        "id": "unit2_fill_429",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "较难",
+        "question": "妈妈下班了! 她回家应先向( )走到银行, 再向( )偏( )( )方向走到超市, 最后向( )偏( )( )方向走回家。![](images/ea1123d0d24ddf2233ddd09615546ccbbb958b0fe116224e3228b56fe1fea952.jpg)",
+        "options": [],
+        "answer": "东东北 $25^{\\circ}$ 南东 $50^{\\circ}$",
+        "solution": "根据\"上北下南，左西右东\"以及图中的角度、妈妈回家的路途，直接描述出回家路线即可。妈妈下班了！她回家应先向东走到银行，再向东偏北 $25^{\\circ}$ 方向（或北偏东 $65^{\\circ}$ 方向）走到超市，最后向南偏东 $50^{\\circ}$ 方向（或东偏南 $40^{\\circ}$ 方向）走回家。![](images/b29f7ecc4da3cf88be10611c8415b90df8ef656120d66482c03e1bbdfbccc23f.jpg)",
+        "knowledge_points": [
+          "根据方向、角度和距离描述路线图"
+        ],
+        "source": "24-25 五年级上·河北石家庄·期中",
+        "merged_id": "unit2_429",
+        "original_id": "unit2_fill_429"
+      },
+      {
+        "id": "unit2_fill_430",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "较难",
+        "question": "周末小华去图书馆看书, 他从家出发向东走 200 米到影视城, 再向东北方向走 120 米到图书馆; 如果按原路回家, 他从图书馆出发向( )方走( )米到影视城, 再向( )方走( )米回到家。",
+        "options": [],
+        "answer": "西南 120 西 200",
+        "solution": "根据方向的相对性，东北对西南，确定从图书馆到影视城的方向，东对西，再确定影视城到家的方向，距离不变，进行填空。如果按原路回家，他从图书馆出发向西南方走120米到影视城，再向西方走200米回到家。",
+        "knowledge_points": [
+          "根据方向和距离描述简单的路线"
+        ],
+        "source": "24-25 六年级上·山东济宁·期中",
+        "merged_id": "unit2_430",
+        "original_id": "unit2_fill_430"
+      },
+      {
+        "id": "unit2_fill_431",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "下面每个小正方形的对角线的长度表示 $3 \\mathrm{~km}$ 。![](images/350b000953867d4e92516c4b17a2b508d59f3163750cbdd1c2648352b88078ea.jpg)(1)一辆货车从（3，2）出发，向东偏北 $45^{\\circ}$ 方向行驶 $6\\mathrm{km}$ 到达( )。(2)一辆客车从（2，4）出发，行驶到（5，1），它向( )偏( )°方向行驶了( )km。",
+        "options": [],
+        "answer": "(1)5, 4 (2) 南东 45 9",
+        "solution": "（1）数对（3，2）表示第3列第2行。每个小正方形的对角线的长度表示$3 \\mathrm{~km}, 6 \\div 3 = 2$ , 根据地图 \"上北下南, 左西右东\" 的规定, 向东偏北 $45^{\\circ}$ 方向 2 个对角线的长度处, 即是到达的位置, 根据数对 \"先列后行\" 的特点即可写出数对。(2) 数对 (2, 4) 表示第 2 列第 4 行, 数对 (5, 1) 表示第 5 列第 1 行, 根据题意, 是向南偏东 $45^{\\circ}$ 方向行驶了 3 个对角线的长度, $3 \\times 3 = 9 \\mathrm{~(km)}$ 。据此解答。如下图所示：![](images/62c2989f03ee75b625f6eca9234b3f6341fd1ac9ea5c1923dfaf4561910d4b94.jpg)（1）通过分析可得：一辆货车从（3，2）出发，向东偏北 $45^{\\circ}$ 方向行驶 $6\\mathrm{km}$ 到达(5，4）。(2) $3 \\times 3 = 9 \\text{ (km)}$ ，一辆客车从（2，4）出发，行驶到（5，1），它向南偏东 $45^{\\circ}$ 方向（或东偏南 $45^{\\circ}$ 方向）行驶了 $9 \\mathrm{~km}$ 。",
+        "knowledge_points": [
+          "用数对表示位置",
+          "根据数对找位置",
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·陕西安康·期中",
+        "merged_id": "unit2_431",
+        "original_id": "unit2_fill_431"
+      },
+      {
+        "id": "unit2_fill_432",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如图, 图书馆在体育馆的( )偏( ) $35^{\\circ}$ 方向上,距离是 1200 米, 体育馆在人民广场的南偏西( )°方向上, 距离是( )米。![](images/3940436315d2b7b52962b3ba99d7bce40d0d9f92712e03ec0dec3a688985315c.jpg)",
+        "options": [],
+        "answer": "北西 60 900",
+        "solution": "图中一小段长度表示实际距离 300 米，以体育馆为观测点，图书馆在北偏西 $35^{\\circ}$ （或西偏北 $55^{\\circ}$ ）方向上，距离是 1200 米；以人民广场为观测点，体育馆在南偏西 $60^{\\circ}$ （或西偏南 $30^{\\circ}$ ）方向上，距离是图中 3 小段长度；据此解答。$3 \\times 300 = 900$ （米）图书馆在体育馆的北偏西 $35^{\\circ}$ 方向上, 距离是 1200 米, 体育馆在人民广场的南偏西 $60^{\\circ}$ 方向上, 距离是 900 米。因此图书馆在体育馆的北偏西 $35^{\\circ}$ 方向上，距离是 1200 米，体育馆在人民广场的南偏西 $60^{\\circ}$ 方向上，距离是 900 米。",
+        "knowledge_points": [
+          "三位数与一位数的乘法口算",
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "22-23 六年级上·陕西安康·期中",
+        "merged_id": "unit2_432",
+        "original_id": "unit2_fill_432"
+      },
+      {
+        "id": "unit2_fill_433",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "较难",
+        "question": "如下图, 小明在小乐的( )偏( ) $45^{\\circ}$ 方向上, 小乐在小明的( )偏( ) $45^{\\circ}$ 方向上。小明![](images/1dfcf31535667ec9c9d6bf0927d900ead22517d20db048a4f54e97fade07e89a.jpg)",
+        "options": [],
+        "answer": "西北东南",
+        "solution": "以图上的\"上北下南，左西右东\"为准，小明在小乐的方向是以小乐为观测点，小乐在小明的方向是以小明为观测点。根据位置的相对性可知，观测点不同，方向相反，夹角的度数相同，距离相同；据此解答。小明在小乐的西偏北（或北偏西） $45^{\\circ}$ 方向上，小乐在小明的东偏南（或南偏东） $45^{\\circ}$ 方向上。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "22-23 六年级上·江西南昌·期中",
+        "merged_id": "unit2_433",
+        "original_id": "unit2_fill_433"
+      }, {
+        "id": "unit2_fill_434",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "小红家在小明家的西偏北 $35^{\\circ}$ 方向上, 那么小明家在小红家的( )方向上。",
+        "options": [],
+        "answer": "东偏南 $35^{\\circ}$ /南偏东 $55^{\\circ}$",
+        "solution": "根据位置的相对性，方向相反，角度相同，距离相等，据此解答即可。根据分析可得：小红家在小明家的西偏北 $35^{\\circ}$ 方向上, 那么小明家在小红家的东偏南 $35^{\\circ}$ (南偏东 $55^{\\circ}$ ) 方向上。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·河南南阳·期中",
+        "merged_id": "unit2_434",
+        "original_id": "unit2_fill_434"
+      },
+      {
+        "id": "unit2_fill_435",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "较难",
+        "question": "学校在少年宫东偏南 $30^{\\circ}$ 方向 $400 \\mathrm{~m}$ 处, 则少年宫在学校( )偏( )( )方向上, 距离学校( ) $\\mathrm{m}$ 。",
+        "options": [],
+        "answer": "西 北 $30^{\\circ}$ 400",
+        "solution": "两个地点的相对位置，方向相反，角度和距离不变。东偏南的相反方向是西偏北。学校在少年宫东偏南 $30^{\\circ}$ 方向 $400 \\mathrm{~m}$ 处, 则少年宫在学校西偏北 $30^{\\circ}$ (或北偏西 $60^{\\circ}$ ) 方向上, 距离学校 $400 \\mathrm{~m}$ 。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·内蒙古通辽·期中",
+        "merged_id": "unit2_435",
+        "original_id": "unit2_fill_435"
+      },
+      {
+        "id": "unit2_fill_436",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "较难",
+        "question": "确定一个具体的位置需要知道方向和____。",
+        "options": [],
+        "answer": "距离和观察点",
+        "solution": "确定一个具体的位置需要知道方向和距离、观察点。如图，台风中心在A市的东偏南 $30^{\\circ}$ 方向上，距离A市600千米处。![](images/057309ff1393f2097e27c2044d0e82c08756f8e31959c8c997a55d30cae3908f.jpg)",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·河南南阳·期中",
+        "merged_id": "unit2_436",
+        "original_id": "unit2_fill_436"
+      },
+      {
+        "id": "unit2_fill_437",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "较难",
+        "question": "小宇家在学校东偏南 $40^{\\circ}$ 方向 600 米处, 则学校在小宇家( )方向( )米处。",
+        "options": [],
+        "answer": "西偏北 $40^{\\circ}$ /北偏西 $50^{\\circ}$ 600",
+        "solution": "一个事物在另一个事物的某个方向一定度数的位置，那么另一个事物在这个事物相对的方向相同度数的位置，据此解答即可。由分析可知：小宇家在学校东偏南 $40^{\\circ}$ 方向 600 米处, 则学校在小宇家西偏北 $40^{\\circ}$ (或北偏西 $50^{\\circ}$ ) 600 米处。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·广东肇庆·期中",
+        "merged_id": "unit2_437",
+        "original_id": "unit2_fill_437"
+      },
+      {
+        "id": "unit2_fill_438",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "较难",
+        "question": "图书馆在剧院的东偏南 $60^{\\circ}$ 方向 500 米处, 那么剧院在图书馆的( )偏( )( )方向上, 距离( )米处。",
+        "options": [],
+        "answer": "西北60 500",
+        "solution": "图书馆在剧院的东偏南 $60^{\\circ}$ 方向 500 米处, 是以剧院为观测点; 那么剧院在图书馆的方向是以图书馆为观测点;根据位置的相对性可知，观测点不同，方向相反，夹角的度数相同，距离相同；由此可知, 东偏南 $60^{\\circ}$ 相对的是西偏北 $60^{\\circ}$ , 西和北之间的夹角是 $90^{\\circ}, 90^{\\circ} - 60^{\\circ} = 30^{\\circ}$ , 所以西偏北 $60^{\\circ}$ 方向, 还可以说成北偏西 $30^{\\circ}$ 方向。图书馆在剧院的东偏南 $60^{\\circ}$ 方向 500 米处, 那么剧院在图书馆的西偏北 $60^{\\circ}$ (或北偏西 $30^{\\circ}$ ) 方向上, 距离 500 米处。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·吉林四平·期中",
+        "merged_id": "unit2_438",
+        "original_id": "unit2_fill_438"
+      },
+      {
+        "id": "unit2_fill_439",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "较难",
+        "question": "少年宫在图书馆西偏北 $30^{\\circ} 200$ 米处, 则图书馆在少年宫( )偏( ) $( )^{\\circ} 200$ 米处。",
+        "options": [],
+        "answer": "东南30",
+        "solution": "根据方向的相对性，西偏北对东偏南，角度和距离不变，进行填空。少年宫在图书馆西偏北 $30^{\\circ} 200$ 米处, 则图书馆在少年宫东偏南 $30^{\\circ}$ 或南偏东 $60^{\\circ} 200$ 米处。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·甘肃兰州·期中",
+        "merged_id": "unit2_439",
+        "original_id": "unit2_fill_439"
+      },
+      {
+        "id": "unit2_fill_440",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "较难",
+        "question": "林林家在冬冬家的北偏东 $35^{\\circ}$ 方向 500 米处, 则冬冬家在林林家( )偏( )( )方向( )米处。",
+        "options": [],
+        "answer": "南西 $35^{\\circ}$ 500",
+        "solution": "林林家在冬冬家的北偏东 $35^{\\circ}$ 方向 500 米处, 是以冬冬家为观测点; 冬冬家在林林家的方向是以林林家为观测点;根据位置的相对性可知，观测点不同，方向相反，夹角的度数相同，距离相同；由此可知, 北偏东 $35^{\\circ}$ 相对的是南偏西 $35^{\\circ}$ , 南和西之间的夹角是 $90^{\\circ}, 90^{\\circ} - 35^{\\circ} = 55^{\\circ}$ , 所以南偏西 $35^{\\circ}$ 方向, 还可以说成西偏南 $55^{\\circ}$ 方向。林林家在冬冬家的北偏东 $35^{\\circ}$ 方向 500 米处, 则冬冬家在林林家南偏西 $35^{\\circ}$ (西偏南 $55^{\\circ}$ ) 方向 500 米处。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "22-23 六年级上·江西南昌·期中",
+        "merged_id": "unit2_440",
+        "original_id": "unit2_fill_440"
+      },
+      {
+        "id": "unit2_fill_441",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "小丽和小明的位置如图所示, 小明在小丽的( )偏( )（ ）°方向上，小丽在小明的( )偏( )（ ）°方向上。![](images/1f84f8f876c74df753c15d8e5755238a68e23fe87827b1fdb3ce02cd2eeb568d.jpg)",
+        "options": [],
+        "answer": "东 南 30 西 北 30",
+        "solution": "小丽看小明，说明小丽是观测点，根据\"上北下南，左西右东\"，即角度信息和距离，可确定小明的位置。小明看小丽，小明就是观测点，根据位置的相对性，方向相反，角度相同，距离相等，即可确定小丽的位置。据此解答。根据分析，可得：小丽和小明的位置如图所示, 小明在小丽的东偏南 $30^{\\circ}$ 方向上, 小丽在小明的西偏北 $30^{\\circ}$ 方向上。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·山东菏泽·期中",
+        "merged_id": "unit2_441",
+        "original_id": "unit2_fill_441"
+      },
+      {
+        "id": "unit2_fill_442",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "小远家在学校北偏西 $30^{\\circ}$ 方向 $400 \\mathrm{~m}$ 处, 则学校在小远家( )偏( )（ ）°方向 $400 \\mathrm{~m}$ 处。",
+        "options": [],
+        "answer": "南东30",
+        "solution": "根据方向的相对性，它们的方向相反，角度相等，距离相等；据此解答。$90^{\\circ} - 30^{\\circ} = 60^{\\circ}$ 小远家在学校北偏西 $30^{\\circ}$ 方向 $400 \\mathrm{~m}$ 处, 则学校在小远家南偏东 $30^{\\circ}$ (或东偏南 $60^{\\circ}$ ) 方向 $400 \\mathrm{~m}$ 处。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·江西赣州·期中",
+        "merged_id": "unit2_442",
+        "original_id": "unit2_fill_442"
+      },
+      {
+        "id": "unit2_fill_443",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "看图回答问题。![](images/8101f4e9694eb42859e797706e986d52117db41de531ec5b15ba50f257f0164d.jpg)若奇奇想乘车去书店购买书籍，要从家出发先向正东方向行驶( )站到汽车站，然后向( )方向行驶2站到( )，再向( )偏( )（ ）°方向行驶( )站到百货大楼，最后向( )方向行驶2站就到书店了。",
+        "options": [],
+        "answer": "3 正南/南 公园 南 东 60 4 正东/东",
+        "solution": "用方向和距离结合来描述路线时，要注意三个要素：一是观测点（即参照物），二是方向，三是距离。地图上按上北下南左西右东确定方向。若奇奇想乘车去书店购买书籍，要从家出发先向正东方向行驶 3 站到汽车站，然后向正南方向行驶 2 站到公园，再向南偏东 $60^{\\circ}$ 或东偏南 $30^{\\circ}$ 方向行驶 4 站到百货大楼，最后向正东方向行驶 2 站就到书店了。",
+        "knowledge_points": [
+          "根据方向、角度和距离描述路线图"
+        ],
+        "source": "23-24五年级上·广西贵港·期中",
+        "merged_id": "unit2_443",
+        "original_id": "unit2_fill_443"
+      },
+      {
+        "id": "unit2_fill_444",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "较难",
+        "question": "乐乐家在欢欢家东偏北 $50^{\\circ}$ 的方向, 距离 $900 \\mathrm{~m}$ , 那么欢欢家在乐乐家( )的方向( )m 处。",
+        "options": [],
+        "answer": "西偏南 $50^{\\circ}$ 900",
+        "solution": "根据方向的相对性，东偏北对西偏南，角度和距离不变，进行填空。乐乐家在欢欢家东偏北 $50^{\\circ}$ 的方向, 距离 $900 \\mathrm{~m}$ , 那么欢欢家在乐乐家西偏南 $50^{\\circ}$ 或南偏西 $40^{\\circ}$ 的方向 $900 \\mathrm{~m}$ 处。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·山东菏泽·期中",
+        "merged_id": "unit2_444",
+        "original_id": "unit2_fill_444"
+      },
+      {
+        "id": "unit2_fill_445",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "小玲从家出发, 先沿 方向走 m 到商场,再沿 方向走 m 到书店。![](images/cf7a3bfaaca76c9863904a1af9dd1ed01cf85f3b55083922bb6349df0308013a.jpg)",
+        "options": [],
+        "answer": "西偏北 $30^{\\circ}$ /北偏西 $60^{\\circ}$； 1000； 西偏南 $45^{\\circ}$ /南偏西 $45^{\\circ}$ ； 400",
+        "solution": "根据图中：上北下南，左西右东，图中每段相当于实际长度为 $200 \\mathrm{~m}$ ，商场在小玲家左上角，结合偏向度数属于西偏北 $30^{\\circ}$ 或北偏西 $60^{\\circ}$ ，一共 5 段，距离是 5 个 $200 \\mathrm{~m}$ ；书店在商场左下角，结合偏向度数属于西偏南 $45^{\\circ}$ 或南偏西 $45^{\\circ}$ ，一共 2 段，距离是 2 个 $200 \\mathrm{~m}$ ；据此解答。$5 \\times 200 = 1000 \\, (\\mathrm{m})$ ； $2 \\times 200 = 400 \\, (\\mathrm{m})$ 小玲从家出发，先沿西偏北 $30^{\\circ}$ 或北偏西 $60^{\\circ}$ 方向走 $1000 \\mathrm{~m}$ 到商场；再沿西偏南 $45^{\\circ}$ 或南偏西 $45^{\\circ}$ 方向走 $400 \\mathrm{~m}$ 到书店。",
+        "knowledge_points": [
+          "根据方向、角度和距离描述路线图"
+        ],
+        "source": "24-25 六年级上·河北秦皇岛·期中",
+        "merged_id": "unit2_445",
+        "original_id": "unit2_fill_445"
+      },
+      {
+        "id": "unit2_fill_446",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如图，依依从家出发，先向( )偏( ) $20^{\\circ}$ 方向走400米到公园，再向正东方向走( )米到医院，最后向南偏东( )°方向走( )米到学校。![](images/33bd364d3d280aada54d2ecff1a756cdfced473822e770a07a84548b8e27b162.jpg)",
+        "options": [],
+        "answer": "北东 600 35 400",
+        "solution": "用方向和距离结合来描述路线时，要注意三个要素：一是观测点（即参照物），二是方向，三是距离。地图上按上北下南左西右东确定方向，图上厘米数 $\\times 200 =$ 实际米数。$3 \\times 200 = 600$ （米）、 $2 \\times 200 = 400$ （米）依依从家出发，先向北偏东 $20^{\\circ}$ 方向走 400 米到公园，再向正东方向走 600 米到医院，最后向南偏东 $35^{\\circ}$ 方向走 400 米到学校。",
+        "knowledge_points": [
+          "根据方向、角度和距离描述路线图"
+        ],
+        "source": "22-23六年级上·陕西安康·期中",
+        "merged_id": "unit2_446",
+        "original_id": "unit2_fill_446"
+      },
+      {
+        "id": "unit2_fill_447",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "东东从家出发去图书馆看书，他先向正东方向走800米到邮政局，接着向西偏北 $30^{\\circ}$ 方向走600米到达图书馆。东东看完书按原路回家，他从图书馆出发，先向( )偏( )（ ）°方向走600米到达邮政局，再向正( )方走800米到家。",
+        "options": [],
+        "answer": "东 南 30 西",
+        "solution": "方向和距离两个条件才能确定物体的位置，根据上北下南左西右东以及位置的相对性，可知两处位置观测点不同，它们的方向相反，角度相等，据此解答。东东从家出发去图书馆看书，他先向正东方向走 800 米到邮政局，接着向西偏北 $30^{\\circ}$ 方向走 600 米到达图书馆。东东看完书按原路回家，他从图书馆出发，先向东偏南 $30^{\\circ}$ （南偏东 $60^{\\circ}$ ）方向走 600 米到达邮政局，再向正西方走 800 米到家。",
+        "knowledge_points": [
+          "根据方向、角度和距离描述路线图"
+        ],
+        "source": "23-24六年级上·河北邢台·期中",
+        "merged_id": "unit2_447",
+        "original_id": "unit2_fill_447"
+      },
+      {
+        "id": "unit2_fill_448",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "小红家在学校的北偏东 $60^{\\circ}$ 的方向上, 距离是 450 米, 那么学校在小红家______偏______度的方向上, 距离是______米。",
+        "options": [],
+        "answer": "南 西 60 450",
+        "solution": "根据方向的相对性：方向相反，角度相同，距离相同。即北相反是南，东相反是西，据此解答即可。据分析可知, 小红家在学校的北偏东 $60^{\\circ}$ 的方向上, 距离是 450 米, 那么学校在小红家南偏西 60 度的方向上, 距离是 450 米。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·山东菏泽·期中",
+        "merged_id": "unit2_448",
+        "original_id": "unit2_fill_448"
+      },
+      {
+        "id": "unit2_fill_449",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "较难",
+        "question": "体育课上, 王艺和李红所站位置如图, 王艺在李红的( )偏( )（ ）°方向上; 李红在王艺的( )偏( )（ ）°方向上。![](images/4fe4d556acc4a3df9bac9aef0b7eb6a3aa63f11226cb07706725f46a9f755cd4.jpg)",
+        "options": [],
+        "answer": "北东55 西南35",
+        "solution": "根据\"上北下南左西右东\"的图上方向，结合图示分析、找准观测点解答即可。体育课上，王艺和李红所站位置如图，王艺在李红的北偏东 $55^{\\circ}$ （东偏北 $35^{\\circ}$ ）方向上；李红在王艺的西偏南 $35^{\\circ}$ （南偏西 $55^{\\circ}$ ）方向上。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 五年级上·河北石家庄·期中",
+        "merged_id": "unit2_449",
+        "original_id": "unit2_fill_449"
+      },
+      {
+        "id": "unit2_fill_450",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "较难",
+        "question": "张亮家在学校的西偏北 $30^{\\circ}$ 方向上, 学校在张亮家的( )方向上。",
+        "options": [],
+        "answer": "东偏南 $30^{\\circ}$",
+        "solution": "张亮家在学校的西偏北 $30^{\\circ}$ 方向上，是以学校为观测点；学校在张亮家的方向是以张亮家为观测点；根据位置的相对性可知，观测点不同，方向相反，夹角的度数相同，距离相同；由此可知, 西偏北 $30^{\\circ}$ 相对的是东偏南 $30^{\\circ}$ , 东和南之间的夹角是 $90^{\\circ}, 90^{\\circ} - 30^{\\circ} = 60^{\\circ}$ , 所以东偏南 $30^{\\circ}$ 方向, 还可以说成南偏东 $60^{\\circ}$ 方向。张亮家在学校的西偏北 $30^{\\circ}$ 方向上，学校在张亮家的东偏南 $30^{\\circ}$ （或南偏东 $60^{\\circ}$ ）方向上。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "22-23 六年级上·湖北襄阳·期中",
+        "merged_id": "unit2_450",
+        "original_id": "unit2_fill_450"
+      },
+      {
+        "id": "unit2_fill_451",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "较难",
+        "question": "图书馆在学校东偏北 $30^{\\circ}$ 方向上, 这是以( )为观测点, 从( )的正东方向开始, 向北偏转( )。",
+        "options": [],
+        "answer": "学校 ；学校 ； $30^{\\circ} / 30$ 度",
+        "solution": "先确定观测点，一般把\"在\"、位于\"后面的场所作为观测点；再根据观测点的\"上北下南，左西右东\"以及角度确定方向，据此解答。图书馆在学校东偏北 $30^{\\circ}$ 方向上，这是以学校为观测点，从学校的正东方向开始，向北偏转 $30^{\\circ}$ 。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "22-23 六年级上·江西南昌·期中",
+        "merged_id": "unit2_451",
+        "original_id": "unit2_fill_451"
+      },
+      {
+        "id": "unit2_fill_452",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "较难",
+        "question": "要比较准确地描述一个物体的位置, ( )和( )这两条件缺一不可。",
+        "options": [],
+        "answer": "方向 距离",
+        "solution": "要确定某一物体的位置，必须先确定参照物，再以参照物为观测点，根据这一物体所在的方向和距离确定它的具体位置，方向、距离这两个条件缺一不可。由分析可知，要比较准确地描述一个物体的位置，方向和距离这两条件缺一不可。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·江西南昌·期中",
+        "merged_id": "unit2_452",
+        "original_id": "unit2_fill_452"
+      },
+      {
+        "id": "unit2_fill_453",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "少年宫在银行的南偏东 $30^{\\circ}$ 方向 2000 米处，从少年宫到银行需要向( )方向走 2000 米。",
+        "options": [],
+        "answer": "北偏西 $30^{\\circ}$",
+        "solution": "根据位置的相对性，方向刚好相反，少年宫在银行的南偏东 $30^{\\circ}$ 方向 2000 米处，则银行在少年宫的北偏西 $30^{\\circ}$ 的方向或者西偏北 $60^{\\circ}$ 方向。则从少年宫到银行就是以少年宫为观测点，即按照这个方向走 2000 米。从少年宫到银行需要向北偏西 $30^{\\circ}$ （西偏北 $60^{\\circ}$ ）方向走2000米。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·河南安阳·期末",
+        "merged_id": "unit2_453",
+        "original_id": "unit2_fill_453"
+      },
+      {
+        "id": "unit2_fill_454",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "较难",
+        "question": "学校在小明家东偏北 $30^{\\circ}$ 方向上, 距离 800 米, 则小明家在学校( )方向上, 距离( )米。",
+        "options": [],
+        "answer": "西偏南 $30^{\\circ}$ 800",
+        "solution": "根据方向的相对性，东偏北对西偏南，角度和距离不变，进行填空。学校在小明家东偏北 $30^{\\circ}$ 方向上, 距离 800 米, 则小明家在学校西偏南 $30^{\\circ}$ 或南偏西 $60^{\\circ}$ 方向上, 距离 800 米。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·河北邯郸·期中",
+        "merged_id": "unit2_454",
+        "original_id": "unit2_fill_454"
+      },
+      {
+        "id": "unit2_fill_455",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "小明从家里出发往南走100米，再往西走100米，这时，他正好在家的( )偏( )（ ）°方向上。",
+        "options": [],
+        "answer": "西南 45",
+        "solution": "可画出相关图帮助理解：以小明家为观测点，先向南走 100 米，再向西走 100 米，此时小明的位置正好在他家的西偏南 $45^{\\circ}$ （或南偏西 $45^{\\circ}$ ）方向上。如图所示：![](images/639199d69bb80fd1518010120da8fdb955b9ab861b99106f747844457db08b4a.jpg)小明从家里出发往南走 100 米, 再往西走 100 米, 这时, 他正好在家的西偏南 $45^{\\circ}$ (或南偏西 $45^{\\circ}$ ) 方向上。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·湖北武汉·期中",
+        "merged_id": "unit2_455",
+        "original_id": "unit2_fill_455"
+      },
+      {
+        "id": "unit2_fill_456",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "较难",
+        "question": "公园在小红家的东偏南 $60^{\\circ}$ 方向, 距离 300 米。那么小红家在公园的( )偏 (_____, _____)°方向, 距离 (_____)米。",
+        "options": [],
+        "answer": "西北 60 300",
+        "solution": "公园在小红家的东偏南 $60^{\\circ}$ 方向上, 是以小红家为观测点; 小红家在公园的方向是以公园为观测点; 根据位置的相对性可知, 观测点不同, 方向相反, 夹角的度数相同,所以东偏南 $60^{\\circ}$ 相对的是西偏北 $60^{\\circ}$ , 由此解答。公园在小红家的东偏南 $60^{\\circ}$ 方向，距离300米。那么小红家在公园的西偏北 $60^{\\circ}$ （或北偏西 $30^{\\circ}$ ）方向，距离300米。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·山东临沂·期中",
+        "merged_id": "unit2_456",
+        "original_id": "unit2_fill_456"
+      },
+      {
+        "id": "unit2_judge_457",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "判断题",
+        "difficulty": "较难",
+        "question": "甲在乙的东偏南 $30^{\\circ}$ 方向上, 那么乙就在甲的西偏北 $60^{\\circ}$ 方向上。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "甲在乙的东偏南 $30^{\\circ}$ 方向上，是以乙为观测点；那么乙就在甲的方向是以甲为观测点；根据位置的相对性可知，观测点不同，方向相反，夹角的度数相同，距离相同；据此解答。根据位置的相对性可知: 甲在乙的东偏南 $30^{\\circ}$ 方向上, 那么乙就在甲的西偏北 $30^{\\circ}$ (或北偏西 $60^{\\circ}$ ) 方向上, 原题说法错误。故答案为： $\\times$",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·江西宜春·期中",
+        "merged_id": "unit2_457",
+        "original_id": "unit2_judge_457"
+      },
+      {
+        "id": "unit2_judge_458",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "目前台风位于 A 市西偏东 $30^{\\circ}$ 方向, 距 A 市 900 千米的海面上。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "确定一个物体的位置时，首先要找到观测点，然后以观测点为中心，根据\"上北下南，左西右东\"结合角度描述方向，最后根据物体和观测点之间的距离确定物体的位置，据此解答。以 A 市为观测点, 台风可能位于 A 市西偏北或西偏南 $30^{\\circ}$ 方向, 东和西是相反的方向, 台风不可能位于 A 市西偏东 $30^{\\circ}$ 方向, 台风在距 A 市 900 千米的海面上。所以原题说法错误。故答案为：×",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·河南南阳·期中",
+        "merged_id": "unit2_458",
+        "original_id": "unit2_judge_458"
+      },
+      {
+        "id": "unit2_judge_459",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "判断题",
+        "difficulty": "较难",
+        "question": "确定一个物体的位置, 只要知道它的方向就可以了。 ( )",
+        "options": [],
+        "answer": "×",
+        "solution": "物体位置对于某一观察点来说，是由一定的方向和距离确定的，只知道方向或距离不能确定物体的位置。据此解答即可。由分析可知：确定一个物体的位置，需要知道它的方向，角度距离才可以确定。原说法错误。故答案为： $\\times$",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "22-23 六年级上·江西南昌·期中",
+        "merged_id": "unit2_459",
+        "original_id": "unit2_judge_459"
+      },
+      {
+        "id": "unit2_judge_460",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "判断题",
+        "difficulty": "较难",
+        "question": "乙在甲的东偏南 $40^{\\circ}$ 方向上, 则甲在乙的西偏北 $50^{\\circ}$ 方向上。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "根据方向的相对性，东偏南对西偏北，角度不变，进行判断。乙在甲的东偏南 $40^{\\circ}$ 方向上，则甲在乙的西偏北 $40^{\\circ}$ 方向上。![](images/1a39a3793a7228c3a5a569f870df8426b9ace671d8a737fb106034339cfaaa0f.jpg)故答案为： $\\times$",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·河北秦皇岛·期中",
+        "merged_id": "unit2_460",
+        "original_id": "unit2_judge_460"
+      },
+      {
+        "id": "unit2_judge_461",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "丁丁在冬冬的南偏东 $40^{\\circ}$ 方向 $500 \\mathrm{~m}$ 处, 那么冬冬在丁丁的北偏西 $40^{\\circ}$ 方向 $500 \\mathrm{~m}$ 处。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "根据方向具有相对性可知: 方向相反, 距离相同, 丁丁在冬冬的南偏东 $40^{\\circ}$ 方向 $500 \\mathrm{~m}$ 处, 冬冬就在丁丁的北偏西 $40^{\\circ}$ 方向 $500 \\mathrm{~m}$ 处。据此判断即可。丁丁在冬冬的南偏东 $40^{\\circ}$ 方向 $500 \\mathrm{~m}$ 处, 那么冬冬在丁丁的北偏西 $40^{\\circ}$ 方向 $500 \\mathrm{~m}$ 处。所以原题说法正确。故答案为： $\\sqrt{}$",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "22-23 六年级上·陕西安康·期中",
+        "merged_id": "unit2_461",
+        "original_id": "unit2_judge_461"
+      },
+      {
+        "id": "unit2_judge_462",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "同一个观测点的北偏东 $30^{\\circ}$ 和东偏北 $30^{\\circ}$ 是一个方向。",
+        "options": [],
+        "answer": "×",
+        "solution": "北偏东 $30^{\\circ}$ 表示从正北方向往东偏 $30^{\\circ}$ , 东偏北 $30^{\\circ}$ 表示从正东方向往北偏 $30^{\\circ}$ 。正北和正东方向的夹角是 $90^{\\circ}, 90^{\\circ} - 30^{\\circ} = 60^{\\circ}$ , 则同一个观测点的北偏东 $30^{\\circ}$ 和东偏北 $60^{\\circ}$ 是一个方向。据此解答。通过分析可得：$ 90^{\\circ} - 30^{\\circ} = 60^{\\circ}$ , 同一个观测点的北偏东 和东偏北 是一个方向, 和东偏北 不是一个方向。原题说法错误。故答案为： $\\times$$90^{\\circ} - 30^{\\circ} = 60^{\\circ}​$ , 同一个观测点的北偏东 $30^{\\circ}​$ 和东偏北 $60^{\\circ}​$ 是一个方向, 和东偏北 $30^{\\circ}​$ 不是一个方向。原题说法错误。",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·陕西安康·期中",
+        "merged_id": "unit2_462",
+        "original_id": "unit2_judge_462"
+      },
+      {
+        "id": "unit2_judge_463",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "判断题",
+        "difficulty": "较难",
+        "question": "东偏南 $45^{\\circ}$ 方向上距离 $800 \\mathrm{~m}$ 处与西偏北 $45^{\\circ}$ 方向上距离 $800 \\mathrm{~m}$ 处相同。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "根据方向的相对性：方向相反，角度相同，距离相同。据此可知，东偏南 $45^{\\circ}$ 方向上距离 $800 \\mathrm{~m}$ 处与西偏北 $45^{\\circ}$ 方向上距离 $800 \\mathrm{~m}$ 处相反。据分析可知, 东偏南 $45^{\\circ}$ 方向上距离 $800 \\mathrm{~m}$ 处与西偏北 $45^{\\circ}$ 方向上距离 $800 \\mathrm{~m}$ 处相反。 原题说法错误。故答案为：×",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·陕西安康·期中",
+        "merged_id": "unit2_463",
+        "original_id": "unit2_judge_463"
+      },
+      {
+        "id": "unit2_judge_464",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "由 \"笑笑家在东偏北 $40^{\\circ}$ 方向上, 距离是 $500 \\mathrm{~m}$ ,可以得出笑笑家的具体位置。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "将方向和距离结合起来描述位置时，要注意三个要素：一是观测点，二是方向，三是距离。\"笑笑家在东偏北 $40^{\\circ}$ 方向上, 距离是 $500 \\mathrm{~m}$ , 这句话没有观测点, 无法得出笑笑家的具体位置, 所以原题说法错误。故答案为：×",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·河南郑州·期中",
+        "merged_id": "unit2_464",
+        "original_id": "unit2_judge_464"
+      },
+      {
+        "id": "unit2_judge_465",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "判断题",
+        "difficulty": "较难",
+        "question": "如图所示, 体育馆在学校的东偏北 $60^{\\circ}$ 方向上。( )![](images/a7c27c718c6f8c9d27a34fe640a0d495fb022cd401e270de0c0a337b5acaf166.jpg)",
+        "options": [],
+        "answer": "×",
+        "solution": "判断体育馆在学校的什么方向，即观测点在学校，站在学校看向体育馆。结合图中的方向标，根据地图上的方向辨别方法：\"上北下南，左西右东\"，即可解决问题。$90^{\\circ} - 60^{\\circ} = 30^{\\circ}$ , 站在学校看向体育馆, 结合图形知: 体育馆在学校的东偏北 $30^{\\circ}$ 。故答案为： $\\times$",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·河南驻马店·期中",
+        "merged_id": "unit2_465",
+        "original_id": "unit2_judge_465"
+      },
+      {
+        "id": "unit2_draw_466",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "根据下面的描述, 在平面图上标出各场所的位置。(1) 文化广场在电视塔的北偏东 $45^{\\circ}$ 方向 $1000 \\mathrm{~m}$ 处。(2) 体育场在电视塔的西偏南 $30^{\\circ}$ 方向 $2500 \\mathrm{~m}$ 处。(3) 动物园在电视塔的西偏北 $20^{\\circ}$ 方向 $2000 \\mathrm{~m}$ 处。(4) 科技馆在电视塔的东偏南 $40^{\\circ}$ 方向 $1500 \\mathrm{~m}$ 处。(5) 电视塔在丁丁家的北偏东 $20^{\\circ}$ 方向 $1000 \\mathrm{~m}$ 处。![](images/65e4f7b404954b8e1ff2daae5fdf9b227c66e1552eee8bee38709d5b7ff70a2e.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "solution": "根据题意，根据实际距离、比例尺及图上距离的关系，可分别求出文化广场、体育场、动物园、科技馆到电视塔的图上距离，以电视塔为观察点，再根据在地图上北下南，左西右东的方向分别确定文化广场、体育场、动物园和和科技馆的方向，电视塔在丁丁家的北偏东 $20^{\\circ}$ 方向 $1000 \\mathrm{~m}$ 处，可以看作丁丁家在电视塔的南偏西 $20^{\\circ}$ 方向 $1000 \\mathrm{~m}$ 处，结合图上所给信息，在平面图上确定出各场所的位置即可。本题考查了位置与方向，确定场所的位置，首先要确定观测中心，抓住方向和距离两个要素，即可解决此类问题。（1）文化广场在电视塔的北偏东 $45^{\\circ}$ 方向 $1000 \\mathrm{~m}$ 处，图上距离为： $1000 \\div 500 =$ 2（厘米）；(2) 体育场在电视塔的西偏南 $30^{\\circ}$ 方向 $2500 \\mathrm{~m}$ 处, 图上距离为: $2500 \\div 500 = 5$ (厘米); (3) 动物园在电视塔的西偏北 $20^{\\circ}$ 方向 $2000 \\mathrm{~m}$ 处, 图上距离为: $2000 \\div 500 = 4$ (厘米); (4) 科技馆在电视塔的东偏南 $40^{\\circ}$ 方向 $1500 \\mathrm{~m}$ 处, 图上距离为: $1500 \\div 500 = 3$ (厘米): (5) 电视塔在丁丁家的北偏东 $20^{\\circ}$ 方向 $1000 \\mathrm{~m}$ 处, 可以看作丁丁家在电视塔的南偏西 $20^{\\circ}$ 方向 $1000 \\mathrm{~m}$ 处, 图上距离为: $1000 \\div 500 = 2$ (厘米)。各场所位置如图所示：![](images/c217275c06745eac57bbeb50e5a697f9624024fd15f839b38a1c05417198f9f9.jpg)",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·河南许昌·期中",
+        "merged_id": "unit2_466",
+        "original_id": "unit2_draw_466"
+      },{
+        "id": "unit2_draw_467",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "以学校为观测点, 回答下面的问题。![](images/62f408f822b7276446de57757f5eae8cf41e79e4cd93f13e2e51f24f4f299404.jpg)(1) 明明家在学校的________方向，距离是________米。(2) 根据描述，在平面图中，标出各场所的位置。亮亮家在学校的东偏南 $40^{\\circ}$ 方向, 400 米处; 童童家在学校的西偏南 $45^{\\circ}$ 方向, 500 米处;红红家在学校的北偏西 $30^{\\circ}$ 方向 300 米处。(3)聪聪从学校应沿着怎样的路线回家？描述一下具体方向和所走路程。",
+        "options": [],
+        "answer": "（1）正东；300 (2）见详解 (3）见详解",
+        "solution": "（1）根据题意可知，1 厘米表示实际距离是 100 米，先计算出明明家到学校的实际距离，再根据地图上方向的规定\"上北下南，左西右东\"，以学校为观测点，确定出明明家的位置；(2) 先计算出亮亮家到学校的图上距离; 童童家到学校的图上距离; 红红家到学校的图上距离, 以学校为观测点, 画出亮亮家、童童家、红红家的位置; (3) 先计算出各个路段的实际距离; 确定线路时, 注意起始点与目的地, 起始点是观测点,再按照地图上 \"上北下南, 左西右东\" 确定方向, 用方向、角度和距离描述聪聪从学校到家行驶路线。(1) $100 \\times 3 = 300$ （米）明明家在学校正东方向，距离是300米。(2) $400 \\div 100 = 4$ （厘米）$5 0 0 \\div 1 0 0 = 5 \\text {(厘 米)} $ $3 0 0 \\div 1 0 0 = 3 (\\text {厘 米}) $如图：![](images/f66dc17b45b1606d8a687ed3f2365c3033fbe08cc31865d0946435efd87a8276.jpg)(3) $100 \\times 4 = 400$ （米）； $100 \\times 1 = 100$ （米）$9 0 ^ {\\circ} - 3 5 ^ {\\circ} = 5 5 ^ {\\circ}$聪聪从学校向东偏北 $35^{\\circ}$ （或北偏东 $55^{\\circ}$ ）方向走400米，再向正东走方向100米到聪聪家。",
+        "knowledge_points": [
+          "根据方向、角度和距离描述路线图",
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·河南南阳·期中",
+        "merged_id": "unit2_467",
+        "original_id": "unit2_draw_467"
+      },
+      {
+        "id": "unit2_draw_468",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "以小明家为观测点, 根据下面提供的信息, 在平面图上标出各场所的位置。(1) 火车站在小明家南偏西 $30^{\\circ}$ 方向 200 米处。(2) 银行在小明家正西方向 400 米处。(3) 超市在小明家北偏西 $45^{\\circ}$ 方向 300 米处。(4) 医院在小明家北偏东 $30^{\\circ}$ 方向 500 米处。![](images/d11ade4de34976be5c72ba633f9db0ec48743b540e94b0f3f7c3e80fa6ffd5e5.jpg)",
+        "options": [],
+        "answer": "图见详解",
+        "solution": "（1）以小明家为观测点，火车站在小明家南偏西 $30^{\\circ}$ 方向，长度是 200 米，即图中 2 段长度；(2) 以小明家为观测点，银行在小明家正西方向，长度是400米，即图中4段长度； (3) 以小明家为观测点, 超市在小明家北偏西 $45^{\\circ}$ 方向, 长度是 300 米, 即图中 3 段长度; (4) 以小明家为观测点, 医院在小明家北偏东 $30^{\\circ}$ 方向, 长度是 500 米, 即图中 5 段长度; 据此作图。(1) (2) (3) (4) 作图如下：![](images/d195835e7e8b65fc901353cf1836f3573a07cf5fcfbf764f53ba19cae16bbce6.jpg)",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·湖北武汉·期中",
+        "merged_id": "unit2_468",
+        "original_id": "unit2_draw_468"
+      },
+      {
+        "id": "unit2_draw_469",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "在平面图中标出各建筑的位置。![](images/a9b296aa06cdbe6849978a5e0ad9732f02f9ab0f3ca7740714d65dadc810385a.jpg)(1) 电子城在文化广场西偏北 $45^{\\circ}$ 方向 200 米处。(2) 美食城在文化广场西偏南 $40^{\\circ}$ 方向 100 米处。(3) 服装城在文化广场北偏东 $30^{\\circ}$ 方向250米处。",
+        "options": [],
+        "answer": "见详解",
+        "solution": "（1）据图可知，图中的 1 厘米表示实际的 50 米，用除法算出各建筑物在图中应该画出的长度；(2) 根据 \"上北下南, 左西右东\" 的方位判断方法并结合给出的角度确定各建筑物的方位即可。$200 \\div 50 = 4$ （厘米）$100 \\div 50 = 2$ （厘米）$250 \\div 50 = 5$ （厘米）(1) (2) (3) 画图如下：![](images/5a1424b37612cc1bbbf0aa72c4d6b8eac149e7e73739a9fd279b86bd4c0fefd7.jpg)",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25六年级上·广东肇庆·期中",
+        "merged_id": "unit2_469",
+        "original_id": "unit2_draw_469"
+      },
+      {
+        "id": "unit2_draw_470",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "根据下面的描述，在平面图上标出各场所的位置。![](images/785bb4491bab168ab3e3c2466c7cbfaa3141114294c785ac17f1606c4cac6829.jpg)(1) 小丽家在广场西偏南 $45^{\\circ}$ 方向 600 米处。(2) 军军在广场东偏北 $40^{\\circ}$ 方向 900 米处。",
+        "options": [],
+        "answer": "(1) (2）见详解",
+        "solution": "（1）根据题意可知，1 厘米表示 300 米，先计算出小丽家到广场的图上距离；再根据地图上方向的规定\"上北下南，左西右东\"，以广场为观测点，画出小丽家的位置；(2) 先计算出军军家到广场的图上距离，再用广场为观测点，画出军军的位置，据此解答。(1) $600 \\div 300 = 2$ （厘米）图如下：(2) $900 \\div 300 = 3$ （厘米）图如下：![](images/3b63f6300d2adf29659d37eaa663a40aed328adcbed64b99d7df4a9cc184c48d.jpg)",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25六年级上·甘肃兰州·期中",
+        "merged_id": "unit2_470",
+        "original_id": "unit2_draw_470"
+      },
+      {
+        "id": "unit2_draw_471",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "星光电器商行接到一份外卖订单。外卖员从星光电器商行出发, 先向东偏南 $45^{\\circ}$ 方向行驶 1000 米到万达广场, 再向西偏南 $30^{\\circ}$ 方向行驶 1500 米到达体育中心, 再向北偏西 $70^{\\circ}$ 方向行驶 1000 米到达电影院, 最后正西方向行驶 500 米到达绿苑小区。根据上面的描述画出外卖员的路线图。![](images/3075ed5ca71364c7108e8cc16a10592a7913f48d875b9891068b3c9d9d59055e.jpg)",
+        "options": [],
+        "answer": "图见详解",
+        "solution": "由题意知：以星光电器商行为参照点建立方向标， $1000 \\div 500 = 2$ （个），从星光电器商行向东偏南 $45^{\\circ}$ 方向画 2 个单位长度，标出万达广场；再以万达广场为参照点建立方向标， $1500 \\div 500 = 3$ （个），从万达广场向西偏南 $30^{\\circ}$ 方向画 3 个单位长度，标出体育中心；再以体育中心为参照点建立方向标，从体育中心向北偏西 $70^{\\circ}$ 方向画出 2 个单位长度，标出电影院；最后以电影院为参照点建立方向标，从电影院向正西方向画 1 个单位长度，标出绿苑小区。据此作图即可。据分析作图如下：![](images/ee385ec03f2e390a4ab63a2f2c0908e4fd5a89d6742ae7230d1216bf0045f5f2.jpg)",
+        "knowledge_points": [
+          "根据方向、角度和距离画线路图"
+        ],
+        "source": "24-25 六年级上·福建福州·期中",
+        "merged_id": "unit2_471",
+        "original_id": "unit2_draw_471"
+      },
+      {
+        "id": "unit2_draw_472",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "根据下面的描述，在平面图上标出各场所的位置。![](images/a21fb34c9dce8df4be8fc05214a29ed359707d963f45a3895de8124c0c34e293.jpg)(1) 医院在学校的西偏北 $30^{\\circ}$ 方向 1200 米处。(2) 政务中心在学校的东偏北 $35^{\\circ}$ 方向 900 米处。(3) 市政府在学校的南偏西 $40^{\\circ}$ 方向 600 米处。",
+        "options": [],
+        "answer": "见详解",
+        "solution": "弄清要标示的物体在哪个方位上, 有多少度, 按要求的方位和度数准确画图; 注意各场所离中心点的距离, 根据要求画出相应的长度。地图上按上北下南左西右东确定方向, 实际米数 $\\div 300 =$ 要画的厘米数。$1200 \\div 300 = 4$ （厘米）$9 0 0 \\div 3 0 0 = 3 (\\text {厘 米}) $ $6 0 0 \\div 3 0 0 = 2 (\\text {厘 米})$![](images/cd0e6457bebe57a602115217c467faa1bfbe82f7b4fd28e731d2ea78fc156495.jpg)",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "22-23六年级上·陕西安康·期中",
+        "merged_id": "unit2_472",
+        "original_id": "unit2_draw_472"
+      },
+      {
+        "id": "unit2_draw_473",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "根据描述，把小敏从家出发去学校的路线图画完整。小敏从家O处出发，先向东偏北 $60^{\\circ}$ 方向，移动 $150\\mathrm{m}$ 到新华书店，再向东偏南 $30^{\\circ}$ 方向移动 $100\\mathrm{m}$ 到公园，然后向正北方向移动 $50\\mathrm{m}$ 到游泳馆，最后东偏北 $45^{\\circ}$ 方向移动 $250\\mathrm{m}$ 到达学校。![](images/fba8d7395feac2050550b28c8c75c26e879df9e323da7532e8acfbd03de7bdd9.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "solution": "用方向和距离结合来画路线时，要注意三个要素：一是观测点（即参照物），二是方向，三是距离。弄清要标示的物体在哪个方位上，有多少度，按要求的方位和度数准确画图；注意各场所离出发点的距离，根据要求画出相应的长度。$150 \\div 50 = 3$ (cm) $1 0 0 \\div 5 0 = 2 (\\mathrm {c m})$ $5 0 \\div 5 0 = 1 (\\mathrm {c m})$ $2 5 0 \\div 5 0 = 5 (\\mathrm {c m})$![](images/5279e2d48d668b5617a03bd7d3b30d49a808905bc39cfb7ad88f31d35e45bf9c.jpg)",
+        "knowledge_points": [
+          "根据方向、角度和距离画线路图"
+        ],
+        "source": "22-23六年级上·陕西安康·期中",
+        "merged_id": "unit2_473",
+        "original_id": "unit2_draw_473"
+      },
+      {
+        "id": "unit2_draw_474",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "2 路公共汽车从起点站向东偏南 $30^{\\circ}$ 方向行驶 6 千米到体育馆, 然后又向东行驶 4 千米到银行, 最后向北偏东 $40^{\\circ}$ 方向行驶 4 千米到达火车站。![](images\\90eef6a6-ecfa-42f6-9ae8-a98aec81deb4.png)(1) 根据上面的描述完成 2 路公共汽车行驶的路线图。(2) 根据路线图说说 2 路公共汽车返回时行驶的方向和路程。",
+        "options": [],
+        "answer": "(1) 见详解 (2）见详解（答案不唯一）",
+        "solution": "（1）图上 1 厘米表示实际距离 2 千米, $6 \\div 2 = 3$ （厘米），以起点站为观测点，根据地图 \"上北下南, 左西右东\" 的规定，在起点站东偏南 $30^{\\circ}$ 方向 3 厘米处标出体育馆的位置。 $4 \\div 2 = 2$ （厘米），再以体育馆为观测点，在正东方向 2 厘米处标出银行的位置。最后以银行为观测点，在北偏东 $40^{\\circ}$ 方向 2 厘米处标出火车站的位置。(2) 根据位置的相对性可知, 描述两个物体之间的相对位置时, 方向相反, 角度相等, 距离相等, 据此解答。![](images/4309ea88c607123e724426a3780e4ef55be82cce7de3208a53ad7e261b9dd95b.jpg)(2) 通过分析可得:2路公共汽车返回时行驶的方向和路程是：从火车站向南偏西 $40^{\\circ}$ 方向行驶4千米到达银行，然后又向西行驶4千米到体育馆，最后向西偏北 $30^{\\circ}$ 方向行驶6千米到起点站。",
+        "knowledge_points": [
+          "根据方向、角度和距离描述路线图",
+          "根据方向、角度和距离画线路图"
+        ],
+        "source": "23-24 六年级上·吉林四平·期中",
+        "merged_id": "unit2_474",
+        "original_id": "unit2_draw_474"
+      },
+      {
+        "id": "unit2_draw_475",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "根据描述，画出路线图。为创建环保社会，倡导低碳生活，小明骑自行车去书店。他先从家向北偏东 $40^{\\circ}$ 方向骑500米到中国银行，再向正东方向骑1000米到万家福超市，最后向东偏南 $30^{\\circ}$ 方向骑1000米到书店。![](images/0f7b538ef55f7ffae3fd14a50e9d3b22d6dbf356d0f8e171e84a1825a2325446.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "solution": "以图上的 \"上北下南, 左西右东\" 为准, 图例表示图上 1 厘米相当于实际距离 250 米。小明先从家向北偏东 $40^{\\circ}$ 方向骑 500 米到中国银行, 即在家的北偏东 $40^{\\circ}$ 方向上画 $500 \\div 250 = 2$ 厘米长的线段, 即是中国银行;再向正东方向骑 1000 米到万家福超市, 即在中国银行的正东方向上画 $1000 \\div 250 = 4$ 厘米长的线段, 即是万家福超市;最后向东偏南 $30^{\\circ}$ 方向骑 1000 米到书店, 即在万家福超市的东偏南 $30^{\\circ}$ 方向上画 $1000 \\div 250 = 4$ 厘米长的线段, 即是书店。根据方向、角度和距离画小明从家去书店的路线图。如图：![](images/6f9ebb871abf997f5f6c3569378f467302e03d63719c2054e8ef6a637a36aafa.jpg)",
+        "knowledge_points": [
+          "根据方向、角度和距离画线路图"
+        ],
+        "source": "22-23六年级上·陕西安康·期中",
+        "merged_id": "unit2_475",
+        "original_id": "unit2_draw_475"
+      },
+      {
+        "id": "unit2_draw_476",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "\"军训\" 最早可追测至夏商周时期, 是学生接受国防教育的基本形式, 是培养 \"四有\" 人才的一项重要措施。某学校同学们参加军训, 从学校出发, 先向北偏东 $60^{\\circ}$ 方向行进 600 米, 然后向正东方向行进 200 米, 最后向南偏东 $35^{\\circ}$ 方向行进 200 米到达指定地点军营。根据描述, 画出同学们从学校到军营的行走路线。![](images/4dc78ea0436839bbca83aacea4047670f90283d22bc817d81579f11cb2f8c8aa.jpg)",
+        "options": [],
+        "answer": "图见详解",
+        "solution": "以图上的\"上北下南，左西右东\"为准，图例表示图上1厘米相当于实际距离200米，结合方向、角度和距离画出同学们从学校到军营的行走路线。$600 \\div 200 = 3$ （厘米） $ 2 0 0 \\div 2 0 0 = 1 (\\text {厘 米})$如图：![](images/aef47c684123a951492bd8af961abf4431112cebb967ff27dc8ce86372b4e650.jpg)",
+        "knowledge_points": [
+          "根据方向、角度和距离画线路图"
+        ],
+        "source": "23-24 六年级上·广东汕尾·期中",
+        "merged_id": "unit2_476",
+        "original_id": "unit2_draw_476"
+      },
+      {
+        "id": "unit2_fill_477",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如图，小志家在学校的（）偏（）（）方向上，学校在小志家的（）偏（）（）方向上；小欣家在学校的（）偏（）（）方向上。![](images/9616882d24a26dc39c48702163d80b638c36851c8fe5261275941858daff7b22.jpg)② 画一画, 小红家在学校的东偏南 $30^{\\circ}$ 距离 $1.5 \\mathrm{~km}$ 处, 请标出小红家。图书馆在学校西偏北 $45^{\\circ}$ 距离 $2 \\mathrm{~km}$ 处, 请标出图书馆。",
+        "options": [],
+        "answer": "①东；北； $40^{\\circ}$ ；西；南； $40^{\\circ}$ ；西；南； $20^{\\circ}$ ② 见详解",
+        "solution": "①将方向和距离结合起来描述位置时，要注意三个要素：一是观测点，二是方向，三是距离。(2) 弄清要标示的物体在哪个方位上, 有多少度, 按要求的方位和度数准确画图; 注意各场所离中心点的距离, 根据要求画出相应的长度。①小志家在学校的东偏北 $40^{\\circ}$ 或北偏东 $50^{\\circ}$ 方向上，学校在小志家的西偏南 $40^{\\circ}$ 或南偏西 $50^{\\circ}$ 方向上；小欣家在学校的西偏南 $20^{\\circ}$ 或南偏西 $70^{\\circ}$ 方向上。![](images/2d0fa881026b8a3369620e1cb261b5691a65e3fee3c43efabce1fce99f801bd4.jpg)",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·福建龙岩·期中",
+        "merged_id": "unit2_477",
+        "original_id": "unit2_fill_477"
+      },
+      {
+        "id": "unit2_fill_478",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "操作。![](images/79333d7685a3c95ccfa1fee9553b1c3f1473c88facdb02462e0baa96a2cabb5c.jpg)![](images/4570660459a4a0f1041f102a3068f6de7ce2cb29ac042276ea2aed0e9328eaf5.jpg)(1) 动物园在游乐场的（）偏（）（）°方向，距离游乐场（）m处。(2) 在图中标出下面地点的位置。迷宫在游乐场南偏西 $30^{\\circ}$ 方向 $1000\\mathrm{m}$ 处。花卉园在游乐场北偏西 $60^{\\circ}$ 方向 $1500\\mathrm{m}$ 处。",
+        "options": [],
+        "answer": "（1）北；东；45；1500 (2）见详解",
+        "solution": "（1）根据上北下南左西右东，以游乐场为观测点，即可确定动物园的方向，游乐场和动物园在图上相距 $3 \\mathrm{~cm}$ ，根据图上 $1 \\mathrm{~cm}$ 表示实际 $500 \\mathrm{~m}$ ，求出游乐场和动物园的实际距离，据此解答。 (2) 以游乐场为观测点, 即可分别确定迷宫和花卉园的方向, 进而求出 $1000 \\mathrm{~m}$ 和 $1500 \\mathrm{~m}$ 的图上距离, 据此作图。(1) $500 \\times 3 = 1500$ (m) 动物园在游乐场的北偏东 $45^{\\circ}$ （东偏北 $45^{\\circ}$ ）方向，距离游乐场 $1500 \\mathrm{~m}$ 处。(2) $1000 \\div 500 = 2$ (cm) $1 5 0 0 \\div 5 0 0 = 3 (\\mathrm {c m})$如图：![](images/a3e94cbf7bfd076203572d3b22c01111cec776ed735603e75a293b0243c9de78.jpg)![](images/ae6c91bbacadb45fd01806a67b09ef58c756f7de44d721432d447e946af1972f.jpg)",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·河南郑州·期中",
+        "merged_id": "unit2_478",
+        "original_id": "unit2_fill_478"
+      },
+      {
+        "id": "unit2_draw_479",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "根据下面的描述，在平台图上标出各场所的位置。![](images/02dabb4e1b1a9bde0d1e3363d7358290a3efbe988a45f96f3059939783390c1b.jpg)(1) 文化广场在电视塔的东偏南 $30^{\\circ}$ 方向 $1 \\mathrm{~km}$ 处。(2) 体育场在电视塔的西偏南 $45^{\\circ}$ 方向 $500 \\mathrm{~m}$ 。(3) 博物馆在电视塔的北偏西 $40^{\\circ}$ 方向 $1 \\mathrm{~km}$ 。(4) 动物园在电视塔的北偏东 $20^{\\circ}$ 方向 $1500 \\mathrm{~m}$ 。",
+        "options": [],
+        "answer": "(1)(2)(3)(4）见详解",
+        "solution": "根据上北下南，左西右东的方向分别确定文化广场、体育场、博物馆和动物园的方向，再根据图上1厘米表示实际距离500米，分别用实际距离除以500，求出它们的图上距离，即可画图。(1) $1\\mathrm{km} = 1000\\mathrm{m}$ ， $1000 \\div 500 = 2$ (cm)(2) $500 \\div 500 = 1$ (cm) (3) $1\\mathrm{km} = 1000\\mathrm{m}$ , $1000 \\div 500 = 2$ (cm) (4) $1500 \\div 500 = 3$ (cm)如图：![](images/04d9255a122ba81c087275e0fc7ff817409e46b07cf0feba932f25c017903ac7.jpg)",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·福建厦门·期中",
+        "merged_id": "unit2_479",
+        "original_id": "unit2_draw_479"
+      },
+      {
+        "id": "unit2_draw_480",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "下面是奇奇写的一篇关于参观消防站的日记。我从大门出发，向东行走100米到达执勤楼，再向正北方向行走100米到达模拟训练场。同行的妙妙则从大门出发，朝东偏北 $45^{\\circ}$ 方向行走，与我在模拟训练场相遇。请你根据奇奇的描述画出他们两人行走的路线。![](images/a35d31eb22985aceed47c474ff38d14b66f941e97bce52476cbb1fc3fb9faac7.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "solution": "根据题意可知, 1 厘米表示实际距离是 50 米, 先用除法计算出图上的长度; 再根据地图上方法的规定 \"上北下南, 左西右东\", 找准观测点, 确定下一个要到的地点的方向和距离，据此解答。$100 \\div 50 = 2$ （厘米）如图：![](images/57fba12964c8000e205566efaa519fb637ae2c260461d0e65f04b409b8a51582.jpg)",
+        "knowledge_points": [
+          "东、南、西、北方向",
+          "根据方向、角度和距离确定物体的位置",
+          "根据方向、角度和距离画线路图"
+        ],
+        "source": "24-25 六年级上·广东东莞·期中",
+        "merged_id": "unit2_480",
+        "original_id": "unit2_draw_480"
+      },
+      {
+        "id": "unit2_draw_481",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "公交车从 A 站开始先向西行驶 2 千米到达 B 站, 再向西偏南 $40^{\\circ}$ 方向行驶 4 千米到达 C 站, 最后向南偏东 $35^{\\circ}$ 方向行驶 3 千米, 到达终点 D 站,请你画出行驶路线图。(用 1 厘米的线段表示 1 千米)![](images/44deecf765d2e337bd7f04ff6b1e25af722630b92630ed5d714483927f84aed5.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "solution": "图上的方向是上北下南、左西右东，据题意图上1厘米表示实际1千米，先确定方向，然后根据实际距离确定图上距离，根据图上的方向、夹角的度数和图上距离画出路线图即可。$2 \\div 1 = 2$ （厘米）$4 \\div 1 = 4 (\\text {厘 米})$ $3 \\div 1 = 3 (\\text {厘 米})$作行驶路线图如下：![](images/542692c8c22f90aa8adb793482e878208ba8b808e10594b9aa3dc4e5cdeb2660.jpg)",
+        "knowledge_points": [
+          "根据方向、角度和距离画线路图"
+        ],
+        "source": "24-25 六年级上·河南南阳·期中",
+        "merged_id": "unit2_481",
+        "original_id": "unit2_draw_481"
+      },
+      {
+        "id": "unit2_fill_482",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "按要求完成下面各题。![](images/54ae19ff4331237e5cdaad0523646328cf82e4f530ad0150831fff5cb5160e66.jpg)(1) 大剧院在学校的东偏北 $45^{\\circ}$ 的方向上，与学校的距离是 800 米。请在平面图上确定并标出大剧院的位置。(2) 从学校出发, 向南偏 ( ) ( )°的方向走 ( ) 米, 到图书馆。(3) 从学校出发，向西偏（）（）°的方向走（）米，到公园。",
+        "options": [],
+        "answer": "(1) 见详解 (2）西；30；600 (3) 北；40；1000",
+        "solution": "以学校为观测点，以图上的\"上北下南，左西右东\"为准，图例表示图上1厘米相当于实际距离200米。(1) 在学校的东偏北 $45^{\\circ}$ 的方向上画 $800 \\div 200 = 4$ 厘米长的线段, 即是大剧院。 (2) 从图中可知, 图书馆与学校相距 3 厘米, 则实际相距 $200 \\times 3 = 600$ 米, 再根据方向、角度和距离描述图书馆与学校的位置关系。 (3) 从图中可知, 公园与学校相距 5 厘米, 则实际相距 $200 \\times 5 = 1000$ 米, 再根据方向、角度和距离描述公园与学校的位置关系。(1) $800 \\div 200 = 4$ （厘米）作图如下：![](images/772cfdf1705cc840c88bb4e11fb28e3bce874c2eb7871a99077a452c4a2365fe.jpg)(2) $200 \\times 3 = 600$ （米）从学校出发，向南偏西 $30^{\\circ}$ 的方向走 600 米，到图书馆。(3) $200 \\times 5 = 1000$ （米）从学校出发，向西偏北 $40^{\\circ}$ 的方向走 1000 米，到公园。",
+        "knowledge_points": [
+          "根据方向、角度和距离描述路线图",
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·河北唐山·期中",
+        "merged_id": "unit2_482",
+        "original_id": "unit2_fill_482"
+      },
+      {
+        "id": "unit2_draw_483",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "根据相关数据, 在平面图上用圆点标出各个飞镖的位置。(最小圆的半径为 4 米, 相邻两个圆之间的距离都是 4 米)| 飞镖 | 与中心点的位置 | 与中心点的距离 || ---- | -------------- | -------------- || A | 北偏西30° | 8米 || B | 东偏南40° | 4米 || C | 北偏东40° | 12米 || D | 西偏南30° | 16米 |![](images/ac1993e12af2f351d690c3c05059510f90781dc793252969c81f445655d32895.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "solution": "将方向和距离结合起来描述位置时，要注意三个要素：一是观测点，二是方向，三是距离。1、弄清要标示的物体在哪个方位上，有多少度，按要求的方位和度数准确画图； 2、注意各场所离中心点的距离，根据要求画出相应的长度。根据最小圆的半径为 4 米, 相邻两个圆之间的距离都是 4 米, A 点距离中心点 8 米, 应该在从里往外数第二个圆上; B 点距离中心点 4 米, 应该在从里往外数第一个圆上; C 点距离中心点 12 米, 应该在从里往外数第三个圆上; D 点距离中心点 16 米, 应该在从里往外数第四个圆上。$8 \\div 4 = 2$ （个） $ 4 \\div 4 = 1 \\quad (\\text {个}) $ $1 2 \\div 4 = 3 (\\text {个})$ $1 6 \\div 4 = 4 (\\text {个})$![](images/3231fc6e33fcb503efc8abeceeeb4c9f80750bfe1aadf01a769f91e964f5ee41.jpg)",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·河北邢台·期中",
+        "merged_id": "unit2_483",
+        "original_id": "unit2_draw_483"
+      },
+      {
+        "id": "unit2_draw_484",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "15路公共汽车从起点站先向正西方向行驶 $2\\mathrm{km}$ 后，接着向西偏北 $45^{\\circ}$ 方向行驶 $4\\mathrm{km}$ ，再向正东方向行驶 $8\\mathrm{km}$ ，最后向东偏南 $30^{\\circ}$ 方向行驶 $6\\mathrm{km}$到达终点站。根据上面的描述，画出15路公共汽车行驶的路线图。![](images/dff6cdb6e807a49e92fe23e62acea8c687ea22e88a3d85bffa76b3d3541d9bee.jpg)![](images/8ba43e16dea4763ff5b822bd94423b087b5de29261690c8606d0f09bf86c595b.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "solution": "用方向和距离结合来画路线时, 要注意三个要素: 一是观测点 (即参照物), 二是方向, 三是距离。地图上按上北下南左西右东确定方向, 图上 $1 \\mathrm{~cm}$ 表示实际 $2 \\mathrm{~km}$ , 实际距离 $\\div 2 =$ 要画的厘米数。$4 \\div 2 = 2$ （厘米）$8 \\div 2 = 4$ （厘米）$6 \\div 2 = 3$ （厘米）![](images/1dcd3e6b35ac888ad243e5d64e7c0ebd8400fd32691f704abcd2c494f7c317cf.jpg)",
+        "knowledge_points": [
+          "根据方向、角度和距离画线路图"
+        ],
+        "source": "23-24六年级上·江西抚州·期中",
+        "merged_id": "unit2_484",
+        "original_id": "unit2_draw_484"
+      },
+      {
+        "id": "unit2_draw_485",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "晓勇从家出发先向东偏北 $30^{\\circ}$ 方向走 300 米到少年宫，再向东走200米到体育馆，最后向东偏南 $45^{\\circ}$ 方向走200米就到了学校，根据描述画出路线图。计算出回家路程：（）。![](images/2051adf26acb62d12546a27c2d508de618bac3441137631c0bbb766fdf71719a.jpg)",
+        "options": [],
+        "answer": "路线图见详解；路程：700米",
+        "solution": "①图中1小段长度代表实际距离是100米，以晓勇家为观测点，先向东偏北 $30^{\\circ}$ 方向走300米到少年宫，也就是图中3小段长度；再向东走200米到体育馆，也就是图中2小段长度；最后向东偏南 $45^{\\circ}$ 方向走200米，也就是图中2小段长度，据此画出线路图。②晓勇从家出发到少年宫的距离是300米，从少年宫到体育馆的距离是200米，从体育馆到学校的距离是200米，因此他回家的总路程也就是这三段距离相加之和，即 $(300 + 200 + 200)$ 米。如图所示：![](images/46bcefd7bf2aede16138b292af51a5c22a31c6a6223aa96fa41208f4bef797b4.jpg)因此晓勇回家的路程是700米。",
+        "knowledge_points": [
+          "根据方向、角度和距离画线路图",
+          "两、三位数的连加运算"
+        ],
+        "source": "24-25 六年级上·新疆喀什·期中",
+        "merged_id": "unit2_485",
+        "original_id": "unit2_draw_485"
+      },
+      {
+        "id": "unit2_draw_486",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "张老师早上从家出发, 沿正东方向走 300 米到医院,再北偏东 $30^{\\circ}$ 方向走了 200 米到学校上课, 根据描述在如图中画出示意图。![](images/201d50b88500688634eefad1cda028416366fd8d446c7ce411bc10ffbaab9fae.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "solution": "以图上的 \"上北下南, 左西右东\" 为准, 图例表示图上 1 厘米相当于实际距离 100 米; 张老师早上从家出发, 在家的正东方向上画 $300 \\div 100 = 3$ 厘米长的线段, 即是医院;在医院的北偏东 $30^{\\circ}$ 方向上画 $200 \\div 100 = 2$ 厘米长的线段, 即是学校; 据此完成路线图。$300 \\div 100 = 3$ （厘米）$200 \\div 100 = 2$ （厘米）如图所示：![](images/a0ad92414025165897955d85a11aaaafd3c2880455d6069bea178184194971e4.jpg)",
+        "knowledge_points": [
+          "根据方向、角度和距离画线路图"
+        ],
+        "source": "23-24 六年级上·贵州遵义·期中",
+        "merged_id": "unit2_486",
+        "original_id": "unit2_draw_486"
+      },
+      {
+        "id": "unit2_fill_487",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "操作。![](images/895ebbf0386b5799be398c00cbf2607a514b0e6ae83f2f83549506b9f6851a5b.jpg)(1) 商场在小丽家的（）偏（）（）方向上，距离是（）米。(2) 银行在小丽家的南偏西 35 度方向上, 1200 米处, 请在平面图上标出银行的位置。",
+        "options": [],
+        "answer": "（1）北；西； $40^{\\circ}$ ；800 (2）见详解",
+        "solution": "（1）以小丽家为观测点，依照图上的方向规定：上北下南左西右东，结合角度，确定商场的方向，再根据图上距离与实际距离的换算，计算出小丽家到商场的实际距离，进而描述商场位置；(2) 根据银行位置的描述, 把实际距离换算成图上距离, 以小丽家为观测点, 结合方向和角度确定银行的位置并在图上表示即可。(1) $400 \\times 2 = 800$ （米）所以商场在小丽家的北偏西 $40^{\\circ}$ （西偏北 $50^{\\circ}$ ）方向上，距离是 800 米。(2) $1200 \\div 400 = 3$ （厘米）如图：![](images/2cf64e1a4ef3c046d1334792ff5df84d64b0292ec78c8288b2ef08af526e664a.jpg)",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·河南郑州·期中",
+        "merged_id": "unit2_487",
+        "original_id": "unit2_fill_487"
+      },
+      {
+        "id": "unit2_draw_488",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "15 路公共汽车从起点站向正西方向行驶 $2 \\mathrm{~km}$ 后, 再向西偏北 $40^{\\circ}$ 方向行驶 $4 \\mathrm{~km}$ , 最后向南偏西 $30^{\\circ}$ 方向行驶 $6 \\mathrm{~km}$ 到达终点站。根据上面的描述, 画出 15 路公共汽车行驶的路线图。![](images/8cb9599a81374bca135ee126ff412ae792f069bb60e2bb208309427c5d44de00.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "solution": "用方向和距离结合来画路线时, 要注意三个要素: 一是观测点 (即参照物), 二是方向, 三是距离。地图上按上北下南左西右东确定方向, 图上 $1 \\mathrm{~cm}$ 表示实际 $2 \\mathrm{~km}$ , 实际距离 $\\div 2 =$ 要画的厘米数。$2 \\div 2 = 1$ （厘米） $4 \\div 2 = 2 \\text {(厘 米)}$ $6 \\div 2 = 3 (\\text {厘 米})$![](images/bb319eb015e52174584e915a57a5578b2c52857421133490d71148a8efc2171b.jpg)",
+        "knowledge_points": [
+          "根据方向、角度和距离画线路图"
+        ],
+        "source": "23-24 六年级上·河南郑州·期中",
+        "merged_id": "unit2_488",
+        "original_id": "unit2_draw_488"
+      },
+      {
+        "id": "unit2_draw_489",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "按要求做题。为创建环保社会，倡导低碳生活，李老师骑自行车去书店，他先从家向北偏东 $40^{\\circ}$ 方向骑 $500\\mathrm{m}$ 到中国银行，再向正东方向骑 $250\\mathrm{m}$ 到万家乐超市，最后向东偏南 $30^{\\circ}$ 方向骑 $1000\\mathrm{m}$ 到书店。根据以上描述，画出李老师的骑行路线图。![](images/cc2797603cdafb89ec379e5b7457e73b4627481f6bc8a2d4e31ff73043df75d9.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "solution": "根据上北下南左西右东，结合距离以及角度画出路线图，根据题意，设定比例尺为 $1: 250$ ，所以李老师先从家向北偏东 $40^{\\circ}$ 方向骑 $500 \\mathrm{~m}$ 到中国银行，图上画出 $(500 \\div 250)$ 段线段，再向正东方向骑 $250 \\mathrm{~m}$ 到万家乐超市，图上画出 $(250 \\div 250)$ 段线段，最后向东偏南 $30^{\\circ}$ 方向骑 $1000 \\mathrm{~m}$ 到书店，图上画出 $(1000 \\div 250)$ 段线段。$500 \\div 250 = 2$ （段） $ 2 5 0 \\div 2 5 0 = 1 \\text {(段)} $ $1 0 0 0 \\div 2 5 0 = 4 \\text {(段)}$如图：![](images/8ba44f54d0cfc1c489740c56f405e3f99b9f816b1a2a6560ffb45a29ebf98bbb.jpg)",
+        "knowledge_points": [
+          "根据方向、角度和距离画线路图"
+        ],
+        "source": "23-24六年级上·河南安阳·期中",
+        "merged_id": "unit2_489",
+        "original_id": "unit2_draw_489"
+      },
+      {
+        "id": "unit2_draw_490",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "201路公交车从起始站出发，先向正南方向行驶了4000米到达人民医院，再向正东方向驶了2000米到达市中心，然后向正南方向行驶了1000米到达科技馆，最后向东偏南 $30^{\\circ}$ 的方向行驶了3000米到达终点站。(1) 根据上面的提示, 请你完成下面的路线示意图。![](images/bdcd6fa681dc1b0cea2efd4040834308048646212d5e67e51c2bcc5ed2a59b7f.jpg)(2) 201 路公交车从起始站到人民医院用了 15 分钟, 由人民医院到市中心用了 10 分钟, 由市中心到科技馆用了 9 分钟, 由科技馆到终点站用了 6 分钟。该路公交车单程的平均速度是多少?",
+        "options": [],
+        "answer": "①图见详解 ②$ 250米",
+        "solution": "①图上的方向是上北下南、左西右东，据题意图上1厘米表示实际1000米，先确定方向，然后根据实际距离确定图上距离，根据图上的方向、夹角的度数和图上距离画出路线图即可。(2) 根据题意, 作路线示意图如下: $4000 + 2000 + 1000 + 3000 = 10000$ 米, 总时间是 $15 + 10 + 9 + 6 = 40$ 分钟。根据速度 $=$ 路程 $\\div$ 时间, 用 $10000 \\div 40$ 即可求出单程的平均速度。① $4000 \\div 1000 = 4$ （厘米）$3000 \\div 1000 = 3$ （厘米）$2000 \\div 1000 = 2$ （厘米）根据分析，作路线示意图如下：![](images/cc13538728ad5d6cf6e8ef6b6aced1250e047c39a4a4d66a996a67e322d03639.jpg)(2) $(4000 + 2000 + 1000 + 3000) \\div (15 + 10 + 9 + 6)$ $= 1 0 0 0 0 \\div 4 0 = 2 5 0 \\text {(米)} $答：该路公交车单程的平均速度是每分钟行250米。",
+        "knowledge_points": [
+          "基础行程问题",
+          "根据方向、角度和距离画线路图"
+        ],
+        "source": "23-24六年级上·福建厦门·期中",
+        "merged_id": "unit2_490",
+        "original_id": "unit2_draw_490"
+      },
+      {
+        "id": "unit2_draw_491",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "根据下面的描述, 画出公交车行驶路线图。一辆公交车从总站出发，先向正东方向行驶5千米到达幸福花园站，再向东偏北 $45^{\\circ}$ 方向行驶8千米到达飞跃电脑城，再向东偏南 $40^{\\circ}$ 方向行驶6千米到达家具城，最后向东偏北 $35^{\\circ}$ 方向行驶4千米到达人民公园。公交总站",
+        "options": [],
+        "answer": "见详解",
+        "solution": "以图上的 \"上北下南, 左西右东\" 为准, 确定图例表示图上 1 厘米相当于实际距离的 1 千米;公交车从总站出发，先向正东方向行驶 5 千米到达幸福花园站，即以总站为观测点，在正东方向上画 $5 \\div 2 = 2.5$ 厘米长的线段，即是幸福花园站；再向东偏北 $45^{\\circ}$ 方向行驶 8 千米到达飞跃电脑城, 是以幸福花园站为观测点, 在东偏北 $45^{\\circ}$ 方向上画 $8 \\div 2 = 4$ 厘米长的线段, 即是飞跃电脑城;再向东偏南 $40^{\\circ}$ 方向行驶 6 千米到达家具城, 是以飞跃电脑城为观测点, 在东偏南 $40^{\\circ}$ 方向画 $6 \\div 2 = 3$ 厘米长的线段, 即是家具城;最后向东偏北 $35^{\\circ}$ 方向行驶 4 千米到达人民公园, 是以家具城为观测点, 在东偏北 $35^{\\circ}$ 方向画 $4 \\div 2 = 2$ 厘米长的线段, 即是人民公园;据此画出公交车行驶路线图。如图：![](images/e6659804242d31e74d19660c1c032f123780f51a02327a5389cb4c576b91eaef.jpg)",
+        "knowledge_points": [
+          "根据方向、角度和距离画线路图"
+        ],
+        "source": "22-23 六年级上·江西南昌·期中",
+        "merged_id": "unit2_491",
+        "original_id": "unit2_draw_491"
+      },
+      {
+        "id": "unit2_draw_492",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "较难",
+        "question": "商店在汽车站东偏北 $60^{\\circ}$ 方向 300 米处, 体育馆在汽车站南偏西 $30^{\\circ}$ 方向 200 米处。(画出它们的位置)![](images/b21479bf6b8c2f3e31d54fa858c8a838519b0673cb132ea321f0358081c373dd.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "solution": "根据比例尺可知：图上1厘米相当于实际100米， $300 \\div 100 = 3$ （厘米）， $200 \\div 100 = 2$ （厘米），将商店和体育馆分别到汽车站的实际距离换算成图上距离。再根据地图方向规定的\"上北下南，左西右东\"，以汽车站为观测点，分别确定商店和体育馆的位置。$300 \\div 100 = 3$ （厘米）， $200 \\div 100 = 2$ （厘米）如图所示：![](images/612412d5e0d499bc8be840d36d860d1369337fbe37e6f6a05384449d534339c7.jpg)",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24 六年级上·湖南岳阳·期中",
+        "merged_id": "unit2_492",
+        "original_id": "unit2_draw_492"
+      },
+      {
+        "id": "unit2_fill_493",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "下面是笑笑周日早晨跑步的路线图。![](images/33aee301d8cacaf001abb70c811c16a4662bbbceb89bf2d9b272e4511f85b830.jpg)(1) 笑笑从家出发，向（ ）（ ）°方向，跑（ ）米到（ ）；再向（ ）（ ）°方向，跑（ ）米到（ ）；最后向（ ）方向，跑（ ）米到东山广场。(2）医院在东山广场东偏北 $40^{\\circ}$ 方向1000米处，请在图上标出医院的位置。",
+        "options": [],
+        "answer": "（1）西偏北；44；1500；西偏南；35；1000；西；1250 (2）见详解",
+        "solution": "（1）确定线路时，注意起始点与目的地，起始点是观测点，再按照地图上\"上北下南，左西右东\"确定方向，用方向、角度和距离描述笑笑跑步的路线。 (2) 1 厘米表示 250 米, 计算出医院到中山广场的图上距离, 再以中山广场为观测点, 画出医院的位置。（1）笑笑家到菜市场的实际距离： $250\\times 6 = 1500$ （米）菜市场到儿童乐园的实际距离$250\\times 4 = 1000$ （米）儿童乐园到东山公园的实际距离： $250\\times 5 = 1250$ （米）$9 0 ^ {\\circ} - 4 4 ^ {\\circ} = 4 6 ^ {\\circ} $ $ 9 0 ^ {\\circ} - 3 5 ^ {\\circ} = 5 5 ^ {\\circ}$ 笑笑从家出发，向西偏北 $44^{\\circ}$ （或北偏西 $46^{\\circ}$ ）方向，跑1500米到菜市场；再向西偏南 $35^{\\circ}$ （南偏西 $55^{\\circ}$ ）方向，跑1000米到儿童乐园；最后向西向，跑1250米到东山公园。(2) $1000 \\div 250 = 4$ （厘米）如图:![](images/2d36a69bc1dafd5fcee9bce839d29e55f1d09f450e9f6fc4a9ec5b6d55096bec.jpg)",
+        "knowledge_points": [
+          "根据方向、角度和距离描述路线图",
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·河南驻马店·期中",
+        "merged_id": "unit2_493",
+        "original_id": "unit2_fill_493"
+      },
+      {
+        "id": "unit2_draw_494",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "根据下面的描述, 在平面图上标出各场所的位置。(1) 少年宫在广场正北方向, 距离 200 米; (2) 人民公园在广场东偏北 $35^{\\circ}$ 方向, 距离 300 米; (3) 宾馆在广场南偏东 $40^{\\circ}$ 方向, 距离 400 米; (4) 医院在广场西偏南 $50^{\\circ}$ 方向, 距离 350 米。![](images/79202866de66eaee92de8dd5128226139825078fd2680aa0b0e5f33dad7ee140.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "solution": "以广场为观测点，以图上的\"上北下南，左西右东\"为准，图例表示图上1厘米相当于实际距离100米。(1) 在广场正北方向上画 $200 \\div 100 = 2$ 厘米长的线段，即是少年宫； (2) 在广场东偏北 $35^{\\circ}$ 方向上画 $300 \\div 100 = 3$ 厘米长的线段, 即是人民公园; (3) 在广场南偏东 $40^{\\circ}$ 方向上画 $400 \\div 100 = 4$ 厘米长的线段, 即是宾馆; (4) 在广场西偏南 $50^{\\circ}$ 方向上画 $350 \\div 100 = 3.5$ 厘米长的线段, 即是医院。如图：![](images/6658806b922174553347f9114581d5f624826214dd4f5502bf3826ad7b433c74.jpg)",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "24-25 六年级上·河北邯郸·期中",
+        "merged_id": "unit2_494",
+        "original_id": "unit2_draw_494"
+      },
+      {
+        "id": "unit2_draw_495",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "\"1路公共汽车从起点站向西偏北 $50^{\\circ}$ 方向行驶 4KM ，然后向西行驶 ，最后向南偏西 方向行驶 到达终点站。\"根据上面的描述，把1路公共汽车行驶的路线图画完整。![](images/f008099f160fdfb2f7fd497ebb8bfa180802b6dc2c08e3c705ee33afbe65c5d5.jpg)$4\\mathrm{km}$ ，然后向西行驶 $3\\mathrm{km}$ ，最后向南偏西 $30^{\\circ}$ 方向行驶 $2\\mathrm{km}$ 到达终点站。\"根据上面的描述，",
+        "options": [],
+        "answer": "见详解",
+        "solution": "图中一小段距离代表 $1 \\mathrm{~km}$ , 以起点站为观测点, 公共汽车先向西偏北 $50^{\\circ}$ 方向行驶 $4 \\mathrm{~km}$ , 也就是 4 段距离; 然后向西行驶 $3 \\mathrm{~km}$ , 也就是 3 段距离; 最后向南偏西 $30^{\\circ}$ 方向行驶 $2\\mathrm{km}$ ，也就是2段距离；据此画出公共汽车行驶的路线图。如图所示：![](images/743d0f4fc932f3779f93a8718e877971fd5e5e315943f530f5026a2525259ddf.jpg)",
+        "knowledge_points": [
+          "根据方向、角度和距离画线路图"
+        ],
+        "source": "23-24六年级上·河南安阳·期末",
+        "merged_id": "unit2_495",
+        "original_id": "unit2_draw_495"
+      },
+      {
+        "id": "unit2_draw_496",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "根据下面的描述, 在平面图上标出各场所的位置。![](images\\aebd06a5-bcaf-401f-9e4e-0493ff69819b.png)(1) 聪聪家在公园北偏西 $30^{\\circ}$ 方向 450 米处。(2) 思思家在公园南偏西 $45^{\\circ}$ 方向 900 米处。(3) 涵涵家在公园南偏东 $60^{\\circ}$ 方向 600 米处。(4) 晶晶家在公园北偏东 $40^{\\circ}$ 方向 300 米处。",
+        "options": [],
+        "answer": "见详解",
+        "solution": "将方向和距离结合起来描述位置时，要注意三个要素：一是观测点，二是方向，三是距离。弄清要标示的物体在哪个方位上，有多少度，按要求的方位和度数准确画图；注意各场所离中心点的距离，根据要求画出相应的长度。$450 \\div 300 = 1.5$ （厘米） $9 0 0 \\div 3 0 0 = 3 (\\text {厘 米})$ $600 \\div 300 = 2$ （厘米）300÷300＝1（厘米）$300 \\div 300 = 1​$ （厘米）![](images/c03f04ce1a79a3e5f2e8b338c67f76e58dc3cf2c3d2fed2ac4786860df54fa4c.jpg)",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "21-22 六年级上·河北保定·期末",
+        "merged_id": "unit2_496",
+        "original_id": "unit2_draw_496"
+      },
+      {
+        "id": "unit2_draw_497",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "作图。![](images/f268cf2dedc5b3728d8b1233862d5c98cf3956c332dc2d6976c8cc55ce825220.jpg)某勘探队在 A 城南偏西 $40^{\\circ}$ 约 60 千米处发现稀有金属矿, 请在图中标出该金属矿的位置。",
+        "options": [],
+        "answer": "见详解",
+        "solution": "根据纸上方向：上北下南，左西右东。由图可知，图上1厘米表示实际距离20千米，用60除以20得可用图上3厘米表示，再根据方向角度和距离在图中标出金属矿的位置。$60 \\div 20 = 3$ （厘米）作图如下：![](images/4b47dea42f11da56e108e36921edebb37d06180e9fb266392da89bb06ea1eac7.jpg)",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·河北邢台·期中",
+        "merged_id": "unit2_497",
+        "original_id": "unit2_draw_497"
+      },
+      {
+        "id": "unit2_draw_498",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "根据下面的描述，画出平面示意图。以学校为观测点，小明家在北偏东 $60^{\\circ}$ 方向400米处；书店在西偏北 $45^{\\circ}$ 方向300米处；邮局在南偏西 $50^{\\circ}$ 方向200米处。![](images/6a48222486893a3c8de76c87e9f964a1629b196fafc3e195af56f26eccc5175a.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "solution": "根据题意可知, 1 厘米表示 100 厘米; 先计算出学校到小明家、到书店、到邮局的图上距离; 再根据地图上方向的规定 \"上北下南, 左西右东\", 以学校为观测点, 画出小明家、书店、邮局的位置。学校到小明家： $400\\div 100 = 4$ （厘米）学校到书店： $300 \\div 100 = 3$ （厘米）学校到邮局： $200 \\div 100 = 2$ （厘米）如下图：![](images/553b88c8659005ce1eb376785bd3190c4684050281eef1e0ed81dfa02e5691d4.jpg)",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·山东临沂·期中",
+        "merged_id": "unit2_498",
+        "original_id": "unit2_draw_498"
+      },
+      {
+        "id": "unit2_draw_499",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "根据下面的描述，在平面图上标出各场所的位置。![](images/ecc7482a55fbc43f4f53ebd97bf94642bff6de5f8e3ec3b8de65d95ac4346ee2.jpg)(1) 教学楼在校标的正北方向 100 米处。(2) 图书馆在校标的北偏东 $45^{\\circ}$ 方向上, 距离是 200 米。(3) 运动场在校标的西偏南 $30^{\\circ}$ 方向上, 距离是 150 米。(4) 实验楼在校标的东偏南 $40^{\\circ}$ 方向上, 距离是 250 米。",
+        "options": [],
+        "answer": "见详解",
+        "solution": "以\"向东偏北 $30^{\\circ}$ 方向 3 米是 A 点\"为例，以坐标交点为顶点，以正东方向为一条边向北作一个 $30^{\\circ}$ 的角，画出这个角的另一条边，再画出 3 米的距离就是 A 点。据此以校标为起点作出所要求的方位角，并画出校标到各场所的距离，据此作图。(1) $100 \\div 50 = 2$ （厘米），以校标为起点向正北方向画出 100 米就是教学楼，如图所示；(2) $200 \\div 50 = 4$ (厘米), 以校标为起点, 以正北方向为一条边向东作一个 $45^{\\circ}$ 的角, 画出这个角的另一条边, 再画出 200 米的距离就是图书馆, 如图所示;(3) $150 \\div 50 = 3$ (厘米), 以校标为起点, 以正西方向为一条边向南作一个 $30^{\\circ}$ 的角, 画出这个角的另一条边, 再画出 150 米的距离就是运动场, 如图所示; (4) $250 \\div 50 = 5$ (厘米), 以校标为起点, 以正东方向为一条边向南作一个 $40^{\\circ}$ 的角, 画出这个角的另一条边, 再画出 250 米的距离就是图书馆, 如图所示。![](images/b6e7b3ebac6db01a19fd0a97ccd092b3ba699f026d602df972096872f2f49e34.jpg)",
+        "knowledge_points": [
+          "根据方向、角度和距离确定物体的位置"
+        ],
+        "source": "23-24六年级上·山东临沂·期中",
+        "merged_id": "unit2_499",
+        "original_id": "unit2_draw_499"
+      },
+      {
+        "id": "unit2_solve_500",
+        "unit": "第二单元：位置与方向",
+        "chapter": "位置与方向综合练习",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "下面是某部队行军路线图, 请写出从军营到 1 号目的地的行军路线。![](images/90cd1149602f8dc37a981823cba19f40eb0cde621353593aab7abb9105ed7fbc.jpg)",
+        "options": [],
+        "answer": "从军营出发，先向西偏北 $50^{\\circ}$ 方向走 3 千米，再向西走 3 千米，然后向南偏西 $30^{\\circ}$ 方向走 1 千米，最后向东偏南 $45^{\\circ}$ 方向走 2 千米到达 1 号目的地",
+        "solution": "先根据\"上北下南，左西右东\"确定行走的方向，再根据给出的角度确定方向对应的角度，最后根据图中给出的距离确定具体的行走路程，据此解答。答：从军营出发，先向西偏北 $50^{\\circ}$ 方向走 3 千米，再向西走 3 千米，然后向南偏西 $30^{\\circ}$ 方向走 1 千米，最后向东偏南 $45^{\\circ}$ 方向走 2 千米到达 1 号目的地。",
+        "knowledge_points": [
+          "根据方向、角度和距离描述路线图"
+        ],
+        "source": "24-25 六年级上·广西河池·期中",
+        "merged_id": "unit2_500",
+        "original_id": "unit2_solve_500"
+      }
+    
+      
+  ]
+  
\ No newline at end of file
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new file mode 100644
index 0000000000000000000000000000000000000000..7b8ecf8e0e92a4916abf646926563fa79f41ba3e
--- /dev/null
+++ "b/\345\205\255\345\271\264\347\272\247\351\242\230\351\233\206/\345\205\255\345\271\264\347\272\247\344\270\212/\345\205\255\345\271\264\347\272\247\344\270\212\347\254\254\344\272\224\345\215\225\345\205\203\345\234\206\346\225\260\346\215\256\351\233\206.json"
@@ -0,0 +1,9409 @@
+[
+  {
+    "id": "unit5_choice_001",
+    "unit": "第五单元：圆",
+    "chapter": "圆的认识",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "下列说法正确的是 ( )。",
+    "options": [
+      "A. π = 3.14",
+      "B. 直径是半径的 2 倍",
+      "C. 直径是圆内最长的线段"
+    ],
+    "answer": "C",
+    "solution": "【分析】判断各选项需结合圆的性质和 π 的特点。π 是无限不循环小数, 直径是半径 2 倍的前提是同圆/等圆, 直径是圆内最长线段, 据此分析。【详解】A. π 是无限不循环小数, 3.14 是近似值, 不是相等, 错误。B. '直径是半径的 2 倍' 需在同圆或等圆中, 题目未提及,错误。C. 直径是圆内最长的线段, 正确。故答案为: C",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "圆的周长"
+    ],
+    "source": "25-26 六年级上·全国·单元测试",
+    "merged_id": "unit5_001",
+    "original_id": "unit5_choice_001"
+  },
+  {
+    "id": "unit5_choice_002",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一个圆桌的直径是 150 cm , 在圆桌边上要求每隔 50 cm 摆一套餐具, 为了保证间距, 最多可以摆餐具 ( ) 套。",
+    "options": [
+      "A. 8",
+      "B. 9",
+      "C. 10",
+      "D. 11"
+    ],
+    "answer": "B",
+    "solution": "【分析】先根据 'C圆形 = πd' 求出圆桌的周长, 可以摆餐具的套数 = 圆桌的周长 ÷ 间距, 最后结果取整数, 据此解答。【详解】150 × 3.14 = 471 (cm)  471 ÷ 50 ≈ 9 (套)所以，最多可以摆餐具9套。故答案为: B",
+    "knowledge_points": [
+      "圆的周长的应用",
+      "小数与整数的乘法"
+    ],
+    "source": "23-24 六年级下·天津河东·期末",
+    "merged_id": "unit5_002",
+    "original_id": "unit5_choice_002"
+  },
+  {
+    "id": "unit5_choice_003",
+    "unit": "第五单元：圆",
+    "chapter": "圆的认识",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "下面图形中，对称轴最多的是（）。",
+    "options": [
+      "A. 长方形",
+      "B. 正方形",
+      "C. 等边三角形",
+      "D. 圆"
+    ],
+    "answer": "D",
+    "solution": "【分析】轴对称图形的意义: 如果一个图形沿着一条直线对折后两部分完全重合, 这样的图形叫做轴对称图形, 这条直线叫做对称轴; 依次进行判断即可。【详解】A.![](images/258ed1da34922b0cd3e24f869ca36f7b530e18582e8a518478586b724f8ba5d8.jpg)长方形有2条对称轴;B.![](images/7120c3fb4c401fb4bef2f2d024d8cc0210bc4c88b849cb5834a2d8edca99fdf7.jpg)正方形有 4 条对称轴;C![](images/007ab50a71636ff799ea621a34cdcc2c16fea034e7d84d3a70768e9a829e855f.jpg)等边三角形有3条对称轴;D.![](images/4b423ba726635098135533779ca5b352b2b1a8603a3ba1335822a2f5bcee0ee4.jpg)圆有无数条对称轴。上面图形中，对称轴最多的是圆，有无数条对称轴。故答案为: D",
+    "knowledge_points": [
+      "正方形的概念及特点",
+      "圆的概念及特点",
+      "长方形的概念及特点",
+      "对称轴的画法及数量"
+    ],
+    "source": "2022·重庆大渡口·小升初真题",
+    "merged_id": "unit5_003",
+    "original_id": "unit5_choice_003"
+  },
+  {
+    "id": "unit5_choice_004",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "我国伟大的数学家 ( )计算出圆周率应在 3.1415926 和 3.1415927 之间，成为世界上第一个把圆周率的值精确到小数点 7 位的人。",
+    "options": [
+      "A. 刘徽",
+      "B. 祖冲之",
+      "C. 杨辉",
+      "D. 朱世杰"
+    ],
+    "answer": "B",
+    "solution": "【分析】中国数学家祖冲之是世界上第一次把圆周率精确到小数点后第七位数字的人，比外国早了近一千年，他推算出圆周率的数值在 3.1415926 到 3.1415927 之间，也就是精确到小数点后第七位。【详解】由分析可知：我国伟大的数学家祖冲之计算出圆周率应在 3.1415926 和3.1415927 之间，成为世界上第一个把圆周率的值精确到小数点 7 位的人。故答案为: B",
+    "knowledge_points": [
+      "圆的周长"
+    ],
+    "source": "2024·河北石家庄·小升初真题",
+    "merged_id": "unit5_004",
+    "original_id": "unit5_choice_004"
+  },
+  {
+    "id": "unit5_choice_005",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "圆周率是一个 ( ) 小数。",
+    "options": [
+      "A. 循环",
+      "B. 有限",
+      "C. 无限不循环"
+    ],
+    "answer": "C",
+    "solution": "【分析】小数分为有限小数和无限小数，无限小数又分为无限循环小数和无限不循环小数。有限小数的小数点后面的小数是有限的、可数的；而无限小数的小数点后面的小数是无限的、不可数的。一个数的小数部分从某一位起，一个数字或几个数字依次不断重复出现，这样的小数叫循环小数。任意一个圆的周长与它的直径的比值是一个固定的数，我们把它叫做圆周率。圆周率用字母 π 表示，π = 3.1415926⋯，它是一个无限不循环小数，但在实际应用中通常只取它的近似值 3.14。【详解】根据分析可知, 圆周率是一个无限不循环小数。故答案为: C",
+    "knowledge_points": [
+      "有限小数和无限小数的认识",
+      "圆的周长",
+      "循环小数的认识与简写"
+    ],
+    "source": "24-25 六年级下·海南海口·单元测试",
+    "merged_id": "unit5_005",
+    "original_id": "unit5_choice_005"
+  },
+  {
+    "id": "unit5_choice_006",
+    "unit": "第五单元：圆",
+    "chapter": "圆的认识",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "《墨经》中记载: '圆, 一中同长也'。以下选项与这句话隐藏的数学道理不相关的是 ( )。",
+    "options": [
+      "A. 丢手绢游戏时同学围成圆形",
+      "B. 车轮做成圆形",
+      "C. 下水道圆形沙井盖",
+      "D. 圆的周长是半径的 2π 倍"
+    ],
+    "answer": "D",
+    "solution": "【分析】《墨经》中记载: '圆, 一中同长也'。由圆的特征可知, 连接圆心和圆上任意一点的线段叫作半径, 同圆的半径都相等, 找出没有应用到圆的这一特征的选项即可。【详解】A．丢手绢游戏时同学围成圆形更公平，利用了圆心到圆上的距离都相等这一特征；B. 车轮做成圆形骑起来比较平稳, 利用了圆心到圆上的距离都相等这一特征;  C. 下水道圆形沙井盖主要利用了同圆的半径都相等这一特征;D. 圆的周长是半径的 2π 倍, 虽然圆的周长与半径有关, 但主要涉及的是圆周率, 而不是同圆的半径都相等这一特征。故答案为: D",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "圆的周长"
+    ],
+    "source": "24-25 六年级上·广东云浮·期末",
+    "merged_id": "unit5_006",
+    "original_id": "unit5_choice_006"
+  },
+  {
+    "id": "unit5_choice_007",
+    "unit": "第五单元：圆",
+    "chapter": "圆的认识",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "数学课上，同学们将做好的硬纸板'车轮'沿直尺的边滚一滚，描出圆心留下的痕迹，如下图所示。为什么圆心留下的痕迹是直线？下面解释最合理的是（）。![](images/92c2079d7ebd4ac15206d40320d2947b6ab866db6201b1da28575057284ae2bb.jpg)",
+    "options": [
+      "A. 圆心决定圆的位置",
+      "B. 圆是曲边图形, 它的边很光滑",
+      "C. 同一个圆中所有的半径都相等",
+      "D. 圆是轴对称图形"
+    ],
+    "answer": "C",
+    "solution": "【分析】圆上的点到圆心的距离都是相等的，这个相等的距离就是圆的半径；当车轮在直尺的边滚动时，车轮的圆心到直尺边上的距离总是等于车轮的半径，因此圆心留下的痕迹是直线。【详解】圆心留下的痕迹的直线的原因是: 车轮的圆心到车轮边上任意一点的距离都是相等的, 这个相等的距离就是车轮的半径, 即同一个圆中所有的半径都相等。故答案为: C",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "24-25六年级上·北京海淀·期末",
+    "merged_id": "unit5_007",
+    "original_id": "unit5_choice_007"
+  },
+  {
+    "id": "unit5_choice_008",
+    "unit": "第五单元：圆",
+    "chapter": "圆的认识",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "学校田径队要选拔实心球男运动员，成绩达到校队标准可以进入校队。李老师用一条与校队标准等长的绳子固定在中心点O点画一条弧线，快速筛选出了入选的运动员。李老师运用了圆的（）性质。![](images/81b9222a90fc5510a09f54c17db61539f0bb3356733c53ecc70e7a950ec1e4b4.jpg)",
+    "options": [
+      "A. 圆是轴对称图形",
+      "B. 同一个圆内, 直径的长度是半径的2倍",
+      "C. 圆的周长与直径的比值相等",
+      "D. 同一个圆的所有半径径相等"
+    ],
+    "answer": "D",
+    "solution": "【分析】圆的特性：圆的位置由圆心决定，圆心的位置决定了圆的位置。在同一个圆内，所有的半径都相等，所有的直径也相等，直径的长度是半径的 2 倍。圆是轴对称图形, 任意一条直径所在的直线就是圆的对称轴。【详解】李老师用一条与校队标准等长的绳子固定在中心点O点画一条弧线，快速筛选出了入选的运动员。李老师运用了圆的（同一个圆的所有半径相等）性质。故答案为: D",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "24-25 六年级上·吉林·期末",
+    "merged_id": "unit5_008",
+    "original_id": "unit5_choice_008"
+  },
+  {
+    "id": "unit5_choice_009",
+    "unit": "第五单元：圆",
+    "chapter": "半圆的周长",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一个半圆形的周长是它半径的 ( ) 倍。",
+    "options": [
+      "A. 2π",
+      "B. 1/2π + 1",
+      "C. π + 2"
+    ],
+    "answer": "C",
+    "solution": "【分析】设半圆形的半径为 r , 根据半圆形的周长 = π × 半径 + 2 × 半径, 求出半圆形的周长, 再用半圆形的周长除以半径, 即可解答。【详解】设半圆形的半径为 r 。(π × r + 2r) ÷ r = r (π + 2) ÷ r = π + 2 一个半圆形的周长是它半径的 (π + 2) 倍。故答案为: C",
+    "knowledge_points": [
+      "半圆的周长"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit5_009",
+    "original_id": "unit5_choice_009"
+  },
+  {
+    "id": "unit5_choice_010",
+    "unit": "第五单元：圆",
+    "chapter": "圆的认识",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "下面说法错误的是 ( )。",
+    "options": [
+      "A. 圆心决定圆的位置",
+      "B. 半径决定圆的大小",
+      "C. 直径是半径的 2 倍"
+    ],
+    "answer": "C",
+    "solution": "【分析】圆心决定圆的位置；半径决定圆的大小；在同一个圆内，直径是半径的2倍。据此判断。【详解】圆心决定圆的位置；半径决定圆的大小，所以A、B 两个选项的说法正确;在同一个圆内，直径是半径的 2 倍，所以 C 选项说法错误。故答案为: C",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit5_010",
+    "original_id": "unit5_choice_010"
+  },
+  {
+    "id": "unit5_choice_011",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "一座台钟, 钟面上分针的长度是 1 dm , 这根分针走 60 分钟, 针尖走过的路程是 ( ) dm 。",
+    "options": [
+      "A. 3.14",
+      "B. 6.28",
+      "C. 37.68",
+      "D. 12.56"
+    ],
+    "answer": "B",
+    "solution": "【分析】分针走 60 分钟就是走了一个圆, 这个圆的半径就是分针的长度, 则针尖走过的路程就是圆的周长, 圆的周长 = 2πr , 代入数据计算得出周长。【详解】2 × 3.14 × 1 = 6.28 (dm)针尖走过的路程是 6.28 dm 。故答案为: B",
+    "knowledge_points": [
+      "圆的周长的应用",
+      "圆的周长"
+    ],
+    "source": "24-25 六年级上·全国·课后作业",
+    "merged_id": "unit5_011",
+    "original_id": "unit5_choice_011"
+  },
+  {
+    "id": "unit5_choice_012",
+    "unit": "第五单元：圆",
+    "chapter": "半圆的周长",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "把一个直径是 10 cm 的圆沿直径分成两个半圆，这两个半圆的周长之和是（）。",
+    "options": [
+      "A. 31.4 cm",
+      "B. 41.4 cm",
+      "C. 51.4 cm",
+      "D. 61.4 cm"
+    ],
+    "answer": "C",
+    "solution": "【分析】根据题意, 把一个圆沿直径分成两个半圆, 根据半圆的周长 = 圆周长的一半 + 直径 = πd ÷ 2 + d , 据此求出一个半圆的周长, 再乘 2 , 即是这两个半圆的周长之和。【详解】3.14 × 10 ÷ 2 + 10 = 15.7 + 10 = 25.7 (cm) 25.7 × 2 = 51.4 (cm)这两个半圆的周长之和是 51.4 cm 。故答案为: C",
+    "knowledge_points": [
+      "半圆的周长"
+    ],
+    "source": "24-25 六年级上·全国·单元测试",
+    "merged_id": "unit5_012",
+    "original_id": "unit5_choice_012"
+  },
+  {
+    "id": "unit5_choice_013",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "若一个圆的半径由 3cm 增加到 6cm ，则它的面积（）。",
+    "options": [
+      "A. 扩大到原来的 3 倍",
+      "B. 扩大到原来的 9 倍",
+      "C. 扩大到原来的 4 倍",
+      "D. 扩大到原来的 8 倍"
+    ],
+    "answer": "C",
+    "solution": "【分析】圆的面积 = 圆周率 × 半径的平方，一个圆的半径扩大到原来的几倍，面积扩大到原来的倍数 × 倍数，据此分析。【详解】6 ÷ 3 = 2  2 × 2 = 4 若一个圆的半径由 3 cm 增加到 6 cm , 则它的面积扩大到原来的 4 倍。故答案为: C",
+    "knowledge_points": [
+      "圆的面积"
+    ],
+    "source": "24-25六年级上·全国·单元测试",
+    "merged_id": "unit5_013",
+    "original_id": "unit5_choice_013"
+  },
+  {
+    "id": "unit5_choice_014",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "时钟的时针长10cm，从下午3时到晚上9时，时针走过钟面的面积是（） cm² 。",
+    "options": [
+      "A. 10 × 10 × π",
+      "B. 10 × 10 × π × 1/2",
+      "C. 10 × 10 × π × 1/3",
+      "D. 10 × 10 × π × 1/4"
+    ],
+    "answer": "B",
+    "solution": "【分析】时针长度相当于圆的半径，从下午3时到晚上9时，时针走过钟面的 1/2 ，时针走过钟面的面积相当于半圆的面积，根据半圆的面积 = 圆周率 × 半径的平方 × 1/2 ，列式即可。【详解】10² × π × 1/2 = 10 × 10 × π × 1/2 与 B 选项一样 = 50π (cm²) 时针走过钟面的面积是 50π cm² 。故答案为: B",
+    "knowledge_points": [
+      "圆的面积的应用"
+    ],
+    "source": "24-25 六年级上·全国·单元测试",
+    "merged_id": "unit5_014",
+    "original_id": "unit5_choice_014"
+  },
+  {
+    "id": "unit5_choice_015",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "两个圆的面积不相等，原因是它们（）。",
+    "options": [
+      "A. 圆心的位置不同",
+      "B. 圆周率不同",
+      "C. 直径不相等",
+      "D. 画法不同"
+    ],
+    "answer": "C",
+    "solution": "【分析】圆心决定圆的位置，半径决定圆的大小，同圆或等圆中直径是半径的 2 倍。据此解答。【详解】A．圆心只决定圆的位置，不能决定圆的面积大小，原因错误；B. 所有圆的圆周率都相同, 原因错误; C. 两个圆的直径不相等, 则它们的半径不相等, 那么两个圆的面积就不相等, 原因正确; D. 圆的画法不同, 不能决定圆的面积大小, 原因错误。故答案为: C",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "圆的面积的应用"
+    ],
+    "source": "24-25 六年级上·全国·单元测试",
+    "merged_id": "unit5_015",
+    "original_id": "unit5_choice_015"
+  },
+  {
+    "id": "unit5_choice_016",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "大圆的周长是小圆周长的 2 倍, 小圆的面积是 6 dm² , 大圆的面积是 ( )。",
+    "options": [
+      "A. 12 dm²",
+      "B. 24 dm²",
+      "C. 36 dm²",
+      "D. 18 dm²"
+    ],
+    "answer": "B",
+    "solution": "【分析】已知大圆的周长是小圆周长的 2 倍, 即大圆与小圆的周长之比是 2: 1 ; 根据圆的周长公式 C = 2πr 可知, 大圆与小圆的半径之比也是 2: 1 ; 根据圆的面积公式 S = πr² 可知,大圆与小圆的面积之比等于它们半径的平方比, 也就是 4: 1 。已知小圆的面积是 6 dm² , 由大圆与小圆的面积之比 4: 1 可知, 大圆的面积是小圆面积的 4 倍, 据此用小圆的面积乘 4 , 即是大圆的面积。【详解】大圆的周长：小圆的周长 = 2:1则大圆的半径：小圆的半径 = 2:1那么大圆的面积：小圆的面积 = 2²: 1² = 4: 16 × 4 = 24 (dm²)大圆的面积是 24 dm² 。故答案为: B",
+    "knowledge_points": [
+      "比的应用",
+      "圆的周长的应用",
+      "圆的面积的应用",
+      "比的意义"
+    ],
+    "source": "24-25 六年级上·全国·单元测试",
+    "merged_id": "unit5_016",
+    "original_id": "unit5_choice_016"
+  },
+  {
+    "id": "unit5_choice_017",
+    "unit": "第五单元：圆",
+    "chapter": "扇形的认识",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "在长方形、正方形、圆、扇形中，只有1条对称轴的图形有（）种。",
+    "options": [
+      "A. 1",
+      "B. 2",
+      "C. 3",
+      "D. 4"
+    ],
+    "answer": "A",
+    "solution": "【分析】一个图形沿一条直线对折，直线两旁的图形完全重合，这样的图形叫做轴对称图形，折痕所在的直线就是对称轴，据此根据轴对称图形的认识以及平面图形的特征进行分析。【详解】长方形有 2 条对称轴, 正方形有 4 条对称轴, 圆有无数条对称轴, 扇形只有 1 条对称轴, 在长方形、正方形、圆、扇形中, 只有 1 条对称轴的图形是扇形, 有 1 种。故答案为: A",
+    "knowledge_points": [
+      "弧、圆心角、扇形的认识",
+      "对称轴的画法及数量"
+    ],
+    "source": "24-25六年级上·全国·单元测试",
+    "merged_id": "unit5_017",
+    "original_id": "unit5_choice_017"
+  },
+  {
+    "id": "unit5_choice_018",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "一个圆的半径由3cm 增加到 8cm, 则它的面积增加了 ( ) cm² 。",
+    "options": [
+      "A. 55",
+      "B. 39",
+      "C. 55π",
+      "D. 39π"
+    ],
+    "answer": "C",
+    "solution": "【分析】圆的面积公式为 S = πr² , 代入公式先算出半径为 3 cm 时小圆的面积, 再计算出半径为 8 cm 时大圆的面积,然后用大圆面积减去小圆面积即可。【详解】π × 8² - π × 3² = π × 64 - π × 9 = 64π - 9π = 55π (平方厘米)所以，它的面积增加了 55π 平方厘米。故答案为: C",
+    "knowledge_points": [
+      "圆的面积"
+    ],
+    "source": "24-25 六年级上·全国·课后作业",
+    "merged_id": "unit5_018",
+    "original_id": "unit5_choice_018"
+  },
+  {
+    "id": "unit5_choice_019",
+    "unit": "第五单元：圆",
+    "chapter": "半圆的周长",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一个半圆的半径是 r, 它的周长是 ( )。",
+    "options": [
+      "A. πr",
+      "B. πr + r",
+      "C. r(π + 2)"
+    ],
+    "answer": "C",
+    "solution": "【分析】如图:![](images/734b7088dba7eba59a1b18b14b68d10776b1149a5abaadbcff86ed6817c2bc02.jpg)半圆的周长 = 圆周长的一半 + 直径，根据圆周长公式：C=2πr，可知半圆的周长 = 2πr ÷ 2 + 2r，再化简即可。【详解】2πr ÷ 2 + 2r = πr + 2r = r(π + 2) 一个半圆的半径是 r , 它的周长是 r(π + 2) 。故答案为: C",
+    "knowledge_points": [
+      "半圆的周长"
+    ],
+    "source": "2022·山东济宁·小升初真题",
+    "merged_id": "unit5_019",
+    "original_id": "unit5_choice_019"
+  },
+  {
+    "id": "unit5_choice_020",
+    "unit": "第五单元：圆",
+    "chapter": "圆的认识",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "同一圆内的所有线段中，（）是最长的。",
+    "options": [
+      "A. 直径",
+      "B. 半径",
+      "C. 不过圆心并且两端都在圆上的线段"
+    ],
+    "answer": "A",
+    "solution": "【分析】连接圆心和圆上任意一点的线段叫做半径。通过圆心并且两端都在圆上的线段都叫做直径。在同一个圆内，所有的半径都相等，所有的直径也相等，直径的长度是半径的 2 倍。在同一个圆内的所有线段中，直径最长。【详解】同一圆内的所有线段中，直径是最长的。故答案为：A",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "24-25 六年级上·全国·课后作业",
+    "merged_id": "unit5_020",
+    "original_id": "unit5_choice_020"
+  },
+  {
+    "id": "unit5_choice_021",
+    "unit": "第五单元：圆",
+    "chapter": "圆的认识",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "车轮做成圆形的, 是因为 ( )。",
+    "options": [
+      "A. 圆的直径是半径的 2 倍",
+      "B. 圆是轴对称图形",
+      "C. 从圆心到圆上任意一点的距离都相等",
+      "D. 圆的半径决定圆的大小"
+    ],
+    "answer": "C",
+    "solution": "【分析】车轮做成圆形，主要是因为从圆心到圆上任意一点的距离都相等。当车轮在平面上滚动时，车轴与平面的距离就能保持不变。这样车辆行驶起来会更加平稳，不会出现颠簸的情况。据此解答。【详解】A．圆的直径是半径的2倍，这是圆的基本性质，但与车轮做成圆形的原因无关。B. 圆是轴对称图形, 这是圆的特征之一, 但不是车轮做成圆形的主要原因。C. 从圆心到圆上任意一点的距离都相等, 这使得车轮在滚动时, 车轴与平面的距离始终保持不变, 保证了车辆行驶的平稳, 所以这是车轮做成圆形的主要原因。D. 圆的半径决定圆的大小, 这是圆的另一个性质, 与车轮做成圆形的直接原因无关。故答案选：C",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "23-24 六年级上·辽宁·期末",
+    "merged_id": "unit5_021",
+    "original_id": "unit5_choice_021"
+  },
+  {
+    "id": "unit5_choice_022",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "圆 A 的直径是6cm , 圆 B 的直径是2dm , 它们的圆周率 ( )。",
+    "options": [
+      "A. 圆 A 大",
+      "B. 圆 B 大",
+      "C. 一样大",
+      "D. 不确定大小"
+    ],
+    "answer": "C",
+    "solution": "【分析】圆周率表示圆的周长与直径的比值，比值是 π ,据此解答。【详解】圆周率表示圆的周长与直径的比值，比值是 π ，与圆的直径是多少无关，所以圆 A 的直径是 6 cm ，圆 B 的直径是 2 dm ，它们的圆周率一样大。故答案为: C",
+    "knowledge_points": [
+      "圆的周长"
+    ],
+    "source": "23-24 六年级上·辽宁·单元测试",
+    "merged_id": "unit5_022",
+    "original_id": "unit5_choice_022"
+  },
+  {
+    "id": "unit5_choice_023",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "约 2000 年前, 中国的古代数学著作《周髀算经》中就有 '周三径一' 的说法,意思是说 ( )。",
+    "options": [
+      "A. 圆的面积约是它周长的 3 倍",
+      "B. 圆的周长约是它半径的3倍",
+      "C. 圆的周长与它直径的比约是 3: 1"
+    ],
+    "answer": "C",
+    "solution": "【分析】周指周长，径指直径，周三径一的意思圆的周长大约是直径的 3 倍，据此解答。【详解】A．题目说的圆的周长大约是直径的3倍，而不是面积，故此选项错误；B. 题目说的圆的周长大约是直径的 3 倍, 而不是半径, 故此选项错误; C. 圆的周长与它直径的比约是 3: 1 , 则圆的周长大约是直径的 3 倍, 这符合 '周三径一'的说法, 故此选项正确。故答案为: C",
+    "knowledge_points": [
+      "圆的周长"
+    ],
+    "source": "23-24六年级下·湖南常德·期末",
+    "merged_id": "unit5_023",
+    "original_id": "unit5_choice_023"
+  },
+  {
+    "id": "unit5_choice_024",
+    "unit": "第五单元：圆",
+    "chapter": "圆的认识",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "在长10厘米、宽6厘米的长方形中画一个最大的半圆，这个半圆的半径是（ ）厘米。",
+    "options": [
+      "A. 10",
+      "B. 6",
+      "C. 5",
+      "D. 3"
+    ],
+    "answer": "C",
+    "solution": "【分析】要在一个长方形里画最大的半圆, 这个半圆的直径应等于长方形的长, 即直径是 10 厘米, 根据半径 = 直径 ÷ 2 , 代入计算, 即可求出半径, 据此解答。【详解】由分析得：10 ÷2=5 (厘米)即这个半圆的半径是 5 厘米。故答案为: C",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "23-24五年级下·湖南邵阳·期末",
+    "merged_id": "unit5_024",
+    "original_id": "unit5_choice_024"
+  },
+  {
+    "id": "unit5_choice_025",
+    "unit": "第五单元：圆",
+    "chapter": "扇形的认识",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "钟面上的时针从 1 走到 5 , 时针扫过的区域是一个圆心角是 ( )°的扇形。",
+    "options": [
+      "A. 150",
+      "B. 120",
+      "C. 50",
+      "D. 40"
+    ],
+    "answer": "B",
+    "solution": "【分析】钟面上共分为 12 个大格, 则每个大格为 360 ÷ 12 = 30 度, 时针从 1 走到 5 共走了 5 - 1 = 4 个大格, 用 1 个大格的度数乘格数即可求出这个扇形的圆心角。【详解】360 ÷ 12 = 30（度）(5 - 1) × 30 = 4 × 30 = 120(度)则时针扫过的区域是一个圆心角是 120° 的扇形。故答案为: B",
+    "knowledge_points": [
+      "弧、圆心角、扇形的认识"
+    ],
+    "source": "23-24 五年级下·湖南邵阳·期末",
+    "merged_id": "unit5_025",
+    "original_id": "unit5_choice_025"
+  },
+  {
+    "id": "unit5_choice_026",
+    "unit": "第五单元：圆",
+    "chapter": "半圆的周长",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一个半径是4分米的半圆，它的周长是（）分米。",
+    "options": [
+      "A. 10.56",
+      "B. 10.28",
+      "C. 20.56",
+      "D. 30.56"
+    ],
+    "answer": "C",
+    "solution": "【分析】半圆的周长由半圆的弧长和一条直径组成，即半圆的周长等于圆周长的一半加上一条直径，根据公式：C= 2πr, d = 2r , 代入数据计算即可, 据此解答。【详解】2 × 3.14 × 4 ÷ 2 = 12.56 （分米）2 × 4 = 8 (分米)12.56 + 8 = 20.56 (分米)即一个半径是 4 分米的半圆, 它的周长是 20.56 分米。故答案为: C",
+    "knowledge_points": [
+      "半圆的周长"
+    ],
+    "source": "23-24六年级上·全国·单元测试",
+    "merged_id": "unit5_026",
+    "original_id": "unit5_choice_026"
+  },
+  {
+    "id": "unit5_choice_027",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "周长相等的长方形、正方形和圆, 面积最小的是 ( )。",
+    "options": [
+      "A. 长方形",
+      "B. 正方形",
+      "C. 圆",
+      "D. 无法确定"
+    ],
+    "answer": "A",
+    "solution": "【分析】要比较周长相等的正方形、长方形和圆形，谁的面积最大，谁面积最小，可以先假设这三种图形的周长是多少，再利用这三种图形的面积公式，分别计算出它们的面积，最后比较这三种图形面积的大小。【详解】假设正方形、长方形和圆形的周长都是 16 。则圆的半径为: 16/(2π) = 8/π , 面积为: π × (8/π)² = 64/π ≈ 20.38 ;正方形的边长为: 16 ÷ 4 = 4 , 面积为: 4 × 4 = 16 ;长方形取长为 5 宽为 3 , 面积为: 5 × 3 = 15 , 当长方形的长和宽最接近时面积也小于 16。所以周长相等的正方形、长方形和圆, 圆面积最大, 长方形面积最小。故答案为：A",
+    "knowledge_points": [
+      "圆的周长的应用",
+      "圆的面积的应用",
+      "长方形的面积",
+      "正方形的面积"
+    ],
+    "source": "23-24 六年级下·甘肃平凉·期末",
+    "merged_id": "unit5_027",
+    "original_id": "unit5_choice_027"
+  },
+  {
+    "id": "unit5_choice_028",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "圆周率是圆的周长与直径的比值，人类对圆周率的研究历史悠久。我国古代数学著作《周髀算经》中就有'周三径一'的记载，东晋时期数学家刘徽用'割圆术'求得圆周率的近似值是3.14，到了南北朝时期数学家祖冲之求得了圆周率的7位小数精确值。如果如图中线段 AF 表示一个圆的周长，那么这个圆的直径可能是（）。![](images/5dc4d9e64038320dede92b481eee18db2bb779e6ccc2819bdfdeb86bafefe7c0.jpg)",
+    "options": [
+      "A. AB",
+      "B. AC",
+      "C. AD",
+      "D. CE"
+    ],
+    "answer": "B",
+    "solution": "【分析】根据圆周率的含义：圆的周长和它直径的比值，叫做圆周率，圆周率用 π 表示，根据圆的周长 i 直径乘 π ，即 C = πd ，直径为 d ，那么周长与直径的比值即是 π 。因为 π 的近似值是 3.14，所以图中线段 AF 代表一个圆的周长的话，那么这个圆的直径大约是周长的三分之一，据此解答即可。【详解】根据圆的周长 d 直径 × π, 即 C = πd, C ÷ d = π, π ≈ 3.14 ,图中线段 AF 代表一个圆的周长, 则圆的直径大约是周长的三分之一, 根据图示线段 AC 最适合。故答案为: B,此题解答关键是明确 π 大约等于 3.14, 周长大约是直径的三倍。",
+    "knowledge_points": [
+      "圆的周长"
+    ],
+    "source": "2024 六年级下·山东·专题练习",
+    "merged_id": "unit5_028",
+    "original_id": "unit5_choice_028"
+  },
+  {
+    "id": "unit5_choice_029",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "用同样长的铁丝围成一个正方形和一个圆，（）。",
+    "options": [
+      "A. 正方形的面积大",
+      "B. 圆的面积大",
+      "C. 一样大",
+      "D. 无法比较"
+    ],
+    "answer": "B",
+    "solution": "【分析】周长相等的正方形和圆, 圆的面积更大。据此解题。【详解】用同样长的铁丝围成一个正方形和一个圆，说明围成的正方形和圆的周长相等，那么圆的面积更大。故答案为: B",
+    "knowledge_points": [
+      "圆的面积",
+      "正方形的面积",
+      "正方形的周长",
+      "圆的周长"
+    ],
+    "source": "2022·湖南岳阳·小升初真题",
+    "merged_id": "unit5_029",
+    "original_id": "unit5_choice_029"
+  },
+  {
+    "id": "unit5_choice_030",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "甲、乙两个圆的半径分别是 5 厘米和 3 厘米。甲圆直径是乙圆直径的( )，乙圆周长是甲圆周长的( )。",
+    "options": [],
+    "answer": "5/3;3/5",
+    "solution": "【分析】（1）先根据直径 = 半径 × 2 求出甲、乙两圆的直径,再用甲圆的直径除以乙圆的直径即可解答;(2) 先根据圆的周长 = πd 求出甲圆和乙圆的周长, 再用乙圆的周长除以甲圆的周长即可解答。【详解】5 × 2 = 10 （厘米）3 × 2 = 6 (厘米)10 ÷ 6 = 5/3  10 × 3.14 = 31.4 (厘米)6 × 3.14 = 18.84 (厘米)   18.84 ÷ 31.4 = 3/5  甲、乙两个圆的半径分别是 5 厘米和 3 厘米。甲圆直径是乙圆直径的 5/3 , 乙圆周长是甲圆周长的 3/5 。",
+    "knowledge_points": [
+      "圆的周长",
+      "求一个数占另一个数几分之几"
+    ],
+    "source": "25-26 六年级上·全国·单元测试",
+    "merged_id": "unit5_030",
+    "original_id": "unit5_choice_030"
+  },
+  {
+    "id": "unit5_choice_031",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的面积",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个圆环, 内圆半径 1 dm , 外圆半径 3 dm , 这个环形的面积是( ) dm² 。",
+    "options": [],
+    "answer": "25.12",
+    "solution": "【分析】分析题目, 圆环的面积 = π(R² - r²) , 据此把 R = 3, r = 1 代入数据列式计算即可。【详解】3.14 × (3² - 1²) = 3.14 × (9 - 1) = 3.14 × 8 = 25.12 (dm²) 一个圆环, 内圆半径 1 dm , 外圆半径 3 dm , 这个环形的面积是 25.12 dm² 。",
+    "knowledge_points": [
+      "圆环的面积"
+    ],
+    "source": "25-26 六年级上·全国·单元测试",
+    "merged_id": "unit5_031",
+    "original_id": "unit5_choice_031"
+  },
+  {
+    "id": "unit5_choice_032",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一根圆木，小玉用绳子绕一周后量得绳长恰好是12.56分米，这根圆木的横截面的半径是( )分米，这根圆木的横截面的面积是( )平方分米。",
+    "options": [],
+    "answer": "2 ;12.56",
+    "solution": "【分析】小玉用绳子绕一周后量得绳长恰好是 12.56 分米,也就是这根圆木的横截面的周长是 12.56 分米; 根据圆的周长公式 C = 2πr , 可求出半径 r = C ÷ (2π) , 再根据面积公式 S = πr² 计算面积。【详解】12.56 ÷ (2 × 3.14) =12.56 ÷ 6.28 = 2 (分米) 3.14 × 2² =3.14 × 4=12.56 (平方分米)因此这根圆木的横截面的半径是2分米，这根圆木的横截面的面积是12.56平方分米。",
+    "knowledge_points": [
+      "圆的周长的应用",
+      "圆的面积"
+    ],
+    "source": "25-26六年级上·全国·单元测试",
+    "merged_id": "unit5_032",
+    "original_id": "unit5_choice_032"
+  },
+  {
+    "id": "unit5_choice_033",
+    "unit": "第五单元：圆",
+    "chapter": "圆的认识",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "在同一个圆中, 直径是半径的( )倍, 半径与直径的比是( )。",
+    "options": [],
+    "answer": "2;1:2",
+    "solution": "【分析】根据圆的特点：在同圆或等圆中，半径是直径的一半，直径是半径的2倍。当直径是2厘米，半径就是1厘米，据此解答。【详解】半径：直径 = 1:2所以在同一个圆中，直径是半径的 2 倍，半径与直径的比是 1: 2 。",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "比的意义"
+    ],
+    "source": "25-26 六年级上·全国·单元测试",
+    "merged_id": "unit5_033",
+    "original_id": "unit5_choice_033"
+  },{
+    "id": "unit5_choice_034",
+    "unit": "第五单元：圆",
+    "chapter": "圆的认识",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "将一个圆形纸片至少对折( )次, 才能找到它的圆心。",
+    "options": [],
+    "answer": "2",
+    "solution": "【分析】用折叠法将圆形纸片对折两次，两次折痕的交点就是圆心的位置，据此解答。【详解】如图：![](images/7da040136f913a40a06ddc2ea2ad6e23a616706d02d6e53ca46e94d01364756a.jpg)将一个圆形纸片至少对折 2 次, 才能找到它的圆心。",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "图形的折叠问题"
+    ],
+    "source": "2024·河北承德·小升初真题",
+    "merged_id": "unit5_034",
+    "original_id": "unit5_choice_034"
+  },
+  {
+    "id": "unit5_choice_035",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "直径为 4 cm 的半圆形，它的周长是 ( ) cm ，面积是 ( ) cm² ，在这个半圆内画一个最大的三角形，这个三角形的面积是 ( ) cm² 。",
+    "options": [],
+    "answer": "10.28;6.28;4",
+    "solution": "【分析】根据半圆的周长 = 圆周长的一半 + 直径, 圆的周长公式 C = πd , 圆的面积除以 2 可得半圆面积, 圆的半径 = 直径 ÷ 2 , 圆的面积公式 S = πr² , 在这个半圆内画一个最大的三角形, 直径就是这个三角形的底, 半径就是它的高, 根据三角形的面积 = 底 × 高 ÷ 2 , 代入数据计算即可。【详解】3.14 × 4 ÷ 2 + 4=6.28+4 =10.28 (cm) 3.14 × (4 ÷ 2)² ÷ 2 = 3.14 × 2² ÷ 2 = 3.14 × 4 ÷ 2 =12.56÷2 =6.28 (cm²) 4 × (4 ÷ 2) ÷ 2 =4×2÷2 =4 cm² 直径为 4 cm 的半圆形, 它的周长是 10.28 cm , 面积是 6.28 cm² , 在这个半圆内画一个最大的三角形, 这个三角形的面积是 4 cm² 。",
+    "knowledge_points": [
+      "三角形面积的计算",
+      "半圆的周长",
+      "圆的面积",
+      "含圆的组合图形的面积"
+    ],
+    "source": "24-25 六年级下·海南海口·单元测试",
+    "merged_id": "unit5_035",
+    "original_id": "unit5_choice_035"
+  },
+  {
+    "id": "unit5_choice_036",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "在一个圆形花坛上每隔 50 厘米摆一盆花, 一共摆了 20 盆, 这个圆形花坛的周长是 米。",
+    "options": [],
+    "answer": "10",
+    "solution": "【分析】封闭图形植树，棵数 = 段数，间距 × 段数 = 花坛的周长，据此列式计算，根据 1 米 = 100 厘米，统一单位即可。【详解】50 × 20 = 1000 （厘米） = 10 （米）这个圆形花坛的周长是 10 米。",
+    "knowledge_points": [
+      "封闭图形上的植树问题",
+      "圆的周长"
+    ],
+    "source": "24-25 六年级下·海南海口·单元测试",
+    "merged_id": "unit5_036",
+    "original_id": "unit5_choice_036"
+  },
+  {
+    "id": "unit5_choice_037",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "在边长是 9 cm 的正方形里画一个最大的圆, 这个圆的面积是 _ cm² 。",
+    "options": [],
+    "answer": "63.585",
+    "solution": "【分析】分析题目，在正方形里画一个最大的圆，则正方形的边长就等于圆的直径，根据圆的面积 = π(d ÷ 2)² 代入数据计算即可。【详解】3.14 × (9 ÷ 2)² = 3.14 × 4.5² = 3.14 × 20.25 = 63.585 (cm²) 在边长是 9 cm 的正方形里画一个最大的圆, 这个圆的面积是 63.585 cm² 。",
+    "knowledge_points": [
+      "方中圆和圆中方的面积问题"
+    ],
+    "source": "24-25 六年级下·海南海口·单元测试",
+    "merged_id": "unit5_037",
+    "original_id": "unit5_choice_037"
+  },
+  {
+    "id": "unit5_choice_038",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长和面积",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "一个圆的半径是 3 cm , 这个圆的周长是 ( ) cm , 面积是 ( ) cm² 。",
+    "options": [],
+    "answer": "18.84;28.26",
+    "solution": "【分析】圆的周长 = 2πr , 圆的面积 = πr² , 据此代入数据列式计算即可。【详解】2 × 3 × 3.14 = 6 × 3.14 = 18.84 (cm) 3.14 × 3² = 3.14 × 9 = 28.26 (cm²) 一个圆的半径是 3 cm , 这个圆的周长是 18.84 cm , 面积是 28.26 cm² 。",
+    "knowledge_points": [
+      "圆的面积",
+      "圆的周长"
+    ],
+    "source": "19-20 六年级上·内蒙古通辽·期末",
+    "merged_id": "unit5_038",
+    "original_id": "unit5_choice_038"
+  },
+  {
+    "id": "unit5_choice_039",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的面积",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "坐落于辽宁省沈抚新区的'生命之环'，无论是高度还是形式都是世界独有的。它近似于一个圆环，它的外直径是170米，内直径是150米，则'生命之环'的面积约是( )平方米。",
+    "options": [],
+    "answer": "5024",
+    "solution": "【分析】直径 ÷ 2 = 半径, 据此分别计算出外圆和内圆半径,根据圆环面积 = 圆周率 × (大圆半径的平方一小圆半径的平方), 列式计算即可。【详解】170 ÷ 2 = 85 （米） 150 ÷ 2 = 75 (米) 3.14 × (85² - 75²) = 3.14 × (7225 - 5625) = 3.14 × 1600 = 5024 (平方米) '生命之环'的面积约是5024平方米。",
+    "knowledge_points": [
+      "圆环的面积"
+    ],
+    "source": "24-25 六年级上·全国·单元测试",
+    "merged_id": "unit5_039",
+    "original_id": "unit5_choice_039"
+  },
+  {
+    "id": "unit5_choice_040",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "白居易的《府西池》中'柳无气力枝先动，池有波纹冰尽开'描述了雨点打在水面上荡开层层的波纹。已知水池是长 6 m 、宽 5 m 的长方形，当波纹到池边时，所形成的最大整圆的周长是( ), 面积是( )。",
+    "options": [],
+    "answer": "15.7m;19.625m²",
+    "solution": "【分析】根据题意，长 6 m 、宽 5 m 的长方形水池中最大的圆的直径等于长方形的宽，根据圆的周长公式 C = πd ，圆的面积公式 S = πr² , 代入数据计算, 求出最大圆的周长和面积。【详解】圆的半径: 5 ÷ 2 = 2.5 (m)圆的周长: 3.14 × 5 = 15.7 (m)圆的面积:3.14 × 2.5² = 3.14 × 6.25 = 19.625 (m²) 当波纹到池边时, 所形成的最大整圆的周长是 15.7 m , 面积是 19.625 m² 。",
+    "knowledge_points": [
+      "圆的面积",
+      "圆的周长"
+    ],
+    "source": "24-25 六年级上·全国·单元测试",
+    "merged_id": "unit5_040",
+    "original_id": "unit5_choice_040"
+  },
+  {
+    "id": "unit5_choice_041",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一幅圆形壁画的边框长是 3.14 m , 这幅壁画的面积是( )。",
+    "options": [],
+    "answer": "0.785 m²或0.785 平方米",
+    "solution": "【分析】圆周长 = 2πr , 那么将圆形壁画的边框长除以 3.14 再除以 2 , 可求出壁画的半径。再根据 '圆面积 = πr²' 列式求出这幅壁画的面积即可。【详解】3.14 ÷ 3.14 ÷ 2 = 0.5 (m) 3.14 × 0.5² = 3.14 × 0.25 = 0.785 (m²) 所以，这幅壁画的面积是 0.785 m² 。",
+    "knowledge_points": [
+      "圆的周长的应用",
+      "圆的面积的应用",
+      "小数与小数的乘法",
+      "除数是小数的小数除法"
+    ],
+    "source": "24-25 六年级上·全国·课后作业",
+    "merged_id": "unit5_041",
+    "original_id": "unit5_choice_041"
+  },
+  {
+    "id": "unit5_choice_042",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的面积",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "若用 R 表示外圆半径, 用 r 表示内圆半径, 则环形面积公式是( )。",
+    "options": [],
+    "answer": "πR² - πr²",
+    "solution": "【分析】我们知道圆的面积公式是 S = πr² 。对于圆环来说，它的面积等于外圆的面积减去内圆的面积。外圆面积为 πR² ，内圆面积为 πr² 。【详解】所以圆环面积就是外圆面积减去内圆面积，即 πR² - πr²",
+    "knowledge_points": [
+      "用字母表示数",
+      "数量关系",
+      "圆环的面积"
+    ],
+    "source": "24-25 六年级上·全国·课后作业",
+    "merged_id": "unit5_042",
+    "original_id": "unit5_choice_042"
+  },
+  {
+    "id": "unit5_choice_043",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一根圆木, 小玉用绳子绕一周后量得绳长恰好是 12.56 分米, 这根圆木的横截面的直径是( )分米。",
+    "options": [],
+    "answer": "4",
+    "solution": "【分析】根据圆周长公式 C = πd , 可知 d = C ÷ π , 代入数据解答。【详解】12.56 ÷ 3.14 = 4 （分米）这根圆木的横截面的直径是 4 分米。",
+    "knowledge_points": [
+      "除数是小数的小数除法",
+      "圆的周长"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit5_043",
+    "original_id": "unit5_choice_043"
+  },
+  {
+    "id": "unit5_choice_044",
+    "unit": "第五单元：圆",
+    "chapter": "圆的认识",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "在同一个圆中, 半径是直径( ), 直径与半径的比是( )。",
+    "options": [],
+    "answer": "1/2;2:1",
+    "solution": "【分析】根据对圆的认识可知，在同一个圆中，直径是半径的2倍，当直径是2厘米，半径就是1厘米，据此解答。【详解】1 ÷ 2 = 1/2 直径：半径 = 2:1所以，在同一个圆中，半径是直径 1/2 ，直径与半径的比是 2: 1 。",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "分数的意义",
+      "比的意义"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit5_044",
+    "original_id": "unit5_choice_044"
+  },
+  {
+    "id": "unit5_choice_045",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长和面积",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "甲圆的半径是 2 分米, 乙圆的半径是 3 分米, 甲圆和乙圆直径之比是( ), 周长之比是( ), 面积之比是( )。",
+    "options": [],
+    "answer": "2:3;2:3;4:9",
+    "solution": "【分析】直径 = 半径 × 2 , 分别求出甲圆和乙圆的直径; 再根据比的意义: 用甲圆的直径 : 乙圆的直径; 根据圆的周长的公式: 周长 = π × 半径 × 2 , 分别求出甲圆的周长和乙圆的周长, 再用甲圆周长 : 乙圆周长; 根据圆的面积公式:面积 = π × 半径 2 , 分别求出甲圆的面积和乙圆的面积, 再用甲圆的面积：乙圆的面积，化简，即可解答。【详解】(2 × 2): (3 × 2) = 4: 6 = (4 ÷ 2): (6 ÷ 2) = 2: 3  (π × 2 × 2) : (π × 3 × 2) = (4π): (6π) = (4π ÷ 2π): (6π ÷ 2π) = 2: 3  (π × 2²) : (π × 3²) = (4π): (9π) = (4π ÷ π): (9π ÷ π) = 4: 9 甲圆的半径是 2 分米, 乙圆的半径是 3 分米, 甲圆和乙圆直径之比是 2: 3 , 周长之比是 2: 3 , 面积之比是 4: 9 。",
+    "knowledge_points": [
+      "圆的面积",
+      "比的化简",
+      "比的意义",
+      "圆的周长"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit5_045",
+    "original_id": "unit5_choice_045"
+  },
+  {
+    "id": "unit5_choice_046",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "圆无论大小, 它的周长总是它直径的( )倍多一些, 这个倍数是个固定不变的数，我们叫它( )，用字母( )表示。",
+    "options": [],
+    "answer": "3;圆周率;π",
+    "solution": "【分析】研究发现任意一个圆的周长与它的直径的比值是一个固定的数，我们把它叫做圆周率，它是一个无限不循环小数, 通常用字母 π 表示, 实际应用中常常只取它的近似数，如 π ≈ 3.14 ，据此解答。【详解】根据分析可知，圆无论大小，它的周长总是它直径的 3 倍多一些, 这个倍数是个固定不变的数, 我们叫它圆周率，用字母 π 表示。",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "圆的周长"
+    ],
+    "source": "24-25 六年级上·全国·单元测试",
+    "merged_id": "unit5_046",
+    "original_id": "unit5_choice_046"
+  },
+  {
+    "id": "unit5_choice_047",
+    "unit": "第五单元：圆",
+    "chapter": "圆的认识",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "用长为 10 dm 、宽为6dm 的长方形纸片剪半径为 5cm 的圆，最多能剪( )个。",
+    "options": [],
+    "answer": "60",
+    "solution": "【分析】如图，![](images/c7fef131edba7e69f66890eb1d7c11da.png)正方形内最大的圆，圆的直径 = 正方形边长，统一单位， 5cm = 0.5dm ，长方形纸片剪半径为 5cm 的圆，相当于在长方形纸片剪边长（ 0.5× 2 ）dm的正方形，沿长能剪 [10÷ (0.5× 2)] 个，沿宽能剪 [6÷ (0.5× 2)] 个，沿长剪的个数 × 沿宽剪的个数 = 能剪的总个数。【详解】5 cm = 0.5 dm [10 ÷ (0.5 × 2)] × [6 ÷ (0.5 × 2)] = [10 ÷ 1] × [6 ÷ 1] = 10 × 6 = 60 (个) 最多能剪 60 个。",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "平面图形的分割"
+    ],
+    "source": "24-25 六年级上·全国·单元测试",
+    "merged_id": "unit5_047",
+    "original_id": "unit5_choice_047"
+  },
+  {
+    "id": "unit5_choice_048",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "在同圆或等圆中,圆的周长是半径的( )倍。",
+    "options": [],
+    "answer": "2π",
+    "solution": "【分析】根据圆的周长公式 C = 2πr , 则可以推导出 C/r = 2π , 即可得解。【详解】据分析可得，在同圆或等圆中，圆的周长是半径的 2π 倍。",
+    "knowledge_points": [
+      "圆的周长"
+    ],
+    "source": "24-25 六年级上·全国·单元测试",
+    "merged_id": "unit5_048",
+    "original_id": "unit5_choice_048"
+  },
+  {
+    "id": "unit5_choice_049",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "两个圆的半径之比是 4: 3 , 它们的面积之比是( )。",
+    "options": [],
+    "answer": "16:9",
+    "solution": "【分析】已知两个圆的半径之比是 4: 3 , 可以设两个圆的半径为 4 和 3 。根据圆的面积公式 S = πr² , 分别求出两个圆的面积, 再根据比的意义写出两个圆的面积之比, 并化简比。【详解】设两个圆的半径分别为4和3。(π × 4²) : (π × 3²) = 16π : 9π = 16: 9 它们的面积之比是 16: 9 。",
+    "knowledge_points": [
+      "圆的面积的应用",
+      "比的化简",
+      "比的意义"
+    ],
+    "source": "24-25 六年级上·全国·单元测试",
+    "merged_id": "unit5_049",
+    "original_id": "unit5_choice_049"
+  },
+  {
+    "id": "unit5_choice_050",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "一个直径是 8 m 的圆形花坛, 它的占地面积是( ) m² 。",
+    "options": [],
+    "answer": "50.24",
+    "solution": "【分析】先根据圆的直径求圆的半径, 8 ÷ 2 = 4 (m), 然后代入公式 'S = πr²'计算即可。【详解】3.14 × (8 ÷ 2)² = 3.14 × 4² = 3.14 × 16 = 50.24 (m²) 所以，圆形花坛的占地面积是 50.24 m² 。",
+    "knowledge_points": [
+      "圆的面积"
+    ],
+    "source": "24-25 六年级上·全国·课后作业",
+    "merged_id": "unit5_050",
+    "original_id": "unit5_choice_050"
+  },
+  {
+    "id": "unit5_choice_051",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "圆的半径扩大到原来的 5 倍, 这个圆的周长就扩大到原来的( )倍。",
+    "options": [],
+    "answer": "5",
+    "solution": "【分析】根据圆的周长 = 2 × π × 半径, π 是一个定值, 则圆的周长是半径的 2π 倍, 当圆的半径扩大到原来的 5 倍, 根据积的变化规律, 当其他因数不变, 一个因数扩大到原来的几倍, 则积也扩大到原来的几倍, 可得周长也扩大到原来的 5 倍, 据此得解。【详解】由分析可知, 圆周长 = 2 × π × 半径, 当半径扩大到原来的 5 倍, 这个圆的周长就扩大到原来的 5 倍。",
+    "knowledge_points": [
+      "圆的周长"
+    ],
+    "source": "24-25 六年级上·全国·课后作业",
+    "merged_id": "unit5_051",
+    "original_id": "unit5_choice_051"
+  },
+  {
+    "id": "unit5_choice_052",
+    "unit": "第五单元：圆",
+    "chapter": "圆的认识",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "在一张边长是 6 cm的正方形纸上剪下一个最大的圆，这个圆的直径是( )cm，半径是( )cm。",
+    "options": [],
+    "answer": "6;3",
+    "solution": "【分析】正方形纸上剪下一个最大的圆，圆的直径=正方形边长，直径÷2=半径，据此分析。【详解】6 ÷ 2 = 3 (cm)这个圆的直径是 6 cm , 半径是 3 cm 。",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "24-25 六年级上·全国·课后作业",
+    "merged_id": "unit5_052",
+    "original_id": "unit5_choice_052"
+  },
+  {
+    "id": "unit5_choice_053",
+    "unit": "第五单元：圆",
+    "chapter": "扇形的认识",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个扇形的面积是它所在圆面积的 1/3 , 这个扇形的圆心角是( )°。",
+    "options": [],
+    "answer": "120",
+    "solution": "【分析】扇形面积占它所在圆面积的 1/3 , 则扇形的圆心角也占整个圆的圆心角的 1/3 , 而整个圆的圆心角是 360° , 即求这个扇形的圆心角是多少, 就是求 360° 的 1/3 是多少, 根据求一个数的几分之几是多少, 用乘法计算, 据此解答。【详解】360° × 1/3 = 120° 即这个扇形的圆心角是 120° 。",
+    "knowledge_points": [
+      "弧、圆心角、扇形的认识",
+      "求一个数的几分之几的问题"
+    ],
+    "source": "23-24 六年级上·全国·课后作业",
+    "merged_id": "unit5_053",
+    "original_id": "unit5_choice_053"
+  },
+  {
+    "id": "unit5_choice_054",
+    "unit": "第五单元：圆",
+    "chapter": "圆中方的面积问题",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "在一个直径是 10 m 的圆里画一个最大的正方形, 正方形的面积是( ) cm² 。",
+    "options": [],
+    "answer": "50",
+    "solution": "【分析】在一个直径是 10 m 的圆里画一个最大的正方形,正方形的对角线的长度等于圆的直径的长度，正方形面积 = 对角线长 × 对角线长的一半，据此求出正方形的面积。【详解】正方形面积：10 × 10 ÷ 2 =100 ÷ 2 = 50 (cm²)所以正方形的面积是 50 cm² 。",
+    "knowledge_points": [
+      "方中圆和圆中方的面积问题"
+    ],
+    "source": "2014 三年级·全国·课后作业",
+    "merged_id": "unit5_054",
+    "original_id": "unit5_choice_054"
+  },
+  {
+    "id": "unit5_choice_055",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "用圆规画一个周长是 18.84 厘米的圆, 那么圆规两脚间的距离是( )厘米。",
+    "options": [],
+    "answer": "3",
+    "solution": "【分析】圆规两脚之间的距离就是圆的半径，已知圆的周长，根据周长公式：r = C ÷ π ÷ 2 ，代入数据计算，即可求出圆规两脚间的距离是多少厘米，据此解答。【详解】18.84 ÷ 3.14 ÷ 2 = 3（厘米）即圆规两脚间的距离是3厘米。",
+    "knowledge_points": [
+      "圆的周长"
+    ],
+    "source": "18-19 五年级下·山东青岛·期中",
+    "merged_id": "unit5_055",
+    "original_id": "unit5_choice_055"
+  },
+  {
+    "id": "unit5_choice_056",
+    "unit": "第五单元：圆",
+    "chapter": "扇形的认识",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "当扇形的圆心角是( )°时, 扇形就是一个圆; 当扇形的圆心角是 180° 时, 扇形就是一个( )圆。",
+    "options": [],
+    "answer": "360;半",
+    "solution": "【分析】由一条弧和经过这条弧两端的两条半径所围成的图形叫做扇形。一个圆的圆心角是 360° , 扇形的圆心角决定了扇形的形状和大小, 据此解答。【详解】因为一个圆的圆心角是 360° , 所以当扇形的圆心角是 360° 时, 扇形的弧长正好构成一个完整的圆。因为 180° 是 360° 的一半, 所以当扇形的圆心角是 180° 时,扇形的弧长正好是圆周长的一半, 此时扇形就是一个半圆。即当扇形的圆心角是 360° 时, 扇形就是一个圆; 当扇形的圆心角是 180° 时, 扇形就是一个半圆。",
+    "knowledge_points": [
+      "弧、圆心角、扇形的认识"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit5_056",
+    "original_id": "unit5_choice_056"
+  },
+  {
+    "id": "unit5_choice_057",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "一个轮子滚动一周前进的距离是 18.84 分米, 则这个轮子的直径是( )分米。",
+    "options": [],
+    "answer": "6",
+    "solution": "【分析】一个轮子滚动一周前进的距离就是这个轮子的周长, 根据公式: d = C ÷ π , 代入数据计算即可解答。【详解】18.84 ÷ 3.14 = 6 （分米）则这个轮子的直径是 6 分米。",
+    "knowledge_points": [
+      "圆的周长"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit5_057",
+    "original_id": "unit5_choice_057"
+  },
+  {
+    "id": "unit5_choice_058",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长和面积",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "御林湾小区门口钟楼上有一个挂钟, 分针长 40 厘米, 1 小时分针针尖经过的路程是( )厘米, 分针走 30 分钟扫过的面积是( )平方厘米。",
+    "options": [],
+    "answer": "251.2;2512",
+    "solution": "【分析】1 小时分针转动 1 圈, 分针长度相当于圆的半径,根据圆的周长 = 2 × 圆周率 × 半径, 求出 1 小时分针针尖经过的路程; 30 分钟分针扫过部分是个半圆, 根据半圆面积 = 圆周率 × 半径的平方 ÷ 2 , 求出分针走 30 分钟扫过的面积。【详解】2 × 3.14 × 40 = 251.2 （厘米）3.14 × 40² ÷ 2 = 3.14 × 1600 ÷ 2 = 2512 (平方厘米) 1 小时分针针尖经过的路程是 251.2 厘米, 分针走 30 分钟扫过的面积是 2512 平方厘米。",
+    "knowledge_points": [
+      "圆的面积",
+      "圆的周长",
+      "小数与整数的乘法"
+    ],
+    "source": "23-24六年级下·内蒙古呼伦贝尔·期末",
+    "merged_id": "unit5_058",
+    "original_id": "unit5_choice_058"
+  },
+  {
+    "id": "unit5_choice_059",
+    "unit": "第五单元：圆",
+    "chapter": "半圆的周长",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个长方形的长是8cm, 宽是 5 cm , 在这个长方形中画一个最大的半圆, 这个半圆的周长是( )cm。",
+    "options": [],
+    "answer": "20.56",
+    "solution": "【分析】如下图，在长 8 cm 、宽 5 cm 的长方形中画一个最大的半圆，那么这个半圆的直径等于长方形的长 8 cm ；根据半圆的周长 = 圆周长的一半 + 直径 ，其中圆的周长公式 C = πd ，代入数据计算求解。![](images/cf2f0dd4f96f99f94cb909c9d8bfb1f171c2401f26a1bba4eff1dba126fb9ca7.jpg)【详解】3.14 × 8 ÷ 2 + 8 = 12.56 + 8 = 20.56 (cm) 这个半圆的周长是 20.56 cm 。",
+    "knowledge_points": [
+      "半圆的周长"
+    ],
+    "source": "23-24 六年级下·四川凉山·期末",
+    "merged_id": "unit5_059",
+    "original_id": "unit5_choice_059"
+  },
+  {
+    "id": "unit5_choice_060",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "把一个圆形纸片切分成若干等份后，拼成一个近似的长方形，周长增加了8 cm , 则圆形纸片的半径是( ) cm , 面积是( ) cm² 。(π取3.14)",
+    "options": [],
+    "answer": "4;50.24",
+    "solution": "【分析】把一个圆形纸片剪拼成一个近似的长方形, 那么长方形的长等于圆的周长的一半, 宽等于圆的半径; 拼成的长方形的周长比原来圆的周长增加了 2 条宽的长度, 即增加了 2 个半径的长度, 用增加的周长除以 2 , 即可求出圆的半径; 然后根据圆的面积公式 S = πr² , 求出这个圆的面积。【详解】圆的半径：8 ÷ 2 = 4 (cm)圆的面积:3.14 × 4² = 3.14 × 16 = 50.24 (cm²)则圆形纸片的半径是 4 cm , 面积是 50.24 cm² 。",
+    "knowledge_points": [
+      "圆的周长的应用",
+      "圆的面积的应用",
+      "长方形的概念及特点",
+      "长、正方形周长的应用"
+    ],
+    "source": "23-24 六年级下·河南南阳·期末",
+    "merged_id": "unit5_060",
+    "original_id": "unit5_choice_060"
+  },
+  {
+    "id": "unit5_choice_061",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长和面积",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "用圆规画一个周长是 12.56 厘米的圆, 圆规两脚间的距离应为( )厘米, 所画圆的面积是( )平方厘米。",
+    "options": [],
+    "answer": "2;12.56",
+    "solution": "【分析】圆规两脚间的距离即为圆的半径, 根据 C = 2πr ,则 r = C ÷ 2 ÷ π , 再根据圆的面积 S = πr² , 代入数据即可算出圆的面积。【详解】12.56 ÷ 2 ÷ 3.14 =6.28 ÷ 3.14 =2 (厘米) 3.14 × 2² =3.14 × 4=12.56 (平方厘米)所以圆规两脚间的距离应为 2 厘米, 所画圆的面积是 12.56 平方厘米。",
+    "knowledge_points": [
+      "圆的面积",
+      "圆的周长"
+    ],
+    "source": "14-15 六年级上·全国·假期作业",
+    "merged_id": "unit5_061",
+    "original_id": "unit5_choice_061"
+  },
+  {
+    "id": "unit5_choice_062",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "用一根长 31.4 厘米的绳子围成一个圆, 这个圆的半径是 厘米。",
+    "options": [],
+    "answer": "5",
+    "solution": "【分析】绳子长度相当于圆的周长，根据圆的半径 = 周长 ÷ 圆周率 ÷ 2 ，列式计算即可。【详解】31.4 ÷ 3.14 ÷ 2 = 5 （厘米）这个圆的半径是 5 厘米。",
+    "knowledge_points": [
+      "圆的周长的应用"
+    ],
+    "source": "23-24 六年级上·河南南阳·期末",
+    "merged_id": "unit5_062",
+    "original_id": "unit5_choice_062"
+  },
+  {
+    "id": "unit5_choice_063",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长和面积",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "千年银杏被誉为'东北树王'，一根长 25.12 米的麻绳刚好可以在这棵银杏树的树干上绕 4 圈。这棵银杏树的树干横截面的半径是( )米,面积是( )平方米。",
+    "options": [],
+    "answer": "1;3.14",
+    "solution": "【分析】根据题意，用 25.12 ÷ 4 ，求出麻绳绕这棵银杏树的树干1圈的长度，也就是这个树干的周长，再根据圆的周长公式：周长 = π × 半径 × 2 ，半径 = 周长 ÷ 2 ÷ π ，代入数据，求出树干横截面的半径；再根据圆的面积公式：面积 =π × 半径 ² 。代入数据，即可解答。【详解】25.12 ÷ 4 = 6.28 （米）6.28 ÷ 2 ÷ 3.14 = 3.14 ÷ 3.14 = 1 (米) 3.14 × 1² = 3.14 × 1 = 3.14 (平方米) 千年银杏被誉为'东北树王'，一根长25.12米的麻绳刚好可以在这棵银杏树的树干上绕4圈。这棵银杏树的树干横截面的半径是1米，面积是3.14平方米。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "2024 六年级下·全国·专题练习",
+    "merged_id": "unit5_063",
+    "original_id": "unit5_choice_063"
+  },
+  {
+    "id": "unit5_choice_064",
+    "unit": "第五单元：圆",
+    "chapter": "圆的认识",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "圆周率 π 等于 3.14。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】圆周率是圆的周长与直径的比值，一般用希腊字母 π 表示。π 是一个常数，它是一个无限不循环小数，π = 3.1415926535… 。在日常生活中，通常都用 3.14 代表圆周率去进行近似计算。据此解答。【详解】通过分析可得: 圆周率 π 是一个无限不循环小数,不等于 3.14。原题说法错误。故答案为: ×",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "2022·甘肃天水·小升初真题",
+    "merged_id": "unit5_064",
+    "original_id": "unit5_choice_064"
+  },
+  {
+    "id": "unit5_choice_065",
+    "unit": "第五单元：圆",
+    "chapter": "扇形的认识",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "圆心角越大，扇形的面积就越大。（ ）（对的打'√'，错的打'×'，并说明理由）理由：________________________________________ 。",
+    "options": [],
+    "answer": "× 根据扇形面积公式: 面积 = n/360 πr² (n 为圆心角的度数), 可知, 扇形的面积由圆心角和半径决定,所以只有圆心角大, 不能确定扇形面积的大小。",
+    "solution": "【分析】一条弧和经过这条弧两端的两条半径所围成的图形叫做扇形；根据扇形的面积公式：面积 = n/360 πr² （n 为圆心角的度数），可知，扇形的面积由圆心角和半径决定的，据此解答。【详解】圆心角越大，扇形的面积就越大。（×）理由: 根据扇形面积公式: 面积 = n/360 πr² (n 为圆心角的度数), 可知, 扇形的面积由圆心角和半径决定, 所以只有圆心角大, 不能确定扇形面积的大小。",
+    "knowledge_points": [
+      "弧、圆心角、扇形的认识",
+      "扇形的周长和面积"
+    ],
+    "source": "23-24六年级上·全国·单元测试",
+    "merged_id": "unit5_065",
+    "original_id": "unit5_choice_065"
+  },
+  {
+    "id": "unit5_choice_066",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "π = 3.14 。（）（对的打'√'，错的打'×'，并说明理由）理由：________________ 。",
+    "options": [],
+    "answer": "×; 圆周率 π 是个无限不循环小数, 3.14 是近似值, 应是 π ≈ 3.14 。",
+    "solution": "【分析】圆周率 π 是个无限不循环小数, 3.14 是近似值, 不可用等号表示。【详解】π ≈ 3.14 理由：圆周率 π 是个无限不循环小数，3.14 是近似值，应是 π ≈ 3.14 。故答案为: ×",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit5_066",
+    "original_id": "unit5_choice_066"
+  },
+  {
+    "id": "unit5_choice_067",
+    "unit": "第五单元：圆",
+    "chapter": "圆的认识",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "圆有无数条对称轴,任意一条直径所在的直线都是圆的对称轴。（）（对的打'√'，错的打'×'，并说明理由）理由：__________________________________ 。",
+    "options": [],
+    "answer": "√ ;一个图形沿着一条直线对折后两边能完全重合, 这个图形就是轴对称图形, 折痕所在的直线就是对称轴。圆有无数条直径, 有无数条对称轴。每条直径所在的直线都是圆的对称轴",
+    "solution": "【分析】轴对称图形的意义: 如果一个图形沿着一条直线对折后两部分完全重合, 这样的图形叫做轴对称图形, 这条直线叫做对称轴; 注意, 语言要严密, 不能说成圆的直径就是圆的对称轴，因为对称轴是一条直线，直径是线段。【详解】圆有无数条对称轴, 任意一条直径所在的直线都是圆的对称轴。(√)理由：一个图形沿着一条直线对折后两边能完全重合，这个图形就是轴对称图形，折痕所在的直线就是对称轴。圆有无数条直径，有无数条对称轴。每条直径所在的直线都是圆的对称轴。",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "轴对称的认识及辨认",
+      "对称轴的画法及数量"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit5_067",
+    "original_id": "unit5_choice_067"
+  },
+  {
+    "id": "unit5_choice_068",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "大圆的圆周率一定比小圆的圆周率大。( ) (对的打 '√', 错的打 '×', 并说明理由)理由：_____________________________________ 。",
+    "options": [],
+    "answer": "×;根据圆周率的意义: 圆周率是任意一个圆的周长与它的直径的比值, 这个比值是一个固定的数; 不管大圆小圆, 圆周率是一定的。",
+    "solution": "【分析】圆周率是任意一个圆的周长与它的直径的比值,这个比值是一个固定的数。据此解答。【详解】大圆的圆周率一定比小圆的圆周率大。（x）理由：根据圆周率的意义：圆周率是任意一个圆的周长与它的直径的比值，这个比值是一个固定的数；不管大圆小圆，圆周率是一定的。",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit5_068",
+    "original_id": "unit5_choice_068"
+  },
+  {
+    "id": "unit5_choice_069",
+    "unit": "第五单元：圆",
+    "chapter": "圆的认识",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "圆上两点间最长的线段就是圆的直径。( ) (对的打 '√', 错的打 '×', 并说明理由)理由：",
+    "options": [],
+    "answer": "√;根据圆的特征可知, 圆内的所有线段中,直径最长, 所以连接圆上两点最长的线段一定是直径。",
+    "solution": "【分析】通过圆心并且两端都在圆上的线段叫做直径。所以两端都在圆上的线段中，直径最长，据此解答。【详解】圆上两点间最长的线段就是圆的直径。（√）理由：根据圆的特征可知，圆内的所有线段中，直径最长，所以连接圆上两点最长的线段一定是直径。",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit5_069",
+    "original_id": "unit5_choice_069"
+  },
+  {
+    "id": "unit5_choice_070",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长和面积",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "半径是 2 厘米的圆, 它的周长和面积相等。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】围成圆的曲线的长叫做圆的周长；围成圆的平面的大小叫做圆的面积。两者意义不同，不能比较大小。【详解】圆的周长和面积不是同类量，无法比较大小。原题说法错误。故答案为: ×",
+    "knowledge_points": [
+      "圆的周长的应用",
+      "圆的面积的应用"
+    ],
+    "source": "2012·全国·小升初真题",
+    "merged_id": "unit5_070",
+    "original_id": "unit5_choice_070"
+  },
+  {
+    "id": "unit5_choice_071",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "若大圆的直径为 d₁ , 小圆的直径为 d₂ , 则两圆的周长相差 π(d₁-d₂) 。( )",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】根据圆的周长公式 C = πd , 则大圆周长为 πd₁ , 小圆周长为 πd₂ , 用大圆周长减小圆周长, 化简后即可得解。【详解】据分析可得，两圆的周长差为：πd₁ - πd₂ = π(d₁ - d₂) 故答案为： √",
+    "knowledge_points": [
+      "圆的周长"
+    ],
+    "source": "24-25 六年级上·全国·单元测试",
+    "merged_id": "unit5_071",
+    "original_id": "unit5_choice_071"
+  },
+  {
+    "id": "unit5_choice_072",
+    "unit": "第五单元：圆",
+    "chapter": "扇形的认识",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "圆心角越大, 扇形的面积越大。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】一条弧和经过这条弧两端的两条半径所围成的图形叫做扇形。根据扇形的面积公式 S扇 = n/360 πr² (n 为圆心角的度数), 可知扇形的面积由圆心角和半径决定。【详解】扇形的面积是由圆心角和半径共同决定。所以，只是圆心角大，不能确定扇形面积的大小。原题说法错误。故答案为: ×",
+    "knowledge_points": [
+      "弧、圆心角、扇形的认识",
+      "扇形的周长和面积"
+    ],
+    "source": "18-19 五年级下·江苏·期末",
+    "merged_id": "unit5_072",
+    "original_id": "unit5_choice_072"
+  },
+  {
+    "id": "unit5_choice_073",
+    "unit": "第五单元：圆",
+    "chapter": "圆的认识",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "圆环也是轴对称图形，它有无数条对称轴。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】一个图形沿一条直线对折后，折痕两旁的部分能够完全重合，这样的图形就是轴对称图形，这条直线就是对称轴。【详解】圆环也是轴对称图形, 圆环的对称轴是每条直径所在的直线, 所以圆环有无数条对称轴。原题说法正确。故答案为: √",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "与圆相关的轴对称图形",
+      "对称轴的画法及数量"
+    ],
+    "source": "24-25 六年级上·全国·单元测试",
+    "merged_id": "unit5_073",
+    "original_id": "unit5_choice_073"
+  },
+  {
+    "id": "unit5_choice_074",
+    "unit": "第五单元：圆",
+    "chapter": "扇形的认识",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "如图, 实线部分是弧, 虚线部分不是弧。( )![](images/9796e8afe6178f72fb75baf1c23ee2f986e0ab9d5a1935e0d4b736309752644c.jpg)",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】圆上任意两点之间的部分叫弧，据此进行分析。【详解】图中, 实线部分是弧, 虚线部分也是弧, 所以原题说法错误。故答案为: ×",
+    "knowledge_points": [
+      "弧、圆心角、扇形的认识"
+    ],
+    "source": "24-25 六年级上·全国·课后作业",
+    "merged_id": "unit5_074",
+    "original_id": "unit5_choice_074"
+  },
+  {
+    "id": "unit5_choice_075",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "半径是 2 cm 的圆, 它的周长和半径相等。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】已知圆的半径是 2 cm , 根据圆的周长公式 C=2πr, 求出这个圆的周长; 据此判断。【详解】周长: 2 × 3.14 × 2 = 12.56 (cm) 12.56 ≠ 2半径是 2 cm 的圆, 它的周长和半径不相等。原题说法错误。故答案为: ×",
+    "knowledge_points": [
+      "圆的周长"
+    ],
+    "source": "2022·湖南长沙·小升初真题",
+    "merged_id": "unit5_075",
+    "original_id": "unit5_choice_075"
+  },
+  {
+    "id": "unit5_choice_076",
+    "unit": "第五单元：圆",
+    "chapter": "圆的认识",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "圆内最长的线段是直径。( )",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】通过圆心并且两端都在圆上的线段叫做直径。所有两端都在圆上的线段中，直径最长。据此判断。【详解】圆内最长的线段是直径。原题说法正确。故答案为： √",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "23-24 六年级上·内蒙古通辽·期末",
+    "merged_id": "unit5_076",
+    "original_id": "unit5_choice_076"
+  },
+  {
+    "id": "unit5_choice_077",
+    "unit": "第五单元：圆",
+    "chapter": "圆的认识",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "圆和半圆都是轴对称图形，它们都有无数条对称轴。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】根据对称轴的定义：如果一个图形沿着一条直线对折，两侧的图形能完全重合，这个图形就是轴对称图形。折痕所在的这条直线叫做对称轴，依此作答。【详解】圆和半圆都是轴对称图形，半圆有1条对称轴，圆有无数条对称轴。故答案为: ×",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "对称轴的画法及数量"
+    ],
+    "source": "2022·四川绵阳·小升初真题",
+    "merged_id": "unit5_077",
+    "original_id": "unit5_choice_077"
+  },
+  {
+    "id": "unit5_choice_078",
+    "unit": "第五单元：圆",
+    "chapter": "扇形的面积",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "圆心角大的扇形的面积不一定大于圆心角小的扇形的面积。( )",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】扇形的面积 = πr² × 圆心角/360度 , 即扇形的面积除了与圆心角的大小有关, 还与半径的长短有关系, 据此分析。【详解】因为扇形的面积除了与圆心角的大小有关，还与半径的长短有关系，所以圆心角大的扇形的面积不一定大于圆心角小的扇形的面积。说法正确。故答案为: √",
+    "knowledge_points": [
+      "扇形的周长和面积"
+    ],
+    "source": "23-24 六年级上·河北邯郸·期末",
+    "merged_id": "unit5_078",
+    "original_id": "unit5_choice_078"
+  },
+  {
+    "id": "unit5_choice_079",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "小圆与大圆的半径比是 2: 5 , 则周长的比也是 2: 5 。( )",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】圆的周长 = 2 × 圆周率 × 半径, 两数相除又叫两个数的比, 由此可知, 两个圆的半径比等于这两个圆的周长比,据此分析。【详解】根据分析, 小圆与大圆的半径比是 2: 5 , 则周长的比也是 2: 5 , 说法正确。故答案为: √",
+    "knowledge_points": [
+      "比的意义",
+      "圆的周长"
+    ],
+    "source": "23-24 六年级上·河南新乡·期末",
+    "merged_id": "unit5_079",
+    "original_id": "unit5_choice_079"
+  },
+  {
+    "id": "unit5_choice_080",
+    "unit": "第五单元：圆",
+    "chapter": "圆的认识",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "用 2 个半圆, 一定可以拼成一个圆。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】半径相同的两个半圆能拼成一个圆，据此解答。【详解】半径相同的两个半圆能拼成一个圆，当两个半圆的半径不相等时就不能拼成一个圆。原题说法错误。故答案为: ×",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "14-15 六年级上·全国·课后作业",
+    "merged_id": "unit5_080",
+    "original_id": "unit5_choice_080"
+  },
+  {
+    "id": "unit5_choice_081",
+    "unit": "第五单元：圆",
+    "chapter": "扇形的认识",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "在同一个圆中, 扇形的大小与这个扇形的圆心角大小无关。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】在同一个圆中，扇形的大小与这个扇形的圆心角大小有关，圆心角越大扇形越大，反之亦然。【详解】根据分析可知：在同一个圆中，扇形的大小与这个扇形的圆心角大小无关，这种说法是错误的。故答案为: ×",
+    "knowledge_points": [
+      "弧、圆心角、扇形的认识"
+    ],
+    "source": "22-23 六年级上·湖北省直辖县级单位·期末",
+    "merged_id": "unit5_081",
+    "original_id": "unit5_choice_081"
+  },
+  {
+    "id": "unit5_choice_082",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长和面积",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "周长相等的两个圆面积一定相等。（ ）",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】根据圆周长公式：C = 2πr ，已知周长相等，可求出两个圆的半径相等，根据圆面积公式：S = πr² ，可确定两个圆的面积相等。据此解答。【详解】根据分析可知, 两个圆的周长相等, 它们的面积一定相等。原题干说法正确。故答案为: √",
+    "knowledge_points": [
+      "圆的面积",
+      "圆的周长"
+    ],
+    "source": "2017·江苏扬州·小升初真题",
+    "merged_id": "unit5_082",
+    "original_id": "unit5_choice_082"
+  },
+  {
+    "id": "unit5_choice_083",
+    "unit": "第五单元：圆",
+    "chapter": "画圆",
+    "question_type": "作图题",
+    "difficulty": "中等",
+    "question": "按下列要求用圆规画圆。以O点为圆心，OA长为半径画圆。![](images/080d62eeb6938f839785baf79410dc0d.png)",
+    "options": [],
+    "answer": "见详解",
+    "solution": "【分析】先确定一点为圆心 O，以该点为定点，圆规的针尖与 O 点重合，圆规两脚间的距离等于 O A，据此画图。【详解】画图如下：![](images/22d1862c7c442500cfeb9885576f02a5c896e0f3b43aa81af64efe4b5fc940ae.jpg)",
+    "knowledge_points": [
+      "画圆"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit5_083",
+    "original_id": "unit5_choice_083"
+  },
+  {
+    "id": "unit5_choice_084",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一种游戏车，前轮直径4分米，后轮直径3分米。前轮转动12周所走过的路程，后轮要转动多少周？",
+    "options": [],
+    "answer": "16周",
+    "solution": "【分析】根据圆的周长公式 C = πd , 前轮和后轮转动的路程相等, 通过计算前轮转动 12 周的路程, 再除以后轮的周长即可得到后轮转动的周数。【详解】(3.14 × 4 × 12) ÷ (3.14 × 3) =150.72÷9.42 =16 (周)答: 后轮要转动 16 周。",
+    "knowledge_points": [
+      "圆的周长的应用",
+      "圆的周长"
+    ],
+    "source": "25-26 六年级上·全国·单元测试",
+    "merged_id": "unit5_084",
+    "original_id": "unit5_choice_084"
+  },
+  {
+    "id": "unit5_choice_085",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一种手榴弹爆炸后，有效杀伤范围的半径是 7 米，有效杀伤面积是多少平方米？",
+    "options": [],
+    "answer": "153.86 平方米",
+    "solution": "【分析】根据圆的面积公式 S = πr² , 代入数据计算即可。【详解】3.14 × 7² =3.14 × 49 =153.86 (平方米)答: 有效杀伤面积是 153.86 平方米。",
+    "knowledge_points": [
+      "圆的面积",
+      "圆的面积的应用"
+    ],
+    "source": "25-26 六年级上·全国·单元测试",
+    "merged_id": "unit5_085",
+    "original_id": "unit5_choice_085"
+  },
+  {
+    "id": "unit5_choice_086",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "小明爸爸要用铁丝网围一个半径是 15 米的圆形鸡栏, 请问: 围这个鸡栏需要多少米的铁丝网？（接头处不计）",
+    "options": [],
+    "answer": "94.2 米",
+    "solution": "【分析】根据圆的周长公式 C = 2πr , 代入数据计算。【详解】2 × 3.14 × 15 =6.28 × 15 =94.2 (米)答: 围这个鸡栏需要 94.2 米的铁丝网。",
+    "knowledge_points": [
+      "圆的周长的应用",
+      "圆的周长"
+    ],
+    "source": "25-26 六年级上·全国·单元测试",
+    "merged_id": "unit5_086",
+    "original_id": "unit5_choice_086"
+  },
+  {
+    "id": "unit5_choice_087",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长和面积",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "某传送带装置侧面如图所示，大轮的面积是 3.14 m² 。（假设绳索与轮子之间没有滑动）(1) 若物体在传送带上, 小轮转动 2 圈, 物体前进多少米? (2) 物体从 A 点传送到 B 点时, 小轮转动的圈数约是大轮转动圈数的多少倍?![](images/1db7959400ebfcdd69a6e1e3ad893298a7f9b72899a376ae207a8ff20d266463.jpg)",
+    "options": [],
+    "answer": "(1) 6.28 米;(2) 2 倍",
+    "solution": "【分析】（1）根据题意可知，物体前进的距离，等于小轮转动2圈的周长，根据圆的周长 = π × 半径 × 2 ，代入数据，即可解答。(2) 根据圆的面积 = π × 半径 ² , 半径 ² = 面积 ÷ π , 据此先求出大轮的半径, 再根据物体从点传送到点的距离, 分别求出小轮和大轮转动的圈数, 进而求出倍数关系。【详解】(1) 3.14 × 0.5 × 2 × 2 = 1.57 × 2 × 2 = 3.14 × 2 = 6.28 (米) 答：物体前进 6.28 米。(2) 3.14 ÷ 3.14 = 1 （平方米）1 × 1 = 1 , 所以大轮半径是 1 米。小轮周长: 3.14 × 0.5 × 2 = 1.57 × 2 = 3.14 (米) 大轮周长: 3.14 × 1 × 2 = 3.14 × 2 = 6.28 (米) 小轮转动圈数: 15.7 ÷ 3.14 = 5 (圈) 大轮转动圈数: 15.7 ÷ 6.28 = 2.5 (圈)5 ÷ 2.5 = 2答: 小轮转动的圈数约是大轮转动圈数的 2 倍。",
+    "knowledge_points": [
+      "圆的周长的应用",
+      "圆的面积的应用"
+    ],
+    "source": "24-25 六年级上·全国·单元测试",
+    "merged_id": "unit5_087",
+    "original_id": "unit5_choice_087"
+  },
+  {
+    "id": "unit5_choice_088",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "下面是一家披萨店的致歉声明：顾客朋友们，很抱歉地通知你们，由于中午客流量大，店内直径是30厘米的批萨已经售空，我们将为您换成相同口味的 2 个直径是 15 厘米的披萨（厚度相同），祝您用餐愉快！如果你是顾客，你觉得这样换有没有吃亏？请说明理由。",
+    "options": [],
+    "answer": "吃亏;理由见详解",
+    "solution": "【分析】先根据圆的面积公式 S = πr² , 分别求出一个直径是 30 厘米批萨的面积, 2 个直径是 15 厘米披萨的面积, 再比较大小，得出结论。【详解】3.14 × (30 ÷ 2)² = 3.14 × 15² = 3.14 × 225 = 706.5 （平方厘米）3.14 × (15 ÷ 2)² × 2 = 3.14 × 7.5² × 2 = 3.14 × 56.25 × 2 = 353.25（平方厘米） 706.5 > 353.25答：这样换吃亏，因为 2 个直径是 15 厘米的披萨的面积小于一个直径是 30 厘米的批萨面积，所以吃亏了。",
+    "knowledge_points": [
+      "圆的面积的应用"
+    ],
+    "source": "24-25 六年级上·全国·单元测试",
+    "merged_id": "unit5_088",
+    "original_id": "unit5_choice_088"
+  },
+  {
+    "id": "unit5_choice_089",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "小明爸爸要用铁丝网围一个半径是 20 米的圆形鸡栏, 请问: 围这个鸡栏需要多少米的铁丝网? (接头处不计)",
+    "options": [],
+    "answer": "125.6 米",
+    "solution": "【分析】根据题意，求这个鸡栏需要铁丝网的长度，就是求半径是 20 米的圆的周长, 根据圆的周长公式: C = 2πr , 代入数据即可解答。【详解】3.14 × 20 × 2 = 62.8 × 2 = 125.6 (米) 答: 围这个鸡栏需要 125.6 米的铁丝网。",
+    "knowledge_points": [
+      "圆的周长的应用"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit5_089",
+    "original_id": "unit5_choice_089"
+  },
+  {
+    "id": "unit5_choice_090",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的面积",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "在一个半径是 3 米的圆形水池周围修一条宽 1 米的石子路, 这条石子路的面积是多少平方米？（π 取 3.14）",
+    "options": [],
+    "answer": "21.98平方米",
+    "solution": "【分析】根据题意，在一个半径是3米的圆形水池周围修一条宽1米的石子路，求这条石子路的面积，就是求圆环的面积；圆环的内圆半径r是3米，外圆半径R是（3+1）米；根据圆环的面积公式 S环 = π(R² - r²) ，代入数据计算即可求解。【详解】3 + 1 = 4 （米）3.14 × (4² - 3²) = 3.14 × (16 - 9) = 3.14 × 7 = 21.98 (平方米) 答: 这条石子路的面积是 21.98 平方米。",
+    "knowledge_points": [
+      "圆环的面积"
+    ],
+    "source": "23-24 五年级下·湖南邵阳·期末",
+    "merged_id": "unit5_090",
+    "original_id": "unit5_choice_090"
+  },
+  {
+    "id": "unit5_choice_091",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一根铁丝恰好围成一个边长是6.28米的正方形，如果用这根铁丝围成一个圆形，这个圆的面积是多少？",
+    "options": [],
+    "answer": "50.24 平方米",
+    "solution": "【分析】据题意可知, 正方形的周长与圆的周长相等。根据正方形的周长 = 边长 × 4 , 计算出正方形的周长即是圆的周长,再根据圆的周长公式 C = 2πr 的逆运算, 根据 r = C ÷ π ÷ 2 求出半径,最后根据圆的面积公式 S = πr² , 代入数据计算即可。【详解】6.28 × 4 ÷ 3.14 ÷ 2 =25.12 ÷ 3.14 ÷ 2 =8 ÷ 2 =4 (米) 3.14 × 4² =3.14 × 16 =50.24 (平方米)答: 这个圆的面积是 50.24 平方米。",
+    "knowledge_points": [
+      "圆的面积",
+      "正方形的周长",
+      "圆的周长"
+    ],
+    "source": "23-24六年级上·全国·单元测试",
+    "merged_id": "unit5_091",
+    "original_id": "unit5_choice_091"
+  },
+  {
+    "id": "unit5_choice_092",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "工人师傅为了能省力地把油桶搬到汽车上，想了一个办法，就是在汽车后面搭一块木板。油桶的半径是30厘米，工人师傅推着油桶沿木板滚了5圈正好把油桶推上了车。这块木板长多少米？",
+    "options": [],
+    "answer": "9.42 米",
+    "solution": "【分析】根据题意, 工人师傅推着油桶沿木板滚了 5 圈正好把油桶推上了车, 说明这块木板的长度等于圆周长的 5 倍;根据圆的周长公式 C = 2πr , 求出油桶滚 1 圈的长度, 再乘 5 即可。注意单位的换算: 1 米 = 100 厘米。【详解】30厘米 = 0.3 米2 × 3.14 × 0.3 = 6.28 × 0.3 = 1.884 (米) 1.884 × 5 = 9.42 (米)答: 这块木板长 9.42 米。",
+    "knowledge_points": [
+      "厘米和米之间的进率与换算",
+      "圆的周长的应用"
+    ],
+    "source": "24-25 六年级上·全国·单元测试",
+    "merged_id": "unit5_092",
+    "original_id": "unit5_choice_092"
+  },
+  {
+    "id": "unit5_choice_093",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的面积",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "沿着直径是 8 米的圆形水池周围铺一条1米宽的石子路，石子路占地面积是多少平方米？",
+    "options": [],
+    "answer": "28.26 平方米",
+    "solution": "【分析】求石子路的占地面积就是求圆环的面积，根据圆环的面积公式：S = πR² - πr² ，据此进行计算即可。【详解】8 ÷ 2 = 4 （米）4 + 1 = 5 (米) 3.14 × 5² - 3.14 × 4² = 3.14 × 25 - 3.14 × 16 = 3.14 × (25 - 16) = 3.14 × 9 = 28.26 (平方米) 答: 石子路占地面积是 28.26 平方米。",
+    "knowledge_points": [
+      "圆环的面积"
+    ],
+    "source": "24-25 六年级上·全国·课后作业",
+    "merged_id": "unit5_093",
+    "original_id": "unit5_choice_093"
+  },
+  {
+    "id": "unit5_choice_094",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "在一个周长为62.8米的圆形草坪中安装自动旋转喷灌装置，现在有射程为 20米、15 米和 10 米的三种装置, 你认为选哪种装置比较合适?安装在什么地方？请写出你的思考过程。",
+    "options": [],
+    "answer": "10米;见详解",
+    "solution": "【分析】已知圆形草坪的周长为 62.8 米, 根据圆的周长公式 C = 2πr 可知, r = C ÷ π ÷ 2 , 由此求出圆的半径, 也就是自动旋转喷灌装置的射程, 据此选择合适的装置。根据圆的特征，在同一个圆内，所有的半径都相等，由此可知自动旋转喷灌装置的安装位置。【详解】62.8 ÷ 3.14 ÷ 2 = 20 ÷ 2 = 10 (米) 答: 选射程为 10 米的装置比较合适, 安装在圆形草坪的圆心处。",
+    "knowledge_points": [
+      "小数的连除运算",
+      "圆的概念及特点",
+      "圆的周长的应用"
+    ],
+    "source": "23-24六年级上·辽宁·单元测试",
+    "merged_id": "unit5_094",
+    "original_id": "unit5_choice_094"
+  },
+  {
+    "id": "unit5_choice_095",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一辆自行车的前轮半径是40厘米，如果车轮每分转100圈，王叔叔骑着它通过一座长2512米的大桥，大约需要几分？",
+    "options": [],
+    "answer": "10分",
+    "solution": "【分析】已知自行车前轮半径是 40 厘米, 根据圆的周长公式 C = 2πr , 求出前轮的周长; 然后用前轮的周长乘车轮每分转的圈数, 即是车轮每分走的距离, 也就是自行车的速度;已知要通过一座长 2512 米的大桥, 根据 '路程÷速度 = 时间', 据此求出骑自行车通过这座大桥需要的时间。注意单位的换算: 1 米 = 100 厘米。【详解】40 厘米 = 0.4 米 2 × 3.14 × 0.4 = 2.512 （米） 2.512 × 100 = 251.2 （米）2512 ÷ 251.2 = 10 （分）答: 大约需要 10 分。",
+    "knowledge_points": [
+      "厘米和米之间的进率与换算",
+      "圆的周长的应用",
+      "除数是小数的小数除法",
+      "基础行程问题"
+    ],
+    "source": "23-24六年级上·辽宁·单元测试",
+    "merged_id": "unit5_095",
+    "original_id": "unit5_choice_095"
+  },
+  {
+    "id": "unit5_choice_096",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "半径分别为1、2、3、4 厘米的四个圆的周长之和是多少厘米？（π 取3.14)",
+    "options": [],
+    "answer": "62.8 厘米",
+    "solution": "【分析】根据圆的周长 = 2 × 圆周率 × 半径, 分别计算四个圆的周长, 相加即可。【详解】2 × 3.14 × 1 + 2 × 3.14 × 2 + 2 × 3.14 × 3 + 2 × 3.14 × 4 = 2 × 3.14 × (1 + 2 + 3 + 4) = 6.28 × 10 = 62.8 (厘米) 答: 四个圆的周长之和是 62.8 厘米。",
+    "knowledge_points": [
+      "圆的周长"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit5_096",
+    "original_id": "unit5_choice_096"
+  },
+  {
+    "id": "unit5_choice_097",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "如图, 有一个圆形花坛，直径为20米，一只小蜜蜂沿着花坛外周飞了一圈，请问它飞了多少米？如果小蜜蜂沿着图中的虚线，飞一个'8'字，路线构成过花坛圆心的两个小圆，那么这次它飞了多少米？（π 取 3.14）![](images/18b6f561a5c4fae081c420c703a8bff004c126be11132643a3250b0f1fc11e04.jpg)",
+    "options": [],
+    "answer": "62.8 米; 62.8 米",
+    "solution": "【分析】蜜蜂沿花坛外周飞时，路线是一个直径为 20 米的圆的周长，计算飞过的路程直接用 C = πd 计算即可；在花坛中飞'8'字路线时，两个小圆的直径等于大圆的半径，也就是10米，此时它飞过的路程等于两个圆的周长，算出一个小圆的周长再乘2即可。通过计算我们会发现两次飞过的路程是相等的，我们可以推导出一个结论：若多个小圆的直径之和等于大圆的直径，那么多个小圆的周长之和等于大圆的周长。πd₁ + πd₂ + πd₃ = π(d₁ + d₂ + d₃)【详解】第一次：3.14 × 20 = 62.8 （米）第二次: 20 ÷ 2 = 10 (米) 3.14 × 10 × 2 = 62.8 （米）答: 沿着花坛外周飞时, 它飞了 62.8 米; 飞一个 '8'字时,它也飞了 62.8 米。",
+    "knowledge_points": [
+      "圆的周长的应用"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit5_097",
+    "original_id": "unit5_choice_097"
+  },
+  {
+    "id": "unit5_choice_098",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长和面积",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "小明和小红经常到公园里的圆形露天舞台散步。一天，他们从圆形露天舞台边沿同一地点同时出发，沿着圆形露天舞台的边沿背向而行，2分钟后，两人相遇，小明每分钟走75米，小红每分钟走82米。(1) 这个圆形露天舞台的周长是多少米? (2) 这个圆形露天舞台占地多少平方米?",
+    "options": [],
+    "answer": "(1) 314 米(2) 7850 平方米",
+    "solution": "【分析】（1）根据题意，圆形露天舞台的周长 = 两人速度和 × 相遇时间，据此解答即可；(2) 根据圆形舞台的周长 C = 2πr , 据此求出其半径, 面积 S = πr² , 代入数据解答即可。【详解】(1) (75 + 82) × 2 = 157 × 2 = 314 (米) 答: 这个圆形露天舞台的周长是 314 米。(2) 314 ÷ 2 ÷ 3.14 = 157 ÷ 3.14 =50(米) 3.14 × 50² = 3.14 × 2500 = 7850 (平方米) 答: 这个圆形舞台的占地面积是 7850 平方米。",
+    "knowledge_points": [
+      "圆的周长的应用",
+      "圆的面积的应用",
+      "相遇问题"
+    ],
+    "source": "23-24五年级下·福建漳州·期末",
+    "merged_id": "unit5_098",
+    "original_id": "unit5_choice_098"
+  },
+  {
+    "id": "unit5_choice_099",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的面积",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一根空心钢管的横截面是环形，测得钢管的外圆直径是3分米，内圆直径是2分米，这根钢管的横截面的面积是多少平方分米？",
+    "options": [],
+    "answer": "3.925平方分米",
+    "solution": "【分析】根据题意, 先分别用外圆和内圆的直径除以 2 , 求出外圆和内圆的半径, 求这根钢管的横截面的面积, 即为环形面积, 根据环形面积公式: S = π(R² - r²) , 代入数据求值即可。【详解】由分析可得:3.14 × [ (3 ÷ 2)² - (2 ÷ 2)² ] = 3.14 × [ 1.5² - 1² ] = 3.14 × [ 2.25 - 1 ] = 3.14 × 1.25 = 3.925 (平方分米) 答: 这根钢管的横截面的面积是 3.925 平方分米。",
+    "knowledge_points": [
+      "小数与小数的乘法",
+      "圆环的面积"
+    ],
+    "source": "23-24六年级上·内蒙古通辽·期末",
+    "merged_id": "unit5_099",
+    "original_id": "unit5_choice_099"
+  },
+  {
+    "id": "unit5_choice_100",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的面积",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一个圆形养鱼池的周长是 100.48 米, 中间有一个圆形小岛, 小岛的半径是 6 米。这个养鱼池的水域面积是多少平方米？",
+    "options": [],
+    "answer": "690.8平方米",
+    "solution": "【分析】由题意可知: 养鱼池的水域是一个环形。根据圆的周长公式 C = 2πr 可知: r = C ÷ 2π , 代入数据求出外圆半径。再带入圆环的面积公式 S = π(R² - r²) , 计算即可。【详解】100.48 ÷ 3.14 ÷ 2 = 32 ÷ 2 = 16 (米) 3.14 × (16² - 6²) = 3.14 × (256 - 36) = 3.14 × 220 = 690.8 (平方米) 答: 这个养鱼池的水域面积是 690.8 平方米。",
+    "knowledge_points": [
+      "圆的周长的应用",
+      "圆环的面积"
+    ],
+    "source": "18- 19 六年级上·全国·期中",
+    "merged_id": "unit5_100",
+    "original_id": "unit5_choice_100"
+  }, {
+    "id": "unit5_choice_101",
+    "unit": "第五单元：圆",
+    "chapter": "圆的认识",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "一张圆形的纸, 要想找到它的圆心, 至少要对折 ( ) 次。",
+    "options": [
+      "A. 1",
+      "B. 2",
+      "C. 4"
+    ],
+    "answer": "B",
+    "solution": "【分析】将一个圆形纸片最少要对折两次，找到两条折痕相交的那个点，即圆心。本题主要考查了圆的认识, 明确圆心到圆上各个点的距离相等是解答本题的关键。【详解】一张圆形的纸, 要想找到它的圆心, 至少要对折 2 次。故答案为: B",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "图形的折叠问题"
+    ],
+    "source": "22-23 六年级上·河南信阳·期末",
+    "merged_id": "unit5_101",
+    "original_id": "unit5_choice_101"
+  },
+  {
+    "id": "unit5_choice_102",
+    "unit": "第五单元：圆",
+    "chapter": "圆周率的认识",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "有不少数学家都对圆周率做出过研究, 中国数学家刘徽在注释《九章算术》(263 年) 时只用圆内接正多边形就求得 π 的近似值, 得出精确到两位小数的 π 值, 他的方法被后人称为割圆术。虽然电子计算机的出现, 使 π 值计算一直到小数点后 5 万亿位。即使我们已经知道 π 是一个（）数，但在工程测量、数学解题过程中，大部分都取前两位数，就是 π ≈ 3.14 ，也产生了圆周率日（3月14日）。",
+    "options": [
+      "A. 无限小数",
+      "B. 有限小数",
+      "C. 循环小数",
+      "D. 较大数"
+    ],
+    "answer": "A",
+    "solution": "【分析】无限小数：一个小数的小数部分的位数是无限的；有限小数: 小数点后的小数个数有限的。一个数的小数部分从某一位起，一个或几个数字依次重复出现的无限小数叫循环小数。根据 π 的位数以及它是否取得近似数, 判断它是一个怎样的数。本题主要考查对有限小数、无限小数和循环小数的认识。【详解】由题干可知: π 的位数已经计算到 5 万亿位, 小数部分没有重复出现某个或某几个数字, 并且它只计算出了近似值, 综合分析, π 是一个无限小数。故答案为: A",
+    "knowledge_points": [
+      "有限小数和无限小数的认识",
+      "循环小数的认识与简写"
+    ],
+    "source": "22-23 五年级上·甘肃金昌·期中",
+    "merged_id": "unit5_102",
+    "original_id": "unit5_choice_102"
+  },
+  {
+    "id": "unit5_choice_103",
+    "unit": "第五单元：圆",
+    "chapter": "方中圆和圆中方",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "在一个正方形内画一个最大的圆, 这个圆的面积是正方形的 ( )。",
+    "options": [
+      "A. 1/4",
+      "B. π/4",
+      "C. 4/π",
+      "D. 3/4"
+    ],
+    "answer": "B",
+    "solution": "【分析】设正方形的边长是 4 分米, 在正方形中画一个最大圆, 其直径和正方形的边长相等, 由此求出半径, 再根据 '圆的面积公式: S = πr²' 求出圆的面积; 再根据正方形的面积 = 边长 × 边长, 求出正方形的面积, 最后根据求一个数占另一个数的几分之几, 用一个数除以另一个数, 则用圆的面积除以正方形的面积即可。此题主要考查的是正方形和圆的面积关系，熟记相关公式是解答本题的关键。【详解】设正方形的边长是4分米。圆的面积:π × (4 ÷ 2)² = π × 2² = π × 4 = 4π (平方分米)正方形的面积: 4 × 4 = 16 (平方分米) 4π ÷ 16 = π/4 这个圆的面积是正方形面积的 π/4 。故答案为: B",
+    "knowledge_points": [
+      "求一个数占另一个数几分之几",
+      "圆的面积及应用",
+      "正方形的面积"
+    ],
+    "source": "2020·山西朔州·小升初真题",
+    "merged_id": "unit5_103",
+    "original_id": "unit5_choice_103"
+  },
+  {
+    "id": "unit5_choice_104",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "两圆半径之比为2:3, 则这两圆面积之比是 ( )。",
+    "options": [
+      "A. 2:3",
+      "B. 4:6",
+      "C. 4:9"
+    ],
+    "answer": "C",
+    "solution": "【分析】已知两圆半径之比为 2:3 , 设小圆的半径为 2 , 则大圆的半径为 3 ; 先根据圆的面积公式 S = πr² , 分别求出两圆的面积, 再根据比的意义写出两圆的面积之比, 并化简。本题考查比的意义以及圆的面积公式的运用，明确两圆的面积之比等于它们半径的平方比。【详解】设小圆的半径为 2 , 则大圆的半径为 3 。小圆面积: π × 2² = 4π 大圆面积: π × 3² = 9π 4π : 9π = 4: 9 两圆半径之比为 2:3 , 则这两圆面积之比是 4:9 。故答案为: C",
+    "knowledge_points": [
+      "圆的面积及应用",
+      "比的化简",
+      "比的意义"
+    ],
+    "source": "18-19 六年级上·四川·单元测试",
+    "merged_id": "unit5_104",
+    "original_id": "unit5_choice_104"
+  },
+  {
+    "id": "unit5_choice_105",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长和面积",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "一个圆和一个正方形的周长相等，他们的面积比较（）。",
+    "options": [
+      "A. 正方形的面积大",
+      "B. 圆的面积大",
+      "C. 一样大",
+      "D. 无法确定"
+    ],
+    "answer": "B",
+    "solution": "【分析】周长相等的正方形和圆，圆的面积比正方形的面积大，可以通过举例证明，设周长是 C，则正方形的边长是 C÷4，圆的半径是 C÷2π；根据它们的面积公式，求出它们的面积，再进行比较，即可解答。此题主要考查周长相等的正方形和圆, 圆的面积比正方形的面积大。【详解】设周长是 C, 则正方形的边长是: C ÷ 4 = C/4 , 圆的半径是: C ÷ 2π = C/(2π) , 则圆的面积为: π(C/(2π))² = C²/(4π) , 正方形的面积为: C/4 × C/4 = C²/16 , 因为 C²/(4π) > C²/16 , 所以圆的面积大于正方形的面积。故答案为：B",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用",
+      "正方形的面积",
+      "正方形的周长"
+    ],
+    "source": "2019·全国·小升初真题",
+    "merged_id": "unit5_105",
+    "original_id": "unit5_choice_105"
+  },
+  {
+    "id": "unit5_choice_106",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "一个圆的周长是25.12 厘米，画这个圆时，圆规两脚间的距离是（）厘米。",
+    "options": [
+      "A. 4",
+      "B. 8",
+      "C. 5"
+    ],
+    "answer": "A",
+    "solution": "【分析】根据圆的周长：C = 2πr 可知，r = C ÷ π ÷ 2 ，用 25.12 ÷ 3.14 ÷ 2 即可求出圆的半径; 用圆规画圆时, 圆规两脚间的距离等于所画圆的半径, 据此解答。本题考查圆周长公式的灵活运用，明确圆规两脚间的距离等于所画圆的半径是解题的关键。【详解】25.12 ÷ 3.14 ÷ 2 = 8 ÷ 2 = 4 (厘米)圆规两脚间的距离是4厘米。故答案为：A",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "19-20 六年级上·全国·单元测试",
+    "merged_id": "unit5_106",
+    "original_id": "unit5_choice_106"
+  },
+  {
+    "id": "unit5_choice_107",
+    "unit": "第五单元：圆",
+    "chapter": "方中圆的面积",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "要想在一张正方形纸上剪出一个面积是 12.56 平方厘米的圆形纸片, 这张正方形纸的面积至少是（）平方厘米。",
+    "options": [
+      "A. 12.56",
+      "B. 14",
+      "C. 16",
+      "D. 25"
+    ],
+    "answer": "C",
+    "solution": "【分析】当正方形的边长等于圆的直径时，圆外面正方形的面积最小，根据 'S = πr²' 先求出半径的平方，再求出圆的半径，最后求出正方形的边长，并根据 'S = a²' 求出正方形的面积，据此解答。理解最小正方形的边长等于圆的直径，并灵活运用圆的面积计算公式是解答题目的关键。【详解】![](images/378177968568853753a7664a99ebed335191383a209cbdd1f8cab0e590f0aa34.jpg)半径的平方: 12.56 ÷ 3.14 = 4 (平方厘米)因为 4 = 2² , 所以圆形纸片的半径是 2 厘米。边长: 2 × 2 = 4 (厘米)正方形的面积: 4 × 4 = 16 (平方厘米)所以，这张正方形纸的面积至少是 16 平方厘米。故答案为: C",
+    "knowledge_points": [
+      "圆的面积及应用",
+      "正方形的面积"
+    ],
+    "source": "22-23 六年级上·河南三门峡·期末",
+    "merged_id": "unit5_107",
+    "original_id": "unit5_choice_107"
+  },
+  {
+    "id": "unit5_choice_108",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "一个圆的半径扩大 4 倍, 面积扩大 ( ) 倍。",
+    "options": [
+      "A. 4",
+      "B. 8",
+      "C. 16",
+      "D. 2"
+    ],
+    "answer": "C",
+    "solution": "【分析】可用设数法解决此题。设原来圆的半径为 1 , 则扩大后圆的半径为 4 。根据圆的面积 S = πr² 分别计算出两圆的面积, 再用扩大后圆的面积除以原来圆的面积, 从而得到两圆面积的倍数关系。如果一个圆的半径扩大到原来的若干倍, 则这个圆的面积就扩大到该倍数的平方倍。【详解】设原来圆的半径为 1 , 则扩大后圆的半径为 4 。原来圆的面积: π × 1² = π × 1 = π 扩大后圆的面积: π × 4² = π × 16 = 16π 16π ÷ π = 16 所以面积扩大 16 倍。故答案为: C",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "14-15 六年级上·全国·课后作业",
+    "merged_id": "unit5_108",
+    "original_id": "unit5_choice_108"
+  },
+  {
+    "id": "unit5_choice_109",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "把一个圆的半径扩大到原来的2倍，它的面积将扩大到原来的（）倍。",
+    "options": [
+      "A. 2",
+      "B. 4",
+      "C. 3",
+      "D. 9"
+    ],
+    "answer": "B",
+    "solution": "【分析】可采用设数法解决此题。不妨先设原来圆的半径为 1 厘米, 则扩大后圆的半径是 2 厘米; 再根据圆的面积 S = πr² 分别求出原来圆的面积和扩大后圆的面积；最后用扩大后圆的面积÷原来圆的面积可求出面积扩大的倍数。如果一个圆的半径扩大到原来的若干倍, 则这个圆的面积就扩大到该倍数的平方倍。【详解】设原来圆的半径为1厘米。原来圆的面积: π × 1² = π × 1 = π (平方厘米)扩大后圆的面积: π × (1 × 2)² = π × 2² = π × 4 = 4π （平方厘米） 4π ÷ π = 4 所以它的面积将扩大到原来的 4 倍。故答案为: B",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "19-20六年级上·河北·单元测试",
+    "merged_id": "unit5_109",
+    "original_id": "unit5_choice_109"
+  },
+  {
+    "id": "unit5_choice_110",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "一个圆的直径等于一个正方形的边长，这个圆的面积（）正方形的面积。",
+    "options": [
+      "A. 大于",
+      "B. 小于",
+      "C. 等于",
+      "D. 无法比较"
+    ],
+    "answer": "B",
+    "solution": "【分析】根据题意, 可设正方形的边长为 4 厘米, 那么圆的半径为 2 厘米, 可根据正方形的面积公式和圆的面积公式进行计算后再比较即可得到答案。此题主要考查的是正方形的面积公式和圆的面积公式的应用。【详解】设正方形的边长为 4 厘米, 则圆的半径为 2 厘米, 正方形的面积为: 4 × 4 = 16 (平方厘米),圆的面积为: 3.14 × 2² = 12.56 (平方厘米),所以正方形的面积大于圆的面积，即这个圆的面积小于正方形的面积。故答案为: B",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "正方形的周长"
+    ],
+    "source": "2021·河北石家庄·小升初真题",
+    "merged_id": "unit5_110",
+    "original_id": "unit5_choice_110"
+  },
+  {
+    "id": "unit5_choice_111",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "周长是 15.7 cm 的圆,画这个圆时圆规两脚间的距离是 ( )。",
+    "options": [
+      "A. 2 cm",
+      "B. 2.5 cm",
+      "C. 4 cm",
+      "D. 5 cm"
+    ],
+    "answer": "B",
+    "solution": "【分析】圆规两脚之间的距离就是圆的半径，根据'r = C ÷ π ÷ 2'求出圆的半径，据此解答。灵活运用圆的周长计算公式是解答题目的关键。【详解】15.7 ÷ 3.14 ÷ 2 = 5 ÷ 2 = 2.5 (cm)所以，画这个圆时圆规两脚间的距离是 2.5 cm 。故答案为: B",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "18-19 六年级上·山东·阶段练习",
+    "merged_id": "unit5_111",
+    "original_id": "unit5_choice_111"
+  },
+  {
+    "id": "unit5_choice_112",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "在推导圆的面积公式时，把一个圆分成若干等份后，拼成一个近似的长方形，这个长方形的长是（）。",
+    "options": [
+      "A. 圆的半径",
+      "B. 圆的直径",
+      "C. 圆的周长",
+      "D. 圆周长的一半"
+    ],
+    "answer": "D",
+    "solution": "【分析】推导圆的面积公式时，沿着半径将圆分成若干份，再拼成一个近似长方形。近似长方形的长是圆周长的一半，宽是圆的半径。据此解题。本题考查了圆的面积, 掌握圆面积的推导过程是解题的关键。【详解】在推导圆的面积公式时, 把一个圆分成若干等份后, 拼成一个近似的长方形, 这个长方形的长是圆周长的一半。故答案为: D",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级上·山东临沂·期末",
+    "merged_id": "unit5_112",
+    "original_id": "unit5_choice_112"
+  },
+  {
+    "id": "unit5_choice_113",
+    "unit": "第五单元：圆",
+    "chapter": "圆的认识",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "把圆规的两脚分开，使两脚之间的距离是 3 cm ，这样画出的圆的直径是（） cm 。",
+    "options": [
+      "A. 3",
+      "B. 4",
+      "C. 6"
+    ],
+    "answer": "C",
+    "solution": "【分析】圆规两脚之间的距离是圆的半径，同圆中，圆的直径是半径的 2 倍，据此答题即可。掌握圆规画圆的方法以及圆的特征是解题的关键。【详解】由分析可知:3 × 2 = 6 (cm)这样画出的圆的直径是 6 cm 。故答案为: C",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "21-22六年级上·河北沧州·期末",
+    "merged_id": "unit5_113",
+    "original_id": "unit5_choice_113"
+  },
+  {
+    "id": "unit5_choice_114",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "小圆的直径等于大圆的半径，则大圆的面积与小圆的面积比是（）。",
+    "options": [
+      "A. 1:2",
+      "B. 2:1",
+      "C. 1:4",
+      "D. 4:1"
+    ],
+    "answer": "D",
+    "solution": "【分析】小圆的直径等于大圆的半径，说明大圆半径是小圆半径的2倍，假设小圆半径是1，大圆半径是2，根据圆的面积 = πr² ，分别表示出大圆和小圆的面积，写出面积比，化简即可。关键是理解比的意义，掌握并灵活运用圆的面积公式。【详解】假设小圆半径是 1 , 大圆半径是 2 。(π × 2²) : (π × 1²) = 2² : 1² = 4: 1 大圆的面积与小圆的面积比是 4:1 。故答案为: D",
+    "knowledge_points": [
+      "圆的面积及应用",
+      "比的化简",
+      "比的意义"
+    ],
+    "source": "2014 六年级·全国·课后作业",
+    "merged_id": "unit5_114",
+    "original_id": "unit5_choice_114"
+  },
+  {
+    "id": "unit5_choice_115",
+    "unit": "第五单元：圆",
+    "chapter": "平面图形的分割",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "李阳把一张长 12 厘米、宽 9 厘米的彩色纸剪成半径是 2 cm 的小圆（不能剪拼），最多能剪（）个小圆。",
+    "options": [
+      "A. 8",
+      "B. 6",
+      "C. 9"
+    ],
+    "answer": "B",
+    "solution": "【分析】因为圆的半径是 2 cm , 则直径是 4 cm , 那么这张彩色纸可以剪 (9 ÷ 4) 排, 每排可以剪 (12 ÷ 4) 个, 用乘法即可解题。找出要剪的圆的直径，是解答此题的关键。【详解】2 × 2 = 4 （厘米） 9 ÷ 4 = 2 (个) ··· 1 (厘米) 12 ÷ 4 = 3 (个) 2 × 3 = 6 （个）所以，最多能剪 6 个圆。故答案为: B",
+    "knowledge_points": [
+      "长方形的概念及特点",
+      "圆的概念及特点",
+      "平面图形的分割"
+    ],
+    "source": "20-21 六年级上·河南新乡·期末",
+    "merged_id": "unit5_115",
+    "original_id": "unit5_choice_115"
+  },
+  {
+    "id": "unit5_choice_116",
+    "unit": "第五单元：圆",
+    "chapter": "圆的认识",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "一张圆形的纸，至少要对折（）次，才得到圆心。",
+    "options": [
+      "A. 1 次",
+      "B. 2 次",
+      "C. 3 次"
+    ],
+    "answer": "B",
+    "solution": "【分析】圆的两条直径相交，得到的交点一定是圆心。如果一个图形沿一条直线对折，直线两侧的部分能够完全重合，这个图形就是轴对称图形，对折一次展开，圆中出现的折痕，就是它的直径，将圆换一个角度，再对折一次，使两个半圆重合，展开后又出现了一条折痕。因此至少要对折两次，才能使两条直径相交，从而得到圆心。据此解答。圆是轴对称图形, 采取对折的方法能确定圆心,注意两次对折的角度是不同的, 否则重合的两条直径上有无数个交点, 不能确定圆心。【详解】由分析得：一张圆形的纸，至少要对折（2）次，才得到圆心。故答案为: B",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "图形的折叠问题",
+      "轴对称的认识及辨认"
+    ],
+    "source": "2014 六年级·全国·课后作业",
+    "merged_id": "unit5_116",
+    "original_id": "unit5_choice_116"
+  },
+  {
+    "id": "unit5_choice_117",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长和面积",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "大圆的半径与小圆半径的比是 2:1 , 则大圆的周长与小圆周长的比是 ( ) ,大圆的面积与小圆面积的比是 ( )。",
+    "options": [
+      "A. 2:1; 8:1",
+      "B. 2:1; 4:1",
+      "C. 8:1; 2:1"
+    ],
+    "answer": "B",
+    "solution": "【分析】此题可采用设数法解决。根据大圆半径与小圆半径的比是 2:1 , 可设大圆半径为 2 , 小圆半径为 1 。根据圆的周长公式 C = 2πr , 分别求出两圆的周长, 再求两圆周长的比。根据圆的面积公式 S = πr² , 分别求出两圆的面积,再求两圆面积的比。（1）圆的半径或直径扩大到原来的几倍，它的周长也扩大到原来的几倍。 (2) 圆的半径（直径或周长）扩大到原来的若干倍，则这个圆的面积就扩大到该倍数的平方倍。【详解】设大圆半径为2，小圆半径为1。大圆周长为: 2π × 2 = 4π ; 小圆周长为: 2π × 1 = 2π ;大圆周长: 小圆周长 = 4π / 2π = 2: 1 。大圆面积: π × 2² = 4π ; 小圆面积: π × 1² = π ;大圆面积: 小圆面积 = 4π / π = 4: 1 。A. 周长的比为 2:1 是正确的; 面积的比是 8:1 是错误的。 B. 周长的比为 2:1 是正确的; 面积的比是 4:1 是正确的。 C. 周长的比为 8:1 是错误的; 面积的比是 2:1 是错误的。故答案为: B",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用",
+      "比的化简"
+    ],
+    "source": "21-22 六年级上·广东广州·期末",
+    "merged_id": "unit5_117",
+    "original_id": "unit5_choice_117"
+  },
+  {
+    "id": "unit5_choice_118",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长和面积",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "周长相等, 面积最大的是 ( )。",
+    "options": [
+      "A. 长方形",
+      "B. 正方形",
+      "C. 圆"
+    ],
+    "answer": "C",
+    "solution": "【分析】根据题意，假设周长都等于 12.56 分米，可利用正方形、长方形、圆的周长公式，分别计算出它们的边长或半径，然后再依据面积公式计算出这些图形的面积，然后再比较大小即可得到答案。在周长一定的情况下，所围成的平面图形的面积从大到小依次是圆、正方形、长方形、三角形，即越接近圆面积越大。【详解】假设周长都等于 12.56 分米,正方形的边长是: 12.56 ÷ 4 = 3.14 (分米)正方形的面积是: 3.14 × 3.14 = 9.8596 (平方分米);长方形的长和宽的和是: 12.56 ÷ 2 = 6.28 (分米)长和宽越接近面积越大，长可为3.15分米，宽为3.13分米，长方形的面积是: 3.15 × 3.13 = 9.8595 (平方分米);圆的半径是: 12.56 ÷ 2 ÷ 3.14 = 2 (分米)圆的面积是: 2 × 2 × 3.14 = 12.56 (平方分米); 9.8595 < 9.8596 < 12.56 ;所以围成的圆的面积最大。故答案为: C",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用",
+      "正方形的面积",
+      "长方形的面积"
+    ],
+    "source": "14-15 六年级·全国·期末",
+    "merged_id": "unit5_118",
+    "original_id": "unit5_choice_118"
+  },
+  {
+    "id": "unit5_choice_119",
+    "unit": "第五单元：圆",
+    "chapter": "圆的对称轴",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "以同一个点为圆心，画两个大小不同的圆，这个图形有（）条对称轴。",
+    "options": [
+      "A. 0",
+      "B. 1",
+      "C. 2",
+      "D. 无数"
+    ],
+    "answer": "D",
+    "solution": "【分析】假设同一点为 A 点，先以 A 点为圆心画一个小圆，再同样以 A 点为圆心画一个较大的圆，据此解答。解答本题的关键要注意该图形是同一个点为圆心。【详解】作图如下：![](images/2569852d90d660bb4b73c12c27feb80a7fd6671d2820f441cbba14932bcb3cf0.jpg)观察图形发现，过圆心 A 的直线都是该图形的对称轴。故答案为: D",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "轴对称的认识及辨认"
+    ],
+    "source": "2022·湖南岳阳·小升初真题",
+    "merged_id": "unit5_119",
+    "original_id": "unit5_choice_119"
+  },
+  {
+    "id": "unit5_choice_120",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "明明用圆规画一个周长是 31.4 cm 的圆，圆规两脚间的距离是（） cm。",
+    "options": [
+      "A. 15.7",
+      "B. 5",
+      "C. 10"
+    ],
+    "answer": "B",
+    "solution": "【分析】根据圆规画圆的方法可知, 圆规两脚间的距离是圆的半径; 根据圆的周长公式 C = 2πr 可知, 圆的半径 r = C ÷ π ÷ 2 , 代入数据计算即可。灵活运用圆的周长公式是解题的关键。【详解】31.4 ÷ 3.14 ÷ 2 = 10 ÷ 2 = 5 (cm)故答案为: B",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "2022·黑龙江哈尔滨·小升初真题",
+    "merged_id": "unit5_120",
+    "original_id": "unit5_choice_120"
+  },
+  {
+    "id": "unit5_choice_121",
+    "unit": "第五单元：圆",
+    "chapter": "圆周率与其他知识",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "下面说法, 正确的有 ( ) 个。(1) 计算 1/3 + 2/7 时, 分子不能直接相加, 是因为分数单位不同 (2) 圆的周长总是直径的 3 倍 (3) 一根钢管长 2 米, 用去 2/3 , 还剩 1/3 (4) 大于 2 的偶数都是合数",
+    "options": [
+      "A. 1",
+      "B. 2",
+      "C. 3",
+      "D. 4"
+    ],
+    "answer": "C",
+    "solution": "【分析】A．根据分数的意义可知，异分母分数分数单位不同，因此不能直接相加减，异分母分数相加减计算时要先进行通分化为同分母的分数后再进行加减；B. 根据教材中关于圆周率的含义的推导可知: 圆的周长总是它直径的 3 倍多一些, 圆的周长和它直径的比值, 叫做圆周率, 圆周率用 'π' 表示;C. 把钢管的全长看成单位 '1', 用 1 减去用去的 2/3 , 即可还剩下全长的 1 - 2/3 ;D. 因为能被 2 整除的数为偶数, 又因为除了 1 和它本身外,还有别的因数的数为合数，所以大于 2 的偶数除了 1 和它本身外至少还有其它因数，符合合数的特点。本题考查了学生对于异分母加减的计算法则, 圆周率的认识, 分数的意义, 合数与偶数的意义。【详解】A．据分数的意义可知，异分母分数分数单位不同，因此 1/3 + 2/7 不能直接相加减，选项说法正确； B. 根据圆周率的定义, 圆的周长总是直径的 3 倍多一些,选项说法错误; C. 由分析可得: 1 - 2/3 = 1/3 , 选项说法正确; D. 因为大于 2 的偶数除了 1 和它本身外至少还有 2 这个约数, 符合合数的特点, 所以大于 2 的偶数都是合数, 选项说法正确;故答案为: C",
+    "knowledge_points": [
+      "异分母分数加、减法",
+      "圆的概念及特点",
+      "质数与合数",
+      "分数的意义"
+    ],
+    "source": "21-22 五年级下·湖南邵阳·期末",
+    "merged_id": "unit5_121",
+    "original_id": "unit5_choice_121"
+  },
+  {
+    "id": "unit5_choice_122",
+    "unit": "第五单元：圆",
+    "chapter": "圆的对称轴",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "下列图形中，对称轴最多的是（）。",
+    "options": [
+      "A. 正方形",
+      "B. 圆",
+      "C. 等边三角形",
+      "D. 长方形"
+    ],
+    "answer": "B",
+    "solution": "【分析】依据轴对称图形的概念, 即在平面内, 如果一个图形沿一条直线折叠, 直线两旁的部分能够完全重合, 这样的图形叫做轴对称图形, 据此先找出对称轴, 从而得出对称轴最多的图形即可进行选择。解答此题的主要依据是：轴对称图形的定义及其对称轴的条数。【详解】A．正方形有4条对称轴； B. 圆有无数条对称轴; C. 等边三角形有 3 条对称轴; D. 长方形有 2 条对称轴。故答案为: B",
+    "knowledge_points": [
+      "正方形的概念及特点",
+      "圆的概念及特点",
+      "等腰三角形和等边三角形的认识及特征",
+      "对称轴的画法及数量"
+    ],
+    "source": "2014·全国·小升初真题",
+    "merged_id": "unit5_122",
+    "original_id": "unit5_choice_122"
+  },
+  {
+    "id": "unit5_choice_123",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "将一个圆平均分成 1000 个完全相同的小扇形，割拼成近似的长方形的周长比原来圆的周长长 20 厘米，这个圆的面积是（）平方厘米。",
+    "options": [
+      "A. 62.8",
+      "B. 1256",
+      "C. 31.4",
+      "D. 314"
+    ],
+    "answer": "D",
+    "solution": "【分析】将一个圆平均分成 1000 个完全相同的小扇形, 割拼成近似的长方形, 周长增加了两条半径, 增加的周长 ÷ 2 = 半径, 再根据圆的面积 = πr² , 列式计算即可。关键是熟悉圆面积公式推导公式，掌握圆的面积公式。【详解】3.14 × (20 ÷ 2)² = 3.14 × 100 = 314 (平方厘米)故答案为: D",
+    "knowledge_points": [
+      "圆的面积及应用",
+      "平面图形的拼接"
+    ],
+    "source": "21-22六年级下·湖北武汉·期末",
+    "merged_id": "unit5_123",
+    "original_id": "unit5_choice_123"
+  },
+  {
+    "id": "unit5_choice_124",
+    "unit": "第五单元：圆",
+    "chapter": "半圆的周长",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "一个半圆, 直径是 d, 它的周长是 ( )。",
+    "options": [
+      "A. (1/2)πd",
+      "B. πd + d",
+      "C. (1/2)πd + d"
+    ],
+    "answer": "C",
+    "solution": "【分析】半圆为周长为圆周长的一半加上直径, 即: (1/2)πd + d; 据此可得出答案。本题主要考查的是扇形的周长, 解题的关键是熟练运用半圆周长公式, 进而得出答案。【详解】半圆, 直径是 d , 它的周长是 (1/2)πd + d 。故答案为: C。",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "21-22 六年级上·江西吉安·期末",
+    "merged_id": "unit5_124",
+    "original_id": "unit5_choice_124"
+  },
+  {
+    "id": "unit5_choice_125",
+    "unit": "第五单元：圆",
+    "chapter": "方中圆的面积",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "在一个边长为 6 cm的正方形中剪一个最大的圆, 则正方形与圆的面积比为 ( )。",
+    "options": [
+      "A. 4:π",
+      "B. π:2",
+      "C. 2:π",
+      "D. π:4"
+    ],
+    "answer": "A",
+    "solution": "【分析】在正方形内剪一个最大的圆，圆的直径和正方形的边长相等, 据此可知圆的半径为 6 ÷ 2 = 3 cm , 再分别求出正方形和圆的面积, 写出它们的比即可。明确在正方形内剪一个最大的圆，圆的直径和正方形的边长的关系是解答本题的关键。【详解】6 ÷ 2 = 3 (cm)正方形: 6 × 6 = 36(cm²)圆: 3² π = 9π(cm²)正方形与圆的面积比为 36: 9π = 4: π故答案为: A。",
+    "knowledge_points": [
+      "圆的面积及应用",
+      "正方形的面积",
+      "比的意义"
+    ],
+    "source": "20-21 六年级上·云南文山·期末",
+    "merged_id": "unit5_125",
+    "original_id": "unit5_choice_125"
+  },
+  {
+    "id": "unit5_choice_126",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "一根铁丝正好围成一个直径是 8 cm 的圆, 如果围成正方形, 它的边长是 ( )。",
+    "options": [
+      "A. 3.14",
+      "B. 6.28",
+      "C. 12.56",
+      "D. 25.12"
+    ],
+    "answer": "B",
+    "solution": "【分析】根据圆的周长公式: C = πd , 把数据代入公式求出这根铁丝的长度, 再根据正方形的周长公式: C = 4a , 那么 a = C ÷ 4 , 把数据代入公式解答。此题主要考查圆的周长公式、正方形的周长公式的灵活运用，关键是熟记公式【详解】3.14 × 8 ÷ 4 = 25.12 ÷ 4 = 6.28故答案为：B",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "正方形的周长"
+    ],
+    "source": "19-20 六年级上·福建南平·期中",
+    "merged_id": "unit5_126",
+    "original_id": "unit5_choice_126"
+  },
+  {
+    "id": "unit5_choice_127",
+    "unit": "第五单元：圆",
+    "chapter": "圆的认识",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "下水道的井盖设计成圆形, 主要是因为 ( )。",
+    "options": [
+      "A. 美观",
+      "B. 周长相等",
+      "C. 直径相等, 怎么放都掉不下去"
+    ],
+    "answer": "C",
+    "solution": "【详解】根据圆的特征，下水道的井盖设计成圆形，主要是因为直径相等，怎么放都掉不下去。故答案为: C",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "20-21 六年级下·山西阳泉·期中",
+    "merged_id": "unit5_127",
+    "original_id": "unit5_choice_127"
+  },
+  {
+    "id": "unit5_choice_128",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的面积",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "一个周长是 43.96 cm 的圆, 半径增加了 5 cm , 面积增加了 ( )。",
+    "options": [
+      "A. 153.86 cm²",
+      "B. 452.16 cm²",
+      "C. 298.3 cm²",
+      "D. 518.1 cm²"
+    ],
+    "answer": "C",
+    "solution": "【分析】圆的半径增加，增加的形状是个圆环，先确定大圆和小圆半径，根据圆环面积 = (R² - r²) ，列式计算即可。关键是掌握圆环面积公式，先确定大圆和小圆半径。【详解】43.96 ÷ 3.14 ÷ 2 = 7 （厘米） 7 + 5 = 12 (厘米)3.14 × (12² - 7²) = 3.14 × (144 - 49) = 3.14 × 95 = 298.3 （平方厘米）故答案为: C",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆环的面积"
+    ],
+    "source": "21-22 六年级上·云南曲靖·期末",
+    "merged_id": "unit5_128",
+    "original_id": "unit5_choice_128"
+  },
+  {
+    "id": "unit5_choice_129",
+    "unit": "第五单元：圆",
+    "chapter": "圆的认识",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "圆规两脚叉开 3 cm , 画出的圆的直径是 ( )。",
+    "options": [
+      "A. 3 cm",
+      "B. 6 cm",
+      "C. 18.84 cm²"
+    ],
+    "answer": "B",
+    "solution": "【分析】画圆时，圆规两角叉开的距离是半径，根据直径 = 半径 × 2 ，计算即可。关键是熟悉圆的特征，掌握画圆的方法。【详解】3 × 2 = 6 （厘米）故答案为: B",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "画圆"
+    ],
+    "source": "21-22 六年级上·四川绵阳·期末",
+    "merged_id": "unit5_129",
+    "original_id": "unit5_choice_129"
+  },
+  {
+    "id": "unit5_choice_130",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "圆的半径扩大到原来的 3 倍，面积扩大到原来的（）倍。",
+    "options": [
+      "A. 3",
+      "B. 6",
+      "C. 9"
+    ],
+    "answer": "C",
+    "solution": "【分析】根据圆的面积公式 S = πr² 可知，其中 π 是一个定值，根据积的变化规律：一个因数不变，另一个因数扩大或缩小为原来的几分之一，积就扩大或缩小为原来的几分之一；如果两个因数都扩大 n 倍，积就扩大到 n² 倍，据此即可解答。此题考查了积的变化规律在圆的面积公式中的灵活应用。【详解】圆的面积 S = πr² , r 扩大 3 倍, 则圆的面积就扩大: 3 × 3 = 9 倍。故答案为: C",
+    "knowledge_points": [
+      "积的变化规律（小数乘法）",
+      "圆的面积及应用"
+    ],
+    "source": "2014·全国·小升初真题",
+    "merged_id": "unit5_130",
+    "original_id": "unit5_choice_130"
+  },
+  {
+    "id": "unit5_choice_131",
+    "unit": "第五单元：圆",
+    "chapter": "圆周率",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "圆的周长同它的直径的比值是一个（）。",
+    "options": [
+      "A. 两位小数",
+      "B. 循环小数",
+      "C. 无限不循环小数"
+    ],
+    "answer": "C",
+    "solution": "【分析】圆周率是圆的周长与直径的比值，一般用希腊字母 π 表示， π 是一个常数（约等于 3.14），它是一个无限不循环小数。在日常生活中，通常都用 3.14 代表圆周率去进行近似计算。【详解】圆的周长同它的直径的比值是一个无限不循环小数。故答案为: C",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "2014 六年级·全国·课后作业",
+    "merged_id": "unit5_131",
+    "original_id": "unit5_choice_131"
+  },
+  {
+    "id": "unit5_choice_132",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长和面积",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "大小两个圆的半径之比是4:3，它们的周长之比是( )，面积之比是( )。",
+    "options": [],
+    "answer": "4:3 16:9",
+    "solution": "【分析】根据'大小两个圆的半径之比是 4:3'，可以设大圆的半径是4，则小圆的半径是3；然后根据圆的周长公式 C = 2πr ，圆的面积公式 S = πr² ，分别求出大圆、小圆的周长和面积，再根据比的意义，写出它们的周长之比和面积之比，并化简比。明确两个圆的周长之比等于它们的半径之比, 两个圆的面积之比等于它们的半径的平方比。【详解】设大圆的半径是 4 , 则小圆的半径是 3 。大圆的周长是: 2π × 4 = 8π 小圆的周长是: 2π × 3 = 6π大圆与小圆的周长之比是 8π: 6π = 4: 3 ;大圆的面积是: 2π × 4² = 32π 小圆的周长是: 2π × 3² = 18π 大圆与小圆的面积之比是 32π: 18π = 16: 9 。大小两个圆的半径之比是 4:3 , 它们的周长之比是 4:3 , 面积之比是 16:9 。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用",
+      "比的化简",
+      "比的意义"
+    ],
+    "source": "18-19六年级上·广东惠州·期末",
+    "merged_id": "unit5_132",
+    "original_id": "unit5_choice_132"
+  },
+  {
+    "id": "unit5_choice_133",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "一个时钟的时针长 5 厘米,这个时针的尖端转动一昼夜所走的路程是( )厘米。",
+    "options": [],
+    "answer": "62.8",
+    "solution": "【分析】根据生活经验可知, 一昼夜有 24 小时, 一圈有 12 小时, 所以一昼夜时针转了 2 圈, 根据圆的周长公式: C = 2πr , 用 2 × 3.14 × 5 × 2 即可解答。此题主要考查圆的周长公式的灵活运用，关键是熟记公式。【详解】2 × 3.14 × 5 × 2 = 31.4 × 2 = 62.8 (厘米)这个时针的尖端转动一昼夜所走的路程是 62.8 厘米。",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "2017·全国·小升初真题",
+    "merged_id": "unit5_133",
+    "original_id": "unit5_choice_133"
+  },
+  {
+    "id": "unit5_choice_134",
+    "unit": "第五单元：圆",
+    "chapter": "扇形的认识",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "( )个圆心角是 90° 的相同大小的扇形可以组成一个圆。",
+    "options": [],
+    "answer": "4",
+    "solution": "【分析】因为圆周角是 360 度, 所以只要求出 360 里面有几个 90 即可, 由此用除法列式解答。关键是明确圆周角是 360 度, 再根据除法的意义解答。【详解】因为 360° ÷ 90° = 4 （个）所以 4 个圆心角是 90° 的相同大小的扇形可以组成一个圆。",
+    "knowledge_points": [
+      "扇形的周长和面积",
+      "圆的概念及特点"
+    ],
+    "source": "22-23 六年级上·广东江门·期末",
+    "merged_id": "unit5_134",
+    "original_id": "unit5_choice_134"
+  },
+  {
+    "id": "unit5_choice_135",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "要画一个周长是25.12cm 的圆，圆规两脚间的距离是( )cm。",
+    "options": [],
+    "answer": "4",
+    "solution": "【分析】求圆规两脚间的距离, 实际上是求圆的半径, 根据圆的周长公式: C = 2πr , 代入数据即可求出圆的半径,即圆规两脚间的距离。此题的解题关键是灵活运用圆的周长公式求解。【详解】25.12 ÷ 2 ÷ 3.14 = 12.56 ÷ 3.14 = 4 (cm) 即圆规两脚间的距离是 4 cm 。",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "2017·云南玉溪·小升初真题",
+    "merged_id": "unit5_135",
+    "original_id": "unit5_choice_135"
+  },
+  {
+    "id": "unit5_choice_136",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长和面积",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "一个圆的半径扩大到原来的3倍，它的周长扩大到原来的( )倍，面积扩大到原来的( )倍。",
+    "options": [],
+    "answer": "3 9",
+    "solution": "【分析】根据圆的周长 = 2πr , 圆的面积 = πr² 以及积的变化规律可得: 一个圆的半径扩大到原来的 n 倍, 这个圆的周长就扩大到原来的 n 倍, 面积就扩大到原来的 n² 倍, 据此解答。本题主要考查圆的周长公式与面积公式的灵活运用。【详解】由分析可得: 一个圆的半径扩大到原来的 3 倍, 它的周长扩大到原来的 3 倍, 面积扩大到原来的 9 倍。",
+    "knowledge_points": [
+      "积的变化规律（小数乘法）",
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "18-19六年级上·全国·课后作业",
+    "merged_id": "unit5_136",
+    "original_id": "unit5_choice_136"
+  },
+  {
+    "id": "unit5_choice_137",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "一个钟表分针长 6 厘米, 经过半小时后, 分针的尖端走过的路程是_____。",
+    "options": [],
+    "answer": "18.84 厘米或18.84cm",
+    "solution": "【分析】根据题意可知, 半小时分针的尖端走过的路程是一个半径是 6 厘米的圆周长的一半, 根据圆周长: C = 2πr ,用 2 × 3.14 × 6 ÷ 2 即可求出分针的尖端走过的路程。本题考查了圆周长公式的灵活应用，明确分针走过的路程和圆周长的关系是解答本题的关键。【详解】2 × 3.14 × 6 ÷ 2 = 3.14 × 6 = 18.84 (厘米) 分针的尖端走过的路程是 18.84 厘米。",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "22-23 六年级上·四川乐山·期末",
+    "merged_id": "unit5_137",
+    "original_id": "unit5_choice_137"
+  },
+  {
+    "id": "unit5_choice_138",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长和面积",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "用圆规画一个周长是 31.4 厘米的圆，圆规两脚间的距离是( )厘米，这个圆的面积是( )平方厘米。",
+    "options": [],
+    "answer": "5 78.5",
+    "solution": "【分析】根据尺规画圆的方法可知，两脚间的距离就是这个圆的半径，由此先利用圆的周长公式求出半径：r = C ÷ π ÷ 2 ，再利用圆的面积公式：S = πr² ，将数值代入计算即可。此题考查圆的周长与面积公式的灵活应用。【详解】用圆规画一个周长是 31.4 厘米的圆, 圆规两脚间的距离是: 31.4 ÷ 3.14 ÷ 2 = 10 ÷ 2 = 5 (厘米) 这个圆的面积是: 3.14 × 5² = 3.14 × 25 = 78.5 (平方厘米)",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "2018·山东临沂·小升初真题",
+    "merged_id": "unit5_138",
+    "original_id": "unit5_choice_138"
+  },
+  {
+    "id": "unit5_choice_139",
+    "unit": "第五单元：圆",
+    "chapter": "圆的认识",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "在同一个圆里, 所有的半径都( )，半径的长度等于直径的( )。",
+    "options": [],
+    "answer": "相等 一半",
+    "solution": "【详解】在同一个圆里，所有的半径都相等，半径的长度等于直径的一半。",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "22-23 六年级上·广东广州·期末",
+    "merged_id": "unit5_139",
+    "original_id": "unit5_choice_139"
+  },
+  {
+    "id": "unit5_choice_140",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "画一个周长是 9.42 cm 的圆，圆规两脚间的距离是( )cm。",
+    "options": [],
+    "answer": "1.5",
+    "solution": "【分析】求圆规两脚间的距离, 就是求所画圆的半径; 根据圆的半径 r = C ÷ π ÷ 2 , 代入数据计算即可。本题考查圆的周长公式的灵活运用，明确圆规两脚间的距离等于所画圆的半径。【详解】圆的半径：9.42 ÷ 3.14 ÷ 2 = 3 ÷ 2 = 1.5 (cm)画一个周长是 9.42 cm 的圆, 圆规两脚间的距离是 1.5 cm 。",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "2014·全国·小升初真题",
+    "merged_id": "unit5_140",
+    "original_id": "unit5_choice_140"
+  },
+  {
+    "id": "unit5_choice_141",
+    "unit": "第五单元：圆",
+    "chapter": "圆的认识和周长",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "同一个圆中, 半径与直径的比是( ), 周长与直径的比值是( )。",
+    "options": [],
+    "answer": "1:2 π",
+    "solution": "【分析】同一个圆中，直径是半径的 2 倍，圆的周长 = πd ，圆的周长 ÷ 直径 = π ，据此分析。关键是理解比的意义，熟悉圆的特征，掌握圆的周长公式。【详解】同一个圆中，半径与直径的比是 1:2 ，周长与直径的比值是 π 。",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "圆的周长及应用",
+      "比的意义"
+    ],
+    "source": "2014 六年级·全国·课后作业",
+    "merged_id": "unit5_141",
+    "original_id": "unit5_choice_141"
+  },
+  {
+    "id": "unit5_choice_142",
+    "unit": "第五单元：圆",
+    "chapter": "方中圆的周长",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "玲玲要在一个长 10 厘米、宽 8 厘米的长方形内画一个最大的圆，这个圆的周长是( )厘米。",
+    "options": [],
+    "answer": "25.12",
+    "solution": "【分析】在长方形内画最大的圆, 圆的直径等于长方形的宽; 根据圆的周长公式: 周长 = π × 直径, 代入数据, 即可解答。本题考查圆的周长公式的应用, 关键明确长方形内画最大的圆, 圆的直径等于长方形的宽。【详解】3.14 × 8 = 25.12 （厘米）玲玲要在一个长 10 厘米、宽 8 厘米的长方形内画一个最大的圆, 这个圆的周长是 25.12 厘米。",
+    "knowledge_points": [
+      "长方形的概念及特点",
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·河北石家庄·期末",
+    "merged_id": "unit5_142",
+    "original_id": "unit5_choice_142"
+  },
+  {
+    "id": "unit5_choice_143",
+    "unit": "第五单元：圆",
+    "chapter": "方中圆的面积",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "在长 8 厘米、宽 6 厘米的长方形内画一个最大的圆，这个圆的面积是( )平方厘米。",
+    "options": [],
+    "answer": "28.26",
+    "solution": "【分析】根据题意可知, 长方形内画最大的圆, 圆的直径等于长方形的宽, 根据圆的面积公式: S = πr² , 代入数据,即可解答。本题考查圆的面积公式的应用, 关键明确长方形内画最大的圆, 圆的直径等于长方形的宽。【详解】3.14 × (6 ÷ 2)² = 3.14 × 3² = 3.14 × 9 = 28.26 (平方厘米) 所以，这个圆的面积是 28.26 平方厘米。",
+    "knowledge_points": [
+      "长方形的概念及特点",
+      "圆的概念及特点",
+      "圆的面积及应用",
+      "画圆"
+    ],
+    "source": "2014·全国·小升初真题",
+    "merged_id": "unit5_143",
+    "original_id": "unit5_choice_143"
+  },
+  {
+    "id": "unit5_choice_144",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "把一个圆形纸片剪成两个相等的半圆，它的周长增加20厘米，这个圆的面积是( )。",
+    "options": [],
+    "answer": "78.5平方厘米",
+    "solution": "【分析】根据半圆周长的意义可知, 半圆的周长 = 圆周长的一半 + 直径, 那么剪成的两个相等半圆的周长比原来一个圆的周长增加了 2 条直径; 用增加的周长除以 2 , 即可求出圆的直径; 再根据圆的面积公式 S = πr² , 求出这个圆的面积。本题考查半圆周长的意义以及圆的面积公式的应用。明确半圆的周长等于圆周长的一半加上一条直径, 求出圆的直径是解题的关键。【详解】圆的直径：20 ÷ 2 = 10 （厘米）圆的面积: 3.14 × (10 ÷ 2)² = 3.14 × 25 = 78.5（平方厘米） 这个圆的面积是 78.5 平方厘米。",
+    "knowledge_points": [
+      "圆的周长的应用",
+      "圆的面积的应用"
+    ],
+    "source": "21-22六年级上·重庆潼南·期末",
+    "merged_id": "unit5_144",
+    "original_id": "unit5_choice_144"
+  },
+  {
+    "id": "unit5_choice_145",
+    "unit": "第五单元：圆",
+    "chapter": "半圆的周长和面积",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "一个半圆的半径是 2 cm , 它的周长是( ) cm , 面积是( ) cm² 。",
+    "options": [],
+    "answer": "10.28 6.28",
+    "solution": "【分析】根据半圆的周长 = πr + 2r , 半圆的面积 = πr² ÷ 2 , 列式计算即可。关键是掌握并灵活运用半圆的周长和面积公式。【详解】3.14 × 2 + 2 × 2 = 6.28 + 4 = 10.28 (cm) 3.14 × 2² ÷ 2 = 3.14 × 4 ÷ 2 = 6.28(cm²) 它的周长是 10.28 cm , 面积是 6.28 cm² 。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "2022·湖南怀化·小升初真题",
+    "merged_id": "unit5_145",
+    "original_id": "unit5_choice_145"
+  },
+  {
+    "id": "unit5_choice_146",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长和面积",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "用一根长 12.56 m 的绳子围成一个圆，这个圆的直径是( )m，面积是()m²。",
+    "options": [],
+    "answer": "4;12.56",
+    "solution": "【分析】这条绳子的长度就是圆的周长, 根据圆的周长公式: C = πd , 据此求出圆的直径, 进而求出圆的半径, 再根据圆的面积公式: S = πr² , 据此求出圆的面积。本题考查圆的周长和面积, 熟记公式是解题的关键。【详解】12.56 ÷ 3.14 = 4 (m) 3.14 × (4 ÷ 2)² = 3.14 × 4 = 12.56(m²) 则这个圆的直径是 4 m , 面积是 12.56 m² 。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "2022·新疆吐鲁番·小升初真题",
+    "merged_id": "unit5_146",
+    "original_id": "unit5_choice_146"
+  },
+  {
+    "id": "unit5_choice_147",
+    "unit": "第五单元：圆",
+    "chapter": "方中圆的面积",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "玲玲拿来一张长方形彩纸, 长为 40 cm , 宽为 30 cm 。用这张彩纸剪一个尽可能大的圆, 这个圆的直径是( )cm, 面积是( )cm²。",
+    "options": [],
+    "answer": "30 ;706.5",
+    "solution": "【分析】长方形内剪一个最大的圆, 圆的直径 = 长方形的宽, 根据圆的面积 = πr² , 列式计算即可。关键是理解圆和长方形之间的关系，掌握圆的面积公式。【详解】3.14 × (30 ÷ 2)² = 3.14 × 15² = 3.14 × 225 = 706.5(cm²) 这个圆的直径是 30 cm , 面积是 706.5 cm² 。",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "22-23 六年级上·辽宁·单元测试",
+    "merged_id": "unit5_147",
+    "original_id": "unit5_choice_147"
+  },
+  {
+    "id": "unit5_choice_148",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长和面积",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "圆规两脚间的距离是 2 厘米, 所画的圆的直径是( ), 周长是( )厘米, 面积是( )平方厘米。",
+    "options": [],
+    "answer": "4 厘米/4cm ;12.56 ; 12.56",
+    "solution": "【分析】圆规两脚间的距离是圆的半径，圆的直径=半径×2，圆的周长=πd，圆的面积=πr²，据此列式计算。关键是熟悉圆的特征，掌握并灵活运用圆的周长和面积公式。【详解】2 × 2 = 4 （厘米） 3.14 × 4 = 12.56 （厘米） 3.14 × 2² = 3.14 × 4 = 12.56 (平方厘米）",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "21-22 六年级上·广东·期末",
+    "merged_id": "unit5_148",
+    "original_id": "unit5_choice_148"
+  },
+  {
+    "id": "unit5_choice_149",
+    "unit": "第五单元：圆",
+    "chapter": "方中圆的面积",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "在一张长 30 cm 、宽 20 cm 的长方形纸上剪一个最大的圆, 圆的面积是( ) cm² 。",
+    "options": [],
+    "answer": "314",
+    "solution": "【分析】长方形纸上剪一个最大的圆，最大的圆的直径是长方形的宽 20 cm ，据此即可求出它的半径是 10 cm ，再利用圆的面积 = πr² ，即可求出圆的面积。此题考查圆的面积公式的计算应用, 关键是明确圆的直径等于长方形的宽。【详解】3.14 × (20 ÷ 2)² = 3.14 × 10² = 3.14 × 100 = 314(cm²)",
+    "knowledge_points": [
+      "平面图形的分割",
+      "圆的面积及应用"
+    ],
+    "source": "21-22 六年级下·山东临沂·期末",
+    "merged_id": "unit5_149",
+    "original_id": "unit5_choice_149"
+  },
+  {
+    "id": "unit5_choice_150",
+    "unit": "第五单元：圆",
+    "chapter": "方中圆的周长和面积",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "在长6厘米，宽4厘米的长方形中画一个最大圆，这个圆的周长是( )厘米，面积是( )平方厘米。",
+    "options": [],
+    "answer": "12.56 12.56",
+    "solution": "【分析】在长 6 厘米, 宽 4 厘米的长方形中画一个最大圆,这个圆的直径最大是 4 厘米, 再根据圆的周长和面积公式进行计算即可。本题考查圆的周长和面积、解答本题的关键是掌握在长方形中画一个最大圆的特征。【详解】3.14 × 4 = 12.56 （厘米） 3.14 × (4 ÷ 2)² = 3.14 × 4 = 12.56 (平方厘米)",
+    "knowledge_points": [
+      "长方形的概念及特点",
+      "圆的概念及特点",
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "2022·湖南邵阳·小升初真题",
+    "merged_id": "unit5_150",
+    "original_id": "unit5_choice_150"
+  },
+  {
+    "id": "unit5_choice_151",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长和面积",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "一个圆形纸片的周长是 12.56 cm , 它的直径是( ) cm , 面积是( ) cm² 。",
+    "options": [],
+    "answer": "4 12.56",
+    "solution": "【分析】根据圆的周长公式：C = πd ，代入周长的数据，即可求出直径。再求出半径，利用圆的面积公式：S = πr² ，代入即可求出圆的面积。此题的解题关键是熟悉圆的周长和圆的面积公式。【详解】12.56 ÷ 3.14 = 4 (cm) 3.14 × (4 ÷ 2)² = 3.14 × 4 = 12.56(cm²)",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级上·安徽宿州·期末",
+    "merged_id": "unit5_151",
+    "original_id": "unit5_choice_151"
+  },
+  {
+    "id": "unit5_choice_152",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长和面积",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "用圆规画一个周长是 50.24 cm 的圆, 圆规两脚间的距离是( )cm, 所画圆的半径为面积是( )cm²。",
+    "options": [],
+    "answer": "8 ;200.96",
+    "solution": "【分析】圆规两脚间的距离就是圆的半径，根据圆的周长公式：C = 2πr ，据此求出圆的半径，然后根据圆的面积公式：S = πr² ，据此代入数值求出圆的面积。本题考查圆的周长和面积，明确圆规两脚间的距离就是圆的半径是解题的关键。【详解】50.24 ÷ 3.14 ÷ 2 = 16 ÷ 2 = 8 (cm) 3.14 × 8² = 200.96 (cm²)",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级上·辽宁本溪·期中",
+    "merged_id": "unit5_152",
+    "original_id": "unit5_choice_152"
+  },
+  {
+    "id": "unit5_choice_153",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "把一个圆形纸片剪成两个相等的半圆, 周长增加了 8 cm 。原来这个圆形纸片的面积是( ) cm² 。",
+    "options": [],
+    "answer": "12.56",
+    "solution": "【分析】根据半圆周长的意义，半圆的周长等于该圆周长的一半加上直径。把一个圆形纸片剪成两个相等的半圆，周长增加了 8 cm 。周长增加的 8 cm 是两个圆的直径的长度，据此可以求出圆的直径；再根据圆的面积公式： S = πr² ，把数据代入公式解答。此题主要考查圆的周长公式、面积公式的灵活运用，关键是熟记公式，重点是明确半圆周长的意义。【详解】圆的直径：8 ÷ 2 = 4 (cm)圆的面积: 3.14 × (4 ÷ 2)² = 3.14 × 4 = 12.56(cm²)",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "18-19 六年级上·全国·单元测试",
+    "merged_id": "unit5_153",
+    "original_id": "unit5_choice_153"
+  },
+  {
+    "id": "unit5_choice_154",
+    "unit": "第五单元：圆",
+    "chapter": "方中圆的面积",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "青青在剪圆, 她用边长 12 cm 的正方形剪了一个最大的圆, 剪去的面积是() cm² 。",
+    "options": [],
+    "answer": "30.96",
+    "solution": "【分析】根据题意可知，在这个正方形内剪一个最大的圆，这个圆的直径等于正方形的边长，根据正方形的面积公式：S = a² ，圆的面积公式：S = πr² ，把数据代入公式求出正方形与圆面积的差即可。此题主要考查正方形和圆面积公式的灵活运用,关键是熟记公式。【详解】12 × 12 - 3.14 × (12 ÷ 2)² = 12 × 12 - 3.14 × 36 = 144 - 113.04 = 30.96 (cm²) 所以，剪去的面积是 30.96 cm² 。",
+    "knowledge_points": [
+      "圆的面积及应用",
+      "正方形的面积",
+      "含圆的组合图形的计算（周长和面积）"
+    ],
+    "source": "21-22 六年级上·河南安阳·期末",
+    "merged_id": "unit5_154",
+    "original_id": "unit5_choice_154"
+  },
+  {
+    "id": "unit5_choice_155",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "一个正方形的边长和圆的半径相等, 已知正方形的面积是 100 cm² , 这个圆的面积是( ) cm² 。",
+    "options": [],
+    "answer": "314",
+    "solution": "【分析】根据正方形的面积，先求出正方形的边长。因为正方形的边长和圆的半径相等，所以圆的半径也求出来了，从而利用圆的面积公式，求出它的面积即可。本题考查了正方形和圆的面积, 正方形面积 = 边长 × 边长, 圆的面积 = 3.14 × 半径。【详解】10 × 10 = 100 (cm²)所以，正方形的边长是 10 cm ，圆的半径也是 10 cm 。3.14 × 10² = 314 (cm²)所以，这个圆的面积是 314 cm² 。",
+    "knowledge_points": [
+      "圆的面积及应用",
+      "正方形的面积"
+    ],
+    "source": "21-22 六年级下·贵州遵义·期末",
+    "merged_id": "unit5_155",
+    "original_id": "unit5_choice_155"
+  },
+  {
+    "id": "unit5_choice_156",
+    "unit": "第五单元：圆",
+    "chapter": "半圆的周长",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "'五角钱' (猜一数学图形) ( )；这个图形的周长用字母表示( )。",
+    "options": [],
+    "answer": "半圆 πr + 2r",
+    "solution": "【分析】一元的一半刚好是五角，则这个图形是半圆，半圆的周长 = 圆周长的一半 + 一条直径，据此解答。掌握圆的周长计算公式，半圆的周长比圆周长的一半多一条直径的长度。【详解】分析可知, 这个图形是半圆, 这个图形的周长用字母表示为: 2πr ÷ 2 + 2r = πr + 2r 。",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级下·湖南娄底·期末",
+    "merged_id": "unit5_156",
+    "original_id": "unit5_choice_156"
+  },
+  {
+    "id": "unit5_choice_157",
+    "unit": "第五单元：圆",
+    "chapter": "方中圆的面积",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "在一个周长是 64 cm 的正方形纸片内, 剪一个最大的圆, 这个圆的半径是( ) cm , 这个圆的面积是( ) cm² 。",
+    "options": [],
+    "answer": "8   200.96",
+    "solution": "【分析】在正方形纸片内剪一个最大的圆, 那么圆的直径是正方形的边长, 根据正方形的边长 = 周长 ÷ 4 , 求出圆的直径; 然后根据 r = d ÷ 2 , 求出圆的半径; 再根据圆的面积公式: S = πr² , 求出圆的面积。掌握正方形内剪一个最大的圆，圆的直径与正方形边长的关系是解题的关键。【详解】圆的直径: 64 ÷ 4 = 16 (cm)圆的半径: 16 ÷ 2 = 8 (cm)圆的面积: 3.14 × 8² = 3.14 × 64 = 200.96(cm²)",
+    "knowledge_points": [
+      "圆的面积及应用",
+      "正方形的周长"
+    ],
+    "source": "21-22 六年级上·云南曲靖·期末",
+    "merged_id": "unit5_157",
+    "original_id": "unit5_choice_157"
+  },
+  {
+    "id": "unit5_choice_158",
+    "unit": "第五单元：圆",
+    "chapter": "方中圆的面积",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "在一块长 5 分米、宽 4 分米的长方形铁板上剪一个最大的圆, 这个圆的面积是 ( )平方分米。",
+    "options": [],
+    "answer": "12.56",
+    "solution": "【分析】长方形铁板上剪一个最大的圆，圆的直径=长方形的宽，根据圆的面积=πr²，列式计算即可。关键是理解圆和长方形之间的关系，掌握圆的面积公式。【详解】3.14 × (4 ÷ 2)² = 3.14 × 4 = 12.56 （平方分米）",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "21-22 六年级上·河南三门峡·期末",
+    "merged_id": "unit5_158",
+    "original_id": "unit5_choice_158"
+  },
+  {
+    "id": "unit5_choice_159",
+    "unit": "第五单元：圆",
+    "chapter": "圆的认识",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "两端都在圆上的线段, 直径最长。( )",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】通过圆心并且两端都在圆上的线段是直径，两端都在圆上的线段，直径是最长的一条，据此判断即可。【详解】如图：![](images/d65c04c4dff2829d021e9c46a307417af38896fde8859a13a6fa1a63a5ad7ef1.jpg)则两端都在圆上的线段，直径最长。说法正确。故答案为: √",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "17-18 五年级下·全国·单元测试",
+    "merged_id": "unit5_159",
+    "original_id": "unit5_choice_159"
+  },
+  {
+    "id": "unit5_choice_160",
+    "unit": "第五单元：圆",
+    "chapter": "圆的认识",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "直径是圆内最长的一条线段。( )",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】通过圆心并且两端都在圆上的线段叫做直径。所有两端都在圆上的线段中，直径最长。【详解】直径是圆内最长的一条线段。原题说法正确。故答案为: √",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "2014 六年级·全国·课后作业",
+    "merged_id": "unit5_160",
+    "original_id": "unit5_choice_160"
+  },
+  {
+    "id": "unit5_choice_161",
+    "unit": "第五单元：圆",
+    "chapter": "圆的认识",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "下水道井盖平面轮廓做成圆形的，圆形井盖怎么放都不会掉到井里，并且能恰好盖住井口，是利用了同一圆中直径都相等的性质。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】同圆或等圆中，所有的直径都相等，因此这就是圆形井盖设计成圆形的原因。关键是熟悉圆的特征，根据圆的特征进行分析。【详解】下水道井盖平面轮廓做成圆形的，圆形井盖怎么放都不会掉到井里，并且能恰好盖住井口，是利用了同一圆中直径都相等的性质，说法正确。故答案为: √",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "22-23六年级上·四川南充·期末",
+    "merged_id": "unit5_161",
+    "original_id": "unit5_choice_161"
+  },
+  {
+    "id": "unit5_choice_162",
+    "unit": "第五单元：圆",
+    "chapter": "圆周率",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "大小不同的两个圆,其周长与它的直径的商相等。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】根据圆周率的意义，任意一个圆的周长与它的直径的比值是一个固定的数，我们把它叫做圆周率，一般用 'π'表示。由此判断。本题考查圆周率的认识, 明确圆周率是一个固定的数, 不随圆的大小而变化。【详解】根据圆周率的意义可知, 每个圆的圆周率都是 π ,所以每个圆的圆周率都相等。周长 ÷ 直径 = π （一定）所以，大小不同的两个圆，其周长与它的直径的商相等。 原题说法正确。故答案为: √",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "22-23 六年级上·陕西宝鸡·期中",
+    "merged_id": "unit5_162",
+    "original_id": "unit5_choice_162"
+  },
+  {
+    "id": "unit5_choice_163",
+    "unit": "第五单元：圆",
+    "chapter": "圆周率",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "大圆的圆周率大,小圆的圆周率小。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】任意一个圆的周长与它的直径的比值是一个固定的数，我们把它叫做圆周率，用字母 π 表示，是一个无限不循环小数；据此判断。本题考查圆周率的认识, 明确圆周率是一个固定的数, 不随圆的大小而变化。【详解】根据圆周率的意义可知, 每个圆的圆周率都是 π ,所以每个圆的圆周率都相等。原题说法错误。故答案为: ×",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "21-22 六年级上·吉林白山·期末",
+    "merged_id": "unit5_163",
+    "original_id": "unit5_choice_163"
+  },
+  {
+    "id": "unit5_choice_164",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长和面积",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "半径是 2 米的圆, 周长和面积相等。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】根据圆的周长和面积公式，先分别求出周长和面积，再根据周长和面积的定义解题即可。本题考查了圆的周长和面积, 掌握周长和面积的公式, 以及周长和面积的意义是解题的关键。【详解】周长: 2 × 3.14 × 2 = 12.56 (米)面积: 3.14 × 2² = 3.14 × 4 = 12.56 (平方米)从数值上看, 12.56 = 12.56 , 但是周长的单位是长度单位,面积的单位是面积单位，周长和面积是两个意义完全不同的量, 所以不能说半径是 2 米的圆, 周长和面积相等。故答案为:×",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "2017·贵州铜仁·小升初真题",
+    "merged_id": "unit5_164",
+    "original_id": "unit5_choice_164"
+  },
+  {
+    "id": "unit5_choice_165",
+    "unit": "第五单元：圆",
+    "chapter": "半圆的周长",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "在同一圆中, 半圆的面积是圆面积的一半，半圆的周长也是圆周长的一半。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】根据半圆面积 = 圆面积 ÷ 2 , 半圆周长 = 圆周长 ÷ 2 + 直径。据此判断即可。本题考查半圆的面积和周长，明确半圆的周长的计算方法是解题的关键。【详解】由分析可知：在同圆或等圆内，半圆的面积是圆面积的一半，半圆的周长不是圆周长的一半。原题干说法错误。故答案为: ×",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "2014·全国·小升初真题",
+    "merged_id": "unit5_165",
+    "original_id": "unit5_choice_165"
+  },
+  {
+    "id": "unit5_choice_166",
+    "unit": "第五单元：圆",
+    "chapter": "扇形的认识",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "用 4 个圆心角都是 90° 的扇形，可以拼成一个圆。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】由圆的两条半径与这两条半径所夹的圆心角所对的弧围成的图形就是扇形。扇形的大小由半径和圆心角决定，据此分析。关键是熟悉扇形的特点，虽然扇形是圆的一部分，但要注意圆心角相同的扇形，大小可能不同。【详解】半径相同, 圆心角都是 90° 的 4 个扇形, 才可以拼成一个圆, 原题说法错误。故答案为: ×",
+    "knowledge_points": [
+      "弧、圆心角、扇形的认识"
+    ],
+    "source": "2020·湖北十堰·小升初真题",
+    "merged_id": "unit5_166",
+    "original_id": "unit5_choice_166"
+  },
+  {
+    "id": "unit5_choice_167",
+    "unit": "第五单元：圆",
+    "chapter": "扇形的面积",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "圆心角是 120° 的扇形面积一定大于圆心角是 90° 的扇形面积。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】扇形的面积 = πr² × (圆心角/360°) , 据此分析。关键是掌握并灵活运用扇形面积公式。【详解】扇形的面积除了与圆心角的大小有关，还与半径的长短有关系，所以原题说法错误。故答案为: ×",
+    "knowledge_points": [
+      "扇形的周长和面积"
+    ],
+    "source": "22-23 六年级上·河南南阳·期末",
+    "merged_id": "unit5_167",
+    "original_id": "unit5_choice_167"
+  },
+  {
+    "id": "unit5_choice_168",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长和面积",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "如果把圆的半径扩大到原来的 2 倍, 周长与面积都扩大到原来的 2 倍。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】根据圆的面积公式 S = πr² , 圆的周长公式 C = 2πr ,假设圆的半径为 1 , 分别求出圆的周长和面积以及半径扩大后圆的周长和面积, 然后进行对比即可。解答此题的关键是根据原来圆的半径，圆的周长公式和圆的面积公式进行计算即可。【详解】假设圆的半径为1，圆的周长: 2 × 3.14 × 1 = 6.28;圆的面积: 3.14 × 1² = 3.14 × 1 = 3.14半径扩大到原来的2倍后，1 × 2 = 2;圆的周长: 2 × 3.14 × 2 = 12.56圆的面积: 3.14 × 2² = 3.14 × 4 = 12.56 12.56 ÷ 6.28 = 2  12.56 ÷ 3.14 = 4 即圆的周长扩大到原来的2倍，圆的面积扩大到原来的4倍。故答案为: ×",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "18-19 五年级下·山东·期中",
+    "merged_id": "unit5_168",
+    "original_id": "unit5_choice_168"
+  },
+  {
+    "id": "unit5_choice_169",
+    "unit": "第五单元：圆",
+    "chapter": "圆的认识",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "两个半圆可以拼成一个圆。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】半径相同的两个半圆能拼成一个圆，据此解答。本题的关键是两个半圆的半径相等时才能拼成一个圆。【详解】因半径相同的两个半圆能拼成一个圆，所以当两个半圆的半径不相等时就不能拼成一个圆，原题说法错误。故答案为：×",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "22-23 六年级上·河南信阳·期末",
+    "merged_id": "unit5_169",
+    "original_id": "unit5_choice_169"
+  },
+  {
+    "id": "unit5_choice_170",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长和面积",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "面积相等的两个圆，它们的周长不一定相等。（ ）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】圆心确定圆的位置，半径决定圆的大小，如果两个圆的面积相等，那么它们的半径一定相等，由 'C = 2πr' 可知，如果两个圆的半径相等，那么它们的周长一定相等，据此解答。根据圆的面积相等得出两个圆的半径相等, 并掌握圆的周长计算公式是解答题目的关键。【详解】分析可知，面积相等的两个圆，它们的半径也相等，所以这两个圆的周长一定相等。故答案为: ×",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "18-19六年级上·山东青岛·期末",
+    "merged_id": "unit5_170",
+    "original_id": "unit5_choice_170"
+  },
+  {
+    "id": "unit5_choice_171",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "如果两个圆的半径相等, 那么它们的面积也一定相等。( )",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】根据圆的面积公式：S = πr² 可知，如果两个圆的半径相等，那么它们的面积也一定相等。解答关键是，熟记圆的面积计算公式。",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "圆的面积及应用"
+    ],
+    "source": "19-20 六年级上·内蒙古通辽·期末",
+    "merged_id": "unit5_171",
+    "original_id": "unit5_choice_171"
+  },
+  {
+    "id": "unit5_choice_172",
+    "unit": "第五单元：圆",
+    "chapter": "圆的认识",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "圆心决定圆的位置，半径决定圆的大小。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】根据圆的概念和特点，直接判断即可。本题考查了圆, 掌握圆的概念和特点是解题的关键。【详解】圆心决定圆的位置，半径决定圆的大小。故答案为: √",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "2017·重庆·小升初真题",
+    "merged_id": "unit5_172",
+    "original_id": "unit5_choice_172"
+  },
+  {
+    "id": "unit5_choice_173",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "圆的周长是它直径的3.14 倍, 半径是它直径的 1/2 。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】圆的周长和它的直径的比值是一个固定的数，我们叫它圆周率（π），也就是说圆的周长是它的直径的π倍，在实际应用时，我们常常取它的近似值，π≈3.14；圆的半径是它的直径的 1/2 。明确圆周率是一个准确值，实际应用时，3.14 只是我们取的 π 的近似值。【详解】由分析可知: 圆的周长是它直径的 π 倍, 半径是它直径的 1/2 , 原题说法错误。故答案为: ×",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "圆的周长及应用"
+    ],
+    "source": "22-23 六年级上·全国·期末",
+    "merged_id": "unit5_173",
+    "original_id": "unit5_choice_173"
+  },
+  {
+    "id": "unit5_choice_174",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长和面积",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "周长相等的正方形、长方形和圆，圆的面积最大。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】根据题意可知, 此题用举例法解答, 先假设正方形、长方形和圆形的周长都是 16 , 分别求出圆、正方形、长方形的面积, 然后比较大小即可。本题考查了长方形、正方形与圆的周长和面积,关键是理解周长相等的正方形、长方形和圆形, 圆的面积最大。【详解】假设正方形、长方形和圆形的周长都是 16 则圆的面积为: π × (16 ÷ 2π)² ≈ 20.38 ;正方形的边长为: 16 ÷ 4 = 4 , 面积为: 4 × 4 = 16 ;长方形取长为 5 宽为 3 , 面积为: 5 × 3 = 15 ;所以周长相等的正方形、长方形和圆形，圆面积最大，原说法正确。故答案为: √",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用",
+      "正方形的面积",
+      "面积认识及大小的比较"
+    ],
+    "source": "2017·甘肃张掖·小升初真题",
+    "merged_id": "unit5_174",
+    "original_id": "unit5_choice_174"
+  },
+  {
+    "id": "unit5_choice_175",
+    "unit": "第五单元：圆",
+    "chapter": "圆周率",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "只要有耐心, π 可以计算出具体数值。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】圆周率用希腊字母 π 表示, 是代表圆周长和直径的比值。它是一个无限不循环小数。据此解答。此题的解题关键是理解圆周率的定义。【详解】根据分析得, π 是一个无限不循环小数, 不可以计算出具体的数值。故答案为: ×",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "21-22 五年级下·河南洛阳·期末",
+    "merged_id": "unit5_175",
+    "original_id": "unit5_choice_175"
+  },
+  {
+    "id": "unit5_choice_176",
+    "unit": "第五单元：圆",
+    "chapter": "扇形的认识",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "用 2 个圆心角都是 90° 的扇形, 一定可以拼成一个半圆。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】根据扇形的面积 = (nπr²)/360 , 扇形的面积大小与圆心角、半径有关系，圆心角相等，半径不一定相等，所以面积无法确定；进而判断即可。此题考查的是扇形面积的大小与哪些量有关系，应注意分析要全面，不能以点代面。【详解】由分析知：扇形的面积大小与圆心角、半径有关，圆心角相等，半径不一定相等，所以面积无法确定，因此两个圆心角均为 90° 的扇形不一定能组成半圆；故答案为：×",
+    "knowledge_points": [
+      "扇形的周长和面积"
+    ],
+    "source": "18-19 六年级上·全国·单元测试",
+    "merged_id": "unit5_176",
+    "original_id": "unit5_choice_176"
+  },
+  {
+    "id": "unit5_choice_177",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的认识",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "在一个大圆里挖去一个小圆就是圆环。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】在一个圆内剪去一个小圆，可以出现很多种情况，只有当大圆和小圆是同一个圆心时，才能成为圆环；进而判断即可。此题考查的是对圆环的认识。掌握圆环的特征是解答的关键。【详解】从一个圆中剪去一个小圆，只有当大圆和小圆是同一个圆心时，从大圆中减去一个小圆，才能成为圆环。所以题干说法错误。故答案为: ×",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "圆环的面积"
+    ],
+    "source": "2022·湖北孝感·小升初真题",
+    "merged_id": "unit5_177",
+    "original_id": "unit5_choice_177"
+  },
+  {
+    "id": "unit5_choice_178",
+    "unit": "第五单元：圆",
+    "chapter": "扇形的面积",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "在同一个圆中，圆心角的度数越大，它所对应的扇形的面积就越大。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】在同一个圆中，扇形的与扇形对应的圆心角度数密切相关，根据圆心角与扇形面积的关系，可以做出判断。本题主要考查扇形面积与圆心角的大小关系, 同一圆内圆心角的大小决定扇形的大小。【详解】在同一个圆中，扇形的大小与这个扇形的圆心角大小有关，圆心角越大扇形越大，圆心角越小扇形越小。故答案为: √",
+    "knowledge_points": [
+      "扇形的周长和面积"
+    ],
+    "source": "20-21 六年级上·贵州安顺·期末",
+    "merged_id": "unit5_178",
+    "original_id": "unit5_choice_178"
+  },
+  {
+    "id": "unit5_choice_179",
+    "unit": "第五单元：圆",
+    "chapter": "圆的认识",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "同一个圆中所有的半径都相等, 所有的直径也都相等。( )",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】根据圆的特征可知: 在同圆或等圆中, 同一圆里,所有的直径都相等, 所有的半径也相等; 进而判断即可。掌握圆的特征是解答本题的关键。【详解】根据圆的特征可得：在同圆或等圆中，所有的半径都相等，所有的直径也都相等。故答案为: √",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "16-17 六年级上·广东深圳·期末",
+    "merged_id": "unit5_179",
+    "original_id": "unit5_choice_179"
+  },
+  {
+    "id": "unit5_choice_180",
+    "unit": "第五单元：圆",
+    "chapter": "轴对称图形",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "三角形、正方形、长方形和圆都是轴对称图形。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】依据轴对称图形的定义判断：一个图形沿一条直线对折，对折后的两部分都能完全重合，这个图形是轴对称图形，则这条直线就是这个图形的对称轴，据此解答即可。解答此题的主要依据是：轴对称图形的概念及特征。【详解】由分析可知：正方形、长方形和圆都是轴对称图形，但三角形中除等腰三角形外，其他的三角形直线两旁的部分不能够完全重合，则除等腰三角形外的三角形不是轴对称图形。所以原题干说法错误。故答案为: ×",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "轴对称的认识及辨认"
+    ],
+    "source": "2014·云南红河·小升初真题",
+    "merged_id": "unit5_180",
+    "original_id": "unit5_choice_180"
+  },
+  {
+    "id": "unit5_choice_181",
+    "unit": "第五单元：圆",
+    "chapter": "圆的对称轴",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "圆中的任一条直径都是它的对称轴。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】依据轴对称图形的定义：一个圆形沿一条直径所在的直线对折，对折后的两部分都能完全重合，所以，任一条直径所在的直线都是它的对称轴。即可作答。解答此题的关键是明白对称轴是直线，而圆的直径是线段。【详解】沿圆的每一条直径所在的直线对折, 对折后的两部分都能完全重合, 所以, 任一条直径所在的直线都是它的对称轴, 因此圆的直径不是其对称轴, 圆的直径所在的直线是其对称轴。故答案为: ×",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "与圆相关的轴对称图形",
+      "轴对称的认识及辨认"
+    ],
+    "source": "16-17 六年级上·四川·期末",
+    "merged_id": "unit5_181",
+    "original_id": "unit5_choice_181"
+  },
+  {
+    "id": "unit5_choice_182",
+    "unit": "第五单元：圆",
+    "chapter": "圆周率",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "大圆的圆周率一定比小圆的圆周率大。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】圆周率是任意一个圆的周长与它的直径的比值,这个比值是一个固定的数。掌握圆周率的意义是解答题目的关键。【详解】任意一个圆的周长都是它的直径的 π 倍, 所以不管是大圆还是小圆, 圆周率都相同。故答案为×",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "14-15 六年级上·全国·课后作业",
+    "merged_id": "unit5_182",
+    "original_id": "unit5_choice_182"
+  },
+  {
+    "id": "unit5_choice_183",
+    "unit": "第五单元：圆",
+    "chapter": "画圆",
+    "question_type": "作图题",
+    "difficulty": "容易",
+    "question": "画半径是 2 厘米的圆。",
+    "options": [],
+    "answer": "见详解",
+    "solution": "【分析】圆的画法：确定圆心，圆规针尖固定在圆心上，拉开圆规的两脚，使两脚之间的距离为 2 厘米，然后旋转一周，即可画出半径为 2 厘米的圆。本题考查学生对圆的画法的掌握。【详解】如图所示:![](images/9680eb3c90500ed8f928215f54742d59eacbf915fb171ac01e5c1a2cac5dcb3c.jpg)",
+    "knowledge_points": [
+      "画圆"
+    ],
+    "source": "2014 六年级·全国·课后作业",
+    "merged_id": "unit5_183",
+    "original_id": "unit5_choice_183"
+  },
+  {
+    "id": "unit5_choice_184",
+    "unit": "第五单元：圆",
+    "chapter": "画圆",
+    "question_type": "作图题",
+    "difficulty": "容易",
+    "question": "用圆规画一个直径为 3 cm 的圆，并用字母 O、r、d 标出它的圆心、半径和直径。",
+    "options": [],
+    "answer": "见详解。",
+    "solution": "【分析】由题意知, 要画一个直径是 3 厘米的圆, 首先确定圆的半径为 3 ÷ 2 = 1.5 厘米, 再依据画圆的方法画一个圆,并用字母标出它的圆心 O、半径 r 和直径 d 即可。解答此题要明确半径是 1.5 厘米, 即画圆时圆规两脚叉开的距离为 1.5 厘米。【详解】半径为: 3 ÷ 2 = 1.5 (厘米), 如图所示:![](images/da3b72ba59f06a469c941c7ef3c8e42cd37321866b49b6acf484f4f926b87363.jpg)",
+    "knowledge_points": [
+      "画圆"
+    ],
+    "source": "2014 六年级·全国·课后作业",
+    "merged_id": "unit5_184",
+    "original_id": "unit5_choice_184"
+  },
+  {
+    "id": "unit5_choice_185",
+    "unit": "第五单元：圆",
+    "chapter": "画圆",
+    "question_type": "作图题",
+    "difficulty": "容易",
+    "question": "用圆规画一个半径是 1.5 cm 的圆，并用字母 O、r、d 标出它的圆心、半径和直径。",
+    "options": [],
+    "answer": "见详解",
+    "solution": "【分析】先确定圆心，用圆规有针的一脚固定在圆心，然后以圆规两脚之间的距离为 1.5 cm 进行旋转一周，得到的图形就是我们要画的圆；圆心用字母 'O'表示；半径用字母 'r'表示，直径用字母 'd'表示，由此求解。本题主要考查了圆的画法, 注意圆规的两脚分开的距离是圆的半径。【详解】作图如下：![](images/00572910d28949a288e3a9992072c8710d5aa2acad0071deb859d9988495e935.jpg)",
+    "knowledge_points": [
+      "画圆"
+    ],
+    "source": "19-20六年级上·河北·课后作业",
+    "merged_id": "unit5_185",
+    "original_id": "unit5_choice_185"
+  },
+  {
+    "id": "unit5_choice_186",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的面积",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "一个半径为20米的圆形喷水池，在它周围修一条宽1米的环形花带，如果每平方米植花25棵，每棵成本为4元，这条环形花带共需投资多少元？",
+    "options": [],
+    "answer": "12874元",
+    "solution": "【分析】根据求环形面积的公式，外圆面积一内圆面积 = 环形面积，已知内圆半径是 20 米，环宽是 1 米，先求出外圆半径，再利用环形面积计算公式 'S = π (R² - r²)' ，求出环形花带的面积；用环形花带的面积乘每平方米植花的数量，得到植花的总数量，再乘每株的单价，即得到一共投资多少钱.此题属于有关圆的应用题, 解题本题的关键是掌握圆环面积的求解方法以及总价、单价、数量三者之间的关系。【详解】20 + 1 = 21 （米） 3.14 × (21² - 20²) = 3.14 × (441 - 400) = 3.14 × 41 = 128.74 (平方米) 128.74 × 25 × 4 = 12874 （元）答：这条环形花带共需投资 12874 元。",
+    "knowledge_points": [
+      "经济问题",
+      "圆环的面积",
+      "小数与整数的乘法"
+    ],
+    "source": "22-23六年级上·河南驻马店·期末",
+    "merged_id": "unit5_186",
+    "original_id": "unit5_choice_186"
+  },
+  {
+    "id": "unit5_choice_187",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "新世纪广场有一个直径为 6 米的圆形花坛，这个圆形花坛的占地面积是多少？",
+    "options": [],
+    "answer": "28.26 平方米",
+    "solution": "【分析】根据题意, 圆的面积 = (d ÷ 2)² × π , 将数据代入公式计算即可。此题考查了圆形的面积计算，关键熟记公式。【详解】(6 ÷ 2)² × 3.14 = 9 × 3.14 = 28.26 （平方米）答: 这个圆形花坛的占地面积是 28.26 平方米。",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "22-23 五年级下·山东·单元测试",
+    "merged_id": "unit5_187",
+    "original_id": "unit5_choice_187"
+  },
+  {
+    "id": "unit5_choice_188",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "一辆自行车轮的半径是0.4米，车轮转动一周，自行车前进多少米？",
+    "options": [],
+    "answer": "2.512 米",
+    "solution": "【分析】根据公式：圆的周长 = 2πr ，将半径 0.4 代入公式计算即可。此题考查了圆的周长计算，关键熟记公式。【详解】0.4 × 2 × 3.14 = 0.8 × 3.14 = 2.512 (米) 答: 自行车前进 2.512 米。",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "22-23 五年级下·山东·单元测试",
+    "merged_id": "unit5_188",
+    "original_id": "unit5_choice_188"
+  },
+  {
+    "id": "unit5_choice_189",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的面积",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "一个直径 10 米的水池，周围有一条环形小路，路面宽3米，这条小路的面积是多少平方米？",
+    "options": [],
+    "answer": "122.46 平方米",
+    "solution": "【分析】先利用圆的半径 = 圆的直径 ÷ 2 , 求出内圆的半径,加上环宽, 就是外圆的半径, 再利用圆环的面积公式: S = π(R² - r²) 即可得解。此题重点理解求小路的面积实际上是求圆环的面积，解题关键是掌握圆环的面积计算方法。【详解】10 ÷ 2 = 5 （米） 5 + 3 = 8 (米) 3.14 × (8² - 5²) = 3.14 × (64 - 25) = 3.14 × 39 = 122.46 (平方米) 答: 这条小路的面积是 122.46 平方米。",
+    "knowledge_points": [
+      "圆环的面积"
+    ],
+    "source": "19-20 六年级上·全国·单元测试",
+    "merged_id": "unit5_189",
+    "original_id": "unit5_choice_189"
+  },
+  {
+    "id": "unit5_choice_190",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长和面积",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "圆的周长为12.56米，那么这个圆的半径是多少米？面积是多少平方米？",
+    "options": [],
+    "answer": "2米；12.56平方米",
+    "solution": "【分析】圆的周长公式为 C = 2πr , 则 r = C ÷ π ÷ 2 , 把圆的周长代入公式求出这个圆的半径, 再利用 S = πr² 求出这个圆的面积, 据此解答。熟记并灵活运用圆的周长和面积计算公式是解答题目的关键。【详解】12.56 ÷ 3.14 ÷ 2 = 4 ÷ 2 = 2 (米) 3.14 × 2² = 12.56 （平方米）答: 这个圆的半径是 2 米, 面积是 12.56 平方米。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "17-18六年级上·吉林·期中",
+    "merged_id": "unit5_190",
+    "original_id": "unit5_choice_190"
+  },
+  {
+    "id": "unit5_choice_191",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "李叔叔用 157 厘米长的铁条做了一个圆形的铁环, 这个铁环的半径是多少厘米?",
+    "options": [],
+    "answer": "25 厘米",
+    "solution": "【分析】根据圆的半径 = 周长 ÷ π ÷ 2 , 列式解答即可。关键是掌握并灵活运用圆的周长公式。【详解】157 ÷ 3.14 ÷ 2 = 25 （厘米）答: 这个铁环的半径是 25 厘米。",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "22-23六年级上·辽宁·单元测试",
+    "merged_id": "unit5_191",
+    "original_id": "unit5_choice_191"
+  },
+  {
+    "id": "unit5_choice_192",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "猫和老鼠在一个直径是50米的圆周上的同一地点向相反方向运动，猫每分钟走21.98米，老鼠每分钟走9.42米。当猫和老鼠第一次相遇时，猫比老鼠多走了多少米？",
+    "options": [],
+    "answer": "62.8米",
+    "solution": "【分析】相遇时间 = 相遇路程 ÷ 速度和, 再根据猫每分钟比老鼠多走 12.56 米, 乘相遇时间, 求出当猫和老鼠第一次相遇时, 猫比老鼠多走了多少米即可。本题考查相遇问题、圆的周长, 解答本题的关键是掌握相遇问题中的数量关系。【详解】3.14 × 50 ÷ (21.98 + 9.42) = 157 ÷ 31.4 = 5(分钟) (21.98 - 9.41) × 5 = 12.56 × 5 = 62.8 (米) 答: 猫和老鼠第一次相遇时, 猫比老鼠多走了 62.8 米。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "相遇问题"
+    ],
+    "source": "20-21六年级下·湖北孝感·期末",
+    "merged_id": "unit5_192",
+    "original_id": "unit5_choice_192"
+  },
+  {
+    "id": "unit5_choice_193",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的面积",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "公园里有一个圆形的花圃，直径是8米，在花圃的周围修一条1米宽的石子路，这条石子路的面积是多少平方米？",
+    "options": [],
+    "answer": "28.26 平方米",
+    "solution": "【分析】公园里有一个圆形的花圃，直径是 8 米，在花圃的周围修一条 1 米宽的石子路，则外圆的直径是 10 米，再用外圆的面积减去内圆的面积，求出这条石子路的面积即可。本题考查圆环的面积, 解答本题的关键是掌握圆环的面积计算公式。【详解】8 ÷ 2 = 4（米） (8 + 1 + 1) ÷ 2 = 10 ÷ 2 = 5 （米） 3.14 × (5² - 4²) = 3.14 × 9 = 28.26 (平方米)答: 这条石子路的面积是 28.26 平方米。",
+    "knowledge_points": [
+      "圆环的面积"
+    ],
+    "source": "21-22 五年级下·湖南邵阳·期末",
+    "merged_id": "unit5_193",
+    "original_id": "unit5_choice_193"
+  },
+  {
+    "id": "unit5_choice_194",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的面积",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "文化广场的喷水池直径是 40 米, 在它的外围有一条宽 4 米的环形路。这条环形路的面积是多少平方米?",
+    "options": [],
+    "answer": "552.64平方米",
+    "solution": "【分析】先根据喷水池的直径计算出内圆半径, 外圆半径 = 内圆半径 + 环宽, 最后根据 S环形 = π × (R² - r²) 求出这条环形路的面积, 据此解答。本题主要考查环形面积公式的应用, 熟记公式是解答题目的关键。【详解】内圆半径: 40 ÷ 2 = 20 (米)    外圆半径: 20 + 4 = 24 (米) 3.14 × (24² - 20²) = 3.14 × (576 - 400) = 3.14 × 176 = 552.64 (平方米) 答: 这条环形路的面积是 552.64 平方米。",
+    "knowledge_points": [
+      "圆环的面积"
+    ],
+    "source": "20-21 五年级下·河南新乡·期末",
+    "merged_id": "unit5_194",
+    "original_id": "unit5_choice_194"
+  },
+  {
+    "id": "unit5_choice_195",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长和植树问题",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "人民广场新建了一个圆柱形花坛，花坛中间有一个底面周长是 6.28 m 的圆柱形喷水池，准备在距喷水池边 1 m 处栽一圈观赏树，如果沿着这一圈每隔 1.57 m 栽一棵，一共要栽多少棵观赏树？",
+    "options": [],
+    "answer": "8棵",
+    "solution": "【分析】根据植树问题可知, 圆形水池的一周种一圈观赏树, 这一圈观赏树形成一个圆形, 它的半径为: 水池的半径 +1 米, 树的棵数 = 间隔数, 用水池一周的长度, 除以每个间隔的长度即可求解。本题考查了植树问题, 关键求出是圆形喷水池的半径, 利用树的棵数 = 间隔数, 解决问题。【详解】6.28 ÷ 3.14 ÷ 2 + 1 = 1 + 1 = 2 (米) 3.14 × 2 × 2 ÷ 1.57 = 6.28 × 2 ÷ 1.57 = 12.56 ÷ 1.57 = 8 (棵) 答: 一共要栽 8 棵观赏树。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "植树问题"
+    ],
+    "source": "2021·陕西榆林·小升初真题",
+    "merged_id": "unit5_195",
+    "original_id": "unit5_choice_195"
+  },
+  {
+    "id": "unit5_choice_196",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长和面积",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "公园里有一个圆形儿童游乐场，周长是75.36米，后来扩建时将它的半径增加了3米，扩建后这个游乐游乐场的面积是多少平方米？",
+    "options": [],
+    "answer": "706.5 平方米",
+    "solution": "【分析】由题意可知, 根据圆的周长公式 C = 2πr , 进而求出圆的半径，然后求出扩建后的半径，最后根据圆的面积公式 S = πr² ，代入数值进行计算即可。本题考查圆的周长和面积, 熟记公式是解题的关键。【详解】75.36 ÷ 3.14 ÷ 2 + 3 = 24 ÷ 2 + 3 = 12 + 3 = 15 (米) 3.14 × 15² = 706.5 （平方米）答: 扩建后这个游乐游乐场的面积是 706.5 平方米。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "21-22六年级上·河北邯郸·期末",
+    "merged_id": "unit5_196",
+    "original_id": "unit5_choice_196"
+  },
+  {
+    "id": "unit5_choice_197",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "人民公园建了一个周长为 62.8 米的圆形草坪，并准备为它安装一个自动旋转式喷水装置进行喷灌，现有射程为20米、15米、10米的三种装置，你认为选哪种射程的装置比较合适？安装在什么地方最好？",
+    "options": [],
+    "answer": "选10米射程的装置比较合适；安装在圆心处最好",
+    "solution": "【分析】根据圆的周长公式：C = 2πr 可知，圆的半径 r = C ÷ π ÷ 2 ，代入数据，求出圆形草坪的半径。因为自动旋转式喷水装置的射程是圆的半径，用草坪的半径与三个射程相比较，选出最合适的射程。灵活运用圆的周长公式是解题的关键。【详解】62.8 ÷ 3.14 ÷ 2 = 20 ÷ 2 = 10 (米) 20 > 10, 15 > 10, 10 = 10 答: 选 10 米射程的装置比较合适; 安装在圆心处最好。",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "圆的周长及应用"
+    ],
+    "source": "20-21六年级上·山东菏泽·期末",
+    "merged_id": "unit5_197",
+    "original_id": "unit5_choice_197"
+  },
+  {
+    "id": "unit5_choice_198",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的面积",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "一个圆形水池的直径是6米，现要在这个水池的周围修一条宽1米的石子路，这条石子路的面积是多少平方米？",
+    "options": [],
+    "answer": "21.98 平方米",
+    "solution": "【分析】先计算出水池的半径, 大圆的半径 = 水池的半径 + 石子路的宽度, 利用圆环的面积公式: S = π(R² - r²) , 即可求得石子路的面积。掌握环形的面积计算公式是解答题目的关键。【详解】大圆半径: 6 ÷ 2 + 1 = 3 + 1 = 4 (米) 3.14 × (4² - 3²) = 3.14 × 7 = 21.98 (平方米) 答: 这条石子路的面积是 21.98 平方米。",
+    "knowledge_points": [
+      "圆环的面积"
+    ],
+    "source": "21-22 六年级上·四川凉山·期末",
+    "merged_id": "unit5_198",
+    "original_id": "unit5_choice_198"
+  },
+  {
+    "id": "unit5_choice_199",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长和面积",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "一张圆桌的直径是18分米，王叔叔给这张圆桌的边上围上一圈铁条，并给桌面刷油漆。王叔叔至少需要准备多长的铁条？需要刷油漆的面积有多大？",
+    "options": [],
+    "answer": "56.52 分米；254.34 平方分米",
+    "solution": "【分析】由题意可知, 计算铁条的长度就是求圆的周长,计算刷油漆的面积就是求圆的面积, 利用圆的周长和面积公式即可求得。掌握圆的周长和面积计算公式是解答题目的关键。【详解】铁条的长度: 3.14 × 18 = 56.52 (分米) 刷油漆的面积: 3.14 × (18 ÷ 2)² = 3.14 × 81 = 254.34 (平方分米) 答: 王叔叔至少需要准备 56.52 分米的铁条, 需要刷油漆的面积是 254.34 平方分米。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级上·广东广州·期末",
+    "merged_id": "unit5_199",
+    "original_id": "unit5_choice_199"
+  },
+  {
+    "id": "unit5_choice_200",
+    "unit": "第五单元：圆",
+    "chapter": "半圆的周长和三角形面积",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "画一个直径是4cm的半圆，在半圆内画一个以半圆直径为边，且这边上的高最大的三角形。分别算出半圆的周长和三角形的面积。",
+    "options": [],
+    "answer": "图见详解；半圆周长 10.28 厘米；三角形面积 4 平方厘米",
+    "solution": "【分析】先确定一个圆心, 再取半径 4 ÷ 2 = 2 (厘米) 为半径, 画一个半圆。再以直径为底, 半径为高, 画一个半圆内的三角形, 使其满足题干要求;根据圆的周长公式，先计算出这个圆的周长，再将其除以2，最后加上直径，求出半圆的周长。根据三角形的面积公式，计算出这个三角形的面积即可。本题考查了画圆、圆的周长以及三角形的面积,掌握画圆方法，圆的周长公式及三角形的面积公式是解题的关键。【详解】![](images/1d18e626db15d9eb1ff8be0a66ddf5636b4b582f06342c4c586d36666d05bcf8.jpg)半圆周长: 3.14 × 4 ÷ 2 + 4 = 6.28 + 4 = 10.28 (厘米)三角形面积： 4 × 2 ÷ 2 = 4 （平方厘米）答: 半圆的周长是 10.28 厘米, 三角形的面积是 4 平方厘米。",
+    "knowledge_points": [
+      "三角形的面积",
+      "圆的周长及应用",
+      "画三角形",
+      "画圆"
+    ],
+    "source": "21-22六年级上·湖北鄂州·期末",
+    "merged_id": "unit5_200",
+    "original_id": "unit5_choice_200"
+  },{
+    "id": "unit5_choice_201",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "用同样长的铁丝, 分别围成长方形、正方形、三角形、圆, ( ) 的面积最大。",
+    "options": [
+      "A. 圆",
+      "B. 长方形",
+      "C. 正方形",
+      "D. 三角形"
+    ],
+    "answer": "A",
+    "solution": "【分析】根据长方形、正方形、三角形、圆的周长一定，圆的面积最大进行选择。有些结论记下来可以节省大量思考和解题时间。\n\n【详解】用同样长的铁丝，说明周长一定，分别围成长方形、正方形、三角形、圆，圆的面积最大。故答案为：A",
+    "knowledge_points": [
+      "三角形的面积",
+      "正方形的面积",
+      "长方形的面积",
+      "圆的面积及应用"
+    ],
+    "source": "21-22 六年级上·云南昭通·期末",
+    "merged_id": "unit5_201",
+    "original_id": "unit5_choice_201"
+  },
+  {
+    "id": "unit5_choice_202",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "在推导圆的面积公式时，把一个圆分成若干（偶数）等份后，拼成一个近似长方形，这个长方形的长近似于（）。",
+    "options": [
+      "A. 圆的半径",
+      "B. 圆的周长",
+      "C. 圆周长的 $\\frac{1}{2}$"
+    ],
+    "answer": "C",
+    "solution": "【分析】推导圆的面积公式时, 将圆转化成一个近似的长方形, 如图: 长方形的长相当于圆周长的一半, 长方形的宽相当于圆的半径, 据此解答即可。![](images/7045dddd078ef49bcfb737d0b95f07b0aac9e9511e1f2e6f32329bfbc8a6c315.jpg)熟练掌握圆的面积推导过程是解答本题的关键。\n\n【详解】在推导圆的面积公式时, 把一个圆分成若干（偶数）等份后, 拼成一个近似长方形, 这个长方形的长近似于圆周长的 $\\frac{1}{2}$ ;故答案为: C。",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "2014 六年级·全国·专题练习",
+    "merged_id": "unit5_202",
+    "original_id": "unit5_choice_202"
+  },
+  {
+    "id": "unit5_choice_203",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "通过圆心并且两端都在圆上的 ( ) 叫直径。",
+    "options": [
+      "A. 直线",
+      "B. 线段",
+      "C. 射线"
+    ],
+    "answer": "B",
+    "solution": "【详解】通过圆心，并且两端都在圆上的线段叫做圆的直径。例如：通过圆心并且两端都在圆上的线段（如 BC）是直径，通常用字母 d 表示。![](images/7b916316f8ce512b0d75d9b679503f6eeee0c98a946905a885d16543b62dc9c1.jpg)故答案选：B",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "14-15 六年级上·全国·课后作业",
+    "merged_id": "unit5_203",
+    "original_id": "unit5_choice_203"
+  },
+  {
+    "id": "unit5_choice_204",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "把一个半径为$6 \\mathrm{~cm}$ 的圆平均方成 9 份, 其中一份的面积是 ( ) $\\mathrm{cm}^{2}$ 。",
+    "options": [
+      "A. 113.04",
+      "B. 12.56",
+      "C. 254.34"
+    ],
+    "answer": "B",
+    "solution": "【分析】根据圆的面积公式: $S = \\pi r^{2}$ , 计算出圆的面积, 将圆平均分成 9 份, 求其中一份的面积, 用圆的面积除以 9 即可解答。本题主要考查圆的面积公式。\n\n【详解】$3.14 \\times 6 \\times 6 \\div 9\\\\ = 1 1 3. 0 4 \\div 9 \\\\ = 1 2. 5 6 (\\mathrm {c m} ^ {2}) \\\\ $\n\n故答案为: B。",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "21-22 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit5_204",
+    "original_id": "unit5_choice_204"
+  },
+  {
+    "id": "unit5_choice_205",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "两个圆的周长不相等,是因为 ( )。",
+    "options": [
+      "A. 圆心的位置不同",
+      "B. 半径大小不同",
+      "C. 圆周率大小不同"
+    ],
+    "answer": "B",
+    "solution": "【分析】根据 'C=2πr''C=πd'可知，圆的周长与圆的直径和半径有关系；两个圆的周长不相等，说明它们的半径和直径不相等，据此解答即可。熟记圆的周长公式并能灵活利用是解答本题的关键。\n\n【详解】两个圆的周长不相等, 是因为半径和直径不相等;故答案为: B。",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_205",
+    "original_id": "unit5_choice_205"
+  },
+  {
+    "id": "unit5_choice_206",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "车轮滚动一周所行的长度是 ( )。",
+    "options": [
+      "A. 车轮的半径",
+      "B. 车轮的直径",
+      "C. 车轮的周长"
+    ],
+    "answer": "C",
+    "solution": "【分析】车轮滚动一周刚好等于一个整圆的周长，求车轮行驶的路程即求圆的周长。理解圆的周长的意义是解答题目的关键。\n\n【详解】分析可知, 车轮滚动一周所行的长度是车轮的周长。故答案为: C",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit5_206",
+    "original_id": "unit5_choice_206"
+  },
+  {
+    "id": "unit5_choice_207",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "在一个长 4 分米，宽 2 分米的长方形中画一个最大的圆，这个圆面积是（）平方分米。",
+    "options": [
+      "A. 3.14",
+      "B. 12.56",
+      "C. 25.12"
+    ],
+    "answer": "A",
+    "solution": "【分析】长方形中画一个最大的圆，圆的直径=长方形的宽，根据圆的面积=πr²，列式计算即可。关键是理解长方形和圆之间的关系，掌握圆的面积公式。\n\n【详解】$3.14 \\times (2 \\div 2)^{2}\\\\= 3. 1 4 \\times 1\\\\= 3.14（平方分米）$\n\n故答案为: A",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit5_207",
+    "original_id": "unit5_choice_207"
+  },
+  {
+    "id": "unit5_choice_208",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "为周长 62.8 米的圆形草坪安装自动旋转喷灌装置，应选择射程为（）米的装置。",
+    "options": [
+      "A. 20",
+      "B. 15",
+      "C. 10"
+    ],
+    "answer": "C",
+    "solution": "【分析】要明确射程, 即圆的半径, 根据 $r = c \\div \\pi \\div 2$ 求出半径, 即射程。熟练掌握圆的周长公式并能灵活利用是解答本题的关键。\n\n【详解】$62.8 \\div 3.14 \\div 2\\\\= 2 0 \\div 2\\\\=10（米）$；\n\n故答案为: C。",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "merged_id": "unit5_208",
+    "original_id": "unit5_choice_208"
+  },
+  {
+    "id": "unit5_choice_209",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一个时钟的分针长 $8 \\mathrm{~cm}$ , 时针长 $6 \\mathrm{~cm}$ , 从 1 时到 4 时, 时针针尖走过的面积是 ( ) $\\mathrm{cm}^{2}$ 。",
+    "options": [
+      "A. 12.56",
+      "B. 50.24",
+      "C. 28.26",
+      "D. 25.12"
+    ],
+    "answer": "C",
+    "solution": "【分析】一个钟表的时针旋转一周是 $360^{\\circ}$ , 平均分成 12 个大格, 每个大格是 $30^{\\circ}$ 从 1 时到 4 时共旋转 3 个大格共 $30^{\\circ} \\times 3 = 90^{\\circ}$ , 求出 $90^{\\circ}$ 占整个钟表的几分之几, 即是钟表面积的几分之几, 根据圆的面积公式进行解答即可。本题考查圆的面积，熟记公式是解题的关键。\n\n【详解】$30^{\\circ} \\times 3 = 90^{\\circ}$\n\n$3.1 4 \\times 6 ^ {2} \\times \\frac {9 0}{3 6 0} \\\\ = 3. 1 4 \\times 3 6 \\times \\frac {1}{4} \\\\ = 3. 1 4 \\times 9 \\\\ = 2 8. 2 6 \\text {(平 方 厘 米)} $故\n\n答案为: C",
+    "knowledge_points": [
+      "圆的面积及应用",
+      "扇形的周长和面积"
+    ],
+    "source": "20-21 六年级上·江西赣州·单元测试",
+    "merged_id": "unit5_209",
+    "original_id": "unit5_choice_209"
+  },
+  {
+    "id": "unit5_choice_210",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "下面各图形中,对称轴最多的是 ( )。",
+    "options": [
+      "A. 正方形",
+      "B. 等边三角形",
+      "C. 扇形",
+      "D. 圆"
+    ],
+    "answer": "D",
+    "solution": "【分析】根据轴对称图形的定义：一个图形沿某条直线对折，直线两旁的部分能够完全重合，这个图形就是轴对称图形，这条直线就是它的一条对称轴，据此即可确定这两个图形的对称轴条数。此题主要考查轴对称图形的定义以及对称轴条数的确定方法。\n\n【详解】由分析可知：正方形有 4 条对称轴, 等边三角形有 3 条对称轴, 扇形有 1 条对称轴, 圆有无数条对称轴。故选：D",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "对称轴的画法及数量",
+      "弧、圆心角、扇形的认识"
+    ],
+    "source": "20-21 六年级上·江西赣州·单元测试",
+    "merged_id": "unit5_210",
+    "original_id": "unit5_choice_210"
+  },
+  {
+    "id": "unit5_choice_211",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "圆周率 $\\pi$ ( ) $3.14$ 。",
+    "options": [
+      "A. 大于",
+      "B. 等于",
+      "C. 小于"
+    ],
+    "answer": "A",
+    "solution": "【分析】圆周率是任意一个圆的周长与它的直径的比值,它是一个无限不循环小数, 但在实际生活中通常取它的近似值 3.14。掌握圆周率的意义和多位小数比较大小的方法是解答题目的关键。\n\n【详解】$\\pi = 3.1415926535 \\ldots$ 3.1415926535...>3.14 则 $\\pi > 3.14$ 故答案为: A",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "多位小数的大小比较"
+    ],
+    "source": "2014 六年级·全国·课后作业",
+    "merged_id": "unit5_211",
+    "original_id": "unit5_choice_211"
+  },
+  {
+    "id": "unit5_choice_212",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一个圆的半径扩大到原来的 4 倍，面积就扩大到原来的（）倍。",
+    "options": [
+      "A. 4",
+      "B. 8",
+      "C. 16",
+      "D. 11"
+    ],
+    "answer": "C",
+    "solution": "【分析】圆的面积 $= \\pi r^2$ ，据此解答即可。本题考查圆的面积, 解答本题的关键是掌握圆的面积公式。\n\n【详解】一个圆的半径扩大到原来的 4 倍, 面积就扩大到原来的 $4 \\times 4 = 16$ 倍。故答案为: C。",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级上·江西宜春·期末",
+    "merged_id": "unit5_212",
+    "original_id": "unit5_choice_212"
+  },
+  {
+    "id": "unit5_choice_213",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "用同样长的绳子围成下面的图形, ( ) 的面积最大。",
+    "options": [
+      "A. 长方形",
+      "B. 等边三角形",
+      "C. 圆 D. 正方形"
+    ],
+    "answer": "C",
+    "solution": "【分析】根据题意可假设铁丝的长为 12.56 米, 那么根据正方形、长方形、圆形、三角形可分别计算出他们的边长,然后再利用他们的面积公式进行计算后再比较即可得到答案。此题主要考查的是：在周长相等的所有图形中，围成的圆的面积最大。\n\n【详解】假设铁丝的长为12.56米，正方形的边长是： $12.56 \\div 4 = 3.14$ （米）正方形的面积是: $3.14 \\times 3.14 = 9.8596$ (平方米);长方形的长和宽的和是: $12.56 \\div 2 = 6.28$ (米)长和宽越接近面积越大，长可为3.15米，宽为3.13米长方形的面积是: $3.15 \\times 3.13 = 9.8595$ (平方米);假设是正三角形，其边长是： $12.56 \\div 3 \\approx 4.2$ （米）三角形的高小于斜边，所以三角形的面积就小于 $4.2 \\times 4.2 \\div 2= 8.82$ （平方米）；圆的半径是： $12.56 \\div 2 \\div 3.14 = 2$ （米）圆的面积是： $2 \\times 2 \\times 3.14 = 12.56$ （平方米）；$8.82 < 9.8595 < 9.8596 < 12.56$ ;所以围成的圆的面积最大。故答案为: C",
+    "knowledge_points": [
+      "三角形的面积",
+      "正方形的面积",
+      "长方形的面积",
+      "圆的面积及应用"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "merged_id": "unit5_213",
+    "original_id": "unit5_choice_213"
+  },
+  {
+    "id": "unit5_choice_214",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "以下对称轴条数最多的是 ( )。",
+    "options": [
+      "A. 正方形",
+      "B. 长方形",
+      "C. 圆",
+      "D. 半圆"
+    ],
+    "answer": "C",
+    "solution": "【分析】一个图形沿一条直线折叠, 直线两旁的部分能够完全重合的图形就叫轴对称图形; 根据轴对称图形的定义,\n\n先找出各个图形的对称轴的条数，再进行对比选择即可。此题主要考查了轴对称图形的定义, 能够正确说出轴对称图形的对称轴是解题关键。\n\n【详解】A．正方形有4条对称轴；B. 长方形有 2 条对称轴; C. 圆有无数条对称轴; D. 半圆有 1 条对称轴;\n\n故对称轴最多的是圆;故答案为: C。",
+    "knowledge_points": [
+      "对称轴的画法及数量"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "merged_id": "unit5_214",
+    "original_id": "unit5_choice_214"
+  },
+  {
+    "id": "unit5_choice_215",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "下面的信息中（）与'圆周率'无直接联系。",
+    "options": [
+      "A. 割圆术",
+      "B. 阿基米德",
+      "C. 黄金分割",
+      "D. 蒲丰投针试验"
+    ],
+    "answer": "C",
+    "solution": "【分析】黄金分割是被公认为最能引起美感的比例，它与'圆周率'没有直接关系。掌握圆周率的相关课外知识是解答题目的关键。\n\n【详解】A. '割圆术'就是不断倍增圆内接正多边形的边数求出圆周率的方法; B. 阿基米德是通过计算边数倍增的圆外切和内接正多边形的周长来求圆周率近似值的; C. 黄金分割是指将整体一分为二, 较大部分与整体部分的比值等于较小部分与较大部分的比值, 比值接近 0.618 ; D. 利用蒲丰投针试验计算圆周率。故答案为: C",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·辽宁·期末",
+    "merged_id": "unit5_215",
+    "original_id": "unit5_choice_215"
+  },
+  {
+    "id": "unit5_choice_216",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "用 $10 \\mathrm{~m}$ 长的绳子,分别围成长方形, 正方形、圆, 面积最大的 ( )。",
+    "options": [
+      "A. 正方形",
+      "B. 长方形",
+      "C. 圆",
+      "D. 一样大"
+    ],
+    "answer": "C",
+    "solution": "【分析】已知用 $10 \\mathrm{~m}$ 长的绳子, 分别围成长方形, 正方形、圆, 则它们的周长都是 $10 \\mathrm{~m}$ , 然后根据长方形、正方形、圆的面积公式分别求出正方形, 长方形, 圆的面积进行比较即可。本题考查了圆, 正方形以及长方形的周长与面积公式。结论: 在周长相等的情况下, 圆的面积最大。\n\n【详解】由分析可知：长方形：设长为 $3 \\mathrm{~m}$ ，宽为 $2 \\mathrm{~m}, 3 \\times 2 = 6$ （平方米）正方形: $10 \\div 4 = 2.5$ （米）， $2.5 \\times 2.5 = 6.25$ （平方米）圆: $10 \\div 3.14 \\div 2 \\approx 1.6$ (米), $3.14 \\times 1.6^{2} = 8.0384$ (平方米)​ ，所以圆的面积最大。\n\n$8.0384 > 6.25 > 6$ ，所以圆的面积最大。故选：C",
+    "knowledge_points": [
+      "求最大面积",
+      "正方形的面积",
+      "长方形的面积",
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级上·湖南怀化·期末",
+    "merged_id": "unit5_216",
+    "original_id": "unit5_choice_216"
+  },
+  {
+    "id": "unit5_choice_217",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个圆形的直径是 8 厘米, 这个圆的周长是( )厘米; 面积是( )平方厘米。",
+    "options": [],
+    "answer": "25.12 50.24",
+    "solution": "【分析】根据圆的周长 $= \\pi d$ , 圆的面积 $= \\pi r^{2}$ , 列式计算即可。关键是掌握圆的周长和面积公式。\n\n【详解】$3.14 \\times 8 = 25.12（厘米）$ \n\n$3.1 4 \\times (8 \\div 2) ^ {2} \\\\ = 3. 1 4 \\times 1 6 \\\\ = 5 0. 2 4 \\text {(平 方 厘 米)} $",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级上·北京·单元测试",
+    "merged_id": "unit5_217",
+    "original_id": "unit5_choice_217"
+  },
+  {
+    "id": "unit5_choice_218",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "从一个边长是 10 分米的正方形纸里剪一个最大的圆，这个圆的周长是( )分米，面积是( )平方分米。",
+    "options": [],
+    "answer": "31.4 78.5",
+    "solution": "【分析】由题意, 在边长为 10 分米的正方形里剪一个最大的圆, 则正方形的边长的长度相当于圆的直径的长度, 即 10 分米。可套用圆的周长公式、面积公式来计算其周长和面积。熟悉圆的周长、面积公式，且能够根据具体题意，已知直径或半径，灵活应用相关公式来解答。\n\n【详解】$C_{\\text{圆}} = \\pi d = 3. 1 4 \\times 1 0 = 3 1. 4 \\text {(分 米)} $\n\n$ S _ {\\text {圆}} = \\pi (d \\div 2) ^ {2} \\\\ = 3. 1 4 \\times (1 0 \\div 2) ^ {2} \\\\ = 3. 1 4 \\times 2 5 \\\\=78.5（平方分米） $",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "15-16 六年级·全国·期末",
+    "merged_id": "unit5_218",
+    "original_id": "unit5_choice_218"
+  },
+  {
+    "id": "unit5_choice_219",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个圆的半径是3厘米，它的周长是( )厘米，面积是( )平方厘米。",
+    "options": [],
+    "answer": "18.84 28.26",
+    "solution": "【分析】根据圆的周长和面积公式，结合题中数据，列式计算出这圆的周长和面积。本题考查了圆的周长和面积, 掌握它的周长和面积公式是解题的关键。\n\n【详解】周长: $2 \\times 3.14 \\times 3 = 18.84$ (厘米) 面积: $3.14 \\times 3^{2} = 28.26$ (平方厘米) 所以，这个圆的周长是 18.84 厘米，面积是 28.26 平方厘米。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "2014·全国·小升初真题",
+    "merged_id": "unit5_219",
+    "original_id": "unit5_choice_219"
+  },
+  {
+    "id": "unit5_choice_220",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "圆的半径是3厘米，直径是( )厘米，周长是( )厘米，面积是( )平方厘米。",
+    "options": [],
+    "answer": "6 18.84 28.26",
+    "solution": "【分析】在同一个圆中直径是半径的 2 倍, 利用 $C = \\pi d$ 求出圆的周长, 利用 $S = \\pi r^{2}$ 求出圆的面积, 据此解答。掌握圆的周长和面积计算公式是解答题目的关键。\n\n【详解】直径: $3 \\times 2 = 6$ (厘米) 周长: $6 \\times 3.14 = 18.84$ (厘米)面积：$3.14\\times 3\\times 3= 9. 4 2 \\times 3=28.26 (平方厘米)$",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "14-15六年级上·宁夏中卫·期中",
+    "merged_id": "unit5_220",
+    "original_id": "unit5_choice_220"
+  },
+  {
+    "id": "unit5_choice_221",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "两端都在圆上的线段中,( )是最长的一条。",
+    "options": [],
+    "answer": "直径",
+    "solution": "【分析】通过圆心并且两端都在圆上的线段是直径，通常用字母 d 表示。关键是熟悉圆的特征。\n\n【详解】两端都在圆上的线段中，直径是最长的一条。",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "2014 六年级·全国·课后作业",
+    "merged_id": "unit5_221",
+    "original_id": "unit5_choice_221"
+  },
+  {
+    "id": "unit5_choice_222",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个长方形的长是 $8 \\mathrm{~cm}$ , 宽是 $6 \\mathrm{~cm}$ 。在这个长方形里画一个尽可能大的圆,这个圆的直径是( )厘米，周长是( )厘米。",
+    "options": [],
+    "answer": "6 18.84",
+    "solution": "【分析】在这个长方形里画一个尽可能大的圆，这个圆的直径等于长方形的宽，再根据圆的周长公式求出周长即可。本题考查长方形、圆的周长, 解答本题的关键是掌握圆的周长计算公式。\n\n【详解】直径：6cm周长: $3.14 \\times 6 = 18.84 \\, (\\mathrm{cm})$",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "长方形的概念及特点"
+    ],
+    "source": "21-22 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit5_222",
+    "original_id": "unit5_choice_222"
+  },
+  {
+    "id": "unit5_choice_223",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个时钟的分针长10厘米，经过1小时后，分针尖端所走的路程是( )厘米，分针扫过的面积是( )平方厘米。",
+    "options": [],
+    "answer": "62.8 314",
+    "solution": "【分析】经过 1 小时后, 分针尖端所走的路程是圆的周长,扫过的面积是圆的面积，根据圆的周长和面积公式解答即可。本题考查圆的周长和面积, 解答本题的关键是掌握圆的面积和周长计算公式。\n\n【详解】$2 \\times 3.14 \\times 10\\\\= 6. 2 8 \\times 1 0 \\\\ = 6 2. 8 \\text {(厘 米)}$\n\n$ 3. 1 4 \\times 1 0 ^ {2} \\\\ = 3. 1 4 \\times 1 0 0 \\\\ = 3 1 4 \\text {(平 方 厘 米)} $",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "17-18六年级上·全国·期末",
+    "merged_id": "unit5_223",
+    "original_id": "unit5_choice_223"
+  },
+  {
+    "id": "unit5_choice_224",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个时钟的分针长 4 厘米, 当它正好走一圈时, 它的尖端走了( )厘米, 分针所扫过的面积是( )平方厘米。",
+    "options": [],
+    "answer": "25.12 50.24",
+    "solution": "【分析】分针尖端走过的路程是以分针长为半径的圆的周长，分针所扫过的面积是以分针长为半径的圆的面积，据此解答。掌握圆周长和面积的计算公式是解答题目的关键。\n\n【详解】(1) $2 \\times 3.14 \\times 4\\\\= 6. 2 8 \\times 4\\\\=25.12 (厘米)$\n\n(2) $3.14 \\times 4^{2}\\\\= 3. 1 4 \\times 1 6\\\\=50.24 (平方厘米)$",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "15-16 六年级上·云南玉溪·期末",
+    "merged_id": "unit5_224",
+    "original_id": "unit5_choice_224"
+  },
+  {
+    "id": "unit5_choice_225",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "经测量一个圆从圆心到圆上一点的距离是 $5 \\mathrm{~cm}$ , 这个圆的面积是( )$\\mathrm {c m} ^ {2} 。$",
+    "options": [],
+    "answer": "78.5",
+    "solution": "【分析】由题意可知, 从圆心到圆上一点的距离是 $5 \\mathrm{~cm}$ , 即半径是 $5 \\mathrm{~cm}$ , 根据圆的面积公式 $S = \\pi r^{2}$ , 据此代入数值进行计算即可。本题考查圆的面积，熟记公式是解题的关键。\n\n【详解】$3.14 \\times 5^{2}\\\\= 3. 1 4 \\times 2 5\\\\=78.5 (平方厘米)$",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "20-21六年级上·湖南怀化·单元测试",
+    "merged_id": "unit5_225",
+    "original_id": "unit5_choice_225"
+  },
+  {
+    "id": "unit5_choice_226",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "男生和女生人数比是 $5: 7$ , 男生占总人数的 $\\frac{()}{()}$ , 女生占总人数 $\\frac{()}{()}$ 。",
+    "options": [],
+    "answer": "$\\frac{5}{12}; \\frac{7}{12}$",
+    "solution": "【分析】总人数平均分成（5+7）份，男生占其中的5份，女生占其中的7份，据此解答。计算出全班人数的总份数是解答题目的关键。\n\n【详解】男生占总人数: $5 \\div (5 + 7) = \\frac{5}{12}$ 女生占总人数: $7 \\div (5 + 7) = \\frac{7}{12}$",
+    "knowledge_points": [
+      "求一个数占另一个数几分之几",
+      "比的意义"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit5_226",
+    "original_id": "unit5_choice_226"
+  },
+  {
+    "id": "unit5_choice_227",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "用圆规画一个周长为 $25.12 \\mathrm{~cm}$ 的圆，圆规两脚间的距离应取( )cm；在圆内画一个圆心角是 $90^{\\circ}$ 的扇形，扇形的面积是圆面积的( )%。",
+    "options": [],
+    "answer": "4 25",
+    "solution": "【分析】由题意可知, 要求圆规两脚间的距离即求圆的半径, 根据圆的周长公式 $C = 2 \\pi r$ , 代入数值进行计算即可;在圆内画一个圆心角是 $90^{\\circ}$ 的扇形, 该扇形的面积等于整圆面积的 $\\frac{1}{4}$ , 据此解答即可。本题考查圆的周长和圆心角是 $90^{\\circ}$ 的扇形, 明确圆心角是 $90^{\\circ}$ 的扇形的面积是整圆面积的 $25 \\text{‰}$\n\n【详解】由分析可知:\n\n$ 2 5. 1 2 \\div 3. 1 4 \\div 2 \\\\ = 8 \\div 2 \\\\=4 (厘米)$\n\n$90 ^ {\\circ} \\div 360 ^ {\\circ} = \\frac {1}{4} = 25 \\%$ 则圆规两脚间的距离应取 $4 \\mathrm{~cm}$ , 扇形的面积是圆面积的$25 \\text{‰}$",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用",
+      "弧、圆心角、扇形的认识"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit5_227",
+    "original_id": "unit5_choice_227"
+  },
+  {
+    "id": "unit5_choice_228",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "$3: 9 = 3 \\div ( )= \\frac {()}{()}; \\quad 3 0: 6 = (\\quad): 1 = 1 5: () $。",
+    "options": [],
+    "answer": "9; $\\frac{1}{3}$ ; 5; 3",
+    "solution": "【分析】(1) 根据比、分数、除法的关系 $a:b = a \\div b = \\frac{a}{b}$ $(b \\neq 0$ 计算即可; (2) 利用比的基本性质计算即可。掌握比、分数、除法之间的关系以及比的基本性质是解答题目的关键。\n\n【详解】(1) $3: 9 = 3 \\div 9 = \\frac{1}{3}$ (2) $30: 6 = (30 \\div 6): (6 \\div 6) = 5: 1 = (5 \\times 3): (1 \\times 3)= 1 5: 3$",
+    "knowledge_points": [
+      "比的基本性质",
+      "比与分数、除法的关系"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit5_228",
+    "original_id": "unit5_choice_228"
+  },
+  {
+    "id": "unit5_choice_229",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个半圆的半径是 $3 \\mathrm{~m}$ , 这个半圆的周长是( ) $\\mathrm{m}$ , 面积是( ) $\\mathrm{m}^{2}$ 。",
+    "options": [],
+    "answer": "15.42 14.13",
+    "solution": "【分析】由题意可知, 半圆的周长 = 整圆的周长的一半 + 一条直径的长度, 根据圆的面积公式 $S = \\pi r^{2}$ , 据此代入数字进行计算即可。本题考查圆的周长和面积, 明确半圆的周长 = 整圆周长的一半 + 一条直径的长度是解题的关键。\n\n【详解】$3.14 \\times (3 \\times 2) \\div 2 + 3 \\times 2\\\\= 3. 1 4 \\times 6 \\div 2 + 6 \\\\ = 9. 4 2 + 6 \\\\= 15.42 (米)$\n\n$3.1 4 \\times 3 ^ {2} \\div 2 \\\\ = 3. 1 4 \\times 9 \\div 2 \\\\ = 2 8. 2 6 \\div 2 \\\\ = 14.13（平方米）$",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit5_229",
+    "original_id": "unit5_choice_229"
+  },
+  {
+    "id": "unit5_choice_230",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "将一个直径是 6 厘米的圆形纸片沿着直径对折一次，得到的一个半圆形纸片的周长是( )厘米，面积是( )平方厘米。",
+    "options": [],
+    "answer": "15.42 14.13",
+    "solution": "【分析】因为半圆的周长 = 圆的周长的一半 + 直径，半圆的面积 = 圆的面积 ÷ 2，利用圆的周长和面积公式即可求解。此题主要考查圆的周长和面积的计算方法, 熟记公式是解题关键。\n\n【详解】$3.14 \\times 6 \\div 2 + 6 = 9. 4 2 + 6 = 1 5. 4 2 \\text {(厘 米)} $\n\n$3.1 4 \\times (6 \\div 2) ^ {2} \\div 2 \\\\ = 3. 1 4 \\times 9 \\div 2 \\\\ = 1 4. 1 3 \\text {(平 方 厘 米)}$",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级上·云南德宏·期末",
+    "merged_id": "unit5_230",
+    "original_id": "unit5_choice_230"
+  },
+  {
+    "id": "unit5_choice_231",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "圆有( )条对称轴, 长方形有( )条对称轴, 正方形有( )条对称轴, 等边三角形有( )对称轴。",
+    "options": [],
+    "answer": "无数 2 4 3",
+    "solution": "【分析】一个图形沿一条直线对折，直线两旁的图形完全重合，这样的图形叫做轴对称图形，折痕所在的直线就是对称轴, 据此解答即可。明确轴对称图形的特点是解答本题的关键。\n\n【详解】![](images/c5fcee9c0ae9ba8de069fe925a27db406523e50f504cf9c3601737ee420da3cf.jpg)![](images/c0523bf9abc127e732fc2754fa82b7a282282342ca8fc0d1ca99537a4e13f0bd.jpg)![](images/c97f977244c416a07df7c45d180f5e962077beb5a241131701701a512bf2be96.jpg)![](images/582c83654b30b2aaf54e7740d3091579ceeb3a208901d916ddd38916488d2391.jpg)圆有条对称轴;长方形有 2 条对称轴;正方形有 4 条对称轴;等边三角形有3对称轴。",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "对称轴的画法及数量"
+    ],
+    "source": "21-22 六年级上·海南省直辖县级单位·阶段练习",
+    "merged_id": "unit5_231",
+    "original_id": "unit5_choice_231"
+  },
+  {
+    "id": "unit5_choice_232",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "用一根长 10.28 米绳子, 围成一个半圆形, 这个半圆的半径是( )米, 面积是( )平方米。",
+    "options": [],
+    "answer": "2 6.28",
+    "solution": "【分析】半圆的周长 = 圆的周长的一半 + 直径，列方程计算出半径的长度，再利用圆的面积公式： $S = \\pi r^2$ ，即可求得。利用圆的周长计算公式求出圆的半径是解答题目的关键。\n\n【详解】（1）解：设圆的半径为 $r$ 米。\n\n$ 2 \\pi r \\div 2 + 2 r = 1 0. 2 8 \\\\ \\pi r + 2 r = 1 0. 2 8 \\\\ (\\pi + 2) \\mathrm {r} = 1 0. 2 8 \\\\ 5. 1 4 \\mathrm {r} = 1 0. 2 8 \\\\ \\mathrm {r} = 1 0. 2 8 \\div 5. 1 4 \\\\ \\mathrm {r} = 2 $\n\n所以，这个半圆的半径是 2 米。\n\n(2) $3.14 \\times 2^{2} \\times \\frac{1}{2}\\\\= 3. 1 4 \\times 4 \\times \\frac {1}{2} \\\\ = 3. 1 4 \\times (4 \\times \\frac {1}{2}) \\\\ = 3. 1 4 \\times 2 \\\\ = 6. 2 8 \\text {(平 方 米)} $",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "14-15 六年级上·全国·课后作业",
+    "merged_id": "unit5_232",
+    "original_id": "unit5_choice_232"
+  },
+  {
+    "id": "unit5_choice_233",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "一个圆的半径扩大 4 倍, 周长扩大( )倍, 面积扩大( )倍。",
+    "options": [],
+    "answer": "4 16",
+    "solution": "【分析】一个圆的半径扩大到原来的几倍，周长也扩大到原来的几倍；面积扩大倍数是半径扩大倍数的平方，据此解答即可。明确半径、周长和面积之间扩大倍数之间的关系是解答本题的关键。\n\n【详解】一个圆的半径扩大到原来的 4 倍, 周长也扩大到原来的 4 倍, 面积扩大到原来的 16 倍。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "15-16 六年级上·全国·课后作业",
+    "merged_id": "unit5_233",
+    "original_id": "unit5_choice_233"
+  },{
+    "id": "unit5_choice_234",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "把一个直径 $8 \\mathrm{~cm}$ 的圆沿直径对折, 剪成两个半圆, 每个半圆的周长是( )cm, 面积是( )cm²。",
+    "options": [],
+    "answer": "20.56 25.12",
+    "solution": "【分析】把一个直径 $8 \\mathrm{~cm}$ 的圆沿直径对折, 剪成两个半圆,每个半圆的周长 = 整圆周长的一半 + 一条直径的长度, 此时半圆的面积等于整圆面积的一半, 据此解答即可。本题考查圆的周长和面积, 明确半圆的周长等于整圆周长的一半加上直径的长度是解题的关键。\n\n【详解】$3.14 \\times 8 \\div 2 + 8\\\\= 2 5. 1 2 \\div 2 + 8 \\\\ = 1 2. 5 6 + 8 \\\\ = 2 0. 5 6 \\text {(厘 米)} \\\\ $\n\n$3.1 4 \\times (8 \\div 2) ^ {2} \\div 2 \\\\ = 3. 1 4 \\times 1 6 \\div 2 \\\\ = 5 0. 2 4 \\div 2 \\\\ = 2 5. 1 2 \\text {(平 方 厘 米)} $",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "18-19六年级上·河北秦皇岛·期末",
+    "merged_id": "unit5_234",
+    "original_id": "unit5_choice_234"
+  },
+  {
+    "id": "unit5_choice_235",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "把一个圆分成若干（偶数）等份，拼成一个近似的长方形，这个长方形的长是 $9.42 \\mathrm{~cm}$ ，它的宽是( )cm，面积是( )cm²。",
+    "options": [],
+    "answer": "3 28.26",
+    "solution": "【分析】根据圆拼成的长方形的过程可知: 近似长方形的长等于圆周长的一半, 宽等于圆的半径, 然后根据长方形的面积公式推导出圆的面积公式。据此解答。本题主要考查了学生利用知识的迁移推导圆面积公式的过程。\n\n【详解】由分析可知:$9.42 \\div 3.14 = 3$ （厘米）$3.14 \\times 3^{2} = 28.26$ （平方厘米） 则它的宽是 $3 \\mathrm{~cm}$ , 面积是 $28.26 \\mathrm{~cm}^{2}$ 。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "长方形的面积"
+    ],
+    "source": "20-21 六年级上·江西赣州·单元测试",
+    "merged_id": "unit5_235",
+    "original_id": "unit5_choice_235"
+  },
+  {
+    "id": "unit5_choice_236",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个圆的直径扩大到原来的 3 倍, 它的周长扩大到原来的( )倍, 面积扩大到原来的( )倍。",
+    "options": [],
+    "answer": "3 9",
+    "solution": "【分析】设出原来圆的半径，表示出直径，以及扩大后圆的半径和直径，再利用圆的周长和面积计算公式计算出周长和面积扩大的倍数。一个圆的直径扩大到原来的 a 倍, 它的周长扩大到原来的 a 倍, 面积扩大到原来的 $a^{2}$ 倍。\n\n【详解】假设原来圆的半径为 $r$ , 周长为 $2 r$ ; 则现在圆的直径为 $6 r$ , 半径为 $3 r$ 。原来圆的周长: $2 \\pi r$现在圆的周长: $6 \\pi r$ , $(6 \\pi r) \\div (2 \\pi r) = 3$ 原来圆的面积： $\\pi r^2$ 现在圆的面积: $(3r)^{2}\\pi = 9\\pi r^{2}$ $(9 \\pi r ^ {2}) \\div (\\pi r ^ {2}) = 9$",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "2014·全国·小升初真题",
+    "merged_id": "unit5_236",
+    "original_id": "unit5_choice_236"
+  },
+  {
+    "id": "unit5_choice_237",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "两个圆的半径分别是 $4 \\mathrm{~cm}$ 和 $6 \\mathrm{~cm}$ , 它们的直径的比是( ), 周长的比是( ), 面积的比是( )。",
+    "options": [],
+    "answer": "2:3 2:3 4:9",
+    "solution": "【分析】圆的半径比 = 圆的直径比 = 圆的周长比，圆的面积比等于圆的半径的平方之比，据此解答。如果两圆的半径之比为 $a:b$ , 直径比和周长比为 $a:b$ , 面积比为 $a^{2}:b^{2}$ 。\n\n【详解】半径比： $4\\mathrm{cm}:6\\mathrm{cm} = 4:6 = (4\\div 2)$ ： $(6\\div 2) =$2:3 面积比: $2^{2}: 3^{2} = 4: 9$两个圆的半径分别是 $4\\mathrm{cm}$ 和 $6\\mathrm{cm}$ ，它们的直径的比是 $2:3$周长的比是 $2: 3$ , 面积的比是 $4: 9$ 。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用",
+      "比的意义"
+    ],
+    "source": "20-21 六年级上·江西赣州·单元测试",
+    "merged_id": "unit5_237",
+    "original_id": "unit5_choice_237"
+  },
+  {
+    "id": "unit5_choice_238",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个圆的半径扩大到原来的 3 倍, 直径扩大到原来的( )倍, 周长扩大到原来的( )倍，面积扩大到原来的( )倍。",
+    "options": [],
+    "answer": "3 3 9",
+    "solution": "【分析】根据半径扩大到原来的几倍，直径就扩大到原来的几倍，周长也扩大到原来的几倍，面积扩大到原来的倍数 $\\times$ 倍数，进行分析。圆的周长 $= \\pi d = 2\\pi r$ ，圆的面积 $= \\pi r^2$ 。\n\n【详解】$3 \\times 3 = 9$ , 一个圆的半径扩大到原来的 3 倍, 直径扩大到原来的 3 倍, 周长扩大到原来的 3 倍, 面积扩大到原来的 9 倍。",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "17-18 六年级上·吉林·单元测试",
+    "merged_id": "unit5_238",
+    "original_id": "unit5_choice_238"
+  },
+  {
+    "id": "unit5_choice_239",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个圆的周长是 78.5 厘米, 它的半径是____厘米, 面积是____平方厘米。",
+    "options": [],
+    "answer": "12.5 490.625",
+    "solution": "【分析】根据圆的周长公式可知: $r = C \\div \\pi \\div 2$ , 然后根据圆的面积公式 $S = \\pi r^{2}$ , 据此解答即可。本题考查圆的周长和面积, 熟记公式是解题的关键。\n\n【详解】$78.5 \\div 3.14 \\div 2\\\\= 2 5 \\div 2\\\\=12.5 (厘米)$\n\n$3.14 \\times 12.5^{2}$ \n$= 3.14 \\times 156.25$ \n=490.625 (平方厘米)\n\n则它的半径是 12.5 厘米，面积是 490.625 平方厘米。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "21-22 六年级上·山东·单元测试",
+    "merged_id": "unit5_239",
+    "original_id": "unit5_choice_239"
+  },
+  {
+    "id": "unit5_choice_240",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个小圆和一个大圆的直径比是 $2:3$ , 小圆和大圆的周长比是 , 面积比是 。",
+    "options": [],
+    "answer": "2:3 4:9",
+    "solution": "【分析】一个小圆和一个大圆的直径比是 $2: 3$ , 说明一个小圆和一个大圆的半径比是 $2: 3$ , 则根据圆的周长公式可得小圆和大圆的周长比是 $2: 3$ , 根据圆的面积公式可得, 面积比是 $4: 9$ 。本题考查比的意义、圆的周长和面积, 解答本题的关键是掌握圆的面积和周长公式。\n\n【详解】小圆和大圆的周长比是 $2: 3$ ，面积比是 $4: 9$ 。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用",
+      "比的意义"
+    ],
+    "source": "21-22 六年级上·山东·单元测试",
+    "merged_id": "unit5_240",
+    "original_id": "unit5_choice_240"
+  },
+  {
+    "id": "unit5_choice_241",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个圆的( )和( )的比值叫做圆周率, 用字母( )表示, 精确到百分位是( )。",
+    "options": [],
+    "answer": "周长 直径 $\\pi$ 3.14",
+    "solution": "【分析】根据圆周率的含义: 圆的周长和它直径的比值叫做圆周率, 用字母 ' $\\pi$ ' 表示, $\\pi$ 是一个无限不循环小数, 计算时一般精确到百分位取它的近似值 $3.14$ 。此题考查的是圆周率的含义。\n\n【详解】根据分析得,圆的周长与直径的比值叫作圆周率，用字母 $\\pi$ 表示，精确到百分位是 $3.14 $",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "21-22 六年级上·全国·课后作业",
+    "merged_id": "unit5_241",
+    "original_id": "unit5_choice_241"
+  },
+  {
+    "id": "unit5_choice_242",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个圆的周长是12.56cm, 它的面积是( )cm²。",
+    "options": [],
+    "answer": "12.56",
+    "solution": "【分析】由 $C = 2 \\pi r$ 的 $r = C \\div \\pi \\div 2$ , 计算出圆的半径, 最后根据 $S = \\pi r^{2}$ 计算出圆的面积即可。掌握圆的周长和面积计算公式是解答本题的关键。\n\n【详解】半径: $12.56 \\div 3.14 \\div 2\\\\= 4 \\div 2\\\\=2 (厘米)$\n\n面积: $3.14 \\times 2 \\times 2\\\\= 6. 2 8 \\times 2\\\\=12.56 (平方厘米)$",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "14-15 六年级·山东菏泽·期末",
+    "merged_id": "unit5_242",
+    "original_id": "unit5_choice_242"
+  },
+  {
+    "id": "unit5_choice_243",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "佳佳在文具店买了个半圆形量角器, 直径是 $8 \\mathrm{~cm}$ , 周长是( ) $\\mathrm{cm}$ , 面积是( ) $\\mathrm{cm}^{2}$ 。",
+    "options": [],
+    "answer": "20.56 25.12",
+    "solution": "【分析】（1）半圆周长 $=$ 圆周长的一半 + 一条直径的长度;(2) 半圆面积 $= \\text{圆的面积} \\div 2$ 。掌握圆的周长和面积的计算公式是解答本题的关键。\n\n【详解】(1) $3.14 \\times 8 \\div 2 + 8\\\\=2 5. 1 2 \\div 2 + 8 \\\\ = 1 2. 5 6 + 8 \\\\ = 2 0. 5 6 \\text {(厘 米)} $\n\n(2) 圆的半径: $8 \\div 2 = 4 \\text{ (cm)}$\n\n$3.1 4 \\times 4 \\times 4 \\div 2 \\\\ = 1 2. 5 6 \\times 4 \\div 2 \\\\ = 5 0. 2 4 \\div 2 \\\\= 25.12 （平方厘米）$",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "merged_id": "unit5_243",
+    "original_id": "unit5_choice_243"
+  },
+  {
+    "id": "unit5_choice_244",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "在同一个圆内,有( )条直径, ( )条半径, 且半径是直径的( )。",
+    "options": [],
+    "answer": "无数 无数 $\\frac{1}{2}$",
+    "solution": "【分析】通过圆心并且两端都在圆上的线段叫做直径，直径是圆中最长的线段，在同一个圆内有无数条直径；连接圆心和圆上任意一点的线段叫做半径，半径决定圆的大小，在同一个圆内有无数条半径在同圆或等圆中, 半径的长度是直径的 $\\frac{1}{2}$ , 直径的长度是半径的 2 倍。掌握圆的特征是解答本题的关键。\n\n【详解】在同一个圆内，有（无数）条直径，（无数）条半径，且半径是直径的 $(\\frac{1}{2})$ 。",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "merged_id": "unit5_244",
+    "original_id": "unit5_choice_244"
+  },
+  {
+    "id": "unit5_choice_245",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "在同一个圆中,扇形的大小与( )有关, 以 $\\frac{1}{6}$ 圆为弧的扇形圆心角是( )度。",
+    "options": [],
+    "answer": "圆心角的大小 60",
+    "solution": "【分析】一条弧和经过这条弧两端的两条半径所围成的图形叫做扇形，在同圆或等圆中，扇形的大小与这个扇形的圆心角的大小有关，圆心角越大扇形就越大； $\\frac{1}{6}$ 圆的圆心角度数一个周角度数的 $\\frac{1}{6}$ ，该圆心角度数表示为： $360 \\times \\frac{1}{6} = 60^{\\circ}$ ，据此解答。熟记扇形的特征是解答本题的关键。\n\n【详解】在同一个圆中，扇形的大小与（圆心角的大小）有关，以 $\\frac{1}{6}$ 圆为弧的扇形圆心角是（60）度。",
+    "knowledge_points": [
+      "弧、圆心角、扇形的认识"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "merged_id": "unit5_245",
+    "original_id": "unit5_choice_245"
+  },
+  {
+    "id": "unit5_choice_246",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "要知道一口水井圆形井口的直径我的办法是（至少写出一条）：________。",
+    "options": [],
+    "answer": "方法一：可以直接用尺子量出井口的直径；方法二：就是量井口的周长然后用周长除以圆周率算直径。(答案不唯一)",
+    "solution": "【分析】第一个办法就是直接用尺子去量井口间最长的线段的长度，这是最直接最简单的。第二个就是量井口的周长然后算直径。此题考查的是圆的直径的认识以及利用周长求直径的方法。\n\n【详解】可以直接用尺子量出井口的直径，第二个就是量井口的周长然后用周长除以圆周率算直径。",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "圆的周长及应用"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "merged_id": "unit5_246",
+    "original_id": "unit5_choice_246"
+  },
+  {
+    "id": "unit5_choice_247",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "一个圆的周长是 12.56m，半径增加3m后，它的面积是( )m²。",
+    "options": [],
+    "answer": "78.5",
+    "solution": "【分析】根据 ' $r = c \\div \\pi \\div 2$ ' 先求出变化前圆的半径，即12.56÷3.14÷2，再加上3即可求出变化后圆的半径，再根据 's = πr²'求出圆的面积即可。熟练掌握圆的周长和面积公式并能灵活利用是解答本题的关键。\n\n【详解】$12.56 \\div 3.14 \\div 2\\\\= 4 \\div 2\\\\= 2$ （米）；\n\n$2 + 3 = 5 \\text {(米)} ;$ $3.14 \\times 5^{2} = 78.5$ （平方米）",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "19-20 六年级上·河南信阳·期末",
+    "merged_id": "unit5_247",
+    "original_id": "unit5_choice_247"
+  },
+  {
+    "id": "unit5_choice_248",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个圆形花坛，直径6米，它的周长是( )米，面积是( )平方米。",
+    "options": [],
+    "answer": "18.84 28.26",
+    "solution": "【分析】(1) $C_{\\text {圆形}} = \\pi d$ ; (2) 根据圆的直径计算圆的半径, 利用 $S_{\\text {圆形}} = \\pi r^{2}$ 计算即可。掌握圆的周长和面积的计算公式是解答本题的关键。\n\n【详解】(1) $6 \\times 3.14 = 18.84$ （米） (2) $3.14 \\times (6 \\div 2)^{2} = 28.26$ （平方米）",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "19-20六年级上·河南三门峡·期末",
+    "merged_id": "unit5_248",
+    "original_id": "unit5_choice_248"
+  },
+  {
+    "id": "unit5_choice_249",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个圆的周长是12.56, 在这个圆内画一个最大的正方形, 这个正方形的面积是()。",
+    "options": [],
+    "answer": "8",
+    "solution": "【分析】先根据圆的周长公式 $C = 2 \\pi r$ , 已知周长是 12.56,求出圆的半径为 $12.56 \\div 2 \\div 3.14 = 2$ , 这个最大正方形的面积就等于 4 个直角边为 2 的等腰直角三角形的面积, 据此可以求出正方形的面积。本题考查了正方形的面积, 关键是先根据周长求出圆的半径。\n\n【详解】$12.56 \\div 3.14 \\div 2\\\\ = 4 \\div 2 \\\\ = 2 $\n\n由分析可知, 正方形的面积为: $2 \\times 2 \\div 2 \\times 4\\\\= 4 \\div 2 \\times 4\\\\ = 8$",
+    "knowledge_points": [
+      "三角形的面积",
+      "圆的周长及应用",
+      "正方形的面积"
+    ],
+    "source": "2014 三年级·全国·课后作业",
+    "merged_id": "unit5_249",
+    "original_id": "unit5_choice_249"
+  },
+  {
+    "id": "unit5_choice_250",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "把一个周长为$314 \\mathrm{~cm}$ 的圆纸片剪成两个半圆形, 每个半圆形纸片的周长是$257 \\mathrm{~cm}$ 。 ( )",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】圆的周长 $= \\pi d$ , 据此求出圆的直径; 圆纸片剪成两个半圆形, 每个半圆的周长等于圆周长的一半加上一条直径, 据此解答即可。本题考查圆的周长, 解答本题的关键是掌握圆的周长的计算公式。\n\n【详解】$314 \\div 3.14 = 100$ （厘米）\n\n$ 3 1 4 \\div 2 + 1 0 0 \\\\ = 1 5 7 + 1 0 0 \\\\ = 2 5 7 \\text {(厘 米)} $\n\n故答案为: √。",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·浙江杭州·期末",
+    "merged_id": "unit5_250",
+    "original_id": "unit5_choice_250"
+  },
+  {
+    "id": "unit5_choice_251",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "圆中两端都在圆上的线段, 一定是圆的直径。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】通过圆心并且两端都在圆上的线段叫做直径，直径是圆中最长的线段，在同一个圆内有无数条直径；据此解答。掌握直径的意义是解答题目的关键。\n\n【详解】圆中两端都在圆上并经过圆心的线段是圆的直径,如图: AB 是圆的直径, CD 不是圆的直径。![](images/3c9d79750f1717ead92e1f49b591fb087e3db23273993c13f92eedc38dc166b2.jpg)故答案为: ×",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "14-15 六年级上·全国·期末",
+    "merged_id": "unit5_251",
+    "original_id": "unit5_choice_251"
+  },
+  {
+    "id": "unit5_choice_252",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "圆是轴对称图形，它有很多条对称轴。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】依据轴对称图形的概念, 即在平面内, 如果一个图形沿一条直线对折, 直线两旁的部分能够完全重合, 这样的图形叫做轴对称图形, 这条直线就是其对称轴, 据此解答即可。明确轴对称图形的特点是解答本题的关键。\n\n【详解】因为圆沿任意一条直径所在的直线对折，对折后的两部分都能完全重合，所以圆是轴对称图形，并且有无数条对称轴，原题说法正确；故答案为： √",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "轴对称的认识及辨认"
+    ],
+    "source": "18-19六年级上·全国·课后作业",
+    "merged_id": "unit5_252",
+    "original_id": "unit5_choice_252"
+  },
+  {
+    "id": "unit5_choice_253",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "所有直径都相等,所有的半径也相等。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】同一圆内，所有直径都相等，所有的半径也相等。本题考查圆的认识, 解答本题的关键是掌握圆的概念。\n\n【详解】根据分析可知：所有直径都相等，所有的半径也相等，说法错误。故答案为: ×。",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "21-22 六年级上·山东·单元测试",
+    "merged_id": "unit5_253",
+    "original_id": "unit5_choice_253"
+  },
+  {
+    "id": "unit5_choice_254",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "所有圆的周长和它的直径的比值都相等。( )",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】圆的周长和它的直径的比值是一个固定的数，叫作圆周率，用字母' $\\pi$ '表示，它是一个无限不循环的小数。明确圆周率的含义是解答本题的关键。\n\n【详解】所有圆的周长和它的直径的比值都相等，叫作圆周率，原题说法正确；故答案为： √。",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "21-22 六年级上·全国·单元测试",
+    "merged_id": "unit5_254",
+    "original_id": "unit5_choice_254"
+  },
+  {
+    "id": "unit5_choice_255",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "在同一圆内, 所有的半径都相等。( )",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】![](images/d3ba16a99f84962ab6d03313183b93a96d6a9a687706f54324d778c622c6db8c.jpg)\n\n【详解】连接圆心和圆上任意一点的线段叫做半径，半径决定圆的大小，在同一个圆内有无数条半径，所有的半径都相等；故答案为: √",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "19-20 六年级下·全国·开学考试",
+    "merged_id": "unit5_255",
+    "original_id": "unit5_choice_255"
+  },
+  {
+    "id": "unit5_choice_256",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "周长相等的圆和正方形，圆的面积大。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】假设圆和正方形的周长是 12.56，确定圆的半径和正方形边长，分别求出面积，比较即可。关键是掌握圆和正方形的周长及面积公式。\n\n【详解】假设圆和正方形的周长是 $12.56$ 。圆的半径: $12.56 \\div 3.14 \\div 2 = 2$；圆的面积: $3.14 \\times 2^{2} = 12.56$ 正方形边长: $12.56 \\div 4 = 3.14$；正方形面积: $3.14 \\times 3.14 = 9.8596$ $12.56 > 9.8596$ 故答案为: √",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "正方形的面积",
+      "圆的面积及应用",
+      "正方形的周长"
+    ],
+    "source": "2014 三年级·全国·课后作业",
+    "merged_id": "unit5_256",
+    "original_id": "unit5_choice_256"
+  },
+  {
+    "id": "unit5_choice_257",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "在长 $12.4 \\mathrm{~cm}$ 、宽7.2cm 的长方形纸中，剪半径是 1cm 的圆，最多只能剪 18个。( )",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】半径是 $1 \\mathrm{~cm}$ 的圆, 则直径为 $2 \\mathrm{~cm}$ , 横着可以剪这样的圆的个数为 $12.4 \\div 2 \\approx 6$ 个, 竖着可以剪这样的圆的个数为 $7.2 \\div 2 \\approx 3$ 个 (用去尾法取近似值, 余下的不能剪 1 个就舍去), 共 $6 \\times 3 = 18$ 个, 据此判断即可。本题考查圆的认识, 解答本题的关键是掌握长方形和圆的特征。\n\n【详解】$1 \\times 2 = 2 \\mathrm{~cm}$ , $12.4 \\div 2 \\approx 6$ （个）， $7.2 \\div 2 \\approx 3$ （个），$6 \\times 3 = 18$ （个），所以最多只能剪 18 个这样的圆。故答案为: √。",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "平面图形的分割",
+      "长方形的概念及特点"
+    ],
+    "source": "22-23 六年级上·全国·假期作业",
+    "merged_id": "unit5_257",
+    "original_id": "unit5_choice_257"
+  },
+  {
+    "id": "unit5_choice_258",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "用两个圆心角是 $180^{\\circ}$ 的扇形, 一定可以拼成一个圆。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】两个圆心角是 $180^{\\circ}$ 的扇形, 半径不一定相等, 当两个扇形的半径相等时, 可以拼成一个圆; 当两个扇形的半径不相等时, 不可以拼成一个圆。掌握扇形的特征是解答题目的关键。\n\n【详解】用两个圆心角是 $180^{\\circ}$ 且半径相等的扇形, 一定可以拼成一个圆。所以答案为: ×",
+    "knowledge_points": [
+      "弧、圆心角、扇形的认识"
+    ],
+    "source": "20-21 六年级上·山西朔州·期末",
+    "merged_id": "unit5_258",
+    "original_id": "unit5_choice_258"
+  },
+  {
+    "id": "unit5_choice_259",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "圆心角的度数越大,扇形的面积就越大。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】在同圆或等圆中，圆心角越大所对应的弧线越长，得到的扇形的面积越大，圆心角越小所对应的弧线越短，扇形的面积越小，据此判断即可。熟练掌握有关圆心角的基础知识是解答本题的关键。\n\n【详解】在同圆或等圆中，圆心角的度数越大，扇形的面积就越大，原题说法错误；故答案为：×。",
+    "knowledge_points": [
+      "弧、圆心角、扇形的认识"
+    ],
+    "source": "17-18 六年级上·全国·单元测试",
+    "merged_id": "unit5_259",
+    "original_id": "unit5_choice_259"
+  },
+  {
+    "id": "unit5_choice_260",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "两个圆的面积相等, 它们的周长也一定相等。( )",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】圆的面积与半径有关，当两个圆的面积相等时，它们的半径和直径都相等，所以周长也相等，据此解答即可。\n\n【详解】两个圆的面积相等，它们的周长也一定相等，原题说法正确；故答案为: √",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "2019·山东潍坊·小升初模拟",
+    "merged_id": "unit5_260",
+    "original_id": "unit5_choice_260"
+  },
+  {
+    "id": "unit5_choice_261",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "圆的对称轴有无数条, 半圆的对称轴也有无数条。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】轴对称图形定义为平面内，一个图形沿一条直线折叠，直线两旁的部分能够完全重合的图形，直线叫做对称轴。关键是熟悉圆和半圆的特征。\n\n【详解】圆的对称轴有无数条，半圆的对称轴只有一条，所以原题说法错误。",
+    "knowledge_points": [
+      "与圆相关的轴对称图形",
+      "对称轴的画法及数量"
+    ],
+    "source": "19-20 六年级上·山西忻州·期末",
+    "merged_id": "unit5_261",
+    "original_id": "unit5_choice_261"
+  },
+  {
+    "id": "unit5_choice_262",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "圆的直径缩小到它的 $\\frac{1}{10}$ , 面积也缩小到它的 $\\frac{1}{10}$ 。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】圆的直径缩小到它的 $\\frac{1}{10}$ , 半径、周长也会缩小到它的 $\\frac{1}{10}$ , 面积会缩小到它的 $\\frac{1}{100}$ , 据此解答即可。半径、直径、周长变化是一致的，面积的变化是它们的平方。\n\n【详解】圆的直径缩小到它的 $\\frac{1}{10}$ , 面积缩小到它的 $\\frac{1}{100}$ , 原题说法错误; 故答案为: ×。",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "圆的面积及应用"
+    ],
+    "source": "19-20 六年级上·湖北十堰·期末",
+    "merged_id": "unit5_262",
+    "original_id": "unit5_choice_262"
+  },
+  {
+    "id": "unit5_choice_263",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "大圆的圆周率等于小圆的圆周率。( )",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】圆周率是任意一个圆的周长与它的直径的比值,这个比值是一个固定的数, 据此解答。掌握圆周率的意义是解答题目的关键。\n\n【详解】任意一个圆的周长都是它的直径的 $\\pi$ 倍, 所以不管是大圆还是小圆, 圆周率都相同。故答案为: √",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "2020·山西忻州·小升初真题",
+    "merged_id": "unit5_263",
+    "original_id": "unit5_choice_263"
+  },
+  {
+    "id": "unit5_choice_264",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "一张纸至少对折两次, 才能找到圆心。( )",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】圆中心的那个点即圆心，所有直径都相交于圆心，将一个圆形纸片最少要对折两次，才能找到两条折痕相交的那个点，即圆心，据此解答即可。本题较易，考查了确定圆心的方法。\n\n【详解】一张纸至少对折两次, 才能找到圆心, 说法正确; 故答案为: √。",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "merged_id": "unit5_264",
+    "original_id": "unit5_choice_264"
+  },
+  {
+    "id": "unit5_choice_265",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "周长相等的圆、正方形和等边三角形中, 正方形的面积最大。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】周长相等时，平面图形越接近圆，面积越大，据此解答即可。熟记周长相等时，面积最大的是圆。\n\n【详解】周长相等的圆、正方形和等边三角形中, 圆的面积最大, 其次是正方形, 原题说法错误;故答案为: ×。",
+    "knowledge_points": [
+      "三角形的面积",
+      "圆的周长及应用",
+      "正方形的面积",
+      "圆的面积及应用"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "merged_id": "unit5_265",
+    "original_id": "unit5_choice_265"
+  },
+  {
+    "id": "unit5_choice_266",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "判断题",
+    "difficulty": "困难",
+    "question": "$\\pi = 3.14$ 。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】圆周率是任意一个圆的周长与它的直径的比值,这个比值是一个固定的数, 它是一个无限不循环小数, 据此解答。掌握圆周率的意义和生活中常用的圆周率的近似值是解答题目的关键。\n\n【详解】$\\pi = 3.1415926535 \\ldots$ , 但在实际生活中常取它的近似值 3.14 , 所以 $\\pi \\approx 3.14$ 。故答案为: ×",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "2020·山西忻州·小升初真题",
+    "merged_id": "unit5_266",
+    "original_id": "unit5_choice_266"
+  },
+  {
+    "id": "unit5_choice_267",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "圆是轴对称图形, 直径就是对称轴。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】依据轴对称图形的意义, 即在平面内, 如果一个图形沿一条直线折叠, 直线两旁的部分能够完全重合, 这样的图形叫做轴对称图形, 这条直线就是其对称轴, 从而作出正确判断。解答此题的主要依据是：轴对称图形的意义及特征，注意对称轴是一条直线。\n\n【详解】因为圆沿任意一条直径所在的直线对折, 对折后的两部分都能完全重合, 则圆是轴对称图形, 任意一条直径所在的直线就是圆的对称轴, 又因为圆的直径是一条线段, 所以原题说法错误。故答案为: ×",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "轴对称的认识及辨认"
+    ],
+    "source": "15-16 五年级上·山东青岛·课后作业",
+    "merged_id": "unit5_267",
+    "original_id": "unit5_choice_267"
+  },
+  {
+    "id": "unit5_choice_268",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "任意一个圆的周长都是它直径的 $\\pi$ 倍。( )",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】根据圆的周长公式，分析解题即可。本题考查了圆的周长, 掌握圆的周长公式是解题的关键。\n\n【详解】圆的周长 $= \\pi \\times$ 直径, 所以任意一个圆的周长都是它直径的 $\\pi$ 倍。所以判断正确。",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "2019·广东·小升初真题",
+    "merged_id": "unit5_268",
+    "original_id": "unit5_choice_268"
+  },
+  {
+    "id": "unit5_choice_269",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "在同一个圆内，最长的线段是直径。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】通过圆心并且两端都在圆上的线段叫做直径。据此解答即可。本题考查圆的直径的定义, 明确直径的定义是解题的关键。\n\n【详解】由分析可知:直径是同一个圆内最长的线段。故题干说法正确。",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "2014·全国·小升初真题",
+    "merged_id": "unit5_269",
+    "original_id": "unit5_choice_269"
+  },
+  {
+    "id": "unit5_choice_270",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "周长与直径的比值就是圆周率。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】圆周率的定义是: 任意一个圆的周长与它的直径的比的比值是一个固定的数, 人们称它为圆周率, 用字母 $\\pi$ 表示。此题考查了圆周率的定义；注意圆周率是一个固定的数值。\n\n【详解】根据圆周率的含义可知：圆的周长与圆的直径的比值叫做圆周率, 本题并没有强调是圆的周长, 故本题说法错误。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "求比值"
+    ],
+    "source": "19-20六年级上·全国·期末",
+    "merged_id": "unit5_270",
+    "original_id": "unit5_choice_270"
+  },
+  {
+    "id": "unit5_choice_271",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "每相邻两条跑道周长相差a米，长、短跑内侧起跑线都要比内侧前移a米。（ ）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】不同的项目、不同的跑道相差也不一样，要使长、短跑的公平，则要求跑道周长相等，如果只要跑一圈，也就是短跑比内侧前移a米，如果是长跑则比内侧前移的就不是a米。据此解答即可。此题属于圆的周长的实际应用, 关键是不同的比赛项目前移的米数也不相同。\n\n【详解】由分析可知：因为赛跑分为长跑和短跑，不同的项目相差也不一样，所以长、短跑内侧起跑线都要根据跑的几圈比内侧前移几个a米，所以这句话错的。故答案为: ×",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "21-22六年级上·江西赣州·单元测试",
+    "merged_id": "unit5_271",
+    "original_id": "unit5_choice_271"
+  },
+  {
+    "id": "unit5_choice_272",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "把下面各比化成最简单的整数比。\n\n64:16   2.7:0.36     $\\frac {5}{7}: \\frac {3}{8}$",
+    "options": [],
+    "answer": "4:1; 15:2; 40:21",
+    "solution": "【分析】（1）整数比的化简方法：比的前项和后项同时除以它们的最大公因数；(2) 小数比的化简方法: 先根据比的基本性质移动小数点的位置, 把小数比化为整数比, 再按照整数比的化简方法计算; (3) 分数比的化简方法: 比的前项和后项同时乘分母的最小公倍数, 把分数比化为整数比, 再按照整数比的化简方法计算, 据此计算。\n\n【详解】(1) $64: 16 = (64 \\div 16): (16 \\div 16) = 4: 1$ \n\n(2) $2.7: 0.36 = (2.7 \\times 100): (0.36 \\times 100) = 270: 36 = (270÷18) : (36÷18) = 15:2$ \n\n(3) $\\frac{5}{7}: \\frac{3}{8} = (\\frac{5}{7} \\times 56): (\\frac{3}{8} \\times 56) = 40: 21$",
+    "knowledge_points": [
+      "比的化简"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit5_272",
+    "original_id": "unit5_choice_272"
+  },
+  {
+    "id": "unit5_choice_273",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "计算下面各题，能简算的要简算。\n\n$ \\frac {7}{2 5} \\times 1 9 9 + \\frac {7}{2 5} \\quad 4 ^ {2} - 3. 1 4 \\times 2 ^ {2} \\quad 5 6. 5 2 \\div 3. 1 4 \\times (\\frac {1}{2} - \\frac {1}{9}) \\\\ \\frac {9}{5} \\times \\frac {2 3}{1 2 1} \\div \\frac {4 6}{5 5} \\quad (\\frac {1}{2} - \\frac {3}{7}) \\times \\frac {7}{5} + \\frac {3}{1 0} \\quad (\\frac {1}{3} - \\frac {1}{4} + \\frac {1}{6}) \\div \\frac {1}{2 4} $",
+    "options": [],
+    "answer": "56；3.44；7 $ \\frac {9}{2 2}; \\frac {2}{5}; 6 $",
+    "solution": "【分析】$\\frac{7}{25} \\times 199 + \\frac{7}{25}$ , 利用乘法分配律进行简算;$4^{2} - 3.14 \\times 2^{2}$ , 先算平方数, 再算乘法, 最后算减法;$56.52 \\div 3.14 \\times (\\frac{1}{2} - \\frac{1}{9})$ , 同时算出除法和减法, 再算乘法;$\\frac{9}{5} \\times \\frac{23}{121} \\div \\frac{46}{55}$ , 将除法改写成乘法, 约分后再计算;$(\\frac{1}{2} - \\frac{3}{7}) \\times \\frac{7}{5} + \\frac{3}{10}$ , 先利用乘法分配律进行简算, 再利用交换律进行简算;$(\\frac{1}{3} - \\frac{1}{4} + \\frac{1}{6}) \\div \\frac{1}{24}$ , 利用乘法分配律进行简算。\n\n【详解】$\\frac{7}{25} \\times 199 + \\frac{7}{25}= \\frac {7}{2 5} \\times (1 9 9 + 1) \\\\ = \\frac {7}{2 5} \\times 2 0 0 \\\\ = 5 6 $\n\n$ 4 ^ {2} - 3. 1 4 \\times 2 ^ {2} \\\\ = 1 6 - 1 2. 5 6 \\\\ = 3. 4 4 $\n\n$ 5 6. 5 2 \\div 3. 1 4 \\times (\\frac {1}{2} - \\frac {1}{9}) \\\\ = 1 8 \\times \\frac {7}{1 8} \\\\ = 7 $\n\n$ \\frac {9}{5} \\times \\frac {2 3}{1 2 1} \\div \\frac {4 6}{5 5} \\\\ = \\frac {9}{5} \\times \\frac {2 3}{1 2 1} \\times \\frac {5 5}{4 6} \\\\ = \\frac {9}{2 2} $\n\n$(\\frac {1}{2} - \\frac {3}{7}) \\times \\frac {7}{5} + \\frac {3}{1 0} \\\\ = \\frac {1}{2} \\times \\frac {7}{5} - \\frac {3}{7} \\times \\frac {7}{5} + \\frac {3}{1 0} \\\\ = \\frac {7}{1 0} + \\frac {3}{1 0} - \\frac {3}{5} \\\\ = 1 - \\frac {3}{5} \\\\ = \\frac {2}{5} $\n\n$ (\\frac {1}{3} - \\frac {1}{4} + \\frac {1}{6}) \\div \\frac {1}{2 4} \\\\ = \\frac {1}{3} \\times 2 4 - \\frac {1}{4} \\times 2 4 + \\frac {1}{6} \\times 2 4 \\\\ = 8 - 6 + 4 \\\\ = 6 $",
+    "knowledge_points": [
+      "分数的四则混合运算",
+      "分数乘法运算律"
+    ],
+    "source": "2021六年级上·全国·专题练习",
+    "merged_id": "unit5_273",
+    "original_id": "unit5_choice_273"
+  },
+  {
+    "id": "unit5_choice_274",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "直接写出得数。\n\n$5 \\times 3. 1 4 =$ $1 2. 5 6 \\div 3. 1 4 =$ $3.1 4 \\times (4 ^ {2} - 3 ^ {2})=$\n\n$3.1 4 \\times 3 =$ $3 1. 4 \\div 3. 1 4 =$ $3.1 4 \\times 2 ^ {2} =$",
+    "options": [],
+    "answer": "15.7; 4; 21.98; 9.42; 10; 12.56",
+    "solution": "【详解】略",
+    "knowledge_points": [
+      "小数与整数的乘法",
+      "除数是小数的小数除法"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "merged_id": "unit5_274",
+    "original_id": "unit5_choice_274"
+  },
+  {
+    "id": "unit5_choice_275",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "作图题",
+    "difficulty": "中等",
+    "question": "请你先画一个长 $3 \\mathrm{~cm}$ 、宽 $2 \\mathrm{~cm}$ 的长方形，再在里面画一个最大的半圆，并画出整个图形的对称轴。",
+    "options": [],
+    "answer": "见详解",
+    "solution": "【分析】先画出一个长 3 厘米, 宽 2 厘米的长方形, 再在这个长方形内, 以一条长的中点为圆心, 以 1.5 厘米为半径在长方形内画一个半圆, 过半圆的圆心画出对称轴。本题考查长方形和圆的特点, 明确它们的特点是解题的关键。\n\n【详解】由分析可知：![](images/366b9f3ebee83e9a35b9f3370693cf659e4dc6f70c89b7a121c488ea30b9219a.jpg)",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "对称轴的画法及数量",
+      "长方形的概念及特点"
+    ],
+    "source": "20-21 六年级上·江西赣州·单元测试",
+    "merged_id": "unit5_275",
+    "original_id": "unit5_choice_275"
+  },
+  {
+    "id": "unit5_choice_276",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "作图题",
+    "difficulty": "中等",
+    "question": "画一个周长是 12.56 厘米的圆，并求出它的面积。",
+    "options": [],
+    "answer": "图见详解，面积 12.56 平方厘米",
+    "solution": "【分析】根据圆的周长公式，先求出圆的半径。确定一个圆心，取半径长度画圆，再根据圆的面积公式求出它的面积。本题考查了圆的周长和面积, 灵活运用圆的周长和面积公式是解题的关键。\n\n【详解】半径: $12.56 \\div 3.14 \\div 2 = 2$ (厘米) 面积: $3.14 \\times 2^{2} = 12.56$ (平方厘米)![](images/ae1062f6904e9b8001ed295747e3f53136dc9e8e582dcee2448eca9b7f702aae.jpg)",
+    "knowledge_points": [
+      "画圆",
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "20-21六年级上·广东广州·期末",
+    "merged_id": "unit5_276",
+    "original_id": "unit5_choice_276"
+  },
+  {
+    "id": "unit5_choice_277",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一颗大树，树干的周长是62.8厘米，这棵大树树干的横截面面积是多少平方厘米？",
+    "options": [],
+    "answer": "314平方厘米",
+    "solution": "【分析】根据圆的半径 $= \\text{周长} \\div \\pi \\div 2$ , 圆的面积 $= \\pi r^{2}$ , 列式解答即可。关键是掌握圆的周长和面积公式。\n\n【详解】$62.8 \\div 3.14 \\div 2 = 10$ （厘米）$3.14 \\times 10^{2} = 314$ （平方厘米）答: 这棵大树树干的横截面面积是 314 平方厘米。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "2021六年级上·全国·专题练习",
+    "merged_id": "unit5_277",
+    "original_id": "unit5_choice_277"
+  },
+  {
+    "id": "unit5_choice_278",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "用一块边长 8 分米的木板加工成一个最大的圆桌，可能浪费了多少木板？",
+    "options": [],
+    "answer": "13.76 平方分米",
+    "solution": "【分析】在正方形内剪一个最大的圆，圆的直径和正方形的边长相等，据此可知圆桌的直径为 8 分米，再根据 's=πr²' 求出圆桌的面积，用木板的总面积减去圆桌的面积即可求出浪费的木板面积。明确在正方形内剪一个最大的圆，圆的直径和正方形的边长相等是解答本题的关键。\n\n【详解】$8 \\times 8 - 3.14 \\times (8 \\div 2)^{2}= 6 4 - 5 0. 2 4= 13.76$（平方分米）答: 可能浪费了 13.76 平方分米木板。",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "2022 六年级上·全国·专题练习",
+    "merged_id": "unit5_278",
+    "original_id": "unit5_choice_278"
+  },
+  {
+    "id": "unit5_choice_279",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "学校围绕一个半径是 7 米的圆形花坛铺一条 1 米宽的石子小路，求小路面积为多少平方米？",
+    "options": [],
+    "answer": "47.1 平方米",
+    "solution": "【分析】用花坛的半径加上小路宽度，求出外圆的半径，再用外圆的面积减去花坛的面积即可。本题考查圆环的面积, 解答本题的关键是掌握圆的面积计算公式。\n\n【详解】$7 + 1 = 8$ (米)\n\n$3.1 4 \\times (8 ^ {2} - 7 ^ {2}) \\\\ = 3. 1 4 \\times 1 5\\\\ = 47.1$ （平方米）\n\n答: 小路面积为 47.1 平方米。",
+    "knowledge_points": [
+      "圆环的面积"
+    ],
+    "source": "2022 六年级上·全国·专题练习",
+    "merged_id": "unit5_279",
+    "original_id": "unit5_choice_279"
+  },
+  {
+    "id": "unit5_choice_280",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一辆自行车的车轮半径是 40 厘米，车轮每分钟转 100 圈，要通过 2512 米的桥，大约需要几分钟？",
+    "options": [],
+    "answer": "10分钟",
+    "solution": "【分析】先求车轮一圈的长度，再求出自行车的速度，用路程除以速度等于时间，据此解答即可。本题考查圆的周长、行程问题, 解答本题的关键是掌握圆的周长计算公式。\n\n【详解】$3.14 \\times 40 \\times 2 \\times 100= 2 5 1. 2 \\times 1 0 0 = 2 5 1 2 0 \\text {(厘 米)} $ 25120 厘米 = 251.2 米 2512÷251.2=10 (分钟)\n\n答: 大约需要 10 分钟。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "基础行程问题"
+    ],
+    "source": "2014·全国·小升初真题",
+    "merged_id": "unit5_280",
+    "original_id": "unit5_choice_280"
+  },
+  {
+    "id": "unit5_choice_281",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "圆形草坪的直径是40米, 每平方米草坪9元。铺满草坪需要多少钱?",
+    "options": [],
+    "answer": "11304元",
+    "solution": "【分析】根据圆的面积公式，先计算出圆的面积，再将其乘9，求出铺满草坪需要多少钱。本题考查了圆的面积, 灵活运用圆的面积公式是解题的关键。\n\n【详解】$3.14 \\times (40 \\div 2)^{2} \\times 9\\\\= 3. 1 4 \\times 4 0 0 \\times 9\\\\= 1 1 3 0 4 \\text {(元)}$\n\n答：铺满草坪需要 11304 元钱。",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "21-22六年级上·黑龙江鹤岗·期末",
+    "merged_id": "unit5_281",
+    "original_id": "unit5_choice_281"
+  },
+  {
+    "id": "unit5_choice_282",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "在一张边长是20厘米的正方形白铁皮上，剪下若干个半径为3厘米的圆片，最多可以剪多少个？",
+    "options": [],
+    "answer": "9 个",
+    "solution": "【分析】在正方形白铁皮上，剪下若干个半径为 3 厘米的圆片，相当于剪边长 $3 \\times 2 = 6$ 厘米的正方形，用正方形白铁皮的边长 $\\div$ 小正方形边长，结果用去尾法保留近似数，求出一条边剪下的个数，个数 $\\times$ 个数即可。关键是熟悉圆的特征，直径=半径×2。\n\n【详解】$3 \\times 2 = 6$ （厘米） $2 0 \\div 6 \\approx 3 (\\text {个})$ $3 \\times 3 = 9 (\\text {个})$ 答: 最多可以剪 9 个。",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "平面图形的分割"
+    ],
+    "source": "21-22六年级上·河南南阳·期末",
+    "merged_id": "unit5_282",
+    "original_id": "unit5_choice_282"
+  },
+  {
+    "id": "unit5_choice_283",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一个圆形花坛的周长是 $50.24 \\mathrm{~m}$ , 在这个花坛里面留出 $\\frac{3}{8}$ 的面积种百合花。百合花的占地面积是多少平方米？",
+    "options": [],
+    "answer": "75.36 平方米",
+    "solution": "【分析】根据花坛的周长计算出圆形的半径，再根据半径计算出圆形花坛的面积，最后用乘法求出花坛面积的 $\\frac{3}{8}$ 即为百合花的占地面积。灵活运用圆的周长和面积计算公式是解答题目的关键。\n\n【详解】半径： $50.24 \\div 3.14 \\div 2= 1 6 \\div 2 = 8 \\text {(米)} $\n\n百合花的占地面积: $3.14 \\times 8^{2} \\times \\frac{3}{8}=3.1 4 \\times 6 4 \\times \\frac {3}{8} \\\\ = 3. 1 4 \\times (6 4 \\times \\frac {3}{8}) \\\\ = 3. 1 4 \\times 2 4 \\\\ = 7 5. 3 6 \\text {(平 方 米)} $\n\n答: 百合花的占地面积是 75.36 平方米。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "21-22 六年级上·山东临沂·期末",
+    "merged_id": "unit5_283",
+    "original_id": "unit5_choice_283"
+  },
+  {
+    "id": "unit5_choice_284",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "公园里有一个周长约为 $25.12 \\mathrm{~m}$ 的圆形假山水池, 这个水池的占地面积约是多少平方米?",
+    "options": [],
+    "answer": "50.24 平方米",
+    "solution": "【分析】圆的周长 $= 2 \\pi r$ , 据此求出圆的半径, 再根据圆的面积 $= \\pi r^{2}$ 解答即可。答: 这个水池的占地面积约是 50.24 平方米。本题考查圆的周长和面积, 解答本题的关键是掌握圆的周长和面积计算公式。\n\n【详解】$25.12 \\div 3.14 \\div 2= 8 \\div2=4 (米)$\n\n$1 4 \\times 4 ^ {2} \\\\ = 3. 1 4 \\times 1 6 \\\\= 50.24（平方米 )$",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "21-22 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit5_284",
+    "original_id": "unit5_choice_284"
+  },
+  {
+    "id": "unit5_choice_285",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "为倡导'节能减排，绿色出行'，李叔叔每天骑自行车上班。李叔叔的自行车车轮的外直径是0.7米，平均每分钟转100圈。李叔叔家离单位有1800米，他上班途中大约需要多长时间？(得数保留整数)",
+    "options": [],
+    "answer": "8 分钟",
+    "solution": "【分析】先根据车轮的一圈长度，求出叔叔骑自行车的速度，再用路程除以速度，求出时间即可。本题考查圆的周长、行程问题, 解答本题的关键是掌握圆的周长公式。\n\n【详解】$3.14 \\times 0.7 \\times 100 \\\\= 2. 1 9 8 \\times 1 0 0 \\\\ = 2 1 9. 8 \\text {(米)} $\n\n$1800 \\div 219.8 \\approx 8$ （分钟）答: 他上班途中大约需要 8 分钟。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "基础行程问题"
+    ],
+    "source": "20-21六年级上·湖南怀化·单元测试",
+    "merged_id": "unit5_285",
+    "original_id": "unit5_choice_285"
+  },
+  {
+    "id": "unit5_choice_286",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "圆圆看一本故事书, 看了 224 页, 看的页数与没看的页数的比是 $2: 3$ , 这本书有多少页?",
+    "options": [],
+    "answer": "560页",
+    "solution": "【分析】把这本书的总页数看作单位 '1', 则已看的 225 页所对应的分率是 $\\frac{2}{2 + 3}$ , 用对应量除以对应分率, 就是这本书的总页数。此题考查的是比的应用，解答此题的关键是：求出225页所对应的分率，利用对应量除以对应分率，即可求出单位'1'对应的量。\n\n【详解】$224 \\div \\frac{2}{2 + 3} \\\\= 2 2 4 \\div \\frac {2}{5} \\\\ = 5 6 0 \\text {(页)} $\n\n答: 这本书有 560 页。",
+    "knowledge_points": [
+      "按比分配问题"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit5_286",
+    "original_id": "unit5_choice_286"
+  },
+  {
+    "id": "unit5_choice_287",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "马戏团猴子表演独轮车走钢丝，车轮的直径为 $40 \\mathrm{~cm}$ ，要骑过 $25.12 \\mathrm{~m}$ 长的钢丝，车轮大约要转动多少周？",
+    "options": [],
+    "answer": "20周",
+    "solution": "【分析】根据圆的周长 $= \\pi d$ , 求出车轮周长, 用钢丝长度: 车轮周长 = 转动周数, 据此列式解答。关键是掌握圆的周长公式，注意统一单位。\n\n【详解】$40 厘米 = 0.4 米 2 5. 1 2 \\div (3. 1 4 \\times 0. 4) \\\\ = 2 5. 1 2 \\div 1. 2 5 6 \\\\ = 2 0 \\text {(周)}$\n\n答: 车轮大约要转动 20 周。",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit5_287",
+    "original_id": "unit5_choice_287"
+  },
+  {
+    "id": "unit5_choice_288",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "从一个长9分米、宽 8 分米的长方形木板上锯下一个最大的圆，剩下的木板是多少平方分米？",
+    "options": [],
+    "answer": "21.76 平方分米",
+    "solution": "【分析】长方形内画一个最大的圆，圆的直径和长方形的宽相等，根据 $s = \\pi r^{2}$ 求出圆的面积，用长方形减去圆的面积即可解答。明确长方形内画一个最大的圆，圆的直径和长方形的宽相等是解答本题的关键。\n\n【详解】$9 \\times 8 - 3.14 \\times (8 \\div 2)^{2}= 7 2 - 5 0. 2 4= 21.76$ （平方分米）答: 剩下的木板是 21.76 平方分米。",
+    "knowledge_points": [
+      "圆的面积及应用",
+      "平面图形的分割"
+    ],
+    "source": "21-22六年级上·江西赣州·单元测试",
+    "merged_id": "unit5_288",
+    "original_id": "unit5_choice_288"
+  },
+  {
+    "id": "unit5_choice_289",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "聪聪家有一张直径 $10 \\mathrm{dm}$ 的圆桌，现要按桌面的大小做一块桌布并在桌布周围镶上花边，请你算一算需要多大面积的桌布？需要花边多少？",
+    "options": [],
+    "answer": "78.5平方分米；31.4分米",
+    "solution": "【分析】计算桌布的面积即计算直径为 $10 \\mathrm{dm}$ 的圆的面积,花边的长度为直径 $10 \\mathrm{dm}$ 圆的周长, 据此计算。灵活运用圆的周长和面积计算公式是解答题目的关键。\n\n【详解】面积： $3.14 \\times (10 \\div 2)^{2}= 3. 1 4 \\times 5 ^ {2} \\\\ = 3. 1 4 \\times 2 5\\\\= 78.5（平方分米) $\n\n周长: $3.14 \\times 10 = 31.4$ (分米)答: 需要 78.5 平方分米的桌布, 需要花边 31.4 分米。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级上·江西赣州·单元测试",
+    "merged_id": "unit5_289",
+    "original_id": "unit5_choice_289"
+  },
+  {
+    "id": "unit5_choice_290",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "张大爷建了一个周长 20.56 米的半圆形羊圈，这个羊圈的面积是多少？",
+    "options": [],
+    "answer": "25.12 平方米",
+    "solution": "【分析】根据半圆的周长 $=$ 整圆的周长的一半 + 直径的长度, 可求出羊圈的半径, 然后再根据圆的面积公式 $S = \\pi r^{2}$ ,据此可求出羊圈的面积。本题考查圆的面积，明确半圆的周长的计算方法是解题的关键。\n\n【详解】$20.56 \\div (3.14 + 2)\\\\= 2 0. 5 6 \\div 5. 1 4 \\\\ = 4 (\\text {米}) $\n\n$3.1 4 \\times 4 ^ {2} \\div 2 \\\\ = 5 0. 2 4 \\div 2 \\\\ = 2 5. 1 2 \\text {(平 方 米)} $\n\n答: 这个羊圈的面积是 25.12 平方米。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级上·江西赣州·单元测试",
+    "merged_id": "unit5_290",
+    "original_id": "unit5_choice_290"
+  },
+  {
+    "id": "unit5_choice_291",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一个直径6米的圆形花坛，在它周围有一条宽1米的环形鹅卵石小路，小路的面积是多少平方米？",
+    "options": [],
+    "answer": "21.98 平方米",
+    "solution": "【分析】此题就是求大圆半径为 $6 \\div 2 + 1 = 4$ 米, 小圆半径为 $6 \\div 2 = 3$ 米的圆环的面积, 利用圆环的面积 $= \\pi (\\mathrm{R}^{2} - \\mathrm{r}^{2})$ ,即可解答。此题考查了圆环的面积公式的灵活应用, 这里关键是把实际问题转化成数学问题中, 并找到对应的数量关系。\n\n【详解】$3.14 \\times [ (6 \\div 2 + 1)^{2} - (6 \\div 2)^{2}]\\\\= 3. 1 4 \\times [ 1 6 - 9 ] \\\\ = 3. 1 4 \\times 7 \\\\ = 2 1. 9 8 \\text {(平 方 米)} $\n\n答: 小路的面积是 21.98 平方米。",
+    "knowledge_points": [
+      "圆环的面积"
+    ],
+    "source": "19-20六年级上·内蒙古巴彦淖尔·期中",
+    "merged_id": "unit5_291",
+    "original_id": "unit5_choice_291"
+  },
+  {
+    "id": "unit5_choice_292",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "在一个长30厘米、宽16厘米的长方形内画一个直径5厘米的圆，最多能画多少个？",
+    "options": [],
+    "answer": "18 个",
+    "solution": "【分析】根据题意知, 长方形的长为 30 厘米, 宽为 16 厘米,所以横向可以画 6 个圆, 纵向可以画 3 个圆, 相乘后即为总共圆的个数。此题考查了图形的拆拼，注意不能用面积解决，最好动手操作分析解答。\n\n【详解】$30 \\div 5 = 6$ （个） $1 6 \\div 5 \\approx 3 (\\text {个})$$6 \\times 3 = 1 8 (\\text {个})$ 答: 最多能画 18 个。",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "平面图形的分割"
+    ],
+    "source": "20-21六年级上·江西赣州·单元测试",
+    "merged_id": "unit5_292",
+    "original_id": "unit5_choice_292"
+  },
+  {
+    "id": "unit5_choice_293",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一个圆形水池的周长是 18.84 米，它的面积是多少平方米？",
+    "options": [],
+    "answer": "28.26 平方米",
+    "solution": "【分析】根据圆形的周长计算出圆形水池的半径，利用 $S = \\pi r^{2}$ 即可计算出水池的面积。灵活运用圆的周长和面积计算公式是解答本题的关键。\n\n【详解】半径： $18.84 \\div 3.14 \\div 2\\\\ = 6 \\div 2 \\\\ = 3 (\\text {米}) \\\\ $\n\n面积: $3.14 \\times 3^{2}\\\\ = 3. 1 4 \\times 9 \\\\ = 2 8. 2 6 \\text {(平 方 米)} $\n\n答: 水池的面积是 28.26 平方米。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "14-15 六年级上·全国·期末",
+    "merged_id": "unit5_293",
+    "original_id": "unit5_choice_293"
+  },
+  {
+    "id": "unit5_choice_294",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一座体育馆的围墙是圆形的，淘气沿着围墙走了一圈，一共是628步，淘气每步长0.6米。这座体育馆的占地面积是多少？",
+    "options": [],
+    "answer": "11304 平方米",
+    "solution": "【分析】因为体育馆的围墙是圆形的，要求这座体育馆的占地面积是多少，应求出圆形体育馆的半径。根据题意，体育馆的周长是 $0.6 \\times 628 = 376.8$ （米），然后求出半径，再运用圆的面积公式解答即可。此题考查了运用圆的周长和面积公式解答实际问题的能力。\n\n【详解】体育馆的周长：$0.6 \\times 6 2 8 = 3 7 6. 8 (\\text {米})$体育馆的占地面积:\n\n$3.1 4 \\times (3 7 6. 8 \\div 3. 1 4 \\div 2) ^ {2} \\\\ = 3. 1 4 \\times 6 0 ^ {2} = 3. 1 4 \\times 3 6 0 0 \\\\ = 1 1 3 0 4 \\text {(平 方 米)} $\n\n答: 这座体育馆的占地面积是 11304 平方米。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "2019·四川自贡·小升初模拟",
+    "merged_id": "unit5_294",
+    "original_id": "unit5_choice_294"
+  },
+  {
+    "id": "unit5_choice_295",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "大本钟——伦敦市的标志性建筑物，钟盘上时针的长度是 $2.75 \\mathrm{~m}$ 。\n\n(1) 大本钟的时针经过 8 小时, 时针的尖端走过的路程是多少米? (得数保留两位小数)\n\n(2) 如果这个时钟的时针尖端走过的路程是 $8.635 \\mathrm{~m}$ , 则时针走过了多少小时?",
+    "options": [],
+    "answer": "(1) $11.51 \\mathrm{~m}$ (2) 6小时",
+    "solution": "【分析】（1）因为时针走一圈是 12 小时, 当大本钟的时针经过 8 小时即是走了圆周长的 $\\frac{8}{12}$ , 求走过的路程即是求以 275 米为半径的圆周长的 $\\frac{8}{12}$ 即可;此题考查的是圆周长在实际生活中的应用, 关键是把实际问题转化成数学问题进行解答。\n\n(2) 如果这个时钟的时针尖端走过的路程是 $8.635 \\mathrm{~m}$ , 则时针走过了几小时, 用走过的路程除以周长, 结果即是时针走过的时间与 12 小时的分率, 据此解答。\n\n【详解】(1) $3.14 \\times 2.75 \\times 2 \\times \\frac{8}{12}\\\\= 8. 6 3 5 \\times 2 \\times \\frac {8}{1 2} \\\\ = 1 7. 2 7 \\times \\frac {8}{1 2} \\\\ \\approx 1 1. 5 1 \\mathrm {(m)} $\n\n答：时针的尖端走过的路程是 11.51 米。\n\n(2) $8.635 \\div (3.14 \\times 2.75 \\times 2) \\times 12 \\\\= 8. 6 3 5 \\div 1 7. 2 7 \\times 1 2 \\\\ = 0. 5 \\times 1 2 \\\\ = 6 \\text {(小 时)} $\n\n答: 时针走过了 6 小时。",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "21-22六年级上·全国·课后作业",
+    "merged_id": "unit5_295",
+    "original_id": "unit5_choice_295"
+  },
+  {
+    "id": "unit5_choice_296",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一辆自行车外轮外直径是 $60 \\mathrm{~cm}$ ，如果自行车车轮每分钟转 100 圈，骑这辆车要经过一座 $1884 \\mathrm{~m}$ 的大桥，需要多少分钟？",
+    "options": [],
+    "answer": "10 分钟",
+    "solution": "【分析】根据题意知道，先求出自行车车轮的周长，进而求出自行车每分钟行驶的速度，再根据路程÷速度 = 时间，即可求出答案。解答此题的关键是，知道自行车过桥时，是自行车的车轮的周长与大桥的长度有关系，理清思路，利用公式和数量关系，即可解答。\n\n【详解】$ 6 0 \\mathrm {c m} = 0. 6 \\mathrm {m} \\quad 1 8 8 4 \\div (3. 1 4 \\times 0. 6 \\times 1 0 0) \\\\ = 1 8 8 4 \\div 1 8 8. 4 \\\\=10 (分钟)$\n\n答：需要10分钟。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "基础行程问题"
+    ],
+    "source": "21-22六年级上·江西赣州·单元测试",
+    "merged_id": "unit5_296",
+    "original_id": "unit5_choice_296"
+  },
+  {
+    "id": "unit5_choice_297",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一种圆形脸盆是用直径为 0.6 米的圆形塑片冲压而成的。生产这种脸盆 10个，需塑片多少平方米？",
+    "options": [],
+    "answer": "2.826 平方米",
+    "solution": "【分析】先根据 $S = \\pi r^{2}$ 计算出一个圆形塑片的面积, 最后结果乘 10 即可。掌握圆的面积计算公式是解答本题的关键。\n\n【详解】$3.14 \\times (0.6 \\div 2)^{2} \\times 10\\\\= 3. 1 4 \\times 0. 3 ^ {2} \\times 1 0\\\\= 3. 1 4 \\times 0. 0 9 \\times 1 0 \\\\ = 0. 2 8 2 6 \\times 1 0 \\\\ = 2. 8 2 6 \\text {(平 方 米)} $\n\n答: 需塑片 2.826 平方米。",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "merged_id": "unit5_297",
+    "original_id": "unit5_choice_297"
+  },
+  {
+    "id": "unit5_choice_298",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一个圆形冰场的周长是94.2米，扩建后半径增加了5米，扩建后的冰场的面积是多少平方米？",
+    "options": [],
+    "answer": "1256 平方米",
+    "solution": "【分析】用周长 $\\div \\pi \\div 2$ , 求出原来的半径, 原来的半径 $+5$ 米,求出扩建后的半径, 再根据圆的面积 $= \\pi r^{2}$ , 列式解答即可。关键是掌握圆的周长和面积公式。\n\n【详解】$94.2 \\div 3.14 \\div 2 = 15$ （米）$3.1 4 \\times (1 5 + 5) ^ {2} \\\\ = 3. 1 4 \\times 2 0 ^ {2} \\\\= 1256（平方米）$\n\n答: 扩建后的冰场的面积是 1256 平方米。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "19-20六年级上·河北·单元测试",
+    "merged_id": "unit5_298",
+    "original_id": "unit5_choice_298"
+  },
+  {
+    "id": "unit5_choice_299",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一个环形零件,内圆半径是 $2 \\mathrm{~cm}$ , 外圆半径是 $3 \\mathrm{~cm}$ , 这个零件的环形面的面积是多少平方厘米?",
+    "options": [],
+    "answer": "15.7 平方厘米",
+    "solution": "【分析】$S_{\\text{环形}} = \\pi (R^{2} - r^{2})$ ，据此解答。掌握环形的面积计算公式是解答题目的关键。\n\n【详解】$\\pi \\times (3^{2} - 2^{2})\\\\= 5 \\pi\\\\= 15.7（平方厘米）$ \n\n答: 这个零件的环形面的面积是 15.7 平方厘米。",
+    "knowledge_points": [
+      "圆环的面积"
+    ],
+    "source": "18-19 六年级上·河南三门峡·期末",
+    "merged_id": "unit5_299",
+    "original_id": "unit5_choice_299"
+  },
+  {
+    "id": "unit5_choice_300",
+    "unit": "第五单元：圆",
+    "chapter": "圆",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "画一个直径是 4 厘米的圆，再在圆中画一个圆心角是 $90^{\\circ}​$ 的扇形，求出这个圆的周长和面积。",
+    "options": [],
+    "answer": "画图见详解，周长 12.56 厘米；面积 12.56 平方厘米",
+    "solution": "【分析】圆心确定圆的位置，半径确定圆的大小，由此以点O为圆心，以 $4 \\div 2 = 2$ 厘米为半径，即可画出这个圆，因为圆周角为 $360^{\\circ}$ ，所以用以圆的任意一条半径为扇形的边，再利用量角器画出圆心角为 $90^{\\circ}$ 的扇形即可。此题考查了圆的画法及扇形的面积公式，抓住圆的两大要素：圆心和半径即可画圆。\n\n【详解】$4 \\div 2 = 2$ （厘米）![](images/b129b1936fc2977844e0801a73623af2beb4e79e26bb2012e95b6240e9be7833.jpg)\n\n圆的周长: $2 \\times 2 \\times \\pi= 4 \\times \\pi= 12.56 (厘米)$\n\n圆的面积: $2^{2} \\times \\pi= 4 \\times \\pi= 12.56$ （平方厘米）\n\n答: 这个圆的周长为 12.56 厘米, 面积为 12.56 平方厘米。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用",
+      "画扇形"
+    ],
+    "source": "15-16 六年级上·湖北黄冈·期末",
+    "merged_id": "unit5_300",
+    "original_id": "unit5_choice_300"
+  },{
+    "id": "unit5_choice_301",
+    "unit": "第五单元：圆",
+    "chapter": "圆的概念及特点",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "(21-22 四年级上·上海·单元测试) 一个圆的直径增加 6 厘米, 它的半径增加 ( )。",
+    "options": [
+      "A. 2 厘米",
+      "B. 3 厘米",
+      "C. 6 厘米",
+      "D. 12 厘米"
+    ],
+    "answer": "B",
+    "solution": "【分析】直径的长度是半径的 2 倍, 因此用增加直径的长度除以 2 即可。熟练掌握直径、半径的关系是解答此题的关键。\n\n【详解】一个圆的直径增加 6 厘米, 它的半径增加:  $6 \\div 2 =$  3 (厘米)故答案为: B",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "21-22 四年级上·上海·单元测试",
+    "merged_id": "unit5_301",
+    "original_id": "unit5_choice_301"
+  },
+  {
+    "id": "unit5_choice_302",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "(18-19 六年级上·全国·单元测试) 一个钟表的分针长 10厘米，分针从下午2点走到下午5点，分针针尖走过了（）厘米。",
+    "options": [
+      "A. 31.4",
+      "B. 62.8",
+      "C. 188.4",
+      "D. 251.2"
+    ],
+    "answer": "C",
+    "solution": "【分析】钟面上，分针走过一周正好是半径为 10 厘米的圆，分针从下午 2 点走到下午 5 点，走了 3 周，根据 'C=2πr' 求出圆的周长，再进一步解答即可。明确分针走过一周正好是半径为 10 厘米的圆是解答本题的关键。\n\n【详解】  $2 \\times 3.14 \\times 10 \\times 3\\\\ = 6 2. 8 \\times 3 \\\\ = 1 8 8. 4 \\text {(厘 米)}; $\n\n故答案为: C。",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "18-19 六年级上·全国·单元测试",
+    "merged_id": "unit5_302",
+    "original_id": "unit5_choice_302"
+  },
+  {
+    "id": "unit5_choice_303",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "(21-22 六年级上·全国·单元测试) 在一个边长是 6 厘米的正方形里, 画一个最大的圆, 这个圆的面积是 ( )。",
+    "options": [
+      "A. 28.26 平方厘米",
+      "B. 18.84 平方厘米",
+      "C. 37.68 平方厘米",
+      "D. 50.24 平方厘米"
+    ],
+    "answer": "A",
+    "solution": "【分析】在正方形内画一个最大的圆, 圆的直径和正方形的边长相等, 据此可知圆的直径为 6 厘米, 再根据  $s = \\pi r^{2}$  求出圆的面积即可。明确在正方形内画一个最大的圆，圆的直径和正形的边长相等是解答本题的关键。\n\n【详解】  $3.14 \\times (6 \\div 2)^{2}= 3. 1 4 \\times 9= 28.26（平方厘米）$故答案为: A",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "21-22 六年级上·全国·单元测试",
+    "merged_id": "unit5_303",
+    "original_id": "unit5_choice_303"
+  },
+  {
+    "id": "unit5_choice_304",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用、圆的面积及应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "(21-22 六年级上·全国·单元测试) 一个圆的周长是31.4cm，那么它的面积是（）。",
+    "options": [
+      "A.  $314 \\mathrm{~cm}^{2}$",
+      "B.  $78.5 \\mathrm{~cm}^{2}$",
+      "C.  $15.7 \\mathrm{~cm}^{2}$",
+      "D.  $50.24 \\mathrm{~cm}^{2}$"
+    ],
+    "answer": "B",
+    "solution": "【分析】先利用圆的周长公式求出半径，再利用圆的面积公式计算即可。本题考查圆的周长、面积, 解答本题的关键是掌握圆的周长、面积计算公式。\n\n【详解】  $31.4 \\div 3.14 \\div 2 \\\\= 1 0 \\div 2 \\\\ = 5 (\\mathrm {c m}) $\n\n$ 3. 1 4 \\times 5 ^ {2} \\\\ = 3. 1 4 \\times 2 5 \\\\ = 7 8. 5 (\\mathrm {c m} ^ {3}) $\n\n所以这个圆的面积是  $78.5 \\mathrm{~cm}^{3}$  。故答案为: B",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "21-22 六年级上·全国·单元测试",
+    "merged_id": "unit5_304",
+    "original_id": "unit5_choice_304"
+  },
+  {
+    "id": "unit5_choice_305",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "（20-21六年级下·河南郑州·期末）如果一个大圆的半径正好等于一个小圆的直径，那么这个大圆的面积是小圆面积的（）倍。",
+    "options": [
+      "A. 2",
+      "B. 4",
+      "C. 3",
+      "D. 5"
+    ],
+    "answer": "B",
+    "solution": "【分析】直径 = 半径  $\\times 2$ , 大圆的半径正好等于小圆的直径,说明大圆半径是小圆半径的 2 倍, 大圆面积是小圆面积的倍数  $\\times$  倍数。关键是掌握圆的面积公式，圆的面积  $= \\pi r^{2}$ 。\n\n【详解】  $2 \\times 2 = 4$故答案为: B",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "20-21六年级下·河南郑州·期末",
+    "merged_id": "unit5_305",
+    "original_id": "unit5_choice_305"
+  },
+  {
+    "id": "unit5_choice_306",
+    "unit": "第五单元：圆",
+    "chapter": "画圆",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "(20-21 六年级下·全国·期末) 用圆规画一个直径为  $10 \\mathrm{~cm}$  的圆, 圆规两脚之间的距离应取 ( ) cm。",
+    "options": [
+      "A. 5",
+      "B. 10",
+      "C. 20"
+    ],
+    "answer": "A",
+    "solution": "【分析】圆规两脚之间的距离即为半径，用直径除以 2 即可求出半径，据此解答即可。明确圆规两脚之间的距离为半径的长度是解答本题的关键。故答案为: A\n\n【详解】  $10 \\div 2 = 5$  (cm)",
+    "knowledge_points": [
+      "画圆"
+    ],
+    "source": "20-21 六年级下·全国·期末",
+    "merged_id": "unit5_306",
+    "original_id": "unit5_choice_306"
+  },
+  {
+    "id": "unit5_choice_307",
+    "unit": "第五单元：圆",
+    "chapter": "弧、圆心角、扇形的认识、对称轴的画法及数量",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "(20-21 六年级下·云南玉溪·期末) 扇形有 ( ) 对称轴。",
+    "options": [
+      "A. 无数条",
+      "B. 一条",
+      "C. 没有"
+    ],
+    "answer": "B",
+    "solution": "【分析】一个图形沿一条直线对折，直线两旁的图形完全重合，这样的图形叫做轴对称图形，折痕所在的直线就是对称轴。明确轴对称图形的特点是解答本题的关键。\n\n【详解】扇形有 1 条对称轴, 如图:![](images/a35138ddaef65f2fd54ede0df13d99dd6015f20015ae7921c451690049ae638d.jpg)故答案为: B。",
+    "knowledge_points": [
+      "弧、圆心角、扇形的认识",
+      "对称轴的画法及数量"
+    ],
+    "source": "20-21 六年级下·云南玉溪·期末",
+    "merged_id": "unit5_307",
+    "original_id": "unit5_choice_307"
+  },
+  {
+    "id": "unit5_choice_308",
+    "unit": "第五单元：圆",
+    "chapter": "扇形的周长和面积",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "(2021·辽宁沈阳·小升初真题) 一个半径是  $r$  的半圆形, 它的周长是 ( )。",
+    "options": [
+      "A.  $2 \\pi \\mathrm{r} \\div 2$",
+      "B.  $r + \\pi r$",
+      "C.  $(2 + \\pi)$",
+      "D.  $\\pi r \\div 2$"
+    ],
+    "answer": "C",
+    "solution": "【分析】根据圆的周长公式:  $C = 2 \\pi r$ , 由此即可求出一半圆弧的长度, 由于它是半圆, 则半圆形的周长  $=$  一半圆的周长 + 直径, 即  $\\pi r + 2 r$ , 由此即可解答。本题主要考查圆的周长公式，要注意半圆的周长要加上直径。\n\n【详解】由分析可知, 半圆周长  $= \\pi r + 2r = (2 + \\pi) r$ , 故答案为: C。",
+    "knowledge_points": [
+      "扇形的周长和面积"
+    ],
+    "source": "2021·辽宁沈阳·小升初真题",
+    "merged_id": "unit5_308",
+    "original_id": "unit5_choice_308"
+  },
+  {
+    "id": "unit5_choice_309",
+    "unit": "第五单元：圆",
+    "chapter": "列方程解含一个未知数的问题、正方形的面积、圆的面积及应用、含圆的组合图形的计算（周长和面积）",
+    "question_type": "选择题",
+    "difficulty": "困难",
+    "question": "（19-20六年级上·全国·期末）从一张圆形纸上剪去一个最大的正方形，剩余部分的面积是6.84平方分米，那么这个圆的面积是（）平方分米。",
+    "options": [
+      "A.  $6 \\pi$",
+      "B.  $9 \\pi$",
+      "C. 9",
+      "D.  $8 \\pi$"
+    ],
+    "answer": "A",
+    "solution": "【分析】可设这个圆的半径是 R 分米，由于在一个圆内剪去一个最大的正方形，就会有一条直径将这个正方形平均分成两个相等的直角三角形，底是直径，高是半径，根据等量关系圆的面积减去正方形的面积等于 6.84 平方分米，结合圆的面积公式和正方形的面积公式列方程解方程即可。本题考查了圆和正方形的面积, 灵活运用圆和正方形的面积公式是解题的关键。\n\n【详解】设圆的半径为 R 分米，则依题意有：$3.14 \\times \\mathrm{R}^{2} - (\\mathrm{R} \\times 2 \\mathrm{R} \\div 2) \\times 2 = 6.84$\n\n$3.14 \\mathrm{R}^{2} - 2 \\mathrm{R}^{2} = 6.84$解得， $\\mathrm{R}^{2} = 6$所以，圆的面积是  $6 \\pi$  。故答案为: A",
+    "knowledge_points": [
+      "列方程解含一个未知数的问题",
+      "正方形的面积",
+      "圆的面积及应用",
+      "含圆的组合图形的计算（周长和面积）"
+    ],
+    "source": "19-20六年级上·全国·期末",
+    "merged_id": "unit5_309",
+    "original_id": "unit5_choice_309"
+  },
+  {
+    "id": "unit5_choice_310",
+    "unit": "第五单元：圆",
+    "chapter": "轴对称的认识及辨认、与圆相关的轴对称图形",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "（2019·北京海淀·小升初真题）用两个大小不同的圆组成图形，对称轴最多可以有（）条。",
+    "options": [
+      "A. 无数",
+      "B. 2",
+      "C. 1"
+    ],
+    "answer": "A",
+    "solution": "【分析】由题意可知, 要想对称轴的条数最多, 那两个大小不同的圆组成同心圆的时候, 对称轴最多, 据此可解答。本题考查圆的相关知识, 因为一个圆的对称轴有无数条, 所以注意同心圆的对称轴是多的。\n\n【详解】由分析可知，两个大小不同的圆组成同心圆的时候，对称轴最多，有无数条。故答案为：A",
+    "knowledge_points": [
+      "轴对称的认识及辨认",
+      "与圆相关的轴对称图形"
+    ],
+    "source": "2019·北京海淀·小升初真题",
+    "merged_id": "unit5_310",
+    "original_id": "unit5_choice_310"
+  },
+  {
+    "id": "unit5_choice_311",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "（2021 五年级下·江苏·专题练习）大圆的半径是小圆的直径，则小圆的周长是大圆周长的（）。",
+    "options": [
+      "A.  $\\frac{1}{2}$",
+      "B. 4 倍",
+      "C. 2 倍",
+      "D.  $\\frac{1}{4}$"
+    ],
+    "answer": "A",
+    "solution": "【分析】根据题意可知，大圆的半径是小圆的直径，设：小圆直径为 d, 则大圆直径为 2d, 根据圆的周长公式, 求出小圆和大圆的周长, 在进行比较, 即可解答。本题考查圆的周长公式的运用, 关键是明确大圆的直径是小圆的直径的 2 倍。\n\n【详解】设：小圆直径为  $d$ ，则大圆直径为 $2d$\n\n$ \\pi \\mathrm {d} \\div (\\pi \\times 2 \\mathrm {d}) \\\\ = 1 \\div 2 \\\\ = \\frac {1}{2} $\n\n故答案选：A",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "2021 五年级下·江苏·专题练习",
+    "merged_id": "unit5_311",
+    "original_id": "unit5_choice_311"
+  },
+  {
+    "id": "unit5_choice_312",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "(2018 六年级下·天津河北·学业考试) 一个圆的直径是$2 \\mathrm{~cm}$ , 这个圆的面积是 ( )。",
+    "options": [
+      "A.  $12.56 \\mathrm{~cm}^{2}$",
+      "B.  $6.28 \\mathrm{~cm}^{2}$",
+      "C.  $3.14 \\mathrm{~cm}^{2}$",
+      "D.  $3.14 \\mathrm{~cm}$"
+    ],
+    "answer": "C",
+    "solution": "【分析】通过直径先求出半径，根据圆的面积  $= \\pi r^{2}$ ，列式计算即可。关键是掌握和熟练运用圆的面积公式。\n\n【详解】  $3.14 \\times (2 \\div 2)^{2}= 3. 1 4 \\times 1= 3.14（平方厘米）$  故答案为: C",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "2018 六年级下·天津河北·学业考试",
+    "merged_id": "unit5_312",
+    "original_id": "unit5_choice_312"
+  },
+  {
+    "id": "unit5_choice_313",
+    "unit": "第五单元：圆",
+    "chapter": "多位小数的大小比较、圆的概念及特点",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "(15-16 六年级上·山东青岛·单元测试) 圆周率  $\\pi$  的值 ( )。",
+    "options": [
+      "A. 等于 3.14",
+      "B. 大于 3.14",
+      "C. 小于 3.14"
+    ],
+    "answer": "B",
+    "solution": "【分析】根据圆周率的值与 3.14 进行比较即可解答。此题主要考查学生对  $\\pi$  的值的掌握以及小数大小比较。\n\n【详解】圆周率  $\\pi$  的值是 3.1415926……, 属于无限小数,大于 3.14。故答案为: B。",
+    "knowledge_points": [
+      "多位小数的大小比较",
+      "圆的概念及特点"
+    ],
+    "source": "15-16 六年级上·山东青岛·单元测试",
+    "merged_id": "unit5_313",
+    "original_id": "unit5_choice_313"
+  },
+  {
+    "id": "unit5_choice_314",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用、植树问题",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "(18-19 六年级下·云南曲靖·期末) 在直径是 10 米的圆形喷水池边上, 每隔 3.14 米放一盆花, 一共可以放 ( ) 盒花。",
+    "options": [
+      "A. 9",
+      "B. 10",
+      "C. 11",
+      "D. 12"
+    ],
+    "answer": "B",
+    "solution": "【分析】先根据'圆周长公式： $C = \\pi d$ '求出圆的周长；再用'圆的周长÷间隔距离=盆数'算出一共可以放多少盆花。此题考查圆的周长与植树问题, 封闭图形中棵树 =间隔数。\n\n【详解】  $3.14 \\times 10 \\div 3.14 = 10$  （盆）故答案为：B。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "植树问题"
+    ],
+    "source": "18-19 六年级下·云南曲靖·期末",
+    "merged_id": "unit5_314",
+    "original_id": "unit5_choice_314"
+  },
+  {
+    "id": "unit5_choice_315",
+    "unit": "第五单元：圆",
+    "chapter": "比的应用、圆的面积及应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "（19-20 六年级下·河南信阳·期末）两个圆的直径分别是 3 厘米和 4 厘米，则大圆与小圆的面积的比是（）。",
+    "options": [
+      "A. 9:16",
+      "B. 4:3",
+      "C.  $3: 4$",
+      "D. 16:9"
+    ],
+    "answer": "D",
+    "solution": "【分析】根据半径  $= \\text{直径} \\div 2$ , 面积  $= \\pi r^{2}$ , 分别求出大小圆的半径和面积, 然后再求出比即可。该类型的题目，计算时应先根据公式分别求出，然后根据题意进行比，最后要化成最简整数比。\n\n【详解】  $4 \\div 2 = 2$  （厘米）  $3 \\div 2 = 1.5$  (厘米)   $(3.14×2^{2}) : (3.14×1.5^{2}) = 4: 2. 2 5 = 1 6: 9 $故答案为: D",
+    "knowledge_points": [
+      "比的应用",
+      "圆的面积及应用"
+    ],
+    "source": "19-20 六年级下·河南信阳·期末",
+    "merged_id": "unit5_315",
+    "original_id": "unit5_choice_315"
+  },
+  {
+    "id": "unit5_choice_316",
+    "unit": "第五单元：圆",
+    "chapter": "圆的概念及特点",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "(19-20 五年级下·天津滨海新·期中) 在一个长  $6dm$ ,宽  $4dm$  的长方形内画一个最大的圆, 圆的半径是 ( )  $dm$ 。",
+    "options": [
+      "A. 6",
+      "B. 4",
+      "C. 3",
+      "D. 2"
+    ],
+    "answer": "D",
+    "solution": "【分析】在一个长方形中画一个最大的圆，圆的直径等于长方形中较短的边，据此解答。此题考查了圆的特征，找出圆与长方形之间的关系是解题关键。\n\n【详解】由分析可知，圆的直径等于长方形的宽，则半径是  $4 \\div 2 = 2$  (分米)故选择：D",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "19-20 五年级下·天津滨海新·期中",
+    "merged_id": "unit5_316",
+    "original_id": "unit5_choice_316"
+  },
+  {
+    "id": "unit5_choice_317",
+    "unit": "第五单元：圆",
+    "chapter": "轴对称的认识及辨认",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "(19-20 三年级下·山东·单元测试) 下面图形中不一定是轴对称图形的是 ( )。",
+    "options": [
+      "A. 长方形",
+      "B. 圆形",
+      "C. 平行四边形"
+    ],
+    "answer": "C",
+    "solution": "【分析】根据对称轴性质逐项进行判断: 如果一个图形沿着某条直线对折, 对折的两部分是完全重的, 我们称这样的图形为轴对称图形, 这条直线叫做这个图形的对称。常见图形的对称轴数量：圆有无数条对称轴，直径所在的直线就是圆的对称轴，正方形有 4 条对称轴，长方形有 2 条对称轴，等腰梯形有 1 条对称轴，等边三角形有 3 条对称轴，等腰三角形有 1 条对称轴，平行四边形没有对称轴。\n\n【详解】A. 长方形存在对称轴使该图形沿着对称轴对折, 对折的两部分是完全重合, 故长方形是轴对称图形。B. 圆形存在对称轴使该图形沿着对称轴对折, 对折的两部分是完全重合。  C. 平行四边形不存在对称轴使该图形沿着对称轴对折, 对折的两部分是完全重合。故答案为: C",
+    "knowledge_points": [
+      "轴对称的认识及辨认"
+    ],
+    "source": "19-20 三年级下·山东·单元测试",
+    "merged_id": "unit5_317",
+    "original_id": "unit5_choice_317"
+  },
+  {
+    "id": "unit5_choice_318",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用、正方形的周长",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "（20-21 六年级上·吉林·期末）一根铁丝，围成一个正方形时，边长为  $3.14 \\mathrm{dm}$ ，若把它围成一个圆，则圆的周长是（）dm。",
+    "options": [
+      "A. 3.14",
+      "B. 9.8596",
+      "C. 12.56"
+    ],
+    "answer": "C",
+    "solution": "【分析】因为是同一根铁丝，所以围成的圆的周长就是正方形的周长，根据正方形周长  $= \\text{边长} \\times 4$  ，计算即可。关键是理解正方形和圆的周长都是这根铁丝的长度, 圆的周长  $= \\pi d$  。\n\n【详解】  $3.14 \\times 4 = 12.56$  （分米）故答案为: C",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "正方形的周长"
+    ],
+    "source": "20-21 六年级上·吉林·期末",
+    "merged_id": "unit5_318",
+    "original_id": "unit5_choice_318"
+  },
+  {
+    "id": "unit5_choice_319",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "（19-20六年级上·河北·单元测试）把一个周长是18.84分米的圆平均分成两个半圆，每个半圆的周长是（）分米。",
+    "options": [
+      "A. 15.42",
+      "B. 9.42",
+      "C. 12.24",
+      "D. 12.56"
+    ],
+    "answer": "A",
+    "solution": "【分析】半圆的周长 = 圆周长的一半 + 直径，据此解答即可。明确半圆的周长公式是解答本题的关键。\n\n【详解】  $18.84 \\div 2 + 18.84 \\div 3.14\\\\= 9. 4 2 + 6\\\\=15.42 (分米)$ ;\n\n故答案为: A。",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "19-20六年级上·河北·单元测试",
+    "merged_id": "unit5_319",
+    "original_id": "unit5_choice_319"
+  },
+  {
+    "id": "unit5_choice_320",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "(18-19 六年级上·江苏·开学考试) 圆的半径有  $3 \\mathrm{~cm}$  增加到  $4 \\mathrm{~cm}$ , 圆的面积增加 ( )  $\\mathrm{cm}^{2}$  。",
+    "options": [
+      "A. 3.14",
+      "B. 6.28",
+      "C. 21.98"
+    ],
+    "answer": "C",
+    "solution": "【分析】根据圆的面积公式:  $S = \\pi r^{2}$ , 可以求出圆的面积,两次求出的面积差就是要求的答案。解答此题的关键是牢记圆的面积公式,  $S = \\pi r^{2}$ , 尤其是半径的平方, 计算时要注意  $r^{2}$  是两个半径相乘。\n\n【详解】  $3.14 \\times 4^{2} - 3.14 \\times 3^{2}\\\\= 3. 1 4 \\times (1 6 - 9)\\\\ = 3. 1 4 \\times 7 \\\\ = 2 1. 9 8 \\text {(平 方 厘 米)} $\n\n故答案为: C。",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "18-19 六年级上·江苏·开学考试",
+    "merged_id": "unit5_320",
+    "original_id": "unit5_choice_320"
+  },
+  {
+    "id": "unit5_choice_321",
+    "unit": "第五单元：圆",
+    "chapter": "比的应用、圆的面积及应用、分数与整数的除法",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·河南驻马店·期末) 甲圆半径是乙圆半径的  $\\frac{2}{3}$ , 甲圆面积是  $120 \\mathrm{~cm}^{2}$ , 乙圆面积是 ( )。",
+    "options": [
+      "A.  $180 \\mathrm{~cm}^{2}$",
+      "B.  $270 \\mathrm{~cm}^{2}$",
+      "C.  $80 \\mathrm{~cm}^{2}$"
+    ],
+    "answer": "B",
+    "solution": "【分析】已知甲圆半径是乙圆半径的  $\\frac{2}{3}$ , 结合两圆面积等于两圆半径平方之比可得, 甲圆面积是乙圆面积的  $\\frac{4}{9}$ , 又因为甲圆面积是  $120 \\mathrm{~cm}^{2}$ , 所以乙圆面积是  $120 \\div \\frac{4}{9}$  。明确两圆面积之比等于两圆半径平方之比是解题关键，在此基础上运用分数除法运算解决问题即可。\n\n【详解】因为甲圆半径是乙圆半径的  $\\frac{2}{3}$ , 所以甲圆面积是乙圆面积的  $\\frac{4}{9}$ ,$1 2 0 \\div \\frac {4}{9} = 2 7 0 (\\mathrm {c m} ^ {2})$故答案为: B。",
+    "knowledge_points": [
+      "比的应用",
+      "圆的面积及应用",
+      "分数与整数的除法"
+    ],
+    "source": "20-21 六年级上·河南驻马店·期末",
+    "merged_id": "unit5_321",
+    "original_id": "unit5_choice_321"
+  },
+  {
+    "id": "unit5_choice_322",
+    "unit": "第五单元：圆",
+    "chapter": "比的应用、圆的面积及应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "（20-21 六年级上·浙江温州·期末）在一个边长为 4 厘米的正方形纸片内剪一个最大的圆，圆面积和正方形面积的比是（）。",
+    "options": [
+      "A.  $\\pi : 2$",
+      "B.  $2: \\pi$",
+      "C.  $4: \\pi$",
+      "D.  $\\pi : 4$"
+    ],
+    "answer": "D",
+    "solution": "【分析】因为是在这个边长为 4 厘米的正方形纸片内剪一个最大的圆, 所以正方形的边长相当于圆的直径, 则圆的面积、正方形的面积都可求，最后再求它们的比。首先确定与圆的面积、正方形的面积相关的要素的具体数值，然后依据公式计算面积，最后求得面积之比。\n\n【详解】由分析得：$S _ {\\text {圆}} = \\pi \\times (4 \\div 2) ^ {2} = 4 \\pi$$S _ {\\text {圆}}: S _ {\\text {正 方 形}} = 4 \\pi : 4 \\times 4 = \\pi : 4$故答案为: D",
+    "knowledge_points": [
+      "比的应用",
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级上·浙江温州·期末",
+    "merged_id": "unit5_322",
+    "original_id": "unit5_choice_322"
+  },
+  {
+    "id": "unit5_choice_323",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用、圆的面积及应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "(19-20 六年级上·山西阳泉·期末) 一个半圆的周长是  $30.84 \\mathrm{~m}$ , 这个半圆的面积是 ( )  $\\mathrm{m}^{2}$  。",
+    "options": [
+      "A. 56.52",
+      "B. 68.52",
+      "C. 113.04"
+    ],
+    "answer": "A",
+    "solution": "【分析】假设半圆的半径为  $x$ , 根据 '圆周长  $\\times \\frac{1}{2}+$  直径 = 半圆周长' 列方程解答, 求出半径的长度, 再根据 ' $s=\\pi r^{2}$ '求出圆的面积，最后除以2即可求出半圆的面积。解答本题时一定明确半圆周长等于圆周长一半加上一条直径的长度, 由此列方程求出半圆的半径, 进而求出面积即可。\n\n【详解】解：设半圆的半径为  $\\mathrm{xm}$ ；\n\n$ 2 \\times 3. 1 4 x \\times \\frac {1}{2} + 2 x = 3 0. 8 4 \\\\ 5. 1 4 x = 3 0. 8 4 \\\\ x = 6 \\\\ 3. 1 4 \\times 6 ^ {2} \\div 2 \\\\ = 1 1 3. 0 4 \\div 2 \\\\ = 5 6. 5 2 (\\mathrm {m} ^ {2}) $\n\n故答案为: A",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "19-20 六年级上·山西阳泉·期末",
+    "merged_id": "unit5_323",
+    "original_id": "unit5_choice_323"
+  },
+  {
+    "id": "unit5_choice_324",
+    "unit": "第五单元：圆",
+    "chapter": "扇形的周长和面积、求比值、比的意义",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "（19-20六年级上·全国·课后作业）一个扇形的圆心角扩大为原来的2倍，半径长缩小为原来的一半，那么变化后所得扇形面积与原来的扇形面积的比值为（）。",
+    "options": [
+      "A.  $\\frac{1}{2}$",
+      "B.  $\\frac{1}{4}$",
+      "C. 2",
+      "D. 1"
+    ],
+    "answer": "A",
+    "solution": "【分析】根据扇形面积  $= \\pi r^{2} \\times \\frac{\\text { 圆心角 }}{360^{\\circ}}$ , 表示出前后扇形面积,根据比的意义写出比, 求比值即可。关键是掌握扇形面积公式，理解比的意义，求比值直接用前项÷后项。\n\n【详解】  $[\\pi (\\frac{1}{2}\\mathrm{r})^{2} \\times \\frac{\\text{圆心角} \\times 2}{360^{\\circ}}]: (\\pi \\mathrm{r}^{2} \\times \\frac{\\text{圆心角}}{360^{\\circ}})\\\\= [ \\frac {1}{4} \\pi r ^ {2} \\times \\frac {\\text {圆 心 角} \\times 2}{3 6 0 ^ {\\circ}} ]: (\\pi r ^ {2} \\times \\frac {\\text {圆 心 角}}{3 6 0 ^ {\\circ}}) \\\\ = [ \\frac {1}{2} \\pi r ^ {2} \\times \\frac {\\text {圆 心 角}}{3 6 0 ^ {\\circ}} ]: (\\pi r ^ {2} \\times \\frac {\\text {圆 心 角}}{3 6 0 ^ {\\circ}}) \\\\ = \\frac {1}{2}: 1 \\\\ = \\frac {1}{2} $\n\n故答案为：A",
+    "knowledge_points": [
+      "扇形的周长和面积",
+      "求比值",
+      "比的意义"
+    ],
+    "source": "19-20六年级上·全国·课后作业",
+    "merged_id": "unit5_324",
+    "original_id": "unit5_choice_324"
+  },
+  {
+    "id": "unit5_choice_325",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "（19-20六年级上·江西宜春·期末）把一个圆平均分成64份，然后剪开，拼成一个近似的长方形，这个转化过程中（）。",
+    "options": [
+      "A. 周长、面积都没变",
+      "B. 周长没变，面不变积变了",
+      "C. 周长变了, 面积没变"
+    ],
+    "answer": "C",
+    "solution": "【分析】把一个圆平均分成 64 份, 然后剪开, 拼成一个近似的长方形, 长方形和圆的面积相等, 周长多了两条半径,由此解答即可。熟练掌握圆的面积推导过程是解答本题的关键。\n\n【详解】把一个圆平均分成 64 份, 然后剪开, 拼成一个近似的长方形, 这个转化过程中周长变了, 面积没变;故答案为: C",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "19-20六年级上·江西宜春·期末",
+    "merged_id": "unit5_325",
+    "original_id": "unit5_choice_325"
+  },
+  {
+    "id": "unit5_choice_326",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "(19-20 六年级上·江西宜春·期末) 钟表的分针长 12 厘米, 它从 5 时走到 6 时, 分针的针尖走了 ( ) 厘米。",
+    "options": [
+      "A.  $12 \\pi$",
+      "B.  $24 \\pi$",
+      "C.  $144 \\pi$",
+      "D.  $6 \\pi$"
+    ],
+    "answer": "B",
+    "solution": "【分析】根据分析可知，分针从 5 时走到 6 时，正好走了一个小时，分针走了一圈，针尖走的距离就是一个圆的周长，分针的长度就是圆的半径，根据圆的周长公式即可解答。此题主要考查是圆的周长公式的应用。\n\n【详解】  $C = 2 \\pi r = 2 4 \\pi$  （厘米）故答案为: B",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "19-20 六年级上·江西宜春·期末",
+    "merged_id": "unit5_326",
+    "original_id": "unit5_choice_326"
+  },
+  {
+    "id": "unit5_choice_327",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "(19-20 六年级上·江西上饶·期末) 一个圆的半径增加  $2 \\mathrm{~cm}$ , 这个圆的周长比原来增加 ( ) cm。",
+    "options": [
+      "A.  $4 \\pi$",
+      "B.  $2 \\pi$",
+      "C.  $\\pi$"
+    ],
+    "answer": "A",
+    "solution": "【分析】根据圆的周长公式，将增加的圆的周长计算出来即可。本题考查了圆的周长, 灵活运用圆的周长公式是解题的关键。\n\n【详解】  $2 \\times \\pi \\times 2 = 4 \\pi$  （厘米），所以这个圆的周长比原来增加  $4 \\pi$  厘米。故答案为: A",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "19-20 六年级上·江西上饶·期末",
+    "merged_id": "unit5_327",
+    "original_id": "unit5_choice_327"
+  },
+  {
+    "id": "unit5_choice_328",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "(21-22 六年级上·全国·单元测试) 把一个圆沿半径平均分成若干份后可以拼成一个近似长方形, 这个长方形的长等于( )，宽等于( )，从而得到圆的面积计算公式是( )。",
+    "options": [],
+    "answer": "圆周长的一半 圆的半径  $s = \\pi r^{2}$",
+    "solution": "【详解】把一个圆沿半径平均分成若干份后可以拼成一个近似长方形，这个长方形的长等于圆周长的一半，宽等于圆的半径，因为长方形的面积  $= \\text{长} \\times$  宽，所以圆的面积  $= \\text{圆}$  周长的一半  $\\times$  半径，即  $s = \\pi r \\times r = \\pi r^2$ 。",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "21-22 六年级上·全国·单元测试",
+    "merged_id": "unit5_328",
+    "original_id": "unit5_choice_328"
+  },
+  {
+    "id": "unit5_choice_329",
+    "unit": "第五单元：圆",
+    "chapter": "圆的概念及特点、圆的周长及应用、比的意义、圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(21-22 六年级上·全国·单元测试) A 圆和 B 圆的半径比是  $5: 3$ , 它们的直径比是 ( ) : ( ), 周长比是 ( ) : ( ), 面积比是 ( ) : ( )。",
+    "options": [],
+    "answer": "5 3 5 3 25 9",
+    "solution": "【分析】半径比和周长比、直径比相等；面积比是半径比的平方，据此解答即可。熟记规律能够提高解答的速度, 也可以分别计算出两圆的直径、周长和面积, 再写出比。\n\n【详解】A 圆和 B 圆的直径比是  $5: 3$ ;A 圆和 B 圆的周长比是  $5: 3$ ,A 圆和 B 圆的面积比是  $25: 9$  。",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "圆的周长及应用",
+      "比的意义",
+      "圆的面积及应用"
+    ],
+    "source": "21-22 六年级上·全国·单元测试",
+    "merged_id": "unit5_329",
+    "original_id": "unit5_choice_329"
+  },
+  {
+    "id": "unit5_choice_330",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用、圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(21-22 六年级上·全国·单元测试) 圆的周长 9.42 分米,它的直径是( )分米, 面积是( )平方分米。",
+    "options": [],
+    "answer": "3 7.065",
+    "solution": "【分析】圆的周长  $= \\pi \\times$  直径, 圆的面积  $= \\pi \\times$  半径  $2$ , 据此解答。本题考查圆的周长和面积公式，计算圆的面积时要切记先求半径。\n\n【详解】直径:  $9.42 \\div 3.14 = 3$  (分米) ;面积:  $3.14 \\times (3 \\div 2)^{2} = 7.065$  (平方分米)",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "21-22 六年级上·全国·单元测试",
+    "merged_id": "unit5_330",
+    "original_id": "unit5_choice_330"
+  },
+  {
+    "id": "unit5_choice_331",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(21-22六年级上·全国·单元测试) 一个圆的直径是6分米，它的面积是( )平方分米。",
+    "options": [],
+    "answer": "28.26",
+    "solution": "【分析】根据公式  $d = 2 r$  和圆的面积公式  $S = \\pi r^{2}$  进行解答即可。本题考查圆的面积, 解答本题的关键是掌握圆的面积计算公式。\n\n【详解】  $3.14 \\times (6 \\div 2)^{2}\\\\= 3. 1 4 \\times 9\\\\= 28.26（平方分米）$\n\n所以这个圆的面积是 28.26 平方分米。",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "21-22六年级上·全国·单元测试",
+    "merged_id": "unit5_331",
+    "original_id": "unit5_choice_331"
+  },
+  {
+    "id": "unit5_choice_332",
+    "unit": "第五单元：圆",
+    "chapter": "圆的概念及特点",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(21-22 六年级上·全国·单元测试) 在一个圆里有( )条直径，直径的长度是半径的( )倍。",
+    "options": [],
+    "answer": "无数 2",
+    "solution": "【分析】据圆的认识及在同一个圆中半径与直径的关系即可作答。本题考查圆的认识, 解答本题的关键是掌握圆的特征。\n\n【详解】在一个圆里有无数条直径, 直径的长度是半径的 2 倍。",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "21-22 六年级上·全国·单元测试",
+    "merged_id": "unit5_332",
+    "original_id": "unit5_choice_332"
+  },
+  {
+    "id": "unit5_choice_333",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用、圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(18-19 六年级上·全国·单元测试) 一个圆形水池, 周长是 25.12 米, 它的面积是( )平方米。",
+    "options": [],
+    "answer": "50.24",
+    "solution": "【分析】先根据圆的周长公式，求圆的半径，再根据圆的面积公式  $s = \\pi r^2$  ，列式解答。此题考查有关圆的应用题, 解决此题关键是理解要求圆形水池的面积, 也就是求水池底面圆的面积, 进而运用公式:  $S = \\pi r^{2}$ , 列式计算即可。\n\n【详解】  $3.14 \\times (25.12 \\div 3.14 \\div 2)^{2}\\\\= 3. 1 4 \\times 1 6 \\\\ = 5 0. 2 4 \\text {(平 方 米)} $",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "18-19 六年级上·全国·单元测试",
+    "merged_id": "unit5_333",
+    "original_id": "unit5_choice_333"
+  },{
+    "id": "unit5_choice_334",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "（2023·四川成都·小升初真题）在一个长10厘米，宽8厘米的长方形内画一个最大的圆。这个圆的面积是( )平方厘米。",
+    "options": [],
+    "answer": "50.24",
+    "solution": "【分析】在长方形内画一个最大的圆，圆的直径等于长方形的宽，根据圆的面积计算即可。关键是确定圆和长方形的关系，圆的面积  $= \\pi r^2$ 。\n\n【详解】  $3.14 \\times (8 \\div 2)^{2}\\\\= 3. 1 4 \\times 1 6\\\\= 50.24（平方厘米）$",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "2023·四川成都·小升初真题",
+    "merged_id": "unit5_334",
+    "original_id": "unit5_choice_334"
+  },
+  {
+    "id": "unit5_choice_335",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的面积",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(19-20 六年级下·湖南郴州·期末) 光盘的银色部分是一个圆环, 外圆直径是  $10 \\mathrm{~cm}$ , 内圆直径是  $6 \\mathrm{~cm}$ , 光盘的面积是( )  $\\mathrm{cm}^{2}$  。",
+    "options": [],
+    "answer": "50.24",
+    "solution": "【分析】要想求出圆环的面积，应该用外圆的面积减去内圆的面积即根据圆环的面积  $= \\pi (\\mathrm{R}^{2} - \\mathrm{r}^{2})$  即可。本题考查圆环的面积公式，掌握圆环的计算方法是解题的关键。\n\n【详解】  $10 \\div 2 = 5$  （厘米）\n\n$ 6 \\div 2 = 3 (\\text {厘 米}) \\\\ 3. 1 4 \\times (5 ^ {2} - 3 ^ {2}) \\\\ = 3. 1 4 \\times 1 6 \\\\ = 5 0. 2 4 \\text {(平 方 厘 米)} $",
+    "knowledge_points": [
+      "圆环的面积"
+    ],
+    "source": "19-20 六年级下·湖南郴州·期末",
+    "merged_id": "unit5_335",
+    "original_id": "unit5_choice_335"
+  },
+  {
+    "id": "unit5_choice_336",
+    "unit": "第五单元：圆",
+    "chapter": "扇形的周长和面积",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "（2021·甘肃武威·小升初模拟）一个钟表的分针长 16 厘米, 15 分钟针尖能走 ( ) 厘米, 分针扫过的面积是 ( ) 平方厘米。",
+    "options": [],
+    "answer": "25.12 ;200.96",
+    "solution": "【分析】已知一个钟表的分针长 16 厘米, 求其 15 分钟时针尖一共走过的距离以及分针扫过的面积; 因为分针相当于半径, 15 分钟又是 1 小时的  $\\frac{1}{4}$ , 即此时分针针尖走过的距离相当于圆心角是整个圆的  $\\frac{1}{4}$  的扇形的弧长, 分针扫过的面积相当于这个扇形的面积; 所以可先求出以分针为半径的圆的周长及面积, 再乘  $\\frac{1}{4}$  即可。注意这里一定要分清楚时针与分针的区别：15分钟，分针走过的路程是整个圆的周长的  $\\frac{1}{4}$ ，扫过的面积也是整个圆的面积的  $\\frac{1}{4}$ ；而对于时针，15分钟只走 1 个大格的  $\\frac{1}{4}$ ，距离和面积都与分针的有很大差异。\n\n【详解】  $3.14 \\times 16 \\times 2 \\times (15 \\div 60)\\\\= 3. 1 4 \\times 3 2 \\times \\frac {1}{4}\\\\= 25.12（厘米）$\n\n$3.1 4 \\times 1 6 ^ {2} \\times (1 5 \\div 6 0) \\\\ = 3. 1 4 \\times 1 6 ^ {2} \\times \\frac {1}{4} \\\\ = 2 0 0. 9 6 \\text {(平 方 厘 米)} $",
+    "knowledge_points": [
+      "扇形的周长和面积"
+    ],
+    "source": "2021·甘肃武威·小升初模拟",
+    "merged_id": "unit5_336",
+    "original_id": "unit5_choice_336"
+  },
+  {
+    "id": "unit5_choice_337",
+    "unit": "第五单元：圆",
+    "chapter": "画圆、圆的周长及应用、圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(2021·江西九江·小升初真题) 用圆规画一个周长为  50.24 厘米的圆，圆规两脚间的距离是( )厘米，这个圆的面积是( )平方厘米。",
+    "options": [],
+    "answer": "8 ;200.96",
+    "solution": "【分析】圆规两脚间的距离是指这个圆的半径，由此利用圆的半径  $=$  圆的周长  $\\div 3.14 \\div 2$  和圆的面积  $S = \\pi r^{2}$  即可解答。此题考查了圆的周长和面积公式的灵活应用。\n\n【详解】  $50.24 \\div 3.14 \\div 2 = 8$  （厘米） $3.14 \\times 8^{2} = 200.96$  （平方厘米）",
+    "knowledge_points": [
+      "画圆",
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "2021·江西九江·小升初真题",
+    "merged_id": "unit5_337",
+    "original_id": "unit5_choice_337"
+  },
+  {
+    "id": "unit5_choice_338",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用、圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(19-20 六年级下·云南曲靖·期末) 画一个直径为  $2 \\mathrm{~cm}$  的半圆, 这个半圆的周长是 ( ) cm, 面积是 ( ) cm²。",
+    "options": [],
+    "answer": "5.14; 1.57",
+    "solution": "【分析】求半圆的周长 = 整圆周长的一半加上一条直径,根据圆面积的公式  $S = \\pi r^{2}$ , 求半圆的面积用整圆的面积除以 2 即可。本题考查圆的周长和面积公式，注意半圆的周长应该等于整圆周长的一半加上直径是解题的关键。\n\n【详解】  $3.14 \\times 2 \\div 2 + 2= 3. 1 4 + 2=5.14 (厘米)$\n\n$3.1 4 \\times (2 \\div 2) ^ {2} \\div 2\\\\ = 3. 1 4 \\div 2 \\\\ = 1. 5 7 \\text {(平 方 厘 米)} $",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "19-20 六年级下·云南曲靖·期末",
+    "merged_id": "unit5_338",
+    "original_id": "unit5_choice_338"
+  },
+  {
+    "id": "unit5_choice_339",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(19-20 五年级下·河南洛阳·期末) 一根绳子长 31.4 厘米, 用它围成一个圆, 圆的直径是( )厘米。",
+    "options": [],
+    "answer": "10",
+    "solution": "【分析】由题意知: 围成的圆的周长就是这根绳子的长度,用圆的周长除以 3.14 , 即得圆的直径。据此解答。灵活运用圆的周长公式求直径是解答本题的关键。\n\n【详解】  $31.4 \\div 3.14 = 10$  （米）",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "19-20 五年级下·河南洛阳·期末",
+    "merged_id": "unit5_339",
+    "original_id": "unit5_choice_339"
+  },
+  {
+    "id": "unit5_choice_340",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "（2021 六年级下·江苏·专题练习）一个圆的周长是 15.7 厘米，将这个圆切成两个半圆，每个半圆的周长是( )厘米。",
+    "options": [],
+    "answer": "12.85",
+    "solution": "【分析】半圆的周长等于整圆周长的一半与直径之和。圆的周长  $= \\pi d$ , 用周长除以  $\\pi$  求出直径后, 再计算半圆的周长。计算半圆的周长，要用整圆周长的一半加上直径。\n\n【详解】  $15.7 \\div 3.14 = 5（厘米）$\n\n$ 1 5. 7 \\div 2 + 5 \\\\ = 7. 8 5 + 5 \\\\ = 1 2. 8 5 \\text {(厘 米)} $",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "2021 六年级下·江苏·专题练习",
+    "merged_id": "unit5_340",
+    "original_id": "unit5_choice_340"
+  },
+  {
+    "id": "unit5_choice_341",
+    "unit": "第五单元：圆",
+    "chapter": "扇形的周长和面积、圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "（2021 五年级下·江苏·专题练习）把一张半径 3 厘米的圆形彩纸对折剪成两个半圆，两个半圆的周长和比原来圆的周长多( )厘米，两个半圆的面积和是( )平方厘米。",
+    "options": [],
+    "answer": "12 28.26",
+    "solution": "【分析】把一张半径为 3 厘米的圆形彩纸对折剪成两个半圆, 则半圆的周长  $=$  一半圆的周长 + 直径, 则两个半圆的周长  $=$  一个圆的周长 +2 个直径, 用两个半圆的周长减去一个圆的周长即可。半圆面积  $=$  圆的面积  $\\div 2$ , 由此可知两个半圆面积  $=$  圆的面积。根据圆的面积公式:  $S = \\pi r^{2}$ , 把数代入即可求解。本题主要考查半圆的周长以及圆的面积公式，熟练掌握圆的面积公式并灵活运用，要注意半圆周长需要加个直径。\n\n【详解】由分析可知，两个半圆的周长比原来的圆的周长多了2个直径\n\n$ 2 \\times 3 \\times 2 \\\\ = 6 \\times 2 \\\\ = 1 2 \\text {(厘 米)} $\n\n两个半圆的面积： $3.14 \\times 3 \\times 3\\\\ = 9. 4 2 \\times 3 \\\\ = 2 8. 2 6 \\text {(平 方 厘 米)} $",
+    "knowledge_points": [
+      "扇形的周长和面积",
+      "圆的面积及应用"
+    ],
+    "source": "2021 五年级下·江苏·专题练习",
+    "merged_id": "unit5_341",
+    "original_id": "unit5_choice_341"
+  },
+  {
+    "id": "unit5_choice_342",
+    "unit": "第五单元：圆",
+    "chapter": "平面图形的拼接、长、正方形周长的应用、圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "（2021·湖南长沙·小升初模拟）把一个圆分成若干等份，拼成一个近似的长方形，周长增加了6厘米，则圆的半径是（ ）厘米，圆的面积是( )平方厘米。",
+    "options": [],
+    "answer": "3 28.26",
+    "solution": "【分析】把一个圆分成若干等份，拼成一个近似的长方形，所以长方形的长为  $\\pi \\mathrm{r}$ ，长方形的宽是  $\\mathrm{r}$ ，此时长方形周长为  $2(\\pi \\mathrm{r} + \\mathrm{r})$ ，增加量为  $2(\\pi \\mathrm{r} + \\mathrm{r}) - 2\\pi \\mathrm{r} = 2 \\mathrm{r}, 2 \\mathrm{r} = 6$  厘米，求出  $\\mathrm{r}$  是多少，再根据圆面积公式  $S = \\pi r^{2}$ ，列式计算即可解答此题。把一个圆分成若干等份，拼成一个近似的长方形，由于拼接而来，所以面积保持不变，周长增加  $2 \\mathrm{r}$  。\n\n【详解】  $6 \\div 2 = 3$  （厘米）\n\n$ 3. 1 4 \\times 3 ^ {2} \\\\ = 3. 1 4 \\times 9 \\\\ = 2 8. 2 6 \\text {(平 方 厘 米)} $\n\n圆的半径是 3 厘米, 圆的面积是 28.26 平方厘米。",
+    "knowledge_points": [
+      "平面图形的拼接",
+      "长、正方形周长的应用",
+      "圆的面积及应用"
+    ],
+    "source": "2021·湖南长沙·小升初模拟",
+    "merged_id": "unit5_342",
+    "original_id": "unit5_choice_342"
+  },
+  {
+    "id": "unit5_choice_343",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(19-20 五年级下·安徽马鞍山·期末)一根长 7.85 米的绳子, 正好可以绕这棵树的树干 5 圈。这棵树树干横截面的直径大约是( )厘米。",
+    "options": [],
+    "answer": "50",
+    "solution": "【分析】绳子的长度÷圈数 = 树干横截面的周长，截面的直径 = 周长÷π，据此解答。此题考查了圆的周长的相关应用, 明确圆的周长  $C = \\pi d$ , 并能灵活运用。\n\n【详解】7.85 米 = 785 厘米\n\n$785 \\div 5 \\div 3.14  \\\\= 157 \\div 3.14  \\\\=50 (厘米)$\n\n这棵树树干横截面的直径大约是50厘米。",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "19-20 五年级下·安徽马鞍山·期末",
+    "merged_id": "unit5_343",
+    "original_id": "unit5_choice_343"
+  },
+  {
+    "id": "unit5_choice_344",
+    "unit": "第五单元：圆",
+    "chapter": "圆的概念及特点、弧、圆心角、扇形的认识",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(19-20 五年级下·河南洛阳·期末) 圆的位置与( )有关系, 圆的( )与半径有关系, 同一圆中扇形的大小与( )有关系。",
+    "options": [],
+    "answer": "圆心 大小 圆心角",
+    "solution": "【详解】圆的位置与圆心有关系，圆的大小与半径有关系，同一圆中扇形的大小与圆心角有关系。",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "弧、圆心角、扇形的认识"
+    ],
+    "source": "19-20 五年级下·河南洛阳·期末",
+    "merged_id": "unit5_344",
+    "original_id": "unit5_choice_344"
+  },
+  {
+    "id": "unit5_choice_345",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(2021 六年级下·全国·竞赛) 神舟飞船绕地球共飞行14 圈, 其中后 10 圈沿离地面 343 千米的圆形轨道飞行。请计算后 10 圈飞船沿圆形轨道飞行了( )千米。(地球半径 6371 千米)",
+    "options": [],
+    "answer": "421639.2",
+    "solution": "【分析】沿离地面 343 千米的圆形轨道飞行, 转一周的轨迹相当于是半径 6714 千米的圆, 求出圆的周长, 乘 10 即可。本题考查的是圆的周长的计算, 求出圆形轨道的半径是解题的关键。\n\n【详解】  $2 \\times 3.14 \\times (343 + 6371) \\times 10\\\\=2 2 \\times 3. 1 4 \\times 6 7 1 4 \\times 1 0 \\\\=421639.2 (千米)$\n\n后 10 圈飞船沿圆形轨道飞行了 421639.2 千米",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "2021 六年级下·全国·竞赛",
+    "merged_id": "unit5_345",
+    "original_id": "unit5_choice_345"
+  },
+  {
+    "id": "unit5_choice_346",
+    "unit": "第五单元：圆",
+    "chapter": "弧、圆心角、扇形的认识、与圆相关的轴对称图形",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "(20-21 五年级下·全国·单元测试) 圆的对称轴是圆的( ), 半圆有( )条对称轴。同一个圆中, 扇形的大小与( )的大小有关。",
+    "options": [],
+    "answer": "直径所在的直线  1    圆心角",
+    "solution": "【分析】轴对称图形的意义：如果一个图形沿着一条直线对折后两部分完全重合，这样的图形叫做轴对称图形，这条直线叫做对称轴，进行解答即可；在同一个圆里，半径相同， $1^{\\circ}$ 的圆心角的扇形面积占圆面积的三百六十分之一， $90^{\\circ}$ 的圆心角的扇形面积占圆面积的四分之一，因此同一圆内圆心角的大小决定扇形的大小；据此判断。此题考查了轴对称图形的意义以及对圆的认识。\n\n【详解】圆的对称轴是圆的直径所在的直线，半圆有 1 条对称轴。同一个圆中，扇形的大小与圆心角的大小有关。",
+    "knowledge_points": [
+      "弧、圆心角、扇形的认识",
+      "与圆相关的轴对称图形"
+    ],
+    "source": "20-21 五年级下·全国·单元测试",
+    "merged_id": "unit5_346",
+    "original_id": "unit5_choice_346"
+  },
+  {
+    "id": "unit5_choice_347",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用、圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "（20-21 六年级下·浙江·期末）把一个周长为 18.84 厘米的圆平均分成两个半圆，每个半圆的周长是( )，每个半圆的面积是( )。",
+    "options": [],
+    "answer": "15.42 厘米 ;14.13 平方厘米",
+    "solution": "【分析】半圆的周长等于圆周长的一半加上直径；半圆的面积等于圆的面积除以 2，据此解答即可。明确半圆的周长和面积公式是解答本题的关键。\n\n【详解】  $18.84 \\div 3.14 = 6$  （厘米）；\n\n$ 1 8. 8 4 \\div 2 + 6 \\\\ = 9. 4 2 + 6 \\\\ = 1 5. 4 2 \\text {(厘 米)} ; $\n\n$ 3. 1 4 \\times (6 \\div 2) ^ {2} \\div 2 \\\\ = 3. 1 4 \\times 9 \\div 2 \\\\ = 2 8. 2 6 \\div 2\\\\ = 14.13（平方厘米）$",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级下·浙江·期末",
+    "merged_id": "unit5_347",
+    "original_id": "unit5_choice_347"
+  },
+  {
+    "id": "unit5_choice_348",
+    "unit": "第五单元：圆",
+    "chapter": "圆的概念及特点、数与形",
+    "question_type": "填空题",
+    "difficulty": "困难",
+    "question": "（2021 六年级·全国·竞赛）圆周上有 8 个点，两点所连的线段叫'弦'，每两点连一条弦，各弦无公共端点，共可连四条弦，各弦互不相交的连法共有 种。",
+    "options": [],
+    "answer": "14",
+    "solution": "【分析】直接考虑 8 个点的情况是比较麻烦的, 可以先从比较简单的情况进行分析, 可以先考虑只有 2 个、4 个、6 个点的情况, 发现并总结规律, 然后应用规律求解问题。归纳递推计数也是计数问题中常用的方法，关键是发现规律，并利用规律求解问题。\n\n【详解】本题可以利用归纳的方法解决。若圆周上只有2个点，只有1种连法；若圆周上只有 4 个点, 先选中 1 个点, 它可以与相邻的两个点相连, 它连好后其它两点只有 1 种连法, 所以此时有$1 \\times 2 = 2$  种连法;若圆周上只有6个点，先选中1个点，此时它可以与相邻的2 个点相连, 也可以相对的 1 个点相连, 若与相邻的点相连,剩下的4个点有2种连法；若与相对的点相连，剩下的4个点只有 1 种连法, 所以此时有  $2 \\times 2 + 1 = 5$  种连法;若圆周上只有8个点，先选中一个点，此时它可以与相邻的2 个点相连，也可以与与它相隔 2 个点的另外两个点相连。若与相邻的点相连，剩下的6个点有5种连法；若与相隔两个点的点相连，剩下的 6 个点被分成两边，一边 2 个点，只有一种连法，一边 4 个点，有 2 种连法。所以此时共有  $5 \\times2+2 \\times2=14$种连法。\n\n$5 \\times 2 + 2 \\times 2 = 14$  种连法。",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "数与形"
+    ],
+    "source": "2021 六年级·全国·竞赛",
+    "merged_id": "unit5_348",
+    "original_id": "unit5_choice_348"
+  },
+  {
+    "id": "unit5_choice_349",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用、圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "(19-20 六年级下·浙江杭州·期末) 推导圆的面积计算公式时，把一个圆分成若干等份后，将它剪拼成一个近似的长方形，若剪拼成的长方形的长是9.42厘米，则这个圆的面积是( )平方厘米。",
+    "options": [],
+    "answer": "28.26",
+    "solution": "【分析】根据题意可知, 剪拼成的长方形的长就是圆周长的一半, 据此求出圆的周长, 再根据 ' $r = c \\div \\pi \\div 2$ ' 求出圆的半径, 进而求出面积即可。熟练掌握圆的面积公式推导过程是解答本题的关键。\n\n【详解】  $9.42 \\times 2 \\div 3.14 \\div 2\\\\= 1 8. 8 4 \\div 3. 1 4 \\div 2\\\\=3 (厘米)$ ;\n\n$3.14 \\times 3^{2} = 28.26（平方厘米）$",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "19-20 六年级下·浙江杭州·期末",
+    "merged_id": "unit5_349",
+    "original_id": "unit5_choice_349"
+  },
+  {
+    "id": "unit5_choice_350",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "(20-21 六年级上·广东河源·期末) 在一块边长是 6 分米的正方形铁片中剪一个最大的圆形，这个圆的面积是()。",
+    "options": [],
+    "answer": "28.26平方分米",
+    "solution": "【分析】在正方形内剪一个最大的圆，圆的直径和正方形的边长相等，再根据 ' $s = \\pi r^2$ '求出圆的面积即可。明确 '在正方形内剪一个最大的圆, 圆的直径和正方形的边长相等' 是解答本题的关键。\n\n【详解】 $3.14 \\times (6\\div2)^{2}= 3. 1 4 \\times 9=28.26(平方分米）$",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级上·广东河源·期末",
+    "merged_id": "unit5_350",
+    "original_id": "unit5_choice_350"
+  },
+  {
+    "id": "unit5_choice_351",
+    "unit": "第五单元：圆",
+    "chapter": "扇形的周长和面积、列方程解含一个未知数的问题、圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "（19-20六年级上·上海闵行·期末）如果一个半径为  $2 \\mathrm{~cm}$  的圆的面积恰好与一个半径为  $4 \\mathrm{~cm}$  的扇形面积相等,那么这个扇形的圆心角为 度。",
+    "options": [],
+    "answer": "90",
+    "solution": "【分析】根据圆的面积公式： $S = \\pi r^2$ ，扇形面积公式：$S_{\\text {扇形}} = \\pi r^{2} \\times \\frac{n}{360}$ , 把数据分别代入公式解答。此题主要考查圆的面积公式、扇形面积公式的灵活应用，关键是熟记公式。\n\n【详解】  $3.14 \\times 2^{2} = 12.56$  （平方厘米）设这个扇形的圆心角度数为  $x$ 度。\n\n$ 3. 1 4 \\times 4 ^ {2} \\times \\frac {x}{3 6 0} = 1 2. 5 6 \\\\ 5 0. 2 4 \\times \\frac {x}{3 6 0} = 1 2. 5 6 \\\\ 1 2. 5 6 \\times \\frac {x}{9 0} = 1 2. 5 6 \\\\ \\frac {x}{9 0} = 1 \\\\ x = 9 0 $",
+    "knowledge_points": [
+      "扇形的周长和面积",
+      "列方程解含一个未知数的问题",
+      "圆的面积及应用"
+    ],
+    "source": "19-20六年级上·上海闵行·期末",
+    "merged_id": "unit5_351",
+    "original_id": "unit5_choice_351"
+  },
+  {
+    "id": "unit5_choice_352",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用、圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(2021 六年级下·全国·专题练习) 王大伯用 9.42 米长的篱笆靠墙围成了一个半圆形菜地。这个菜地的面积是( )平方米。",
+    "options": [],
+    "answer": "14.13",
+    "solution": "【分析】由题意可知: 圆周长的一半是 9.42 米, 则圆的周长是  $9.42 \\times 2 = 18.84$  米。带入周长公式:  $C = 2 \\pi r$ , 求出半径,再将半径值带入圆的面积公式求出面积, 最后用圆的面积  $\\div 2$  即可。本题主要考查圆的周长、面积公式的实际应用,求出圆的半径是解题的关键。\n\n【详解】  $9.42 \\times 2 \\div 3.14 \\div 2 \\\\= 1 8. 8 4 \\div 3. 1 4 \\div 2 \\\\ = 6 \\div 2 \\\\ = 3 (\\text {米}) $\n\n$ 3. 1 4 \\times 3 ^ {2} \\div 2 \\\\ = 2 8. 2 6 \\div 2 \\\\ = 1 4. 1 3 \\text {(平 方 米)} $",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "2021 六年级下·全国·专题练习",
+    "merged_id": "unit5_352",
+    "original_id": "unit5_choice_352"
+  },
+  {
+    "id": "unit5_choice_353",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "（2021 六年级下·全国·专题练习）钟面上时针长 6 厘米，分针长 10 厘米，从 8:00 到 8:45 分钟，针尖走过的路程是( )厘米。",
+    "options": [],
+    "answer": "47.1",
+    "solution": "【分析】在钟面上分针 1 小时转一圈, 从  $8: 00$  到  $8: 45$ ,经过了 45 分钟, 分针转了  $\\frac{45}{60}$  圈, 针尖走过的路程是以分针为半径的圆, 分针半径  $= 10$  厘米, 圆的周长 =  $\\pi \\times (10 \\times 2)$ , 再乘  $\\frac{45}{60}$ , 就是针尖走过的路程, 即可解答问题。本题考查圆的周长公式，熟练应用圆的周长公式解决实际问题。\n\n【详解】从  $8: 00$  到  $8: 45$ , 经过 45 分钟, 也就转了  $\\frac{45}{60} \\times 3.14 \\times (2 \\times 10) \\times \\frac{45}{60}\\\\= 3. 1 4 \\times 2 0 \\times \\frac {4 5}{6 0}\\\\= 47.1$  （厘米）",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "2021 六年级下·全国·专题练习",
+    "merged_id": "unit5_353",
+    "original_id": "unit5_choice_353"
+  },
+  {
+    "id": "unit5_choice_354",
+    "unit": "第五单元：圆",
+    "chapter": "圆的概念及特点、三角形的面积",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(19-20 六年级上·湖北武汉·期末) 一个圆的直径是$20 \\mathrm{~cm}$ , 以这个圆的直径为三角形的底, 顶点在圆周上运动,那么这个三角形的面积最大是( )cm²。",
+    "options": [],
+    "answer": "100",
+    "solution": "【分析】三角形的底一定，当高是半径时，三角形的面积最大。掌握圆的特点以及三角形的面积计算公式，找出三角形的高是解题关键。\n\n【详解】  $20 \\div 2 = 10$  （厘米）\n\n$ 2 0 \\times 1 0 \\div 2 \\\\ = 2 0 0 \\div 2 \\\\ = 1 0 0 \\text {(平 方 厘 米)} $\n\n这个三角形的面积最大是 100 平方厘米。",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "三角形的面积"
+    ],
+    "source": "19-20 六年级上·湖北武汉·期末",
+    "merged_id": "unit5_354",
+    "original_id": "unit5_choice_354"
+  },
+  {
+    "id": "unit5_choice_355",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·安徽宿州·期末) 在一个长  $16 \\mathrm{~cm}$ , 宽  $8 \\mathrm{~cm}$  的长方形中画一个最大的圆, 这个圆的半径是()cm, 周长是( )cm。",
+    "options": [],
+    "answer": "4 25.12",
+    "solution": "【分析】长方形里面画出的最大圆的直径与长方形的宽相等，由此用长方形的宽除以2即可求出圆的半径长度，根据圆的周长  $= 2 \\pi r$  求出周长。本题主要考查圆的周长公式，解题的关键是明确圆的直径等于长方形的宽。\n\n【详解】  $8 \\div 2 = 4$  （厘米）$3.14 \\times 4 \\times 2 = 25.12$  （厘米）",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·安徽宿州·期末",
+    "merged_id": "unit5_355",
+    "original_id": "unit5_choice_355"
+  },
+  {
+    "id": "unit5_choice_356",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "（20-21 六年级上·湖北孝感·期末）在一个边长是 4 分米的正方形内剪下一个最大的圆，这个圆的面积是( )平方分米。",
+    "options": [],
+    "answer": "12.56",
+    "solution": "【分析】这个圆的直径就是正方形的边长，先计算出圆的半径（直径  $\\div 2$ ），再依据圆的面积公式  $(S = \\pi r^{2})$  即可求其面积。此题主要考查正方形及圆的面积公式，关键是明白圆的直径即为正方形的边长。\n\n【详解】圆的面积：\n\n$ 3. 1 4 \\times (4 \\div 2) ^ {2} \\\\ = 3. 1 4 \\times 4 \\\\ = 1 2. 5 6 \\text {(平 方 分 米)} $",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级上·湖北孝感·期末",
+    "merged_id": "unit5_356",
+    "original_id": "unit5_choice_356"
+  },
+  {
+    "id": "unit5_choice_357",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的面积",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "(20-21 六年级上·内蒙古巴彦淖尔·期末) 一个圆环,内圆半径是  $2 \\mathrm{dm}$ , 外圆半径是  $3 \\mathrm{dm}$ , 圆环面积是( )  $\\mathrm{dm}^{2}$ 。",
+    "options": [],
+    "answer": "15.7",
+    "solution": "【分析】圆环的面积  $S = \\pi (R^{2} - r^{2})$ , 由此代入数据即可解决问题。此题考查了圆环的面积公式的计算应用, 熟记公式是关键。\n\n【详解】  $3.14 \\times (3^{2} - 2^{2})= 3. 1 4 \\times 5=15.7 (平方分米)$",
+    "knowledge_points": [
+      "圆环的面积"
+    ],
+    "source": "20-21 六年级上·内蒙古巴彦淖尔·期末",
+    "merged_id": "unit5_357",
+    "original_id": "unit5_choice_357"
+  },
+  {
+    "id": "unit5_choice_358",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用、圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(18-19六年级上·河南商丘·期末)在长8厘米、宽6厘米的长方形中画一个最大的半圆，这个半圆的周长是( )厘米，面积是( )平方厘米。",
+    "options": [],
+    "answer": "20.56;25.12",
+    "solution": "【分析】在长 8 厘米、宽 6 厘米的长方形中画一个最大的半圆, 半圆直径是这个长方形的长, 根据半圆周长 = 圆周长的一半 + 直径, 半圆面积 = 圆的面积  $\\div 2$  进行计算。关键是确定半圆的直径，半圆周长  $= \\pi r + 2r$  ，半圆面积  $= \\pi r^2 \\div 2$  。\n\n【详解】  $8 \\div 2 = 4$  （厘米）\n\n$ 3. 1 4 \\times 4 + 8 \\\\ = 1 2. 5 6 + 8 \\\\ = 2 0. 5 6 \\text {(厘 米)} \\\\ 3. 1 4 \\times 4 ^ {2} \\div 2 \\\\ = 5 0. 2 4 \\div 2 \\\\ = 2 5. 1 2 \\text {(平 方 厘 米)} $",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "18-19六年级上·河南商丘·期末",
+    "merged_id": "unit5_358",
+    "original_id": "unit5_choice_358"
+  },
+  {
+    "id": "unit5_choice_359",
+    "unit": "第五单元：圆",
+    "chapter": "平面图形的分割、长方形的概念及特点、圆的面积及应用、长方形的面积",
+    "question_type": "填空题",
+    "difficulty": "困难",
+    "question": "（19-20六年级上·山东临沂·期末）在一块长10分米、宽5分米的长方形铁板上，最多能截取( )个直径是2分米的圆形铁板。剩下部分的面积是( )。",
+    "options": [],
+    "answer": "10   18.6平方分米",
+    "solution": "【分析】求出长和宽分别能截取多少个直径为 2 分米的圆, 再相乘求出, 最多能截取几个直径是 2 分米的圆形铁板; 用长方形铁板的面积减去所有圆形铁板的面积, 求出剩下部分的面积。本题考查长方形、圆的面积, 解答本题的关键是掌握在长方形中截取圆形的方法。\n\n【详解】  $10 \\div 2 = 5$  （个）\n\n$5 \\div 2 = 2 \\text {(个)} \\dots \\dots 1 \\text {(分 米)}$   $5 \\times 2 = 1 0 (\\text {个})\\quad   10 \\times 5 - 1 0 \\times 3. 1 4 \\times (2 \\div 2) ^ {2}\\quad 50-31.4=18.6 (平方分米)$",
+    "knowledge_points": [
+      "平面图形的分割",
+      "长方形的概念及特点",
+      "圆的面积及应用",
+      "长方形的面积"
+    ],
+    "source": "19-20六年级上·山东临沂·期末",
+    "merged_id": "unit5_359",
+    "original_id": "unit5_choice_359"
+  },
+  {
+    "id": "unit5_choice_360",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用、比的意义、圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "（19-20六年级上·山东临沂·期末）有大小两个圆，小圆的半径是3厘米，大圆的半径是6厘米，它们的周长之比是( )，面积之比是( )。",
+    "options": [],
+    "answer": "$1: 2 \\quad 1: 4$",
+    "solution": "【分析】已知小圆的半径是 3 厘米, 大圆的半径是 6 厘米, 则它们的半径之比为  $3: 6 = 1: 2$ , 然后再依据周长之比等于半径之比; 面积之比等于半径的平方之比来解答。解答本题时，明确两个圆的周长之比和面积之比同半径之比的关系是关键。\n\n【详解】由分析得:3:6=1:2   $1^{2}: 2^{2} = 1: 4$   所以周长之比为  $1: 2$ ; 面积之比为  $1: 4$ 。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "比的意义",
+      "圆的面积及应用"
+    ],
+    "source": "19-20六年级上·山东临沂·期末",
+    "merged_id": "unit5_360",
+    "original_id": "unit5_choice_360"
+  },
+  {
+    "id": "unit5_choice_361",
+    "unit": "第五单元：圆",
+    "chapter": "圆的概念及特点",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "(20-21 六年级上·陕西西安·期中) 同一个圆中, 只能画一条半径和一条直径。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【详解】连接圆心和圆上任意一点的线段叫做圆的半径,通过圆心且两个端点都在圆周上的线段叫做圆的直径,在一个圆中, 有无数条半径和直径, 原题说法错误;故答案为: ×  。",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "20-21 六年级上·陕西西安·期中",
+    "merged_id": "unit5_361",
+    "original_id": "unit5_choice_361"
+  },
+  {
+    "id": "unit5_choice_362",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "(21-22 六年级上·全国·单元测试) 圆的周长是半径的6.28 倍。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】根据'圆的周长  $= 2 \\pi r$ '可知，圆的周长  $\\div r = 2 \\pi$ ，即圆的周长是它半径的  $2 \\pi$  倍，由此判断即可。解答此题应根据圆的周长和半径之间的关系进行解答, 注意  $\\pi$  与 3.14 的区别;\n\n【详解】同一个圆的周长一定是它半径的  $2 \\pi$  倍, 不是 6.28 倍, 原题说法错误; 故答案为: ×  。",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "21-22 六年级上·全国·单元测试",
+    "merged_id": "unit5_362",
+    "original_id": "unit5_choice_362"
+  },
+  {
+    "id": "unit5_choice_363",
+    "unit": "第五单元：圆",
+    "chapter": "正方形的面积、圆的面积及应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级下·河北张家口·期末) 若一个圆的直径和一个正方形的边长相等, 则圆的面积大。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】令圆的直径为 2 , 据此分别计算出圆和正方形的面积, 再判断哪个图形的面积大。本题考查了圆和正方形的面积, 灵活运用圆和正方形的面积公式是解题的关键。\n\n【详解】正方形:  $2 \\times 2 = 4$  圆:  $3.14 \\times (2 \\div 2)^{2} = 3.14$  所以，若一个圆的直径和一个正方形的边长相等，则正方形的面积大，所以原题说法错误。故答案为:  ×",
+    "knowledge_points": [
+      "正方形的面积",
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级下·河北张家口·期末",
+    "merged_id": "unit5_363",
+    "original_id": "unit5_choice_363"
+  },
+  {
+    "id": "unit5_choice_364",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用、圆的面积及应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "(21-22 六年级上·全国·单元测试) 半圆的面积是它所在圆面积的  $\\frac{1}{2}$ , 半圆的周长也是它所在圆周长的  $\\frac{1}{2}$ 。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】首先理解半圆的周长的意义，半圆的周长是这个圆周长的一半加上它的直径，由此解答。本题考查圆的周长、面积, 解答本题的关键是掌握圆的周长、面积的计算公式。\n\n【详解】半圆的面积是这个圆面积的  $\\frac{1}{2}$ , 而半圆的周长是这个圆周长的一半加上它的直径。因此半圆的面积是这个圆面积的  $\\frac{1}{2}$ , 所以半圆的周长也是这个圆周长的  $\\frac{1}{2}$ , 这种说法是错误的。故答案为:  ×",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "21-22 六年级上·全国·单元测试",
+    "merged_id": "unit5_364",
+    "original_id": "unit5_choice_364"
+  },
+  {
+    "id": "unit5_choice_365",
+    "unit": "第五单元：圆",
+    "chapter": "对称轴的画法及数量、与圆相关的轴对称图形",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "(20-21 六年级下·河南许昌·期末) 正方形、长方形、半圆和等边三角形中, 对称轴条数最多的是正方形。( )",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】先判断出正方形、长方形、半圆和等边三角形的对称轴数量，再判断出对称轴中数量最多的图形即可。本题考查了对称轴, 明确常见图形有几条对称轴是解题的关键。\n\n【详解】正方形有 4 条对称轴, 长方形有 2 条对称轴, 半圆有 1 条对称轴, 等边三角形有 3 条对称轴。所以, 这四个图形中, 对称轴数量最多的是正方形。故答案为:  √",
+    "knowledge_points": [
+      "对称轴的画法及数量",
+      "与圆相关的轴对称图形"
+    ],
+    "source": "20-21 六年级下·河南许昌·期末",
+    "merged_id": "unit5_365",
+    "original_id": "unit5_choice_365"
+  },
+  {
+    "id": "unit5_choice_366",
+    "unit": "第五单元：圆",
+    "chapter": "圆的概念及特点、平面图形的分割",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级下·河南商丘·期末) 在长  $15.4 \\mathrm{~cm}$ , 宽  $12 \\mathrm{~cm}$  的长方形纸中, 剪半径是  $1 \\mathrm{~cm}$  的圆, 能剪 13 个。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】圆的半径是 1 厘米, 则直径是  $1 \\times 2 = 2$  (厘米) 。沿长方形纸的长边来剪, 可以剪  $15.4 \\div 2 \\approx 7$  （个）; 沿宽边剪, 可以剪  $12 \\div 2 = 6$  （行）。每行剪 7 个, 一共剪 6 行, 则一共可以剪  $7 \\times 6 = 42$  （个）圆。求长方形能剪几个圆, 要沿长和宽分别计算每行剪几个、能剪几行, 再把每行的个数乘行数即可求出圆的总数。因为圆不能密铺, 则不能用长方形的面积除以圆的面积求出圆的个数。\n\n【详解】  $1 \\times 2 = 2$  （厘米） 15.4÷2≈7 (个)  $12 \\div 2 = 6$  (行)  $7 \\times 6 = 42$  （个） 这张长方形纸能剪 42 个圆。故答案为:  ×",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "平面图形的分割"
+    ],
+    "source": "20-21 六年级下·河南商丘·期末",
+    "merged_id": "unit5_366",
+    "original_id": "unit5_choice_366"
+  },{
+    "id": "unit5_choice_367",
+    "unit": "第五单元：圆",
+    "chapter": "扇形的周长和面积",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "(19-20 六年级上·全国·单元测试) 一个扇形的圆心角是  $120^{\\circ}$ , 它的面积是它所在圆的面积的  $\\frac{1}{3}$ 。( )",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】扇形面积  $= \\frac{n^{\\circ}}{360^{\\circ}} \\times \\pi r^{2}$ , 用扇形的圆心角的度数除以  $360^{\\circ}$  即可求出扇形的面积是它所在圆面积的几分之几。本题考查扇形的面积。扇形的圆心角是  $360^{\\circ}$  的几分之几, 扇形的面积就是其所在圆面积的几分之几。\n\n【详解】  $120^{\\circ} \\div 360^{\\circ} = \\frac{1}{3}$则扇形的面积是它所在圆的面积的  $\\frac{1}{3}$  。故答案为: √",
+    "knowledge_points": [
+      "扇形的周长和面积"
+    ],
+    "source": "19-20 六年级上·全国·单元测试",
+    "merged_id": "unit5_367",
+    "original_id": "unit5_choice_367"
+  },
+  {
+    "id": "unit5_choice_368",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "（2014·贵州安顺·小升初真题）大小两个圆，大圆周长与直径的比值和小圆周长与直径的比值相等。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】大圆周长等于  $\\pi \\times$  直径, 大圆的周长与直径的比值等于  $\\pi$ , 同样, 小圆的周长与直径的比值等于  $\\pi$ , 即可判断对错。本题考查熟练运用圆的周长公式解答问题。\n\n【详解】大圆的周长  $= \\pi \\times$  大圆的直径, 大圆的周长: 直径  $= \\pi$ ;小圆周长  $= \\pi \\times$  小圆直径, 小圆周长: 小圆直径  $= \\pi$ ;大小两个园，大圆周长与直径的比值和小圆周长与直径的比值相等，说法是正确的。故答案为:  √",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "2014·贵州安顺·小升初真题",
+    "merged_id": "unit5_368",
+    "original_id": "unit5_choice_368"
+  },
+  {
+    "id": "unit5_choice_369",
+    "unit": "第五单元：圆",
+    "chapter": "弧、圆心角、扇形的认识",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "(17-18 五年级下·全国·单元测试) 在同一个圆中, 扇形的大小与这个扇形的圆心角的大小有关。( )",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】在同一个圆中，扇形的大小与这个扇形的圆心角大小有关，圆心角越大扇形越大，反之亦然。如果两个扇形不是在同一个圆中, 那么即使圆心角相同, 也不能保证这两个扇形完全一样。因此, 在同一个圆中, 是必要的前提。\n\n【详解】因此，在同一个圆中，扇形的大小与这个扇形的圆心角的大小有关，这种说法是正确的。故答案为:  √",
+    "knowledge_points": [
+      "弧、圆心角、扇形的认识",
+      "扇形的周长和面积"
+    ],
+    "source": "17-18 五年级下·全国·单元测试",
+    "merged_id": "unit5_369",
+    "original_id": "unit5_choice_369"
+  },
+  {
+    "id": "unit5_choice_370",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·黑龙江齐齐哈尔·期末) 两个圆半径的比是  $2: 3$ , 那么它们面积的比也是  $2: 3$ 。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】根据圆的面积计算公式可得：平方以后的半径比是面积比，进行分析。关键是理解比的意义，掌握圆的面积公式，圆的面积  $= \\pi r^{2}$ 。\n\n【详解】  $2^{2}: 3^{2} = 4: 9$ , 两个圆半径的比是  $2: 3$ , 那么它们面积的比是  $4: 9$ , 所以原题说法错误。",
+    "knowledge_points": [
+      "比的意义",
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级上·黑龙江齐齐哈尔·期末",
+    "merged_id": "unit5_370",
+    "original_id": "unit5_choice_370"
+  },
+  {
+    "id": "unit5_choice_371",
+    "unit": "第五单元：圆",
+    "chapter": "扇形的周长和面积",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "（20-21 六年级上·河北张家口·期末）一个扇形的圆心角是  $90^{\\circ}$ ，则这个扇形的面积是同半径圆面积的  $\\frac{1}{4}$ 。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】周角的度数为  $360^{\\circ}$ , 扇形的圆心角度占总角度的  $\\frac{1}{4}$ ,因此面积也是同半径圆面积的  $\\frac{1}{4}$ , 由此解答即可解答本题的关键是抓住扇形圆心角和周角的关系，进而判断面积关系。\n\n【详解】一个扇形的圆心角是  $90^{\\circ}$ , 则这个扇形的面积是同半径圆面积的  $\\frac{1}{4}$ , 原题说法正确;故答案为: √",
+    "knowledge_points": [
+      "扇形的周长和面积",
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级上·河北张家口·期末",
+    "merged_id": "unit5_371",
+    "original_id": "unit5_choice_371"
+  },
+  {
+    "id": "unit5_choice_372",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "(2014 六年级·全国·课后作业) 圆周率是圆的周长与直径的比值。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】根据圆的周长  $= \\pi d$ , 圆周率  $= \\text{周长} \\div$  直径, 进行分析。圆周率是圆的周长与直径的比值，一般用希腊字母  $\\pi$  表示，是一个在数学及物理学中普遍存在的数学常数。\n\n【详解】圆周率是圆的周长与直径的比值，说法正确。故答案为： √",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "2014 六年级·全国·课后作业",
+    "merged_id": "unit5_372",
+    "original_id": "unit5_choice_372"
+  },
+  {
+    "id": "unit5_choice_373",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "(20-21 六年级上·全国·期末) 圆的直径变大, 其面积依然不变。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】圆的直径变大，半径就会变大；根据' $s = \\pi r^2$ '可知，半径变大，面积也会变大，由此解答即可。直径的大小直接影响到半径的大小，因此也影响面积的大小。\n\n【详解】圆的直径变大, 其面积也会变大, 原题说法错误; 故答案为: × 。",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级上·全国·期末",
+    "merged_id": "unit5_373",
+    "original_id": "unit5_choice_373"
+  },
+  {
+    "id": "unit5_choice_374",
+    "unit": "第五单元：圆",
+    "chapter": "画圆",
+    "question_type": "作图题",
+    "difficulty": "容易",
+    "question": "(21-22 六年级上·全国·单元测试) 按要求画圆。d=3cm 的圆。",
+    "options": [],
+    "answer": "见详解",
+    "solution": "【分析】画圆的步骤：把圆规的两脚分开，定好两脚的距离，即半径；把有针尖的一只脚固定在一点上，即圆心；把装有铅笔尖的一只脚旋转一周，就画出一个圆。熟练掌握画圆的方法是解答本题的关键。\n\n【详解】如图：![](images/767ccd787eca2366bc68aa3fdfbbd51bf71aada7117107e97d718383586dde38.jpg)",
+    "knowledge_points": [
+      "画圆"
+    ],
+    "source": "21-22 六年级上·全国·单元测试",
+    "merged_id": "unit5_374",
+    "original_id": "unit5_choice_374"
+  },
+  {
+    "id": "unit5_choice_375",
+    "unit": "第五单元：圆",
+    "chapter": "画圆",
+    "question_type": "作图题",
+    "difficulty": "容易",
+    "question": "(21-22 六年级上·全国·单元测试) 按要求画圆。r=2cm 的圆。![](images/22fa778623ee795d4b95816a7ad5a6daec9b4f922964e8db4a4fc3c2726b6cf8.jpg)",
+    "options": [],
+    "answer": "见详解",
+    "solution": "【分析】画圆的步骤：把圆规的两脚分开，定好两脚的距离，即半径；把有针尖的一只脚固定在一点上，即圆心；把装有铅笔尖的一只脚旋转一周，就画出一个圆。熟练掌握画圆的方法是解答本题的关键。\n\n【详解】如图:![](images/c14e603236f8ed4688515eb6832e1548915481960b6b87224e77c176d74e7112.jpg)",
+    "knowledge_points": [
+      "画圆"
+    ],
+    "source": "21-22 六年级上·全国·单元测试",
+    "merged_id": "unit5_375",
+    "original_id": "unit5_choice_375"
+  },
+  {
+    "id": "unit5_choice_376",
+    "unit": "第五单元：圆",
+    "chapter": "画圆",
+    "question_type": "作图题",
+    "difficulty": "中等",
+    "question": "（20-21 六年级上·湖南郴州·期末）在下面空白处用圆规画一个半径是 1.5 厘米的圆，并标出圆心和半径。",
+    "options": [],
+    "answer": "见详解",
+    "solution": "【分析】画一条长 1.5 厘米的线段, 以这条线段的一个端点为圆心, 以圆规的另一个脚到线段另一个端点的长为半径画圆即可。本题考查了学生画圆的作图能力。\n\n【详解】根据题意, 以  $\\mathrm{O}$  为圆心, 以 1.5 厘米为半径, 画圆。如图所示:![](images/e60da7835899467536068bce225730adaa4b7cd49d5e93aabcf359a2675c24c4.jpg)",
+    "knowledge_points": [
+      "画圆"
+    ],
+    "source": "20-21 六年级上·湖南郴州·期末",
+    "merged_id": "unit5_376",
+    "original_id": "unit5_choice_376"
+  },
+  {
+    "id": "unit5_choice_377",
+    "unit": "第五单元：圆",
+    "chapter": "画扇形",
+    "question_type": "作图题",
+    "difficulty": "中等",
+    "question": "(19-20 六年级上·山东临沂·期末) 画一个半径是  $1 \\mathrm{~cm}$  的圆, 再在圆中画一个圆心角是  $100^{\\circ}$  的扇形。",
+    "options": [],
+    "answer": "见详解",
+    "solution": "【分析】画圆：把圆规的两脚分开，定好两脚的距离（1厘米）；有针尖的一只脚固定在一点上；把装有铅笔尖的一只脚旋转一周，就画出一个圆；扇形：先画出一条半径，用量角器量出100度，再画出另一条半径即可。熟练掌握圆和扇形的画法是解答本题的关键。\n\n【详解】如图：![](images/903a34c2baf63da99128a78ab71df4e1c3d54771624b0569e809d05863c23602.jpg)",
+    "knowledge_points": [
+      "画圆",
+      "画扇形"
+    ],
+    "source": "19-20 六年级上·山东临沂·期末",
+    "merged_id": "unit5_377",
+    "original_id": "unit5_choice_377"
+  },
+  {
+    "id": "unit5_choice_378",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的面积",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(21-22 六年级上·全国·单元测试) 校园圆形花池的半径是 6 米, 在花池的周围修一条 2 米宽的水泥路。(1) 水泥路的面积是多少平方米？(2) 沿水泥路的外边缘每隔 3.14 米栽一棵柳树, 最多可以栽几棵柳树?",
+    "options": [],
+    "answer": "(1) 87.92 平方米    (2) 16 棵",
+    "solution": "【分析】（1）由题意可知，就是求圆环的面积，先求出大圆的半径, 即  $6 + 2$ , 再根据 'S 环形  $= \\pi$ $(\\mathrm{R}^{2} - \\mathrm{r}^{2})$  ' 进行解答即可;(2) 封闭图形植树属于只栽一端的情况, 棵数等于间隔数;用周长  $\\div$  间隔长度即可求出间隔数, 也就是植树的棵数。熟记求圆环的面积公式以及封闭图形植树的特点是解答本题的关键。\n\n【详解】 (1)  $3.14 \\times [6 + 2)^{2} - 6^{2}]\\\\= 3. 1 4 \\times 2 8 \\\\ = 8 7. 9 2 \\text {(平 方 米)}; $\n\n答: 水泥路的面积是 87.92 平方米;\n\n(2)  $2 \\times 3.14 \\times (6 + 2) \\div 3.14 \\\\= 5 0. 2 4 \\div 3. 1 4 \\\\ = 1 6 \\text {(棵)}; $\n\n答: 最多可以栽 16 棵柳树。",
+    "knowledge_points": [
+      "圆环的面积",
+      "植树问题"
+    ],
+    "source": "21-22 六年级上·全国·单元测试",
+    "merged_id": "unit5_378",
+    "original_id": "unit5_choice_378"
+  },
+  {
+    "id": "unit5_choice_379",
+    "unit": "第五单元：圆",
+    "chapter": "正方形的面积",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(21-22 六年级上·全国·单元测试) 在一个正方形纸板上画了一个最大的圆，已知圆的周长是 37.68 厘米，这张正方形纸板的面积是多少？",
+    "options": [],
+    "answer": "144平方厘米",
+    "solution": "【分析】在正方形内画一个最大的圆, 圆的直径和正方形的边长相等; 根据 '  $d = c \\div \\pi$  ' 求出圆的直径, 即正方形的边长, 再用边长乘边长即可求出正方形的面积。明确在正方形内画一个最大的圆，圆的直径和正方形的边长相等是解答本题的关键。\n\n【详解】  $37.68 \\div 3.14 = 12$  （厘米）； $12 \\times 12 = 144$  （平方厘米）；答: 这张正方形纸板的面积是 144 平方厘米。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "正方形的面积"
+    ],
+    "source": "21-22 六年级上·全国·单元测试",
+    "merged_id": "unit5_379",
+    "original_id": "unit5_choice_379"
+  },
+  {
+    "id": "unit5_choice_380",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（18-19六年级上·全国·课后作业）有一个水缸，缸壁厚5厘米，从里面量，缸口直径是50厘米，要制作一个缸盖，使它正好盖住缸口的外沿，这个缸盖的面积是多少平方厘米？如果在缸盖的边沿贴上一圈金属条（接头处不计），这圈金属条长多少厘米？",
+    "options": [],
+    "answer": "2826平方厘米；188.4厘米",
+    "solution": "【分析】根据题意可知, 缸壁厚 5 厘米。从里面量, 缸口直径是 50 厘米。这个缸盖的半径等于缸口里面的半径加上缸壁的厚度, 根据圆的面积公式:  $S = \\pi r^{2}$ , 圆的周长公式:  $C = \\pi d$ , 把数据代入公式解答。此题主要考查圆的面积公式、周长公式在实际生活中的应用，关键是熟记公式。\n\n【详解】  $3.14 \\times (50 \\div 2 + 5)^{2}\\\\ = 3. 1 4 \\times 3 0 ^ {2} \\\\ = 3. 1 4 \\times 9 0 0 \\\\ = 2 8 2 6 \\text {(平 方 厘 米)} $\n\n答: 这个缸盖的面积是 2826 平方厘米。\n\n3.14× (50+5×2)  \n$= 3.14 \\times 60$  \n=188.4 (厘米)\n\n答: 这个金属条长 188.4 厘米。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "18-19六年级上·全国·课后作业",
+    "merged_id": "unit5_380",
+    "original_id": "unit5_choice_380"
+  },
+  {
+    "id": "unit5_choice_381",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的面积",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(21-22 六年级上·全国·单元测试) 一个直径为  $14 \\mathrm{~m}$  的圆形水池, 周围有一条宽  $1 \\mathrm{~m}$  的环形小路, 小路的面积是多少?",
+    "options": [],
+    "answer": "47.1 平方米",
+    "solution": "【分析】这条小路的面积就是这个外圆半径为  $14 \\div 2 + 1 = 8$  米, 内圆半径为  $14 \\div 2 = 7$  米的圆环的面积, 由此利用圆环的面积公式即可计算。本题考查圆环的面积, 解答本题的关键是掌握圆环的面积计算公式。\n\n【详解】![](images/0bfcfda9a6bf6c0fb3f42838f1f62f921440c446aaace6382b7342286131c15b.jpg)$14 \\div 2 = 7$  （米）， $7 + 1 = 8$  （米），所以小路的面积为:\n\n\n$ 3. 1 4 \\times (8 ^ {2} - 7 ^ {2}) \\\\ = 3. 1 4 \\times (6 4 - 4 9) \\\\ = 3. 1 4 \\times 1 5\\\\= 47.1(平方米) $\n答: 小路的面积是 47.1 平方米。",
+    "knowledge_points": [
+      "圆环的面积"
+    ],
+    "source": "21-22 六年级上·全国·单元测试",
+    "merged_id": "unit5_381",
+    "original_id": "unit5_choice_381"
+  },
+  {
+    "id": "unit5_choice_382",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(21-22 六年级上·全国·单元测试) 画一个直径为 4 厘米的圆。要求：①要标明圆心。②要标明半径是多少。③要计算出这个圆周长是多少厘米。",
+    "options": [],
+    "answer": "![](images/e28cffd983c1bf20ea7fca677f7de6dc9235f60eb26ed6614246384c4cb6c19e.jpg)半径是 2 厘米; 周长是 12.56 厘米",
+    "solution": "【分析】由题意知, 要画一个直径是 4 厘米的圆, 首先确定圆的半径为  $4 \\div 2 = 2$  厘米, 再依据画圆的方法画一个圆, 并用字母标出它的圆心 O、半径 r 和直径 d; 求圆的周长, 根据: C=πd, 由此解答即可。本题考查圆的画法、圆的周长, 解答本题的关键是掌握圆的画法。\n\n【详解】 (1) 如图所示:\n\n![](images/552a919d4ecc9f472f2a062a28e983e906417bbe53d224b66e96a11727f9566a.jpg)半径:  $4 \\div 2 = 2$  (厘米)(2)  $3.14 \\times 4 = 12.56$  (厘米)答: 半径是 2 厘米, 周长是 12.56 厘米。",
+    "knowledge_points": [
+      "画圆",
+      "圆的周长及应用"
+    ],
+    "source": "21-22 六年级上·全国·单元测试",
+    "merged_id": "unit5_382",
+    "original_id": "unit5_choice_382"
+  },
+  {
+    "id": "unit5_choice_383",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的面积",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（21-22六年级上·全国·单元测试）一个圆形花坛，直径为10米，经过扩建，直径增加到16米，花坛占地面积增加了多少平方米？",
+    "options": [],
+    "answer": "122.46 平方米",
+    "solution": "【分析】根据题意可知，扩建后直径增加到 16 米，求面积增加了多少平方米，也就是求这个环形的面积，已知内圆直径，首先求出内圆半径，根据环形面积 = 外圆面积 - 内圆面积，由此列式解答。本题考查圆环的面积, 解答本题的关键是掌握圆环的面积计算公式。\n\n【详解】内圆半径: $10 \\div 2 = 5  (米)外圆半径:  $16 \\div 2 = 8$  (米)  增加的面积:\n\n$ 3. 1 4 \\times (8 ^ {2} - 5 ^ {2}) \\\\ = 3. 1 4 \\times (6 4 - 2 5) \\\\ = 3. 1 4 \\times 3 9 \\\\ = 1 2 2. 4 6 \\text {(平 方 米)} $\n\n答: 花坛占地面积增加了 122.46 平方米。",
+    "knowledge_points": [
+      "圆环的面积"
+    ],
+    "source": "21-22六年级上·全国·单元测试",
+    "merged_id": "unit5_383",
+    "original_id": "unit5_choice_383"
+  },
+  {
+    "id": "unit5_choice_384",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(20-21 五年级下·江苏苏州·期末) 一根长 188.4 厘米的绳子, 正好在一棵树上绕了 10 圈。这棵树的横截面的直径约是多少厘米? 面积呢?",
+    "options": [],
+    "answer": "6 厘米；28.26 平方厘米",
+    "solution": "【分析】先用  $188.4 \\div 10$  计算出绕树的树干 1 圈的长度, 即树干的周长, 根据圆的周长公式:  $C = 2 \\pi r$ , 求出半径进而得出直径。在根据圆的面积公式:  $S = \\pi r^{2}$ , 把数据代入公式解答。解答此题的关键是先计算出树的树干 1 圈的长度,继而根据圆的直径和圆周率和周长的关系进行解答。\n\n【详解】  $188.4 \\div 10 \\div 3.14= 1 8. 8 4 \\div 3. 1 4 = 6  (厘米)$\n\n$  3. 1 4 \\times (6 \\div 2) ^ {2} \\\\ = 3. 1 4 \\times 9\\\\=28.26 (平方厘米) $\n\n答: 这棵树的横截面的直径约是 6 厘米, 面积是 28.26 平方厘米。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "20-21 五年级下·江苏苏州·期末",
+    "merged_id": "unit5_384",
+    "original_id": "unit5_choice_384"
+  },
+  {
+    "id": "unit5_choice_385",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（19-20五年级下·湖南邵阳·期末）魏源公园有一个直径是 20 米的圆形花坛, 想全部用来栽杜鹃花, 其中每 0.5平方米栽一棵，请你帮忙计算一下要买多少棵杜鹃花？",
+    "options": [],
+    "answer": "628 棵",
+    "solution": "【分析】由题意可知, 求出这个圆形花坛的面积, 每 0.5 平方米栽一棵, 那这个圆形花坛里面有几个 0.5 就有几棵树杜鹃花, 用除法计算即可。本题考查圆的面积, 掌握圆的面积计算公式是关键。\n\n【详解】  $\\pi \\times (20 \\div 2)^{2} \\div 0.5$\n\n$ = 3. 1 4 \\times 1 0 0 \\div 0. 5 \\\\ = 3 1 4 \\div 0. 5 \\\\ = 6 2 8 \\text {(棵)} $\n\n答: 要买 628 棵杜鹃花。",
+    "knowledge_points": [
+      "除数是小数的小数除法",
+      "圆的面积及应用"
+    ],
+    "source": "19-20五年级下·湖南邵阳·期末",
+    "merged_id": "unit5_385",
+    "original_id": "unit5_choice_385"
+  },
+  {
+    "id": "unit5_choice_386",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的面积",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（20-21五年级下·江苏·期末）学校修建一个直径为20米的圆形花坛，并在花坛的四周修一条2米宽的彩砖小路。修建的彩砖小路占地多少平方米？",
+    "options": [],
+    "answer": "138.16 平方米",
+    "solution": "【分析】由题意可知: 彩砖小路的占地面积是内圆半径是  $20 \\div 2 = 10$  米, 外圆半径是  $10 + 2$  米的圆环的面积, 代入数据计算即可。本题主要考查圆环面积的实际应用, 明确内、外圆的半径是解题的关键。\n\n【详解】  $3.14 \\times (20 \\div 2 + 2)^{2} - 3.14 \\times (20 \\div 2)^{2}$\n\n$ = 3. 1 4 \\times 1 2 ^ {2} - 3. 1 4 \\times 1 0 ^ {2} \\\\ = 3. 1 4 \\times (1 4 4 - 1 0 0) \\\\ = 3. 1 4 \\times 4 4 \\\\ = 1 3 8. 1 6 \\quad (\\text {平 方 米}) $\n答: 修建的彩砖小路占地 138.16 平方米。",
+    "knowledge_points": [
+      "圆环的面积"
+    ],
+    "source": "20-21五年级下·江苏·期末",
+    "merged_id": "unit5_386",
+    "original_id": "unit5_choice_386"
+  },
+  {
+    "id": "unit5_choice_387",
+    "unit": "第五单元：圆",
+    "chapter": "工程问题",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（2022·安徽合肥·小升初真题）加工一批零件，王师傅每小时加工36个，与李师傅每小时加工个数的比是4:5，两人共同加工5小时，可以加工多少个零件？",
+    "options": [],
+    "answer": "405 个",
+    "solution": "【分析】根据两位师傅的工作效率比，用王师傅每小时加工个数  $\\div$  对应份数  $\\times$  李师傅每小时加工个数的对应份数  $=$  李师傅每小时加工个数，两位师傅每小时加工个数和  $\\times$  工作时间  $=$  加工的总个数。关键是理解比的意义，掌握工作效率、工作时间、工作总量之间的关系。\n\n【详解】  $36 \\div 4 \\times 5 = 45$  （个）\n\n$ (4 5 + 3 6) \\times 5 \\\\ = 8 1 \\times 5 \\\\= 405（个）$\n\n答: 可以加工 405 个零件。",
+    "knowledge_points": [
+      "按比分配问题",
+      "工程问题"
+    ],
+    "source": "2022·安徽合肥·小升初真题",
+    "merged_id": "unit5_387",
+    "original_id": "unit5_choice_387"
+  },
+  {
+    "id": "unit5_choice_388",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（20-21六年级上·广东湛江·期末）一台压路机的滚轮直径是1.5米，向前滚动50圈压路面的长度是多少米？",
+    "options": [],
+    "answer": "235.5 米",
+    "solution": "【分析】压路机的滚轮就是一个圆柱体，向前滚动一圈就是这个圆柱的底面的周长，也就是底面的圆的周长，根据圆的周长，算出圆柱底面圆的周长，即  $3.14 \\times 1.5$  ，向前滚动50圈，就是50个圆柱的底面的圆向前滚了50圈，长是多少，就有一个圆柱底面圆的周长  $\\times 50$  ，即：  $3.14 \\times 1.5 \\times 50$  ，即可算出。本意考查圆柱体底面的特点和圆的周长，根据圆的周长公式解答问题。\n\n【详解】  $3.14 \\times 1.5 \\times 50\\\\= 4. 7 1 \\times 5 0\\\\= 235.5$  （平方米）\n\n答: 向前滚动 50 圈压路机面的长度是 235.5 米。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆柱的认识及特征"
+    ],
+    "source": "20-21六年级上·广东湛江·期末",
+    "merged_id": "unit5_388",
+    "original_id": "unit5_choice_388"
+  },
+  {
+    "id": "unit5_choice_389",
+    "unit": "第五单元：圆",
+    "chapter": "含圆的组合图形的计算",
+    "question_type": "解答题",
+    "difficulty": "困难",
+    "question": "(19-20 六年级上·辽宁·期中) 一块正方形的草地, 边长是 3 米, 在两个对角的顶点处各种一棵树, 树上各拴一只羊，拴羊的绳子都是3米。这两只羊都能吃到的草的面积有多大？",
+    "options": [],
+    "answer": "5.13 平方米",
+    "solution": "【分析】根据所画图形可知，两只羊都能吃到的草的面积  $=$  （圆的面积的  $\\frac{1}{4} -$  正方形面积的一半）  $\\times 2$  ，其中圆的半径是 3 米，据此解答。此题考查了组合图形的面积计算, 明确问题所求,找准面积之间的关系是解题关键。\n\n【详解】  $(3.14 \\times 3^{2} \\times \\frac{1}{4} - 3 \\times 3 \\div 2) \\times 2\\\\= (7. 0 6 5 - 4. 5) \\times 2 \\\\ = 2. 5 6 5 \\times 2\\\\= 5.13$  （平方米）\n\n答: 这两只羊都能吃到的草的面积有 5.13 平方米。",
+    "knowledge_points": [
+      "含圆的组合图形的计算（周长和面积）"
+    ],
+    "source": "19-20 六年级上·辽宁·期中",
+    "merged_id": "unit5_389",
+    "original_id": "unit5_choice_389"
+  },
+  {
+    "id": "unit5_choice_390",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的面积",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（20-21 六年级上·福建泉州·阶段练习）李爷爷家有一块圆形菜地，周长是 12.56 米，他想在菜地周围加宽 2 米，加宽后的菜地比原来的面积多了多少平方米？",
+    "options": [],
+    "answer": "37.68 平方米",
+    "solution": "【分析】先根据圆的周长求出原来菜地的半径，再求出原来菜地的面积；加宽后菜地的半径是原来菜地半径加 2 ，再用加宽后的面积减去原来菜地的面积即可。本题考查圆环的面积, 解答本题的关键是掌握圆的面积公式。\n\n【详解】  $12.56 \\div 2 \\div 3.14\\\\ = 6. 2 8 \\div 3. 1 4 \\\\ = 2 (\\text {米}) $\n\n$2 + 2 = 4 \\text {(米)}$\n\n$ 3. 1 4 \\times (4 ^ {2} - 2 ^ {2}) \\\\ = 3. 1 4 \\times 1 2 \\\\ = 3 7. 6 8 \\text {(平 方 米)} $\n\n答: 加宽后的菜地比原来的面积多了 37.68 平方米。",
+    "knowledge_points": [
+      "圆环的面积",
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级上·福建泉州·阶段练习",
+    "merged_id": "unit5_390",
+    "original_id": "unit5_choice_390"
+  },
+  {
+    "id": "unit5_choice_391",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的面积",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（20-21六年级上·黑龙江齐齐哈尔·期末）为美化校园环境，学校准备在直径是6米的花坛外围铺一条1米宽的环形小路，这条小路的面积是多少平方米？",
+    "options": [],
+    "answer": "21.98 平方米",
+    "solution": "【分析】小路面积是一个圆环, 先求出内外圆的半径, 根据圆环面积  $= \\pi (\\mathrm{R}^{2} - \\mathrm{r}^{2})$ , 列式解答。关键是先求出内外圆半径，掌握圆环面积公式。\n\n【详解】  $6 \\div 2 = 3$  （米）\n\n$ 3 + 1 = 4 \\text {(米)} $\n\n$3.1 4 \\times (4 ^ {2} - 3 ^ {2}) \\\\ = 3. 1 4 \\times (1 6 - 9) \\\\ = 3. 1 4 \\times 7 \\\\= 21.98（平方米）$\n\n答: 这条小路的面积是 21.98 平方米。",
+    "knowledge_points": [
+      "圆环的面积"
+    ],
+    "source": "20-21六年级上·黑龙江齐齐哈尔·期末",
+    "merged_id": "unit5_391",
+    "original_id": "unit5_choice_391"
+  },
+  {
+    "id": "unit5_choice_392",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·内蒙古呼和浩特·期末) 现在有一根长 125.6 米的绳子, 要围成一块尽量大的土地, 你认为怎样围，围成的是什么图形？面积是多少？",
+    "options": [],
+    "answer": "圆形面积最大；1256平方米",
+    "solution": "【分析】根据在所有的平面图形中，周长一定围成了圆的面积最大，所以可以把这根绳子围成一个圆形，然后再根据圆的周长公式  $C = 2 \\pi r$ , 得出圆的半径  $r = C \\div \\pi \\div 2$ , 最后再根据圆的面积公式:  $S = \\pi r^{2}$  进行计算即可得到答案。此题主要考查的是在所有的平面图形中, 周长一定围成的圆的面积最大, 然后再灵活利用圆的周长公式和圆的面积公式进行计算即可。\n\n【详解】  $125.6 \\div 3.14 \\div 2\\\\= 4 0 \\div 2\\\\=20 (米)  $\n\n$3.14 \\times 20^{2}$  \n$= 3.14 \\times 400$  \n$= 1256$  （平方米）\n\n答: 把绳子围成圆形面积最大, 面积是 1256 平方米。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级上·内蒙古呼和浩特·期末",
+    "merged_id": "unit5_392",
+    "original_id": "unit5_choice_392"
+  },
+  {
+    "id": "unit5_choice_393",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（20-21六年级上·广东广州·期末）一个圆形餐桌的直径是  $2 \\mathrm{~m}$  。(1) 如果在这张餐桌的中央放一个半径是  $0.5 \\mathrm{~m}$  的圆形转盘,剩下的桌面面积是多少?(2) 如果一个人需要  $0.5 \\mathrm{~m}$  宽的位置就餐, 这张桌子大约能坐多少人?",
+    "options": [],
+    "answer": "(1)  $2.355 \\mathrm{~m}^{2}$   (2) 12人",
+    "solution": "【分析】（1）根据圆的面积公式' $s = \\pi r^2$ '，分别求出圆形转盘和桌面的面积，再相减即可；(2) 根据 'C=πd' 求出圆形餐桌的周长, 再除以每个人需要的位置宽度, 即可求出这张桌子大约能坐多少人。熟练掌握圆的周长和面积公式是解答本题的关键,尤其在最后问大约能坐多少人时，根据实际情况商的近似数要采用'去尾法'。\n\n【详解】 (1)  $3.14 \\times (2 \\div 2)^{2} - 3.14 \\times 0.5^{2}\\\\ = 3. 1 4 - 0. 7 8 5 \\\\ = 2. 3 5 5 \\text {(平 方 米)}; $\n\n答: 剩下的桌面面积是 2.355 平方米;\n\n(2)  $3.14 \\times 2 \\div 0.5\\\\ = 6. 2 8 \\div 0. 5 \\\\ \\approx 1 2 \\text {(人)} ; $\n\n答: 这张桌子大约能坐 12 人。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "20-21六年级上·广东广州·期末",
+    "merged_id": "unit5_393",
+    "original_id": "unit5_choice_393"
+  },
+  {
+    "id": "unit5_choice_394",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（20-21六年级上·广东广州·期末）给一把圆形的扇子镶边，共用去  $62.8 \\mathrm{~cm}$  长的布条（接头忽略不计）。这把扇子的面积约是多少平方厘米？",
+    "options": [],
+    "answer": "314平方厘米",
+    "solution": "【分析】根据圆的周长公式，先求出扇面的半径，再根据圆的面积  $= \\pi r^{2}$ ，列式解答。关键是掌握圆的周长和面积公式，圆的周长  $= \\pi d = 2\\pi r$ 。\n\n【详解】  $62.8 \\div 3.14 \\div 2 = 10$  （厘米）$3.14 \\times 10^{2} = 314$  （平方厘米） 答: 这把扇子的面积约是 314 平方厘米。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "20-21六年级上·广东广州·期末",
+    "merged_id": "unit5_394",
+    "original_id": "unit5_choice_394"
+  },
+  {
+    "id": "unit5_choice_395",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的面积",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（20-21 六年级上·湖南郴州·期末）校园圆形花池的半径是 6 米，在花池的周围修一条 2 米宽的水泥路。水泥路的面积是多少平方米？",
+    "options": [],
+    "answer": "87.92平方米",
+    "solution": "【分析】由题意可知, 水泥路的面积就是一个圆环的面积,  $S_{\\text {圆环}} = \\pi (R^{2} - r^{2})$ , 其中  $r = 6$  米,  $R = r + 2$ , 代入数据计算即可。此题主要考查圆环面积的计算，掌握圆环面积公式，并找出两个圆的半径是解题关键。\n\n【详解】  $6 + 2 = 8$  （米），\n\n$ 3. 1 4 \\times (8 ^ {2} - 6 ^ {2}) \\\\ = 3. 1 4 \\times 2 8 \\\\ = 8 7. 9 2 \\text {(平 方 米)} $\n\n答: 水泥路的面积是 87.92 平方米。",
+    "knowledge_points": [
+      "圆环的面积"
+    ],
+    "source": "20-21 六年级上·湖南郴州·期末",
+    "merged_id": "unit5_395",
+    "original_id": "unit5_choice_395"
+  },
+  {
+    "id": "unit5_choice_396",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·吉林通化·期末) 在一块长 36 米, 宽30 米的空地上，修建一个周长为 62.8 米的圆形花坛，剩下的空地种草坪，草坪的面积是多少平方米？",
+    "options": [],
+    "answer": "766 平方米",
+    "solution": "【分析】利用圆的周长，先计算圆的半径，再用长方形面积减去圆的面积，得到草坪的面积。半径决定圆的大小, 已知半径可以确定周长、面积, 已知周长、面积, 也能确定半径。\n\n【详解】圆的半径：\n\n$62.8 \\div 2 \\div 3.14\\\\=31.4÷3.14\\\\=10 (米)$\n\n$36 \\times 30 - 3.14 \\times 10^{2}\\\\=1080-314\\\\=76$  (平方米)\n\n答: 草坪的面积是 766 平方米。",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级上·吉林通化·期末",
+    "merged_id": "unit5_396",
+    "original_id": "unit5_choice_396"
+  },
+  {
+    "id": "unit5_choice_397",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的面积",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（18-19 五年级下·江苏·单元测试）一个圆形花坛的直径是 12 米，在它的周围铺一条 1 米宽的小路，这条小路的面积是多少平方米？",
+    "options": [],
+    "answer": "40.82 平方米",
+    "solution": "【分析】大圆面积减去小圆面积，得到小路的面积。 $S = \\pi (R^{2} - r^{2})$ , 掌握圆环的面积公式是解题关键。\n\n【详解】  $12 \\div 2 = 6$  （米） $6 + 1 = 7 \\text {(米)}$  $3.1 4 \\times (7 ^ {2} - 6 ^ {2})-3.14 \\times 1 3= 4 0. 8 2 \\text {(平 方 米)}$\n\n答: 这条小路的面积是 40.82 平方米。",
+    "knowledge_points": [
+      "圆环的面积",
+      "圆的面积及应用"
+    ],
+    "source": "18-19 五年级下·江苏·单元测试",
+    "merged_id": "unit5_397",
+    "original_id": "unit5_choice_397"
+  },
+  {
+    "id": "unit5_choice_398",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的面积",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（19-20六年级上·江西宜春·期末）有一个圆形花坛，直径是  $16 \\mathrm{~m}$ ，在它的周围修建一条  $2 \\mathrm{~m}$  宽的小路。(1) 这条小路的面积是多少？(2) 沿环形小路的（靠内侧）边缘每隔  $5 \\mathrm{~m}$  装一盏灯，一共约可安装多少盏灯？",
+    "options": [],
+    "answer": "(1)  $113.04 \\mathrm{~m}^{2}$   (2) 10 盏",
+    "solution": "【分析】（1）由题意可知，就是求圆环的面积，先求出大圆的半径，即  $16 \\div 2 + 2$ ，再根据 ' $S_{\\text {环形}} = \\pi (R^{2} - r^{2})$ ' 进行解答即可；(2) 封闭图形植树属于只栽一端的情况, 棵数等于间隔数;用周长  $\\div$  间隔长度即可求出间隔数, 也就是植树的棵数, 据此解答即可。熟练掌握圆环的面积公式以及封闭图形植树的特点是解答本题的关键。\n\n【详解】 (1)  $16 \\div 2 = 8$  （米）；8 + 2 = 10（米）；  $3.14× (10^{2}-8^{2})  = 3.14 \\times 36  = 113.04 (平方米)$答: 这条小路的面积是 113.04 平方米。(2)  $3.14 \\times 16 \\div 5  = 50.24 \\div 5\\approx 10$（盏）；答: 一共约可安装 10 盏灯。",
+    "knowledge_points": [
+      "圆环的面积",
+      "圆的周长及应用",
+      "植树问题"
+    ],
+    "source": "19-20六年级上·江西宜春·期末",
+    "merged_id": "unit5_398",
+    "original_id": "unit5_choice_398"
+  },
+  {
+    "id": "unit5_choice_399",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的面积",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（2021 六年级上·全国·专题练习）如果将一根  $6 \\mathrm{~m}$  长的拴马绳加长  $2 \\mathrm{~m}$ ，马儿可多吃多大面积的草？",
+    "options": [],
+    "answer": "$87.92 \\mathrm{~m}^{2}$",
+    "solution": "【分析】多吃的面积是一个圆环, 求出内外圆半径, 根据圆环面积  $= \\pi (\\mathrm{R}^{2} - \\mathrm{r}^{2})$ , 列式解答即可。关键是掌握圆环面积公式，栓马绳的长度相当于圆的半径。\n\n【详解】  $6 + 2 = 8$  （米）\n\n$ 3. 1 4 \\times (8 ^ {2} - 6 ^ {2}) \\\\ = 3. 1 4 \\times 2 8 \\\\ = 8 7. 9 2 \\text {(平 方 米)} $\n答: 马儿可多吃  $87.92 \\mathrm{~m}^{2}$  的草。",
+    "knowledge_points": [
+      "圆环的面积"
+    ],
+    "source": "2021 六年级上·全国·专题练习",
+    "merged_id": "unit5_399",
+    "original_id": "unit5_choice_399"
+  },
+  {
+    "id": "unit5_choice_400",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（20-21六年级上·全国·期末）梅园新村新建一个圆形喷水池，其直径是  $4 \\mathrm{~m}$ ，小齐绕着喷水池走了 3 圈，共走多少米？这喷水池的占地面积是多少平方米？",
+    "options": [],
+    "answer": "37.68 米；12.56 平方米",
+    "solution": "【分析】要求 3 圈是多少米, 可以先求出一圈的长度, 即圆的周长。求这个喷水池的占地面积可以直接套用圆面积公式。本题综合考查了有关圆的周长、圆的面积的应用,牢记其相关公式是解题关键。\n\n【详解】  $3.14 \\times 4 \\times 3\\\\= 1 2. 5 6 \\times 3\\\\=37.68 (米)$\n\n$3.1 4 \\times (4 \\div 2) ^ {2}\\\\= 3. 1 4 \\times 4\\\\= 12.56（平方米）$  \n\n答: 走了 3 圈, 共走 37.68 米; 这喷水池的占地面积是 12.56 平方米。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "20-21六年级上·全国·期末",
+    "merged_id": "unit5_400",
+    "original_id": "unit5_choice_400"
+  },{
+    "id": "unit5_choice_401",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的面积",
+    "question_type": "选择题",
+    "difficulty": "困难",
+    "question": "（2021 六年级上·全国·专题练习）一个圆环的内圆半径是外圆半径的  $\\frac{1}{2}$ , 这个圆环的面积是内圆面积的 ( )。",
+    "options": [
+      "A. 4 倍",
+      "B. 2",
+      "C. 3 倍"
+    ],
+    "answer": "C",
+    "solution": "【分析】通过内圆半径是外圆半径的  $\\frac{1}{2}$ , 确定内圆半径和外圆半径的比是  $1: 2$ , 根据平方以后的半径比是面积比, 确定内外圆的面积比, 用 (外圆面积 - 内圆面积) ÷ 内圆面积即可。关键是理解比的意义，掌握圆环面积  $=$  外圆面积一内圆面积。\n\n【详解】通过分析, 内外圆面积比是  $1: 2^{2}, \\quad (2^{2} - 1) \\div 1 = 3 \\div 1 = 3$ , 圆环的面积是内圆面积的 3 倍。故答案为: C",
+    "knowledge_points": [
+      "圆环的面积",
+      "比的意义"
+    ],
+    "source": "2021 六年级上·全国·专题练习",
+    "merged_id": "unit5_401",
+    "original_id": "unit5_choice_401"
+  },
+  {
+    "id": "unit5_choice_402",
+    "unit": "第五单元：圆",
+    "chapter": "半圆的周长",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "（19-20六年级上·河北·单元测试）把一个直径是  $4 \\mathrm{~cm}$  的圆形纸片剪成两个半圆后每个半圆形纸片的周长是（）  $\\mathrm{cm}$  。",
+    "options": [
+      "A. 12.56",
+      "B. 6.28",
+      "C. 10.28"
+    ],
+    "answer": "C",
+    "solution": "【分析】半圆形纸片的周长等于圆周长的一半加上一条直径, 由此解答即可。明确半圆的周长公式是解答本题的关键。\n\n【详解】  $3.14 \\times 4 \\div 2 + 4\\\\=6.2 8 + 4 \\\\ = 1 0. 2 8 \\text {(厘 米)}; $\n\n故答案为: C。",
+    "knowledge_points": [
+      "半圆的周长"
+    ],
+    "source": "19-20六年级上·河北·单元测试",
+    "merged_id": "unit5_402",
+    "original_id": "unit5_choice_402"
+  },
+  {
+    "id": "unit5_choice_403",
+    "unit": "第五单元：圆",
+    "chapter": "求最大面积",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "(19-20 六年级上·四川成都·期末) 用同样长的铁丝围成一个面积最大的形状, 应当围成 ( )。",
+    "options": [
+      "A. 等边三角形",
+      "B. 正方形",
+      "C. 长方形",
+      "D. 圆"
+    ],
+    "answer": "D",
+    "solution": "【分析】等边三角形，正方形，长方形，圆的周长相等，要找出面积最大的一个，可两两比较，确定出面积最大的图形。正 n 变形的周长相等，n 越大，图形越趋近于圆，面积也就越大。\n\n【详解】周长相等的长方形和正方形，正方形面积较大，排除C选项；周长相等的正多边形，边的条数越多，面积越大，即正方形面积大于等边三角形的面积，排除A选项；圆可以看做是正无数边形，在周长相等的条件下，圆的面积大于正方形的面积，排除B选项；故在周长相等的条件下, 圆的面积最大, 故答案选  $\\mathrm{D}$  。",
+    "knowledge_points": [
+      "求最大面积"
+    ],
+    "source": "19-20 六年级上·四川成都·期末",
+    "merged_id": "unit5_403",
+    "original_id": "unit5_choice_403"
+  },
+  {
+    "id": "unit5_choice_404",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "(20-21 六年级上·全国·期末) 用 18.84 米篱包围成一块菜地, 围成 ( ) 的面积最大。",
+    "options": [
+      "A. 长方形",
+      "B. 正方形",
+      "C. 圆",
+      "D. 平行四边形"
+    ],
+    "answer": "C",
+    "solution": "【分析】周长相等的平面图形中, 圆的面积最大, 据此解题即可。本题考查了平面图形的面积, 明确周长一致的情况下圆的面积最大是解题的关键。\n\n【详解】围成圆的面积最大。故答案为: C。",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级上·全国·期末",
+    "merged_id": "unit5_404",
+    "original_id": "unit5_choice_404"
+  },
+  {
+    "id": "unit5_choice_405",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的面积",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·全国·单元测试) 一个圆环, 外圆半径是内圆半径的 2 倍, 则圆环面积是内圆面积的 ( )。",
+    "options": [
+      "A.  $\\frac{1}{2}$",
+      "B. 3 倍",
+      "C.  $\\frac{1}{3}$"
+    ],
+    "answer": "B",
+    "solution": "【分析】假设内圆半径为 1 , 则外圆半径是 2 , 根据 '  $s_{\\text {环}} = \\pi (r_{1}^{2} - r_{2}^{2})$  ' 求出圆环面积, 进而求出圆环面积是内圆面积的几分之几即可。熟练掌握圆环的面积公式是解答本题的关键。\n\n【详解】假设内圆半径为 1 , 则外圆半径是 2 ;\n\n$3.1 4 \\times (2 ^ {2} - 1 ^ {2}) \\div (3. 1 4 \\times 1 ^ {2}) \\\\ = 3. 1 4 \\times 3 \\div 3. 1 4 \\\\ = 3$\n\n故答案为: B",
+    "knowledge_points": [
+      "圆的面积及应用",
+      "圆环的面积"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_405",
+    "original_id": "unit5_choice_405"
+  },
+  {
+    "id": "unit5_choice_406",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的面积",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "（20-21 六年级上·全国·单元测试）小明画了一个半径是 2 厘米的圆，老师让他把半径增加到 5 厘米，它的面积比原来增加了（）平方厘米。",
+    "options": [
+      "A.  $9 \\pi$",
+      "B.  $21 \\pi$",
+      "C.  $25 \\pi$"
+    ],
+    "answer": "B",
+    "solution": "【分析】由题意可知, 就是求圆环的面积, 根据 ' $\\mathrm{s}_{\\text {环}} = \\pi (\\mathrm{r}_{1}^{2} - \\mathrm{r}_{2}^{2})$ ' 进行解答即可。明确本题就是圆环的面积是解答本题的关键。\n\n【详解】  $(5^{2} - 2^{2})\\pi = 21\\pi ;$故答案为: B。",
+    "knowledge_points": [
+      "圆环的面积"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_406",
+    "original_id": "unit5_choice_406"
+  },
+  {
+    "id": "unit5_choice_407",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "(20-21 六年级上·全国·单元测试) 小明买了一个台玩具越野车，前轮直径是后轮的一半，后轮滚动6圈，前轮要滚动（）圈。",
+    "options": [
+      "A. 3",
+      "B. 12",
+      "C. 9"
+    ],
+    "answer": "B",
+    "solution": "【分析】假设后轮直径为 2 , 则前轮直径是 1 , 求出后轮滚动 6 圈的路程再除以前轮的周长即可解答。本题采用了假设法，假设法使题目变得具体化，简单化。\n\n【详解】  $3.14 \\times 2 \\times 6 \\div (3.14 \\times 1)\\\\= 3 7. 6 8 \\div 3. 1 4\\\\= 12（圈）$\n\n故答案为: B。",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_407",
+    "original_id": "unit5_choice_407"
+  },
+  {
+    "id": "unit5_choice_408",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·全国·单元测试) 下列说法正确的是 ( ) 。",
+    "options": [
+      "A. 半径是直径的一半",
+      "B. 圆的半径扩大到原来的1.5倍，其面积也扩大到原来的1.5倍",
+      "C. 圆的周长与半径的比是  $2 \\pi: 1$",
+      "D.  $2 \\pi \\mathrm{r}$  与  $\\pi \\mathrm{r}^{2}$  表示的意义相同"
+    ],
+    "answer": "C",
+    "solution": "【分析】A．同圆或等圆内，半径是直径的一半，由此解答即可；B. 假设圆的面积为 1 , 则扩大后的半径为 1.5 , 分别计算出前后的面积再进行判断即可;C. 根据 'C=2πr'求出圆的周长与半径的关系;D.  $2 \\pi r$  表示圆的周长,  $\\pi r^{2}$  表示圆的面积, 由此判断即可。本题综合性较强，熟练掌握有关圆的基础知识是关键。\n\n【详解】A．同圆或等圆内，半径是直径的一半，原题说法错误；  B. 假设圆的面积为 1 , 则扩大后的半径为 1.5 ; \n$3.14 \\times 1.5^{2} \\div 3.14 \\times 1^{2} = 1.5^{2} \\div 1^{2} = 2.25$ , 原题说法错误;  C. 根据 ' $C = 2 \\pi r$ ' 可知, 圆的周长与半径的比是  $2 \\pi: 1$ , 原题说法正确;  D.  $2 \\pi r$  表示圆的周长,  $\\pi r^{2}$  表示圆的面积, 它们的意义不同。故答案为: C。",
+    "knowledge_points": [
+      "圆的面积及应用",
+      "圆的概念及特点",
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_408",
+    "original_id": "unit5_choice_408"
+  },
+  {
+    "id": "unit5_choice_409",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "(20-21 六年级上·全国·单元测试) 小圆的直径是 5 厘米,大圆的直径是 10 厘米, 大圆面积是小圆面积的 ( ) 倍。",
+    "options": [
+      "A. 8",
+      "B. 4",
+      "C. 2"
+    ],
+    "answer": "B",
+    "solution": "【分析】根据  $s = \\pi r^{2}$  求出两圆的面积, 再进行解答即可。面积比是直径比的平方。\n\n【详解】  $3.14 \\times (5 \\div 2)^{2}= 3. 1 4 \\times 6. 2 5=19.625（平方厘米）；$$3.1 4 \\times (1 0 \\div 2) ^ {2}=3.1 4 \\times 2 5= 78.5（平方厘米）$；$7 8. 5 \\div 1 9. 6 2 5 = 4;$故答案为: B。",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_409",
+    "original_id": "unit5_choice_409"
+  },
+  {
+    "id": "unit5_choice_410",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "（20-21六年级上·全国·单元测试）小明画了两个圆，它们的面积不相等，是因为它们的（）。",
+    "options": [
+      "A. 圆心位置不一样",
+      "B. 直径不相等",
+      "C. 圆周率不一样"
+    ],
+    "answer": "B",
+    "solution": "【分析】圆心决定圆的位置，半径和直径决定圆的大小，由此解答即可；本题属于基础性题目, 一定要熟练掌握有关圆的基础知识。\n\n【详解】小明画了两个圆，它们的面积不相等，是因为它们的半径和直径不相等；故答案为：B。",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "20-21六年级上·全国·单元测试",
+    "merged_id": "unit5_410",
+    "original_id": "unit5_choice_410"
+  },
+  {
+    "id": "unit5_choice_411",
+    "unit": "第五单元：圆",
+    "chapter": "扇形的周长和面积",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·全国·单元测试) 同圆中扇形甲的弧长是扇形乙的弧长的  $\\frac{1}{6}$ , 那么扇形乙的面积是扇形甲面积的 ( )。",
+    "options": [
+      "A. 36 倍",
+      "B. 12 倍",
+      "C. 6 倍",
+      "D. 3 倍"
+    ],
+    "answer": "C",
+    "solution": "【分析】根据扇形面积  $= \\pi r^{2} \\times \\frac{\\text { 圆心角 }}{360^{\\circ}}$ , 同圆中扇形甲的弧长是扇形乙的弧长的  $\\frac{1}{6}$ , 说明扇形甲的圆心角是扇形乙的圆心角的  $\\frac{1}{6}$ , 扇形甲的面积是扇形乙的  $\\frac{1}{6}$ , 反过来确定扇形乙的面积是扇形甲的倍数即可。关键是掌握扇形的面积公式，扇形面积根据圆的面积来求。\n\n【详解】根据分析, 同圆中扇形甲的弧长是扇形乙的弧长的  $\\frac{1}{6}$ , 那么扇形乙的面积是扇形甲面积的 6 倍。",
+    "knowledge_points": [
+      "扇形的周长和面积"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_411",
+    "original_id": "unit5_choice_411"
+  },
+  {
+    "id": "unit5_choice_412",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "（20-21 六年级上·全国·单元测试）一辆自行车车轮直径是 1 米，如果平均每分钟转 100 周。通过一座 628 米的大桥，经过大桥大约需要（）。",
+    "options": [
+      "A. 1 分钟",
+      "B. 0.5 分钟",
+      "C. 2 分钟",
+      "D. 3 分钟"
+    ],
+    "answer": "C",
+    "solution": "【分析】根据  $C = \\pi d$  先求出自行车车轮的周长, 周长  $\\times 100 =$  速度, 再根据: 路程  $\\div$  速度  $=$  时间, 即可求出答案。灵活运用圆的周长公式和路程、速度、时间的关系解决问题。\n\n【详解】  $628 \\div (3.14 \\times 1 \\times 100)\\\\= 6 2 8 \\div 3 1 4\\\\= 2\\text {(分 钟)}$\n\n故选: C。",
+    "knowledge_points": [
+      "基础行程问题",
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_412",
+    "original_id": "unit5_choice_412"
+  },
+  {
+    "id": "unit5_choice_413",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "(20-21 六年级上·全国·单元测试)一台拖拉机的前轮半径是后轮半径的  $\\frac{1}{2}$ , 后轮转 4 圈, 前轮转 ( ) 圈。",
+    "options": [
+      "A. 2",
+      "B. 4",
+      "C. 8"
+    ],
+    "answer": "C",
+    "solution": "【分析】假设后轮的半径为 2 , 则前轮的半径为 1 , 求出后轮转到 4 圈走过的路程再除以前轮的周长即可求出前轮转多少圈。熟练掌握圆的周长公式是解答本题的关键。\n\n【详解】假设后轮的半径为2，则前轮的半径为1；\n\n$2 \\times 3. 1 4 \\times 2 \\times 4 \\div (2 \\times 3. 1 4 \\times 1) \\\\ = 1 2. 5 6 \\times 4 \\div 6. 2 8 \\\\ = 8 \\text {(圈)} $\n\n故答案为: C。",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_413",
+    "original_id": "unit5_choice_413"
+  },
+  {
+    "id": "unit5_choice_414",
+    "unit": "第五单元：圆",
+    "chapter": "圆的概念及特点",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "（20-21 六年级上·全国·单元测试）在长  $12 \\mathrm{~cm}$  、宽  $8 \\mathrm{~cm}$  的长方形最多能剪出（）个直径是  $4 \\mathrm{~cm}$  的圆。",
+    "options": [
+      "A. 8",
+      "B. 6",
+      "C. 9"
+    ],
+    "answer": "B",
+    "solution": "【分析】用  $12 \\div 4$  求出沿长可以剪多少个, 用  $8 \\div 4$  求出沿宽可以剪多少个, 进而求出总共剪多少个即可。明确沿长和沿宽各剪多少个是解答本题的关键。\n\n【详解】  $(12 \\div 4) \\times (8 \\div 4)= 3 \\times 2= 6(个)$故答案为: B。",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_414",
+    "original_id": "unit5_choice_414"
+  },
+  {
+    "id": "unit5_choice_415",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·全国·单元测试) 两个圆直径的比是  3:2，那么它们周长的比是（）。",
+    "options": [
+      "A.  $3: 2$",
+      "B. 6:4",
+      "C.  $8: 4$",
+      "D. 9:4"
+    ],
+    "answer": "A",
+    "solution": "【分析】圆的周长  $= \\pi d$ , 两个圆的直径之比等于两圆的周长之比, 据此解答即可。本题考查圆的周长、比的概念, 解答本题的关键是掌握圆的周长公式。\n\n【详解】根据分析可得，两个圆直径的比是  $3:2$  ，那么它们周长的比是  $3: 2$  。故答案为: A。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "比的意义"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_415",
+    "original_id": "unit5_choice_415"
+  },
+  {
+    "id": "unit5_choice_416",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "（20-21 六年级上·全国·单元测试）一个圆的周长和它的直径的比是（）。",
+    "options": [
+      "A.  $3: 1$",
+      "B. 2:1",
+      "C.  $\\pi : 1$",
+      "D.  $1: \\pi$"
+    ],
+    "answer": "C",
+    "solution": "【分析】圆的周长为  $C = \\pi d$ , 据此写出圆的周长和它的直径的比即可。熟练掌握圆的周长公式是解答本题的关键。\n\n【详解】一个圆的周长和它的直径的比是  $\\pi d: d = \\pi: 1$ ; 故答案为: C。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "比的意义"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_416",
+    "original_id": "unit5_choice_416"
+  },
+  {
+    "id": "unit5_choice_417",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·全国·单元测试) 两根都是长 6.28 厘米的铁丝分别围成一个正方形和一个圆，比较围成的这两个图形的面积，（）。",
+    "options": [
+      "A. 正方形的大",
+      "B. 圆的大",
+      "C. 它们同样大",
+      "D. 无法比较"
+    ],
+    "answer": "B",
+    "solution": "【分析】由题意可知, 正方形和圆形的周长都为 6.28 厘米,则围成正方形的边长为:  $6.28 \\div 4 = 1.57$  (厘米); 面积:$1.57 \\times 1.57$ ; 围成圆的半径:  $r = 6.28 \\div 3.14 \\div 2 = 1$  (厘米);面积:  $3.14 \\times 1^{2}$ ; 由此分别计算出它们的面积再进行比较即可。明确正方形和圆形的周长是解答本题的关键。\n\n【详解】  $6.28 \\div 4 = 1.57$  （厘米）；$1.57 \\times 1.57 = 2.4649$  （平方厘米）；$6.28 \\div 3.14 \\div 2  = 2 \\div 2  = 1（厘米)$；$3.14 \\times 1^{2} = 3.14$  （平方厘米）；  $3.14 > 2.4649$ ;故答案为: B。",
+    "knowledge_points": [
+      "圆的面积及应用",
+      "正方形的面积",
+      "圆的周长及应用",
+      "正方形的周长"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_417",
+    "original_id": "unit5_choice_417"
+  },
+  {
+    "id": "unit5_choice_418",
+    "unit": "第五单元：圆",
+    "chapter": "圆的概念及特点",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "(18-19 二年级上·浙江·课后作业) 在一块长  $10 \\mathrm{dm}$ , 宽  $6 \\mathrm{dm}$  的长方形铁板上, 最多能截取( )个直径是  $2 \\mathrm{dm}$  的圆形铁板。",
+    "options": [],
+    "answer": "15",
+    "solution": "【分析】用  $10 \\div 2$  求出一行可以截取几个, 再用  $6 \\div 2$  求出可以截取几行，再相乘即可。明确一行可以截取几个，可以截取几行是解答本题的关键。\n\n【详解】  $(10 \\div 2) \\times (6 \\div 2)\\\\=5 \\times 3 \\\\ = 1 5 (\\text {个}) $",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "18-19 二年级上·浙江·课后作业",
+    "merged_id": "unit5_418",
+    "original_id": "unit5_choice_418"
+  },
+  {
+    "id": "unit5_choice_419",
+    "unit": "第五单元：圆",
+    "chapter": "圆的概念及特点",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "(20-21 六年级上·河北·单元测试) 圆是____图形, 它有____条对称轴。",
+    "options": [],
+    "answer": "轴对称 无数",
+    "solution": "【分析】一个图形沿一条直线对折，直线两旁的图形完全重合，这样的图形叫做轴对称图形，折痕所在的直线就是对称轴，由此解答即可。明确圆的特征是解答本题的关键。\n\n【详解】圆是轴对称图形，它有无数条对称轴。",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "轴对称的认识及辨认",
+      "对称轴的画法及数量"
+    ],
+    "source": "20-21 六年级上·河北·单元测试",
+    "merged_id": "unit5_419",
+    "original_id": "unit5_choice_419"
+  },
+  {
+    "id": "unit5_choice_420",
+    "unit": "第五单元：圆",
+    "chapter": "圆的概念及特点",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "(2014 六年级·全国·课后作业) 一个圆有( )条半径, ( )条直径。",
+    "options": [],
+    "answer": "无数 无数",
+    "solution": "【详解】直径是通过圆心并且两端都在圆上的线段；半径是连接圆心和圆上任意一点的线段。一个圆有无数条半径，无数条直径。",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "2014 六年级·全国·课后作业",
+    "merged_id": "unit5_420",
+    "original_id": "unit5_choice_420"
+  },
+  {
+    "id": "unit5_choice_421",
+    "unit": "第五单元：圆",
+    "chapter": "圆的概念及特点",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·全国·单元测试) 若一个正方形的面积是 64 平方厘米, 那么在这个正方形内可以画( )个直径是 2 厘米的圆。",
+    "options": [],
+    "answer": "16",
+    "solution": "【分析】由题意可知, 正方形的边长为 8 厘米, 用正方形的边长除以圆的直径即可求出一排可以画几个, 可以画几排,进而求出总共可以画多少个。明确一排可以画几个，可以画几排是解答本题的关键。\n\n【详解】由题意可知，正方形的边长为8厘米； $8 \\div 2 = 4$  （个）； $4 \\times 4 = 16$  （个）",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_421",
+    "original_id": "unit5_choice_421"
+  },
+  {
+    "id": "unit5_choice_422",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "(20-21 六年级上·全国·单元测试) 在一个长  $16 \\mathrm{~cm}$ , 宽  $12 \\mathrm{~cm}$  的长方形内, 画一个最大的圆, 其面积是( )  $\\mathrm{cm}^{2}$ 。",
+    "options": [],
+    "answer": "113.04",
+    "solution": "【分析】长方形内画一个最大的圆，圆的直径和长方形的宽相等，根据  $s = \\pi r^{2}$  求出圆的面积即可。明确长方形内画一个最大的圆，圆的直径和长方形的宽相等是解答本题的关键。\n\n【详解】  $3.14 \\times (12 \\div 2)^{2}\\\\= 3. 1 4 \\times 3 6\\\\=113.04 (平方厘米)$",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_422",
+    "original_id": "unit5_choice_422"
+  },
+  {
+    "id": "unit5_choice_423",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·全国·单元测试) 在一个圆中, 其周长是  $28.26 \\mathrm{~cm}$ , 它的直径是( )  $\\mathrm{cm}$ , 半径是( )  $\\mathrm{cm}$ , 面积是()  $\\mathrm{cm}^{2}$  。",
+    "options": [],
+    "answer": "9    4.5     63.585",
+    "solution": "【分析】根据圆的周长公式以及这个圆的周长  $28.26 \\mathrm{~cm}$ , 先后求出圆的半径和直径, 再求出圆的面积即可。本题考查了圆的面积和周长, 熟练掌握公式并会用其计算相应的半径、周长和面积是解题的关键。\n\n【详解】半径:  $28.26 \\div (2 \\times 3.14) = 4.5 \\, (\\mathrm{cm})$   直径:  $4.5 \\times 2 = 9 \\text{ (cm)}$    面积:  $3.14 \\times 4.5^{2} = 63.585 (\\mathrm{~cm}^{2})$",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_423",
+    "original_id": "unit5_choice_423"
+  },
+  {
+    "id": "unit5_choice_424",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "(20-21 六年级上·全国·单元测试) ( )和( )的比值叫圆周率, 用字母( )表示, 它的近似值是( )。",
+    "options": [],
+    "answer": "周长  直径   $\\pi$    3.14",
+    "solution": "【详解】根据圆周率的含义解答。周长和直径的比值叫圆周率，用字母  $\\pi$  表示，它的近似值是  $3.14$ 。",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_424",
+    "original_id": "unit5_choice_424"
+  },
+  {
+    "id": "unit5_choice_425",
+    "unit": "第五单元：圆",
+    "chapter": "圆的概念及特点",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "（20-21六年级上·全国·单元测试）小明画了一个圆，在这个圆中所有半径都( )，所有的直径都( )，而且直径的长度是半径的( )倍。",
+    "options": [],
+    "answer": "相等  相等 2",
+    "solution": "【详解】小明画了一个圆，在这个圆中所有半径都相等，所有的直径都相等，而且直径的长度是半径的2倍。",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "20-21六年级上·全国·单元测试",
+    "merged_id": "unit5_425",
+    "original_id": "unit5_choice_425"
+  },
+  {
+    "id": "unit5_choice_426",
+    "unit": "第五单元：圆",
+    "chapter": "圆的概念及特点",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "(20-21 六年级上·全国·单元测试) 王同学在一个圆中可以画( )条直径和半径, 且这些直径都( ), 半径也( ), 直径是半径的( )。",
+    "options": [],
+    "answer": "无数   相等   相等   2倍",
+    "solution": "【详解】根据圆的特征及在同一个圆中半径与直径的关系即可作答。王同学在一个圆中可以画无数条直径和半径，且这些直径都相等，半径也相等，直径是半径的2倍。",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_426",
+    "original_id": "unit5_choice_426"
+  },
+  {
+    "id": "unit5_choice_427",
+    "unit": "第五单元：圆",
+    "chapter": "圆的概念及特点",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "（20-21 六年级上·全国·单元测试）连接圆心和圆上任意一个点的线段称为圆的（ ），两端都在圆上且通过圆心的线段是（ ）。",
+    "options": [],
+    "answer": "半径 直径",
+    "solution": "【详解】如图所示：![](images/182217d9547a868485067d595faf3d09286a33f6db72587da375d5dd4b16d6b8.jpg)圆心: 圆中心的一点叫做圆心;半径：连接圆心到圆上任意一点的线段叫做半径；直径：通过圆心，并且两端都在圆上的线段叫做圆的直径。",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_427",
+    "original_id": "unit5_choice_427"
+  },
+  {
+    "id": "unit5_choice_428",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "(20-21 六年级上·全国·单元测试) 圆的 ( ) 与 ( )的比值是一个固定的数，我们把它叫做圆周率。",
+    "options": [
+      "A. 面积，直径",
+      "B. 面积, 半径",
+      "C. 周长, 半径径",
+      "D. 周长, 直径"
+    ],
+    "answer": "D",
+    "solution": "【详解】圆的周长与直径的比值是一个固定的数，我们把它叫做圆周率。故答案为: D。",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_428",
+    "original_id": "unit5_choice_428"
+  },
+  {
+    "id": "unit5_choice_429",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "(20-21 六年级上·全国·单元测试) 一个车轮的外直径是 1.5 米, 它向前滚动 10 周能前进( )米。",
+    "options": [],
+    "answer": "47.1",
+    "solution": "【分析】用  $3.14 \\times 1.5$  求出车轮的周长, 再乘 10 周即可求出向前滚动 10 周前进的距离。熟练掌握圆的周长公式是解答本题的关键。\n\n【详解】  $3.14 \\times 1.5 \\times 10\\\\=4.7 1 \\times 1 0 \\\\ = 4 7. 1 \\text {(米)} $",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_429",
+    "original_id": "unit5_choice_429"
+  },
+  {
+    "id": "unit5_choice_430",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "（20-21 六年级上·全国·单元测试）一个半圆形花坛的直径是 4 米，这个花坛的周长是( )米，面积是( )平方米。",
+    "options": [],
+    "answer": "10.28 6.28",
+    "solution": "【分析】半圆形的周长等于圆周长的一半加上直径；半圆形的面积等于圆的面积除以 2，由此解答即可。熟练掌握半圆形的周长和面积的计算方法是解答本题的关键。\n\n【详解】  $3.14 \\times 4 \\div 2 + 4\\\\=6.2 8 + 4 \\\\ = 1 0. 2 8 \\text {(米)} \\\\ $\n\n$3.1 4 \\times (4 \\div 2) ^ {2} \\div 2 \\\\ = 3. 1 4 \\times 4 \\div 2 \\\\ = 6. 2 8 \\text {(平 方 米)}  $",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_430",
+    "original_id": "unit5_choice_430"
+  },
+  {
+    "id": "unit5_choice_431",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的面积",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "（20-21 六年级上·全国·单元测试）一个环形纸片，它的内圆直径是 10 厘米，外圆直径 12 厘米，它的面积是( )平方厘米。",
+    "options": [],
+    "answer": "34.54",
+    "solution": "【分析】由题意可知, 就是求圆环的面积根据 'S 环形  $=$ $\\pi (\\mathrm{R}^{2}-\\mathrm{r}^{2})$ ' 进行解答即可,熟练掌握求圆环面积的公式是解答本题的关键。\n\n【详解】  $10 \\div 2 = 5$  （厘米）； $12 \\div 2 = 6$  （厘米）；  $3.14 \\times  (6^{2}-5^{2}) =3.14 \\times 11 = 34.54（平方厘米）$",
+    "knowledge_points": [
+      "圆环的面积"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_431",
+    "original_id": "unit5_choice_431"
+  },
+  {
+    "id": "unit5_choice_432",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "(20-21 六年级上·全国·单元测试) 圆的半径由 4 厘米增加到 7 厘米, 面积增加了( )  $\\mathrm{cm}^{2}$  。",
+    "options": [],
+    "answer": "103.62",
+    "solution": "【分析】根据圆的面积公式  $s = \\pi r^{2}$ , 分别计算出两次面积,再比较即可。熟练掌握圆的面积公式是解答本题的关键。\n\n【详解】  $3.14 \\times 4^{2}\\\\=3.1 4 \\times 1 6 \\\\ = 5 0. 2 4 \\text {(平 方 厘 米)}$ \n\n$ 3. 1 4 \\times 7 ^ {2} \\\\ = 3. 1 4 \\times 4 9 \\\\ = 1 5 3. 8 6 \\text {(平 方 厘 米)}$\n\n$1 5 3. 8 6 - 5 0. 2 4 = 1 0 3. 6 2 \\text {(平 方 厘 米)}  $",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_432",
+    "original_id": "unit5_choice_432"
+  },
+  {
+    "id": "unit5_choice_433",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "（20-21 六年级上·全国·单元测试）把一个圆平均分成若干（偶数）等份，分得份数越多，拼成的图形越接近长方形，拼成的长方形与原图形比较，周长( )，面积( )。",
+    "options": [],
+    "answer": "变大 不变",
+    "solution": "【分析】把圆形拼成一个长方形后，周长多了两条半径，面积不变，由此解答即可。熟练掌握圆的面积推导过程是解答本题的关键。\n\n【详解】把一个圆平均分成若干（偶数）等份，分得份数越多，拼成的图形越接近长方形，拼成的长方形与原图形比较，周长变大，面积不变。",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_433",
+    "original_id": "unit5_choice_433"
+  },
+  {
+    "id": "unit5_choice_434",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "（20-21 六年级上·全国·单元测试）小冰家里的一张圆形的饭桌，饭桌面的周长是 37.68 分米，饭桌面的面积是( )平方分米。",
+    "options": [],
+    "answer": "113.04",
+    "solution": "【分析】先根据圆的周长求出饭桌的半径，再根据圆的面积公式求出饭桌的面积。本题考查圆的周长、面积, 解答本题的关键是掌握圆的周长、面积公式。\n\n【详解】半径:  $37.68 \\div 3.14 \\div 2\\\\= 1 2 \\div 2\\\\=6 (分米)$\n\n面积： $3.14 \\times 6 \\times 6= 3. 1 4 \\times 3 6= 113.04（平方分米）$",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_434",
+    "original_id": "unit5_choice_434"
+  },
+  {
+    "id": "unit5_choice_435",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "（20-21 六年级上·全国·单元测试）用三根同样长的铁丝分别围成一个圆、一个长方形和一个正方形，其中____的面积最大。",
+    "options": [],
+    "answer": "圆",
+    "solution": "【分析】根据题意可知, 铁丝的长度是围成图形的周长,此题用举例法解答, 先假设正方形、长方形和圆形的周长都是 16 米, 分别求出圆、正方形、长方形的面积, 然后比较大小即可。所以周长相等，圆面积  $>$  正方形面积  $>$  长方形面积。\n\n【详解】假设正方形、长方形和圆形的周长都是 16 米, 则圆的面积为:  $\\pi \\times (\\frac{16}{2 \\times 3.14})^{2} \\approx 20.38$  (平方米); 正方形的边长为:  $16 \\div 4 = 4$  (米), 面积为:  $4 \\times 4 = 16$  (平方米) ;长方形长、宽越接近，面积越大，就取长为5米、宽为3米，面积为:  $5 \\times 3 = 15$  (平方米),当长方形的长和宽最接近时面积也小于 16 平方米;所以周长相等的正方形、长方形和圆形，圆面积最大。",
+    "knowledge_points": [
+      "圆的面积及应用",
+      "求最大面积"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_435",
+    "original_id": "unit5_choice_435"
+  },
+  {
+    "id": "unit5_choice_436",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·全国·单元测试) 一个正方形的边长和一个圆的半径相等，已知正方形的面积是20平方分米，圆的面积是____平方分米。",
+    "options": [],
+    "answer": "62.8",
+    "solution": "【分析】根据正方形和圆的面积公式，直接用  $\\pi \\times$  正方形的面积  $=$  圆的面积。正方形面积  $= \\text{边长}^{2}$ , 圆的面积  $= \\pi r^{2}$  。\n\n【详解】  $3.14 \\times 20 = 62.8$  （平方分米）",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_436",
+    "original_id": "unit5_choice_436"
+  },
+  {
+    "id": "unit5_choice_437",
+    "unit": "第五单元：圆",
+    "chapter": "半圆的周长",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·全国·单元测试) 一个长方形长 10 厘米, 宽 8 厘米, 在里面剪一个最大的半圆, 这个半圆的周长是____厘米。",
+    "options": [],
+    "answer": "25.7",
+    "solution": "【分析】在一个长方形长 10 厘米, 宽 8 厘米中剪一个最大的半圆, 则半圆的直径为 10 厘米, 半圆的周长 = 圆的周长的一半 ( $\\pi \\times$  直径  $\\div 2$ ) + 直径。关键是确定半圆的直径。\n\n【详解】圆的直径为 10 厘米,$3.1 4 \\times 1 0 \\div 2 + 1 0 \\\\ = 1 5. 7 + 1 0 \\\\ = 2 5. 7 \\text {(厘 米)} $所以半圆的周长为 25.7 厘米。",
+    "knowledge_points": [
+      "半圆的周长"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_437",
+    "original_id": "unit5_choice_437"
+  },
+  {
+    "id": "unit5_choice_438",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·全国·单元测试) 一个半圆的周长是  25.7cm，这个半圆的面积是____$\\mathrm{cm}^2$ 。",
+    "options": [],
+    "answer": "39.25",
+    "solution": "【分析】设半圆半径是  $r$ , 根据半圆周长  $= \\pi r + + 2 r$ , 列出方程, 求出  $r$  的值, 再根据半圆的面积  $=$  圆的面积  $\\div 2$ , 计算即可。关键是掌握半圆周长和面积的求法，半圆周长包括圆周长的一半和一条直径。\n\n【详解】解：设半圆半径是  $r$ 。$3.14 \\mathrm{r} + 2 \\mathrm{r} = 25.7\\\\5.14 \\mathrm{r} = 25.7\\\\r=5$\n\n$3.14 \\times 5^{2} \\div 2 = 39.25$  （平方厘米）",
+    "knowledge_points": [
+      "圆的面积及应用",
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_438",
+    "original_id": "unit5_choice_438"
+  },
+  {
+    "id": "unit5_choice_439",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·全国·单元测试) 在一个长  $3 \\mathrm{dm}$  、宽  $2 \\mathrm{dm}$  的长方形里画一个最大的圆, 这个圆的面积是( )周长是( )。",
+    "options": [],
+    "answer": "3.14 平方分米 6.28 分米",
+    "solution": "【分析】在长方形里剪一个最大的圆, 圆的直径是长方形的宽, 据此根据圆的面积和周长公式计算即可。关键是掌握圆的周长和面积公式，圆的周长  $= \\pi d$ ，圆的面积  $= \\pi r^2$ 。\n\n【详解】  $2 \\div 2 = 1$  (分米)\n\n$3.14 \\times 1^{2} = 3.14$  （平方分米）  \n$3.14 \\times 2 = 6.28$  (分米)",
+    "knowledge_points": [
+      "圆的面积及应用",
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_439",
+    "original_id": "unit5_choice_439"
+  },
+  {
+    "id": "unit5_choice_440",
+    "unit": "第五单元：圆",
+    "chapter": "半圆的周长",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·全国·单元测试) 一个半圆形的半径是 3 米, 它的周长是( )米。",
+    "options": [],
+    "answer": "15.42",
+    "solution": "【分析】根据半圆周长  $= \\pi r + 2r$ , 列式计算即可。关键是想清楚半圆的周长包括圆周长的一半，还有一条直径。\n\n【详解】  $3.14 \\times 3 + 2 \\times 3\\\\=9.4 2 + 6 \\\\ = 1 5. 4 2 \\text {(米)} $",
+    "knowledge_points": [
+      "半圆的周长"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_440",
+    "original_id": "unit5_choice_440"
+  },
+  {
+    "id": "unit5_choice_441",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "（2014·全国·小升初真题）圆的周长与这个圆的直径的比是( )。",
+    "options": [],
+    "answer": "$\\pi : 1$",
+    "solution": "【分析】圆的周长是直径的  $\\pi$  倍, 据此解答即可。本题考查圆的周长、比的意义, 解答本题的关键是掌握圆的直径与周长的倍数关系。\n\n【详解】圆的周长与这个圆的直径的比是  $\\pi:1$ 。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "比的意义"
+    ],
+    "source": "2014·全国·小升初真题",
+    "merged_id": "unit5_441",
+    "original_id": "unit5_choice_441"
+  },
+  {
+    "id": "unit5_choice_442",
+    "unit": "第五单元：圆",
+    "chapter": "比的意义",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "（20-21 六年级上·全国·单元测试）把  $1.8: 0.6$  化成最简单的整数比是( )，比值是( )。",
+    "options": [],
+    "answer": "3:1 3",
+    "solution": "【分析】比的性质：比的前项和比的后项乘或除以同一个不为 0 的数，比值不变，据此解答即可。本题考查比的化简和求比值, 解答本题的关键是掌握比的性质。\n\n【详解】$1.8:0.6 \\\\=(1. 8 \\div 0. 6): (0. 6 \\div 0. 6) \\\\ = 3: 1 \\\\ = 3$",
+    "knowledge_points": [
+      "比的意义"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_442",
+    "original_id": "unit5_choice_442"
+  },
+  {
+    "id": "unit5_choice_443",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "(20-21 六年级上·全国·单元测试) 在一个圆中, 其半径是  $8 \\mathrm{~cm}$ , 它的直径是( )  $\\mathrm{cm}$ , 周长是( )  $\\mathrm{cm}$ , 面积是( )  $\\mathrm{cm}^{2}$  。",
+    "options": [],
+    "answer": "16 50.24 200.96",
+    "solution": "【分析】根据  $d = 2 r 、 C = 2 \\pi r 、 s = \\pi r^{2}$  进行解答即可。熟练圆的半径和直径的关系、周长和面积公式是解答本题的关键。\n\n【详解】  $8 \\times 2 = 16$  （厘米）；\n\n$2 \\times 3. 1 4 \\times 8 \\\\ = 6. 2 8 \\times 8 \\\\ = 5 0. 2 4 \\text {(厘 米)}$\n\n$3.1 4 \\times 8 ^ {2} \\\\ = 3. 1 4 \\times 6 4 \\\\ = 2 0 0. 9 6 \\text {(平 方 厘 米)} $",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_443",
+    "original_id": "unit5_choice_443"
+  },
+  {
+    "id": "unit5_choice_444",
+    "unit": "第五单元：圆",
+    "chapter": "圆的概念及特点",
+    "question_type": "填空题",
+    "difficulty": "困难",
+    "question": "(20-21 六年级上·全国·单元测试) 张玲在一个圆中可以画出( )条半径和( )条直径。",
+    "options": [],
+    "answer": "无数 无数",
+    "solution": "【分析】在一个圆中，半径和直径都有无数条，由此解答即可。明确圆的特点是解答本题的关键。\n\n【详解】张玲在一个圆中可以画出无数条半径和无数条直径。",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_444",
+    "original_id": "unit5_choice_444"
+  },
+  {
+    "id": "unit5_choice_445",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "（20-21 六年级上·全国·单元测试）张奶奶买了一个钟，分针长  $1 \\mathrm{dm}$ ，从 1 时到 2 时，分针针尖走过了（ ）cm；从 1 时到 2 时，分针扫过的面积是（ ）cm²。",
+    "options": [],
+    "answer": "62.8   314",
+    "solution": "【分析】由题意可知, 从 1 时到 2 时, 分针的针尖走过一个半径为 1 分米的圆, 根据 '  $C = 2 \\pi r$  ' 求出周长即可; 分针扫过的面积根据  $s = \\pi r^2$  求出面积即可。明确从 1 时到 2 时, 分针正好走一圈, 能够正确判断出针尖走过的轨迹是求圆的周长, 扫过的面积是求圆的面积是解答本题的关键。\n\n【详解】1分米  $= 10$  厘米$ 2 \\times 3. 1 4 \\times 1 0 \\\\ = 6. 2 8 \\times 1 0\\\\=62.8 (厘米) \\\\ 3. 1 4 \\times 1 0 ^ {2} \\\\ = 3. 1 4 \\times 1 0 0 \\\\ =314 (平方厘米)$",
+    "knowledge_points": [
+      "圆的面积及应用",
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_445",
+    "original_id": "unit5_choice_445"
+  },
+  {
+    "id": "unit5_choice_446",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的面积",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "（20-21六年级上·全国·单元测试）一个内圆半径是4dm，外圆直径是14dm的圆环，这个圆环的面积是( )dm²。",
+    "options": [],
+    "answer": "103.62",
+    "solution": "【分析】圆环的面积 = 大圆面积 - 小圆面积，根据圆的面积公式计算即可。本题考查圆环的面积, 解答本题的关键是掌握求圆环面积的方法。\n\n【详解】  $3.14 \\times (14 \\div 2)^{2} - 3.14 \\times 4^{2}\\\\=153.86-50.24\\\\=103.62 (平方分米)$",
+    "knowledge_points": [
+      "圆环的面积"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_446",
+    "original_id": "unit5_choice_446"
+  },
+  {
+    "id": "unit5_choice_447",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "（20-21 六年级上·全国·单元测试）一个圆的直径扩大到原来的 3 倍，半径扩大到原来的( )倍，周长扩大到原来的( )倍，面积扩大到原来的( )倍。",
+    "options": [],
+    "answer": "3  3   9",
+    "solution": "【分析】根据直径扩大的倍数等于半径扩大的倍数，等于周长扩大的倍数，平方以后的倍数是面积扩大的倍数，进行填空。关键是熟悉圆的特征，掌握圆的周长和面积公式。\n\n【详解】  $3^{2} = 9$ , 一个圆的直径扩大到原来的 3 倍, 半径扩大到原来的 3 倍, 周长扩大到原来的 3 倍, 面积扩大到原来的 9 倍。",
+    "knowledge_points": [
+      "圆的面积及应用",
+      "圆的概念及特点",
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_447",
+    "original_id": "unit5_choice_447"
+  },
+  {
+    "id": "unit5_choice_448",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·全国·单元测试) 小张画了一个周长为  $25.12 \\mathrm{~cm}$  的圆, 他画的圆的半径是( ) cm, 面积是( ) cm²。",
+    "options": [],
+    "answer": "4 50.24",
+    "solution": "【分析】圆的周长  $62 \\pi r$ , 据此求出圆的半径, 再根据圆的面积公式求出圆的面积即可。本题考查圆的周长、面积, 解答本题的关键是掌握圆的周长、面积公式。\n\n【详解】半径： $25.12 \\div 2 \\div 3.14\\\\=12.56÷3.14\\\\=4  (厘米)$\n\n面积： $3.14 \\times 4^{2}= 3.14 \\times 16=50.24 (平方厘米)$",
+    "knowledge_points": [
+      "圆的面积及应用",
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_448",
+    "original_id": "unit5_choice_448"
+  },
+  {
+    "id": "unit5_choice_449",
+    "unit": "第五单元：圆",
+    "chapter": "与圆相关的轴对称图形",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "(20-21 六年级上·全国·单元测试) 圆环有( )条对称轴, 半圆有( )条对称轴。",
+    "options": [],
+    "answer": "无数  1",
+    "solution": "【分析】一个图形沿一条直线对折，直线两旁的图形完全重合，这样的图形叫做轴对称图形，折痕所在的直线就是对称轴,明确轴对称图形的特点是解答本题的关键。\n\n【详解】圆环有无数条对称轴，半圆有 1 条对称轴。![](images/211a695c08e5c83630a1088244ecc508464a7ed689bcacabbfabb5d3ed0fd798.jpg)![](images/05944da9814607f63a8eaa8bbf07166df10e3dc867d49039a2dbfad9b6931c2b.jpg)",
+    "knowledge_points": [
+      "与圆相关的轴对称图形"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_449",
+    "original_id": "unit5_choice_449"
+  },
+  {
+    "id": "unit5_choice_450",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "（20-21 六年级上·全国·单元测试）小白要画一个半径是  $6 \\mathrm{~cm}$  的圆，圆规两脚之间的距离是( )cm，这个圆的半径是( )cm，周长是( )cm，面积是( )cm²。",
+    "options": [],
+    "answer": "6 6 37.68 113.04",
+    "solution": "【分析】画圆时, 圆规两脚之间的距离是圆的半径; 周长  $= 2 \\pi r$ , 面积  $= \\pi \\times r \\times r$ , 代入数据求解即可。此题考查圆的周长以及面积的求法, 了解圆规两脚之间的距离是圆的半径是解题的关键。\n\n【详解】小白要画一个半径是  $6 \\mathrm{~cm}$  的圆, 圆规两脚之间的距离是  $6 \\mathrm{~cm}$ , 这个圆的半径是  $6 \\mathrm{~cm}$ , 周长是  $2 \\times 3.14 \\times 6 =37.68 \\mathrm{~cm}$ , 面积是  $3.14 \\times 6 \\times 6 = 113.04 \\mathrm{~cm}^{2}$  。",
+    "knowledge_points": [
+      "圆的面积及应用",
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_450",
+    "original_id": "unit5_choice_450"
+  },
+  {
+    "id": "unit5_choice_451",
+    "unit": "第五单元：圆",
+    "chapter": "画圆",
+    "question_type": "填空题",
+    "difficulty": "困难",
+    "question": "(20-21 六年级上·全国·单元测试) 李小二画圆时先用圆规的针尖确定好圆的( ), 一般用字母( )表示, 它的作用是确定圆的( )。",
+    "options": [],
+    "answer": "圆心   O    位置",
+    "solution": "【分析】画圆时先用圆规的针尖确定好圆的圆心, 一般用字母 O 表示, 它的作用是确定圆的位置。本题考查圆的画法，解答本题的关键是掌握圆的画法。\n\n【详解】根据分析可得, 李小二画圆时先用圆规的针尖确定好圆的圆心, 一般用字母 O 表示, 它的作用是确定圆的位置。",
+    "knowledge_points": [
+      "画圆"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_451",
+    "original_id": "unit5_choice_451"
+  },
+  {
+    "id": "unit5_choice_452",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "（20-21 六年级上·全国·单元测试）用圆规画一个周长是  $12.56 \\mathrm{dm}$  的圆，圆规两脚之间的距离是  $\\_ \\mathrm{dm}$ ，这个圆的面积是  $\\_ \\mathrm{dm}^{2}$ 。",
+    "options": [],
+    "answer": "2 12.56",
+    "solution": "【分析】画圆时, 圆规两脚之间的距离是半径的长度, 根据圆的周长求出半径, 再根据圆的面积公式求出面积即可。关键是掌握圆的周长和面积公式，圆的周长  $= \\pi d = 2 \\pi r$  ，圆的面积  $= \\pi r^{2}$  。\n\n【详解】  $12.56 \\div 3.14 \\div 2 = 2$  （分米） $3.14 \\times 2^{2} = 12.56$  （平方分米）",
+    "knowledge_points": [
+      "圆的面积及应用",
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_452",
+    "original_id": "unit5_choice_452"
+  },
+  {
+    "id": "unit5_choice_453",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的面积",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "（2022·河南信阳·小升初真题）一个圆环的外圆半径是  $10 \\mathrm{~cm}$ , 内圆半径是  $8 \\mathrm{~cm}$ , 它的面积是_____。",
+    "options": [],
+    "answer": "113.04平方厘米",
+    "solution": "【分析】根据圆环的面积  $= \\pi$  （ $\\mathrm{R}^2 - \\mathrm{r}^2$ ），列式计算即可。关键是掌握和运用圆环的面积公式。\n\n【详解】  $3.14 \\times (10^{2} - 8^{2})\\\\=3.1 4 \\times 3 6 \\\\ = 1 1 3. 0 4 \\text {(平 方 厘 米)}  $",
+    "knowledge_points": [
+      "圆环的面积"
+    ],
+    "source": "2022·河南信阳·小升初真题",
+    "merged_id": "unit5_453",
+    "original_id": "unit5_choice_453"
+  },
+  {
+    "id": "unit5_choice_454",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "(2014 六年级·全国·课后作业) 当大、小两圆的半径比为  $3: 2$ , 则两圆的周长比是( ), 面积比是( )。",
+    "options": [],
+    "answer": "3:2 9:4",
+    "solution": "【分析】根据圆的周长公式  $C = 2 \\pi r$ , 求出大、小两圆的周长, 再写出它们的比即可; 根据圆的面积公式  $s = \\pi r^{2}$ , 分别计算大、小两圆的面积, 再写出它们的比即可。周长比和半径比相等，面积比是周长比的平方。\n\n【详解】大圆周长:  $2 \\pi \\times 3 = 6 \\pi$ ;小圆周长:  $2 \\pi \\times 2 = 4 \\pi$ ;大、小两圆的周长比是  $6 \\pi: 4 \\pi = 3: 2$ ;大圆面积:  $\\pi 3^{2} = 9 \\pi$ ;小圆面积:  $\\pi 2^{2} = 4 \\pi$ ;大、小两圆的面积比为  $9 \\pi: 4 \\pi = 9: 4$  。",
+    "knowledge_points": [
+      "圆的面积及应用",
+      "圆的周长及应用",
+      "比的意义"
+    ],
+    "source": "2014 六年级·全国·课后作业",
+    "merged_id": "unit5_454",
+    "original_id": "unit5_choice_454"
+  },
+  {
+    "id": "unit5_choice_455",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "(20-21 六年级上·全国·单元测试) 明明要画一个直径$4 \\mathrm{~cm}$  的圆, 圆规的两脚应叉开( )cm, 画出的这个圆的周长是长是( )cm，面积是( )cm²。",
+    "options": [],
+    "answer": "2   12.56  12.56",
+    "solution": "【分析】圆规两脚叉开的距离表示的是圆的半径，由此解答即可; 再根据 'C=πd'、'S=πr²'求出圆的周长和面积即可。明确两脚叉开的距离表示的是圆的半径是解答本题的关键，一定要熟练掌握圆的周长和面积公式。\n\n【详解】  $4 \\div 2 = 2$  （厘米）；$3.14 \\times 4 = 12.56$  （厘米）；  $3.14 \\times 2^{2} = 3.14 \\times 4 = 12.56$  （平方厘米）",
+    "knowledge_points": [
+      "圆的面积及应用",
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_455",
+    "original_id": "unit5_choice_455"
+  },
+  {
+    "id": "unit5_choice_456",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "（20-21 六年级上·全国·单元测试）一种小汽车的轮子的直径是 40 厘米。小汽车在行驶过程中轮子每分钟大约转 1000 圈，这样这辆小汽车每小时大约走（）千米（取整千米数）。",
+    "options": [],
+    "answer": "75",
+    "solution": "【分析】求根据圆的周长公式求出一周的长度, 再算出每分钟走的长度, 最后算 1 小时走的长度。本题考查圆的周长, 解答本题的关键是掌握圆的周长公式。\n\n【详解】40 厘米  $= 0.4$  米   1时=60分\n\n$3.14 \\times 0.4 \\times 1000 \\times 60\\\\=1256×60\\\\=75360 (米)$\n\n75360 米 = 75.36 千米 ≈ 75 千米",
+    "knowledge_points": [
+      "小数的近似数和改写",
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_456",
+    "original_id": "unit5_choice_456"
+  },
+  {
+    "id": "unit5_choice_457",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·全国·单元测试) 在一个边长 8 厘米的正方形里画一个最大的圆, 这个圆的周长是( )厘米,面积是( )平方厘米; 如果在这正方形中画 4 个最大的圆,这些圆的周长的和是( )厘米, 面积的和是( )平方厘米。",
+    "options": [],
+    "answer": "25.12   50.24    50.24    50.24",
+    "solution": "【分析】在一个边长 8 厘米的正方形里画一个最大的圆, 则圆的直径是正方形的边长; 如果在这正方形中画 4 个最大的圆, 这 4 个圆的直径是正方形边长的一半, 据此解答即可。本题考查圆的周长、面积, 解答本题的关键是掌握圆的周长、面积公式。\n\n【详解】  $3.14 \\times 8 = 25.12$  （厘米）\n\n$3.1 4 \\times (8 \\div 2) ^ {2}=3.1 4 \\times 1 6=50.24 (平方厘米)$\n\n$3.14 \\times (8 \\div 2) \\times 4\\\\=12.56 \\times 4\\\\=50.24 (厘米)$\n\n$3.1 4 \\times (8 \\div 2 \\div 2) ^ {2} \\times 4\\\\=3.1 4 \\times 4 \\times 4\\\\=50.24 (平方厘米)$",
+    "knowledge_points": [
+      "圆的面积及应用",
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_457",
+    "original_id": "unit5_choice_457"
+  },
+  {
+    "id": "unit5_choice_458",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·全国·单元测试) 一个圆的周长是125.6 厘米, 半径是( )厘米, 面积是( )平方厘米。",
+    "options": [],
+    "answer": "20 1256",
+    "solution": "【分析】根据圆的周长公式，先求出圆的半径，再根据圆的面积公式求出面积即可。圆的周长  $= \\pi d = 2 \\pi r$ , 圆的面积  $= \\pi r^{2}$  。\n\n【详解】  $125.6 \\div 3.14 \\div 2 = 20$  （厘米）$3.14 \\times 20^{2} = 1256$  （平方厘米）",
+    "knowledge_points": [
+      "圆的面积及应用",
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_458",
+    "original_id": "unit5_choice_458"
+  },
+  {
+    "id": "unit5_choice_459",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·全国·单元测试) 圆的面积比圆外的正方形面积( )，比圆内的正方形面积( )。\n\n(1) 圆形桌面的直径为 1.2 米, 这个桌面的面积是( )。  \n\n(2) 圆形花坛的半径是 4.5 米, 这个花坛的占地面积是 ( )。  \n(3) 一圆形物体的周长为 18.84 厘米, 其面积是( )。",
+    "options": [],
+    "answer": "小    大    1.1304平方米      63.585 平方米     28.26平方厘米",
+    "solution": "【分析】如图 ![](images/cf76b1dbbaf7eff8f9e0a4f031d271f0404a749bf869f86a0bc784d483f29833.jpg)![](images/3c17e9cf99fa536716a8b5e1952904f1bb21b502b6d9738a6b439529852aa106.jpg)可知，圆外和圆内正方形与圆的面积之间的大小关系；根据圆的面积  $= \\pi r^{2}$ ，列式计算即可，第（1）（3）小题先求出圆的半径，再根据圆的面积公式进行计算。关键是掌握圆的周长和面积公式，圆的周长  $= 2\\pi r$ 。\n\n【详解】圆的面积比圆外的正方形面积小, 比圆内的正方形面积大。\n\n(1)  $3.14 \\times (1.2 \\div 2)^{2}\\\\= 3. 1 4 \\times 0. 3 6\\\\= 1.1304（平方米）$  \n\n(2)  $3.14 \\times 4.5^{2} = 63.585$  （平方米）  \n(3)  $3.14 \\times (18.84 \\div 3.14 \\div 2)^{2}\\\\=3.1 4 \\times 9 \\\\ = 2 8. 2 6 \\text {(平 方 厘 米)} $",
+    "knowledge_points": [
+      "圆的面积及应用",
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_459",
+    "original_id": "unit5_choice_459"
+  },
+  {
+    "id": "unit5_choice_460",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·全国·单元测试) 要在底面半径是 14厘米的圆柱形水桶外面打上一个铁丝箍，接头部分是6厘米，需用铁丝( )厘米。",
+    "options": [],
+    "answer": "93.92",
+    "solution": "【分析】圆柱形水桶的外围是一个圆，所以这个铁丝箍的长度等于圆柱底面周长和接头部分之和。本题考查了对圆柱形的图形理解, 明确 '圆柱内每一个平行于底面的截面都是圆' 是解题的关键。\n\n【详解】  $3.14 \\times 2 \\times 14 + 6\\\\=8 7. 9 2 + 6 \\\\ = 9 3. 9 2 \\text {(厘 米)} $\n\n所以，需要铁丝 93.92 厘米。",
+    "knowledge_points": [
+      "圆柱的认识及特征",
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_460",
+    "original_id": "unit5_choice_460"
+  },
+  {
+    "id": "unit5_choice_461",
+    "unit": "第五单元：圆",
+    "chapter": "圆的概念及特点",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "(2014·全国·小升初真题) 在等圆中, 所有的直径都( ), 所有的半径都( ) , 直径长度是半径的( )。",
+    "options": [],
+    "answer": "相等   相等   2倍",
+    "solution": "【分析】根据圆的直径、半径的概念解答即可。本题考查圆的认识, 解答本题的关键是掌握圆的概念。\n\n【详解】在等圆中，所有的直径都相等，所有的半径都相等，直径长度是半径的 2 倍。",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "2014·全国·小升初真题",
+    "merged_id": "unit5_461",
+    "original_id": "unit5_choice_461"
+  },
+  {
+    "id": "unit5_choice_462",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·全国·期中) 一个圆的周长是 18.84 米,它的半径是( )，面积是( )。",
+    "options": [],
+    "answer": "3 米    28.26 平方米",
+    "solution": "【分析】根据圆的半径  $= \\text{周长} \\div \\pi \\div 2$ , 求出半径, 再根据圆的面积  $= \\pi r^{2}$ , 求出面积即可。关键是掌握圆的周长和面积公式，圆的周长  $= \\pi d = 2\\pi r$ 。\n\n【详解】  $18.84 \\div 3.14 \\div 2 = 3$（米）$3.14 \\times 3^{2} = 28.26$  （平方米）",
+    "knowledge_points": [
+      "圆的面积及应用",
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·全国·期中",
+    "merged_id": "unit5_462",
+    "original_id": "unit5_choice_462"
+  },
+  {
+    "id": "unit5_choice_463",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·全国·单元测试) 一个周长是  $12.56 \\mathrm{~cm}$  圆, 这个圆的直径是( )  $\\mathrm{cm}$ , 半径是( )  $\\mathrm{cm}$ , 面积是( ) $\\mathrm{cm}^{2}$  。",
+    "options": [],
+    "answer": "4 2    12.56",
+    "solution": "【分析】根据周长公式先求出圆的半径，后求出圆的直径。再利用圆的面积公式求出圆的面积即可。本题考查了圆的面积公式, 能够根据半径长求出面积是解题的关键。\n\n【详解】半径:  $12.56 \\div (2 \\times 3.14)\\\\=1 2. 5 6 \\div 6. 2 8 \\\\ = 2 \\quad (\\mathrm {c m}) $\n\n直径:  $2 \\times 2 = 4 \\text{ (cm)}$\n\n面积:  $3.14 \\times 2^{2}\\\\=3.1 4 \\times 4 \\\\ = 1 2. 5 6 (\\mathrm {c m} ^ {2})  $",
+    "knowledge_points": [
+      "圆的面积及应用",
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_463",
+    "original_id": "unit5_choice_463"
+  },
+  {
+    "id": "unit5_choice_464",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "（20-21 六年级上·全国·期末）圆的半径扩大 3 倍，圆的面积就扩大 3 倍。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】根据圆的半径扩大几倍，面积扩大倍数的平方倍，进行分析。关键是掌握圆的面积公式，圆的面积  $= \\pi r^{2}$ 。\n\n【详解】  $3^{2} = 9$ , 圆的半径扩大 3 倍, 圆的面积就扩大 9 倍,所以原题说法错误。故答案为:  ×",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级上·全国·期末",
+    "merged_id": "unit5_464",
+    "original_id": "unit5_choice_464"
+  },
+  {
+    "id": "unit5_choice_465",
+    "unit": "第五单元：圆",
+    "chapter": "弧、圆心角、扇形的认识",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "（19-20六年级上·河南南阳·期末）两个扇形相比，圆心角大的不一定就大。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】扇形面积等于扇形所在圆的面积乘圆心角度数除以 360, 所以决定扇形面积大小的有两个因素, 一个是半径,一个是圆心角的大小。据此判断。此题考查的是对扇形的认识, 要牢固掌握基础性知识。\n\n【详解】两个扇形相比，圆心角大的不一定就大，还跟扇形的半径有关。故答案为:  √。",
+    "knowledge_points": [
+      "弧、圆心角、扇形的认识"
+    ],
+    "source": "19-20六年级上·河南南阳·期末",
+    "merged_id": "unit5_465",
+    "original_id": "unit5_choice_465"
+  },
+  {
+    "id": "unit5_choice_466",
+    "unit": "第五单元：圆",
+    "chapter": "弧、圆心角、扇形的认识",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "(19-20 六年级上·河南南阳·期末) 扇形的大小只与它的圆心角的度数有关。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】扇形：由圆心角的两条边（即两条半径）和圆心角所对的弧围成的图形叫做扇形。扇形可看作圆的一部分，圆的大小主要是半径来决定的，所以扇形的大小既受圆半径的制约，又受圆心角大小的制约，因此，扇形大小与圆心角的度数和半径的长短有关。结合扇形的概念可知, 它的大小是由半径和圆心角的度数来决定的。\n\n【详解】扇形的大小与它的圆心角大小和半径长短二者都有关。故答案为:  ×",
+    "knowledge_points": [
+      "弧、圆心角、扇形的认识"
+    ],
+    "source": "19-20 六年级上·河南南阳·期末",
+    "merged_id": "unit5_466",
+    "original_id": "unit5_choice_466"
+  },
+  {
+    "id": "unit5_choice_467",
+    "unit": "第五单元：圆",
+    "chapter": "弧、圆心角、扇形的认识",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "(18-19 六年级上·全国·期末) 圆心角相等的扇形, 其面积都相等。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】扇形是由两条半径和圆上一段弧组成的，扇形的大小除了与圆心角有关外，还与半径有关系，由此解答即可。明确扇形的概念是解答本题的关键。\n\n【详解】圆心角相等的扇形，半径不一定相等，所以面积也不一定相等。故答案为:  ×  。",
+    "knowledge_points": [
+      "画扇形"
+    ],
+    "source": "18-19 六年级上·全国·期末",
+    "merged_id": "unit5_467",
+    "original_id": "unit5_choice_467"
+  },
+  {
+    "id": "unit5_choice_468",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "(20-21 六年级上·全国·单元测试) 圆周率就是 3.1415926。()",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】圆周率是一个无限不循环小数,  $\\pi = 3.1415926\\ldots$ ,据此解答即可。本题考查圆周率, 解答本题的关键是掌握圆周率的概念。\n\n【详解】根据分析可得，本题说法错误，故答案为：×。",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_468",
+    "original_id": "unit5_choice_468"
+  },
+  {
+    "id": "unit5_choice_469",
+    "unit": "第五单元：圆",
+    "chapter": "圆的概念及特点",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "（2014 六年级·全国·课后作业）在同一个圆上，连接圆上任意两点的线段中，直径最长。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "【详解】根据圆的特征，在同一个圆上，连接圆上任意两点的线段中，直径最长，说法正确。故答案为:  √",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "2014 六年级·全国·课后作业",
+    "merged_id": "unit5_469",
+    "original_id": "unit5_choice_469"
+  },
+  {
+    "id": "unit5_choice_470",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "(20-21 六年级上·全国·单元测试) 式子  $\\pi r^{2}$  可看成圆的面积, 也可看成是以  $r$  为边长的正方形面积的  $\\pi$  倍。()。",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】以  $r$  为边长的正方形的面积为  $r^{2}$ , 由此进一步解答即可。明确正方形的面积是解答本题的关键。\n\n【详解】  $\\pi r^{2} \\div r^{2} = \\pi$ ;故答案为: √。",
+    "knowledge_points": [
+      "圆的面积及应用",
+      "正方形的面积"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_470",
+    "original_id": "unit5_choice_470"
+  },
+  {
+    "id": "unit5_choice_471",
+    "unit": "第五单元：圆",
+    "chapter": "圆的概念及特点",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "(20-21 六年级上·全国·单元测试) 从圆内到圆上的任意一点的线段叫半径。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】从圆心到圆上的任意一点的线段叫半径，据此判断即可。本题考查圆的认识, 解答本题到的关键是掌握半径的概念。\n\n【详解】根据分析可得，本题说法错误。故答案为:× 。",
+    "knowledge_points": [
+      "圆的概念及特点"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_471",
+    "original_id": "unit5_choice_471"
+  },
+  {
+    "id": "unit5_choice_472",
+    "unit": "第五单元：圆",
+    "chapter": "与圆相关的轴对称图形",
+    "question_type": "作图题",
+    "difficulty": "容易",
+    "question": "（20-21 六年级上·全国·单元测试）两个圆能组成许多有趣的图形，请按下面的要求画一画。\n\n(1) 只有一条对称轴 (2) 只有两条对称轴 (3) 有无数条对称轴",
+    "options": [],
+    "answer": "见详解",
+    "solution": "【分析】一个图形沿一条直线对折，直线两旁的图形完全重合，这样的图形叫做轴对称图形，折痕所在的直线就是对称轴，由此画图即可。明确轴对称图形的特点是解答本题的关键。\n\n【详解】①![](images/292daa1486b504224f57d84ad01e91d67cd367253a6ba6c3288468e9c666e84e.jpg)\n\n② ![](images/fa98edd522173de8c17918eaa99f943809b7a8d3f5f0563a8e5689407a82750b.jpg)\n\n③![](images/bd6017c6c66183f05bf52c0f5b6c73ba2f1aca6f5e1e2b4a361d4d54313961a9.jpg)",
+    "knowledge_points": [
+      "与圆相关的轴对称图形"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_472",
+    "original_id": "unit5_choice_472"
+  },
+  {
+    "id": "unit5_choice_473",
+    "unit": "第五单元：圆",
+    "chapter": "画圆",
+    "question_type": "作图题",
+    "difficulty": "容易",
+    "question": "(20-21 六年级上·全国·单元测试) 请用圆规画一个半径是 2.5 厘米的圆，并分别用 O、r 表示圆心和半径。",
+    "options": [],
+    "answer": "见详解",
+    "solution": "【分析】画圆的步骤如下1、把圆规的两脚分开，定好两脚的距离，即半径。  2、把有针尖的一只脚固定在一点上，即圆心。  3、把装有铅笔尖的一只脚旋转一周，就画出一个圆。关键是在掌握画圆的方法，半径决定圆的大小，圆心决定圆的位置。\n\n【详解】作图如下:![](images/91a38c7872f51cb26601f7add72d446f.png)",
+    "knowledge_points": [
+      "画圆"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_473",
+    "original_id": "unit5_choice_473"
+  },
+  {
+    "id": "unit5_choice_474",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的面积",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(19-20 六年级上·新疆省直辖县级单位·期末) 锦绣小区的维修队要在小区广场修建一个周长为62.8米的圆形喷水池，绕着这个喷水池再修一条宽2米的水泥路。这条水泥路的面积是多少平方米？",
+    "options": [],
+    "answer": "138.16平方米",
+    "solution": "【分析】根据圆的周长公式，先求出喷水池的半径，再用喷水池半径  $+$  水泥路的宽，求出大圆半径，根据圆环面积  $= \\pi (R^{2} - r^{2})$  ，列式解答即可。关键是掌握圆的周长和圆环面积公式。\n\n【详解】  $62.8 \\div 3.14 \\div 2 = 10$（米）\n\n$1 0 + 2 = 1 2 \\text {(米)}$\n\n$3.14 \\times (12 ^ {2} - 10 ^ {2}) \\\\ =3.14 \\times 4 4\\\\ =138.16 (\\text {平 方 米})$\n\n答: 这条水泥路的面积是  $138.16$  平方米。",
+    "knowledge_points": [
+      "圆环的面积",
+      "圆的周长及应用"
+    ],
+    "source": "19-20 六年级上·新疆省直辖县级单位·期末",
+    "merged_id": "unit5_474",
+    "original_id": "unit5_choice_474"
+  },
+  {
+    "id": "unit5_choice_475",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的面积",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "(19-20 六年级上·山东临沂·期末) 一个直径是 6 米的圆形花坛，在它的周围铺2米宽的石子路，这条石子路的面积是多少平方米？",
+    "options": [],
+    "answer": "50.24 平方米",
+    "solution": "【分析】由题意可知, 就是求圆环的面积, 先求出大圆的半径, 即  $6 \\div 2 + 2$ , 再根据 'S 环形  $= \\pi (\\mathrm{R}^{2} - \\mathrm{r}^{2})$  ' 进行解答即可。熟练掌握求圆环面积的公式是解答本题的关键。\n\n【详解】  $6 \\div 2 + 2\\\\= 3 + 2 \\\\ = 5 \\text {(米)} $\n\n$6 \\div 2 = 3 (\\text {米})$;\n\n$3.14 \\times (5 ^ {2} - 3 ^ {2}) \\\\ = 3. 1 4 \\times 1 6 \\\\ = 5 0. 2 4 \\text {(平 方 米)} $\n\n答: 这条石子路的面积是 50.24 平方米。",
+    "knowledge_points": [
+      "圆环的面积"
+    ],
+    "source": "19-20 六年级上·山东临沂·期末",
+    "merged_id": "unit5_475",
+    "original_id": "unit5_choice_475"
+  },
+  {
+    "id": "unit5_choice_476",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·全国·期末) 一个圆形餐桌的周长是6.28m，要在餐桌中间放一个旋转玻璃盘，餐桌边到这个旋转玻璃盘的宽度是  $30 \\mathrm{~cm}$  。那么放在中间的旋转玻璃盘的面积是多少平方米？",
+    "options": [],
+    "answer": "$1.5386 \\mathrm{~m}^{2}$",
+    "solution": "【分析】由题意可知, 用餐桌的半径减去餐桌边到这个旋转玻璃盘的宽度即可得到玻璃盘的宽度（半径）, 再根据 ' $s = \\pi r^{2}$ ' 求出面积即可。解答本题的关键是明确玻璃盘的宽度（半径）就是用餐桌的半径减去餐桌边到这个旋转玻璃盘的宽度。\n\n【详解】  $6.28 \\div 3.14 \\div 2= 2 \\div 2=1(米)$；1米=100厘米;\n\n$3.1 4 \\times (1 0 0 - 3 0) ^ {2} \\\\ = 3. 1 4 \\times 4 9 0 0 \\\\ = 1 5 3 8 6 \\text {(平 方 厘 米)};  $\n\n15386 平方厘米 = 1.5386 平方米; 答: 放在中间的旋转玻璃盘的面积是 1.5386 平方米。",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级上·全国·期末",
+    "merged_id": "unit5_476",
+    "original_id": "unit5_choice_476"
+  },
+  {
+    "id": "unit5_choice_477",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "(20-21 六年级上·全国·单元测试) 小美把一个周长是  $28.26 \\mathrm{~m}$  的圆形纸片截成两个半圆, 求每个半圆的周长。",
+    "options": [],
+    "answer": "23.13 米",
+    "solution": "【分析】半圆的周长等于圆周长的一半加上直径的长度,由此解答即可。\n\n【 详 解 】$ 2 8. 2 6 \\div 2 + 2 8. 2 6 \\div 3. 1 4 \\\\ = 1 4. 1 3 + 9 \\\\=23.13 (米)$\n\n答: 每个半圆的周长是 23.13 米。明确半圆的周长等于圆周长的一半加上直径的长度是解答本题的关键。",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_477",
+    "original_id": "unit5_choice_477"
+  },
+  {
+    "id": "unit5_choice_478",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（20-21六年级上·全国·单元测试）王云阅读科普杂志，发现大卡车车轮直径是1.5米，如果每分钟转300圈，行驶28.26千米需要多少时间？",
+    "options": [],
+    "answer": "20 分钟",
+    "solution": "【分析】用  $3.14 \\times 1.5$  求出大卡车车轮的周长, 再乘 300 圈即可求出每分钟行驶的路程, 再根据 '路程÷速度 = 时间' 解答即可。求出大卡车每分钟行驶的路程是解答本题的关键。\n\n【详解】28.26千米  $= 28260$  米;  \n\n$28260 \\div (3.14 \\times 1.5 \\times 300)\\\\= 2 8 2 6 0 \\div 1 4 1 3 \\\\ = 2 0 \\text {(分 钟)}  $\n\n答: 行驶 28.26 千米需要 20 分钟。",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "20-21六年级上·全国·单元测试",
+    "merged_id": "unit5_478",
+    "original_id": "unit5_choice_478"
+  },
+  {
+    "id": "unit5_choice_479",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "(20-21 六年级上·全国·单元测试) 在公园里有一圆形花坛, 其周长是  $18.84 \\mathrm{~m}$ , 它的面积是多少  $\\mathrm{m}^{2}$ ?",
+    "options": [],
+    "answer": "28.26 平方米",
+    "solution": "【分析】用  $18.84 \\div 3.14 \\div 2$  求出圆形花坛的半径, 再根据 ' $s = \\pi r^{2}$ ' 求出面积即可。求出圆形花坛的半径是解答本题的关键。\n\n【详解】  $3.14 \\times (18.84 \\div 3.14 \\div 2)^{2}\\\\=3.1 4 \\times 9 \\\\ = 2 8. 2 6 \\text {(平 方 米)}  $\n\n答: 它的面积是 28.26 平方米。",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_479",
+    "original_id": "unit5_choice_479"
+  },
+  {
+    "id": "unit5_choice_480",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·全国·单元测试) 一个圆形的周长为37.68m 的小花园，施工方计划用  $\\frac{1}{6}$  的面积种植玫瑰花，种玫瑰花的面积是多少？",
+    "options": [],
+    "answer": "$18.84 \\mathrm{~m}^{2}$",
+    "solution": "【分析】根据圆的周长先求出圆的半径，再根据圆的面积公式求出小花园面积，将小花园面积看作单位'1'，用小花园面积×玫瑰花对应分率即可。关键是掌握圆的周长和面积公式，求一个数的几分之几用乘法。\n\n【详解】  $37.68 \\div 3.14 \\div 2 = 6$  （米）$3.14 \\times 6^{2} \\times \\frac{1}{6} = 113.04 \\times \\frac{1}{6} = 18.84$  (平方米)答: 种玫瑰花的面积是  $18.84 \\mathrm{~m}^{2}$  。",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_480",
+    "original_id": "unit5_choice_480"
+  },
+  {
+    "id": "unit5_choice_481",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（20-21 六年级上·全国·单元测试）小明的妈妈骑车去大桥对面的外婆家，自行车轮胎的半径是  $3 \\mathrm{dm}$ ，如果每分钟转 100 圈，通过 1000 米大桥要多少分钟？（结果保留整数)",
+    "options": [],
+    "answer": "5 分钟",
+    "solution": "【分析】根据圆的周长  $= 2 \\pi r$ , 求出车轮周长, 用车轮周长  $\\times$ 每分钟转的圈数, 求出每分钟行驶距离, 用大桥长度  $\\div$  每分钟行驶距离即可。关键是掌握圆的周长公式，用四舍五入法保留近似数，本题注意需要统一单位。\n\n【详解】  $3.14 \\times 2 \\times 3 \\times 100 = 1884$  （分米） 1884 分米=188.4 米 $1000 \\div 188.4 \\approx 5$  （分钟）\n\n答: 通过 1000 米大桥要 5 分钟。",
+    "knowledge_points": [
+      "商的近似数",
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_481",
+    "original_id": "unit5_choice_481"
+  },
+  {
+    "id": "unit5_choice_482",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的面积",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "（20-21六年级上·全国·单元测试）赣县人民广场有一个直径是4米的圆形花坛，在它的周围有一条宽1米的环形鹅卵石小路，小路的面积是多少？",
+    "options": [],
+    "answer": "15.7 平方米",
+    "solution": "【分析】由题意可知, 就是求圆环的面积, 先求出大圆的半径, 即 $4 \\div 2 + 1$, 再根据 'S 环形 $= \\pi (R^{2} - r^{2})$  进行解答即可。熟练掌握求圆环面积的公式是解答本题的关键。\n\n【详解】  $4 \\div 2 + 1 = 3$  （米）$4 \\div 2 = 2 (\\text {米})$\n\n$3.1 4 \\times (3 ^ {2} - 2 ^ {2}) \\\\ = 3. 1 4 \\times 5 \\\\ = 1 5. 7 \\text {(平 方 米)}; $\n\n答: 小路的面积是 15.7 平方米。",
+    "knowledge_points": [
+      "圆环的面积"
+    ],
+    "source": "20-21六年级上·全国·单元测试",
+    "merged_id": "unit5_482",
+    "original_id": "unit5_choice_482"
+  },
+  {
+    "id": "unit5_choice_483",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "（20-21六年级上·全国·单元测试）在一片草地上拴着一只小羊，如果拴羊的绳子长2米，这只羊最多可以吃到多大面积的草？",
+    "options": [],
+    "answer": "12.56 平方米",
+    "solution": "【分析】由题意可知, 羊吃草的范围就是一个半径为 2 米的圆, 根据 ' $s = \\pi r^{2}$ ' 求出圆的面积即可。熟练掌握圆的面积公式是解答本题的关键。\n\n【详解】  $3.14 \\times 2^{2}\\\\=3.1 4 \\times 4 \\\\ = 1 2. 5 6 \\text {(平 方 米)};$\n\n答: 这只羊最多可以吃到草的面积为 12.56 平方米。",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "20-21六年级上·全国·单元测试",
+    "merged_id": "unit5_483",
+    "original_id": "unit5_choice_483"
+  },
+  {
+    "id": "unit5_choice_484",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的面积",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "(20-21 六年级上·全国·单元测试) 沿着直径是 6 米的圆形花坛周围铺1米宽的石子路，石子路的面积是多少平方米？",
+    "options": [],
+    "answer": "21.98 平方米",
+    "solution": "【分析】由题意可知, 就是求圆环的面积, 先求出大圆的半径, 即  $6 \\div 2 + 1$ , 再根据 'S 环形  $= \\pi$ $(\\mathrm{R}^{2} - \\mathrm{r}^{2})$  ' 进行解答即可。熟练掌握求圆环面积的公式是解答本题的关键。\n\n【详解】  $6 \\div 2 + 1 = 4$  （米）；  $6 \\div 2 = 3$  （米）；\n\n$3.14× (4^{2}-3^{2})  \\\\= 3.14 \\times 7 \\\\ = 21.98（平方米）$；\n\n答: 石子路的面积是 21.98 平方米。",
+    "knowledge_points": [
+      "圆环的面积"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_484",
+    "original_id": "unit5_choice_484"
+  },
+  {
+    "id": "unit5_choice_485",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "（20-21六年级上·全国·单元测试）要从一个长5分米，宽4分米的长方形木板上锯下一个最大的圆，剩下的木板占圆面积的百分之几？",
+    "options": [],
+    "answer": "$59.24\\%$",
+    "solution": "【分析】由题意可知，长方形木板上锯下一个最大的圆，圆的直径和长方形的宽相等，根据圆的面积公式  $s = \\pi r^{2}$  求出圆的面积，用长方形的面积减去圆的面积即可求出剩下的木板面积，进而求出剩下的木板占圆面积的百分之几。明确长方形木板上锯下一个最大的圆，圆的直径和长方形的宽相等是解答本题的关键。\n\n【详解】  $3.14 \\times (4 \\div 2)^{2}\\\\= 3. 1 4 \\times 4\\\\= 12.56（平方分米)$ ；\n\n$5 \\times 4 - 1 2. 5 6\\\\= 2 0 - 1 2. 5 6\\\\= 7.44$  （平方分米）；\n\n$7.44 \\div 12.56 \\approx 59.24 \\%;$答：剩下的木板占圆面积的  $59.24 \\%$ 。",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "20-21六年级上·全国·单元测试",
+    "merged_id": "unit5_485",
+    "original_id": "unit5_choice_485"
+  },
+  {
+    "id": "unit5_choice_486",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的面积",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（20-21六年级上·全国·单元测试）公园一个圆形草坪，量得它的周长是50.24米。\n\n(1) 这个草坪的占地面积是多少？  \n(2) 公园要在草坪的四周铺一条宽 1 米的小路, 这条小路的面积是多少?  \n(3) 如果要给这条小路铺上地砖, 大约每平方米需要用地砖 50 块, 这样大约需要多少块地砖?",
+    "options": [],
+    "answer": "(1) 200.96 平方米;  (2) 53.38 平方米;    (3) 2669 块",
+    "solution": "【分析】（1根据圆的周长公式' $C = 2\\pi r$ '，求出圆的半径，再根据' $s = \\pi r^2$ '求出圆的面积即可；(2) 由题意可知, 就是求圆环的面积, 根据 ' $\\mathrm{s}_{\\text {环}} = \\pi (\\mathrm{r}_{1}^{2} - \\mathrm{r}_{2}^{1})$ ' 进行解答即可;  (3) 用圆环的面积乘 50 块即可求出需要的砖块数。熟练掌握圆周长和面积公式以及圆环面积公式,并能灵活运用是解答本题的关键。\n\n【详解】 (1)  $50.24 \\div 3.14 \\div 2\\\\= 1 6 \\div 2 \\\\ = 8 \\text {(米)} \\\\ 3.14 \\times 8^{2} = 3.14 \\times 64 = 200.96 （平方米）$ \n\n答: 这个草坪的占地面积是 200.96 平方米。\n\n(2)  $8 + 1 = 9$  (米)\n\n$3.1 4 \\times (9 ^ {2} - 8 ^ {2}) \\\\ = 3. 1 4 \\times 1 7 \\\\ = 5 3. 3 8 \\text {(平 方 米)} $\n\n答: 这条小路的面积是 53.38 平方米。\n\n(3)  $53.38 \\times 50 = 2669$  （块）；答: 这样大约需要 2669 块地砖。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆的面积及应用",
+      "圆环的面积"
+    ],
+    "source": "20-21六年级上·全国·单元测试",
+    "merged_id": "unit5_486",
+    "original_id": "unit5_choice_486"
+  },
+  {
+    "id": "unit5_choice_487",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "(20-21 六年级上·全国·单元测试) 一个圆形水池的直径是  $20 \\mathrm{~m}$ , 沿水池走一圈, 至少要走多少米?",
+    "options": [],
+    "answer": "62.8米",
+    "solution": "【分析】沿水池走一圈，相当于沿圆形水池周长走一圈，根据圆的周长  $= \\pi d$ ，列式解答即可。关键是理解题意，掌握圆的周长公式。\n\n【详解】  $3.14 \\times 20 = 62.8$  （米）答：至少要走62.8米。",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_487",
+    "original_id": "unit5_choice_487"
+  },
+  {
+    "id": "unit5_choice_488",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（20-21 六年级上·全国·单元测试）一辆自行车轮胎的外直径为  $72 \\mathrm{~cm}$ ，如果平均每分钟转 100 周，骑行通过一座2260.8m 的大桥，需要几分钟？",
+    "options": [],
+    "answer": "10分钟",
+    "solution": "【分析】先根据圆的周长公式计算出圆的周长，再算 100 周的长度，最后根据时间 = 路程 ÷ 速度，求出时间即可。本题考查圆的周长、行程问题, 解答本题的关键是掌握圆的周长公式。\n\n【详解】$72 厘米 = 0.72米$    $3.1 4 \\times 0. 7 2 \\times 1 0 0 \\\\ = 2. 2 6 0 8 \\times 1 0 0 \\\\ =226.08 (米)$\n\n2260.8÷226.08=10 (分钟)答: 需要 10 分钟。",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_488",
+    "original_id": "unit5_choice_488"
+  },
+  {
+    "id": "unit5_choice_489",
+    "unit": "第五单元：圆",
+    "chapter": "圆环的面积",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（20-21 六年级上·全国·单元测试）游乐园里有一项游乐设施，在半径是 32 米的圆形水池中央有一个圆形小岛，小岛的直径是 10 米，这个水池能放水的区域面积是多少  $\\mathrm{m}^{2}$ ？",
+    "options": [],
+    "answer": "3136.86 平方米",
+    "solution": "【分析】水池能放水的区域面积实际上是内圆直径为 10 米,外圆半径为 32 米的圆环面积, 根据圆环面积 = 外圆面积 - 内圆面积解答即可。本题考查圆环的面积, 解答本题的关键是掌握求圆环面积的方法。\n\n【详解】  $3.14 \\times 32^{2} - 3.14 \\times (10 \\div 2)^{2}=3215.36-78.5=3136.86 (平方米)$答: 这个水池能放水的区域面积是 3136.86 平方米。",
+    "knowledge_points": [
+      "圆环的面积"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_489",
+    "original_id": "unit5_choice_489"
+  },
+  {
+    "id": "unit5_choice_490",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（20-21 六年级上·全国·单元测试）小崔有一个直径是  $50 \\mathrm{~cm}$  的铁圈，她至少要在地上滚多少圈，才能使这铁圈通过一条 15.7 米的小沟？",
+    "options": [],
+    "answer": "10 圈",
+    "solution": "【分析】根据圆的周长公式求出一圈的长度, 再用总长除以一圈的长度得到圈数即可。本题考查圆的周长, 解答本题的关键是掌握圆的周长公式。\n\n【详解】50 厘米  $= 0.5$  米  $15.7 \\div (3. 1 4 \\times 0. 5)=15.7÷1.57=10$  (圈) 答: 她至少要在地上滚 10 圈。",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_490",
+    "original_id": "unit5_choice_490"
+  },
+  {
+    "id": "unit5_choice_491",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·全国·单元测试) 一辆自行车的车轮外直径约 6 分米, 如果车轮每分钟转 100 周, 从学校到公园的路程是 3.768 千米, 行完这段路程需要几分钟?",
+    "options": [],
+    "answer": "20 分钟",
+    "solution": "【分析】先用  $3.14 \\times 6$  求出自行车车轮的周长, 再乘 100 周,即可求出自行车一分钟行驶的路程, 再根据 '路程÷速度 =时间' 进行解答即可。求出自行车一分钟行驶的路程是解答本题的关键。\n\n【详解】  $3.14 \\times 6 \\times 100\\\\=18.84 \\times 1 0 0 \\\\ = 1 8 8 4 \\text {(分 米)} $\n\n3.768千米 = 37680分米   $37680 \\div 1884 = 20（分钟）$答：行完这段路程需要20分钟。",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_491",
+    "original_id": "unit5_choice_491"
+  },
+  {
+    "id": "unit5_choice_492",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（18-19六年级上·辽宁·单元测试）一台压路机前轮半径是 0.4 米, 如果前轮每分钟转动 6 周, 十分钟可以从路的一端转到另一端，这条路约长多少米？",
+    "options": [],
+    "answer": "150.72 米",
+    "solution": "【分析】用  $2 \\times 3.14 \\times 0.4$  求出压路机前轮旋转一周的长度,再乘 6 周即可求出旋转一分钟走过的路程, 再根据 '速度  $x$时间  $=$  路程'进行解答即可。求出压路机前轮旋转一分钟走过的路程是解答本题的关键。\n\n【详解】  $2 \\times 3.14 \\times 0.4 \\times 6 \\times 10\\\\=6.28 \\times 0. 4 \\times 6 \\times 1 0 \\\\ = 2. 5 1 2 \\times \\times 6 \\times 1 0 \\\\ = 1 5 0. 7 2 (\\text {米})$\n\n答: 这条路约长 150.72 米。",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "18-19六年级上·辽宁·单元测试",
+    "merged_id": "unit5_492",
+    "original_id": "unit5_choice_492"
+  },
+  {
+    "id": "unit5_choice_493",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（20-21 六年级上·全国·单元测试）小文的自行车轮子的直径是 0.6 米，如果平均每分钟转 125 圈，她从家到学校需 10 分钟，那么小文家到学校有多远？",
+    "options": [],
+    "answer": "2355米",
+    "solution": "【分析】根据圆的周长公式，先求出车轮转一圈的距离，再用转一圈的距离  $\\times$  每分钟转的圈数  $\\times$  总时间即可。关键是掌握圆的周长公式，圆的周长  $= \\pi d_{0}$\n\n【详解】  $0.6 \\times 3.14 \\times 125 \\times 10 = 2355 (米)$答: 小文家到学校有 2355 米远。",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_493",
+    "original_id": "unit5_choice_493"
+  },
+  {
+    "id": "unit5_choice_494",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（20-21 六年级上·全国·单元测试）一个圆与一个长方形面积相等，圆周长是 18.84 厘米，长方形长 6 厘米，宽是多少厘米？",
+    "options": [],
+    "answer": "4.71 厘米",
+    "solution": "【分析】根据圆的周长先计算出圆的半径，后计算出圆的面积。由于圆的面积等于长方形的面积，从而求出长方形的宽。本题考查了圆的面积、周长以及长方形的面积公式, 属于基础题, 解题时细心即可。\n\n【详解】圆的半径：\n\n$1 8. 8 4 \\div (2 \\times 3. 1 4) \\\\ = 1 8. 8 4 \\div 6. 2 8 \\\\= 3$  (厘米)\n\n圆的面积:\n\n$3 \\times 3 \\times 3. 1 4\\\\= 9 \\times 3. 1 4\\\\=28.26 (平方厘米)$\n\n长方形的宽:  $28.26 \\div 6 = 4.71$  (厘米)答: 长方形的宽是  $4.71$  厘米。",
+    "knowledge_points": [
+      "圆的面积及应用",
+      "长方形的面积",
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_494",
+    "original_id": "unit5_choice_494"
+  },
+  {
+    "id": "unit5_choice_495",
+    "unit": "第五单元：圆",
+    "chapter": "圆的面积及应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（20-21 六年级上·全国·单元测试）把一个圆平均分成若干等份后，能拼成一个周长为 20.7 分米的长方形，这个圆形的面积是多少平方分米？",
+    "options": [],
+    "answer": "19.625平方分米",
+    "solution": "【分析】把一个圆平均分成若干等份拼成长方形，长方形的长是圆周长的一半，长方形的宽是圆的半径，根据圆的周长  $+$  半径  $\\times 2 =$  长方形周长, 求出半径, 再根据圆的面积公式求出圆的面积即可。关键是熟悉圆的面积推导过程, 圆的周长  $= \\pi d =$ $2 \\pi r$ , 圆的面积  $= \\pi r^{2}$  。\n\n【详解】解：设圆的半径是  $r$  分米。$3.14 \\times 2 \\times \\mathrm{r} + 2 \\mathrm{r} = 20.7$  $6.28 \\mathrm{r} + 2 \\mathrm{r} = 20.7$  $8.28 \\mathrm{r} = 20.7$  $\\mathrm{r} = 2.5$  \n\n$3.14 \\times 2.5^{2} = 19.625(平方分米)$答: 这个圆形的面积是 19.625 平方分米。",
+    "knowledge_points": [
+      "圆的面积及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_495",
+    "original_id": "unit5_choice_495"
+  },
+  {
+    "id": "unit5_choice_496",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（20-21 六年级上·全国·单元测试）一只钟的时针长 40 毫米, 这根时钟的尖端一天 (24 小时) 所走过的路程是多少分米?",
+    "options": [],
+    "answer": "5.024分米",
+    "solution": "【分析】一个时钟的时针一天会走完钟表的两圈，所以它的尖端所走过的路程为两个圆的周长。据此解题即可。本题考查了圆的周长。解题时，除了要会使用圆的周长公式以外，还应注意单位的换算，将题中的40毫米换算成分米。\n\n【详解】40毫米  $= 0.4$  分米\n\n$2 \\times 3. 1 4 \\times 0. 4 \\times 2 \\\\ = 4 \\times 1. 2 5 6 \\\\ = 5. 0 2 4 \\text {(分 米)}  $\n\n答: 这根时钟的尖端一天 (24 小时) 所走过的路程是 5.024 分米。",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_496",
+    "original_id": "unit5_choice_496"
+  },
+  {
+    "id": "unit5_choice_497",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（20-21 六年级上·全国·单元测试）一辆自行车轮胎的外直径是 70 厘米，如果每分转 120 周，一小时能行多少千米？（保留正千米数）",
+    "options": [],
+    "answer": "16千米",
+    "solution": "【分析】根据圆的周长  $= \\pi d$ , 求出轮胎转一周的距离, 用轮胎转一周的距离  $\\times$  每分钟转的周数  $\\times 60$  分钟即可, 最后将结果换算成以千米为单位的数, 四舍五入成整数即可。关键是掌握圆的周长公式，熟记单位间的进率。\n\n【详解】1小时  $= 60$  分钟  $3.14 \\times 70 \\times 120 \\times 60 = 1582560$ （厘米）1582560 厘米 = 15.82560 千米 ≈ 16 千米   答：一小时能行 16 千米。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "时、分的认识及换算"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_497",
+    "original_id": "unit5_choice_497"
+  },
+  {
+    "id": "unit5_choice_498",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（20-21六年级上·全国·单元测试）刘老师提倡绿色出行，她每天骑车去学校。已知刘老师家距离学校  $2198 \\mathrm{~m}$ ，车轮外直径为  $70 \\mathrm{~cm}$ ，车轮每分钟可转 100 圈，骑车到学校需要几分钟？",
+    "options": [],
+    "answer": "10分钟",
+    "solution": "【分析】已知车轮的外直径, 则车轮的周长可求; 又因为车轮每分钟可转 100 圈, 则每分钟自行车行驶的距离可求;最后结合刘老师家与学校的距离, 来计算刘老师骑车到学校需要几分钟。解题思路就是先求出速度, 即每分钟车轮行进的距离, 要套用圆的周长公式; 再依据速度、路程、时间三者之间的关系, 计算求得所需时间。特别注意长度单位的转化。\n\n【详解】  $70 \\mathrm{~cm} = 0.7 \\mathrm{~m}$ \n\n$ 2 1 9 8 \\div (3. 1 4 \\times 0. 7 \\times 1 0 0) \\\\ = 2 1 9 8 \\div 2 1 9. 8 \\\\ = 1 0 \\text {(分 钟)}  $\n\n答: 刘老师骑车到学校需要 10 分钟。",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "20-21六年级上·全国·单元测试",
+    "merged_id": "unit5_498",
+    "original_id": "unit5_choice_498"
+  },
+  {
+    "id": "unit5_choice_499",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（20-21六年级上·全国·单元测试）毕爸爸开一辆汽车路过大桥，若车轮要滚动800圈才能下桥。已知车轮的直径为  $8 \\mathrm{dm}$ , 则这座大桥有多长?",
+    "options": [],
+    "answer": "$2009.6 \\mathrm{~m}$",
+    "solution": "【分析】已知车轮的直径为  $8 \\mathrm{dm}$ , 且车轮要滚动 800 圈才能下桥, 滚动一圈行驶了车轮一周的长度, 那么滚动 800 圈, 就是行驶了 800 圈车轮的长度。则先求圆的周长, 再乘 800 就是所求。本题需要注意的地方：化单位；车轮的直径以分米为单位，而一般大桥长度是以米为单位，故计算结果要转化单位。\n\n【详解】  $8 \\times 3.14 \\times 800\\\\= 2 5. 1 2 \\times 8 0 0 \\\\ = 2 0 0 9 6 (\\mathrm {d m}) \\\\ = 2 0 0 9. 6 \\quad (\\mathrm {m})  $\n\n答: 这座大桥有  $2009.6 \\mathrm{~m}$  长。",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "20-21六年级上·全国·单元测试",
+    "merged_id": "unit5_499",
+    "original_id": "unit5_choice_499"
+  },
+  {
+    "id": "unit5_choice_500",
+    "unit": "第五单元：圆",
+    "chapter": "圆的周长及应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（20-21 六年级上·全国·单元测试）公园里有一个圆形小花园，其直径为  $24 \\mathrm{~m}$ ，沿着它的周长每隔  $3 \\mathrm{~m}$  种一棵树苗，要种植多少棵？",
+    "options": [],
+    "answer": "25 棵",
+    "solution": "【分析】已知圆的直径, 则周长可求; 又因为是沿着它的周长每隔  $3 \\mathrm{~m}$  种一棵树苗, 在沿着封闭环形植树时, 棵树 =间隔数, 而间隔数可以用周长  $\\div$  间距, 据此解答。本题属于植树问题的第三种类型，封闭图形。在生活中，把树、花沿着各种封闭图形种植，是很常见的，只要灵活应用其规律解答即可。\n\n【详解】  $24 \\times 3.14 \\div 3\\\\=7 5. 3 6 \\div 3 \\\\ = 2 5. 1 2 \\text {(棵)} \\\\ \\approx 2 5 \\text {(棵)}  $\n\n答: 要种植 25 棵。",
+    "knowledge_points": [
+      "圆的周长及应用"
+    ],
+    "source": "20-21 六年级上·全国·单元测试",
+    "merged_id": "unit5_500",
+    "original_id": "unit5_choice_500"
+  }
+]
\ No newline at end of file
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new file mode 100644
index 0000000000000000000000000000000000000000..c68fa9e6ae163740678c467dafe0ec169be4b8c9
--- /dev/null
+++ "b/\345\205\255\345\271\264\347\272\247\351\242\230\351\233\206/\345\205\255\345\271\264\347\272\247\344\270\212/\345\205\255\345\271\264\347\272\247\344\270\212\347\254\254\345\205\253\345\215\225\345\205\203\346\225\260\344\270\216\345\275\242\346\225\260\346\215\256\351\233\206.json"
@@ -0,0 +1,8766 @@
+[
+    {
+      "id": "001",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "如图,按照此规律,图形⑥需要( )个○。\n\n![](images/b173363edfc760b50cb074b74c3ff050.png)",
+      "options": [
+        "A. 15",
+        "B. 21",
+        "C. 28"
+      ],
+      "answer": "B",
+      "solution": "【分析】第一个图形有 1 个 ○, 第二个图形有 3 个 ○, 第三个图形有 6 个 ○,\n\n1 = 1\n3 = 1 + 2\n\n6 = 1 + 2 + 3\n\n那么第 n 个图形中 ○ 的数量就是 1 + 2 + 3 + ... + n, 据此解答即可。\n\n【详解】图形⑥的○数：1 + 2 + 3 + 4 + 5 + 6 = 21(个)故答案为: B",
+      "knowledge_points": [
+        "图形的变化规律",
+        "数与形(探索规律)"
+      ],
+      "source": "23-24 三年级下·湖南·期末",
+      "merged_id": "unit8_001",
+      "original_id": "001"
+    },
+    {
+      "id": "002",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "珊珊用石子摆出了下图中的图案, 根据规律判断第 6 个图案中石子总数为 ( )个。\n\n![](images/50067eaa84cb93614c8458559c1c50003bd1b0ed7ab22ff390825379af7b16f0.jpg)",
+      "options": [
+        "A. 16",
+        "B. 20",
+        "C. 24"
+      ],
+      "answer": "C",
+      "solution": "【分析】第 1 个图案需要石子 4 个, 第 2 个图案需要石子 8个, 第 3 个图案需要石子 12 个, 第 4 个图案需要石子 16 个,由此可知，下一个图案比上一个图案多4个石子;第1个图案需要石子4个，可以写成： 4 × 1 ; 第2个图案需要石子8个，可以写成： 4 × 2 ; 第3个图案需要石子12个，可以写成： 4 × 3 ; 第4个图案需要石子16个，可以写成： 4× 4 由此可知, 第 n 个图案需要石子 4 n 个, 当 n = 6 时, 求出石子的数量, 据此解答。\n\n【详解】根据分析可知，第 n 个图案需要石子 4n 个。当 n = 6 时:4 × 6 = 24(个)珊珊用石子摆出了下图中的图案，根据规律判断第 6 个图案中石子总数为 24 个。\n\n![](images/50067eaa84cb93614c8458559c1c50003bd1b0ed7ab22ff390825379af7b16f0.jpg)\n\n故答案为: C",
+      "knowledge_points": [
+        "数与形(探索规律)"
+      ],
+      "source": "23-24 五年级下·湖南邵阳·期末",
+      "merged_id": "unit8_002",
+      "original_id": "002"
+    },
+    {
+      "id": "003",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "选择题",
+      "difficulty": "困难",
+      "question": "给定一列按规律排列的数： 1/2 、 2/5 、 3/10 、 4/17 …则这列数的第 8 个数是( )。",
+      "options": [
+        "A. 8/63",
+        "B. 8/65",
+        "C. 7/65",
+        "D. 9/64"
+      ],
+      "answer": "B",
+      "solution": "【分析】由前 4 个是 1/2 、 2/5 、 3/10 、 4/17 , 可知: 分子是1, 2, 3, 4, 第几个数分子就是几, 所以第 8 个数的分子是 8; 分母是 2, 5, 10, 17, 相邻两个数之间的差分别是 3, 5, 7…，由此求出第 8 个数的分母。观察各个分数的分子、分母，分析题意找出题中几个分数之间的规律。\n\n【详解】第8个数的分子是8，分母是:\n17 + 9 + 11 + 13 + 15= 26 + 11 + 13 + 15\n= 37 + 13 + 15\n= 50 + 15\n= 65\n则这列数的第 8 个数是 8/65 。故答案为: B",
+      "knowledge_points": [
+        "数字排列的规律"
+      ],
+      "source": "2024 六年级下·辽宁·专题练习",
+      "merged_id": "unit8_003",
+      "original_id": "003"
+    },
+    {
+      "id": "004",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "根据下面一组有规律的算式, 可以推出下一个算式是 ( )。\n\n6 × 7 = 42\n\n66 × 67 = 4422\n\n666 × 667 = 444222",
+      "options": [
+        "A. 6666 × 667 = 4446222",
+        "B. 666 × 6667 = 4440222",
+        "C. 6666 × 6677 = 44508882",
+        "D. 6666 × 6667= 44442222"
+      ],
+      "answer": "D",
+      "solution": "【分析】第一个算式 1 个 6 和 1 个 7 相乘等于 42，第二个算式 2 个 6 和 67 相乘等于 4422，第三个算式 3 个 6 和 667 相乘等于 444222，第四个算式应该是 4 个 6 和 6667 相乘等于 44442222，据此选择即可。\n\n【详解】可以推出下一个算式是 6666 × 6667 = 44442222 。故答案为：D",
+      "knowledge_points": [
+        "算式的规律"
+      ],
+      "source": "23-24 四年级上·陕西汉中·期末",
+      "merged_id": "unit8_004",
+      "original_id": "004"
+    },
+    {
+      "id": "005",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "如图, 第 8 个点子图的点子数是 ( )。\n\n![](images/63ff8ef3f80b1523ddef9869f5b050a892ec183e6668c02089e689bb148ba4ac.jpg)",
+      "options": [
+        "A. 26",
+        "B. 27",
+        "C. 28",
+        "D. 29"
+      ],
+      "answer": "D",
+      "solution": "【分析】观察图形可知，第一个图形点子数是 1，可以写成：4 × 1 - 3;第二个图形点子数是5，可以写成： 4 × 2 - 3 ;第三个图形点子数是9，可以写成： 4 × 3 - 3 ;........由此可知, 第 n 个图形的点子数是 (4 n - 3) , 当 n = 8 时,求出图中的点数。\n\n【详解】根据分析可知，第 n 个图形的点子数： (4n - 3) 个点;当 n = 8 时:\n\n4 × 8 - 3\n= 32 - 3\n= 29(个)\n\n如图，第8个点子图的点子数是 29 。\n\n![](images/a63759741c9e9761fdb20546e893dd786e7d22cd5606b9f858bc77f27dddbbdf.jpg)\n\n故答案为: D",
+      "knowledge_points": [
+        "数与形",
+        "图形的规律"
+      ],
+      "source": "22-23 五年级上·广东深圳·期末",
+      "merged_id": "unit8_005",
+      "original_id": "005"
+    },
+    {
+      "id": "006",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "根据 15 × 15 = 225 ,25 × 25 = 625, 35 × 35 = 1225, 45 × 45 = 2025 , 可推算出 85 × 85 = ( )。",
+      "options": [
+        "A. 4225",
+        "B. 5625",
+        "C. 6425",
+        "D. 7225"
+      ],
+      "answer": "D",
+      "solution": "【分析】当个位数字是 5 的两个相同的两位数相乘时, 积的后两位数是 25 , 前两位数是因数中十位的数字和比它大 1 的数字的乘积, 由此解答。\n\n【详解】因为 15 × 15 = 225 所以 85 × 85 = 7225 故答案为: D",
+      "knowledge_points": [
+        "算式的规律"
+      ],
+      "source": "22-23 四年级上·四川成都·期末",
+      "merged_id": "unit8_006",
+      "original_id": "006"
+    },
+    {
+      "id": "007",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "与 1 + 3 + 5 + 7 + 9 + 5 + 3 + 1 表示相同结果的算式是 ( )。",
+      "options": [
+        "A. 5 + 3",
+        "B. 4²",
+        "C. 5² + 3²",
+        "D. 5² - 3²"
+      ],
+      "answer": "C",
+      "solution": "【分析】把算式 1 + 3 + 5 + 7 + 9 + 5 + 3 + 1 看作两部分: 1 + 3 + 5 + 7 + 9 和 5 + 3 + 1 , 根据 '连续奇数的和等于奇数个数的平方' 可得, 1 + 3 + 5 + 7 + 9 = 5², 5 + 3 + 1 = 3² , 据此解答。本题是找规律的题型，从已知的数据中找到规律，并按规律解题。\n\n【详解】 1 + 3 + 5 + 7 + 9 + 5 + 3 + 1\n= (1 + 3 + 5 + 7 + 9) + (5 + 3 + 1)\n= 5² + 3²\n= 25 + 9\n= 34\n\n所以, 与 1 + 3 + 5 + 7 + 9 + 5 + 3 + 1 表示相同结果的是 5²+ 3² 。故答案为: C",
+      "knowledge_points": [
+        "算式的规律(整数)"
+      ],
+      "source": "16-17 六年级·全国·期末",
+      "merged_id": "unit8_007",
+      "original_id": "007"
+    },
+    {
+      "id": "008",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "有一组图，它的排列规律如下图，第 7 个图形由( )个●组成。\n\n![](images/ddc4370f4445a0a65527fcdfe757af84a4577c52136ff79f3eade2500dc4aa90.jpg)",
+      "options": [
+        "A. 21",
+        "B. 25",
+        "C. 28",
+        "D. 32"
+      ],
+      "answer": "C",
+      "solution": "【分析】第1个图形有 ●: 1 个; 第2个图形有 ●: 3 个, 3 = 1 + 2 ; 第3个图形有 ●: 6 个, 6 = 1 + 2 + 3 ;\n\n规律：第 n 个图形中 ● 的个数 = 1 + 2 + 3 + ... +n = n× (n+1) ÷ 2 ;据此求出第 7 个图形中 ● 的个数。通过数与形的结合，从已知的图形和数据中找到规律，并按规律解题。\n\n【详解】第7个图形：1 + 2 + 3 + 4 + 5 + 6 + 7\n= 7 × 8 ÷ 2\n= 56 ÷ 2\n= 28(个)故答案为: C",
+      "knowledge_points": [
+        "数与形",
+        "用字母表示数"
+      ],
+      "source": "2022 六年级下·全国·专题练习",
+      "merged_id": "unit8_008",
+      "original_id": "008"
+    },
+    {
+      "id": "009",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "按△△☆☆◇◇△△☆☆◇◇……的规律排序，第50个图形是( )。",
+      "options": [
+        "A. △",
+        "B. ☆",
+        "C. ◇"
+      ],
+      "answer": "A",
+      "solution": "【分析】看图形规律， △△☆☆◇◇◇ 为一组重复出现，所以用50除以7，求出商和余数，余数是几，第50个图形就是第一组图形的第几个。本题考查了图形的排列规律, 有一定观察力是解题的关键。\n\n【详解】 50 ÷ 7 = 7 ... 1, '△△☆☆◇◇◇'中第一个图形是 △ , 所以第 50 个图形也是 △ 。故答案为：A",
+      "knowledge_points": [
+        "数与形",
+        "图形的规律"
+      ],
+      "source": "15-16六年级下·广西玉林·期中",
+      "merged_id": "unit8_009",
+      "original_id": "009"
+    },
+    {
+      "id": "010",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "在一个扇形统计图中, 黄花朵数对应的圆心角是 90° , 那么黄花朵数占各种颜色花的总数的 ( )。",
+      "options": [
+        "A. 1/4",
+        "B. 1/2",
+        "C. 1/8"
+      ],
+      "answer": "A",
+      "solution": "【分析】在扇形统计图中, 圆周角是 360° , 计算黄花朵数对应的圆心角是 90° 占 360° 的分率即可。在扇形统计图中计算部分占整体的分率, 可以求出部分对应的圆心角占 360° 的分率。\n\n【详解】 90° ÷ 360° = 1/4 故答案为：A",
+      "knowledge_points": [
+        "扇形统计图的特点及绘制",
+        "弧、圆心角、扇形的认识",
+        "求一个数占另一个数几分之几"
+      ],
+      "source": "21-22 六年级上·湖南怀化·单元测试",
+      "merged_id": "unit8_010",
+      "original_id": "010"
+    },
+    {
+      "id": "011",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "图中有 ( ) 个角。\n\n![](images/afede37a2c0a52f6895f2206c0f07dab0626f34e056f4c72646fa329dd65cc7f.jpg)",
+      "options": [
+        "A. 6",
+        "B. 4",
+        "C. 3"
+      ],
+      "answer": "A",
+      "solution": "【分析】从一点引出两条射线所形成的图形叫做角。据此可知, 图中单独的角共 3 个, 由两个单独的角组成的角共 2\n\n个, 有三个单独的角组成的角共 1 个, 则图中共 (3 + 2 + 1) 个角。熟练掌握角的定义。数角时, 要按照顺序数, 才能做到不重不漏。\n\n【详解】 3 + 2 + 1 = 6 （个）;则图中有 6 个角。 故答案为：A",
+      "knowledge_points": [
+        "角的初步认识及辨认",
+        "数与形"
+      ],
+      "source": "21-22 四年级上·湖南岳阳·期中",
+      "merged_id": "unit8_011",
+      "original_id": "011"
+    },
+    {
+      "id": "012",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "观察101001000100001...的排列，第5和第6个1之间应排( )个 0。",
+      "options": [
+        "A. 1",
+        "B. 2",
+        "C. 4",
+        "D. 5"
+      ],
+      "answer": "D",
+      "solution": "【分析】观察数字的排列发现：第1和第2个1之间有1个0, 第 2 和第 3 个 1 之间有 2 个 0, 第 3 和第 4 个 1 之间有 3 个 0, 依次类推, 第 n 和 n+1 个 1 之间应该有 n 个 0。据此解答即可。本题主要考查了学生通过特例分析从而归纳总结出一般结论的能力。对于找规律的题目首先应找出哪些部分发生了变化，是按照什么规律变化的，通过分析找到各部分的变化规律后直接利用规律求解。\n\n【详解】由分析可知, 第 5 和第 6 个 1 之间应排 5 个 0 。故答案为: D",
+      "knowledge_points": [
+        "数字排列的规律"
+      ],
+      "source": "21-22六年级上·全国·单元测试",
+      "merged_id": "unit8_012",
+      "original_id": "012"
+    },
+    {
+      "id": "013",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "已知: 3 × 9 = 27, 3 × 99 = 297, 3 × 999 = 2997, 3 × 9999 = 29997 , 那么: 3 × 999999 = ( )。",
+      "options": [
+        "A. 299997",
+        "B. 3000007",
+        "C. 2999997",
+        "D. 29999997"
+      ],
+      "answer": "C",
+      "solution": "【分析】根据已知的算式可得积的规律：固定数字不变的是 2、7;如果第二个因数 9 的个数是 n 个，那么在积的数字 '2' 和 '7' 的之间就排 n-1 个 9;据此解答。主要考查了学生通过特例分析从而归纳总结出一般结论的能力。对于找规律的题目首先应找出哪些部分发生了变化，是按照什么规律变化的，通过分析找到各部分的变化规律后直接利用规律求解。\n\n【详解】根据分析可得,3 × 999999 = 2999997;故答案为: C。",
+      "knowledge_points": [
+        "算式的规律"
+      ],
+      "source": "19-20 四年级上·全国·期末",
+      "merged_id": "unit8_013",
+      "original_id": "013"
+    },
+    {
+      "id": "014",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "一组图形按'☆☆☆○○△☆☆☆○○△……'的规律排列，第36个图形是( )。",
+      "options": [
+        "A. △",
+        "B. ○",
+        "C. ☆"
+      ],
+      "answer": "A",
+      "solution": "【分析】观察图形可知, 这组图形是 6 个图形一个循环周期,分别按照 ☆☆☆○○△ 的顺序依次循环排列, 据此求出第 36 个图形是第几个图形中的第几个图形即可解答问题。根据题干得出这组图形的排列周期规律是解决此类问题的关键。\n\n【详解】 36 ÷ 6 = 6 所以第 36 个图形是第 6 周期的最后一个是 △。故答案为: A。",
+      "knowledge_points": [
+        "周期问题"
+      ],
+      "source": "2021 三年级下·全国·专题练习",
+      "merged_id": "unit8_014",
+      "original_id": "014"
+    },
+    {
+      "id": "015",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "某商场要表示2012年到2019年销售额的增长变化情况，应选用( )。",
+      "options": [
+        "A. 条形统计图",
+        "B. 折线统计图",
+        "C. 扇形统计图"
+      ],
+      "answer": "B",
+      "solution": "【分析】条形统计图的特点：用直条的长短表示数量的多少，从图中直观地看出数量的多少，便于比较。折线统计图的特点：不仅能看清数量的多少，还能通过折线的上升和下降表示数量的增减变化情况。扇形统计图清楚地看出各部分数量与总数量之间，部分与部分之间的关系。一定要明确各种统计图的特点，进而选择出合适的统计方式。\n\n【详解】某商场要表示 2012 年到 2019 年销售额的增长变化情况，应选用折线统计图;故答案为: B",
+      "knowledge_points": [
+        "统计图的选择"
+      ],
+      "source": "17-18六年级上·河南周口·期末",
+      "merged_id": "unit8_015",
+      "original_id": "015"
+    },
+    {
+      "id": "016",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "选择题",
+      "difficulty": "困难",
+      "question": "现有 200 根相同的钢管，把它们堆放成正三角形垛，使剩余的钢管尽可能的少，那么剩余的钢管有( )根。",
+      "options": [
+        "A. 9",
+        "B. 10",
+        "C. 11",
+        "D. 12"
+      ],
+      "answer": "B",
+      "solution": "【分析】\n\n| 钢管层数/(层) | 钢管根数/(根) |\n| ------------- | ------------- |\n| 2             | 1+2           |\n| 3             | 1+2+3         |\n| 4             | 1+2+3+4       |\n| n             | n(n+1)/2      |\n\n本题应用了有关数字排列的知识, 只要找出其中的规律, 并结合题意恰当运用相关公式, 就能够解答出来。\n\n【详解】由分析得：只要依据公式 n(n + 1)/2 确定出三角形钢管垛在 200 根以内最多的排列方式, 并求出具体根数, 再用 200 减去这个根数即可。当 n = 18 时, 共有 18 × (18 + 1)/2 = 171 (根);当 n = 19 时, 共有 19 × (19 + 1)/2 = 190 (根);当 n = 20 时, 共有 20 × (20 + 1)/2 = 210 (根)即摆放成三角形钢管垛，最多摆19层，共有190根，200 - 190 = 10 （根）故答案为 B。",
+      "knowledge_points": [
+        "数与形",
+        "数字排列的规律"
+      ],
+      "source": "2020 六年级·全国·竞赛",
+      "merged_id": "unit8_016",
+      "original_id": "016"
+    },{
+      "id": "017",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "先找规律, 再填一填。\n\n① 2 = 1 × 2\n② 2 + 4 = ()×()\n③ 2 + 4 + 6 = () ×()④ 2 + 4 + 6 + 8 = () ×()\n\n根据上面的规律写一写。\n\n2 + 4 + 6 + 8 + 10 = ()× () =()\n2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 = ()× ()=()",
+      "answer": "2;3;3;4;4;5;5;6;30;8;9;72",
+      "solution": "【分析】通过观察算式可知, 从 2 起的几个连续偶数的和等于偶数的个数与偶数的个数加 1 的积, 即第 n 个式子中, 等式左侧从 2 起有连续 n 个偶数相加, 对应的乘法算式为 n × (n + 1)。\n\n【详解】根据②式中，算式左边有从2起2个连续偶数的和等于偶数的个数2与3的积2 + 4 = 2 × 3\n\n③ 式中, 算式左边有从 2 起 3 个连续偶数的和等于偶数的个数 3 与 4 的积2 + 4 + 6 = 3 × 4\n\n④式中, 算式左边有从 2 起 4 个连续偶数的和等于偶数的个数 4 与 5 的积2 + 4 + 6 + 8 = 4 × 5\n\n则 2 + 4 + 6 + 8 + 10 这是前 5 个偶数的和, 对应乘法算式为 5 × 6 = 302 + 4 + 6 + 8 + 10 = 5 × 6 = 30\n\n2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 , 这是前8个偶数的和, 对应乘法算式为8 × 9 = 72  2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 = 8 × 9 = 72",
+      "knowledge_points": [
+        "算式的规律(整数)"
+      ],
+      "source": "25-26 六年级上·全国·单元测试",
+      "merged_id": "unit8_017",
+      "original_id": "017"
+    },
+    {
+      "id": "018",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "困难",
+      "question": "在以下数列：\n\n1/1, 2/1, 1/2, 3/1, 2/2, 1/3, 4/1, 3/2, 2/3, 1/4 , 中 9/17 位于第___项。",
+      "options": [],
+      "answer": "317",
+      "solution": "【分析】首先发现第一个数的分子分母的和为 2 , 第二、第三个数的分子分母的和为 3 , 第四、五、六个数的分子分母的和为 4 , 由此将分子与分母之和相等的归于同一组, 算出 9/17 在 7 + 19 - 1 = 25 组, 再算出在 25 组的位置, 由此找出规律解决问题。解决这类数列找项数的问题, 关键是通过分组找出分子分母的和、每组项数的规律, 再利用等差数列求和等知识计算。要注意分析每组内数的分子分母变化特点,准确确定目标数所在组及组内位置，从而算出总项数。\n\n【详解】因为，在这一列数中，第 1 组: 分子、分母的和为 2 的有 1/1 , 共 1 项,第2组：分子、分母的和为3的有 2/1, 1/2 ，共2项，第3组: 分子、分母的和为4的有 3/1, 2/2, 1/3 , 共3项,依此类推就会发现, 分子分母和为 n 的组, 有 (n - 1) 项,即第 (n - 1) 组, 且每组中分子从 n - 1 开始依次递减 1 ,分母从 1 开始依次递增 1 ;9/17 的分子与分母和是 26 , 那么该分子所在的组数就是:26 - 1 = 25 (组)根据前面找到的规律，分子分母和为 n 的组有 n - 1 项，那么前 24 组的项数之和是\n\n1 + 2 + 3 + ... + 24\n= (24 + 1) × 24 ÷ 2\n= 25 × 24 ÷ 2\n= 25 × 12\n= 300\n\n而 9/17 在分子、分母和为 26 一组中, 前面还有25/1, 24/2, 23/3, 22/4, 21/5, ..., 11/15, 10/16 , 共 16 个数, 即 9/17 在第 17 项, 所以 9/17 在这一整列数的项数为 300 + 17 = 317 (项)。",
+      "knowledge_points": [
+        "分数排列的规律"
+      ],
+      "source": "25-26六年级上·全国·单元测试",
+      "merged_id": "unit8_018",
+      "original_id": "018"
+    },
+    {
+      "id": "019",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "困难",
+      "question": "有许多等式:\n\n2 + 4 + 6 = 1 + 3 + 5 + 3;\n8 + 10 + 12 + 14 = 7 + 9 + 11 + 13 + 4;\n16 + 18 + 20 + 22 + 24 = 15 + 17 + 19 + 21 + 23 + 5;\n\n那么第 10 个等式的和是____。",
+      "options": [],
+      "answer": "1668",
+      "solution": "【分析】分析等式的规律，第 n 个等式左边有 n + 2 个连续偶数，确定第10个等式左边的第一个数，第1个等式左边的第一个数为 1 × (1 + 3) - 2 = 2 ，第2个等式左边的第一个数为 2 × (2 + 3) - 2 = 8 ，第3个等式左边的第一个数为 3 × (3 + 3) - 2 = 16 ，那么第 n 个等式左边的第一个数为 n × (n + 3) - 2 ;右侧有 n + 3 项，其中前 n + 2 项为连续奇数，右侧第一个数为 n × (n + 3) - 3 ，最后一个数为 n + 2(n + 2)(n² + 4n - 1) 。本题考查的对于等式的规律的探索, 分析等式两边的规律进而求和。\n\n【详解】第10个等式等号左边的第一个数字是:\n\n10 × (10 + 3) - 2\n=10 × 13 - 2\n=128\n\n所以第 10 个等式是:128 + 130 + 132 + 134 + 136 + 138 + 140 + 142 + 144 + 146 + 148 + 150 = 127 + 129 + 131 + 133 + 135 + 137 + 139+141+143+145+147+12所以第 10 个等式的结果是:128 + 130 + 132 + 134 + 136 + 138 + 140 + 142 + 144 + 146 + 148 + 150=(128 + 150) + (130 + 148) + (132 + 146) + (134 + 144) + (136 + 142) + (138 + 140)\n=278+278+278+278+278\n=278×6\n=1668",
+      "knowledge_points": [
+        "算式的规律(整数)"
+      ],
+      "source": "25-26 六年级上·全国·单元测试",
+      "merged_id": "unit8_019",
+      "original_id": "019"
+    },
+    {
+      "id": "020",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "按规律填空:7, 9, 12, 16, _, 27。",
+      "options": [],
+      "answer": "21",
+      "solution": "【分析】观察可知，第一个数 +2 = 第二个数，第二个数 +3 = 第三个数，第三个数 +4 = 第四个数，由此可知，第几个数就是前一个数加几。\n\n【详解】 16 + 5 = 21\n\n7, 9, 12, 16, 21, 27。",
+      "knowledge_points": [
+        "整数排列的规律"
+      ],
+      "source": "24-25 六年级下·海南海口·单元测试",
+      "merged_id": "unit8_020",
+      "original_id": "020"
+    },
+    {
+      "id": "021",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "聪聪用小棒摆了 4 个树状图, 以下是树状图变化的规律:\n\n![](images/96fdc7d98c382f8855ed7062d9e18d98.png)\n\n(1)按此规律继续摆下去，第5个树状图要用( )根小棒。\n(2)按此规律继续摆下去，第( )个树状图要用 1023 根小棒。",
+      "options": [],
+      "answer": "(1)31;(2)10",
+      "solution": "【分析】（1）看图可知，第1个图用（2-1）根小棒，第2个图用 (2× 2 - 1) 根小棒，第3个图用 (2× 2× 2 - 1) 根小棒，第4个图用 (2× 2× 2× 2 - 1) 根小棒，那么第5个图需要用 (2× 2× 2× 2× 2 - 1) 根小棒。\n\n(2) 根据 (1) 可知, 第 n 个图需要 (2^n - 1) 根小棒。令 2^n - 1 = 1023 , 求出 n , 即可求出第几个树状图要用 1023 根小棒。\n\n【详解】 (1) 2 × 2 × 2 × 2 × 2 - 1= 32 - 1  = 31(根)  所以第 5 个树状图要用 31 根小棒。\n\n(2) 2^n - 1 = 1023 , 那么 2^n = 1024 。2^10 = 1024, 那么 n = 10 。所以第 10 个树状图要用 1023 根小棒。",
+      "knowledge_points": [
+        "数与形(探索规律)"
+      ],
+      "source": "24-25 六年级上·浙江嘉兴·期末",
+      "merged_id": "unit8_021",
+      "original_id": "021"
+    },
+    {
+      "id": "022",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "先观察下面算式的规律, 再按要求做一做。\n\n1 = 1² , 1 + 3 = 2² , 1 + 3 + 5 = 3² , 1 + 3 + 5 + 7 = 4²\n\n利用上面的规律直接写一写。\n\n1 + 3 + 5 + 7 + 9 = ( )²\n1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = ( )²",
+      "options": [],
+      "answer": "5²;8²",
+      "solution": "【分析】发现规律为：从 1 开始的连续奇数之和，等于数字个数的平方，据此直接写结果。\n\n【详解】 1 + 3 + 5 + 7 + 9 ，一共是5个数字和，即 1 + 3 + 5 + 7 + 9 = 5²\n\n1 + 3 + 5 + 7 + 9 + 11 + 13 + 15, 一共是 8 个数字和，即 1 + 3 +5 + 7 + 9 + 11 + 13 + 15 = 8²",
+      "knowledge_points": [
+        "数与形(归纳递推)"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit8_022",
+      "original_id": "022"
+    },
+    {
+      "id": "023",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "按下面的方式摆正方形，摆1个正方形需要4根小棒，摆2个正方形需要7根小棒，摆20个正方形需要( )根小棒。\n\n![](images/bbf888dc0c7485bb1fe1d59f8c6a9e3f.png)",
+      "options": [],
+      "answer": "61",
+      "solution": "【分析】观察图形可知, 摆 1 个正方形需要 4 = 1 × 3 + 1 根小棒, 摆 2 个正方形需要 7 = 2 × 3 + 1 根小棒, 摆 3 个正方形需要 10 = 3 × 3 + 1 根小棒, 由此可知摆 n 个正方形需要 (3 n + 1) 根小棒。据此解答即可。\n\n【详解】 3 × 20 + 1\n= 60 + 1\n= 61(根)\n\n则摆 20 个正方形需要 61 根小棒。",
+      "knowledge_points": [
+        "图形的变化规律",
+        "数与形(探索规律)"
+      ],
+      "source": "23-24 六年级上·全国·课后作业",
+      "merged_id": "unit8_023",
+      "original_id": "023"
+    },
+    {
+      "id": "024",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "找规律: 2/3, 4/9, 8/27 ,( )，( )， 64/729 。",
+      "options": [],
+      "answer": "16/81;32/243",
+      "solution": "【分析】观察可知, 前一个分数的分母 × 3 = 后一个分数的分母, 前一个分数的分子 × 2 = 后一个分数的分子, 据此分别确定后面两个分数的分母和分子, 写出后面的两个分数即可。\n\n【详解】 27 × 3 = 81 、 8 × 2 = 16\n\n81 × 3 = 243, 16 × 2 = 32  2/3, 4/9, 8/27, 16/81, 32/243, 64/729 。",
+      "knowledge_points": [
+        "分数排列的规律",
+        "数字排列的规律"
+      ],
+      "source": "23-24 六年级上·四川泸州·期中",
+      "merged_id": "unit8_024",
+      "original_id": "024"
+    },
+    {
+      "id": "025",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "找规律, 写答案。\n\n0.1 ÷ 11 = 0.009090909...\n0.2 ÷ 11 = 0.018181818......\n0.3 ÷ 11 = 0.027272727......\n0.5 ÷ 11 = ( )\n\n找一找，这个商的小数点后第2022个数字是( )。",
+      "options": [],
+      "answer": "0.0454545…… 4",
+      "solution": "【分析】仔细观察题中的式子可以发现：除数不变，被除数乘几，商也乘几，据此求出 0.5 ÷ 11 = 0.0454545... ;这个商的小数点后第 2022 个数字（总数位数 -1）÷循环节的位数。能发现题中的规律并利用规律求出 0.5 ÷ 11 的结果以及熟练掌握求循环小数第几位上数字是多少的方法是解题的关键。\n\n【详解】 0.5 ÷ 0.1 = 5  0.5 ÷ 11 的商是一个循环小数，这个循环小数的循环节是0.5 ÷ 11 = 0.009090909...... × 5 = 0.0454545......\n45,\n(2022 - 1) ÷ 2\n= 2021 ÷ 2=101(组)……1(位)余数 1 表示一个循环里的第 1 个数即 4 ; 所以这个商的小数点后第 2022 个数字是 4。",
+      "knowledge_points": [
+        "循环小数的认识与简写",
+        "算式的规律",
+        "除数是整数的小数除法"
+      ],
+      "source": "21-22 五年级上·浙江台州·期末",
+      "merged_id": "unit8_025",
+      "original_id": "025"
+    },
+    {
+      "id": "026",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "困难",
+      "question": "1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ... + 1/256 = ( )。",
+      "options": [],
+      "answer": "255/256",
+      "solution": "【分析】如图所示, 把整个正方形的面积看作单位 '1', 依次表示出 1/2 、 1/4 、 1/8 、 1/16 、 1/32 ... 1/256 , 式子中所有分数的和等于整体减去最后一个分数的差, 据此解答。\n\n![](images/d840e8391303e81a90fea0adeee15618ffe4bdd210db875c5ac632885bf1a472.jpg)\n\n【详解】\n\n1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ... + 1/256\n= 1 - 1/256\n= 255/256",
+      "knowledge_points": [
+        "数与形"
+      ],
+      "source": "20-21 六年级下·贵州遵义·期末",
+      "merged_id": "unit8_026",
+      "original_id": "026"
+    },
+    {
+      "id": "027",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "找规律, 写得数。\n\n1 × 5 + 4 = 9 = 3 × 3; 2 × 6 + 4 = 16 = 4 × 4; 3 × 7 + 4 = 25 = 5 × 5; 4 × 8 + 4 = 36 = 6 × 6; ...; 10 × ( ) + 4 = ( ) = ( ) × ( ); ...; ( ) × ( ) + 4 = ( ) = ( ) × 32 。",
+      "options": [],
+      "answer": "14;144;12;12;30;34;1024;32",
+      "solution": "【分析】注意观察，在每个算式中，第二个数比第一个数大 4 ，第四个数等于第二个数减去 2 的差的平方数，比如1 × 5 + 4 = 9 = 3 × 3 中, 5 = 1 + 4 , (5-2)² = 9 , 据此解答。此题主要通过观察，找到数字之间的规律，解决实际问题。\n\n【详解】 10 + 4 = 14 ， (14 - 2)² = 144;所以 10 × 14 + 4 = 144 = 12 × 12 ;32 × 32 = 1024, 32 + 2 = 34, 34 - 4 = 30;所以 30 × 34 + 4 = 1024 = 32 × 32 。",
+      "knowledge_points": [
+        "数与形",
+        "算式的规律"
+      ],
+      "source": "19-20 五年级下·湖南娄底·期末",
+      "merged_id": "unit8_027",
+      "original_id": "027"
+    },
+    {
+      "id": "028",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "1/2, 1/4, 1/8, 1/16, ... 这组数字的第 6 个数是( )。",
+      "options": [],
+      "answer": "1/64",
+      "solution": "【分析】 4 ÷ 2 = 2, 8 ÷ 4 = 2, 16 ÷ 8 = 2 , 规律: 分子都是 1 , 前一个分母依次乘 2 得到后一个分母。通过观察，分析、归纳并发现其中的规律，并应用发现的规律解决问题是应该具备的基本能力。\n\n【详解】 16 × 2 = 32   32 × 2 = 64  所以，这组数字的第 6 个数是 1/64 。",
+      "knowledge_points": [
+        "数与形",
+        "数字排列的规律"
+      ],
+      "source": "18-19 六年级下·四川遂宁·期末",
+      "merged_id": "unit8_028",
+      "original_id": "028"
+    },
+    {
+      "id": "029",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "按规律填数。\n\n1，4，9，16，( )，36，49，( )。\n6.25, 2.5, 1, ( ), ( ), 0.064。",
+      "options": [],
+      "answer": "25;64;0.4;0.16",
+      "solution": "【分析】（1）通过观察发现：第一个数是 1² ，第二个数是 2² ，第三个数是 3² ，第 n 个数是 n² ;(2) 前一个数除以 2.5 得后一个数, 如第二个数 2.5 ÷ 2.5 = 1 , 据此解答。此题考查的是找规律，解答此题关键是正确找出规律并运用规律解决问题。\n\n【详解】由分析得，1, 4, 9, 16, 25, 36, 49, 64。6.25, 2.5, 1, 0.4, 0.16, 0.064。",
+      "knowledge_points": [
+        "数字排列的规律"
+      ],
+      "source": "20-21五年级上·湖南怀化·单元测试",
+      "merged_id": "unit8_029",
+      "original_id": "029"
+    },
+    {
+      "id": "030",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "请你根据规律直接写一写。\n\n2 + 4 + 6 + 8 = ( ) × ( )\n\n2 + 4 + 6 + 8 + 10 + 12 + 14 = ( ) × ( )\n () = 10 × 11",
+      "options": [],
+      "answer": "4;5;7;8;2 + 4 + 6 + 8 + 10 + 12 + 14 + 16+18 + 20",
+      "solution": "【分析】\n\n2 = 1 × 2\n2 + 4 = 2 × 3\n2 + 4 + 6 = 3 × 4\n2 + 4 + 6 + 8 = 4 × 5\n2 + 4 + 6 + 8 + 10 = 5 × 6\n2 + 4 + 6 + 8 + 10 + 12 = 6 × 7\n2 + 4 + 6 + 8 + 10 + 12 + 14 = 7 × 8\n··················\n\n从 2 开始的连续偶数相加的和正好等于相加的偶数的个数与 '个数 +1' 的乘积。\n\n由上可知, 等式左边是从从 2 开始的连续偶数相加的和, 等式右边是等式左边偶数的个数与 '个数 +1' 的积。\n\n【详解】 2 + 4 + 6 + 8 = (4) × (5)\n\n2 + 4 + 6 + 8 + 10 + 12 + 14 = (7) × (8)\n\n2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 = 10 × 11",
+      "knowledge_points": [
+        "数与形"
+      ],
+      "source": "21-22 六年级上·江西赣州·单元测试",
+      "merged_id": "unit8_030",
+      "original_id": "030"
+    },
+    {
+      "id": "031",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "百分之零点八七写作( )， 21.58% 读作( )。",
+      "options": [],
+      "answer": "0.87%;百分之二十一点五八",
+      "solution": "【分析】写百分数时，先写分子，再写百分号。一个百分数，%前面的数是几，就读作百分之几。百分数常常不写成分母是 100 的分数形式，而是在原来的分子后面添加上百分号 '%'来表示。\n\n【详解】百分之零点八七写作 0.87% , 21.58% 读作百分之二十一点五八。",
+      "knowledge_points": [
+        "百分数的读法和写法"
+      ],
+      "source": "20-21 六年级上·江西赣州·单元测试",
+      "merged_id": "unit8_031",
+      "original_id": "031"
+    },
+    {
+      "id": "032",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "按规律填数： 1/7 , 2/7 , 4/7 , 1又1/7 , ( )，( )。",
+      "options": [],
+      "answer": "2又2/7;4又4/7",
+      "solution": "【分析】观察数列发现: 1/7 × 2 = 2/7, 2/7 × 2 = 4/7, 4/7 × 2 = 1又1/7 , 相邻两数的商是 2 , 据此解答即可。本题考查数字之间的规律，发现规律，利用规律是解题的关键。\n\n【详解】由分析可知：1又1/7 × 2 = 2又2/7, 16/7 × 2 = 4又4/7",
+      "knowledge_points": [
+        "数字排列的规律"
+      ],
+      "source": "20-21 六年级上·福建厦门·单元测试",
+      "merged_id": "unit8_032",
+      "original_id": "032"
+    },
+    {
+      "id": "033",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "按规律填数: 1/4 、 3/4 、 2又1/4 、( )、 20又1/4 。",
+      "options": [],
+      "answer": "6又3/4",
+      "solution": "【分析】根据题意可知, 分母都为 4 , 分子的规律: 后一个数是前一个数的 3 倍, 据此解答即可。根据题意找到数字排列的规律是解答本题的关键,再根据规律解答问题。\n\n【详解】 1/4 、 3/4 、 2又1/4 （ 9/4 ）、 6又3/4 （ 27/4 ）、 20又1/4 （ 81/4 ）",
+      "knowledge_points": [
+        "数字排列的规律"
+      ],
+      "source": "20-21 六年级上·福建厦门·单元测试",
+      "merged_id": "unit8_033",
+      "original_id": "033"
+    }, {
+      "id": "034",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "1 + 3 + 5 + 7 + 9= ( )² = ( )\n\n1/2 + 1/4 + 1/8 + 1/16 + 1/32 = ( )",
+      "options": [],
+      "answer": "5;25;31/32",
+      "solution": "【分析】（1）n个连续从1相加的奇数和等于n的平方。 1 + 3 + 5 + 7 + 9 共5个奇数相加，所以得 5² = 25(2) 根据分子是 1 的分数相加, 且从第二个数开始, 每个数是前一个数的一半, 继续加下去, 和越来越接近 1 , 可得,相加的和的分子比分母小 1 , 据此解答。此类试题属于难度一般的试题, 解答此类试题时一定要主要找出解答的规律, 进而求解出此类试题的规律。\n\n【详解】 1 + 3 + 5 + 7 + 9 = 5² = 25  1/2 + 1/4 + 1/8 + 1/16 + 1/32 = 31/32",
+      "knowledge_points": [
+        "异分母分数加、减法"
+      ],
+      "source": "21-22 六年级上·江西赣州·单元测试",
+      "merged_id": "unit8_034",
+      "original_id": "034"
+    },
+    {
+      "id": "035",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "某校今年新招收的一年级新生中，男女生人数比是 3:2 ，如果要制作成扇形统计图，得算出女生占总人数的( )%，男生占总人数的( )%。",
+      "options": [],
+      "answer": "40;60",
+      "solution": "【分析】根据男女生人数比可知，男生人数有 3 份，女生人数有 2 份，总人数有 5 份。据此利用除法分别求出女生占总人数的百分之几，男生占总人数的百分之几即可。本题考查了百分数, 求一个数是另一个数的百分之几, 用除法。\n\n【详解】 2 ÷ (3 + 2) = 40% , 3 ÷ (3 + 2) = 60% , 所以, 女生占总人数的 40% , 男生占总人数的 60% 。",
+      "knowledge_points": [
+        "求一个数是另一个数的百分之几(百分率问题)",
+        "比的应用"
+      ],
+      "source": "21-22六年级上·江西赣州·单元测试",
+      "merged_id": "unit8_035",
+      "original_id": "035"
+    },
+    {
+      "id": "036",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "用小棒按下列方式摆图形，照这样方式摆下去，第5个图形用( )小棒，第10个图形用( )根小棒，第n个图形用( )根小棒。\n\n![](images/a470ed752a4e9e78fac63bb64132dcca271923dd10b15da2aa7d18260215a435.jpg)\n\n![](images/06117b7b43401a6f2756078fa16ffb0dfd3e03f31bfc42baab3918adcbb93c23.jpg)\n\n![](images/9b0e14c16d6372e230a8b565efa6a7e4eb065321b4509161bd303d0008a1ba17.jpg)",
+      "options": [],
+      "answer": "26;51;5n + 1",
+      "solution": "【分析】第一个图形用了 6 根小棒, 5 × 1 + 1 = 6 (根); 第二个图形用了 11 根小棒, 5 × 2 + 1 = 11 (根); 第三个图形用了 16 根小棒, 5 × 3 + 1 = 16 (根)。根据这个规律, 推断出第 5 个、第 10 个以及第 n 个图形需要的小棒数量即可。本题考查了找规律, 有一定逻辑推理和抽象概括能力是解题的关键。\n\n【详解】 5 × 5 + 1 = 26 （根），所以第 5 个图形用 26 小棒; 5 × 10 + 1 = 51 （根），所以第 10 个图形用 51 根小棒; 5 × n + 1 = 5n + 1 ，所以第 n 个图形用 5n + 1 根小棒。",
+      "knowledge_points": [
+        "用字母表示数",
+        "图形的规律",
+        "含有字母式子的化简与求值"
+      ],
+      "source": "21-22六年级上·江西赣州·单元测试",
+      "merged_id": "unit8_036",
+      "original_id": "036"
+    },
+    {
+      "id": "037",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "接着写。19，5，18，6，17，7，_、_。",
+      "options": [],
+      "answer": "16;8",
+      "solution": "【分析】这个题，我们来找规律，我们发现第一个数是 19，第三个数是 18，第五个数是 17，也就是奇数位置上得数依次都是减 1，所以第七个数应该是 16;同理可得，发现偶数位置上得数分别是 5、6、7，依次加 1，所以最后一个数应该是 8.这个题型主要是找规律，找出对应得数之后，一定要在检查一下，看是否满足。\n\n【详解】这组数得奇数位置上得数分别是:19、18、17、16这组数得偶数位置上得数分别是：5、6、7、8所以横线上分别填 16 和 8。",
+      "knowledge_points": [
+        "数字排列的规律"
+      ],
+      "source": "21-22 六年级上·全国·单元测试",
+      "merged_id": "unit8_037",
+      "original_id": "037"
+    },
+    {
+      "id": "038",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "根据规律, 在横线上画图。\n\n![](images/a433169f15d8c73f1ca72cc1796ac34e2c7e9e01761aca6de492c161230c9d70.jpg)",
+      "options": [],
+      "answer": "见详解",
+      "solution": "【分析】观察图形发现，第一个图形有一个正方形，第二个图形的第一层由2个正方形，第二层有1个，第三个图形的第一层有3个正方形，第二层有2个，第三层有1个，依次递减1，据此解答即可。本题考查图形的规律，发现规律，利用规律是关键。\n\n【详解】有分析可知, 第四个图形的第一层应该有 4 个正方形, 第二层有 3 个, 第三层有 2 个, 最上面的一层有 1 个, 如图所示:\n\n![](images/ef2b25a3225466da61960e5c1aa416a79511edec5033777136a14f8ec9e48c5a.jpg)",
+      "knowledge_points": [
+        "图形的规律"
+      ],
+      "source": "21-22 六年级上·全国·单元测试",
+      "merged_id": "unit8_038",
+      "original_id": "038"
+    },
+    {
+      "id": "039",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "找规律填数。\n\n100、80、60、（ ）、（ ）、（ ）。",
+      "options": [],
+      "answer": "40;20;0",
+      "solution": "【分析】由题目可知后一个数都比前一个数少 20 , 据此填空。解决本题的关键是根据所给数列得出规律，再利用这个规律写数。\n\n【详解】100、80、60、(40)、(20)、(0)。",
+      "knowledge_points": [
+        "数字排列的规律"
+      ],
+      "source": "21-22 六年级上·全国·单元测试",
+      "merged_id": "unit8_039",
+      "original_id": "039"
+    },
+    {
+      "id": "040",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "观察下面的算式看看你有什么发现？\n\n1³ + 2³ = 9 , (1 + 2)² = 9\n1³ + 2³ + 3³ = 36 , (1 + 2 + 3)² = 36\n1³ + 2³ + 3³ + 4³ = 100 , (1 + 2 + 3 + 4)² = 100\n......\n\n通过你的发现计算： 1³ + 2³ + 3³ + 4³ + ... +15³ = ( )",
+      "options": [],
+      "answer": "14400",
+      "solution": "【分析】观察算式发现，由 1³ = 1² ， 1³ + 2³ = (1 + 2)² ， 1³ + 2³ + 3³ = (1 + 2 + 3)²可知， 1³ + 2³ + 3³ + 4³ + ... + n³ = (1 + 2 + 3 + 4 + ... + n)²,主要考查了学生通过特例分析从而归纳总结出一般结论的能力。对于找规律的题目首先应找出哪些部分发生了变化，是按照什么规律变化的，通过分析找到各部分的变化规律后直接利用规律求解。\n\n【详解】由分析可知,\n\n1 + 2 + 3 + 4 + ... + 15\n= 16 × 7 + 8\n= 112 + 8\n= 120\n\n120² = 14400\n1³ + 2³ + 3³ + 4³ + ... + 15³ = (1 + 2 + 3 + 4 + ... + 15)² = 14400",
+      "knowledge_points": [
+        "算式的规律"
+      ],
+      "source": "21-22六年级上·全国·单元测试",
+      "merged_id": "unit8_040",
+      "original_id": "040"
+    },
+    {
+      "id": "041",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "困难",
+      "question": "瑞士数学教师巴尔末成功地从光谱数据 9/5 、 16/12 、 25/21 、 36/32 ，…中得到巴尔末公式，从而打开了光谱奥妙的大门，按这种规律写出的第7个数是( )。",
+      "options": [],
+      "answer": "81/77",
+      "solution": "【分析】由前面四个数可知, 分子是序数与 2 的和的平方,分母比分子小 4 , 可得第 7 个数。本题考查了数字排列的规律, 关键是要从前面的几个数找出规律从而进行解答。\n\n【详解】由题目可得： 9/5 = 3²/(3² - 4) ; 16/12 = 4²/(4² - 4) ; 25/21 = 5²/(5² - 4) ; 36/32 =6²/(6² - 4);所以第 7 个数为: 9²/(9² - 4) = 81/77 。",
+      "knowledge_points": [
+        "数字排列的规律"
+      ],
+      "source": "2021·辽宁沈阳·小升初真题",
+      "merged_id": "unit8_041",
+      "original_id": "041"
+    },
+    {
+      "id": "042",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "找规律, 填数。\n\n(1) 1, 2, 4, 7, 11, 16, ( ), 29, ( )。\n(2) 0.1, 0.01, 0.001, 0.0001, ( ), …这列数的每一项越来越小, 越来越接近( )。",
+      "options": [],
+      "answer": "22;37;0.00001;0",
+      "solution": "【分析】（1）后一位数减前一位数的差是逐项加 1，据此规律填空;(2) 由题目可知, 每个数都比前一个数多一位小数, 并且末位都是 1 , 其他数位上的数都是 0 , 这列数的每一项越来越小, 越来越接近 0 。本题考查的是探索规律的知识, 找到题目的规律是解答本题的关键。\n\n【详解】（1）由分析可得：1，2，4，7，11，16,(22) , 29, (37) ;  (2) 0.1, 0.01, 0.001, 0.0001, (0.00001), …这列数的每一项越来越小, 越来越接近 (0)。",
+      "knowledge_points": [
+        "数字排列的规律"
+      ],
+      "source": "21-22 六年级上·全国·单元测试",
+      "merged_id": "unit8_042",
+      "original_id": "042"
+    },
+    {
+      "id": "043",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "找规律:\n1, 3, 2, 6, 4, 9, 8, ( ), 16, 15, …",
+      "options": [],
+      "answer": "12",
+      "solution": "【分析】如图![](images/9399e6f909923cb0e39bd5603acecd20bad9bf7d77c8d698f9fb09cb415e788d.jpg)，提取出奇数项和偶数项, 奇数项的规律是前一个数 × 2 = 下一个数,偶数项的规律是前一个数 +3 = 下一个数。寻找数字排列中的规律，平时要注重多积累，培养数感。\n\n【详解】 9 + 3 = 12",
+      "knowledge_points": [
+        "数字排列的规律"
+      ],
+      "source": "20-21 六年级下·河南三门峡·期末",
+      "merged_id": "unit8_043",
+      "original_id": "043"
+    },
+    {
+      "id": "044",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "困难",
+      "question": "按规律填数。\n\n2、4、8、14、22、( )、44...",
+      "options": [],
+      "answer": "32",
+      "solution": "【分析】观察数列发现 4 - 2 = 2 、 8 - 4 = 4 、 14 - 8 = 6 、 22 - 14 = 8 后面一个数字减前面一个数字的结果依次是2、4、6、8、10、12、……，那么 22 后面的数字就是 22 + 10 = 32,根据给出的数列，找出数与数之间的变化规律，再由规律解决问题。\n\n【详解】因为 4 - 2 = 2 、 8 - 4 = 4 、 14 - 8 = 6 、 22 - 14 = 8 后面一个数字减前面一个数字的结果依次是2、4、6、8、10、12、……那么22后面的数字就是： 22 + 10 = 32",
+      "knowledge_points": [
+        "数字排列的规律"
+      ],
+      "source": "21-22 六年级上·全国·单元测试",
+      "merged_id": "unit8_044",
+      "original_id": "044"
+    },
+    {
+      "id": "045",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "找规律, 填一填。\n\n①0.2 、0.6、1.8、( )、16.2、( )、( )。\n② 0.1 、 0.2 、 0.3 、 0.5 、 0.8 、 ( )、 2.1 、 ( ) 。",
+      "options": [],
+      "answer": "5.4;48.6;145.8;1.3;3.4",
+      "solution": "【分析】①规律：每次扩大3倍，据此解答即可;② 规律: 从第三个数开始, 每个数是它前两个数的和, 据此解答即可。根据题干中的数字找到规律，再根据规律解答问题是解答本题的关键。\n\n【详解】① 1.8 × 3 = 5.4 ;\n\n16.2 × 3 = 48.6;\n48.6 × 3 = 145.8;\n② 0.5 + 0.8 = 1.3;\n\n1.3 + 2.1 = 3.4 。",
+      "knowledge_points": [
+        "数字排列的规律"
+      ],
+      "source": "21-22 六年级上·全国·单元测试",
+      "merged_id": "unit8_045",
+      "original_id": "045"
+    },
+    {
+      "id": "046",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "根据下列数的规律,填上适当的数。\n\n5，12，（ ），26，33，40，（ ），54，61。",
+      "options": [],
+      "answer": "19;47",
+      "solution": "【分析】观察题中的数字发现, 12 - 5 = 7, 33 - 26 = 7, 40 - 33 = 7, 61 - 54 = 7 , 据此推出, 前后两个数的差是 7 , 这是一组差为 7 的有规律数字。据此得解。解答本题的关键是根据已有的数字找到规律，再根据规律解决问题。\n\n【详解】 12 + 7 = 19 ;40 + 7 = 47",
+      "knowledge_points": [
+        "数字排列的规律"
+      ],
+      "source": "21-22 六年级上·全国·单元测试",
+      "merged_id": "unit8_046",
+      "original_id": "046"
+    },
+    {
+      "id": "047",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "找规律, 填一填。\n\n0，3，8，15，24，（ ）。",
+      "options": [],
+      "answer": "35",
+      "solution": "【分析】这组数的排列规律是：后一个数依次比前一个数多 3、5、7、9、11...，据此解答。本题考查数字的排列规律，需通过认真观察、分析、计算等方法，找出数字排列的规律。\n\n【详解】 24 + 11 = 35",
+      "knowledge_points": [
+        "数字排列的规律"
+      ],
+      "source": "20-21 六年级下·辽宁·期末",
+      "merged_id": "unit8_047",
+      "original_id": "047"
+    },
+    {
+      "id": "048",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "找规律填数:\n\n64、49、( )、25、16、9、4。",
+      "options": [],
+      "answer": "36",
+      "solution": "【分析】经过观察发现，第一个数 64 是 8 的平方、第二个数 49 是 7 的平方、第四个数 25 是 5 的平方……因此，第三个数就应该是 6 的平方 36。数字从右至左逐渐变大，应用的计算可能有乘法、加法或平方;再经过进一步计算验证，可得答案。\n\n【详解】 64 = 8² , 49 = 7² , 36 = 6² , 25 = 5²",
+      "knowledge_points": [
+        "数字排列的规律"
+      ],
+      "source": "19-20 六年级下·浙江金华·期末",
+      "merged_id": "unit8_048",
+      "original_id": "048"
+    },
+    {
+      "id": "049",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "把一根绳对折1次，得到2段，且有一个折痕;再对折1次，得到4段，且有3个折痕，如此下去，对折5次后，得到( )段，且有( )个折痕。",
+      "options": [],
+      "answer": "32;31",
+      "solution": "【分析】把一根绳对折 1 次, 得到 2 段, 且有一个折痕; 再对折 1 次, 得到 4 段, 且有 3 个折痕。则所得的段数减 1 得到对应折痕数。折几次, 得到的段数就为几个 2 相乘的积。本题考查数和形中的找规律问题。找到共同特征解决问题即可。\n\n【详解】 2 × 2 × 2 × 2 × 2 = 32 （段）32 - 1 = 31 （个）则对折 5 次后, 得到 32 段, 且有 31 个折痕。",
+      "knowledge_points": [
+        "数与形"
+      ],
+      "source": "2021三年级下·全国·专题练习",
+      "merged_id": "unit8_049",
+      "original_id": "049"
+    },
+    {
+      "id": "050",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "有黑白颜色的珠子共84个，按照5个黑珠子，2个白珠子的顺序排列，白色的珠子占总数的( )，第58个珠子是( )色。",
+      "options": [],
+      "answer": "2/7;黑",
+      "solution": "【分析】根据题干，可得这组珠子是7个一个循环周期，分别按照5黑2白的顺序排列， 84 ÷ 7 = 12 ，即经历了12个循环周期，据此即可求出白色珠子的个数，再除以总个数即可求出白色的珠子占总数的几分之几。用58除以7，求出第58颗是第几个循环周期的第几个即可解答。根据题干得出这组珠子的排列规律是解决此类问题的关键。\n\n【详解】 5 + 2 = 7 （个）84 ÷ 7 = 12   12 × 2 ÷ 84 = 2/7      58 ÷ 7 = 8 .... 2所以第 58 个珠子是第 9 循环周期的第 2 个, 是黑珠子。",
+      "knowledge_points": [
+        "间隔问题"
+      ],
+      "source": "2021三年级下·全国·专题练习",
+      "merged_id": "unit8_050",
+      "original_id": "050"
+    }, {
+      "id": "051",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "困难",
+      "question": "2020 个 8 连乘, 乘积的个位上的数字是( )。",
+      "options": [],
+      "answer": "6",
+      "solution": "【分析】要求 n 个 8 连乘的积的个位上的数字, 只需看个位上的数字与 8 相乘的情况, 即 8 × 8 = 64, 4 × 8 = 32, 2 × 8 = 16,6 × 8 = 48, 8 × 8 = 64, 4 × 8 = 32, ... 由此可以发现, n个 8 相乘的积的个位上的数字依次按 8 , 4 , 2 , 6 这四个数字进行循环, 因此可以用解决周期问题的方法来解答。 2020 ÷ 4 = 505 ,所以 2020 个 8 连乘的积的个位上的数字是 6 。因为是 n 个 8 相乘，在乘的过程中，乘积末尾的数字必然会按一定规律排列，只要找到这列数字排列的规律，问题就可迎刃而解。只是注意第一个乘积末尾数字是 8 本身,而不是 8 × 8 = 64 中的数字 4。\n\n【详解】由分析可知: n 个 8 相乘的积的个位上的数字依次按 8, 4, 2, 6 这四个数字进行循环。2020 ÷ 4 = 505 , 所以 2020 个 8 连乘, 乘积的个位上的数字是 6。",
+      "knowledge_points": [
+        "间隔问题",
+        "数字排列的规律"
+      ],
+      "source": "20-21 六年级下·全国·课后作业",
+      "merged_id": "unit8_051",
+      "original_id": "051"
+    },
+    {
+      "id": "052",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "2, 4, 6, 8_, 14, 14, _。",
+      "options": [],
+      "answer": "10;12;16",
+      "solution": "【分析】偶数：能被2整除的自然数叫偶数。如：2、4、6、8、10…。首先依据数字排列规律大胆猜测, 然后写出后面的数字, 为防止出错, 还可以进行验证, 看看填入的数字是否符合这一组数字的规律。\n\n【详解】观察前几个数可知, 这是一组偶数, 且分别是 2 的 1 倍、 2 倍、 3 倍、 4 倍……, 则后面数就是 2 的 5 倍、 6 倍、 7 倍、 8 倍。",
+      "knowledge_points": [
+        "数字排列的规律"
+      ],
+      "source": "20-21 六年级上·全国·单元测试",
+      "merged_id": "unit8_052",
+      "original_id": "052"
+    },
+    {
+      "id": "053",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "按规律填数。\n\n(1) 21, 32, 43, _, _, _。\n(2) _, 31, 38, 45, _, _, 66。",
+      "options": [],
+      "answer": "54;65;76;24;52;59",
+      "solution": "【分析】后面的数比前面的数每次递增 11, 43 + 11 = 54, 54 + 11 = 65, 65 + 11 = 76 。后面的数比前面的数每次递增 7, 31 - 7 = 24, 45 + 7 = 52, 52 + 7 = 59 。对于找规律的题目首先应找出哪些部分发生了变化，是按照什么规律变化的，通过分析找到各部分的变化规律后直接利用规律求解。\n\n【详解】 (1) 21, 32, 43, 54, 65, 77。(2) 24, 31, 38, 45, 52, 57, 66。",
+      "knowledge_points": [
+        "数字排列的规律"
+      ],
+      "source": "20-21 六年级上·全国·单元测试",
+      "merged_id": "unit8_053",
+      "original_id": "053"
+    },
+    {
+      "id": "054",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "根据所给出的算式的规律，写出其他算式的得数。\n\n11 × 99 = 1089\n111 × 999 = 110889\n1111 × 9999 = 11108889\n11111 × 99999 =\n111111 × 999999 =",
+      "options": [],
+      "answer": "1111088889;11110888889",
+      "solution": "【分析】观察上面的算式，先找出规律第几个算式就有几个 1 和 8, 1 与 8 之间有一个 0 , 最后一位是 9。本题考查数的规律, 关键是找出已知算式中的规律即可填空。\n\n【详解】 11111 × 99999 = 1111088889   111111 × 999999 = 11110888889",
+      "knowledge_points": [
+        "算式的规律"
+      ],
+      "source": "18-19六年级上·全国·单元测试",
+      "merged_id": "unit8_054",
+      "original_id": "054"
+    },
+    {
+      "id": "055",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "笑笑在一座宝塔的底层，发现有一张字条告诉她再登多少个台阶有藏宝图：\n\n'比125大，小于180;5个5个地数，没有剩余;又是4和8的倍数。'笑笑再登____台阶才能找到藏宝图。",
+      "options": [],
+      "answer": "160个",
+      "solution": "【分析】因为台阶数是 4、5、8 的倍数，所以从 4、5、8 的最小公倍数中判断台阶数即可。此题从 4、5、8 的最小公倍数入手, 找出适合的倍数。\n\n【详解】 4 × 8 = 32, 32 × 5 = 160, 160 既是 4 和 8 的倍数, 又是 5 的倍数, 125 < 160 < 180 , 所以台阶数是 160 个。",
+      "knowledge_points": [
+        "最小公倍数的应用"
+      ],
+      "source": "20-21 六年级上·全国·单元测试",
+      "merged_id": "unit8_055",
+      "original_id": "055"
+    },
+    {
+      "id": "056",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "如果要反映一天温度的变化情况，我们应该绘制的统计图是( )。",
+      "options": [],
+      "answer": "折线统计图",
+      "solution": "【分析】折线统计图能直观地反映事物变化情况，由此解答即可。明确折线统计图的特点是解答本题的关键。\n\n【详解】如果要反映一天温度的变化情况，我们应该绘制的统计图是折线统计图。",
+      "knowledge_points": [
+        "单式折线统计图"
+      ],
+      "source": "20-21 六年级上·全国·单元测试",
+      "merged_id": "unit8_056",
+      "original_id": "056"
+    },
+    {
+      "id": "057",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "条形统计图能直观地表示出每个项目的( );折线统计图能直观地反映事物( );扇形统计图能直观地表示出各部分在总体中所占的( )。",
+      "options": [],
+      "answer": "具体数目;变化情况;百分比",
+      "solution": "【详解】条形统计图能直观地表示出每个项目的具体数目;折线统计图能直观地反映事物变化情况;扇形统计图能直观地表示出各部分在总体中所占的百分比。",
+      "knowledge_points": [
+        "统计图的特点"
+      ],
+      "source": "20-21 六年级上·全国·单元测试",
+      "merged_id": "unit8_057",
+      "original_id": "057"
+    },
+    {
+      "id": "058",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "在第 28 届奥运会上,为了统计在奥运会上各国奖牌数所占的百分比，应选用( )统计图。",
+      "options": [],
+      "answer": "扇形",
+      "solution": "【分析】扇形统计图能直观地表示出各部分在总体中所占的百分比，由此解答即可。本题主要考查了扇形统计图的特点，一定要牢记。\n\n【详解】在第 28 届奥运会上, 为了统计在奥运会上各国奖牌数所占的百分比, 应选用扇形统计图。",
+      "knowledge_points": [
+        "统计图的选择"
+      ],
+      "source": "20-21 六年级上·全国·单元测试",
+      "merged_id": "unit8_058",
+      "original_id": "058"
+    },
+    {
+      "id": "059",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "困难",
+      "question": "1 + 3 + 5 + 7 + 9 = ( )² = ( )\n\n4² = 1 + 3 + ( ) + ( )",
+      "options": [],
+      "answer": "5;25;5;7",
+      "solution": "【分析】①首先计算出原题的结果，是25，而25是5的平方，所以先填入数字5，再填入数字25;② 先计算出 4 的平方是 16, 再用 16 减去 1 与 3 的和, 是\n\n12, 因为 1 和 3 都是奇数, 所以接下来填入的数字也必须是\n\n奇数，这样，就想到 12 = 5 + 7 ，解答完毕。\n\n【详解】 1 + 3 + 5 + 7 + 9 = 25\n\n25 = 5²\n4² = 16\n16 - 1 - 3 = 12\n12 = 5 + 7",
+      "knowledge_points": [
+        "数与形"
+      ],
+      "source": "20-21 六年级上·全国·单元测试",
+      "merged_id": "unit8_059",
+      "original_id": "059"
+    },
+    {
+      "id": "060",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "1 + 3 + 5 + 7 + 9 + 11+13 + 11 + 9 + 7 + 5 + 3 + 1 = ( )² + ( )² = ( ) 。",
+      "options": [],
+      "answer": "7;6;85",
+      "solution": "【分析】通过观察发现，此题即 1 + 3 + 5 + 7 + 9 + 11 + 13 和 11 + 9 + 7 + 5 + 3 + 1; 1 + 3 + 5 + 7 + 9 + 11 + 13 = 7², 11 + 9 + 7 + 5 + 3 + 1 = 6² ，由此解答即可。本题主要考查了学生分析、总结、归纳的能力。\n\n【详解】 1 + 3 + 5 + 7 + 9 + 11 + 13 + 11 + 9 + 7 + 5 + 3 + 1 = 7² + 6²= 85",
+      "knowledge_points": [
+        "算式的规律"
+      ],
+      "source": "20-21六年级上·全国·单元测试",
+      "merged_id": "unit8_060",
+      "original_id": "060"
+    },
+    {
+      "id": "061",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "找规律填数：1，8，27，（ ），125。",
+      "options": [],
+      "answer": "64",
+      "solution": "【分析】观察数据发现: 1 = 1³, 8 = 2³, 27 = 3³, 125 = 5³ , 括号里的数应是 4³ = 64 , 据此解答即可。本题考查找规律, 解答本题的关键是找到数据排列的特点。\n\n【详解】这列数的排列规律是 1³ , 2³ , 3³ , 4³ , 5³ , 所以括号里的数为 64。",
+      "knowledge_points": [
+        "数字排列的规律"
+      ],
+      "source": "2020·河南·小升初真题",
+      "merged_id": "unit8_061",
+      "original_id": "061"
+    },
+    {
+      "id": "062",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "(9, 8),(6, 12), (4, 18), (3, _), (_, 36)。",
+      "options": [],
+      "answer": "24;2",
+      "solution": "【分析】观察数据发现，每组数据中，第一个数与第二个数的乘积都相等。9 × 8 = 72, 6 × 12 = 72, 4 × 18 = 72，因此 3 × ? = 72，? = 24;? × 36 = 72，? = 2。\n\n【详解】观察可知，每组两个数的乘积均为72，所以 3 × 24 = 72，2 × 36 = 72。",
+      "knowledge_points": [
+        "数字排列的规律"
+      ],
+      "source": "19-20 六年级下·四川·单元测试",
+      "merged_id": "unit8_062",
+      "original_id": "062"
+    },
+    {
+      "id": "063",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "一张方桌最多坐8人，2 张这样的方桌拼在一起排放，可以坐 16 人。（）",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】根据并排方桌的特点，当两张桌子拼在一起，则方桌各有一条边不能坐人，据此用一张方桌最多坐的人数乘2，再减去并在一起时的两边坐的人数即可判断。\n\n【详解】 8 ÷ 4 = 2 （人） 8 × 2 - 2 × 2\n= 16 - 4\n= 12(人)\n\n所以一张方桌最多坐 8 人, 2 张这样的方桌拼在一起排放,可以坐 12 人。所以原题说法错误。故答案为: ×改正：错在两张方桌并在一起，没有考虑两边不能坐人，一张方桌最多坐 8 人, 2 张这样的方桌拼在一起排放, 可以坐12人。",
+      "knowledge_points": [
+        "数与形(归纳递推)"
+      ],
+      "source": "23-24五年级上·全国·单元测试",
+      "merged_id": "unit8_063",
+      "original_id": "063"
+    },
+    {
+      "id": "064",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "根据 99 × 99 = 9801, 999 × 999 = 998001 , 9999 × 9999 = 99980001 , 可以直接得出 99999 × 99999 = 99998000001 。（）",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】算式中两个因数都相同，且每个数位上的数都是 9，通过观察发现，积前面几位 9 的个数比因数 9 的个数少一个，积写完几个 9 之后，就是数字 8，接着是 0，积中 0 的个数比因数 9 的个数少 1 个，积最后一位是数字 1，据此解答。\n\n【详解】根据 99 × 99 = 9801, 999 × 999 = 998001, 9999 × 9999 = 99980001 , 可以直接得出 99999 × 99999 = 9999800001 , 原题说法错误。故答案为: ×",
+      "knowledge_points": [
+        "算式的规律"
+      ],
+      "source": "23-24 四年级上·陕西西安·期中",
+      "merged_id": "unit8_064",
+      "original_id": "064"
+    },
+    {
+      "id": "065",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "在 1 + 3 + 5+7 + 9 + ... 中, 从数 '1' 到数 '15' 的和是 8² 。( )",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】连续几个奇数的和等于奇数的个数的平方。据此判断即可。本题考查连续几个奇数的和, 明确求连续的奇数的和的计算方法是解题的关键。\n\n【详解】由分析可知：1 + 3 + 5 + 7 + 9 + ... 15 = 8² = 8 × 8 = 64 。原题干说法正确。故答案为: √",
+      "knowledge_points": [
+        "数与形",
+        "算式的规律"
+      ],
+      "source": "22-23 六年级上·内蒙古呼伦贝尔·期末",
+      "merged_id": "unit8_065",
+      "original_id": "065"
+    },
+    {
+      "id": "066",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 13 + 11 + 9 + 7 + 5 + 3 + 1 = 8² + 7² ( )",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 , 一共有 8 个相邻的奇数, 13 + 11 + 9 + 7 + 5 + 3 + 1 , 一共有 7 个相邻的奇数,从 1 开始 n 个相邻奇数的和 = n² , 据此计算。关键是理解并明确从 1 开始 n 个相邻奇数的总和等于奇数个数的平方。\n\n【详解】 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 8²   13 + 11 + 9 + 7 + 5 + 3 + 1 = 7²所以 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 13 + 11 + 9 + 7 + 5 + 3 + 1\n= 8² + 7²\n= 64 + 49\n= 113\n\n故答案为: √",
+      "knowledge_points": [
+        "数与形"
+      ],
+      "source": "20-21 六年级上·四川德阳·期末",
+      "merged_id": "unit8_066",
+      "original_id": "066"
+    },
+    {
+      "id": "067",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "![](images/264783a46a7f75de80981c9aa427bc85.png)(按规律往下画，第19个图形是![](images/4528272204e8a5b0b6259bee95567919.png)",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】观察这组图形可得 3 个图形是一个周期, 求第 n 个图形是什么, 则用 n ÷ 3 , 得出的余数是 1 时则与第一个图形相同; 得出的余数是 2 时则与第二个图形相同; 没有余数时即与第三个图形相同。此题考查是的找规律, 正确找出规律并用规律解决问题是解题关键。\n\n【详解】 19 ÷ 3 = 6 ... 1 所以第 19 个图形与第一个图形相同, 是 ★ , 即正确。故答案为: √",
+      "knowledge_points": [
+        "周期问题"
+      ],
+      "source": "2022 六年级下·全国·专题练习",
+      "merged_id": "unit8_067",
+      "original_id": "067"
+    },{
+      "id": "068",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "1 + 3 + 5 + 7 + 9 + 7 + 5 + 3 + 1 = 5² + 4² ( )",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】从 1 开始的连续奇数相加, 和等于奇数个数的平方,据此解答即可。本题考查数与形, 解答本题的关键是掌握题中相加数字的和的规律。\n\n【详解】 1 + 3 + 5 + 7 + 9 + 7 + 5 + 3 + 1\n=(1 + 3 + 5 + 7 + 9) + (7 + 5 + 3 + 1)\n=5² + 4²\n\n故答案为: √ 。",
+      "knowledge_points": [
+        "数与形"
+      ],
+      "source": "21-22 六年级上·山东临沂·期末",
+      "merged_id": "unit8_068",
+      "original_id": "068"
+    },
+    {
+      "id": "069",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "判断题",
+      "difficulty": "困难",
+      "question": "6 个点连成线段, 最多可以连成 25 条线段。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】一个点和另外 5 个点最多可以连成 5 条线段, 在不重复的情况下, 第 2 个点最多可以连成 4 条线段, 第 3 个点最多可以连成 3 条线段, 第 4 个点最多可以连成 2 条线段,第 5 个点最多可以连成 1 条线段, 最后计算出它们的和即可。把点的个数看作 n , 则 n 个点最多可以连成线段的条数就等于从 1 到 (n - 1) 的连续自然数的和。\n\n【详解】 5 + 4 + 3 + 2 + 1\n\n= 9 + 3 + 2 + 1\n= 12 + 2 + 1\n= 14 + 1\n= 15(条)\n\n故答案为: ×",
+      "knowledge_points": [
+        "数与形"
+      ],
+      "source": "21-22 六年级上·江西赣州·单元测试",
+      "merged_id": "unit8_069",
+      "original_id": "069"
+    },
+    {
+      "id": "070",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ... = 1 。()",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】根据算式可知, 后面一个数是前一个数的 1/2 , 如果把一条线段看作 1 , 先取它的一半表示 1/2 , 再取余下的一半的一半表示 1/4 , 这样不断地取下去, 最终相当于取了整条线段。所以 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ... = 1 , 据此解答即可。![](images/05703f090f5775efc07bf673b0bfe0e16a8e6a3d1ca8bbb2b5363f1f525950c3.jpg)本题采用了数形结合的思想，使题目形象化，再利用极限思想得到结果。\n\n【详解】 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ... = 1 , 说法正确; 故答案为: √ 。",
+      "knowledge_points": [
+        "数与形"
+      ],
+      "source": "20-21 六年级上·湖北荆门·期末",
+      "merged_id": "unit8_070",
+      "original_id": "070"
+    },
+    {
+      "id": "071",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "判断题",
+      "difficulty": "困难",
+      "question": "1 - 1/2 - 1/4 - 1/8 - 1/16 - 1/32 = 1/32 。",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】可从减数是前 1 个分数、前 2 个分数、前 3 个分数这样稍简单的式子推理出一般规律，再计算整个式子。如果计算整个式子，计算量将非常大，且容易马虎做错。像这样由易到难，推理出算式的规律，再加以应用, 十分易于理解和计算。本题也可以通过作图进行简便计算。\n\n【详解】由分析得:\n\n1 - 1/2 = 1/2\n=1 - 1/2 - 1/4\n= 1 - 2/4 - 1/4\n= 1/4\n\n1 - 1/2 - 1/4 - 1/8\n= 1 - 4/8 - 2/8 - 1/8\n= 1/8\n\n因此: 1 - 1/2 - 1/4 - 1/8 - 1/16 - 1/32 = 1/32 。故答案为: √。",
+      "knowledge_points": [
+        "数与形"
+      ],
+      "source": "19-20 六年级下·安徽铜陵·期末",
+      "merged_id": "unit8_071",
+      "original_id": "071"
+    },
+    {
+      "id": "072",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "8 个点可以连成 20 条线段。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】3 个点连成线段的条数: 1 + 2 = 3 (条), 4 个点连成线段的条数: 1 + 2 + 3 = 6 (条), 5 个点连成线段的条数: 1 + 2 + 3 + 4 = 10 (条), 由此得出规律: n 个点的线段数是: 1 + 2 + 3 + 4 + ...... + n - 1 条线段, 据此规律解答即可。本题考查探索规律, 解答本题的关键是根据数形结合的思想, 在草稿纸上画出几个点, 然后得出规律, 根据规律解答问题。\n\n【详解】8 个点可以连成线段数为 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28（条），所以本题说法错误。故答案为: ×",
+      "knowledge_points": [
+        "数与形"
+      ],
+      "source": "2018·山东临沂·小升初真题",
+      "merged_id": "unit8_072",
+      "original_id": "072"
+    },
+    {
+      "id": "073",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "在扇形统计图中,所有扇形的百分比之和为 1。( )",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】扇形统计图中，将整个圆看作单位'1'，则所有扇形的百分比之和为 1 。本题主要考查扇形统计图的概念和意义, 圆代表整体, 即单位 '1', 各个扇形代表部分, 掌握扇形统计图的特点是关键。\n\n【详解】在扇形统计图中，所有扇形的百分比之和为 1 ，题目描述正确。故答案为: √",
+      "knowledge_points": [
+        "扇形统计图的特点及绘制"
+      ],
+      "source": "19-20 六年级下·江苏·单元测试",
+      "merged_id": "unit8_073",
+      "original_id": "073"
+    },
+    {
+      "id": "074",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "在数列 1/1, 3/4, 5/9, 7/16, 9/25, 11/36, ... 中，第10个数是 19/100 。",
+      "options": [],
+      "answer": "√",
+      "solution": "【详解】这个数列中从左到右分别是1、3、5、7、9、11、13、15、17、19……分母从左到右分别是 1², 2², 3², 4², 5², ...由此可知: 第 10 个数分子是 19 , 分母是 100 。故答案为: √",
+      "knowledge_points": [
+        "数与形"
+      ],
+      "source": "19-20六年级上·全国·期末",
+      "merged_id": "unit8_074",
+      "original_id": "074"
+    },
+    {
+      "id": "unit8_calc_075",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "计算题",
+      "difficulty": "困难",
+      "question": "观察下列等式: 第一个等式: a₁ = 1/(1 × 3) = 1/2 × (1 - 1/3) ;第二个等式: a₂ = 1/(3 × 5) = 1/2 × (1/3 - 1/5) 第三个等式: a₃ = 1/(5 × 7) = 1/2 × (1/5 - 1/7) 第四个等式: a₄ = 1/(7 × 9) = 1/2 × (1/7 - 1/9) 请回答下列问题:\n\n(1) 按以上规律列出第 5 个等式: a₅ = ;\n(2) 用含 n 的代数式表示第 n 个等式: aₙ = (n 为正整数);\n(3) 求 a₁ + a₂ + a₃ + a₄ + ... + a₁₀₀ 的值。",
+      "options": [],
+      "answer": "(1) 1/2 × (1/9 - 1/11);(2) 1/2 × (1/(2n - 1) - 1/(2n + 1));(3) 100/201",
+      "solution": "【分析】（1）（2）通过观察可知，第一个等号后面的式子规律是分子不变，分母是两个连续奇数的乘积，它们与式子序号之间的关系为序号的2倍减1和序号2倍加1;第二个等号后面的式子则是 1/2 乘以第一个奇数分数减第二个奇数分数，根据规律列式即可解得。(3) 将具体各项根据分数列项公式, 括号外提取 1/2 , 将括号内依次相消进行计算即可解得。本题的难点在于通过观察等式规律写出新的等式,再通过提取公因数将复杂的等式进行化简即可解得。\n\n【详解】（1）通过观察可知，第一个等号后面的分子为 1,分母为序号的 2 倍减 1 乘以序号 2 倍加 1 ,\n\n第二个等号后面为 1/2 乘以第一个奇数分数减第二个奇数分数, 故 a₅ = 1/(9 × 11) = 1/2 × (1/9 - 1/11) 。\n\n(2) 通过 (1) 的规律可知,aₙ = 1/((2 × n - 1) × (2 × n + 1)) = 1/2 × (1/(2 × n - 1) - 1/(2 × n + 1)) (n为正整数) 。\n\n(3) 由题, a₁ + a₂ + a₃ + a₄ + ⋯ + a₁₀₀\n\n1/2 × (1 - 1/3) + 1/2 × (1/3 - 1/5) + 1/2 × (1/5 - 1/7) + ... + 1/2 × (1/199 - 1/201)\n=1/2 (1 - 1/3 + 1/3 - 1/5 + ... + 1/199 - 1/201)\n= 100/201",
+      "knowledge_points": [
+        "算式的规律（分数）",
+        "分数的四则混合运算"
+      ],
+      "source": "25-26 六年级上·全国·单元测试",
+      "merged_id": "unit8_075",
+      "original_id": "unit8_calc_075"
+    },
+    {
+      "id": "unit8_calc_076",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "计算。\n\n1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729",
+      "options": [],
+      "answer": "364/729",
+      "solution": "【分析】根据分数的意义作图如下：\n\n图一![](images/b1cb4852b3ec3997a502e41bc543d23f5b685cf993596af65e86cc49730934f5.jpg) 图二 ![](images/855fa3e18e3e9e9a6f90be4aa4cfcf675930e8d98d88e8ae8c84fb22387b98e1.jpg)\n\n图三![](images/f19acb17d32d36c63802613c0e9ad8d3e4b4601253dfe34f78839df304b291a4.jpg)\n\n从图一可知: 1/3 = (1 - 1/3) ÷ 2 从图二可知: 1/3 + 1/9 = 4/9 = (1 - 1/9) ÷ 2 从图三可知: 1/3 + 1/9 + 1/27 = 13/27 = (1 - 1/27) ÷ 2 根据此规律， 1/3 + 1/9 + 1/27 + 1/81 = (1 - 1/81) ÷ 2 = 40/81因此计算 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729 , 只需 (1 - 1/729) ÷ 2 即可。\n\n【详解】 1/3 +1/9 +1/27 +1/81 +1/243 +1/729\n= (1 - 1/729) ÷ 2\n= 728/729 ÷ 2\n= 364/729",
+      "knowledge_points": [
+        "异分母分数加、减法",
+        "分数与整数的除法",
+        "数与形(归纳递推)"
+      ],
+      "source": "23-24 六年级上·全国·课后作业",
+      "merged_id": "unit8_076",
+      "original_id": "unit8_calc_076"
+    },
+    {
+      "id": "unit8_calc_077",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "计算。\n\n(1) 1 + 3 + 5 + 7 + 9 + 11 + 13\n\n(2) 103 + 105 + 107 + ...+203 + 205",
+      "options": [],
+      "answer": "(1)49;(2)8008",
+      "solution": "【分析】（1）计算规律：从 1 开始，几个连续奇数相加，和等于加数个数的平方，据此计算即可;(2) 103 + 205 = 105 + 203 ... , 一共有 (52 ÷ 2) 组, 据此解答即可。\n\n【详解】 (1) 1 + 3 + 5 + 7 + 9 + 11 + 13=7²=7 × 7=49\n\n(2) 103 + 105 + 107 + ... + 203 + 205\n=(103 + 205) × (52 ÷ 2)\n=308 × 26\n=8008",
+      "knowledge_points": [
+        "数与形（探索规律）"
+      ],
+      "source": "23-24 六年级上·全国·单元测试",
+      "merged_id": "unit8_077",
+      "original_id": "unit8_calc_077"
+    },
+    {
+      "id": "unit8_calc_078",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "计算题",
+      "difficulty": "困难",
+      "question": "找规律，算一算。\n\n(1) 1 - 1/2 - 1/4 - 1/8 - 1/16 - 1/32    (2) 5 + 7 + 9 + 11 + 13 + ... + 25",
+      "options": [],
+      "answer": "(1) 1/32;(2) 165",
+      "solution": "【分析】（1）将 1/2 拆成 (1 - 1/2) ， 1/4 拆成 (1/2 - 1/4) ， 1/8 拆成 (1/4 - 1/8) ， 1/16 拆成 (1/8 - 1/16) ， 1/32 拆成 (1/16 - 1/32) ，再去括号，括号前边是减号，去掉括号，括号里的减号变加号，前边抵消后，只剩下 1/32 ，据此计算;(2) 可以将数列分为 5 对数, 每对数的和都是 30, 即 (5 + 25) + (7 + 23) + (9 + 21) + (11 + 19) + (13 +\n\n17), 除了中间的 15, 先计算出 5 对数的和, 再加上 15 即可。较复杂的分数加运算可以利用裂项以及转化法解决计算是解答本题的关键。\n\n【详解】 (1) 1 - 1/2 - 1/4 - 1/8 - 1/16 - 1/32\n= 1 - (1 - 1/2) - (1/2 - 1/4) - (1/4 - 1/8) - (1/8 - 1/16) - (1/16 - 1/32)\n= 1 - 1 + 1/2 - 1/2 + 1/4 - 1/4 + 1/8 - 1/8 + 1/16 - 1/16 + 1/32\n= 1/32\n\n(2) 5 + 7 + 9 + 11 + 13 + ... + 25\n= (5 + 25) + (7 + 23) + (9 + 21) + (11 + 19) + (13 + 17) + 15\n= 30 × 5 + 15\n= 150 + 15\n= 165",
+      "knowledge_points": [
+        "数与形(归纳递推)"
+      ],
+      "source": "23-24六年级上·全国·单元测试",
+      "merged_id": "unit8_078",
+      "original_id": "unit8_calc_078"
+    },
+    {
+      "id": "unit8_calc_079",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "计算题",
+      "difficulty": "容易",
+      "question": "直接写出得数。\n\n28.5 - 0.85=  1.25 × 2/5 =  2.4 + 1.2 ÷ 4 =  3 + 3/4 - 3 + 3/4 =\n0.5² =  9 × 70% =  5/8 ÷ 1/2 =  1 + 3 + 5 + 7 + 9 + ... + 21 =",
+      "options": [],
+      "answer": "27.65; 0.5; 2.7; 3/2; 0.25; 6.3; 5/4; 121",
+      "solution": "【详解】略。",
+      "knowledge_points": [
+        "分数与分数的除法",
+        "小数与小数的乘法",
+        "含百分数的运算",
+        "除数是整数的小数除法"
+      ],
+      "source": "22-23 六年级上·江西吉安·期末",
+      "merged_id": "unit8_079",
+      "original_id": "unit8_calc_079"
+    },
+    {
+      "id": "unit8_calc_080",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "计算题",
+      "difficulty": "困难",
+      "question": "求从 1 开始的连续自然数的和公式为: 1 + 2 + 3 + ... + n = (1 + n) × n ÷ 2 ,\n\n例如：\n\n1 + 2 + 3 + 4 + ... + 50\n= (1 + 50) × 50 ÷ 2\n= 51 × 50 ÷ 2\n= 1275\n\n根据方法计算。\n\n(1) 1 + 2 + 3 + ... + 100\n(2) 10 + 11 + 12 + ... + 50",
+      "options": [],
+      "answer": "(1) 5050;(2) 1230",
+      "solution": "【分析】 1 + 2 + 3 + ... + n = (1 + n) × n ÷ 2 ，求从 1 开始的 n 个连续自然数之和，可以利用梯形面积公式简便计算，梯形面积 = （上底 + 下底） × 高 ÷ 2 ， 1 + 2 + 3 + ... + n 可以看作上底为 1 ，下底为 n ，高也为 n 的梯形面积;(1) 1 + 2 + 3 + ... + 100 , 即梯形的上底为 1, 下底为 100,高为 100 , 求梯形面积可列式; (1 + 100) × 100 ÷ 2 ; (2) 10 + 11 + 12 + ... + 50 = (1 + 2 + 3 + ... + 50) - (1 + 2 + 3 + ... + 9) , 即上底 1, 下底 50, 高为 50 的大梯形面积减去上底为 1, 下底为 9, 高为 9 的小梯形面积。\n\n【详解】 (1) 1 + 2 + 3 + ... + 100\n= (1 + 100) × 100 ÷ 2\n= 101 × 100 ÷ 2= 10100 ÷ 2\n= 5050\n\n(2) 10 + 11 + 12 + ... + 50\n= (1 + 2 + 3 + ... + 50) - (1 + 2 + 3 + ... + 9)\n= (1 + 50) × 50 ÷ 2 - (1 + 9) × 9 ÷ 2\n= 51 × 50 ÷ 2 - 10 × 9 ÷ 2\n= 1275 - 45\n= 1230",
+      "knowledge_points": [
+        "数与形",
+        "梯形的面积"
+      ],
+      "source": "23-24 五年级上·江苏盐城·期末",
+      "merged_id": "unit8_080",
+      "original_id": "unit8_calc_080"
+    },
+    {
+      "id": "unit8_calc_081",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "计算题",
+      "difficulty": "困难",
+      "question": "根据你发现的规律, 直接写出下面各题的得数。\n\n1 - 1/2 - 1/4 - 1/8 - 1/16 - 1/32 - 1/64 =()\n1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512 + 1/1024 + ... =()",
+      "options": [],
+      "answer": "1/64;1",
+      "solution": "【分析】 1 - 1/2 - 1/4 - 1/8 - 1/16 - 1/32 - 1/64 , 通过观察发现, 1/2 = 1 - 1/2, 1/4= 1/2 -1/4,1/8 = 1/4 -1/8... ，依此类推，将原式改写为：1 - 1/2 - 1/4 - 1/8 - 1/16 - 1/32 - 1/64 = 1 - (1 - 1/2) - (1/2 - 1/4) - (1/4 - 1/8)(1/8 - 1/16) - (1/16 - 1/32) - (1/32 - 1/64) ，化简即可;令 n =1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512 + 1/1024 + ⋯ ，等式两边同时乘2得：\n\n2n = 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512 + ⋯ , 两式相减化简即可。本题的关键是观察算式，寻找规律，运用减法性质进行化简，最后得出得数。\n\n【详解】 1 - 1/2 -1/4 -1/8 -1/16 -1/32 -1/64\n= 1 - (1 - 1/2) - (1/2 - 1/4) - (1/4 - 1/8) - (1/8 - 1/16) - (1/16 - 1/32) - (1/32 - 1/64)\n= 1 - 1 + 1/2 - 1/2 + 1/4 - 1/4 + 1/8 - 1/8 + 1/16 - 1/16 + 1/32 - 1/32 + 1/64\n= 1/64令 n = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512 + 1/1024 + ...等式两边同时乘2得：2n = 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512 + ...两式相减得:n = 1",
+      "knowledge_points": [
+        "异分母分数加、减法",
+        "算式的规律"
+      ],
+      "source": "18-19六年级下·浙江·期末",
+      "merged_id": "unit8_081",
+      "original_id": "unit8_calc_081"
+    },
+    {
+      "id": "unit8_calc_082",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "计算题",
+      "difficulty": "困难",
+      "question": "(1) 1 + 3 + 5 + 11+9 + 7 + 5 + 3 + 1 =\n\n(2) 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 =",
+      "options": [],
+      "answer": "45    127/128",
+      "solution": "【分析】（1）利用加法交换结合律进行计算即可，方法不唯一;(2) 1/2 = 1 - 1/2, 1/4 = 1/2 - 1/4, 1/8 = 1/4 - 1/8 , 据此将每个加数进行分解, 中间全部抵消, 再计算即可。\n\n【详解】 (1) 1 + 3 + 5 + 11 + 9 + 7 + 5 + 3 + 1\n= (1 + 1 + 3) + (5 + 5) + (11 + 9) + (7 + 3)\n= 5 + 10 + 20 + 10\n= 45\n\n(2) 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128\n= (1 - 1/2) + (1/2 - 1/4) + (1/4 - 1/8) + ...... + (1/64 - 1/128)\n= 1 - 1/128\n= 127/128",
+      "knowledge_points": [
+        "整数加法交换律",
+        "分数加、减简便运算",
+        "整数加法结合律"
+      ],
+      "source": "20-21 六年级上·全国·单元测试",
+      "merged_id": "unit8_082",
+      "original_id": "unit8_calc_082"
+    },
+    {
+      "id": "unit8_calc_083",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "计算题",
+      "difficulty": "中等",
+      "question": "解方程。\n\n60%x + 25 = 40  24 - 120%x = 18\n35%x - 25%x = 1  (1 - 25%)x = 45",
+      "options": [],
+      "answer": "25;5;10;60",
+      "solution": "【分析】利用等式的解方程即可。\n\n【详解】 60%x + 25 = 40\n\n解: 0.6x = 40 - 25\n0.6x = 15\nx = 25\n\n24 - 120%x = 18\n\n解: 24 - 1.2x = 18\n24 - 18 = 1.2x\n2x = 6\nx = 5\n\n35%x - 25%x= 1\n\n解: 0.35x - 0.25x = 1\n0.1x = 1\nx = 10\n\n(1 - 25%)x = 45\n\n解: 0.75x= 45\nx = 60",
+      "knowledge_points": [
+        "应用等式的性质1和2解方程"
+      ],
+      "source": "20-21 六年级上·全国·单元测试",
+      "merged_id": "unit8_083",
+      "original_id": "unit8_calc_083"
+    },
+    {
+      "id": "unit8_draw_084",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "作图题",
+      "difficulty": "中等",
+      "question": "找规律, 画一画。\n\n![](images/dc59eedabe63b9b3e0d50c04952b3efdd22b2236162e1a320b9af7f0ef46ff03.jpg)",
+      "options": [],
+      "answer": "见详解",
+      "solution": "【分析】依次是 5 个 △ , 4 个 △ , 3 个 △ , 2 个 △ , 1 个 △ ; 据此解决即可。\n\n【详解】由题意分析得:\n\n![](images/bf6c8de59e4239a19e3635fc444ba23b0082d5e06e9c9c3f63893b9bccf957eb.jpg)",
+      "knowledge_points": [
+        "数一数",
+        "图形的变化规律"
+      ],
+      "source": "24-25一年级上·辽宁·课前预习",
+      "merged_id": "unit8_084",
+      "original_id": "unit8_draw_084"
+    },
+    {
+      "id": "unit8_draw_085",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "作图题",
+      "difficulty": "中等",
+      "question": "在每个□里画一个图形, 每四个图形为一组, 每组中有 ○★△ 这三种图形,按一定的规律排列, 并且使第 23 个图形为 △ 。\n\n□□□□□□□□□□□□□□□□□□□□□□□",
+      "options": [],
+      "answer": "☆O△☆☆O△☆☆O△☆☆O△☆☆O△☆☆O△☆",
+      "solution": "【分析】每 4 个图形一循环, 第 23 个图形是 △ , 可以按 ☆○△☆ 的顺序画。先找到规律，再根据规律求解。\n\n【详解】 23 ÷ 4 = 5 （组）……3（个）如图:☆〇△☆☆〇△☆☆〇△☆☆〇△☆☆〇△☆☆〇△☆（答案不唯一）。",
+      "knowledge_points": [
+        "图形的规律"
+      ],
+      "source": "22-23 四年级上·福建宁德·期中",
+      "merged_id": "unit8_085",
+      "original_id": "unit8_draw_085"
+    },
+    {
+      "id": "086",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "观察下面各式:\n\n(1) 1 × 3 = 2² - 1 ,\n\n2 × 4 = 3² - 1\n\n3 × 5 = 4² - 1,\n\n4 × 6 =\n\n10 × 12 =\n\n(2) 将你猜到的规律用只含有一个字母 n 的式子表示出来。",
+      "options": [],
+      "answer": "(1) 5² - 1; 11² - 1 ;(2) 这个式子是: (n - 1) × (n + 1) = n² - 1",
+      "solution": "【分析】（1）给出的等式左边为差为 2 的两个数相乘，等式右边为左侧两个数的一半的平方再减 1，则 4 × 6 = ((4 + 6)/2)² - 1 ， 10 × 12 = ((10 + 12)/2)² - 1 ，整理即可得出结果;(2) 根据所列出的算式, 总结可知等式左侧为两个乘数 n - 1 与 n + 1 的乘积, 等式右侧为左侧两个乘数和的一半的平方减 1。\n\n【详解】 (1) 4 × 6 = 5² - 1 ; 10 × 12 = 11² - 1; (2) 这个式子是: (n - 1) × (n + 1) = ((n + 1 + n - 1)/2)² - 1 = n² - 1",
+      "knowledge_points": [
+        "用字母表示数",
+        "数量关系",
+        "算式的规律（整数）"
+      ],
+      "source": "25-26 六年级上·全国·单元测试",
+      "merged_id": "unit8_086",
+      "original_id": "086"
+    },
+    {
+      "id": "087",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "解答题",
+      "difficulty": "困难",
+      "question": "阅读理解应用。\n\n(1) 实验: 20 名同学排成一排。\n\n① 从排头到排尾按 1 至 3 的规律报数, 报 3 的同学共有_人;\n\n② 从排头到排尾按 1 至 4 的规律报数, 报 3 的同学共有_人。\n\n(2) 探究: 50 名同学排成一排。\n\n① 从排头到排尾按 1 至 3 的规律报数, 再从排头到排尾按 1 至 4 报数, 那么两次报数都报 3 的同学共有人;\n② 从排头到排尾按 1 至 3 的规律报数, 再从排尾向排头按 1 至 4 报数, 那么两次报数都报 3 的同学共有人。\n\n(3) 应用:\n\n2015 名同学排成一排, 从排头到排尾按 1 至 4 的规律报数,再从排尾向排头按 1 至 6 报数, 则两次报数都报 3 的共有多少人？",
+      "options": [],
+      "answer": "(1) ①6; ②5  (2) ①4; ②4   (3) 168 人",
+      "solution": "【分析】(1) ①从排头到排尾按 1 至 3 的规律报数, 每组有 3 人, 每组里有 1 人报 3 , 求出 20 里面有几个 3 即可知道有几组, 20 ÷ 3 = 6 (组) ……2 (人), 余下的 2 人里没有报数报 3 的, 所以有 6 人报 3 ;②从排头到排尾按1至4的规律报数，每组有4人，每组里有1人报3，求出20里面有几个4即可知道重复出现几组，再根据商的情况判断有几人报数报3;(2) ①都是从排头到排尾报数, 3 和 4 的最小公倍数是12，从排头到排尾 12 人为一组报数情况重复出现，每组第 3 个人两次都报 3，求出 50 里有多少个 12，即可知道重复出现几组，再根据商和余数的情况判断有多少人两次报数都报 3;② 50 ÷ 3 = 16 （组）……2（人），余数2人中没有人报数报3，所以从排头到排尾按1至3的规律报数，报数报3的同学有16人; 50 ÷ 4 = 12 （组）……2（人），余数2人中没有人报数报3，所以从排尾向排头按1至4报数，报数报3的同学有12人;3和4的最小公倍数是12，即每12人为一组。报数情况重复一次，通过列举可知，每组中第12人两次都报3。求出50里面有多少个12，再根据商和余数的情况判断有多少人两次都报3。(3) 2015 ÷ 4 = 503 (组) ... 3 (人), 余下的 3 人里有 1 人报 3 , 所以从排头到排尾按 1 至 4 的规律报数, 报数报 3 的同学有 (503 + 1) 人; 2015 ÷ 6 = 335 (组) ... 5 (人),从排尾向排头按 1 至 6 报数，余下的 5 人里有 1 人报数报 3，所以报数报 3 的同学有 (335 + 1) 人;4 和 6 的最小公倍数是 12，即每 12 人为一组报数情况重复一次，通过列举可知，每组中第 3 人两次都报 3。 2015 ÷ 12 = 167 （组）……11（人），余下的 11 人里的第 3 个人报数报 3，所以用 167 + 1 即可求出有多少人两次都报 3。解题关键是通过除法运算确定按固定周期报数时特定数字出现的次数;利用最小公倍数找出两种报数规则下的重复周期，根据总人数除以周期所得组数与余数，计算两次都报特定数字的人数。\n\n【详解】 (1) ① 20 ÷ 3 = 6 （组）……2（人）从排头到排尾按 1 至 3 的规律报数, 报 3 的同学共有 6 人; 20 ÷ 4 = 5 (组)从排头到排尾按 1 至 4 的规律报数, 报 3 的同学共有 5 人。(2) ①[3, 4] = 1250 ÷ 12 = 4 （组）……2（人）从排头到排尾 12 人为一组报数情况重复出现, 每组第 3 个人两次都报 3 , 共有 4 组, 即有 4 人每次都报 3 。从排头到排尾按 1 至 3 的规律报数, 再从排头到排尾按 1 至4 报数, 那么两次报数都报 3 的同学共有 4 人;(2) 50 ÷ 3 = 16 (组) ...... 2 (人)50 ÷ 4 = 12 （组）……2（人）\n\n![](images/9ba012337749d8b95c16d9d1c1d8d2eb1c29b227a075994ec22f2d50f16270ad.jpg)\n\n[3, 4] = 12每 12 人为一组报数情况重复一次, 每组中第 12 人两次都报 3。50 ÷ 12 = 4 （组）……2（人）\n\n所以共有 4 人两次都报 3 。从排头到排尾按 1 至 3 的规律报数, 再从排尾向排头按 1 至 4 报数, 那么两次报数都报 3 的同学共有 4 人。(3) 2015 ÷ 4 = 503 (组) ... 3 （人）2015÷6=335 (组) ……5 (人)\n\n![](images/fcc5a692055507dabaffacb5b8a3f25358a9bd9e7134d2196c38c42c71ce4fac.jpg)\n\n[4, 6] = 12每 12 人为一组报数情况重复一次, 每组中第 3 人两次都报 3。2015÷12=167（组）……11（人）167 + 1 = 168 （人）答: 两次报数都报 3 的共有 168 人。",
+      "knowledge_points": [
+        "周期问题",
+        "整数排列的规律",
+        "公倍数与最小公倍数"
+      ],
+      "source": "25-26 六年级上·全国·单元测试",
+      "merged_id": "unit8_087",
+      "original_id": "087"
+    },
+    {
+      "id": "088",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "解答题",
+      "difficulty": "困难",
+      "question": "有一列数是2, 9, 8, 2, 6, 2...从第3个数起，每一个数都是它前面的两个数乘积的个位数字。这一列数的第 2023 个数是多少？",
+      "options": [],
+      "answer": "2",
+      "solution": "【分析】通过列举2, 9, 8, 2, 6, 2, 2, 4, 8, 2, 6, 2, 2, 4...可以发现，从第三个数 8 开始每 6 个数一组重复出现, 用 2023 先减去不参加重复出现的 2 个数字, 然后除以 6 可得有多少组, 如果整除了那么就是这一组的最后一个数，如果有余数，那么余数是几，就是下一组的第几个;据此可解此题。本题关键是通过列举找出这一列数的规律，从而进行解答。\n\n【详解】 (2023 - 2) ÷ 6\n= 2021 ÷ 6\n= 336 .... 5\n\n说明第 2023 个就是第 337 组中的第 5 个，所以是 2。答: 这一列数的第 2023 个数是 2。",
+      "knowledge_points": [
+        "整数排列的规律",
+        "带有小括号的混合运算",
+        "有余数除法的实际应用"
+      ],
+      "source": "23-24 四年级上·辽宁·课后作业",
+      "merged_id": "unit8_088",
+      "original_id": "088"
+    },
+    {
+      "id": "089",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "'黑洞'是宇宙空间中一种神秘的天体，能把接近它的物质吸引进去。在数学中也有神秘的黑洞现象。四位数的黑洞是6174，即：任意一个四位数，把各个数位上的数按从大到小排列，组成一个新数，再按从小到大排列组成另一个新数，这两个数相减，得到的差再按上面的步骤做，若干次后，得到的差始终是6174，6174就是四位数的黑洞。举例说明，例如4321 - 1234 = 3087   8730 - 378 = 83528532 - 2358 = 6174  7641 - 1467 = 6174  7641 - 1467 = 6174按此继续思考，你能找出三位数的黑洞吗？任选一个三位数试试吧。",
+      "options": [],
+      "answer": "495",
+      "solution": "【分析】任意写一个三位数，把各个数位上的数按从大到小排列，组成一个新数，再按从小到大排列组成另一个新数，这两个数相减，得到的差再按上面的步骤做，若干次后，得到的差是几，三位数的黑洞就是几。\n\n【详解】例如：576   765 - 567 = 198  981 - 189 = 792 792 - 279 = 513  531 - 153 = 378  873 - 378 = 495   954 - 459 = 495  954 - 459 = 495答: 三位数的黑洞是 495 。",
+      "knowledge_points": [
+        "算式的规律（整数）"
+      ],
+      "source": "23-24四年级下·湖北襄阳·期末",
+      "merged_id": "unit8_089",
+      "original_id": "089"
+    },
+    {
+      "id": "090",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "已知一列数按294736294736294……排列，那么前 40 个数字之和是多少？",
+      "options": [],
+      "answer": "208",
+      "solution": "【分析】294736294736294……这一列数字是按照2、9、4、7、3、6这6个数字为一组进行循环出现的，求出40里面有多少个这样的一组，还余几;求出每组和，进而求出前40个数字的和。解决这类问题往往是把重复出现的部分看成一组,先找出排列的周期性规律，再根据规律求解。\n\n【详解】2、9、4、7、3、6这6个数字为一组进行循环出现2 + 9 + 4 + 7 + 3 + 6 = 31 40 ÷ 6 = 6(组) ... 4(个)6组还余4个数字，余下的4个是2，9，4，7 2 + 9 + 4 + 7 = 22 31 × 6 + 22 = 208答: 前 40 个数字之和是 208。",
+      "knowledge_points": [
+        "间隔问题",
+        "数字排列的规律"
+      ],
+      "source": "2022六年级下·全国·专题练习",
+      "merged_id": "unit8_090",
+      "original_id": "090"
+    },
+    {
+      "id": "091",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "先找规律填数，再计算每相邻两个数的比的比值，比值用小数表示。（除不尽的保留三位小数）你能发现什么规律？2, 3, 5, 8, 13, 21, 34, ( ), ( ).......",
+      "options": [],
+      "answer": "55;89",
+      "solution": "【分析】根据题意可知，依据数据的变化，可以发现：前两个数据相加等于后一个数据，据此解答，求比值时，用前项÷后项=比值，据此解答。此题考查是的找规律, 正确找出规律并用规律解决问题是解题关键。\n\n【详解】2、3、5、8、13、21、34、(55)、(89)……比值分别为:0.667, 0.6, 0.625, 0.615, 0.619, 0.618, 0.618, 0.618...#我发现：前两项之和等于后一项。",
+      "knowledge_points": [
+        "数字排列的规律",
+        "求比值"
+      ],
+      "source": "2022六年级下·全国·专题练习",
+      "merged_id": "unit8_091",
+      "original_id": "091"
+    },
+    {
+      "id": "092",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "下面的算式是按照某种规律排列的：\n\n1 + 1, 2 + 3, 3 + 5, 4 + 7, 1 + 9, 2 + 11, 3 + 13, 4 + 15, 1 + 17 ...\n\n(1) 第 13 个算式的得数是多少?  (2) 第 2019 个算式是什么?",
+      "options": [],
+      "answer": "(1) 26; (2) 3 + 4037",
+      "solution": "【分析】根据题目中的式子可知，第一个加数是1、2、3、4这样按照周期来循环，即一个周期4个数;第二个加数分别是 1、3、5、7……，属于连续的奇数，即 3 =1 + 1 × 2, 5 = 1 + 2 × 2, 7 = 1 + 3 × 2 , 即第 n 个式子的第二个加数: 1 + (n - 1) × 2 = 1 + 2n - 2 = 2n - 1 , 由此即可解答。本题主要考查算式的规律，找准两个加数的规律是解题的关键。\n\n【详解】（1）由分析可知：第13个式子的第一个加数：13 ÷ 4 = 3 ... 1 , 由此即可知道第 13 个算式的第一个加数是:1;第二个加数: 2 × 13 - 1\n= 26 - 1\n= 25\n\n即 1 + 25 = 26答: 第 13 个算式的得数是 26。(2) 2019 ÷ 4 = 504 ... 3即第 2019 个算式的第一个加数是: 3第二个加数：\n\n 2019 × 2 - 1\n= 4038 - 1\n= 4037\n\n所以第 2019 个算式是: 3 + 4037答: 第 2019 个算式是 3 + 4037 。",
+      "knowledge_points": [
+        "周期问题",
+        "算式的规律"
+      ],
+      "source": "21-22六年级上·甘肃金昌·期末",
+      "merged_id": "unit8_092",
+      "original_id": "092"
+    },
+    {
+      "id": "093",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "刘老师带了 41名同学去北海公园划船，共租了10条船。每条大船坐6人，每条小船坐4人，问大船、小船各租了几条？",
+      "options": [],
+      "answer": "大船租 1 条; 小船租 9 条",
+      "solution": "【分析】根据题干分析可得，一共有 41 + 1 = 42 人，假设全部租大船, 10 条船能坐 6 × 10 = 60 人, 比实际多算了: 60-42=18 人, 因为把小船看作了大船, 每条小船多算了 6-4 =2 人, 所以小船的条数是: 18 ÷ 2 = 9 条, 那么大船的条数就是: 10 - 9 = 1 条, 据此解答。解答鸡兔同笼问题一般用假设法, 也就是假设全部为某种量, 和实际的总量相比较, 就会出现矛盾, 然后利用这个矛盾求出另一个量, 继而求出假设的量。\n\n【详解】 (10 × 6 - 41 - 1) ÷ (6 - 4)\n= 18 ÷ 2= 9 (条)\n\n10 - 9 = 1 (条)答: 大船租 1 条, 小船租用 9 条。",
+      "knowledge_points": [
+        "鸡兔同笼"
+      ],
+      "source": "20-21 六年级上·湖南怀化·单元测试",
+      "merged_id": "unit8_093",
+      "original_id": "093"
+    },
+    {
+      "id": "094",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "小明用牙签搭六边形, 如下图。\n\n![](images/45044f99b9e8d78cfd5498d0b84dd57414aa6de588cbb6eecad4b69e479ad98d.jpg)\n\n(1) 数一数, 上面四幅图每幅各用了多少根牙签?  (2) 接着画下去, 第五幅图将用多少根牙签? 第八幅图呢?\n(3) 你能利用规律直接写成第 n 幅图一共要用多少根吗?",
+      "options": [],
+      "answer": "（1）6根;11根;16根;21根;(2) 26 根; 41 根;  (3) (5n + 1) 根",
+      "solution": "【分析】分析图形可知，每增加一个六边形就增加 5 根牙签，第 1 个图形一共用了 6 根牙签，第 2 个图形一共用了 (6 + 5) 根牙签，第 3 个图形一共用了 (6 + 5 × 2) 根牙签，第 4 个图形一共用了 (6 + 5 × 3) 根牙签……则第 n 个图形一共用了 [6 + 5 × (n - 1)] 根牙签，据此解答。用含有字母的式子表示出图形变化的规律是解答题目的关键。\n\n【详解】（1）第 1 幅图用了 6 根，第 2 幅图用了 11 根，第 3 幅图用了 16 根，第 4 幅图用了 21 根。\n\n(2) 第5幅图: 6 + 5 × (5 - 1)\n= 6 + 5 × 4\n= 6 + 20\n= 26(根)\n\n第8幅图: 6 + 5 × (8 - 1)\n= 6 + 40 - 5\n= 46 - 5\n= 41(根)\n\n答: 第五幅图将用 26 根牙签, 第八幅图将用 41 根牙签。\n\n(3) 6 + 5 × (n - 1)\n= 6 + 5n - 5\n= (5n + 1)根\n\n答: 第 n 幅图一共要用 (5n + 1) 根。",
+      "knowledge_points": [
+        "数与形",
+        "含有字母式子的化简与求值"
+      ],
+      "source": "21-22 六年级上·江西赣州·单元测试",
+      "merged_id": "unit8_094",
+      "original_id": "094"
+    },
+    {
+      "id": "095",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "观察下列顺序排列的等式，猜想第21个等式应该是多少？\n\n9 × 0 + 1 = 1\n9 × 1 + 2 = 11\n9 × 2 + 3 = 21\n9 × 4 + 5 = 41",
+      "options": [],
+      "answer": "201",
+      "solution": "【分析】观察下列等式，发现 9 × 0 + 1 = 10 × 0 + 1 = 1, 9 × 1 + 2 = 10 × 1 + 1 = 11, 9 × 2 + 3 = 10 × 2 + 1 = 21, 9 × 4 + 5 = 10 × 3 + 1 = 41 ，发现第 n 个等式为： 9 × (n - 1) + n = 10 × (n - 1) + 1 。据此解答即可。此题考查的找规律，先分别观察左右两边的式子，根据特点找出规律，用规律计算。对于归纳递推计数问题，通常从最简单的情况展开分析，然后找规律求解问题。\n\n【详解】由分析得，第21个等式应为 9 × (21 - 1) + 21\n= 10 × (21 - 1) + 1\n= 201",
+      "knowledge_points": [
+        "算式的规律"
+      ],
+      "source": "2021 六年级上·全国·专题练习",
+      "merged_id": "unit8_095",
+      "original_id": "095"
+    },
+    {
+      "id": "096",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "辨析选优, 我有主见。1, 3, 6, 10 ( ) ……聪聪发现的规律是依次增加 2、3、4、5……括号里填 15;明明发现求第几个数就是从 1 连续加到几，第五个数为： 1 + 2 + 3 + 4 + 5 = 15 ;你认为哪种规律较好？说一说你的理由。",
+      "options": [],
+      "answer": "我认为明明发现的规律较好，对于较大的数更好计算。",
+      "solution": "【分析】聪聪和明明发现的规律都很正确，但是聪聪的计算相较而言麻烦一点，随着要求数位的增加，明明的计算更加简单一点。本题主要考查找规律。\n\n【详解】 1 = 1 = (1 × (1 + 1))/2\n=1 + 2 = 3 = (2 × (2 + 1))/2\n1 + 2 + 3 = 6 = (3 × (3 + 1))/2\n\n故 1 + 2 + 3 + ... + n = (n × (n + 1))/2所以对于较大的数，相比明明发现的规律更好计算。答: 我认为明明发现的规律较好, 对于较大的数更好计算。",
+      "knowledge_points": [
+        "数字排列的规律"
+      ],
+      "source": "21-22 六年级上·江西赣州·单元测试",
+      "merged_id": "unit8_096",
+      "original_id": "096"
+    },
+    {
+      "id": "097",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "解答题",
+      "difficulty": "困难",
+      "question": "在一大块面包上切 6 刀最多能将面包切成多少块。（注：面包是一个立体几何图形，切面可以是任何方向)",
+      "options": [],
+      "answer": "42块",
+      "solution": "【分析】题目相当于 6 个平面能将空间划分为多少个部分。通过找规律来寻找递推关系，如图，显然的 1 个平面能将空间划分成 2 块，2 个平面能将空间划分成 4 块，3 个平面能将空间划分成 8 个平面，当增加到第四个平面时，第四个平面这能将原来空间中的 8 个部分中的其中几个划分。\n\n【详解】如图所示：\n\n![](images/c5027e75ffb1324d040838a98838b61a999211642f46b5b771c4fd95aa878299.jpg)\n\n![](images/ab18b28ef7e6a75a7d3182fd917fdd34cf64164783e56500346f016454b7b3d7.jpg)\n\n注意到第四个平面与其他三个平面相交形成 3 条直线, 这三条直线将第四个平面分割成 7 个部分, 而每一部分将原来三个平面划分的 8 个空间中的 7 个划分成两份, 所以 4 个平面能将空间划分成 8 + 7 = 15 个部分。同样的第五个平面与前四个平面分别相交成4条直线，这四条直线能将第5个平面分割成 1 + 1 + 2 + 3 + 4 = 11 个部分，每一部分都划分原空间中的某一区域，所以第五个平面能使空间中的区域增加到 15 + 11 = 26 个部分。当增加到6个平面时，第六个平面共被划分成 1 + 1 + 2 + 3 + 4 + 5 = 16 个部分，所以第6个平面能将空间中的区块数增加到 26 + 16 = 42 个部分。答: 6 刀能将面包切成 42 块。",
+      "knowledge_points": [
+        "递推法计数"
+      ],
+      "source": "2021 六年级·全国·竞赛",
+      "merged_id": "unit8_097",
+      "original_id": "097"
+    },
+    {
+      "id": "098",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "解答题",
+      "difficulty": "困难",
+      "question": "有 8 个选手进行乒乓球单循环赛，结果每人获胜局数各不相同，那么冠军胜了几局？",
+      "options": [],
+      "answer": "7局",
+      "solution": "【分析】8 个选手进行乒乓球单循环赛, 每个选手都要参加 7 场比赛, 而且每人获胜局数各不相同, 所以每人获胜的局数分别为 0 ~ 7 局, 那么冠军胜了 7 局。本题考查的是体育比赛中的数学问题, 单循环比赛, n 个队伍, 每个队伍要比 n-1 场。\n\n【详解】单循环赛，每个人都要和其余的每个人进行比赛; 8 - 1 = 7 （场）每个选手都要参加 7 场比赛, 既然每人获胜局数各不相同,那么只能是 0 ~ 7 ;冠军的胜场数肯定是最大的，所以冠军胜了7局;答: 冠军胜了 7 局。",
+      "knowledge_points": [
+        "数与形"
+      ],
+      "source": "2021 六年级·全国·竞赛",
+      "merged_id": "unit8_098",
+      "original_id": "098"
+    },
+    {
+      "id": "099",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "解答题",
+      "difficulty": "困难",
+      "question": "小猴子去小猫家参加生日聚会，它进门后和屋里的每一只小动物都握了一次手，一共握了9次，小猫家里一共有多少只小动物？",
+      "options": [],
+      "answer": "10只",
+      "solution": "【详解】略。",
+      "knowledge_points": [
+        "搭配问题"
+      ],
+      "source": "20-21二年级上·全国·单元测试",
+      "merged_id": "unit8_099",
+      "original_id": "099"
+    },
+    {
+      "id": "100",
+      "unit": "第八单元：数与形",
+      "chapter": "数与形",
+      "question_type": "解答题",
+      "difficulty": "困难",
+      "question": "仔细观察下面的数列，说说你发现它有什么特点和规律，并按照数列的规律，写出 2011 个分数。1, 3/4, 2/3, 5/8, 3/5, 7/12, 4/7, ...",
+      "options": [],
+      "answer": "1006/2011",
+      "solution": "【分析】把 1换成2/2 , 利用分数的基本性质, 分子分母同时乘一个数, 值不变, 把 2/3 化成 4/6, 3/5 化成 6/10, 4/7 化成 8/14 , 即可看出规律, 对应第 n 项, 分子等于 n + 1 , 分母等于 2n 因此得解。主要考查探索规律的实际能力，关键是先找出规律，然后根据规律写出第 2011 个分数，根据这个可以直接解答，注意解答的准确性。\n\n【详解】将原数列变为 2/2, 3/4, 4/6, 5/8, 6/10, 7/12, 8/14, ⋯ 由此得出分子为 n + 1 , 分母为 2n ( n 为第几个数), 所以第 2011 个分数为 2012/4022 , 化简得 1006/2011 。",
+      "knowledge_points": [
+        "分数的基本性质及应用",
+        "数字排列的规律"
+      ],
+      "source": "2020 六年级下·江苏·专题练习",
+      "merged_id": "unit8_100",
+      "original_id": "100"
+    }, {
+        "id": "101",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "1、2、3、2、3、4、3、4、5、4、5、6...，照这样从左数第50个数是（ ）",
+        "options": [
+          "A. 50",
+          "B. 17",
+          "C. 18",
+          "D. 19"
+        ],
+        "answer": "C",
+        "solution": "这个数列每三个数看成一组，那么：$5 0 \\div 3 = 1 6 \\dots 2;$50 就是第 17 组数中的第二个;它是:  $17 + 1 = 18$ ;故选：C.",
+        "knowledge_points": [
+          "数字排列的规律",
+          "数与形"
+        ],
+        "source": "19-20六年级上·全国·单元测试",
+        "merged_id": "unit8_101",
+        "original_id": "101"
+      },
+      {
+        "id": "102",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "$2.504$  小数部分的第100 位数字是 ( )",
+        "options": [
+          "A. 5",
+          "B. 0",
+          "C. 4"
+        ],
+        "answer": "A",
+        "solution": "因为  $2.504$  的循环节是 504,  $100 \\div 3 = 33 \\dots 1$ ,所以第 100 个数字是 5 .故选: A .",
+        "knowledge_points": [
+          "小数的读写、意义及分类",
+          "数与形"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_102",
+        "original_id": "102"
+      },
+      {
+        "id": "103",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "仔细观察  $37 \\times 3 =111$;$37 \\times 6 = 222$ ;  $37 \\times 9 = 333$ ; 请你推算出  $37 \\times 21$  的得数是 ( )",
+        "options": [
+          "A. 444",
+          "B. 555",
+          "C. 777"
+        ],
+        "answer": "C",
+        "solution": "通过观察发现：第二个因数是3的几倍积就是3个几。故答案为: C",
+        "knowledge_points": [
+          "算式的规律",
+          "数与形"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_103",
+        "original_id": "103"
+      },
+      {
+        "id": "104",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "根据  $3 \\times 4 =$ 12、 $33 \\times 34 = 1122$ 、 $333 \\times 334 = 111222$ ，推测  $3333 \\times 3334 =$  （）",
+        "options": [
+          "A. 11111222",
+          "B. 11122222",
+          "C. 11112222",
+          "D. 11111112"
+        ],
+        "answer": "C",
+        "solution": "根据观察知：因数有 3 的个数与积中 1 的个数和 2 的个数相同.$3333 \\times 3334 = 11112222$故选: C.",
+        "knowledge_points": [
+          "数与形"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_104",
+        "original_id": "104"
+      },
+      {
+        "id": "105",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "一列分数的前 4 个是$\\frac{1}{2} 、 \\frac{2}{5} 、 \\frac{3}{10} 、 \\frac{4}{17}$ . 根据这 4 个分数的规律可知, 第 8 个分数是 ( )",
+        "options": [
+          "A.  $\\frac{8}{61}$",
+          "B.  $\\frac{8}{63}$",
+          "C.  $\\frac{8}{65}$"
+        ],
+        "answer": "C",
+        "solution": "略",
+        "knowledge_points": [
+          "数字排列的规律"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_105",
+        "original_id": "105"
+      },
+      {
+        "id": "106",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "4 8 12 ( ) 20。",
+        "options": [
+          "A. 16",
+          "B. 17",
+          "C. 15"
+        ],
+        "answer": "A",
+        "solution": "略",
+        "knowledge_points": [
+          "数字排列的规律"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_106",
+        "original_id": "106"
+      },
+      {
+        "id": "107",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "求  $2 + 6 + 10 + 14$  的和,下面算式错误的是 ( )",
+        "options": [
+          "A.  $16 \\times 2$",
+          "B.  $(14 + 2) \\times 4$",
+          "C.  $30 + 2$",
+          "D.  $(14 + 2) \\times 2$"
+        ],
+        "answer": "B",
+        "solution": "略",
+        "knowledge_points": [
+          "数与形"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_107",
+        "original_id": "107"
+      },
+      {
+        "id": "108",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "根据你发现的规律，算式  $1234567 \\times 8 + 7$  的得数是（）\\n$ 1 \\times 8 + 1 = 9 \\quad 1 2 \\times 8 + 2 = 9 8 \\quad 1 2 3 \\times 8 + 3 = 9 8 7 $",
+        "options": [
+          "A. 9876",
+          "B. 98765",
+          "C. 987654",
+          "D. 9876543"
+        ],
+        "answer": "D",
+        "solution": "$1 + 9 = 2 + 8 = 3 + 7 = 4 + 6 = 5 + 5 = 6 + 4 = 7 + 3,$算式  $1234567 \\times 8 + 7 = 9876543$ .故选D",
+        "knowledge_points": [
+          "算式的规律"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_108",
+        "original_id": "108"
+      },
+      {
+        "id": "109",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "0.123412341234…,小数点后第 100 个数字是 ( )",
+        "options": [
+          "A. 1",
+          "B. 2",
+          "C. 3",
+          "D. 4"
+        ],
+        "answer": "D",
+        "solution": "小数 0.123412341234... 循环节为 1234，共 4 位数。 $100 \\div 4 = 25$小数点后第 100 个数字是 4 故选 D",
+        "knowledge_points": [
+          "周期问题"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_109",
+        "original_id": "109"
+      },
+      {
+        "id": "110",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "观察已给数列，括号中应填入所缺的数为：1，1，2，3，5，8，13（ ），34，……",
+        "options": [
+          "A. 15",
+          "B. 17",
+          "C. 21",
+          "D. 30"
+        ],
+        "answer": "C",
+        "solution": "要填的数是：$8 + 1 3 = 2 1;$故选 C",
+        "knowledge_points": [
+          "数字排列的规律"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_110",
+        "original_id": "110"
+      },
+      {
+        "id": "111",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "一组数据按下面顺序依次排列：1, 3, 2014, 2, 4, 2012, 3, 5, 2010, 4, 6, 2008…第2016个数是（）",
+        "options": [
+          "A. 672",
+          "B. 674",
+          "C. 670",
+          "D. 676"
+        ],
+        "answer": "A",
+        "solution": "根据观察发现，这组数据每 3 个数一组：第一个数字为从 1 开始的自然数排列;第二个数为从 3 开始的自然数排列;第3个数为从2014开始，每组减2.第2016个数包含几组：2016÷3=672 (组)所以第 2016 个数为:$ 2 0 1 4 - (6 7 2 - 1) \\times 2 \\\\ = 2 0 1 4 - 1 3 4 2 \\\\ = 6 7 2 $\\n答: 第 2016 个数为 672 .故选 A.",
+        "knowledge_points": [
+          "整数排列的规律",
+          "简单周期问题"
+        ],
+        "source": "19-20六年级上·全国·单元测试",
+        "merged_id": "unit8_111",
+        "original_id": "111"
+      },
+      {
+        "id": "112",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "笑笑用小棒以下面的方式摆六边形.1个    2个   3个\\n![](images/af67a202ee18d067388673c419abd0d4c28083bb2241a97343faef375749e711.jpg)\\n\\n摆  $\\mathrm{n}$  个六边形要用小棒（）根.",
+        "options": [
+          "A. 6n",
+          "B.  $5 \\mathrm{n} + 1$",
+          "C.  $4 \\mathrm{n} + 3$"
+        ],
+        "answer": "B",
+        "solution": "【分析】第一个图形需要六根小棒，以后每加一个六边形，只需要加五根小棒即可，进而找出规律。本题考查数与形的知识，根据图形找出规律。\\n【详解】1个六边形，需要  $5 + 1 = 6$  根小棒。2 个六边形, 需要  $5 \\times 2 + 1 = 11$  根小棒。3个六边形，需要  $5 \\times 3 + 1 = 16$  根小棒。n 个六边形, 需要  $5 \\times n + 1$  根小棒。故答案为: B",
+        "knowledge_points": [
+          "数与形"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_112",
+        "original_id": "112"
+      },
+      {
+        "id": "113",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "$\\frac{1}{2}, \\frac{3}{4}, \\frac{5}{6}, \\frac{7}{8}, \\cdots$ , 则这一行数中的第 10 个数应该是 ( )。",
+        "options": [
+          "A.  $\\frac{9}{10}$",
+          "B.  $\\frac{17}{18}$",
+          "C.  $\\frac{19}{20}$",
+          "D.  $\\frac{19}{22}$"
+        ],
+        "answer": "C",
+        "solution": "【分析】通过观察题目发现, 这个分数的分子是从 1 开始的. 相邻奇数:  $1 = 2 \\times 1 - 1 、 3 = 2 \\times 2 - 1 、 5 = 2 \\times 3 - 1 、 7 = 2 \\times 4 -1 \\cdots$.第 n 项是  $2 \\mathrm{n} - 1$ ; 这个分数的分母由左到右是从 2 开始的相邻偶数,  $2 = 2 \\times 1 、 4 = 2 \\times 2 、 6 = 2 \\times 3 、 8 = 2 \\times 4 \\dots$ . 第 n 项是 2n, 据此即可写出第 10 个分数。解答此题的关键是找出这列分数的分子、分母与序数之间的关系，然后再根据这一关系写出第 10 个分数。\\n【详解】由分析可知：第10个数的分子是：\\n$ 2 \\times 1 0 - 1 \\\\ = 2 0 - 1 \\\\ = 1 9 $\\n第 10 个数的分母是:  $2 \\times 10 = 20$   第 10 个数的是:  $\\frac{19}{20}$故答案为: C",
+        "knowledge_points": [
+          "数字排列的规律"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_113",
+        "original_id": "113"
+      },
+      {
+        "id": "114",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "一列数$1，\\frac{1}{2},   \\frac{1}{2}, \\frac{1}{3}, \\frac{1}{3}, \\frac{1}{3}, \\frac{1}{4}, \\frac{1}{4}, \\frac{1}{4}, \\frac{1}{4}$  ……中的第27 个数是 ( )",
+        "options": [
+          "A.  $\\frac{1}{6}$",
+          "B.  $\\frac{1}{7}$",
+          "C. $\\frac{1}{8}$",
+          "D. $\\frac{1}{9}$"
+        ],
+        "answer": "B",
+        "solution": "解：根据规律，设第27个数为  $\\frac{1}{n}$ ，则  $1 + 2 + 3 + \\ldots + n - 1 < 27 < 1 + 2 + 3 + \\ldots + n,$\\n$\\frac {7 \\times (7 - 1)}{2} <   2 7 <   \\frac {7 \\times (7 + 1)}{2};$所以  $n = 7$ , 则第 27 个数是  $\\frac{1}{7}$ .故选 B.",
+        "knowledge_points": [
+          "数字排列的规律"
+        ],
+        "source": "19-20六年级上·全国·单元测试",
+        "merged_id": "unit8_114",
+        "original_id": "114"
+      },
+      {
+        "id": "115",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "不计算, 用发现的规律直接写出后两题的结果。\\n$3 \\times 0. 7 = 2. 1$\\n$3.3 \\times 6. 7 = 22. 11$\\n$3.33 \\times 66.7 = 222.111$  \\n$3.333 \\times 666.7 =$  \\n$3.3333×6666.7=$\\n用发现的规律，请接着写一个算式",
+        "options": [],
+        "answer": "2222.1111 ;2222.1111; 3.33333×66666.7=22222.111111",
+        "solution": "略",
+        "knowledge_points": [
+          "算式的规律"
+        ],
+        "source": "19-20 五年级上·全国·期末",
+        "merged_id": "unit8_115",
+        "original_id": "115"
+      },
+      {
+        "id": "116",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "先用计算器计算出横线里是几，再根据规律直接写出后两道题的结果。\\n$1 \\times 9 = 9 \\\\ 2 \\times 9 9 = \\_ 9\\_ \\\\ 3 \\times 9 9 9 = \\_ 9 9\\_ \\\\ 4 \\times 9 9 9 9 = \\_ 9 9 9\\_ \\\\ 5 \\times 9 9 9 9 9 =\\_\\_\\_\\_\\_ \\\\ 6 \\times 9 9 9 9 99 =\\_\\_\\_\\_\\_$",
+        "options": [],
+        "answer": "1   8   2    7    3   6   499995    5999994",
+        "solution": "略",
+        "knowledge_points": [],
+        "source": "19-20六年级上·全国·单元测试",
+        "merged_id": "unit8_116",
+        "original_id": "116"
+      },
+      {
+        "id": "117",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "按规律填数\\n(1) 1、2、5、_、17、26、37、50.  \\n(2) 1、1、2、3、5、8、_、21、34.",
+        "options": [],
+        "answer": "10 13",
+        "solution": "(1)  $5 + 5 = 10$ ; (2)  $5 + 8 = 13$  故答案为: 10、13.",
+        "knowledge_points": [],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_117",
+        "original_id": "117"
+      },
+      {
+        "id": "118",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "按照一定规律列出的加法算式： $1 + 6, 2 + 7, 3 + 8, 4 + 9 \\ldots$ ，第 7 个算式是____，这 7 个算式的和是____。",
+        "options": [],
+        "answer": "$7 + 12$  91",
+        "solution": "【分析】观察算式可得规律：每个算式第 1 个加数为第几个算式的序数，两个加数的差是 5，据此规律写出第 7 个算式，然后把 7 个算式相加，利用高斯求和方法求出算式的和。\\n【详解】第7个算式为  $7 + 12$\\n7 个算式的和:\\n$1 + 6 + 2 + 7 + 3 + 8 + 4 + 9 + 5 + 1 0 + 6 + 1 1 + 7 + 1 2 \\\\ = (1 + 1 2) + (6 + 7) + (2 + 1 1) + (7 + 6) + (3 + 1 0) + (8 + 5)+(4 + 9) \\\\ = 1 3 \\times 7 \\\\ = 9 1 $\\n故答案为:  $7 + 12; 91$  。",
+        "knowledge_points": [
+          "算式的规律"
+        ],
+        "source": "2019 六年级上·全国·专题练习",
+        "merged_id": "unit8_118",
+        "original_id": "118"
+      },
+      {
+        "id": "119",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "下面这些分数是按一定规律排列的： $\\frac{1}{2}$ 、 $\\frac{2}{3}$ 、 $\\frac{3}{4}$ 、 $\\frac{4}{5}$ …… $\\frac{15}{16}$ 、 $\\frac{16}{17}$ 。\\n(1) 一共有____个分数。\\n(2) 这些分数相乘所得的积是____。",
+        "options": [],
+        "answer": "16  $\\frac{1}{17}$",
+        "solution": "【分析】（1）观察分数可得规律：每个分数的分子比分母少 1 , 这组分数的分子是从 1 开始的连续自然数, 一直到 16 结束, 一共有 16 个分数;  (2) 要求这些分数的乘积, 直接用乘法计算, 先约分后计算, 据此解答\\n【详解】（1）根据分析可得，这组分数分别是： $\\frac{1}{2}$ 、 $\\frac{2}{3}$ 、 $\\frac{3}{4}$ 、 $\\frac{4}{5}$ 、 $\\frac{5}{6}$ 、 $\\frac{6}{7}$ 、 $\\frac{7}{8}$ 、 $\\frac{8}{9}$ 、 $\\frac{9}{10}$ 、 $\\frac{10}{11}$ 、 $\\frac{11}{12}$ 、 $\\frac{12}{13}$ 、 $\\frac{13}{14}$ 、 $\\frac{14}{15}$ 、 $\\frac{15}{16}$ 、 $\\frac{16}{17}$ ，一共有 16 个分数。\\n(2)  $\\frac{1}{2} \\times \\frac{2}{3} \\times \\frac{3}{4} \\times \\frac{4}{5} \\times \\frac{5}{6} \\times \\frac{6}{7} \\times \\frac{7}{8} \\times \\frac{8}{9} \\times \\frac{9}{10} \\times \\frac{10}{11} \\times \\frac{11}{12} \\times \\frac{12}{13} \\times \\frac{13}{14} \\times \\frac{14}{15} \\times \\frac{15}{16} \\times \\frac{16}{17} = \\frac{1}{17}$ 。\\n故答案为：(1) 16;(2)  $\\frac{1}{17}$ 。",
+        "knowledge_points": [
+          "分数乘法",
+          "数与形"
+        ],
+        "source": "2019 六年级上·全国·专题练习",
+        "merged_id": "unit8_119",
+        "original_id": "119"
+      },
+      {
+        "id": "120",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "按规律填一填.\\n(1) 4001, 5002, 6003, _, 8005.  \\n(2) 8600, 8400, 8200, 8000, _____.",
+        "options": [],
+        "answer": "7004 7800",
+        "solution": "根据分析可得，(1)  $6003 + 1001 = 7004$  (2)  $8000 - 200 = 7800$故答案为: 7004, 7800.",
+        "knowledge_points": [
+          "数字排列的规律"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_120",
+        "original_id": "120"
+      },
+      {
+        "id": "121",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "按规律在括号里填上适当的数.\\n(1)  $9, 3, \\_ , \\_ , \\frac{1}{9}, \\frac{1}{27}$ .  \\n(2)  $\\frac{1}{3}, \\frac{3}{6}, \\frac{5}{9}, \\frac{7}{12}, ,\\_\\_,\\frac{11}{18}, \\_\\_\\dots$",
+        "options": [],
+        "answer": "$1\\quad\\frac{1}{3} \\quad\\frac{9}{15} \\quad\\frac{13}{21}$",
+        "solution": "根据分析可得,\\n(1)  $3 \\div 3 = 1$  $1 \\div 3 = \\frac {1}{3}$\\n(2)  $\\frac{7 + 2}{3 \\times 5} = \\frac{9}{15}$   $\\frac {1 1 + 2}{3 \\times 7} = \\frac {1 3}{2 1}$\\n故答案为:  $1, \\frac{1}{3}; \\frac{9}{15}, \\frac{13}{21}$ .",
+        "knowledge_points": [
+          "数字排列的规律"
+        ],
+        "source": "19-20六年级上·全国·单元测试",
+        "merged_id": "unit8_121",
+        "original_id": "121"
+      },
+      {
+        "id": "122",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "按规律填空\\n$\\frac {1}{6}, \\frac {1}{3}, \\frac {1}{2}, \\_\\_, \\frac {4}{3}, \\_\\_, \\cdots$",
+        "options": [],
+        "answer": "$\\frac{5}{6} \\cdot \\frac{13}{6}$",
+        "solution": "$\\frac{1}{3} = \\frac{2}{6}, \\frac{1}{2} = \\frac{3}{6}, \\frac{4}{3} = \\frac{8}{6},$ $\\frac {2 + 3}{6} = \\frac {5}{6}$   $\\frac {5 + 8}{6} = \\frac {1 3}{6}$  故答案为:  $\\frac{5}{6}; \\frac{13}{6}$ .",
+        "knowledge_points": [
+          "数字排列的规律"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_122",
+        "original_id": "122"
+      },
+      {
+        "id": "123",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "找规律填数:\\n$100，50，25，\\_\\_,\\_\\_.$",
+        "options": [],
+        "answer": "12.5 6.25",
+        "solution": "$25 \\times 0.5 = 12.5$  $12.5 \\times 0.5 = 6.25$ 即100，50，25，12.5，6.25.故答案为：12.5，6.25.",
+        "knowledge_points": [],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_123",
+        "original_id": "123"
+      },
+      {
+        "id": "124",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如下图, 第 20 行第2个数是( ).\\n1  \\n2 3  \\n4 5 6  \\n7 8 9 10\\n.......",
+        "options": [],
+        "answer": "192",
+        "solution": "略",
+        "knowledge_points": [
+          "数字排列的规律"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_124",
+        "original_id": "124"
+      },
+      {
+        "id": "125",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "张老师把38张明信片按顺序依次发给张华、刘明、许兵，最后一张发给（ ），（ ）和（ ）发的一样多，（ ）发的最少。",
+        "options": [],
+        "answer": "刘明 张华 刘明 许兵",
+        "solution": "略",
+        "knowledge_points": [],
+        "source": "19-20六年级上·全国·单元测试",
+        "merged_id": "unit8_125",
+        "original_id": "125"
+      },
+      {
+        "id": "126",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "选找规律再填空.\\n0.2，0.7，1.2，1.7， ，___ ，",
+        "options": [],
+        "answer": "2.2 2.7 3.2",
+        "solution": "略",
+        "knowledge_points": [
+          "数字排列的规律"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_126",
+        "original_id": "126"
+      },
+      {
+        "id": "127",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "找出下面一行数的排列规律，填上合适的数。\\n4，8，16，32，_____，_______",
+        "options": [],
+        "answer": "64    128",
+        "solution": "略",
+        "knowledge_points": [
+          "数字排列的规律"
+        ],
+        "source": "19-20六年级上·全国·单元测试",
+        "merged_id": "unit8_127",
+        "original_id": "127"
+      },
+      {
+        "id": "128",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "想一想, 在括号里填上正确的数字.\\n① $1,2,4,8,16,32,$_______;  \\n②$21, 1, 2, 3, 5, 8, 13, \\_\\_\\_;  $\\n③ 625, 125, 25, _, _;  \\n④   $9 \\times 0 + 1 = 1; 9 \\times 1 + 2 = 11; 9 \\times 2 + 3 = 21; 9 \\times 3 + 4 = 31;$\\n猜想：第  $\\mathrm{n}$  个等式（n为正整数）应为____.",
+        "options": [],
+        "answer": "64 21 5 1  $9(\\mathrm{n - 1}) + \\mathrm{n}$",
+        "solution": "略",
+        "knowledge_points": [
+          "数字排列的规律"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_128",
+        "original_id": "128"
+      },
+      {
+        "id": "129",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "找规律填空\\n(1)  $5^{2} = 1 + 3 + 5 + (\\quad) + (\\quad)$  \\n(2)  $1 + 0.9 + 0.09 + 0.009 + \\dots = ()$  \\n(3)  $2 + 4 + 6 + 8 + 10 =$  ( )  \\n(4)  $1 + 3 + 5 + 7 + 9 + 11 + 13 + 11 + 9 + 7 + 5 + 3 + 1 = (\\quad)$",
+        "options": [],
+        "answer": "7 9 1.9999……… 30 85",
+        "solution": "略",
+        "knowledge_points": [
+          "数与形"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_129",
+        "original_id": "129"
+      },
+      {
+        "id": "130",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "要表示一个病人在一天中不同时间体温的变化情况, 用( )统计图更合适.",
+        "options": [],
+        "answer": "折线",
+        "solution": "略",
+        "knowledge_points": [
+          "统计图的选择"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_130",
+        "original_id": "130"
+      },
+      {
+        "id": "131",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "小明今年 11 岁, 属相是'猴'，小明的姑姑比他大一轮，小明姑姑今年( )岁;妈妈比小明大两轮，小明妈妈今年( )岁。",
+        "options": [],
+        "answer": "23 35",
+        "solution": "略",
+        "knowledge_points": [
+          "间隔问题"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_131",
+        "original_id": "131"
+      },
+      {
+        "id": "132",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "今天是星期三, 第 47 天是星期( );第 99 天是星期( )。",
+        "options": [],
+        "answer": "一 四",
+        "solution": "略",
+        "knowledge_points": [
+          "周期问题"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_132",
+        "original_id": "132"
+      },
+      {
+        "id": "133",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "○□△◆○□△◆○□△◆……第50个是( )，第80个是( )，第2019 个是( ).",
+        "options": [],
+        "answer": "□   ◆  △",
+        "solution": "略",
+        "knowledge_points": [
+          "间隔问题"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_133",
+        "original_id": "133"
+      },{
+        "id": "134",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "有80面小旗，按  $\\upharpoonright \\upharpoonright \\upharpoonright \\upharpoonright \\upharpoonright \\upharpoonright \\dots$  的规律排列，那么'P'有( )面，'P'有( )面.",
+        "options": [],
+        "answer": "54 26",
+        "solution": "略",
+        "knowledge_points": [
+          "数与形"
+        ],
+        "source": "19-20六年级上·全国·单元测试",
+        "merged_id": "unit8_134",
+        "original_id": "134"
+      },
+      {
+        "id": "135",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "按规律填数.\\n(1) -3、3、9、15、( )  \\n(2)  $1 、 3 、 7 、 1 5 、 (\\quad)$",
+        "options": [],
+        "answer": "21 31",
+        "solution": "略",
+        "knowledge_points": [
+          "数字排列的规律"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_135",
+        "original_id": "135"
+      },
+      {
+        "id": "136",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "△○△○△○△○△左图中一共有( )个  △ , ( )个  $\\circ$ ,○ 的个数比  △ 少( )\\n; 象这样一共摆 20 个  △ , 那么中间一共要摆 ( )个 $\\circ$.",
+        "options": [],
+        "answer": "5    4    1   19",
+        "solution": "略",
+        "knowledge_points": [
+          "数与形"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_136",
+        "original_id": "136"
+      },
+      {
+        "id": "137",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "把1、2、3、4、5、6、7、8、9这九个数字每相邻的两个数字组成两位数，可以组成( )个不同的两位数;如果每相邻的三个数字组成三位数，可以组成( )个不同的三位数.",
+        "options": [],
+        "answer": "16 42",
+        "solution": "略",
+        "knowledge_points": [
+          "排列组合问题"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_137",
+        "original_id": "137"
+      },
+      {
+        "id": "138",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "找规律填数。\\n(1) 1、4、9、（ ）、（ ）  \\n(2)  $\\frac{1}{2} + \\frac{1}{4} + \\frac{1}{8} + \\frac{1}{16} + \\frac{1}{32} = 1 - (\\quad)$",
+        "options": [],
+        "answer": "16 25  $\\frac{1}{32}$",
+        "solution": "略",
+        "knowledge_points": [
+          "算式的规律",
+          "数字排列的规律"
+        ],
+        "source": "19-20 六年级上·福建福州·单元测试",
+        "merged_id": "unit8_138",
+        "original_id": "138"
+      },
+      {
+        "id": "139",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "数列 2, 3,  $\\frac{9}{2}$ ,  $\\frac{27}{4}$ , ……, 则其中第 6 个数是____.",
+        "options": [],
+        "answer": "$\\frac{243}{16}$",
+        "solution": "这组数据的规律为：后面的数等于前面的数乘  $\\frac{3}{2}$  所以，第 6 个数为：$2 \\times \\frac {3}{2} \\times \\frac {3}{2} \\times \\frac {3}{2} \\times \\frac {3}{2} \\times \\frac {3}{2} = \\frac {2 4 3}{1 6}$\\n答：其中第6个数是  $\\frac{243}{16}$故答案为 $\\frac{243}{16}$",
+        "knowledge_points": [],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_139",
+        "original_id": "139"
+      },
+      {
+        "id": "140",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "找规律:  $\\frac{2}{1} 、 \\frac{4}{4}$\\n$\\frac {6}{9}  ,  \\frac {8}{1 6} , \\_\\_、\\_\\_\\quad \\dots  , \\quad \\frac {1 4}{4 9}$",
+        "options": [],
+        "answer": "$\\frac{10}{25} \\quad \\frac{12}{36}$",
+        "solution": "第一空分子是  $8 + 2 = 10$  ，分母是  $52 = 25$  第二空分子是  $10 + 2 = 12$ , 分母是  $62 = 36$\\n$ \\frac {2}{1} 、\\frac {4}{4}  \\frac {6}{9} 、\\frac {8}{1 6}、\\frac {1 0}{2 5} 、  \\frac {1 2}{3 6} 、 \\frac {1 4}{4 9}$故答案为  $\\frac{10}{25}$ ,  $\\frac{12}{36}$",
+        "knowledge_points": [],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_140",
+        "original_id": "140"
+      },
+      {
+        "id": "141",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "$2 \\div 11$  的商用简便方法记作____，小数点后面第100位上的数字是____.",
+        "options": [],
+        "answer": "$0.\\dot{1}\\dot{8}$  8",
+        "solution": "$2 \\div 11 = 0.\\dot{1}\\dot{8}$循环节是 18 两个数字; $100 \\div 2 = 50$ , 说明到第 100 位数字出现了 50 个循环节, 所以\\n100 位上的数字是 8 ;故答案为$0.\\dot{1}\\dot{8} $, 8",
+        "knowledge_points": [
+          "周期问题",
+          "小数的读写、意义及分类"
+        ],
+        "source": "19-20六年级上·全国·单元测试",
+        "merged_id": "unit8_141",
+        "original_id": "141"
+      },
+      {
+        "id": "142",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "先计算前三题，再根据发现的规律直接写出其他算式的结果。\\n$1 + 3 = \\_\\_\\_= 2 ^ {2}$\\n$1 + 3 + 5 = \\_\\_\\_ = 3 ^ {2}$\\n$1 + 3 + 5 + 7 = \\_ \\_\\_=\\_\\_\\_$\\n$\\dots\\dots$\\n$1 + 3 + 5 + 7 + \\dots + 1 5 = \\_\\_ =\\_\\_$\\n$1 + 3 + 5 + 7 + \\dots + 2 0 1 7 = \\_\\_ =\\_\\_$",
+        "options": [],
+        "answer": "4    9    16    4²   64   8²   1018081   1009²",
+        "solution": "$1 + 3 = 4 = 2^{2}$\\n$1 + 3 + 5 = 9 = 3 ^ {2}$\\n$1 + 3 + 5 + 7 = 1 6 = 4 ^ {2}$\\n$\\dots\\dots$\\n$1 + 3 + 5 + 7 + \\dots + 1 5 = 6 4 = 8 ^ {2}$\\n$1 + 3 + 5 + 7 + \\dots + 2 0 1 7 = 1 0 1 6 0 6 4 = 1 0 0 8 ^ {2}$\\n故答案为 4, 9, 16,  $4^{2}, 64, 8^{2}, 1016064, 1008^{2}$",
+        "knowledge_points": [
+          "算式的规律"
+        ],
+        "source": "19-20六年级上·全国·单元测试",
+        "merged_id": "unit8_142",
+        "original_id": "142"
+      },
+      {
+        "id": "143",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "按规律填 1,  $\\frac{3}{4}$ ,  $\\frac{5}{9}$ ,  $\\frac{7}{16}$ , _,_\\_\\_ ……",
+        "options": [],
+        "answer": "$\\frac {9}{2 5} \\quad \\frac {1 1}{3 6}$",
+        "solution": "【分析】这一列数的分子分别是 1,3,5,7，每次比前一个多2，据此，可写出后面两个分子分别为9和11;分母分别是1,4,9,16，分别加3，加5，加7，按此规律，后面两个分母分别为  $16 + 9 = 25; 25 + 11 = 36$ .把这一列数分成分子和分母两部分, 分别找出规律, 再求解。\\n【详解】  $\\frac{1}{1};\\frac{1 + 2}{1 + 3} = \\frac{3}{4};\\frac{3 + 2}{4 + 5} = \\frac{5}{9};\\frac{5 + 2}{9 + 7} = \\frac{7}{16};\\frac{7 + 2}{16 + 9} = \\frac{9}{25};$   $\\frac {9 + 2}{2 5 + 1 1} = \\frac {1 1}{3 6}$   故答案为 $\\frac{9}{25}$  $\\frac{11}{36}$",
+        "knowledge_points": [],
+        "source": "19-20 六年级上·全国·期中",
+        "merged_id": "unit8_143",
+        "original_id": "143"
+      },
+      {
+        "id": "144",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "神奇的 9 找出规律, 写出结果。\\n$9 = 1 \\times 9$    $1 0 8 = 1 2 \\times 9$     $1 1 0 7 = 1 2 3 \\times 9$       $1 1 1 0 6 = 1 2 3 4 \\times 9$     $\\_ = 1 2 3 4 5 \\times 9$   $9 \\times 9 = 8 1$\\n$9 8 \\times 9 = 8 8 2$     $9 8 7 \\times 9 = 8 8 8 3$     $9 8 7 6 \\times 9 = 8 8 8 8 4$      $9 8 7 6 5 \\times 9 =$",
+        "options": [],
+        "answer": "111105    888885",
+        "solution": "略",
+        "knowledge_points": [
+          "算式的规律"
+        ],
+        "source": "19-20 四年级上·吉林长春·期中",
+        "merged_id": "unit8_144",
+        "original_id": "144"
+      },
+      {
+        "id": "145",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "将正整数按如图所示的规律排列下去，若有序实数对（n，m）表示第  $n$ 排，从左到右第  $m$  个数, 如  $(4, 2)$  表示实数 9 , 则表示实数 17 的有序实数对是____.\\n![](images/41bec866f73ae3539540f4178558424e.png)",
+        "options": [],
+        "answer": "(6, 5)",
+        "solution": "寻找规律，然后解答，每排的数字个数就是排数：且奇数排从左到右，从小到大，而偶数排从左到右，从大到小。观察图表可知：每排的数字个数就是排数;且奇数排从左到右，从小到大，而偶数排从左到右，从大到小。实数  $15 = 1 + 2 + 3 + 4 + 5$ 。则 17 在第 6 排，第 5 个位置，即其坐标为 (6, 5)。故答案填：(6, 5)。",
+        "knowledge_points": [
+          "数与形"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_145",
+        "original_id": "145"
+      },
+      {
+        "id": "146",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "找规律填一填:3，5，9，17，\\_\\_\\_\\_，\\_\\_\\_\\_，129.",
+        "options": [],
+        "answer": "33   65",
+        "solution": "略",
+        "knowledge_points": [
+          "数字排列的规律"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_146",
+        "original_id": "146"
+      },
+      {
+        "id": "147",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "$1 - \\frac{9}{10} - \\frac{9}{100} - \\frac{9}{1000} - \\frac{9}{10000} - \\dots = (\\quad)$",
+        "options": [],
+        "answer": "0",
+        "solution": "因为  $\\frac{9}{10} + \\frac{9}{100} + \\frac{9}{1000} + \\frac{9}{10000} + \\dots = 1$  ，所以 $1 - \\frac{9}{10} - \\frac{9}{100} - \\frac{9}{1000} - \\dots = 1 - (\\frac{9}{10} + \\frac{9}{100} + \\frac{9}{1000} + \\frac{9}{10000}) + \\dots = 0$ 。",
+        "knowledge_points": [],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_147",
+        "original_id": "147"
+      },
+      {
+        "id": "148",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "$1 + \\frac{9}{10} + \\frac{9}{100} + \\frac{9}{1000} + \\frac{9}{10000} + \\dots = (\\quad)$",
+        "options": [],
+        "answer": "2",
+        "solution": "因为  $\\frac{9}{10} + \\frac{9}{100} + \\frac{9}{1000} + \\frac{9}{10000} + \\dots = 1$  ，所以  $1 + \\frac{9}{10} + \\frac{9}{100} + \\frac{9}{1000} + \\dots = 2$ .",
+        "knowledge_points": [],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_148",
+        "original_id": "148"
+      },
+      {
+        "id": "149",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "已知  $121 = 11^{2}, 12321 = 111^{2}, 1234321 = 1111^{2}, \\ldots$ , 那么 12345678987654321 = ( ).",
+        "options": [],
+        "answer": "$111111111^{2}$",
+        "solution": "略",
+        "knowledge_points": [
+          "数与形"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_149",
+        "original_id": "149"
+      },
+      {
+        "id": "150",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "找出下面一组数的排列规律，然后填上适当的数。4，7，10，13，16，____，____.",
+        "options": [],
+        "answer": "19 22",
+        "solution": "第一个数 4 加上 3 得第二个数 7 , 第二个数 7 加上 3 得第三个数 10 , 第三个数 10 加上 3 得第四个数 13 ,第四个数 13 加上 3 得第五个数 16 ; 第六个数是第五个数加上 3 , 第七个数是第六个数加上 3 .",
+        "knowledge_points": [
+          "数字排列的规律"
+        ],
+        "source": "19-20六年级上·全国·单元测试",
+        "merged_id": "unit8_150",
+        "original_id": "150"
+      },
+      {
+        "id": "151",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "找规律.\\n9876,8765,7654,\\_\\_\\_\\_,\\_\\_\\_\\_,\\_\\_\\_\\_3210.",
+        "options": [],
+        "answer": "6543  5432  4321",
+        "solution": "【分析】观察数据可知, 每相邻的两个数据之间相差 1111 , 前一个数  $-1111 =$  后一个数, 据此解答.\\n【详解】7654- 1111=6543，6543- 1111=5432，5432- 1111=4321.",
+        "knowledge_points": [
+          "数字排列的规律"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_151",
+        "original_id": "151"
+      },
+      {
+        "id": "152",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "1,1,2,3,5,8,( ),21,( ). . .",
+        "options": [],
+        "answer": "13  34",
+        "solution": "略",
+        "knowledge_points": [],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_152",
+        "original_id": "152"
+      },
+      {
+        "id": "153",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "学校的教学楼有四层。六年级一班的学生第一节课到二楼上数学课，第二节课到三楼上美术课，第三节课到二楼上语文课，第四节课到四楼上音乐课，中午到一楼食堂吃饭。下面图( )描述的是六年级一班学生的这一活动过程。\\n![](images/9e8731eb9ab883c4ed32af7ecb66143679734e24aed09232c3812d82af37518c.jpg)\\n①\\n![](images/448b607b8a9d5da982de76a96c248cb5d5310a32d30f0fa4b3365d58d7b1c443.jpg)\\n②\\n![](images/5a5c895ee43a37461075fbae8ac85a763df3ea6e275505fa17b0324f4bee79b3.jpg)\\n③",
+        "options": [
+          "①",
+          "②",
+          "③"
+        ],
+        "answer": "③",
+        "solution": "略",
+        "knowledge_points": [
+          "数与形"
+        ],
+        "source": "19-20六年级上·全国·单元测试",
+        "merged_id": "unit8_153",
+        "original_id": "153"
+      },
+      {
+        "id": "154",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "按规律填数.\\n(1)  $\\frac{1}{2}, \\frac{3}{4}, \\frac{5}{6}, \\frac{7}{8},$  ( ).  \\n(2) 1, 9, 25, 49, (), (), 169.",
+        "options": [],
+        "answer": "$\\frac{9}{10}$     81   121",
+        "solution": "略",
+        "knowledge_points": [
+          "数字排列的规律"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_154",
+        "original_id": "154"
+      },
+      {
+        "id": "155",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "要统计小红、小明、小丽、小东四个人的身高情况,应绘制()比较合适;要统计小红从一年级到四年级的身高变化的情况,应绘制()比较合适.",
+        "options": [],
+        "answer": "条形统计图  折线统计图",
+        "solution": "略",
+        "knowledge_points": [
+          "统计图的选择"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_155",
+        "original_id": "155"
+      },
+      {
+        "id": "156",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "小明的爸爸把一个月的工资做了一次消费计划,要想清楚地看到各部分消费与工资的关系,要选用( )统计图.",
+        "options": [],
+        "answer": "扇形",
+        "solution": "略",
+        "knowledge_points": [
+          "统计图的选择"
+        ],
+        "source": "19-20六年级上·全国·单元测试",
+        "merged_id": "unit8_156",
+        "original_id": "156"
+      },
+      {
+        "id": "157",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "乐乐今年初中毕业考高中,妈妈想知道他各科成绩占总分的百分比,应选用( )统计图;杨校长想知道2014年上半年学校的支出情况,王会计应给杨校长绘制一张( )统计图;明明发烧住院,李医生想绘制一张统计图来了解他的体温变化情况,李医生应选用( )统计图.",
+        "options": [],
+        "answer": "扇形 条形 折线",
+        "solution": "略",
+        "knowledge_points": [
+          "统计图的选择"
+        ],
+        "source": "19-20六年级上·全国·单元测试",
+        "merged_id": "unit8_157",
+        "original_id": "157"
+      },
+      {
+        "id": "158",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "找规律填数,6.877、6.872、6.867、\\_\\_\\_\\_、\\_\\_\\_\\_、\\_\\_\\_\\_.",
+        "options": [],
+        "answer": "6.862   6.857 6.852",
+        "solution": "$6.867 - 0.005 = 6.8626 \\quad 6.8 6 2 - 0. 0 0 5 = 6. 8 5 7 \\quad 6. 8 5 7 - 0. 0 0 5 = 6. 8 5 2 $\\n即 6.877、6.872、6.867、6.862、6.857、6.852。故答案为 6.862、6.857、6.852。",
+        "knowledge_points": [],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_158",
+        "original_id": "158"
+      },
+      {
+        "id": "159",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "①  $1^{3} + 2^{3} = 9$ , $(1 + 2) ^ {2} = 9;$\\n$②1 ^ {3} + 2 ^ {3} + 3 ^ {3} = 3 6, \\quad (1 + 2 + 3) ^ {2} = 3 6; \\\\ ③ 1 ^ {3} + 2 ^ {3} + 3 ^ {3} + 4 ^ {3} = 1 00, \\quad (1 + 2 + 3 + 4) ^ {2} = 1 0 0; $\\n$\\dots \\dots$\\n通过观察发现:  $1^{3} + 2^{3} + 3^{3} + 4^{3} + 5^{3} + 6^{3} = \\underline{\\quad}$ . (填得数)",
+        "options": [],
+        "answer": "441",
+        "solution": "$1^{3} + 2^{3} + 3^{3} + 4^{3} + 5^{3} + 6^{3}\\\\ = (1 + 2 + 3 + 4 + 5 + 6) ^ {2} \\\\ = 2 1 ^ {2} = 4 4 1$\\n即通过观察发现:  $1^{3} + 2^{3} + 3^{3} + 4^{3} + 5^{3} + 6^{3} = 441$ .故答案为 441.",
+        "knowledge_points": [
+          "算式的规律"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_159",
+        "original_id": "159"
+      },
+      {
+        "id": "160",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "根据各式的规律填空：\\n$1 = 1 ^ {2}$\\n$1 + 3 = 2 ^ {2}$\\n$1 + 3 + 5 = 3 ^ {2}$\\n$1 + 3 + 5 + 7 = 4 ^ {2}$\\n(1)  $1 + 3 + 5 + 7 + 9 + 11 + 13 = \\underline{}^{2}$ .  \\n(2) 从 1 开始, _个连续奇数相加的和是  $20^{2}$",
+        "options": [],
+        "answer": "7   20",
+        "solution": "(1)  $1 + 3 + 5 + 7 + 9 + 11 + 13 = 7^{2}$ ;(2) 观察图形可知, 从 1 开始的连续奇数的和等于奇数个\\n数的平方，所以从 1 开始, 20 个连续奇数相加的和是  $20^{2}$ .故答案为 (1) 7; (2) 20.",
+        "knowledge_points": [
+          "数与形"
+        ],
+        "source": "19-20六年级上·全国·单元测试",
+        "merged_id": "unit8_160",
+        "original_id": "160"
+      },
+      {
+        "id": "161",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "1 除以 111 的商的小数部分第 15 位数字是 0。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "$1 \\div 111 = 0.009009 \\dots$ , 循环节是 009, 三位,  $15 \\div 3 = 5$ , 所以商的小数部分第 15 位数字是 9。故答案为:  ×",
+        "knowledge_points": [
+          "周期问题",
+          "除数是整数的小数除法"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_161",
+        "original_id": "161"
+      },
+      {
+        "id": "162",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "根据几个乘法算式找出的得数的规律，适用于所有具有同一特征算式的结果。",
+        "options": [],
+        "answer": "√",
+        "solution": "如\\n$ 1 \\times 9 = 9 \\quad 1 2 \\times 9 = 1 0 8 \\quad 1 2 3 \\times 9 = 1 1 0 7  \\quad\\dots$\\n根据几个乘法算式找出的得数的规律，适用于所有具有同一特征算式的结果，这种说法正确。故答案为： √",
+        "knowledge_points": [
+          "算式的规律"
+        ],
+        "source": "19-20六年级上·全国·单元测试",
+        "merged_id": "unit8_162",
+        "original_id": "162"
+      },
+      {
+        "id": "163",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "19. 325 小数点后第 80 位上的数字是 2. _____.",
+        "options": [],
+        "answer": "√",
+        "solution": "19. 325 可以看出：循环从小数点后第一位就开始了，循环节是325，共3位;$80 \\div 3 = 26 \\dots 2$ , 那么第 80 位的数字就是 2.",
+        "knowledge_points": [
+          "周期问题",
+          "小数的读写、意义及分类"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_163",
+        "original_id": "163"
+      },
+      {
+        "id": "164",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "判断题",
+        "difficulty": "困难",
+        "question": "$30 \\div 11 = 2.$  72 , 小数点后 100 位上的数字是 7 .",
+        "options": [],
+        "answer": "×",
+        "solution": "循环节是72两个数字;$100 \\div 2 = 50$ , 说明到第 100 位数字出现了 50 个循环节, 所以100 位上的数字是 2 ;所以原题错误，故答案为×",
+        "knowledge_points": [
+          "周期问题",
+          "小数的读写、意义及分类"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_164",
+        "original_id": "164"
+      },
+      {
+        "id": "165",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "判断题",
+        "difficulty": "困难",
+        "question": "一个数列为:1，2，3，1，2，3，…按这样的顺序排下去，第20个数是3.（）",
+        "options": [],
+        "answer": "×",
+        "solution": "$20 \\div 3 = 6$  （组）…2（个）每组中的第 2 个是 2 , 所以第 20 个数是 2 .故答案为×",
+        "knowledge_points": [
+          "数字排列的规律"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_165",
+        "original_id": "165"
+      },
+      {
+        "id": "166",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "判断题",
+        "difficulty": "困难",
+        "question": "$44 \\times 9 = 396, 444 \\times 9 = 3996$ , 由此可得  $44444 \\times 9 = 399996$ .",
+        "options": [],
+        "answer": "√",
+        "solution": "因为  $44 \\times 9 = 396$  $4 4 4 \\times 9 = 3 9 9 6$  所以  $44444 \\times 9 = 399996$故答案为  √",
+        "knowledge_points": [
+          "算式的规律"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_166",
+        "original_id": "166"
+      },{
+        "id": "167",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "判断题",
+        "difficulty": "困难",
+        "question": "在数列 '  $\\frac{1}{1}, \\frac{3}{4}$ ,$\\frac{5}{9}$ ,  $\\frac{7}{16}$ ,  $\\frac{9}{25}$ ,  $\\frac{11}{36}$ , …'中，第10个数是  $\\frac{19}{100}$ 。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "这个数列中从左到右分别是1、3、5、7、9、11、13、15、17、19…分母是  $10^{2} = 100$因此，在数列'  $\\frac{1}{1},\\frac{3}{4},\\frac{5}{9},\\frac{7}{16},\\frac{9}{25},\\frac{11}{36},…'$中，第10个数是  $\\frac{19}{100}$故答案为  √",
+        "knowledge_points": [
+          "数字排列的规律"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_167",
+        "original_id": "167"
+      },
+      {
+        "id": "168",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "将  $\\frac{1}{7}$  化成小数以后,小数点后第 2008 位上的数字是 7。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "$\\frac{1}{7} = 1 \\div 7 = 0.142857142857 \\dots$一直重复 142857，所以小数点后的数字周期为  $6_{\\circ}$$2 0 0 8 \\div 6 = 3 3 4 \\dots 4$故小数点后第 2008 位上的数字是 8。故答案为: ×",
+        "knowledge_points": [
+          "周期问题",
+          "循环小数的认识与简写"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_168",
+        "original_id": "168"
+      },
+      {
+        "id": "169",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "若一列数为:2, 4, 6, 8, 10, ……96, 98, 100, 则这列数的和是2550。 ( )",
+        "options": [],
+        "answer": "√",
+        "solution": "解：  $2 + 4 + 6 + 8 + 10 + \\dots +100\\\\ = \\frac {5 0 \\times (2 + 1 0 0)}{2} \\\\ = \\frac {5 1 0 0}{2} \\\\ = 2 5 5 0 $\\n所以原题计算正确。故答案为: √",
+        "knowledge_points": [
+          "数与形"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_169",
+        "original_id": "169"
+      },
+      {
+        "id": "170",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "判断题",
+        "difficulty": "困难",
+        "question": "在  $1 + 3 + 5 + 7 + 9 + \\ldots$  中,从 '1' 到数 '13' 的和是 49. ( )",
+        "options": [],
+        "answer": "√",
+        "solution": "在  $1 + 3 + 5 + 7 + 9 + \\ldots$  中，从'1'到数'13'的项数为：$(13 - 1)  \\div 2 + 1 = 12 \\div 2 + 1 = 6 + 1 = 7$ . 前 7 项的和为:$(1 3 + 1) \\times \\frac {7}{2} = 1 4 \\times 3. 5 = 4 9$因此，在  $1 + 3 + 5 + 7 + 9 + \\ldots$  中，从'1'到数'13'的和是49,原题的说法是正确的.故答案为   √。",
+        "knowledge_points": [
+          "数与形"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_170",
+        "original_id": "170"
+      },
+      {
+        "id": "unit8_calc_171",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "计算题",
+        "difficulty": "困难",
+        "question": "$(1) 1 + 3 + 5 + 11 + 9+7 + 5 + 3 + 1 = (\\quad)$\\n$(2) \\frac{1}{2} + \\frac{1}{4} + \\frac{1}{8} + \\frac{1}{16} + \\frac{1}{32} + \\frac{1}{64} + \\frac{1}{128} = (\\quad)$ .",
+        "options": [],
+        "answer": "45",
+        "solution": "略",
+        "knowledge_points": [
+          "分数的简便计算",
+          "整数的运算律与简便计算"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_171",
+        "original_id": "unit8_calc_171"
+      },
+      {
+        "id": "unit8_calc_172",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "看算式, 找规律,填答案.\\n(1)  $88.2 \\div 9 = 9.8$\\n(2)  $88.83 \\div 9 = 9.87$\\n(3)  $88.884 \\div 9 = 9.876$\\n(4)  $88.8885 \\div 9 =$\\n(5)  $88.88886 \\div 9 =$\\n(6)  $88.888887 \\div 9 =$",
+        "options": [],
+        "answer": "9.8765 9.87654 9.876543",
+        "solution": "【分析】通过观察发现，除数不变，被除数整数部分不变。小数部分的变化规律为：2、83、884、8885、88886、888887;商的变化规律为：9.8、9.87、9.876、9.8765、9.87654、9.876543;根据这个规律解答。本题主要考查算术中的变化规律, 解题的关键在于找到题中算式的规律。\\n【详解】④  $88.8885 \\div 9 = 9.8765$ ⑤  $88.88886 \\div 9 = 9.87654$   ⑥ $88.888887 \\div 9 = 9.876543$",
+        "knowledge_points": [
+          "算式的规律"
+        ],
+        "source": "19-20 五年级上·湖南张家界·期中",
+        "merged_id": "unit8_172",
+        "original_id": "unit8_calc_172"
+      },
+      {
+        "id": "unit8_calc_173",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "巧算下面各题。\\n(1)  $1 + 3 + 5 + 7 + 9 + 7 + 5 + 3 + 1$  \\n(2)$\\frac {1}{2} + \\frac {1}{4} + \\frac {1}{8} + \\frac {1}{1 6} + \\frac {1}{3 2} + \\frac {1}{6 4}$",
+        "options": [],
+        "answer": "(1) 41; (2)  $\\frac{63}{64}$",
+        "solution": "(1)  $1 + 3 + 5 + 7 + 9 + 7 + 5 + 3 + 1\\\\ = 1 0 \\times 4 + 1 \\\\ = 4 1$\\n(2)  $\\frac{1}{2} + \\frac{1}{4} + \\frac{1}{8} + \\frac{1}{16} + \\frac{1}{32} + \\frac{1}{64}\\\\= 1 - \\frac {1}{2} + \\frac {1}{2} - \\frac {1}{4} + \\frac {1}{4} - \\frac {1}{8} + \\frac {1}{8} - \\frac {1}{1 6} + \\frac {1}{1 6} - \\frac {1}{3 2} + \\frac {1}{3 2} - \\frac {1}{6 4} \\\\ = 1 - \\frac {1}{6 4} \\\\ = \\frac {6 3}{6 4} $",
+        "knowledge_points": [
+          "异分母分数加、减法",
+          "数与形"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_173",
+        "original_id": "unit8_calc_173"
+      },
+      {
+        "id": "unit8_calc_174",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "计算下面各题.\\n(1)  $\\frac{1}{2} + \\frac{1}{4} + \\frac{1}{8} + \\frac{1}{16}$  \\n(2)  $\\frac{1}{8} + \\frac{1}{16} + \\frac{1}{32} + \\frac{1}{64} + \\frac{1}{128}$  \\n(3)$1 - \\frac {1}{2} - \\frac {1}{4} - \\frac {1}{8} - \\frac {1}{1 6} - \\frac {1}{3 2}$",
+        "options": [],
+        "answer": "$(1)\\frac{15}{16};(2)\\frac{31}{128};(3)\\frac{1}{32}$",
+        "solution": "$(1)\\frac{1}{2} +\\frac{1}{4} +\\frac{1}{8} +\\frac{1}{16}\\\\ = 1 - \\frac {1}{2} + \\frac {1}{2} - \\frac {1}{4} + \\frac {1}{4} - \\frac {1}{8} + \\frac {1}{8} - \\frac {1}{1 6} \\\\ = 1 - \\frac {1}{1 6} \\\\ = \\frac {1 5}{1 6} $\\n(2)  $\\frac{1}{8} + \\frac{1}{16} + \\frac{1}{32} + \\frac{1}{64} + \\frac{1}{128}\\\\ = \\frac {1}{4} - \\frac {1}{8} + \\frac {1}{8} - \\frac {1}{1 6} + \\frac {1}{1 6} - \\frac {1}{3 2} + \\frac {1}{3 2} - \\frac {1}{6 4} + \\frac {1}{6 4} - \\frac {1}{1 2 8} \\\\ = \\frac {1}{4} - \\frac {1}{1 2 8}\\\\=\\frac{31}{128} $\\n(3)$1-\\frac{1}{2}-\\frac{1}{4}-\\frac{1}{8}-\\frac{1}{16}-\\frac{1}{32}\\\\=\\frac {1}{2} - \\frac {1}{4} - \\frac {1}{8} - \\frac {1}{1 6} - \\frac {1}{3 2} \\\\ = \\frac {1}{4} - \\frac {1}{8} - \\frac {1}{1 6} - \\frac {1}{3 2} \\\\ = \\frac {1}{8} - \\frac {1}{1 6} - \\frac {1}{3 2} \\\\ = \\frac {1}{1 6} - \\frac {1}{3 2} \\\\ = \\frac {1}{3 2} $",
+        "knowledge_points": [
+          "运算定律与简便运算",
+          "数与形"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_174",
+        "original_id": "unit8_calc_174"
+      },
+      {
+        "id": "unit8_calc_175",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "计算.\\n(1)  $1 - \\frac{1}{2} - \\frac{1}{4} - \\frac{1}{8} - \\frac{1}{16} - \\frac{1}{32}$  \\n(2)  $\\frac{1}{2} + \\frac{1}{6} + \\frac{1}{12} + \\frac{1}{20} + \\frac{1}{30}$",
+        "options": [],
+        "answer": "(1)  $\\frac{1}{32}$  (2)  $\\frac{5}{6}$",
+        "solution": "【分析】根据式子进行计算得出结果即可。本题考查了异分母分数的加减法, 关键是要掌握在进行异分母分数相加减时, 要先通分, 转化为分母一样时再进行计算。\\n【详解】 (1)  $1 - \\frac{1}{2} - \\frac{1}{4} - \\frac{1}{8} - \\frac{1}{16} - \\frac{1}{32} = \\frac{1}{2} - \\frac{1}{4} - \\frac{1}{8} - \\frac{1}{16} - \\frac{1}{32} =\\frac {1}{4} - \\frac {1}{8} - \\frac {1}{1 6} - \\frac {1}{3 2} = \\frac {1}{8} - \\frac {1}{1 6} - \\frac {1}{3 2} = \\frac {1}{1 6} - \\frac {1}{3 2} = \\frac {1}{3 2}$\\n(2)  $\\frac{1}{2} + \\frac{1}{6} + \\frac{1}{12} + \\frac{1}{20} + \\frac{1}{30} = \\frac{2}{3} + \\frac{1}{12} + \\frac{1}{20} + \\frac{1}{30} = \\frac{3}{4} + \\frac{1}{20} + \\frac{1}{30} = \\frac{4}{5} + \\frac{1}{30} = \\frac{5}{6}$\\n故答案为:  $\\frac{1}{32}; \\frac{5}{6}$",
+        "knowledge_points": [
+          "分数的加、减法",
+          "分数的简便计算"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_175",
+        "original_id": "unit8_calc_175"
+      },
+      {
+        "id": "unit8_draw_176",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "六年级一班40名同学上学期期末数学成绩考试得优的有10人，良有20人，及格的和待及格的都是5人。请用圆表示上学期期末数学测试的结果。",
+        "options": [],
+        "answer": "![](images/4f55cde5ef0b99208940e9c1887de130ae68d357a9ea820f8e0126436cee1b06.jpg)",
+        "solution": "根据题意, 考试得优的 10 人占总人数的  $\\frac{1}{4}$ , 得良的 20 人, 占总人数的  $\\frac{1}{2}$ , 及格和待及格的占总人数的  $\\frac{1}{8}$ , 因此画图得到上图.",
+        "knowledge_points": [],
+        "source": "19-20六年级上·全国·单元测试",
+        "merged_id": "unit8_176",
+        "original_id": "unit8_draw_176"
+      },
+      {
+        "id": "177",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "小兔想做一个正方形的画框, 可是它的细木条长短都不一样, 有1, 2, 3, 4, 5, 6, 7, 8, 9 厘米长的细木条各一根.\\n(1) 如果不能锯断, 但可以连接 (接头处不计损耗), 小兔可以做一个怎样的正方形画框?  \\n(2) 要做边长是 9 厘米的正方形的画框, 可以怎么做?",
+        "options": [],
+        "answer": "（1）边长是 10 厘米的正方形画框。(2) 取 1 厘米与 8 厘米连接、取 2 厘米与 7 厘米连接、取 3 厘米与 6 厘米连接、取 9 厘米的木条依次连接即可做出。",
+        "solution": "【分析】根据题意, 9 根细木条可以连接成边长为 10 厘米的正方形或边长为 9 厘米的正方形, 利用整数加法的运算法则计算即可。本题主要考查数与形结合的规律，关键根据图示发现这组图形的规律，并运用规律做题。\\n【详解】 (1) 根据题意: $1 + 9 = 10  (厘米)   2 + 8 = 10 (厘米)   3 + 7 = 10  (厘米)  4 + 6 = 10  (厘米)$答: 小兔可以做边长是 10 厘米的正方形画框。（答案不唯一，合理即可。）\\n(2) 要做边长是 9 厘米的正方形的画框:\\n$1 + 8 = 9  (厘米)\\quad 2 + 7 = 9  (厘米)\\quad 3 + 6 = 9  (厘米)$最后取9厘米的木条。",
+        "knowledge_points": [
+          "正方形的概念及特点"
+        ],
+        "source": "19-20 六年级上·全国·期末",
+        "merged_id": "unit8_177",
+        "original_id": "177"
+      },
+      {
+        "id": "178",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "如图是用小棒摆出的正方形, 观察图形中的规律, 画出后面的图形, 再填一填。\\n![](images/354754f732f117e155747d60d6b15722.png)",
+        "options": [],
+        "answer": "![](images/cd6e3473fdad9fbb71b5a341b2b302a5.png)",
+        "solution": "【分析】根据图示，发现这组图形的规律：摆1个正方形需要小棒：4根;摆2个正方形需要小棒： $4 + 3 = 7$ （根）;摆3个正方形需要小棒： $4 + 3 + 3 = 10$ （根）;……摆n个正方形需要小棒： $4 + 3$ （n-1）=（3n+1）根。据此解答并完成作图。本题主要考查数与形结合的规律, 关键根据所给图示发现这组图形的规律, 然后利用规律做题。\\n【详解】如图：![](images/cd6e3473fdad9fbb71b5a341b2b302a5.png)摆1个正方形需要小棒：4根  \\n摆 2 个正方形需要小棒:  $4 + 3 = 7$  (根)  摆 3 个正方形需要小棒:  $4 + 3 + 3 = 10$  (根) $\\dots \\dots$摆  $n$  个正方形需要小棒:  $4 + 3(n - 1) = (3n + 1)$  根所以，摆 4 个正方形需要小棒:\\n$ 3 \\times 4 + 1 \\\\ = 1 2 + 1 \\\\ = 1 3 (\\text {根}) $",
+        "knowledge_points": [
+          "图形的规律"
+        ],
+        "source": "19-20 六年级上·全国·期末",
+        "merged_id": "unit8_178",
+        "original_id": "178"
+      },
+      {
+        "id": "179",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "张伟在纸上画了四个点，如果把这四个点彼此连接成一个图形，这个图形中可能有几个三角形？",
+        "options": [],
+        "answer": "8个   4个   3个   0个",
+        "solution": "略",
+        "knowledge_points": [],
+        "source": "19-20六年级上·全国·单元测试",
+        "merged_id": "unit8_179",
+        "original_id": "179"
+      },
+      {
+        "id": "180",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "A、B、C、D、E、F六位同学见面，每两人之间都要握一次手。到现在为止，A 握了 5 次，B 握了 4 次，最少的也握了2次．这时6个人至少握了多少次？",
+        "options": [],
+        "answer": "18次",
+        "solution": "略",
+        "knowledge_points": [
+          "数与形"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_180",
+        "original_id": "180"
+      },
+      {
+        "id": "181",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "一条马路长 180 米,小明和他的小狗分别以均匀的速度同时从马路的起点出发.当小明走到马路的  $\\frac{1}{3}$  时, 小狗已经到达马路的终点。然后返回与小明相向而行，遇到小明以后再跑向终点，达到终点以后再与小明相向而行……直到小明到达终点。小狗从出发开始，一共跑了多少米？",
+        "options": [],
+        "answer": "540 米",
+        "solution": "$180 \\div \\frac{1}{3} = 540$  （米）",
+        "knowledge_points": [
+          "数与形"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_181",
+        "original_id": "181"
+      },
+      {
+        "id": "182",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "小明、小强、小丽、小芳、和小琼进行象棋比赛，每两人之间都要下一盘。小明、小强都已经下了 4 盘, 小丽、小芳都已经下了 3 盘, 小琼下的盘数最少．请问：小琼下了几盘？分别和谁下的？",
+        "options": [],
+        "answer": "2 盘，与小明、小强各下一盘",
+        "solution": "小明与小强、小丽、小芳、小琼各下一盘小强与小明、小丽、小芳、小琼各下一盘小丽与小明、小强、小丽各下一盘小丽与小明、小强、小丽各下一盘小琼与小明、小强各下一盘答: 小琼下了 2 盘, 小琼与小明、小强各下一盘",
+        "knowledge_points": [],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_182",
+        "original_id": "182"
+      },
+      {
+        "id": "183",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "欢欢和乐乐带着一条狗在郊外玩，欢欢与乐乐相距  $1800\\mathrm{m}$ ，两人相向而行，这时狗从欢欢跑往乐乐，遇到乐乐后狗立即返回跑往欢欢，狗在两人之间不停在往返跑动。乐乐每分钟行  $55\\mathrm{m}$ ，欢欢每分钟行  $65\\mathrm{m}$ ，狗的速度是160米/分。当欢欢和乐乐相遇时狗共跑了多少米？",
+        "options": [],
+        "answer": "2400 米",
+        "solution": "$1800 \\div (55 + 65) = 15$（分） $1 6 0 \\times 1 5 = 2 4 0 0 (\\text {米})$",
+        "knowledge_points": [],
+        "source": "19-20六年级上·全国·单元测试",
+        "merged_id": "unit8_183",
+        "original_id": "183"
+      },
+      {
+        "id": "184",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "下图是百世超市一月份统计的营业额中各种支付方式情况统计图。\\n![](images/27e04d9b8ed97300a025e3d7f620f519c8f6e384b8c2285215dab7a9e41b69b4.jpg)\\n(1)财付通支付占百分之几？哪种支付方式所占百分比最多？  \\n(2)你还了解到了哪些信息？  \\n(3)看到上面的统计图，你想到了什么？",
+        "options": [],
+        "answer": "(1)  $1 - 40\\% - 16\\% - 20\\% - 18\\% = 6\\%$  支付宝(2)使用财付通支付的最少等。(答案不唯一)  (3)现在购物的支付方式越来越多，用现金支付的人逐渐减少。（答案不唯一，合理即可）",
+        "solution": "略",
+        "knowledge_points": [
+          "统计图表的综合应用"
+        ],
+        "source": "19-20六年级上·全国·单元测试",
+        "merged_id": "unit8_184",
+        "original_id": "184"
+      },
+      {
+        "id": "185",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "下面是六(1)班的李明同学一天作息时间安排情况统计图。\\n![](images/5e33d47a9e23e8f32742fcd43f71590503a769cc599ee016c18781b4d403c790.jpg)\\n李明的作息时间安排合理吗？请你提出几条合理化的建议。",
+        "options": [],
+        "answer": "不合理.建议：应安排上自主学习的时间、写作业的时间等。（答案不唯一，合理即可）",
+        "solution": "略",
+        "knowledge_points": [
+          "统计图表的综合应用"
+        ],
+        "source": "19-20六年级上·全国·单元测试",
+        "merged_id": "unit8_185",
+        "original_id": "185"
+      },
+      {
+        "id": "186",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "在式子  $\\square \\div 8 = 9$  。○中, 可以出现多少种不同的形式, 你能把它们都写出来吗?",
+        "options": [],
+        "answer": "七种.\\n$7 3 \\div 8 = 9 \\dots \\dots 1; 7 4 \\div 8 = 9 \\dots \\dots 2; 7 5 \\div 8 = 9 \\dots \\dots 3; 7 6 \\div 8 =9 \\dots \\dots 4;\\\\7 7 \\div 8 = 9 \\dots \\dots . 5; 7 8 \\div 8 = 9 \\dots \\dots . 6; 7 9 \\div 8 = 9 \\dots \\dots . 7。$",
+        "solution": "略",
+        "knowledge_points": [],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_186",
+        "original_id": "186"
+      },
+      {
+        "id": "187",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "$1 + 2 + 3 + \\dots \\dots + 99 =？$说一说你的计算方法.",
+        "options": [],
+        "answer": "$495049 \\times 100 + 50 = 4950$",
+        "solution": "1 和 99 相加，2 和 98 相加……49 和 51 相加，最剩 50，所以结果是  $49 \\times 100 + 50 = 4950$",
+        "knowledge_points": [],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_187",
+        "original_id": "187"
+      },
+      {
+        "id": "188",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "一张桌子坐 6 人,两张桌子并起来坐10人，三张桌子并起来坐14人，照这样，10 张桌子并成一排可以坐多少人？如果一共有 38 人，需要并多少张桌子才能坐下？",
+        "options": [],
+        "answer": "42人;9张",
+        "solution": "$9 \\times 4 + 6 = 42$  （人） $3 8 - 1 0 = 2 8 \\text {(人)}$  $2 8 \\div 4 = 7 \\text {(张)}$ $7 + 2 = 9$  (张) 答: 并成一排可以坐 42 人; 如果一共有 38 人, 需要并 9 张桌子才能坐下。",
+        "knowledge_points": [
+          "数与形"
+        ],
+        "source": "19-20 五年级上·全国·单元测试",
+        "merged_id": "unit8_188",
+        "original_id": "188"
+      },
+      {
+        "id": "189",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "一条马路全长420米，小张步行，小刘骑自行车，他们同时以均匀的速度从起点出发，当小张走到这条马路的  $\\frac{1}{6}$  时，小刘到了马路的  $\\frac{1}{2}$  处。当小刘到达终点后立即返回，遇到小张又掉头骑向终点，到达终点后在返回，---，直到小张也到达终点。小刘从出发开始，一共骑了多少米？",
+        "options": [],
+        "answer": "1260 米",
+        "solution": "$420 \\div (\\frac{1}{2} \\div \\frac{1}{6}) = 1260$  （米）",
+        "knowledge_points": [
+          "行程问题"
+        ],
+        "source": "19-20六年级上·全国·单元测试",
+        "merged_id": "unit8_189",
+        "original_id": "189"
+      },
+      {
+        "id": "190",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "有一列数:  $\\frac{1}{1}$ ,$\\frac{1}{2}, \\frac{2}{2}, \\frac{1}{3}, \\frac{2}{3}, \\frac{3}{3}, \\frac{1}{4}, \\cdots$  它的前 2015 个数的和是多少？",
+        "options": [],
+        "answer": "1007.5",
+        "solution": "以 1 为分母的数有 1 个, 相加和  $S_{1} = 1$ ,以2为分母的数有2个，相加和  $S_{2} = \\frac{1}{2} +\\frac{2}{2} = \\frac{3}{2}$以3为分母的数有3个，相加和  $S_{3} = \\frac{1}{3} +\\frac{2}{3} +\\frac{3}{3} = 2,\\dots$以  $\\mathrm{n}$  为分母的数有  $\\mathrm{n}$  个，相加和  $S_{n} = \\frac{1}{n} +\\frac{2}{n} +\\dots +\\frac{n}{n} =$\\n$\\frac {n (n + 1)}{2 n} = \\frac {n + 1}{2},$求前 2015 个数的和, 先确定第 2015 个数分母是什么, 即求满足  $1 + 2 + 3 + 4 \\ldots + m = \\frac{n(n + 1)}{2} \\geqslant 2015$  的最小整数  $n$ , 易得  $n$$= 6 3, 6 2 \\times 6 3 \\div 2 = 1 9 5 3,$分母为 63 的数有  $2015 - 1953 = 62$  个, 即  $\\frac{1}{63}  、\\frac{2}{63}  \\frac {3}{6 3}  \\dots  \\frac {6 2}{6 3},$则前 2015 个数的和是:$S = S _ {1} + S _ {2} + \\dots S _ {6 2} + \\frac {1}{6 3} + \\frac {2}{6 3} + \\frac {3}{6 3} + \\dots + \\frac {6 2}{6 3}\\\\ = (1 + 2 + 3 + \\dots 6 2) \\div 2 + (1 + 2 + 3 + \\dots + 6 2) \\div 6 3 \\\\ = (1 + 6 2) \\times 6 2 \\div 2 \\div 2 + (1 + 6 2) \\times 6 2 \\div 2 \\div 6 3 \\\\ = 9 7 6. 5 + 3 1 \\\\ = 1 0 0 7. 5 $\\n答: 它的前 2015 个数的和是 1007.5 .",
+        "knowledge_points": [],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_190",
+        "original_id": "190"
+      },
+      {
+        "id": "191",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "龙一鸣、黄霏霏、苹苹、淘淘和依依5个好朋友聚会，每两人之间都要握一次手．已知龙一鸣已经握了4次手，黄霏霏已经握了3次手，依依已经握了2次手，苹苹已经握了1次手．淘淘一共握了几次手？分别与谁握的？(可画图连一连)![](images/d5cfddf85af15b7f7817c6f791dc2259.png)",
+        "options": [],
+        "answer": "![](images/1a90e7c9b4ce323f799cbc825ddb2af0299c2bfccc6e3e9966e2cb9ac6716924.jpg)由图可以看出淘淘一共握了两次手，分别是与龙一鸣和黄\\n霏霏握的.",
+        "solution": "略",
+        "knowledge_points": [
+          "数与形"
+        ],
+        "source": "19-20六年级上·全国·单元测试",
+        "merged_id": "unit8_191",
+        "original_id": "191"
+      },
+      {
+        "id": "192",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "用所学知识解决下列问题.\\n$\\frac {2}{3} + \\frac {2}{9} + \\frac {2}{2 7} + \\frac {2}{8 1} + \\frac {2}{2 4 3} + \\dots \\dots =$",
+        "options": [],
+        "answer": "1",
+        "solution": "![](images/13e9c32082838d0e15d646825ec75084.png)从图上可以看出这些分数不断加下去，总和就是一",
+        "knowledge_points": [],
+        "source": "19-20六年级上·全国·单元测试",
+        "merged_id": "unit8_192",
+        "original_id": "192"
+      },
+      {
+        "id": "193",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "有一次篮球比赛，共有10支球队参加比赛，如果每两支球队之间进行一场比\\n赛，一共要比赛多少场？",
+        "options": [],
+        "answer": "45场",
+        "solution": "【分析】根据题意每两支球队之间都要进行一场比赛，可以画图为：![](images/c86a619ebdad5f9ca79c99a52cd8a5d3.png)\\n从图中看出第一队要打 9 场, 第二队要 8 场, 第三队 7 场,……\\n【详解】  $9 + 8 + 7 + 6 + 5 + \\dots \\dots 1 = 45$  场.答: 一共要比赛 45 场.",
+        "knowledge_points": [
+          "搭配问题"
+        ],
+        "source": "19-20六年级上·全国·单元测试",
+        "merged_id": "unit8_193",
+        "original_id": "193"
+      },
+      {
+        "id": "194",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "张林带着他心爱的小狗去外婆家。他步行的速度是每分钟 80 米, 小狗以每分\\n钟200米的速度奔向了外婆家。小狗过了8分钟到达外婆家后又立刻返回，路上遇见张林后它又返回奔向外婆家……就这样它不停来回跑。当张林到达外婆家时，小狗总共跑了多少米？",
+        "options": [],
+        "answer": "$200 \\times 8 \\div 80 \\times 200 = 4000$  （米）",
+        "solution": "略",
+        "knowledge_points": [
+          "数与形"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_194",
+        "original_id": "194"
+      },
+      {
+        "id": "195",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "甲、乙、丙、丁与小明五人进入象棋比赛的决赛，每两人要比赛一场，每胜一场得2分，和一场得1分，输一场得0分．到现在为止，甲赛了4场，得2分;乙赛了3场，得4分;丙赛了2场，得1分;丁赛了1场，得2分．那么小明现在已经赛了几场？得多少分？",
+        "options": [],
+        "answer": "赛了 2 场, 得 3 分",
+        "solution": "略",
+        "knowledge_points": [
+          "数与形"
+        ],
+        "source": "19-20六年级上·全国·单元测试",
+        "merged_id": "unit8_195",
+        "original_id": "195"
+      },
+      {
+        "id": "196",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "某仓库堆放一批圆木，一共10层，第一层3根，每往下一层多1根，问这堆圆木一共有多少根？",
+        "options": [],
+        "answer": "75 根",
+        "solution": "【分析】根据题意, 把第一层的根数看作梯形的上底, 最下层的根数看作梯形的下底, 层数看作梯形的高, 由梯形的面积公式就可以求出结果。\\n【详解】根据题意可得最下面的一层的根数是:  $10 + 3-1 = 12$  (根)\\n$(1 2 + 3) \\times 1 0 \\div 2\\\\ = 1 5 \\times 1 0 \\div 2 \\\\ = 1 5 0 \\div 2 \\\\ = 7 5 \\text {(根)} $\\n答: 这堆圆木一共有 75 根。",
+        "knowledge_points": [
+          "梯形面积的应用",
+          "数与形"
+        ],
+        "source": "19-20六年级上·全国·单元测试",
+        "merged_id": "unit8_196",
+        "original_id": "196"
+      },
+      {
+        "id": "197",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "某种细菌在培养过程中，每  $\\frac{1}{2}$  小时分裂 1 次，每次一分为二。若这种细菌由 1 个分裂到 16 个，这个过程要多长时间？",
+        "options": [],
+        "answer": "2 小时",
+        "solution": "略",
+        "knowledge_points": [
+          "分数与整数的乘法"
+        ],
+        "source": "19-20六年级上·全国·单元测试",
+        "merged_id": "unit8_197",
+        "original_id": "197"
+      },
+      {
+        "id": "198",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "3支球队进行单循环赛（参赛的每支队都与其他所有各队比赛一场），总的比赛场数是多少？4支球队呢？12支球队呢？",
+        "options": [],
+        "answer": "3666",
+        "solution": "略",
+        "knowledge_points": [
+          "搭配问题"
+        ],
+        "source": "19-20六年级上·全国·单元测试",
+        "merged_id": "unit8_198",
+        "original_id": "198"
+      },
+      {
+        "id": "199",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "1、4、7、10、13、…这个数列中，有6个连续数字的和是159，那么这6个数中最小的是几？",
+        "options": [],
+        "answer": "19",
+        "solution": "解：设这6个数中最小的数为  $x$ ，据题意可得方程：\\n$x +  (x + 3)  + \\dots +  (x + 1 5) = 1 5 9\\\\6 x + (3 + 6 + \\dots + 1 5) = 1 5 9\\\\6 x + 4 5 = 1 5 9\\\\6 x = 1 1 4\\\\x = 1 9$\\n答: 这 6 个数中最小的是  $19$  。",
+        "knowledge_points": [
+          "数字排列的规律",
+          "列方程解含一个未知数的问题"
+        ],
+        "source": "19-20六年级上·全国·单元测试",
+        "merged_id": "unit8_199",
+        "original_id": "199"
+      },
+      {
+        "id": "200",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "通过计算发现规律。\\n$6543 - 2345 =\\_\\_\\_$   $9876 - 5678 =\\_\\_\\_$   $7654-3456=\\_\\_\\_$\\n按找到的规律，再写两个算式。",
+        "options": [],
+        "answer": "4198;4198;4198   $8765 - 4567 = 4198$    $5432 - 1234 = 4198$",
+        "solution": "$6543 - 2345 = 4198$   $9876 - 5678 = 4198$   $7654 - 3456 = 4198$另外两个算式:$8765 - 4567 = 4198$$5432 - 1234 = 4198$",
+        "knowledge_points": [
+          "数与形"
+        ],
+        "source": "19-20 六年级上·全国·单元测试",
+        "merged_id": "unit8_200",
+        "original_id": "200"
+      },
+      {
+        "id": "201",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "瑞士的巴尔末从测量光谱的数据  9/5 、 16/12 、 25/21 、 36/32… 中得到了巴尔末公式, 请你按这种规律写出第 6 个数据, 这个数据为 ( )。",
+        "options": [
+          "A. 81/77",
+          "B. 79/77",
+          "C. 64/60",
+          "D. 81/60"
+        ],
+        "answer": "C",
+        "solution": "【分析】观察所给数据的分子，可得：  9 = 3² ，  16 = 4² ，25 = 5² ，  36 = 6² ，可以发现分子是从3开始的连续自然数的平方，  3 - 1 = 2 ，即第  n  个数据的分子为  (n + 2)² 。分母依次为  5 = 3² - 4 ，  12 = 4² - 4 ，  21 = 5² - 4 ，  32 = 6² - 4 ，可以发现分母是从3开始的连续自然数的平方减4，即第  n  个数据的分母为  (n + 2)² - 4 。当  n = 6  时，分子为  (6 + 2)² = 8² =64, 分母为  (6 + 2)² - 4 = 64 - 4 = 60 , 所以第 6 个数据为64/60。【详解】由分析可知:3 - 1 = 2  第  n  个数据的分子:  (n + 2)²   第  n  个数据的分母:  (n + 2)² - 4。当  n = 6  时;分子: (6 + 2)² = 8² = 64。分母: (6 + 2)² - 4 = 8² - 4 = 64 - 4 = 60。所以第 6 个数据为  64/60  。故答案为: C",
+        "knowledge_points": [
+          "分数排列的规律"
+        ],
+        "source": "24-25 六年级上·湖南湘西·期末",
+        "merged_id": "unit8_201",
+        "original_id": "201"
+      },
+      {
+        "id": "202",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "古希腊著名的毕达哥拉斯学派对'形数'的研究最为突出，有效印证了'凡物皆数'的观点。观察下图的点阵图形，依次排列下去，根据点数的变化规律，则第9个图形中的点数为（）。![](images/f6d95cbb40af8508a5bbd46d8731fa54cb898bb22cb2818747b12994f7afae64.jpg)",
+        "options": [
+          "A. 25",
+          "B. 29",
+          "C. 33",
+          "D. 37"
+        ],
+        "answer": "C",
+        "solution": "【分析】由图可知, 第 1 个图形中的点数为 1 , 可表示为  4 × 1 - 3 ;第2个图形中的点数为5，可表示为  4 × 2 - 3 ;第3个图形中的点数为9，可表示为  4 × 3 - 3 ;第 4 个图形中的点数为 13，可表示为  4 × 4 - 3 ;由此可推出, 第  n  个图形中的点数为  (4 n - 3)  。据此解答。【详解】分析可知，第  n  个图形中的点数为（4n-3）。当  n = 9  时, 4 n - 3 = 4 × 9 - 3 = 36 - 3 = 33。所以第 9 个图形中的点数为 33。故答案为: C",
+        "knowledge_points": [
+          "含有字母式子的化简与求值",
+          "用字母表示数、数量关系",
+          "数与形（归纳递推）"
+        ],
+        "source": "24-25六年级上·湖北襄阳·期末",
+        "merged_id": "unit8_202",
+        "original_id": "202"
+      },
+      {
+        "id": "203",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "如图，是一组有规律的图案，它们是由边长相同的小正方形组成的，其中部分小正方形涂有阴影，按照这样的规律，第2025个图案中涂有阴影的小正方形有（）个。![](images/a58632e3ee37bee840e5d4a97e1b85e453dd9aeb7d12abc1e72948f7217d3946.jpg)第1个![](images/923b5a5a0857229414784b6f6a99f32e1a07dd0ced01101b4595335c609d7b6e.jpg)第2个![](images/064ac66b2bf7e6ad67c6c4faf583a30580bcc018b769125887bd1e3cedd2d80f.jpg)第3个![](images/4a3bd56ff2aa35b8e4463b3f9054396b295b96c2527bba141b1b22312d301e2e.jpg)",
+        "options": [
+          "A. 8101",
+          "B. 8103",
+          "C. 4051",
+          "D. 4053"
+        ],
+        "answer": "A",
+        "solution": "【分析】由图可知, 第 1 个图案中涂有阴影的小正方形有 5 个, 可表示为  1 + 4 × 1 ;第 2 个图案中涂有阴影的小正方形有 9 个, 可表示为  1 + 4 × 2 ;第3个图案中涂有阴影的小正方形有13个，可表示为  1 + 4 × 3 ;发现规律：第  n  个图案中涂有阴影的小正方形有  (1 + 4n)个; 最后将  n = 2025  代入  1 + 4 n  中计算出第 2025 个图案中涂有阴影的小正方形个数。【详解】分析可知，第  n  个图案中涂有阴影的小正方形有 (1 + 4n)  个。当  n = 2025  时, 1 + 4 n = 1 + 4 × 2025 = 1 + 8100 = 8101。所以第 2025 个图案中涂有阴影的小正方形有 8101 个。故答案为：A",
+        "knowledge_points": [
+          "含有字母式子的化简与求值",
+          "用字母表示数、数量关系",
+          "数与形（归纳递推）"
+        ],
+        "source": "24-25六年级下·黑龙江哈尔滨·期末",
+        "merged_id": "unit8_203",
+        "original_id": "203"
+      },
+      {
+        "id": "204",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "晓东用'□'按下面的规律拼图形，第6个图形中，'□'的数量为（）个。![](images/14a67b36ac743c7662d6dc9ff2faf979ad8420bbab90c75770ee50b6061e7edc.jpg)(1)![](images/3e39d379261e70b8447af1e852cca7d393c818a9a9b904313d6b5c6f9463bee0.jpg)(2）![](images/22fd0e2dbdfc279e9bf43bcebbae748cdc91032565c5e949bda3522b31b870a0.jpg)(3）![](images/f7c445c42550baa4606d048e3649ddd079ac26165553562d4a9aec238f1404bf.jpg)(4)",
+        "options": [
+          "A. 40",
+          "B. 35",
+          "C. 27",
+          "D. 20"
+        ],
+        "answer": "C",
+        "solution": "【分析】看图可知, 第 1 个图形有 2 个  □ ; 第二个图形有 5 个  □, 5 = 2 + 3 ; 第 3 个图形有 9 个  □, 9 = 2 + 3 + 4... 由此可知, ' □ ' 的数量 = 第几个图形就从 2 依次加几个数。【详解】  2 + 3 + 4 + 5 + 6 + 7 = 27  （个）第 6 个图形中，'□'的数量为 27 个。故答案为: C",
+        "knowledge_points": [
+          "数与形（归纳递推）"
+        ],
+        "source": "24-25六年级下·广东东莞·期末",
+        "merged_id": "unit8_204",
+        "original_id": "204"
+      },
+      {
+        "id": "205",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "如图, 三个杯子叠起来高 13 厘米, 五个杯子叠起来高 17 厘米。n 个杯子叠起来的高度是 ( ) 厘米。![](images/642fc6678ecc2ee066a630802dd1b8636873e73ce542028f6905f619b0b860ae.jpg)![](images/2b3781a78acf0baae89d2563c7f0aec483a5b8b00b398693c81933b30d9deb8b.jpg)",
+        "options": [
+          "A. 6 + 2n",
+          "B. 7 + 2n",
+          "C. 8 + 2n",
+          "D. 9 + 2n"
+        ],
+        "answer": "B",
+        "solution": "【分析】已知 3 个杯子叠起来高 13 厘米, 5 个杯子叠起来高 17 厘米, 那么 (5-3) 个杯子叠起来的高度是 (17-13) 厘米, 用  (17 - 13) ÷ (5 - 3) = 2  厘米, 求出每多叠一个杯子增加的高度为 2 厘米;从图中可知, 3 个杯子叠起来时高 13 厘米, 有 2 个重叠部分高  2 × 2 = 4  厘米, 则一个杯子的高度为  13 - 4 = 9  厘米。由 3 个杯子叠起来有 2 个重叠部分, 5 个杯子叠起来有 4 个重叠部分, 可得出: n 个杯子叠起来有 (n-1) 个重叠部分,那么 n 个杯子叠起来的高度 = 一个杯子的高度 + (n-1) × 每多叠一个杯子增加的高度, 据此得出规律。【详解】每多叠一个杯子增加的高度为: (17 - 13) ÷ (5 - 3) = 4 ÷ 2 = 2（厘米）。一个杯子的高度为:13 - 2 × 2 = 13 - 4 =9 (厘米)。n 个杯子叠起来的高度是:9 + (n-1) ×2 = 9 + 2n - 2 = (7 + 2n) 厘米。所以，n个杯子叠起来的高度是（7+2n）厘米。故答案为: B",
+        "knowledge_points": [
+          "含有字母式子的化简与求值",
+          "用字母表示稍复杂的数量关系",
+          "数与形（探索规律）"
+        ],
+        "source": "24-25 六年级下·云南曲靖·期末",
+        "merged_id": "unit8_205",
+        "original_id": "205"
+      },
+      {
+        "id": "206",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "有这样一组数:  8、12、16、20、…，第  n  个数是（）。",
+        "options": [
+          "A. 4n + 4",
+          "B. 4n",
+          "C. 3n",
+          "D. 2n"
+        ],
+        "answer": "A",
+        "solution": "【分析】观察这组数列：8、12、16、20、…。计算相邻两个数的差值:  12 - 8 = 4, 16 - 12 = 4, 20 - 16 = 4 , 可以得出相邻两个数相差 4 。以第 2 个数 12 为例, 可以表示为:  8 + (2 - 1) × 4 = 8 + 1 × 4 = 8 + 4 = 12  。由此可知第 n 个数表示为:  8 + (n - 1) × 4 , 然后计算这个关系式即可。【详解】  12 - 8 = 4   16 - 12 = 4。20 - 16 = 4。8 + (n - 1) × 4 = 8 + 4 n - 4 = 4 n + 4。所以第  n  个数是  4 n + 4  。故答案为: A",
+        "knowledge_points": [
+          "整数排列的规律",
+          "用字母表示数、数量关系"
+        ],
+        "source": "24-25 六年级下·四川凉山·期末",
+        "merged_id": "unit8_206",
+        "original_id": "206"
+      },
+      {
+        "id": "207",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "古希腊著名的毕达哥拉斯学派把1、3、6、10...这样的数称为'三角形数'，把1、4、9、16...这样的数称为'正方形数'。从下图中可以发现，任何一个大于 1 的'正方形数'都可以看作是两个相邻'三角形数'之和。下列等式中符合这一规律的是（）。![](images/340272b9185a55cef60526a455dfc0bc8773d2a9206f7d3ca7b008a933b272c4.jpg)",
+        "options": [
+          "A. 13 = 3 + 10",
+          "B. 25 = 9 + 16",
+          "C. 25 = 10 + 15"
+        ],
+        "answer": "C",
+        "solution": "【分析】'三角形数'的规律是  1, 1 + 2 = 3, 1 + 2 + 3 = 6, 1 + 2 + 3 + 4 = 10, 1 + 2 + 3 + 4 + 5 = 15… ，即第  n  个'三角形数'为  1 + 2 + … + n = n(n + 1)/2 。'正方形数'的规律是 1² = 1, 2² = 4, 3² = 9, 4² = 16, 5² = 25… ，即第  n  个'正方形数'为  n² 。据此分析各选项，进而得出正确答案。【详解】A.'三角形数'有 1, 3, 6, 10, 15…，13 不是'三角形数'，所以 A 不符合。B.9 和 16 都是'正方形数'，而不是'三角形数'，所以 B 不符合。C.10 和 15 是相邻的两个'三角形数'，25 是'正方形数'，符合规律。故答案为: C",
+        "knowledge_points": [
+          "数与形（归纳递推）"
+        ],
+        "source": "24-25 五年级下·重庆万州·期末",
+        "merged_id": "unit8_207",
+        "original_id": "207"
+      },
+      {
+        "id": "209",
+        "unit": "数与形",
+        "question_type": "选择题",
+        "difficulty": 0.65,
+        "question": "如下图：正方形纸片按规律拼成如下图形，第( )个图案恰好有37个纸片。![](images/32957f1ee1c846ed1e93c0f57ee7330a2f28b66b7b684ec8cde0c4987ce93749.jpg)第1个 ![](images/49b8cded392faa14141654b5527dc7c340f50653689aba2904ea8341c34b5812.jpg)第2个 ![](images/1578f32b43af465a3e79f6fa6850a440a636ea3954056fafff5ab1354e6625ce.jpg)第3个",
+        "options": [
+          "A. 7",
+          "B. 8",
+          "C. 9",
+          "D. 10"
+        ],
+        "answer": "C",
+        "solution": "【分析】观察可知，第1个图形有5个纸片，5=1×4+1;第2个图形有9个纸片，9=2×4+1;第3个图形有13个纸片，13=3×4+1……由此可知纸片个数=第几个图形×4+1。反过来求第几个图形，用(纸片个数−1)÷4即可。\n【详解】(37−1)÷4=36÷4=9（个）。第9个图案恰好有37个纸片。故答案为：C",
+        "knowledge_points": [
+          "数与形 (归纳递推)"
+        ],
+        "source": "24-25 六年级下·湖南永州·期末"
+      },
+      {
+        "id": "210",
+        "unit": "数与形",
+        "question_type": "选择题",
+        "difficulty": 0.65,
+        "question": "古希腊数学家毕达哥拉斯用石子在沙滩上画画，发现了数与形的规律。按照如图中图形的排列规律，第⑩个图形中直角三角形的个数是（）。![](images/324f19ef6c7807a3ffe27b7d4b62b015.png)",
+        "options": [
+          "A. 38",
+          "B. 40",
+          "C. 42",
+          "D. 44"
+        ],
+        "answer": "B",
+        "solution": "【分析】根据图示可知：①幅图直角三角形个数为4个;②幅图为8个，8=2×4;③幅图为12个，12=3×4;……第n幅图直角三角形个数为4n个。据此解答。\n【详解】当n=10时，4n=4×10=40。第⑩个图形中直角三角形的个数是40。故答案为：B",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "24-25 六年级下·重庆秀山·期末"
+      },
+      {
+        "id": "211",
+        "unit": "数与形",
+        "question_type": "选择题",
+        "difficulty": 0.85,
+        "question": " 下列说法不正确的是( )。![](images/f83b2d184ea059545ad37c040f1a93f2.png)",
+        "options": [
+          "A. 亮亮面向东站立，他向右转30°所面对的方向是南偏东60°",
+          "B. 在同一平面内，两条直线只有两种位置关系——平行和相交",
+          "C. 如图，在一根铁丝AB的C、D处各剪一刀，得到的三根铁丝能围成三角形（见图①）",
+          "D. 图中共有6个角（见图②）"
+        ],
+        "answer": "C",
+        "solution": "【分析】可根据方向与角度、同一平面内直线位置关系、三角形三边关系、角的计数方法，对各选项逐一分析，进而找出正确答案。\n【详解】A. 亮亮面向东站立，向右转30°，此时所面对的方向是南偏东60°（或东偏南30°），该说法正确。B. 在同一平面内，不重合的两条直线只有平行和相交两种位置关系，该说法正确。C. 图①中D是AB中点，则DB=AD。根据三角形三边关系‘任意两边之和大于第三边’，因为AC+CD=DB，不满足三边关系，所以不能围成三角形，该说法错误。D. 数图②中的角，单个小角有3个，两个小角组成的角有2个，三个小角组成的角有1个，总共3+2+1=6个角，该说法正确。只有选项C的说法错误，其它选项说法正确。故答案为：C",
+        "knowledge_points": [
+          "平行的特征及性质",
+          "三角形三边关系",
+          "根据方向、角度和距离确定物体的位置",
+          "角的初步认识及辨认"
+        ],
+        "source": "24-25 六年级下·湖北随州·期末"
+      },
+      {
+        "id": "212",
+        "unit": "数与形",
+        "question_type": "选择题",
+        "difficulty": 0.65,
+        "question": "下图中能大致反映龟兔赛跑的路程s随时间t变化情况的是( )。A.![](images/1fabf1a4556d4b4fc6bca9b130f7ca8164993c035cfd3d3b0666001cd9834a8c.jpg) B.![](images/fe6ac6859dd9ee58896fe704874e82d902f2c68d5e43f243b590c3fa2788211b.jpg) C.![](images/a277048078c7ee3b10b14ab50a957e8c5f6154245498b3543aaddf0063fd0269.jpg)",
+        "options": [
+          "A. 选项A所示图像",
+          "B. 选项B所示图像",
+          "C. 选项C所示图像"
+        ],
+        "answer": "B",
+        "solution": "【分析】乌龟一直坚持不懈地向终点爬着，速度较均匀，所以路程s随时间t均匀增加，图像为从原点出发的倾斜直线;兔子一开始更快，路程增加更快，图像更陡;中途睡觉路程不变，图像出现水平线段;醒来后继续增加，但最终乌龟先到终点。据此逐项分析。\n【详解】A. 情节基本符合，但图像表示兔子先到终点，不符合题意;B. 兔子先快后水平（睡觉）再上升，且乌龟先到终点，符合题意;C. 一开始乌龟路程应小于兔子，该图不符合。故答案为：B",
+        "knowledge_points": [
+          "复式折线统计图"
+        ],
+        "source": "23-24 六年级下·重庆酉阳·期末"
+      },
+      {
+        "id": "213",
+        "unit": "数与形",
+        "question_type": "选择题",
+        "difficulty": 0.65,
+        "question": "小明从家出发去超市购物，慢走了一段路后发现会员卡落家里了，马上小跑回家取卡，5分钟后找到会员卡继续出发，购物若干时间后再散步回家。能够比较准确地描述离家距离与经过时间的关系的是( )。A.![](images/53d0a6566be8da4d5d2ead4a2565248d8dc00eaf940fa5372f7e00a8834f666c.jpg) B.![](images/f8bf010698047e05592433cfe173319390b0600d53ef9532bab695ae9a3a23e8.jpg) C.![](images/ecb5a19b04081f16bf1995268874c48341612c4cc3b01c030127c335cef14856.jpg) D.![](images/25d3be866f36da422e1557a521e1417ba9c881d6bed540f331afc2e91aad4f7b.jpg)",
+        "options": [
+          "A. 选项A所示图像",
+          "B. 选项B所示图像",
+          "C. 选项C所示图像",
+          "D. 选项D所示图像"
+        ],
+        "answer": "C",
+        "solution": "【分析】①先慢走离家距离增加（上升线段）;②小跑回家距离减小直至0（下降线段到横轴）;③在家找卡5分钟距离为0（横轴上水平线段）;④继续出发距离增加（上升线段且更长）;⑤购物距离不变（水平线段）;⑥散步回家距离减小且较缓（下降线段至0）。据此解答。\n【详解】符合上述过程的图像为选项C所示。故答案为：C",
+        "knowledge_points": [
+          "单式折线统计图"
+        ],
+        "source": "23-24 六年级下·浙江温州·期末"
+      },
+      {
+        "id": "214",
+        "unit": "数与形",
+        "question_type": "选择题",
+        "difficulty": 0.65,
+        "question": " 用同样长的小棒摆出如下图形，照这样继续摆，第⑥个图形用( )根小棒。![](images/fe97664a7ec495cb254228595f5462521ff8120a89744a66072c7c05b855b18e.jpg)![](images/99be301c639866006755835abcfe751244314d2912d079c91b41988c5749156f.jpg)![](images/adeda341d0014bb3bbfe450844d56dc6e303485e7699a0541a35ac13755f7e4f.jpg)",
+        "options": [
+          "A. 7",
+          "B. 11",
+          "C. 13"
+        ],
+        "answer": "C",
+        "solution": "【分析】第1个图形小棒数为3根，即2×1+1;第2个为5根，即2×2+1;第3个为7根，即2×3+1……第⑥个图形用小棒数为2×6+1，据此解答。\n【详解】2×6+1=12+1=13（根）。第⑥个图形用13根小棒。故答案为：C",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "24-25 五年级下·河南信阳·期中"
+      },
+      {
+        "id": "215",
+        "unit": "数与形",
+        "question_type": "选择题",
+        "difficulty": 0.65,
+        "question": "餐厅里一张桌子可坐6人，按照下图的摆放规律，2张桌子可坐10人，n张桌子能坐（ ）人。![](images/1171844a6c65e68ad438608ab0068818e65ecd96ccc7afbeb279d568a221f3cc.jpg)![](images/da49e0ff43d2f2559609bf41391a4511759e2d082ffeeb141d88ce805e65bea3.jpg)![](images/b0306b880e571e18e9484163e71e3c6c997e7a5e830cc20acaab405a58e9df92.jpg)",
+        "options": [
+          "A. 4n+2",
+          "B. 4n+4",
+          "C. 6n+4",
+          "D. 6n+6"
+        ],
+        "answer": "A",
+        "solution": "【分析】观察可知：1张桌子可坐4×1+2=6人，2张桌子可坐4×2+2=10人，由此规律为4×桌子数+2=可坐人数，即n张桌子能坐(4n+2)人。\n【详解】按照题图规律，n张桌子能坐(4n+2)人。故答案为：A",
+        "knowledge_points": [
+          "用字母表示数",
+          "数量关系",
+          "数与形（探索规律）"
+        ],
+        "source": "2024·福建·小升初真题"
+      },
+      {
+        "id": "216",
+        "unit": "数与形",
+        "question_type": "选择题",
+        "difficulty": 0.65,
+        "question": "观察如图，按规律画下去，当某幅图中O的个数有25个时，□的个数为（）。![](images/1da3e7272c163e64fadda470e74e530b.png)",
+        "options": [
+          "A. 144",
+          "B. 121",
+          "C. 100",
+          "D. 81"
+        ],
+        "answer": "A",
+        "solution": "【分析】第1个图中O有1个，没有□;第2个图中O有3个，□有1个(1×1);第3个图中O有5个，□有4个(2×2);第4个图中O有7个，□有9个(3×3)……可见□的个数等于：O的个数去掉左下角1个后，剩余O的一半作边长的平方。\n【详解】O有25个时，去掉左下角1个O：25−1=24（个），24÷2=12（个），所以□的个数为12×12=144（个）。故答案为：A",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "2024·浙江宁波·小升初真题"
+      },
+      {
+        "id": "217",
+        "unit": "数与形",
+        "question_type": "选择题",
+        "difficulty": 0.65,
+        "question": "小雨遇到一道数与形的问题，她想到老师教的'化繁为简'的思想，就画表格从简单问题开始研究（如下表）。按照规律，序号是( )的图形中一共有48个三角形。![](images/4df216bef7ba94024b869e43b86fa132.png)",
+        "options": [
+          "A. 12",
+          "B. 8",
+          "C. 5",
+          "D. 6"
+        ],
+        "answer": "B",
+        "solution": "【分析】序号1三角形个数是(1+2+3)×1;序号2是(1+2+3)×2;……序号n是(1+2+3)×n。若共有48个三角形，则(1+2+3)×n=48，求n。\n【详解】(1+2+3)×n=48，即6n=48，n=8。所以序号是8的图形中一共有48个三角形。故答案为：B",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "24-25 六年级下·福建厦门·期中"
+      },
+      {
+        "id": "218",
+        "unit": "数与形",
+        "question_type": "填空题",
+        "difficulty": 0.65,
+        "question": "苗族千人长桌宴是苗族宴席的最高形式与隆重礼仪，已有几千年的历史。通常用于接亲嫁女、满月酒以及村寨联谊宴饮活动。在设定座位时如下图设计，认真观察，完成后面的问题。![](images/e07106b02ba9a9b502e93551d0592415fc26d5cb84f1945c638369d667222682.jpg)![](images/559533d7fe8e33e0e3f3cab94f349fe17fc31315a76d55cbae05a6255287c203.jpg)![](images/f28d89e38a36248d5d0bba3bf49f3f0a7d55087f2a52e2b7302968437f78e847.jpg)(1)按照上图所示规律，安排5张桌子最多可以坐_人。(2)如果有42人，需要安排_张桌子。",
+        "options": [],
+        "answer": "(1)22(2)10",
+        "solution": "【分析】(1)观察图形：1张桌子最多坐6人，6=4×1+2;2张桌子最多坐10人，10=4×2+2;3张桌子最多坐14人，14=4×3+2……规律：n张桌子最多坐(4n+2)人。(2)若有42人，则4n+2=42，解方程求n。\n【详解】(1)当n=5时，4n+2=4×5+2=20+2=22(人)。\n(2)4n+2=42，解：4n=40，n=10。所以安排5张桌子最多可以坐22人;如果有42人，需要安排10张桌子。",
+        "knowledge_points": [
+          "含有字母式子的化简与求值",
+          "用字母表示稍复杂的数量关系",
+          "数与形（探索规律）"
+        ],
+        "source": "2025·河南信阳·小升初真题"
+      },
+      {
+        "id": "219",
+        "unit": "数与形",
+        "question_type": "填空题",
+        "difficulty": 0.65,
+        "question": "动脑筋，填一填。![](images/b561245515987207c02c33f7ee2c1150e6444cc8aa23d371b7463f0f3cf6a185.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】通过观察可以发现：下面一行两个相邻白色三角形的和等于上面一行与这两个三角形相邻的白色三角形，根据这个规律填一填。\n【详解】![](images/164cf347eeeff8eb2669f7a917539582c38b8989c587608e80e5b4b68f49d529.jpg)观察图片中这个部分：![](images/72af94e538e044dea2305da7527aee4dea50f487ef2cfdcd1da158aa02a4f7c9.jpg)最下面一行的两个白色三角形是2和3，上面一行与这两个三角形相邻的白色三角形是5。以此类推：①=7−4=3，③=16−7=9，②=9−4=5;右下方的三角形：⑤=2+5=7，⑥=14−7=7，④=7−5=2。",
+        "knowledge_points": [
+          "数表中的规律"
+        ],
+        "source": "25-26 六年级上·全国·单元测试"
+      },
+      {
+        "id": "220",
+        "unit": "数与形",
+        "question_type": "填空题",
+        "difficulty": 0.4,
+        "question": " 找规律填空。![](images/d76ce82fb7137b96e83f3ff31da27378.png)(1)如图1所示，一个正方体，用刀截去一个角后，所得的几何体有个顶点。(2)如图2所示，各图中每个正方体的棱长都是a厘米，各图的表面积分别是多少？（按图形顺序依次将答案填在对应的横线内）①_平方厘米;②_平方厘米;③_平方厘米;④_平方厘米。(3)观察下面的几个算式：1+2+1=4;1+2+3+2+1=9;1+2+3+4+3+2+1=16;1+2+3+4+5+4+3+2+1=25。根据你所发现的规律，直接写出：1+2+3+…+99+100+99+…+3+2+1=_。",
+        "options": [],
+        "answer": "(1)7或8或9或10 (2)6a^2;10a^2;14a^2;206a^2 (3)10000",
+        "solution": "【分析】(1)正方体截去一个角后会有多种情况，根据不同情况分析顶点的变化。(2)第1个图形有6个面，表面积为6a^2;第2个图形有10个面，表面积为10a^2;第3个图形有14个面，表面积为14a^2;第51个图形有206个面，表面积为206a^2。(3)由算式可见：1+2+…+n+(n−1)+…+2+1=n^2，当n=100时结果为100^2。\n【详解】(1)截角后如图①顶点7个;图②顶点8个;图③顶点9个;图④顶点10个。\n![](images/2383214d83b0c40de515623e222561d5270add60645dfc288f9ecef9af7b3159.jpg)① ![](images/7ed17743ebd36012bc08e86e9b8be82be764af6a4749203fa924977b6d84895b.jpg)② ![](images/5e9dadb4c7c158376fd9beeb36373249f8ccf874df2bffaa43e8756c0eb50b72.jpg)③ ![](images/ec32765131cc1f7fe0ddaeb1f824af6e6305a7fee0ab3f0e5dd44affd958d9e2.jpg)④。\n(2)图(1)表面积6a^2;图(2)表面积=6a^2×2−a^2×2×1=10a^2;图(3)表面积=6a^2×3−a^2×2×2=14a^2;图(4)表面积=6a^2×51−a^2×2×50=206a^2。\n(3)1+2+3+…+99+100+99+…+3+2+1=100^2=10000。",
+        "knowledge_points": [
+          "正方体表面积的计算",
+          "算式的规律（整数）",
+          "整数排列的规律"
+        ],
+        "source": "25-26 六年级上·全国·单元测试"
+      },
+      {
+        "id": "221",
+        "unit": "数与形",
+        "question_type": "填空题",
+        "difficulty": 0.65,
+        "question": " 找规律。1，3/4，5/9，7/16，…第n个数是( )。",
+        "options": [],
+        "answer": "(2n-1)/n^2",
+        "solution": "【分析】1可看作1/1。第2个数3/4：3=2×2−1，4=2^2;第3个数5/9：5=2×3−1，9=3^2;第4个数7/16：7=2×4−1，16=4^2。由此可知分子为2n−1，分母为n^2。\n【详解】第n个数的分子为2n−1，分母为n^2，所以第n个数是(2n−1)/n^2。",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "2025·西藏·小升初真题"
+      },
+      {
+        "id": "222",
+        "unit": "数与形",
+        "question_type": "填空题",
+        "difficulty": 0.65,
+        "question": " 如图，按这样的规律摆下去，第8个图形有( )个•，第n个图形有( )个•。![](images/f620051ab9f6ff8ad2cfb394179d8b6f63c82d1d18505612942d066e92d9e4dd.jpg)",
+        "options": [],
+        "answer": "32;4n",
+        "solution": "【分析】观察发现每增加一个图形就增加4个点：第1个4个点=4×1;第2个8个点=4×2;第3个12个点=4×3;第4个16个点=4×4……据此第8个为4×8，第n个为4n。\n【详解】第8个图形有4×8=32个点;第n个图形有4n个点。",
+        "knowledge_points": [
+          "用字母表示数",
+          "数量关系",
+          "数与形（探索规律）"
+        ],
+        "source": "24-25 六年级下·浙江杭州·期末"
+      },
+      {
+        "id": "223",
+        "unit": "数与形",
+        "question_type": "填空题",
+        "difficulty": 0.65,
+        "question": "用小棒摆成一组有规律的图案如下图所示，第1个图案需要4根小棒，第2个图案需要10根小棒……按此规律摆下去，第7个图案需要( )根小棒，第n个图案需要( )根小棒。![](images/e7ab0e1602756afb60a28a0538afef20d9a4a2b40494d35a9452d6770241b5e8.jpg)",
+        "options": [],
+        "answer": "40;6n-2",
+        "solution": "【分析】由图可知：第1个图案4根;第2个图案4+6根;第3个图案4+6×2根;第4个图案4+6×3根……每次增加6根，所以第n个图案需要4+6×(n−1)根，化简得6n−2;再代入n=7求第7个。\n【详解】4+6×(n−1)=4+6n−6=6n−2（根）。当n=7时，6×7−2=42−2=40（根）。所以第7个图案需要40根小棒，第n个图案需要(6n−2)根小棒。",
+        "knowledge_points": [
+          "数与形（探索规律）",
+          "数与形（归纳递推）"
+        ],
+        "source": "2025·云南西双版纳·小升初模拟"
+      },
+      {
+        "id": "224",
+        "unit": "数与形",
+        "question_type": "填空题",
+        "difficulty": 0.65,
+        "question": "（如图）用长度相等的火柴棒按照图中的方式拼摆正方形，第4幅图需要( )根火柴棒，第n幅图需要( )根火柴棒。![](images/482833ca1ecb43a1a18c68255989142e75812ce699e9edf088c20ee6a85a1db3.jpg)第1幅图 ![](images/02c5cc119df00f883709ccd1c40b944856ec1330466edf4c9d1d954e2fdaf716.jpg)第2幅图 ![](images/8cb545c3e94abe56ab38f1530ff8c4224ea6bd82aff9d8d8a37d2fa458d4add3.jpg)第3幅图",
+        "options": [],
+        "answer": "13;3n+1",
+        "solution": "【分析】看图可知：第1幅图4根=1×3+1;第2幅图7根=2×3+1;第3幅图10根=3×3+1……规律：火柴棒根数=幅数×3+1。据此求第4幅与第n幅。\n【详解】第4幅图：4×3+1=12+1=13（根）。第n幅图需要n×3+1=(3n+1)（根）。",
+        "knowledge_points": [
+          "用字母表示数",
+          "数量关系",
+          "数与形（探索规律）"
+        ],
+        "source": "24-25 六年级下·河南郑州·期末"
+      },
+      {
+        "id": "225",
+        "unit": "数与形",
+        "question_type": "填空题",
+        "difficulty": 0.65,
+        "question": "用小棒摆三角形（如下图），照这样摆5个三角形要用( )根小棒，摆n个三角形需要用( )根小棒。![](images/2a90e84d721141ecedf40129a928fe89e23975906fdc0d583c3e483cb25ee279.jpg)1个三角形 ![](images/6805cbbd630c834133eff706c7ca15434b09fe56f88fb2353e1a3ecfcbbb162a.jpg)2个三角形 ![](images/6af4a8141ae04ca14c8f5b6d3f1cfd86f83144884abb0517ac12996e9f37258e.jpg)3个三角形",
+        "options": [],
+        "answer": "11;2n+1",
+        "solution": "【分析】摆1个三角形用3根;摆2个用5根（比前一个多2根）;摆3个用7根（比前一个多2根）。因此每增加1个三角形多2根，小棒数为等差规律，可表示为2n+1根;代入n=5求摆5个。\n【详解】摆n个三角形要用(2n+1)根。摆5个：2×5+1=10+1=11（根）。",
+        "knowledge_points": [
+          "用字母表示数",
+          "数量关系",
+          "数与形（探索规律）"
+        ],
+        "source": "24-25 六年级上·浙江绍兴·期末"
+      },
+      {
+        "id": "226",
+        "unit": "数与形",
+        "question_type": "填空题",
+        "difficulty": 0.65,
+        "question": "《庄子·天下篇》中有一句话：'一尺之棰，日取其半，万世不竭。'照此说法，第( )天取得的长度是1/32尺。",
+        "options": [],
+        "answer": "5",
+        "solution": "【分析】'日取其半'表示每天取得的长度是前一天的1/2，则第n天取得的长度为1/2^n（n为正整数）。1/32=1/2^5，所以n=5。\n【详解】因为1/32=1/2^5，所以第5天取得的长度是1/32尺。",
+        "knowledge_points": [
+          "连续求一个数的几分之几是多少的问题",
+          "数与形（探索规律）"
+        ],
+        "source": "24-25 六年级上·浙江绍兴·期末"
+      },
+      {
+        "id": "227",
+        "unit": "数与形",
+        "question_type": "填空题",
+        "difficulty": 0.65,
+        "question": " 同学们用小棒玩游戏，如图：搭第一个图形要8根小棒，搭第二个图形要15根小棒，照这样的方法，搭第6个图形要( )根小棒，搭第n个图形要( )根小棒。![](images/8abe8de3976067a7cbacd6509decee8a9300396ffb7028947b0133880cec7fb5.jpg)① ![](images/14901ddcda656b3a7a6354e38d1770867afe703f811a7a66812d1e2dc96c00c9.jpg)② ![](images/d383f31870aca27fd9d591104bf6b4f67d5a426a3128ed0992131fc5caff7e7a.jpg)③",
+        "options": [],
+        "answer": "43;7n+1",
+        "solution": "【分析】观察图形可知：搭第1、2、3个图形分别需要8、15、22根小棒，每增加一个正八边形，小棒数量增加7根。规律：搭第n个图形要7n+1根。代入n=6求第6个。\n【详解】7n+1中代入n=6：7×6+1=42+1=43（根）。所以搭第6个图形要43根小棒，搭第n个图形要(7n+1)根小棒。",
+        "knowledge_points": [
+          "含有字母式子的化简与求值",
+          "用字母表示稍复杂的数量关系",
+          "数与形（探索规律）"
+        ],
+        "source": "24-25 六年级上·湖南永州·期末"
+      },
+      {
+        "id": "228",
+        "unit": "数与形",
+        "question_type": "填空题",
+        "difficulty": 0.65,
+        "question": " 如图，小明用小棒搭房子，搭7间房子要用( )根小棒，搭n间房子需要( )根小棒。![](images/abec1c39c3a2b641c99338ca8c6ff13a341f4e7d002f3f39e01e7574b222c3fc.jpg)",
+        "options": [],
+        "answer": "29;4n+1",
+        "solution": "【分析】搭1间房子用5根，可表示为4×1+1;搭2间用9根=4×2+1;搭3间用13根=4×3+1……规律：搭n间房子需要(4n+1)根。代入n=7求搭7间。\n【详解】4×7+1=28+1=29（根）。所以搭7间房子要用29根小棒，搭n间房子需要(4n+1)根小棒。",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "24-25 六年级下·江西九江·期末"
+      },
+      {
+        "id": "229",
+        "unit": "数与形",
+        "question_type": "填空题",
+        "difficulty": 0.65,
+        "question": "(24-25 六年级下·广东江门·期末) 画直线把一个圆分成尽可能多的块数（如图）请仔细观察，找出其中的规律。按照你发现的规律，6条直线最多能把圆分成( )块。![](images/6f3d3e19b9620000d022d39dc634a01f282c4f0edd717d6d8b4e038985995f5a.jpg)1条直线 ![](images/5f73ff7b8dc89b51232cbe4f761881cdb43657801cd319e4178ee08411f56c9a.jpg)2条直线 ![](images/d856d403e31344d9f2ff1f7eef37ed3dda5c4067dd1806038770c6fbfbf430a5.jpg)3条直线",
+        "options": [],
+        "answer": "22",
+        "solution": "【分析】观察可得：0条直线分成1块;1条直线分成(1+1)块;2条直线分成(1+1+2)块;3条直线分成(1+1+2+3)块……n条直线分成(1+1+2+3+…+n)块。据此求n=6。\n【详解】当n=6时：1+1+2+3+4+5+6=1+(1+6)×6÷2=1+7×6÷2=1+21=22（块）。所以6条直线最多能把圆分成22块。",
+        "knowledge_points": [
+          "数与形 (归纳递推)"
+        ],
+        "source": "24-25 六年级下·广东江门·期末"
+      },
+      {
+        "id": "230",
+        "unit": "数与形",
+        "question_type": "填空题",
+        "difficulty": 0.65,
+        "question": "(24-25 六年级下·贵州黔西·期末) 老师制作教具，如图①所示，搭一个长方体框架用了12根小棒。如果搭如图(2)所示的2个长方体框架，需要( )根小棒。如果像这样搭n个长方体框架，需要( )根小棒。![](images/db8a4f88876b8f8dec2bde982dad5879.png)",
+        "options": [],
+        "answer": "20;8n+4",
+        "solution": "【分析】搭1个长方体框架用12根;搭2个时有4根小棒重合，因此需要12×2−4=20根。搭n个时，第1个用12根，从第2个开始每个与前一个重合4根，所以每增加一个只需12−4=8根新小棒，总数为12+8(n−1)=8n+4。\n【详解】2个框架：12×2−4=20（根）。n个框架：12+8(n−1)=12+8n−8=8n+4（根）。",
+        "knowledge_points": [
+          "长方体有关棱长的应用",
+          "图形的变化规律"
+        ],
+        "source": "24-25 六年级下·贵州黔西·期末"
+      },
+      {
+        "id": "231",
+        "unit": "数与形",
+        "question_type": "填空题",
+        "difficulty": 0.65,
+        "question": " 如图所示，用小木棒摆正方形。按这样的规律，摆8个正方形需要小木棒( )根，摆n个正方形需要小木棒( )根。（用含字母'n'的式子表示）![](images/67662be679a37d6bc60752cc5224582f.png)",
+        "options": [],
+        "answer": "25;3n+1",
+        "solution": "【分析】看图可知：1个正方形4根=1×3+1;2个正方形7根=2×3+1;3个正方形10根=3×3+1……规律：小木棒根数=正方形个数×3+1。\n【详解】摆8个：3×8+1=24+1=25（根）。摆n个：n×3+1=(3n+1)（根）。",
+        "knowledge_points": [
+          "数与形 (归纳递推)"
+        ],
+        "source": "24-25 六年级下·贵州安顺·期末"
+      },
+      {
+        "id": "232",
+        "unit": "数与形",
+        "question_type": "填空题",
+        "difficulty": 0.65,
+        "question": "如图，下面每个三角形都是由若干个小三角形组成的。按照这样的规律，摆第5个图形需要( )个小三角形。![](images/48d8a8c3e1303aa6357c9ddccb6ee5198ce10c3df295af65a04a694f65a11e39.jpg)第1个 ![](images/92cdd0b4f89ed4095a6365f24e2ce541c976d95bd86233ee67c1c340346088e9.jpg)第2个 ![](images/1707ade33727c5ed042701d31241a75fe91004a7c6e020c5edfec48ec0142167.jpg)第3个 ![](images/3b2702eb0860228b60424449d663d11609dccbbf9a841970ae24b0435cf200cc.jpg)",
+        "options": [],
+        "answer": "25",
+        "solution": "【分析】第1个图形小三角形数为1，可写成1^2;第2个为4=2^2;第3个为9=3^2……规律：第n个图形小三角形个数为n^2。代入n=5求解。\n【详解】当n=5时，5^2=25（个）。摆第5个图形需要25个小三角形。",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "24-25 六年级下·河南省直辖县级单位·期末"
+      },
+      {
+        "id": "233",
+        "unit": "数与形",
+        "question_type": "填空题",
+        "difficulty": 0.65,
+        "question": "将一些五角星如下图摆放，照这样的规律继续摆下去，第7幅图中共有( )颗五角星，第n幅图中共有( )颗五角星。![](images/4e03e1b931f09cea1741d75e0ad6ddc65987e16eed442365bc47786c76578421.jpg)① ![](images/efc9eab3ebbfdfa5d52c9548817f4950fd8256f827ab880329c6bb975f2906c0.jpg)② ![](images/486baefc5c8dd19c0f4eba850d53f4b8ffda48af568e505755a869bebbe18575.jpg)③ ![](images/2ef724d6bdcf613b667ae3e4854682645b1f5167665595d36a42da22b914bee8.jpg)④",
+        "options": [],
+        "answer": "25;4n-3",
+        "solution": "【分析】第1幅图共有1颗五角星;第2幅图共有(1+4)颗;第3幅图共有(1+4×2)颗;第4幅图共有(1+4×3)颗……每次增加4颗五角星。第n幅图共有1+4×(n−1)颗，化简为4n−3。代入n=7求第7幅。\n【详解】1+4×(n−1)=1+4n−4=4n−3（颗）。当n=7时：4×7−3=28−3=25（颗）。所以第7幅图共有25颗五角星，第n幅图共有(4n−3)颗五角星。",
+        "knowledge_points": [
+          "数与形（探索规律）",
+          "数与形（归纳递推）"
+        ],
+        "source": "24-25 六年级下·湖南邵阳·期末"
+      },
+    
+
+      {
+        "id": "234",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "把数和形结合起来思考, 是数学学习经常使用的方法。认真观察如图所示的点阵图, 按这样的规律, 第 6 幅图一共有( )个圆点, 第 n 幅图一共有( )个圆点。![](images/63609e9ac1267b12a0a87ed7d4cb7e6cd0c760d777a7086e0cabdf71a2b5fba5.jpg)![](images/a48616c9d190510c0380eca4cbc1205c5654f0085cec134a0994e661c871a2a3.jpg)![](images/f66da5eeff1baa10751bc9898903d806af9da18197555c00552b31f75dd2b847.jpg)![](images/512f0c96626fd080cd135a4bb44039b08cca220d74716c1b4066fc3833e8bb7a.jpg)",
+        "options": [],
+        "answer": "21;3 + 3n",
+        "solution": "【分析】由图可知:第①幅图中圆点数等于  1 + 2 + 3  第②幅图中圆点数等于  2 + 3 + 4  第③幅图中圆点数等于  3 + 4 + 5第⑥幅图中圆点数等于  6 + 7 + 8 = 21  （个）圆点第  n  幅图中圆点数等于  n + (n + 1) + (n + 2) = (3 +3n) 个圆点, 由此解答即可。【详解】第6幅点阵图中的圆点数：6 + 7 + 8 = 13 + 8 = 21 (个)。第  n  幅点阵图中的圆点数:  n + (n + 1) + (n + 2) = (3 + 3n)  个。第 6 幅图一共有 21 个圆点, 第 n 幅图一共有 (3+3n) 个圆点。",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "24-25 六年级下·湖南郴州·期末",
+        "merged_id": "unit8_234",
+        "original_id": "234"
+      },
+      {
+        "id": "235",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(24-25 六年级下·云南西双版纳·期末) 观察下面的点阵图形规律, 第 8 个图形中有( )个点, 第 n 个图形中有( )个点。![](images/8edca0658cc3d32d68810e0e46f16e342cd53bdc8d3e050febd8e76049965d72.jpg)",
+        "options": [],
+        "answer": "64 n²",
+        "solution": "【分析】由图可知，第1个图形中有1个点，表示为  1² ;第2个图形中有4个点，表示为  2² ;第3个图形中有9个点，表示为  3² ;第 4 个图形中有 16 个点，表示为  4² ;发现规律：第  n  个图形中有  n²  个点。据此解答。【详解】  8² = 8 × 8 = 64  （个）所以，第8个图形中有64个点，第n个图形中有n²个点。",
+        "knowledge_points": [
+          "用字母表示数、数量关系",
+          "数与形（探索规律）"
+        ],
+        "source": "24-25 六年级下·云南西双版纳·期末",
+        "merged_id": "unit8_235",
+        "original_id": "235"
+      },
+      {
+        "id": "236",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "'黄金螺旋'线'是一种优美的曲线，它是由长度不一、但圆心角都是90° 的弧组成的。如图是形形尝试画它的步骤, 第一步中弧所在扇形的半径是 1 厘米，第二步中弧所在扇形的半径是 1厘米，第三步中弧所在扇形的半径是2厘米，按照这样的方法继续画下去，第_步中的弧所在扇形的半径是21厘米。![](images/715ccd2a600d724d5ef00aff809da2b57f2a574eb9908e8da7641365669e978c.jpg)第一步![](images/cd044861a8ecdf0ad94bdd1bda4e49fb047e4d48f58178b4a1ecdacdc403b0c3.jpg)第二步![](images/a3b4186f634a84631b6776f4c08d62dbd88f7fa231d80ecbc54fb0f5180af384.jpg)第三步![](images/f0ed070dc5fd987751c19aa485a140159610fd6f541f6fc3489c282dcaa436d6.jpg)第四步![](images/623826c9a8b09ab198da1c98c97834342d258388742a8f54e0fca63d33240742.jpg)第五步",
+        "options": [],
+        "answer": "八",
+        "solution": "【分析】观察图形可知，第一步，弧所在扇形的半径是1厘米;第二步，弧所在扇形的半径是1厘米;第三步，弧所在扇形的半径是  1 + 1 = 2  厘米;第四步，弧所在扇形的半径是  1 + 2 = 3  厘米;第五步，弧所在扇形的半径是  2 + 3 = 5  厘米……以此类推，从第三步开始，弧所在扇形的半径依次是前两步中弧所在扇形的半径之和，据此解答。【详解】分析可知：第一步：1 厘米 ;第二步：1厘米;第三步： 1 + 1 = 2  （厘米）;第四步:  1 + 2 = 3  (厘米);第五步： 2 + 3 = 5  （厘米）;第六步： 3 + 5 = 8  （厘米）;第七步:  5 + 8 = 13  (厘米);第八步:  13 + 8 = 21  (厘米)所以，第八步中的弧所在扇形的半径是 21 厘米。",
+        "knowledge_points": [
+          "数与形（归纳递推）"
+        ],
+        "source": "24-25 六年级下·黑龙江哈尔滨·期末",
+        "merged_id": "unit8_236",
+        "original_id": "236"
+      },
+      {
+        "id": "237",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "用小棒按下图方式搭图形，第6个图形需要( )根小棒，第n个图形需要( )根大棒。![](images/739e721c3655fb8e9f6cb67b3a9ec8b5.png)",
+        "options": [],
+        "answer": "26 4n + 2",
+        "solution": "【分析】由图可知，第1个图形需要6根小棒，第2个图形需要  (6 + 4)  根小棒，第3个图形需要  (6 + 4 × 2)  根小棒……每次增加4根小棒，以此类推，第n个图形需要  [6 + 4 × (n - 1)]  根小棒，最后求出  n = 6  时含有字母式子的值，据此解答。【详解】  6 + 4 × (n - 1) = 6 + 4n - 4 × 1 = 6 + 4 n - 4 = 4 n + 6 - 4 = (4 n + 2) 根。当  n = 6  时。4 n + 2 = 4 × 6 + 2 = 24 + 2 =26 (根)。所以，第 6 个图形需要 26 根小棒，第 n 个图形需要  (4n + 2)  根小棒。",
+        "knowledge_points": [
+          "数与形（探索规律）",
+          "数与形（归纳递推）"
+        ],
+        "source": "24-25 六年级下·广西百色·期末",
+        "merged_id": "unit8_237",
+        "original_id": "237"
+      },
+      {
+        "id": "238",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "古希腊毕达哥拉斯学派把 1、3、6、10…这样的数称为 '三角形数', 把 1、4、9、16...这样的数称为'正方形数'。从图中可以发现，任何一个大于 1 的'正方形数'可以看作两个相邻'三角形数'之和，比如  4 = 1 + 3, 9 = 3 + 6 ...那么  49 = () +( )。![](images/f3d0ea2eb27fe61616ecf67371e1406c00d23418370ed79b7be8a7b7406f9704.jpg)",
+        "options": [],
+        "answer": "21 28",
+        "solution": "【分析】1、3、6、10...这样的数称为'三角形数'， 3 = 1+2, 6 = 1 + 2 + 3, 10 = 1 + 2 + 3 + 4…  第  n  个三角形数是  1 + 2 + 3 + … + n = n(n + 1)/2 ; 正方形数是  1 = 1², 4 = 2², 9 = 3²  …第  n  个正方形数是  n² ; 任何一个大于 1 的 '正方形数' 可以看作两个相邻 '三角形数' 之和, 正方形数  49 = 7² , 所以正方形数是第 7 个, 根据第  n  个三角形数是  n(n + 1)/2  计算即可解答。【详解】由分析可知：第  n  个三角形数是  n(n + 1)/2。49 = 7²。所以正方形数是第 7 个, 则两个三角形数分别是第 7 个和第 6 个, 把 7 和 6 分别代入  n(n + 1)/2 , 得:[7 × (7 + 1)]/2 = [7 × 8]/2 = 7 × 4 = 28。[6 × (6 + 1)]/2 = [6 × 7]/2 = 3 × 7 = 21。所以  49 = 21 + 28",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "23-24 六年级下·浙江杭州·期末",
+        "merged_id": "unit8_238",
+        "original_id": "238"
+      },
+      {
+        "id": "239",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "如图, 下列图案均正确的是 ( )是长度相同的火柴按一定的规律拼搭而成：第1个图案中有两个正方形，第2个图案中有4个正方形，…，依此规律，第10个图案中有( )个正方形。![](images/d531c7852232ce165fa53d45db4bc937056aedc6a99f64dd7318fb412a88a7b2.jpg)第1个![](images/76796b4730c659188e4a99f8e1c5206f428d02eeeac074bdba9bf37f23558c1c.jpg)第2个![](images/4f81e7cea88117555ada4077d6939bb4593fc3f565433fd420ba921700c26cc0.jpg)第3个![](images/3f268abf260ba97ba2adfa0b7b43c53ff7685af49b53737f8a2827ef0613307c.jpg)第4个",
+        "options": [],
+        "answer": "56",
+        "solution": "【分析】第 1 个图案中有 2 个正方形，第 2 个图案中有  (1 + 2 + 1)  个正方形，第 3 个图案中有  (1 + 2 + 3 + 1)  个正方形，…，依此规律，第  n  个图案中有  (1 + 2 + 3 + … + n + 1)  个正方形。即  1 + 2 + 3 + … + n + 1 = n × (1 + n) ÷ 2 + 1 。此题主要考查了图形的变化, 关键是根据题目中给出的图形, 通过观察思考, 归纳总结出规律, 再利用规律解答。【详解】由分析可知, 第  n  个图案中有:  n × (1 + n) ÷ 2 + 1  个正方形。当  n = 10  时, 10 × (1 + 10) ÷ 2 + 1 = 10 × 11 ÷ 2 + 1 = 110 ÷ 2 + 1 = 55 + 1 = 56 (个)。所以第 10 个图案中有 56 个正方形。",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "24-25 六年级下·江苏南京·期末",
+        "merged_id": "unit8_239",
+        "original_id": "239"
+      },
+      {
+        "id": "240",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如图, 每个图形都是用边长  1cm  的小正方形纸片摆出来的, 照这样摆下去,第 4 个图形用了( )个小正方形纸片, 第 4 个图形的周长是( )cm。![](images/8c34dc730f175e7d9d4f28ae352192ef0dd49b71ad498ba683ed052b1684e0ce.jpg)第1个![](images/482f9240e3ea69a9b85a54b32995c0387b910dffd220afee3445e0f53f6c8261.jpg)第2个![](images/6de86c1884fe080e82ac2db71284472ffc80b8f0cbca161ac2fae107d58be6ac.jpg)第3个",
+        "options": [],
+        "answer": "16 22",
+        "solution": "【分析】第 1 个图形: 小正方形纸片数量为  1² = 1  个。第 2 个图形: 小正方形纸片数量为  2² = 4  个。第 3 个图形: 小正方形纸片数量为  3² = 9  个。由此可推出, 第 n 个图形小正方形纸片数量为  n²  个。那么第 4 个图形小正方形纸片数量为  4² = 16  个。第1个图形：周长为  4 × 1 = 4cm  。第2个图形：周长为  6 × 2 - 2 = 12 - 2 = 10cm  。第3个图形：周长为  6 × 3 - 2 = 18 - 2 = 16cm  。可以发现，可得第  n  个图形的周长为：（6n-2）  cm  。那么第4个图形的周长为：  6 × 4 - 2 = 24 - 2 = 22cm  。【详解】由分析可知，第  n  个图形小正方形纸片数量为  n²  个。4² = 4 × 4 = 16 (个)。第  n  个图形的周长为:  (6 n - 2) cm。6 × 4 - 2 = 24 - 2 = 22 (cm)。第 4 个图形用了 16 个小正方形纸片, 第 4 个图形的周长是  22cm  。",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "24-25 六年级下·湖北武汉·期末",
+        "merged_id": "unit8_240",
+        "original_id": "240"
+      },
+      {
+        "id": "241",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如图, 像这样有规律地排列, 摆第 8 个图形需要( )根小棒。![](images/ad41135d759699d10d40089ddd1c7a4ca4e28e99787eabc2377264964f4061ed.jpg)①![](images/a361962880a38837289182c6f046e6619b0183fc640ffd617aafa00641e8eb72.jpg)②![](images/c2fb4b668db36ff9f953c243bf12d4f0001e6eb0f0fc2c60d5900c2ad45ac5db.jpg)③![](images/6a279deec309d2afc85bcb9e5d9d1e669a51af0d19fdda8199eec8e0521521c3.jpg)④",
+        "options": [],
+        "answer": "17",
+        "solution": "【分析】摆第 1 个图形（三角形）需要 3 根小棒，可表示为  2 × 1 + 1 = 3  根。摆第 2 个图形需要 5 根小棒，可表示为  2 × 2 + 1 = 5  根。摆第 3 个图形需要 7 根小棒，可表示为  2 × 3 + 1 = 7  根。摆第 4 个图形需要 9 根小棒，可表示为  2 × 4 + 1 = 9  根。由此可推出，摆第 n 个图形需要的小棒数量为  2n + 1  根。据此把数字 8 代入计算即可。【详解】由分析可知, 摆第  n  个图形需要的小棒数量为  2n + 1  根。2 × 8 + 1 = 16 + 1 = 17 (根)。所以摆第 8 个图形需要 17 根小棒。",
+        "knowledge_points": [
+          "用字母表示数、数量关系",
+          "数与形（探索规律）"
+        ],
+        "source": "24-25 六年级下·河南新乡·期末",
+        "merged_id": "unit8_241",
+        "original_id": "241"
+      },
+      {
+        "id": "242",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "A、B、C、D、E、F六人赛棋, 采用单循环制。现在知道A、B、C、D、E五人已经分别赛过 5 盘、4 盘、3 盘、2 盘、1 盘。问: 这时  F  已赛过( )盘。",
+        "options": [],
+        "answer": "3",
+        "solution": "【分析】单循环赛是每两个人之间都要比赛一场，所以每个人最多比赛 5 盘（因为有 6 个人，除去自己，要和另外 5 个人比赛）。A 已经赛过 5 盘, 这说明 A 和 B、C、D、E、F 都比赛过了。E 只赛过 1 盘, 而 A 已经和所有人都比赛过了, 所以 E 这 1盘只能是和 A 比赛的, E 没有和 B、C、D、F 比赛过。B 赛过 4 盘, 因为  E  没有和  B  比赛过, 所以  B  是和A、C、D、F 比赛的。D 赛过 2 盘, 由前面的分析可知, D 是和 A、B 比赛的, D没有和 C、F 比赛过。C 赛过 3 盘, 因为 E 没有和 C 比赛过, D 也没有和 C 比赛过,所以 C 是和 A、B、F 比赛的。F和A、B、C比赛过，所以F已赛过3盘。本题根据每个人最多只能比赛 5 盘作为突破口, 进行逐个推理, 找出  F  进行比赛的次数。【详解】每个人最多比赛5盘。A 已经赛过 5 盘, A 和 B、C、D、E、F 都比赛过了。E 只赛过 1 盘, 所以 E 这 1 盘只能是和 A 比赛的, E 没有和 B、C、D、F 比赛。B 赛过 4 盘, B 是和 A、C、D、F 比赛。D 赛过 2 盘, D 是和 A、B 比赛的, 没有和 C、F 比赛。C 赛过 3 盘, C 是和 A、B、F 比赛的。F 和 A、B、C 比赛过。所以 F 已赛过 3 盘。",
+        "knowledge_points": [
+          "推理问题",
+          "比赛问题"
+        ],
+        "source": "24-25 六年级下·湖南长沙·期末",
+        "merged_id": "unit8_242",
+        "original_id": "242"
+      },
+      {
+        "id": "243",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": " 小亮按下面的摆法摆六边形。摆 1 个六边形用 6 根小棒（如图①），摆 2 个六边形用了 11 根小棒（如图②）……![](images/b67cbc77182fc9000ac2ccbe9b79787c532914050afa28382031270d4990a9f7.jpg)①![](images/9deeda69f37a150deca4cf1c096d792084d8e725d3ea35ac3de2aab97b433ab0.jpg)②![](images/a95ff72ca0a1c9b8f8871662e4fc23f1b8608bc198e690c2c8aa40771926e8a2.jpg)③。按照这个摆法，摆4个六边形要用( )根小棒，摆n个六边形要用( )根小棒, 用 36 根小棒能摆出( )个六边形。",
+        "options": [],
+        "answer": "21 5n + 1 7",
+        "solution": "【分析】由图可知，摆 1 个六边形用 6 根小棒，摆 2 个六边形用  (6 + 5)  根小棒，摆 3 个六边形用  (6 + 5 × 2)  根小棒……以此类推，每次增加 5 根小棒，摆 n 个六边形用  [6 + 5 × (n - 1)]  根小棒，最后求出  n = 4  时含有字母式子的值，以及式子的值为 36 时字母的值，据此解答。【详解】  6 + 5 × (n - 1) = 6 + (5 n - 5 × 1) = 6 + (5 n - 5) = 6 + 5 n - 5 = 5 n + 6 - 5 = (5 n + 1) 根。当  n = 4  时。5 n + 1 = 5 × 4 + 1 = 20 + 1 = 21 (根)。5 n + 1 = 36。解:  5n + 1 - 1 = 36 - 1。5 n = 35。5 n ÷ 5 = 35 ÷ 5。n = 7。所以，摆 4 个六边形要用 21 根小棒, 摆 n 个六边形要用  (5n + 1)  根小棒, 用 36 根小棒能摆出 7 个六边形。",
+        "knowledge_points": [
+          "应用等式的性质 1 和 2 解方程",
+          "用字母表示数、数量关系",
+          "数与形（探索规律）"
+        ],
+        "source": "24-25 五年级下·江苏南通·期中",
+        "merged_id": "unit8_243",
+        "original_id": "243"
+      },
+      {
+        "id": "244",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如图由同样大小的圆按一定规律排列所组成，其中第1个图形中有4个圆，第2个图形中有8个圆，第3个图形中有14个圆，第4个图形中有22个圆……，按此规律排列下去，第30个图形有____个圆。![](images/c52b043e99e434d823624ad1b161510a9c868c3eaf376c78da2dbd9860f11ea7.jpg)第1个图形![](images/153a0aa506ba346d4bade851abb29f8012e6f950019e47d89272abfa5e4f87b7.jpg)第2个图形![](images/3c9749987278f8b43810bc0ddd7c5b6b95123b4552bca0d0e4798974fcad991e.jpg)第3个图形![](images/36231d174ca6fbf5c3e432439009378b5547970f35d571757d4eb13346c33eb5.jpg)第4个图形![](images/2a1972425f48e1dfd438c872a96f47eab708fb6c60d7bdcd1be6354850263258.jpg)",
+        "options": [],
+        "answer": "932",
+        "solution": "【分析】根据题意可知，第一个图形一共有圆：  1 × (1 + 1) + 2 = 4  个;第二个图形一共有圆：  2 × (2 + 1) + 2 = 8  个;第三个图形一共有圆：  3 × (3 + 1) + 2 = 14  个;第四个图形一共有圆：4 × (4 + 1) + 2 = 22  个;圆的个数等于图形序号与序号数加多 1 数的积再加上上面圆的个数 2 , 根据图形得出第  n  个图形中圆的个数是  n × (n + 1) + 2 , 据此进行解答。【详解】第1个图形中一共有  1 × (1 + 1) + 2 = 1 × 2 + 2 = 2 + 2 = 4  （个）圆第2个图形中一共有  2 × (2 + 1) + 2 = 2 × 3 + 2 = 6 + 2 = 8(个) 圆第3个图形中一共有  3 × (3 + 1) + 2 = 3 × 4 + 2 = 12 + 2 =14（个）圆第 4 个图形中一共有  4 × (4 + 1) + 2 = 4 × 5 + 2 = 20 + 2 =22（个）圆可得第  n  个图形中圆的个数是： n(n + 1) + 2所以第 30 个图形中圆的个数:  30 × (30 + 1) + 2 = 30 × 31 + 2 = 930 + 2 = 932  （个）第30个图形有932个圆。",
+        "knowledge_points": [
+          "图形的变化规律",
+          "用字母表示稍复杂的数量关系"
+        ],
+        "source": "24-25六年级下·湖南长沙·期末",
+        "merged_id": "unit8_244",
+        "original_id": "244"
+      },
+      {
+        "id": "245",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(24-25 六年级下·北京顺义·期末) 如图是用小棒拼摆的 3 个不同的图形, 按照这个规律, 第五个图形需要( )根小棒, 第 n 个图形需要( )根小棒。![](images/6bf6d2a65278288db8ca81f35d3082d58cdd1b925564f12091a4506281487e46.jpg)1![](images/66c841d93d744374ccab2bd900c88a314bf02241fdc6cd68a6453e81f79561b2.jpg)2![](images/99e5c634b833c38094d1be0ae17f8a61ab7ffe7a3c03500c63b40dccee180f60.jpg)3",
+        "options": [],
+        "answer": "60 2n(n + 1)",
+        "solution": "【分析】根据图示可知:第1幅图小棒根数：4根第2幅图小棒根数：12根，  12 = 4 + 8 = 2× 2× (2 + 1)第3幅图小棒根数：24根，  24 = 4 + 8 + 12 = 2 × 3 × (3 + 1)中第  n  幅图小棒根数:  2 n(n + 1) , 据此解答。【详解】根据分析，第  n  幅图小棒根数： 2n (n + 1)当  n = 5, 2n(n + 1) = 2 × 5 × (5 + 1) = 60第五个图形需要 60 根小棒, 第 n 个图形需要  2n(n + 1)  根小棒。",
+        "knowledge_points": [
+          "图形的变化规律",
+          "数与形（探索规律）"
+        ],
+        "source": "24-25 六年级下·北京顺义·期末",
+        "merged_id": "unit8_245",
+        "original_id": "245"
+      },
+      {
+        "id": "246",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(2025·重庆江北·小升初真题) 按如图规律铺黑白砖,第 49 幅图形中有( )块黑瓷砖。![](images/c002169af19c858613d5e24f7821ee6a4f49498a45bc8449d915173f6e925248.jpg)![](images/cbb07acb705936c514710b1a0e913fd4d3aaff458eb2eaca47e6afc550c7c33a.jpg)![](images/d1ea696ef1d537fae5e03c7ec342dc2acf31a68283e368ceefe6d9ffe9ef0440.jpg)",
+        "options": [],
+        "answer": "148",
+        "solution": "【分析】根据图形可知, 第一个图形中, 黑瓷砖有 4 块, 可以写成  3 × 1 + 1 ;第二个图形中, 黑瓷砖有 7 块, 可以写成  3 × 2 + 1 ;第三个图形中，黑瓷砖有 10 块，可以写成  3 × 3 + 1; …由此可以得出一般规律, 第  n  个图形中黑瓷砖有  (3 n + 1) 块, 据此解答。【详解】第  n  幅图形中有  (3 n + 1)  块黑瓷砖, 所以第 49 幅图形中有: 3 × 49 + 1 = 147 + 1 = 148 (块)。因此，第49幅图形中有148块黑瓷砖。",
+        "knowledge_points": [
+          "数与形（归纳递推）"
+        ],
+        "source": "2025·重庆江北·小升初真题",
+        "merged_id": "unit8_246",
+        "original_id": "246"
+      },
+
+      {
+        "id": "247",
+        "unit": "第八单元 数与形",
+        "question_type": "填空题",
+        "difficulty": 0.65,
+        "question": "如图，用黑白两种颜色的正方形纸片，按黑色纸片逐渐加1的规律拼成如图图案。按照这样的规律，如果一个图案白色纸片有70张，那么这个图案中有黑色纸片( )张。![](images/87a8ec52be6d3d33c9a3f7853d9d858169944541906391b68b18883a3e8b1fb5.jpg) ![](images/4f62e1df8a7f51a92e2b462ee51b9b875066c00524127a24d2280f6679615117.jpg) ![](images/49dd72bbc89176f4c1ad61ccdcdabf1b3b3f1d4572c002db0f453c6ffefaecac.jpg) ![](images/7c0e404d425df19cd8de761714cf3e38c2e49723f15d88e78f3bc9045b916280.jpg)",
+        "options": [],
+        "answer": "23",
+        "solution": "【分析】第1个图案黑色1张、白色4张;第2个图案黑色2张、白色7张;第3个图案黑色3张、白色10张。可得第n个图案黑色n张、白色(3n+1)张。已知白色70张，即3n+1=70。\n【详解】3n+1=70，3n=69，n=23。所以黑色纸片有23张。",
+        "knowledge_points": ["图形的变化规律", "数与形（探索规律）"],
+        "source": "2025·福建龙岩·小升初真题"
+      },
+      {
+        "id": "248",
+        "unit": "第八单元 数与形",
+        "question_type": "填空题",
+        "difficulty": 0.65,
+        "question": "找规律：0.1，4，0.2，16，0.3，36，0.4，（ ），0.5。",
+        "options": [],
+        "answer": "64",
+        "solution": "【分析】奇数项是0.1、0.2、0.3、0.4、0.5，依次增加0.1;偶数项是4、16、36，分别是2²、4²、6²，所以第8项应是8²。\n【详解】8²=64，所以括号里填64。",
+        "knowledge_points": ["整数排列的规律", "小数排列的规律"],
+        "source": "24-25 六年级下·贵州铜仁·期中"
+      },
+      {
+        "id": "249",
+        "unit": "第八单元 数与形",
+        "question_type": "填空题",
+        "difficulty": 0.4,
+        "question": "古希腊的毕达哥拉斯喜欢用小石子摆数，他发现当小石子的数量是1、3、6、10...这些数时，都能摆成正三角形，于是把这样的数称为‘三角形数’。如下图，第10个三角形数是( )，第100个三角形数是( )。![](images/c449cb5f287afb0eb881cdb0d879b8d85d972526b45469792ecaeffe386d115a.jpg)",
+        "options": [],
+        "answer": "55;5050",
+        "solution": "【分析】第n个三角形数为1+2+...+n。分别代入n=10、n=100计算。\n【详解】1+2+...+10=55;1+2+...+100=5050。",
+        "knowledge_points": ["含有字母式子的化简与求值", "用字母表示稍复杂的数量关系", "数与形（探索规律）"],
+        "source": "24-25 六年级下·山东济南·期中"
+      },
+      {
+        "id": "250",
+        "unit": "第八单元 数与形",
+        "question_type": "填空题",
+        "difficulty": 0.65,
+        "question": "如图，1张桌子可坐4人，2张桌子拼起来可坐6人，3张桌子拼起来可坐8人。像这样10张桌子可坐( )人，( )张桌子拼起来可坐38人。![](images/c45bdb3aa8b2dd6cd5cc428abb228544a6c63514197bbf67202a51481b0f576b.jpg) ![](images/4c124f8fc0be3365e7396141ad973c2aa94d730027704361b7300026b28f6e71.jpg) ![](images/f2b7e1fef7343ddf3715e80ded8fe55fdbf9fbe501d23e7bc30ccf081f218da9.jpg)",
+        "options": [],
+        "answer": "22;18",
+        "solution": "【分析】规律：人数=2×桌子张数+2。\n【详解】10张桌子：2×10+2=22（人）;38人对应桌子数：(38-2)÷2=18（张）。",
+        "knowledge_points": ["数与形 (归纳递推)"],
+        "source": "24-25 六年级下·湖北黄冈·期末"
+      },
+      {
+        "id": "251",
+        "unit": "第八单元 数与形",
+        "question_type": "填空题",
+        "difficulty": 0.65,
+        "question": "按下面图形的排列规律，摆到第9个图形共需要小三角形个数是( )个。![](images/98315d09e677770869f2678b4ce98419d1a732758d0269ad37ae6073345e12bb.jpg) ![](images/468f4bd3b5f806b33a55e28d36508dd3a27f5d72cf277fdd0ac4f6b828b276e1.jpg) ![](images/136521cf16b356ccafd41c193fa507819dc17b3f57a4f66c10d29314b2a1d1b5.jpg) ![](images/11c27ef143b9ecaf03e46836fc3c72cc8bd422330996bf8b079d32bb8841ee30.jpg)",
+        "options": [],
+        "answer": "81",
+        "solution": "【分析】第1、2、3个图分别对应1²、2²、3²，可得摆到第n个图形需要n²个小三角形。\n【详解】n=9时，9²=81。",
+        "knowledge_points": ["数与形（探索规律）"],
+        "source": "24-25 六年级下·湖南邵阳·期中"
+      },
+      {
+        "id": "252",
+        "unit": "第八单元 数与形",
+        "question_type": "填空题",
+        "difficulty": 0.65,
+        "question": "用边长是1厘米的正方形分别摆出如图的图形。按照规律，第5个图形的周长是( )厘米。![](images/27fe8a8f91379e906dbff237919f9cf6fa56c868eb3994fe6236241f4ad79b8c.jpg)第1个 ![](images/f90dcfbdbdfbb5bb333efd65d33c759d233c06c6f3366e5081e6b0ce779f758e.jpg)第2个 ![](images/b4fb53c6d47f0cc4187dfbf081f2ddd1f4a74424fea674944c08ae3b04924a7d.jpg)第3个",
+        "options": [],
+        "answer": "28",
+        "solution": "【分析】第1、2、3个图周长分别为4、10、16，可归纳为：周长=层数×6-2。\n【详解】第5个图：5×6-2=28（厘米）。",
+        "knowledge_points": ["数与形 (归纳递推)"],
+        "source": "24-25 六年级下·湖南湘西·期末"
+      },
+      {
+        "id": "253",
+        "unit": "第八单元 数与形",
+        "question_type": "填空题",
+        "difficulty": 0.65,
+        "question": "用白色和黑色的小正方形按照下面的方法摆图形。按照这样的方法继续摆下去，第4个图形中，黑色小正方形有( )个;第n个图形中，黑色小正方形有( )个。![](images/c60bc459c663ef5ce0cddceea086cc439a3d75a6986251aa2f9b2aa5b2ac11e2.jpg)第1个 ![](images/32967ee8c7c32fdffcaae3af47af5e69ab17108095a0aeb8078e5efb155b3929.jpg)第2个 ![](images/31da9b430440669ef424661655f56d83d0ed961c831126e0a4d408a1fdefdfd8.jpg)第3个 ![](images/c9eb626df859a606c4c11aefc81e9073378fadce48ebbaba950008aeec40eb13.jpg)",
+        "options": [],
+        "answer": "17;4n+1",
+        "solution": "【分析】第1、2、3个图黑色小正方形分别为5、9、13，可写作1×4+1、2×4+1、3×4+1，所以第n个图为4n+1。\n【详解】第4个图：4×4+1=17。第n个图：4n+1。",
+        "knowledge_points": ["数与形 (归纳递推)"],
+        "source": "24-25 六年级下·江西南昌·期末"
+      },
+      {
+        "id": "254",
+        "unit": "第八单元 数与形",
+        "question_type": "填空题",
+        "difficulty": 0.65,
+        "question": "如图，用‘+’字形分割正方形，分割一次分成4个小正方形，分割两次分成7个小正方形。请思考分割次数和正方形个数的关系：如果分成了361个正方形，共用‘+’字形分割了( )次。![](images/f07dac4aa6e4bd30d0d2bf950acbb30356128c824138af546a6fedc989e7d1ca.jpg) ![](images/da60f524ea7ae3d3e1a8cb103749070767e8f318104a27f1449c268da8d9a1b8.jpg) ![](images/aa9c2f65180126f90a120d6c1d3123dda462e7e65ea2f0c551c32066f3d641a5.jpg) ![](images/a3feeb3bd4b0e80bdc0c3e26a1459296be9d0248fd1796851f4c15d903ec58e6.jpg)",
+        "options": [],
+        "answer": "120",
+        "solution": "【分析】初始1个正方形;每分割一次增加3个正方形，所以总数=1+3×分割次数。\n【详解】(361-1)÷3=360÷3=120（次）。",
+        "knowledge_points": ["平面图形的分割", "数与形 (归纳递推)"],
+        "source": "24-25 六年级下·天津南开·期末"
+      },
+      {
+        "id": "255",
+        "unit": "第八单元 数与形",
+        "question_type": "填空题",
+        "difficulty": 0.65,
+        "question": "找规律填一填。![](images/6cdc8b235a2b5aadd8e95e854d3b1cbea616e9171797eb1838e781710be3f130.jpg) ![](images/78415b13c0d12950f924061f6fc64695bded7fa9985a410a729a1e2524a3e45d.jpg) ![](images/b1081d5cda64c4cb1f46db87621559d9bb44237c81c619948929a64f5eab9089.jpg) ![](images/943116916b5a440e0b6b748da5cd907ccaa034002c1f30f90d5f47bb54622540.jpg) 依次摆下去，第8个图形是( )形。摆第10个图形需要( )根小棒，摆n个图形需要( )根小棒。",
+        "options": [],
+        "answer": "平行四边形;21;2n+1",
+        "solution": "【分析】图形类型从第2个起平行四边形与梯形交替出现，双数位置是平行四边形;小棒数依次为3、5、7、9...，每次+2，可写为1+2n。\n【详解】第8个是双数位，所以是平行四边形。第10个小棒数：1+2×10=21。第n个：1+2n。",
+        "knowledge_points": ["数与形（探索规律）"],
+        "source": "24-25 六年级下·湖南常德·期末"
+      },
+      {
+        "id": "256",
+        "unit": "第八单元 数与形",
+        "question_type": "填空题",
+        "difficulty": 0.65,
+        "question": "观察下图，依次排下去，第n幅图有( )个棋子。![](images/de2795696e7213043372361c94279a41b6286bfc3a879d6a5d33be9a8f0d9eba.jpg) ![](images/ba029af254f946582a5a0cc8eb1d06785a583517f4089a3ef6ae8b1ceb8bf69e.jpg) ![](images/9f0d7999de2b0395e674bf13c3168360e703bbd790ee66e7be402d6145c98a32.jpg) ![](images/3e8cd4c20f96e33784431f03ce35d1293d0d593911b5389a18cfbf675b756c24.jpg)",
+        "options": [],
+        "answer": "n²",
+        "solution": "【分析】第1、2、3、4幅图棋子数分别为1、4、9、16，对应1²、2²、3²、4²。\n【详解】所以第n幅图有n²个棋子。",
+        "knowledge_points": ["数与形（探索规律）"],
+        "source": "24-25 六年级下·云南文山·期末"
+      },
+      {
+        "id": "257",
+        "unit": "第八单元 数与形",
+        "question_type": "填空题",
+        "difficulty": 0.65,
+        "question": "用小棒摆五边形，如下图所示。按照这样的方法继续摆下去，摆第20幅图需要( )根小棒。![](images/98b4f573e83d4b6d825d1b6d6f07309b0213af300a2890c7975271ce49199aad.jpg)第1幅 ![](images/e7ab76d3c58d2f46bbabcce5070d2b0d418ab1ead8cbd7eb0df4670b96f8116b.jpg)第2幅 ![](images/6f53f676c5869c4e46e47527d28e2060f270400f53308105b91e1c6f19da87d9.jpg)第3幅",
+        "options": [],
+        "answer": "81",
+        "solution": "【分析】第1、2、3幅分别为5、9、13根，公差为4，可得第n幅为4n+1。\n【详解】第20幅：4×20+1=81（根）。",
+        "knowledge_points": ["数与形（探索规律）"],
+        "source": "24-25六年级下·山西吕梁·期末"
+      },
+      {
+        "id": "258",
+        "unit": "第八单元 数与形",
+        "question_type": "填空题",
+        "difficulty": 0.65,
+        "question": "我们经常采用数形结合的方法探索规律：例如，计算1/2+1/4+1/8+1/16+1/32+...，可根据下面两种图探索计算结果。如果这样一直加下去，加到1/1024结果是（ ），还发现这样不断加下去，结果越来越接近（ ）。![](images/65f993aa34264c161b84a02a57feb45b7c791d6e3385f4610e28ffd52472c657.jpg) ![](images/c2a322e21d153e49ea472107f1e4854f7794a4cc8e7f0f3910215b1357a2b3ca.jpg)",
+        "options": [],
+        "answer": "1023/1024;1",
+        "solution": "【分析】该和是等比序列前n项和，可看成1减去最后一项。\n【详解】加到1/1024时和为1-1/1024=1023/1024;继续加下去极限趋近1。",
+        "knowledge_points": ["异分母分数加、减法", "数与形（探索规律）"],
+        "source": "24-25六年级下·云南昭通·期末"
+      },
+      {
+        "id": "259",
+        "unit": "第八单元 数与形",
+        "question_type": "填空题",
+        "difficulty": 0.65,
+        "question": "著名的毕达哥拉斯学派把1、3、6、10...称为‘三角形数’，把1、4、9、16...称为‘正方形数’。从图中可发现任何一个大于1的正方形数都可看作两个相邻三角形数之和。把正方形数100写成两个相邻三角形数之和：100=( )。![](images/b94e837f242eafbea8593aa4b73de78f7fd53bc23b77b6e122ce30f08bf4e745.jpg) ![](images/935c5a36fb94c6fda315e715220b01aec684e5172ebf7e1d19b956555bb9d568.jpg) ![](images/4af2dd53707e3f68c3a5ff6f60d9e9606ca1950bcce304575c9d7a6867cfd5c9.jpg)",
+        "options": [],
+        "answer": "45+55",
+        "solution": "【分析】列出相邻三角形数：...36、45、55、66...，寻找和为100的一对相邻数。\n【详解】45和55是相邻三角形数，且45+55=100。",
+        "knowledge_points": ["数与形（探索规律）"],
+        "source": "24-25六年级下·湖南岳阳·期末"
+      },
+      {
+        "id": "260",
+        "unit": "第八单元 数与形",
+        "question_type": "填空题",
+        "difficulty": 0.65,
+        "question": "将黑色棋子按照一定规律排列成一系列如图所示的图案。第1个图中有6枚黑棋子，第2个图中有11枚黑棋子，第3个图中有16枚黑棋子……按照此规律，第n个图中有( )枚黑棋子。![](images/238fed06e89264a41fe4b5498854b9e92115eade8889efd086dbcb3164f2caa3.jpg)第1个 ![](images/e04959e9fbd193db682d6d7f67586e9dbe1eedbce3cbdd7aa400806993d532de.jpg)第2个 ![](images/2a6d3fc22cce48ea05a7303018bc3c05d3c57daeb6b3301e59b33985a7729083.jpg)第3个",
+        "options": [],
+        "answer": "5n+1",
+        "solution": "【分析】6、11、16...每次增加5，可写为6+5(n-1)。\n【详解】6+5(n-1)=1+5n，所以第n个图有5n+1枚黑棋子。",
+        "knowledge_points": ["数与形（探索规律）"],
+        "source": "24-25 六年级下·湖南长沙·期末"
+      },
+      {
+        "id": "261",
+        "unit": "第八单元 数与形",
+        "question_type": "填空题",
+        "difficulty": 0.65,
+        "question": "如图，一张正方形桌子能坐4人，两张桌子拼成一张长桌子能坐6人……(1)按规律填写下表，10张桌子能坐( )人;(2)n张正方形桌子拼成一张长桌子能坐（ ）人。![](images/fd25d8b52ccd515808a652f9486980ba46c6b3046b9c6e7147a3333d089a9ecd.jpg) ![](images/b3d37e58b3493b5e5ea938fd52acc4ef250148f9754bdedc3cef8a59f854143a.jpg) ![](images/ec5f72c93bf608e105e041924cb2f82213bd1826dbc0652f43bb65db39783408.jpg)",
+        "options": [],
+        "answer": "22;2n+2",
+        "solution": "【分析】每增加1张桌子增加2个座位，故人数是等差规律。\n【详解】10张：2×10+2=22。n张：4+(n-1)×2=2n+2。",
+        "knowledge_points": ["用字母表示数", "数量关系", "数与形（探索规律）"],
+        "source": "24-25 六年级下·河北邢台·期末"
+      },
+      {
+        "id": "262",
+        "unit": "第八单元 数与形",
+        "question_type": "填空题",
+        "difficulty": 0.65,
+        "question": "探究规律：观察图形和算式，完成填空。![](images/4d4204baa4e52f064a855babdc43b2ba53a55bc22ac04054f894a4df8a7f1ce3.jpg) ![](images/d05af02c929f48ab1fa58e9117495876033edbdfe465a5e8178044ae0ff62dfe.jpg) (1)1=( )²;1+3=( )²;1+3+5=( )²。 (2)1+3+5+7+9+11=( )²。 (3)8²=( )。",
+        "options": [],
+        "answer": "(1)1，2，3;(2)6;(3)1+3+5+7+9+11+13+15",
+        "solution": "【分析】从1开始的连续奇数和：有几个奇数，和就是几的平方。\n【详解】(1)分别是1²、2²、3²;(2)6个连续奇数之和是6²;(3)8²可表示为前8个奇数之和，即1+3+5+7+9+11+13+15。",
+        "knowledge_points": ["数与形 (归纳递推)"],
+        "source": "24-25六年级下·湖南株洲·期末"
+      },
+      {
+        "id": "263",
+        "unit": "第八单元 数与形",
+        "question_type": "填空题",
+        "difficulty": 0.65,
+        "question": "观察下列等式：1³=1，1³+2³=9，1³+2³+3³=36，依此类推，第5个等式是：( )，结果等于( )。",
+        "options": [],
+        "answer": "1³+2³+3³+4³+5³;225",
+        "solution": "【分析】规律：1³+2³+...+n³=(1+2+...+n)²。\n【详解】第5个为1³+2³+3³+4³+5³=(1+2+3+4+5)²=15²=225。",
+        "knowledge_points": ["算式的规律（整数）", "数与形（归纳递推）"],
+        "source": "24-25 六年级下·湖南株洲·期末"
+      },
+      {
+        "id": "264",
+        "unit": "第八单元 数与形",
+        "question_type": "填空题",
+        "difficulty": 0.65,
+        "question": "观察图的规律，第6个图中有( )个点，第n个图中有( )个点。![](images/68eabab7b1d1d7b086bc45ccdcf1a4780e07fb5cd0275274f0fe69e5578114cd.jpg) ![](images/d032f657e7db06c98498e6926588e5ae5e0a768586497f42fc172611c79e9ddc.jpg) ![](images/db67186eeae8ac1cae25e2eb64fe9fc6c2ba419e4118ba55f1c7f0b883885caa.jpg) ![](images/215e75c83430ce04562d949dfc4e46cc04076589474664af1d833091ec96d692.jpg)",
+        "options": [],
+        "answer": "21;4n-3",
+        "solution": "【分析】第1、2、3、4个图点数可写为1、1+4×1、1+4×2、1+4×3，故第n个图为1+4(n-1)=4n-3。\n【详解】n=6时，4×6-3=21。",
+        "knowledge_points": ["数与形（探索规律）"],
+        "source": "24-25 六年级下·江西吉安·期末"
+      },
+      {
+        "id": "265",
+        "unit": "第八单元 数与形",
+        "question_type": "填空题",
+        "difficulty": 0.65,
+        "question": "下面图形都是由面积为1的正方形组成的，观察图形变化规律，第⑧个图形中正方形的数量是( )个。![](images/cc6ff32f91539867cd2bf46c4632d9c8d0a609f599e9e43dc339dba98d05613f.jpg)① ![](images/0c96c7e78cc5f2470e94321b1803a50a9da95e39cb388cdf6f84bd9a103dfdc5.jpg)② ![](images/2c1b7cd446167d089ee809ba092f5251c62825a9ce875fbee3279bf2af04370e.jpg)③ ![](images/7121faeaa2bce76339cd4c69c4a61a953cda5f1f1c50524762b2d5d951035405.jpg)④",
+        "options": [],
+        "answer": "44",
+        "solution": "【分析】第①②③④个分别对应2、2+3、2+3+4、2+3+4+5，故第⑧个是2到9的和。\n【详解】2+3+4+5+6+7+8+9=44。",
+        "knowledge_points": ["数与形 (归纳递推)"],
+        "source": "24-25 六年级下·天津·期末"
+      },
+      {
+        "id": "266",
+        "unit": "第八单元 数与形",
+        "question_type": "填空题",
+        "difficulty": 0.65,
+        "question": "按下图这样的规律摆下去，第10个图中有( )个涂成阴影的小正方形，第n个图中有( )个涂成阴影的小正方形。![](images/c2542837f9f258b0921f9d1df17e3b191f1d772ea4f2419c6774f8e4d528240d.jpg) ![](images/4f39f48567c36b644e36d7f83cca94752aa1e0c5f301936b0bc31bb04cde40f9.jpg) ![](images/4823db133885b79446ede4dc3210905fb71fd2d1c9be4f186cb7b3225c37cbd8.jpg)",
+        "options": [],
+        "answer": "19;2n-1",
+        "solution": "【分析】第1、2、3个图阴影小正方形分别为1、3、5，呈奇数序列，规律为第n个图有2n-1个。\n【详解】当n=10时，2×10-1=19。故第10个图有19个，第n个图有2n-1个。",
+        "knowledge_points": ["含有字母式子的化简与求值", "用字母表示稍复杂的数量关系", "数与形（探索规律）"],
+        "source": "24-25六年级下·湖南张家界·期末"
+      },
+      
+      {
+        "id": "267",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如图，用小棒摆一个正六边形，摆4个正六边形需要()根小棒;摆51个小正六边形需要()根小棒;用1001根小棒，可以摆个()小正六边形。![](images/b2491631f4249ef178ba61ceebbfa0fe3ed47a816964a5d5a69a0f9feaa0b3ed.jpg)![](images/30c77ee0e8790223bdef5abb6ec7941b8399995803813825d89701c1eef27bf0.jpg)![](images/648cc2441107d81d0bec89c6c3679c3033fe03d3bcb68dfc6882455067120dee.jpg)",
+        "options": [],
+        "answer": "21;256;200",
+        "solution": "【分析】摆1个正六边形时，需要6根小棒，可表示为$5 \\times 1 + 1$;摆2个正六边形时，第二个正六边形与第一个正六边形共用1根小棒，所以需要$6 + 5 = 11$根小棒，可表示为$5 \\times 2 + 1$;摆3个正六边形时，第三个正六边形与第二个正六边形共用1根小棒，所以需要$11 + 5 = 16$根小棒，可表示为$5 \\times 3 + 1$;由此发现，摆几个正六边形所需小棒的数量就为5乘几再加1。【详解】摆4个正六边形所需小棒的数量就为5乘4再加$5 \\times 4 + 1 = 20 + 1 = 21$(根)。所以摆4个正六边形需要21根小棒;摆51个正六边形所需小棒的数量就为5乘51再加1：$5 \\times 51 + 1 = 255 + 1 = 256$(根)。所以摆51个小正六边形需要256根小棒;用1001减1，再除以5就是所摆小正方形的个数：$(1001 - 1) \\div 5 = 1000 \\div 5 = 200$(个)。所以用1001根小棒，可以摆200个小正六边形。",
+        "knowledge_points": [
+          "数与形(归纳递推)"
+        ],
+        "source": "2024·甘肃兰州·小升初真题",
+        "merged_id": "unit8_267",
+        "original_id": "267"
+      },
+      {
+        "id": "268",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "把边长为1厘米的正方形纸片，按下面的规律拼成长方形：![](images/e3c85574dddfb6bbcdeb52eb8ef2adac9b3e0a5dd22f7d462da3b41242ba74ee.jpg)![](images/a7be0bd8503f9929251e7658481706d1d7ea08cc014f5c1116092431291059e8.jpg)![](images/94b3765421c940d06539f3b769938e3538b67d2d1957db94a07cd39bb99b901e.jpg)用4个正方形拼成的长方形的周长是( )厘米，用$n$个正方形拼成的长方形的周长是( )厘米。",
+        "options": [],
+        "answer": "10;$(2n+2)$",
+        "solution": "【分析】每增加1个正方形，周长增加了2厘米，则n个正方形周长为上下的$2n$厘米，再加上左右的2厘米，即$(2n + 2)$厘米。【详解】由分析可知，用$n$个正方形拼成的长方形的周长是(2n+2)厘米。当$n = 4$时，$2 + 2 \\times 4 = 2 + 8 = 10$(厘米)。所以用4个正方形拼成的长方形的周长是10厘米，用$n$个正方形拼成的长方形的周长是$(2n + 2)$厘米。",
+        "knowledge_points": [
+          "图形的变化规律"
+        ],
+        "source": "2024·河南郑州·小升初真题",
+        "merged_id": "unit8_268",
+        "original_id": "268"
+      },
+      {
+        "id": "269",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如下图，用棱长1cm的正方体排成一排排并成长方体。像这样，用n个正方体拼成的一个长方体的表面积，比原来n个正方体表面积之和减少了( )cm²。![](images/302634849f980f7b5eefd06bae543b7da0b3cea12b1bedd62fa962e6d42b7877.jpg)",
+        "options": [],
+        "answer": "$2n - 2$",
+        "solution": "【分析】用2个正方体拼成一个长方体，减少了2个面;用3个正方体拼成一个长方体，减少了4个面;用4个正方体拼成一个长方体，减少了6个面……以此类推，用n个正方体拼成一个长方体，减少了（2n-2）个面;正方体的棱长为1cm，则正方体每个面的面积为$1cm^2$，因此拼成的一个长方体比原来n个正方体表面积之和减少了的表面积=减少的面×每个面的面积，据此解答。【详解】用2个正方体拼成一个长方体，减少了2个面;用3个正方体拼成一个长方体，减少了4个面;用4个正方体拼成一个长方体，减少了6个面……用$n$个正方体拼成一个长方体，减少了$2 \\times (n - 1) = (2n - 2)$个面。每个面的面积：$1 \\times 1 = 1(cm^{2})$。$(2n - 2) \\times 1 = (2n - 2)cm^{2}$。因此用$n$个正方体拼成的一个长方体的表面积，比原来$n$个正方体表面积之和减少了$(2n - 2)cm^{2}$。",
+        "knowledge_points": [
+          "含有字母式子的化简与求值",
+          "立体图形的切拼（长方体、正方体的表面积）",
+          "用字母表示数",
+          "数量关系"
+        ],
+        "source": "23-24五年级下·重庆璧山·期末",
+        "merged_id": "unit8_269",
+        "original_id": "269"
+      },
+      {
+        "id": "270",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "观察下列图形的构成规律，按照此规律，第10个图形中★的个数为( )个;第$n$幅图中★的个数为( )个。![](images/c22e111bba8783d6026cea0fdfce130f5597878cc36c13c5422e8f656156a24c.jpg)图形1![](images/e56d667f7f433b09609b4fac834c8e63f38ba9d8689f5204b8e99f9cf9b1c3e7.jpg)图形2![](images/da60d244cfa4412b1d376a04413c038d3161cc24e1016dd45ada67de83225a7b.jpg)图形3",
+        "options": [],
+        "answer": "31;$1 + 3n$或$3n + 1$",
+        "solution": "【分析】图形1：数出★的个数为4个，可写成$3 \\times 1 + 1$。图形2：数出★的个数为7个，可写成$3 \\times 2 + 1$。图形3：数出★的个数为10个，可写成$3 \\times 3 + 1$。通过观察上述规律，发现第几个图形中★的个数就是3乘几加1。【详解】第10个图形中★的个数就是3乘10加1，即$3 \\times 10 + 1 = 30 + 1 = 31$个;第$n$幅图中$\\star$的个数就是3乘$n$加1，即$3 \\times n + 1 = (3n + 1)$个;所以按照此规律，第10个图形中★的个数为31个;第n幅图中★的个数为$(3n + 1)$个。",
+        "knowledge_points": [
+          "用字母表示稍复杂的数量关系",
+          "数与形(归纳递推)"
+        ],
+        "source": "23-24六年级下·广东江门·期末",
+        "merged_id": "unit8_270",
+        "original_id": "270"
+      },
+      {
+        "id": "271",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如图，摆6个同样的正方形需要小棒( )根。![](images/e306208cc73093c3142f111ad576560f47d452d9d293a4ad3f3727ff5f8c379a.jpg)",
+        "options": [],
+        "answer": "19",
+        "solution": "【分析】仔细观察可以发现，每增加一个正方形增加3根小棒，摆1个正方形需要4根小棒，摆2个正方形需要$(4 + 3 \\times 1)$根小棒，摆3个正方形需要$(4 + 3 \\times 2)$根小棒……则摆6个正方形需要$(4 + 3 \\times 5)$根小棒。据此解答。【详解】$4 + 3 \\times 5 = 4 + 15 = 19$(根)。所以摆6个同样的正方形需要小棒19根。",
+        "knowledge_points": [
+          "数与形(探索规律)"
+        ],
+        "source": "23-24六年级上·新疆乌鲁木齐·期末",
+        "merged_id": "unit8_271",
+        "original_id": "271"
+      },
+      {
+        "id": "272",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "科科学家研究发现：植物的花瓣、萼片、果实的数目、特征都吻合于一种奇特的数列：1，1，2，3，5，8，13，21……请你仔细观察此数列，它的第9个数应该是( )。",
+        "options": [],
+        "answer": "34",
+        "solution": "【分析】首先，从已知数列观察出特点：$1 + 1 = 2$;$1 + 2 = 3$;$2 + 3 = 5$;$3 + 5 = 8$。由此可知：在已知数列中，从第三项开始每一项是前两项的和;第9项就是第7项与第8项的和，据此解答。【详解】$13 + 21 = 34$。科科学家研究发现：植物的花瓣、萼片、果实的数目、特征都吻合于一种奇特的数列：1，1，2，3，5，8，13，21……请你仔细观察此数列，它的第9个数应该是34。",
+        "knowledge_points": [
+          "整数排列的规律",
+          "间隔问题"
+        ],
+        "source": "2024·贵州安顺·小升初真题",
+        "merged_id": "unit8_272",
+        "original_id": "272"
+      },
+      {
+        "id": "273",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "用小棒摆八边形，如下所示。按这个规律摆6个八边形需要( )根小棒，摆n个八边形需要( )根小棒。![](images/3635e4cef11d06f1a06205f97f43dc51.png)",
+        "options": [],
+        "answer": "43;$1 + 7n$",
+        "solution": "【分析】从图中可知：一个八边形需要8根小棒，2个八边形需要$(8 + 7)$根小棒，3个八边形需要$(8 + 7 + 7)$根小棒，据此可知：每增加一个八边形，就要增加7根小棒，若将一个八边形的8根小棒改写成$8 = 1 + 7$，这样，n个八边形就要$(1 + 7n)$根小棒。据此计算摆6个八边形需要小棒根数即可。【详解】当$n = 6$时，$1 + 7 \\times 6 = 1 + 42 = 43$(根)。按这个规律摆6个八边形需要43根小棒，摆n个八边形需要$(1 + 7n)$根小棒。",
+        "knowledge_points": [
+          "含有字母式子的化简与求值",
+          "用字母表示稍复杂的数量关系",
+          "数与形(探索规律)"
+        ],
+        "source": "2024·辽宁盘锦·小升初真题",
+        "merged_id": "unit8_273",
+        "original_id": "273"
+      },
+      {
+        "id": "274",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "用黑白两种颜色的正六边形地面砖按如图所示的规律，拼成若干个图案：第4个图案中有白色地面砖( )块，第100个图案中有白色地面砖( )块，第n个图案中有白色地面砖( )块。![](images/d27965e8aef3e71de0d2948dfd171fa5c4997fac2bd07117438697de230270c4.jpg)第1个![](images/2121d26231af70cf8f3d1b94ab74e6924572cecad37d043a70848d8ea6df2fad.jpg)第2个![](images/a1c57c60eda456a281efd72b80cd7d781f0fb8dfe11dd916c29407be948beac4.jpg)第3个",
+        "options": [],
+        "answer": "18;402;$(4n+2)$",
+        "solution": "【分析】观察图形：第1个图形：白色的地面砖有6块;第2个图形：白色的地面砖有10块，$10 = 4 \\times 2 + 2$;第3个图形：白色的地面砖有14块，$14 = 4\\times 3 + 2$。第4个图形：白色的地面砖有18块，$18 = 4 \\times 4 + 2$;第$n$个图形：白色的地面砖有$(4n + 2)$块;【详解】$4 \\times 4 + 2 = 16 + 2 = 18$(块)。$4 \\times 100 + 2 = 400 + 2 = 402$(块)。则4个图案中有白色地面砖18块，第100个图案中有白色地面砖402块，第n个图案中有白色地面砖$(4n + 2)$块。",
+        "knowledge_points": [
+          "数与形(探索规律)"
+        ],
+        "source": "2024·湖北武汉·小升初真题",
+        "merged_id": "unit8_274",
+        "original_id": "274"
+      },
+      {
+        "id": "275",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如图，它是由火柴棒组成的三角形图案，拼9个三角形图案用()根火柴棒，拼n个三角形图案用()根火柴棒。![](images/6927749cf66d64a0713653f8078625e0073f61e9703bcc4532a12345ac46e265.jpg)",
+        "options": [],
+        "answer": "19;$2n + 1$或$1 + 2n$",
+        "solution": "【分析】拼1个三角形需3根火柴棒，拼2个三角形需$(3 + 2)$根火柴棒，拼3个三角形需$(3 + 2 + 2)$根火柴棒，……拼9个三角形需$[3 + 2 \\times (9 - 8)]$个火柴棒，拼n个三角形需$[3 + 2 \\times (n - 1)]$个火柴棒即可解答。【详解】$3 + 2 \\times (9 - 8) = 3 + 2 \\times 8 = 3 + 16 = 19$(根)。拼n个三角形需火柴棒：$3 + 2 \\times (n - 1) = 3 + 2n - 2 = 2n + 1$(根)。如图，它是由火柴棒组成的三角形图案，拼9个三角形图案用19根火柴棒，拼n个三角形图案用$(2n + 1)$根火柴棒。",
+        "knowledge_points": [
+          "用字母表示数",
+          "数量关系",
+          "数与形(探索规律)"
+        ],
+        "source": "2024·河南开封·小升初真题",
+        "merged_id": "unit8_275",
+        "original_id": "275"
+      },
+      {
+        "id": "276",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "用白色和灰色圆形按照如图所示的方法摆图形。按照这样的方法摆下去，第5个图形中，共有____个圆形;当一个图形中有6个灰色圆形时，白色的圆形有____个。![](images/dcb5d6f6232729c5ad9d5e8f56e2f4a0c7af0cfc6984dc5571602f029ad207bf.jpg)",
+        "options": [],
+        "answer": "25;30",
+        "solution": "【分析】观察图形可知：第1个图中有圆形1个，$1 = 1^{2}$，有灰色圆形1个;有白色圆形0个。第2个图中有圆形4个，$4 = 2^{2}$，有灰色圆形2个;有白色圆形2个，$2 = 2^{2} - 2$。第3个图中有圆形9个，$9 = 3^{2}$，有灰色圆形3个;有白色圆形6个，$6 = 3^{2} - 3$。第4个图中有圆形16个，$16 = 4^{2}$，有灰色圆形4个;有白色圆形12个，$12 = 4^{2} - 4$。规律：第n个图中有圆形的个数为$n^{2}$个，有灰色圆形n个，有白色圆形$(n^{2} - n)$个。据此规律解答。【详解】规律：第n个图中有圆形的个数为$n^{2}$个，有灰色圆形n个，有白色圆形$(n^{2} - n)$个。当$n = 5$时，$n^{2} = 5 \\times 5 = 25$(个)。当一个图形中有6个灰色圆形时，即$n = 6$时，则白色圆形有：$6 \\times 6 - 6 = 36 - 6 = 30$(个)。第5个图形中，共有（25）个圆形;当一个图形中有6个灰色圆形时，白色的圆形有（30）个。",
+        "knowledge_points": [
+          "含有字母式子的化简与求值",
+          "用字母表示稍复杂的数量关系",
+          "数与形(探索规律)"
+        ],
+        "source": "2024·河北秦皇岛·小升初真题",
+        "merged_id": "unit8_276",
+        "original_id": "276"
+      },
+      {
+        "id": "277",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "按图中的方式摆棋子，摆第5个图案需要()枚棋子，摆第n个图案需要()枚棋子。![](images/eff2b4b2036cea5c0e737870e015bb97e76d91586cb698836d819b31d6f086f5.jpg)图案1![](images/5faa7cd90d0d6d0fb7cd6c426179d880f0a7746864dfbfae527bbaeba7fdc425.jpg)图案2![](images/37c6942b85a03738b7ed0066adc7cf0d2032cff5cc66197195f51713df00bb87.jpg)图案3![](images/98037243f8abf615791c9615467d5a701206f3d905ed8fb76e715c589d0ed6ed.jpg)",
+        "options": [],
+        "answer": "17;$2 + 3n$",
+        "solution": "【分析】从图中可知：第一个图案需要5枚棋子;第二个图案需要8枚棋子，比第一个图案增加了3枚棋子;第三个图案需要11枚棋子，比第二个图案增加了3枚棋子，据此可以发现，每个图案需要的棋子都比前一个图案增加3枚棋子。若第一个图案表示为$5 = 2 + 3 = 2 + 3 \\times 1$，则第二个图案表示为$8 = 2 + 3 + 3 = 2 + 3 \\times 2$。第三个图案表示为$11 = 2 + 3 + 3 + 3 = 2 + 3 \\times 3$;第$n$个图案可以表示为：$2 + 3n$。据此解答即可。【详解】第五个图案：$2 + 3 \\times 5 = 2 + 15 = 17$(枚)。摆第5个图案需要17枚棋子，摆第n个图案需要$(2 + 3n)$枚棋子。",
+        "knowledge_points": [
+          "数与形(探索规律)"
+        ],
+        "source": "2024·河北衡水·小升初真题",
+        "merged_id": "unit8_277",
+        "original_id": "277"
+      },
+      {
+        "id": "278",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "观察下图，照这样画下去，第5幅图中有( )个互不重叠的三角形，第n幅图中有( )个互不重叠的三角形(用含n的式子表示)。![](images/d1f224e7d12d03a1580cf6309caffe3bfcd1661800e1ae61e1055c71dd5d780c.jpg)第1幅![](images/f7e587fcc9ea107b454fecb0bdf1354ecaa35fe4e776833ffecbf1fd76bfbf74.jpg)第2幅![](images/5be0075dff5481b55a3ed8debebd47d6df9134409df68ee5143311c158fe39d4.jpg)第3幅",
+        "options": [],
+        "answer": "16;$3n + 1$或$1 + 3n$",
+        "solution": "【分析】观察可知，第1幅图中有（3+1）个互不重叠的三角形，第2幅图中有（3×2+1）个互不重叠的三角形，第3幅图中有（3×3+1）个互不重叠的三角形……以此类推，第n幅图中有（3n+1）个互不重叠的三角形，最后求出$n = 5$时含有字母式子的值，据此解答。【详解】规律：第$n$幅图中有$(3n + 1)$个互不重叠的三角形。当$n = 5$时，$3n + 1 = 3 \\times 5 + 1 = 15 + 1 = 16$(个)。所以，第5幅图中有（16）个互不重叠的三角形，第n幅图中有$(3n + 1)$个互不重叠的三角形。",
+        "knowledge_points": [
+          "含有字母式子的化简与求值",
+          "用字母表示稍复杂的数量关系",
+          "数与形(探索规律)"
+        ],
+        "source": "23-24六年级下·浙江温州·期末",
+        "merged_id": "unit8_278",
+        "original_id": "278"
+      },
+      {
+        "id": "279",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "探索规律。请你观察图形规律，当涂色正方形的个数是6时，空白正方形的个数是( );若涂色正方形的个数是a，空白正方形的个数是( )（用含有字母的式子表示）。![](images/5363aab073525451a71015fc5a687c856966d1a9684da00c61acd2e4d608bd15.jpg)![](images/aaaa11453a1c001267bd05dc346745a5173814e87feea2198471c5fc1f87bedf.jpg)![](images/4200e4716d34ae707185aa65ae776a174e2b21ccad5d210c0d8ab4cac278ed17.jpg)",
+        "options": [],
+        "answer": "33;$5a + 3$或$3 + 5a$",
+        "solution": "【分析】由图可知，涂色正方形的个数为1时，正方形的总个数为9，空白正方形的个数为$(9 - 1)$;涂色正方形的个数为2时，正方形的总个数为$(9 + 6)$，空白正方形的个数为$(9 + 6 - 2)$;涂色正方形的个数为3时，正方形的总个数为$(9 + 6 \\times 2)$，空白正方形的个数为$(9 + 6 \\times 2 - 3)$……以此类推，涂色正方形的个数为a时，正方形的总个数为$[9 + 6 \\times (a - 1)]$，空白正方形的个数为$[9 + 6 \\times (a - 1) - a]$，最后求出$a = 6$时含有字母式子的值，据此解答。【详解】$9 + 6 \\times (a - 1) - a = 9 + 6a - 6 - a = (6a - a) + (9 - 6) = 5a + 3$。当$a = 6$时，$5a + 3 = 5 \\times 6 + 3 = 30 + 3 = 33$。所以，当涂色正方形的个数是6时，空白正方形的个数是33，若涂色正方形的个数是$a$，空白正方形的个数是$(5a + 3)$。",
+        "knowledge_points": [
+          "数与形(探索规律)",
+          "数与形(归纳递推)"
+        ],
+        "source": "2024·河北邯郸·小升初真题",
+        "merged_id": "unit8_279",
+        "original_id": "279"
+      },
+      {
+        "id": "280",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "有一组算式如：$4 + 2$，$5 + 8$，$6 + 14$，$7 + 20$，……那么，第10个算式的得数是( )。",
+        "options": [],
+        "answer": "69",
+        "solution": "【分析】观察第一个加数序列4，5，6，7，……，起始为4，后一个数比前一个数大1，第n个算式的第一个加数，是在起始数4的基础上，增加了(n-1)个1，所以表达式为$4 + (n - 1) \\times 1$;第二个加数序列2，8，14，20，……，起始为2，后一个数比前一个数大6，第n个算式的第二个加数，是在起始数2的基础上，增加了(n-1)个6，表达式为$2 + (n - 1) \\times 6$。【详解】根据分析可知：第10个算式的前1个加数为：$4 + (10 - 1) \\times 1 = 4 + 9 \\times 1 = 4 + 9 = 13$。第10个算式的后1个加数为：$2 + (10 - 1) \\times 6 = 2 + 9 \\times 6 = 2 + 54 = 56$。第10个算式为：$13 + 56 = 69$。第10个算式的得数是69。",
+        "knowledge_points": [
+          "带有小括号的混合运算",
+          "数与形(归纳递推)"
+        ],
+        "source": "2024·重庆沙坪坝·小升初真题",
+        "merged_id": "unit8_280",
+        "original_id": "280"
+      },
+      {
+        "id": "281",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "$\\frac{1}{2}$，$\\frac{1}{4}$，$\\frac{1}{8}$，$\\frac{1}{16}$，$\\frac{1}{32}$，……这列数的每一项越来越小，越来越接近( )。",
+        "options": [],
+        "answer": "0",
+        "solution": "【分析】分子都是1，分母依次乘2，这样分母越来越大，分母越大，分数值越小，所以这个分数越来越小，越来越接近0，据此解答。【详解】根据分析可知，$\\frac{1}{2}$，$\\frac{1}{4}$，$\\frac{1}{8}$，$\\frac{1}{16}$，$\\frac{1}{32}$，……这列数的每一项越来越小，越来越接近0。",
+        "knowledge_points": [
+          "分数排列的规律"
+        ],
+        "source": "2024·河南南阳·小升初真题",
+        "merged_id": "unit8_281",
+        "original_id": "281"
+      },
+      {
+        "id": "282",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "摆1个三角形需要3根小棒，摆2个三角形需要5根小棒，摆3个三角形需要7根小棒。摆n个三角形，需要()根小棒。摆50个三角形，需要()根小棒。![](images/cbd1965ec6ea25ca2715160301ec0c13e61f8d16c6cd3cdd54e28f083b591bbd.jpg)![](images/e8f9b906c9e1b29ebbd41a7752f00f365d7103f45477d9111f145e0d4affd3bd.jpg)![](images/261d8576d9396ee106c98310cefeaa7ef2b89213d03f539e5ef5f0bac9bd9a63.jpg)",
+        "options": [],
+        "answer": "(1 + 2n)或(2n + 1);101",
+        "solution": "【分析】观察摆三角形的情况，摆1个三角形用3根小棒，即$1 + 2$根小棒;摆2个三角形用$1 + 2 + 2 = 5$根，即$1 + 2 \\times 2$根小棒，摆3个三角形用$1 + 2 + 2 + 2 = 7$根，即$1 + 2 \\times 3$根小棒;由此归纳：摆$n$个三角形时，需要（$1 + 2n$）根小棒;计算摆50个时，把$n = 50$代入$1 + 2n$中，就能求出小棒数。【详解】摆1个三角形：$3 = 1 + 2\\times 1$根小棒;摆2个三角形：$5 = 1 + 2\\times 2$根小棒;摆3个三角形：$7 = 1 + 2 \\times 3$根小棒;规律：摆$n$个三角形时，需要$(1 + 2n)$根小棒。当$n = 50$时，$1 + 2n = 1 + 2 \\times 50 = 1 + 100 = 101$(根)。摆n个三角形，需要$(1 + 2n)$根小棒。摆50个三角形，需要101根小棒。",
+        "knowledge_points": [
+          "含有字母式子的化简与求值",
+          "用字母表示数",
+          "数量关系",
+          "数与形(探索规律)"
+        ],
+        "source": "2024·河南漯河·小升初真题",
+        "merged_id": "unit8_282",
+        "original_id": "282"
+      },
+      {
+        "id": "283",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "找出下面算式的规律：$2^{2} - 1^{2} = 2 + 1$;$4^{2} - 3^{2} = 4 + 3$;$6^{2} - 5^{2} = 6 + 5$。(1)请你再写个这样的算式：( )。(2)运用规律计算：$50^{2} - 49^{2} + 48^{2} - 47^{2} + 46^{2} - 45^{2} + ... + 2^{2} - 1^{2} = ()$。",
+        "options": [],
+        "answer": "(1)$7^{2} - 6^{2} = 7+6$ (2)1275",
+        "solution": "【分析】（1）观察算式，找出规律：相邻两个数的平方差，等于这两个数的和;可以表示为$(n + 1)^{2} - n^{2} = n + 1 + n$，据此规律再写出一个算式即可。(2)运用规律，将算式改写成$50 + 49 + 48 + 47 + 46 + 45+……+2+1$，再计算出结果即可。【详解】(1)这样的算式：$7^{2} - 6^{2} = 7 + 6$。（答案不唯一）(2)$50^{2} - 49^{2} + 48^{2} - 47^{2} + 46^{2} - 45^{2} + ... + 2^{2} - 1^{2} = 50 + 49 + 48 + 47 + 46 + 45 + ... + 2 + 1 = (50 + 1) \\times 50 \\div 2 = 51 \\times 50 \\div 2 = 1275$。",
+        "knowledge_points": [
+          "算式的规律(整数)",
+          "数与形(探索规律)"
+        ],
+        "source": "2025六年级下·全国·专题练习",
+        "merged_id": "unit8_283",
+        "original_id": "283"
+      },
+      {
+        "id": "284",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "找规律：1，1，2，3，5，8，13，（ ）。",
+        "options": [],
+        "answer": "21",
+        "solution": "【分析】观察数列，得出数列规律：每一个数字都是前两个数字之和。根据规律计算缺失数字即可。【详解】$8 + 13 = 21$。所以括号中填21。",
+        "knowledge_points": [
+          "整数排列的规律"
+        ],
+        "source": "24-25六年级下·湖南怀化·期中",
+        "merged_id": "unit8_284",
+        "original_id": "284"
+      },
+      {
+        "id": "285",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "观察下列等式：解答问题：$2^{1} + 2^{2} + 2^{3} + 2^{4} + 2^{5} + 2^{6} + ... + 2^{2015}$的末位数字是( )。",
+        "options": [],
+        "answer": "4",
+        "solution": "【分析】根据题意，可以分别计算算式中每个加数的个位的和是多少，看末位就是所求;由$2^{n}$，$2^{n + 1}$，$2^{n + 2}$，$2^{n + 3}$的个位依次是2，4，8，6，可知次数每4的倍数一循环，用2015除以4计算出商和余数，商表示2，4，8，6的个数，再余数中包含的数字相加即可。解答此题关键是将算式的末位数字转化为求每个加数末位的数字之和。【详解】根据分析，$2015 \\div 4 = 503...3$。$(2 + 4 + 8 + 6) \\times 503 + (2 + 4 + 8) = 20 \\times 503 + 14 = 10060 + 14 = 10074$。所以，算式$2^{1} + 2^{2} + 2^{3} + 2^{4} + 2^{5} + 2^{6} + ... + 2^{2015}$的末位数字是$4$。",
+        "knowledge_points": [
+          "算式的规律(整数)",
+          "间隔问题"
+        ],
+        "source": "2024·新疆乌鲁木齐·小升初真题",
+        "merged_id": "unit8_285",
+        "original_id": "285"
+      },
+      {
+        "id": "286",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "用小棒摆五边形，如图：按这个规律，摆$n$个五边形( )根小棒;照这样摆，用97根小棒能摆( )个五边形。![](images/39ecd53a969f71983dff7b86bdfe2663c902c21700b841aecf055e0b7b4aed0d.jpg)![](images/357fcb167d3ae3c8bb6715d26ad4fd514dabe5e817e2bfc31f0cce545c1e2ac1.jpg)![](images/3607ef93afeb0b5e0108a0a625523c363c83bb11333e1bc43b71a84afa3edf99.jpg)",
+        "options": [],
+        "answer": "$4n + 1$;24",
+        "solution": "【分析】看图可知，摆1个五边形需要5根小棒，$5 = 1 \\times 4 + 1$;摆2个五边形需要9根小棒，$9 = 2 \\times 4 + 1$;摆3个五边形需要13根小棒，$13 = 3 \\times 4 + 1$。由此可知，小棒根数=摆几个五边形就用几$\\times 4 + 1$，五边形个数=(小棒根数-1)$\\div 4$，据此分析。【详解】摆1个五边形小棒根数：$1 \\times 4 + 1 = 4 + 1 = 5$(根)。摆2个五边形小棒根数：$2\\times 4 + 1 = 8 + 1 = 9$(根)。摆3个五边形小棒根数：$3 \\times 4 + 1 = 12 + 1 = 13$(根)。第$n$个图形小棒的根数是：$n \\times 4 + 1 = (4n+ 1)$根。$(97 - 1) \\div 4 = 96 \\div 4 = 24$(个)。按这个规律，摆$n$个五边形（$4n + 1$）根小棒;照这样摆，用97根小棒能摆24个五边形。",
+        "knowledge_points": [
+          "数与形(归纳递推)"
+        ],
+        "source": "2024·重庆丰都·小升初真题",
+        "merged_id": "unit8_286",
+        "original_id": "286"
+      },
+      {
+        "id": "287",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "一张方桌可坐4人，2张方桌并在一起可以坐6人，6张方桌可坐____人，n张方桌可坐____人。![](images/a62e0327c728581305a66a28361c5b7c1a4bdd4721952ac7d5638b0812bbd5f1.jpg)![](images/084f89f1b00434af8afaf33c54f903643c3f5f702800138a2c7ab1abe33b60b2.jpg)![](images/eaf598e55feb30fd262ba064212309770f381767dca6dc9118a749d52ba6bd1b.jpg)",
+        "options": [],
+        "answer": "14;(2n+2)",
+        "solution": "【分析】1张方桌可坐4人，即$4 = 2 \\times 1 + 2$，2张方桌并在一起可坐6人，即$6 = 2 \\times 2 + 2$，3张方桌可以坐$(2 \\times 3 + 2)$人，……n张方桌可以坐$(2n + 2)$人。【详解】$2 \\times 6 + 2 = 12 + 2 = 14$(人)。n张方桌可坐(2n+2)个人。所以6张方桌可坐14人，n张方桌可坐$(2n + 2)$人。",
+        "knowledge_points": [
+          "数与形(探索规律)"
+        ],
+        "source": "2024·河北保定·小升初真题",
+        "merged_id": "unit8_287",
+        "original_id": "287"
+      },
+      {
+        "id": "288",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "用完全一样的火柴棍拼图形（如图）。![](images/3358d67d1f0ddb3897e398e673831fd49360a79031e93b106372916f8f4c7daf.jpg)第1个![](images/2d80a73b113ab9bce51c6ae2f8f71db37dd90f0c02ebcbfc89b8a77fdeaf32c0.jpg)第2个![](images/f910e59436ff921976cbaf3d67a5df695077b527cd0e69032a9557f2cb8e9985.jpg)第3个![](images/8e83623b6b47bdf7349489ef2d0983fc852d06598211bdbc172aa773335f136a.jpg)。按照这样的方法拼成第4个图形需要火柴棍()根，拼成第$n$个图形需要火柴棍()根。",
+        "options": [],
+        "answer": "34;(8n + 2)或(2 + 8n)",
+        "solution": "【分析】根据题图可知：第1个图形需要的火柴棍为：10根;第2图形需要的火柴棍为：18根，$18 = 10 + 8 \\times 1$。第3个图形需要的火柴棍为：26根，$26 = 10 + 8 \\times 2$。第4个图形需要的火柴棍为：34根，$26 = 10 + 8 \\times 3$;第$n$个图形需要的火柴棍根数为：$10 + 8 \\times (n - 1) = 8n +2$。据此解答。【详解】下午第$n$个图形需要的火柴棍根数为：$10 + 8 \\times (n - 1) = 8n + 2$。当$n = 4$时，$8 \\times 4 + 2 = 32 + 2 = 34$(根)。拼成第4个图形需要火柴棍34根，拼成第n个图形需要火柴棍$(8n + 2)$或$(2 + 8n)$根。",
+        "knowledge_points": [
+          "图形的变化规律",
+          "数与形(探索规律)"
+        ],
+        "source": "2024·北京西城·小升初真题",
+        "merged_id": "unit8_288",
+        "original_id": "288"
+      },
+      {
+        "id": "289",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "用黑、白两种颜色的正方形纸片按如图中的规律拼图案。![](images/138237013e3b37e119a4dc7ffc5768c36bd17f47ab6012f126d0839ec6346e85.jpg)第1个图![](images/b3807d160621436e00c429e75314189a5e90b0753c7e8615c8fd103e41edc903.jpg)第2个图![](images/79b260cb8dc4bc6850596760360c9f4aca290abe9c6045304cc77be86ebd2110.jpg)第3个图。第15个图案有白色纸片( )张，第n个图案中有白色纸片121张，则$n = ()$。",
+        "options": [],
+        "answer": "46;40",
+        "solution": "【分析】观察图案可知，第1个图案中有白色纸片4张，即$(3 \\times 1 + 1)$张;第2个图案中有白色纸片7张，即$(3 \\times 2 + 1)$张;第3个图案中有白色纸片10张，即$(3 \\times 3 + 1)$张。由此可推出第n个图案中有白色纸片$(3n + 1)$张;把$n = 15$代入$3n + 1$计算即可求出第15个图案有白色纸片的张数，令$3n + 1 = 121$，方程两边同时减去1，再同时除以3即可解答。【详解】(1)$3 \\times 15 + 1 = 45 + 1 = 46$(张)。所以第15个图案有白色纸片46张。(2)由分析可得：$3n + 1 = 121$，$3n + 1 - 1 = 121 - 1$，$3n = 120$，$3n \\div 3 = 120 \\div 3$，$n = 40$。所以第40个图案中有白色纸片121张，则$n = 40$。",
+        "knowledge_points": [
+          "应用等式的性质1和2解方程",
+          "用字母表示数",
+          "数量关系",
+          "数与形(探索规律)"
+        ],
+        "source": "2024·江西景德镇·小升初真题",
+        "merged_id": "unit8_289",
+        "original_id": "289"
+      },
+      {
+        "id": "290",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如图，第①个图中有1个小三角形，第②个图中有4个小三角形，第⑤个图形中有( )个小三角形;第$n$个图形中有( )个小三角形。![](images/cc9b4deef4571757c0885840e925e913607233ad5d16fe5eeaab3b5047aba7cd.jpg)①![](images/9e9f39c5263b09937011fd35f17b3f5f403d43e3158185b03c778431b2c3d311.jpg)②![](images/202877cd23fd738c51a913f96113b001549125a841db79ff711d6be59a361644.jpg)③",
+        "options": [],
+        "answer": "25;$n^{2}$",
+        "solution": "【分析】观察可知，第①个图中有$1^{2}$个小三角形，第②个图中有$2^{2}$个小三角形，第⑤个图形中有$5^{2}$个小三角形，第n个图形中有$n^{2}$个小三角形。【详解】$5^{2} = 25$(个)。$n\\times n = n^{2}$(个)。第①个图中有1个小三角形，第②个图中有4个小三角形，第⑤个图形中有25个小三角形;第n个图形中有$n^{2}$个小三角形。",
+        "knowledge_points": [
+          "图形的变化规律",
+          "用字母表示数",
+          "数量关系",
+          "数与形(探索规律)"
+        ],
+        "source": "2024·江西九江·小升初真题",
+        "merged_id": "unit8_290",
+        "original_id": "290"
+      },
+      {
+        "id": "291",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "有一列数：$\\frac{1}{1}$，$\\frac{1}{2}$，$\\frac{2}{2}$，$\\frac{1}{2}$，$\\frac{1}{3}$，$\\frac{2}{3}$，$\\frac{3}{3}$，$\\frac{2}{3}$，$\\frac{1}{3}$，$\\frac{1}{4}$，$\\frac{2}{4}$，$\\frac{3}{4}$，$\\frac{4}{4}$，$\\frac{3}{4}$，$\\frac{2}{4}$，$\\frac{1}{4}$……从左开始数，第111个分数是$\\frac{11}{11}$。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】这一列数中，分母是1的分数有1个，分子是1;分母是2的分数有3个，分子是1，2，1;分母是3的分数有5个，分子是1，2，3，2，1;分母是4的分数有7个;分子是1，2，3，4，3，2，1。分数的个数分别是1，3，5，7...，当分母是n时有$2n - 1$个分数;由此求出从分母是1的分数到分母是11的分数一共有多少个;分子是自然数，先从1增加，到和分母相同时再减少到1;所以$\\frac{11}{11}$还有10个分母是11的分数，由此求解。【详解】分母是11的分数一共有：$2 \\times 11 - 1 = 21$(个)。从分母是1的分数到分母是11的分数一共：$1 + 3 + 5 + 7... + 21 = (1 + 21) \\times 11 \\div 2 = 22 \\times 11 \\div 2 = 121$(个)。$\\frac{11}{11}$还有10个分母是11的分数。$121 - 10 = 111$。有一列数：$\\frac{1}{1}$，$\\frac{1}{2}$，$\\frac{2}{2}$，$\\frac{1}{2}$，$\\frac{1}{3}$，$\\frac{2}{3}$，$\\frac{3}{3}$，$\\frac{2}{3}$，$\\frac{1}{4}$，$\\frac{1}{4}$，$\\frac{2}{4}$，$\\frac{3}{4}$，$\\frac{4}{4}$，$\\frac{3}{4}$，$\\frac{2}{4}$，$\\frac{1}{4}$……从左开始数，$\\frac{11}{11}$是第111个数。原题说法正确。故答案为：√",
+        "knowledge_points": [
+          "分数排列的规律"
+        ],
+        "source": "2025·四川南充·小升初真题",
+        "merged_id": "unit8_291",
+        "original_id": "291"
+      },
+      {
+        "id": "292",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "已知$\\frac{1}{3 \\times 5} = (\\frac{1}{3} - \\frac{1}{5}) \\times \\frac{1}{2}$，$\\frac{1}{5 \\times 7} = (\\frac{1}{5} - \\frac{1}{7}) \\times \\frac{1}{2}$，按这样的规律，请计算：$\\frac{1}{3 \\times 5} + \\frac{1}{5 \\times 7} + \\frac{1}{7 \\times 9} + ... + \\frac{1}{19 \\times 21}$。",
+        "options": [],
+        "answer": "$\\frac{1}{7}$",
+        "solution": "【分析】已知$\\frac{1}{3 \\times 5} = (\\frac{1}{3} - \\frac{1}{5}) \\times \\frac{1}{2}$，$\\frac{1}{5 \\times 7} = (\\frac{1}{5} - \\frac{1}{7}) \\times \\frac{1}{2}$，可以发现规律：对于$\\frac{1}{n \\times (n + 2)}$（n为正整数），可转化为$(\\frac{1}{n} - \\frac{1}{n + 2}) \\times \\frac{1}{2}$。即$\\frac{1}{19 \\times 21}$可转化为$(\\frac{1}{19} - \\frac{1}{21}) \\times \\frac{1}{2}$，则原式变为$(\\frac{1}{3} - \\frac{1}{5}) \\times \\frac{1}{2} + (\\frac{1}{5} - \\frac{1}{7}) \\times \\frac{1}{2} + (\\frac{1}{7} - \\frac{1}{9}) \\times \\frac{1}{2} + ... + (\\frac{1}{19} - \\frac{1}{21}) \\times \\frac{1}{2}$。然后利用乘法分配律计算即可。【详解】$\\frac{1}{3 \\times 5} + \\frac{1}{5 \\times 7} + \\frac{1}{7 \\times 9} + ... + \\frac{1}{19 \\times 21} = (\\frac{1}{3} - \\frac{1}{5}) \\times \\frac{1}{2} + (\\frac{1}{5} - \\frac{1}{7}) \\times \\frac{1}{2} + (\\frac{1}{7} - \\frac{1}{9}) \\times \\frac{1}{2} + ... + (\\frac{1}{19} - \\frac{1}{21}) \\times \\frac{1}{2} = \\frac{1}{3} \\times \\frac{1}{2} - \\frac{1}{5} \\times \\frac{1}{2} + \\frac{1}{5} \\times \\frac{1}{2} - \\frac{1}{7} \\times \\frac{1}{2} + \\frac{1}{7} \\times \\frac{1}{2} - \\frac{1}{9} \\times \\frac{1}{2} + ... + \\frac{1}{19} \\times \\frac{1}{2} - \\frac{1}{21} \\times \\frac{1}{2} = (\\frac{1}{3} - \\frac{1}{5} + \\frac{1}{5} - \\frac{1}{7} + \\frac{1}{7} - \\frac{1}{9} + ... + \\frac{1}{19} - \\frac{1}{21}) \\times \\frac{1}{2} = (\\frac{1}{3} - \\frac{1}{21}) \\times \\frac{1}{2} = \\frac{1}{3} \\times \\frac{1}{2} - \\frac{1}{21} \\times \\frac{1}{2} = \\frac{1}{6} - \\frac{1}{42} = \\frac{7}{42} - \\frac{1}{42} = \\frac{1}{7}$。所以$\\frac{1}{3 \\times 5} + \\frac{1}{5 \\times 7} + \\frac{1}{7 \\times 9} + ... + \\frac{1}{19 \\times 21} = \\frac{1}{7}$。",
+        "knowledge_points": [
+          "整数乘法运算定律推广到分数乘法",
+          "算式的规律(分数)"
+        ],
+        "source": "2025·福建龙岩·小升初真题",
+        "merged_id": "unit8_292",
+        "original_id": "292"
+      },
+      {
+        "id": "293",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "观察$5^{*}2 = 5 + 55 = 60$，$7^{*}4 = 7 + 77 + 777 + 7777 = 8638$，求$9^{*}5$。",
+        "options": [],
+        "answer": "111105",
+        "solution": "【分析】由$5^{*}2 = 5 + 55 = 60$，$7^{*}4 = 7 + 77 + 777 + 7777 = 8638$，推测出$a * b = a + aa + aaa + ...$一直加到b个a为止，据此分析。【详解】$9^{*}5 = 9 + 99 + 999 + 9999 + 99999 = 111105$。",
+        "knowledge_points": [
+          "算式的规律(整数)",
+          "数与形(归纳递推)"
+        ],
+        "source": "2024·重庆云阳·小升初真题",
+        "merged_id": "unit8_293",
+        "original_id": "293"
+      },
+      {
+        "id": "294",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "在分数$\\frac{1}{1}$、$\\frac{1}{2}$、$\\frac{2}{2}$、$\\frac{1}{2}$、$\\frac{1}{3}$、$\\frac{2}{3}$、$\\frac{3}{3}$、$\\frac{2}{3}$、$\\frac{1}{3}$、$\\frac{1}{4}$、$\\frac{2}{4}$、$\\frac{3}{4}$、$\\frac{4}{4}$、$\\frac{3}{4}$、$\\frac{2}{4}$、$\\frac{1}{4}$……中，$\\frac{5}{12}$是第几个分数？第2024个分数是多少？",
+        "options": [],
+        "answer": "126或140;$\\frac{2}{45}$",
+        "solution": "【分析】分母是1、2、3、……、n的分数分别有1、3、5、…、$(2n - 1)$个，再加上分母是12的第5个分数或分母是12的倒数第5个分数就可知$\\frac{5}{12}$是第几个分数。连续的奇数$1 + 3 + 5 + ...$加到多少接近2024，就可知道第2024个分数是多少。【详解】分母是1、2、3、…、11的分数共有：$1 + 3 + 5 + ... + 21 = (1 + 21) + (3 + 19) + (5 + 17) + (7 + 15) + (9 + 13) + 11 = 22 + 22 + 22 + 22 + 22 + 11 = 22 \\times 5 + 11 = 110 + 11 = 121$。因为按照规律，$\\frac{5}{12}$是分母为12的分数中的第5个或第19个，所以$\\frac{5}{12}$是第$121 + 5 = 126$(个)分数，或第$121 + 19 = 140$(个)分数。因为$1 + 3 + 5 + 7 + ... + 89 = (1 + 89) + (3 + 87) + ... (43 + 47) + 45 = 90 + 90 + ... + 90 + 45 = 90 \\times 22 + 45 = 1980 + 45 = 2025$。分母是45的分数有89个，所以需要再减去1个分母是45的分数$\\frac{1}{45}$，则第2024个分数是$\\frac{2}{45}$。答：$\\frac{5}{12}$是第126或140个分数，第2024个分数是$\\frac{2}{45}$。",
+        "knowledge_points": [
+          "分数排列的规律"
+        ],
+        "source": "25-26六年级上·全国·单元测试",
+        "merged_id": "unit8_294",
+        "original_id": "294"
+      },
+      {
+        "id": "295",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "下面是用牙签拼成的图形，观察下图的规律，按要求回答问题。![](images/c26938893456ecf51ecbb11739127aab038b02428eb5cf15ca7a6ed13bd2360e.jpg)①![](images/7762638b800b71f6098d6b0d0ea052094685a3be86f8aa551f388108a404865a.jpg)②![](images/f8d41bf0dfa6f856b6636fc5812539c285928963f65ad85198b031b7d0eebbf2.jpg)③![](images/f6d7a93d0a5c3fac9e2c9f8b7831c77b06ab46b291dc3ea25c015eab65d44625.jpg)④。(1)看图填写下表中的相关数据。(2)根据(1)中信息，你发现了什么规律？(3)按这种拼图方法拼出的第10个图中三角形和正方形各有多少个？共需要牙签多少根？",
+        "options": [],
+        "answer": "(1)见详解 (2)第$n$个图中，三角形有$n$个，正方形有$(2n + 1)$个，所用牙签总根数为$(8n + 4)$根。(3)三角形有10个，正方形有21个;牙签84根",
+        "solution": "【分析】（1）观察图形，三角形的个数：图①有1个三角形，图②有2个三角形，图③有3个三角形，图④有4个三角形，因此图序为$n$时，三角形个数为$n$;正方形的个数：图①有3个正方形，图②有5个正方形，图③有7个正方形，图④有9个正方形，通过观察可知正方形的个数比三角形个数的2倍还多1，因此图序为$n$时，正方形个数为$2n + 1$;牙签总数：图①有12根牙签，图②有20根牙签，图③有28根牙签，图④有36根牙签;(2)通过观察可知牙签的总数比三角形个数的8倍还多4，因此图序为$n$时，牙签的总数为$8n + 4$;(3)根据规律可以求出对应值。【详解】(1)见表格。(2)第$n$个图中，三角形有$n$个，正方形有$(2n + 1)$个，所用牙签总根数为$(8n + 4)$根。(3)$2 \\times 10 + 1 = 20 + 1 = 21$(个);$8 \\times 10 + 4 = 80 + 4 = 84$(根)。答：第10个图中三角形有10个，正方形有21个，共需要牙签84根。",
+        "knowledge_points": [
+          "图形的变化规律",
+          "用字母表示数",
+          "数量关系",
+          "乘加、乘减混合运算"
+        ],
+        "source": "25-26六年级上·全国·单元测试",
+        "merged_id": "unit8_295",
+        "original_id": "295"
+      },
+      {
+        "id": "296",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "现将连续自然数1至2009按下图中的方式排列成一个长方形队列，用正方形方框可以任意框出这个队列中的16个数，如图所示。(1)设任意一个这样的正方形框中的最小数为$n$，请求出这16个数的和。(用$n$的代数式表示)(2)在图中，要使一个正方形框出的16个数之和分别等于832、2000、2010是否可能？若不可能，请说明理由;若可能，请求出该正方形框出的16个数中的最小数和最大数。",
+        "options": [],
+        "answer": "(1)$16n + 192$ (2)832是正方形框内16个数的和，这16个数中的最小数是40，最大数是64。2000是正方形框内16个数的和，这16个数中的最小数是113，最大数是137。由于n应为自然数，2010不可能是正方形框内16个数的和。",
+        "solution": "【分析】（1）正方形框中的16个数构成一个$4 \\times 4$的矩阵，横向的数从左向右依次加1，纵向的数从上向下依次加7，据此找出框中16个数的和与最小数n的关系;(2)框中最小数是$n$，最大数是框中第四列第四行位置上的数，是$n + 3 + 7 \\times 3 = n + 24$，通过给定的和值，反向求解出$n$的值，根据$n$是自然数的特征判断和的可能值，求出16个数中的最小数和最大数。本题考查观察数阵数字排列规律，用含有字母的代数式表示框内16个数的和;并通过列方程求解n，判断能否框出满足条件的16个数。【详解】（1）假设左上角的数为$n$，那么框中的16个数依次为$n$，$n+ 1$，$n + 2$，$n + 3$，$n + 7$，$n + 8$，$n + 9$，$n + 10$，$n + 14$，$n + 15$，$n + 16$，$n + 17$，$n + 21$，$n + 22$，$n + 23$，$n + 24$。$[n + (n + 1) + (n + 2) + (n + 3)] + [(n + 7) + (n + 8) + (n + 9) + (n + 10)] + [(n + 14) + (n + 15) + (n + 16) + (n + 17)] + [(n + 21) + (n + 22) + (n + 23) + (n + 24)] = [n + (n + 1) + (n + 2) + (n + 3)] + [n + (n + 1) + (n + 2) + (n + 3) + 7 \\times 4] + [n + (n + 1) + (n + 2) + (n + 3) + 14 \\times 4] + [n + (n + 1) + (n + 2) + (n + 3) + 21 \\times 4] = (4n + 6) + [(4n + 6) + 28] + [(4n + 6) + 56] + [(4n + 6) + 84] = 16n + 192$。故16个数的和是$16n + 192$。(2)$16n + 192 = 832$，$16n + 192 - 192 = 832 - 192$，$16n = 640$，$16n \\div 16 = 640 \\div 16$，$n = 40$，$n + 24 = 40 + 24 = 64$。所以832是正方形框内16个数的和，这16个数中的最小数是40，最大数是64。$16n + 192 = 2000$，$16n + 192 - 192 = 2000 - 192$，$16n = 1808$，$16n \\div 16 = 1808 \\div 16$，$n = 113$，$n + 24 = 113 + 24 = 137$。所以2000是正方形框内16个数的和，这16个数中的最小数是113，最大数是137。$16n + 192 = 2010$，$16n + 192 - 192 = 2010 - 192$，$16n = 1818$，$16n \\div 16 = 1818 \\div 16$，$n = 113.625$。由于n应为自然数，所以2010不可能是正方形框内16个数的和。",
+        "knowledge_points": [
+          "列方程解含一个未知数的问题",
+          "用字母表示稍复杂的数量关系",
+          "数与形(探索规律)"
+        ],
+        "source": "25-26六年级上·全国·单元测试",
+        "merged_id": "unit8_296",
+        "original_id": "296"
+      },
+      {
+        "id": "297",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "我们把'$n$个相同的数$a$相乘'记为$a^n$，例如$2^3 = 2 \\times 2 \\times 2 = 8$。(1)请计算：$2^{6} = $，$5^{4} = $。(2)观察下列等式：$(x - 1) \\times (x + 1) = x^{2} - 1$，$(x - 1) \\times (x^{2} + x + 1) = x^{3} - 1$，$(x - 1) \\times (x^{3} + x^{2} + x + 1) = x^{4} - 1$。由以上规律，我们可以猜：$(x - 1) \\times (x^{n} + x^{n - 1} + ... + x + 1) = ()$。(3)计算：$3^{2011} + 3^{2010} + ... + 3 + 1$。",
+        "options": [],
+        "answer": "(1)64;625 (2)$x^{n + 1} - 1$ (3)$\\frac{3^{2012} - 1}{2}$",
+        "solution": "【分析】（1）根据'$n$个相同的数$a$相乘'记为$a^n$，将$2^6$写成6个2相乘，$5^4$写成4个5相乘，计算即可;(2)观察$(x - 1) \\times (x + 1) = x^{2} - 1$，$(x - 1) \\times (x^{2} + x + 1) = x^{3} - 1$，$(x - 1) \\times (x^{3} + x^{2} + x + 1) = x^{4} - 1$可知算式左边为$(x - 1)$乘$x^{n}$到1的连续次方数减1的和，结果为$(x^{n + 1} - 1)$;(3)利用(2)的规律，$3^{2011} + 3^{2010} + ... + 3 + 1$符合$x^n + x^{n-1} + ... + x + 1$的形式，其中$x = 3$，$n = 2011$，根据结论$(x - 1) \\times (x^{n} + x^{n - 1} + ... + x + 1) = x^{n + 1} - 1$，可得$(3 - 1)(3^{2011} + 3^{2010} + ... + 3 + 1) = 3^{2012} - 1$，根据等式的性质2，两边同时÷(3-1)，计算即可。关键是理解'$n$个相同的数$a$相乘'记为$a^n$，总结出第(2)题的规律，根据总结出的规律，运用等式的性质，求出第(3)题的结果。【详解】(1)$2^{6} = 2 \\times 2 \\times 2 \\times 2 \\times 2 = 64$，$5^{4} = 5 \\times 5 \\times 5 \\times 5 = 625$。(2)$(x - 1) \\times (x^{n} + x^{n - 1} + ... + x + 1) = x^{n + 1} - 1$。(3)根据分析，可得：(3-1)$(3^{2011} + 3^{2010} + ... + 3 + 1) = 3^{2012} - 1$。解：$(3 - 1)(3^{2011} + 3^{2010} + ... +3 + 1)\\div (3 - 1) = (3^{2012} - 1)\\div (3 - 1)$，$3^{2011} + 3^{2010} + ... + 3 + 1 = \\frac{3^{2012} - 1}{2}$。",
+        "knowledge_points": [
+          "算式的规律(整数)",
+          "用字母表示数",
+          "数量关系",
+          "等式的性质2"
+        ],
+        "source": "2025·重庆渝北·小升初真题",
+        "merged_id": "unit8_297",
+        "original_id": "297"
+      },
+      {
+        "id": "298",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "用小棒摆五边形，如下图所示。![](images/e44c8c0a858661f5e9995b27f7f74a0dbf23e4b6fb00865a6e302faab35c6c33.jpg)。(1)填表。(2)照这样摆120个五边形，需要多少根小棒？",
+        "options": [],
+        "answer": "(1)$5 + 4 + 4 + 4$;$4n + 1$ (2)481根",
+        "solution": "【分析】（1）观察图形可知，摆1个五边形需要5根小棒，摆2个五边形需要$(5 + 4)$根小棒，摆3个五边形需要$(5 + 4 + 4)$根小棒，摆4个五边形需要$(5 + 4 + 4 + 4)$根小棒……则摆n个五边形需要$[5 + 4 \\times (n - 1)]$根小棒，据此解答即可;(2)把$n = 120$代入(1)中所得出的规律中求值即可解答。【详解】(1)$5 + 4 \\times (n - 1) = 5 + 4n - 4 = (4n + 1)$根。填表如下：五边形个数1、2、3、4、…、n，小棒根数5、5+4、5+4+4、5+4+4+4、…、4n+1。(2)$4 \\times 120 + 1 = 480 + 1 = 481$(根)。答：需要481根小棒。",
+        "knowledge_points": [
+          "含有字母式子的化简与求值",
+          "用字母表示数",
+          "数量关系",
+          "数与形(探索规律)"
+        ],
+        "source": "24-25六年级下·江西九江·期末",
+        "merged_id": "unit8_298",
+        "original_id": "298"
+      },
+      {
+        "id": "299",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "如下图，像这样摆下去，摆n个六边形需要多少根小棒？用自己喜欢的方法表达解答的过程。![](images/fef2383bd59fb446c9d4e12384918adc3189c6e6effb044a8f23f2d2fdb32b82.jpg)",
+        "options": [],
+        "answer": "$1 + 5n$(或$5n + 1$)，解答见详解",
+        "solution": "【分析】观察可发现规律，1个六边形：$1 + 5 \\times 1$根小棒;2个六边形：$(1 + 5 \\times 2)$根小棒;3个六边形：$(1 + 5 \\times 3)$根小棒……按照这样的方法分析解答。【详解】1个六边形：$(1 + 5 \\times 1)$根小棒;2个六边形：$(1 + 5 \\times 2)$根小棒;3个六边形：$(1 + 5 \\times 3)$根小棒……按照这样的方法摆下去，摆n个六边形需要$(1 + 5n)$或$(5n + 1)$根小棒。",
+        "knowledge_points": [
+          "数与形(探索规律)"
+        ],
+        "source": "2025五年级上·海南海口·专题练习",
+        "merged_id": "unit8_299",
+        "original_id": "299"
+      },
+      {
+        "id": "300",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "一辆小轿车加满油后，记录了小轿车行驶路程和对应的剩余油量;制作出图象。(如图)剩余油量/升。![](images/7cfa5461ae672947a3f728bf0bc9656ce21b9565b3155f87d63b14900f283cf9.jpg)。(1)行驶130千米时，油箱剩余油量是( )升。(2)这辆车的耗油量是每千米多少升？",
+        "options": [],
+        "answer": "(1)21 (2)$0.06$升",
+        "solution": "【分析】（1）图象的横轴表示轿车行驶的千米数，纵轴表示剩余油量，找到汽车行驶130千米时在图像上的点，横向与纵轴交点就是剩余油量;(2)汽车行驶380千米，剩余油量6升，结合(1)可知行驶250千米，用掉15升油，据此用除法可求出行驶时每千米的耗油量。【详解】（1）由图象可知：汽车行驶130千米，剩余油量21升;(2)$(21 - 6) \\div (380 - 130) = 15 \\div 250 = 0.06$(升)。答：这辆车的耗油量是每千米0.06升。",
+        "knowledge_points": [
+          "统计图表的综合应用"
+        ],
+        "source": "2024·重庆石柱·小升初真题",
+        "merged_id": "unit8_300",
+        "original_id": "300"
+      },{
+        "id": "301",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "如图摆 1 个三角形用 3 根小棒，摆 2 个三角形用 5 根小棒，摆 3 个三角形用 7 根小棒。照这样，摆 12 个三角形用（）根小棒。![](images/04dbf72294c7a8b624a2c05303a504c08761d2879b0dfeebba06617c3acf3415.jpg)![](images/987ee6ee540bfb98f7f509b93628dd6a53787518eaa2448eb14eb9bba38dd367.jpg)![](images/babc6f427bd413ecfcfbff00c14178db9242d6c9d6e20cf77a15d066be515851.jpg)········",
+        "options": [
+          "A. 25",
+          "B. 24",
+          "C. 36"
+        ],
+        "answer": "A",
+        "solution": "【分析】根据题意, 摆 1 个三角形用 3 根小棒, 摆 2 个三角形用 5 根小棒, 摆 3 个三角形用 7 根小棒……发现: 每增加一个三角形, 小棒的数量增加 2 根, 据此找出规律, 并按规律解答。【详解】观察图形可知：摆 1 个三角形用 3 根小棒, $3 = 2 \\times 1 + 1$ ;摆 2 个三角形用 5 根小棒, $5 = 2 \\times 2 + 1$ ;摆 3 个三角形用 7 根小棒, $7 = 2 \\times 3 + 1$ ;规律：摆 $n$ 个三角形用 $(2n + 1)$ 根小棒。当 $n = 12$ 时$ 2 n + 1 \\ = 2 \\times 1 2 + 1 \\ = 2 4 + 1 \\ = 2 5 \\text {(根)}$摆 12 个三角形用 25 根小棒。故答案为: A",
+        "knowledge_points": [
+          "用字母表示稍复杂的数量关系",
+          "数与形（探索规律）",
+          "含有字母式子的化简与求值"
+        ],
+        "source": "2022·河北石家庄·小升初真题",
+        "merged_id": "unit8_301",
+        "original_id": "301"
+      },
+      {
+        "id": "302",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "与 $1 + 3 + 5 + 7 + 9 + 5$ $+ 3 + 1$ 表示结果相同的算式是 ( )。",
+        "options": [
+          "A. (5+3)²",
+          "B. $5^{2} + 3^{2}$",
+          "C. $5^{2}$"
+        ],
+        "answer": "B",
+        "solution": "【分析】$1 + 3 = 4 = 2^{2}, 1 + 3 + 5 = 9 = 3^{2}, 1 + 3 + 5 + 7 = 16$ $= 4^{2}, 1 + 3 + 5 + 7 + 9 = 25 = 5^{2}$ , 故将 $1 + 3 + 5 + 7 + 9 + 5 + 3$ +1 看作 $1 + 3 + 5 + 7 + 9$ 与 $5 + 3 + 1$ 相加, 即可求解。此题考查了算式的加法的计算技巧, 弄清算式中的计算规律是解决本题的关键。【详解】$1 + 3 + 5 + 7 + 9 + 5 + 3 + 1\\ = (1 + 3 + 5 + 7 + 9) + (5 + 3 + 1) \\ = 5 ^ {2} + 3 ^ {2}$与 $1 + 3 + 5 + 7 + 9 + 5 + 3 + 1$ 表示相同结果的算式是 $5^{2} + 3^{2}$ 。故答案为: B",
+        "knowledge_points": [
+          "算式的规律（整数）"
+        ],
+        "source": "24-25 六年级上·河南南阳·期末",
+        "merged_id": "unit8_302",
+        "original_id": "302"
+      },
+      {
+        "id": "303",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "$1 + 3 + 5 + 7 + \\dots +$ $29 + 31 =$ ( )",
+        "options": [
+          "A. 144",
+          "B. 256",
+          "C. 265",
+          "D. 326"
+        ],
+        "answer": "B",
+        "solution": "【分析】$1 + 3 = 4 = 2^{2}$ ， $1 + 3 + 5 = 9 = 3^{2}$ ， $1 + 3 + 5 + 7 = 4^{2}$ 可得规律：从 1 开始的连续奇数之和，等于数字个数的平方，据此解答即可。【详解】根据分析可得：$1 + 3 + 5 + 7 + \\dots + 29 + 31$ , 一共是 16 个数字的和。$ 1 + 3 + 5 + 7 + \\dots + 2 9 + 3 1 \\ = 1 6 ^ {2} \\ = 2 5 6$故答案为: B",
+        "knowledge_points": [
+          "数与形 (归纳递推)"
+        ],
+        "source": "24-25 六年级上·贵州黔东南·期末",
+        "merged_id": "unit8_303",
+        "original_id": "303"
+      },
+      {
+        "id": "304",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "平面上有5个三角形，这些三角形最多将平面划分成（）个部分。",
+        "options": [
+          "A. 45",
+          "B. 54",
+          "C. 62",
+          "D. 72"
+        ],
+        "answer": "C",
+        "solution": "【分析】平面本身是 1 部分, 一个三角形将平面分成三角形内外 2 部分, 即增加了 1 部分。两个三角形不相交时将平面分成 3 部分, 相交时, 交点越多, 分成的部分越多, 新增加的部分数与增加的交点数相同, 据此解答即可。本题考查找规律, 解答本题的关键是掌握增加部分数与三角形边相交时的交点个数有关系。【详解】1个三角形将平面分成2个部分2 个三角形不相交时将平面分成 3 部分, 相交时, 交点越多分成的部分越多 (见下图),![](images/0ed3d26019ec5fb7262837017d98d1d53d1434ee62dd8f2d8ce00061eb782132.jpg) 2个交点 增加2部分![](images/88ddbe3872246d7eafe590228aefe95254dc8eb49aba2c9e373ecadaf3726576.jpg) 3个交点增加3部分![](images/2f21b1545d0c7e1ce3bbeda3a514d1c0135a4ad3041fe8378990bbe41082e218.jpg) 4个交点 增加4部分![](images/8f45edc57e4e147cdeba5e42effa55be94b17661ac2cd7b62069d0dadae4389e.jpg) 5个交点 增加5部分![](images/78c9ec98836facd8a398bfad65df5cc14866b6eb6833ded15c2dfa98d1c6802d.jpg) 6个交点 增加6部分2 个三角形将平面最多分成 $1 + 1 + 6 = 7 + 1 = 8$ 个部分;由上图可以看出，新增加的部分数与增加的交点数相同，则再画第3个三角形时，应使每条边的交点尽量多;对于每个三角形，因为1条直线最多与三角形的两条边相交，所以第3个三角形的每条边最多与前面2个三角形的各两条边相交，共可产生 $3 \\times (2 \\times 2) = 3 \\times 4 = 12$ 个交点，即增加12个部分;因此3个三角形最多可以把平面分成 $1 + 1 + 6 + 12 = 8 + 12$ $= 20$ 个部分;由上面的分析, 当画第 4 个三角形时, 每条边最多与前面 3 个三角形的各两条边相交, 共可产生 $3 \\times (3 \\times 2) = 3 \\times 6 = 18$个交点, 即增加 18 个部分; 当画第 5 个三角形时, 每条边最多与前面 4 个三角形的各两条边相交, 共可产生$3 \\times (4 \\times 2) = 3 \\times 8 = 24$ 个交点, 即增加 24 个部分;所以 5 个三角形最多将平面划分成: $1 + 1 + 6 + 12 + 18 + 24$ $= 8 + 12 + 18 + 24 = 20 + 18 + 24 = 38 + 24 = 62$ 个部分故答案为: C",
+        "knowledge_points": [
+          "平面图形的分割",
+          "数与形（探索规律）"
+        ],
+        "source": "23-24六年级下·湖北武汉·开学考试",
+        "merged_id": "unit8_304",
+        "original_id": "304"
+      },
+      {
+        "id": "305",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "仔细观察, 第 20 个宝塔的最底层有 ( ) 个小三角形。![](images/59250c7d0cab2bdfdf9712fe1e16b04d6eaa3d3286c106ce246cee6cf2810bad.jpg)",
+        "options": [
+          "A. 39",
+          "B. 41",
+          "C. 43",
+          "D. 45"
+        ],
+        "answer": "A",
+        "solution": "【分析】观察图形可知，第1个图形最底层有1个小三角形，第2个图形最底层有 $(1 + 2)$ 个小三角形，第3个图形最底层有 $(1 + 2 \\times 2)$ 个小三角形，第4个图形最底层有 $(1 + 2 \\times 3)$ 个小三角形，第5个图形最底层有 $(1 + 2 \\times 4)$ 个小三角形……以此类推，每次宝塔的最底层增加2个小三角形，第 $n$ 个图形最底层有 $[1 + 2 \\times (n - 1)]$ 个小三角形, 最后求出 $n = 20$ 时含有字母式子的值, 据此解答。【详解】第 $\\mathrm{n}$ 个图形最底层小三角形的数量: $1 + 2 \\times (n - 1)\\= 1 + 2 n - 2 \\ = 2 n - 2 + 1 \\ = 2 n - (2 - 1) \\ = (2 n - 1) \\text {个}$当 $n = 20$ 时。$2 n - 1 \\ = 2 \\times 2 0 - 1 \\ = 4 0 - 1 \\ = 3 9 (\\text {个})$所以，第20个宝塔的最底层有39个小三角形。故答案为: A",
+        "knowledge_points": [
+          "数与形 (归纳递推)"
+        ],
+        "source": "24-25 六年级上·江西吉安·期末",
+        "merged_id": "unit8_305",
+        "original_id": "305"
+      },
+      {
+        "id": "306",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "照下图中的方法继续摆圆片，摆到第5个图案时，需要（）个圆片。![](images/c5291fdd6e563cb403d0a27a1bc722acc1f1a83924d5a861dee363a1d253a3a7.jpg)第1个![](images/5312f770f8c39c1583e599a848d4c54eb83b66b1b1593f45330d422648c370b7.jpg)第2个![](images/33ae5f015da1d6dd8fbc633817398ad997826ecf116488da6dcd70fe091cdb44.jpg)第3个··",
+        "options": [
+          "A. 11",
+          "B. 14",
+          "C. 17",
+          "D. 20"
+        ],
+        "answer": "C",
+        "solution": "【分析】第 1 个图案需要 5 个圆片，可以写成： $3 \\times 1 + 2$ ; 第 2 个图案需要 8 个圆片，可以写成： $3 \\times 2 + 2$ ; 第 3 个图案需要 11 个圆片，可以写成： $3 \\times 3 + 2$ ;由此可知，第 $n$ 个图案需要 $(3 n + 2)$ 个圆片，据此求出第 5 个图案需要圆片的个数，据此解答。【详解】根据分析可知, 第 $n$ 个图案需要 $(3 n + 2)$ 个圆片;第 5 个图案需要圆片:$ 3 \\times 5 + 2 \\ = 1 5 + 2 \\ = 1 7 \\text {(个)}$摆到第 5 个图案时, 需要 17 个圆片。故答案为: C",
+        "knowledge_points": [
+          "用字母表示数",
+          "数量关系",
+          "数与形（探索规律）"
+        ],
+        "source": "24-25 六年级上·广东东莞·期末",
+        "merged_id": "unit8_306",
+        "original_id": "306"
+      },
+      {
+        "id": "307",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "如图，图 1 有 1 个小正方形，图 2 有 3 个小正方形，图 3 有 6 个小正方形。按照规律，图 8 共有（）个小正方形。![](images/8ffb42b2732e112dce6482decd3a30d7a3172bd51aa76deb27ddd9d3e272a4c4.jpg)",
+        "options": [
+          "A. 28",
+          "B. 36",
+          "C. 45",
+          "D. 48"
+        ],
+        "answer": "B",
+        "solution": "【分析】通过分析找出规律：图1: 1个图2：1+2=3（个） 图 $3: 1 + 2 + 3 = 6$ （个） 图 $4: 1 + 2 + 3 + 4 = 10$ （个）··········图 $\\mathrm{n}:1 + 2 + 3 + \\dots \\dots +n$通过规律，将 8 代入得出结果。【详解】$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36$ （个）则图 8 共有 36 个小正方形。故答案为: B",
+        "knowledge_points": [
+          "有趣的找规律"
+        ],
+        "source": "2019 五年级·全国·竞赛",
+        "merged_id": "unit8_307",
+        "original_id": "307"
+      },
+      {
+        "id": "308",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "观察分析淘气跑步的时间和速度关系图，下面说法错误的是（）。![](images/3e73b9abce01af624b906c8780dfbc6fc111c37a8132adc4d6183c9099edd2c4.jpg)",
+        "options": [
+          "A. 在第 1 分钟内, 淘气跑步的速度从 0 米/分提高到 150 米/分",
+          "B. 从第 1 分钟到第 4 分钟, 淘气一共跑了 600 米",
+          "C. 从第 1 分钟到第 4 分钟, 淘气跑步的速度保持不变",
+          "D. 从第 5 分钟到第 6 分钟, 淘气跑步的速度在下降"
+        ],
+        "answer": "B",
+        "solution": "【分析】根据折线统计图，淘气在 1 分钟里面从速度 0 米/分提高了 150 米/分。然后开始匀速跑 3 分钟后，开始在后面的 2 分钟里面降速，从 150 米/分降到 0 米/分，据此分析每个选项。【详解】A. 根据折线统计图, 淘气跑步的速度在第 1 分钟内, 从 0 米/分提高到 150 米/分, 故正确; B. 从第 1 分钟到第 4 分钟, 淘气以 150 米/分的速度跑了 3 分钟, 根据路程 = 速度 $\\times$ 时间, 用 $3 \\times 150 = 450$ (米), 淘气一共跑了 450 米, 故错误;C. 从第 1 分钟到第 4 分钟, 折线统计图中这一段是一条直线, 即淘气跑步的速度保持不变, 故正确; D. 淘气跑步的过程中, 从第 4 分钟开始降速, 降到第 6 分钟停下来, 则第 5 分钟到第 6 分钟, 淘气跑步的速度在下降,故正确。故答案为: B",
+        "knowledge_points": [
+          "单式折线统计图",
+          "基础行程问题"
+        ],
+        "source": "2024 六年级上·全国·专题练习",
+        "merged_id": "unit8_308",
+        "original_id": "308"
+      },
+      {
+        "id": "309",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "与 $1 + 3 + 5 + 7 + 9 + 11 + 13 + 7 + 5 + 3 + 1$ 表示相同结果的算式是（）。",
+        "options": [
+          "A. $4^{2}$",
+          "B. $7^{2} - 4^{2}$",
+          "C. $7^{2}$",
+          "D. $7^{2} + 4^{2}$"
+        ],
+        "answer": "D",
+        "solution": "【分析】把 $1 + 3 + 5 + 7 + 9 + 11 + 13 + 7 + 5 + 3 + 1$ 中的 $1 + 3 + 5 + 7$ 和 $7 + 5 + 3 + 1$ 相加，用 $(1 + 7) \\cdot (3 + 5) \\cdot (5 + 3) \\cdot (7 + 1)$ ，变成 $8 \\times 4$ ，再加 $9 + 11 + 13$ ，然后与各个算式的结果进行比较，据此即可解答。【详解】$1 + 3 + 5 + 7 + 9 + 11 + 13 + 7 + 5 + 3 + 1\\ = (1 + 7) + (3 + 5) + (5 + 3) + (7 + 1) + (9 + 1 1) + 1 3 \\= 8 \\times 4 + 2 0 + 13 = 3 2 + 3 3 \\ = 6 5$A. $4^{2} = 16$ B. $7^{2} - 4^{2} = 49 - 16 = 33$ C. $7^{2} = 49$ D. $7^{2} + 4^{2} = 49 + 16 = 65$所以与 $1 + 3 + 5 + 7 + 9 + 11 + 13 + 7 + 5 + 3 + 1$ 表示相同结果的算式是 $7^{2} + 4^{2}$ 。故答案为: D",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "2024 六年级·全国·竞赛",
+        "merged_id": "unit8_309",
+        "original_id": "309"
+      },
+      {
+        "id": "310",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "用小棒摆三角形 $\\triangle$ ,摆 10 个三角形共需要 ( ) 根小棒。![](images/9742b27f7f49b46513bd3e9ee20fcb1a5c2a415b959f4d1255531e96434a220a.jpg)![](images/9c72551b7861d64582668dc19a3a54127258c68111f5d26c862a78f4852df08a.jpg)![](images/6c4200165431ea2bbee4748eb6c23a15a0759504c6dca92c5a99486caccc6a7e.jpg)![](images/684710c9c0519afa1304abb73328c9715bf163e906efd745d96b75cba730cc00.jpg)![](images/25cda18bccd1f395a25d420b0a2e35b0bc488308593c3118bd13b7ea2fcfd72b.jpg)",
+        "options": [
+          "A. 20",
+          "B. 21",
+          "C. 30"
+        ],
+        "answer": "B",
+        "solution": "【分析】每个三角形有 3 条边, 需要 3 根小棒, 摆 1 个三角形需要 3 根小棒, 后边每多 1 个三角形只需要 2 根小棒就可以了, 摆 10 个三角形需要再多 $(10 - 1) \\times 2$ 根小棒, 最后相加即可求解。【详解】$(10 - 1) \\times 2\\= 9 \\times 2\\=18 (根)$$18 + 3 = 21$ (根)摆 10 个三角形需要 21 根小棒。故答案为: B",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "23-24 三年级下·河北保定·期末",
+        "merged_id": "unit8_310",
+        "original_id": "310"
+      },
+      {
+        "id": "311",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "观察下列图形: 第 1 个图形有 6 根小棒, 第 2 个图形有 11 根小棒, 第 3 个图形有 16 根小棒……, 第 10 个图形有 ( ) 根小棒。![](images/be3a039fccb8940a4fd8d4548936ae7f112fbd1b345bd67215017049abc7824d.jpg)",
+        "options": [
+          "A. 45",
+          "B. 51",
+          "C. 60"
+        ],
+        "answer": "B",
+        "solution": "【分析】观察图形可知，如果以最左边的 1 根小棒为基础，第 1 个图形有 6 根小棒， $6 = 1 + 5$ ;第 2 个图形有 11 根小棒， $11 = 1 + 5 \\times 2$ ;第 3 个图形有 16 根小棒， $16 = 1 + 5 \\times 3$ 。由此可知：小棒的根数 $= 1 + 5 \\times$ 图形的序数，据此求出第 10 个图形有多少根小棒。【详解】通过分析可得: 小棒的根数 $= 1 + 5 \\times$ 图形的序数$1 + 5 \\times 10\\= 1 + 5 0 \\ = 5 1 \\text {(根)}$则第 10 个图形有 51 根小棒。故答案为: B",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "23-24 六年级上·广西玉林·期末",
+        "merged_id": "unit8_311",
+        "original_id": "311"
+      },
+      {
+        "id": "312",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "按照图中的规律,第8个图形中共有（）个点。![](images/61e4daaa9e7b806825d0dd7493f8502820446eee2a3ad68ea8eb9d5cb3e14a84.jpg)",
+        "options": [
+          "A. 16",
+          "B. 28",
+          "C. 32",
+          "D. 36"
+        ],
+        "answer": "D",
+        "solution": "【分析】观察图形可知:第1个图形有1个点;第2个图形有3个点， $3 = 1 + 2$第3个图形有6个点， $6 = 1 + 2 + 3$第4个图形有10个点， $10 = 1 + 2 + 3 + 4$··规律：第 n 个图形有点的个数为 $(1 + 2 + 3 + \\dots + n)$ 个。据此规律解答。【详解】规律：第 n 个图形有点的个数为 $(1 + 2 + 3 + \\dots + n)$ 个。当 $n = 8$ 时$ 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 \\ = (1 + 8) \\times 8 \\div 2 \\ = 9 \\times 8 \\div 2 \\ = 3 6 (\\text {个})$第 8 个图形中共有 36 个点。故答案为: D",
+        "knowledge_points": [
+          "用字母表示稍复杂的数量关系",
+          "数与形（探索规律）",
+          "含有字母式子的化简与求值"
+        ],
+        "source": "23-24 五年级上·河南洛阳·期中",
+        "merged_id": "unit8_312",
+        "original_id": "312"
+      },
+      {
+        "id": "313",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "观察下面的图形,想一想, 第 5 个图形有 ( ) 个黑点。![](images/f2979c715516f156ff4adaa3e616abe20665929e21afcebea2d998dcb1f859f5.jpg)",
+        "options": [
+          "A. 45",
+          "B. 46",
+          "C. 47",
+          "D. 48"
+        ],
+        "answer": "A",
+        "solution": "【分析】第 1 个图形有 5 个黑点; 可以分成 2 部分: 上部分是 1 个黑点, 下部分是 4 个黑点; 上部分可以写成: $2 \\times 1 - 1$ ; 下部分可以写成: $(1 + 1)^{2}$ ; 合起来是: $2 \\times 1 - 1 + (1 + 1)^{2}$ ;第2个图形有12个黑点;可以分成2部分：上部分是3个黑点，下部分是9个黑点;上部分可以写成： $2 \\times 2 - 1$ ;下部分可以写成： $(2 + 1)^{2}$ ;合起来是： $2 \\times 2 - 1 + (2 + 1)^{2}$ ;第3个图形有21个黑点;可以分成2部分：上部分是5个黑点，上部分可以写成： $2 \\times 3 - 1$ ;下部分是16个黑点，可以写成： $(3 + 1)^{2}$ ;合起来是： $2 \\times 3 - 1 + (3 + 1)^{2}$由此可知，第5个图形上部分黑点个数是： $2 \\times 5 - 1$ 个;下部分个数： $(5 + 1)^{2}$ ，合起来是： $2 \\times 5 - 1 + (5 + 1)^{2}$ 。据此解答。解答部分的关键是把这个图形分成2部分，再找出上部分规律和下部分规则，再合起来解答。【详解】根据分析可知，第5个图形小黑点的个数是：$2 \\times 5 - 1 + (5 + 1) ^ {2} \\ = 1 0 - 1 + 6 ^ {2} \\ = 9 + 3 6 \\ = 4 5 (\\text {个})$第5个图形有45个黑点。故答案为: A",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "24-25 六年级上·全国·单元测试",
+        "merged_id": "unit8_313",
+        "original_id": "313"
+      },
+      {
+        "id": "314",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "奇奇用大小相同的棋子按如图规律摆放图案。照这样摆下去，第2024个图案中有（）个棋子。![](images/72696ccfcc8276acefc34020031feceb2a18868b013a8c35fd7ae6b5e57506c1.jpg)第1个![](images/9efaa5c1acc9c052c36c5a606458631bd18a57ab402c869a1a2fb40daf41fb0e.jpg)第2个![](images/0645c0f13602e5ed2bd1550d0815246dae6a00f1bbac0e4fc118c16b65369976.jpg)第3个![](images/a0dddbf4c86f5cb82faa40d5a388227e03db9b79d5f2d2809e6b4021607716e9.jpg)第4个",
+        "options": [
+          "A. 6072",
+          "B. 6075",
+          "C. 6078"
+        ],
+        "answer": "B",
+        "solution": "【分析】观察图形可知，第一个图形的棋子数有 $(3 + 3 \\times 1)$ 个，第二个图形的棋子数有 $(3 + 3 \\times 2)$ 个，第三个图形有 $(3 + 3 \\times 3)$ 个，……可发现规律是：第 $n$ 个图形的棋子数有 $(3 + 3 n)$ 个。据此解答。【详解】$3 + 2024 \\times 3\\= 3 + 6 0 7 2\\= 6075（个）$ 所以，第 2024 个图案中有 6075 个棋子。故答案为: B。",
+        "knowledge_points": [
+          "图形的变化规律",
+          "数与形（探索规律）"
+        ],
+        "source": "23-24六年级下·湖南衡阳·期末",
+        "merged_id": "unit8_314",
+        "original_id": "314"
+      },
+      {
+        "id": "315",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "一组数3、5、7、9、…中，第n个数是（）。",
+        "options": [
+          "A. n",
+          "B. $2 \\mathrm{~n}$",
+          "C. $2 \\mathrm{n} + 1$"
+        ],
+        "answer": "C",
+        "solution": "【分析】观察这组数可知: 第 1 个数是 3, $3 = 1 \\times 2 + 1$ ; 第 2 个数是 5, $5 = 2 \\times 2 + 1$ ; 第 3 个数是 7, $7 = 3 \\times 2 + 1$ ; 第 4 个数是 9, $9 = 4 \\times 2 + 1$ ; 由此可知: 第 n 个数就是 $n \\times 2 + 1$ , 即 $2n + 1$ 。【详解】根据分析可得：一组数 3、5、7、9、…中，第 n 个数是 $2 n + 1$ 。",
+        "knowledge_points": [
+          "整数排列的规律"
+        ],
+        "source": "2022·甘肃陇南·小升初真题",
+        "merged_id": "unit8_315",
+        "original_id": "315"
+      },
+      {
+        "id": "316",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "将小正方体按如图的规律摆放：摆 1 个小正方体有 5 个面露在外面，摆 2 个小正方体有 8 个面露在外面，摆 6 个小正方体有（）个面露在外面，摆 n 个小正方体有（）个面露在外面。![](images/0375dc316397a66080c83510d9a489951242d62e3b0426ffa342ce3ba535c0a4.jpg)",
+        "options": [],
+        "answer": "20 3n + 2",
+        "solution": "【分析】从图中可知，摆 1 个、2 个、3 个小正方体分别有 5 个、8 个、11 个面露在外面，发现每增加一个小正方体，露在外面的面就会增加 3 个，据此找出规律，并按规律解答。【详解】观察图形可知:摆 1 个小正方体有 5 个面露在外面, $5 = 1 \\times 3 + 2$ ; 摆 2 个小正方体有 8 个面露在外面, $8 = 2 \\times 3 + 2$ ; 摆 3 个小正方体有 11 个面露在外面, $11 = 3 \\times 3 + 2$ ;········摆 6 个小正方体露在外面的面有:$ 3 \\times 6 + 2 \\ = 1 8 + 2 \\ = 2 0 (\\text {个})$······规律：摆 $n$ 个小正方体露在外面的面有 $(3 n + 2)$ 个。填空如下:摆 6 个小正方体有（20）个面露在外面，摆 n 个小正方体有 $(3n + 2)$ 个面露在外面。",
+        "knowledge_points": [
+          "用字母表示稍复杂的数量关系",
+          "数与形（探索规律）",
+          "含有字母式子的化简与求值"
+        ],
+        "source": "2024·浙江宁波·小升初真题",
+        "merged_id": "unit8_316",
+        "original_id": "316"
+      },
+      {
+        "id": "317",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如图中第 5 个正方形有( )个点。如果某个正方形每边上的点子数用 a 表示, 则这个正方形的点子总数可表示为( )。![](images/86a26beddfd3723a6c3c5e046386e57b48e22c95a43989d7db5d899c2d964cda.jpg)",
+        "options": [],
+        "answer": "20 4a-4",
+        "solution": "【分析】观察前三个图形，每个正方形每边的点子数分别是2、3、4，正方形中点子的总个数分别是4、8、12，发现后一个图形中的点子数比前一个图形中的多4个，据此得出点子的总数与每边上的点子数之间的关系：点子的总数 $=$ 每边点数 $\\times 4 - 4$ ，用含字母的式子表示数量关系，据此解答。【详解】（1）观察图形可知：第1个正方形每边有2个点，共有4个点， $4 = 2 \\times 4 - 4$ ;第2个正方形每边有3个点，共有8个点， $8 = 3 \\times 4 - 4$ ;第3个正方形每边有4个点，共有12个点， $12 = 4 \\times 4 - 4$ ;·······规律：第（a-1）个正方形每边有a个点，共有 $(4a - 4)$ 个点。第5个正方形，每边的点数是： $5 + 1 = 6$ （个）;当 $a = 6$ 时$4 a - 4 \\ = 4 \\times 6 - 4 \\ = 2 4 - 4 \\ = 2 0 (\\text {个})$填空如下：第 5 个正方形有（20）个点。如果某个正方形每边上的点子数用 a 表示，则这个正方形的点子总数可表示为 $(4a - 4)$ 。",
+        "knowledge_points": [
+          "用字母表示稍复杂的数量关系",
+          "数与形（探索规律）",
+          "含有字母式子的化简与求值"
+        ],
+        "source": "2024·广东广州·小升初真题",
+        "merged_id": "unit8_317",
+        "original_id": "317"
+      },
+      {
+        "id": "318",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "下面图形是由若下个圆按规律组成的，第6个图形中有____个圆，第n个图形中有____个圆。第1个![](images/16525c1251449bca5f8bacc585c0203a04136d7b585fdf250e08c58769f4f0ab.jpg)第2个![](images/4845547f60824021f458a6dbd3cadb6635fd5d1ded0bc76537e89238f2e35015.jpg)第3个![](images/2bf406c51a714c9661e6a1668e4c9356c574cf14879590db00b7799ddf414238.jpg)第4个![](images/4300b3d1d117c7eb06e113e235e9749d73859cf841e4fd7e474aca9c161f3981.jpg)",
+        "options": [],
+        "answer": "37 (n^{2} + 1)",
+        "solution": "【分析】先观察前几个图形中圆的数量，尝试找出数量与图形序号之间的联系，进而得出第6个图形中圆的数量以及第n个图形中圆的数量的表达式。【详解】第 1 个图形: 有 $1 + 1 = 2$ 个圆, 这里前面的 1 可以看作是 $1^{2}$ , 即 $1^{2} + 1 = 2$ 。第2个图形：有 $4 + 1 = 5$ 个圆，其中4恰好是 $2^{2}$ ，也就是 $2^{2} + 1 = 5$ 。第3个图形：有 $9 + 1 = 10$ 个圆，9是 $3^{2}$ ，即 $3^{2} + 1 = 10$ 。第4个图形：有 $16 + 1 = 17$ 个圆，16是 $4^{2}$ ，即 $4^{2} + 1 = 17$ 。·······第 6 个图形圆的个数有: $6^{2} + 1 = 37$ (个)发现规律: 第 $\\mathrm{n}$ 个图形中圆的数量是 $(\\mathrm{n}^{2} + 1)$ 个。填空如下:所以第 6 个图形中有 (37) 个圆, 第 n 个图形中有 $(n^{2}+1)$ 个圆。",
+        "knowledge_points": [
+          "用字母表示稍复杂的数量关系",
+          "数与形（探索规律）"
+        ],
+        "source": "2024·广东广州·小升初真题",
+        "merged_id": "unit8_318",
+        "original_id": "318"
+      },
+      {
+        "id": "319",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "1 张餐桌可坐 4 人, 2张餐桌拼在一起可坐 6 人, 3 张餐桌拼在一起可坐 8 人, 如图。![](images/eee794e2e35a96430ea253962adb4d64e3519d75329fa0d3bf1dbd818c9ed9ff.jpg)![](images/c6758a086096041da61929e544bf7d6ff05beeb40b155a1b443db3bcbd105fd0.jpg)![](images/d96c4ec0fc455821c2037a5af209718d70c13979ee43b8095c52282d4e54e25a.jpg)(1)观察思考，发现规律：每增加 1 张桌子，可坐人数就增加( )人。 (2)如果 10 张餐桌拼在一起, 一共可以坐( )人。 (3)按这样拼下去，m 张餐桌可坐( )人。（用含有字母的式子表示）",
+        "options": [],
+        "answer": "(1)2 (2)22 (3) 2m + 2",
+        "solution": "【分析】（1）一张餐桌可坐 4 人，2 张餐桌拼在一起可坐 6 人，增加了 2 人，3 张餐桌拼在一起可坐 8 人，增加了 2 人，因此每增加 1 张桌子，可坐人数就增加 2 人。(2) 由 (1) 可知, 每增加 1 张桌子, 可坐人数就增加 2 人,如果 10 张餐桌拼在一起, 也就是增加了 $(10 - 1 = 9)$ 张餐桌, 那么增加的人数为 $(2 \\times 9 = 18)$ 人, 一共可以坐的有 $(4 + 18)$ 人。 (3) 观察三个图形得到 1 张正方形桌子可坐 4 人, 2 张正方形桌子可坐 $(4 + 2 \\times 1)$ 人, 则每增加一张桌子就可多坐两个人, 于是得到 $\\mathrm{m}$ 张正方形桌子可坐 $[4 + 2 \\times (m - 1))$ 人, 据此解答。【详解】（1）发现规律：每增加 1 张桌子，可坐人数就增加 2 人。(2) 10 张桌子拼在一起, 相当于增加了 $10 - 1 = 9$ (张)桌子。$ 4 + 2 \\times 9 \\ = 4 + 1 8 \\ = 2 2 \\text {(人)}$因此如果 10 张餐桌拼在一起，一共可以坐 22 人。(3) $4 + 2 \\times (m- 1)\\ = 4 + 2 m - 2 \\ = (2 m + 2) \\text {人}$因此按这样拼下去，m 张餐桌可坐（2m+2）人。",
+        "knowledge_points": [
+          "用字母表示数",
+          "数量关系",
+          "数与形（探索规律）",
+          "乘加、乘减混合运算",
+          "含有字母式子的化简与求值"
+        ],
+        "source": "2024·重庆垫江·小升初真题",
+        "merged_id": "unit8_319",
+        "original_id": "319"
+      },
+      {
+        "id": "320",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "用边长为 1 厘米的正方形纸片分别摆出下面的图形，按这样摆下去，第6个图形的周长是( )厘米。第1个![](images/065f525a6ba322bd312240b0ac8bba3a80b742e422569f1b4cb2fa2c5736a864.jpg) 第2个![](images/0496178350d464e609e082ea6050cba8976c00b177778d58dbd754e55f47d25d.jpg) 第3个![](images/146863c9ac14aba80063e0f4347143165b2314c167dff754c672af31c3f62674.jpg) ···········",
+        "options": [],
+        "answer": "34",
+        "solution": "【分析】由图可知, $\\square$ 第 1 个图形的周长是 4 厘米, $\\square$ 第2个图形的周长是（4+6）厘米，第3个图形的周长是（4+6×2）厘米，……，以此类推，每次周长增加6厘米，那么第n个图形的周长是[4+6×（n-1）]厘米，求出 n = 6时含有字母式子的值，据此解答。【详解】第 $\\mathrm{n}$ 个图形的周长： $4 + 6\\times (n-1)\\ = 4 + 6 n - 6 \\ = 6 n - 6 + 4 \\ = 6 n - (6 - 4) \\ = (6 n- 2) \\text {厘 米}$当 $n = 6$ 时。$ 6 n - 2 \\ = 6 \\times 6 - 2 \\ = 3 6 - 2 \\ = 3 4 \\text {(厘 米)}$所以，第 6 个图形的周长是 34 厘米。",
+        "knowledge_points": [
+          "数与形（归纳递推）"
+        ],
+        "source": "2024·重庆涪陵·小升初真题",
+        "merged_id": "unit8_320",
+        "original_id": "320"
+      },
+      {
+        "id": "321",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如图：![](images/c255a3080c5a634def9c83a7799effbef17ded99348d0bb892419df2e32717c0.jpg)60个这样的小梯形拼出的图形是( )，周长是( )cm。",
+        "options": [],
+        "answer": "平行四边形 182",
+        "solution": "【分析】看图可知，1个小梯形是梯形，周长 $= 1 \\times$ 上底 $+ 1 \\times$ 下底 $+ 1 \\times 2$ ;2个小梯形拼出的是平行四边形，周长 $= 2 \\times$ 上底 $+ 2 \\times$ 下底 $+ 1 \\times 2$ ;3个小梯形拼出的是梯形，周长 $= 3 \\times$ 上底 $+ 3 \\times$ 下底 $+ 1 \\times 2$ ;4个小梯形拼出的是平行四边形，周长 $= 4 \\times$ 上底 $+ 4 \\times$ 下底 $+ 1 \\times 2$ ……由此可知，单数个小梯形拼出的是梯形，双数个小梯形拼出的是平行四边形，周长 $=$ 几个梯形就用几 $\\times$ 上底 + 几个梯形就用几 $\\times$ 下底 $+ 1 \\times 2$ ;据此解答。【详解】根据分析：60 是双数, 拼出的图形是平行四边形$6 0 \\times 1 + 6 0 \\times 2 + 1 \\times 2 \\ = 6 0 + 1 2 0 + 2 \\ = 1 8 2 (c m)$60 个这样的小梯形拼出的图形是平行四边形, 周长是 $182 (cm)$ 。",
+        "knowledge_points": [
+          "用同样的图形拼接",
+          "数与形（归纳递推）",
+          "梯形的周长",
+          "平行四边形的周长"
+        ],
+        "source": "24-25六年级上·重庆长寿·期末",
+        "merged_id": "unit8_321",
+        "original_id": "321"
+      },
+      {
+        "id": "322",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "观察下图的图形规律, 第 5 个图形是由( )个小正方形拼成的。![](images/789414f4c3dce8bda775e5fa1c30f319799b938b960c5b59e1038381861109fa.jpg)",
+        "options": [],
+        "answer": "25",
+        "solution": "【分析】观察可知，第1幅图由1个小正方形拼成，第2幅图由 $(1 + 3)$ 个小正方形拼成，第3幅图由 $(1 + 3 + 5)$ 个小正方形拼成，可知第4幅图由 $(1 + 3 + 5 + 7)$ 个小正方形拼成，第5幅图由 $(1 + 3 + 5 + 7 + 9)$ 个小正方形拼成。【详解】$1 + 3 + 5 + 7 + 9 = 25$ （个）第 5 个图形是由 25 个小正方形拼成的。",
+        "knowledge_points": [
+          "图形的变化规律",
+          "数与形（探索规律）"
+        ],
+        "source": "24-25 六年级上·青海西宁·期末",
+        "merged_id": "unit8_322",
+        "original_id": "322"
+      },
+      {
+        "id": "323",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "相同的小棒按如图所示方式摆图形。摆第6个图形需要( )根小棒，摆n个需要( )根小棒。![](images/a4f26c047600ec0f67a98718772223c854320f39b31f4b99ef158de453893102.jpg)",
+        "options": [],
+        "answer": "31 5n + 1",
+        "solution": "【分析】观察图形可知，摆第 1 个图形需要 6 根小棒，摆第 2 个图形需要 11 根小棒，摆第 3 个图形需要 16 根小棒，…发现：每增加一个六边形，小棒的数量增加 5 根，据此找到规律，按规律解答。【详解】观察图形可知:摆第 1 个图形需要 6 根小棒, $6 = 1 \\times 5 + 1$ ;摆第 2 个图形需要 11 根小棒, $11 = 2 \\times 5 + 1$ ;摆第 3 个图形需要 16 根小棒, $16 = 3 \\times 5 + 1$ ;·······规律：摆第 $\\mathrm{n}$ 个图形需要（5n+1）根小棒;当 $n = 6$ 时$5 n + 1 \\ = 5 \\times 6 + 1 \\ = 3 0 + 1 \\ = 3 1 \\text {(根)}$摆第 6 个图形需要 (31) 根小棒, 摆 n 个 需要 $(5 \\mathrm{n} +$ 1) 根小棒。",
+        "knowledge_points": [
+          "数与形（探索规律）",
+          "含有字母式子的化简与求值"
+        ],
+        "source": "2024·湖北襄阳·小升初真题",
+        "merged_id": "unit8_323",
+        "original_id": "323"
+      },
+      {
+        "id": "324",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "用火柴棒按下图方式搭三角形：搭5个三角形需要火柴棒()根，23根小棒可以搭()个三角形。![](images/23269eac9c7f3f22202dffbcadfe3a1ffe77adf2fd624ace57f46bd41e4551d3.jpg)![](images/b393a8a274ba9c093af5a1a620e74477a7062ae53becc2980f1997e8ec2c44e0.jpg)![](images/2c32b2fc2a5e317cf3ecf507ffc36cc95b1360cf461b6c0f0037b48624654e3f.jpg)![](images/bf202434a2dccdc8f1ac2c849c6ef0fb4f91baca19a6efec5d997b970350fcbe.jpg)",
+        "options": [],
+        "answer": "11 11",
+        "solution": "【分析】观察图形可知：搭1个三角形要3根火柴棒， $2\\times 1 + 1 = 2 + 1 = 3$ （根）; 搭2个三角形要5根火柴棒， $2\\times 2 + 1 = 4 + 1 = 5$ （根）; 搭 3 个三角形要 7 根火柴棒, $2 \\times 3 + 1 = 6 + 1 = 7$ (根); 搭 4 个三角形要 9 根火柴棒, $2 \\times 4 + 1 = 8 + 1 = 9$ (根) ;按此规律搭下去, 搭 n 个三角形要 $2 \\times \\mathrm{n} + 1 = (2 \\mathrm{n} + 1)$ 根火柴棒; 据此解答。【详解】根据分析：规律：搭 n 个三角形要 $(2n + 1)$ 根火柴棒。当 $n = 5$ 时, $2 \\times 5 + 1 = 10 + 1 = 11$ (根)解：设23根小棒可以搭 $n$ 个三角形。$2 n + 1 = 2 3$$2 n + 1 - 1 = 2 3 - 1 \\ 2 n = 2 2 \\ 2 n \\div 2 = 2 2 \\div 2 \\ n = 1 1$搭 5 个三角形需要火柴棒 11 根, 23 根小棒可以搭 11 个三角形。",
+        "knowledge_points": [
+          "应用等式的性质 1 和 2 解方程",
+          "数与形（探索规律）",
+          "列方程解含一个未知数的问题",
+          "含有字母式子的化简与求值"
+        ],
+        "source": "2023·广东韶关·小升初真题",
+        "merged_id": "unit8_324",
+        "original_id": "324"
+      },
+      {
+        "id": "325",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如图, 4 根火柴可拼成一个正方形, 7 根火柴可拼成两个正方形……照这样的规律, 拼 100 个正方形需要( )根火柴, 1000 根火柴最多可以拼成( )个正方形。![](images/8bb61b333a96f20ce66cd22940ff608836f65f3389363248e78652f3a59eec9f.jpg)![](images/554f348434c5b384c8a80131fe2e9e1b86deb9b20c6e832f10d80cc6a6e0135b.jpg)![](images/06013928daa102f8f36c77afaaad5407c83595fd02796601f9757533db9679c3.jpg)",
+        "options": [],
+        "answer": "301 333",
+        "solution": "【分析】观察图形可知：拼1个正方形需要4根火柴， $4 = 3\\times 1 + 1$拼2个正方形需要7根火柴， $7 = 3 \\times 2 + 1$ ;拼 3 个正方形需要 10 根火柴, $10 = 3 \\times 3 + 1$ ;规律：拼 n 个正方形需要（ $3n + 1$ ）根火柴，据此规律解答。【详解】规律：拼 n 个正方形需要 $(3n + 1)$ 根火柴。当 $n = 100$ 时$3 n + 1 \\ = 3 \\times 1 0 0 + 1 \\ = 3 0 0 + 1 \\ = 3 0 1 \\text {(根)}$$3 n + 1 = 1 0 0 0$解: $3 \\mathrm{n} = 1000 - 1\\3 n = 9 9 9\\n = 9 9 9 \\div 3 \\ n = 3 3 3$照这样的规律，拼 100 个正方形需要（301）根火柴，1000根火柴最多可以拼成（333）个正方形。",
+        "knowledge_points": [
+          "应用等式的性质 1 和 2 解方程",
+          "用字母表示稍复杂的数量关系",
+          "数与形（探索规律）",
+          "含有字母式子的化简与求值"
+        ],
+        "source": "24-25 六年级上·河南新乡·期末",
+        "merged_id": "unit8_325",
+        "original_id": "325"
+      },
+      {
+        "id": "326",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "图中第 5 个正方形有____个点。如果某个正方形每边上的点数用 $n$ 表示,那么这个正方形点子总数可以表示为_____。![](images/89d4cadf7ef27ab641f640bcae7a1228099d2ae0adc0ae014ee4f87f63ec6ce5.jpg)![](images/365092bce9b46c46f405c70adae0f089bd9e9cbc3f30e293dc66d2b220619bed.jpg)![](images/5fe3ad679fe000993ea38bf4f088f5d3aa5ccad1295b5e1ac5a71dd04aac7133.jpg)![](images/b10228150ff483b534e21b512fa1c7d83cf68fd515018a5f7634a097178db89d.jpg)",
+        "options": [],
+        "answer": "20 4n - 4",
+        "solution": "【分析】注意观察前三个图形中圆点的个数可以发现分别为: 4, 8, 12, 后一个图形中的圆点个数比前一个图形中圆点多 4 , 所以可得正方形点子总数 S 与每边上的点数 n 之间的关系用字母表示为: $S = 4(n - 1) = 4n - 4$ , 据此解答即可。【详解】第一个图形有 $4 = 4 \\times (2 - 1)$ 个点子;第二个图形有 $8 = 4 \\times (3 - 1)$ 个点子;第三个图形有 $12 = 4 \\times (4 - 1)$ 个点子;第四个图形有 $4 \\times (5 - 1) = 16$ 个点子;第五个图形有 $4 \\times (6 - 1) = 20$ 个点子，则如果某个正方形每边上的点数用 $n$ 表示, 这个正方形的点子总数为 $4(n - 1) = 4n - 4$ 。因此，图中第5个正方形有20个点。如果某个正方形每边上的点数用 $\\mathrm{n}$ 表示，那么这个正方形点子总数可以表示为 $4 \\mathrm{n} - 4$ 。",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "24-25 六年级下·海南海口·单元测试",
+        "merged_id": "unit8_326",
+        "original_id": "326"
+      },
+      {
+        "id": "327",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "照这样的规律, 第 6 个图形有( )个涂色小正方形, 第 $n$ 个图形有( )个涂色小正方形。![](images/e76b832b880a9746782808e5838383f5839137967d456c63f651d14f97b64db5.jpg)![](images/9c317297536c591c189c7753b88cbf41b3d38ed5cec03ab8f6e4ff709b1725f6.jpg)![](images/9f997e8bc757d7918bed77d64ccc0d3244b3e78475bf28c74b3cc6a4cf084822.jpg)$2 ^ {2} - 1 ^ {2} = 3 \\quad 3 ^ {2} - 2 ^ {2} = 5 \\quad 4 ^ {2} - 3 ^ {2} = 7$",
+        "options": [],
+        "answer": "13 (n + 1)^{2} - n^{2}",
+        "solution": "【分析】根据题中给的规律，第几个图形的涂色数量，就是两个相邻的数的平方相减即可;第几个图形，第一个数就是几加 1 的平方，第二个数就是几的平方，即第 n 个图形的涂色小正方形的个数： $(n + 1)^{2} - n^{2}$ ，据此即可填空。【详解】第6个图形：$7 ^ {2} - 6 ^ {2} \\ = 4 9 - 3 6 \\ = 1 3 (\\text {个})$第 6 个图形有 13 个涂色小正方形, 第 $n$ 个图形有 $(n + 1)^{2} - n^{2}$ 个涂色小正方形。",
+        "knowledge_points": [
+          "用字母表示数",
+          "数量关系",
+          "数与形（探索规律）"
+        ],
+        "source": "24-25 六年级上·湖北孝感·期末",
+        "merged_id": "unit8_327",
+        "original_id": "327"
+      },
+      {
+        "id": "328",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "我国宋代数学家杨辉在公元1261年撰写了《详解九章算法》，他在这本著作中给出了一个由数构成的三角形图，我们称之为'杨辉三角'。仔细观察规律，从上到下，第2024行有( )个数字，倒数第2个数字是( )。![](images/80ccf54a405f364f8f8f44a2dcea66f9.png)",
+        "options": [],
+        "answer": "2024 2023",
+        "solution": "【分析】观察'杨辉三角'可知，第几行就有几个数，左右两边都是1，从第三行开始，中间的每一个数都等于它上方相邻的两个数字之和，如图 $1 \\quad 5 \\quad 10 \\quad 5$ ，第2024行的倒数第2个数字上方右侧的数是1，据此确定第2024行的倒数第2个数字。【详解】$2024 - 1 = 2023$仔细观察规律，从上到下，第2024行有2024个数字，倒数第2个数字是2023。",
+        "knowledge_points": [
+          "整数排列的规律"
+        ],
+        "source": "24-25六年级上·山东济南·期末",
+        "merged_id": "unit8_328",
+        "original_id": "328"
+      },
+      {
+        "id": "329",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "看图填空。![](images/f9bd87bd4f7f5ed4aa49db8cc4de2e07227abbe94dd057534dd2f1a30632f4d2.jpg)![](images/6fa04b68b6b191527539ff8d6d805d1fb21b72bf0014ebf5dd62c6887c37ab94.jpg)![](images/b6dd30385e7ed0ced2a49983adf0ccbf0dc4d0c3548eada21e2be362d7c203c3.jpg)黑色：1块 2块 3块白色：8块 13块 18块照这样画下去，第6个图形中黑色有( )块，白色有( )块。第n个图形中白色有( )块。（用含n的式子表示）",
+        "options": [],
+        "answer": "6 33 5n + 3",
+        "solution": "【分析】根据图形可知，黑色：第一个图形有 1 块;第二个图形有 2 块;第三个图形有 3 块;所以第 n 个图形，有 n 块。白色：第一个图形有 8 块，第二个图形有 13 块，第三个图形有 18 块，由此可知，后一个图形比前一个图形多 5 块;第一个图形有 8 块, 可以写成: $5 \\times 1 + 3$ ;第二个图形有 13 块, 可以写成: $5 \\times 2 + 3$ ;第三个图形有 18 块, 可以写成: $5 \\times 3 + 3$ ;由此可知，第 $n$ 个图形有 $(5 n + 3)$ 块，据此解答。【详解】根据分析可知：黑色：第6个图形中黑色有6块。白色: 第 $n$ 个图形: $5 \\times n + 3 = (5 n + 3)$ 块。第6个图形白色:$ 5 \\times 6 + 3 \\ = 3 0 + 3 \\ = 3 3 \\text {(块)}$第 6 个图形中黑色有 6 块, 白色有 33 块。第 n 个图形中白色有 $(5 n + 3)$ 块。",
+        "knowledge_points": [
+          "用字母表示数",
+          "数量关系",
+          "数与形（探索规律）"
+        ],
+        "source": "24-25六年级下·江西南昌·开学考试",
+        "merged_id": "unit8_329",
+        "original_id": "329"
+      },
+      {
+        "id": "330",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "观察下面点阵图的规律, 根据规律填一填。![](images/650a3835583173713ae3aa83d3ae43851eaa9361c79aea6d34ebd6123bb9ca97.jpg) 图1![](images/bbb0524ce7d72e3e8ffc1055c00513c313c9cdfbcd0a7de1ebfaa0837e80955e.jpg) 图2![](images/205bd5eddc527e3f29f29518aa98d7d22540ec202187655aac2dff0dfd410bfa.jpg) 图3![](images/1e0ef641569a0e95270f5ead1971fa24d0db85e96117d4434d4eb819ef1b64ec.jpg) 图4(1)按照规律在第四幅图中应该画( )个圆点。 (2)按照这个规律还可以知道第 $\\mathrm{n}$ 个图形的点阵中, 一共应画( )个圆点。",
+        "options": [],
+        "answer": "(1)15 (2) 3 n + 3",
+        "solution": "【分析】图1有6个圆点，图2有9个圆点，图3有12个圆点，由此可知，后一个图形比前一个图形增加3个圆点。图1画6个圆点，可以写成 $3 \\times 1 + 3$ ; 图2画9个圆点，可以写成： $3 \\times 2 + 3$ 图3画12个圆点，可以写成： $3\\times 3 + 3$ ······由此可知, 图 n 画 $(3n + 3)$ 个圆点, 由此求出图 4 画多少个圆点, 据此解答。【详解】（1）根据分析可知，图 n 画 $(3n + 3)$ 个圆点; $3n + 3$图4有:$ 3 \\times 4 + 3 \\ = 1 2 + 3 \\ = 1 5 (\\text {个})$按照规律在第四幅图中应该画15个圆点。(2) 根据分析可知, 按照这个规律还可以知道第 $\\mathrm{n}$ 个图形的点阵中, 一共应画 (3n+3) 个圆点。",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "24-25 六年级下·河南郑州·开学考试",
+        "merged_id": "unit8_330",
+        "original_id": "330"
+      },
+      {
+        "id": "331",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "明明是个善于观察,乐于思考的好孩子。他通过数形结合 (如图), 发现了求两个连续自然数平方差的规律。请你根据明明发现的规律,直接写出下面算式的结果: $6^{2} - 5^{2} = i( \\quad)$ ; $2025^{2} - 2024^{2} = ( \\quad)$ 。![](images/6c313670afd43e57876d74b046653f9d704d4e34259d6b2928482d88ca04f5c3.jpg)",
+        "options": [],
+        "answer": "11 4049",
+        "solution": "【分析】由图可知: 每个图形阴影部分的面积 = 大正方形面积 - 空白部分正方形面积。假设大正方形边长为 $a$ ( $a$ 为整数), 空白正方形边长则为 $(a - 1)$ , 阴影部分面积 $S = a^{2} - (a - 1)^{2}$ , 即两个相邻自然数的平方差; 由算式可知:两个相邻自然数的平方差其结果为两个相邻自然数的和,据此解答。【详解】$6^{2} - 5^{2} = 16 + 5 = 11; 2025^{2} - 2024^{2} = 12025 + 2024 =$ 4049。",
+        "knowledge_points": [
+          "数与形 (归纳递推)"
+        ],
+        "source": "24-25 六年级上·山西晋中·期末",
+        "merged_id": "unit8_331",
+        "original_id": "331"
+      },
+      {
+        "id": "332",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "我们用小棒摆正方形（如下图），按照这样的规律，摆 10 个正方形需要( )根小棒;49 根小棒可以摆( )个这样的正方形。![](images/1999125bafc7ee9511430ee7c3662c359fe9ca1aaadccf5371ba092909290f4b.jpg)",
+        "options": [],
+        "answer": "31 16",
+        "solution": "【分析】摆一个正方形需要 4 个小棒, 可以写成: $3 \\times 1 + 1$ ;摆二个正方形需要 7 个小棒, 可以写成: $3 \\times 2 + 1$ ; 摆 3 个正方形需要 10 个小棒, 可以写成: $3 \\times 3 + 1 \\dots$ . 摆 n 个正方形需要 $3n + 1$ 个小棒, 由此求出当 $n = 10$ 时, 求出需要小棒的数量, 以及 49 根小棒可以摆多少个正方形。【详解】根据分析可知，摆10和正方形需要小棒：$ 3 \\times 1 0 + 1 \\ = 3 0 + 1 \\ = 3 1 \\text {(根)}$$ (4 9 - 1) \\div 3 \\ = 4 8 \\div 3 \\ = 1 6 (\\text {个})$摆 10 个正方形需要 31 根小棒; 49 根小棒可以摆 16 个这样的正方形。",
+        "knowledge_points": [
+          "图形的变化规律"
+        ],
+        "source": "24-25 五年级上·广东韶关·期末",
+        "merged_id": "unit8_332",
+        "original_id": "332"
+      },
+      {
+        "id": "333",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如下图, 按规律继续画下去, 第 (6) 个图形共有( )个 □ 。(1)![](images/14bd7571df752f54001e78723a1093ca7273e8c99d91d3e7421da47959cbeb94.jpg) (2）![](images/2e07467fae160dc4281aa11bd3b2bd992da3108e7ade45ce27b820695e93d7fa.jpg) (3）![](images/8da5bf889e0995de33bcc8062d378986a5dd8d350d0595c369bf155714d2abf7.jpg) (4)![](images/4b891938d08ed7228ddf466aa7d2cf2535888e90d96e4facf6f7b48e24cf5ede.jpg)",
+        "options": [],
+        "answer": "25",
+        "solution": "【分析】(1)![](images/b29722fe5a15ab976979b842ab966de34f2a9e1005c72c2bc21bd6737cef9627.jpg) (2）![](images/82fd34f109f73107e9916f489daca2eeedb24d07360b6a0977c1263b4adadae6.jpg) (3）![](images/501ded526e43cbc00b2fbbbc99a21c342e2996f0ed49da0cde7fa0cead99f667.jpg) (4)![](images/a2e94f688a4f4d22ce8a1e309b3ad41515074ca96c6e7f06259d9635ba2cf8fe.jpg) 从图中可知：每一个图形，中间对角线处，分别有2个、3个、4个、5个，即的个数 $=$ 序号 $+1$ ，第（6）个图形中间对角线处就有 $6 + 1 = 7$ 个;第（1）个图形有2个 $\\square$ ，第（2）个图形有（ $1 + 3 + 1$ ）个 $\\square$ ，第(3）个图形有（ $2 + 4 + 2$ ）个 $\\square$ ，第（4）个图形有（ $1 + 3 + 5 + 3 + 1$ ）个 $\\square$ ，据此第（6）个图形共有（ $1 + 3 + 5 + 7 + 5 + 3 + 1$ ）个 $\\square$ 。再根据从1开始的连续奇数的和等于奇数个数的平方。计算即可。【详解】$6 + 1 = 7$ （个）$ 1 + 3 + 5 + 7 + 5 + 3 + 1 \\ = (1 + 3 + 5 + 7) + (5 + 3 + 1) \\ = 4 ^ {2} + 3 ^ {2} \\ = 1 6 + 9 \\ = 2 5 (\\text {个})$",
+        "knowledge_points": [
+          "数与形 (归纳递推)"
+        ],
+        "source": "24-25 六年级上·广东云浮·期末",
+        "merged_id": "unit8_333",
+        "original_id": "333"
+      }, 
+      {
+        "id": "334",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "观察下面图形的规律, 第 9 个框里有( )个点。![](images\\c41deb7c746994feb0b2406d6b795bb6.png)",
+        "options": [],
+        "answer": "33",
+        "solution": "【分析】根据题图可知, 每增加一个框就增加 4 个点, 第 $n$ 个框里有 $4(n - 1) + 1 = (4n - 3)$ 个点, 再将 $n = 9$ 代入含字母的式子解答即可。【详解】由分析可知: 第 $\\mathrm{n}$ 个框里有 $4(n - 1) + 1 = (4n - 3)$ 个点, 把 $n = 9$ 代入 $4n - 3$ , 得:$ 4 \\times 9 - 3 \\ = 3 6 - 3 \\ = 3 3 (\\text {个})$所以第 9 个框里有 33 个点。",
+        "knowledge_points": [
+          "数与形（探索规律）",
+          "含有字母式子的化简与求值"
+        ],
+        "source": "24-25 六年级上·河南焦作·期末",
+        "merged_id": "unit8_334",
+        "original_id": "334"
+      },
+      {
+        "id": "335",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如图, 聪聪将黑、白两种颜色的正方形自上而下一层层地排列下去，每层从左到右排列, 第 8 层是( )色的正方形, 8 层共有( )个正方形。![](images/7050aaab36717a937aa93a3c006fee395efaa5859f1835b79ec1294b12b34d18.jpg)",
+        "options": [],
+        "answer": "黑 64",
+        "solution": "【分析】自上而下一层层排列下去，奇数层排列的是白色的正方形，偶数层排列的黑色的正方形，据此确定第8层排列正方形的颜色;第一层有1个正方形，第2层有3个正方形，第3层有5个正方形，以此类推，下面一层均比它的上面一层增加2个正方形，也就是将 $(1 + 3 + 5 + 7 + 9 + 11 + 13 + 15)$ 依次加起来，所得结果即为8层共有的正方形个数。【详解】第8层是偶数层，因此第8层是黑色的正方形。$1 + 3 + 5 + 7 + 9 + 1 1 + 1 3 + 1 5 = 6 4 \\text {(个)}$因此8层共有64个正方形。",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "24-25 六年级上·广西南宁·期末",
+        "merged_id": "unit8_335",
+        "original_id": "335"
+      },
+      {
+        "id": "336",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "用同样大小的黑白两色正方形摆图形, 如图所示, 按照这样的规律摆第 7 个图形需要( )个白色正方形, 摆 n 个图形需要( )个黑色正方形。![](images/d03b4a2672a117f26c1b5a217dd867a37acd57cb3344d447d1741058ceb690c1.jpg)第1个![](images/4d664b61d215c21816d43b4163f0a9b649cd52910b90e3c597d0b2135dd036e6.jpg)第2个![](images/dba70d904bac3dd2cb5b0a62c601604a026dd3b029ae78284c2bcd01afb478b5.jpg)第3个···",
+        "options": [],
+        "answer": "22  $3 \\mathrm{n} + 2$",
+        "solution": "【分析】仔细观察图形可以发现，每增加一个图形，白色正方形增加3个，第1个图形有4个白色正方形，第2个图形有 $(4 + 3)$ 个白色正方形，第3个图形有 $(4 + 3 \\times 2)$ 个白色正方形……第7个图形有 $(4 + 3 \\times 6)$ 个白色正方形;第一个图形有5个黑色正方形，每增加1个图形，黑色正方形增加3个，摆 $n$ 个图形，增加 $(n - 1)$ 个图形，即摆 $n$ 个图形需要 $[5 + 3 \\times (n - 1)]$ 个黑色正方形。据此解答。【详解】由分析可知：摆第7个图形需要白色正方形：$ 4 + 3 \\times 6 \\ = 4 + 1 8 \\ = 2 2 \\text {(个)}$摆 $n$ 个图形需要黑色正方形:$ 5 + 3 \\times (n - 1) \\ = 5 + 3 \\mathrm {n} - 3 \\ = (3 n + 2) \\text {个}$所以摆第 7 个图形需要 22 个白色正方形，摆 n 个图形需要 $(3n + 2)$ 个黑色正方形。",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "24-25 六年级上·湖北十堰·期末",
+        "merged_id": "unit8_336",
+        "original_id": "336"
+      },
+      {
+        "id": "337",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "古希腊著名的毕达哥拉斯学派把1、3、6、10、15、21...这样的数称为三角形数，而把1、4、9、16、25、36...这样的数称为'正方形数'。从下图中可以发现，任何一个大于1的'正方形数'都可以看作两个相邻的'三角形数'之和。如果把'正方形数'64写成两个相邻的'三角形数'之和，这两个'三角形数'分别是( )和( )。![](images/c6675bb55c3afa43da0591289a6001f35dc70cf829168bf61d627d103866f5d2.jpg)![](images/d7124e05da947c21fd14638024da9761c6a0fa6077afa26de6c07a75e484e821.jpg)![](images/74148af231fa7b1a2b61913c2dc596b46091ea6bf1997e86babb2756770d1ff1.jpg)",
+        "options": [],
+        "answer": "28   36",
+        "solution": "【分析】观察图形可知:第1个正方形数是4（ $4 = 2 \\times 2$ ）， $4 = 1 + 3$ ，其中第一个加数是1，第二个加数是 $3 = 1 + 2$ ;第2个正方形数是9（ $9 = 3 \\times 3$ ）， $9 = 3 + 6$ ，其中第一个加数是 $3 = 1 + 2$ ，第二个加数是 $6 = 1 + 2 + 3$ ;第3个正方形数是16（ $16 = 4 \\times 4$ ）， $16 = 6 + 10$ ，其中第一个加数是 $6 = 1 + 2 + 3$ ，第二个加数是 $10 = 1 + 2 + 3 + 4$ ;……发现规律：如果正方形数=n×n，则这个正方形数=（1+2+3+…+n-1）+（1+2+3+…+n），据此规律解答。通过数与形的结合，从已知的图形和数据中找到规律，并按规律解题。【详解】 $64 = 8 \\times 8$$1 + 2 + 3 + \\dots + 6 + 7 \\ = (1 + 7) \\times 7 \\div 2 \\ = 8 \\times 7 \\div 2\\ = 2 8$ $ 1 + 2 + 3 + \\dots + 7 + 8 \\ = (1 + 8) \\times 8 \\div 2 \\ = 9 \\times 8 \\div 2 \\ = 3 6$所以， $64 = 28 + 36$ 。这两个'三角形数'分别是（28）和（36）。",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "24-25六年级上·湖北随州·期末",
+        "merged_id": "unit8_337",
+        "original_id": "337"
+      },
+      {
+        "id": "338",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如图, 3 个同样的杯子叠在一起高 20 厘米, 5 个这样的高 24 厘米。9 个这样的杯子叠起来高度是( )厘米。 $n$ 个这样的杯子叠在一起的高度是( )厘米。![](images/c4d433dae5789bbaba61571ac23ae73e36da28d4451d93a8c86e07bf97b2dc8a.jpg)",
+        "options": [],
+        "answer": "32  $2n + 14$",
+        "solution": "【分析】观察可知，20 厘米的高度等于 $(3 - 1)$ 个杯口高度加 1 个杯子高度，24 厘米等于 $(5 - 1)$ 个杯口高度加 1 个杯子高度，因为用 24 减 20 可得 2 个杯口高度，再除以 2 得 1 个杯口高度，再用 20 减 $(3 - 1)$ 乘 1 个杯口高度的积，得到 1 个杯子的高度。当 9 个杯子叠起来时会有 $(9 - 1)$ 个杯口高度加 1 个杯子高度，据此计算即可。 $n$ 个这样的杯子叠在一起的高度就可用 1 个杯口高度乘 $(n - 1)$ 加 1 个杯子的高度，据此列式并化简解答。【详解】 $(24 - 20)\\div [(5 - 1) - (3 - 1))\\=4 \\div [ 4 - 2)\\= 4 \\div 2\\=2 (厘米)$$2 0 - (3 - 1) \\times 2\\= 2 0 - 2 \\times 2\\=2 0 - 4\\=16 (厘米)$$2 \\times (9 - 1) + 1 6\\=2 \\times 8 + 1 6\\= 16 + 1 6\\=32 (厘米)$$2 (n - 1) + 1 6\\= 2 n - 2 + 1 6\\=2 n + 14（厘米）$ 3个同样的杯子叠在一起高20厘米，5个这样的高24厘米。 9个这样的杯子叠起来高度是32厘米。n个这样的杯子叠在一起的高度是(2n+14)或(14+2n)厘米。",
+        "knowledge_points": [
+          "用字母表示数",
+          "数量关系",
+          "数与形（探索规律）",
+          "含有字母式子的化简与求值"
+        ],
+        "source": "24-25 六年级上·广东东莞·期末",
+        "merged_id": "unit8_338",
+        "original_id": "338"
+      },
+      {
+        "id": "339",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如图, 拼一个三角形用3根小棒，拼2个三角形用5根小棒，拼3个三角形用( )根小棒，拼10个三角形用( )根小棒，拼n个三角形用( )根小棒。(1)![](images/dba115e80b1153e61f1e4ec65a1f3050faf96e1eaa4a700f0c48b42bd7b4e808.jpg)(2)![](images/9f5a01a18e9a9ee0d4545f0c14107905bdd6579742ecc76088ac9e4ad427dd7e.jpg)(3)![](images/418ebcdf3462d0b3351577ee8c7e1cc9cc6c768223791c572148423d619cc8f5.jpg)",
+        "options": [],
+        "answer": "7 21  $1 + 2\\mathrm{n}$",
+        "solution": "【分析】观察图形可知：搭一个三角形需要3根小棒，每增加1个三角形，需要增加2根小棒，搭两个三角形需要（3+2）根小棒，搭三个三角形需要 $(3 + 2 + 2)$ 根小棒。将第一个三角形的3根小棒表示成 $(1 + 2)$ 根，则搭两个三角形表示成 $(1 + 2 + 2)$ 根，搭三个三角形表示成 $(1 + 2 + 2 + 2)$ 根小棒，可知搭n个三角形需要 $(1 + 2\\mathrm{n})$ 根小棒，据此即可解答问题。【详解】拼3个三角形： $3 + 2 + 2\\= 1 + 2 + 2 + 2 \\ = 1 + (2 + 2 + 2) \\ = 1 + 2 \\times 3 \\ = 7 \\text {(根)}$拼 10 个三角形:$ 1 + 2 \\times 1 0 \\ = 1 + 2 0 \\ = 2 1 \\text {(根)}$拼 3 个三角形用 7 根小棒, 拼 10 个三角形用 21 根小棒, 拼 n 个三角形用 $(1 + 2 n)$ 根小棒。",
+        "knowledge_points": [
+          "数与形 (归纳递推)"
+        ],
+        "source": "24-25 六年级上·广东韶关·期末",
+        "merged_id": "unit8_339",
+        "original_id": "339"
+      },
+      {
+        "id": "340",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如图, 摆 1 个六边形需要 6 根小棒, 摆 3 个六边形需要( )根小棒, 摆 n 个六边形, 需要( )根小棒。![](images/4f2e1fc5c4349cb149fee09704fa7b4b9f0208dc6692cb6e335e34afdc041148.jpg)![](images/de905d41445e7798ac6e073239454da94e2cb1a141aae00cc4e52751f1dbd699.jpg)![](images/6e1012295675d16bd338cef6d90900d34e339e3034e637069f3088bfad7da1ac.jpg)",
+        "options": [],
+        "answer": "16   $1 + 5\\mathrm{n}$",
+        "solution": "【分析】通过观察图形，得出规律：摆 1 个六边形需要 $(1 + 5)$ 根小棒,摆 2 个六边形需要 $(1 + 5 \\times 2)$ 根小棒摆 3 个六边形需要 $(1 + 5 \\times 3)$ 根小棒,摆 4 个六边形需要 $(1 + 5 \\times 4)$ 根小棒,··摆 $n$ 个六边形需要 $(1 + 5 \\times n)$ 根小棒。据此解答。【详解】由分析得:$1 + 5 \\times 3 \\ = 1 + 1 5 \\ = 1 6 \\text {(根)}$即摆 3 个六边形需要 16 根小棒。$1 + 5 \\times n = 1 + 5 n$ 即摆 $n$ 个六边形, 需要 $(1 + 5 n)$ 根小棒。",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "23-24 六年级下·四川宜宾·期末",
+        "merged_id": "unit8_340",
+        "original_id": "340"
+      },
+      {
+        "id": "341",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如下图，用小棒摆正五边形，摆10个正五边形需要( )根小棒，摆n个正五边形需要( )根小棒。![](images/862814297181fcc9cd91a54062af1ad1af9f2cada09ede84288fd7ca55c616ec.jpg)",
+        "options": [],
+        "answer": "41  $4\\mathrm{n} + 1$",
+        "solution": "【分析】观察图形可知，摆1个五边形需要5根小棒，摆2个五边形需要 $(5 + 4)$ 根小棒，摆3个五边形需要 $(5+$ $4 \\times 2$ ) 根小棒，摆4个五边形需要 $(5 + 4 \\times 3$ ) 根小棒……则摆 $\\mathrm{n}$ 个五边形需要 $[5 + 4 \\times (n - 1)]$ 根小棒，据此解答即可。【详解】 $5 + 4 \\times (10 - 1)\\ = 5 + 4 \\times 9 \\ = 5 + 3 6\\= 41$ (根)$ 5 + 4 \\times (n - 1) \\ = 5 + 4 n - 4 \\ = (4 n + 1) \\text {根}$用小棒摆正五边形，摆 10 个正五边形需要 41 根小棒，摆 n 个正五边形需要 $(4 n + 1)$ 根小棒。",
+        "knowledge_points": [
+          "用字母表示数",
+          "数量关系",
+          "数与形（探索规律）",
+          "含有字母式子的化简与求值"
+        ],
+        "source": "24-25六年级上·重庆万州·期末",
+        "merged_id": "unit8_341",
+        "original_id": "341"
+      },
+      {
+        "id": "342",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如下图，1张桌子可以坐6人，2张桌子拼起来可以坐10人，5张这样的桌子拼起来能够坐下( )人;要想坐下38人，至少需要( )张这样的桌子拼起来。![](images/908763298cfd0fd2824be727236539e457da5801336e7f3add9e18416d3c1ee3.jpg)![](images/c1c96bb304389e14dd5a53faa3bec3b0ca2e1917b20ef1ce0f538db065e00b98.jpg)![](images/a07ec1540e3a24e2ba7d68875d43407859de93761849db9165fcbb6e80d43c16.jpg)![](images/9894f005f7a22ff59703ac0c6566ec59b9af20e1d1be4ddab944aa3a0bae3bfc.jpg)",
+        "options": [],
+        "answer": "22    9",
+        "solution": "【分析】1 张桌子可以坐 6 人, $6 = 1 \\times 4 + 2$ ; 2 张桌子拼起来可以坐 10 人, $10 = 2 \\times 4 + 2 \\dots$ 由此可知坐下的人数 = 桌子数 $\\times 4 + 2$ ; 桌子数 = (坐下的人数 - 2) $\\div 4$ , 据此列式计算。【详解】 $5 \\times 4 + 2\\= 2 0 + 2\\=22$$ (3 8 - 2) \\div 4 \\ = 3 6 \\div 4 \\ = 9 \\text {(张)}$$= 22$ （人）5 张这样的桌子拼起来能够坐下 22 人; 要想坐下 38 人, 至少需要 9 张这样的桌子拼起来。",
+        "knowledge_points": [
+          "数与形 (归纳递推)"
+        ],
+        "source": "24-25六年级上·江西赣州·期末",
+        "merged_id": "unit8_342",
+        "original_id": "342"
+      },
+      {
+        "id": "343",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "按规律填数: $1, \\frac{3}{4}, \\frac{7}{9}, \\frac{15}{16}, \\frac{31}{25}, (\\quad), (\\quad)$ 。",
+        "options": [],
+        "answer": "$\\frac{63}{36} \\cdot \\frac{127}{49}$",
+        "solution": "【分析】观察分数的分子与分母，可知分子的规律依次是：$2 ^ {1} - 1、2 ^ {2} - 1、2 ^ {3} - 1、2 ^ {4} - 1………2 ^ {n} - 1$，分母的规律是$1 ^ {2}、2 ^ {2}、3 ^ {2}、4 ^ {2}………n ^ {2}$，据此解答。【详解】 $2^{6} - 1\\= 6 4 - 1 \\ = 6 3$ $6 ^ {2} = 6 \\times 6 = 3 6 $ $2 ^ {7} - 1 \\ = 1 2 8 - 1 \\ = 1 2 7 $$ 7 ^ {2} = 7 \\times 7 = 4 9$所以第 6 个数是 $\\frac{63}{36}$ , 第 7 个数是 $\\frac{127}{49}$ 。",
+        "knowledge_points": [
+          "分数排列的规律"
+        ],
+        "source": "24-25 六年级上·四川成都·期末",
+        "merged_id": "unit8_343",
+        "original_id": "343"
+      },
+      {
+        "id": "344",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "活动课上, 同学们合作用白色和灰色的圆片按照下面的方法摆图形。照这样摆下去, 第 6 个图形中, 一共有( )个圆片; 当一个图形中有 8 个灰色圆片时, 白色圆片有( )个。O第1个![](images/51a0b60d2625ab22225f67980af3490613d9a64d4a59bcc2962d9fef52b9b474.jpg)![](images/c646f494cff0f5eea6ab4f7df1e63c2bb215dae5ed0c692c1145cd6135b0331b.jpg)![](images/3ea81c6e3d684839b2f616c71ff001108c544999932f17ce3959d3be07b53c65.jpg)",
+        "options": [],
+        "answer": "36 56",
+        "solution": "【分析】通过观察图可知：第1个图形：一共有（ $1 \\times 1$ ）个圆片，有0个灰色圆片;第2个图形：一共有 $(2\\times 2)$ 个圆片，有2个灰色圆片;第3个图形：一共有 $(3 \\times 3)$ 个圆片，有3个灰色圆片;第 4 个图形: 一共有 $(4 \\times 4)$ 个圆片, 有 4 个灰色圆片。由此可以得出规律: 第 $n$ 个图形一共有 $(n \\times n)$ 个圆片; 从第二个图形开始，第几个图形就对应有几个灰色圆片，且每个图形的白色圆片个数等于一共有的圆片个数减去灰色圆片个数，据此解答。【详解】由分析得:$6 \\times 6 = 3 6 (\\text {个})$第 6 个图形中, 一共有 36 个圆片。当一个图形中有 8 个灰色圆片时, 说明这个图形是第 8 个图形。$8\\times8=64(个)$ $64 - 8 = 56$ （个）即当一个图形中有 8 个灰色圆片时，白色圆片有 56 个。$8 \\times 8 = 64$ （个）",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "24-25 六年级上·山东菏泽·期末",
+        "merged_id": "unit8_344",
+        "original_id": "344"
+      },
+      {
+        "id": "345",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "古希腊毕达哥拉斯学派信奉'万物皆数'，将'数'排列成三角形、正方形等美丽的图形。如下图，排列成三角形的数叫作三角形数。照这样排列下去，第8个图形所表示的数是( )。![](images/90f533cb1680909e7ce88465eac8d7bdb5582806e7cbbd747042785407a94f13.jpg)",
+        "options": [],
+        "answer": "36",
+        "solution": "【分析】根据题意，第一个图形有1个点，表示1;第二个图形有 3 个点, 表示 3 , 可以写成: $1 + 2 = 3$ ; 第三个图形有 6 个点, 表示 6 , 可以写成: $1 + 2 + 3 = 6$ ; 第四个图形有 10 个点, 表示 10 , 可以写成: $1 + 2 + 3 + 4 = 10$ ; ……由此可知, 第 n 个图形有 $(1 + 2 + 3 + \\dots \\dots + n)$ 个点, 表示 $(1 + 2 + 3 + \\dots \\dots + n)$ , 由此当 $n = 8$ 时, 把 8 代入算式计算即可。【详解】根据分析可知, 第 $n$ 个图形有 $(1 + 2 + 3 + \\dots + n)$ 个点。当 $n = 8$ 时，表示的数是：$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 3 6$古希腊毕达哥拉斯学派信奉'万物皆数'，将'数'排列成三角形、正方形等美丽的图形。如下图，排列成三角形的数叫作三角形数。照这样排列下去，第8个图形所表示的数是36。",
+        "knowledge_points": [
+          "图形的变化规律",
+          "数与形（探索规律）"
+        ],
+        "source": "24-25 六年级上·四川凉山·期末",
+        "merged_id": "unit8_345",
+        "original_id": "345"
+      },
+      {
+        "id": "346",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "先找规律再填空。$1 ^ {2} = 1; 2 ^ {2} = 1 + 3; 3 ^ {2} = 1 + 3 + 5; 4 ^ {2} = (\\quad); (\\quad) ^ {2} = 1 + 3 + 5 + 7 + 9 _ {\\circ}$",
+        "options": [],
+        "answer": "$1 + 3 + 5 + 7$   5",
+        "solution": "【分析】观察 $1^{2} = 1, 2^{2} = 1 + 3, 3^{2} = 1 + 3 + 5$ , 发现规律:连续奇数的和等于奇数个数的平方, 据此规律解答。【详解】 $1^{2} = 1$ ;$ 2 ^ {2} = 1 + 3; \\ 3 ^ {2} = 1 + 3 + 5; \\ 4 ^ {2} = 1 + 3 + 5 + 7; \\ 5 ^ {2} = 1 + 3 + 5 + 7 + 9 。$",
+        "knowledge_points": [
+          "算式的规律（整数）"
+        ],
+        "source": "24-25 六年级上·湖北襄阳·期末",
+        "merged_id": "unit8_346",
+        "original_id": "346"
+      },
+      {
+        "id": "347",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如图, ①号图有 1个小三角形, ⑥号图是由( )个小三角形组成的: ⑩号图最底层有( )个小三角形。①![](images/363613a406c78dfb053bcbf6e1d2e6ea8a0cea1df408c7c32b16f53a5e4928bf.jpg)②![](images/bc98ae3d48e7986b745ca40897b4d1488518e48c00d8686292d0a98063fa2d6e.jpg)③ ![](images/68bd00fce7308e5f4521aa7d5a54d1e191c697b51358144704fe88b70d290588.jpg)④![](images/0bd12d4655263e2806ce10f97d7e41bb55db67726d86782b0f607117ec526d92.jpg)",
+        "options": [],
+        "answer": "36 19",
+        "solution": "【分析】第1个图形：1个三角形， $1 = 1^{2}$ 第2个图形：4个三角形， $1 + 3 = 4 = 2^{2}$ 第3个图形：9个三角形， $1 + 3 + 5 = 9 = 3^{2}$ 第4个图形：16个三角形 $1 + 3 + 5 + 7 = 16 = 4^{2}$......第 $\\mathrm{n}$ 个图形: $1 + 3 + 5 + 7 + \\dots \\dots + (2 \\mathrm{n} - 1) = \\mathrm{n}^{2}$ 个三角形, 每个图形的最底层有 $(2n - 1)$ 个小三角形, 据此解答。【详解】⑥号图是由 $6^{2} = 6 \\times 6 = 36$ （个）小三角形组成的;把10代入 $2 \\mathrm{n} - 1$ , 得:$2 \\times 1 0 - 1\\ = 2 0 - 1 \\ = 1 9 \\text {(个)}$所以⑥号图是由36个小三角形组成的，⑩号图最底层有19个小三角形。",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "24-25 五年级上·内蒙古包头·期末",
+        "merged_id": "unit8_347",
+        "original_id": "347"
+      },
+      {
+        "id": "348",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "用算式表示点阵中点的个数。①![](images/e4b8ebf5a4875af1a0a6172e6fedb3db483f42b4cf6d2cb92c718772463c067b.jpg)②![](images/44772959c685bba2fe202885bbbe544ecd06867b2fa29506a4b2d93e2948b332.jpg)③ ![](images/4eb2428210b6ab94adbdfb21bc68e7f7fb39deacdaa231f289081411bf9c237c.jpg)①$1 + 2 = 3$ ; ②$2 + 3 = 5$ ; ③()④()",
+        "options": [],
+        "answer": "$3 + 4 = 7 \\quad 4 + 5 = 9$",
+        "solution": "【分析】用③的上层点数加上下层点数即可，不难发现，点阵的排列规律为：后面一个图形每层的个数都比前面一个图形每层的个数多1个，据此写出④。【详解】由图形易知③: $3 + 4 = 7$ , 根据点阵的排列规律可知④中每行的点数都比③多一个，由此可得④： $4 + 5 = 9$ 。",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "24-25 五年级上·广东河源·期末",
+        "merged_id": "unit8_348",
+        "original_id": "348"
+      },
+      {
+        "id": "349",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "用 6 根同样的小棒可以摆一个正六边形，照图这样摆下去，摆7个正六边形需要( )根小棒，摆 $n$ 个正六边形需要( )根小棒。![](images/85f4e8f7a2a3a28d91a7ed8ae77532f848f2c00b6a3f30a8b14ac31f34fd22f6.jpg)",
+        "options": [],
+        "answer": "36    $5 n + 1 / 1 + 5 n$",
+        "solution": "【分析】观察可知，从第二个正六边形开始每增加一个正六边形就需要增加 5 根小棒，如下图所示，第一个正六边形也有这 5 根小棒，可知 1 个正六边形就有 1 个 5 根再加 1 根，2 个正六边形就有 2 个 5 根再加 1 根，3 个正六边形就有 3 个 5 根再加 1 根……7 个正六边形就有 7 个 5 根再加 1 根小棒， $n$ 个正六边形就有 $n$ 个 5 根再加 1 根小棒，据此解答。![](images/ad7c47f1abdfc4685b930847f89986afed693e0bb0b7b5c0e483b6863911cfbe.jpg)【详解】 $7 \\times 5 + 1\\= 3 5 + 1\\=36 (根)$$5 n + 1$ (或 $1 + 5 n$ )用 6 根同样的小棒可以摆一个正六边形, 照图这样摆下去,摆7个正六边形需要36根小棒，摆 $n$ 个正六边形需要 $5n+1 $(或 $1 + 5 \\mathrm{n}$ ) 根小棒。",
+        "knowledge_points": [
+          "图形的变化规律",
+          "数与形（探索规律）"
+        ],
+        "source": "24-25 六年级上·浙江宁波·期末",
+        "merged_id": "unit8_349",
+        "original_id": "349"
+      },
+      {
+        "id": "350",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "$1 + 3 + 5 + 7 + \\dots +4 7 = (\\quad), \\frac {1}{4} + \\frac {1}{8} + \\frac {1}{1 6} + \\frac {1}{3 2} + \\frac {1}{6 4} = (\\quad) 。$",
+        "options": [],
+        "answer": "576  $\\frac{31}{64}$",
+        "solution": "【分析】![](images/efe5a0c12201c0ec16d8fcbfa44e1f59db62cb41a679838f676f9a084387dbfa.jpg)![](images/7423c26f53e1ee610f2e70fdf7d25bde433bf871a1877b77647aed2005a56fcb.jpg)![](images/444eb7b68f5fcf65a6c6b219736a8690b2dd667ce848ef0f57e6cea4d757bef0.jpg)从 1 开始, n 个连续奇数的和等于奇数个数的平方;![](images/58ad4ec1e9232bde98d6385818f9b74e94544a69de8b57f74fb3c0ace397983b.jpg)由图可知$ \\frac {1}{4}+ \\frac {1}{8} + \\frac {1}{1 6}+ \\frac {1}{3 2}+ \\frac {1}{6 4}$，这些分数不断加下去，总和就是 $\\frac{1}{2}$ ，题目中算式的和就等于 $\\frac{1}{2} -\\frac{1}{64}$ 。据此解答。【详解】根据分析：$ 1 + 3 + 5 + 7 \\dots + 4 7 \\ = [ (1 + 4 7) \\div 2 ] ^ {2} \\ = [ 4 8 \\div 2 ] ^ {2} \\ = [ 2 4 ] ^ {2} \\ = 5 7 6$$\\frac {1}{4} + \\frac {1}{8} + \\frac {1}{1 6} + \\frac {1}{3 2} + \\frac {1}{6 4}\\ = \\frac {1}{2} - \\frac {1}{6 4} \\ = \\frac {3 1}{6 4}$所以, $1 + 3 + 5 + 7 \\dots + 47 = 576, \\frac{1}{4} + \\frac{1}{8} + \\frac{1}{16} + \\frac{1}{32} + \\frac{1}{64} = \\frac{31}{64}$ 。",
+        "knowledge_points": [
+          "数与形（探索规律）",
+          "异分母分数加、减法"
+        ],
+        "source": "24-25 六年级上·江西吉安·期末",
+        "merged_id": "unit8_350",
+        "original_id": "350"
+      },
+      {
+        "id": "351",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如图, 按这样的规律, 第 6 幅图对应的分数是( ), $\\frac{1}{81}$ 是第( )幅图对应的分数, 第( )幅图中涂色部分周长是大正方形的 $\\frac{1}{100}$ 。![](images/928002067f80d5231d99068d2c783c73111c238fe1d469bbdb2d4ffae516c82c.jpg)1第1幅图![](images/82551563303acd040d4e57f1f022876b569b1a83abd58d36ee7ee03f8af53695.jpg)$\\frac{1}{4}$第2幅图![](images/42ad1c75881b08c9ac4ffdefc15d82ae88cba11a448745e50a7065c6f0845600.jpg)$\\frac{1}{9}$第3幅图![](images/7f1535e0610d83ec0a5f3235fa1ae705af1eecc1a8df91f9e0c1bdf86e47b5be.jpg)$\\frac{1}{16}$ 第4幅图",
+        "options": [],
+        "answer": "$\\frac{1}{36} $    9    100",
+        "solution": "【分析】观察图形可知, 第 1 幅图对应的数是 $\\frac{1}{1^{2}}$ , 第 2 幅图对应的数是 $\\frac{1}{2^{2}}$ , 第 3 幅图对应的数是 $\\frac{1}{3^{2}}$ ……, 因此第 6 幅图对应的数是 $\\frac{1}{6^{2}}$ , 把 $\\frac{1}{81}$ 化成 $\\frac{1}{9^{2}}$ , 所以 $\\frac{1}{81}$ 是第 9 幅图对应的分数,观察第一幅图的阴影部分的周长是大正方形周长的 $\\frac{1}{1}$ , 第二幅图的阴影部分的周长是大正方形周长的 $\\frac{1}{2}$ , 第三幅图的阴影部分的周长是大正方形周长的 $\\frac{1}{3}$ , ……第 n 幅图的阴影部分的周长是大正方形周长的 $\\frac{1}{n}$ , 所以第 100 幅图中涂色部分周长是大正方形的 $\\frac{1}{100}$ 。【详解】由分析可知：第6幅图对应的分数是 $\\frac{1}{6^2} = \\frac{1}{36}$ $\\frac{1}{81} = \\frac{1}{9^2}$ ，所以 $\\frac{1}{81}$ 是第9幅图对应的分数第100幅图中涂色部分周长是大正方形的 $\\frac{1}{100}$ 。",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "24-25 六年级上·江苏宿迁·期末",
+        "merged_id": "unit8_351",
+        "original_id": "351"
+      },
+      {
+        "id": "352",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "'数形结合'的思想可以帮助我们进行思考，观察下列图形的规律，在（）里填上合适的数。![](images/77463b5ff793e35dd87bcc2871b9e5c81b9c1753d9766c5312b46e454bc614df.jpg)![](images/d64456ceb1cd3d667a1f76e8daae781e26e0b14fb534adf4449fb4ae8bb04fe4.jpg)![](images/704c6a432c3addb2e1d0a6603d566323ecb53bb9356a331367f4d86e7535306e.jpg)![](images/0b75f434da27cb197aa8ab5fdbc88bfa0fbd987706e474c2d152cd8bd38238e0.jpg)1  $1 + 3$ $1 + 3 + (\\quad)$ $1 + 3 + (\\quad) + (\\quad)$",
+        "options": [],
+        "answer": "5   5    7",
+        "solution": "【分析】观察图形，第一幅图：1个三角形;第二幅图：（1+3）个三角形;第三幅图： $(1 + 3 + 5)$ 个三角形;第四幅图： $(1 + 3 + 5 + 7)$ 个三角形;第 $n$ 幅图： $(1 + 3 + 5 + \\dots + 2n - 1)$ 个三角形;【详解】![](images/83db8843cc82a5ab22f5d7f7e1267afb430fa3e4fb880a1c25a4577b5e1f2990.jpg)1![](images/008228cfe6a2865265919c118c1dc5b0538ce5cebe0695bc223adb78ad4bc999.jpg)1+3![](images/9871b8f3645dcd70bf20c99ca01e1b54bf519c3c26052b7fe473cb6e5226dceb.jpg)$1 + 3 + 5$![](images/9a14b0b50170ac372163ad4dc7e417e4f96c299c00c102385ecded711bce811d.jpg)$1 + 3 + 5 + 7$",
+        "knowledge_points": [
+          "图形的变化规律",
+          "数与形（探索规律）"
+        ],
+        "source": "24-25 六年级上·青海果洛·期末",
+        "merged_id": "unit8_352",
+        "original_id": "352"
+      },
+      {
+        "id": "353",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "下面图形中每个小三角形边长都为 1 , 认真观察, 填出后面三角形的个数和图形的周长。![](images/f3f5dceb8e47ce406ddca1d1646f7f7c170070be627f4d524fe070d3a1602c7a.jpg)……第10个图形……第n个图形三角形的个数：1 4 9 16 ( ) ( )图形的周长：3 6 9 12 ( ) ( )",
+        "options": [],
+        "answer": "$100 \\quad \\mathrm{n}^{2} \\quad 30 \\quad 3 \\mathrm{n}$",
+        "solution": "【分析】观察可知, 第 1 个图形有 1 个小三角形,  $1 = 1^{2}$ ,周长是 3,  $3 = 1 \\times 3$ ; 第 2 个图形有 4 个小三角形,  $4 = 2^{2}$ , 周长是 6,  $6 = 2 \\times 3$ ; 第 3 个图形有 9 个小三角形,  $9 = 3^{2}$ , 周长是 9,  $9 = 3 \\times 3$ . 由此可知, 三角形的个数 = 第几个图形就是几的平方; 图形的周长 = 第几个图形就用几  $\\times 3$ , 据此分析。【详解】  $10^{2} = 10 \\times 10 = 100$  （个）、 $10 \\times 3 = 30$   $n \\times 3 = 3 n$![](images/88bda0d5873d3d8902edc3a800124ca91016d8703891377674f96c541a5a36ff.jpg)……第10个图形……第n个图形三角形的个数:  1   4   9    16   100  n²图形的周长:3   6   9    12   30   3n",
+        "knowledge_points": [
+          "数与形 (归纳递推)"
+        ],
+        "source": "24-25 六年级上·河北保定·期末",
+        "merged_id": "unit8_353",
+        "original_id": "353"
+      },
+      {
+        "id": "354",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "明明是个善于观察,乐于思考的好孩子，他通过数形结合（如图），发现了'求两个连续自然数平方差'的规律。![](images/23c7513854c786bdaad32870a8781bf9f2a61f1387325bbcff31a5a325360dfb.jpg)$2 ^ {2} - 1 ^ {2} = 2 + 1$![](images/ce4b3b16a3efb1fb87b2a24364aa16788c279e4130609c12442e79c285197d15.jpg)$3 ^ {2} - 2 ^ {2} = 3 + 2$![](images/d35f49f03cd149a07474fe14e68f0e222bb5e860234609413e2273573515af67.jpg)$4 ^ {2} - 3 ^ {2} = 4 + 3$![](images/a5b590654caa497289d507aaa5670431facc3c8adad204c6530b386ebc494482.jpg)$5 ^ {2} - 4 ^ {2} = 5 + 4$···请你根据明明发现的规律，直接写出下面算式的结果：$6 ^ {2} - 5 ^ {2} = ( \\quad) \\quad 2 0 2 4 ^ {2} - 2 0 2 3 ^ {2} =  ( \\quad)$",
+        "options": [],
+        "answer": "11  4047",
+        "solution": "【分析】由图可知：每个图形阴影部分的面积  $= \\text{大正方形}$面积一空白部分正方形面积。假设大正方形边长为  $a$  (  $a$  为整数), 空白正方形边长则为  $(a - 1)$ , 阴影部分面积  $S = a^{2} - (a - 1)^{2}$ , 即两个相邻自然数的平方差;由算式可知: 两个相邻自然数的平方差其结果为两个相邻自然数的和, 据此解答。解答此题的关键是要将图形与算式相结合，算式含有两个数的平方，所以利用图形中面积关系来找到图形与算式的联系，平方差的计算结合根据多个算式的规律来总结。【详解】结合图和算式：$2 ^ {2} - 1 ^ {2} = 2 + 1$   $3 ^ {2} - 2 ^ {2} = 3 + 2$     $4 ^ {2} - 3 ^ {2} = 4 + 3$      $5 ^ {2} - 4 ^ {2} = 5 + 4$综上可知:  $a^{2} - (a - 1)^{2} = a + (a - 1)$  。所以， $6^{2} - 5^{2} = 6 + 5 = 11$ ;$2 0 2 4 ^ {2} - 2 0 2 3 ^ {2} = 2 0 2 4 + 2 0 2 3 = 4 0 4 7 。$",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "24-25 六年级上·广东东莞·期末",
+        "merged_id": "unit8_354",
+        "original_id": "354"
+      },
+      {
+        "id": "355",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "$1 + 3 + 5 + 7 + 9 + 1 1 + 1 3 + 1 5 + 1 7 + 1 9 + 1 7 + 1 5 + 1 3 + 1 1 + 9 + 7 + 5 + 3 + 1 = ( \\quad) _ {\\circ}$",
+        "options": [],
+        "answer": "181",
+        "solution": "【分析】从 1 开始的连续奇数的和等于奇数个数的平方, 据此分别数出  $1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19$  和  $17 + 15 + 13 + 11 + 9 + 7 + 5 + 3 + 1$  中奇数的个数即可解答。【详解】  $1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 10^{2} =1 0 \\times 1 0 = 1 0 0$$1 7 + 1 5 + 1 3 + 1 1 + 9 + 7 + 5 + 3 + 1 = 9 ^ {2} = 9 \\times 9 = 8 1$所以  $1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 17 + 15 + 13+1 1 + 9 + 7 + 5 + 3 + 1  = 1 0 0 + 8 1= 1 8 1$",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "24-25 六年级上·吉林四平·期末",
+        "merged_id": "unit8_355",
+        "original_id": "355"
+      },
+      {
+        "id": "356",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "由下图可以看出，每多搭一个三角形，就要增加( )根小棒，照这样搭下去,第8个图形一共需要( )根小棒。①![](images/3fa8fbf28d09b268eae552e95a652a5e85915b508b2a9197d435cdf243c327c4.jpg)  ②![](images/e6a32bfdfd099732ffe62023147ebefe844d966ef7620e32a80d4d13faba89db.jpg)  ③![](images/47a2902b87ac538256a9dacbab694199ea92e2b120e8951e98cd0d385b8cf9c2.jpg)  ④  ![](images/8787da5dfc8fdb8c31ee2437793e680d544840d9283f8f82b8749eea6d03c8b2.jpg)·······",
+        "options": [],
+        "answer": "2   17",
+        "solution": "【分析】由图可知, 第 1 个三角形用 3 根小棒, 后续每增加 2 根小棒就可以多 1 个三角形; 第 8 个图形, 在第 1 个三角形的基础上, 再增加 (8-1) 组 2 根的小棒; 据此解答。【详解】由图可知:第②个图形： $(3 + 2)$  根小棒，第③个图形:  $(3 + 2 + 2)$  根小棒，第④个图形:  $(3 + 2 + 2 + 2)$  根小棒，，则第  $\\mathrm{n}$  个图形:  $3 + 2 \\times (\\mathrm{n} - 1) = 3 + 2 \\mathrm{n} - 2 \\times 1 = 3 + 2 \\mathrm{n} - 2 =2 \\mathrm{n} + 3 - 2 = 2 \\mathrm{n} + (3 - 2) = (2 \\mathrm{n} + 1)$  根小棒则第 8 个图形:  $2 \\times 8 + 1 = 16 + 1 = 17$  (根)所以每多搭一个三角形, 就要增加 2 根小棒, 照这样搭下去,第 8 个图形一共需要 17 根小棒。",
+        "knowledge_points": [
+          "表内乘加、乘减",
+          "数与形（探索规律）",
+          "含有字母式子的化简与求值"
+        ],
+        "source": "24-25六年级上·新疆喀什·期末",
+        "merged_id": "unit8_356",
+        "original_id": "356"
+      },
+      {
+        "id": "357",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "$1 + 3 + 5 + 7 + 5 + 3+1 = (\\quad) ^ {2} $。",
+        "options": [],
+        "answer": "5",
+        "solution": "【分析】先计算出  $1 + 3 + 5 + 7 + 5 + 3 + 1$  后面 3 个数的和是  $5 + 3 + 1 = 9$ , 算式变成  $1 + 3 + 5 + 7 + 9$ , 根据 '连续奇数的和等于奇数个数的平方', 据此解答。【详解】  $1 + 3 + 5 + 7 + 5 + 3 + 1\\ = 1 + 3 + 5 + 7 + (5 + 3 + 1) \\ = 1 + 3 + 5 + 7 + 9 \\ = 5 ^ {2} \\ = 2 5$所以,  $1 + 3 + 5 + 7 + 5 + 3 + 1 = 5^{2}$  。",
+        "knowledge_points": [
+          "算式的规律（整数）",
+          "数与形（探索规律）"
+        ],
+        "source": "24-25 六年级上·吉林白城·期末",
+        "merged_id": "unit8_357",
+        "original_id": "357"
+      },
+      {
+        "id": "358",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "用边长为  $1 \\mathrm{~cm}$  的小正方形摆成如下图所示的图形，则第4个图形的面积是( ) $\\mathrm{cm}^2$  ，第8个图形的面积是( )  $\\mathrm{cm}^2$  。![](images/4b1d9e8c67a1f0aab96750ac0320825c9a19cb99962961d27aaefa7565e45294.jpg)![](images/4c266f89da0441586b07ba82fac0a996b04b286c53ce725675fdc23d8fee6179.jpg)![](images/180cfe8b7808450d2b359752a62a481a49fe715d2d718afc00bb2b78cc8496cf.jpg)",
+        "options": [],
+        "answer": "16  64",
+        "solution": "【分析】由图可知, 第一个图形的面积为  $1 \\times 1 = 1^{2}=1(cm^{2}) $ ，第二个图形的面积为  $2 \\times 2 = 2^{2} = 4$ （ $\\mathrm{cm}^{2}$ ），第三个图形的面积为  $3 \\times 3 = 3^{2} = 9（ cm^{2} )$，所以第四个图形的面积为  $4^{2}$ , 第八个图形的面积为  $8^{2}$ , 据此解答。【详解】  $4^{2} = 16 (\\mathrm{cm}^{2})$$8 ^ {2} = 6 4 (c m ^ {2})$则第 4 个图形的面积是  $16 \\mathrm{~cm}^{2}$ , 第 8 个图形的面积是  $64 \\mathrm{~cm}^{2}$  。",
+        "knowledge_points": [
+          "图形的变化规律"
+        ],
+        "source": "24-25 六年级上·吉林松原·期末",
+        "merged_id": "unit8_358",
+        "original_id": "358"
+      },
+      {
+        "id": "359",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "A、B、C、D、E 五支足球队进行循环赛（每两队之间要比赛1场）。到现在为止，A队已经比了4场，B队已经比了3场，C队已经比了2场，E队已经比了1场，则D队已经比了( )场。",
+        "options": [],
+        "answer": "2",
+        "solution": "【分析】由题意可知, A 队已经赛了 4 场, A 队分别与 B 队、C 队、D 队、E 队进行了比赛; E 队赛了 1 场, E 队只与 A队进行了比赛; B 队赛了 3 场, B 队分别与 A 队、C 队、D队进行了比赛，最后求出D队比赛的场数，据此结合画图解答。【详解】如图：![](images/755c71bcd277c4b7f9fffbf2af5acc521f089f0f2f8e4ea80c2402c3d04d05d9.jpg)由图可知: D 队与 A 队进行了 1 场比赛, 又与 B 队进行了 1 场比赛, 所以 D 队已经比了 2 场比赛。",
+        "knowledge_points": [
+          "数与形 (归纳递推)"
+        ],
+        "source": "24-25 六年级上·湖南衡阳·期末",
+        "merged_id": "unit8_359",
+        "original_id": "359"
+      },
+      {
+        "id": "360",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "找规律回答问题。![](images/bc322a7ce23698036c5c242e0d7838807ac4aa4c78a7940cbb6f56a5876ad532.jpg)第⑤个图形中共有( )个点, 如果其中一个图形有 295 个点, 则这个图形是第( )个图形。",
+        "options": [],
+        "answer": "16 98",
+        "solution": "【分析】根据图形，得出点的个数：第一个图：4个第二个图：  $4 + 3 = 7$  （个）第三个图：  $4 + 3 + 3 = 10$  （个）第四个图:  $4 + 3 + 3 + 3 = 13$  (个)可以得出每增加一个图形, 就增加 3 个点。即第  $n$  个图形,点的个数为:  $4 + 3 \\times (n - 1)$  。图形有 295 个点, 先减去开始的 4 个点, 再除以 3 , 则可以得出增加几个 3 , 再加上 1 就是第几个图形。【详解】  $4 + (5 - 1) \\times 3\\ = 4 + 4 \\times 3 \\ = 4 + 1 2 \\ = 1 6 (\\text {个})$$  (2 9 5 - 4) \\div 3 + 1 \\ = 2 9 1 \\div 3 + 1 \\ = 9 7 + 1\\= 98(个)$  第⑤个图形中共有 16 个点, 如果其中一个图形有 295 个点,则这个图形是第 98 个图形。",
+        "knowledge_points": [
+          "图形的变化规律",
+          "数与形（探索规律）"
+        ],
+        "source": "23-24 六年级下·四川内江·期末",
+        "merged_id": "unit8_360",
+        "original_id": "360"
+      },
+      {
+        "id": "361",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "按  $1, \\frac{1}{3}, \\frac{1}{9}, \\frac{1}{27},$  ……中的规律, 第 5 个数应该是( ), 这一列数越来越接近( )。",
+        "options": [],
+        "answer": "$\\frac{1}{81}$ 0",
+        "solution": "【分析】把 1 看作  $\\frac{1}{1}$ , 即这列数的规律是: 分子是 1、分母依次乘 3; 这一列分数的分子不变, 分母越来越大, 即分数值越来越小, 越来越接近 0 ; 据此解答即可。【详解】由分析可知:$2 7 \\times 3 = 8 1$这列数为:  $1, \\frac{1}{3}, \\frac{1}{9}, \\frac{1}{27}, \\frac{1}{81}, \\dots$所以第 5 个数应该是  $\\frac{1}{81}$ ; 这一列数越来越接近 0。",
+        "knowledge_points": [
+          "分数排列的规律"
+        ],
+        "source": "24-25 六年级上·海南海口·期末",
+        "merged_id": "unit8_361",
+        "original_id": "361"
+      },
+      {
+        "id": "362",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "根据下面的规律，算一算。$1 = 1 ^ {2}$![](images/9a112f5e0e2d08009cb6c89e103ba3e012e9975833445305a5039a1b2ad0f5f9.jpg)$1 + 3 = 2 ^ {2}$![](images/4d5fff1ce089ba665e5af82608c0bad135ad0d81a8a6d56596aba7434f8b2639.jpg)$1 + 3 + 5 = 3 ^ {2}$![](images/21350813afe071a21cf9638ad8a53cc7c188b418ebb723fd0993d4609ced55ca.jpg)$1 + 3 + 5 + 7 + 9 + 7 + 5 + 3 + 1 = (\\quad) _ {\\circ}$",
+        "options": [],
+        "answer": "41",
+        "solution": "【分析】根据规律，从 1 开始连续的几个奇数相加，就等于几的平方，即  $1 + 3 + 5 + \\dots + n = (n + 1) \\div 2^{2}$ ;据此解答。【详解】根据分析：将  $1 + 3 + 5 + 7 + 9 + 7 + 5 + 3 + 1$  分为  $1 + 3 + 5 + 7 + 9$  和$7 + 5 + 3 + 1;$$1 + 3 + 5 + 7 + 9\\=[ (9 + 1) \\div 2 ] ^ {2}\\= [ 1 0 \\div 2 ] ^ {2} \\=5 ^ {2}\\= 2 5$$7 + 5 + 3 + 1\\= [ (7 + 1) \\div 2 ] ^ {2}\\= [ 8 \\div 2) ^ {2}\\= 4 ^ {2}\\=1 6$$2 5 + 1 6 = 4 1$所以  $1 + 3 + 5 + 7 + 9 + 7 + 5 + 3 + 1 = 41$  。",
+        "knowledge_points": [
+          "数与形 (归纳递推)"
+        ],
+        "source": "2024 六年级上·全国·专题练习",
+        "merged_id": "unit8_362",
+        "original_id": "362"
+      },
+      {
+        "id": "363",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "古希腊的数学家毕达哥拉斯在没有纸笔的时代，用沙子在沙滩上画画，后来发现了数与形的规律。如果按照下面的方式用小棒摆出五边形。照这样的规律接着摆下去，第20个图形需要( )根小棒。![](images/1ea76f14327c5230a505cca692a5b55f1c8ee2505e7e553de67af921a37c02ea.jpg)![](images/21c6e100c21c11343f6ab9179774d97596bcac3839a6e8808c367bf795117feb.jpg)![](images/4206c8b67f3277b393f1b059ac76656f67edad5d85d5a837d810d04e1be431f8.jpg)",
+        "options": [],
+        "answer": "81",
+        "solution": "【分析】规律：每多1个五边形就多4根小棒;第1个图形里共有5根小棒，即  $4 \\times 1 + 1$第2个图形里共有9根小棒，即  $4 \\times 2 + 1$第3个图形里共有13根小棒，即  $4 \\times 3 + 1$ ;··第  $\\mathrm{n}$  个图形里需要的小棒数为:  $4\\mathrm{n} + 1$  。【详解】根据分析可知，第  $\\mathrm{n}$  个图形里需要的小棒数为：4n+1, 当  $n = 20$  时,$ 4 n + 1 \\ = 4 \\times 2 0 + 1 \\ = 8 0 + 1 \\ = 8 1$即第20个图形需要81根小棒。",
+        "knowledge_points": [
+          "图形的变化规律"
+        ],
+        "source": "24-25 六年级上·全国·单元测试",
+        "merged_id": "unit8_363",
+        "original_id": "363"
+      },
+      {
+        "id": "364",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "运用规律算一算。$1 + 3 + 5 + 7 + 9 + 1 1 + 1 3 + 1 1 + 9 + 7 + 5 + 3 + 1 = (\\quad) \\ 7 + 9 + 1 1 + 1 3 + 1 5 = (\\quad)$",
+        "options": [],
+        "answer": "85 55",
+        "solution": "【分析】把算式  $1 + 3 + 5 + 7 + 9 + 11 + 13 + 11 + 9 + 7 + 5 + 3+1$ 看作两部分:  $1 + 3 + 5 + 7 + 9 + 11 + 13$  和  $11 + 9 + 7 + 5+3+1$，'根据连续奇数的和等于奇数个数的平方'可得，$1 + 3 + 5 + 7 + 9 + 11 + 13 = 7^{2}, 11 + 9 + 7 + 5 + 3 + 1 = 6^{2}$ , 再两部分相加即可;$7 + 9 + 1 1 + 1 3 + 1 5$ 可 以 看 作 (1 + 3 + 5 + 7 + 9 + 1 1 + 1 3 +15)  $-(1 + 3 + 5)$ , 即  $8^{2} - 3^{2}$  。据此解答。【详解】  $1 + 3 + 5 + 7 + 9 + 11 + 13 + 11 + 9 + 7 + 5 + 3 + 1\\= (1 + 3 + 5 + 7 + 9 + 1 1 + 1 3) + (1 1 + 9 + 7 + 5 + 3 + 1)\\ = 7 ^ {2} + 6 ^ {2} \\ = 4 9 + 3 6 \\ = 8 5$$ 7 + 9 + 1 1 + 1 3 + 1 5 \\ = (1 + 3 + 5 + 7 + 9 + 1 1 + 1 3 + 1 5) - (1 + 3 + 5) \\ = 8 ^ {2} - 3 ^ {2} \\ = 6 4 - 9 \\ = 5 5$$ 1 + 3 + 5 + 7 + 9 + 1 1 + 1 3 + 1 1 + 9 + 7 + 5 + 3 + 1 = 8 5 \\ 7 + 9 + 1 1 + 1 3 + 1 5 = 5 5$",
+        "knowledge_points": [
+          "算式的规律（整数）",
+          "数与形（归纳递推）"
+        ],
+        "source": "2024 六年级上·全国·专题练习",
+        "merged_id": "unit8_364",
+        "original_id": "364"
+      },
+      {
+        "id": "365",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "一个细胞，一分钟后变成 2 个，10 分钟后细胞的个数是____。",
+        "options": [],
+        "answer": "1024",
+        "solution": "【分析】据题意可知，一个细胞，一分钟后变成 2 个，每分钟的数量是前面一分钟的 2 倍, 也就是 2 分钟变成  $(2 \\times 2)$ 个, 3 分钟变成  $(2 \\times 2 \\times 2)$  个, …以此类推, n 分钟后变成的数量就是 n 个 2 相乘的结果, 据此解答。完成本题的关键是据题意推理其分裂的个数构成的算式的规律，由此进行解答即可。【详解】  $2 \\times 2 \\times 2 \\times 2 \\times 2 \\times 2 \\times 2 \\times 2\\ = (2 \\times 2 \\times 2 \\times 2 \\times 2) \\times (2 \\times 2 \\times 2 \\times 2 \\times 2) \\ = 3 2 \\times 3 2 \\ = 1 0 2 4 (\\text {个})$一个细胞, 一分钟后变成 2 个, 10 分钟后细胞的个数是  $1024_\\circ$",
+        "knowledge_points": [
+          "求一个数的几倍是多少",
+          "算式的规律（整数）"
+        ],
+        "source": "2024 六年级·全国·竞赛",
+        "merged_id": "unit8_365",
+        "original_id": "365"
+      },
+      {
+        "id": "366",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "'成都欢迎你成都欢迎你成都……'按这样的规律排下去, 第 2016 个汉字是( )。",
+        "options": [],
+        "answer": "成",
+        "solution": "【分析】由题意可知, 每 5 个字一循环, 则 5 个字一组, 可先用除法计算 2016 个字有几组, 如刚好, 则是一组中最后一个字, 如有余数, 则看是一组中的第几个汉字, 即可得解。【详解】  $2016 \\div 5 = 403$  （组）……1（个）第2016个汉字是成。",
+        "knowledge_points": [
+          "有余数除法的实际应用",
+          "周期问题"
+        ],
+        "source": "2024·四川绵阳·小升初真题",
+        "merged_id": "unit8_366",
+        "original_id": "366"
+      },
+      {
+        "id": "367",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "找规律填数：  $\\frac{1}{2}$  ，  $\\frac{3}{5}$  ， $\\frac{5}{8}$  ，  $\\frac{7}{11}$  ，（ ）...。",
+        "options": [],
+        "answer": "$\\frac{9}{14}$",
+        "solution": "【分析】观察发现：分子是按 1、3、5、7...的顺序排列，即前一个分子加 2 等于后一个分子;分母是按 2、5、8、11...的顺序排列，即前一个分母加 3 等于后一个分母;据此规律解答。【详解】按规律可得:  $\\frac{7 + 2}{11 + 3} = \\frac{9}{14}$填空如下:$\\frac {1}{2}, \\frac {3}{5}, \\frac {5}{8}, \\frac {7}{1 1}, \\quad (\\frac {9}{1 4}) \\dots 。$",
+        "knowledge_points": [
+          "分数排列的规律"
+        ],
+        "source": "23-24六年级上·河北唐山·期中",
+        "merged_id": "unit8_367",
+        "original_id": "367"
+      },
+      {
+        "id": "368",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "用边长为  $1 \\mathrm{~cm}$  的小三角形按下图的方式摆图形，摆5个图形需要( )个小三角形，第9个图形的周长是( )cm。![](images/1f5be3b635ad691430938fd9ede11aeecf7c517dfd00852a949d19222ac31cf9.jpg)![](images/b818d21f8a5f2a4d21276fa8ed770d0afdc4149ca56e6ca7eb58876483e88af3.jpg)![](images/7e940abfd45809919ef2ae61bbe31e762833329b9df95c330459b775afe2a4b6.jpg)",
+        "options": [],
+        "answer": "25   27",
+        "solution": "【分析】观察图形可知，摆1个图形需要1个小三角形，1=1²;摆2个图形需要4个小三角形，4=2²;摆3个图形需要 9 个小三角形,  $9 = 3^{2}$  。由此可知, 小三角形的个数 = 图形序数的平方。据此求出摆 5 个图形的小三角形的个数。第1个图形的边长是  $1 \\mathrm{~cm}$ , 周长是  $1 \\times 3 = 3(\\mathrm{~cm})$ ; 第2个图形的边长是  $2 \\mathrm{~cm}$ , 周长是  $2 \\times 3 = 6$  (cm); 第 3 个图形的边长是  $3 \\mathrm{~cm}$ , 周长是  $3 \\times 3 = 9$  (cm) 。由此可知, 图形的周长 = 图形的序数  $\\times 3$  。据此求出第 9 个图形的周长。【详解】通过分析可得: 小三角形的个数 = 图形序数的平方,  $5^{2} = 25$  (个), 则摆 5 个图形需要 25 个小三角形;图形的周长 = 图形的序数  $\\times 3, 9 \\times 3 = 27 \\text{（cm）}$ , 则第 9 个图形的周长是  $27 \\mathrm{~cm}$  。",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "21-22 六年级上·河南信阳·期末",
+        "merged_id": "unit8_368",
+        "original_id": "368"
+      },
+      {
+        "id": "369",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "①![](images/b38eb6d3b9ee631a24a2eccb4d24b9963536332b1f8953800c5819f8df3fdbd5.jpg)  ②![](images/2ad37069e87d0b27d999f85c3581d4e9c8851e3656f949398a36c9f18307a25a.jpg)  ③![](images/11b2db4cf5e6e56293c9acabdb829c02063a89bdc1dbc6c8a45c7a9a7a463f27.jpg)  ④ ![](images/42e0e5c7b3833eb8282a2341ab6766a2793f1890f67c1dd3818f3a5c455cfb55.jpg)根据如图的排列规律，摆第6个图形需要用( )根同样长的小棒。继续摆下去，第100个图形用( )根同样长的小棒。",
+        "options": [],
+        "answer": "13 201",
+        "solution": "【分析】根据图意，第1个图形需要3根小棒;第2个图形需要  $3 + 2 = 5$  （根）小棒;第3个图形需要  $3 + 2 + 2 = 7(根) $小棒; 第 4 个图形需要  $3 + 2 + 2 + 2 = 9$  (根) 小棒…由此可见, 每增加 1 个三角形就增加 2 根小棒, 小棒的总根数可以用 3 加若干个 2, 2 的个数比图形个数少 1 ; 由此可以计算 6 个图形的小棒根数是  $3 + (6 - 1) \\times 2; 100$  个图形的小棒根数是  $3 + (100 - 1) \\times 2$ , 最后根据四则运算顺序进行计算。据此解答。【详解】根据分析可知:$ 3 + (6 - 1) \\times 2 \\ = 3 + 5 \\times 2\\ = 3 + 1 0 \\ = 1 3 (\\text {根})$$3 + (1 0 0 - 1) \\times 2 \\ = 3 + 9 9 \\times 2 \\ = 3 + 1 9 8 \\ = 2 0 1 \\text {(根)}$所以，摆第 6 个图形需要用 13 根同样长的小棒。继续摆下去，第 100 个图形用 201 同样长的小棒。",
+        "knowledge_points": [
+          "数与形（探索规律）",
+          "带有小括号的混合运算"
+        ],
+        "source": "23-24 四年级下·河南平顶山·期末",
+        "merged_id": "unit8_369",
+        "original_id": "369"
+      },
+      {
+        "id": "370",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "按规律填数：1, 3, 5, 7, ( ), 11, ( )...。",
+        "options": [],
+        "answer": "9   13",
+        "solution": "【分析】这组数的规律： $3 - 1 = 2, 5 - 3 = 2, 7 - 5 = 2$ ，发现后一个数比前一个数大 2，据此解答。【详解】  $7 + 2 = 9$   $1 1 + 2 = 1 3$   按规律填数: 1, 3, 5, 7, 9, 11, 13。",
+        "knowledge_points": [
+          "整数排列的规律"
+        ],
+        "source": "23-24六年级下·黑龙江哈尔滨·期末",
+        "merged_id": "unit8_370",
+        "original_id": "370"
+      },
+      {
+        "id": "371",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如下图, 如果一个小正三角形的边长为  $1 \\mathrm{~cm}$ , 第 5 个图形周长是( )  $\\mathrm{cm}$ , 按此规律拼下去, 第 52 个图形的周长是( )cm。![](images/4aec518c77b36ef02124b3717d3cdcb9f7fa1cd5d4910c7c27b5fabdd7df0926.jpg)",
+        "options": [],
+        "answer": "7   54",
+        "solution": "【分析】封闭图形一周的长度是周长。看图可知, 第 1 个图形的周长是  $3 \\mathrm{~cm}, 3 = 1 + 2$ ; 第 2 个图形的周长是  $4 \\mathrm{~cm}, 4 =$ $2 + 2$ , 第 3 个图形的周长是  $5 \\mathrm{~cm}, 5 = 3 + 2 \\dots \\dots$ , 由此可知,第几个图形的周长  $=$  第几个图形就用几 +2 , 据此分析。【详解】  $5 + 2 = 7$  (cm)   $5 2 + 2 = 5 4 \\text {(c m)}$  第5个图形周长是  $7 \\mathrm{~cm}$ , 第52个图形的周长是  $54 \\mathrm{~cm}$  。",
+        "knowledge_points": [
+          "数与形 (归纳递推)"
+        ],
+        "source": "21-22 六年级上·福建莆田·期末",
+        "merged_id": "unit8_371",
+        "original_id": "371"
+      },
+      {
+        "id": "372",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "观察下列数据, 按某种规律在括号填上适当的数:$1, \\frac {3}{4}, \\frac {5}{9}, \\frac {7}{1 6}, (\\quad), (\\quad), \\dots$",
+        "options": [],
+        "answer": "$\\frac{9}{25} \\cdot \\frac{11}{36}$",
+        "solution": "【分析】先观察分子的规律，分子依次是1、3、5、7，可以发现这些数依次增加2;再观察分母的规律，分母依次是1、4、9、16，可以看出  $1 = 1 \\times 1, 4 = 2 \\times 2, 9 = 3 \\times 3, 16 = 4 \\times 4$ ，即分母是连续自然数的平方。【详解】第五个数的分子是  $7 + 2 = 9$ , 分母是  $5 \\times 5 = 25$ , 所以这个数是  $\\frac{9}{25}$ ;第六个数的分子是  $9 + 2 = 11$ , 分母是  $6 \\times 6 = 36$ , 所以这个数是  $\\frac{11}{36}$ 。$1, \\frac {3}{4}, \\frac {5}{9}, \\frac {7}{1 6}, \\frac {9}{2 5}, \\frac {1 1}{3 6}$",
+        "knowledge_points": [
+          "分数排列的规律"
+        ],
+        "source": "24-25 六年级上·全国·课后作业",
+        "merged_id": "unit8_372",
+        "original_id": "372"
+      },
+      {
+        "id": "373",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "按下图的规律摆下去。(1)![](images/afcd000ba8655401be707c367fd5745c73729fb217cae8d7e4f326e6667ed790.jpg)  (2）![](images/bc0f1be423717893b3bc3987bbdccb22b2c183c9337f87fca66eef458a557614.jpg)  (3）![](images/4f6c13301a9d2c63f93d8a40e5c58f871fdb4a0faf082b942cc6c9745a09551a.jpg)    ..........第7幅图需要( )个点，第n幅图需要( )个点。",
+        "options": [],
+        "answer": "24    $3 + 3n$",
+        "solution": "【分析】第一幅图点的数量： $3 \\times 2 = 6$ （个），第二幅图点的数量： $3 \\times 3 = 9$ （个），第三幅图点的数量： $3 \\times 4 = 12$ （个），……，第  $n$  幅图点的数量： $3(n + 1)$ 个，据此解答即可。【详解】第7幅图需要：  $3\\times (7 + 1) = 3\\times 8 = 24$  （个）第  $n$  幅图需要:  $3(n + 1) = 3 + 3n$  (个)所以第 7 幅图需要 24 个点, 第 n 幅图需要  $(3 + 3 n)$  个点。",
+        "knowledge_points": [
+          "用字母表示数",
+          "数量关系",
+          "数与形（探索规律）"
+        ],
+        "source": "23-24 六年级下·辽宁鞍山·期末",
+        "merged_id": "unit8_373",
+        "original_id": "373"
+      },
+      {
+        "id": "374",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如图中的每个图形都是由若干个小三角形组成的，已知小三角形的边长为  $1$ 。根据图形的排列规律，第5个图形中包含( )个小三角形，第  $\\mathrm{n}$  个图形的周长为( )。![](images/ba5aee44ee4c1243608bd8450d5eac95df53f2e415ea5f4aaa1cfcd2d5c39d5a.jpg)![](images/4cea6e1ce31b439452f065b6e2ea27e15aa1cf788efb6ea81c771647470db01b.jpg)![](images/542a733748ca233a2cb20e2b832db6c3ed78aef01bc7247ff29fdb62b6c18826.jpg)![](images/7013a62f22d8471c57fa509e23378c03564d2c3dfe7e18bae61568db2fde510d.jpg)",
+        "options": [],
+        "answer": "25   (3n)",
+        "solution": "【分析】看图可知, 第 1 个图形有 1 个三角形,  $1 = 1 \\times 1$ ; 第 2 个图形有 4 个三角形,  $4 = 2 \\times 2$ ; 第 3 个图形有 9 个小三角形,  $9 = 3 \\times 3$ , 由此可知, 小三角形的个数 = 第几个图形就用几  $\\times$  几;看图可知, 第 1 个等边三角形的边长是 1 , 第 2 个等边三角形的边长是 2 , 第 3 个等边三角形的边长是 3 , 由此可知,第几个图形就是边长是几的等边三角形，根据等边三角形的周长  $= \\text{边长} \\times 3$  ，列式计算即可。【详解】  $5 \\times 5 = 25$  （个） $3 \\times n =$  (3n) 厘米   第 5 个图形中包含 25 个小三角形, 第 n 个图形的周长是 (3n) 厘米。",
+        "knowledge_points": [
+          "数与形 (归纳递推)"
+        ],
+        "source": "23-24 六年级下·河南许昌·期末",
+        "merged_id": "unit8_374",
+        "original_id": "374"
+      },
+      {
+        "id": "375",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "观察下面各图中小正方形的变化规律，据图可知第⑧个图形中有( )个小正方形。![](images/147490391ad716589c8738b784aff3fb1f868184a303629d428ffd2f440281cc.jpg)![](images/9564c20fcd9919ad37648e1fe563950af0dd5672f4d9220c96f0b43f24d87315.jpg)![](images/25ca88dc342e75c024cf3055e7950e97eb076e57b042abe6fe8a4a102da34648.jpg)",
+        "options": [],
+        "answer": "45",
+        "solution": "【分析】观察可知, 第①个图形有  $1 + 2 = 3$  (个) 小正方形,第②个图形有  $1 + 2 + 3 = 6$  （个）小正方形，第③个图形有  $1 + 2 + 3 + 4 = 10$  （个）小正方形，由此可知，小正方形的个数 = 第几个图形就从 1 依次加到（几 + 1），因此第⑧个图形有  $(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)$  个小正方形。【详解】  $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$  （个）第⑧个图形中有45个小正方形。",
+        "knowledge_points": [
+          "数与形 (归纳递推)"
+        ],
+        "source": "23-24六年级下·广东汕尾·期末",
+        "merged_id": "unit8_375",
+        "original_id": "375"
+      },
+      {
+        "id": "376",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "根据规律填空。$\\frac {1}{2}, \\frac {2}{3}, \\frac {3}{5}, \\frac {5}{8}, \\frac {8}{1 3}, (\\quad), (\\quad) 。$",
+        "options": [],
+        "answer": "$\\frac{13}{21}$ $\\frac{21}{34}$",
+        "solution": "【分析】观察可知, 下一个分数的分子等于前一个分数的分母, 下一个分数的分母等于前一个分数的分子加分母的和, 据此分析解答。【详解】  $8 + 13 = 21$  ，则第6个分数是  $\\frac{13}{21}$   $13 + 21 = 34$ , 则第 7 个分数是  $\\frac{21}{34}$   即  $\\frac{1}{2}, \\frac{2}{3}, \\frac{3}{5}, \\frac{5}{8}, \\frac{8}{13}, \\frac{13}{21}, \\frac{21}{34}$ 。",
+        "knowledge_points": [
+          "分数排列的规律"
+        ],
+        "source": "23-24 六年级下·天津南开·期末",
+        "merged_id": "unit8_376",
+        "original_id": "376"
+      },
+      {
+        "id": "377",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "下列图形都是由相同大小的正方形构成，按照图中的规律，第十个图形中有( )(1)个相同大小的正方形。第一个![](images/a8748da12db4f2fe48a6c728c7cf0be64e87b45c6cf114616b1ec146927fa904.jpg)  第二个![](images/b7a99ed20e52149b52857efcebef35196581939e061663a4253bf59b259dfc75.jpg)第三个![](images/1c19738eac148e1965dcf5605ed9e8cca5bb1f6bdaf3b5e52cf73aacaed97e64.jpg)",
+        "options": [],
+        "answer": "110",
+        "solution": "【分析】观察图形可得：![](images/95add3b6fe825525e89c2513b6c85425aad40c6648e64691360a64333b92e72a.jpg)第一个:  $2 = 1 + 1 \\times 1 = 1 + 1^{2}$![](images/fa9c2ea8766aa1ba5c93dcac8ae88b53e58140ddcca53166ed6b33cac62b7534.jpg)第二个:  $6 = 2 + 4 = 2 + 2 \\times 2 = 2 + 2^{2}$![](images/51dba530fcf7ebd38b42a2c271767da2bc6af6a8ffed3b523a9d7afef68ba3a4.jpg)第三个:  $12 = 2 + 4 + 6 = 3 + 3 \\times 3 = 3 + 3^{2}$![](images/91e7ec54da1e97eb75c19f75fede5882f4495ccc82ed932472f70f6b891d9ec5.jpg)按这样的规律，第四个： $20 = 2 + 4 + 6 + 8 = 4 + 4 \\times 4 = 4 + 4^{2}$第  $\\mathrm{n}$  个图形中就有  $( {\\mathrm{n} + {\\mathrm{n}}^{2}})$  个小正方形。根据规律即可计算出第十个图形中的小正方形的个数。【详解】根据分析可得：$1 0 + 1 0 ^ {2} \\ = 1 0 + 1 0 0 \\ = 1 1 0 (\\text {个})$第十个图形中有 110 个相同大小的正方形。",
+        "knowledge_points": [
+          "图形的变化规律",
+          "数与形（归纳递推）"
+        ],
+        "source": "23-24六年级上·全国·课后作业",
+        "merged_id": "unit8_377",
+        "original_id": "377"
+      },
+      {
+        "id": "378",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "![](images/293f97215a8e0a53cf91bfa85e102213a92c48957db496cbf4275558479e2532.jpg)![](images/0c42803f330df3d9ef7c8e063d9180b83415a03b2642d7dcdcc280d4eb27cc9a.jpg)![](images/084a86899a609eaf9ac6bc8774f3ebfcd9837a34640bb760409cb96f14fdaa10.jpg)摆10张桌子可以坐（）人。",
+        "options": [],
+        "answer": "22",
+        "solution": "【分析】观察图形可知，桌子的左右两边人数始终是2人，另外两个方向每增加1张桌子，就餐人数增加2人。1张桌子可以坐4人， $4 = 2 + 2$ ;2张桌子坐6人， $6 = 2 \\times 2 + 2$ ;3张桌子坐8人， $6 = 3 \\times 2 + 2$ ;……，由此可得：就餐人数=桌子数  $\\times 2 + 2$ ，据此解答。【详解】通过分析可得：就餐人数  $=$  桌子数  $\\times 2 + 2$ $4 \\times 2 + 2\\ = 8 + 2 \\ = 1 0 (\\text {人})$$5 \\times 2 + 2 \\ = 1 0 + 2 \\ = 1 2 \\text {(人)}$$1 0 \\times 2 + 2 \\ = 2 0 + 2 \\ = 2 2 \\text {(人)}$则摆 10 张桌子可以坐 22 人。",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "23-24六年级上·全国·课后作业",
+        "merged_id": "unit8_378",
+        "original_id": "378"
+      },
+      {
+        "id": "379",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "照这样把边长 1 厘米的正方形拼成长方形，用 5 个这样的正方形拼成的长方形周长是( )厘米;用  $\\mathrm{n}$  个这样的正方形拼成的长方形的周长是( )厘米。",
+        "options": [],
+        "answer": "12 $2(n + 1)$",
+        "solution": "【分析】根据长方形的周长  $=$  （长  $+$  宽）  $\\times 2$  。从题意可知：2个正方形拼成□□的长是2厘米，宽是1厘米，周长：（2+1）  $\\times 2 = 3\\times 2 = 6$  厘米;3个正方形拼成□□□的长是3厘米，宽是1厘米，周长：  $(3 + 1)\\times 2 = 4\\times 2 = 8$  厘米;按此规律，用5个这样的正方形拼成的长方形，长是5厘米，宽是1厘米，周长：  $(5 + 1)\\times 2 = 5\\times 2 = 10$  厘米……;由此可知，n个这样的正方形拼成的长方形，长是n厘米，宽是1厘米，周长：  $(\\mathrm{n} + 1)\\times 2 = 2(\\mathrm{n} + 1)$  厘米;据此解答。【详解】  $(5 + 1) \\times 2\\= 5 \\times 2\\= 10  (厘米)$$(n + 1) \\times 2= 2(n + 1)  厘米$![](images/64d4a744063f7d651042364948712c4c976597e9cdb56d5c8ec30afb1ab8ccb9.jpg)……照这样把边长 1 厘米的正方形拼成长方形，用 5 个这样的正方形拼成的长方形周长是 10 厘米;用  $\\mathrm{n}$  个这样的正方形拼成的长方形的周长是  $2(n + 1)$  厘米。",
+        "knowledge_points": [
+          "数与形（探索规律）",
+          "长、正方形周长的应用"
+        ],
+        "source": "2018·浙江杭州·小升初模拟",
+        "merged_id": "unit8_379",
+        "original_id": "379"
+      },
+      {
+        "id": "380",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如图, 用同样规格的正方形铺设地面，按规律依次铺下去，铺设第8个长方形地面共用( )块黑色地砖。![](images/13078ee791617f88206a4bd4c28665611001d0e20cc700b6cc5c0e4c03396ce5.jpg)![](images/67ff410b5e8bb5f7c9d5faa7bb45ee2c9189e14accd044dc27cbbe887be1ec23.jpg)![](images/b227005b0dc3c47763f1e4b9c6d1864a5bb0b26efe617862be4cb2ac8e83f78e.jpg)",
+        "options": [],
+        "answer": "72",
+        "solution": "【分析】通过观察图形，可得规律：第1个长方形地面共用黑色地砖：  $1\\times 2 = 2$  （块）;  第2个长方形地面共用黑色地砖：  $2 \\times 3 = 6$  （块）;  第3个长方形地面共用黑色地砖： $3 \\times 4 = 12$ （块）;第  $n$  个长方形地面共用黑色地砖:  $[n \\times (n + 1)]$  块。根据所得规律，代入数据计算，即可求出铺设第 8 个长方形地面一共用的黑色地砖的块数，据此解答。【详解】由分析得:$ 8 \\times (8 + 1) \\ = 8 \\times 9 \\ = 7 2 \\text {(块)}$即铺设第8个长方形地面共用72块黑色地砖。",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "23-24 六年级下·湖南常德·期末",
+        "merged_id": "unit8_380",
+        "original_id": "380"
+      },
+      {
+        "id": "381",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "图中可看出每多摆 1 个三角形，需增加( )根小棒。搭10个这样的三角形要( )根小棒。![](images/e4ea75a113c6c02081c3ffc6287be2d1.png)",
+        "options": [],
+        "answer": "2 21",
+        "solution": "【分析】摆 1 个三角形需要 3 根小棒, 摆 2 个三角形需要 5 个小棒, 摆 3 个三角形需要 7 根小棒, 摆 4 个三角形需要 9 根小棒, 由此可知, 每多摆 1 个三角形, 需要增加 2 个小棒;摆1个三角形需要3根小棒，可以写成：  $2\\times 1 + 1$摆2个三角形需要5根小棒，可以写成：  $2\\times 2 + 1$摆3个三角形需要7根小棒，可以写成：  $2\\times 3 + 1$.........由此可知，摆  $n$  三角形需要小棒： $(2n + 1)$  根，据此求出n=10时，需要小棒的根数，据此解答。【详解】根据分析可知, 每增加 1 个三角形, 需要增加 2 个小棒;摆  $n$  个三角形需要  $(2n + 1)$  根。当  $n = 10$  时:$ 2 \\times 1 0 + 1 \\ = 2 0 + 1 \\ = 2 1 \\text {(根)}$图中可看出每多摆 1 个三角形, 需增加 2 根小棒. 搭 10 个这样的三角形要 21 根小棒。![](images/ca27f603b374a8cc224a2e2a613b4c92af3fc91b6ffa9a24ca08c82f5ea116e9.jpg)",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "2024·湖南衡阳·小升初真题",
+        "merged_id": "unit8_381",
+        "original_id": "381"
+      },
+      {
+        "id": "382",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如图，隆隆在摆圆片，他在第一层摆了2个圆片，在第二层摆了4个圆片，照这样摆下去，在第5层需要摆( )个圆片，在第7层需要摆( )个圆片。![](images/8c2e0e60e090b6b9a50eb55401b559fc6b48103240cffa9792779a6d3ca846ee.jpg)",
+        "options": [],
+        "answer": "10 14",
+        "solution": "【分析】根据题意可知，第一层有  $2 \\times 1 = 2$  （个）圆片，第二层有  $2 \\times 2 = 4$  （个）圆片，第三层有  $2 \\times 3 = 6$  （个）圆片，第四层有  $2 \\times 4 = 8$  （个）圆片，用层数乘2即可求出该层需要摆的圆片个数，据此填空即可。【详解】  $5 \\times 2 = 10$  （个） $7 \\times 2 = 1 4 (\\text {个})$  在第 5 层需要摆 10 个圆片，在第 7 层需要摆 14 个圆片。",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "23-24四年级下·湖北襄阳·期末",
+        "merged_id": "unit8_382",
+        "original_id": "382"
+      },
+      {
+        "id": "383",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "找规律填空。36, 6, (),  $\\frac{1}{36}, \\frac{1}{216}, \\cdots$",
+        "options": [],
+        "answer": "1 $\\frac{1}{6}$",
+        "solution": "【分析】  $36 \\div 6 = 6$  、  $\\frac{1}{36} \\div 6 = \\frac{1}{36} \\times \\frac{1}{6} = \\frac{1}{216}$  ，由此可知，$前一个数\\div 6 =$  后一个数，据此填空。【详解】  $6 \\div 6 = 1$  、  $1 \\div 6 = \\frac{1}{6}$    36, 6, 1,  $\\frac{1}{6}, \\frac{1}{36}, \\frac{1}{216}, \\cdots$",
+        "knowledge_points": [
+          "分数排列的规律"
+        ],
+        "source": "23-24 六年级下·四川凉山·期末",
+        "merged_id": "unit8_383",
+        "original_id": "383"
+      },
+      {
+        "id": "384",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "用小棒按照如下方式摆图形, 用 2024 根小棒可摆( )个正八边形。![](images/d6b3fdbe8e7e6bc8c0da16083e96364fc2685a31af973f5083ac01231c5b6a8d.jpg)![](images/174bdf0b2baea43b921f48bf37f55f1846d9eeab6652a2bbdc34194ba88da15b.jpg)![](images/206ec14b263f6b4f4bba17b40cae1c62a0556905ce01453e27517a4571e207e2.jpg)",
+        "options": [],
+        "answer": "289",
+        "solution": "【分析】摆 1 个正八边形需要的小棒数为 8 根, 即  $7 \\times 1+$  1 ; 摆 2 个正八边形需要的小棒数为 15 根, 即  $7 \\times 2 + 1$ ; 摆 3 个正八边形需要的小棒数为 22 根, 即  $7 \\times 3 + 1$ ; ……摆 n 个正八边形需要的小棒数为:  $7 \\mathrm{n} + 1$  。求用 2024 根小棒可摆多少个正八边形, 就是当  $7 \\mathrm{n} + 1 = 2024$  时, 求  $\\mathrm{n}$  的值。【详解】摆  $n$  个正八边形需要的小棒数为:  $7 n + 1$ , 由此列方程得:$7 n + 1 = 2 0 2 4$解:  $7 \\mathrm{n} + 1 - 1 = 2024 - 1\\\\7 \\mathrm {n} = 2 0 2 3\\\\7 \\mathrm {n} \\div 7 = 2 0 2 3 \\div 7\\\\n = 2 8 9$所以用 2024 根小棒可摆 289 个正八边形。",
+        "knowledge_points": [
+          "数与形 (归纳递推)"
+        ],
+        "source": "23-24 六年级下·黑龙江鸡西·期末",
+        "merged_id": "unit8_384",
+        "original_id": "384"
+      },
+      {
+        "id": "385",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "找规律:1，2，5，13，34，89，",
+        "options": [],
+        "answer": "233",
+        "solution": "【分析】通过观察可知,  $2 \\times 2 + 1 = 5, 5 \\times 2 + 1 + 2 =13,  13 \\times 2 + 1 + 2 + 5 = 34$ . 以此类推, 从第二项开始, 每项乘以 2 再加上前面所有的项, 得后一项。据此解答。本题属于较难题，需要善于从复杂的数据中发现规律。【详解】  $89 \\times 2 + 1 + 2 + 5 + 13 + 34\\\\ = 1 7 8 + 1 + 2 + 5 + 1 3 + 3 4 \\\\ = 2 3 3$1, 2, 5, 1 3, 3 4, 8 9, 2 3 3 。",
+        "knowledge_points": [
+          "整数排列的规律"
+        ],
+        "source": "2015 六年级·全国·竞赛",
+        "merged_id": "unit8_385",
+        "original_id": "385"
+      },
+      {
+        "id": "386",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "按规律填空:  $\\frac{1}{3}, \\frac{4}{4}, \\frac{9}{8}, \\frac{16}{17}, (\\quad), \\frac{36}{58}$ 。",
+        "options": [],
+        "answer": "$\\frac{25}{33}$",
+        "solution": "【分析】观察可知, 分子分别是 1 的平方、2 的平方、3 的平方…，前一个分数的分子和分母的和是后一个分数的分母，据此分别确定分子和分母，写出这个分数即可。【详解】  $5^{2} = 5 \\times 5 = 25$    $1 6 + 1 7 = 3 3$",
+        "knowledge_points": [
+          "分数排列的规律"
+        ],
+        "source": "23-24 六年级下·湖南永州·期末",
+        "merged_id": "unit8_386",
+        "original_id": "386"
+      },
+      {
+        "id": "387",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "![](images/e0fad959772d4576ffabf95ef87647ec.png)图中一共有10条线段。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】根据题意可知：![](images/29b4c2e38813929afc9706e4ce34075e62010a194a02fdfa4d1930153486eb69.jpg)两点间有 1 条线段;![](images/ce922a64a44e0ff0433de09e0e6ef2e0f50300b9636dc49b12ba1e4a7fe2772d.jpg)三点间有  $1 + 2 = 3$  条线段;![](images/d66c73d6a72d45623f99154ba34c2bec8eb0307b6c5a83901a70a8bdaf33efb6.jpg)四点间有  $1 + 2 + 3 = 6$  条线段;![](images/a8ed4d253e5cb373ad7367eb55fb6efbc514c61ea619e05b125ff93249dbca6a.jpg)五点间有  $1 + 2 + 3 + 4 = 10$  条线段;由此可知: 线段总数等于从 1 开始依次加到 (端点数一 1)。据此判断即可。【详解】根据分析可得：$1 + 2 + 3 + 4 = 10$  (条)即图中共有 10 条线段，原说法正确。故答案为:  √",
+        "knowledge_points": [
+          "数与形 (归纳递推)"
+        ],
+        "source": "2023·广东韶关·小升初真题",
+        "merged_id": "unit8_387",
+        "original_id": "387"
+      },
+      {
+        "id": "388",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 9 ^ {2} $",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】分别计算等号左边式子的结果和等号右边式子的结果，再判断大小是否相等。【详解】  $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9\\\\ = 3 + 3 + 4 + 5 + 6 + 7 + 8 + 9 \\\\ = 6 + 4 + 5 + 6 + 7 + 8 + 9 \\\\ = 1 0 + 5 + 6 + 7 + 8 + 9 \\\\ = 1 5 + 6 + 7 + 8 + 9\\\\ = 2 1 + 7 + 8 + 9 \\\\ = 2 8 + 8 + 9 \\\\ = 3 6 + 9 \\\\ = 4 5$$9 ^ {2} = 9 \\times 9 = 8 1$因此，  $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 \\neq 9^{2}$故答案为:  ×",
+        "knowledge_points": [
+          "算式的规律（整数）"
+        ],
+        "source": "24-25 六年级上·湖北荆州·期末",
+        "merged_id": "unit8_388",
+        "original_id": "388"
+      },
+      {
+        "id": "389",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "计算。$\\frac {1}{1 \\times 2} + \\frac {1}{2 \\times 3} + \\frac {1}{3 \\times 4} + \\dots + \\frac {1}{2 0 1 7 \\times 2 0 1 8} =$",
+        "options": [],
+        "answer": "$\\frac{2017}{2018}$",
+        "solution": "【分析】观察每个分数，其分子都是 1，分母是两个相邻自然数相乘的形式; 先把每个分数写成两个分数相减的形式,即  $(\\frac{1}{1 \\times 2} = 1 - \\frac{1}{2}, \\frac{1}{2 \\times 3} = \\frac{1}{2} - \\frac{1}{3})$ ; 中间的分数可以相互抵消, 据此解答。【详解】  $\\frac{1}{1\\times 2} +\\frac{1}{2\\times 3} +\\frac{1}{3\\times 4} +\\dots +\\frac{1}{2017\\times 2018}$$ (1 - \\frac {1}{2}) + (\\frac {1}{2} - \\frac {1}{3}) + (\\frac {1}{3} - \\frac {1}{4}) + \\dots + (\\frac {1}{2 0 1 7} - \\frac {1}{2 0 1 8}) \\\\ = 1 - \\frac {1}{2} + \\frac {1}{2} - \\frac {1}{3} + \\frac {1}{3} - \\frac {1}{4} + \\dots + \\frac {1}{2 0 1 7} - \\frac {1}{2 0 1 8} \\\\=1 - \\frac {1}{2 0 1 8} \\\\ =\\frac {2 0 1 7}{2 0 1 8}$",
+        "knowledge_points": [
+          "分数加、减简便运算",
+          "用转化法解决分数计算问题（数形结合）"
+        ],
+        "source": "23-24 六年级上·湖南永州·期中",
+        "merged_id": "unit8_389",
+        "original_id": "389"
+      },
+      {
+        "id": "390",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "计算。$\\frac {1}{3} + \\frac {1}{9} + \\frac {1}{2 7} + \\frac {1}{81}$",
+        "options": [],
+        "answer": "$\\frac{40}{81}$",
+        "solution": "【分析】可先原式扩大到原来的 3 倍, 即原式乘 3 , 再减去原式, 得到原式的 2 倍, 再乘  $\\frac{1}{2}$ , 得到原式的结果。计算时,运用乘法分配律和加法交换律、结合律，进行简便运算。【详解】  $\\frac{1}{3} +\\frac{1}{9} +\\frac{1}{27} +\\frac{1}{81} \\\\=[ (\\frac {1}{3} + \\frac {1}{9} + \\frac {1}{2 7} + \\frac {1}{8 1}) \\times 3 - (\\frac {1}{3} + \\frac {1}{9} + \\frac {1}{2 7} + \\frac {1}{8 1})] \\times \\frac {1}{2} \\\\ = [ 1 + \\frac {1}{3} + \\frac {1}{9} + \\frac {1}{2 7} - \\frac {1}{3} - \\frac {1}{9} - \\frac {1}{2 7} - \\frac {1}{8 1}) \\times \\frac {1}{2} \\\\ = [ 1 + (\\frac {1}{3} - \\frac {1}{3}) + (\\frac {1}{9} - \\frac {1}{9}) + (\\frac {1}{2 7} - \\frac {1}{2 7}) - \\frac {1}{8 1}] \\times \\frac {1}{2} \\\\ = [ 1 - \\frac {1}{8 1}] \\times \\frac {1}{2} \\\\= \\frac {8 0}{8 1} \\times \\frac {1}{2} \\\\ = \\frac {4 0}{8 1}$",
+        "knowledge_points": [
+          "分数的四则混合运算",
+          "异分母分数加、减法"
+        ],
+        "source": "23-24 六年级上·全国·课后作业",
+        "merged_id": "unit8_390",
+        "original_id": "390"
+      },
+      {
+        "id": "391",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "计算:  $\\frac{1}{3} + \\frac{1}{9} + \\frac{1}{27}$",
+        "options": [],
+        "answer": "$\\frac{13}{27}$",
+        "solution": "【分析】从第二个数起，每个数是前一个数的  $\\frac{1}{3}$ ，可以通分来计算，但如果再按算式中的规律增加几个分数，用通分的方法计算就比较麻烦，可以画图来表示这道算式，先画了一个正方形表示'1'，平均分成3份，涂色  $\\frac{1}{3}$ ，再把另一个  $\\frac{1}{3}$  平均分成3份，涂色表示  $\\frac{1}{9}$ ，……如图。![](images/3f16390d9a53c7dfc8d39c989d930f62f466982d8212c866db23051038aa149f.jpg)求  $\\frac{1}{3} + \\frac{1}{9} + \\frac{1}{27}$  的和, 就是看图中涂色部分表示多少。从图中不难看出, 涂色部分接近 '1' 的一半, 把图中间的  $\\frac{1}{27}$  平均分成 2 份, 每份是  $\\frac{1}{54}$ , 因此直接计算  $\\frac{1}{2} - \\frac{1}{54}$  即可。【详解】  $\\frac{1}{3} +\\frac{1}{9} +\\frac{1}{27}\\\\=\\frac {1}{2} - \\frac {1}{5 4} \\\\= \\frac {2 6}{5 4} \\\\ =\\frac {1 3}{2 7}$",
+        "knowledge_points": [
+          "数与形（探索规律）",
+          "分数的加、减法混合运算"
+        ],
+        "source": "23-24 六年级上·全国·课后作业",
+        "merged_id": "unit8_391",
+        "original_id": "391"
+      },
+      {
+        "id": "392",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "小明带着一只狗在马路边散步，这条马路长800米，小明和狗都以匀速向前走。当小明走到马路的一半时，小狗已经到达了马路的终点，然后小狗又返回与小明相向而行，遇到小明后又跑向终点，到达终点后再与小明相向而行……直到小明到达终点。小狗从出发开始，一共跑了多少米？",
+        "options": [],
+        "answer": "1600 米",
+        "solution": "【分析】根据题意，当小明走到马路的一半时，小狗已经到达了马路的终点，说明小狗的速度是小明的 2 倍。因为小明和狗都以各自的速度匀速行走，所以在相同的时间内，小狗跑的路程是小明的 2 倍。小明到达终点时走了 800 米,则小狗跑了  $(800 \\times 2)$  米。【详解】  $800 \\times 2 = 1600$  （米）答: 小狗从出发开始, 一共跑了 1600 米。",
+        "knowledge_points": [
+          "相遇问题"
+        ],
+        "source": "24-25六年级上·湖北随州·期末",
+        "merged_id": "unit8_392",
+        "original_id": "392"
+      },
+      {
+        "id": "393",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "观察下面三幅点阵图，按照这样的规律，第10幅图中有多少个点，第n幅图中有多少个点？请说明理由。第1幅图![](images/aa5b06acbcff276a03404299b4a221eb4259ef42c1ee6349ce047c1a97a22d68.jpg)  第2幅图![](images/87f1a29f35c25a52bc5d0a6c8155479018776d07b61c0f338e430d9f22732304.jpg)  第3幅图![](images/1e52f4861062ecd98258300945b79aec598fc29278cefe784e71320e7f879bf5.jpg)  ........",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】观察图形，第一个图形需要  $(1 + 2 + 3)$  个点，第二个图形需要  $(2 + 3 + 4)$  个点，第三个图形需要  $(3 + 4 + 5)$  个点，依次类推，算出第 10 个图形需要的点数。第  $n$  个图形需要  $n + (n + 1) + (n + 2)$  个点，据此解答。【详解】  $10 + 11 + 12\\\\ = 2 1 + 1 2 \\\\ = 3 3 (\\text {个})$第 10 幅图中有 33 个点, 第 n 个这样的点阵图中有 n + (n+ 1) + (n+2) = (3n+3) 个点。",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "24-25 六年级上·黑龙江齐齐哈尔·期末",
+        "merged_id": "unit8_393",
+        "original_id": "393"
+      },
+      {
+        "id": "394",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "一棵树干第一年长出 2 个枝丫，以后每一年每个枝丫上会再长出 2 个新的枝丫，第 3 年这棵树可以长出多少个新枝丫？",
+        "options": [],
+        "answer": "18个",
+        "solution": "【分析】把树干也看作 1 个枝丫，第一年长出 2 个枝丫，第二年  $(1 + 2)$  个枝丫可以长出  $(1 + 2) \\times 2 = 6$  （个）枝丫，第三年  $(1 + 2 + 6)$  个枝丫可以长出  $(1 + 2 + 6) \\times 2 = 18（个$枝丫，据此即可解答。每年的枝数等于前面各年长出的枝数加当年长出的枝数, 再加上树干 (树干也要看作 1 枝) 的和。【详解】第一年：  $1 \\times 2 = 2$  （个）第二年：  $(1 + 2)\\times 2= 3 \\times 2= 6$  （个）第三年：  $(1 + 2 + 6)\\times 2 = 9 \\times 2  = 1 8 \\text {(个)}$答: 第 3 年这棵树可以长 18 个新枝丫。",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "2024 六年级·全国·竞赛",
+        "merged_id": "unit8_394",
+        "original_id": "394"
+      },
+      {
+        "id": "395",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "有 3 个细胞，在自然状态下每天每个细胞由 1 个分裂为 2 个，分裂后新旧细胞每天死去 2 个，1 天后有细胞 4 个，2 天后有细胞 6 个，依此类推，10 天后有多少个细胞？",
+        "options": [],
+        "answer": "1026 个",
+        "solution": "【分析】第一天后： $3 \\times 2 - 2 = 4$ （个） $= (2^{1} + 2)$ 个第二天后：  $4 \\times 2 - 2 = 6$  （个）  $= (2^{2} + 2)$  个第三天后： $6 \\times 2 - 2 = 10$  （个） $= (2^3 + 2)$  个第四天后： $10 \\times 2 - 2 = 18$  （个） $= (2^4 + 2)$  个1··········第10天后：  $(2^{9} + 2)\\times 2 - 2 = (2^{10} + 2)$  个【详解】  $2^{10} + 2\\\\ = 2 \\times 2 \\times 2 \\times 2 \\times 2 \\times 2 \\times 2 \\times 2 \\times 2 + 2 \\\\ = 4 \\times 4 \\times 4 \\times 4 \\times 4 + 2 \\\\ = 1 6 \\times 1 6 \\times 4 + 2 \\\\ = 2 5 6 \\times 4 + 2 \\\\ = 1 0 2 4 + 2 \\\\ = 1 0 2 6 \\text {(个)}$答: 10 天后有 1026 个细胞。",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "2024 六年级·全国·竞赛",
+        "merged_id": "unit8_395",
+        "original_id": "395"
+      },
+      {
+        "id": "396",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "爸爸开车带明明去动物园，在去的路上，明明画出了汽车的速度随时间的变化情况，如图所示。![](images/52d2cf69abc71af6f43f36bfdae16cf4ae24167aec3d1b211ec40f4741a95b20.jpg)(1) 汽车行驶了（）分钟，它行驶的最大速度是（）千米/时。(2) 出发后 8 分钟到 10 分钟这段时间可能出现什么情况?(3) 用自己的语言描述这辆车的行驶情况。",
+        "options": [],
+        "answer": "(1) 16; 60(2) 见详解  (3) 见详解",
+        "solution": "【分析】（1）从图中可知，0 分钟出发，18 分钟到达动物园，途中  $8 \\sim 10$  分钟这段时间停车，所以汽车的行驶时间要用总时间减去停车的 2 分钟;折线的最高点表示汽车行驶的最大速度。(2) 图中 8 分钟到 10 分钟这段时间, 路程为 0 , 表示汽车停车, 没有行驶, 结合生活实际, 得出可能出现的情况。(3) 结合图象, 可以分时间段描述这辆车的行驶情况, 合理即可。【详解】 (1)  $18 - (10 - 8)\\\\ = 1 8 - 2 \\\\ = 1 6 \\text {(分)}$汽车行驶了 16 分钟, 它行驶的最大速度是 60 千米/时。(2) 出发后 8 分钟到 10 分钟这段时间可能出现的情况: 汽车加油或堵车等。(答案不唯一)(3)  $0 \\sim 2$  分时汽车加速行驶, 速度达到 60 千米/时,  $2 \\sim 6$  分时匀速行驶,  $6 \\sim 8$  分时减速行驶, 直到停下, 10 分时又开始加速, 12 分时达到 60 千米/时的速度,  $12 \\sim 16$  分匀速行驶,  $16 \\sim 18$  分开始减速直到停下, 到达目的地。（答案不唯一)",
+        "knowledge_points": [
+          "看图找关系",
+          "单式折线统计图"
+        ],
+        "source": "24-25六年级上·全国·课后作业",
+        "merged_id": "unit8_396",
+        "original_id": "396"
+      },
+      {
+        "id": "397",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "有三堆围棋子, 每堆有 60 枚, 第一堆黑子与第二堆的白子同样多, 第三堆有 10 枚白子。这三堆棋子中, 白子比黑子少多少枚?",
+        "options": [],
+        "answer": "40 枚",
+        "solution": "【分析】由题可得：第一堆黑子  $+$  第一堆白子  $= 60$  枚，第二堆黑子  $+$  第二堆白子  $= 60$  枚，因为第一堆黑子与第二堆的白子同样多，所以第二堆的白子  $+$  第一堆白子  $= 60$  枚，第一堆黑子  $+$  第二堆黑子  $= 60$  枚，因此可知第一二堆的黑子数量和等于第一二堆的白子数量和。在第三堆中可求出白子比黑子少多少枚，就是这三堆棋子中白子比黑子少多少枚。第三堆有10枚白子，即黑子有：  $60 - 10 = 50$  （枚），最后用黑子数减去白子数，即可求出白子比黑子少多少枚，据此解答。【详解】  $60 - 10 = 50$  （枚）$50 - 10 = 40$  (枚)  答: 这三堆棋子中, 白子比黑子少 40 枚。",
+        "knowledge_points": [
+          "等量代换"
+        ],
+        "source": "23-24六年级上·全国·课后作业",
+        "merged_id": "unit8_397",
+        "original_id": "397"
+      },
+      {
+        "id": "398",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "学校一年一度的艺术节即将开幕，五（2）班的节目是一个团体操表演。在排练时，同学们排成了下面的队形。廖老师觉得阵容不够大，所以她决定再增加一些人参加团体操表演，但是要保持队伍形状不变，至少应该增加多少人？![](images/c01992320ec454ee6a87164fe091ae9c524e0ea1c32b8a650762150e6d3d0031.jpg)",
+        "options": [],
+        "answer": "9人",
+        "solution": "【分析】观察队形可知，原来的队形第一排有 1 人，第二排有 2 人。有2人，……，总人数是  $1 + 2 + 3 + 4 + 3 + 2 + 1 = 4 \\times 4 =$ 16（人），要保持队形不变，人数增加最少，第五排应有5  人, 总人数可以变成  $1 + 2 + 3 + 4 + 5 + 4 + 3 + 2 + 1 = 5 \\times 5 =$ 25（人），增加人数就是  $25 - 16 = 9$  （人）。【详解】  $1 + 2 + 3 + 4 + 3 + 2 + 1\\\\ = 4 \\times 4 \\\\ = 1 6 (\\text {人})$$ 1 + 2 + 3 + 4 + 5 + 4 + 3 + 2 + 1 \\\\ = 5 \\times 5 \\\\ = 2 5 \\text {(人)}$$25 - 16 = 9$  (人)答: 至少应该增加 9 人。",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "23-24 五年级上·辽宁·课后作业",
+        "merged_id": "unit8_398",
+        "original_id": "398"
+      },
+      {
+        "id": "399",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "古希腊著名的毕达哥拉斯学派经常把数与形联系在一起，下图是用形来表示数，请你认真观察：第1幅图的点数为1，第2幅图的点数为5，第3幅图的点数为9。![](images/e62ce435d30e0a90b333c03a7bf6c8c011c3e99c7fb9f748d136a944b919f834.jpg)(1)按这个规律排下去，第5幅图的点数是( )。(2)第 20 幅图的点数是( )。(3)第  $n$  幅图的点数是( )。",
+        "options": [],
+        "answer": "(1)17 (2)77    (3)  $4 \\mathrm{n} - 3$",
+        "solution": "【分析】 (1) (2) (3) 通过观察图，发现规律：第1幅图的点数为  $1 + 0 \\times 4 = 1 + 0 = 1$  第2幅图的点数为  $1 + 1 \\times 4 = 1 + 4 = 5$  第3幅图的点数为  $1 + 2 \\times 4 = 1 + 8 = 9$第4幅图的点数为  $1 + 3 \\times 4 = 1 + 12 = 13$·············第  $n$  幅图的点数为:  $1 + (n - 1) \\times 4 = 1 + 4n - 4 = 4n - 3$通过发现的规律，算出第 5 幅图和第 20 幅图的点数是多少,据此解答。【详解】 (1)  $4 \\times 5 - 3 = 20 - 3 = 17$  （个）按这个规律排下去，第5幅图的点数是  $17$  。(2)  $4 \\times 20 - 3 = 80 - 3 = 77$  （个）第20幅图的点数是77。(3) 由分析得: 第  $n$  幅图的点数是  $4 n - 3$  。",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "23-24 六年级下·湖南株洲·期末",
+        "merged_id": "unit8_399",
+        "original_id": "399"
+      },
+      {
+        "id": "400",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "甲乙两人从相距  $50 \\mathrm{~km}$  的两地同时相向而行, 甲每小时走  $7 \\mathrm{~km}$ , 乙每小时走  $3 \\mathrm{~km}$ , 甲带一只狗每小时走  $9 \\mathrm{~km}$ , 当狗一遇到乙时又返回甲处, 一遇到甲时又返回乙处, 直到两人相遇, 求小狗走的路程。",
+        "options": [],
+        "answer": "45千米",
+        "solution": "【分析】此题为相遇问题, 先根据相遇时间  $=$  路程  $\\div$  速度和,把甲和乙的速度相加求出速度和, 再用甲乙两地的路程除以这个和即可求出相遇的时间, 相遇的时间即为小狗跑的时间, 狗 1 小时行驶 9 千米, 根据路程  $=$  速度  $\\times$  时间, 代入数据, 即可求出小狗所走的路程。【详解】  $50 \\div (7 + 3)\\\\= 5 0 \\div 1 0\\\\=5 (小时)$$9 \\times 5 = 4 5 (\\text {千 米})$答: 小狗走的路程是 45 千米。",
+        "knowledge_points": [
+          "相遇问题",
+          "除数是整十数的口算除法"
+        ],
+        "source": "23-24 四年级下·江苏宿迁·期末",
+        "merged_id": "unit8_400",
+        "original_id": "400"
+      },
+      {
+        "id": "401",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "(23-24 六年级下·河南周口·期末) 将黑色棋子按照一定规律排列成一系列如右图所示的图案, 按照此规律, 图 n 中有 ( ) 枚黑棋子。![](images/10d39f23252a605e03016e42f3cdc014.png)",
+        "options": [
+          "A. 5n + 3",
+          "B. 5n - 2",
+          "C. 4n+ 3"
+        ],
+        "answer": "A",
+        "solution": "【分析】观察图形可知, 图 1、图 2、图 3……中分别有黑棋子 8 枚、13 枚、18 枚……由此发现: 后一个图比前一个图的黑棋子数量多5枚，据此找到规律。【详解】观察图形可知：图1中有8枚黑棋子， 8 = 5 × 1 + 3 ;  图2中有13枚黑棋子，  13 = 5× 2 + 3  图3中有18个枚黑棋子， 18 = 5 × 3 + 3 ;·······按照此规律：图 n 中有  (5 n + 3)  枚黑棋子。故答案为: A",
+        "knowledge_points": [
+          "用字母表示稍复杂的数量关系",
+          "数与形（探索规律）"
+        ],
+        "source": "23-24 六年级下·河南周口·期末",
+        "merged_id": "unit8_401",
+        "original_id": "401"
+      },
+      {
+        "id": "402",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "(22-23 六年级上·山西阳泉·期末) 已知: 3 × 9 =27 ，3 × 99 = 297 , 3 × 999 = 2997 , 3 × 9999 = 29997 , 那么:3 × 999999 = ( )。",
+        "options": [
+          "A. 299997",
+          "B. 3000007",
+          "C. 2999997"
+        ],
+        "answer": "C",
+        "solution": "【分析】根据算式的规律可以看出，从第二个算式开始,每个算式的结果的第一个数是 2 , 最后一个数是 7 , 中间是 9 , 乘数里面有几个 9 , 积里面 9 的个数比乘数里 9 的个数少 1 , 据此规律即可选择。【详解】由分析可知：3 × 9 9 9 9 9 9 = 2 9 9 9 9 9 7 故答案为: C",
+        "knowledge_points": [
+          "算式的规律（整数）"
+        ],
+        "source": "22-23 六年级上·山西阳泉·期末",
+        "merged_id": "unit8_402",
+        "original_id": "402"
+      },
+      {
+        "id": "403",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "（2024 六年级下·四川·专题练习）聪聪用小木棒搭三角形（如图），他搭 n 个这样的三角形要用（）根小棒。![](images/c51c0da2f16d21d60a6a213a7d44e0234ef0a6c012afd6f54b07c7c6ce120e23.jpg) ![](images/4225de6ff14a385240dfb1379097d84eadde6157932caeeb1018203189c933be.jpg) ![](images/67edc15888b71355930f6b4d2544a36c8ce562a72566941b2dc41bd463b60fb7.jpg) ![](images/e11f74891f4bd36778441c603327268bca30398eb5d15f737f1e6263ad390d0e.jpg) ···",
+        "options": [
+          "A. n",
+          "B. 2 n - 1",
+          "C. 2 n",
+          "D. 2 n + 1"
+        ],
+        "answer": "D",
+        "solution": "【分析】根据图意, 1 个三角形用 (1 × 2 + 1) 根火柴, 2 个三角形用 (2 × 2 + 1) 根火柴, 3 个三角形用 (3 × 2 + 1) 根火柴, 则 n 个三角形用 (n × 2 + 1) 根火柴，据此解答即可。【详解】分析可知, 用小木棒搭三角形 (如图), 他搭 n 个这样的三角形要用 (2 n + 1) 根小棒。故答案为: D",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "2024 六年级下·四川·专题练习",
+        "merged_id": "unit8_403",
+        "original_id": "403"
+      },
+      {
+        "id": "404",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "（2022·四川绵阳·小升初真题）如果有 2019 名学生排成一列，按 1，2，3，4，3，2，1，2，3，4，3，2，1... 的规律报数，那么第 2019 名学生所报的数是（）。",
+        "options": [
+          "A. 2",
+          "B. 1",
+          "C. 3",
+          "D. 4"
+        ],
+        "answer": "C",
+        "solution": "【分析】观察这组数的特点, 每 6 个数为一轮,1、2、3、4、3、2，再用 2019 除以 6，看余数，即可确定答案。本题是一道找规律的题目，对于此类题目首先应找出哪些部分发生了变化，是按照什么规律变化的。【详解】根据观察，每 6 个数为一轮。2 0 1 9 ÷ 6 = 3 3 6 ... 3则第 2019 名学生所报的数是 3故答案为: C",
+        "knowledge_points": [
+          "周期问题"
+        ],
+        "source": "2022·四川绵阳·小升初真题",
+        "merged_id": "unit8_404",
+        "original_id": "404"
+      },
+      {
+        "id": "405",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "(23-24 六年级上·河南安阳·期末) 照下图的样子摆下去,如果一个小三角形的边长为 1 cm , 第 8 个图形的周长是 ( ) cm。![](images/84e11a38112efe04f5ea7852481f08878fc3df2184c5f040a1d69086c2cb602d.jpg)",
+        "options": [
+          "A. 8",
+          "B. 10",
+          "C. 17",
+          "D. 24"
+        ],
+        "answer": "B",
+        "solution": "【分析】第一个图形的周长是 3 cm , 可以写成: (1 + 2) cm , 第二个图形的周长是 4 cm , 可以写成: (2 +2) cm, 第三个图形的周长是 5 cm , 可以写成: (3+2) cm, …, 由此可知, 每增加一个三角形周长就多 1 cm ,所以第 n 个图形的周长 = (n + 2) cm 。【详解】根据分析可知, 第 8 个图形的周长是: 8 + 2 = 10 (cm)照下图的样子摆下去, 如果一个小三角形的边长为1 cm , 第 8 个图形的周长是 10 cm 。![](images/d7ec913e0052a5f5571a63ac0ad5f07be5dafe4f2793365603b969aa587990d3.jpg)故答案为: B",
+        "knowledge_points": [
+          "数与形"
+        ],
+        "source": "23-24 六年级上·河南安阳·期末",
+        "merged_id": "unit8_405",
+        "original_id": "405"
+      },
+      {
+        "id": "406",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "(22-23 六年级上·福建漳州·期末) 根据下图排列规律,拼出第四个图形要用 ( ) 根小棒。![](images/ef1980256d2466b43fee201df4658c1924af3ec5984d7ac68f5ab8148ab3e8fc.jpg)![](images/10ac62ce2f335746151c5406a63722e8cda6230dddb1f2a376b5a16dc53b2301.jpg)![](images/1fc8180a52ee9d0d1979dea7cbb5f5569656cb95a773b3a72808f8447624ba8b.jpg)",
+        "options": [
+          "A. 21",
+          "B. 22",
+          "C. 23",
+          "D. 24"
+        ],
+        "answer": "A",
+        "solution": "【分析】观察图形可知，第一个图形用了 1 × 5 + 1 = 6 根小棒，第二个图形用了 2 × 5 + 1 = 11 根小棒，第三个图形用了 3 × 5 + 1 = 16 根小棒，则第 n 个图形用了（ 5n + 1 ）根小棒。据此解答即可。【详解】第四个图形用小棒的个数为：5 n + 1 = 5 × 4 + 1 = 2 0 + 1 = 2 1。则拼出第四个图形要用 21 根小棒。故答案为：A",
+        "knowledge_points": [
+          "数与形",
+          "图形的规律"
+        ],
+        "source": "22-23 六年级上·福建漳州·期末",
+        "merged_id": "unit8_406",
+        "original_id": "406"
+      },
+      {
+        "id": "407",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "(22-23 六年级上·重庆丰都·期末) 算式 1 + 3 + 5 + 7 + 5 + 3 + 1 的和等于 ( )。",
+        "options": [
+          "A. 7²",
+          "B. 4² + 4²",
+          "C. 4² + 3²"
+        ],
+        "answer": "C",
+        "solution": "【分析】把算式 1 + 3 + 5 + 7 + 5 + 3 + 1 看作两部分: 1 + 3 + 5 + 7 和 5 + 3 + 1 , 根据 '连续奇数的和等于奇数个数的平方' 可得, 1 + 3 + 5 + 7 = 4², 5 + 3 + 1 = 3² , 据此解答。本题是找规律的题型，从已知的数据中找到规律，并按规律解题。【详解】1 + 3 + 5 + 7 + 5 + 3 + 1 = 4 ² + 3 ² = 1 6 + 9 = 2 5。算式 1 + 3 + 5 + 7 + 5 + 3 + 1 的和等于 4² + 3² 。故答案为: C",
+        "knowledge_points": [
+          "算式的规律"
+        ],
+        "source": "22-23 六年级上·重庆丰都·期末",
+        "merged_id": "unit8_407",
+        "original_id": "407"
+      },
+      {
+        "id": "408",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "仔细分析, 摆第 5 个图形需要用 ( ) 根小棒。![](images/c5b4fa43bf5a5b69bc75211998cd5ee819a57960982988e371d9e99f1d26fcb7.jpg)image.png",
+        "options": [
+          "A. 20",
+          "B. 30",
+          "C. 26"
+        ],
+        "answer": "C",
+        "solution": "【分析】通过观察发现：每增加 1 个正六边形，就增加 5 根小棒。即第 1 个图形需要用 1 + 5 = 6 （根）小棒;第 2 个图形需要用 1 + 5 × 2 = 11 （根）;第 3 个图形需要用 1 + 5 × 3 = 16 （根）小棒;……由此发现规律：第 n 个图形需要用（1 +5n）根小棒。解答数形结合类规律探索问题时，要仔细观察数与形之间的关系，先找到数与形之间隐含的数学规律，再利用规律进行解答即可。【详解】1 + 5 × 5 = 1 + 2 5 =26 (根)。所以摆第5个图形需要用26根小棒。故答案为: C",
+        "knowledge_points": [
+          "数与形"
+        ],
+        "source": "22-23 六年级上·山东临沂·期末",
+        "merged_id": "unit8_408",
+        "original_id": "408"
+      },
+      {
+        "id": "409",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "（22-23六年级上·河南周口·期末）用小棒按照如下方式摆图形，摆n个八边形需要（）根小棒。![](images/f236d22264e2b95425dc94cfb5eae4956aa1043b8cd00239a950d39917f87e8d.jpg)![](images/0f2c0f02d07669cd28e2b17a7433d3b6a0e9520be15edafd0fc95cf1d5db3f10.jpg)",
+        "options": [
+          "A. 8n",
+          "B. 8 n - 1",
+          "C. 7 n + 1",
+          "D. 8 (n-1)"
+        ],
+        "answer": "C",
+        "solution": "【分析】根据图示发现：摆 1 个八边形需要小棒：8 根;摆 2 个八边形需要小棒（8+7）根;摆 3 个八边形需要小棒 (8 + 7 + 7) 根;……摆 n 个八边形需要小棒的根数是 8+ 7 (n-1)。据此解答。本题主要考查数与形结合的规律, 关键根据图示发现这组图形的规律, 并运用规律做题。【详解】8 + 7 (n - 1) = 8 + 7 n - 7 = (7 n + 1) 根。摆 n 个八边形需要 (7 n + 1) 根小棒。故答案为: C",
+        "knowledge_points": [
+          "数与形",
+          "图形的规律"
+        ],
+        "source": "22-23六年级上·河南周口·期末",
+        "merged_id": "unit8_409",
+        "original_id": "409"
+      },
+      {
+        "id": "410",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "(21-22 六年级下·山东济宁·期末) 按照下图规律, 摆第 7 个图形需要 ( ) 根小棒。![](images/80f248f45492c2e928d8acd9c684c394d3b8938cda8b2c1cf5454e00c5281ab6.jpg)",
+        "options": [
+          "A. 21",
+          "B. 22",
+          "C. 28"
+        ],
+        "answer": "B",
+        "solution": "【分析】第 1 个图形需要 1 + 3 = 4 （根）小棒，第 2 个图形需要 1 + 3 × 2 = 7 （根）小棒，第 3 个图形需要 1 + 3 × 3 = 10 （根）小棒，……由此发现规律：第 n 个图形需要（1+3n) 根小棒。在运用数形结合的方法探究数学规律时, 一定要把图形和数一一对应。【详解】1 + 3 × 7 = 1 + 2 1 = 2 2 (根)。所以摆第 7 个图形需要 22 根小棒。故答案为: B",
+        "knowledge_points": [
+          "数与形"
+        ],
+        "source": "21-22 六年级下·山东济宁·期末",
+        "merged_id": "unit8_410",
+        "original_id": "410"
+      },
+      {
+        "id": "411",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "（22-23六年级下·陕西咸阳·期末）下面是由完全相同的正方形按规律摆成的图形，第1个图中有1个正方形，第2个图中有3个正方形，第3个图中有6个正方形，第4个图中有10个正方形，……，按此规律，第8个图中有（）个正方形。![](images/7877c0cbdb4960cf03d753ba24d92a99.png)",
+        "options": [
+          "A. 42",
+          "B. 40",
+          "C. 38",
+          "D. 36"
+        ],
+        "answer": "D",
+        "solution": "【分析】观察可得规律, 第 1 个图中有 1 个正方形, 第 2 个图中有 1 + 2 = 3 个正方形, 第 3 个图中有 1 + 2 + 3 = 6 个正方形, 第 4 个图中有 1 + 2 + 3 + 4 = 10 个正方形, 按此规律,第 8 个图中有 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 个正方形, 据此解答。仔细观察，比较总结出规律是解决本题的关键。【详解】1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36 （个）按此规律，第 8 个图中有 36 个正方形。故答案为: D",
+        "knowledge_points": [
+          "数与形"
+        ],
+        "source": "22-23六年级下·陕西咸阳·期末",
+        "merged_id": "unit8_411",
+        "original_id": "411"
+      },
+      {
+        "id": "412",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "(22-23 六年级上·广东珠海·期末) 如下图, 第 5 个图形是由 ( ) 个小正方形拼成的。![](images/24d3281b01d521f6cb8abde257b22ae3b3cfa932dfe90efc8633c372715a5267.jpg)",
+        "options": [
+          "A. 16",
+          "B. 20",
+          "C. 25",
+          "D. 36"
+        ],
+        "answer": "C",
+        "solution": "【分析】观察图形, 第 1 个图形是由 1 × 1 = 1 = 1² 个小正方形拼成的, 第 2 个图形是由 2 × 2 = 4 = 2² 个小正方形拼成的,第 3 个图形是由 3 × 3 = 9 = 3² 个小正方形拼成的, 依次类推,可以看出第 5 个图形是由 5² 个小正方形拼成的。据此解答。此题的解题关键是利用数与形的结合，通过观察图形，把图形中变化的规律转化成数字，多多练习，培养数感。【详解】5² = 5 × 5 = 25 （个）即第 5 个图形是由 25 个小正方形拼成的。故答案为: C",
+        "knowledge_points": [
+          "数与形",
+          "图形的规律"
+        ],
+        "source": "22-23 六年级上·广东珠海·期末",
+        "merged_id": "unit8_412",
+        "original_id": "412"
+      },
+      {
+        "id": "413",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "(22-23 五年级下·四川绵阳·期末) 在100，95，90，85，80，…，20，15，10，5这一列数中，第12个数是（）。",
+        "options": [
+          "A. 45",
+          "B. 60",
+          "C. 65"
+        ],
+        "answer": "A",
+        "solution": "【分析】根据数排列的规律：前一个数比后一个数大 5，第二个数加 5 得到 100，第三个数加 2 个 5 得到 100，第 12 个数加 11 个 5 得到 100，据此可得出答案。本题主要考查的是数字排列的规律，解题的关键是掌握两数字之间相差 5 ，进而可得出答案。【详解】第 12 个数是用 100 减去 11 个 5 , 即 100 - 11 × 5 = 45。故答案为: A",
+        "knowledge_points": [
+          "数字排列的规律"
+        ],
+        "source": "22-23 五年级下·四川绵阳·期末",
+        "merged_id": "unit8_413",
+        "original_id": "413"
+      },
+      {
+        "id": "414",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(23-24 六年级下·重庆万州·期末) 下图中第 6 个正方形有( )个点, 第 n 个正方形的点子数表示为( )个。![](images/715d3b107aae18529f577978908e92289e53c4128a8f67afac9c059421249e32.jpg)",
+        "options": [],
+        "answer": "24;4n",
+        "solution": "【分析】观察图形可知：第1个正方形的点子数为4个， 4 = 4 × 1 ;第2个正方形比第1个正方形多了4个点，点子数为8个， 8 = 4 × 2 ;第3个正方形比第2个正方形又多了4个点，点子数为12个， 12 = 4 × 3 .由此可知，点子数 = 4 × 正方形的序数。据此解答。【详解】通过分析可得：4 × 6 = 2 4 (个)第 6 个正方形有 24 个点;第 n 个正方形的点子数表示为 4 n 个。",
+        "knowledge_points": [
+          "数与形（探索规律）"
+        ],
+        "source": "23-24 六年级下·重庆万州·期末",
+        "merged_id": "unit8_414",
+        "original_id": "414"
+      },
+      {
+        "id": "415",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "观察下面的点阵图,照这样画下去, 第 (8) 个点阵图中有( )个圆点, 第( )个点阵图中有 90 个圆点。![](images/95228c70d33ae1e300edfe0796ed098662feb80f4fcead4ad1fac677264fd2b2.jpg) ",
+        "options": [],
+        "answer": "27;29",
+        "solution": "【分析】根据题意得：第（1）个点阵图中有6个圆点，第(2) 个点阵图有 9 个圆点, 第 (3) 个点阵图有 12 个圆点。则可得出规律：后一个点阵图圆点数量比前一个点阵图圆点数量多 3 个, 即第 (1) 个圆点数量为: 3 × 1 + 3 , 第 (2)个圆点数量为: 3 × 2 + 3 , 第 (3) 个圆点数量为: 3 × 3 + 3 , 依次类推得出规律, 据此可得出答案。【详解】根据题意得，点阵图圆点变化规律为：后一个点阵图圆点数量比前一个点阵图圆点数量多3个，即第（1）个圆点数量为： 3 × 1 + 3 = 6 ，第（2）个圆点数量为： 3 × 2 + 3 = 9 ，第（3）个圆点数量为： 3 × 3 + 3 = 12 。则第n个点阵图的圆点数量为： (3n + 3) 个。第（8）个点阵图中有圆点：3 × 8 + 3 =24 + 3 =27 (个)。(90-3)÷3 =87÷3 =29(个)。即第 (29) 个点阵图中有 90 个圆点。",
+        "knowledge_points": [
+          "用字母表示数、数量关系",
+          "数与形（探索规律）"
+        ],
+        "source": "21-22 六年级下·湖南长沙·期末",
+        "merged_id": "unit8_415",
+        "original_id": "415"
+      },
+      {
+        "id": "416",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "根据规律填数。1/2、 1/4 、 1/7、 1/1 1、 1/1 6 、1/2 2 、 1/3 7  、1/2 9 、___",
+        "options": [],
+        "answer": "1/37",
+        "solution": "【分析】2 + 2 = 4, 4 + 3 = 7, 7 + 4 = 11, 11 + 5 = 16, 16 + 6 = 22, 22 + 7 = 28 ，通过观察可知，这列分数的分子都是1.分母从左往右依次增加 2, 3, 4, 5, 6, 7...，据此可知,第 8 个数的分子是 1 , 分母是 (29+8) , 据此解答。【详解】29 + 8 = 37。1/2 、 1/4、 1/7 、 1/1 1 、1/1 6 、1/2 2 、 1/2 9 、1/3 7",
+        "knowledge_points": [
+          "分数排列的规律"
+        ],
+        "source": "2012 六年级·全国·竞赛",
+        "merged_id": "unit8_416",
+        "original_id": "416"
+      }, {
+        "id": "417",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": " 如图, 3个同样的杯子叠在一起高20厘米, 5个这样的杯子叠起来高24厘米, 7个这样的杯子叠起来的高度是____厘米, n个这样的杯子叠在一起的高度是____厘米。![](images/13dfb68700e56236e346e41864bf5626ba9f032658dd77db91e0822a6bb6f826.jpg)",
+        "options": [],
+        "answer": "28;14 + 2n",
+        "solution": "【分析】（1）具体思路是先求出单个杯子的高度，再根据与已知条件的关系逐步计算其他情况的高度。求单个杯子的高度：5个杯子比3个杯子多了2个杯子，高度多了 24 - 20 = 4 厘米，所以一个杯子的高度是 4 ÷ 2 = 2 厘米。求7个杯子的高度：3个杯子高20厘米，那么7个杯子比3个杯子多4个杯子，多 4 × 2 = 8 厘米，所以7个杯子叠起来的高度是 20 + 8 = 28 厘米。(2) 已知杯子总高度为 20 厘米, 第一个杯子的高度是: 杯子总高度-间隔高度即为: 20 - 4 = 16 厘米, 已知杯子与杯子间隔高度是 2 厘米, 3 个杯子有 2 个间隔, 以此类推, n 个杯子就有 n-1 个间隔, 每个间隔的高度是 2 厘米, 有 n-1 个间隔, 所以间隔的总高度就是 (n - 1) × 2 , 最后可列式为: 16 + (n - 1) × 2 , 算出结果即可。【详解】（1）两个杯子的高度为： 24 - 20 = 4 （厘米）一个杯子的高度为: 4 ÷ 2 = 2 (厘米) 3 个杯子叠一起 20 厘米, 那么最下面一个杯子的高度为 20 -2 × 2 = 16 (厘米) 7 个杯子叠起来的高度为 16 + 2 × (7 - 1) = 16 + 12 = 28 (厘米) 7 个这样的杯子叠起来的高度是 28 厘米。(2) 16 + (n - 1) × 2 = 1 6 + 2 n - 2 = 1 4 + 2 n。n 个这样的杯子叠在一起的高度是 14 + 2 n 厘米",
+        "knowledge_points": [
+          "用字母表示数、数量关系",
+          "数与形（归纳递推)"
+        ],
+        "source": "23-24六年级下·浙江台州·期末",
+        "merged_id": "unit8_417",
+        "original_id": "417"
+      },
+      {
+        "id": "418",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "在圆周上任意给定 2011 个点，在圆内再选 2009 个点，以这 4020 个点为顶点构成的彼此不重叠的三角形最多有____个。",
+        "options": [],
+        "answer": "6027",
+        "solution": "【分析】![](images/e76b4c9abcd1634b8ed3882699671a29ffedb2c0129b123ab68a805b19e7e4eb.jpg)如图，在圆周上任意给定7个点，连成如图所示的5个三角形，为了彼此不重叠，再在这5个三角形的内部分别选择5个点，将每个三角形又分成了3个彼此不重叠的三角形。由此可得到规律：在圆上任意给定n个点，连成（n-2）个三角形，在圆内再选（n-2）个点，分别将每个三角形分成3个彼此不重叠的三角形，那么就有3（n-2）个彼此不重叠的三角形。解答本题可从数量较少的点进行推理, 得到相应的规律, 再利用规律进行解答。【详解】根据分析可知，在圆上任意给定 2011 个点，在圆内再选 2009 个点，使得以这 2009 个点为顶点构成尽可能多的彼此不重叠的三角形，三角形最多有（ 3 × 2009 ）个。3 × 2 0 0 9 = 6 0 2 7 (个)以这 4020 个点为顶点构成的彼此不重叠的三角形最多有 6027 个。",
+        "knowledge_points": [
+          "图形的变化规律"
+        ],
+        "source": "2012 六年级·全国·竞赛",
+        "merged_id": "unit8_418",
+        "original_id": "418"
+      },
+      {
+        "id": "419",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "按如图所示的规律, 第 8 个图形中有( )个阴影三角形, 第 14 个图形中有( )个空白三角形。![](images/808572f850a80893c0665806020401cc952ef59f916e6cb1a2cb8bff52314202.jpg)![](images/f22de4c20cf2a43dbda7a7fbcbaa11405fa7e5d490a44c34745e642c32754716.jpg)![](images/b6b32692500cf7d85890bea3ef75df0b2873cfcd77d77ff5794a8a3975b5a9c9.jpg)![](images/547a01062c2d78298da6303bc9580db361e4c23c870d583afbfb63f71b0b3335.jpg)",
+        "options": [],
+        "answer": "36  91",
+        "solution": "【分析】观察图形, 第一幅图有 1 个阴影三角形, 第二幅图有 3 个阴影三角形, 3 = 2 + 1 , 第三幅图有 6 个阴影三角形, 6 = 3 + 2 + 1 , 阴影三角形的个数 = 第几个图形就从几开始依次加到 1 ;第一幅图没有空白三角形，第二幅图有1个空白三角形，1= (1 - 1) + 1 , 第三幅图有 3 个空白三角形, 3 = 2 + 1 , 第4 幅图有 6 个空白三角形, 6 = 3 + 2 + 1 , 空白三角形的个数=第几个图形就从几减 1 开始依次加到 0。【详解】8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 36 （个） 1 4 - 1 = 1 3。1 3 + 1 2 + 1 1 + 1 0 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 + 0 = 9 1(个)第 8 个图形中有 36 个阴影三角形, 第 14 个图形中有 91 个空白三角形。",
+        "knowledge_points": [
+          "数与形 (归纳递推)"
+        ],
+        "source": "22-23 六年级上·河南驻马店·期末",
+        "merged_id": "unit8_419",
+        "original_id": "419"
+      },
+      {
+        "id": "420",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": " 用火柴棒搭平行四边形, 搭 1 个平行四边形用 4 根火柴棒, 搭 2 个平行四边形用7 根火柴棒, 搭 3 个平行四边形用 10 根火柴棒, 照这样的规律搭 40 个这样的平行四边形要( )根火柴棒。",
+        "options": [],
+        "answer": "121",
+        "solution": "【分析】由题意可知:搭 1 个平行四边形用 4 根火柴棒, 4 = 3 × 1 + 1 ; 搭 2 个平行四边形用 7 根火柴棒, 7 = 3 × 2 + 1 ; 搭 3 个平行四边形用 10 根火柴棒, 10 = 3 × 3 + 1 ;按此规律摆下去, 搭 n 个平行四边形要 (3 n + 1) 根火柴棒,据此解答。【详解】规律：搭 n 个平行四边形要 (3n + 1) 根火柴棒。当 n = 40 时。3 × 4 0 + 1 = 1 2 0 + 1 = 1 2 1 (根)。照这样的规律搭 40 个这样的平行四边形要 121 根火柴棒。",
+        "knowledge_points": [
+          "含有字母式子的化简与求值",
+          "用字母表示稍复杂的数量关系",
+          "数与形（探索规律）"
+        ],
+        "source": "2024 六年级下·四川·专题练习",
+        "merged_id": "unit8_420",
+        "original_id": "420"
+      },
+      {
+        "id": "421",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "将2004名同学排成一行，先从左到右依次按1，2，3，1，2，3，1，2，3...的顺序报数，再从右到左依次按1，2，3，4，1，2，3，4，1，2，3，4...的顺序报数。那么两次都报 '1'的同学共有____人。",
+        "options": [],
+        "answer": "167",
+        "solution": "【分析】根据题意可知, 2004 是 4 的倍数, 所以同学从右到左依次按 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4... 的顺序报数，相当于从左到右按 4，3，2，1，4，3，2，1... 的顺序报数;报数情况从左到右如下：1，2，3，1，2，3，1，2，3，1，2，3，1，2，3，1，2,3…4, 3, 2, 1, 4, 3, 2, 1, 4, 3, 2, 1, 4, 3, 2, 1, 4, 3, 2, 1...通过观察发现，从左到右第 4 位同学报两次 '1'，之后每12 位出现报两次 '1', 所以用 2004-4 求出第 5 位到第2004 位有多少位同学, 也就是 2000 位同学, 再用 2000 ÷ 12即可求出 2000 里面有几个 12，商是 166，余数是 8，所以166 加上 1 即可求出多少位同学报两次 '1'。本题可通过列举从左往右报的数, 找出它们出现的规律进行解答即可。【详解】2004 - 3 = 2001 （人）2001 ÷ 12 = 166 （个）……9（人）166+1=167(人)。166 + 1 = 167 （人）。两次都报 '1'的同学共有 167 人。",
+        "knowledge_points": [
+          "整数排列的规律"
+        ],
+        "source": "2007六年级·全国·竞赛",
+        "merged_id": "unit8_421",
+        "original_id": "421"
+      },
+      {
+        "id": "422",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "自然数的平方按从小到大的顺序排成一个多位数：14916253649…，第 190 个数位上的数字是_____。",
+        "options": [],
+        "answer": "3",
+        "solution": "【分析】平方数是一位数的有 1 ~ 3 , 共 3 个, 平方数是两位数的有 4 ~ 9 , 共 6 个, 平方数是三位数的有 10 ~ 31 , 共 22 个,平方数是四位数的有 32 ~ 99 , 共 68 个, 平方数少于四位数的一共有 3 × 1 + 6 × 2 + 22 × 3 = 81 (位) 数字, 81 < 190 , 68 × 4 = 272 , 272 > 190 , 所以平方数等于四位数的有 (190-81) 位数字, 也就是 109 个数字, 先用 109 ÷ 4 求出有几个四位数, 商是 27 , 余数是 1 , 所以一共有 27 完整的四位数,第 190 个数位上的数字是第 28 个平方数是四位数的首位数字, 即 (31 + 28)² 的首位数字。分别找到一位数、两位数、三位数、四位数……有几个是解答本题的关键。【详解】3 × 1 + 6 × 2 + 22 × 3 = 3 + 1 2 + 6 6 = 8 1 (个)。1 9 0 - 8 1 = 1 0 9 (个) 1 0 9 ÷ 4 = 2 7 ... 1。第 190 个数位上的数字是第 28 个平方数是四位数的首位数字, 即 (32 + 27) ² 的首位数字。(3 1 + 2 8) ² = 5 9 ² = 3 4 8 1。第 190 个数位上的数字是 3。",
+        "knowledge_points": [
+          "整数排列的规律"
+        ],
+        "source": "2008 六年级下·全国·竞赛",
+        "merged_id": "unit8_422",
+        "original_id": "422"
+      },
+      {
+        "id": "423",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（2024 六年级下·四川·专题练习）按如图用小棒摆正六边形，摆第 n 组正六边形需要( )根小棒。![](images/67c8d20a63d17a8caed67e0614b39497c2875ec441c5782b7cb8ae79c88ef8fb.jpg)",
+        "options": [],
+        "answer": "5n+1",
+        "solution": "【分析】摆 1 个六边形需要 6 根小棒, 可以写作: 5 × 1 + 1 ;摆 2 个需要 11 根小棒, 可以写作: 5 × 2 + 1 ; 摆 3 个需要 16 根小棒, 可以写成: 5 × 3 + 1 ; …由此可以推理得出一般规律:5 × 正六边形的个数 +1 = 6 小棒的根数, 然后解答即可。【详解】当 n = 1 时，需要小棒 1 × 5 + 1 = 6 （根）当 n = 2 时, 需要小棒 2 × 5 + 1 = 11 (根)当 n = 3 时, 需要小棒 3 × 5 + 1 = 16 (根)摆第 n 组正六边形需要 (5 n + 1) 根小棒。",
+        "knowledge_points": [
+          "数与形（探索规律）",
+          "图形的运动规律"
+        ],
+        "source": "2024 六年级下·四川·专题练习",
+        "merged_id": "unit8_423",
+        "original_id": "423"
+      },
+      {
+        "id": "424",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(22-23 六年级上·山西忻州·期末) 1/2 + 1/4 + 1/8 + ... + 1/256 + 1/512 = .",
+        "options": [],
+        "answer": "511/512",
+        "solution": "【分析】观察算式，发现规律： 1/2 = 1 - 1/2, 1/4 = 1/2 - 1/4, 1/8 = 1/4 - 1/8, ，据此规律把算式进行简算。【详解】1/2 +1/4 +1/8 +... +1/256 +1/512 =1 - 1/2 + 1/2 - 1/4 + 1/4 - 1/8 + ... + 1/1 2 8 - 1/2 5 6 + 1/2 5 6 - 1/5 1 2 =1 - 1/5 1 2 = 5 1 1/5 1 2",
+        "knowledge_points": [
+          "算式的规律（分数）"
+        ],
+        "source": "22-23 六年级上·山西忻州·期末",
+        "merged_id": "unit8_424",
+        "original_id": "424"
+      },
+      {
+        "id": "425",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "（2007 六年级·全国·竞赛）有一列数，第一个数是 5，第二个数是 2，从第三个数起每个数都等于它前面两个数中较大数减去较小数的差。则这列数中前 100 个数之和等于",
+        "options": [],
+        "answer": "77",
+        "solution": "【分析】通过观察可知，前几个数是5、2、3、1、2、1、1、0、1、1、0、1、1、0…从第6个数开始每3个数一个循环，先用 100 - 5 ，然后然后求出（100-5）除以3即可求出有几个循环，商是31，余数是2，也就是31个循环，剩余循环中的前2个数，所以用 (5 + 2 + 3 + 1 + 2) 求出前5个数的和，然后用 (1 + 1 + 0) 求出一个循环的和，再乘31即可求出即可求出31个循环的和，最后两个数是1、1;据此将所有数的和相加即可。首先通过列举找出这列数的规律，再根据它的循环周期进行解答即可。【详解】(100 - 5) ÷ 3 = 9 5 ÷ 3 = 3 1 .... 2。有 31 个循环，剩余循环中前 2 个数，5 + 2 + 3 + 1 + 2 = 1 3。(1 + 1 + 0) × 3 1 = 2 × 3 1 = 6 2。1 3 + 6 2 + 1 + 1 = 7 7这列数中前 100 个数之和等于 77。",
+        "knowledge_points": [
+          "周期问题",
+          "整数排列的规律"
+        ],
+        "source": "2007 六年级·全国·竞赛",
+        "merged_id": "unit8_425",
+        "original_id": "425"
+      },
+      {
+        "id": "426",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "（2008 六年级下·全国·竞赛）有 140 个学生排成一排，从左到右进行 1，2，3 报数，从右到左进行 1 至 4 报数。两次所报数的和为 5 的学生共有_个。",
+        "options": [],
+        "answer": "36",
+        "solution": "【分析】每个学生从左往右的报数情况如下：第一次报数: 123123123123...12第二次报数：432143214321...21每个学生报两次数为 1 组, 把两次情况看作 140 组数据来看,每 12 组一个循环, 每个循环中和为 5 的数一共有 3 组, 且是每个循环的前 3 组; 据此先用 140 ÷ 12 即可求出有多少个循环, 已知商是 11, 余数是 8, 说明有 11 个循环, 剩余 8 组, 剩余前 3 组的和也是 5 , 所以用 (11 + 1) × 3 即可求出一共有几组和为 5 。本题可通过列举从左往右报的数，找出它们出现的规律进行解答即可。【详解】140 ÷ 12 = 11 ... 8。(1 1 + 1) × 3 = 1 2 × 3 = 3 6。两次所报数的和为5的学生共有36个。",
+        "knowledge_points": [
+          "间隔问题",
+          "整数排列的规律"
+        ],
+        "source": "2008 六年级下·全国·竞赛",
+        "merged_id": "unit8_426",
+        "original_id": "426"
+      },
+      {
+        "id": "427",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(2023·福建莆田·小升初真题) 有一列由两个数组成的数组：（1，1），（2，1），（2，2），（3，1），(3, 2), (3, 3), (4, 1), (4, 2), (4, 3),(4, 4), (5, 1) …(1)第72组的两个数之和是( )。(2)在前55组中，'5'这个数出现了( )次。",
+        "options": [],
+        "answer": "(1)18   (2)11",
+        "solution": "【分析】①观察数组的规律，第一个数是 1 的有 1 组，第一个数是 2 的有 2 组，第一个数是 3 的有 3 组，第一个数是 4 的有 4 组，……，因为 1 + 2 + 3 + 4 + ... + 11 = 66 组，所以从第 67 组开始，每组的第一个数是 12，第 67 组是(12, 1), 依此类推, 第 72 组是 (12, 6), 两个数的和是 12 + 6 = 18 ;(2) 因为 1 + 2 + 3 + ... + 10 = 55 组, 所以第 55 组恰好是(10, 10), 第一个数是 5 的有 5 组, 即 (5, 1)、(5, 2)、(5、3)、(5, 4)、(5, 5)。第二个数是 5 的只能是 (5, 5)、(6, 5)、(7, 5)、(8, 5)、(9, 5)、(10, 5) 出现了 6 次, 所以 '5'这个数出现了 11 次。据此解答。【详解】（1）观察数组的规律，可知：第一个数是 1 的有 1 组,第一个数是2的有2组，第一个数是3的有3组，第一个数是4的有4组，……又 1 + 2 + 3 + 4 + ... + 11 = 66 组,所以从第 67 组开始，每组的第一个数是 12，第 67 组是(12, 1), 第68组是 (12, 2), 第72组是 (12, 6),两个数的和: 12 + 6 = 18 ;答: 第 72 组的两个数之和是 18。(2) 因为 1 + 2 + 3 + ... + 10 = 55 组, 所以第 55 组恰好是(10, 10)第一个数是 5 的有 5 组，即（5，1）、（5，2）、(5、3) 、 (5, 4) 、 (5, 5)，出现了 5 次;第二个数是 5 的只能是 (5, 5)、(6, 5)、(7, 5)、(8, 5)、(9, 5)、(10, 5) 出现了6次;所以 '5'这个数出现: 5 + 6 = 11 (次)在前 55 组中，'5'这个数出现了 11 次。",
+        "knowledge_points": [
+          "数字排列的规律"
+        ],
+        "source": "2023·福建莆田·小升初真题",
+        "merged_id": "unit8_427",
+        "original_id": "427"
+      },
+      {
+        "id": "428",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "（2009六年级下·全国·竞赛）字母A，B，C，D与数字1，2，3，…按如下规律排列：| A | B | C | D | D | C | B | A | A | B | C | D··· || ---- | ---- | ---- | ---- | ---- | ---- | ---- | ---- | ---- | ---- | ---- | ----- || 1 | 2 | 3 | 6 | 5 | 4 | 7 | 8 | 9 | 12 | 11 | 10··· |则数字 106 在字母__下。",
+        "options": [],
+        "answer": "D",
+        "solution": "【分析】观察题意可知, 数字排列每 3 个为 1 组, 奇数组从小到大排列, 偶数组从大到小排列, 先用 106 ÷ 3 即可求出前106 个数有几组，商是 35，余数是 1，所以第 106 个数字是在第 36 组的第 1 个数字，偶数组从大到小排列，第 36 组也就是 108、107、106，第 105 个数字是 105，所以第 106 个数字是 108，第 108 个数字是 106;字母以 8 个为 1 组，用 108 ÷ 8 即可求出前 108 个字母有几组，商是 13，余数是 4，所以第 108 个字母是第 14 组里面第 4 个数字，也就是 D。找到周期的规律和特点是解答本题的关键。【详解】数字排列每3个为1组，106 ÷ 3 = 35...第106个数字是在第36组的第1个数字，第36组也就是108、107、106，第108个数字是106,字母以 8 个为 1 组,108 ÷ 8 = 13 ... 4第 108 个字母是第 14 组里面第 4 个数字, 也就是 D。数字106在字母D下。",
+        "knowledge_points": [
+          "周期问题"
+        ],
+        "source": "2009六年级下·全国·竞赛",
+        "merged_id": "unit8_428",
+        "original_id": "428"
+      },
+      {
+        "id": "429",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(2018 六年级·全国·竞赛) 在数列 2/1, 1/2, 3/1, 2/2, 1/3, 4/1, 3/2, 2/3, 1/4, ... 中, 12/23 是第_个数。",
+        "options": [],
+        "answer": "583",
+        "solution": "【分析】观察这组数据发现分子和分母的和是3、4、5、6…，则和是3的分数有2个，和是4的分数有3个，和是5的分数有4的，那么 12/23 的分子和分母的和是35，那么和是34的分数有33个，则这个时候的分数的个数是 2 + 3 + 4 + ... + 33 ，是一组等差数列相加，等差数列的和 = （首项+末项）×项数÷2得出是560个，分子和分母的和是35的时候分母是从1开始的，则第23个数就是 12/23 。【详解】2 + 3 + 4 + ... + 33 =(2 + 3 3) × 3 2 ÷ 2 = 3 5 × 3 2 ÷ 2 = 1 12 0 ÷ 2 =5 6 0 (个)。5 6 0 + 2 3 = 5 8 3(个)则 12/23 是第 583 个。",
+        "knowledge_points": [
+          "数字排列的规律"
+        ],
+        "source": "2018 六年级·全国·竞赛",
+        "merged_id": "unit8_429",
+        "original_id": "429"
+      },
+      {
+        "id": "430",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(23-24 六年级下·全国·课后作业) 1/7, 1/14, 1/21, ( ), ( ) 。",
+        "options": [],
+        "answer": "1/28  1/35",
+        "solution": "【分析】观察可知, 这些分数的分子都是 1 , 前一个分数的分母 +7 =后一个分数的分母, 据此填空。【详解】21 + 7 = 28 、 28 + 7 = 35。1/7, 1/1 4, 1/2 1, 1/2 8, 1/3 5 。",
+        "knowledge_points": [
+          "数字排列的规律"
+        ],
+        "source": "23-24 六年级下·全国·课后作业",
+        "merged_id": "unit8_430",
+        "original_id": "430"
+      },
+      {
+        "id": "431",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(23-24 六年级下·全国·课后作业）1，3，2，6，4，9，8，（ ），（ ），15,( )，18。",
+        "options": [],
+        "answer": "12   16   32",
+        "solution": "【分析】单数项依次乘2;双数项依次加3，据此填空。【详解】9 + 3 = 12。8 × 2 = 1 6。1 6 × 2 = 3 2。这组数是：1，3，2，6，4，9，8，12，16，15，32，18。",
+        "knowledge_points": [
+          "数字排列的规律"
+        ],
+        "source": "23-24 六年级下·全国·课后作业",
+        "merged_id": "unit8_431",
+        "original_id": "431"
+      },
+      {
+        "id": "432",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(23-24六年级上·福建莆田·期末）古希腊毕达哥拉斯学派信奉'万物皆数'，将'数'排列成三角形、正方形等美丽的图形。如下图，排列成三角形的数叫作三角形数。照这样排列下去，第10个图形所表示的数是( )。![](images/b44c854dacc4a2a118da408dcbd6c13e.png)",
+        "options": [],
+        "answer": "55",
+        "solution": "【分析】根据题意，第一个图形有1个点，表示1;第二个图形有 3 个点, 表示 3 , 可以写成: 1 + 2 = 3 ;第三个图形有 6 个点, 表示 6 , 可以写成: 1 + 2 + 3 = 6 ;第四个图形有 10 个点, 表示 10 , 可以写成: 1 + 2 + 3 + 4 = 10;由此可知, 第 n 个图形有 (1 + 2 + 3 + ...... + n) 个点, 表示 (1 + 2 + 3 + ...... + n) , 由此当 n = 10 时,求出图形所表示的数。【详解】根据分析可知, 第 n 个图形有 (1 + 2 + 3 + ... + n) 个点。当 n = 10 时;1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 1 0 = 5 5古希腊毕达哥拉斯学派信奉'万物皆数'，将'数'排列成三角形、正方形等美丽的图形。如下图，排列成三角形的数叫作三角形数。照这样排列下去, 第 10 个图形所表示的数是 55。![](images/98fc29ecf922fc45e2a696fe49a5d8da365f988990c9e5b93a272740e96cfd93.jpg)",
+        "knowledge_points": [
+          "数与形"
+        ],
+        "source": "23-24六年级上·福建莆田·期末",
+        "merged_id": "unit8_432",
+        "original_id": "432"
+      },
+      {
+        "id": "433",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（2019 六年级·全国·竞赛）已知数列 1/3, 1/15, 1/63 ...，则第 10 个数是____。",
+        "options": [],
+        "answer": "1/399",
+        "solution": "【分析】通过观察可知, 1/3 = 1/(1 × 3), 1/15 = 1/(3 × 5), 1/63 = 1/(5 × 7) ...,以此类推, 第 n 个数是 1/((2 n - 1) × (2 n + 1)) , 据此解答。【详解】根据分析可知，第10个数是1/((2 × 1 0 - 1) × (2 × 1 0 +1)) = 1/((2 0 - 1) × (2 0 + 1)) = 1/(1 9 × 2 1) = 1/3 9 9",
+        "knowledge_points": [
+          "数字排列的规律"
+        ],
+        "source": "2019 六年级·全国·竞赛",
+        "merged_id": "unit8_433",
+        "original_id": "433"
+      },
+      {
+        "id": "434",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(23-24 五年级上·福建福州·期末) 小青在计算中发现了这样的一类算式:32 × 1 = 32, 32 × 1 0 1 = 3 2 3 2, 32 × 1 0 1 0 1 = 3 2 3 2 3 2, 32 × 1 0 1 0 1 0 1 = ( )。根据我的发现，( )×( )=41414141。",
+        "options": [],
+        "answer": "32323232    41    1010101",
+        "solution": "【分析】由题可知：其中一个乘数不变，从第二个算式开始，另一个乘数的数字每增加一组'01'，积就并排着重复写一个不变的乘数，据此解答。【详解】由分析可知：32 × 1 0 1 0 1 = 3 2 3 2 3 2所以， 32 × 1010101 = 32323232。4 1 × 1 = 4 1。4 1 × 1 0 1 = 4 1 4 1所以， 41 × 1010101 = 41414141",
+        "knowledge_points": [
+          "算式的规律",
+          "积的变化规律（小数乘法）"
+        ],
+        "source": "23-24 五年级上·福建福州·期末",
+        "merged_id": "unit8_434",
+        "original_id": "434"
+      },
+      {
+        "id": "435",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(23-24 六年级上·江西宜春·期末) 一组图形按下面规律排列: ![](images/d8a2c91fe1d0ddef82fb978c441bbf94.png)那么第 n 个图形由( )根小棒组成。",
+        "options": [],
+        "answer": "2 n + 1",
+        "solution": "【分析】观察分图形可知，第1个图形由3根小棒组成，3=2+1;第2个图形由5根小棒组成，5=2×2+1;第3个图形由 7 根小棒组成, 7 = 2 × 3 + 1 ; 第 4 个图形由 9 根小棒组成, 9 = 2 × 4 + 1 。由此可得: 小棒的根数 = 2 × 图形的序数 +1。据此解答。【详解】通过分析可得：小棒的根数 = 2 × 图形的序数 +1, 则第 n 个图形由 (2 n + 1) 根小棒组成。",
+        "knowledge_points": [
+          "数与形"
+        ],
+        "source": "23-24 六年级上·江西宜春·期末",
+        "merged_id": "unit8_435",
+        "original_id": "435"
+      },
+      {
+        "id": "436",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(23-24 六年级上·四川绵阳·期末) 摆 1 个八边形需要 8 根小棒, 如图, ![](images/03a722593ea04c299b403e60226c2f9e.png)想一想, 摆 10 个八边形需要( )根小棒; 摆 n 个八边形需要( )根小棒。",
+        "options": [],
+        "answer": "71  7n + 1",
+        "solution": "【分析】根据图示发现：摆 1 个八边形需要小棒：8 根;摆2 个八边形需要小棒 (8+7) 根; 摆 3 个八边形需要小棒 (8 + 7 + 7) 根; ……摆 n 个八边形需要小棒的根数是 8+ 7 (n - 1) 。据此解答。【详解】根据分析可知，摆 n 个八边形需要小棒：8 + 7 (n - 1) = 8 + 7 n - 7 = (7 n + 1) 根。摆 n 个八边形需要 (7 n + 1) 根小棒。当 n = 10 时,7 n + 1 = 7 × 1 0 + 1 = 7 0 + 1 = 7 1 (根)。摆 10 个八边形需要 71 根小棒; 摆 n 个八边形需要 (7 n + 1) 根小棒。",
+        "knowledge_points": [
+          "用字母表示数、数量关系",
+          "含有字母式子的化简与求值",
+          "数与形（探索规律）",
+          "图形的变化规律"
+        ],
+        "source": "23-24 六年级上·四川绵阳·期末",
+        "merged_id": "unit8_436",
+        "original_id": "436"
+      },
+      {
+        "id": "437",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(23-24 六年级上·湖南永州·期末) 1/2 + 1/4 + 1/8 + 1/16 + 1/16 =( ); 1 + 3 + 5 + 7 + ... + 9 7 + 9 9 = ( ) 。",
+        "options": [],
+        "answer": "1   2500",
+        "solution": "【分析】把 1/2 看作 (1 - 1/2) ，把 1/4 看作 (1/2 - 1/4) ，把 1/8 看作 (1/4 - 1/8) ，把 1/16 看作 (1/8 - 1/16) ，再去掉小括号化简即可;把头尾两个数分别相加，即 (1 + 99) ， (3 + 97) ，……，其中 (1 + 3 + 5 + 7 + ... + 97 + 99) 一共有 (50 ÷ 2) 组这样的数相加，用 (1 + 99) 乘 (50 ÷ 2) 计算，据此解答。【详解】1/2 +1/4 +1/8 +1/16 +1/16。(1 - 1/2) + (1/2 - 1/4) + (1/4 - 1/8) + (1/8 - 1/1 6) + 1/1 6 =1 - 1/2 + 1/2 - 1/4 + 1/4 - 1/8 + 1/8 - 1/1 6 + 1/1 6 = 1 + 3 + 5 + 7 + ... + 9 7 + 9 9 = (1 + 9 9) × (5 0 ÷ 2) = 1 0 0 × 2 5 = 2 5 0 0。因此 1/2 + 1/4 + 1/8 + 1/16 + 1/16 = 1 ; 1 + 3 + 5 + 7 + ... + 97 + 99 = 2500 。",
+        "knowledge_points": [
+          "算式的规律",
+          "异分母分数加、减法"
+        ],
+        "source": "23-24 六年级上·湖南永州·期末",
+        "merged_id": "unit8_437",
+        "original_id": "437"
+      },
+      {
+        "id": "438",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(23-24 六年级上·山东济宁·期末) 3 + 5 + 7 + 9 + 11+1 3 + 1 5 + 1 7 + 1 9 = ( ) 。",
+        "options": [],
+        "answer": "99",
+        "solution": "【分析】观察算式发现是从 3 开始的 9 个连续奇数相加, 给这个算式补上一个加数 1 , 改写成 (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19) - 1 , 这样括号里面有 10 个连续奇数相加;根据 '连续奇数的和等于奇数个数的平方', 可知括号里面 10 个数的和是 10² , 再减去 1 , 即是原式的计算结果。【详解】3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = (1 + 3 + 5 + 7 + 9 + 1 1 + 1 3 + 1 5 + 1 7 + 1 9) - 1 = 1 0 ² - 1 = 1 0 0 - 1 = 9 9。所以， 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 99 。",
+        "knowledge_points": [
+          "算式的规律"
+        ],
+        "source": "23-24 六年级上·山东济宁·期末",
+        "merged_id": "unit8_438",
+        "original_id": "438"
+      },
+      {
+        "id": "439",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(23-24 六年级上·湖南永州·期末) 找规律 4/3 、 1 、 10/11 、 13/15 、 16/19 、 ..., 这串数中第 7 个数是( )。",
+        "options": [],
+        "answer": "19/23",
+        "solution": "【分析】根据 10/11 、 13/15 、 16/19 可得相邻两个分数的分子之差为 3 , 分母之差为 4 , 又因为 4/3 和 10/11 的分子之差为 6 , 分母之差为 8 , 即两数之间应该是 7/7, 7/7 = 1 。所以这串数中第 7 个数应该是 16/19 的分子加上 3 , 分母加上 4 之后所得的数。【详解】16 + 3 = 19。19 + 4 = 23所以这串数中第 7 个数是 19/23 。",
+        "knowledge_points": [
+          "假分数与带分数或整数的互化",
+          "数字排列的规律"
+        ],
+        "source": "23-24 六年级上·湖南永州·期末",
+        "merged_id": "unit8_439",
+        "original_id": "439"
+      },
+      {
+        "id": "440",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "(23-24 六年级上·湖北省直辖县级单位·期末) 学校进行乒乓球比赛，每两个人打一场，比赛前有8个人报名，正式比赛时有 1 名同学因病退赛。这样比赛的总场次减少了( )场。",
+        "options": [],
+        "answer": "7",
+        "solution": "【分析】每两个人打一场，如果 8 人报名，每人都要与其余的 (8-1) 人打一场, 需要比 8 × (8 - 1) 场, 这样重复计算了一遍，除以2是总场数;正式比赛时有1名同学因病退赛，总人数变成7人，每人都要与其余的（7-1）人打一场，需要比 7 × (7 - 1) 场，这样重复计算了一遍，除以 2 是总场数。据此分别计算出前后总场数，求差即可。本题是典型的握手问题，如果人数比较少，可以用枚举法解答;如果人数比较多，将人数看作n，可以用公式：n（n-1）÷2解答。【详解】8 × (8 - 1) ÷ 2 = 8 × 7 ÷ 2 = 28 （场）。8 - 1 = 7 （人）7 × (7 - 1) ÷ 2 = 7 × 6 ÷ 2 = 21（场）。28 - 21 = 7 （场）。这样比赛的总场次减少了7场。",
+        "knowledge_points": [
+          "数与形",
+          "搭配问题"
+        ],
+        "source": "23-24 六年级上·湖北省直辖县级单位·期末",
+        "merged_id": "unit8_440",
+        "original_id": "440"
+      },
+      {
+        "id": "441",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(2023·四川成都·小升初真题) 有一串数如下:1, 2, 4, 7, 11, 16…它的规律是由 1 开始，加 1，加 2，加 3…依次逐个产生这串数，直到产生 50 个数为止。那么这50个数中，被3除余1的数有( )个。",
+        "options": [],
+        "answer": "33",
+        "solution": "【分析】用 1, 2, 4, 7, 11, 16... 分别除以 3 , 所得余数为 1, 2, 1, 1, 2, 1... 每三个数字为一组, 每组数里面有 2 个数的余数是 1 。这 50 组数据里面每 3 个为一组, 一共有 16 组, 余下 2 个数。每组有 2 个数的余数是 1 。剩下的两个数的余数是 1 和 2 。【详解】1 ÷ 3 = 0 ... 1。2 ÷ 3 = 0 .... 2。4 ÷ 3 = 1 ... 1。7 ÷ 3 = 2 ... 1。1 1 ÷ 3 = 3 ... 2。1 6 ÷ 3 = 5 ... 1。2 2 ÷ 3 = 7 .... 1。2 9 ÷ 3 = 9 .... 2。5 0 ÷ 3 = 1 6 (组) .... 2 (个)。1 6 × 2 + 1 = 3 2 + 1 = 3 3 (个)。那么这50个数中，被3除余1的数有33个。",
+        "knowledge_points": [
+          "有余数除法的实际应用",
+          "周期问题"
+        ],
+        "source": "2023·四川成都·小升初真题",
+        "merged_id": "unit8_441",
+        "original_id": "441"
+      },
+      {
+        "id": "442",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(22-23 六年级上·河南新乡·期末)1/2 + 1/4 + 1/8 + 1/1 6 + 1/3 2 + 1/6 4 + 1/1 2 8 + ... = ()",
+        "options": [],
+        "answer": "1",
+        "solution": "【分析】通过画图发现: 把整个图形看作单位 '1', 先平均分成 2 份, 取其中的 1 份, 即 1/2 ; 再把剩下的图形平均分成 2 份, 取其中的 1 份, 即 1/4 ……以此类推, 发现加数越多,涂色部分即它们的和就越来越接近 1。![](images/dbaa57c641535f858ae4f2692ac43f65e2395714085fc4b030615303f09d9b14.jpg)【详解】1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + ... = 1",
+        "knowledge_points": [
+          "数与形"
+        ],
+        "source": "22-23 六年级上·河南新乡·期末",
+        "merged_id": "unit8_442",
+        "original_id": "442"
+      },
+      {
+        "id": "443",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(23-24 六年级上·青海西宁·期末) 观察下图的图形规律，第8个图形一共有( )个小黑点。![](images/825b13a7fe4cc46d95cffbd96c9c03f4857cd74c9412c7ed68dc9381043aba01.jpg)",
+        "options": [],
+        "answer": "36",
+        "solution": "【分析】观察图形可知:第1个图形有1个小黑点;第2个图形有3个小黑点， 3 = 1 + 2 ;第3个图形有6个小黑点， 6 = 1 + 2 + 3 ;第 4 个图形有 10 个小黑点, 10 = 1 + 2 + 3 + 4 ;··第 n 个图形有 (1 + 2 + 3 + 4 + ... + n) 个小黑点。【详解】规律：第 n 个图形有 (1 + 2 + 3 + 4 + ... + n) 个小黑点。当 n = 8 时。1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = (1 + 8) × 8 ÷ 2 = 9 × 8 ÷ 2 = 3 6 (个)。第 8 个图形一共有 36 个小黑点。",
+        "knowledge_points": [
+          "数与形",
+          "图形的规律"
+        ],
+        "source": "23-24 六年级上·青海西宁·期末",
+        "merged_id": "unit8_443",
+        "original_id": "443"
+      },
+      {
+        "id": "444",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(23-24六年级上·河南南阳·期末）观察下面的图形，按照图形与数的规律排列下去，第5个数是( )，第20个数是( )。![](images/a71cb9e17bd179c521343822eae07b04.png)···",
+        "options": [],
+        "answer": "15 210",
+        "solution": "【分析】观察图形可知：第 1 个图形中有 1 个圆，表示的数是 1;第2个图形中有3个圆，表示的数是3， 3 = 1 + 2 ;第3个图形中有6个圆，表示的数是6， 6 = 1 + 2 + 3 ;第 4 个图形中有 10 个圆, 表示的数是 10, 10 = 1 + 2 + 3 + 4;....... 第 n 个图形中有圆的个数与表示的数是 (1 + 2 + 3 + 4 + ... + n）;据此解答。【详解】规律：第 n 个图形表示的数是 (1 + 2 + 3 + 4 + ... + n)。当 n = 5 时， 1 + 2 + 3 + 4 + 5 = 15当 n = 20 时1 + 2 + 3 + 4 + ... + 2 0 = (1 + 2 0) × 2 0 ÷ 2 = 2 1 × 2 0 ÷ 2 = 2 1 0。按照图形与数的规律排列下去，第5个数是15，第20个数是 210 。",
+        "knowledge_points": [
+          "数与形",
+          "含有字母式子的化简与求值",
+          "用字母表示稍复杂的数量关系"
+        ],
+        "source": "23-24六年级上·河南南阳·期末",
+        "merged_id": "unit8_444",
+        "original_id": "444"
+      },
+      {
+        "id": "445",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(23-24 六年级上·湖南岳阳·期末) 摆一个图形用 6 根小木棒, 像这样摆 20 个图形要用( )根小棒。![](images/8e3c4bad7fb9dad1ff934ebc3e6f5187.png)",
+        "options": [],
+        "answer": "101",
+        "solution": "【分析】摆 1 个图形需要 5 × 1 + 1 = 6 （根）小棒，摆 2 个图形需要 5 × 2 + 1 = 11 （根）小棒，摆3个图形需要 5 × 3 + 1 = 16 （根）小棒，摆4个图形需要 5 × 4 + 1 = 21 （根）小棒，每多摆1个图形，需要多用5根小棒。那么，摆20个图形需要 (5 × 20 + 1) 根小棒。【详解】5 × 20 + 1 = 1 0 0 + 1 = 1 0 1 (根)。所以，像这样摆 20 个图形要用 101 根小棒。",
+        "knowledge_points": [
+          "数与形"
+        ],
+        "source": "23-24 六年级上·湖南岳阳·期末",
+        "merged_id": "unit8_445",
+        "original_id": "445"
+      },
+      {
+        "id": "446",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（23-24六年级上·湖北黄石·期末）我国宋代数学家杨辉在公元1261年撰写了《详解九章算法》，他在这本著作中画了一个由数构成的三角形（见图），根据它的规律，第六行数字应该是( )。![](images/366f1556a2498a77568a653a086d5f24d50285195ba26bb26ac16d4f5f0a9acc.jpg)",
+        "options": [],
+        "answer": "1 - 5 - 10 - 10 - 5 - 1",
+        "solution": "【分析】观察'杨辉三角'可知，第几行就有几个数，左右两边都是1，从第三行开始，中间的每一个数都等于它上方相邻的两个数字之和，由此求出第六行数字。【详解】如图：![](images/f036d1775604cf65b9a62304f3500af84f96f6bb778a35b4756225c895bebad4.jpg)第六行数字应该是 1 - 5 - 10 - 10 - 5 - 1 。我国宋代数学家杨辉在公元 1261 年撰写了《详解九章算法》，他在这本著作中画了一个由数构成的三角形（见图），根据它的规律，第六行数字应该是 1 - 5 - 10 - 10 - 5 - 1 。",
+        "knowledge_points": [
+          "数字排列的规律"
+        ],
+        "source": "23-24六年级上·湖北黄石·期末",
+        "merged_id": "unit8_446",
+        "original_id": "446"
+      },
+      {
+        "id": "447",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(23-24 六年级上·广东汕头·期末) 5 + 7 + 9 + 11 + 13+1 5 + 1 7 = () ² - () ² = ( )²",
+        "options": [],
+        "answer": "9    2    77",
+        "solution": "【分析】1 = 1², 1 + 3 = 2², 1 + 3 + 5 = 3², ... 据此可知, 从 1 开始的连续奇数的和等于奇数个数的平方, 所以 5 + 7 + 9 + 11 + 13 + 15 + 17 = (1 + 3 + 5 + ... + 17) - (1 + 3), 1 + 3 + 5 + ... + 17 = 9², 1 + 3 = 2² , 据此解答。【详解】5 + 7 + 9 + 11 + 13 + 15 + 17 = (1 + 3 + 5 + ... + 1 7) - (1 + 3) = 9 ² - 2 ² = 8 1 - 4 = 7 7",
+        "knowledge_points": [
+          "算式的规律"
+        ],
+        "source": "23-24 六年级上·广东汕头·期末",
+        "merged_id": "unit8_447",
+        "original_id": "447"
+      },
+      {
+        "id": "448",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(23-24六年级上·新疆喀什·期末）由下图可以看出，每多搭一个正方形，就要增加( )根小棒，照这样搭下去，搭15个这样的正方形一共需要( )根小棒。![](images/9f3abc8c2d4490f7d1841b2a672b954b83b4f4636e217b5851dd5b592bd1cd78.jpg)",
+        "options": [],
+        "answer": "3 46",
+        "solution": "【分析】观察图形可知, 搭 1 个正方形要用 4 根小棒, 搭 2 个正方形要用 7 根小棒, 搭 3 个正方形要用 10 根小棒……发现: 每多搭一个正方形, 小棒的数量增加 3 根, 据此找到规律, 按此规律解答。【详解】搭1个正方形要用4根小棒， 4 = 3× 1 + 1搭 2 个正方形要用小棒: 4 + 3 = 7 (根), 7 = 3 × 2 + 1 ;搭 3 个正方形要用小棒: 7 + 3 = 10 (根), 10 = 3 × 3 + 1 ;......所以，每多搭一个正方形，就要增加3根小棒。规律：搭 n 个这样的正方形要用 (3n + 1) 根小棒。当 n = 15 时3 n + 1 = 3 × 1 5 + 1 = 4 5 + 1 = 4 6 (根)。照这样搭下去，搭15个这样的正方形一共需要46根小棒。",
+        "knowledge_points": [
+          "数与形",
+          "用字母表示数",
+          "含有字母式子的化简与求值",
+          "图形的规律"
+        ],
+        "source": "23-24六年级上·新疆喀什·期末",
+        "merged_id": "unit8_448",
+        "original_id": "448"
+      },
+      {
+        "id": "449",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(21-22 六年级上·湖南张家界·期末) 9 + 11 + 13 + 15... + 2 9 = ( ) 。",
+        "options": [],
+        "answer": "209",
+        "solution": "【分析】1 = 1², 1 + 3 = 2², 1 + 3 + 5 = 3², ... 据此可知, 从 1 开始的连续奇数的和等于奇数个数的平方, 所以 9 + 11 + 1 3 + 1 5 + ... + 2 9 = (1 + 3 + 5 + ... + 2 9) - (1 + 3 + 5 +7) , 1 + 3 + 5 + ... + 29 = 15², 1 + 3 + 5 + 7 = 4² , 据此解答。【详解】9 + 1 1 + 1 3 + 1 5 + ... + 2 9 = (1 + 3 + 5 + ... + 2 9) - (1 + 3 + 5 + 7) = 1 5 ² - 4 ² = 2 2 5 -1 6 = 2 0 9",
+        "knowledge_points": [
+          "算式的规律"
+        ],
+        "source": "21-22 六年级上·湖南张家界·期末",
+        "merged_id": "unit8_449",
+        "original_id": "449"
+      },
+      {
+        "id": "450",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(23-24 六年级上·吉林·期末) 1 + 3 + 5 + 7 + 9 = ()²。",
+        "options": [],
+        "answer": "5",
+        "solution": "【分析】1 + 3 = 4 = 2²。1 + 3 + 5 = 9 = 3 ²。1 + 3 + 5 + 7 = 1 6 = 4 ²。观察各等式，可得：从 1 开始的连续奇数的和等于奇数的个数的平方。据此解答即可。【详解】由分析可知， 1 + 3 + 5 + 7 + 9 ，是从1开始的连续5 个奇数的和, 所以 1 + 3 + 5 + 7 + 9 = 5² 。",
+        "knowledge_points": [
+          "数与形",
+          "算式的规律"
+        ],
+        "source": "23-24 六年级上·吉林·期末",
+        "merged_id": "unit8_450",
+        "original_id": "450"
+      },
+      {
+        "id": "451",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（22-23六年级下·湖北省直辖县级单位·期末）找规律：摆1个六边形用6根小棒，照这样摆下去，摆第17个图形需要( )根小棒，摆第 n 个图形需要( )根小棒。![](images/168266548e715bf08583b755be207e63.png)",
+        "options": [],
+        "answer": "86  1 + 5 n",
+        "solution": "【分析】通过观察发现: 摆 1 个六边形小棒的根数是 6 根,6 = 1 + 5 ; 摆 2 个六边形小棒的根数是 11 根, 11 = 1 + 5 × 2 ;摆3个六边形小棒的根数是16根， 16 = 1 + 5× 3 ;由此发现规律：摆第几个图形需要的小棒的根数就是 1 加上 5 乘几，即摆第 n 个图形需要（1+5n）根小棒。【详解】1 + 5 × 17 = 1 + 8 5 =86 (根)。所以，摆第 17 个图形需要 86 根小棒，摆第 n 个图形需要 (1 + 5 n) 根小棒。",
+        "knowledge_points": [
+          "数与形",
+          "用字母表示数"
+        ],
+        "source": "22-23六年级下·湖北省直辖县级单位·期末",
+        "merged_id": "unit8_451",
+        "original_id": "451"
+      },
+      {
+        "id": "452",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(22-23 六年级上·四川南充·期末)![](images/dbdd54629f315787f9b533934cf5d4b66b798bacabab5fa17d760bbe9160ad87.jpg)按照这样的规律摆下去，摆出10个这样的正六边形一共需要( )根小棒。",
+        "options": [],
+        "answer": "51",
+        "solution": "【分析】看图可知, 小棒根数 = 正六边形个数 × 5 + 1 , 据此列式计算。【详解】10 × 5 + 1 = 5 0 + 1 = 5 1 (根)。摆出 10 个这样的正六边形一共需要 51 根小棒。",
+        "knowledge_points": [
+          "数与形"
+        ],
+        "source": "22-23 六年级上·四川南充·期末",
+        "merged_id": "unit8_452",
+        "original_id": "452"
+      },
+      {
+        "id": "453",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（2024 六年级下·全国·专题练习）如下图，表 2 是从表 1 中截取的一部分，根据表 1 中数的规律，可以得到 n 的值为( )。表1 | 1 | 2 | 3 | 4 | … || ---- | ---- | ---- | ---- | ---- || 2 | 4 | 6 | 8 | … || 3 | 6 | 9 | 12 | … || 4 | 8 | 12 | 16 | … || … | … | … | … | … |表2 | 20 | 24 || ---- | ---- || 25 | n |",
+        "options": [],
+        "answer": "30",
+        "solution": "【分析】通过表1可知，第1行相邻两数相差1，第2行相邻两数相差2，第3行相邻两数相差3，则第n行相邻两数相差n，由表2可知 24 - 20 = 4 ，20、24是第4行的数，则25、n是第5行的数，所以n比25多5，据此解答。本题考查数表中的规律。通过给出一部分数据，让学生根据已知的规律来推理出未知的部分。既可以提升学生的观察能力，又可以发展学生的推理判断能力。【详解】25 + 5 = 30可以得到 n 的值为 30 。",
+        "knowledge_points": [
+          "数与形"
+        ],
+        "source": "2024 六年级下·全国·专题练习",
+        "merged_id": "unit8_453",
+        "original_id": "453"
+      },
+      {
+        "id": "454",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(23-24 六年级上·湖南永州·期中) 按数的规律填空:1/1 2, 1/6, 1/4, 1/3, ( ), 1/2 。",
+        "options": [],
+        "answer": "5/12",
+        "solution": "【分析】根据题意, 把分数通过通分, 化为分母是 12 的分数, 1/6 = 2/12, 1/4 = 3/12, 1/3 = 4/12; 1/2 = 6/12 , 由此可知, 从 1/12 到 6/12 ,这一列数的分母都是 12 , 分子是从 1 开始的连续的自然数,据此求出要求的数, 即可解答。【详解】根据分析可知, 这一列数排列为 1/12, 2/12, 3/12, 4/12, 5/12, 6/12 。按数的规律填空: 1/12, 1/6, 1/4, 1/3, 5/12, 1/2 。",
+        "knowledge_points": [
+          "通分的认识及应用",
+          "数字排列的规律"
+        ],
+        "source": "23-24 六年级上·湖南永州·期中",
+        "merged_id": "unit8_454",
+        "original_id": "454"
+      },
+      {
+        "id": "455",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（22-23六年级下·湖北恩施·开学考试）按下图所示的方式摆放餐桌和椅子，按照这样的规律，八张餐桌可以坐( )个人，n张餐桌可以坐( )个人。![](images/79b214e7391b2b3cd58cfd02d8e9b637562284937682b405caac0d355fc65f1b.jpg)",
+        "options": [],
+        "answer": "34    4 n + 2",
+        "solution": "【分析】观察图形可知, 一张餐桌可以坐 6 个人, 两张餐桌可以坐 10 人, 三张餐桌可以坐 14 人, 进一步观察发现: 多一张餐桌, 多放 4 把椅子, 就多坐 4 个人, 第 n 张餐桌可以坐 6 + 4(n - 1) = (4n + 2) 个人。【详解】八张餐桌可以坐的人数：4 n + 2 = 4 × 8 + 2 = 3 2 + 2 =34。则按照这样的规律, 八张餐桌可以坐 34 个人, n 张餐桌可以坐 (4n + 2) 个人。",
+        "knowledge_points": [
+          "数与形",
+          "用字母表示数",
+          "含有字母式子的化简与求值",
+          "图形的规律"
+        ],
+        "source": "22-23六年级下·湖北恩施·开学考试",
+        "merged_id": "unit8_455",
+        "original_id": "455"
+      },
+      {
+        "id": "456",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（22-23六年级上·宁夏固原·期末）甲、乙、丙、丁和小强五位同学聚会，见面时，每2个人要握一次，到现在为止甲握了4次手，乙握了3次手，丙握了2次手，丁握了1次手，那么到现在为止小强还没有和( )握手。",
+        "options": [],
+        "answer": "丙、丁",
+        "solution": "【分析】用 5 个点表示甲、乙、丙、丁、小强, 用两点之间相连的线段表示两人握过手, 和这个人相连的有几条线段,就表示他握过几次手; 题中告诉我们, 甲和四人都握了手,那甲分别和另外四人握了手, 乙和三人握了手, 则乙和甲握了一次, 另外两次是和丙、丁、小强中的两人握了手,而丙和二人握了手，丙和甲握了一次，另外一次是和乙或丁或小强握的，丁和一人握了手，是和甲握的，所以乙和甲、丙、小强握了手，丙和甲、乙两人握的手，所以小强和甲、乙两人握过手，即没有和丙、丁握手。【详解】如图：![](images/17406f08b5f6c94dd764de8d88646d6d7e7bb4fe83f653c706a071c0860d3b71.jpg)则到现在为止小强还没有和丙、丁握手。",
+        "knowledge_points": [
+          "搭配问题",
+          "推理问题"
+        ],
+        "source": "22-23六年级上·宁夏固原·期末",
+        "merged_id": "unit8_456",
+        "original_id": "456"
+      },
+      {
+        "id": "457",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "(22-23 六年级上·四川达州·期中) 若1 + 3 + 5 + 7 + 9 + ... + n = 2025 ，则 n = ( ) 。",
+        "options": [],
+        "answer": "89",
+        "solution": "【分析】1 + 3 = [ (1 + 3) ÷ 2 ] ² = 2 ² = 4。1 + 3 + 5 = [ (1 + 5) ÷ 2 ] ² = 3 ² = 9。1 + 3 + 5 + 7 = [ (1 + 7) ÷ 2 ] ² = 4 ² = 1 6。··。1 + 3 + 5 + 7 + 9 + ... + n = [(1 + n)/2]² = 2025, 由此可知, (1 + n)² = 2025 × 4, 据此求出n的值。【详解】根据分析可知， [(1 + n)/2]² = 2025。(1 + n) ² = 2 0 2 5 × 4。(1 + n) ² = 8 1 0 0。(1 + n) ² = 9 0 ²。1 + n = 9 0。n = 9 0 - 1。n = 8 9。若 1 + 3 + 5 + 7 + 9 + ... + n = 2025 , 则 n = 89 。",
+        "knowledge_points": [
+          "算式的规律"
+        ],
+        "source": "22-23 六年级上·四川达州·期中",
+        "merged_id": "unit8_457",
+        "original_id": "457"
+      },
+      {
+        "id": "458",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(22-23 五年级上·江苏徐州·期末) 用若干张边长 1 厘米的正方形纸片, 像下图这样依次摆出 1 层, 2 层, 3 层……的图形。![](images/0ae3e35bc7d087f4b8fb0ac5b54c9474.png)(1) 每个图形的周长分别是多少厘米? 先算一算, 再填一填。(2) 像这样摆 n 层, 摆成的图形周长是 ( ) 厘米; 当 n = 15 时, 摆成图形的周长是 ( ) 厘米。",
+        "options": [],
+        "answer": "4; 8; 12; 4n; 60",
+        "solution": "【分析】观察图形可知，第一个图形的周长边长为 1 厘米的正方形的周长，是 1 × 4 = 4 厘米，第二个图形右上方的两条小正方形的边长分别向上、向右平移，则可得这个图形的周长是边长是 2 厘米的正方形的周长，是 2 × 4 = 8 厘米;同样，第三个图形的周长等于边长是 3 厘米的正方形的周长，是 3 × 4 = 12 厘米;……，摆成 n 层，摆成的图像的周长是4n; 当 n = 15 时, 求出摆成图形的面积.本题考查了通过特例分析从而归纳总结出一般结论的能力。对于找规律的题目首先应找出哪些部分发生了变化，是按照什么规律变化的，通过分析找到各部分的变化规律后直接利用规律求解。【详解】根据分析可知, 第一个图形周长是: 1 × 4 = 4 (厘米)第二个图形周长是： 2 × 4 = 8 （厘米）第三个图形周长是： 3 × 4 = 12 （厘米）![](images/d3e68ed12c07571cb4e6c49eb64d1543e620b717f8c00892690dd096a4e4612b.jpg)若摆成 n 层, 则周长是: n × 4 = 4 n (厘米)当 n = 15 数, 4 × 15 = 60 (厘米)像这样摆 n 层, 摆成的图形周长是 4 n 厘米; 当 n = 15 时,摆成图形的周长是 60 厘米。",
+        "knowledge_points": [
+          "数与形",
+          "含多边形的组合图形的周长"
+        ],
+        "source": "22-23 五年级上·江苏徐州·期末",
+        "merged_id": "unit8_458",
+        "original_id": "458"
+      },
+      {
+        "id": "459",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（21-22六年级上·湖北武汉·期末）用同样大小的黑色棋子按下图所示的方式摆图案。按照这样的规律摆下去，第10个图案需要( )枚棋子，用56枚棋子摆的图案是第( )个，摆第 n 个图案需要( )枚棋子。图案1![](images/80a51960ed70467c0b24335de1570ff3427128de48fa14968f5ebddde1176777.jpg)  图案2![](images/571c5ffc40c1a0422fe1dea9591ee6d8c1d06abef491cf1f1d4af20daf0ba234.jpg)  图案3![](images/5716b3a0c6862a23b9e1c0f171d12b434e60118786ff59d2c8cd2d5237b30877.jpg)",
+        "options": [],
+        "answer": "32   18    3 n + 2 / 2 + 3 n",
+        "solution": "【分析】看图，第一个图案需要 3 × 1 + 2 = 5 （枚）棋子，第二个图案需要 3 × 2 + 2 = 8 （枚）棋子，第三个图案需要 3 × 3 + 2 = 11 （枚）棋子，据此类推第 10 个图案需要 (3 × 10 + 2) 枚棋子，第 n 个图案需要 (3n + 2) 枚棋子。将 56 枚棋子减去 2，将差除以 3，即可求出用 56 枚棋子摆的图案是第几个。本题考查了数与形, 有一定观察总结能力是解题的关键。【详解】3 × 10 + 2 = 3 0 + 2 = 32 (枚)。(5 6 - 2) ÷ 3 = 5 4 ÷ 3 = 1 8 (个)。所以，第 10 个图案需要 32 枚棋子，用 56 枚棋子摆的图案是第 18 个，摆第 n 个图案需要 (3n + 2) 枚棋子。",
+        "knowledge_points": [
+          "数与形",
+          "图形的规律"
+        ],
+        "source": "21-22六年级上·湖北武汉·期末",
+        "merged_id": "unit8_459",
+        "original_id": "459"
+      },
+      {
+        "id": "460",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(22-23 六年级上·重庆梁平·期末) 找规律填空。1/2, 3/4, 5/6, 7/1 0, ( ), ( ) 。",
+        "options": [],
+        "answer": "9/16   11/26",
+        "solution": "【分析】观察数列可知，各个分数的分子分别是按照奇数的大小进行排列的，即 1、3、5、7、9、11;第一个分数的分母加上第二个分数的分母即可得到第三个分数的分母，第二个分数的分母加上第三个分数的分母即可得到第四个分数的分母，依次类推，据此计算即可。本题考查数字排列的规律，发现规律，利用规律是解题的关键。【详解】由分析可知:第五个分数的分子应是 9 , 分母为 6 + 10 = 16 , 则这个分数是 9/16 ;第六个分数的分子应是 11, 分母为 10 + 16 = 26 , 则这个分数是 11/26 ;1/2, 3/4, 5/6, 7/1 0, 9/1 6, 1 1/2 6 。",
+        "knowledge_points": [
+          "数字排列的规律"
+        ],
+        "source": "22-23 六年级上·重庆梁平·期末",
+        "merged_id": "unit8_460",
+        "original_id": "460"
+      },
+      {
+        "id": "461",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(22-23 六年级上·四川南充·期末)1 + 3 + 5 + 7 + 9 + ... + 2 5 + 2 7 + 2 9 = ( )² 。",
+        "options": [],
+        "answer": "15",
+        "solution": "【分析】看题可知, 等号左边是求从 1 开始到 29 结束所有奇数的和, 且这些数都是连续的奇数, 根据从 1 开始的 n 个奇数和等于 n 的平方。所以等号的右边其实是求等号右边数据个数的平方, 据此解答。此题考查了学生的理解分析能力, 关键是找出规律。【详解】1 到 29 之间共有 15 个奇数。所以 1 + 3 + 5 + 7 + 9 + ... + 25 + 27 + 29 = 15² = 225 。",
+        "knowledge_points": [
+          "算式的规律"
+        ],
+        "source": "22-23 六年级上·四川南充·期末",
+        "merged_id": "unit8_461",
+        "original_id": "461"
+      },
+      {
+        "id": "462",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(22-23 六年级上·天津南开·期末) 用小棒按下面的方法摆图形, 第 10 个图形用了( )根小棒。![](images/8af0418ad45439c72cc9cba90db196e00e08d2de98e9ec7998bc71e4d5cd79fc.jpg)",
+        "options": [],
+        "answer": "21",
+        "solution": "【分析】根据图示发现：摆1个三角形需要小棒：3根;摆2个三角形需要小棒（ 3 + 2 ）根;摆3个三角形需要小棒 (3 + 2 + 2) 根;……摆 m 个三角形需要小棒的根数是 3+ 2（ m - 1 ）。据此解答。本题主要考查数与形结合的规律，关键根据图示发现这组图形的规律，并运用规律做题。【详解】摆 m 个三角形需要小棒：3 + 2 (m - 1) = 3 + 2 m - 2 = (2 m + 1) 根。当 m = 10 时,2 × 1 0 + 1 = 2 0 + 1 = 2 1 (根)。用小棒按下面的方法摆图形, 第 10 个图形用了 21 根小棒。",
+        "knowledge_points": [
+          "数与形"
+        ],
+        "source": "22-23 六年级上·天津南开·期末",
+        "merged_id": "unit8_462",
+        "original_id": "462"
+      },
+      {
+        "id": "463",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(22-23 六年级上·四川南充·期末) 4² = 1 + 3+( )+( )。",
+        "options": [],
+        "answer": "5 7",
+        "solution": "【分析】看题可知, 等号的左边是等号右边数据个数的平方, 等号右边是四个数的和, 且前两个数是连续的奇数,根据从 1 开始的 n 个奇数和等于 n 的平方可得出 4² = 1 + 3 + 5 + 7 = 16 。此题考查了学生的理解分析能力，关键是找出规律。【详解】根据题意, 结合从 1 开始的 n 个奇数和等于 n 的平方, 可得出 4² = 1 + 3 + 5 + 7 。",
+        "knowledge_points": [
+          "数与形",
+          "算式的规律"
+        ],
+        "source": "22-23 六年级上·四川南充·期末",
+        "merged_id": "unit8_463",
+        "original_id": "463"
+      },
+      {
+        "id": "464",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(21-22 六年级上·天津南开·期末) 观察下面的点阵图规律，第9个点阵图中有( )个点，第( )个点阵图中有300个点。![](images/8f39046bffe9b40a359b9ef2cc9676ec530addfbd18c01632ca125cc082c4dec.jpg)",
+        "options": [],
+        "answer": "30  99",
+        "solution": "【分析】通过观察，第1幅点阵图中有（3+3）个点，第2幅点阵图中有（3+3+3）个点;第3幅点阵图中（3+3+3+3）个点……以此类推，可得出第n幅点阵图中有（3+3n）个点，据此解答。本题主要考查数与形结合的规律, 关键根据图示发现这组图形的规律, 并运用规律做题。【详解】据分析可知，第 n 幅点阵图中有 (3 + 3n) 个点;第9幅点阵图中有3 + 3 × 9 = 3 + 2 7 =30(个)。第9幅点阵图中有30个点。3 + 3n = 300。解: 3 + 3n - 3 = 300 - 3。3 n = 297。3 n ÷ 3 = 297 ÷ 3。n=99。第99个点阵图中有300个点。",
+        "knowledge_points": [
+          "数与形",
+          "应用等式的性质 1 和 2 解方程",
+          "用字母表示数"
+        ],
+        "source": "21-22 六年级上·天津南开·期末",
+        "merged_id": "unit8_464",
+        "original_id": "464"
+      },
+      {
+        "id": "465",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(22-23 六年级上·浙江温州·期末) '2022 北京冬奥会'某国代表团安排入住奥运村C区宿舍，若每间2人，则有 12 人没有床位: 若每间 3 人, 则多出 2 个空床位, C区共有宿舍( )间，运动员共有( )。",
+        "options": [],
+        "answer": "14 40",
+        "solution": "【分析】根据题意，当每个房间增加 3 - 2 = 1 个人的时候，原来 12 个没有床位的人有了床位，还多出 2 个床位来，也就是说，每个房间增加一个床位，就会多出 12 + 2 = 14 个床位，所以一共有 (12 + 2) ÷ (3 - 2) 间房，再根据题意可求出总人数。此题考查了学生的理解分析能力和整数的四则运算。【详解】根据题意可得宿舍的间数：(12 + 2)÷ (3 - 2) = 1 4 ÷ 1 = 1 4 (间)。那么运动员的人数为: 14 × 2 + 12 = 2 8 + 1 2 = 4 0 (人)。'2022 北京冬奥会'某国代表团安排入住奥运村 C 区宿舍,若每间 2 人，则有 12 人没有床位：若每间 3 人，则多出 2个空床位，C区共有宿舍14间，运动员共有 40 人 。",
+        "knowledge_points": [
+          "搭配问题"
+        ],
+        "source": "22-23 六年级上·浙江温州·期末",
+        "merged_id": "unit8_465",
+        "original_id": "465"
+      },
+      {
+        "id": "466",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(22-23 六年级上·河南郑州·期末) 认真观察 '数形'规律并填空。![](images/715b87915b20001a97c543865a27152e93f7f90fb51a714df8eb3739bac20015.jpg)1 + 3 = 4 = 2 ²。1 + 3 + 5 = 9 = 3 ²。1 + 3 + 5 + 7 = 1 6 = 4 ²。1 + 3 + 5 + 7 + 9 = 6 ( ) = ( ) ² 。..........。1 + 3 + 5 + 7 + ... + 9 7 + 9 9 = ( ) = ( ) ²",
+        "options": [],
+        "answer": "25  5  2500 50",
+        "solution": "【分析】通过观察发现：每个![](images/f23295b98c0a004adc4e89e47d452bff.png)中点的个数都是奇数，分别是1、3、5、7、9、11、……;![](images/da05b7e193534178d906cbccc4c95de7.png) 中的点有2行，每行2个点，即 1 + 3 = 2 × 2 = 2² ;![](images/860ab4c834ddc32a39a3f018b56c3919.png) 中的点有3行，每行 3 个点, 即 1 + 3 + 5 = 3 × 3=3²![](images/9e832237f91224dcb87cdd0faae9a169.png) ; 中的点有 4 行,每行 4 个点, 即 1 + 3 + 5 + 7 = 4 × 4 = 4² ; ……由此发现规律:从 1 开始的 n 个连续奇数相加的和可以用 n² 表示, 即 1 + 3+5 + 7 + 9 + ... + (2 n - 1) = n ² 。解答数形结合类规律探索问题时，要仔细观察数与形之间的关系，先找到数与形之间隐含的数学规律，再利用规律进行解答即可。【详解】算式 1 + 3 + 5 + 7 + 9 是从 1 开始的 5 个连续奇数相加, 和是 5² = 25 , 所以 1 + 3 + 5 + 7 + 9 = 25 = 5² 。算式 1 + 3 + 5 + 7 + ... + 97 + 99 是从 1 开始的 50 个连续奇数相加，和是 50² = 2500 ，所以 1 + 3 + 5 + 7 + ... + 97 + 99 = 2500 = 50² 。",
+        "knowledge_points": [
+          "数与形"
+        ],
+        "source": "22-23 六年级上·河南郑州·期末",
+        "merged_id": "unit8_466",
+        "original_id": "466"
+      }, {
+        "id": "467",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如下图, 1 张桌子可以坐 8 人, 2 张桌子拼起来可以坐 12 人。8 张这样的桌子拼起来能够坐下( )人; 要想坐下 48 人, 需要( )张桌子拼起来。![](images/682d05e2bdc3d0030c5568fda0c604f7db074c962ff8c8bd4ec583585b96b104.jpg)",
+        "options": [],
+        "answer": "36 11",
+        "solution": "【分析】通过观察图形可知, 1 张桌子可以坐 8 人, 2 张桌子可以坐 (8 + 4) 人, 3 张桌子可以坐 (8 + 4 + 4) 人, ……多一张桌子多坐 4 人, 以此类推, n 张桌子可以坐 [8+4 (n-1) ]人。据此解答。本题主要考查数与形结合的规律, 关键根据图示发现这组图形的规律, 并运用规律做题。【详解】n 张桌子可以坐：8 + 4 (n - 1) = 8 + 4 n - 4 = (4 + 4 n) 人。当 n = 8 时,4 × 8 + 4 = 3 2 + 4 = 3 6 (人)。4 + 4 n = 4 8。解: 4 + 4n - 4 = 48 - 4。4 n = 4 4。4 n ÷ 4 = 4 4 ÷ 4。n = 1 1。8 张这样的桌子拼起来能够坐下 36 人; 要想坐下 48 人, 需要 11 张桌子拼起来。",
+        "knowledge_points": [
+          "数与形",
+          "图形的规律"
+        ],
+        "source": "20-21 六年级上·河南安阳·期末",
+        "merged_id": "unit8_467",
+        "original_id": "467"
+      },
+      {
+        "id": "468",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": " 如图, 小三角形的边长是 1 cm , 第五个图形的周长是( ) cm 。![](images/338f5bb018fbec5a007909c93c0a14eabc2c04a52a8cc05fa030f87750ee5465.jpg)",
+        "options": [],
+        "answer": "7",
+        "solution": "【分析】由题意可知, 小三角形的边长是 1 cm , 则第一个图形的周长是 3 cm = 1 + 2 , 第二个图形的周长是 4 cm = 2 + 2 ,第三个图形的周长是 5 cm = 3 + 2 , 则第 n 个图形的周长是 (n + 2) cm 。据此计算即可。本题考查图形的变化规律，发现规律，利用规律是解题的关键。【详解】第五个图形的周长是:n + 2 = 5 + 2 = 7则第五个图形的周长是 7 cm 。",
+        "knowledge_points": [
+          "数与形",
+          "用字母表示数",
+          "图形的规律"
+        ],
+        "source": "21-22 五年级上·安徽黄山·期中",
+        "merged_id": "unit8_468",
+        "original_id": "468"
+      },
+      {
+        "id": "469",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "找规律。6, 3, 3/2 , ( ), 3/8 , ( ), 3/32 。",
+        "options": [],
+        "answer": "3/4 3/16",
+        "solution": "【分析】根据题意可知, 6 × 1/2 = 3, 3 × 1/2 = 3/2, ... 依次类推,每一个数都是前一个数的 1/2 ; 据此解答。解答此题的关键是根据所给出的数字，找出规律，再根据规律解决问题。【详解】3/2 × 1/2 = 3/4。3/8 × 1/2 = 3/1 6。6, 3, 3/2, 3/4, 3/8, 3/16, 3/32 。",
+        "knowledge_points": [
+          "数字排列的规律"
+        ],
+        "source": "22-23 六年级上·河南信阳·期末",
+        "merged_id": "unit8_469",
+        "original_id": "469"
+      },
+      {
+        "id": "470",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": " 下面图形按一定规律排列，这样第⑥幅图中一共有( )个小正方形。①![](images/44864b08da7ad163010e95dd87efcd24088652f89d702fa83977087b1ae7758e.jpg)②![](images/ad7ee659154ca67b88062fd94e2aeb7e56e2aede3f4b6caa8887101518c92006.jpg)③![](images/6d4f64fa293c38c9185f9819910a6fec500a5c4c09190b52dfeb4be6b2de8667.jpg)④![](images/6ea4c00cf2638ebfd8dc852768a738db1a8573b867c06132afacc1a214a34e03.jpg)···",
+        "options": [],
+        "answer": "42",
+        "solution": "【分析】观察图形可知：第①幅图中的小正方形一共有1行，每行2个，第②幅图中的小正方形一共有2行，每行3个，第③幅图中的小正方形一共有3行，每行4个，第④幅图中的小正方形一共有4行，每行5个，据此类推，第n幅图中小正方形一共有n行，每行有（n+1）个，把n=6代入，算出第⑥幅图中一共有多少小正方形即可。本题重点考查数与形的规律，明确行数和列数之间的关系是解答本题的关键。【详解】由分析可知:6 × (6 + 1) = 6 × 7 = 4 2 (个)。所以第⑥幅图中一共有42个小正方形。",
+        "knowledge_points": [
+          "数与形",
+          "图形的规律"
+        ],
+        "source": "22-23 六年级上·福建莆田·期末",
+        "merged_id": "unit8_470",
+        "original_id": "470"
+      },
+      {
+        "id": "471",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(23-24 六年级上·全国·随堂练习) 算一算。1 + 3 + 5 + 7 + 5 + 3 + 1 = ( ); 1 + 3 + 5 + 7 + 9 + 1 1 + 1 3 + 1 1 + 9 + 7 + 5 + 3 + 1 = ( ) 。",
+        "options": [],
+        "answer": "25;85",
+        "solution": "【分析】由相加的数字规律可得：第一式子中最中间的数是7，左右两侧的数都是1、3、5，可运用左侧三个数之和的2倍加上7得出答案;第二个式子中最中间的数是13，左右两侧的数都是1、3、5、7、9、11，可运用左侧6个数字之和的 2 倍, 再加上 13 , 得出答案。本题主要考查的是找出相加的数规律进行简便运算, 解题的关键是掌握中间数的两边数相同的规律, 进而计算得出答案。【详解】1 + 3 + 5 + 7 + 5 + 3 + 1 = (1 + 3 + 5) × 2 + 7 = 1 8 + 7 = 2 5。1 + 3 + 5 + 7 + 9 + 1 1 + 1 3 + 1 1 + 9 + 7 + 5 + 3 + 1 = (1 + 3 + 5 + 7 + 9 + 1 1) × 2 + 1 3 = 7 2 + 1 3 = 8 5",
+        "knowledge_points": [
+          "算式的规律",
+          "数字排列的规律"
+        ],
+        "source": "23-24 六年级上·全国·随堂练习",
+        "merged_id": "unit8_471",
+        "original_id": "471"
+      },
+      {
+        "id": "472",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(22-23 六年级上·黑龙江哈尔滨·期末) 按规律填空:1, 1, 2, 3, 5, 8, ( ), 2 1 。",
+        "options": [],
+        "answer": "13",
+        "solution": "【分析】我们能够发现，从第一个数开始，相邻两个数字的和等于下一个数字，依照这个规律，缺数的位置前面两个数分别是5和8，故所缺数就是5与8的和13，据此解答。仔细观察这列数，找出数字排列的规律是解答题目的关键，可以用已经排列好的数来验证。【详解】分析可知， 5 + 8 = 13 。按规律填空: 1, 1, 2, 3, 5, 8, 13, 21。",
+        "knowledge_points": [
+          "黄金螺旋线、斐波那契数列",
+          "数字排列的规律"
+        ],
+        "source": "22-23 六年级上·黑龙江哈尔滨·期末",
+        "merged_id": "unit8_472",
+        "original_id": "472"
+      },
+      {
+        "id": "473",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(22-23 六年级下·云南德宏·期末)1/2 + 1/4 + 1/8 + 1/1 6 + 1/3 2 + 1/6 4 + 1/1 2 8 + 1/2 5 6 + ... = ()",
+        "options": [],
+        "answer": "1",
+        "solution": "【分析】![](images/d14236ac53f118ca9368f10bb7e3fc6c97e433c3fcfb1a48785cf2a97cac219a.jpg)如图所示，用一个圆表示 '1', 从第 2 个数开始, 每个数都是前一个数的 1/2 , 从图上可以看出, 这些分数不断加下去, 总和就是 1 , 据此解答。本题主要考查数形结合思想的应用, 根据图形求出算式的结果是解答题目的关键。【详解】分析可知， 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + ... = 1 。",
+        "knowledge_points": [
+          "数与形",
+          "异分母分数加、减法"
+        ],
+        "source": "22-23 六年级下·云南德宏·期末",
+        "merged_id": "unit8_473",
+        "original_id": "473"
+      },
+      {
+        "id": "474",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "(22-23 六年级下·浙江绍兴·期末) 李叔叔早上跑步去离家 6 千米远的皂李湖锻炼, 请根据折线统计图回答问题:![](images/074a778718ccba7af6cb3aab27f9cd312bc4fef786134f11a2a6ee0bf960c48e.jpg)(1)如果李叔叔中途不休息，用( )分钟就可以跑到皂李湖。(2)李叔叔在皂李湖休息了( )分钟才返回的。(3)返回时，李叔叔是骑公共自行车的，他返回时骑车的平均速度是每小时( )千米。",
+        "options": [],
+        "answer": "(1)40 (2)40  (3)18",
+        "solution": "【分析】（1）观察折线统计图，李叔叔早上6:00出发，7:00 到达皂李湖, 共花费 1 小时, 即 60 分钟, 中途休息了 60 ÷ 3 = 20 （分钟），所以如果中途不休息，用 60 - 20 = 40 分钟就可以跑到皂李湖。(2) 7: 00 到达皂李湖, 然后休息一段时间后返回, 返回时间是 8: 00 , 共花费 1 小时, 即 60 分钟, 所以在皂李湖休息了 60 ÷ 3 × 2 = 40 (分钟) 才返回的。(3) 返回所花的时间是 60 - 40 = 20 分钟, 路程是 6 千米,根据路程 ÷ 时间 = 速度, 代入数据即可得解。此题考查的目的是理解掌握折线统计图的特点及作用，并且能够根据统计图提供的信息，解决有关的实际问题。【详解】(1) 7: 00 - 6: 00 = 1 （小时） 1小时=60分钟。60 ÷ 3 = 20 (分钟)。60 - 20 = 40 (分钟)即如果李叔叔中途不休息，用40分钟就可以跑到皂李湖。(2) 8: 00 - 7: 00 = 1 (小时) 1小时=60分钟。60 ÷ 3 × 2 = 40 （分钟）即李叔叔在皂李湖休息了40分钟才返回的。(3) 60 - 40 = 20 (分钟)。20 分钟 = 1/3 小时。6 ÷ 1/3 = 6 × 3 = 18 (千米/时) 即他返回时骑车的平均速度是每小时 18 千米。",
+        "knowledge_points": [
+          "单式折线统计图",
+          "基础行程问题"
+        ],
+        "source": "22-23 六年级下·浙江绍兴·期末",
+        "merged_id": "unit8_474",
+        "original_id": "474"
+      },
+      {
+        "id": "475",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(22-23 六年级下·广东河源·期末)六(1)班9名同学进行乒乓球比赛, 如果每2名同学之间都进行一场比赛, 一共要比赛( )场。",
+        "options": [],
+        "answer": "36",
+        "solution": "【分析】每两个人之间都要比赛一场，即进行循环赛，共有 9 人, 则每人都要与另外 8 人进行比赛, 每人要参赛 8 场, 9 人共参赛 9 × 8 = 72 场。由于比赛是在两人之间进行的, 所以一共要进行比赛 72 ÷ 2 = 36 场。循环赛中，参赛人数与比赛场数之间的关系为：参赛人数 × （参赛人数 - 1） ÷ 2 = 比赛总场数。【详解】9 × 8 = 72 （场）。72 ÷ 2 = 36 (场)",
+        "knowledge_points": [
+          "排列组合问题"
+        ],
+        "source": "22-23 六年级下·广东河源·期末",
+        "merged_id": "unit8_475",
+        "original_id": "475"
+      },
+      {
+        "id": "476",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(22-23 六年级下·山东济南·期末) 如图所示, 照这样摆 8 个三角形要小棒( )根, 摆 n 个三角形要小棒( )根。![](images/543081aa90fa2e97f957104a57511349b67f0bbb4849810431a19c29f9c187e9.jpg)![](images/8f55bb57b6e70b7614c8436022a2b7437b48d92ca3377cab84dac8addb8b1b4b.jpg)![](images/3ca7f476b770783742880d7a6477bef7f5d3a04744df6feb30ddc3b5b6a46a72.jpg)",
+        "options": [],
+        "answer": "17;2n+1",
+        "solution": "【分析】观察图形, 摆 1 个三角形需要 3 根小棒, 摆 2 个三角形需要 (3 + 2) 根小棒, 摆 3 个三角形需要 (3 + 2 × 2) 根小棒, 依次类推, 算出摆 8 个三角形需要的小棒数。从而得出摆 n 个三角形需要 [3 + 2 × (n - 1)] 根小棒。据此解答。根据题干中已知的图形的排列特点及其数量关系,推理得出一般的结论进行解答, 是此类问题的关键。【详解】3 + 2 × 7 = 3 + 1 4 = 17 (根)。3 + 2 × (n - 1) = 3 + 2 n - 2 =(2n + 1) 根。即照这样摆 8 个三角形要小棒 17 根, 摆 n 个三角形要小棒 (2 n + 1) 根。",
+        "knowledge_points": [
+          "数与形"
+        ],
+        "source": "22-23 六年级下·山东济南·期末",
+        "merged_id": "unit8_476",
+        "original_id": "476"
+      },
+      {
+        "id": "477",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(22-23 六年级下·辽宁鞍山·期末) 一个叫巴尔默的中学数学老师成功的从光谱数据 9/5, 16/12, 25/21, 36/32, ... 中得到巴尔默公式，从而打开了光谱奥秘大门。请你按照这种规律，写出第 8 个数据是____。",
+        "options": [],
+        "answer": "100/96",
+        "solution": "【分析】根据观察可知，第一个数的分子部分是 3 的平方，第二个数的分子部分是 4 的平方，第三个数的分子部分是 5 的平方，按顺序数下去，第 8 个数的分子部分应该是 10 的平方;分母部分是不断加上奇数，第一个数的分母是 5，第二个数的分母加 7，第三个数的分母继续加 9，第四个数的分母继续加 11，第 5 个数的分母继续加 13，第 6 个数的分母继续加 15，第 7 个数的分母继续加 17，第 5 个数的分母继续加 19。仔细观察分别找到分子和分母的排列规律是解题的关键。【详解】分子： 10 × 10 = 100。分母: 32 + 13 + 15 + 17 + 19 = 96。则第 8 个数据是 100/96 。",
+        "knowledge_points": [
+          "数与形"
+        ],
+        "source": "22-23 六年级下·辽宁鞍山·期末",
+        "merged_id": "unit8_477",
+        "original_id": "477"
+      },
+      {
+        "id": "478",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "(22-23 六年级上·山东菏泽·期末) 1 - 1/2 - 1/4 - 1/8 - ... - 1/64 = ( ) 。",
+        "options": [],
+        "answer": "1/64",
+        "solution": "【分析】观察下图可知：一个连加算式，第一个加数是 1/2 ，之后每个加数都是前一个加数的一半，这个算式的结果就是 1 减去最后一个加数所得的差。即 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ... 的和等于 1 减去最后一个分数。据此先将 1 - 1/2 - 1/4 - 1/8 - ... - 1/64 利用减法的性质化为 1 - (1/2 + 1/4 + 1/8 + ... + 1/64) ，再根据以上规律先算括号里面的，再算括号外面的。![](images/989d25e618394ad61225f3bc16523e7570b273a5ac8f20f4e64d3502eb0dec91.jpg)明确 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ... 的计算方法是解决此题的关键。可借助图示法理解 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ... 的计算方法。【详解】1 - 1/2 -1/4 -1/8 -... -1/64 = 1 - (1/2 + 1/4 + 1/8 + ... + 1/6 4) = 1 - (1 - 1/6 4) = 1 - 6 3/6 4 = 1/6 4",
+        "knowledge_points": [
+          "数与形",
+          "异分母分数加、减法"
+        ],
+        "source": "22-23 六年级上·山东菏泽·期末",
+        "merged_id": "unit8_478",
+        "original_id": "478"
+      },
+      {
+        "id": "479",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(22-23 六年级下·甘肃兰州·期末) 如图, 下列图形都是由面积为 1 的正方形按一定的规律组成，其中，第（1）个图形中面积为 1 的正方形有 2 个, 第 (2) 个图形中面积为 1 的正方形有 5 个, 第 (3) 个图形中面积为 1 的正方形有 9 个, …, 按此规律, 则第 (6) 个图形中面积为 1 的正方体的体积是方形有( )个。（1）![](images/63bbca77b06cdfd3d3750b83c4876b346a655d0c27d64810885ffd23d0ceb709.jpg)（2）![](images/15a10f55bc1d1495c1f36722ca11ae443b45588e405e67dbea7a15bbc303964b.jpg)（3）![](images/0082d4b9f2d76aea27a47998311e2b348d29138e4d8b57ea19cec5f1aeb22f55.jpg)（4）![](images/fa90b8fbcff459228f70d89f76b122a389f7c8d899545e561688145bb29ab75b.jpg)",
+        "options": [],
+        "answer": "27",
+        "solution": "【分析】按照从下往上的顺序观察每行小正方形的个数,第（1）个图形有2个小正方形，第（2）个图形有 (2 + 3)个小正方形，第（3）个图形有 (2 + 3 + 4) 个小正方形，第（4）个图形有 (2 + 3 + 4 + 5) 个小正方形，那么第(5) 个图形有 (2 + 3 + 4 + 5 + 6) 个小正方形, 第 (6) 个图形有 (2 + 3 + 4 + 5 + 6 + 7) 个小正方形，据此解答。本题主要考查数形结合思想的应用, 根据图形找出小正方形的个数变化的规律是解答题目的关键。【详解】2 + 3 + 4 + 5 + 6 + 7 = 5 + 4 + 5 + 6 + 7 = 9 + 5 + 6 + 7 = 1 4 + 6 + 7 = 2 0 + 7 = 2 7 (个)。所以，第（6）个图形中面积为1的正方形有27个。",
+        "knowledge_points": [
+          "数与形"
+        ],
+        "source": "22-23 六年级下·甘肃兰州·期末",
+        "merged_id": "unit8_479",
+        "original_id": "479"
+      },
+      {
+        "id": "480",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(22-23 六年级下·河南南阳·期末) 在同一平面内, 10个点最多可以连( )条线段; 5 ÷ 7 商的小数点后第 100 位上的数字是( ); 2022 个 7 连乘, 乘积个位上的数字是( )。",
+        "options": [],
+        "answer": "45 2 9",
+        "solution": "【分析】在同一平面内, 10 个点的每个点都可以和另外 9 个点连成 9 条线段, 共能连成 9 × 10 = 90 (条) 线段, 由于每条线段重复计算了一次，所以共能连成 90 ÷ 2 = 45 （条）线段;先计算出 5 ÷ 7 的商是 0.714285714285... 的循环节是 714285,小数部分从第 1 个数字开始, 以这 6 个数字为一个周期, 求小数点后第 100 位的数字, 则用 100 除以 7 , 商表示 100 里面有几个周期, 如果结果没有余数, 则小数点后第 100 个数字是一个周期的最后一个数字, 如果有余数, 余数是几,则小数点后第 100 个数字是一个周期的第几个数字;要求乘积个位上的数字是多少，则探索几个7相乘的结果末尾规律，1个7相乘的结果个位是7，2个7相乘的结果个位是9，3个7相乘的结果个位是3，4个7相乘的结果个位是1, 5 个 7 相乘的结果个位是 7, 6 个 7 相乘的结果个位是9...据此可知，结果每4个为1个周期，则先用 2022 ÷ 4 计算出2022里面有几个周期，如果结果没有余数，则2022个7连乘的结果个位是周期里面的最后1个数字，如果有余数，余数是几, 则 2022 个 7 连乘的结果个位是周期里面的第几个数字。本题考查了线段的计数问题、循环小数的认识以及周期问题。【详解】10 × (10 - 1) ÷ 2 =1 0 × 9 ÷ 2 =45(条)。在同一平面内，10个点最多可以连45条线段;5 ÷ 7 = 0.714285714285...。1 0 0 ÷ 6 = 1 6 ... 4。5 ÷ 7 商的小数点后第 100 位上的数字是 2 ;几个 7 相乘的结果末尾规律是 7、9、3、1、7、9、3、1、... ,2 0 2 2 ÷ 4 = 5 0 5 .... 2。2022 个 7 连乘，乘积个位上的数字是 9。",
+        "knowledge_points": [
+          "循环小数的认识与简写",
+          "周期问题",
+          "排列组合问题"
+        ],
+        "source": "22-23 六年级下·河南南阳·期末",
+        "merged_id": "unit8_480",
+        "original_id": "480"
+      },
+      {
+        "id": "481",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "(22-23 六年级下·江西宜春·期末) 找规律, 填数。10，14，22，38，70，134，（ ），（ ）。",
+        "options": [],
+        "answer": "262;518",
+        "solution": "【分析】数列中用后项减去前项, 14 - 10 = 4 = 2 × 2, 22 - 14 = 8 = 2 × 2 × 2 , 依次验证, 发现规律就是第几项减去其前项的差就是几个 2 的乘积, 据此解答。考查探索数的排列规律。【详解】14 - 10 = 4 = 2 × 2。2 2 - 1 4 = 8 = 2 × 2 × 2。3 8 - 2 2 = 1 6 = 2 × 2 × 2 × 2。7 0 - 3 8 = 3 2 = 2 × 2 × 2 × 2 × 2。1 3 4 - 7 0 = 6 4 = 2 × 2 × 2 × 2 × 2。规律：数列的第2项减去第1项的差等于2个2的积，数列的第 3 项减去第 2 项的差等于 3 个 2 的积……数列的第 7 项减去第 6 项的差等于 7 个 2 的积, 即 (262)1 3 4 = 2 × 2 × 2 × 2 × 2 × 2 = 1 2 8;数列的第8项减去第7项的差等于8个2的积，即（518）2 6 2 = 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2 5 6 。",
+        "knowledge_points": [
+          "数字排列的规律"
+        ],
+        "source": "22-23 六年级下·江西宜春·期末",
+        "merged_id": "unit8_481",
+        "original_id": "481"
+      },
+      {
+        "id": "482",
+        "unit": "第八单元：数与形",
+        "chapter": "数与形",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（18-19六年级下·河南南阳·期末）按下图用小棒摆正六边形，摆10个正六边形需要( )根小棒;摆n个正六边形需要( )根小棒。![](images/d4bbd9fb93edda222ca3feabb7daf44a9739bbcd7a0209dc20a90f15dcfe7c90.jpg)",
+        "options": [],
+        "answer": "51;5n+1",
+        "solution": "【分析】由图可知, 摆 1 个正六边形需要 6 根小棒, 摆 2 个正六边形需要 (6 + 5) 根小棒, 摆 3 个正六边形需要 (6 + 5 × 2) 根小棒……每增加一个正六边形就增加 5 根小棒, 以此类推, 摆 n 个正六边形需要 [6 + 5 × (n - 1)] 根小棒, 最后求出 n = 10 时式子的值，据此解答。本题主要考查数形结合思想的应用，观察图形找出正六边形数量和小棒数量的变化规律是解答题目的关键。【详解】摆 n 个正六边形需要小棒的根数： 6 + 5× (n-1) = 6 + 5 n - 5 = 5 n + 6 - 5 =(5n+1) 根当 n = 10 时。5 n + 1 = 5 × 1 0 + 1 = 5 0 + 1 = 5 1 (根)。所以，摆 10 个正六边形需要 51 根小棒，摆 n 个正六边形需要 (5 n + 1) 根小棒。",
+        "knowledge_points": [
+          "数与形"
+        ],
+        "source": "18-19六年级下·河南南阳·期末",
+        "merged_id": "unit8_482",
+        "original_id": "482"
+      },
+      {
+        "id": "483",
+        "unit": "第八单元 数与形",
+        "question_type": "填空题",
+        "difficulty": 0.85,
+        "question": "图中从左往右观察，我发现后面的一个正方形的点数比前一个正方形的点数多( )；第5个正方形有( )个点。(21-22 六年级下·湖南常德·期末) ![](images/fc47409baa83e6b83473f7b3831ae25751e5b019d535af5a407d361d02ea4e57.jpg)",
+        "options": [],
+        "answer": "4;20",
+        "solution": "【分析】点数依次为4、8、12…每次增加4。\n【详解】每次多4；第5个=4+4×4=20。",
+        "knowledge_points": ["数与形"],
+        "source": "21-22 六年级下·湖南常德·期末"
+      },
+      {
+        "id": "484",
+        "unit": "第八单元 数与形",
+        "question_type": "填空题",
+        "difficulty": 0.65,
+        "question": "找规律填空。(18-19 六年级下·河南南阳·期末) 1，2，4，7，( )，16，22，( )。",
+        "options": [],
+        "answer": "11；29",
+        "solution": "【分析】从第二项开始依次加1、2、3、4、5、6、7…\n【详解】7+4=11，22+7=29。",
+        "knowledge_points": ["数字排列的规律"],
+        "source": "18-19 六年级下·河南南阳·期末"
+      },
+      {
+        "id": "485",
+        "unit": "第八单元 数与形",
+        "question_type": "填空题",
+        "difficulty": 0.65,
+        "question": "按规律填空。(22-23 六年级下·浙江温州·期末) 2，5，10，17，26，37…第n个数是( )。",
+        "options": [],
+        "answer": "n²+1",
+        "solution": "【分析】2=1²+1，5=2²+1，10=3²+1，17=4²+1…\n【详解】所以第n个数是n²+1。",
+        "knowledge_points": ["数与形", "数字排列的规律"],
+        "source": "22-23 六年级下·浙江温州·期末"
+      },
+      {
+        "id": "486",
+        "unit": "第八单元 数与形",
+        "question_type": "判断题",
+        "difficulty": 0.65,
+        "question": "同一平面内的6条直线，最多有15个交点。( )(23-24 六年级下·湖北鄂州·期末)",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】2条直线最多1个交点；3条最多1+2=3；4条最多1+2+3=6；5条最多1+2+3+4=10；6条最多1+2+3+4+5=15。\n【详解】1+2+3+4+5=15，所以原说法正确。",
+        "knowledge_points": ["数与形 (归纳递推)"],
+        "source": "23-24 六年级下·湖北鄂州·期末"
+      },
+      {
+        "id": "487",
+        "unit": "第八单元 数与形",
+        "question_type": "判断题",
+        "difficulty": 0.65,
+        "question": "找规律：1/2，1/3，2/3，1/4，2/4，3/4，1/5，( )，括号里应填2/5。( )(23-24 六年级上·河南周口·期末)",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】分母固定时分子从1递增到分母-1，再分母加1后分子重新从1开始。\n【详解】所以1/5后应为2/5，原说法正确。",
+        "knowledge_points": ["数字排列的规律"],
+        "source": "23-24 六年级上·河南周口·期末"
+      },
+      {
+        "id": "488",
+        "unit": "第八单元 数与形",
+        "question_type": "判断题",
+        "difficulty": 0.65,
+        "question": "1+3+5…+13+15+13+11…+3+1=113。( )(20-21 六年级上·湖南张家界·期末)",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】1+3+5+…+15为前8个奇数和=8²；1+3+5+…+13为前7个奇数和=7²。\n【详解】8²+7²=64+49=113，所以原说法正确。",
+        "knowledge_points": ["算式的规律"],
+        "source": "20-21 六年级上·湖南张家界·期末"
+      },
+      {
+        "id": "489",
+        "unit": "第八单元 数与形",
+        "question_type": "判断题",
+        "difficulty": 0.85,
+        "question": "有一组数：1，2，5，10，17，26…根据这组数的排列规律，第8个数应是50。( )(22-23 六年级上·河南信阳·期末)",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】相邻差为1、3、5、7、9、11、13…依次为连续奇数。\n【详解】第7个=26+11=37，第8个=37+13=50，所以原说法正确。",
+        "knowledge_points": ["数字排列的规律"],
+        "source": "22-23 六年级上·河南信阳·期末"
+      },
+      {
+        "id": "490",
+        "unit": "第八单元 数与形",
+        "question_type": "解答题",
+        "difficulty": 0.4,
+        "question": "把一张纸剪成6块，从中取几块，将每一块又剪成6块，再任取几块，又将每一块剪成6块……如此剪下去，经过有限次后，能否恰好剪成2009块？说明理由。(2009 六年级下·全国·竞赛)",
+        "options": [],
+        "answer": "不能经过有限次操作，将一张纸恰好剪成2009块，因为2009被5除余4，余数不是1。",
+        "solution": "【分析】每进行一次操作，纸张总数会在原基础上增加5的整数倍，因此总块数始终是被5除余1的数。\n【详解】2009÷5=401…4，余数不是1，所以不能经过有限次操作恰好剪成2009块。",
+        "knowledge_points": ["平面图形的分割", "图形的变化规律"],
+        "source": "2009 六年级下·全国·竞赛"
+      },
+      {
+        "id": "491",
+        "unit": "第八单元 数与形",
+        "question_type": "解答题",
+        "difficulty": 0.65,
+        "question": "从甲城到乙城共1200千米，一辆汽车从甲城出发，每小时行75千米，行了6小时，共行驶了多少千米？距乙城还有多少千米？(22-23 三年级下·河北邢台·期中)",
+        "options": [],
+        "answer": "450千米;750千米",
+        "solution": "【分析】路程=速度×时间，先求已行路程，再用总路程减去已行路程。\n【详解】75×6=450（千米）；1200-450=750（千米）。",
+        "knowledge_points": ["基础行程问题"],
+        "source": "22-23 三年级下·河北邢台·期中"
+      },
+      {
+        "id": "492",
+        "unit": "第八单元 数与形",
+        "question_type": "解答题",
+        "difficulty": 0.65,
+        "question": "将1～5填入第一行的五个○中，将相邻两个○中的数之和填入上一行的○中，如此下去直到第五行（如图）。要使第五行○中的数最大，第一行中的数应以怎样的顺序填写？最大值是多少？(2010 六年级下·全国·竞赛) 图1![](images/c733623f2619cbaf96e45a9dee647e9f3a52b7efba99306b3cd6f0752e6ae20f.jpg) 图2![](images/d577b8c5e1dfd2653d828da148c1419ed8bf47d0fd4e7a83293e24fcb0d9aa30.jpg)",
+        "options": [],
+        "answer": "第一行可填1，3，5，4，2（满足最大值即可）；最大值是61。",
+        "solution": "【分析】从第一行到第五行的加法次数分别为1次、4次、6次、4次、1次。为使最终和最大，应让最大的数被加的次数最多。\n【详解】把5放在加6次的位置，把3和4放在加4次的位置，把1和2放在加1次的位置，可取顺序1，3，5，4，2。最大值=5×6+3×4+4×4+1+2=30+12+16+3=61。",
+        "knowledge_points": ["整数排列的规律"],
+        "source": "2010 六年级下·全国·竞赛"
+      },
+      {
+        "id": "493",
+        "unit": "第八单元 数与形",
+        "question_type": "解答题",
+        "difficulty": 0.65,
+        "question": "用A、B、C、D、E代表5人的姓名，这5人进行乒乓球比赛，每2人之间都要打一局。A已打了4局，B打了3局，C打了2局，D打了1局。E一共打了几局？分别和谁打的？(23-24 六年级上·湖南娄底·期末)",
+        "options": [],
+        "answer": "2局；分别和A、B。",
+        "solution": "【分析】5人每人最多打4局。根据A、B、C、D已打局数用连线法统计缺少的对局即可确定E参与的对局。\n【详解】如图：![](images/f52ea2c05faf0a486495f48a1860dc18cd2c2e4ac7da14834c7f36a001e395ff.jpg) 可知E共打2局，分别和A、B打。",
+        "knowledge_points": ["推理问题"],
+        "source": "23-24 六年级上·湖南娄底·期末"
+      },
+      {
+        "id": "494",
+        "unit": "第八单元 数与形",
+        "question_type": "解答题",
+        "difficulty": 0.65,
+        "question": "下面图形中各有多少个三角形？有什么规律？(23-24 四年级下·全国·课后作业) ![](images/781177f9d384f400bd7ce375cd2b4c1d3ec07cb7016042a199325bb45f6561e2.jpg)",
+        "options": [],
+        "answer": "1个；3个；6个；10个；规律示例：第n个图形有(1+2+…+n)个三角形（规律不唯一）。",
+        "solution": "【分析】图①为1个；图②为1+2；图③为1+2+3；图④为1+2+3+4…\n【详解】图①：1；图②：1+2=3；图③：1+2+3=6；图④：1+2+3+4=10。可归纳：第n个图形有(1+2+…+n)个三角形。",
+        "knowledge_points": ["用字母表示稍复杂的数量关系", "图形的规律", "三角形的概念及表示方法"],
+        "source": "23-24 四年级下·全国·课后作业"
+      },
+      {
+        "id": "495",
+        "unit": "第八单元 数与形",
+        "question_type": "解答题",
+        "difficulty": 0.65,
+        "question": "有5个人，每2个人握一次手，一共要握多少次手？(23-24 三年级下·全国·随堂练习)",
+        "options": [],
+        "answer": "10次",
+        "solution": "【分析】第1个人握(5-1)次，第2个人握(4-1)次…相加即可。\n【详解】(5-1)+(4-1)+(3-1)+(2-1)=4+3+2+1=10（次）。",
+        "knowledge_points": ["搭配问题"],
+        "source": "23-24 三年级下·全国·随堂练习"
+      },
+      {
+        "id": "496",
+        "unit": "第八单元 数与形",
+        "question_type": "解答题",
+        "difficulty": 0.65,
+        "question": "下列图形共用小棒61根，按规律排列，一共拼成了多少个连续正方形？(23-24 四年级上·江苏常州·期中) ![](images/2a542d004f807f2868d4a327c8b1373f9c7debb1bf62b1204201781dc886733a.jpg)",
+        "options": [],
+        "answer": "20个",
+        "solution": "【分析】连续拼n个正方形用棒数=3n+1。已知3n+1=61。\n【详解】(61-1)÷3=60÷3=20（个）。",
+        "knowledge_points": ["数与形"],
+        "source": "23-24 四年级上·江苏常州·期中"
+      },
+      {
+        "id": "497",
+        "unit": "第八单元 数与形",
+        "question_type": "解答题",
+        "difficulty": 0.4,
+        "question": "用绳子测量一口井的深度，把绳子三折来量，井外每折余16米；把绳子四折来量，井外每折余4米。井深和绳长各是多少？(23-24 六年级上·北京大兴·期中)",
+        "options": [],
+        "answer": "绳长144米；井深32米",
+        "solution": "【分析】三折余16米表示绳长比井深的3倍多48米；四折余4米表示绳长比井深的4倍多16米。用盈亏求井深，再求绳长。\n【详解】井深=(48-16)÷(4-3)=32（米）；绳长=(32+16)×3=144（米）。",
+        "knowledge_points": ["盈亏问题"],
+        "source": "23-24 六年级上·北京大兴·期中"
+      },
+      {
+        "id": "498",
+        "unit": "第八单元 数与形",
+        "question_type": "解答题",
+        "difficulty": 0.65,
+        "question": "用4根小棒可以摆一个正方形，照下图摆下去，摆20个正方形一共要多少根小棒？(22-23 三年级上·江苏淮安·期中) ![](images/418c6ae5bebc144b9d047efd0d159af22d6ec4b965ea95f53a42bf5dc2ac1a82.jpg)",
+        "options": [],
+        "answer": "61根",
+        "solution": "【分析】摆n个连续正方形用棒数=3n+1。\n【详解】3×20+1=61（根）。",
+        "knowledge_points": ["数与形 (归纳递推)"],
+        "source": "22-23 三年级上·江苏淮安·期中"
+      },
+      {
+        "id": "499",
+        "unit": "第八单元 数与形",
+        "question_type": "解答题",
+        "difficulty": 0.65,
+        "question": "李师傅从家开车去县城买家具，去的时候用了3小时，速度是80千米/时，沿原路返回时用了4小时。返回时平均每小时行多少千米？(2023 四年级上·全国·专题练习)",
+        "options": [],
+        "answer": "60千米/时",
+        "solution": "【分析】先求去时路程=80×3，再用返回速度=路程÷4。\n【详解】路程=80×3=240（千米）；返回速度=240÷4=60（千米/时）。",
+        "knowledge_points": ["基础行程问题"],
+        "source": "2023 四年级上·全国·专题练习"
+      },
+      {
+        "id": "500",
+        "unit": "第八单元 数与形",
+        "question_type": "解答题",
+        "difficulty": 0.65,
+        "question": "照这样的规律接着画下去，第5个图形中有多少个○？第8个图形呢？(23-24 六年级上·全国·课后作业) ![](images/a105903b91c020f91e64fa70cf4ff0f91222c06497c7371cd8cac3f0eabb6d3e.jpg)",
+        "options": [],
+        "answer": "24个；36个",
+        "solution": "【分析】第n个图中○和●共有(n+2)²个，●有n²个，所以○有(n+2)²-n²个。\n【详解】第5个：(5+2)²-5²=7²-25=49-25=24；第8个：(8+2)²-8²=10²-64=100-64=36。并可参考图示：![](images/c54e5d10b64a78b948270182a8019b52670c5436dbc78e59b859829c0d31f248.jpg) ![](images/63e540c9e52f420bc894090c00e1407827d9090a0a49ef48b7ca0cc4d3d0823c.jpg) ![](images/dd4048bea0159f69f3f7db3a5561346de78486dfda3e30cad7e17fe3b02d65d1.jpg)",
+        "knowledge_points": ["数与形"],
+        "source": "23-24 六年级上·全国·课后作业"
+      }
+  ]
+  
\ No newline at end of file
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new file mode 100644
index 0000000000000000000000000000000000000000..ca02413237a1dc3b39ac753256fcb60d07b929a0
--- /dev/null
+++ "b/\345\205\255\345\271\264\347\272\247\351\242\230\351\233\206/\345\205\255\345\271\264\347\272\247\344\270\212/\345\205\255\345\271\264\347\272\247\344\270\212\347\254\254\345\205\255\345\215\225\345\205\203\347\231\276\345\210\206\346\257\224\346\225\260\346\215\256\351\233\206.json"
@@ -0,0 +1,9273 @@
+[
+  {
+    "id": "unit6_choice_001",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "（25-26 六年级上·海南海口·单元测试）25克糖全部化在 100 克水中，得到糖水的含糖率是（）。",
+    "options": [
+      "A. $25\\%$",
+      "B. $20\\%$",
+      "C. $\\frac{1}{5}$"
+    ],
+    "answer": "B",
+    "solution": "【分析】先用糖的重量 + 水的重量，求出糖水的重量，再用糖的重量 $\\div$ 糖水的重量，再乘 $100\\%$，即可解答。\n\n【详解】$25 \\div (25 + 100) \\times 100\\%\\\\= 25 \\div 125 \\times 100 \\% \\\\ = 0.2 \\times 100 \\% \\\\ = 20 \\% $\n\n25 克糖全部化在 100 克水中，得到糖水的含糖率是 $20\\%$。故答案为：B",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "25-26 六年级上·海南海口·单元测试",
+    "merged_id": "unit6_001",
+    "original_id": "unit6_choice_001"
+  },
+  {
+    "id": "unit6_choice_002",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "(25-26 六年级上·海南海口·单元测试) 一袋糖 4 千克,吃了 $\\frac{1}{2}$ 千克, 吃了 ( )。",
+    "options": [
+      "A. $125\\%$",
+      "B. $12.5\\%$",
+      "C. $3 \\frac{1}{2}$ 千克"
+    ],
+    "answer": "B",
+    "solution": "【分析】分析题目，用吃了的质量除以这袋糖的总质量即可得到吃了百分之几。\n\n【详解】$\\frac{1}{2} \\div 4 \\times 100\\%\\\\= 0.5 \\div 4 \\times 100 \\% \\\\ = 0.125 \\times 100 \\% \\\\ = 12.5\\% $\n\n一袋糖 4 千克, 吃了 $\\frac{1}{2}$ 千克, 吃了 $12.5 \\%$。故答案为: B",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "25-26 六年级上·海南海口·单元测试",
+    "merged_id": "unit6_002",
+    "original_id": "unit6_choice_002"
+  },
+  {
+    "id": "unit6_choice_003",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "(25-26 六年级上·海南海口·单元测试) 甲班有 160 本图书, 比乙班多 20 本, 甲班比乙班多百分之几的算式是 ()。",
+    "options": [
+      "A. $20 \\div 160 \\times 100 \\%$",
+      "B. $20 \\div (160 - 20) \\times 100 \\%$",
+      "C. $20 \\div (160 + 20) \\times 100\\%$"
+    ],
+    "answer": "B",
+    "solution": "【分析】先用甲班有图书的本数一比乙班多的本数，即 $160 - 20 = 140$ 本，求出乙班有图书的本数；再用甲班比乙班多的本数 $\\div$ 乙班有图书的本数，再乘 $100\\%$，即可求出甲班比乙班多百分之几，据此解答\n\n【详解】$20 \\div (160 - 20) \\times 100\\%\\\\= 20 \\div 140 \\times 100 \\%\\\\ \\approx 0.143 \\times 100 \\% \\\\ = 14.3 \\% $\n\n甲班有 160 本图书, 比乙班多 20 本, 甲班比乙班多百分之几的算式是 $20 \\div (160 - 20) \\times 100\\%$ .故答案为: B",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "25-26 六年级上·海南海口·单元测试",
+    "merged_id": "unit6_003",
+    "original_id": "unit6_choice_003"
+  },
+  {
+    "id": "unit6_choice_004",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的意义",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "（25-26六年级上·海南海口·单元测试）下面各数能改写成百分数的是（）。",
+    "options": [
+      "A. 甲绳子比乙绳子长 $\\frac{3}{5}$ 米",
+      "B. 课桌高 0.8 米",
+      "C. 小明比小华高 $\\frac{1}{20}$"
+    ],
+    "answer": "C",
+    "solution": "【分析】百分数不表示具体的量，没有单位，据此可知选项中没有单位的小数或分数可以改写成百分数，据此解答。\n\n【详解】$\\frac{1}{20} = 0.05 = 5\\%$根据分析可知: 小明比小华高 $\\frac{1}{20}$ 中的 $\\frac{1}{20}$ 可以改写成百分数$5 \\%$。故答案为: C",
+    "knowledge_points": [
+      "百分数的意义",
+      "百分数、小数和分数的互化"
+    ],
+    "source": "25-26六年级上·海南海口·单元测试",
+    "merged_id": "unit6_004",
+    "original_id": "unit6_choice_004"
+  },
+  {
+    "id": "unit6_choice_005",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的互化",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "（25-26六年级上·海南海口·单元测试）把2.8化成百分数写作（）。",
+    "options": [
+      "A. $28\\%$",
+      "B. $280\\%$",
+      "C. $2800\\%$"
+    ],
+    "answer": "B",
+    "solution": "【分析】小数化成百分数：先把小数的小数点向右移动两位，再在后面加上百分号即可。\n\n【详解】$2.8 = 280\\%$ 把2.8化成百分数写作 $280 \\%$。故答案为: B",
+    "knowledge_points": [
+      "百分数、小数和分数的互化"
+    ],
+    "source": "25-26六年级上·海南海口·单元测试",
+    "merged_id": "unit6_005",
+    "original_id": "unit6_choice_005"
+  },
+  {
+    "id": "unit6_choice_006",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "（24-25六年级下·全国·单元测试）乘公交车可以投币，也可以刷卡，刷卡每人每次只需1元，比投币便宜 $50\\%$，那么投币每人每次需要（）元。",
+    "options": [
+      "A. 0.5",
+      "B. 1.5",
+      "C. 2",
+      "D. 2.5"
+    ],
+    "answer": "C",
+    "solution": "【分析】把每人每次投币的钱数看作单位 '1', 刷卡每人每次的钱数是投币的 $(1 - 50 \\%$ , 对应的是刷卡每人每次的钱数 1 元, 求单位 '1', 用刷卡每人每次的钱数 $\\div$ $(1 - 50 \\%$ , 即可解答。\n\n【详解】$1 \\div (1 - 50\\%)\\\\= 1 \\div 50 \\% \\\\ = 2 \\text {(元)} $\n\n乘公交车可以投币，也可以刷卡，刷卡每人每次只需1元，比投币便宜 $50 \\%$ , 那么投币每人每次需要 2 元。故答案为: C",
+    "knowledge_points": [
+      "已知比一个数多/少百分之几是多少，求这个数"
+    ],
+    "source": "24-25六年级下·全国·单元测试",
+    "merged_id": "unit6_006",
+    "original_id": "unit6_choice_006"
+  },
+  {
+    "id": "unit6_choice_007",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "(24-25 六年级下·全国·单元测试) '田'字中的每一小格占整个字的（）。",
+    "options": [
+      "A. $40\\%$",
+      "B. $25\\%$",
+      "C. $50\\%$"
+    ],
+    "answer": "B",
+    "solution": "【分析】将整个'田'字看作单位'1'，共有4个小格，$1 \\div$ 小格总数 $=$ 每一小格占整个字的百分之几。\n\n【详解】$1 \\div 4 = 0.25 = 25\\%$'田'字中的每一小格占整个字的 $25 \\%$。故答案为: B",
+    "knowledge_points": [
+      "百分数的意义",
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "24-25 六年级下·全国·单元测试",
+    "merged_id": "unit6_007",
+    "original_id": "unit6_choice_007"
+  },
+  {
+    "id": "unit6_choice_008",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "(24-25 六年级上·湖北襄阳·期末) 如果 $m$ 的 $\\frac{1}{5}$ 等于 $n$ 的 $25\\%$ （ $m, n$ 均不为 0），那么（）。",
+    "options": [
+      "A. $m < n$",
+      "B. $m = n$",
+      "C. $m > n$",
+      "D. 无法确定"
+    ],
+    "answer": "C",
+    "solution": "【分析】m 的 $\\frac{1}{5}$ 表示为 $m \\times \\frac{1}{5}$ , n 的 $25 \\%$ 表示为 $n \\times 25 \\%$ , 则 $m \\times \\frac{1}{5} = n \\times 25 \\%$ , 当乘法算式的乘积相等且不为 0 时, 如果已知因数越小, 那么与它相乘的另一个因数越大; 相反地,已知因数越大, 与它相乘的另一个因数就越小, 据此解答。\n\n【详解】分析可知, $m \\times \\frac{1}{5} = n \\times 25\\%$ , 则 $m \\times \\frac{1}{5} = n \\times \\frac{1}{4}$ , 因为 $\\frac{1}{5}< \\frac{1}{4}$ , 所以 $m > n$。故答案为: C",
+    "knowledge_points": [
+      "求一个数的几分之几的问题",
+      "求一个数的百分之几是多少",
+      "因数和积的大小关系（分数乘法）"
+    ],
+    "source": "24-25 六年级上·湖北襄阳·期末",
+    "merged_id": "unit6_008",
+    "original_id": "unit6_choice_008"
+  },
+  {
+    "id": "unit6_choice_009",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "（23-24六年级上·全国·单元测试）一件商品，先涨价 $10\\%$ 后，又降价 $10\\%$，这件商品价格与原价相比（）。",
+    "options": [
+      "A. 提高了",
+      "B. 降低了",
+      "C. 没有变化"
+    ],
+    "answer": "B",
+    "solution": "【分析】 一件商品涨价 $10\\%$，是在原来的价格上涨的10%，把原来的价格看作单位'1'，那么涨价后的价格是$(1 + 10\\%)$，又降价 $10\\%$，是在涨价的基础上降低了10%，此时的单位'1'是涨价后的价格，据此求出现价与原价比较即可。\n\n【详解】由分析可知，现价是：$ (1 + 10 \\%) \\times (1 - 10 \\%) \\\\ = 1. 1 \\times 0. 9 \\\\ = 0. 9 9 \\\\ = 99 \\% $\n\n商品现在的价格低于原价。故答案为: B",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "23-24六年级上·全国·单元测试",
+    "merged_id": "unit6_009",
+    "original_id": "unit6_choice_009"
+  },
+  {
+    "id": "unit6_choice_010",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "(23-24 六年级上·全国·单元测试) 甲班有 160 本图书,比乙班多20本，甲班比乙班多百分之几的算式是（）。",
+    "options": [
+      "A. $20 \\div 160 \\times 100 \\%$",
+      "B. $20 \\div (160 - 20) \\times 100 \\%$",
+      "C. $(160 - 20) \\div 160\\times 100\\%$"
+    ],
+    "answer": "B",
+    "solution": "【分析】用甲班有图书本数减去 20 本, 求出乙班有图书的本数; 再把乙班有图数的本数看作单位 '1', 用甲班有图书本数与乙班有图书本数的差, 除以乙班有图书本数, 再乘 $100 \\%$ , 即 $20 \\div (160 - 20) \\times 100\\%$ ; 据此解答。\n\n【详解】根据分析可知，甲班有 160 本图书，比乙班多 20 本，甲班比乙班多百分之几的算式是 $20 \\div (160 - 20) \\times 100\\%$。故答案为: B",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_010",
+    "original_id": "unit6_choice_010"
+  },
+  {
+    "id": "unit6_choice_011",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的意义",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "（23-24六年级上·全国·单元测试）下面各数不能改写成百分数的是（）。",
+    "options": [
+      "A. 甲绳子比乙绳子长 $\\frac{3}{5}$",
+      "B. 课桌高 $0.8$ 米",
+      "C. 小明比小华高 $\\frac{1}{20}$"
+    ],
+    "answer": "B",
+    "solution": "【分析】根据百分数表示一个数是另一个数的百分之几,能改写成百分数的数, 其形式应该是表示两个数之间的比例关系即可。本题考查了百分数的概念, 百分数的表示方法,熟练掌握百分数的概念是解题的关键。\n\n【详解】A. 甲绳子比乙绳子长 $\\frac{3}{5}$ , 表示的是两个量之间的比例关系, 可以改写成百分数, 即 $\\frac{3}{5}$ 可以改写成 $60 \\%$ , 所以 A 项不符合题意;B. 课桌高0.8米是一个具体的长度数值，不能改写成百分数，所以B项符合题意；C. 小明比小华高 $\\frac{1}{20}$ , 表示的是两个量之间的比例关系, 可以改写成百分数，即 $\\frac{1}{20}$ 可以改写成 $5 \\%$，所以C项不符合题意；故答案为：B",
+    "knowledge_points": [
+      "百分数的意义",
+      "百分数、小数和分数的互化"
+    ],
+    "source": "23-24六年级上·全国·单元测试",
+    "merged_id": "unit6_011",
+    "original_id": "unit6_choice_011"
+  },
+  {
+    "id": "unit6_choice_012",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "（24-25六年级上·全国·单元测试） $20\\mathrm{m}$ 增加 $10\\%$ 后，再减少 $10\\%$，结果是（）。",
+    "options": [
+      "A. $20 \\mathrm{~m}$",
+      "B. $22 \\mathrm{~m}$",
+      "C. $19.8 \\mathrm{~m}$",
+      "D. $21 \\mathrm{~m}$"
+    ],
+    "answer": "C",
+    "solution": "【分析】由题意可知, 第一个 $10 \\%$ 是把 $20 \\mathrm{~m}$ 看作单位 ' 1', 根据求一个数的百分之几是多少, 用乘法计算, 用 $20 \\times (1 + 10 \\%)$ 求出增加后的长度, 第二个 $10 \\%$ 是把增加后的长度看作单位 ' 1', 用增加后的长度乘 $(1 - 10 \\%)$ , 即可得解。\n\n【详解】$20 \\times (1 + 10\\%) \\times (1 - 10\\%)\\\\=20 \\times 110 \\% \\times 90 \\% \\\\ = 2\\times 90\\% \\\\ = 1 9. 8  (\\mathrm {m})  $\n\n结果是 $19.8 \\mathrm{~m}$。故答案为: C",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "24-25六年级上·全国·单元测试",
+    "merged_id": "unit6_012",
+    "original_id": "unit6_choice_012"
+  },
+  {
+    "id": "unit6_fill_013",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(25-26 六年级上·海南海口·单元测试) 50 千米增加 $10 \\%$ 后是( )千米; 60 米比( )米少 $20 \\%$。",
+    "options": [],
+    "answer": "55;75",
+    "solution": "【分析】求 50 千米增加 $10 \\%$ 后是多少, 就是把 50 千米看作单位 '1', 用 $50 \\times (1 + 10 \\%)$ 即可。求出 60 米比谁少 $20 \\%$ , 就是把未知长度看作单位 '1', 60 米是这个未知长度的 $(1 - 20\\%)$ , 用 $60 \\div (1 - 20\\%)$ 即可。\n\n【详解】$50 \\times (1 + 10\\%)= 5 0 \\times 1. 1=55 (千米)$  $60 \\div (1 - 20 \\%)  = 6 0 \\div 0. 8 = 75 (米)$50千米增加 $10\\%$ 后是 55 千米；60 米比 75 米少 $20\\%$。",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少",
+      "已知比一个数多/少百分之几是多少，求这个数"
+    ],
+    "source": "25-26 六年级上·海南海口·单元测试",
+    "merged_id": "unit6_013",
+    "original_id": "unit6_fill_013"
+  },
+  {
+    "id": "unit6_fill_014",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "（25-26六年级上·海南海口·单元测试）某超市10月份鸡蛋的零售价比9月份涨了 $8\\%$，11月份又比10月份涨了8%。11月份的零售价是9月份的( )%，与9月份相比，涨价的幅度是( )%。",
+    "options": [],
+    "answer": "116.64;16.64",
+    "solution": "【分析】本题涉及连续两次涨价，均为 $8 \\%$。将9月份价格看作单位 '1', 10月份价格为 $1 \\times (1 + 8 \\%) ,$ 用10月份价格再乘 $(1 + 8 \\%)$ 求出11月份价格。据此解答。\n\n【详解】设9月份价格为'1'（即 $100\\%$ )。10月份价格：$1 \\times (1 + 8\\%) = 108\\% = 1.08$11月份价格：$1.08 \\times (1 + 8\\%) = 1.08 \\times 108\\% = 1.08 \\times 1.08 = 1.1664 = 116.64\\%$9月份相比，涨价的幅度：$116.64\\% - 100\\% = 16.64\\%$因此，11月份价格是9月份的 $116.64\\%$，涨价幅度为 $16.64\\%$。",
+    "knowledge_points": [
+      "含百分数的运算"
+    ],
+    "source": "25-26六年级上·海南海口·单元测试",
+    "merged_id": "unit6_014",
+    "original_id": "unit6_fill_014"
+  },
+  {
+    "id": "unit6_fill_015",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "（25-26 六年级上·海南海口·单元测试）一次计算测试,小明做对了 19 道题, 做错了 1 道题, 正确率是( )%。",
+    "options": [],
+    "answer": "95",
+    "solution": "【分析】已知小明做对了 19 道题, 做错了 1 道题, 先用做对的题数加上做错的题数计算出总题数为 $19 + 1 = 20$ 道; 再用做对的题数除以总题数乘 $100\\%$ 即可计算出正确率。\n\n【详解】$19 \\div (19 + 1) \\times 100\\%\\\\= 19 \\div 20 \\times 100 \\% \\\\ = 0.95 \\times 100 \\% \\\\ = 95 \\% $\n\n因此，正确率是 $95\\%$。",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "25-26 六年级上·海南海口·单元测试",
+    "merged_id": "unit6_015",
+    "original_id": "unit6_fill_015"
+  },
+  {
+    "id": "unit6_fill_016",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(25-26 六年级上·海南海口·单元测试)六(1)班男生占全班人数的 $\\frac{4}{9}$ , 男生比女生人数少( )%。",
+    "options": [],
+    "answer": "20",
+    "solution": "【分析】根据题意，六（1）班男生占全班人数的 $\\frac{4}{9}$ ，即把全班人数分成 9 份，男生占其中的 4 份，则女生占其中的 9 -4=5 份，用 5-4，求出男生比女生少的份数，再除以女生份数，再乘 $100 \\%$ , 即可求出男生比女生人数少百分之几,据此解答。\n\n【详解】$9 - 4 = 5$ （份）$ (5 - 4) \\div 5 \\times 100 \\% \\\\ = 1 \\div 5 \\times 100 \\% \\\\ = 0.2 \\times 100 \\% \\\\ = 20 \\% $\n\n六 (1) 班男生占全班人数的 $\\frac{4}{9}$ , 男生比女生人数少 $20 \\%$。",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "25-26 六年级上·海南海口·单元测试",
+    "merged_id": "unit6_016",
+    "original_id": "unit6_fill_016"
+  },
+  {
+    "id": "unit6_fill_017",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(25-26 六年级上·海南海口·单元测试) '一种花生仁的出油率是 $28 \\%$ '。 $200 \\mathrm{~kg}$ 这样的花生仁可榨油( ) $\\mathrm{kg}$ , 要榨 $7 \\mathrm{~kg}$ 油, 需要( ) $\\mathrm{kg}$ 花生仁。",
+    "options": [],
+    "answer": "56;25",
+    "solution": "【分析】已知花生仁质量为 $200 \\mathrm{~kg}$ , 出油率为 $28 \\%$ , 求一个数的百分之几是多少, 用乘法计算, 因此用花生仁的质量 乘出油率即可计算出榨出的油的质量;已知榨出的油的质量为 $7 \\mathrm{~kg}$ , 出油率为 $28 \\%$ , 已知一个数的百分之几是多少, 求这个数用除法计算, 因此用油的质量除以出油率即可计算出所需花生仁的质量。\n\n【详解】$200 \\times 28\\%\\\\= 2 0 0 \\times 0. 2 8 \\\\ = 5 6 (\\mathrm {k g})$  \n\n $ 7 \\div 28 \\% \\\\ = 7 \\div 0. 2 8 \\\\ = 2 5 (\\mathrm {kg}) $\n\n综上， $200 \\mathrm{~kg}$ 这样的花生仁可榨油 $56 \\mathrm{~kg}$ ，要榨 $7 \\mathrm{~kg}$ 油，需要 $25 \\mathrm{~kg}$ 花生仁。",
+    "knowledge_points": [
+      "求一个数的百分之几是多少",
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "25-26 六年级上·海南海口·单元测试",
+    "merged_id": "unit6_017",
+    "original_id": "unit6_fill_017"
+  },
+  {
+    "id": "unit6_fill_018",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(25-26 六年级上·海南海口·单元测试) 把 8 米长的绳子截成每段长是 $\\frac{4}{5}$ 米的小段, 可截( )段, 每段占全长的()%。",
+    "options": [],
+    "answer": "10;10",
+    "solution": "【分析】求可截成的段数, 需将绳子总长度除以每段长度,即用 8 米除以 $\\frac{4}{5}$ 米; 每段占全长的百分比等于每段数除以总段数后转化为百分数。\n\n【详解】$8 \\div \\frac{4}{5} = 8 \\times \\frac{5}{4} = 10$ （段）$1 \\div 10 = 0.1 = 10 \\%$ 所以可截 10 段, 每段占全长的 $10 \\%$ 。",
+    "knowledge_points": [
+      "分数与整数的除法",
+      "百分数、小数和分数的互化"
+    ],
+    "source": "25-26 六年级上·海南海口·单元测试",
+    "merged_id": "unit6_018",
+    "original_id": "unit6_fill_018"
+  },
+  {
+    "id": "unit6_fill_019",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(25-26 六年级上·海南海口·单元测试) 60 的 $40 \\%$ 是( ); ( )吨的 $75 \\%$ 是 75 吨。",
+    "options": [],
+    "answer": "24;100",
+    "solution": "【分析】求一个数的百分之几是多少，用乘法；已知一个数的百分之几是多少，求这个数用除法。据此解答。\n\n【详解】$60 \\times 40\\% = 24$   $75 \\div 75\\% = 100$ （吨）   所以 60 的 $40 \\%$ 是 24 ; 100 吨的 $75 \\%$ 是 75 吨。",
+    "knowledge_points": [
+      "求一个数的百分之几是多少",
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "25-26 六年级上·海南海口·单元测试",
+    "merged_id": "unit6_019",
+    "original_id": "unit6_fill_019"
+  },
+  {
+    "id": "unit6_fill_020",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "（25-26六年级上·海南海口·单元测试）今年的销售额比去年减少了 $10\\%$，今年的销售额相当于去年的( )%。",
+    "options": [],
+    "answer": "90",
+    "solution": "【分析】把去年的销售额看作单位'1'，今年的销售额比去年减少 $10\\%$ ，即今年的销售额相当于去年的 $1 - 10\\%$ ，据此解答。\n\n【详解】$1 - 10\\% = 90\\%$  今年的销售额比去年减少了 $10\\%$，今年的销售额相当于去年的 $90\\%$。",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几",
+      "单位'1'的认识与确定"
+    ],
+    "source": "25-26六年级上·海南海口·单元测试",
+    "merged_id": "unit6_020",
+    "original_id": "unit6_fill_020"
+  },
+  {
+    "id": "unit6_fill_021",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(25-26 六年级上·海南海口·单元测试) 15 是 40 的( )%；0.4 吨比 0.8 吨少( )%；8 米比 5 米多( )%。",
+    "options": [],
+    "answer": "37.5;50;60",
+    "solution": "【分析】求15是40的百分之几，用15除以40再乘100%;求0.4吨比0.8吨少百分之几，先算两者的差值，再用差值除以0.8吨，最后乘100%;求8米比5米多百分之几，先算两者的差值，再用差值除以5米，最后乘100%。\n\n【详解】$15÷40×100\\%\\\\=0.375×100\\%\\\\=37.5\\%$   \n\n$(0.8-0.4) ÷0.8×100\\%\\\\=0.4÷0.8×100\\%\\\\=0.5×100\\%\\\\=50\\%$\n\n$(8-5) ÷5×100\\%=3÷5×100\\%=0.6×100\\%=60\\%$因此，15是40的 $37.5\\%$ ；0.4吨比0.8吨少 $50\\%$ ；8米比5米多 $60\\%$ 。",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几",
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "25-26 六年级上·海南海口·单元测试",
+    "merged_id": "unit6_021",
+    "original_id": "unit6_fill_021"
+  },
+  {
+    "id": "unit6_fill_022",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的互化",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(25-26 六年级上·海南海口·单元测试) 在 $\\frac{3}{5}$ 、0.67、666.6%、6.066 这四个数中, 最大的数是( ), 最小的数是()。",
+    "options": [],
+    "answer": "666.6%;$\\frac{3}{5}$",
+    "solution": "【分析】根据分数与除法的关系 $\\frac{a}{b} = a \\div b (b \\neq 0)$ , 将 $\\frac{3}{5}$ 化为小数;将百分数化为小数：去掉百分号，再将小数的小数点向左移动两位;再比较这些小数的大小，从而确定最大数和最小数。\n\n【详解】$\\frac{3}{5} = 3 \\div 5 = 0.6$  666.6%去掉百分号为 666.6，再将小数点向左移动两位得 6.666;因为 $0.6 < 0.67 < 6.066 < 6.666$ , 所以 $\\frac{3}{5} < 0.67 < 6.066<$666.6%。因此，这四个数中，最大的是 $666.6\\%$，最小的是 $\\frac{3}{5}$。",
+    "knowledge_points": [
+      "多位小数的大小比较",
+      "百分数、小数和分数的互化"
+    ],
+    "source": "25-26 六年级上·海南海口·单元测试",
+    "merged_id": "unit6_022",
+    "original_id": "unit6_fill_022"
+  },
+  {
+    "id": "unit6_fill_023",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的意义",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(25-26 六年级上·海南海口·单元测试) 一车间上月生产合格产品占 $96 \\%$ , 这句话的意思是把( )看作 100 份, ( )占它的 $96 \\%$ , 不合格的占( )$\\%$。",
+    "options": [],
+    "answer": "产品总数量;合格产品数量;4",
+    "solution": "【分析】百分数表示一个数是另一个数的百分之几，已知合格产品占 $96\\%$，这里 $96\\%$ 表示合格产品数量占产品总数量的 $96\\%$，因此，将产品总数量看作单位'1'（即100份），合格产品数量占其中的96份，不合格产品占比则为 $100\\% - 96\\% = 4\\%$。据此解答。\n\n【详解】$100\\% - 96\\% = 4\\%$一车间上月生产合格产品占 $96 \\%$ , 这句话的意思是把产品总数量看作 100 份, 合格产品数量占它的 $96 \\%$ , 不合格的占 $4 \\%$",
+    "knowledge_points": [
+      "百分数的意义",
+      "含百分数的运算"
+    ],
+    "source": "25-26 六年级上·海南海口·单元测试",
+    "merged_id": "unit6_023",
+    "original_id": "unit6_fill_023"
+  },
+  {
+    "id": "unit6_fill_024",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "（25-26六年级上·海南海口·单元测试）商场降价促销，全场降价百分之十五，横线上的数写作( )，小玉买一件漂亮的衣服，实际花了原价的( )%。",
+    "options": [],
+    "answer": "$15\\%$;85",
+    "solution": "【分析】第一个空需要将'百分之十五'写成百分数形式,即 $15 \\%$。第二个空需要计算实际花费占原价的百分比, 由于降价 $15 \\%$ , 实际花费为原价的 $1 - 15 \\% = 85 \\%$。\n\n【详解】'百分之十五'写作百分数： $15 \\% $ 。$1 - 15\\% = 85\\%$ 所以横线上的数写作 $15 \\%$ , 实际花费是原价的 $85 \\%$。",
+    "knowledge_points": [
+      "含百分数的运算"
+    ],
+    "source": "25-26六年级上·海南海口·单元测试",
+    "merged_id": "unit6_024",
+    "original_id": "unit6_fill_024"
+  },
+  {
+    "id": "unit6_fill_025",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(2022·广东惠州·小升初真题) 5 千克是 4 千克的( )$\\%$, 4 千克比 5 千克少( )$\\%$。",
+    "options": [],
+    "answer": "125;20",
+    "solution": "【分析】求一个数是另一个数的百分之几用除法计算，再将结果转化为百分数；第二空是求一个数比另一个数少百分之几，先算出两数差值，再用差值除以被比的数，最后转化为百分数。\n\n【详解】$5 \\div 4 = 1.25 = 125\\%$\n\n$(5 - 4) \\div 5 \\\\ = 1 \\div 5 \\\\ = 20 \\% $\n\n所以 5 千克是 4 千克的 $125 \\%$ , 4 千克比 5 千克少 $20 \\%$。",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几",
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "2022·广东惠州·小升初真题",
+    "merged_id": "unit6_025",
+    "original_id": "unit6_fill_025"
+  },
+  {
+    "id": "unit6_fill_026",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的互化",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(24-25 六年级下·海南海口·单元测试) 在 $\\frac{6}{7}$ 、0.855、 $85\\%$ 和 $0.85$ 中，最大的数是____，最小的数是____。",
+    "options": [],
+    "answer": "$\\frac{6}{7}$;$85\\%$",
+    "solution": "【分析】把分数、百分数转化为小数，小数比较大小的方法：先看它们的整数部分，整数部分大的那个数就大；整数部分相同，十分位上的数大的那个数就大；十分位上的数相同，百分位上的数大的那个数就大……以此类推。据此比较。\n\n【详解】$\\frac{6}{7} \\approx 0.8571$   $85 \\% = 0.85$  因为 $0.8571 > 0.855 > 0.85 = 0.85$   所以 $\\frac{6}{7} > 0.855 > 0.85 = 85\\%$\n\n所以最大的数是 $\\frac{6}{7}$ , 最小的数是 $85 \\%$",
+    "knowledge_points": [
+      "循环小数的认识与简写",
+      "多位小数的大小比较",
+      "百分数、小数和分数的互化"
+    ],
+    "source": "24-25 六年级下·海南海口·单元测试",
+    "merged_id": "unit6_026",
+    "original_id": "unit6_fill_026"
+  },
+  {
+    "id": "unit6_fill_027",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "(24-25 六年级下·海南海口·单元测试) 15 是 40 的( ) %; 150 是 120 的( )$\\%$。",
+    "options": [],
+    "answer": "37.5;125",
+    "solution": "【分析】根据求一个数是另一个数的百分之几，用一个数÷另一个数 $\\times 100 \\%$ , 求 15 是 40 的百分之几, 用 $15 \\div 40 \\times 100 \\%$ 解答; 求 150 是 120 的百分之几, 用 $150 \\div 120 \\times 100 \\%$ 解答。\n\n【详解】$15 \\div 40 \\times 100\\%\\\\= 0.375 \\times 100 \\% \\\\ = 37.5 \\% $\n\n$ 150 \\div 120 \\times 100 \\% \\\\ = 1.25 \\times 100 \\% \\\\ = 125\\% $\n\n15是40的 $37.5 \\%$ ; 150是120的 $125 \\%$ .",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "24-25 六年级下·海南海口·单元测试",
+    "merged_id": "unit6_027",
+    "original_id": "unit6_fill_027"
+  },
+  {
+    "id": "unit6_fill_028",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(24-25 六年级下·海南海口·单元测试) $\\frac{2}{5}$ 吨比 0.8 吨少( ) \\% 0.8 吨比 $\\frac{2}{5}$ 吨多( ) %。",
+    "options": [],
+    "answer": "50;100",
+    "solution": "【分析】求 $\\frac{2}{5}$ 吨比 0.8 吨少百分之几, 先用减法求出少的吨数, 再除以 0.8 吨即可;求 $0.8 \\text{ 吨比 }\\frac{2}{5}$ 吨多百分之几，先用减法求出多的吨数，再除以 $\\frac{2}{5}$ 吨即可。\n\n【详解】$(0.8 - \\frac{2}{5}) \\div 0.8 \\times 100\\%\\\\= (0.8 - 0.4) \\div 0.8 \\times 100 \\% \\\\ = 0.4 \\div 0.8 \\times 100 \\% \\\\ = 0.5 \\times 100 \\% \\\\ = 50 \\% $\n\n$ (0.8 - \\frac{2}{5})\\div \\frac{2}{5}\\times 100\\% \\\\ = (0.8 - 0.4) \\div 0.4 \\times 100 \\% \\\\ = 0.4 \\div 0.4 \\times 100 \\% \\\\ = 1 \\times 100 \\% \\\\ = 100 \\% $\n\n$\\frac{2}{5}$ 吨比 0.8 吨少 (50) %, 0.8 吨比 $\\frac{2}{5}$ 吨多 (100) %。",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "24-25 六年级下·海南海口·单元测试",
+    "merged_id": "unit6_028",
+    "original_id": "unit6_fill_028"
+  },
+  {
+    "id": "unit6_fill_029",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的互化",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(24-25 六年级上·全国·单元测试) $\\frac{3}{8} = (\\quad): 16 = (\\quad) \\% = 75 \\div (\\quad) = (\\quad)$ 。 (填小数)",
+    "options": [],
+    "answer": "6;37.5;200;0.375",
+    "solution": "【分析】分数与比的关系：分数的分子相当于比的前项，分母相当于后项，分数值相当于比值；比基本性质：比的前项和后项同时乘或除以相同的数（0 除外），比值不变；用分数的分子除以分母，结果用小数表示，小数化成百分数的方法：小数点向右移动两位，添上百分号；分数与除法的关系：分数的分子相当于被除数，分母相当于除数；商不变的性质：被除数和除数都乘或除以一个相同的数（0除外），商不变。据此解答。\n\n【详解】$\\frac{3}{8} = 3:8 = (3\\times 2)$ ： $(8\\times 2) = 6:16$\n\n$\\frac{3}{8} = 3\\div 8 = 0.375 = 37.5\\%$ $3 \\div 8 = (3 \\times 2 5) \\div (8 \\times 2 5) = 7 5 \\div 2 0 0$ 所以 $\\frac{3}{8} = 6:16 = 37.5\\% = 75 \\div 200 = 0.375$ 。",
+    "knowledge_points": [
+      "商不变的规律及应用",
+      "百分数、分数、小数和比的互化",
+      "比的基本性质",
+      "分数化小数"
+    ],
+    "source": "24-25 六年级上·全国·单元测试",
+    "merged_id": "unit6_029",
+    "original_id": "unit6_fill_029"
+  },
+  {
+    "id": "unit6_fill_030",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(23-24 六年级上·全国·单元测试) 40 吨增加 $10\\%$ 后是( )吨; 60 米比( )米少 $20 \\%$",
+    "options": [],
+    "answer": "44;75",
+    "solution": "【分析】把40吨看作单位'1'，求40吨增加 $10\\%$ 后是多少吨，就是求40吨的 $(1 + 10\\%)$ 是多少，根据百分数乘法的意义，用40吨乘 $(1 + 10\\%)$ 即可求出结果；把第二个括号看作单位'1'，求60米比多少米少 $20\\%$，就是求60米是多少米的 $(1 - 20\\%)$ ，根据百分数除法的意义，用60米除以 $(1 - 20\\%)$ 即可求出结果。\n\n【详解】$40 \\times (1 + 10\\%)$\n\n$ 4 0 \\times 1. 1 \\\\ = 4 4 (\\text {吨}) $\n\n$60 \\div (1 - 20 \\%) \\\\ = 60 \\div 80 \\% \\\\ = 7 5 \\text {(米)} $\n\n40吨增加10%后是44吨；60米比75米少20%。",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少",
+      "已知比一个数多/少百分之几是多少，求这个数"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_030",
+    "original_id": "unit6_fill_030"
+  },
+  {
+    "id": "unit6_fill_031",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的意义",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "(24-25 六年级上·江苏·课后作业) 求百分率实质就是求一个数是另一个数的____。",
+    "options": [],
+    "answer": "百分之几",
+    "solution": "【分析】百分率是指一个数占另一个数的百分比，例如达标率、发芽率等都是求部分数量占总数量的百分之几。所以求百分率实质就是求一个数是另一个数的百分之几。据此解答。\n\n【详解】求百分率实质就是求一个数是另一个数的百分之几。",
+    "knowledge_points": [
+      "百分数的意义",
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "24-25 六年级上·江苏·课后作业",
+    "merged_id": "unit6_031",
+    "original_id": "unit6_fill_031"
+  },
+  {
+    "id": "unit6_judge_032",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的意义",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "（23-24六年级上·全国·单元测试）一袋大米 $10 \\mathrm{~kg}$ , 吃了 $20\\%$ 后, 还剩 $80\\% \\mathrm{~kg}$ 。 ( ) (对的打 ' √ ', 错的打 '× ', 并说明理由)   理由：____________________ 。",
+    "options": [],
+    "answer": "×;一袋大米 $10 \\mathrm{~kg}$ , 吃了 $20\\%$ 后, 剩余这袋大米的 $80\\%$ , 重量是 $8 \\mathrm{~kg}$。",
+    "solution": "【分析】百分数表示两个数之间的比率关系，不表示具体的数量，无单位名称，据此解答。\n\n【详解】剩下的大米是 $10 \\mathrm{~kg}$ 的 $(1 - 20 \\%$ ） $= 80 \\%$。重量是 $10 \\times 80 \\% = 8(\\mathrm{~kg})$。因此题目叙述错误。一袋大米 $10 \\mathrm{~kg}$ , 吃了 $20\\%$ 后, 还剩 $80\\% \\mathrm{~kg} 。 (× )$理由：一袋大米 $10 \\mathrm{~kg}$ ，吃了 $20\\%$ 后，剩余这袋大米的80%, 重量是 $8 \\mathrm{~kg}$。",
+    "knowledge_points": [
+      "百分数的意义",
+      "求一个数的百分之几是多少"
+    ],
+    "source": "23-24六年级上·全国·单元测试",
+    "merged_id": "unit6_032",
+    "original_id": "unit6_judge_032"
+  },
+  {
+    "id": "unit6_fill_033",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(23-24 六年级上·全国·单元测试) 某年 8 月初每千克猪肉的价格比7月份初上涨了 $12\\%$，9月初又比8月初回落了 $10\\%$，9月初猪肉价格与7月初猪肉的价格相比，（ ）(填'涨了'或'降了'),幅度是( )$\\%$。",
+    "options": [],
+    "answer": "涨了;0.8",
+    "solution": "【分析】把 7 月初的价格看作单位 '1', 则 8 月初的价格是7月初的 $(1 + 12\\%)$ ；再把8月初的价格看作单位'1'，则9月初的价格是8月初的（1一10%）；根据百分数乘法的意义, 则 9 月初的价格是 7 月初的 $(1 + 12\\%) \\times (1 - 10\\%)$ ；然后再比较上涨了还是下降了。\n\n【详解】$1 \\times (1 + 12\\%) \\times (1 - 10\\%)\\\\= 1 \\times 1. 1 2 \\times 0. 9\\\\=1.0 0 8 \\\\ = 100.8\\% \\\\ 100.8\\% > 1 \\\\ 100.8\\% - 1 = 0.8\\% $\n\n9月初猪肉价格与7月初猪肉的价格相比上涨了，幅度是 $0.8\\%$ 。",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_033",
+    "original_id": "unit6_fill_033"
+  },
+  {
+    "id": "unit6_fill_034",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(23-24 六年级上·全国·单元测试) 某班男生是女生的 $\\frac{4}{5}$ , 男生比女生少( )$\\%$。",
+    "options": [],
+    "answer": "20",
+    "solution": "【分析】把题目里的女生看作单位 '1', 则男生是 $\\frac{4}{5}$ , 那么男生比女生少 $1 - \\frac{4}{5} = \\frac{1}{5}$ , 再用 $\\frac{1}{5}$ 除以 1 解答即可。\n\n【详解】$(1 - \\frac{4}{5}) \\div 1 \\times 100\\%\\\\=\\frac{1}{5}\\div 1\\times 100\\% \\\\ = 0.2 \\times 100 \\% \\\\= 20 \\%$\n\n故男生比女生少 $20$$\\%$",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_034",
+    "original_id": "unit6_fill_034"
+  },
+  {
+    "id": "unit6_fill_035",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(23-24 六年级上·全国·单元测试) 一次计算测试, 小明做对了 30 道题, 做错了 3 道题, 正确率约是( )。",
+    "options": [],
+    "answer": "$91\\%$",
+    "solution": "【分析】用做对的题的数量加上做错的题的数量，求出做题的总数，正确率 = 做题的正确数 ÷ 做题的总数 $\\times 100\\%$，据此代入数据解答即可。\n\n【详解】$30 \\div (30 + 3) \\times 100\\%\\\\=30 \\div 33 \\times 100 \\% \\\\ \\approx 0.909 \\times 100 \\% \\\\ \\approx 91 \\% $\n\n所以正确率约是 $91 \\%$",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_035",
+    "original_id": "unit6_fill_035"
+  },
+  {
+    "id": "unit6_fill_036",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(23-24 六年级上·全国·单元测试) 把 16 米长的绳子截成每段长是 $\\frac{4}{5}$ 米的小段, 可截( )段, 每段占长的( )%。",
+    "options": [],
+    "answer": "20;5",
+    "solution": "【分析】据题意可知, 求可截几段, 是把 16 米平均分, 用除法计算, 用绳子的长度除以每段的长度, 即可得段数; 可分几段, 每段就占全长的几分之一, 再把分数转化为百分数即可。\n\n【详解】$16 \\div \\frac{4}{5} = 16 \\times \\frac{5}{4} = 20$ （段）$\\frac{1}{20} = 0.05 = 5\\%$把 16 米长的绳子截成每段长是 $\\frac{4}{5}$ 米的小段, 可截 20 段, 每段占长的 $5 \\%$",
+    "knowledge_points": [
+      "分数与除法的关系",
+      "百分数、小数和分数的互化"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_036",
+    "original_id": "unit6_fill_036"
+  },
+  {
+    "id": "unit6_fill_037",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的互化",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(23-24 六年级上·全国·单元测试) 在 $\\frac{2}{3} 、 0.6 、 6 \\%$ 、 $\\frac{3}{5}$ 这四个数中，最大的数是( )，最小的数是( )。",
+    "options": [],
+    "answer": "$\\frac{2}{3}$;$6\\%$",
+    "solution": "【分析】先把分数、百分数化成小数，百分数化成小数：把小数点向左移动两位，同时去掉百分号。小数比较大小的方法：先看它们的整数部分，整数部分大的那个数就大；整数部分相同，十分位上的数大的那个数就大；十分位上的数相同，百分位上的数大的那个数就大……以此类推。据此判断。\n\n【详解】$\\frac{2}{3} \\approx 0.67$  $6 \\% = 0.06$   $\\frac {3}{5} = 0. 6$   因为 $0.06 < 0.6 < 0.67$    所以最大的数是 $\\frac{2}{3}$ , 最小的数是 $6 \\%$",
+    "knowledge_points": [
+      "多位小数的大小比较",
+      "百分数、小数和分数的互化"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_037",
+    "original_id": "unit6_fill_037"
+  },
+  {
+    "id": "unit6_fill_038",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(23-24 六年级上·全国·单元测试) '一种花生仁的出油率是 $28 \\%$ '。 $300 \\mathrm{~kg}$ 这样的花生仁可榨油( )kg, 要榨 $70 \\mathrm{~kg}$ 油, 需要( )kg 花生仁。",
+    "options": [],
+    "answer": "84;250",
+    "solution": "【分析】根据求一个数的百分之几是多少，用乘法计算，用花生仁的质量乘出油率即可得出榨油的质量；再根据已知一个数的百分之几，求这个数用除法计算，用榨油的质量除以出油率，即可得花生仁的质量。据此解答。\n\n【详解】$300 \\times 28\\% = 84 (\\mathrm{~kg})$  $70 \\div 28 \\% = 250 (\\mathrm {kg})$  $300 \\mathrm{~kg}$ 这样的花生仁可榨油 $84 \\mathrm{~kg}$ , 要榨 $70 \\mathrm{~kg}$ 油, 需要250$\\mathrm{~kg}$花生仁。",
+    "knowledge_points": [
+      "求一个数的百分之几是多少",
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_038",
+    "original_id": "unit6_fill_038"
+  },
+  {
+    "id": "unit6_fill_039",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "（23-24六年级上·全国·单元测试）今年的销售额比去年减少了 $15\\%$，今年的销售额相当于去年的( )%。",
+    "options": [],
+    "answer": "85",
+    "solution": "【分析】根据'今年的销售额比去年减少了 $15 \\%$ '可知，把去年的销售额看作单位'1'，今年的销售额相当于去年的 $(1 - 15\\%)$。据此解答。\n\n【详解】$1 - 15\\% = 85\\%$ 今年的销售额比去年减少了 $15 \\%$ , 今年的销售额相当于去年的 $85 \\%$。",
+    "knowledge_points": [
+      "含百分数的运算",
+      "单位'1'的认识与确定"
+    ],
+    "source": "23-24六年级上·全国·单元测试",
+    "merged_id": "unit6_039",
+    "original_id": "unit6_fill_039"
+  },
+  {
+    "id": "unit6_fill_040",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(23-24 六年级上·全国·单元测试) 80 的 $40\\%$ 是( )；( )吨的 $75\\%$ 是 75 吨。",
+    "options": [],
+    "answer": "32;100",
+    "solution": "【分析】把 80 看作单位 '1', 根据百分数乘法的意义, 用 80 乘 $40 \\%$ 即可求出 80 的 $40 \\%$ 是多少; 把第二个括号看作单位 '1', 根据百分数除法的意义, 用 75 吨除以 $75 \\%$ 即可求出单位 '1'。\n\n【详解】$80 \\times 40\\% = 32$  $75 \\div 75\\% = 100$ （吨）80 的 $40 \\%$ 是 32; 100 吨的 $75 \\%$ 是 75 吨。",
+    "knowledge_points": [
+      "求一个数的百分之几是多少",
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_040",
+    "original_id": "unit6_fill_040"
+  },
+  {
+    "id": "unit6_fill_041",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(23-24 六年级上·全国·单元测试) 在括号里填上适当的百分数。$\\frac{2}{5}$ 吨比 0.8 吨少( )；0.8 吨比 $\\frac{2}{5}$ 吨多( )。",
+    "options": [],
+    "answer": "$50\\%$;$100\\%$",
+    "solution": "【分析】把另一个数看作单位 '1', 求一个数比另一个数多或少百分之几, 用两个数的差除以另一个数解答。据此列式解答。\n\n【详解】$(0.8 - \\frac{2}{5}) \\div 0.8\\\\=(0. 8 - 0. 4) \\div 0. 8 \\\\ = 0. 4 \\div 0. 8 \\\\ = 50 \\% $\n\n$ (0. 8 - \\frac {2}{5}) \\div \\frac {2}{5} \\\\ = (0. 8 - 0. 4) \\div 0. 4 \\\\ = 0. 4 \\div 0. 4 \\\\ = 100 \\% $\n\n所以 $\\frac{2}{5}$ 吨比 0.8 吨少 $50 \\%$ , 0.8 吨 $\\frac{2}{5}$ 吨多 $100 \\%$。",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_041",
+    "original_id": "unit6_fill_041"
+  },
+  {
+    "id": "unit6_fill_042",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "（23-24六年级上·全国·单元测试）在括号里填上适当的百分数。15是40的（）；150是120的（ ）。",
+    "options": [],
+    "answer": "$37.5\\%$;$125\\%$",
+    "solution": "【分析】求一个数是另一个数的百分之多少，用这个数除以另一个数，据此列式解答。\n\n【详解】$15 \\div 40 = 0.375 = 37.5\\%$  $150 \\div 120 = 1.25 = 125 \\%$  所以15是40的 $37.5 \\%$ , 150是120的 $125 \\%$ .",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "23-24六年级上·全国·单元测试",
+    "merged_id": "unit6_042",
+    "original_id": "unit6_fill_042"
+  },
+  {
+    "id": "unit6_fill_043",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的读写",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(23-24 六年级上·全国·单元测试) $26 \\%$ 读作( )；百分之十二点五写作( )。",
+    "options": [],
+    "answer": "百分之二十六;12.5%",
+    "solution": "【分析】读百分数时，百分号前面的数按照整数、小数的读法来读，百分号读作'百分之'，先读'百分之'，再读百分号前面的数；写百分数时，百分之写作'%'，后面的数按照整数、小数的写法来写，先写整数或小数，再写百分号：'%'。\n\n【详解】$26 \\%$ 读作：百分之二十六百分之十二点五写作： $12.5\\%$ 所以第一空填百分之二十六，第二空填 $12.5\\%$。",
+    "knowledge_points": [
+      "百分数的读法和写法"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_043",
+    "original_id": "unit6_fill_043"
+  },
+  {
+    "id": "unit6_fill_044",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "（24-25六年级上·全国·单元测试）李师傅加工200个零件，有2个不合格，合格率为( )%，不合格率为( )%。",
+    "options": [],
+    "answer": "99;1",
+    "solution": "【分析】（产品总数一不合格数量）÷产品总数×100%=合格率，不合格率 $= 1 -$ 合格率，据此解答。\n\n【详解】$(200 - 2) \\div 200 \\times 100\\%\\\\=198 \\div 200 \\times 100 \\% \\\\ = 0.99 \\times 100 \\% \\\\ = 99 \\% $\n\n$1 - 99 \\% = 1 \\%$所以合格率为 $99 \\%$ , 不合格率为 $1 \\%$",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "24-25六年级上·全国·单元测试",
+    "merged_id": "unit6_044",
+    "original_id": "unit6_fill_044"
+  },
+  {
+    "id": "unit6_fill_045",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(24-25六年级上·全国·单元测试) 甲、乙两车的速度比是 $5: 4$ , 甲车的速度比乙车快 $(\\quad) \\%$。如果甲车的速度是80千米/时，那么乙车行驶 $288 \\mathrm{~km}$ 的路程需( )小时。",
+    "options": [],
+    "answer": "25;4.5",
+    "solution": "【分析】根据甲、乙两车的速度比是 $5: 4$ , 可将甲车速度看作5，乙车速度看作4，甲车速度比乙车速度快：（甲车速度一乙车速度）÷乙车速度×100%；已知甲车的速度，可根据比的基本性质求出乙车的速度，再根据时间=路程÷速度得出答案。\n\n【详解】甲车速度比乙车快:$ (5 - 4) \\div 4 \\times 100 \\% \\\\ =1 \\div 4 \\times 100 \\% \\\\ = 25 \\% $\n\n甲车速度为 80 千米/小时, 则 $5: 4 = (5 \\times 16): (4 \\times 16) =$ 80:64, 即乙车速度为: 64 千米/小时。则乙车行驶 288 千米需要: $288 \\div 64 = 4.5$ (小时) 。",
+    "knowledge_points": [
+      "比的基本性质",
+      "求一个数比另一个数多少百分之几"
+    ],
+    "source": "24-25六年级上·全国·单元测试",
+    "merged_id": "unit6_045",
+    "original_id": "unit6_fill_045"
+  },
+  {
+    "id": "unit6_fill_046",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的互化",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(24-25 六年级上·全国·单元测试) $5 \\div ()= 0.25 = \\frac{()}{4} = (\\quad) \\%$",
+    "options": [],
+    "answer": "20;1;25",
+    "solution": "【分析】小数化成分数, 两位小数先化成分母为 $100$ 的分数, 再化简成最简分数;分数的基本性质: 分数的分子和分母同时乘或除以相同的数 (0 除外), 分数的大小不变;分数与除法的关系：分子相当于被除数，分母相当于除数，分数线相当于除号；小数化成百分数，小数点向右移动两位，同时在数的后面添上百分号。\n\n【详解】$0.25 = \\frac{25}{100} = \\frac{1}{4}$     $\\frac {1}{4} = \\frac {1 \\times 5}{4 \\times 5} = \\frac {5}{2 0}, \\frac {5}{2 0} = 5 \\div 2  \\quad0.25 = 25 \\%$ 即 $5 \\div 20 = 0.25 = \\frac{1}{4} = 25 \\%$",
+    "knowledge_points": [
+      "分数的基本性质的应用",
+      "分数与除法的关系",
+      "百分数、小数和分数的互化"
+    ],
+    "source": "24-25 六年级上·全国·单元测试",
+    "merged_id": "unit6_046",
+    "original_id": "unit6_fill_046"
+  },
+  {
+    "id": "unit6_fill_047",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的意义",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(24-25 六年级上·全国·单元测试) $75 \\%$ 读作( )，它的计数单位是( )，再加( )个这样的单位，就是最小的奇数。",
+    "options": [],
+    "answer": "百分之七十五;$1\\%$;25",
+    "solution": "【分析】百分数的读法：先读百分号，读作百分之，再读百分号前面的数字；百分数的计数单位是 $1 \\%$ ，最小的奇数是 1 , 用 1 减去 $75 \\%$ , 即 $1 - 75 \\% = 25 \\%$ , 看 $25 \\%$ 里有几个计数单位就是再加几个这样的单位, 就是最小的奇数。\n\n【详解】$75 \\%$ 读作：百分之七十五；它的计数单位是 $1 \\%$ $1 - 75 \\% = 25 \\%$所以 $75 \\%$ 读作百分之七十五, 它的计数单位是 $1 \\%$ , 再加 25 个这样的单位, 就是最小的奇数。",
+    "knowledge_points": [
+      "百分数的意义",
+      "百分数的读法和写法"
+    ],
+    "source": "24-25 六年级上·全国·单元测试",
+    "merged_id": "unit6_047",
+    "original_id": "unit6_fill_047"
+  },
+  {
+    "id": "unit6_fill_048",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(24-25 六年级上·全国·单元测试) 六年级女生人数与男生人数的比是 $7: 8$ , 女生人数相当于男生人数的( ),男生人数相当于全班人数的( )。",
+    "options": [],
+    "answer": "$87.5\\%或\\frac{7}{8}$;$53.3\\%$",
+    "solution": "【分析】六年级女生人数与男生人数的比是 $7: 8$ , 将女生人数看作 7 , 男生人数看作 8 , 男生人数看作单位 '1', 女生人数 $\\div$ 男生人数 = 女生人数相当于男生人数的百分之几或几分之几; 全班人数看作单位 ' 1 , 男生人数 $\\div$ 全班人数 = 男生人数相当于全班人数的百分之几或几分之几。\n\n【详解】$7 \\div 8 = 0.875 = 87.5\\%$\n\n$8 \\div (7 + 8) \\\\ = 8 \\div 1 5 \\\\ \\approx 0. 5 3 3 \\\\ = 53.3 \\% $\n\n女生人数相当于男生人数的 $87.5 \\%$ , 男生人数相当于全班人数的 $53.3 \\%$。",
+    "knowledge_points": [
+      "比的意义",
+      "求一个数是另一个数的百分之几（百分率问题）",
+      "求一个数占另一个数几分之几"
+    ],
+    "source": "24-25 六年级上·全国·单元测试",
+    "merged_id": "unit6_048",
+    "original_id": "unit6_fill_048"
+  },
+  {
+    "id": "unit6_judge_049",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "（2014·全国·小升初真题）一种商品先提价 $10\\%$，再降价 $10\\%$，现价与原价相等。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】假设商品原价为 100 元，提价 $10 \\%$ , 就是在原价的基础上增加 $10 \\%$ 的价格, 因此提价后的价格 = 原价 $\\times (1 + 10 \\%)$ , 据此计算出提价后的价格; 再降价 $10 \\%$ , 此时是在提价后的价格基础上降低 $10 \\%$ , 则降价后的价格 = 提价后的价格 $\\times (1 - 10 \\%)$ , 据此计算出降价后的价格, 即现价;最后比较现价与原价作出判断。\n\n【详解】假设商品原价为100元，\n\n提价 $10\\%$ 后的价格为：\n\n$ 100 \\times (1 + 10 \\%) \\\\ = 100 \\times 110 \\% \\\\ = 1 0 0 \\times 1. 1 \\\\ = 1 1 0 (\\text {元}) $再降价 $10\\%$ \n\n后的价格为:$ 110 \\times (1 - 10 \\%) \\\\ = 110 \\times 90 \\% \\\\ = 1 1 0 \\times 0. 9 \\\\ = 9 9 \\text {(元)} $\n\n因为 $99 < 100$ , 所以现价低于原价, 原题说法错误。故答案为: ×",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "2014·全国·小升初真题",
+    "merged_id": "unit6_049",
+    "original_id": "unit6_judge_049"
+  },
+  {
+    "id": "unit6_judge_050",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "（2021·河北石家庄·小升初真题）一批零件，经检验有100个合格，合格率为 $100\\%$ （ ）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】合格率一般用百分数来表示，百分数是表示一个数是另一个数的百分之几的数，也叫百分率或百分比。所以合格率为（合格数÷合数零件数）×100%。在本题中，合部零件为 100 个，但全部零件数不明确，如果全部零件数为 100 个，则合格率为 100%，如果全部零件数多于 100 个，则合格率达不到 100%。\n\n【详解】如果全部零件数为 100 个, 则合格率为 $100 \\%$ , 如果全部零件数多于 100 个, 则合格率达不到 $100 \\%$。所以一批零件，经检验有 100 个合格，合格率为 $100\\%$ 的说法是错误的。故答案为: ×",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "2021·河北石家庄·小升初真题",
+    "merged_id": "unit6_050",
+    "original_id": "unit6_judge_050"
+  },
+  {
+    "id": "unit6_judge_051",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "（24-25六年级下·海南海口·单元测试）六（1）班种了105棵树苗，成活100棵，成活率是 $100\\%$ 。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】成活率 = 成活棵数 ÷ 种植棵数 × 100%, 代入数据,求出成活率, 再进行比较, 即可解答。\n\n【详解】$100 \\div 105 \\times 100\\% \\approx 0.952 \\times 100 \\% = 95.2 \\% $六（1）班种了 105 棵树苗，成活 100 棵，成活率是 $95.2\\%$。原题干说法错误。故答案为: ×",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "24-25六年级下·海南海口·单元测试",
+    "merged_id": "unit6_051",
+    "original_id": "unit6_judge_051"
+  },
+  {
+    "id": "unit6_judge_052",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的意义",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "(24-25 六年级下·海南海口·单元测试) 一个整数 A (A>0) 添上 '%', 这个数就缩小到原数的 100 倍。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】整数 A 添加百分号后变为 A%, 相当于将原数 A 除以 100 , 即缩小到原数的 $\\frac{1}{100}$。题目中 '缩小到原数的 100 倍'的说法错误, 因为 '100 倍' 会导致数值扩大而非缩小。因此, 正确答案应打叉。\n\n【详解】一个整数 A（A>0）添上'%'，这个数就缩小到原数的 $\\frac{1}{100}$ 。例如：25 添上 '%'，25 就变成了 0.25。\n\n所以原题说法错误。故答案为: ×",
+    "knowledge_points": [
+      "百分数的意义"
+    ],
+    "source": "24-25 六年级下·海南海口·单元测试",
+    "merged_id": "unit6_052",
+    "original_id": "unit6_judge_052"
+  },
+  {
+    "id": "unit6_judge_053",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "(24-25 六年级下·海南海口·单元测试) 成语 '十拿九'稳'中的'九稳'用百分数表示是 $90\\%$。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】成语 '十拿九稳' 表示 9 占 10 的百分之几, 用除法计算。\n\n【详解】$9 \\div 10 \\times 100\\%\\\\=0.9 \\times 100 \\% \\\\ = 90 \\% $\n\n成语'十拿九稳'中的'九稳'用百分数表示是 $90 \\%$。原题说法正确。故答案为: √",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "24-25 六年级下·海南海口·单元测试",
+    "merged_id": "unit6_053",
+    "original_id": "unit6_judge_053"
+  },
+  {
+    "id": "unit6_judge_054",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的意义",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "(19-20六年级上·河北·单元测试) 在50的后面添上'%', 就表示这个数缩小到原来 $\\frac{1}{100}$ ( )",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】在一个非零自然数的后面添上'%'，这个数就变成了一个百分数，化成小数时小数点向左移动两位，即这个数就缩小到原来的 $\\frac{1}{100}$ ，据此解答。\n\n【详解】根据分析可知，在50的后面添上'%'，就表示这个数缩小到原来 $\\frac{1}{100}$ 。原题干说法正确。故答案为: √",
+    "knowledge_points": [
+      "百分数、小数和分数的互化"
+    ],
+    "source": "19-20六年级上·河北·单元测试",
+    "merged_id": "unit6_054",
+    "original_id": "unit6_judge_054"
+  },
+  {
+    "id": "unit6_judge_055",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "(23-24 六年级上·全国·单元测试) 一根绳子剪成两段,甲段长 80 米, 乙段占全长的 $40 \\%$ , 甲段比乙段长。( )",
+    "options": [],
+    "answer": "√ 乙段占全长的 $40 \\%$ , 甲段占全长的 $60 \\%$ , $60 \\% > 40 \\%$ , 所以甲段比乙段长。",
+    "solution": "【分析】由题意可知，把绳子全长看做单位'1'，绳子剪成两段, 乙段占全长的 $40 \\%$ , 则甲段占全长的 $(1 - 40 \\%)$ , 再比较百分数的大小。据此解答。\n\n【详解】$1 - 40\\% = 60\\%$  $60 \\% >40 \\%$  甲段比乙段长。故答案为: √  理由：乙段占全长的 $40 \\%$ , 甲段占全长的 $60 \\%$ , $60 \\% > 40 \\%$ , 所以甲段比乙段长。",
+    "knowledge_points": [
+      "含百分数的运算",
+      "单位'1'的认识与确定"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_055",
+    "original_id": "unit6_judge_055"
+  },
+  {
+    "id": "unit6_judge_056",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "(23-24 六年级上·全国·单元测试) 甲数的 $25 \\%$ 和乙数的 $\\frac{1}{4}$ 相等。( ) (对的打 ' √ ', 错的打 ' × ', 并说明理由)",
+    "options": [],
+    "answer": "×;$25 \\% = 0.25 ; \\frac {1}{4} = 0.25$ ；甲数的 $25 \\%$ 就是甲数 $\\times 25\\%$ ；即甲数 $\\times 0.25$ ；乙数的 $\\frac{1}{4}$ 就是乙数 $\\times \\frac{1}{4}$ ；即乙数 $\\times 0.25$如果甲数 $=$ 乙数，则甲数的 $25 \\%$ 和乙数的 $\\frac{1}{4}$ 相等。如果甲数≠乙数，则甲数的 $25\\%$ 和乙数的 $\\frac{1}{4}$ 不相等。",
+    "solution": "【分析】把甲数看作单位 '1', 用乘法求出甲数的 $25 \\%$ 是多少; 把乙数看作单位 '1', 用乘法求出乙数的 $\\frac{1}{4}$ 是多少,由于甲数和乙数的具体数量不知道, 所以无法确定甲数的 $25 \\%$ 和乙数的 $\\frac{1}{4}$ 相等。据此解答。\n\n【详解】甲数的 $25 \\%$ 和乙数的 $\\frac{1}{4}$ 相等。（x）理由: $25\\% = 0.25; \\frac{1}{4} = 0.25;$甲数的 $25 \\%$ 就是甲数 $\\times 25 \\%$ ; 即甲数 $\\times 0.25$ ; 乙数的 $\\frac{1}{4}$ 就是乙数 $\\times \\frac{1}{4}$ ; 即乙数 $\\times 0.25$ ;如果甲数 = 乙数, 则甲数的 $25 \\%$ 和乙数的 $\\frac{1}{4}$ 相等。如果甲数 $\\neq$ 乙数, 则甲数的 $25 \\%$ 和乙数的 $\\frac{1}{4}$ 不相等。",
+    "knowledge_points": [
+      "求一个数的几分之几的问题",
+      "百分数、小数和分数的互化"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_056",
+    "original_id": "unit6_judge_056"
+  },
+  {
+    "id": "unit6_judge_057",
+    "unit": "第六单元：百分数",
+    "chapter": "圆的认识",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "(23-24 六年级上·全国·单元测试) 在一个圆中, 半径长度是它直径长度的 $50 \\%$ ( ) (对的打 ' √ ', 错的打 '× ', 并说明理由)理由：________________________。",
+    "options": [],
+    "answer": "√;同一个圆中每条直径的长度都是半径长度的 2 倍, $1 \\div 2 \\times 100 \\% = 50 \\%$ , 可知同一个圆中, 半径长度是它直径长度的 $50 \\%$",
+    "solution": "【分析】连接圆周上两点并通过圆心的线段叫做直径；半径指连接圆心和圆上任意一点的线段，同一个圆中每条直径的长度都是半径长度的2倍，假设半径为1，直径为2，根据求一个数是另一个数的百分之几，用一个数除以另一个数再乘 $100\\%$，据此求出同一个圆中，半径长度是它直径长度的百分之几。\n\n【详解】在一个圆中，半径长度是它直径长度的 $50\\%$ 。(√)  理由：假设半径为 1，直径为 2，$1 \\div 2 \\times 100 \\% = 50 \\%$同一个圆中每条直径的长度都是半径长度的 2 倍, 根据百分数的意义, 可知同一个圆中, 半径长度是它直径长度的 $50 \\%$ .",
+    "knowledge_points": [
+      "圆的概念及特点",
+      "求一个数是另一个数的百分之几 (百分率问题)"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_057",
+    "original_id": "unit6_judge_057"
+  },
+  {
+    "id": "unit6_judge_058",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的意义",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "(23-24 六年级上·全国·单元测试) 分母是 100 的分数都是百分数。（ ）（对的打'√'，错的打'×'，并说明理由）理由：___________________________________________。",
+    "options": [],
+    "answer": "×;分母是 100 的分数可能是百分数, 也可能是分数, 只有当表示一个数是另一个数的百分之多少时, 才是百分数。",
+    "solution": "【分析】表示一个数是另一个数的百分之多少的数叫做百分数，百分数也叫做百分率或者百分比，根据百分数的意义可知，百分数是分数，百分数的分母是100，但是分母是100的分数有可能表示百分数，也有可能表示分数。\n\n【详解】由分析可知, 分母是 100 的分数不一定都是百分数,也有可能是分数。所以原题说法错误。故答案为: ×理由：百分数表示一个数是另一个数的百分之多少，也叫百分率或者百分比。",
+    "knowledge_points": [
+      "百分数的意义"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_058",
+    "original_id": "unit6_judge_058"
+  },
+  {
+    "id": "unit6_judge_059",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "（24-25六年级上·全国·单元测试）今年产量比去年增加了 $30\\%$，就是今年的产量相当于去年产量的 $130\\%$。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】今年产量比去年增加了 $30 \\%$ , 是把去年产量看作单位 '1', 今年产量占去年产量的 $(1 + 30 \\%)$ , 据此解答。\n\n【详解】$1 + 30\\% = 130\\%$因此，今年产量比去年增加了 $30 \\%$ , 就是今年的产量相当于去年产量的 $130 \\%$。故答案为: √",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "24-25六年级上·全国·单元测试",
+    "merged_id": "unit6_059",
+    "original_id": "unit6_judge_059"
+  },
+  {
+    "id": "unit6_judge_060",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "(24-25 六年级上·全国·单元测试) 1 千克糖, 小红吃了它的 $50 \\%$ , 还剩下 0.5 千克。( )",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】据题意可知, 把 1 千克糖看作单位 '1', 吃了它的 $50 \\%$ , 则剩下的占 1 千克糖的 $(1 - 50 \\%)$ , 根据求一个数的百分之几是多少, 用乘法计算即可分析判断。\n\n【详解】$1 \\times (1 - 50\\%)=1  \\times   50\\%=0.5 (千克) $ 还剩下 0.5 千克。 故答案为: √",
+    "knowledge_points": [
+      "求一个数的百分之几是多少"
+    ],
+    "source": "24-25 六年级上·全国·单元测试",
+    "merged_id": "unit6_060",
+    "original_id": "unit6_judge_060"
+  },
+  {
+    "id": "unit6_judge_061",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "(24-25 六年级上·全国·单元测试) 若苹果树棵数的80%和梨树棵数的76%一样多，则苹果树比梨树少。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】由'苹果树棵数的 $80 \\%$ 和梨树棵数的 $76 \\%$ 一样多'可知，苹果树的棵数 $\\times 80 \\% =$ 梨树的棵数 $\\times 76 \\%$ , 即两个乘法算式的积相等, 根据 '积一定时, 一个因数乘的数越大, 这个数就越小', 比较两个百分数的大小, 即可判断哪种树少。\n\n【详解】苹果树的棵数 $\\times 80 \\% =$ 梨树的棵数 $\\times 76 \\%$ $80 \\% > 76 \\%$ , 则苹果树的棵数 $<$ 梨树的棵数; 若苹果树棵数的 $80 \\%$ 和梨树棵数的 $76 \\%$ 一样多, 则苹果树比梨树少。原题说法正确。故答案为: √",
+    "knowledge_points": [
+      "求一个数的百分之几是多少",
+      "积的变化规律(小数乘法)"
+    ],
+    "source": "24-25 六年级上·全国·单元测试",
+    "merged_id": "unit6_061",
+    "original_id": "unit6_judge_061"
+  },
+  {
+    "id": "unit6_judge_062",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "（24-25六年级上·全国·单元测试）栽103棵树成活100棵，成活率为 $100\\%$ 。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】树的成活率 = 成活棵数 ÷ 栽种总棵数 $\\times 100 \\%$ , 据此计算得出答案。\n\n【详解】成活率 $= 100 \\div 103 \\times 100 \\% \\approx 97.1 \\%$ , 题干中成活率为 $100 \\%$ , 表述错误。故答案为: ×",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "24-25六年级上·全国·单元测试",
+    "merged_id": "unit6_062",
+    "original_id": "unit6_judge_062"
+  },
+  {
+    "id": "unit6_judge_063",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "(24-25 六年级上·全国·单元测试) $5 \\mathrm{~kg}$ 的 $\\frac{1}{8}$ 与 $1 \\mathrm{~kg}$ 的62.5%同样重。( )",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】根据分数乘法的意义：求一个数的几分之几是多少，用乘法；根据百分数乘法的意义：求一个数的百分之几是多少，用乘法；分别求出 $5 \\mathrm{~kg}$ 的 $\\frac{1}{8}$ 是多少 $\\mathrm{kg}$ 、 $1 \\mathrm{~kg}$ 的62.5%是多少kg即可判断。\n\n【详解】$5 \\times \\frac{1}{8} = 0.625 (\\mathrm{~kg})$$1 \\times 62.5 \\% = 0.625 (\\mathrm {kg})$ $0.6 2 5 = 0. 6 2 5$ 所以 $5 \\mathrm{~kg}$ 的 $\\frac{1}{8}$ 与 $1 \\mathrm{~kg}$ 的 $62.5 \\%$ 同样重。 原题说法正确。 故答案为: √",
+    "knowledge_points": [
+      "求一个数的几分之几的问题",
+      "求一个数的百分之几是多少"
+    ],
+    "source": "24-25 六年级上·全国·单元测试",
+    "merged_id": "unit6_063",
+    "original_id": "unit6_judge_063"
+  },
+  {
+    "id": "unit6_calc_064",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的运算",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "(25-26 六年级上·海南海口·单元测试) 用简便方法计算。\n\n$① \\frac {2}{5} \\times \\frac {4}{9} + \\frac {5}{9} \\times 40 \\%$\n\n$② 36 \\times (75 \\% - \\frac {2}{9})$",
+    "options": [],
+    "answer": "①$\\frac{2}{5}$; ②19",
+    "solution": "【分析】①先把 $40\\%$ 化成 $\\frac{2}{5}$ , 再逆用乘法分配律 $a \\times c + b \\times c = (a + b) \\times c$ 把算式写成 $\\frac{2}{5} \\times (\\frac{4}{9} + \\frac{5}{9})$ , 再进一步计算即可; ②先把 $75\\%$ 化成 $\\frac{3}{4}$ , 再根据乘法分配律 $(a + b) \\times c = a \\times c+b\\times c$ 把算式写成 $36 \\times \\frac{3}{4} - 36 \\times \\frac{2}{9}$ , 再进一步计算即可。\n\n【详解】$ ① \\frac { 2 } { 5 } \\times \\frac { 4 } { 9 } + \\frac { 5 } { 9 } \\times 4 0 \\%\\\\=\\frac {2}{5} \\times \\frac {4}{9} + \\frac {5}{9} \\times \\frac {2}{5} \\\\ = \\frac {2}{5} \\times (\\frac {4}{9} + \\frac {5}{9}) \\\\ = \\frac {2}{5} \\times 1 \\\\ = \\frac {2}{5} $\n\n$ ② 36 \\times (75 \\% - \\frac {2}{9}) \\\\ = 3 6 \\times (\\frac {3}{4} - \\frac {2}{9}) \\\\ = 3 6 \\times \\frac {3}{4} - 3 6 \\times \\frac {2}{9} \\\\ = 2 7 - 8 \\\\ = 1 9 $",
+    "knowledge_points": [
+      "含百分数的运算",
+      "整数乘法运算定律推广到分数乘法"
+    ],
+    "source": "25-26 六年级上·海南海口·单元测试",
+    "merged_id": "unit6_064",
+    "original_id": "unit6_calc_064"
+  },
+  {
+    "id": "unit6_calc_065",
+    "unit": "第六单元：百分数",
+    "chapter": "解方程",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "(25-26 六年级上·海南海口·单元测试) 解方程。\n\n$\\frac {5}{6} x = \\frac {5}{9} \\quad x + \\frac {4}{5} = \\frac {9}{10} \\quad x + 60 \\% x = 16$",
+    "options": [],
+    "answer": "$x =\\frac{2}{3};x=\\frac{1}{10};x = 10$",
+    "solution": "【分析】 $\\frac{5}{6} x = \\frac{5}{9}$ , 根据等式的性质 2 , 两边同时 $\\div \\frac{5}{6}$ 即可; $x + \\frac{4}{5} = \\frac{9}{10}$ , 根据等式的性质 1 , 两边同时 $-\\frac{4}{5}$ 即可; $x + 60 \\% x = 16$ , 将左边合并成 $1.6 x$ , 根据等式的性质 2 , 两边同时 $\\div 1.6$ 即可。\n\n【详解】$\\frac{5}{6} x = \\frac{5}{9}$\n\n解: $\\frac{5}{6} x \\div \\frac{5}{6} = \\frac{5}{9} \\div \\frac{5}{6}\\\\x = \\frac {5}{9} \\times \\frac {6}{5}\\\\x = \\frac {2}{3}$\n\n$x + \\frac {4}{5} = \\frac {9}{1 0}$\n解: $x + \\frac{4}{5} - \\frac{4}{5} = \\frac{9}{10} - \\frac{4}{5}\\\\x = \\frac {9}{1 0} - \\frac {8}{1 0}\\\\x = \\frac {1}{1 0}$\n\n$x + 60 \\% x = 16$\n\n解: $1.6x = 16$\n\n$1.6 x \\div 1. 6 = 1 6 \\div 1. 6$\n\n$x = 1 0$",
+    "knowledge_points": [
+      "应用等式的性质 2 解方程",
+      "解百分数方程",
+      "解分数方程"
+    ],
+    "source": "25-26 六年级上·海南海口·单元测试",
+    "merged_id": "unit6_065",
+    "original_id": "unit6_calc_065"
+  },
+  {
+    "id": "unit6_calc_066",
+    "unit": "第六单元：百分数",
+    "chapter": "四则混合运算",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "(24-25 六年级下·海南海口·单元测试) 脱式计算。\n\n$3 6 1 8 \\div 4 5 + 1 6 2 0 \\quad \\frac {5}{6} + \\frac {1}{9} - \\frac {2}{3}  \\quad \\frac {9}{1 1} \\times \\frac {1}{4} \\times \\frac {2 2}{9} \\\\ 21\\times (1 + 25\\%) \\div \\frac{3}{8}\\quad \\frac{3}{8}\\times [6\\div (\\frac{2}{5} -\\frac{1}{4}))\\quad \\frac{5}{16}\\times [\\frac{1}{4}\\div (\\frac{7}{16} -\\frac{1}{4})) $",
+    "options": [],
+    "answer": "1700.4;$\\frac{5}{18}$; $\\frac{1}{2}$;70;15;$\\frac {5}{12}$",
+    "solution": "【分析】根据四则混合运算的顺序，先算除法，再算加法；异分母分数加减法，先通分转化为同分母分数再按照同分母分数计算方法计算；分数的连乘，可以将能约分的先约分，再计算；先将百分数转化为分数，再根据四则混合运算的顺序，先算括号里面的加法，再算括号外面的乘法，最后算除法，在计算分数除法时，除以一个分数相当于乘这个分数的倒数；先算中括号里面的减法，再算中括号里面的除法，最后算括号外面的乘法；先算中括号里面的减法，再算中括号里面的除法，将分数除法转化为分数乘法，可以去掉中括号后得出是分数的连乘，可以约分的要约分可以简便计算。\n\n【详解】$3618 \\div 45 + 1620\\\\=8 0. 4 + 1 6 2 0\\\\=1700.4$\n\n$\\frac {5}{6} + \\frac {1}{9} - \\frac {2}{3}\\\\= \\frac {1 5}{1 8} + \\frac {2}{1 8} - \\frac {1 2}{1 6}\\\\=\\frac {1 7}{1 8} - \\frac {1 2}{1 8}\\\\=\\frac {5}{1 8}$\n\n$\\frac {9}{1 1} \\times \\frac {1}{4} \\times \\frac {2 2}{9}\\\\= \\frac {9}{11} \\times \\frac {22}{9} \\times \\frac {1}{4} \\\\ =\\frac {1}{2}$\n\n$ 21 \\times (1 + 25 \\%) \\div \\frac {3}{8} \\\\ =2 1 \\times (1 + \\frac {1}{4}) \\div \\frac {3}{8} \\\\= 2 1 \\times \\frac {5}{4} \\div \\frac {3}{8} \\\\ = \\frac {1 0 5 }{4 } \\times \\frac {8 }{3} \\\\ =7 0$\n\n$ \\frac {3}{8} \\times [ 6 \\div (\\frac {2}{5} - \\frac {1}{4})) \\\\ = \\frac {3}{8} \\times [ 6 \\div \\frac {3}{2 0}] \\\\ = \\frac {3}{8} \\times [ 6 ^ {2} \\times \\frac {2 0}{3 _ {1}}) \\\\ = \\frac {3}{8} \\times 4 0 \\\\ = 1 5$ \n\n$\\frac {5}{1 6} \\times [ \\frac {1}{4} \\div (\\frac {7}{1 6} - \\frac {1}{4}) ] \\\\ = { \\frac {5}{1 6} \\times [ \\frac {1}{4} \\div \\frac {3}{1 6})} {\\frac {5}{1 6} \\times [ \\frac {1}{4} \\times \\frac {1 6}{3}) .} \\\\ = \\frac {5}{1 6} \\times \\frac {1}{4} \\times \\frac {1 6 }{3} \\\\= \\frac {5}{1 2} $",
+    "knowledge_points": [
+      "分数的四则混合运算",
+      "带有中括号的混合运算",
+      "含百分数的运算",
+      "异分母分数加、减法"
+    ],
+    "source": "24-25 六年级下·海南海口·单元测试",
+    "merged_id": "unit6_066",
+    "original_id": "unit6_calc_066"
+  },{
+    "id": "unit6_solve_067",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(25-26 六年级上·海南海口·单元测试) 甲、乙两个粮仓共存粮 420 吨, 现在从甲仓运走 $\\frac{2}{5}$ , 从乙仓运走 $35\\%$ , 两仓剩下的粮食相等。甲、乙两仓原来各有粮食多少吨?",
+    "options": [],
+    "answer": "甲仓 180 吨,乙仓 240 吨",
+    "solution": "【分析】设甲仓原来有粮食 $x$ 吨, 则乙仓有 $(420 - x)$ 吨, 根据两仓剩下的粮食相等, 列出方程: $x \\times (1 - \\frac{2}{5}) = (420 - x) \\times (1 - 35\\%)$ , 解方程求出 $x$ 即可。\n\n【详解】解: 设甲仓原来有粮食 $x$ 吨, 则乙仓有 $(420 - x)$ 吨。\n\n$x \\times (1 - \\frac{2}{5}) = (420 - x) \\times (1 - 35 \\%)$\n\n$x \\times \\frac{3}{5} = (420 - x) \\times 65 \\%$\n\n$0.6x = (420 - x) \\times 0.65$\n\n$0.6 x = 273 - 0.65 x$\n\n$0.6 x + 0.65 x = 273$\n\n$1.25 x = 273$\n\n$x = 218.4 \\approx 180$ (吨)\n\n$420 - 180 = 240$ (吨)\n\n答: 甲仓原来有 180 吨, 乙仓原来有 240 吨。",
+    "knowledge_points": [
+      "含百分数的运算",
+      "列方程解含有百分数的应用题"
+    ],
+    "source": "25-26 六年级上·海南海口·单元测试",
+    "merged_id": "unit6_067",
+    "original_id": "unit6_solve_067"
+  },
+  {
+    "id": "unit6_solve_068",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(25-26 六年级上·海南海口·单元测试) 一个果园里有苹果树 100 棵, 梨树 80 棵。\n\n(1) 梨树比苹果树少百分之几?\n\n(2) 苹果树比梨树多百分之几?",
+    "options": [],
+    "answer": "(1) $20\\%$;(2)$25\\%$",
+    "solution": "【分析】（1）把苹果树的棵数看作单位 '1', 用苹果树的棵数减去梨树的棵数, 求出梨树比苹果树少多少棵, 再除以苹果树的棵数, 即可求出梨树比苹果树少百分之几; （2）把梨树的棵数看作单位 '1', 用苹果树的棵数减去梨树的棵数, 求出苹果树比梨树多多少棵, 再除以梨树的棵数, 即可求出苹果树比梨树多百分之几。\n\n【详解】(1) $(100 - 80) \\div 100 \\times 100 \\%\\\\= 20 \\div 100 \\times 100 \\% \\\\ = 0.2 \\times 100 \\% \\\\ = 20 \\% $\n\n答: 梨树比苹果树少 $20 \\%$ 。\n\n(2) $(100 - 80) \\div 80 \\times 100 \\%\\\\= 20 \\div 80 \\times 100 \\% \\\\ = 0.25 \\times 100 \\% \\\\ = 25 \\% $\n\n答: 苹果树比梨树多 $25 \\%$ 。",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "25-26 六年级上·海南海口·单元测试",
+    "merged_id": "unit6_068",
+    "original_id": "unit6_solve_068"
+  },
+  {
+    "id": "unit6_solve_069",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(25-26 六年级上·海南海口·单元测试) 我校图书馆买来一批新书，其中文艺书占 $40\\%$ , 科技书占 $\\frac{1}{4}$ , 其余的是连环画，共有 70 本。这批新书一共有多少本?",
+    "options": [],
+    "answer": "200本",
+    "solution": "【分析】将这批新书的总本数看作单位'1'，连环画占总本数的 $(1 - 40 \\% - \\frac{1}{4})$ , 根据已知一个数的百分之几是多少, 求这个数, 用除法计算, 用连环画的本数除以对应的百分率, 即可求出这批新书的总本数。\n\n【详解】$70 \\div (1 - 40 \\% - \\frac{1}{4})\\\\= 70 \\div (1 - 0.4 - 0.25) \\\\ = 70 \\div 0.35 \\\\ = 200$ (本)\n\n答: 这批新书一共有 200 本。",
+    "knowledge_points": [
+      "已知一部分量占总量的百分之几及另一部分量, 求总量"
+    ],
+    "source": "25-26 六年级上·海南海口·单元测试",
+    "merged_id": "unit6_069",
+    "original_id": "unit6_solve_069"
+  },
+  {
+    "id": "unit6_solve_070",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(25-26 六年级上·海南海口·单元测试) 一套运动服降价 $10\\%$ 后是 180 元，这套运动服原价是多少元?",
+    "options": [],
+    "answer": "200元",
+    "solution": "【分析】把这套运动服的原价看作单位 '1', 降价 $10 \\%$ 后, 现价是原价的 $(1 - 10 \\%)$ , 根据已知一个数的百分之几是多少, 求这个数, 用除法计算, 用现价除以对应的百分率, 即可求出原价。\n\n【详解】$180 \\div (1 - 10 \\%)\\\\= 180 \\div 90 \\% \\\\ = 180 \\div 0.9 \\\\ = 200$ (元)\n\n答: 这套运动服原价是 200 元。",
+    "knowledge_points": [
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "25-26 六年级上·海南海口·单元测试",
+    "merged_id": "unit6_070",
+    "original_id": "unit6_solve_070"
+  },
+  {
+    "id": "unit6_solve_071",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(25-26 六年级上·海南海口·单元测试) 一个工厂九月份用电 6000 千瓦时, 十月份比九月份节约用电 $\\frac{1}{5}$ , 十月份用电多少千瓦时?",
+    "options": [],
+    "answer": "4800千瓦时",
+    "solution": "【分析】把九月份的用电量看作单位 '1', 十月份比九月份节约用电 $\\frac{1}{5}$ , 则十月份用电量是九月份的 $(1 - \\frac{1}{5})$ , 根据求一个数的几分之几是多少, 用乘法计算, 用九月份的用电量乘对应的分率, 即可求出十月份的用电量。\n\n【详解】$6000 \\times (1 - \\frac{1}{5})\\\\= 6000 \\times \\frac{4}{5} \\\\ = 4800$ (千瓦时)\n\n答: 十月份用电 4800 千瓦时。",
+    "knowledge_points": [
+      "比一个数多/少几分之几的数是多少"
+    ],
+    "source": "25-26 六年级上·海南海口·单元测试",
+    "merged_id": "unit6_071",
+    "original_id": "unit6_solve_071"
+  },
+  {
+    "id": "unit6_solve_072",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(24-25 六年级下·海南海口·单元测试) 小丽看一本 180 页的故事书, 第一天看了全书的 $25\\%$ , 第二天看了全书的 $\\frac{1}{3}$ , 两天一共看了多少页?",
+    "options": [],
+    "answer": "105页",
+    "solution": "【分析】把全书的页数看作单位 '1', 根据求一个数的百分之几是多少, 用乘法计算, 用全书的页数乘 $25 \\%$ , 求出第一天看的页数; 根据求一个数的几分之几是多少, 用乘法计算, 用全书的页数乘 $\\frac{1}{3}$ , 求出第二天看的页数, 再把两天看的页数相加, 即可求出两天一共看了多少页。\n\n【详解】$180 \\times 25 \\% + 180 \\times \\frac{1}{3}\\\\= 45 + 60 \\\\ = 105$ (页)\n\n答: 两天一共看了 105 页。",
+    "knowledge_points": [
+      "求一个数的百分之几是多少",
+      "求一个数的几分之几的问题"
+    ],
+    "source": "24-25 六年级下·海南海口·单元测试",
+    "merged_id": "unit6_072",
+    "original_id": "unit6_solve_072"
+  },
+  {
+    "id": "unit6_solve_073",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(24-25 六年级下·海南海口·单元测试) 一个服装厂计划做 2400 套服装, 已经做了 $40\\%$ , 已经做了多少套?",
+    "options": [],
+    "answer": "960套",
+    "solution": "【分析】把计划做的总套数看作单位 '1', 根据求一个数的百分之几是多少, 用乘法计算, 用总套数乘 $40 \\%$ , 即可求出已经做了多少套。\n\n【详解】$2400 \\times 40 \\%\\\\= 2400 \\times 0.4 \\\\ = 960$ (套)\n\n答: 已经做了 960 套。",
+    "knowledge_points": [
+      "求一个数的百分之几是多少"
+    ],
+    "source": "24-25 六年级下·海南海口·单元测试",
+    "merged_id": "unit6_073",
+    "original_id": "unit6_solve_073"
+  },
+  {
+    "id": "unit6_solve_074",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(24-25 六年级下·海南海口·单元测试) 一个长方形的长是 $8 \\mathrm{~cm}$ , 宽比长短 $25 \\%$ , 求这个长方形的周长。",
+    "options": [],
+    "answer": "$28\\mathrm{~cm}$",
+    "solution": "【分析】把长方形的长看作单位 '1', 宽比长短 $25 \\%$ , 则宽是长的 $(1 - 25 \\%)$ , 根据求一个数的百分之几是多少, 用乘法计算, 用长乘对应的百分率, 求出宽, 再根据长方形的周长公式: 周长 $= ($ 长 $+$ 宽 $) \\times 2$ , 代入数据计算即可。\n\n【详解】$[8 + 8 \\times (1 - 25 \\%)] \\times 2\\\\= [8 + 8 \\times 75 \\%] \\times 2 \\\\ = [8 + 6] \\times 2 \\\\ = 14 \\times 2 \\\\ = 28 (\\mathrm{~cm})$\n\n答: 这个长方形的周长是 $28 \\mathrm{~cm}$ 。",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少",
+      "长方形的周长"
+    ],
+    "source": "24-25 六年级下·海南海口·单元测试",
+    "merged_id": "unit6_074",
+    "original_id": "unit6_solve_074"
+  },
+  {
+    "id": "unit6_solve_075",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(24-25 六年级下·海南海口·单元测试) 工厂有男工人 240 人, 女工人 160 人, 男工人比女工人多百分之几?",
+    "options": [],
+    "answer": "$50\\%$",
+    "solution": "【分析】把女工人的人数看作单位 '1', 用男工人的人数减去女工人的人数, 求出男工人比女工人多多少人, 再除以女工人的人数, 即可求出男工人比女工人多百分之几。\n\n【详解】$(240 - 160) \\div 160 \\times 100 \\%\\\\= 80 \\div 160 \\times 100 \\% \\\\ = 0.5 \\times 100 \\% \\\\ = 50 \\% $\n\n答: 男工人比女工人多 $50 \\%$ 。",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "24-25 六年级下·海南海口·单元测试",
+    "merged_id": "unit6_075",
+    "original_id": "unit6_solve_075"
+  },
+  {
+    "id": "unit6_solve_076",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(24-25 六年级下·海南海口·单元测试) 学校十月份用水 200 吨, 比九月份节约 $\\frac{1}{5}$ , 九月份用水多少吨?",
+    "options": [],
+    "answer": "250吨",
+    "solution": "【分析】把九月份的用水量看作单位 '1', 十月份比九月份节约 $\\frac{1}{5}$ , 则十月份用水量是九月份的 $(1 - \\frac{1}{5})$ , 根据已知一个数的几分之几是多少, 求这个数, 用除法计算, 用十月份的用水量除以对应的分率, 即可求出九月份的用水量。\n\n【详解】$200 \\div (1 - \\frac{1}{5})\\\\= 200 \\div \\frac{4}{5} \\\\ = 200 \\times \\frac{5}{4} \\\\ = 250$ (吨)\n\n答: 九月份用水 250 吨。",
+    "knowledge_points": [
+      "已知比一个数多/少几分之几的数是多少，求这个数"
+    ],
+    "source": "24-25 六年级下·海南海口·单元测试",
+    "merged_id": "unit6_076",
+    "original_id": "unit6_solve_076"
+  },
+  {
+    "id": "unit6_solve_077",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(23-24 六年级上·全国·单元测试) 某玩具厂生产一批玩具，原计划每天生产 120 个，30 天完成。实际提前 6 天完成，实际每天生产多少个?",
+    "options": [],
+    "answer": "150个",
+    "solution": "【分析】根据工作量 = 工作效率 × 工作时间，用原计划每天生产的个数乘天数，求出这批玩具的总个数，再根据工作效率 = 工作量 ÷ 工作时间，用这批玩具的总个数除以实际生产的天数，即可求出实际每天生产多少个。\n\n【详解】$120 \\times 30 \\div (30 - 6)\\\\= 3600 \\div 24 \\\\ = 150$ (个)\n\n答: 实际每天生产 150 个。",
+    "knowledge_points": [
+      "工程问题"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_077",
+    "original_id": "unit6_solve_077"
+  },
+  {
+    "id": "unit6_solve_078",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(23-24 六年级上·全国·单元测试) 某商店出售一批苹果, 第一天售出总数的 $\\frac{2}{5}$ , 第二天售出总数的 $40\\%$ , 两天共售出 400 千克。这批苹果有多少千克?",
+    "options": [],
+    "answer": "500千克",
+    "solution": "【分析】把这批苹果的总重量看作单位 '1', 两天共售出总数的 $(\\frac{2}{5} + 40 \\%)$ , 根据已知一个数的几分之几（百分之几）是多少, 求这个数, 用除法计算, 用两天售出的重量除以对应的分率（百分率）, 即可求出这批苹果的总重量。\n\n【详解】$400 \\div (\\frac{2}{5} + 40 \\%)\\\\= 400 \\div (\\frac{2}{5} + \\frac{2}{5}) \\\\ = 400 \\div \\frac{4}{5} \\\\ = 400 \\times \\frac{5}{4} \\\\ = 500$ (千克)\n\n答: 这批苹果有 500 千克。",
+    "knowledge_points": [
+      "已知一部分量占总量的百分之几及另一部分量, 求总量"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_078",
+    "original_id": "unit6_solve_078"
+  },
+  {
+    "id": "unit6_solve_079",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(23-24 六年级上·全国·单元测试) 小明家养的鹅和鸡共 110 只, 其中鹅的只数是鸡的 $20 \\%$ , 鹅和鸡各有多少只? (用方程解)",
+    "options": [],
+    "answer": "鹅 $\\frac{220}{11}$只, 鸡$\\frac{990}{11}$ 只",
+    "solution": "【分析】设鸡有 $x$ 只, 则鹅有 $20 \\% x$ 只, 根据鹅和鸡共 110 只, 列出方程: $x + 20 \\% x = 110$ , 解方程求出鸡的只数, 再用 110 减去鸡的只数, 即可求出鹅的只数。\n\n【详解】解: 设鸡有 $x$ 只, 则鹅有 $20 \\% x$ 只。\n\n$x + 20 \\% x = 110$\n\n$1.2 x = 110$\n\n$x = 110 \\div 1.2$\n\n$x = \\frac{275}{3} \\approx 91.67$\n\n$110 - \\frac{275}{3} = \\frac{55}{3} \\approx 18.33$\n\n答: 鸡有约 92 只, 鹅有约 18 只。",
+    "knowledge_points": [
+      "列方程解含有百分数的应用题"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_079",
+    "original_id": "unit6_solve_079"
+  },
+  {
+    "id": "unit6_solve_080",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(23-24 六年级上·全国·单元测试) 学校美术组有 45 人, 比书法组的人数多 $12.5 \\%$ , 书法组有多少人?",
+    "options": [],
+    "answer": "40人",
+    "solution": "【分析】把书法组的人数看作单位 '1', 美术组比书法组的人数多 $12.5 \\%$ , 则美术组的人数是书法组的 $(1 + 12.5 \\%)$ , 根据已知一个数的百分之几是多少, 求这个数, 用除法计算, 用美术组的人数除以对应的百分率, 即可求出书法组的人数。\n\n【详解】$45 \\div (1 + 12.5 \\%)\\\\= 45 \\div 112.5 \\% \\\\ = 45 \\div 1.125 \\\\ = 40$ (人)\n\n答: 书法组有 40 人。",
+    "knowledge_points": [
+      "已知比一个数多/少百分之几的数是多少，求这个数"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_080",
+    "original_id": "unit6_solve_080"
+  },
+  {
+    "id": "unit6_solve_081",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "困难",
+    "question": "(23-24 六年级上·全国·单元测试) 一瓶盐水重 500 克, 其中含盐 $4 \\%$ 。如果要使盐水中含盐 $10 \\%$ , 需要加盐多少克?",
+    "options": [],
+    "answer": "$\\frac{100}{3}$克",
+    "solution": "【分析】根据求一个数的百分之几是多少，用乘法计算，用盐水的重量乘 $4 \\%$ ，求出盐的重量；设需要加盐 $x$ 克，则加盐后盐水的总重量为 $(500 + x)$ 克，盐的重量为 $(500 \\times 4 \\% + x)$ 克，根据含盐率 $= 10 \\%$ ，列出方程: $(500 \\times 4 \\% + x) \\div (500 + x) = 10 \\%$ ，解方程即可。\n\n【详解】解: 设需要加盐 $x$ 克。\n\n$(500 \\times 4 \\% + x) \\div (500 + x) = 10 \\%$\n\n$(20 + x) \\div (500 + x) = 0.1$\n\n$20 + x = 0.1 \\times (500 + x)$\n\n$20 + x = 50 + 0.1x$\n\n$x - 0.1 x = 50 - 20$\n\n$0.9 x = 30$\n\n$x = 30 \\div 0.9$\n\n$x = \\frac{100}{3} \\approx 33.33$\n\n答: 需要加盐 $\\frac{100}{3}$ 克或约 33.33 克。",
+    "knowledge_points": [
+      "浓度问题",
+      "列方程解含有百分数的应用题"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_081",
+    "original_id": "unit6_solve_081"
+  },
+  {
+    "id": "unit6_solve_082",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(24-25 六年级上·全国·单元测试) 一种商品, 先涨价 $10 \\%$ , 又降价 $10 \\%$ , 现在的价格与原价一样吗? 请说明理由。",
+    "options": [],
+    "answer": "不一样，现在的价格比原价低",
+    "solution": "【分析】假设原价为 100 元，把原价看作单位'1'，先涨价 $10 \\%$ ，则涨价后的价格为 $100 \\times (1 + 10 \\%)$ ；再把涨价后的价格看作单位'1'，降价 $10 \\%$ ，则现价为涨价后的价格 $\\times (1 - 10 \\%)$ ，比较现价与原价即可。\n\n【详解】假设原价为 100 元。\n\n$100 \\times (1 + 10 \\%) \\times (1 - 10 \\%)\\\\= 100 \\times 110 \\% \\times 90 \\% \\\\ = 100 \\times 1.1 \\times 0.9 \\\\ = 99$ (元)\n\n因为 $99 < 100$ , 所以现在的价格比原价低。\n\n答: 现在的价格与原价不一样, 现在的价格比原价低。",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "24-25 六年级上·全国·单元测试",
+    "merged_id": "unit6_082",
+    "original_id": "unit6_solve_082"
+  },
+  {
+    "id": "unit6_solve_083",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(24-25 六年级上·全国·单元测试) 学校图书室原有图书 2400 本, 今年图书本数增加了 $15 \\%$ 。今年图书室有图书多少本?",
+    "options": [],
+    "answer": "2760本",
+    "solution": "【分析】把原有图书的本数看作单位 '1', 今年图书本数增加了 $15 \\%$ , 则今年图书本数是原有的 $(1 + 15 \\%)$ , 根据求一个数的百分之几是多少, 用乘法计算, 用原有图书的本数乘对应的百分率, 即可求出今年图书室有图书多少本。\n\n【详解】$2400 \\times (1 + 15 \\%)\\\\= 2400 \\times 115 \\% \\\\ = 2400 \\times 1.15 \\\\ = 2760$ (本)\n\n答: 今年图书室有图书 2760 本。",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "24-25 六年级上·全国·单元测试",
+    "merged_id": "unit6_083",
+    "original_id": "unit6_solve_083"
+  },
+  {
+    "id": "unit6_solve_084",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(24-25 六年级上·全国·单元测试) 小明看一本书, 第一天看了全书的 $25 \\%$ , 第二天看了 24 页, 还剩 56 页没有看。这本书一共有多少页?",
+    "options": [],
+    "answer": "160页",
+    "solution": "【分析】把这本书的总页数看作单位 '1', 第二天看的页数加上剩下的页数, 就是第一天看后剩下的页数, 占总页数的 $(1 - 25 \\%)$ , 根据已知一个数的百分之几是多少, 求这个数, 用除法计算, 用第一天看后剩下的页数除以对应的百分率, 即可求出这本书的总页数。\n\n【详解】$(24 + 56) \\div (1 - 25 \\%)\\\\= 80 \\div 75 \\% \\\\ = 80 \\div 0.75 \\\\ = 160$ (页)\n\n答: 这本书一共有 160 页。",
+    "knowledge_points": [
+      "已知一部分量占总量的百分之几及另一部分量, 求总量"
+    ],
+    "source": "24-25 六年级上·全国·单元测试",
+    "merged_id": "unit6_084",
+    "original_id": "unit6_solve_084"
+  },
+  {
+    "id": "unit6_solve_085",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(24-25 六年级上·全国·单元测试) 把 5000 元钱存入银行, 定期两年, 年利率是 $3.75 \\%$ , 到期时可以取回多少元? (免收利息税)",
+    "options": [],
+    "answer": "5375元",
+    "solution": "【分析】根据利息 = 本金 × 利率 × 时间, 用本金乘利率再乘时间, 求出利息, 再用本金加上利息, 即可求出到期时可以取回多少元。\n\n【详解】$5000 \\times 3.75 \\% \\times 2 + 5000\\\\= 5000 \\times 0.0375 \\times 2 + 5000 \\\\ = 375 + 5000 \\\\ = 5375$ (元)\n\n答: 到期时可以取回 5375 元。",
+    "knowledge_points": [
+      "利息、利率问题"
+    ],
+    "source": "24-25 六年级上·全国·单元测试",
+    "merged_id": "unit6_085",
+    "original_id": "unit6_solve_085"
+  },
+  {
+    "id": "unit6_solve_086",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(24-25 六年级上·全国·单元测试) 一张课桌的价钱是 80 元，一把椅子的价钱比一张课桌便宜 $15 \\%$ 。一把椅子的价钱是多少元?",
+    "options": [],
+    "answer": "68元",
+    "solution": "【分析】把一张课桌的价钱看作单位 '1', 一把椅子的价钱比一张课桌便宜 $15 \\%$ , 则一把椅子的价钱是一张课桌的 $(1 - 15 \\%)$ , 根据求一个数的百分之几是多少, 用乘法计算, 用一张课桌的价钱乘对应的百分率, 即可求出一把椅子的价钱。\n\n【详解】$80 \\times (1 - 15 \\%)\\\\= 80 \\times 85 \\% \\\\ = 80 \\times 0.85 \\\\ = 68$ (元)\n\n答: 一把椅子的价钱是 68 元。",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "24-25 六年级上·全国·单元测试",
+    "merged_id": "unit6_086",
+    "original_id": "unit6_solve_086"
+  },
+  {
+    "id": "unit6_solve_087",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(24-25 六年级上·全国·单元测试) 一辆汽车从甲地开往乙地, 已经行了 120 千米, 还剩全程的 $40 \\%$ 没有行。甲、乙两地相距多少千米?",
+    "options": [],
+    "answer": "200千米",
+    "solution": "【分析】把甲、乙两地的全程看作单位 '1', 已经行了 120 千米, 还剩全程的 $40 \\%$ 没有行, 则已经行的占全程的 $(1 - 40 \\%)$ , 根据已知一个数的百分之几是多少, 求这个数, 用除法计算, 用已经行的千米数除以对应的百分率, 即可求出甲、乙两地相距多少千米。\n\n【详解】$120 \\div (1 - 40 \\%)\\\\= 120 \\div 60 \\% \\\\ = 120 \\div 0.6 \\\\ = 200$ (千米)\n\n答: 甲、乙两地相距 200 千米。",
+    "knowledge_points": [
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "24-25 六年级上·全国·单元测试",
+    "merged_id": "unit6_087",
+    "original_id": "unit6_solve_087"
+  },
+  {
+    "id": "unit6_solve_088",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(24-25 六年级上·全国·单元测试) 一件上衣打八折后是 240 元, 这件上衣原价是多少元?",
+    "options": [],
+    "answer": "300元",
+    "solution": "【分析】打八折就是现价是原价的 $80 \\%$ , 把这件上衣的原价看作单位 '1', 根据已知一个数的百分之几是多少, 求这个数, 用除法计算, 用现价除以 $80 \\%$ , 即可求出原价。\n\n【详解】$240 \\div 80 \\%\\\\= 240 \\div 0.8 \\\\ = 300$ (元)\n\n答: 这件上衣原价是 300 元。",
+    "knowledge_points": [
+      "折扣问题",
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "24-25 六年级上·全国·单元测试",
+    "merged_id": "unit6_088",
+    "original_id": "unit6_solve_088"
+  },
+  {
+    "id": "unit6_solve_089",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(24-25 六年级上·全国·单元测试) 一条公路长 1200 米, 修路队第一周修了全长的 $30 \\%$ , 第二周修了全长的 $\\frac{2}{5}$ 。两周一共修了多少米?",
+    "options": [],
+    "answer": "840米",
+    "solution": "【分析】把公路的全长看作单位 '1', 根据求一个数的百分之几是多少, 用乘法计算, 用全长乘 $30 \\%$ , 求出第一周修的米数; 根据求一个数的几分之几是多少, 用乘法计算, 用全长乘 $\\frac{2}{5}$ , 求出第二周修的米数, 再把两周修的米数相加, 即可求出两周一共修了多少米。\n\n【详解】$1200 \\times 30 \\% + 1200 \\times \\frac{2}{5}\\\\= 360 + 480 \\\\ = 840$ (米)\n\n答: 两周一共修了 840 米。",
+    "knowledge_points": [
+      "求一个数的百分之几是多少",
+      "求一个数的几分之几的问题"
+    ],
+    "source": "24-25 六年级上·全国·单元测试",
+    "merged_id": "unit6_089",
+    "original_id": "unit6_solve_089"
+  },
+  {
+    "id": "unit6_solve_090",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(23-24 六年级上·全国·单元测试) 红星小学有学生 1200 人, 其中男生占 $48 \\%$ 。男生有多少人?",
+    "options": [],
+    "answer": "576人",
+    "solution": "【分析】把学生总人数看作单位 '1', 根据求一个数的百分之几是多少, 用乘法计算, 用学生总人数乘 $48 \\%$ , 即可求出男生有多少人。\n\n【详解】$1200 \\times 48 \\%\\\\= 1200 \\times 0.48 \\\\ = 576$ (人)\n\n答: 男生有 576 人。",
+    "knowledge_points": [
+      "求一个数的百分之几是多少"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_090",
+    "original_id": "unit6_solve_090"
+  },
+  {
+    "id": "unit6_solve_091",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "困难",
+    "question": "(23-24 六年级上·全国·单元测试) 一桶油, 连桶重 125 千克。用去油的 $60 \\%$ 后, 连桶重 65 千克。这个桶重多少千克?",
+    "options": [],
+    "answer": "25千克",
+    "solution": "【分析】把油的重量看作单位 '1', 用去油的 $60 \\%$ 后, 剩下的油占油的重量的 $(1 - 60 \\%)$ , 用去前连桶的重量减去用去后连桶的重量, 求出用去油的重量, 再除以 $60 \\%$ , 求出油的总重量, 再用用去前连桶的重量减去油的总重量, 即可求出桶的重量。\n\n【详解】$(125 - 65) \\div 60 \\%\\\\= 60 \\div 0.6 \\\\ = 100$ (千克)\n\n$125 - 100 = 25$ (千克)\n\n答: 这个桶重 25 千克。",
+    "knowledge_points": [
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_091",
+    "original_id": "unit6_solve_091"
+  },
+  {
+    "id": "unit6_solve_092",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(23-24 六年级上·全国·单元测试) 水泥厂 8 个月完成全年生产计划的 $75 \\%$ , 照这样, 如果再生产 1800 吨水泥就可以超产 300 吨, 水泥厂计划生产多少吨水泥?",
+    "options": [],
+    "answer": "6000吨",
+    "solution": "【分析】将 1800 吨减去超产的 300 吨, 求出再生产多少吨恰好完成计划。将全年生产计划看作单位 '1', 用单位 '1' 减去 $75 \\%$ , 求出余下没有生产的是全年计划的百分之几。单位 '1' 未知, 将余下没有生产的除以对应的百分率, 即可求出全年生产计划。\n\n【详解】$(1800 - 300) \\div (1 - 75 \\%)\\\\= 1500 \\div 25 \\% \\\\ = 1500 \\div 0.25 \\\\ = 6000$ (吨)\n\n答: 水泥厂计划生产 6000 吨水泥。",
+    "knowledge_points": [
+      "含百分数的运算",
+      "已知一部分量占总量的百分之几及另一部分量，求总量"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_092",
+    "original_id": "unit6_solve_092"
+  },
+  {
+    "id": "unit6_solve_093",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(23-24 六年级上·全国·单元测试) 某市公交车的正常票价为每人 1 元。如果刷卡付费, 那么每人 0.8 元。刷卡付费比正常票价便宜百分之几?",
+    "options": [],
+    "answer": "$20\\%$",
+    "solution": "【分析】根据求一个数比另一个数多（少）百分之几，用相差数除以另一个数再乘 $100 \\%$ , 则用正常票价减去刷卡付费的价格的差，除以正常票价再乘 $100 \\%$ ，即可求出刷卡付费比正常票价便宜百分之几。\n\n【详解】$(1 - 0.8) \\div 1 \\times 100 \\%\\\\= 0.2 \\div 1 \\times 100 \\% \\\\ = 0.2 \\times 100 \\% \\\\ = 20 \\% $\n\n答: 刷卡付费比正常票价便宜 $20 \\%$ 。",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_093",
+    "original_id": "unit6_solve_093"
+  },
+  {
+    "id": "unit6_solve_094",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(23-24 六年级上·安徽亳州·期中) 一条水渠, 第一周修了全长的 $\\frac{2}{5}$ , 第二周修了全长的 $40 \\%$ , 还剩 200 米, 水渠全长多少米?",
+    "options": [],
+    "answer": "1000米",
+    "solution": "【分析】把全长看作单位 '1', 剩下的长度是全长的 $(1 - \\frac{2}{5} - 40 \\%)$ , 根据分数除法的意义, 用剩下的长度除以 $(1 - \\frac{2}{5} - 40 \\%)$ 即可求出全长。\n\n【详解】$200 \\div (1 - \\frac{2}{5} - 40 \\%)\\\\= 200 \\div (1 - \\frac{2}{5} - \\frac{2}{5}) \\\\ = 200 \\div \\frac{1}{5} \\\\ = 200 \\times 5 \\\\ = 1000$ (米)\n\n答: 水渠全长 1000 米。",
+    "knowledge_points": [
+      "已知一部分量占总量的百分之几及另一部分量, 求总量"
+    ],
+    "source": "23-24 六年级上·安徽亳州·期中",
+    "merged_id": "unit6_094",
+    "original_id": "unit6_solve_094"
+  },
+  {
+    "id": "unit6_solve_095",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(24-25 六年级上·全国·单元测试) 某美食城进行促销, 饭菜一律降价 $5 \\%$ , 在此基础上, 美食城又返还消费额 $4 \\%$ 的现金。此时到该美食城就餐, 相当于降价百分之几?",
+    "options": [],
+    "answer": "$8.8\\%$",
+    "solution": "【分析】根据题意得：将饭菜价格看作'1'，降价 $5 \\%$ 得到 $95 \\%$ ；此时美食城又返还消费额得的 $4 \\%$ ，即在 $95 \\%$ 的基础上减少 $4 \\%$ , 可计算得出现在的价格是原价的百分之几, 再用 1 减去这个百分数, 得出答案。\n\n【详解】将饭菜的原价看作单位'1'，则降价：\n\n$1 \\times (1 - 5 \\%) \\times (1 - 4 \\%)\\\\= 1 \\times 95 \\% \\times 96 \\% \\\\ = 1 \\times 0.95 \\times 0.96 \\\\ = 0.912 \\\\ = 91.2 \\% $\n\n$1 - 91.2 \\% = 8.8 \\%$\n\n答: 相当于降价 $8.8 \\%$ 。",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少",
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "24-25 六年级上·全国·单元测试",
+    "merged_id": "unit6_095",
+    "original_id": "unit6_solve_095"
+  },
+  {
+    "id": "unit6_solve_096",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(24-25 六年级上·全国·单元测试) 某厂 5 月份用水 280 吨, 6 月份比 5 月份节约用水 $15 \\%$ , 5、6 月份共用水多少吨?",
+    "options": [],
+    "answer": "518 吨",
+    "solution": "【分析】由题意可知, 把 5 月份用水量看作单位 '1', 已知 6 月份比 5 月份节约用水 $15 \\%$ , 则 6 月份用水量是 5 月份的 $(1 - 15 \\%)$ , 根据求一个数的百分之几是多少, 用乘法计算, 用 5 月份的用水量乘 6 月份对应的百分率, 得到 6 月份的用水量, 再把 5、6 月份的用水量加起来, 即可得解。\n\n【详解】$280 \\times (1 - 15 \\%) + 280\\\\= 280 \\times 85 \\% + 280 \\\\ = 238 + 280 \\\\ = 518$ (吨)\n\n答: 5、6 月份共用水 518 吨。",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "24-25 六年级上·全国·单元测试",
+    "merged_id": "unit6_096",
+    "original_id": "unit6_solve_096"
+  },
+  {
+    "id": "unit6_solve_097",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(24-25 六年级上·全国·单元测试) 一辆汽车从甲地开往乙地, 每小时行 45 千米, $\\frac{6}{5}$ 小时到达。如果把速度提高 $20 \\%$ , 那么几小时可以到达?",
+    "options": [],
+    "answer": "1小时",
+    "solution": "【分析】根据'速度 × 时间 = 路程'，用汽车每小时行的 45 千米乘 $\\frac{6}{5}$ 小时，求出甲地到乙地的路程，再根据'路程 ÷ 速度 = 时间'，用甲地到乙地的路程除以汽车提速 $20 \\%$ 后的速度：$[45 \\times (1 + 20 \\%)]$ 千米/小时即可解答。\n\n【详解】$45 \\times \\frac{6}{5} \\div [45 \\times (1 + 20 \\%)]\\\\= 54 \\div [45 \\times 1.2] \\\\ = 54 \\div 54 \\\\ = 1$ (小时)\n\n答: 1 小时可以到达。",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少",
+      "含百分数的运算",
+      "基础行程问题"
+    ],
+    "source": "24-25 六年级上·全国·单元测试",
+    "merged_id": "unit6_097",
+    "original_id": "unit6_solve_097"
+  },
+  {
+    "id": "unit6_solve_098",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(24-25 六年级上·全国·单元测试) 一台冰箱, 原来售价是 2500 元, 促销活动中比原来便宜 300 元, 降价后的价格是原价的百分之几?",
+    "options": [],
+    "answer": "$88\\%$",
+    "solution": "【分析】根据题意，先用原来的售价减去 300，求出降价后的价格；再用降价后的价格除以原来的售价，即是降价后的价格是原价的百分之几。\n\n【详解】$(2500 - 300) \\div 2500 \\times 100 \\%\\\\= 2200 \\div 2500 \\times 100 \\% \\\\ = 0.88 \\times 100 \\% \\\\ = 88 \\% $\n\n答: 降价后的价格是原价的 $88 \\%$ 。",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "24-25 六年级上·全国·单元测试",
+    "merged_id": "unit6_098",
+    "original_id": "unit6_solve_098"
+  },
+  {
+    "id": "unit6_solve_099",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "困难",
+    "question": "(24-25 六年级上·全国·单元测试) 用 200 粒种子做发芽试验, 结果有 4 粒没有发芽, 求发芽率。",
+    "options": [],
+    "answer": "$98\\%$",
+    "solution": "【分析】根据题意，先用种子的总数减去没有发芽的种子数，求出发芽的种子数；再根据'发芽率 = 发芽的种子数 ÷ 种子总数 × $100 \\%$'，代入数据计算，求出发芽率。\n\n【详解】$(200 - 4) \\div 200 \\times 100 \\%\\\\= 196 \\div 200 \\times 100 \\% \\\\ = 0.98 \\times 100 \\% \\\\ = 98 \\% $\n\n答: 发芽率是 $98 \\%$ 。",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "24-25 六年级上·全国·单元测试",
+    "merged_id": "unit6_099",
+    "original_id": "unit6_solve_099"
+  },
+  {
+    "id": "unit6_solve_100",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(24-25 六年级上·全国·课后作业) 学校购买吊扇 60 台, 购买落地式电扇比吊扇少 $50 \\%$ , 吊扇和落地扇共购买了多少台?",
+    "options": [],
+    "answer": "90台",
+    "solution": "【分析】已知落地式电扇比吊扇少 $50 \\%$ , 把吊扇的台数看作单位 '1', 则落地式电扇的台数是吊扇的 $(1 - 50 \\%)$ , 单位 '1' 已知, 用吊扇的台数乘 $(1 - 50 \\%)$ , 求出落地式电扇的台数, 再加上吊扇的台数, 即是吊扇和落地扇的总台数。\n\n【详解】$60 \\times (1 - 50 \\%)\\\\= 60 \\times (1 - 0.5) \\\\ = 60 \\times 0.5 \\\\ = 30$ (台)\n\n$60 + 30 = 90$ (台)\n\n答: 吊扇和落地扇共购买了 90 台。",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "24-25 六年级上·全国·课后作业",
+    "merged_id": "unit6_100",
+    "original_id": "unit6_solve_100"
+  },  {
+    "id": "unit6_choice_101",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "如果a: $\\frac{3}{5} = b \\times 62.5\\% = c \\div 62.5\\%$ (a、b、c均不为0)，那么（）。",
+    "options": [
+      "A. a 最大",
+      "B. b 最大",
+      "C. c 最大",
+      "D. 无法确定"
+    ],
+    "answer": "B",
+    "solution": "【分析】假设 $a: \\frac{3}{5} = b \\times 62.5\\% = c \\div 62.5\\% = 1$ ，分别求出 a、b、c 的值，再比较大小即可。\n\n【详解】假设 $a: \\frac{3}{5} = b \\times 62.5\\% = c \\div 62.5\\% = 1$ 。则 $a: \\frac{3}{5} = 1, a = \\frac{3}{5} \\times 1 = \\frac{3}{5} = 0.6$$b = 1 \\div 62.5 \\% = 1.6$$c = 1 \\times 62.5 \\% = 0.625$因为 $0.6 < 0.625 < 1.6$所以 $a < c < b$所以最大的是 b 。故答案为: B",
+    "knowledge_points": [
+      "比的意义",
+      "含百分数的运算"
+    ],
+    "source": "24-25 六年级上·全国·单元测试",
+    "merged_id": "unit6_101",
+    "original_id": "unit6_choice_101"
+  },
+  {
+    "id": "unit6_choice_102",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "某食品厂原计划年产值达到 40 万元，实际产值达到 50 万元，超产（）。",
+    "options": [
+      "A. $25\\%$",
+      "B. $20\\%$",
+      "C. $80\\%$",
+      "D. $40\\%$"
+    ],
+    "answer": "A",
+    "solution": "【分析】超产百分之几，意思是实际产值比原计划年产值多百分之几。根据求一个数比另一个数多百分之几，用除法计算，用实际与原计划产值的差除以原计划年产值再乘百分之百，即可得解。\n\n【详解】$(50 - 40) \\div 40 \\times 100\\%$\n\n$ 10 \\div 40 \\times 100 \\% \\\\= 0.25 \\times 100 \\%\\\\=25\\%$\n\n因此，实际产值超产 $25\\%$故答案为：A",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "24-25 六年级上·全国·单元测试",
+    "merged_id": "unit6_102",
+    "original_id": "unit6_choice_102"
+  },
+  {
+    "id": "unit6_choice_103",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "某单位共有职工2000人，今天的出勤率是 $95\\%$ ，今天缺勤（）人。",
+    "options": [
+      "A. 100",
+      "B. 200",
+      "C. 10",
+      "D. 20"
+    ],
+    "answer": "A",
+    "solution": "【分析】出勤率表示的是出勤人数占职工总人数的百分率,由此可知, 把职工总人数看作单位 '1', 缺勤人数占职工总人数的 $(1 - 95 \\%)$ , 根据求一个数的百分之几是多少, 用乘法计算, 用职工总人数乘缺勤人数对应的百分率, 即可得解。\n\n【详解】$2000 \\times (1 - 95\\%)= 2000 \\times 5 \\%=100 (人)$今天缺勤 100 人。故答案为：A",
+    "knowledge_points": [
+      "求一个数的百分之几是多少",
+      "含百分数的运算"
+    ],
+    "source": "24-25六年级上·全国·单元测试",
+    "merged_id": "unit6_103",
+    "original_id": "unit6_choice_103"
+  },
+  {
+    "id": "unit6_choice_104",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "下面情况中, 百分数可能超过 $100\\%$ 的是 ( )。",
+    "options": [
+      "A. 商场今年的营业额占去年的百分比",
+      "B. 酒瓶上标注的酒精度",
+      "C. 人类可利用的淡水资源占水资源总量的百分比"
+    ],
+    "answer": "A",
+    "solution": "【分析】表示一个数是另一个数的百分之几的数叫百分数，如果部分量超过单位'1'的量，百分数可能超过 $100\\%$ ，据此分析。\n\n【详解】A．去年营业额是单位'1'，今年的营业额有可能超过去年营业额，商场今年的营业额占去年的百分比可能超过 $100\\%$ ;B. 酒的质量是单位 '1', 酒精的质量不可能超过酒的质量,酒瓶上标注的酒精度不可能超过 $100 \\%$ ;C. 水资源总量是单位 '1', 淡水资源不可能超过水资源总量, 人类可利用的淡水资源占水资源总量的百分比不可能超过 $100 \\%$ .百分数可能超过 $100\\%$ 的是商场今年的营业额占去年的百分比。故答案为: A",
+    "knowledge_points": [
+      "百分数的意义"
+    ],
+    "source": "24-25 六年级上·辽宁·课后作业",
+    "merged_id": "unit6_104",
+    "original_id": "unit6_choice_104"
+  },
+  {
+    "id": "unit6_choice_105",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "下面的百分率可能大于 $100\\%$ 的是（）。",
+    "options": [
+      "A. 成活率",
+      "B. 增长率",
+      "C. 合格率"
+    ],
+    "answer": "B",
+    "solution": "【分析】一般来讲，出勤率、成活率、发芽率、及格率、合格率、正确率、达标率能达到 $100\\%$ ，增长率能超过 $100\\%$ ；出米率、出粉率、出油率达不到 $100\\%$ 。据此解答。\n\n【详解】A．成活率能达到 $100\\%$ 。 B. 增长率可能大于 $100 \\%$ C. 合格率达到 $100 \\%$ .百分率可能大于 $100\\%$ 的是增长率。故答案为: B",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）",
+      "百分数的意义"
+    ],
+    "source": "24-25六年级上·江苏·单元测试",
+    "merged_id": "unit6_105",
+    "original_id": "unit6_choice_105"
+  },
+  {
+    "id": "unit6_choice_106",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "金星小学有 100 人参加数学竞赛, 其中有 30 人获奖, 获奖人数占参赛总人数的 ( )。",
+    "options": [
+      "A. $20\\%$",
+      "B. $25\\%$",
+      "C. $30\\%$"
+    ],
+    "answer": "C",
+    "solution": "【分析】已知 100 人参加数学竞赛, 其中有 30 人获奖, 用获奖人数除以参赛总人数, 即是获奖人数占参赛总人数的百分之几。\n\n【详解】$30 \\div 100 \\times 100\\%$ $ 0.3 \\times 100 \\% = 30 \\% $ 获奖人数占参赛总人数的 $30\\%$ 故答案为: C",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "24-25 六年级上·全国·课后作业",
+    "merged_id": "unit6_106",
+    "original_id": "unit6_choice_106"
+  },
+  {
+    "id": "unit6_choice_107",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "有含糖 $25 \\%$ 的糖水 $60 \\mathrm{~kg}$ , 蒸发 ( ) kg 水后浓度变为 $40 \\%$ 。",
+    "options": [
+      "A. 37.5",
+      "B. 40",
+      "C. 25",
+      "D. 22.5"
+    ],
+    "answer": "D",
+    "solution": "【分析】根据糖的质量 $=$ 糖水的质量 $\\times$ 含糖率, 用 $60 \\times 25 \\%$ 求出糖水中糖的质量, 浓度变为 $40 \\%$ 后, 糖的质量不变,已知一个数的百分之几是多少, 求这个数用除法解答, 据此用糖的质量除以 $40 \\%$ ，求出浓度变为 $40 \\%$ 后糖水的质量,再用原来糖水的质量减去浓度变为 $40 \\%$ 后糖水的质量即可解答。\n\n【详解】$60 - 60 \\times 25\\% \\div 40\\%\\\\= 6 0 - 1 5 \\div 0. 4 \\\\ = 6 0 - 3 7. 5 \\\\ = 2 2. 5 \\mathrm {(k g)} $\n\n故答案为: D",
+    "knowledge_points": [
+      "已知一个数的百分之几是多少，求这个数",
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_107",
+    "original_id": "unit6_choice_107"
+  },
+  {
+    "id": "unit6_choice_108",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "甲数比乙数多 $30\\%$ ，甲数与乙数的比是（）。",
+    "options": [
+      "A. 13:10",
+      "B. $7: 3$",
+      "C. $10: 7$",
+      "D. $7: 10$"
+    ],
+    "answer": "A",
+    "solution": "【分析】根据'甲数比乙数多 $30\\%$ '，把乙数看作单位'1', 则甲数是 $(1 + 30 \\%)$ , 然后去求甲数与乙数的比, 据此解答。\n\n【详解】$(1 + 30\\%):1$ 1.3:1 13:10 故答案为: A",
+    "knowledge_points": [
+      "比的化简",
+      "比的应用",
+      "含百分数的运算"
+    ],
+    "source": "23-24六年级上·全国·单元测试",
+    "merged_id": "unit6_108",
+    "original_id": "unit6_choice_108"
+  },
+  {
+    "id": "unit6_choice_109",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一条裙子原价360元，降价 $15\\%$ 后，又涨价 $5\\%$ ，现在售价是（）元。",
+    "options": [
+      "A. 321.3",
+      "B. 290.7",
+      "C. 306",
+      "D. 288"
+    ],
+    "answer": "A",
+    "solution": "【分析】把这条裙子的原价看作单位'1'，先降价 $15 \\%$ ,则降价后的价格是原价的 $(1 - 15 \\%)$ , 单位'1'已知, 用原价乘 $(1 - 15 \\%)$ , 求出降价后的价格;又涨价 $5 \\%$ , 是把降价后的价格看作单位 '1', 涨价后的价格是降价后价格的 $(1 + 5 \\%)$ , 单位 '1'已知, 用降价后的价格乘 $(1 + 5 \\%)$ , 求出现价。\n\n【详解】$360 \\times (1 - 15\\%) \\times (1 + 5\\%)\\\\= 3 6 0 \\times (1 - 0. 1 5) \\times (1 + 0. 0 5) \\\\ = 3 6 0 \\times 0. 8 5 \\times 1. 0 5 \\\\ = 3 0 6 \\times 1. 0 5 \\\\ = 3 2 1. 3 \\text {(元)} $\n\n现在售价是 321.3 元。\n\n故答案为: A",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "23-24六年级上·全国·单元测试",
+    "merged_id": "unit6_109",
+    "original_id": "unit6_choice_109"
+  },
+  {
+    "id": "unit6_choice_110",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "( ) 比 $32 \\mathrm{~kg}$ 多$25.5\\%$",
+    "options": [
+      "A. $4.2 \\mathrm{~kg}$",
+      "B. $34.55 \\mathrm{~kg}$",
+      "C. $40.16 \\mathrm{~kg}$",
+      "D. $57.5 \\mathrm{~kg}$"
+    ],
+    "answer": "C",
+    "solution": "【分析】求比 $32 \\mathrm{~kg}$ 多 $25.5 \\%$ 的数是多少，就是求 $32 \\mathrm{~kg}$ 的(1+25.5%) 是多少，根据求比一个数多百分之几是多少，用乘法解答即可。\n\n【详解】$32 \\times (1 + 25.5\\%)\\\\=3 2 \\times 1. 2 5 5 \\\\ = 4 0. 1 6 \\mathrm {(k g)} $\n\n故答案为: C",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_110",
+    "original_id": "unit6_choice_110"
+  },
+  {
+    "id": "unit6_choice_111",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "下面与 $75 \\%$ 相等的数是 ( )。",
+    "options": [
+      "A. 7.5",
+      "B. 0.75",
+      "C. $\\frac{3}{5}$",
+      "D. $\\frac{7}{10}$"
+    ],
+    "answer": "B",
+    "solution": "【分析】百分数化成小数的方法：去掉百分号，把小数点向左移动两位；把 $75 \\%$ 写成分数 $\\frac{75}{100}$ ，再根据分数的基本性质，分数的分子和分母同时乘或除以相同的数（0 除外），分数的大小不变，把分数 $\\frac{75}{100}$ 化成最简分数；据此选择。\n\n【详解】$75\\% = 0.75\\\\=75 \\% = \\frac {75}{100} \\\\= \\frac {75 \\div 25}{100 \\div 25} \\\\= \\frac {3}{4}$\n\n故答案为: B",
+    "knowledge_points": [
+      "百分数、小数和分数的互化"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_111",
+    "original_id": "unit6_choice_111"
+  },
+  {
+    "id": "unit6_choice_112",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "下面情形中的百分数可能超过 $100\\%$ 的是（）。",
+    "options": [
+      "A. 六年级学生社会实践活动的参与率",
+      "B. 小黄 5 分钟跳绳成绩的提高率",
+      "C. 五年级学生眼睛的近视率",
+      "D. 一批零件的合格率"
+    ],
+    "answer": "B",
+    "solution": "【分析】百分数是表示一个数是另一个数的百分之几，依此分别计算出选项中所要求的最大参与率、提高率、近视率和合格率，然后再进行选择。\n\n【详解】A．六年级学生社会实践活动的参与率，参与的人数最多等于总人数，所以参与率最大是 $100 \\%$ 。 B. 小黄 5 分钟跳绳成绩的提高率, 成绩可以不断提高, 所以提高率可以超过 $100 \\%$ . C. 五年级学生眼睛的近视率, 近视的人数最多等于总人数,所以近视率最大是 $100 \\%$ . D. 一批零件的合格率, 合格的零件数最多等于零件总数,所以合格率最大是 $100 \\%$ .故答案为: B",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "23-24六年级上·全国·单元测试",
+    "merged_id": "unit6_112",
+    "original_id": "unit6_choice_112"
+  },
+  {
+    "id": "unit6_choice_113",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "$45\\mathrm{m}$ 增加 $20\\%$ 后，再减少 $20\\%$ ，结果是（） $\\mathrm{m}_{\\circ}$",
+    "options": [
+      "A. 54",
+      "B. 64.8",
+      "C. 45",
+      "D. 43.2"
+    ],
+    "answer": "D",
+    "solution": "【分析】增加 $20\\%$ ，以 $45 \\mathrm{~m}$ 为单位 '1'，增加之后是：$45 \\times (1 + 20\\%) = 45 \\times 1.2 = 54 \\mathrm{~m}$ , 再减少 $20 \\%$ , 是以 $54 \\mathrm{~m}$ 为单位'1'，则减少了： $54 \\times 20\\% = 10.8\\mathrm{m}$ ，结果是： $54 - 10.8 = 43.2\\mathrm{m}$据此解答即可。\n\n【详解】增加之后是： $45 \\times (1 + 20\\%)=45 \\times 1. 2=54 (m) $减少之后是： $54 - 54 \\times 20\\%\\\\=5 4 - 1 0. 8\\\\=43.2 (m) $\n\n故答案为: D",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "23-24六年级上·全国·单元测试",
+    "merged_id": "unit6_113",
+    "original_id": "unit6_choice_113"
+  },
+  {
+    "id": "unit6_choice_114",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "在含糖 $25\\%$ 的糖水中，糖比水大约少（）。(百分号前保留一位小数)",
+    "options": [
+      "A. $66.7\\%$",
+      "B. $50\\%$",
+      "C. $33.3\\%$"
+    ],
+    "answer": "A",
+    "solution": "【分析】把糖水的质量看作单位 '1', 糖水中糖的含量为 $25 \\%$ , 所以水的含量为 $(1 - 25 \\%)$ , 用糖水中糖的含量和水的含量的差除以水的含量, 即可求出糖比水大约少百分之几, 据此解答。\n\n【详解】$1 - 25\\% = 75\\%$\n\n$ (75 \\% -25 \\%) \\div 75 \\% \\\\ = 0. 5 \\div 0. 7 5 \\\\ \\approx 66.7 \\% $\n\n即糖比水大约少 $$66.7 %$\n\n故答案为: A",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "23-24六年级上·全国·单元测试",
+    "merged_id": "unit6_114",
+    "original_id": "unit6_choice_114"
+  },
+  {
+    "id": "unit6_choice_115",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "千分数表示一个数是另一个数的千分之几，千分数也叫千分率，用'‰'表示。初步核算，忠县 2023 年全年实现地区生产总值(GDP) 比上年增长 $67 \\%​$ , 用百分率表示应是 ( )。",
+    "options": [
+      "A. $0.67\\%$",
+      "B. $6.7\\%$",
+      "C. $67\\%$",
+      "D. $670\\%$"
+    ],
+    "answer": "B",
+    "solution": "【分析】根据千分数的意义，先把 $67 \\%$ 化成小数，再把小数化成百分数即可。千分数化成小数，把千分号去掉，同时把小数点向左移动三位;小数化成百分数，小数点向右移动两位，同时在数的后面添上百分号。\n\n【详解】$67 \\%= 0.067$ $0.067 = 6.7\\%$ 初步核算，忠县 2023 年全年实现地区生产总值（GDP）比上年增长 $67 \\%$ ，用百分率表示应是 $6.7 \\%$ 。故答案为: B",
+    "knowledge_points": [
+      "百分数、小数和分数的互化"
+    ],
+    "source": "23-24六年级下·重庆忠县·期末",
+    "merged_id": "unit6_115",
+    "original_id": "unit6_choice_115"
+  },
+  {
+    "id": "unit6_choice_116",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "选择题",
+    "difficulty": "困难",
+    "question": "一次数学竞赛, 所有选手的平均分是 70 分, 其中女选手人数与男选手人数的比是 $3: 5$ , 男选手的平均分比女选手平均分高 $40 \\%$ 。那么男选手的平均分是 ( ) 分。",
+    "options": [
+      "A. 56",
+      "B. 75.6",
+      "C. 78.4",
+      "D. 81.2"
+    ],
+    "answer": "C",
+    "solution": "【分析】女选手人数与男选手人数的比是 $3: 5$ , 假设女生人数为 3 人, 男生人数为 5 人, 所有选手的平均分是 70 分,则 8 人总分是 $70 \\times 8 = 560$ 分; 设女选手平均分是 x 分, 则男选手平均分是 $(1 + 40\\%) x$ 分, 再根据男选手总分 + 女选手总分 = 560 分, 列出方程, 求出女选手平均分, 进而求出男选手平均分, 据此解答即可。本题考查按比分配、平均数、百分数，解答本题的关键是掌握题中的数量关系。\n\n【详解】设女选手平均分是 $x$ 分, 则男选手平均分是 $(1 + 40\\%) x$ 分。\n\n$ 3x + 5(1 + 40\\%)x = 70\\times (3 + 5) \\\\ 3 x + 7 x = 7 0 \\times 8 \\\\ 1 0 x = 5 6 0 \\\\ 1 0 x \\div 1 0 = 5 6 0 \\div 1 0 \\\\ x = 5 6 $\n\n所以男选手平均分是: $56 \\times (1 + 40\\%) = 56 \\times 1.4 = 78.4$ （分）故答案为: C",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少",
+      "按比分配问题",
+      "列方程解含一个未知数的问题",
+      "平均数的意义及求法"
+    ],
+    "source": "23-24六年级下·重庆渝中·期末",
+    "merged_id": "unit6_116",
+    "original_id": "unit6_choice_116"
+  },
+  {
+    "id": "unit6_choice_117",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "不能用算式 ' $9 \\div 15$ '解决的问题是 ( )。",
+    "options": [
+      "A. 老师买了 9 米的红绸带, 平均分给表演节目的 15 名女生。平均每人分几米?",
+      "B. 王叔叔是骑行爱好者, 他 15 分钟骑行 9 千米, 每骑行 1 千米需要多少分钟?",
+      "C. 第 24 届冬奥会上, 中国体育代表团获得 9 枚金牌,是奖牌总数 15 枚的几分之几?",
+      "D. 明明投篮 15 个, 投中 9 个, 命中率是多少?"
+    ],
+    "answer": "B",
+    "solution": "【分析】A. 将绸带总长 9 米除以女生人数, 求出平均每人分几米; B. 将 15 分钟除以 9 千米, 求出每骑行 1 千米需要多少分钟; C. 求一个数是另一个数的几分之几, 用这个数除以另一个数; D. 将命中的数量除以投篮总数量, 求出命中率。\n\n【详解】A．求平均每人分几米，列式为： $9 \\div 15$ B. 每骑行 1 千米需要多少分钟, 列式为: $15 \\div 9$ ; C. 求金牌是奖牌总数 15 枚的几分之几, 列式为: $9 \\div 15$ ; D. 求命中率是多少, 列式为: $9 \\div 15$ 。故答案为: B",
+    "knowledge_points": [
+      "求一个数占另一个数几分之几",
+      "基础行程问题",
+      "求一个数是另一个数的百分之几（百分率问题）",
+      "平均数的意义及求法"
+    ],
+    "source": "2024·甘肃兰州·小升初真题",
+    "merged_id": "unit6_117",
+    "original_id": "unit6_choice_117"
+  },
+  {
+    "id": "unit6_choice_118",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "甲数的 $\\frac{1}{5}$ 与乙数的 $\\frac{1}{4}$ 相等，甲数的 $25 \\%$ 与丙数的 $20 \\%$ 相等。比较甲、乙、丙三个数的大小，下列结果正确的是哪一个？（）（甲、乙、丙均不为0）",
+    "options": [
+      "A. 甲>乙>丙",
+      "B. 丙>乙>甲",
+      "C. 甲>丙>乙",
+      "D. 丙>甲>乙"
+    ],
+    "answer": "D",
+    "solution": "【分析】本题可以采用设数法进行分析: 将甲数设为 20,分别计算出乙与丙表示的数的大小, 再进行比较。设甲数为 20, 求甲数的 $\\frac{1}{5}$ 用 $20 \\times \\frac{1}{5}$ 等于 4 , 求甲数的 $25 \\%$ 用 $20 \\times 25 \\%$ 等于 5 ;乙数的 $\\frac{1}{4}$ 是 4 , 求乙数用 $4 \\div \\frac{1}{4}$ 等于 16 ;丙数的 $20\\%$ 是 5, 求丙数用 $5 \\div 20\\%$ 等于 25 ;\n\n【详解】假设甲数是 20,$2 0 \\times \\frac {1}{5} = 4$ $20 \\times 25 \\% = 5$ $4 \\div \\frac {1}{4} = 4 \\times 4 = 1 6$ $5 \\div 20\\% = 25$ $25 > 20 > 16$ , 所以丙 $>$ 甲 $>$ 乙。故答案为: D",
+    "knowledge_points": [
+      "已知一个数的百分之几是多少，求这个数",
+      "求一个数的几分之几的问题",
+      "求一个数的百分之几是多少",
+      "已知一个数的几分之几是多少，求这个数"
+    ],
+    "source": "2016·全国·小升初真题",
+    "merged_id": "unit6_118",
+    "original_id": "unit6_choice_118"
+  },
+  {
+    "id": "unit6_choice_119",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "下面的百分率中, ( ) 可能大于 $100 \\%$ .",
+    "options": [
+      "A. 合格率",
+      "B. 出油率",
+      "C. 增长率",
+      "D. 中奖率"
+    ],
+    "answer": "C",
+    "solution": "【分析】合格率：合格人数占总人数的百分率；出油率：油的质量占菜籽质量的百分率，增长率：增长部分占原来总量的百分率；中奖率：表示中奖的概率；据此解答。\n\n【详解】A. 当全部产品都合格时, 合格的数量等于产品的总量, 此时合格率为 $100 \\%$ ;B. 油的质量不可能大于出油种子的质量, 则出油率一般小于 $100 \\%$ ; C. 增长部分有可能比原来的总量多, 则增长率可能大于 $100 \\%$ ; D. 一般来说, 中奖率能达到 $100 \\%$ , 但不会大于 $100 \\%$ 。故答案为: C",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "2014·湖南长沙·小升初真题",
+    "merged_id": "unit6_119",
+    "original_id": "unit6_choice_119"
+  },
+  {
+    "id": "unit6_choice_120",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "据省统计局发布， 2019年第四季度我省GDP总值约为m万亿元，2020年第一季度比2019年第四季度降低a%，2020年第二季度比 2020年第一季度增长 $b \\%$ , 则我省2020年上半年GDP总值可表示为（）。",
+    "options": [
+      "A. $(1 - a\\% + b\\%)$ m",
+      "B. $(1 - a\\%) \\quad (1 + b\\%) \\mathrm{~m}$",
+      "C. $(1 - a\\%)m + (1 + b\\%)$m",
+      "D. $(1 - a\\%)m + (1+b \\%)$m"
+    ],
+    "answer": "D",
+    "solution": "【分析】把2019年第四季度我省GDP总值看作单位'1'，根据题意可知，2020年第一季度我省GDP总值是2019年第四季度的（ $1 - a\\%$ ），根据百分数乘法的意义，可知2020年第一季度的GDP总值为（ $1 - a\\%$ ）m；再把2020年第一季度我省GDP总值看作单位'1'，2020年第二季度我省GDP总值是2020年第一季度的（ $1 + b\\%$ ），根据百分数乘法的意义，用（ $1 - a\\%$ ）（ $1 + b\\%$ ）m即可求出2020年第二季度的GDP总值；然后将2020年第一季度和第二季度的GDP总值相加即可。\n\n【详解】根据分析可知，我省2020年上半年GDP总值可表示为 $(1 - a\\%)m + (1 - a\\%) (1 + b\\%)m$ 。故答案为: D",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "2023六年级下·全国·专题练习",
+    "merged_id": "unit6_120",
+    "original_id": "unit6_choice_120"
+  },
+  {
+    "id": "unit6_choice_121",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "2020年12月1日23时11分嫦娥五号着陆后，开始了为期两天的月面工作，实际采集样本时间19小时，采取样品重量设定为2千克。同样以无人采样返回的方式，苏联曾成功开展过三次月球任务，一共带回约300克样品。嫦娥五号一次设定带回的样品质量比苏联三次带回的样品质量约多（）。",
+    "options": [
+      "A. $566.7\\%$",
+      "B. $85\\%$",
+      "C. $15\\%$"
+    ],
+    "answer": "A",
+    "solution": "【分析】因为 1 千克 = 1000 克, 则 2 千克 = 2000 克, 然后求出嫦娥五号一次设定带回的样品质量比苏联三次带回的样品质量多多少, 再除以苏联三次带回的样品质量, 最后再乘 $100\\%$ 即可。\n\n【详解】2千克 $= 2000$ 克\n\n$(2000 - 300) \\div 300 \\times 100 \\%\\\\= 1700 \\div 300 \\times 100 \\% \\\\ \\approx 5.667 \\times 100 \\% \\\\ = 566.7 \\% $\n\n则嫦娥五号一次设定带回的样品质量比苏联三次带回的样品质量约多 $566.7 \\%$ .故答案为：A",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "2024六年级下·全国·专题练习",
+    "merged_id": "unit6_121",
+    "original_id": "unit6_choice_121"
+  },
+  {
+    "id": "unit6_choice_122",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "如图是某通信公司发送给用户的一条提醒短信。请你仔细阅读后思考：计算移动套餐流量总量，下列选项错误的是（）。尊敬的客户，您好！截至2021年12月5日19时10分，您套餐包含的移动数据流量已使用2.5GB，剩余7.5GB。剩余流量占套餐总量的 $75\\%$ ，请您放心使用。以上内容仅供参考，具体以月结账单为准。",
+    "options": [
+      "A. $2.5 + 7.5$",
+      "B. $7.5 \\div 75\\%$",
+      "C. $2.5 \\div (1 - 75\\%)$",
+      "D.$2.5\\div75\\%$"
+    ],
+    "answer": "D",
+    "solution": "【分析】用已使用的流量加剩余流量，即可得移动套餐流量总量；用剩余流量除以 $75 \\%$ , 即可得移动套餐流量总量;用已使用的流量除以 $(1 - 75 \\%)$ , 即可得移动套餐流量总量。\n\n【详解】A．用已使用的流量加剩余流量，计算移动套餐流量总量可以列式为：$2.5 + 7. 5 = 1 0 \\quad (\\mathrm {G B})$B. 用剩余流量除以 $75 \\%$ , 计算移动套餐流量总量可以列式为:$7.5 \\div 75 \\% = 10 (\\mathrm {GB})$C. 用已使用的流量除以 $(1 - 75 \\%)$ ，计算移动套餐流量总量可以列式为：$2.5 \\div (1 - 75 \\%)$$2.5 \\div 25 \\% = 1 0 \\mathrm {(G B)} $D. 该选项不符合计算移动套餐流量总量的列式。故答案为: D",
+    "knowledge_points": [
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "2024 六年级下·全国·专题练习",
+    "merged_id": "unit6_122",
+    "original_id": "unit6_choice_122"
+  },
+  {
+    "id": "unit6_choice_123",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "为创建优质教育均衡发展区，我区计划投资540万元更新一批教学设备，实际比计划节约了 $24 \\%$ , 实际投资 ( ) 万元。",
+    "options": [
+      "A. $540 \\times 24\\%$",
+      "B. $540 \\div 24\\%$",
+      "C. $540 \\times (1 - 24\\%)$",
+      "D. $540 \\div (1 - 24\\%)$"
+    ],
+    "answer": "C",
+    "solution": "【分析】把计划投资的钱数看作单位'1'，实际投资的钱数是计划的 $(1 - 24\\%)$ ，根据百分数乘法的意义，用 $540 \\times (1 - 24\\%)$ 即可求出实际投资的钱数。\n\n【详解】$540 \\times (1 - 24\\%)\\\\=540\\times 76\\% \\\\ = 4 1 0. 4 \\text {(万 元)} $\n\n实际投资410.4万元。故答案为: C",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "18-19六年级下·山东枣庄·期中",
+    "merged_id": "unit6_123",
+    "original_id": "unit6_choice_123"
+  },
+  {
+    "id": "unit6_fill_124",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "《中国睡眠研究报告2022》数据显示，目前中国只有 $35\\%$ 的人每天睡眠时间够8时。 $35\\%$ 表示( )。",
+    "options": [],
+    "answer": "中国人每天睡眠时间够8时的人数占中国总人数的 $35 \\%$",
+    "solution": "【分析】根据百分数的意义：一个数是另一个数的百分之几，又叫百分率或百分比；把中国总人数看作单位'1'，中国人每天睡眠时间够8时的人数占中国总人数的 $35\\%$ ，据此解答。\n\n【详解】根据分析可知，《中国睡眠研究报告2022》数据显示，目前中国只有 $35\\%$ 的人每天睡眠时间够8时。 $35\\%$ 表示中国人每天睡眠时间够8时的人数占中国总人数的 $35\\%$ 。",
+    "knowledge_points": [
+      "百分数的意义"
+    ],
+    "source": "24-25六年级上·辽宁·课后作业",
+    "merged_id": "unit6_124",
+    "original_id": "unit6_fill_124"
+  },
+  {
+    "id": "unit6_fill_125",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "港珠澳大桥是中国境内一座连接香港、珠海和澳门的大桥，用钢量是42万吨，法国埃菲尔铁塔的用钢量只占港珠澳大桥的百分之一点六七。横线上的数写作( )。",
+    "options": [],
+    "answer": "$1.67\\%$",
+    "solution": "【分析】百分数的写法：先写出'百分之'后面的数，然后在这个数的后面加'%'。\n\n【详解】法国埃菲尔铁塔的用钢量只占港珠澳大桥的百分之一点六七。横线上的数写作 $1.67 \\%$",
+    "knowledge_points": [
+      "百分数的读法和写法"
+    ],
+    "source": "24-25六年级上·辽宁·课后作业",
+    "merged_id": "unit6_125",
+    "original_id": "unit6_fill_125"
+  },
+  {
+    "id": "unit6_fill_126",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "'学习强国'平台报道：2021年我国粮食总量是2020年的 $102\\%$ 。 $102\\%$ 表示( )。",
+    "options": [],
+    "answer": "2021年我国粮食总量比2020年粮食总量增长了 $2 \\%$",
+    "solution": "【分析】2021年我国粮食总量是2020年的 $102 \\%$ , 以2020年的粮食总量为单位 '1', 相当于把2020年的粮食总量平均分成100份, 2021年粮食总量是102份, 即2021年粮食总量比2020年增长多了2份, $102 \\%$ 表示2021年我国粮食总量比2020年粮食总量增长了 $2 \\%$ 。\n\n【详解】'学习强国'平台报道：2021年我国粮食总量是2020年的 $102 \\%$ 。 $102 \\%$ 表示2021年我国粮食总量比2020年粮食总量增长了 $2 \\%$ 。",
+    "knowledge_points": [
+      "百分数的意义"
+    ],
+    "source": "24-25 六年级上·辽宁·课后作业",
+    "merged_id": "unit6_126",
+    "original_id": "unit6_fill_126"
+  },
+  {
+    "id": "unit6_fill_127",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "2022年北京冬奥会上，中国体育代表团参赛运动员共176人，其中女运动员约占 $49.43\\%$ 。 $49.43\\%$ 读作( )。",
+    "options": [],
+    "answer": "百分之四十九点四三",
+    "solution": "【分析】百分数的读法：先读分母（即%），再读分子，读作'百分之……'。\n\n【详解】$49.43\\%$ 读作百分之四十九点四三。",
+    "knowledge_points": [
+      "百分数的读法和写法"
+    ],
+    "source": "24-25六年级上·辽宁·课后作业",
+    "merged_id": "unit6_127",
+    "original_id": "unit6_fill_127"
+  },
+  {
+    "id": "unit6_fill_128",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "科研人员培育了一种治沙植物'红柳'，在离沙漠边缘40千米处种了10000株红柳，成活了8500株，这批红柳的成活率是( )。",
+    "options": [],
+    "answer": "$85\\%$",
+    "solution": "【分析】成活率表示成活的棵数占总棵数的百分之几，计算方法：成活率 $=$ 成活的棵数 $\\div$ 总棵数 $\\times 100\\%$ ，代入数据计算即可。\n\n【详解】$8500 \\div 10000 \\times 100\\%\\\\= 0.85 \\times 100 \\% \\\\ = 85 \\% $\n\n这批红柳的成活率是 $85\\%$ 。",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题)"
+    ],
+    "source": "23-24六年级下·新疆乌鲁木齐·期末",
+    "merged_id": "unit6_128",
+    "original_id": "unit6_fill_128"
+  },
+  {
+    "id": "unit6_fill_129",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一种什锦糖是由奶糖、水果糖和软糖按 $5: 3: 2$ 的比例混合而成的，那么奶糖占 $(\\quad) \\%$ ，软糖占 $(\\quad) \\%$ 。",
+    "options": [],
+    "answer": "50 20",
+    "solution": "【分析】由题意可知, 把什锦糖平均分成 $5 + 3 + 2 = 10$ (份), 其中水果糖、奶糖与软糖分别占 5 份、3 份、2 份, 根据求一个数是另一个数的百分之几, 分别用奶糖、软糖占的份数除以总份数解答即可。\n\n【详解】$5 + 3 + 2$ $= 8 + 2$ $= 10$（份） $5\\div 10 = 50\\%$ $2\\div 10 = 20\\%$ 所以奶糖占 $50\\%$ ，软糖占 $20\\%$ 。",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_129",
+    "original_id": "unit6_fill_129"
+  },
+  {
+    "id": "unit6_fill_130",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "张师傅缝制了 180 条裤子, 经过质检, 5 条不合格。张师傅缝制的这批裤子的合格率是( )。(百分号前保留一位小数。)",
+    "options": [],
+    "answer": "$97.2\\%$",
+    "solution": "【分析】合格率表示合格的裤子条数占张师傅缝制的裤子总条数的百分之几。先算出合格的裤子条数，再根据合格率 = 合格的裤子条数 ÷ 缝制的裤子总条数 × 100%，代入数据计算，即可求出张师傅缝制的这批裤子的合格率。据此解答。\n\n【详解】$180 - 5 = 175$ （条）$175 \\div 180 \\times 100 \\% \\approx 97.2 \\%$即张师傅缝制的这批裤子的合格率是 $97.2\\%$ 。",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_130",
+    "original_id": "unit6_fill_130"
+  },
+  {
+    "id": "unit6_fill_131",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "9 比( )少 $10 \\%$ , ( )比 12 多 $30 \\%$ 。",
+    "options": [],
+    "answer": "10 $\\frac{78}{5}$ 或 $15\\frac{3}{5}$ 或 15.6",
+    "solution": "【分析】（1）已知一个数的百分之几是多少，求这个数，用除法计算，用已知数 $\\div$ 其对应的分率；(2) 求一个数的百分之几是多少, 用乘法计算, 用这个数 $\\times$ 所求数对应的分率。据此计算即可。\n\n【详解】$9 \\div (1 - 10\\%)\\\\= 9 \\div 90 \\%\\\\= 9 \\div 0. 9\\\\= 1 0$\n\n$12 \\times (1 + 30) \\%\\\\=1 2\\times 130\\%\\\\=1 2 \\times 1.3\\\\=15. 6$\n\n$1 5. 6 = \\frac {7 8}{5} = 1 5 \\frac {3}{5}$ 9比10少 $10\\%$ ，15.6比12多 $30\\%$ 。",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少",
+      "已知比一个数多/少百分之几是多少，求这个数"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_131",
+    "original_id": "unit6_fill_131"
+  },
+  {
+    "id": "unit6_fill_132",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "修一条公路，实际比原计划多修 $15 \\%$ 。是把( )看作单位'1'，实际所修的相当于原计划所修的( )%。",
+    "options": [],
+    "answer": "原计划所修的 115",
+    "solution": "【分析】在这类题目中，通常把'比'字后面的量看作单位'1'，即把原计划所修的看作单位'1'，实际比原计划多修 $15\\%$ ，则实际所修的是原计划所修的 $(1 + 15\\%)$ ，据此解答。\n\n【详解】由分析可得：$1 + 15 \\% = 115 \\%$ 则修一条公路，实际比原计划多修 $15 \\%$ 。是把原计划所修的看作单位 '1'，实际所修的相当于原计划所修的 $115 \\%$ 。",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）",
+      "单位'1'的认识与确定"
+    ],
+    "source": "23-24六年级上·全国·单元测试",
+    "merged_id": "unit6_132",
+    "original_id": "unit6_fill_132"
+  },
+  {
+    "id": "unit6_fill_133",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "1 里面有( )个 $10\\%$ , 有( )个 $20\\%$ , 有( )个 $5\\%$ 。",
+    "options": [],
+    "answer": "10 5 20",
+    "solution": "【分析】求 1 里面有几个 $10 \\%$ , 1 里面有几个 $20 \\%$ , 1 里面有几个 $5 \\%$ , 根据除法的意义, 分别用 1 除以 $10 \\%$ 、 $20 \\%$ 、 $5 \\%$ 解答。\n\n【详解】$1 \\div 10\\%\\\\= 1 \\div 0. 1 \\\\ = 1 0 $\n\n$ 1 \\div 20 \\% \\\\ = 1 \\div 0. 2 \\\\ = 5 $\n\n$1 \\div 5 \\% \\\\ = 1 \\div 0. 0 5\\\\ = 2 0$\n\n1里面有10个 $10\\%$ ，有5个 $20\\%$ ，有20个 $5\\%$ 。",
+    "knowledge_points": [
+      "含百分数的运算"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_133",
+    "original_id": "unit6_fill_133"
+  },{
+    "id": "unit6_fill_134",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "写出下面各百分数。百分之五十四写作( ) 百分之一点八五写作( ) 百分之百二十五写作( ) 百分之六点九写作( )",
+    "options": [],
+    "answer": "$54\\%$ $1.85\\%$ $125\\%$ $6.9\\%$",
+    "solution": "【分析】在写百分数时，先写分子，再写百分号。\n\n【详解】百分之五十四写作 $54\\%$ 百分之一点八五 写作 $1.85\\%$ 百分之一百二十五 写作 $125\\%$ 百分之六点九 写作 $6.9\\%$",
+    "knowledge_points": [
+      "百分数的读法和写法"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_134",
+    "original_id": "unit6_fill_134"
+  },
+  {
+    "id": "unit6_fill_135",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "读出下面各百分数。37%读作( ) 90.75%读作( ) 234%读作( ) 0.96%读作( )",
+    "options": [],
+    "answer": "百分之三十七 百分之九十点七五 百分之二百三十四 百分之零点九六",
+    "solution": "【分析】百分数的读法和一般分数的读法相同，都是先读分母再读分子，即读作'百分之几'；注意不要出现阿拉伯数字，要用汉字。据此依次读出即可。\n\n【详解】$37 \\%$ 读作百分之三十七; 90.75%读作百分之九十点七五; 234%读作百分之二百三十四; 0.96%读作百分之零点九六。",
+    "knowledge_points": [
+      "百分数的读法和写法"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_135",
+    "original_id": "unit6_fill_135"
+  },
+  {
+    "id": "unit6_fill_136",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "蜿蜒雄伟的长城凝结了中国古代人民的智慧结晶，是中华民族精神和文化的重要物质载体。根据国家文物局 2012 年宣布的调查结果，中国历代长城总长度为 21196.18 千米，其中明长城 8851.8 千米。(1)横线上的数读作( )千米。 (2)估算：明长城长度占历代长城总长度的( )%。",
+    "options": [],
+    "answer": "(1)二万一千一百九十六点一八 (2) $41.76$",
+    "solution": "【分析】（1）小数的读法，整数部分根据整数的读法，从高位到低位，一级一级地读，每一级末尾的0都不读出来，其余数位连续几个0都只读一个零，小数点直接读点，小数点后面的数直接读出即可。(2) 用明长城长度除以历代长城总长度乘 $100 \\%$ , 就是明长城长度占历代长城总长度的百分数。\n\n【详解】 (1) 21196.18 读作: 二万一千一百九十六点一八 (2) $8851.8 \\div 21196.18 \\times 100\\% \\\\\\approx 0.4176\\times 100\\% \\\\ = 41.76 \\% $\n\n则明长城长度占历代长城总长度的 $41.76\\%$ 。",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）",
+      "小数的读法和写法"
+    ],
+    "source": "2024 六年级下·四川·专题练习",
+    "merged_id": "unit6_136",
+    "original_id": "unit6_fill_136"
+  },
+  {
+    "id": "unit6_fill_137",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "某订餐APP推出新用户满减活动，一份套餐实际支付24元，比原价少付了6元，降低了( )%。",
+    "options": [],
+    "answer": "20",
+    "solution": "【分析】用实际支付的钱数十少付的钱数，求出套餐的原价，再用少付的钱数 $\\div$ 原价 $\\times 100\\%$ ，即可求出降低了百分之几，据此解答。\n\n【详解】$6 \\div (24 + 6) \\times 100\\%\\\\= 6 \\div 30 \\times 100 \\%\\\\= 0.2 \\times 100 \\% \\\\ = 20 \\% $\n\n某订餐APP推出新用户满减活动，一份套餐实际支付24元，比原价少付了6元，降低了 $20\\%$ 。",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "23-24六年级下·四川·单元测试",
+    "merged_id": "unit6_137",
+    "original_id": "unit6_fill_137"
+  },
+  {
+    "id": "unit6_fill_138",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "第31届世界大学生运动会于2023年8月8日在四川省成都市举行。中国队以178枚奖牌位居奖牌榜第一，其中获得金牌103枚，这是中国队参加历届大运会以来，所获金牌数的最高纪录，金牌占比高达( )%；我国第30届世界大学生运动会获得162枚奖牌，第31届获得奖牌比第30届多( )%。（得数在百分号前保留整数)",
+    "options": [],
+    "answer": "58 10",
+    "solution": "【分析】根据求一个数占另一个数的百分之几，用一个数除以另一个数再乘 $100\\%$ ，则用 $103 \\div 178 \\times 100\\%$ 即可求出第31届金牌占的百分比；根据求一个数比另一个数多百分之几，用相差数除以另一个数再乘 $100\\%$ ，用（178一 162) $\\div 162 \\times 100 \\%$ 即可求出第 31 届获得奖牌比第 30 届多百分之几。\n\n【详解】$103 \\div 178 \\times 100\\% \\approx 58\\%$\n\n$(178 - 162) \\div 162 \\times 100 \\% \\\\ = 16 \\div 162 \\times 100 \\% \\\\ \\approx 10 \\% $\n\n第31届获得金牌占比高达 $58\\%$ ；第31届获得奖牌比第30届多 $10\\%$ 。",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）",
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "23-24六年级上·四川绵阳·期末",
+    "merged_id": "unit6_138",
+    "original_id": "unit6_fill_138"
+  },
+  {
+    "id": "unit6_fill_139",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "2023年杭州亚运会是我国第三次举办亚洲最高规格的国际综合性体育赛事，比赛共设42个竞赛项目，其中有26个奥运项目，其余均是非奥运项目，非奥运项目和竞赛项目数量的最简比是( )，奥运项目比非奥运项目的数量多( )%。",
+    "options": [],
+    "answer": "8:21 62.5",
+    "solution": "【分析】由题意可知，比赛共设42个竞赛项目，其中有26个奥运项目，其余均是非奥运项目，即非奥运项目有（42一26）个，据此写出非奥运项目和竞赛项目数量的比，再化简即可；根据求一个数比另一个数多百分之几，用相差数除以另一个数再乘 $100\\%$ ，即可解答。\n\n【详解】$42 - 26 = 16$ （个）\n\n$ 1 6: 4 2 \\\\ = (1 6 \\div 2): (4 2 \\div 2) \\\\ = 8: 2 1 $\n\n$ (26 - 16) \\div 16 \\times 100 \\% \\\\ = 10 \\div 16 \\times 100 \\% \\\\ = 0.625 \\times 100 \\% \\\\ = 62.5 \\% $\n\n非奥运项目和竞赛项目数量的最简比是 $8: 21$ , 奥运项目比非奥运项目的数量多 $62.5 \\%$",
+    "knowledge_points": [
+      "比的化简",
+      "比的意义",
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "23-24六年级上·山西大同·期末",
+    "merged_id": "unit6_139",
+    "original_id": "unit6_fill_139"
+  },
+  {
+    "id": "unit6_fill_140",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "填空题",
+    "difficulty": "困难",
+    "question": "有浓度是 $20\\%$ 的盐水溶液若干千克，如果再加入 20 千克盐，那么盐水的浓度就变为 $30\\%$ ，原来的盐水是( )千克。",
+    "options": [],
+    "answer": "140",
+    "solution": "【分析】假设原来的盐水有 $\\mathrm{x}$ 千克, 根据盐水的质量 $\\times$ 含盐率 $=$ 盐的质量, 可知原来的盐有 $20 \\% \\mathrm{x}$ 千克, 再加入 20 千克盐, 现在的盐有 $(20 \\% \\mathrm{x} + 20)$ 千克, 现在的盐水有 $(x+20) $千克, 盐水的浓度就变为 $ 30 \\%$ , 据此列方程为 $(x + 20) \\times 30\\% = 20\\% x + 20$ ，然后解出方程即可。本题主要考查了浓度问题, 可用列方程解决问题,找到相应的数量关系是解答本题的关键。\n\n【详解】解：设原来的盐水有 $\\mathrm{x}$ 千克。\n\n$(x + 20) \\times 30 \\% = 20 \\% x + 20=0.3 x + 6 = 0. 2 x + 2 0 \\\\ 0. 3 x + 6 - 0. 2 x = 2 0 \\\\ 0. 1 x + 6 = 2 0 \\\\ 0. 1 x = 2 0 - 6 \\\\ 0. 1 x = 1 4 \\\\ x = 1 4 \\div 0. 1 \\\\ x = 1 4 0 $\n\n原来的盐水是 140 千克。",
+    "knowledge_points": [
+      "浓度问题",
+      "列方程解含一个未知数的问题"
+    ],
+    "source": "2023·四川·小升初真题",
+    "merged_id": "unit6_140",
+    "original_id": "unit6_fill_140"
+  },
+  {
+    "id": "unit6_fill_141",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "2020年12月17日，探月工程'嫦娥五号'任务取得圆满成功。'嫦娥五号'从出发奔月到回归地球共经历了23天，其中8天完成从地球到月球的旅程，在月球工作2天，其余的时间用于返回地球。(1)'嫦娥五号'在月球工作的天数占它从出发奔月到回归地球的总天数的( )。 (2)'嫦娥五号'返回地球用的时间比从地球到月球用的时间多( )‰。",
+    "options": [],
+    "answer": "$(1)\\frac{2}{23}$ (2) $62.5 \\mathrm{~g}$",
+    "solution": "【分析】（1）根据求一个数占另一个数的几分之几，用一个数除以另一个数，则用 $2 \\div 23$ 即可求出'嫦娥五号'在月球工作的天数占它从出发奔月到回归地球的总天数的几分之几;(2) 根据题意可知，'嫦娥五号'返回地球用了（23-8 -2）天，再根据求一个数比另一个数多百分之几，用相差数除以另一个数，则用返回的天数减去从地球到月球的天数的差除以从地球到月球的天数即可求出'嫦娥五号'返回地球用的时间比从地球到月球用的时间多百分之几。\n\n【详解】 (1) $2 \\div 23 = \\frac{2}{23}$'嫦娥五号'在月球工作的天数占它从出发奔月到回归地球的总天数的 $\\frac{2}{23}$ 。(2) $23 - 8 - 2 = 13$ (天) $ (1 3 - 8) \\div 8 = 5 \\div 8 = 62.5 \\% $'嫦娥五号'返回地球用的时间比从地球到月球用的时间多 $62.5 \\%$ .",
+    "knowledge_points": [
+      "求一个数占另一个数几分之几",
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "2024六年级下·全国·专题练习",
+    "merged_id": "unit6_141",
+    "original_id": "unit6_fill_141"
+  },
+  {
+    "id": "unit6_fill_142",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "中国向外传播文化的途经之一是孔子学院。某一年底，全球已有146个国家（地区）建立了525所孔子学院，其中'一带一路'沿线有53国设立140所孔子学院。此时，'一带一路'沿线设立的孔子学院占全球的( )%。（百分号前保留一位小数）",
+    "options": [],
+    "answer": "26.7",
+    "solution": "【分析】百分比表示一个数是另一个数的百分之几；求'一带一路'沿线设立的孔子学院占全球的百分比，即'一带一路'沿线设立孔子学院140所 $\\div$ 全球孔子学院525所×100%，列式计算即可。注意百分号前保留一位小数，则计算 $140 \\div 525$ 保留3位小数。\n\n【详解】$140 \\div 525 \\times 100\\%\\\\\\approx 0.267 \\times 100 \\%\\\\\\approx 26.7\\%$\n\n所以'一带一路'沿线设立的孔子学院占全球的 $26.7\\%$ 。",
+    "knowledge_points": [
+      "百分数、小数和分数的互化",
+      "百分数的意义"
+    ],
+    "source": "2024六年级下·全国·专题练习",
+    "merged_id": "unit6_142",
+    "original_id": "unit6_fill_142"
+  },
+  {
+    "id": "unit6_judge_143",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "某班男生人数比女生人数多 $20 \\%$ , 女生人数与男生人数的比是 $6: 5$ 。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】某班男生人数比女生人数多 $20 \\%$ , 将女生的人数看成单位 '1', 则男生的人数是女生人数的 $(1 + 20 \\%)$ ,将 $120 \\%$ 转化为分数, 为 $\\frac{6}{5}$ , 根据分数、除法以及比的关系,则女生人数与男生人数的比是 $5: 6$ 。\n\n【详解】$1 + 20\\% = 120\\%$ $120 \\% = \\frac {6}{5} = 6 \\div 5 = 6: 5$ 则女生人数与男生人数的比是 $5: 6$ 。故答案为: ×",
+    "knowledge_points": [
+      "比的意义",
+      "百分数、分数、小数和比的互化"
+    ],
+    "source": "23-24 六年级下·山东济宁·期末",
+    "merged_id": "unit6_143",
+    "original_id": "unit6_judge_143"
+  },
+  {
+    "id": "unit6_judge_144",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "把 $48\\%$ 的百分号去掉，这个数就扩大为原来的 100 倍。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】去掉 $48 \\%$ 的百分号，这个数就变成了 48 ，求出 48 是 $48 \\%$ 的多少倍即可。\n\n【详解】$48 \\div 48\\% = 48 \\times \\frac{100}{48} = 100$ 48是 $48\\%$ 的100倍。故答案为:√",
+    "knowledge_points": [
+      "百分数、小数和分数的互化",
+      "求一个数是另一个数的几倍",
+      "百分数的意义"
+    ],
+    "source": "23-24六年级下·湖南益阳·期末",
+    "merged_id": "unit6_144",
+    "original_id": "unit6_judge_144"
+  },
+  {
+    "id": "unit6_judge_145",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "购物中心一双皮鞋的价格比原来降低 $15\\%$ ，现价相当于原价的 $85\\%$ 。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】把皮鞋的原价看作单位'1'，现在价格比原来降低 $15\\%$ ，现价是 $1 - 15\\% = 85\\%$ ，即现价相当于原价的 $85\\%$ ，据此解答。\n\n【详解】$1 - 15\\% = 85\\%$购物中心一双皮鞋的价格比原来降低 $15 \\%$ , 现价相当于原价的 $85 \\%$ 。原题干说法正确。 故答案为: √",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "23-24六年级上·全国·课后作业",
+    "merged_id": "unit6_145",
+    "original_id": "unit6_judge_145"
+  },
+  {
+    "id": "unit6_judge_146",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "$\\frac{50}{200}, \\frac{25}{100}, 25 \\%$ 三个数的数值相等，意义也相同。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】先把三个数化成小数，再进行比较；再根据分数可以表示一个分率，也可表示一个具体数量，即它后面可以不带计量单位（表示分率），也可带计量单位（表示一个具体数量）；百分数只表示求一个数是另一个数的百分之几，它后面不能带计量单位。据此解答。\n\n【详解】$\\frac{50}{200} = 0.25\\quad\\frac {2 5}{1 0 0} = 0. 2 5$ $25 \\% = 0.25$ $0.25 = 0.25 = 0.25$ ，三个数的数值相等，但意义不相同。\n\n$\\frac{50}{200}, \\frac{25}{100}, 25 \\%$ 三个数的数值相等，但意义不相同。原题干说法错误。 故答案为: ×",
+    "knowledge_points": [
+      "百分数、小数和分数的互化",
+      "分数的意义",
+      "百分数的意义"
+    ],
+    "source": "18-19 六年级上·浙江·课后作业",
+    "merged_id": "unit6_146",
+    "original_id": "unit6_judge_146"
+  },
+  {
+    "id": "unit6_judge_147",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "$3 \\mathrm{~m}$ 的 $10 \\%$ 与 $10 \\mathrm{~m}$ 的$3 \\%$ 一样长。（)",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】根据求一个数的百分之几是多少，用乘法解答，\n\n分别求出 $3 \\mathrm{~m}$ 的 $10 \\%$ 与 $10 \\mathrm{~m}$ 的 $3 \\%$ 是多少 $\\mathrm{m}$ , 据此判断。\n\n【详解】$3 \\times 10\\% = 0.3 \\text{ (m)}$ $10 \\times 3 \\% = 0.3 (\\mathrm {m})$ $0.3 = 0. 3$ 所以 $3 \\mathrm{~m}$ 的 $10 \\%$ 与 $10 \\mathrm{~m}$ 的 $3 \\%$ 一样长。原题说法正确。故答案为: √",
+    "knowledge_points": [
+      "求一个数的百分之几是多少"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_147",
+    "original_id": "unit6_judge_147"
+  },
+  {
+    "id": "unit6_judge_148",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "一筐苹果重$45 \\% \\mathrm{kg}$ 。 ( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】百分数表示一个数是另一个数的百分之多少，如 $14 \\%$ 表示一个数是另一个数的 $\\frac{14}{100}$ ，百分数不能带单位。\n\n【详解】由分析可知, 百分数不能带单位, 所以一筐苹果重 $45 \\mathrm{~kg}$ 的说法错误。故答案为: ×",
+    "knowledge_points": [
+      "百分数的意义"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_148",
+    "original_id": "unit6_judge_148"
+  },
+  {
+    "id": "unit6_judge_149",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "15 分钟是 1 小时的 $25 \\%$ ( )",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】1小时是60分钟，15分钟是60分钟的百分之几用除法。\n\n【详解】1小时=60分钟 $15 \\div 60 = 0.25 = 25 \\%$ 故 15 分钟是 1 小时的 $25 \\%$ 。故答案为: √",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_149",
+    "original_id": "unit6_judge_149"
+  },
+  {
+    "id": "unit6_judge_150",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "一种冰箱先涨价20%, 再降价 20%, 现价和原价一样。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】设该冰箱的原价为 1 , 把该冰箱的原价看作单位'1'，先涨价 $20\\%$ ，则涨价后的价格是原价的（1+20%），单位'1'已知，用原价乘 $(1 + 20\\%)$ ，求出涨价后的价格;再降价 $20 \\%$ , 是把涨价后的价格看作单位 '1', 降价后的价格是涨价后价格的（1-20%）；单位'1'已知，用涨价后的价格乘（1-20%），求出现价；最后把该冰箱的现价与原价进行比较，得出结论。\n\n【详解】设该冰箱的原价为 1。冰箱的现价为：$ 1 \\times (1 + 20 \\%) \\times (1 - 20 \\%) \\\\ = 1 \\times 1. 2 \\times 0. 8 \\\\ = 0. 9 6 $\n\n$0.9 6 \\neq 1$现价和原价不一样。原题说法错误。故答案为: ×",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_150",
+    "original_id": "unit6_judge_150"
+  },
+  {
+    "id": "unit6_judge_151",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "百分数一定比 1 小。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】表示一个数是另一个数的百分之几的数是百分数,百分数可以比 1 小, 也可以比 1 大, 或等于 1 。可举例解答。\n\n【详解】例如： $50\\%<1$ $120 \\%> 100%=1$ 所以原题说法错误。 故答案为: ×",
+    "knowledge_points": [
+      "百分数的意义"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_151",
+    "original_id": "unit6_judge_151"
+  },
+  {
+    "id": "unit6_judge_152",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "若我旗今年玉米的产量比去年增长 $25 \\%$ , 那么今年产量是去年的 $\\frac{5}{4}$ 。( )",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】把去年的玉米的产量看作单位 '1', 因为今年玉米的产量比去年增长 $25 \\%$ , 说明今年玉米的产量是去年玉米的产量的 $(1 + 25 \\%)$ , 据此解答。\n\n【详解】$1 + 25\\% = 1 + \\frac{1}{4} = \\frac{5}{4}$ ,即今年产量是去年的 $\\frac{5}{4}$ 。 故答案为: √",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "23-24 六年级下·内蒙古呼伦贝尔·期末",
+    "merged_id": "unit6_152",
+    "original_id": "unit6_judge_152"
+  },
+  {
+    "id": "unit6_judge_153",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "$\\frac{1}{100}$ 和 $1 \\%$ 都是分母为100的分数，它们表示的意义相同。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】根据分数和百分数的意义: $\\frac{1}{100}$ 表示是把单位 '1'平均分成 100 份, 取其中一份的数, 加上单位可以表示具体的数量; $1 \\%$ 表示的是一个数是另一个数的百分之一, 百分数不能带单位名称, 即不能表示具体的数量; 据此可判断。此题考查了分数和百分数的意义区别, 解答此题关键是注意分数既能表示份数也能表示具体的数量, 而百分数不能表示具体的数量。\n\n【详解】根据分数和百分数的意义可知: $\\frac{1}{100}$ 和 $1 \\%$ 都是分母为 100 的分数, 它们表示的意义不同。故答案为: ×",
+    "knowledge_points": [
+      "分数的意义",
+      "百分数的意义"
+    ],
+    "source": "2022·新疆吐鲁番·小升初真题",
+    "merged_id": "unit6_153",
+    "original_id": "unit6_judge_153"
+  },
+  {
+    "id": "unit6_judge_154",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "把一批玉米种子做试验，发芽的有98粒，未发芽的有2粒，发芽率是 $98\\%$ 。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】根据发芽率 = 发芽种子数 ÷ 种子总数 × 100%, 列式计算即可。\n\n【详解】$98 \\div (98 + 2) \\times 100\\%\\\\=98 \\div 100 \\times 100 \\% \\\\ = 0.98 \\times 100 \\% \\\\ = 98 \\% $\n\n把一批玉米种子做试验，发芽的有98粒，未发芽的有2粒，发芽率是 $98 \\%$ , 说法正确。故答案为: √",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "23-24六年级上·海南省直辖县级单位·期末",
+    "merged_id": "unit6_154",
+    "original_id": "unit6_judge_154"
+  },
+  {
+    "id": "unit6_judge_155",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "一种商品先降价 $20 \\%$ ,再提价 $20 \\%$ , 现价与原价相等。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】设商品的原价是 100 元; 降价后的价格是原价的 $(1 - 20 \\%$ , 用商品的原价 $\\times$ $(1 - 20 \\%$ ), 求出降价后的价格; 再把降价后的价格看作单位 '1', 提价后的价格是降价后价格的 $(1 + 20 \\%$ , 用降价后的价格 $\\times$ $(1 + 20 \\%$ ), 求出提价后的价格, 再进行比较, 即可解答。\n\n【详解】设商品的原价是100元。\n\n$100 \\times (1 - 20 \\%) \\times (1 + 20 \\%) \\\\ = 1 0 0 \\times 0. 8 \\times 1. 2 \\\\= 8 0 \\times 1. 2 \\\\ = 9 6 (\\text {元}) $\n\n$96 < 100$ , 现价比原价低了。一种商品先降价 $20 \\%$ , 再提价 $20 \\%$ , 现价比原价低了。原题干说法错误。故答案为: ×",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "2012·贵州安顺·小升初真题",
+    "merged_id": "unit6_155",
+    "original_id": "unit6_judge_155"
+  },
+  {
+    "id": "unit6_judge_156",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "一种商品提价 $10\\%$ 后，又降价 $10\\%$ ，这种商品的价格不变。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】把这种商品的原价看作单位'1'，提价 $10\\%$ 后的价格为 $1 \\times (1 + 10\\%)$ ；再将提价后的价格看作单位'1'，又降价 $10\\%$ 后的价格为 $1 \\times (1 + 10\\%) \\times (1 - 10\\%)$ ，据此可求出现价，最后再与原价对比即可。\n\n【详解】$1 \\times (1 + 10\\%) \\times (1 - 10\\%)$ $= 1 \\times 1.1 \\times 0.9$ $= 1.1 \\times 0.9$ $= 0.99$ $0.99 < 1$ 则一种商品提价 $10\\%$ 后，又降价 $10\\%$ ，这种商品的价格此时比原价低。原说法错误。 故答案为：×",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "2019·湖南郴州·小升初真题",
+    "merged_id": "unit6_156",
+    "original_id": "unit6_judge_156"
+  },
+  {
+    "id": "unit6_judge_157",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "生产120个零件，全部合格，合格率是 $120\\%$ 。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】合格率 $=$ 合格零件的数量÷生产零件的总数量$\\times 100\\%$ ，代入数据，求出合格率，再进行比较，即可解答。\n\n【详解】$120 \\div 120 \\times 100\\%\\\\=1 \\times 100 \\% \\\\ = 100 \\% $\n\n生产 120 个零件，全部合格，合格率是 $100\\%$ 。原题干说法错误故答案为: ×",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问 题）"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit6_157",
+    "original_id": "unit6_judge_157"
+  },
+  {
+    "id": "unit6_judge_158",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "甲数的 $\\frac{3}{4}$ 与乙数的 $75\\%$ 一定相等。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】求一个数的几分之几或百分之几是多少用乘法，甲数和乙数如果相等，则甲数的 $\\frac{3}{4}$ 与乙数的 $75\\%$ 一定相等；甲数和乙数如果不相等, 则甲数的 $\\frac{3}{4}$ 与乙数的 $75 \\%$ 不相等,举例说明即可。\n\n【详解】如果甲数和乙数都是 $100$ 。$1 0 0 \\times \\frac {3}{4} = 7 5$ $100 \\times 75 \\% = 100 \\times 0.75 = 75$ 如果甲数是 100, 乙数是 $200$ 。 $1 0 0 \\times \\frac {3}{4} = 7 5$ $200 \\times 75 \\% = 200 \\times 0.75 = 150$甲数的 $\\frac{3}{4}$ 与乙数的 $75 \\%$ 可能相等也可能不相等, 所以原题说法错误。故答案为: ×",
+    "knowledge_points": [
+      "求一个数的几分之几的问题",
+      "求一个数的百分之几是多少"
+    ],
+    "source": "19-20 三年级上·全国·课后作业",
+    "merged_id": "unit6_158",
+    "original_id": "unit6_judge_158"
+  },
+  {
+    "id": "unit6_judge_159",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "$\\frac{37}{100}$ 写作 $37 \\%$ ，读作：百分之三十七。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】百分数的读法：先读百分号，再读出百分号前的部分；百分数的写法：通常不写成分数形式，而在原来的分子后面加上百分号'%'，由此解决问题。此题比较简单，主要考查的是百分数的读法和写法，百分数的读法：先读百分号，再读出百分号前的部分；百分数的写法：通常不写成分数形式，而在原来的分子后面加上百分号'%'，同时应注意基础知识的积累。\n\n【详解】根据百分数的读写方法可知: $\\frac{37}{100}$ 写作 $37 \\%$ , 读作:百分之三十七, 原说法正确。故答案为: √",
+    "knowledge_points": [
+      "百分数的读法和写法"
+    ],
+    "source": "21-22 六年级上·广东惠州·期中",
+    "merged_id": "unit6_159",
+    "original_id": "unit6_judge_159"
+  },
+  {
+    "id": "unit6_judge_160",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "在 100 克水中放入 10 克盐，盐的质量占盐水质量的 $10\\%$ 。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】盐水的质量 = 盐的质量 + 水的质量，盐的质量占盐水质量的百分率 = 盐的质量 ÷ 盐水的质量 × 100%，据此解答。本题主要考查百分率问题, 掌握求一个数占另一个数百分之几的计算方法是解答题目的关键。\n\n【详解】$10 \\div (100 + 10) \\times 100\\%\\\\= 10 \\div 110 \\times 100 \\% \\\\ \\approx 0.091 \\times 100 \\% \\\\ = 9.1\\% $\n\n所以，盐的质量占盐水质量的 $9.1 \\%$ 。故答案为: ×",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题)"
+    ],
+    "source": "2018·山东临沂·小升初真题",
+    "merged_id": "unit6_160",
+    "original_id": "unit6_judge_160"
+  },
+  {
+    "id": "unit6_calc_161",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "解方程。\n\n$84 \\% \\mathrm {x} = 42 \\quad 12 \\% \\mathrm {x} + 7 = 7.24 \\quad 55 \\% \\mathrm {x} - 20 \\% \\mathrm {x} = 280$",
+    "options": [],
+    "answer": "$x = 50; x = 2; x = 800$",
+    "solution": "【分析】等式的性质 1: 等式两边加上或减去同一个数, 左右两边仍然相等。等式的性质 2: 等式两边乘或除以同一个不为 0 的数, 左右两边仍然相等。(1) 方程等号左右两边同时除以 $84 \\%$ , 即可解出方程。 (2) 方程等号左右两边同时减去 7 , 等号左右两边同时除以 $12 \\%$ , 即可解出方程; (3) 先化简方程得到 $35\\% \\mathrm{x} = 280$ ，等号左右两边同时除以 $35\\%$ ，即可解出方程。\n\n【详解】 (1) $84\\% \\mathrm{x} = 42$\n\n解: $x = 42 \\div 84\\%$\n\n$x = 5 0$\n\n(2) $12\\% \\mathrm{x} + 7 = 7.24$\n\n解： $12\\% \\mathrm{x} = 7.24 - 7$\n\n$12 \\% \\mathrm {x} = 0.24$\n\n$\\mathrm{x} = 0.24 \\div 12 \\%$\n\n$x = 2$\n\n(3) $55\\% \\mathrm{x} - 20\\% \\mathrm{x} = 280$\n\n解: $35\\% \\mathrm{x} = 280$\n\n$x = 280 \\div 35 \\% \\\\ x = 8 0 0 $",
+    "knowledge_points": [
+      "应用等式的性质1和2解方程",
+      "解百分数方程"
+    ],
+    "source": "23-24六年级上·辽宁·课后作业",
+    "merged_id": "unit6_161",
+    "original_id": "unit6_calc_161"
+  },
+  {
+    "id": "unit6_calc_162",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "一个数减少它的15%后是 5.1，这个数是多少？（用方程解）",
+    "options": [],
+    "answer": "6",
+    "solution": "【分析】求一个数的百分之几是多少用乘法，设这个数是x，根据这个数一这个数 $\\times 15 \\% = 5.1$ ，列出方程求出 x 的值即可。\n\n【详解】解：设这个数是 $x$\n\n$x - 15 \\% \\mathrm{x} = 5.1\\\\0.8 5x = 5. 1\\\\0.8 5 \\mathrm {x} \\div 0. 8 5 = 5. 1 \\div 0. 8 5\\\\x = 6$\n\n这个数是 6。",
+    "knowledge_points": [
+      "已知比一个数多/少百分之几是多少，求这个数",
+      "求一个数的百分之几是多少",
+      "列方程解含一个未知数的问题"
+    ],
+    "source": "23-24 六年级下·四川凉山·期末",
+    "merged_id": "unit6_162",
+    "original_id": "unit6_calc_162"
+  },
+  {
+    "id": "unit6_calc_163",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "将下列各小数化为百分数。\n\n(1) $0.35 =$ (2) $1.266 =$ (3) $5.001 =$",
+    "options": [],
+    "answer": "(1) $35\\%$ ;(2) $126.6\\%$ ;(3) $500.1\\%$",
+    "solution": "【分析】把小数化成百分数，小数点向右移动两位，加上百分号即可。\n\n【详解】 (1) $0.35 = 35\\%$ (2) $1.266 = 126.6\\%$ (3) $5.001 = 500.1\\%$",
+    "knowledge_points": [
+      "百分数、分数、小数和比的互化"
+    ],
+    "source": "23-24 六年级上·全国·假期作业",
+    "merged_id": "unit6_163",
+    "original_id": "unit6_calc_163"
+  },
+  {
+    "id": "unit6_solve_164",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "王奶奶家 5 月份的用电量比4月份增加了 $5 \\%$ ，6月份又比5月份增加了8%，6月份与4月份相比，增加了百分之几？",
+    "options": [],
+    "answer": "$13.4\\%$",
+    "solution": "【分析】设 4 月份的用电量为 1 , 先把 4 月份的用电量看作单位 '1', 5 月份的用电量比 4 月份增加了 $5 \\%$ , 则 5 月份的用电量是 4 月份的 ( $1 + 5 \\%$ ), 单位 '1'已知, 用乘法求出5月份的用电量；再把5月份的用电量看作单位'1'，6月份又比5月份增加了 $8\\%$ ，则6月份的用电量是5月份的 $(1 + 8\\%)$ ，单位'1'已知，用乘法求出6月份的用电量；最后用减法求出6月份比4月份增加的用电量，再除以4月份的用电量，即是6月份比4月份增加了百分之几。\n\n【详解】设4月份的用电量为 $1_{\\circ}$\n\n$1 \\times (1 + 5 \\%) \\times (1 + 8 \\%) \\\\ = 1 \\times (1 + 0. 0 5) \\times (1 + 0. 0 8) \\\\ = 1 \\times 1. 0 5 \\times 1. 0 8 \\\\ = 1. 1 3 4 $\n\n$ (1. 1 3 4 - 1) \\div 1 \\\\ = 0. 1 3 4 \\div 1 \\\\ = 0. 1 3 4 \\\\ = 13.4 \\% $\n\n答：6月份与4月份相比，增加了 $13.4\\%$ 。",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少",
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "24-25 六年级上·全国·课后作业",
+    "merged_id": "unit6_164",
+    "original_id": "unit6_solve_164"
+  },
+  {
+    "id": "unit6_solve_165",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "某工厂扩建厂房实际投资4.76万元，比原计划节约了2.04万元，实际比原计划节约了百分之几？",
+    "options": [],
+    "answer": "$30\\%$",
+    "solution": "【分析】根据题意，实际投资4.76万元，比原计划节约了2.04万元，则原计划投资（ $4.76 + 2.04$ ）万元；求实际比原计划节约了百分之几，用节约的钱数除以原计划投资的钱数，即是实际比原计划节约了百分之几。\n\n【详解】$2.04 \\div (4.76 + 2.04) \\times 100\\%\\\\=2.04 \\div 6.8 \\times 100 \\% \\\\ = 0.3 \\times 100 \\% \\\\ = 30 \\% $\n\n答：实际比原计划节约了 $30 \\%$ 。",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "24-25六年级上·辽宁·期中",
+    "merged_id": "unit6_165",
+    "original_id": "unit6_solve_165"
+  },
+  {
+    "id": "unit6_solve_166",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "春蕾小学的一项调查表明，有牙病的学生人数占全校人数的 $20\\%$ 。春蕾小学共有750名学生，有牙病的学生有多少人？",
+    "options": [],
+    "answer": "150人",
+    "solution": "【分析】把春雷小学总人数看作单位 '1', 有牙病的学生人数占全校人数的 $20 \\%$ , 求有牙病的学生人数, 用春雷小学的总人数 $\\times 20 \\%$ , 即可解答。\n\n【详解】$750 \\times 20\\% = 150$ （人）\n\n答: 有牙病的学生有 150 人。",
+    "knowledge_points": [
+      "求一个数的百分之几是多少"
+    ],
+    "source": "24-25六年级上·辽宁·课后作业",
+    "merged_id": "unit6_166",
+    "original_id": "unit6_solve_166"
+  },  {
+    "id": "unit6_solve_167",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "陈叔叔家今年收花生1500千克，比去年增产 $25\\%$ 。陈叔叔家去年收花生多少千克？",
+    "options": [],
+    "answer": "1200千克",
+    "solution": "【分析】已知今年收花生 1500 千克, 比去年增产 $25 \\%$ , 把去年收花生的质量看作单位 '1', 则今年收花生的质量是去年的 $(1 + 25 \\%) ,$ 单位 '1' 未知, 用今年收花生的质量除以 $(1 + 25$ ), 求出去年收花生的质量。\n\n【详解】$1500 \\div (1 + 25\\%)\\\\=1 5 0 0 \\div (1 + 0. 2 5) \\\\ = 1 5 0 0 \\div 1. 2 5 \\\\ = 1 2 0 0 \\text {(千 克)} $\n\n答: 陈叔叔家去年收花生 1200 千克。",
+    "knowledge_points": [
+      "已知比一个数多/少百分之几是多少，求这个数"
+    ],
+    "source": "24-25六年级上·江苏·单元测试",
+    "merged_id": "unit6_167",
+    "original_id": "unit6_solve_167"
+  },
+  {
+    "id": "unit6_solve_168",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "某面粉厂6月生产面粉2400吨，六月份比7月多生产 $\\frac{1}{7}$ ，7月份生产面粉多少吨？",
+    "options": [],
+    "answer": "2100吨",
+    "solution": "【分析】把 7 月份生产面粉的重量看作单位 '1', 6 月份的生产面粉的重量是 7 月份的 $(1 + \\frac{1}{7})$ , 对应的是 6 月份生产面粉的重量, 求单位 '1', 用 6 月份生产面粉的重量 $\\div$ $(1 + \\frac{1}{7})$ 解答。\n\n【详解】$2400 \\div (1 + \\frac{1}{7})\\\\= 2 4 0 0 \\div \\frac {8}{7} \\\\ = 2 4 0 0 \\times \\frac {7}{8} \\\\ = 2 1 0 0 (\\text {吨}) $\n\n答: 7月份生产面粉2100吨。",
+    "knowledge_points": [
+      "已知比一个数多/少几分之几是多少，求这个数"
+    ],
+    "source": "23-24六年级上·全国·单元测试",
+    "merged_id": "unit6_168",
+    "original_id": "unit6_solve_168"
+  },
+  {
+    "id": "unit6_solve_169",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "学校开展'读好书'活动，小明三天读完一本书，第一天读了全书的 $\\frac{1}{5}$ ，第二天读了全书的 $50\\%$ ，第三天读了 60 页，这本书一共有多少页？",
+    "options": [],
+    "answer": "200页",
+    "solution": "【分析】将全书页数看作单位 '1', 1- 第一天读的对应分率 - 第二天读的对应百分率 = 第三天读的对应分率或百分率, 第三天读的页数 ÷ 对应分率或百分率 = 全书页数, 据此列式解答。\n\n【详解】$60 \\div (1 - \\frac{1}{5} - 50\\%)\\\\=6 0 \\div 0. 3 \\\\ = 2 0 0 \\text {(页)} $\n\n答: 这本书一共有 200 页。",
+    "knowledge_points": [
+      "已知一部分量占总量的几分之几及另一部分量,求总量",
+      "已知一部分量占总量的百分之几及另一部分量,求总量"
+    ],
+    "source": "23-24六年级下·河北保定·期末",
+    "merged_id": "unit6_169",
+    "original_id": "unit6_solve_169"
+  },
+  {
+    "id": "unit6_solve_170",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "小亮和小静去买文具, 共带了 60 元, 小亮用去了自己钱数的 $\\frac{4}{5}$ , 小静用去了自己钱数的 $60 \\%$ , 两人剩下的钱数正好一样多, 那么两人原来各带多少钱?",
+    "options": [],
+    "answer": "小亮40元；小静20元",
+    "solution": "【分析】从题意可知：以小亮原来的钱数为单位'1'，小亮还剩下自己钱数的 $1 - \\frac{4}{5} = \\frac{1}{5}$ ；以小静原来的钱数为单位'1'，小静还剩下自己钱数的 $1 - 60\\% = 40\\%$ ；根据'两人剩下的钱数正好一样多'，可得：小亮原来的钱数 $\\times \\frac{1}{5} =$ 小静原来的钱数 $\\times 40\\%$ 。设小亮原来有 $x$ 元，小静就有 $(60 - x)$ 元，根据等量关系式，列出方程，并求出 $x$ 和 $(60 - x)$ 的值，即可求出两人原来各带多少钱。据此解答。\n\n【详解】解：设小亮原来有 $x$ 元，小静有 $(60 - x)$ 元。\n\n$(1 - \\frac{4}{5}) x = (60 - x) \\times (1 - 60\\%)\\\\(1 - 0. 8) x = (6 0 - x) \\times (1 - 0. 6) \\\\ 0. 2x = (6 0 - x) \\times 0. 4 \\\\ 0. 2 x = 6 0 \\times 0. 4 - 0. 4 ^ {x} \\\\ 0. 2 x + 0. 4 x = 6 0 \\times 0. 4 \\\\ 0. 6 x = 2 4 \\\\ x = 2 4 \\div 0. 6 \\\\ x = 4 0 \\\\ 6 0 - 4 0 = 2 0 \\text {(元)} $\n\n答: 小亮原来有 40 元, 小静原来有 20 元。",
+    "knowledge_points": [
+      "百分数、小数和分数的互化",
+      "单位 '1'的认识与确定",
+      "列方程解含两个未知数的问题",
+      "含百分数的运算"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_170",
+    "original_id": "unit6_solve_170"
+  },
+  {
+    "id": "unit6_solve_171",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "仓库里有 20 吨大米,第一次运走总数的 $30 \\%$ , 第二次运走剩下的 $15 \\%$ , 两次共运走多少吨大米？",
+    "options": [],
+    "answer": "8.1 吨",
+    "solution": "【分析】从题意可知：以总数为单位'1'，第一次运走总数的 $30\\%$ ，用 $20 \\times 30\\% = 6$ 吨，即求出了第一次运走的吨数；第一次运走后，剩下总数的 $1 - 30\\% = 70\\%$ ，用 $20 \\times (1 - 30\\%) = 14$ 吨，即剩下的吨数；以剩下的吨数为单位'1'，用 $14 \\times 15\\%$ 即可求出第二次运走的吨数；将两次的吨数相加，即可求出两次一共运走的吨数。据此解答。\n\n【详解】$20 \\times 30\\% + 20 \\times (1 - 30\\%) \\times 15\\%$ \n\n$ 6 + 20 \\times 70 \\% \\times 15 \\% \\\\ = 6 + 2. 1 \\\\ = 8. 1 \\text {(吨)} $答: 两次共运走 8.1 吨大米",
+    "knowledge_points": [
+      "求一个数的百分之几是多少",
+      "单位'1'的认识与确定",
+      "含百分数的运算"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_171",
+    "original_id": "unit6_solve_171"
+  },
+  {
+    "id": "unit6_solve_172",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "某家具厂去年计划生产桌椅 38000 套，实际比计划少生产了 5000 套。实际完成了计划的百分之多少？（百分号前保留一位小数。）",
+    "options": [],
+    "answer": "$86.8\\%$",
+    "solution": "【分析】先求出实际生产的桌椅套数，再用实际生产的套数除以计划生产的套数，最后乘 $100\\%$ 并保留一位小数，据此解答。\n\n【详解】$(38000 - 5000) \\div 38000 \\times 100\\%\\\\= 33000 \\div 38000 \\times 100 \\% \\\\ \\approx 86.8 \\% $\n\n答：实际完成了计划的 $86.8\\%$ 。",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "23-24六年级上·全国·单元测试",
+    "merged_id": "unit6_172",
+    "original_id": "unit6_solve_172"
+  },
+  {
+    "id": "unit6_solve_173",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "修一条公路，已经修了 $35\\%$ ，距离中点还有 16.2 千米，这条公路全长多少千米？",
+    "options": [],
+    "answer": "108千米",
+    "solution": "【分析】把这条公路的全长看作单位'1'，已经修了 $35\\%$ ，距离中点即离全长的 $50\\%$ 还有 16.2 千米，那么 16.2 千米占全长的 $(50\\% - 35\\%)$ ，单位'1'未知，根据百分数除法的意义解答。\n\n【详解】$16.2 \\div (50\\% - 35\\%)\\\\=16.2 \\div (0. 5 - 0. 3 5) \\\\ = 1 6. 2 \\div 0. 1 5 \\\\ = 1 0 8 \\text {(千 米)} $\n\n答: 这条公路全长 108 千米。",
+    "knowledge_points": [
+      "已知一部分量占总量的百分之几及另一部分量,求总量"
+    ],
+    "source": "23-24六年级上·全国·单元测试",
+    "merged_id": "unit6_173",
+    "original_id": "unit6_solve_173"
+  },
+  {
+    "id": "unit6_solve_174",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "很多人喜欢用玉米油炒菜，而玉米的出油率仅为 $25\\%$ ，要得到35千克玉米油，至少需要多少千克玉米？",
+    "options": [],
+    "answer": "140千克",
+    "solution": "【分析】已知玉米的出油率为 $25 \\%$ , 即玉米油的质量占玉米质量的 $25 \\%$ , 把玉米的质量看作单位 '1', 单位 '1'未知,用玉米油的质量除以 $25 \\%$ , 即可求出玉米的质量。\n\n【详解】$35 \\div 25\\%\\\\= 3 5 \\div 0. 2 5 \\\\ = 1 4 0 (\\text {千 克}) $\n\n答: 至少需要 140 千克玉米。",
+    "knowledge_points": [
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "23-24六年级上·全国·单元测试",
+    "merged_id": "unit6_174",
+    "original_id": "unit6_solve_174"
+  },
+  {
+    "id": "unit6_solve_175",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "植树造林活动中,共植柳树85棵，杨树65棵，有9棵没有成活，这次植树的成活率是多少？",
+    "options": [],
+    "answer": "$94\\%$",
+    "solution": "【分析】成活率表示成活的棵数占种植的总棵数的百分之几，先算出种植的总棵数，再算出成活的棵数，根据成活率 = 成活的棵数 ÷ 种植的总棵数 × 100%，代入数据计算即可，据此解答。\n\n【详解】$(85 + 65 - 9) \\div (85 + 65) \\times 100\\% \\\\= 141 \\div 150 \\times 100 \\% \\\\ = 0.94 \\times 100 \\% \\\\ = 94 \\% $\n\n答: 这次植树的成活率是 $94 \\%$",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题)"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_175",
+    "original_id": "unit6_solve_175"
+  },
+  {
+    "id": "unit6_solve_176",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "某校六年级共有学生195人，从中选出男生的 $15\\%$ 和10名女生参加数学竞赛，剩下的男生人数与女生人数相等，那么，该校六年级参加数学竞赛的学生有多少人？",
+    "options": [],
+    "answer": "25人",
+    "solution": "【分析】设六年级男生有 $x$ 人，则女生有（ $195 - x$ ）人；选出男生的 $15 \\%$ 和 10 名女生后，男生还有（ $1 - 15\\%$ ） $x$ 人，女生还有（ $195 - x - 10$ ）人；根据数量关系：剩下的男女生人数正好相等，据此列方程，解方程求出六年级男生有多少人；最后用男生人数乘 $15 \\%$ 计算参加数学竞赛的男生人数，再加上 10 ，所得结果即为参加数学竞赛的学生人数。\n\n【详解】解：设六年级男生有 $\\mathrm{x}$ 人，则女生有 $(195 - \\mathrm{x})$ 人。\n\n$(1 - 15 \\%) x = 195 - x - 10\\\\85 \\% x = 185 - x\\\\0.8 5 x + x = 1 8 5 - x + x\\\\1.8 5 x = 1 8 5\\\\1.8 5 \\mathrm {x} \\div 1. 8 5 = 1 8 5 \\div 1. 8 5\\\\x = 1 0 0$\n\n$ 100 \\times 15 \\% + 10 \\\\ = 1 5 + 1 0 \\\\ = 2 5 \\text {(人)} $\n\n答: 该校六年级参加数学竞赛的学生有 25 人。",
+    "knowledge_points": [
+      "求一个数的百分之几是多少",
+      "列方程解含两个未知数的问题",
+      "应用等式的性质 1 和 2 解方程"
+    ],
+    "source": "23-24六年级上·全国·单元测试",
+    "merged_id": "unit6_176",
+    "original_id": "unit6_solve_176"
+  },
+  {
+    "id": "unit6_solve_177",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "有浓度为 $7 \\%$ 的盐水 600 克, 为得到浓度为 $10 \\%$ 的盐水, 应向盐水中加入多少克盐?",
+    "options": [],
+    "answer": "20克",
+    "solution": "【分析】盐水浓度 $=$ 盐的质量 $\\div$ 盐水的质量 $\\times 100 \\%$ , 往盐水中加盐时盐和盐水的质量都发生变化, 不变的是水的质量,根据原来的浓度可以算出水的质量, 再用水的质量除以 $(1-10\\% $就可以算出新的盐水总质量，两次盐水总质量作差即可。本题考查百分数, 解答本题的关键是掌握盐水浓度的计算方法。\n\n【详解】水： $600 \\times (1 - 7\\%)\\\\= 600 \\times 93 \\% \\\\ = 5 5 8 \\text {(克)} $\n\n现盐水： $558 \\div (1 - 10\\%)\\\\= 558 \\div 90 \\% \\\\ = 6 2 0 \\text {(克)} $\n\n加盐: $620 - 600 = 20$ (克)答: 应向盐水中加入 20 克盐。",
+    "knowledge_points": [
+      "求一个数的百分之几是多少",
+      "已知一部分量占总量的百分之几及另一部分量，求总量"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_177",
+    "original_id": "unit6_solve_177"
+  },
+  {
+    "id": "unit6_solve_178",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "小明看完一本书用了 30 天, 小红看书的速度比小明快 $20 \\%$ , 小亮看书的速度是小红的 $\\frac{5}{3}$ , 小亮看完这本书需要多少天?",
+    "options": [],
+    "answer": "15天",
+    "solution": "【分析】把这本书的总页数看作单位 '1', 已知小明看完一本书用了 30 天, 用 '1'除以小明看的天数, 求出小明每天看这本书的几分之几, 也就是小明看书的速度;已知小红看书的速度比小明快 $20 \\%$ , 把小明看书的速度看作单位 '1', 则小红看书的速度是小明的 $(1 + 20 \\%)$ , 单位 '1' 已知, 用小明看书的速度乘 $(1 + 20 \\%)$ , 即可求出小红看书的速度;已知小亮看书的速度是小红的 $\\frac{5}{3}$ , 把小红看书的速度看作单位 '1', 单位 '1' 已知, 用小红看书的速度乘 $\\frac{5}{3}$ , 即可求出小亮看书的速度;最后用 '1'除以小亮看书的速度, 即是小亮看完这本书需要的天数。\n\n【详解】小明看书的速度：$1 \\div 3 0 = \\frac {1}{3 0}$\n\n小红看书的速度:$\\frac{1}{30}\\times (1 + 20\\%) \\\\ = \\frac {1}{3 0} \\times (1 + \\frac {1}{5}) \\\\ = \\frac {1}{3 0} \\times \\frac {6}{5} \\\\ = \\frac {1}{2 5} $\n\n小亮看书的速度:$\\frac {1}{2 5} \\times \\frac {5}{3} = \\frac {1}{1 5}$\n\n小亮看的天数:$ 1 \\div \\frac {1}{1 5} \\\\ = 1 \\times 1 5 \\\\ = 1 5 \\text {(天)} $\n\n答: 小亮看完这本书需要 15 天。",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少",
+      "分数与整数的除法",
+      "求一个数的几分之几的问题",
+      "分数乘分数"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_178",
+    "original_id": "unit6_solve_178"
+  },
+  {
+    "id": "unit6_solve_179",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "乘坐火车从 A 城到B城，2016年需要19.5小时，2017年这列火车提速30%，2018年第二次提速25%，2019年第三次提速20%，经过这三次提速后，从A城到B城乘火车需要多少时间？",
+    "options": [],
+    "answer": "10小时",
+    "solution": "【分析】将 A 城到 B 城的总路程看成 '1', 2016 年火车的速度是 $1 \\div 19.5 = \\frac{2}{39}$ ; 2017 年的速度是 2016 年速度的 (1+30%), 2018 年的速度是 2017 年速度的 (1+25%), 2019 年的速度是 2018 年速度的 (1+20%) ; 求比一个数多百分之几的数是多少用乘法计算, 求出 2019 年的速度, 再用总路程 '1'除以 2019 年的速度, 即可求出经过这三次提速后,从 A 城到 B 城乘火车需要多少时间。\n\n【详解】$1 \\div 19.5 \\times (1 + 30\\%) \\times (1 + 25\\%) \\times (1 + 20\\%)\\\\= \\frac {2}{3 9} \\times 1. 3 \\times 1. 2 5 \\times 1. 2 \\\\ = \\frac {1}{1 5} \\times 1. 2 5 \\times 1. 2 \\\\ = \\frac {1}{1 2} \\times 1. 2 \\\\ = \\frac {1}{1 0} $\n\n$ 1 \\div \\frac {1}{1 0} \\\\ = 1 \\times 1 0 \\\\ = 1 0 \\text {(小 时)} $\n\n答: 经过三次提速后, 需要 10 小时。",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少",
+      "含百分数的运算"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_179",
+    "original_id": "unit6_solve_179"
+  },
+  {
+    "id": "unit6_solve_180",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "某商品按定价卖出可获利润100元，如果按定价的 $90\\%$ 出售，则亏损20元。则商品的原价是多少元？",
+    "options": [],
+    "answer": "1100元",
+    "solution": "【分析】根据题意可得到关系式：商品原价 $+100 =$ 定价，\n\n原价-定价 $\\times 90 \\% = 20$ ; 可设商品原价为未知数 $x$ , 根据关系式可列出方程, 可得出答案。\n\n【详解】解：设商品原价为 $x$ 元，可列出方程：\n\n$ x - (100 + x) \\times 90 \\% = 20 \\\\ x - (1 0 0 \\times 0. 9 + 0. 9 x) = 2 0 \\\\ x - (9 0 + 0. 9 x) = 2 0 \\\\ x - 9 0 - 0. 9 x = 2 0 \\\\ 0. 1 x - 9 0 + 9 0 = 2 0 + 9 0 \\\\ 0. 1 x = 1 1 0 \\\\ 0. 1 x \\div 0. 1 = 1 1 0 \\div 0. 1 \\\\ x = 1 1 0 0 $\n\n答: 商品原价是 1100 元。",
+    "knowledge_points": [
+      "已知一个数的百分之几是多少，求这个数",
+      "求一个数的百分之几是多少",
+      "列方程解含一个未知数的问题"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_180",
+    "original_id": "unit6_solve_180"
+  },
+  {
+    "id": "unit6_solve_181",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一台收录机如果按原售价打九折出售可获利70元，如果按原售价打九五折出售可获利100元，这台收录机原价是多少元？",
+    "options": [],
+    "answer": "600元",
+    "solution": "【分析】把这台收录机的原售价看作单位'1'。如果按原售价打九折出售可获利70元，即原售价 $\\times 90\\%$ —成本价 $= 70$ 元；如果按原售价打九五折出售可获利100元，即原售价 $\\times 95\\%$ —成本价 $= 100$ 元；那么获利相差（ $100 - 70$ ）元是原售价的（ $95\\% - 90\\%$ ），单位'1'未知，根据百分数除法的意义解答，求出这台收录机的原售价。\n\n【详解】$(100 - 70) \\div (95\\% - 90\\%)\\\\=30 \\div 0. 0 5 \\\\ = 6 0 0 \\text {(元)} $\n\n答: 这台收录机原价是 600 元。",
+    "knowledge_points": [
+      "求原价（折扣问题）",
+      "已知一个数的百分之几是多少，求这个数",
+      "经济问题"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_181",
+    "original_id": "unit6_solve_181"
+  },
+  {
+    "id": "unit6_solve_182",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "解答题",
+    "difficulty": "困难",
+    "question": "一件商品出售，如果按原价降低一成，仍可盈利200元；如果降价二成，则亏损220元。这件商品原价多少元？",
+    "options": [],
+    "answer": "4200元",
+    "solution": "【分析】把这件商品的原价看成单位'1'。降低一成，仍可盈利200元，即现价比原价降低 $10\\%$ ，则现价是原价的 $90\\%$ ，可得出：原价 $\\times 90\\% -$ 成本 $= 200$ 元；如果降价二成，则亏损220元，即现价比原价降低 $20\\%$ ，则现价是原价的 $80\\%$ ，可得出：成本一原价 $\\times 80\\% = 220$ 元；那么盈利 200 元与亏损 220 元相差的（ $200 + 220$ ）元占原价的（ $90\\% - 80\\%$ ），单位'1'未知，根据百分数除法的意义解答，求出这件商品的原价。掌握原价、现价、成本、盈利、亏损之间的关系，分析出盈利与亏损相差的金额占原价的百分之几是解题的关键。\n\n【详解】一成 $= 10\\%$ ，二成 $= 20\\%$\n\n$(1 - 10 \\%) - (1 - 20 \\%)\\\\=90 \\% -80 \\% \\\\ = 10 \\% $\n\n$(200 + 220) \\div 10 \\% \\\\ = 4 2 0 \\div 0. 1 \\\\ = 4 2 0 0 (\\text {元}) $\n\n答：这件商品原价4200元。",
+    "knowledge_points": [
+      "已知一个数的百分之几是多少，求这个数",
+      "根据成数反求单位'1'",
+      "经济问题"
+    ],
+    "source": "23-24六年级上·全国·单元测试",
+    "merged_id": "unit6_182",
+    "original_id": "unit6_solve_182"
+  },
+  {
+    "id": "unit6_solve_183",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "解答题",
+    "difficulty": "困难",
+    "question": "体育组有一筐球,其中足球占 $45 \\%$ , 如果再放入 5 个篮球, 足球就只占$36\\%$,那么，这筐球中，足球有多少个？",
+    "options": [],
+    "answer": "9 个",
+    "solution": "【分析】放入篮球后篮球数量与球的总数都发生变化，我们可以把不变量：足球数量看作单位'1'，原来足球占 $45\\%$ ，则篮球占 $55\\%$ ，所以原来的篮球数量就是足球数量的 $\\frac{11}{9}$ ，同理可知放入 5 个篮球后的篮球数量是足球数量的 $\\frac{16}{9}$ 那么 5 个篮球就对应了 $\\frac{16}{9} - \\frac{11}{9} = \\frac{5}{9}$ ，用 9 除以对应分率就能算出足球数量。据此解答即可。解答此题的关键是根据题意，判断出足球数量不变，并把足球看成单位'1'，再求出5个篮球所对应的分率，列式解答即可。\n\n【详解】$(1 - 45\\%) \\div 45\\%\\\\=0.5 5 \\div 0. 4 5 \\\\ = \\frac {1 1}{9}$\n\n $(1 - 36 \\%) \\div 36 \\% \\\\ = 0. 6 4 \\div 0. 3 6 \\\\ = \\frac {1 6}{9} $\n\n足球: $5 \\div (\\frac{16}{9} - \\frac{11}{9})\\\\= 5 \\div \\frac {5}{9}\\\\= 5 \\times \\frac {9}{5} \\\\ = 9 (\\text {个}) $\n\n答: 足球有 9 个。",
+    "knowledge_points": [
+      "已知一个数的几分之几是多少，求这个数",
+      "含百分数的运算"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_183",
+    "original_id": "unit6_solve_183"
+  },
+  {
+    "id": "unit6_solve_184",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "解答题",
+    "difficulty": "困难",
+    "question": "书店里有男生和女生共 90 人, 其中女生占 $60 \\%$ , 走了一部分女生后, 女生占现在总数的 $28 \\%$ , 走了多少个女生?",
+    "options": [],
+    "answer": "40 个",
+    "solution": "【分析】根据'男生和女生共90人，其中女生占 $60\\%$ '，可以列式 $90 - 90 \\times 60\\%$ 得出男生人数。再根据'走了一部分女生后，女生占现在总数的 $28 \\%$ , 得出男生的人数占走一部分女生后现在人数的 $(1 - 28 \\%)$ , 由此求出现在人数, 再用原来的总人数减去现在人数即可求出走了多少个女生。据此解答。解答此题的关键是根据题意，判断出男生人数不变，再找出对应量，列式解答即可。\n\n【详解】$(90 - 90 \\times 60\\%) \\div (1 - 28\\%)\\\\= (90 - 54) \\div 72 \\% \\\\ = 3 6 \\div 0. 7 2 \\\\ = 5 0 (\\text {个}) $\n\n$9 0 - 5 0 = 4 0 (\\text {个})$ 答: 走了 40 个女生。",
+    "knowledge_points": [
+      "求一个数的百分之几是多少",
+      "已知一部分量占总量的百分之几及另一部分量，求总量"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_184",
+    "original_id": "unit6_solve_184"
+  },
+  {
+    "id": "unit6_solve_185",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "体育用品商店有篮球和排球共45个，其中篮球占 $60\\%$ ，当卖出一批篮球后，篮球占现存总数的 $25\\%$ ，卖出的篮球多少个？",
+    "options": [],
+    "answer": "21个",
+    "solution": "【分析】排球的数量是不变的，求一个数的百分之几是多少用乘法计算，则用 $45 \\times 60\\%$ 即可求出原来篮球的数量，用45减去篮球的数量就是排球的数量；排球的数量还是剩余总数的（ $1 - 25\\%$ ），已知一个数的百分之几是多少求这个数用除法计算，则用排球的数量除以（ $1 - 25\\%$ ）即可求出剩余的总数，用原来的总数45减去现在剩余的总数即可求出卖出多少个篮球。\n\n【详解】原来的篮球数量：\n\n$ 45 \\times 60 \\% \\\\ = 4 5 \\times 0. 6 \\\\ = 2 7 \\text {(个)} $\n\n原来的排球数量： 45-27=18（个）\n\n$45 - 27 = 18$ （个）\n\n卖出一批篮球后剩余的总球数:\n\n$ 18 \\div (1 - 25 \\%) \\\\ = 18 \\div 75 \\% \\\\ = 1 8 \\div 0. 7 5 \\\\ = 2 4 \\text {(个)} $\n\n卖出的篮球数量: $45 - 24 = 21$ （个）答: 卖出篮球 21 个。",
+    "knowledge_points": [
+      "求一个数的百分之几是多少",
+      "已知一部分量占总量的百分之几及另一部分量，求总量"
+    ],
+    "source": "23-24六年级上·全国·单元测试",
+    "merged_id": "unit6_185",
+    "original_id": "unit6_solve_185"
+  },
+  {
+    "id": "unit6_solve_186",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "体育用品商店有篮球和排球共50个，其中篮球占 $60\\%$ ，当卖出一批篮球后，篮球占现存总数的 $37.5\\%$ ，卖出了多少个篮球？",
+    "options": [],
+    "answer": "18 个",
+    "solution": "【分析】先把原来的总数 50 个看作单位 '1', 算出排球的数量, 卖出一批篮球后, 把新的总数看作单位 '1', 此时排球数量不变, 用排球数量除以其现在所对应的分率, 也就是 $1 - 37.5 \\%$ , 求出现在的总数, 两次总数相减就是卖出的篮球数量。据此解答即可。解答此题的关键是根据题意，判断出排球数量不变, 再找出对应量, 列式解答即可。\n\n【详解】排球： $50 \\times (1 - 60\\%)\\\\= 5 0 \\times 0. 4 \\\\ = 2 0 \\text {(个)} $\n\n现总： $20 \\div (1 - 37.5\\%)\\\\=2 0 \\div 0. 6 2 5 \\\\ = 3 2 (\\text {个}) $\n\n卖出篮球： $50 - 32 = 18$ （个）答: 卖出了 18 个篮球。",
+    "knowledge_points": [
+      "已知一个数的百分之几是多少，求这个数",
+      "求一个数的百分之几是多少"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_186",
+    "original_id": "unit6_solve_186"
+  },
+  {
+    "id": "unit6_solve_187",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "学校去年春季栽树500 棵，成活率 $85 \\%$ , 去年秋季的成活率为 $90 \\%$ 。已知去年春季比秋季多活了20棵数，去年秋季学校共种了多少棵树？",
+    "options": [],
+    "answer": "450 棵",
+    "solution": "【分析】学校去年春季栽树 500 棵, 成活率 $85 \\%$ , 用去年春季栽树乘成活率, 求出去年春季成活棵数; 去年春季成活棵树减去 20 棵, 求出去年秋季成活棵数, 再用去年秋季的成活棵数除以去年秋季的成活率, 求出去年秋季学校共种了多少棵树即可。本题考查百分数，解答本题的关键是掌握成活率 $=$ 成活棵数 $\\div$ 栽树棵数 $\\times 100 \\%$ 。\n\n【详解】去年春季成活棵数： $500 \\times 85\\% = 425$ （棵）去年秋季成活棵数： $425 - 20 = 405$ （棵）去年秋季学校栽树： $405 \\div 90\\% = 450$ （棵）答: 去年秋季学校共种了 450 棵树。",
+    "knowledge_points": [
+      "已知一个数的百分之几是多少，求这个数",
+      "求一个数的百分之几是多少"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_187",
+    "original_id": "unit6_solve_187"
+  },
+  {
+    "id": "unit6_solve_188",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "解答题",
+    "difficulty": "困难",
+    "question": "陈大爷第一次植树200 棵，成活率为 $85\\%$ ；第二次植树成活率为 $90\\%$ 。第一次植树比第二次植树多死了8棵。第二次植树多少棵？",
+    "options": [],
+    "answer": "220 棵",
+    "solution": "【分析】陈大爷第一次植树 200 棵, 成活率为 $85 \\%$ , 则死亡的棵数是第一次植树的 $(1 - 85 \\%)$ , 用第一次植树棵数乘死亡棵数占第一次植树棵数的分率, 求出第一次植树的死亡棵数;用第一次植树的死亡棵数减去 8 棵, 求出第二次植树死亡棵数, 用第二次植树死亡棵数除以第二次植树死亡棵数占第二次植树棵数的分率, 求出第二次植树棵数。本题考查百分数, 解答本题的关键是掌握题中的数量关系。\n\n【详解】第二次死亡棵数： $200 \\times (1 - 85\\%) - 8\\\\= 2 0 0 \\times 0. 1 5 - 8\\\\=3 0 - 8\\\\=22 (棵)$\n\n第二次植树棵数： $22 \\div (1 - 90\\%)\\\\= 2 2 \\div 0. 1\\\\=220 (棵)$\n\n答: 第二次植树 220 棵。",
+    "knowledge_points": [
+      "求一个数的百分之几是多少",
+      "已知一部分量占总量的百分之几及另一部分量，求总量"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_188",
+    "original_id": "unit6_solve_188"
+  },
+  {
+    "id": "unit6_solve_189",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "东乡去年春季植树450 棵，成活率为 $80 \\%$ , 去年秋季植树的成活率为 $90 \\%$ , 已知去年春季比秋季多死了 18 棵, 去年一共种活了多少棵树?",
+    "options": [],
+    "answer": "1008 棵",
+    "solution": "【分析】根据春季的植树数量与成活率可以算出春季种活的树的数量，从总数中减去成活的树等于春季死掉的树的数量，再减去18算出秋季死去的树的数量，这个数量对应的分率是 $1 - 90\\%$ 即 $10\\%$ ，用秋季树死去的数量除以 $10\\%$ 算出秋季植树总量再乘 $90\\%$ 算出成活数量，最后将春秋成活数量相加即可。\n\n【详解】春季种活：\n\n$450 \\times 80 \\% \\\\ = 4 5 0 \\times 0. 8 \\\\ = 3 6 0 \\text {(棵)} $\n\n春季没种活:\n\n$4 5 0 - 3 6 0 = 9 0 \\text {(棵)}$\n\n秋季没种活：\n\n$90 - 18 = 72$ (棵)\n\n秋季种活：\n\n$ 72 \\div (1 - 90 \\%) \\times 90 \\% \\\\ = 7 2 \\div 0. 1 \\times 0. 9 \\\\ = 7 2 0 \\times 0. 9 \\\\ = 6 4 8 \\text {(棵)} $\n\n共种活: 360+648=1008(棵)答: 去年一共种活了 1008 棵。\n\n$360 + 648 = 1008$ （棵）",
+    "knowledge_points": [
+      "已知一个数的百分之几是多少，求这个数",
+      "求一个数的百分之几是多少"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_189",
+    "original_id": "unit6_solve_189"
+  },
+  {
+    "id": "unit6_solve_190",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "《世界奥数专刊》的标价是 40 元/本，大赛组委会决定减价 $10\\%$ 出售给参加'世奥赛'的学生，但打了折扣后需附加 $5 \\%$ 作为税金。那么，参赛学生购买专刊实际需要多少元？",
+    "options": [],
+    "answer": "37.8元",
+    "solution": "【分析】已知标价是 40 元/本，减价 $10 \\%$ 出售，把原价看作单位 '1'，则售价是原价的 $(1 - 10\\%)$ ，单位 '1'已知，用原价乘 $(1 - 10\\%)$ ，求出售价；但打了折扣后需附加 $5 \\%$ 作为税金，即需交售价的 $5 \\%$ 作为税金，根据求一个数的百分之几是多少，用售价乘 $5 \\%$ , 即可求出税金, 再加上售价, 就是实际购买需付的钱数。\n\n【详解】售价：\n\n$ 40 \\times (1 - 10 \\%) \\\\ = 4 0 \\times (1 - 0. 1) \\\\ = 4 0 \\times 0. 9 \\\\ = 3 6 \\text {(元)} $\n\n税金：\n\n$36 \\times 5 \\%\\\\=3 6 \\times 0. 0 5 \\\\ = 1. 8 \\text {(元)} $\n\n实际应付：\n\n$3 6 + 1. 8 = 3 7. 8 \\text {(元)}$答: 参赛学生购买专刊实际需要 37.8 元。",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少",
+      "求一个数的百分之几是多少"
+    ],
+    "source": "23-24六年级上·全国·单元测试",
+    "merged_id": "unit6_190",
+    "original_id": "unit6_solve_190"
+  },
+  {
+    "id": "unit6_solve_191",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "甲、乙两家商店出售同一款兔宝宝玩具，每只原售价都是25元，为了促销，甲店先提价 $10 \\%$ , 再降价 $20 \\%$ ; 乙店则直接降价 $10 \\%$ 。那么，调价后对于这款兔宝宝玩具，哪家店的售价更便宜，便宜多少元？",
+    "options": [],
+    "answer": "甲；0.5元",
+    "solution": "【分析】甲商店：先提价 $10\\%$ ，是把原价看作单位'1'，则提价后的价格是原价的 $(1 + 10\\%)$ , 单位 '1' 已知, 用原价乘 $(1 + 10\\%)$ , 求出提价后的价格;再降价 $20 \\%$ , 是把提价后的价格看作单位 '1', 则降价后的价格是提价后的（1一20%），单位'1'已知，用提价后的价格乘（1一20%），即可求出甲商店的现价；乙商店：降价 $10 \\%$ , 把原价看作单位 '1', 则降价后的价格是原价的（1一10%），单位'1'已知，用原价乘（1一10%），即可求出乙商店的现价；最后比较两家商店的现价，得出哪家商店更便宜，再用减法求出便宜的钱数。\n\n【详解】甲商店：\n\n$ 25 \\times (1 + 10 \\%) \\times (1 - 20 \\%) \\\\ = 2 5 \\times 1. 1 \\times 0. 8 \\\\ = 2 2 \\text {(元)} $\n\n乙商店:\n\n$25 \\times (1 - 10 \\%)\\\\=2 5 \\times 0. 9 \\\\ = 2 2. 5 \\text {(元)} $\n\n$22< 22.5$ , 甲商店便宜;便宜了: $22.5 - 22 = 0.5$ (元)答：甲商店的售价更便宜，便宜0.5元。",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "23-24 六年级上·全国·单元测试",
+    "merged_id": "unit6_191",
+    "original_id": "unit6_solve_191"
+  },
+  {
+    "id": "unit6_solve_192",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "解答题",
+    "difficulty": "困难",
+    "question": "一辆汽车以每小时45千米的速度从甲地开往乙地，行了全程的 $20\\%$ 后，又行了1小时，这时未行路程与已行路程的比是 $3:1$ ，甲、乙两地相距多少千米？",
+    "options": [],
+    "answer": "900千米",
+    "solution": "【分析】将全程看作单位'1'，开始行了全程的 $20\\%$ ，又行了 1 小时, 即行驶了 45 千米, 这时未行路程与已行路程的比是 $3: 1$ , 则现在已行驶路程是全程的 $\\frac{1}{3+1}$ , 这 1 小时行了全程的 $(\\frac{1}{3+1} -20 \\%)$ ，这 1 小时行驶路程÷对应分率或百分率 = 全程, 据此列式解答。关键是确定单位 '1', 根据比的意义, 确定现在已行驶路程的对应分率或百分率, 从而确定已知行驶路程的对应分率或百分率, 根据部分数量 $\\div$ 对应分率或百分率 = 整体数量, 进行解答。\n\n【详解】$45 \\div (\\frac{1}{3 + 1} -20\\%)$\n\n$4 5 \\div (\\frac {1}{4} - 0. 2) \\\\ = 4 5 \\div (0. 2 5 - 0. 2) \\\\ = 4 5 \\div 0. 0 5 \\\\ = 9 0 0 \\text {(千 米)} $\n\n答: 甲、乙两地相距 900 千米。",
+    "knowledge_points": [
+      "已知一个数的百分之几是多少，求这个数",
+      "已知一个数的几分之几是多少，求这个数",
+      "比的意义",
+      "比的应用"
+    ],
+    "source": "23-24六年级上·海南·期末",
+    "merged_id": "unit6_192",
+    "original_id": "unit6_solve_192"
+  },
+  {
+    "id": "unit6_solve_193",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "阅读下列信息回答问题。\n\n《山东省中小学校园园林绿化管理办法》中规定：新建学校绿地率不得低于 $35\\%$ ，绿化覆盖率应该在 $50\\%$ 及以上。山东省小学规范化学校标准中对绿化方面的要求：绿地面积每块不小于50平方米，学生人均绿化用地不少于0.5平方米。\n\n某小学的校园是长方形，长200米，宽150米，目前在校生有1500人。该学校的绿地面积是12000平方米，学校绿化覆盖面积为18000平方米。\n\n绿地率 $= \\frac{\\text{绿地面积}}{\\text{用地总面积}} \\times 100\\%$ 绿化覆盖率 $= \\frac{\\text{绿化覆盖面积}}{\\text{用地总面积}}$ $x100\\%$ 人均绿化用地面积 $= \\frac{\\text{绿地面积}}{\\text{总人数}}$\n\n请同学们计算一下该学校的绿地率、绿地覆盖率和学生人均绿化用地面积是否符合要求？",
+    "options": [],
+    "answer": "该学校的绿地率、绿地覆盖率和学生人均绿化用地面积均符合要求",
+    "solution": "【分析】根据长方形的面积 $= \\text{长} \\times$ 宽，代入数据计算校园的用地总面积；再根据绿地率 $= \\frac{\\text{绿地面积}}{\\text{用地总面积}} \\times 100\\%$ 、绿化覆盖率 $= \\frac{\\text{绿化覆盖面积}}{\\text{用地总面积}} \\times 100\\%$ 、人均绿化用地面积 $=$ 绿地面积 总人数，代入数据计算绿地率、绿化覆盖率和学生人均绿化用地面积；最后根据规定判断各项数据是否符合要求。\n\n【详解】$200 \\times 150 = 30000$ （平方米）\n\n$12000 \\div 30000 \\times 100 \\% \\\\ = 0.4 \\times 100 \\% \\\\ = 40 \\% $\n\n$18000 \\div 30000 \\times 100 \\% \\\\ = 0.6 \\times 100 \\% \\\\ = 60 \\% $\n\n$12000 \\div 1500 = 8$ （平方米）40%>35% 60%>50% 8平方米 $>0.5$ 平方米 答：该学校的绿地率、绿地覆盖率和学生人均绿化用地面积均符合要求。",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "22-23 六年级下·山东德州·期末",
+    "merged_id": "unit6_193",
+    "original_id": "unit6_solve_193"
+  },
+  {
+    "id": "unit6_solve_194",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "有两盘葡萄, 每盘都是 42 颗, 从甲盘中拿出 6 颗放进乙盘后, 甲盘葡萄数是乙盘的几分之几?",
+    "options": [],
+    "answer": "$\\frac{3}{4}$",
+    "solution": "【分析】每盘原有颗数 $-6 =$ 甲盘现在颗数, 每盘原有颗数 $+6 =$ 乙盘现在颗数, 甲盘现在颗数 $\\div$ 乙盘现在颗数 = 现在甲盘葡萄数是乙盘的几分之几, 根据分数与除法的关系表示出结果, 约分即可。分数的分子相当于被除数, 分母相当于除数。\n\n【详解】$(42 - 6) \\div (42 + 6)\\\\=3 6 \\div 4 8 \\\\ =\\frac {3 6}{4 8} \\\\ = \\frac {3}{4} $\n\n答: 甲盘葡萄数是乙盘的 $\\frac{3}{4}$ 。",
+    "knowledge_points": [
+      "约分的认识及应用",
+      "求一个数占另一个数几分之几"
+    ],
+    "source": "23-24 五年级下·四川·单元测试",
+    "merged_id": "unit6_194",
+    "original_id": "unit6_solve_194"
+  },
+  {
+    "id": "unit6_solve_195",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "'新世界七大奇迹'之一的港珠澳大桥，于2018年10月24日上午9点正式通行。从此，珠海、香港、澳门三地口岸实现了24小时通车和通关。\n\n(1) 港珠澳大桥小型客车的收费标准是每辆车每次 150 元,比大型客车少 $\\frac{1}{4}$ , 大型客车每辆车每次收费多少元? \n(2) 港珠澳大桥总长 55 千米, 一辆巴士从珠海口岸前往香港口岸，先行驶了全程的 $\\frac{1}{4}$ ，又行驶了全程的 $40\\%$ 。这辆巴士距离香港口岸还有多远？",
+    "options": [],
+    "answer": "(1) 200 元 (2) 19.25 千米",
+    "solution": "【分析】（1）把大型客车每次收费的钱数看作单位'1'，已知小型客车的收费比大型客车少 $\\frac{1}{4}$ ，即小型客车收费是大型客车的收费的 $(1 - \\frac{1}{4})$ ，求单位'1'，用小型客车收费除以 $(1 - \\frac{1}{4})$ 即可解答；(2) 把港珠澳大桥总长看作单位 '1', 先行驶了全程的 $\\frac{1}{4}$ ,又行驶了全程的 $40 \\%$ , 即还剩全程的 $(1 - \\frac{1}{4} - 40 \\%)$ , 再用港珠澳大桥总长乘剩下路程的分率即可解答。\n\n【详解】 (1) $150 \\div (1 - \\frac{1}{4})\\\\=1 5 0 \\div \\frac {3}{4} \\\\ = 1 5 0 \\times \\frac {4}{3} \\\\ = 2 0 0 \\text {(元)} $\n\n答: 大型客车每辆车每次收费 200 元。\n\n(2) $55 \\times (1 - \\frac{1}{4} - 40\\%)\\\\= 55 \\times (\\frac{3}{4} -40\\%) \\\\ = 5 5 \\times 0. 3 5 \\\\ = 1 9. 2 5 \\text {(千 米)} $\n\n答: 这辆巴士距离香港口岸还有 19.25 千米。",
+    "knowledge_points": [
+      "已知比一个数多/少几分之几是多少，求这个数",
+      "分数的四则混合运算",
+      "含百分数的运算"
+    ],
+    "source": "23-24六年级上·河南驻马店·期末",
+    "merged_id": "unit6_195",
+    "original_id": "unit6_solve_195"
+  },
+  {
+    "id": "unit6_solve_196",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "某校六年级有 140名师生去参观博物馆，有两种车可供选择：(1) 限乘 40 人的大客车, 每人票价 5 元, 若坐满, 票价是原价的 $\\frac{4}{5}$ ; (2) 限乘 10 人的面包车, 每人票价 6 元, 若坐满, 票价可按 $75\\%$ 优惠。请你根据以上信息为六年级师生设计一种最省钱的租车方案，并算出总租金。",
+    "options": [],
+    "answer": "租3辆大客车，2辆面包车；570元",
+    "solution": "【分析】（1）如果选择大客车，需要大客车： $140 \\div 40 \\approx 4$ （辆），求出租大客车需要的钱数；(2) 如果选择面包车, 需要面包车: $140 \\div 10 = 14$ (辆),求出租面包车需要的钱数; (3) 选择租大客车和面包车, 用 $140 \\div 40$ , 求出租大客车的辆数, 剩下的人数租面包车, 求出租大客车的钱数; 求出租面包车的钱数, 再相加, 求出租大客车和面包车的钱数,再进行比较, 即可解答。\n\n【详解】（1）如果全部租大客车：$140 \\div 40 = 3$ （辆）……20（人），需要租4辆大客车；其中\n\n有 3 辆票价按原价的 $\\frac{4}{5}$ 。\n\n$ 4 0 \\times 3 \\times 5 \\times \\frac {4}{5} + 2 0 \\times 5 \\\\ = 1 2 0 \\times 5 \\times \\frac {4}{5} + 1 0 0 \\\\ = 6 0 0 \\times \\frac {4}{5} + 1 0 0 \\\\ = 4 8 0 + 1 0 0 \\\\ = 5 8 0 \\text {(元)} $\n\n(2) 如果全租面包车:\n\n$1 4 0 \\div 1 0 = 1 4 (\\text {辆}) \\\\ 140 \\times 6 \\times 75 \\% \\\\ = 840 \\times 75 \\% \\\\ = 6 3 0 (\\text {元}) $\n\n(3) $140 \\div 40 = 3$ （辆）……20（人） $2 0 \\div 1 0 = 2 \\text {(辆)}$ 租3辆大客车，2辆面包车。\n\n$ 40 \\times 3 \\times 5 \\times \\frac{4}{5} + 20 \\times 6 \\times 75 \\% \\\\ = 120 \\times 5 \\times \\frac{4}{5} + 120 \\times 75 \\% = 6 0 0 \\times \\frac {4}{5} + 9 0 \\\\ = 4 8 0 + 9 0 \\\\ = 5 7 0 \\text {(元)} $\n\n$570 < 580 < 630$ , 租 3 辆大客车和 2 辆面包车最省钱。答: 租 3 辆大客车和 2 辆面包车最省钱, 租金是 570 元。",
+    "knowledge_points": [
+      "购票问题",
+      "求一个数的几分之几的问题",
+      "求一个数的百分之几是多少"
+    ],
+    "source": "22-23六年级上·江西吉安·期末",
+    "merged_id": "unit6_196",
+    "original_id": "unit6_solve_196"
+  },
+  {
+    "id": "unit6_solve_197",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "同学们在围棋社团学习围棋，磨炼自己的毅力。围棋老师为了提升同学们的棋艺，准备在网上购买一些相关书籍，刚好赶上店铺做优惠活动，'满300元优惠 $10\\%$ '，最后付了360元。围棋老师购买的这些书籍的原价一共是多少元？（请你列方程解答）",
+    "options": [],
+    "answer": "400元",
+    "solution": "【分析】设这些书籍的原价一共是 $x$ 元, 最后付了 360 元可知, 已参加了店铺的优惠活动, 原价 $\\times (1 - 10 \\%) =$ 现价,据此解答。\n\n【详解】解：设这些书籍的原价一共是 $x$ 元，\n\n$(1 - 10 \\%) x = 360\\\\90 \\% \\mathrm {x} = 360 \\\\ 0. 9 \\mathrm {x} = 3 6 0 \\\\ x = 3 6 0 \\div 0. 9 \\\\ x = 4 0 0 $\n\n答: 围棋老师购买的这些书籍的原价一共是 400 元。",
+    "knowledge_points": [
+      "已知比一个数多/少百分之几是多少，求这个数",
+      "列方程解含一个未知数的问题"
+    ],
+    "source": "22-23六年级上·河南郑州·期末",
+    "merged_id": "unit6_197",
+    "original_id": "unit6_solve_197"
+  },
+  {
+    "id": "unit6_solve_198",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "同学们在国学社团里诵读国学经典，传承国学文化精髓。喜欢读《三字经》的有 24 人, 喜欢读《百家姓》的人数比喜欢读《三字经》的人数多 $50 \\%$ , 喜欢读《百家姓》的有多少人?",
+    "options": [],
+    "answer": "36人",
+    "solution": "【分析】已知喜欢读《百家姓》的人数比喜欢读《三字经》的人数多 $50 \\%$ , 把喜欢读《三字经》的人数看作单位 '1', 喜欢读《百家姓》的人数是喜欢读《三字经》的 (1 +50%), 又已知喜欢读《三字经》的有 24 人, 根据百分数乘法的意义, 用 $24 \\times (1 + 50 \\%)$ 即可求出喜欢读《百家姓》的人数。\n\n【详解】$24 \\times (1 + 50\\%)\\\\ 2 4 \\times 1. 5 \\\\ = 3 6 (\\text {人}) $\n\n答: 喜欢读《百家姓》的有 36 人。",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "22-23 六年级上·河南郑州·期末",
+    "merged_id": "unit6_198",
+    "original_id": "unit6_solve_198"
+  },
+  {
+    "id": "unit6_solve_199",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "李老师使用笔记本电脑时，他突然听到电量不足的声音，并看到了'剩余电量 $10\\%$ ，还能使用24分'的提示。照这样计算，电池满电时，这个笔记本电脑能工作多长时间？",
+    "options": [],
+    "answer": "240 分钟",
+    "solution": "【分析】将笔记本电脑工作总时间看作单位 '1', 剩余使用时间 $\\div$ 对应百分率 $=$ 笔记本电脑工作总时间, 据此列式解答。\n\n【详解】$24 \\div 10\\% = 24 \\div 0.1 = 240$ （分钟）答: 电池满电时, 这个笔记本电脑能工作 240 分钟。",
+    "knowledge_points": [
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "23-24六年级上·甘肃定西·期中",
+    "merged_id": "unit6_199",
+    "original_id": "unit6_solve_199"
+  },
+  {
+    "id": "unit6_solve_200",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数",
+    "question_type": "解答题",
+    "difficulty": "困难",
+    "question": "已知某种商品每件定价为10元，网购这种商品的数量不满100件，则按定价付款，另外每件还要加付定价的 $10\\%$ 作为快递费；网购的数量达到或超过100件，则每件按定价的九折付款，而且免付快递费。某公司两次共网购这种商品200件，其中第一次的数量不满100件（第二次超过），两次网购总计付款1960元，求第一次、第二次分别网购多少件？",
+    "options": [],
+    "answer": "80 件; 120 件",
+    "solution": "【分析】第一次网购的每件定价看作单位'1'，算上快递费实际价格应为单位'1'的（ $1 + 10\\%$ ）；第二次网购件数 = 200 件 - 第一次网购件数，第二次网购的每件实际价格 = 每件定价 $\\times 90\\%$ 。根据数量 $\\times$ 单价 $=$ 总价, 分别求出两次网购所付金额, 再根据等量关系: 第一次网购付款金额 + 第二次网购付款金额 =1960 元, 据此列方程, 解方程。本题可以看作分段计费的问题，关键是找到两次购物的单价和数量，根据'单价、数量、总价'之间的关系，分别求出两次网购的所付金额。才能进一步根据等量关系列方程、解方程。\n\n【详解】解：设第一次网购 $x$ 件，则第二次网购 $(200 - x)$ 件。\n\n$(1 + 10 \\%) \\times 10 x + 10 \\times 90 \\% \\times (200 - x) = 1960\\\\1 1 x + 1 8 0 0 - 9 x = 1 9 6 0\\\\2 x = 1 9 6 0 - 1 8 0 0\\\\2 x \\div 2 = 1 6 0 \\div 2\\\\x = 8$\n\n$200 - 80 = 120$ （件）答：第一次网购80件，第二次网购120件。",
+    "knowledge_points": [
+      "列方程解含两个未知数的问题",
+      "含百分数的运算"
+    ],
+    "source": "2024六年级下·全国·专题练习",
+    "merged_id": "unit6_200",
+    "original_id": "unit6_solve_200"
+  }, {
+    "id": "unit6_choice_201",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "同学们种花, 活了 100 棵, 有 5 棵没活, 成活率约是 ( )。(保留一位小数)",
+    "options": [
+      "A. 100%",
+      "B. 95%",
+      "C. 95.2%"
+    ],
+    "answer": "C",
+    "solution": "【分析】成活率 = 成活的棵数 ÷ 种的总棵数 × 100%, 由题干可知, 成活的棵数是 100, 栽种的棵数是 100 + 5 = 105 棵,代入公式计算即可。此题属于百分率问题, 解答时注意明确栽种的总棵数。\n【详解】100 ÷ (100 + 5) × 100% = 100 ÷ 105 × 100% ≈ 0.952 × 100% = 95.2%。成活率是 95.2%",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）",
+      "小数的近似数和改写"
+    ],
+    "source": "21-22 六年级上·吉林白山·期末",
+    "merged_id": "unit6_201",
+    "original_id": "unit6_choice_201"
+  },
+  {
+    "id": "unit6_choice_202",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "在含盐率为 20% 的盐水中，盐和水各加入 20 克，这时的含盐率同原来相比（）。",
+    "options": [
+      "A. 增加了",
+      "B. 减少了",
+      "C. 不变"
+    ],
+    "answer": "A",
+    "solution": "【分析】含盐率 = 盐的质量 / (盐的质量 + 水的质量) × 100%，据此先求出新加入的盐水的含盐率，即 20 ÷ (20 + 20) × 100%；再把新加入的盐水的含盐率与原来盐水的含盐率作比较，若新加入的盐水的含盐率比原来盐水的含盐率高，则混合后的盐水的含盐率比原来的盐水的含盐率高。求各种百分率的实质就是求一个数是另一个数的百分之几，只是在计算时要乘 100%，把结果化成百分数。\n【详解】20 ÷ (20 + 20) × 100% = 20 ÷ 40 × 100% = 0.5 × 100% = 50%。50% > 20% 所以这时的含盐率比原来增加了。",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "22-23六年级下·河北保定·期末",
+    "merged_id": "unit6_202",
+    "original_id": "unit6_choice_202"
+  },
+  {
+    "id": "unit6_choice_203",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "困难",
+    "question": "万达商场以 100 元的价格卖出两套不同的服装，结果一套赚 20%，一套亏本 20%，那么对于商场老板整体是（）。",
+    "options": [
+      "A. 亏本",
+      "B. 赚钱",
+      "C. 不亏也不赚",
+      "D. 无法确定"
+    ],
+    "answer": "A",
+    "solution": "【分析】根据题意，商场以 100 元的价格卖出两套不同的服装，即两套服装的售价都是 100 元；把第一套服装的原价看作单位'1', 第一套赚了 20%, 即售价比原价高 20%, 那么售价是原价的 (1 + 20%), 单位'1'未知, 用除法求出第一套的原价;把第二套服装的原价看作单位'1', 第二套亏本 20%, 即售价比原价低 20%, 那么售价是原价的 (1 - 20%), 单位'1'未知, 用除法求出第二套的原价;分别用加法求出两套服装的原价之和与售价之和，再比较，如果售价大于原价，则赚钱；如果售价小于原价，则亏本；如果售价等于原价，则不亏也不赚。\n【详解】第一套的原价：100 ÷ (1 + 20%) = 100 ÷ 1.2 ≈ 83.33(元)。第二套的原价：100 ÷ (1 - 20%) = 100 ÷ 0.8 = 125(元)。两套服装的售价: 100 + 100 = 200(元)。两套服装的原价：83.33 + 125 = 208.33(元)。200 < 208.33",
+    "knowledge_points": [
+      "经济问题",
+      "已知比一个数多/少百分之几是多少，求这个数"
+    ],
+    "source": "2018·全国·小升初模拟",
+    "merged_id": "unit6_203",
+    "original_id": "unit6_choice_203"
+  },
+  {
+    "id": "unit6_choice_204",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一种纺织品的合格率是 98%。照这样计算，300 件产品中有（）件不合格。",
+    "options": [
+      "A. 2",
+      "B. 6",
+      "C. 294"
+    ],
+    "answer": "B",
+    "solution": "【分析】把产品的总数量看作单位'1'，合格产品的数量占产品总数量的 98%，则不合格产品的数量占产品总数量的 (1 - 98%)，不合格产品的数量 = 产品的总数量 × (1 - 98%)，据此解答。找出题目中的单位'1'并表示出不合格产品数量占产品总数量的百分率是解答题目的关键。\n【详解】300 × (1 - 98%) = 300 × 0.02 = 6(件)。所以，300件产品中有6件不合格。",
+    "knowledge_points": [
+      "求一个数的百分之几是多少"
+    ],
+    "source": "22-23五年级下·山东济南·期中",
+    "merged_id": "unit6_204",
+    "original_id": "unit6_choice_204"
+  },
+  {
+    "id": "unit6_choice_205",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一件上衣现在的售价是 200 元，比原价便宜了 50 元，比原价便宜了（）。",
+    "options": [
+      "A. 20%",
+      "B. 80%",
+      "C. 25%"
+    ],
+    "answer": "A",
+    "solution": "【分析】先用现价加上降低的钱数求出原价，再用降低的钱数除以原价，就是降低了百分之几。本题是求一个数是另一个数的百分之几, 关键是看把谁当成了单位'1', 单位'1'的量为除数。\n【详解】50 ÷ (200 + 50) × 100% = 50 ÷ 250 × 100% = 0.2 × 100% = 20%。比原价便宜了 20%。",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "2022·河南三门峡·小升初真题",
+    "merged_id": "unit6_205",
+    "original_id": "unit6_choice_205"
+  },
+  {
+    "id": "unit6_choice_206",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一件商品，先提价 10%，又降价 10%，现价与原来相比，（）。",
+    "options": [
+      "A. 提高了",
+      "B. 降低了",
+      "C. 不变",
+      "D. 无法确定"
+    ],
+    "answer": "B",
+    "solution": "【分析】把商品原价看作单位'1'，提价 10% 后价格占原价的 (1 + 10%)，又降价 10% 后的价格占原价的 (1 + 10%) × (1 - 10%)，据此解答。理解两个百分数的单位'1'不相同, 并求出现在价格占原价的百分率是解答题目的关键。\n【详解】假设商品原价为1；(1 + 10%) × (1 - 10%) = 1.1 × 0.9 = 0.99 = 99%。99% < 1，所以现价比原价降低了。",
+    "knowledge_points": [
+      "经济问题",
+      "含百分数的运算",
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "2022·江西景德镇·小升初真题",
+    "merged_id": "unit6_206",
+    "original_id": "unit6_choice_206"
+  },
+  {
+    "id": "unit6_choice_207",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "某体育场大约能容纳3万名观众，在一次足球比赛中，上座率 68%。估一估，大约有（）名观众现场观看了比赛。",
+    "options": [
+      "A. 6800",
+      "B. 15000",
+      "C. 20000",
+      "D. 28000"
+    ],
+    "answer": "C",
+    "solution": "【分析】结合题意，求一个数的百分之几是多少，用乘法计算，用总人数乘上座率即可解答。此题的解题关键是掌握求一个数的百分之几是多少的计算方法。\n【详解】30000 × 68% = 20400(名)。20400≈20000，即大约有 20000 名观众观看了比赛。",
+    "knowledge_points": [
+      "求一个数的百分之几是多少"
+    ],
+    "source": "2018·山东临沂·小升初真题",
+    "merged_id": "unit6_207",
+    "original_id": "unit6_choice_207"
+  },
+  {
+    "id": "unit6_choice_208",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "一种商品先涨价10%，再降价10%，现价（）原价。",
+    "options": [
+      "A. 高于",
+      "B. 低于",
+      "C. 等于",
+      "D. 不能确定"
+    ],
+    "answer": "B",
+    "solution": "【分析】第一个 10% 的单位'1'是原价，设原价是 1，涨价后的价格是原价的 (1 + 10%), 用乘法求出涨价后的价格;再把涨价后的价格看成单位'1', 现价是涨价后价格的 (1 - 10%), 用乘法求出现价, 然后与原价比较即可。掌握求比一个数多（少）百分之几的数是多少的计算方法是解答题目的关键。\n【详解】假设原价是1，1 × (1 + 10%) × (1 - 10%) = 1 × (1 + 0.1) × (1 - 0.1) = 1.1 × 0.9 = 0.99。0.99 < 1，现价比原价便宜。",
+    "knowledge_points": [
+      "单位'1'的认识与确定",
+      "比一个数多少/多少之几的数是多少"
+    ],
+    "source": "15-16 六年级下·全国·课后作业",
+    "merged_id": "unit6_208",
+    "original_id": "unit6_choice_208"
+  },
+  {
+    "id": "unit6_choice_209",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "一本书, 已经看了总页数的 60%, 没看的页数与全书总页数的比是 ( )。",
+    "options": [
+      "A. 5:3",
+      "B. 3:5",
+      "C. 2:3",
+      "D. 2:5"
+    ],
+    "answer": "D",
+    "solution": "【分析】把这本书的总页数看作单位'1', 已经看了总页数的 60%, 没看的页数占总页数的 (1-60%), 根据比的意义, 即可求出没看的页数与全书总页数的比。此题的解题关键是把这本书的总页数看作单位'1'，利用百分数的减法以及比的意义，解决问题。\n【详解】根据分析得，把这本书的总页数看作单位'1'，1 - 60% = 40%，即没看的页数占总页数的 40%。40%:1 = 0.4:1 = (0.4 × 10):(1 × 10) = 4:10 = 2:5。即没看的页数与全书总页数的比是 2:5。",
+    "knowledge_points": [
+      "比的意义",
+      "单位'1'的认识与确定",
+      "含百分数的运算"
+    ],
+    "source": "2022·山东德州·小升初真题",
+    "merged_id": "unit6_209",
+    "original_id": "unit6_choice_209"
+  },
+  {
+    "id": "unit6_choice_210",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一杯糖水的含糖率是 10%, 那么糖与水的比是 ( )。",
+    "options": [
+      "A. 1:10",
+      "B. 1:11",
+      "C. 1:9",
+      "D. 无法确定"
+    ],
+    "answer": "C",
+    "solution": "【分析】把这杯糖水的质量看作单位'1', 已知含糖率是 10%, 根据含糖率的意义可知, 糖的质量占糖水质量的 10%, 那么水的质量占糖水质量的 (1 - 10%); 根据比的意义写出糖与水的比，并化简比。本题考查比的意义及化简比, 关键是理解含糖率的意义。\n【详解】10%:(1 - 10%) = 0.1:0.9 = (0.1 × 10):(0.9 × 10) = 1:9。糖与水的比是 1:9。",
+    "knowledge_points": [
+      "比的化简",
+      "求一个数是另一个数的百分之几（百分率问题）",
+      "比的意义"
+    ],
+    "source": "2020·湖北武汉·小升初真题",
+    "merged_id": "unit6_210",
+    "original_id": "unit6_choice_210"
+  },
+  {
+    "id": "unit6_choice_211",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一种商品的价格，经过两次调价后，现价与原价相同的是（）。",
+    "options": [
+      "A. 先降价 20%, 再涨价 20%",
+      "B. 先降价 20%, 再涨价 25%",
+      "C. 先涨价 20%, 再降价 25%",
+      "D. 先涨价 25%, 再降价 25%"
+    ],
+    "answer": "B",
+    "solution": "【分析】A．把这件商品的原价看作单位'1'，用1乘（1-20%），求出降价20%后的价格是多少；然后把降价20%的价格看作单位'1'，用降价20%的价格乘（1+20%），求出现价是多少，再把现价和原价比较大小；B. 用1乘(1-20%),求出降价20%后的价格是多少;然后把降价20%的价格看作单位'1',用降价20%的价格乘(1+25%),求出现价是多少,再把现价和原价比较大小;C. 用1乘(1+20%),求出涨价20%后的价格是多少;然后把涨价20%的价格看作单位'1',用涨价20%的价格乘(1-25%),求出现价是多少,再把现价和原价比较大小;D. 用1乘(1+25%),求出涨价25%后的价格是多少;然后把涨价25%的价格看作单位'1',用涨价25%的价格乘(1-25%),求出现价是多少,再把现价和原价比较大小。\n【详解】A. 1 × (1 - 20%) × (1 + 20%) = 1 × 0.8 × 1.2 = 0.96。因为 0.96 < 1，所以现价比原价低。B. 1 × (1 - 20%) × (1 + 25%) = 1 × 0.8 × 1.25 = 1。因为 1 = 1，所以现价和原价相等。C. 1 × (1 + 20%) × (1 - 25%) = 1 × 1.2 × 0.75 = 0.9。因为 0.9 < 1, 所以现价比原价低。D. 1 × (1 + 25%) × (1 - 25%) = 1 × 1.25 × 0.75 = 0.9375。因为 0.9375 < 1，所以现价比原价低。",
+    "knowledge_points": [
+      "单位'1'的认识与确定",
+      "比一个数多少/多少之几的数是多少"
+    ],
+    "source": "19-20六年级上·全国·期中",
+    "merged_id": "unit6_211",
+    "original_id": "unit6_choice_211"
+  },
+  {
+    "id": "unit6_choice_212",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "小刚以前乘火车去奶奶家要16小时，现在火车提速了，14小时就能到。现在火车速度比原来提高百分之几？可用下面（）列式解答。",
+    "options": [
+      "A. (16 - 14) ÷ 14",
+      "B. (16-14) ÷ 16",
+      "C. 1 -(14÷16)",
+      "D. (1/14 - 1/16) ÷ 1/14"
+    ],
+    "answer": "A",
+    "solution": "【分析】方法一：将时间比反过来就是速度比，前后速度差÷原来速度=提高百分之几;方法二：将总路程看作单位'1'，时间分之一可以看作速度，前后速度差÷原来速度=提高百分之几，据此分析。两数相除又叫两个数的比, 差 ÷ 较小数 = 增加或提高百分之几。\n【详解】方法一：(16 - 14)÷ 14 = 2 ÷ 14 ≈ 14.3%。方法二:(1/14 - 1/16) ÷ 1/16 = 1/112 ÷ 1/16 ≈ 14.3%。",
+    "knowledge_points": [
+      "比的意义",
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "2022·湖北武汉·小升初真题",
+    "merged_id": "unit6_212",
+    "original_id": "unit6_choice_212"
+  },
+  {
+    "id": "unit6_choice_213",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "下面的百分率可能大于 100% 的是（）。",
+    "options": [
+      "A. 增长率",
+      "B. 出勤率",
+      "C. 成活率"
+    ],
+    "answer": "A",
+    "solution": "【分析】出勤率 = 出勤人数 / 应出勤人数 × 100%。成活率 = 成活的植物（动物）数量 / 种植（养殖）的总数量 × 100%。增长率 = 增长量 / 原来量 × 100%, 分数的分子大于分母时, 分数值大于 1, 分数对应的百分数就大于 100%, 所以只要分别比较分子有没有可能大于分母即可解答。本题考查生活中各种率的计算, 以及率值的范围的判断。\n【详解】增长量可能大于原来量，所以增长率可能大于 100%；出勤人数最大可以等于应出勤人数，出勤率最大是 100%；同理，成活率最大也是 100%。",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_213",
+    "original_id": "unit6_choice_213"
+  },
+  {
+    "id": "unit6_choice_214",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一件商品，先提价 20%, 又降价 20%, 现在的价格与原来相比, ( )。",
+    "options": [
+      "A. 提高了",
+      "B. 降低了",
+      "C. 不变",
+      "D. 无法确定"
+    ],
+    "answer": "B",
+    "solution": "【分析】把商品原价看作单位'1'，提价 20% 后价格占原价的 (1 + 20%), 降价 20% 的价格占原价的 (1+ 20%)× (1 - 20%)，据此解答。理解两个百分数的单位'1'不相同, 并求出现在价格占原价的百分率是解答题目的关键。\n【详解】假设商品原价为1。1 × (1 + 20%) × (1 - 20%) = 1 × 1.2 × 0.8 = 0.96。因为 0.96 < 1, 所以现价比原价降低了。",
+    "knowledge_points": [
+      "经济问题",
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "2014·全国·小升初真题",
+    "merged_id": "unit6_214",
+    "original_id": "unit6_choice_214"
+  },
+  {
+    "id": "unit6_choice_215",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一件衣服进价80元，卖100元，赚了（）。",
+    "options": [
+      "A. 20%",
+      "B. 25%",
+      "C. 80%",
+      "D. 125%"
+    ],
+    "answer": "B",
+    "solution": "【分析】用赚的钱除以进价，求出赚了百分之几即可。本题考查百分数, 解答本题的关键是掌握求一个数占另一个数的百分之几的计算方法。\n【详解】(100 - 80) ÷ 80 = 20 ÷ 80 = 25%",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "20-21六年级下·广东中山·期末",
+    "merged_id": "unit6_215",
+    "original_id": "unit6_choice_215"
+  },
+  {
+    "id": "unit6_choice_216",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "把 2 米长的绳子平均剪成 5 段, 每段绳长是全长的 ( )。",
+    "options": [
+      "A. 2/5",
+      "B. 40%",
+      "C. 20%",
+      "D. 5/2"
+    ],
+    "answer": "C",
+    "solution": "【分析】将绳子长度看作单位'1'，求每段是全长的几分之几或百分之几，1 ÷ 段数即可。求一个数占另一个数的几分之几或百分之几，用除法，此类问题一般用表示单位'1'的量作被除数。\n【详解】1 ÷ 5 = 1/5 = 20%",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）",
+      "单位'1'的认识与确定"
+    ],
+    "source": "21-22 六年级下·湖北襄阳·期末",
+    "merged_id": "unit6_216",
+    "original_id": "unit6_choice_216"
+  },
+  {
+    "id": "unit6_choice_217",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "小明每天上学要用 8 分钟, 放学回家要用 10 分钟, 上学比放学速度提高了 ()。",
+    "options": [
+      "A. 20%",
+      "B. 25%",
+      "C. 80%",
+      "D. 120%"
+    ],
+    "answer": "B",
+    "solution": "【分析】把家到学校的距离看成单位'1', 根据路程÷时间 = 速度, 小明每天上学的速度是 1/8, 小明每天放学回家的速度是 1/10, 求出二者的速度差, 然后用速度差除以放学回家的速度即可。本题把路程看成单位'1', 把速度用分数表示出来,然后根据求一个数比另一个数多百分之几的方法求解。\n【详解】根据分析得，1 ÷ 8 = 1/8，1 ÷ 10 = 1/10。(1/8 - 1/10) ÷ 1/10 = (5/40 - 4/40) ÷ 1/10 = 1/40 ÷ 1/10 = 1/4 = 25%",
+    "knowledge_points": [
+      "单位'1'的认识与确定",
+      "求一个数比另一个数多/少百分之几",
+      "基础行程问题"
+    ],
+    "source": "21-22 六年级下·江西吉安·期末",
+    "merged_id": "unit6_217",
+    "original_id": "unit6_choice_217"
+  },
+  {
+    "id": "unit6_choice_218",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "在 100 克水中加入 25 克盐, 那么盐水中的含盐率是 ( )。",
+    "options": [
+      "A. 10%",
+      "B. 15%",
+      "C. 20%",
+      "D. 25%"
+    ],
+    "answer": "C",
+    "solution": "【分析】含盐率是指盐占盐水的百分率，盐有 25 克，盐水有 (100 + 25) 克，用盐除以盐水再乘 100% 即可得解。此题的解题关键是理解含盐率的含义。\n【详解】25 ÷ (100 + 25) × 100% = 25 ÷ 125 × 100% = 0.2 × 100% = 20%",
+    "knowledge_points": [
+      "百分数的意义",
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "2020·湖南永州·小升初真题",
+    "merged_id": "unit6_218",
+    "original_id": "unit6_choice_218"
+  },
+  {
+    "id": "unit6_choice_219",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "'一种产品原价50元，现价比原价降低了5元，求降低了百分之几？'解决这一问题的正确列式是（）。",
+    "options": [
+      "A. 5/50 × 100%",
+      "B. 5/(50+5) × 100%",
+      "C. 45/(50+5) × 100%"
+    ],
+    "answer": "A",
+    "solution": "【分析】求降低了百分之几，意思是求现价比原价降低了百分之几，用降低的钱数除以原价即可。明确求一个数比另一个数多或少百分之几，用两者的差值除以另一个数。\n【详解】5 ÷ 50 × 100% = 5/50 × 100%",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "19-20六年级上·陕西·期末",
+    "merged_id": "unit6_219",
+    "original_id": "unit6_choice_219"
+  },
+  {
+    "id": "unit6_choice_220",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "根据'五年级今天有3人请假，出勤率是 98.75%'这两条信息，可以解决的问题是（）。",
+    "options": [
+      "A. 五年级有多少人",
+      "B. 五年级男生有多少人",
+      "C. 六年级有多少人",
+      "D. 五年级女生有多少人"
+    ],
+    "answer": "A",
+    "solution": "【分析】由题意可知, 出勤率是 98.75%, 则缺勤率是 1 - 98.75% = 1.25%, 根据五年级总人数 = 缺勤的人数 ÷ 缺勤率,据此解答即可。本题考查已知一个数的百分之几是多少，求这个数，明确缺勤率是解题的关键。\n【详解】由分析可知:根据题干中的信息可以解决求五年级的总人数是多少。",
+    "knowledge_points": [
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "21-22六年级上·山西长治·期末",
+    "merged_id": "unit6_220",
+    "original_id": "unit6_choice_220"
+  },
+  {
+    "id": "unit6_choice_221",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "某电器，先涨价 10%，再降价 10%，现价与原价相比（）。",
+    "options": [
+      "A. 比原价高",
+      "B. 比原价低",
+      "C. 一样",
+      "D. 不确定"
+    ],
+    "answer": "B",
+    "solution": "【分析】假设原价为 100 元，现价 = 原价 × (1 + 10%) × (1 - 10%)，由此计算出现价，再与原价进行比较即可。本题考查了'求比一个数多（少）百分之几的数是多少'，关键是找准单位'1'。\n【详解】假设原价为100元。100 × (1 + 10%) × (1 - 10%) = 100 × 1.1 × 0.9 = 99(元)。99 < 100",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "20-21 六年级上·云南文山·期末",
+    "merged_id": "unit6_221",
+    "original_id": "unit6_choice_221"
+  },
+  {
+    "id": "unit6_choice_222",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "2020年冬季体检结果显示，第一小学的近视率是 15%，第二小学的近视率是17%。两个小学的近视人数相比（）。",
+    "options": [
+      "A. 第一小学多",
+      "B. 第二小学多",
+      "C. 两个学校一样多",
+      "D. 无法确定"
+    ],
+    "answer": "D",
+    "solution": "【分析】单位'1'不确定, 也就是说第一小学和第二小学的人数不一定相等, 所以无法比较, 据此解答即可。\n【详解】第一小学的近视率是 15%, 单位'1'是第一小学的人数; 第二小学的近视率是 17%, 单位'1'是第二小学的人数, 而近视人数 = 总人数 × 近视率, 所以无法确定具体的近视人数。此题考查了百分数的实际应用, 熟练掌握百分数的意义是解决此题的关键。",
+    "knowledge_points": [
+      "求一个数的百分之几是多少"
+    ],
+    "source": "20-21六年级上·福建南平·期末",
+    "merged_id": "unit6_222",
+    "original_id": "unit6_choice_222"
+  },
+  {
+    "id": "unit6_choice_223",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "下面的数可以写成百分数的是 ( )。",
+    "options": [
+      "A. 面粉 35 袋",
+      "B. 大米的袋数是面粉的 4/5",
+      "C. 面粉比大米多50袋"
+    ],
+    "answer": "B",
+    "solution": "【分析】百分数是表示一个数是另一个数的百分之几，它不能表示一个具体的数，所以百分数不能有单位名称；据此即可判断。此题考查百分数的意义：百分数只表示一个数占另一个数的百分比，不能用来表示具体的数。\n【详解】A．面粉35袋，不能写出百分数；B. 大米的袋数是面粉的 4/5, 4/5 = 80%;C. 面粉比大米多 50 袋, 不能写成百分数;",
+    "knowledge_points": [
+      "百分数的意义"
+    ],
+    "source": "20-21 六年级上·河北邯郸·期中",
+    "merged_id": "unit6_223",
+    "original_id": "unit6_choice_223"
+  },
+  {
+    "id": "unit6_choice_224",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "按糖与水的比是 1:9 配制一种糖水, 现在有糖 5 克, 水 50 克, 要求把糖全部用完, 则水 ( )。",
+    "options": [
+      "A. 多 5 克",
+      "B. 少 5 克",
+      "C. 无法判断"
+    ],
+    "answer": "A",
+    "solution": "【分析】按糖和水的比为 1:9 配制一种糖水, 这种糖水的含糖和糖水的比例是 1:10, 含糖率就是 1 除以 10, 再化成百分数; 现有糖 5 克, 用 5 除以 10%, 求出可配制这种糖水多少克, 再减去 5 克糖得出需要的水的质量和 50 比较即可。此题考查的是百分数的应用, 解答此题关键是根据比不变有糖 5 克, 求出需配制糖水的质量。\n【详解】1 ÷ 10 = 10%。5 ÷ 10% = 50(克)。50 - 5 = 45(克)。要求把糖全部用完，则水多 50 - 45 = 5(克)",
+    "knowledge_points": [
+      "比的应用",
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "20-21 六年级上·河南新乡·期中",
+    "merged_id": "unit6_224",
+    "original_id": "unit6_choice_224"
+  },
+  {
+    "id": "unit6_choice_225",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "一本书共200页，看了20页，还剩（）%没有看。",
+    "options": [
+      "A. 180",
+      "B. 80",
+      "C. 90",
+      "D. 20"
+    ],
+    "answer": "C",
+    "solution": "【分析】求还剩百分之几没有看, 就是求还剩下的页数占总页数的百分之几, 先用总页数减去看了的页数, 求出还剩下的页数, 再除以总页数即可。明确求一个数占另一个数的百分之几，用除法计算。\n【详解】(200 - 20) ÷ 200 = 180 ÷ 200 = 0.9 = 90%",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "18-19六年级上·山东青岛·单元测试",
+    "merged_id": "unit6_225",
+    "original_id": "unit6_choice_225"
+  },
+  {
+    "id": "unit6_choice_226",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一本书有 160 页, 已经看了 128 页, 还剩下 ( ) 没看。",
+    "options": [
+      "A. 80%",
+      "B. 20%",
+      "C. 25%"
+    ],
+    "answer": "B",
+    "solution": "【分析】一本书共有 160 页，已经看了 128 页，则还剩下(160-128) 页没有看, 用剩下的页数除以总页数乘百分之百, 即可得知没看的百分率。完成本题要注意是求没有看的部分占全部的分率,而不是具体数量。\n【详解】(160 - 128) ÷ 160 × 100% = 32 ÷ 160 × 100% = 0.2 × 100% = 20%",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）",
+      "含百分数的运算"
+    ],
+    "source": "21-22 六年级下·河南信阳·开学考试",
+    "merged_id": "unit6_226",
+    "original_id": "unit6_choice_226"
+  },
+  {
+    "id": "unit6_choice_227",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一份文件，原计划3小时打完，实际2.5小时就完成了任务。实际工作效率比计划提高了（）%。",
+    "options": [
+      "A. 13.3",
+      "B. 20",
+      "C. 25",
+      "D. 50"
+    ],
+    "answer": "B",
+    "solution": "【分析】根据实际工作效率比计划提高的百分率 = （实际的工作效率一计划的工作效率）÷计划的工作效率，据此解答即可。本题考查一个数比另一个多百分之几，明确单位'1'是解题的关键。\n【详解】1 ÷ 3 = 1/3，1 ÷ 2.5 = 2/5。(2/5 - 1/3) ÷ 1/3 = 1/15 ÷ 1/3 = 20%",
+    "knowledge_points": [
+      "工程问题",
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "2022六年级下·全国·专题练习",
+    "merged_id": "unit6_227",
+    "original_id": "unit6_choice_227"
+  },
+  {
+    "id": "unit6_choice_228",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "一批纺织品的合格率是 98%，300件产品中有（ ）件不合格。",
+    "options": [
+      "A. 2",
+      "B. 4",
+      "C. 6",
+      "D. 294"
+    ],
+    "answer": "C",
+    "solution": "【分析】把这批纺织品的总数看作单位'1'，合格率是 98%，即合格产品占总数的 98%，那么不合格产品占总数的 (1 - 98%)，单位'1'已知，用总数乘 (1 - 98%) 即可求出不合格产品数。找准单位'1', 理解百分率的含义, 明确求一个数的百分之几是多少, 用乘法计算。\n【详解】300 × (1 - 98%) = 300 × 0.02 = 6(件)",
+    "knowledge_points": [
+      "求一个数的百分之几是多少",
+      "单位'1'的认识与确定"
+    ],
+    "source": "14-15六年级·全国·期末",
+    "merged_id": "unit6_228",
+    "original_id": "unit6_choice_228"
+  },
+  {
+    "id": "unit6_choice_229",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一台冰箱先涨价 10%，再降价 10%, 这台冰箱现在的售价与原价相比, ( )。",
+    "options": [
+      "A. 提高了",
+      "B. 降低了",
+      "C. 不变"
+    ],
+    "answer": "B",
+    "solution": "【分析】把这台冰箱的原价看作单位'1'，这台冰箱的售价 = (1 + 10%) × (1 - 10%)，最后原价和售价比较大小即可。第一个 10% 的单位'1'是冰箱的原价，第二个10%\n【详解】假设这台冰箱的原价为 1。售价为: 1 × (1 + 10%) × (1 - 10%) = 1.1 × 0.9 = 0.99。因为 1 > 0.99, 所以这台冰箱现在的售价比原价降低了。",
+    "knowledge_points": [
+      "单位'1'的认识与确定",
+      "经济问题",
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "2015·全国·小升初真题",
+    "merged_id": "unit6_229",
+    "original_id": "unit6_choice_229"
+  },
+  {
+    "id": "unit6_fill_230",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一段路甲车 5 小时行完，乙车 4 小时行完，那么乙车的速度比甲车快( )%。",
+    "options": [],
+    "answer": "25",
+    "solution": "【分析】把这段路看作单位'1', 根据路程÷时间 = 速度,据此分别求出甲车的速度是 1/5, 乙车的速度是 1/4, 然后求出乙车的速度比甲车快多少, 再除以甲车的速度, 最后再乘 100% 即可。本题考查求一个数比另一个数多百分之几, 求出甲车和乙车的速度是解题的关键。\n【详解】(1/4 - 1/5) ÷ 1/5 × 100% = 1/20 ÷ 1/5 × 100% = 0.25 × 100% = 25%。则乙车的速度比甲车快 25%",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "2015 六年级上·全国·专题练习",
+    "merged_id": "unit6_230",
+    "original_id": "unit6_fill_230"
+  },
+  {
+    "id": "unit6_fill_231",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "0.6 = ( )/25 = 12:( ) = ( )%。",
+    "options": [],
+    "answer": "15; 20; 60",
+    "solution": "【分析】根据小数与分数的关系，把0.6化为分数形式，即 0.6 = 3/5；根据分数的基本性质，分子和分母同时乘5就是 3/5 = 15/25；根据分数与比的关系就是 3:5，根据比的基本性质，比的前项和后项同时乘4就是 3:5 = 12:20；把0.6的小数点向右移动两位，再加上百分号即可化为百分数。本题考查小数、分数、比和百分数的互化，明确它们之间的关系是解题的关键。\n【详解】由分析可知:0.6 = 15/25 = 12:20 = 60%",
+    "knowledge_points": [
+      "百分数、分数、小数和比的互化"
+    ],
+    "source": "17-18 六年级上·重庆永川·期末",
+    "merged_id": "unit6_231",
+    "original_id": "unit6_fill_231"
+  },
+  {
+    "id": "unit6_fill_232",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "儿童的负重最好不要超过自身体重的 15%，如果长期背负过重物体，会妨碍骨骼生长。红红的体重是 50 千克，红红的书包最好不要超过____千克。",
+    "options": [],
+    "answer": "7.5",
+    "solution": "【分析】把红红的体重看作单位'1', 书包的重量最好不要超过她的体重的 15%, 用红红的体重乘 15%, 就是红红的书包最好不要超过的重量。熟练掌握求一个是的百分之几的计算方法是解答本题的关键。\n【详解】50 × 15% = 7.5(千克)。儿童的负重最好不要超过自身体重的 15%, 如果长期背负过重物体, 会妨碍骨骼生长。红红的体重是 50 千克, 红红的书包最好不要超过 7.5 千克。",
+    "knowledge_points": [
+      "求一个数的百分之几是多少"
+    ],
+    "source": "2022·广东惠州·小升初真题",
+    "merged_id": "unit6_232",
+    "original_id": "unit6_fill_232"
+  },
+  {
+    "id": "unit6_fill_233",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一种大豆的出油率是30%, 300 千克大豆可出油( )千克, 要榨 300 千克豆油需大豆( )千克。",
+    "options": [],
+    "answer": "90; 1000",
+    "solution": "【分析】将大豆质量看作单位'1', 大豆质量 × 出油率 = 可出油的质量; 豆油质量 ÷ 出油率 = 需要的大豆质量, 据此列式计算。关键是确定单位'1'，理解百分率的意义。\n【详解】300 × 30% = 300 × 0.3 = 90(千克)。300 ÷ 30% = 300 ÷ 0.3 = 1000(千克)。一种大豆的出油率是 30%, 300 千克大豆可出油 90 千克,要榨 300 千克豆油需大豆 1000 千克。",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）",
+      "求一个数的百分之几是多少",
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "2016·全国·小升初真题",
+    "merged_id": "unit6_233",
+    "original_id": "unit6_fill_233"
+  },{
+    "id": "unit6_fill_234",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "某服装城再一次清仓大甩卖活动中，某品牌服装现在只卖80元，比原价降低了 60%，这个品牌服装原价( )元。",
+    "options": [],
+    "answer": "200",
+    "solution": "【分析】将原价看作单位'1'，单位'1'未知，利用除法求解。将现价80元除以（1 - 60%），求出这个服装的原价。本题考查了含百分数的运算, 已知比一个数少百分之几的数是多少, 求这个数用除法。\n【详解】80 ÷ (1 - 60%) = 80 ÷ 40% = 200(元)。所以，这个品牌服装的原价是200元。",
+    "knowledge_points": [
+      "含百分数的运算",
+      "已知比一个数多/少百分之几,是多少, 求这个数"
+    ],
+    "source": "22-23六年级上·吉林·期末",
+    "merged_id": "unit6_234",
+    "original_id": "unit6_fill_234"
+  },
+  {
+    "id": "unit6_fill_235",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "20 的 5% 是( )，40是50的( )%，50比40多( )%，30是20的( )%, 40 比 50 少()%, 比 35 米少 20% 是( )米, 20 吨比( )多 20%。",
+    "options": [],
+    "answer": "1; 80; 25; 150; 20; 28; 50/3",
+    "solution": "【分析】求 20 的 5% 是多少, 用 20 × 5% 即可; 求 40 是 50 的百分之几, 用 40 ÷ 50 × 100% 即可;用 50 减去 40, 求出 50 比 40 多多少, 再用多的量除以 40 即可;求 30 是 20 的百分之几，用 30 ÷ 20 × 100% 即可；要求40比50少百分之几，就是求40比50少的部分占50的百分比，用 (50 - 40) ÷ 50 即可;把35米看成单位'1'，要求的数量是35米的（1一20%），由此用乘法求出要求的数量；把要求数看作单位'1'，20吨是单位'1'的（1 + 20%），求单位'1', 用除法解答。解答此题的关键是分清两个单位'1'的区别，求单位'1'的百分之几用乘法；已知单位'1'的百分之几是多少，求单位'1'用除法。\n【详解】20 × 5% = 1。20 的 5%是1; 40 ÷ 50 × 100% = 0.8 × 100% = 80%。40是50的 80%; (50 - 40) ÷ 40 × 100% = 10 ÷ 40 × 100% = 0.25 × 100% = 25%。50比40多 25%。30 ÷ 20 × 100% = 1.5 × 100% = 150。30是20的 150%; (50 - 40) ÷ 50 × 100% = 10 ÷ 50 × 100% = 0.2 × 100% = 20%。40比50少20%; 35 × (1 - 20%) = 35 × 0.8 = 28(米)。比35米少 20% 是28米; 20 ÷ (1 + 20%) = 20 ÷ 1.2 = 50/3(吨)。20吨比 50/3 多 20%。",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）",
+      "求一个数的百分之几是多少",
+      "求一个数比另一个数多/少百分之几",
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "22-23 六年级下·全国·单元测试",
+    "merged_id": "unit6_235",
+    "original_id": "unit6_fill_235"
+  },
+  {
+    "id": "unit6_fill_236",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "学校四月份付水费是2000元，五月份比四月份节约500元，节约了( )%。",
+    "options": [],
+    "answer": "25",
+    "solution": "【分析】学校四月份付水费是 2000 元，五月份比四月份节约 500 元，根据百分数除法的意义，用节约钱数除以四月份付的水费，即得节约了百分之几。求一个数比另一个数多或少百分之几，用除法。\n【详解】500 ÷ 2000 × 100% = 0.25 × 100% = 25%。节约了 25%",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "22-23六年级下·全国·单元测试",
+    "merged_id": "unit6_236",
+    "original_id": "unit6_fill_236"
+  },
+  {
+    "id": "unit6_fill_237",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "绿丰农场购进一批大豆种子，技术员在这批种子中随机抽取了150粒进行发芽实验，实验结果是发芽粒数与没有发芽粒数的比是 4:1，则发芽率为( )。",
+    "options": [],
+    "answer": "80%",
+    "solution": "【分析】发芽率 = 发芽种子数量 ÷ 种子总数量 × 100%，发芽粒数与没有发芽粒数的比是 4:1, 说明发芽种子占 4 份, 种子总数量为 4 + 1 = 5 份, 用 (4 ÷ 5) × 100% 即可求解。明确发芽率的含义是解题的关键, 也可按比分分配求出发芽种子数量, 再根据发芽率的计算方法求解即可。\n【详解】(4÷ 5)× 100% = 0.8 × 100% = 80%。发芽率为 80%",
+    "knowledge_points": [
+      "按比分配问题",
+      "求一个数是另一个数的百分之几 (百分率问题)"
+    ],
+    "source": "21-22 六年级上·福建福州·期末",
+    "merged_id": "unit6_237",
+    "original_id": "unit6_fill_237"
+  },
+  {
+    "id": "unit6_fill_238",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "六年一班期末检测时，有 2 名学生测试没及格，及格率是 95%, 六年一班一共有( )人。",
+    "options": [],
+    "answer": "40",
+    "solution": "【分析】先用'1一95%'求出不及格率，然后根据'已知一个数的百分之几是多少，求这个数用除法计算'即可求出六年一班的学生总数。此题属于百分率问题，重点找出 2 人所占全班人数的对应百分率，这是解题的关键。\n【详解】2 ÷ (1 - 95%) = 2 ÷ 5% = 40(人)",
+    "knowledge_points": [
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "20-21 六年级下·黑龙江伊春·期末",
+    "merged_id": "unit6_238",
+    "original_id": "unit6_fill_238"
+  },
+  {
+    "id": "unit6_fill_239",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "六 (3) 班学生今天实到 49 人, 有 1 人请假, 六 (3) 班学生今天的出勤率为( )。",
+    "options": [],
+    "answer": "98%",
+    "solution": "【分析】出勤率 = 出勤人数 ÷ 总人数 × 100%，据此解答。掌握出勤率的计算方法是解答题目的关键。\n【详解】49 ÷ (49 + 1) × 100% = 49 ÷ 50 × 100 = 0.98 × 100% = 98%",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "21-22六年级下·湖北襄阳·期末",
+    "merged_id": "unit6_239",
+    "original_id": "unit6_fill_239"
+  },
+  {
+    "id": "unit6_fill_240",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "9 ÷ ( )= 0.75 = 3/()= ( ): 20 = ( )",
+    "options": [],
+    "answer": "12；4；15；75",
+    "solution": "【分析】先把小数化为最简分数并把小数的小数点向右移动两位，再添上百分号'%'，把小数化为百分数；然后根据 a/b = a ÷ b = a : b (b ≠ 0) 利用商不变的规律和比的基本性质求出除数和比的前项；据此解答。掌握比、分数、除法之间的关系以及小数、分数、百分数互相转化的方法是解答题目的关键。\n【详解】0.75 = 3/4 = 3÷ 4 = 3:4 = 75%。3 ÷ 4 = (3 × 3) ÷ (4 × 3) = 9 ÷ 12。3: 4 = (3 × 5): (4 × 5) = 15: 20",
+    "knowledge_points": [
+      "百分数、小数和分数的互化",
+      "比与分数、除法的关系"
+    ],
+    "source": "21-22 六年级下·广东汕尾·期末",
+    "merged_id": "unit6_240",
+    "original_id": "unit6_fill_240"
+  },
+  {
+    "id": "unit6_fill_241",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "强强做数学题, 做对了 47 题, 做错了 3 题, 强强做题的错误率是( ), 正确率是()。",
+    "options": [],
+    "answer": "6%; 94%",
+    "solution": "【分析】根据公式：错误率 = 错误的题数 ÷ 总题数 × 100%，正确率 = 正确的题数 ÷ 总题数 × 100%，把数代入公式即可求解。本题主要考查百分率的求法，要注意先求出题目的总数是解题的关键。\n【详解】47 + 3 = 50(道)。3÷ 50× 100% = 0.06× 100% = 6%。47÷ 50× 100% = 0.94× 100% = 94%",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "21-22 六年级下·湖南永州·期末",
+    "merged_id": "unit6_241",
+    "original_id": "unit6_fill_241"
+  },
+  {
+    "id": "unit6_fill_242",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "一件衣服，原价是250元，为了尽快销售，价格降为200元，价格降低了( )%，现价是原价的( )%。",
+    "options": [],
+    "answer": "20; 80",
+    "solution": "【分析】用原价减去现价，等于降低的价格，用这个价格除以原价，即可计算出价格降低的百分比。用现价除以原价再乘 100% 即可计算出现价是原价的百分之几。此题的解题关键是掌握求一个数比另一个数少百分之几和求一个数是另一个数的百分之几的计算方法。\n【详解】(250-200)÷ 250 = 50÷ 250 = 0.2 = 20%。200÷ 250× 100% = 0.8× 100% = 80%",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）",
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "19-20六年级上·陕西·期末",
+    "merged_id": "unit6_242",
+    "original_id": "unit6_fill_242"
+  },
+  {
+    "id": "unit6_fill_243",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "72 的 1/9 是( ), ( )吨的 20% 是 36 吨。",
+    "options": [],
+    "answer": "8; 180",
+    "solution": "【分析】求一个数的几分之几是多少，用乘法，72 乘 1/9 即可得解；已知一个数的百分之几是多少，求这个数，用除法，36 吨除以 20% 即可得解。此题的解题关键是掌握求一个数的几分之几是多少以及已知一个数的百分之几是多少，求这个数的计算方法。\n【详解】72 × 1/9 = 8。36 ÷ 20% = 180(吨)",
+    "knowledge_points": [
+      "求一个数的几分之几的问题",
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "21-22 六年级上·宁夏石嘴山·期末",
+    "merged_id": "unit6_243",
+    "original_id": "unit6_fill_243"
+  },
+  {
+    "id": "unit6_fill_244",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "做了 20 道口算题, 错了 2 道题, 他这次口算的正确率是( )%。",
+    "options": [],
+    "answer": "90",
+    "solution": "【分析】根据正确率 = 正确的题数 ÷ 总题数 × 100%，据此计算即可。本题考查正确率，明确正确率的求法是解题的关键。\n【详解】(20 - 2) ÷ 20 × 100% = 18 ÷ 20 × 100% = 0.9 × 100% = 90%",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "21-22 六年级上·内蒙古乌海·期末",
+    "merged_id": "unit6_244",
+    "original_id": "unit6_fill_244"
+  },
+  {
+    "id": "unit6_fill_245",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "在 3/8、3.75、35%、0.37、0.375 这五个数中，最大的数是( )，最小的数是( )。",
+    "options": [],
+    "answer": "3.75; 35%",
+    "solution": "【分析】根据题目要求，应把 3/8、35% 化成小数后再比较大小，最后得出最大的数和最小的数各是什么。在有分数、小数和百分数的数中找出最大和最小的数，先化成小数，通过比较大小找出最大和最小的数。\n【详解】3/8 = 0.375。35% = 0.35。0.35 < 0.37 < 0.375 = 0.375 < 3.75。35% < 0.37 < 3/8 = 0.375 < 3.75",
+    "knowledge_points": [
+      "百分数、小数和分数的互化"
+    ],
+    "source": "20-21 六年级上·江西·期中",
+    "merged_id": "unit6_245",
+    "original_id": "unit6_fill_245"
+  },
+  {
+    "id": "unit6_fill_246",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "5/8 = ( )% = ( )÷ 48 = 45:( ) = ( )(填小数)。",
+    "options": [],
+    "answer": "62.5; 30; 72; 0.625",
+    "solution": "【分析】把分数 5/8 转化成小数，分子除以分母可得 5/8 = 5 ÷ 8 =0.625;小数转化成百分数，小数点向右移动两位后，加上百分号，可得 0.625 = 62.5%；根据分数与除法的关系以及商的变化规律，可得 5/8 = 5 ÷ 8 = (5 × 6) ÷ (8 × 6) = 30 ÷ 48;根据比与除法的关系以及比的基本性质，可得 5/8 = 5 ÷ 8 = 5: 8 = (5 × 9) : (8 × 9) = 45: 72。此题的解题关键是掌握分数、百分数、小数和比之间的互化，根据比、分数与除法的关系以及商的变化规律，利用比的基本性质，求出结果。\n【详解】根据分析得，5/8 = 62.5% = 30 ÷ 48 = 45:72 = 0.625(填小数)。",
+    "knowledge_points": [
+      "商的变化规律及应用",
+      "百分数、分数、小数和比的互化",
+      "比的基本性质",
+      "比与分数、除法的关系"
+    ],
+    "source": "21-22 六年级下·河南三门峡·期末",
+    "merged_id": "unit6_246",
+    "original_id": "unit6_fill_246"
+  },
+  {
+    "id": "unit6_fill_247",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一批 200 千克的花生能榨出花生油 96 千克。这批花生的出油率是( )，如果需要榨出 36 千克的花生油，需要( )千克的花生。",
+    "options": [],
+    "answer": "48%; 75",
+    "solution": "【分析】用花生油的质量除以花生的质量乘上 100% 即可;把花生的质量看成单位'1', 它的 48% 就是要榨出油的质量 36 千克, 由此用除法求出需要花生的质量即可。解决本题注意理解出油率，出油率 = 油的质量 ÷ 花生的质量 × 100%，能根据其中的两个量求出第三个量。\n【详解】96 ÷ 200 × 100% = 0.48 × 100% = 48%。36 ÷ 48% = 75(千克)",
+    "knowledge_points": [
+      "含百分数的运算"
+    ],
+    "source": "20-21 六年级上·福建南平·期末",
+    "merged_id": "unit6_247",
+    "original_id": "unit6_fill_247"
+  },
+  {
+    "id": "unit6_fill_248",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "某种商品的进价为 70 元,零售价为 80 元, 该商品的销售利润率为( )。",
+    "options": [],
+    "answer": "14.29%",
+    "solution": "【分析】利用利润率 = (售价一进价)÷进价，代入数据进行解答即可。\n【详解】(80 - 70) ÷ 70 × 100% = 10 ÷ 70 × 100% ≈ 14.29%",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）",
+      "经济问题"
+    ],
+    "source": "21-22 六年级下·全国·期末",
+    "merged_id": "unit6_248",
+    "original_id": "unit6_fill_248"
+  },
+  {
+    "id": "unit6_fill_249",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "为庆贺春节，学校买来'中国结'20个，'福'字10个，'福'字是中国结的()%，'中国结'比'福'字多( )%。",
+    "options": [],
+    "answer": "50; 100",
+    "solution": "【分析】根据百分数意义：百分数表示一个数是另一个数的百分之几；把 10 个'福'字看作单位'1'，用它们的差除以 10 即可。此题考查的是百分数的意义, 掌握用两数的差除以单位'1'的量 = 多 (少) 的分率是解题关键。\n【详解】10 ÷ 20 = 50%。(20 - 10) ÷ 10 = 10 ÷ 10 = 100%",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）",
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "21-22六年级下·河南信阳·开学考试",
+    "merged_id": "unit6_249",
+    "original_id": "unit6_fill_249"
+  },
+  {
+    "id": "unit6_fill_250",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "每套原价a元的衣服，降价 20% 的价格是( )元。",
+    "options": [],
+    "answer": "80%a",
+    "solution": "【分析】本题的单位'1'是原价, 现价是原价的 (1- 80%), 然后根据求单位'1'的百分之几是多少用乘法计算即可。本题是简单的百分数乘法应用题, 求单位'1'的百分之几是多少用乘法, 注意这里是用字母表示数。\n【详解】a × (1 - 20%) = 80%a = 0.8a。降价 20% 的价格是 0.8a 元。",
+    "knowledge_points": [
+      "求一个数的百分之几是多少",
+      "用字母表示数"
+    ],
+    "source": "2021·宁夏石嘴山·小升初真题",
+    "merged_id": "unit6_250",
+    "original_id": "unit6_fill_250"
+  },
+  {
+    "id": "unit6_judge_251",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "在一杯 100 克水中加入 10 克糖, 这杯糖水的含糖率是 10%( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】含糖率 = 糖的质量 ÷ 糖水的质量 × 100%, 据此求出含糖率, 进行判断。掌握含糖率的计算方法是解答此题的关键。\n【详解】10 ÷ (100 + 10) × 100% = 10 ÷ 110 × 100% ≈ 0.09 × 100% = 9%。这杯糖水的含糖率大约是 9%。故答案为: ×",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "19-20 六年级下·四川·课后作业",
+    "merged_id": "unit6_251",
+    "original_id": "unit6_judge_251"
+  },
+  {
+    "id": "unit6_judge_252",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "王师傅加工98个零件，合格98个，合格率是 98%。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】合格率是指合格的零件数量占加工零件的总数量的百分比，计算方法是：合格的零件数量÷加工零件的总数量×100%=合格率，代入数据求解即可。本题考查合格率，明确合格率的计算方法是解题的关键。\n【详解】98 ÷ 98 × 100%= 1 × 100% = 100%。即合格率是 100%。故答案为: ×",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "2014·全国·小升初真题",
+    "merged_id": "unit6_252",
+    "original_id": "unit6_judge_252"
+  },
+  {
+    "id": "unit6_judge_253",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "栽了 102 棵树，全部成活，成活率是 102%。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】成活率 = 成活棵数 ÷ 栽种总棵数 × 100%，据此解答。关键是理解百分率的意义，出油率、出勤率、及格率等不可能超过 100%。\n【详解】102 ÷ 102 × 100% = 1 × 100% = 100%。栽了 102 棵树, 全部成活, 成活率是 100%; 原题说法错误;故答案为: ×",
+    "knowledge_points": [
+      "百分数的意义",
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "2014 五年级·全国·课后作业",
+    "merged_id": "unit6_253",
+    "original_id": "unit6_judge_253"
+  },
+  {
+    "id": "unit6_judge_254",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "一批零件里100件正品，4件次品，正品率为 100%。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】根据题意，正品率 = 正品数量 ÷ 总数量 × 100%，由此代入数据判断。此题属于百分率问题，都是用一部分数量（或全部数量）除以全部数量乘 100%。\n【详解】100 ÷ (100 + 4) × 100% = 100 ÷ 104 × 100% ≈ 96.2%。96.2% < 100%。一批零件里 100 件正品，4 件次品，正品率约 96.2%, 为原题说法错误;故答案为: ×",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "2022·河南焦作·小升初真题",
+    "merged_id": "unit6_254",
+    "original_id": "unit6_judge_254"
+  },
+  {
+    "id": "unit6_judge_255",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "甲乙两班今天的出勤率都是 98%，那么甲乙两班今天的出勤人数相等。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】出勤率是指出勤人数占总人数的百分比，那么98%的单位'1'分别是两个班的人数，出勤人数的多少就由班级的总人数决定。出勤率 = 出勤人数 ÷ 全部人数 × 100%, 分为三种情况: 当甲乙两班人数相等、甲班大于乙班人数、甲班人数小于乙班人数。分类分析即可得出答案。本题主要考查的是百分率问题, 解题的关键是熟练运用百分数运算, 进而得出答案。\n【详解】①当甲班<乙班时 甲班 × 98% <乙班 × 98% 甲班出勤人数比乙班出勤人数少;② 当甲班 = 乙班时,甲班 × 98% = 乙班 × 98% 两班出勤人数相等;③ 当甲班 > 乙班时 甲班 × 98% > 乙班 ×98% 甲班出勤人数 > 乙班出勤人数。则原题干说法错误。故答案为: ×",
+    "knowledge_points": [
+      "求一个数的百分之几是多少"
+    ],
+    "source": "2020·湖北武汉·小升初真题",
+    "merged_id": "unit6_255",
+    "original_id": "unit6_judge_255"
+  },
+  {
+    "id": "unit6_judge_256",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "男生比女生多 25%, 也就是女生比男生少 25%。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】由题意可知, 设女生的人数为 1, 则男生的人数有 1 × (1 + 25%)，然后求出女生比男生少多少，再除以男生的人数即可。本题考查求比一个数多百分之几的数是多少，明确单位'1'是解题的关键。\n【详解】假设女生的人数为 1。1 × (1 + 25%) = 1 × 1.25 = 1.25。(1.25 - 1) ÷ 1.25 = 0.25 ÷ 1.25 = 20%。则女生比男生少 20%。原题干说法错误。故答案为: ×",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几",
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "15-16六年级上·全国·期末",
+    "merged_id": "unit6_256",
+    "original_id": "unit6_judge_256"
+  },
+  {
+    "id": "unit6_judge_257",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "把20克盐溶解在80克水中，盐水的含盐率是 20%。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】根据含盐率 = 盐的质量 ÷ 盐水的质量 × 100%，已知盐的质量是 20 克, 盐水的质量是 (20 + 80) 克, 代入到公式中, 即可求出盐水的含盐率。此题的解题关键是理解含盐率的意义, 掌握求一个数占另一个数的百分之几的计算方法。\n【详解】20 ÷ (20 + 80) × 100% = 20 ÷ 100 × 100% = 0.2 × 100% = 20%。即盐水的含盐率是 20%。故答案为: √",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "16-17六年级·江苏·单元测试",
+    "merged_id": "unit6_257",
+    "original_id": "unit6_judge_257"
+  },
+  {
+    "id": "unit6_judge_258",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "生产99个产品，全部合格，合格率是 99%。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】根据'合格率 = 合格的产品 ÷ 产品总数 × 100%'可知，全部合格时，合格率是 100%，据此判断即可。明确合格率的含义是解答本题的关键。\n【详解】99 ÷ 99 × 100% = 1 × 100% = 100%。生产 99 个产品，全部合格，合格率是 100%, 原题说法错误。故答案为: ×",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "2017·湖南长沙·小升初真题",
+    "merged_id": "unit6_258",
+    "original_id": "unit6_judge_258"
+  },
+  {
+    "id": "unit6_judge_259",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "生产的一批产品中，有 100 件合格品，1 件次品，这批产品的合格率是 100%。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】先用合格产品的件数加上不合格产品的件数，求出这批产品的总件数，然后根据'合格率 = 合格产品的件数 ÷ 产品的总件数 × 100%'，代入数据计算即可这批产品的合格率。本题考查百分率问题, 掌握合格率的计算方法是解题的关键。\n【详解】100 ÷ (100 + 1) × 100% = 100 ÷ 101 × 100% ≈ 0.99 × 100% = 99%。这批产品的合格率是 99%, 原题说法错误。故答案为: ×",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "2014 五年级·全国·课后作业",
+    "merged_id": "unit6_259",
+    "original_id": "unit6_judge_259"
+  },
+  {
+    "id": "unit6_judge_260",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "102个零件，全部合格，这些零件的合格率是 102%。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】首先理解合格率的意义，合格率是指合格产品数占产品总数的百分之几，计算方法为：合格产品数÷产品总数×100%=合格率，由此解答。此题属于百分率问题，计算的结果最大值为 100%，都是用一部分数量（或全部数量）除以全部数量乘百分之百，解题的时候不要被表面数字困惑。\n【详解】102 ÷ 102 × 100% = 1 × 100% = 100%。合格率是 100%, 原题说法错误。故答案为: ×",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "2017·云南楚雄·小升初真题",
+    "merged_id": "unit6_260",
+    "original_id": "unit6_judge_260"
+  },
+  {
+    "id": "unit6_judge_261",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "在含盐率 10% 的盐水中,加入10克盐和10克水后，盐水的含盐率没有改变。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】先求出后来加入部分的盐水的含盐率，然后与原来的含盐率 10% 比较，如果与 10% 相等，那么混合后含盐率就是 10%，否则就不是 10%。理解含盐率的含义，明确两种含盐率不同的盐水混合后的含盐率和原来的都不相等。\n【详解】10 ÷ (10 + 10) × 100% = 10 ÷ 20 × 100% = 0.5 × 100% = 50%。50%的盐水与 10%的盐水混合后含盐率要大于 10%。故答案为: ×",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "2014 五年级·全国·课后作业",
+    "merged_id": "unit6_261",
+    "original_id": "unit6_judge_261"
+  },
+  {
+    "id": "unit6_judge_262",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "23/100 吨可以写成 23% 吨。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】百分数是'表示一个数是另一个数百分之几的数。'它只能表示两数之间的倍数关系，不能表示某一具体数量，所以 23% 吨的表示方法是错误的，据此判断。百分数不能表示具体的数量是百分数与分数的区别之一。\n【详解】根据百分数的意义可知, 百分数不能表示某一具体数量, 23/100 吨不可以写成 23% 吨, 原题说法错误。故答案为: ×",
+    "knowledge_points": [
+      "百分数的意义"
+    ],
+    "source": "22-23 六年级上·黑龙江鸡西·期末",
+    "merged_id": "unit6_262",
+    "original_id": "unit6_judge_262"
+  },
+  {
+    "id": "unit6_judge_263",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "在 100 克水中加入5克糖，糖占糖水的 5%。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】求一个数占另一个数的百分之几，用除法，糖的质量是 5 克，糖水的质量是 (100 + 5) 克，用糖的质量除以糖水的质量，即可得解。此题的解题关键是掌握求一个数占另一个数的百分之几的计算方法。\n【详解】5 ÷ (100 + 5) × 100% = 5 ÷ 105 × 100% ≈ 0.048 × 100% = 4.8%。即糖约占糖水的 4.8%。故答案为: ×",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "22-23 六年级上·黑龙江鸡西·期末",
+    "merged_id": "unit6_263",
+    "original_id": "unit6_judge_263"
+  },
+  {
+    "id": "unit6_judge_264",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "种98棵树全部活了，成活率是 98%。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】根据成活率是指成活的棵数占栽的总棵数的百分之几，因为成活树的棵数和栽的总棵数都是98棵，用成活树的棵数除以栽的总棵数即可求出成活率。此题的解题关键是理解百分数的意义以及成活率的含义。\n【详解】98 ÷ 98 × 100% = 1× 100% = 100%。即成活率是 100%。故答案为: ×",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "2017·江苏扬州·小升初真题",
+    "merged_id": "unit6_264",
+    "original_id": "unit6_judge_264"
+  },
+  {
+    "id": "unit6_judge_265",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "祥和社区今年植树99 棵，全部成活，成活率为 99%（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】根据成活率 = 成活的棵数 ÷ 植树的棵数 × 100%, 据此代入数值进行计算即可。本题考查成活率，明确成活率的计算方法是解题的关键。\n【详解】99 ÷ 99 × 100% = 1 × 100% = 100%。则成活率为 100%。所以原题干说法错误。故答案为: ×",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "16-17六年级上·全国·单元测试",
+    "merged_id": "unit6_265",
+    "original_id": "unit6_judge_265"
+  },
+  {
+    "id": "unit6_judge_266",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "如果科技书与文艺书的本数比是 3:5, 那么文艺书比科技书多 40%。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】由'科技书与文艺书的比是 3:5'可知, 科技书是 3 份, 文艺书 5 份, 文艺书比科技书多 (5-3) 份, 所以, 文艺书比科技书多 (5 - 3) ÷ 3 × 100% ≈ 67%,若求文艺书比科技书多百分之几, 就是求文艺书比科技书多的本数是科技书的百分之几。\n【详解】根据分析可得：(5 - 3) ÷ 3 × 100% = 2 ÷ 3 × 100% ≈ 0.67× 100% = 67%。所以，如果科技书与文艺书的本数比是 3:5，那么文艺书比科技书多 67%；故原题干说法错误。故答案为: ×",
+    "knowledge_points": [
+      "比的应用",
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "21-22 六年级上·河南安阳·期末",
+    "merged_id": "unit6_266",
+    "original_id": "unit6_judge_266"
+  },{
+    "id": "unit6_judge_267",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "用 101 粒种子做发芽试验, 全部发芽了, 这些种子的发芽率为 101%( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】发芽率 = 发芽种子的数量 ÷ 种子总数量 × 100%, 据此解答。掌握发芽率的计算方法是解答题目的关键。\n【详解】101 ÷ 101 × 100% = 1 × 100% = 100%。所以，这些种子的发芽率为 100%。故答案为: ×",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "2022·湖南永州·小升初真题",
+    "merged_id": "unit6_267",
+    "original_id": "unit6_judge_267"
+  },
+  {
+    "id": "unit6_judge_268",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "2 米的 1/5 是 20% 米。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】求一个数的几分之几是多少，用乘法，求出结果是 0.2 米；但根据百分数的定义，百分数表示一个数是另一个数的百分之几，也叫百分比，或百分率。它只是表示两者之间的关系，是一种分率的关系。分率后面不能加单位名称。据此解答。此题的解题关键是理解百分数的意义。\n【详解】2 × 1/5 = 0.2(米)。0.2 米≠20%米。百分数后面不能加单位名称。故答案为: ×",
+    "knowledge_points": [
+      "百分数的意义",
+      "求一个数的几分之几的问题"
+    ],
+    "source": "22-23 六年级上·全国·课后作业",
+    "merged_id": "unit6_268",
+    "original_id": "unit6_judge_268"
+  },
+  {
+    "id": "unit6_judge_269",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "如果甲班人数的50%比乙班人数的 50%多，那么甲班总人数一定比乙班总人数多。（ ）",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】甲班人数的 50% = 甲班总人数 × 50%, 乙班人数的 50% = 乙班总人数 × 50%, 根据题意, 可知甲班总人数 × 50% > 乙班人数 × 50%。根据一个因数相同，另一个因数越大，积越大，可知甲班总人数多于乙班总人数，据此解答即可。解答本题的关键是明确'甲班总人数 × 50% > 乙班总人数 × 50%', 再进一步解答。\n【详解】如果甲班人数的 50% 比乙班人数的 50% 多, 那么甲班总人数一定比乙班总人数多，说法正确。故答案为: √",
+    "knowledge_points": [
+      "求一个数的百分之几是多少"
+    ],
+    "source": "20-21 六年级下·山东临沂·期末",
+    "merged_id": "unit6_269",
+    "original_id": "unit6_judge_269"
+  },
+  {
+    "id": "unit6_judge_270",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "今年小麦喜获丰收，总产量比去年增加 30% 万吨。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】百分数是表示一个数是另一个数的百分之几的数，百分数不能带单位。本题考查百分数的意义。\n【详解】据分析可知：今年小麦喜获丰收，总产量比去年增加 30% 万吨，说法错误。故答案为: ×",
+    "knowledge_points": [
+      "百分数的意义"
+    ],
+    "source": "20-21 六年级上·湖北荆州·期末",
+    "merged_id": "unit6_270",
+    "original_id": "unit6_judge_270"
+  },
+  {
+    "id": "unit6_judge_271",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "所有的百分数都小于1。（ ）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】百分数是表示一个数是另一个数的百分之几的数,可以举例说明。此题的解题关键是认识百分数，理解百分数的意义。\n【详解】举例：100% 等于1，102%，120% 等都大于1。故答案为：×",
+    "knowledge_points": [
+      "百分数的意义"
+    ],
+    "source": "2014 六年级·全国·课后作业",
+    "merged_id": "unit6_271",
+    "original_id": "unit6_judge_271"
+  },
+  {
+    "id": "unit6_judge_272",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "若甲比乙多 15%，那么乙就比甲少 15%。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】根据题意，甲比乙多 15%，把乙看作单位'1'，则甲是乙的 (1 + 15%), 求乙比甲少百分之几, 用甲乙相差的 15% 除以甲即可, 据此判断。明确求一个数比另一个数多或少百分之几，用两数的差值除以另一个数。\n【详解】15% ÷ (1 + 15%) × 100% = 0.15 ÷ 1.15 × 100% ≈ 0.13 × 100% = 13%。原题说法错误。故答案为: ×",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "19-20六年级上·辽宁·期末",
+    "merged_id": "unit6_272",
+    "original_id": "unit6_judge_272"
+  },
+  {
+    "id": "unit6_calc_273",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "只列式不计算。180 比一个数的 50% 多 10，这个数是多少？",
+    "options": [],
+    "answer": "340",
+    "solution": "【分析】180 比一个数的 50% 多 10, 那么 180 减去 10 就是这个数的 50%, 已知一个数的百分之几是多少求这个数用除法。\n【详解】(180 - 10) ÷ 50% = 170 ÷ 0.5 = 340。这个数是 340。",
+    "knowledge_points": [
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "2018·浙江杭州·小升初模拟",
+    "merged_id": "unit6_273",
+    "original_id": "unit6_calc_273"
+  },
+  {
+    "id": "unit6_calc_274",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "计算下面各题, 怎样算简便就怎样算。(1) 63 × 75% + 3/4 × 37 (2) 50% × 2.5 × 1/3 × 0.4 (3) (3.7 × 40% + 6.3 × 40%) ÷ 3 (4) 35.5+(2.8-0.8)÷ 80% (5) 80 × (1 - 10%) × (1 + 10%) (6) 500 × 75% × (1 ÷ 25%)",
+    "options": [],
+    "answer": "(1) 75; (2) 1/6; (3) 4/3; (4) 38; (5) 79.2; (6) 1500",
+    "solution": "【分析】（1）把 75% 和 3/4 化成小数，再根据乘法分配律，把式子转化为 (63 + 37) × 0.75 进行简算；(2) 根据乘法交换律和乘法结合律, 把式子转化为 0.5 × 1/3 × (2.5 × 0.4) 进行简算; (3) 根据乘法分配律, 把式子转化为 (3.7 + 6.3) × 0.4 ÷ 3 进行简算; (4) 根据运算顺序, 先计算括号里的减法, 再计算括号外的除法, 最后计算括号外的加法; (5) 根据运算顺序, 先计算括号里的加法和减法, 再计算括号外的乘法;(6) 根据运算顺序, 先计算括号外的乘法和括号里的除法,最后计算括号外的乘法。\n【详解】(1) 63 × 75% + 3/4 × 37 = 63 × 0.75 × 0.75 × 37 = (63 + 37) × 0.75 = 100 × 0.75 = 75。(2) 50% × 2.5 × 1/3 × 0.4 = 0.5 × 2.5 × 1/3 × 0.4 = 0.5 × 1/3 × (2.5 × 0.4) = 1/6 × 1 = 1/6。(3) (3.7 × 40% + 6.3 × 40%) ÷ 3 = (3.7 × 0.4 + 6.3 × 0.4) ÷ 3 = (3.7 + 6.3) × 0.4 ÷ 3 = 10 × 0.4 ÷ 3 = 4 ÷ 3 = 4/3。(4) 35.5 + (2.8 - 0.8) ÷ 80% = 35.5 + 2 ÷ 0.8 = 35.5 + 2.5 = 38。(5) 80 × (1 - 10%) × (1 + 10%) = 80 × 0.9 × 1.1 = 72 × 1.1 = 79.2。(6) 500 × 75% × (1 ÷ 25%) = 375 × 4 = 1500",
+    "knowledge_points": [
+      "分数的四则混合运算",
+      "含百分数的运算",
+      "小数乘法运算律"
+    ],
+    "source": "22-23 六年级下·全国·单元测试",
+    "merged_id": "unit6_274",
+    "original_id": "unit6_calc_274"
+  },
+  {
+    "id": "unit6_calc_275",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "计算题",
+    "difficulty": "容易",
+    "question": "直接写出得数。7 × 60% = 45 ÷ 62.5% = 0.67 + 1.33 = 20 × 70% = 60 ÷ 1.5 = (0.18 + 9) ÷ 9 = 10 - 0.01 = 45 ÷ 90% = 200 × (1 - 40%)=",
+    "options": [],
+    "answer": "4.2; 72; 2; 14; 40; 1.02; 9.99; 50; 120",
+    "solution": "【详解】略",
+    "knowledge_points": [
+      "除数是小数的小数除法",
+      "小数的四则运算及法则",
+      "含百分数的运算",
+      "多位小数的进位加法、退位减法"
+    ],
+    "source": "22-23 六年级下·全国·单元测试",
+    "merged_id": "unit6_275",
+    "original_id": "unit6_calc_275"
+  },
+  {
+    "id": "unit6_calc_276",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "直接写出得数。1 ÷ 40% = 8 ÷ 50% = 75% - 3/5 = 2 ÷ 1% = 1 - 25% = 5 × 60% = 20 × 20% = 16 × 25%= 3 ÷ 75% = 30 ÷ 60% = 40% ÷ 4 = 3/8 ÷ 37.5% =",
+    "options": [],
+    "answer": "2.5; 16; 0.15; 200; 0.75; 3; 4; 4; 4; 50; 0.1; 1",
+    "solution": "【详解】略",
+    "knowledge_points": [
+      "含百分数的运算"
+    ],
+    "source": "22-23 六年级下·全国·单元测试",
+    "merged_id": "unit6_276",
+    "original_id": "unit6_calc_276"
+  },
+  {
+    "id": "unit6_calc_277",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "化简比并求比值。80平方米：3.2公顷 0.4²:25% 3.5小时：24分",
+    "options": [],
+    "answer": "1:400, 1/400; 16:25, 16/25; 35:4, 35/4",
+    "solution": "【分析】根据比的基本性质作答, 即比的前项和后项同时乘或除以一个相同的数 (0 除外) 值不变; 用比的前项除以后项求比值。\n【详解】80平方米：3.2公顷 = 80平方米：32000 平方米 = (80 ÷ 80):(32000 ÷ 80) = 1:400 = 1/400。0.4²:25% = 0.16:0.25 = (0.16 × 100):(0.25 × 100) = 16:25 = 16/25。3.5小时：24分 = 210分:24分 = (210 ÷ 6):(24 ÷ 6) = 35:4 = 35/4",
+    "knowledge_points": [
+      "比的化简",
+      "求比值",
+      "百分数、小数和分数的互化"
+    ],
+    "source": "20-21 六年级下·云南文山·期末",
+    "merged_id": "unit6_277",
+    "original_id": "unit6_calc_277"
+  },
+  {
+    "id": "unit6_solve_278",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一辆玩具车，按照获利 50% 定价后，恰逢某宝'双十一'进行促销活动，商家又返还售价 20% 的现金。此时卖这辆玩具车，商家赚了还是亏了？请说明理由。",
+    "options": [],
+    "answer": "因为最后结果大于 1, 即售价大于进价, 所以商家赚了。",
+    "solution": "【分析】先把玩具车进价看作单位'1'，算出这辆玩具车定价为 1 × (1 + 50%)，再把玩具车售价看作单位'1'，算出卖这辆玩具车赚到的钱为：1 × (1 + 50%) × (1 - 20%)，最后判断商家赚了还是亏了。此题考查了学生的分析理解能力以及分数乘法。\n【详解】1 × (1 + 50%) × (1 - 20%) = 1 × 150% × 80% = 1 × 3/2 × 4/5 = 6/5。6/5 > 1, 所以商家赚了。答: 因为最后结果大于 1, 即售价大于进价, 所以商家赚了。",
+    "knowledge_points": [
+      "含百分数的运算",
+      "折扣问题"
+    ],
+    "source": "22-23 六年级上·福建莆田·期末",
+    "merged_id": "unit6_278",
+    "original_id": "unit6_solve_278"
+  },
+  {
+    "id": "unit6_solve_279",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一瓶洗衣液，第一周用了这瓶洗衣液的 2/5，第二周用了这瓶洗衣液的 20%，还剩 0.8 升，这瓶洗衣液原有多少升？",
+    "options": [],
+    "answer": "2 升",
+    "solution": "【分析】把这瓶洗液的总容量看作单位'1'，第一周用了这瓶洗衣液的 2/5，第二周用了这瓶洗衣液的 20%，则还剩下 (1 - 2/5 - 20%)，已知还剩0.8升，对应着分率 (1 - 2/5 - 20%)，根据量÷对应的分率 = 单位'1'的量，据此求出这瓶洗衣液原有多少升。本题考查了分数、百分数复合应用题, 关键是确定单位'1', 找到部分对应分率。\n【详解】0.8 ÷ (1 - 2/5 - 20%) = 0.8 ÷ (1 - 0.4 - 0.2) = 0.8 ÷ 0.4 = 2(升)。答：这瓶洗衣液原有 2 升。",
+    "knowledge_points": [
+      "含百分数的运算",
+      "百分数的其他问题"
+    ],
+    "source": "2024·甘肃兰州·小升初真题",
+    "merged_id": "unit6_279",
+    "original_id": "unit6_solve_279"
+  },
+  {
+    "id": "unit6_solve_280",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "杂交水稻，是中国式现代化的'特色名片'。众所周知，'杂交水稻之父'袁隆平院士用'一粒种子改变世界'，既让中国人民远离了饥饿，也为解决全球粮食短缺作出了重要贡献。2021年袁隆平院士指导的杂交水稻示范片'双季稻'年平均产量达到了每公顷24.06吨，比攻关目标提高了 7%。攻关目标每公顷约是多少吨？（得数保留一位小数）",
+    "options": [],
+    "answer": "22.5吨",
+    "solution": "【分析】把攻关目标年平均每公顷水稻的产量看作单位'1'，杂交水稻示范片'双季稻'年平均产量相当于攻关目标年平均每公顷水稻的产量的（1 + 7%），已知一个数的百分之几是多少，求这个数，用除法，用24.06除以（1 + 7%），即可求出攻关目标每公顷约是多少吨。此题的解题关键是掌握已知比一个数多百分之几的数是多少，求这个数的计算方法。\n【详解】24.06 ÷ (1 + 7%) = 24.06 ÷ 1.07 ≈ 22.5(吨)。答: 攻关目标每公顷约是 22.5 吨。",
+    "knowledge_points": [
+      "已知比一个数多/少百分之几是多少，求这个数"
+    ],
+    "source": "22-23六年级下·甘肃兰州·期末",
+    "merged_id": "unit6_280",
+    "original_id": "unit6_solve_280"
+  },
+  {
+    "id": "unit6_solve_281",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一种阿克苏苹果，原来每箱售价为36元。现在由于运输成本提高，单价提高了 25%。原来买10箱的钱，现在能买多少箱？",
+    "options": [],
+    "answer": "8箱",
+    "solution": "【分析】把原来每箱的价格看作单位'1'，则提高后的价格是原来的（1 + 25%），根据求一个数的百分之几是多少，用乘法计算，则提高后的价格是 36 × (1 + 25%)，根据单价 × 数量 = 总价，用36乘10求出原来的钱数，再根据总价÷单价 = 数量，据此求解即可。本题考查求比一个数多百分之几的数是多少，求出提高后的价格是解题的关键。\n【详解】36 × 10 ÷ [36 × (1 + 25%)] = 360 ÷ [36 × 1.25] = 360 ÷ 45 = 8(箱)。答：现在能买 8 箱。",
+    "knowledge_points": [
+      "经济问题",
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "22-23六年级下·重庆潼南·期末",
+    "merged_id": "unit6_281",
+    "original_id": "unit6_solve_281"
+  },
+  {
+    "id": "unit6_solve_282",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "学校图书馆科技书占图书总数的 25%，故事书占图书总数的 15%，科技书比故事书多1280本，学校图书馆共有图书多少本？",
+    "options": [],
+    "answer": "12800 本",
+    "solution": "【分析】由题意可知: 图书总数看作单位'1', 单位'1'是未知的, 关键是求出 1280 本占图书总数的百分之几, 然后根据已知一个数的百分之几是多少, 求这个数, 用除法解答。此题考查百分数的实际应用，找准单位'1'，数量除以对应的分率可求出单位'1'。\n【详解】1280 ÷ (25% - 15%) = 1280 ÷ 10% = 1280 ÷ 0.1 = 12800(本)。答：学校图书馆共有图书 12800 本。",
+    "knowledge_points": [
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "22-23六年级下·广东湛江·期末",
+    "merged_id": "unit6_282",
+    "original_id": "unit6_solve_282"
+  },
+  {
+    "id": "unit6_solve_283",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一本 120 页的书, 小红用三天看完。第一天看了全书总页数的 25%, 第二天与第三天看的页数之比是 3:2, 第二天和第三天各看了多少页?",
+    "options": [],
+    "answer": "第二天看了54页；第三天看了36页。",
+    "solution": "【分析】先求出第一天看的页数，再把剩下的页数分成5等份，求出每份多少页，再解答即可。先求出第一天看的页数，再把剩下的页数分成5等份，求出每份多少页，这是解答此题的关键。\n【详解】120 × 25% = 30(页)。120 - 30 = 90(页)。90 ÷ (3 + 2) = 90 ÷ 5 =18(页)。18 × 3 = 54(页)。18 × 2 = 36(页)。答: 第二天看了 54 页, 第三天看了 36 页。",
+    "knowledge_points": [
+      "按比分配问题",
+      "求一个数的百分之几是多少"
+    ],
+    "source": "2023 六年级下·浙江嘉兴·专题练习",
+    "merged_id": "unit6_283",
+    "original_id": "unit6_solve_283"
+  },
+  {
+    "id": "unit6_solve_284",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "玩具厂生产一种电动玩具，原来每件成本96元，革新技术后，每件成本降低了84元，每件成本降低了百分之几？",
+    "options": [],
+    "answer": "12.5%",
+    "solution": "【分析】先求出每件的成本降低了多少元，用降低了的成本除以原来的成本，就是每件成本降低了多少百分之几。本题是求一个数是另一个数的百分之几, 关键是看把谁当成了单位'1', 单位'1'的量为除数。\n【详解】(96-84) ÷ 96 × 100% = 12 ÷ 96 × 100% = 0.125 × 100% = 12.5%。答: 每件成本降低了 12.5%。",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "18-19六年级上·全国·课后作业",
+    "merged_id": "unit6_284",
+    "original_id": "unit6_solve_284"
+  },
+  {
+    "id": "unit6_solve_285",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "只列式, 不计算。少年宫在元旦期间举行联欢晚会，实到46人，事假2人，病假有2人，参加元旦晚会的出勤率是多少？",
+    "options": [],
+    "answer": "46 ÷ (46 + 2 + 2) × 100%",
+    "solution": "【分析】先表示出参加元旦晚会的总人数，再根据'出勤率 = 出勤人数 ÷ 总人数 × 100%'求出参加元旦晚会的出勤率,据此解答。掌握一个数占另一个数百分之几的计算方法是解答题目的关键。\n【详解】46 ÷ (46 + 2 + 2) × 100% = 46 ÷ 50 × 100% = 0.92 × 100% = 92%。答：参加元旦晚会的出勤率是 92%。",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "22-23 六年级上·湖南怀化·期末",
+    "merged_id": "unit6_285",
+    "original_id": "unit6_solve_285"
+  },
+  {
+    "id": "unit6_solve_286",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "困难",
+    "question": "洪水灾情发生后,某校师生踊跃捐款，六年级（2）班女生捐款数比全班的45%还多 150 元, 男生捐款数是女生捐款数的 2/3, 这个班一共为灾区捐款多少元？",
+    "options": [],
+    "answer": "1000元",
+    "solution": "【分析】将全班捐款数看作单位'1', 全班捐款数 × 45% + 150 元 = 女生捐款数, 再将女生捐款数看作单位'1', 女生捐款数 × 2/3 = 男生捐款数, 设这个班一共为灾区捐款 x 元,根据全班捐款数 - 男生捐款数 = 女生捐款数, 列出方程解答即可。整体数量 × 部分对应分率或百分数 = 部分数量, 用方程解决问题的关键是找到等量关系。\n【详解】解：设这个班一共为灾区捐款 x 元。x - (45%x + 150) × 2/3 = 45%x + 150。x - 3/10x - 100 = 9/20x + 150。7/10x - 9/20x = 150 + 100。1/4x × 4 = 250 × 4。x = 1000。答: 这个班一共为灾区捐款 1000 元。",
+    "knowledge_points": [
+      "列方程解含一个未知数的问题",
+      "求一个数的几分之几的问题",
+      "求一个数的百分之几是多少"
+    ],
+    "source": "19-20 六年级下·重庆黔江·期末",
+    "merged_id": "unit6_286",
+    "original_id": "unit6_solve_286"
+  },
+  {
+    "id": "unit6_solve_287",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "天然气公司铺设一条长2800米的天然气管道，第一周铺设了全长的 45%，第二周铺设了余下的 7/10，还剩多少米没有铺设？",
+    "options": [],
+    "answer": "462 米",
+    "solution": "【分析】根据题意，先将第二周铺的占总长的百分之几计算出来，再利用减法求出剩下的占总长的百分之几。最后，利用乘法求出还剩多少米没有铺设即可。本题考查了百分数的应用, 能根据题意求出剩下的占总长的百分之几是解题的关键。\n【详解】2800 × [1 - 45% - (1 - 45%) × 7/10] = 2800 × [0.55 - 0.55 × 7/10] = 2800 × [0.55 - 0.385] = 2800 × 0.165 = 462(米)。答: 还剩 462 米没有铺设。",
+    "knowledge_points": [
+      "含百分数的运算"
+    ],
+    "source": "21-22六年级下·湖北襄阳·期末",
+    "merged_id": "unit6_287",
+    "original_id": "unit6_solve_287"
+  },
+  {
+    "id": "unit6_solve_288",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "在'庆祝中华人民共和国成立70周年大会'阅兵式上，受阅部队人数约1.5万人，比群众游行人数少了 85%，参加群众游行人数有多少万人？",
+    "options": [],
+    "answer": "10万人",
+    "solution": "【分析】把群众游行人数看作单位'1', 受阅部队人数是群众游行人数的 (1 - 85%), 已知一个数的百分之几是多少,求这个数, 用除法, 列式 1.5 ÷ (1 - 85%) 即可求出参加群众游行的人数。此题的解题关键是掌握已知比一个数少百分之几的数是多少，求这个数的计算方法。\n【详解】1.5 ÷ (1 - 85%) = 1.5 ÷ 15% = 10(万人)。答: 参加群众游行人数有 10 万人。",
+    "knowledge_points": [
+      "已知比一个数多/少百分之几是多少，求这个数"
+    ],
+    "source": "2022·广东广州·小升初真题",
+    "merged_id": "unit6_288",
+    "original_id": "unit6_solve_288"
+  },
+  {
+    "id": "unit6_solve_289",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "甲、乙、丙三个修路队共同修完了一条公路。甲队长说：'我们完成了全部任务的一半。'；乙队长说：'我们修了360米。'；丙队长说：'我们承担了全长的 30%。'请算一算这条公路长多少米？（用方程解）",
+    "options": [],
+    "answer": "1800 米",
+    "solution": "【分析】设这条公路长 x 米, 再根据全长的 (1 - 1/2 - 30%) 是 360 米, 列出方程解答即可。本题考查列方程解决问题、百分数，解答本题的关键是掌握题中的等量关系式。\n【详解】解：设这条公路长 x 米。x × (1 - 1/2 - 30%) = 360。0.2x = 360。x = 1800。答: 这条公路全长 1800 米。",
+    "knowledge_points": [
+      "列方程解含一个未知数的问题",
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "21-22 六年级下·江西抚州·期末",
+    "merged_id": "unit6_289",
+    "original_id": "unit6_solve_289"
+  },
+  {
+    "id": "unit6_solve_290",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "水果批发店运来苹果和梨共180箱，其中苹果的箱数占 60%，由于苹果的销售量较大，所以后来又运来一些苹果，这时苹果的箱数占总数的 70%，求又运进多少箱苹果？",
+    "options": [],
+    "answer": "60箱",
+    "solution": "【分析】求一个数的百分之几是多少，用乘法，180 × 60% 求出原来苹果的箱数; 假设又运来 x 箱苹果, 现在苹果的箱数是 (x + 180 × 60%) 箱, 总的箱数是 (x + 180) 箱, 现在苹果的箱数占总数的 70%, 列出方程, 求解即可。此题主要考查百分数的实际应用, 根据数量关系,列出方程, 解决实际问题。\n【详解】解：设又运来x 箱苹果，x + 180 × 60% = (x + 180) × 70%。x + 108 = 0.7x + 180 × 0.7。x - 0.7x = 126 - 108。0.3x = 18。x = 18 ÷ 0.3。x = 60。答: 又运进 60 箱苹果。",
+    "knowledge_points": [
+      "列方程解含一个未知数的问题",
+      "求一个数的百分之几是多少"
+    ],
+    "source": "2020·湖南永州·小升初真题",
+    "merged_id": "unit6_290",
+    "original_id": "unit6_solve_290"
+  },
+  {
+    "id": "unit6_solve_291",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "青园小学六年级有男生 240 人, 女生人数与男生人数的比是 5:6, 六年级数恰好占全校人数的 25%，青园小学全校共有学生多少人？",
+    "options": [],
+    "answer": "1760 人",
+    "solution": "【分析】先把六年级学生人数看作单位'1'，其中男生占六年级学生人数的 6/(5 + 6)，又知男生有240人，根据分数除法的意义，用男生人数除以 6/(5 + 6) 就是六年级学生人数；再把全校学生总人数看作单位'1'，根据百分数除法的意义，用六年级人数除以 25% 就是全校学生总人数。解答此题的关键是把比转化成分数，再根据分数、百分数除法的意义解答。\n【详解】240 ÷ 6/(5 + 6) ÷ 25% = 240 ÷ 6/11 ÷ 25% = 440 ÷ 25% = 1760(人)。答: 青园小学全校共有学生 1760 人。",
+    "knowledge_points": [
+      "比的应用",
+      "已知一个数的百分之几是多少, 求这个数"
+    ],
+    "source": "21-22 六年级下·湖南永州·期末",
+    "merged_id": "unit6_291",
+    "original_id": "unit6_solve_291"
+  },
+  {
+    "id": "unit6_solve_292",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "某小学去年学生人数是1500人，新修学校后扩招，增加了300人，今年人数比去年增加了百分之几？",
+    "options": [],
+    "answer": "20%",
+    "solution": "【分析】用增加的人数除以去年的人数即可求出今年人数比去年增加了百分之几，据此解答即可。解答本题的关键是读懂问题, 明确今年人数比去年增加了去年的百分之几。\n【详解】300 ÷ 1500 = 20%。答：今年人数比去年增加了 20%。",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "20-21六年级上·云南文山·期末",
+    "merged_id": "unit6_292",
+    "original_id": "unit6_solve_292"
+  },
+  {
+    "id": "unit6_solve_293",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "大珑小学六（1）班50名同学参加体质健康测试，达标率为 92%。班上有多少名同学达标？",
+    "options": [],
+    "answer": "46人",
+    "solution": "【分析】达标率为 92%, 根据公式: 达标率 = 达标人数 ÷ 总人数 × 100%, 即总人数 × 达标率 = 达标人数。本题主要考查百分率问题，需熟练掌握。\n【详解】50 × 92% = 46(名)。答：班上有46名同学达标。",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）",
+      "求一个数的百分之几是多少",
+      "含百分数的运算"
+    ],
+    "source": "2022六年级下·全国·专题练习",
+    "merged_id": "unit6_293",
+    "original_id": "unit6_solve_293"
+  },
+  {
+    "id": "unit6_solve_294",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一根绳子，第一次用去全长的 20%，第二次用去余下长度的 20%，两次所用绳子长度相差 2m。这根绳子原来长( ) m。",
+    "options": [],
+    "answer": "50",
+    "solution": "【分析】根据题意，可以把这根绳子原来的长看作单位'1'，第一次用去全长的 20%，第二次用去剩下的 20% 即 (1 - 20%) × 20% = 16%，两次相差全长的 20% - 16%，两次相差 2m，即可求出原来的长。解答此题的关键是找单位'1'，进一步发现比单位'1'多或少百分之几，由此解决问题。\n【详解】(1 - 20%) × 20% = 80% × 20% = 16%。2 ÷ (20% - 16%) = 2 ÷ 4% = 50(m)。",
+    "knowledge_points": [
+      "含百分数的运算",
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "2021·河北石家庄·小升初真题",
+    "merged_id": "unit6_294",
+    "original_id": "unit6_solve_294"
+  },
+  {
+    "id": "unit6_solve_295",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "园林工人种植一批树苗，结果未成活树苗的棵数占成活树苗的 1/9，求这批树苗的成活率是多少？",
+    "options": [],
+    "answer": "90%",
+    "solution": "【分析】1/9的单位'1'是成活的棵数，设成活的棵数是9棵，那么未存活的棵数是1棵，把成活的棵数和未存活的棵数相加，求出总棵数，再根据成活率 = 成活的棵数 ÷ 总棵数 × 100%，代入数据计算即可。本题设出成活的棵数, 再把总棵数表示出来, 根据求百分率的方法计算。\n【详解】9 ÷ (1 + 9) × 100% = 9 ÷ 10 × 100% = 0.9 × 100% = 90%。答: 这批树苗的成活率是 90%.",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "20-21 六年级上·江西·期中",
+    "merged_id": "unit6_295",
+    "original_id": "unit6_solve_295"
+  },
+  {
+    "id": "unit6_solve_296",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一批货物分三次运完，第一次运了这批货物的 28%，第二次运了这批货物的 3/10，其余的第三次运完。已知前两次共运货物 87 吨，第三次运货物多少吨？",
+    "options": [],
+    "answer": "63吨",
+    "solution": "【分析】把这批货物的总质量看成单位'1', 前两次运走了总质量的 (28% + 3/10), 它对应的数量是 87 吨, 由此用除法求出总质量, 然后用总质量减去前两次运走的质量, 即可解决问题。本题的关键是找出单位'1', 并找出单位'1'的百分之几对应的数量, 用除法就可以求出单位'1'的量, 进而解决问题。\n【详解】87 ÷ (28% + 3/10) = 87 ÷ 0.58 =150(吨)。150 - 87 = 63(吨)。答: 第三次运货物 63 吨。",
+    "knowledge_points": [
+      "含百分数的运算",
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "20-21 六年级上·江西·期中",
+    "merged_id": "unit6_296",
+    "original_id": "unit6_solve_296"
+  },
+  {
+    "id": "unit6_solve_297",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "有浓度是 15% 的盐水300克，要配制成浓度是 40% 的盐水可以采取两种方法：(1) 如果要用加盐的方法, 可以加多少盐? (2) 如果要用蒸发水分的方法, 要蒸发掉多少克水?",
+    "options": [],
+    "answer": "(1) 125 克 (2) 187.5克",
+    "solution": "【分析】（1）根据公式：溶质 = 溶液 × 浓度，由此即可求出盐水里面盐的质量，即 300 × 15% = 45(克），水的质量：300 - 45 = 255(克），由于加盐的方法，即水的量不变，浓度变为 40%，即此时水的含量占了盐水的：1 - 40% = 60%，单位'1'，未知，用除法，即 255 ÷ 60% = 425(克），用425减300即可求出需要加盐的质量。(2) 如果蒸发水分的方法, 那么此时里面盐的质量不变,根据公式: 对应量 ÷ 对应百分率 = 单位'1', 即此时的盐水: 45 ÷ 40% = 112.5(克), 用原有盐水的质量减现在的盐水质量即可求出蒸发掉的水分。本题主要考查浓度问题以及百分数的应用，找准单位'1'，单位'1'未知，用除法。\n【详解】(1) 盐的质量: 300 × 15% = 45(克)。水的质量: 300 - 45 = 255(克)。255 ÷ (1 - 40%) = 255 ÷ 60% = 425(克)。425 - 300 = 125(克)。答: 可以加 125 克盐。(2) 45 ÷ 40% = 112.5(克)。300 - 112.5 = 187.5(克)。答: 要蒸发掉 187.5 克水。",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题)",
+      "已知一个数的百分之几是多少，求这个数",
+      "百分数的其他问题"
+    ],
+    "source": "21-22六年级下·辽宁·期末",
+    "merged_id": "unit6_297",
+    "original_id": "unit6_solve_297"
+  },
+  {
+    "id": "unit6_solve_298",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "小洁原来平均每分钟写 25 个字, 现在平均每分钟写 36 个字, 她现在平均每分钟写字的数量比原来增加了百分之几?",
+    "options": [],
+    "answer": "44%",
+    "solution": "【分析】根据题意，把原来平均每分钟写字的个数看作单位'1', 根据求一个数比另一个数多百分之几, 先求出现在比原来每分钟多写几个字, 用除法解答。\n【详解】(36 - 25) ÷ 25 × 100% = 11 ÷ 25 × 100% = 0.44 × 100% = 44%。答：她现在平均每分钟写字的数量比原来增加了 44%。",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "20-21 六年级上·福建南平·期末",
+    "merged_id": "unit6_298",
+    "original_id": "unit6_solve_298"
+  },
+  {
+    "id": "unit6_solve_299",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "某种子公司用360粒种子做发芽试验，其中有52粒种子没发芽。求发芽率。",
+    "options": [],
+    "answer": "85.6%",
+    "solution": "【分析】根据发芽率的意义，发芽率 = 发芽种子数 / 实验种子总数 × 100%，发芽种子 = 360 - 52(粒）；代入数据，即可解答。根据求一个数是另一个数的百分之几（百分率问题）的知识进行解答，关键是明确发芽的种子 = 实验种子总数 - 没发芽种子数。\n【详解】(360 - 52) / 360 × 100% = 308/360 × 100% ≈ 0.856 × 100% = 85.6%。答: 发芽率为 85.6%",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "20-21六年级上·河北唐山·期中",
+    "merged_id": "unit6_299",
+    "original_id": "unit6_solve_299"
+  },
+  {
+    "id": "unit6_solve_300",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一份稿件，打字员第一天打了总数的 1/4，第二天打了总数的 40%，还剩70页未打，这份稿件有多少页？",
+    "options": [],
+    "answer": "200页",
+    "solution": "【分析】把一份稿件的总页数看作单位'1', 打字员第一天打了总数的 1/4, 第二天打了总数的 40%, 还剩 (1 - 1/4 - 40%) 未打, 用未打的 70 页除以对应的分率即可解答。此题考查的是分数除法的应用，明确单位'1'未知用除法是解题关键。\n【详解】70 ÷ (1 - 1/4 - 40%) = 70 ÷ 0.35 = 200(页)。答: 这份稿件有 200 页。",
+    "knowledge_points": [
+      "分数除法的应用",
+      "含百分数的运算"
+    ],
+    "source": "20-21六年级下·河南新乡·期末",
+    "merged_id": "unit6_300",
+    "original_id": "unit6_solve_300"
+  }, {
+    "id": "unit6_choice_301",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一本书按原价的 $80\\%$ 购买可便宜 3 元, 这本书的原价是 ( ) 元。",
+    "options": [
+      "A. 15",
+      "B. 12",
+      "C. 10",
+      "D. 3.75"
+    ],
+    "answer": "A",
+    "solution": "【分析】把原价看作单位'1'，一本书按原价的 $80\\%$ 购买可便宜3元，即现价是原价的 $80\\%$，便宜的3元是原价的 $(1 - 80\\%)$，求原价，用除法计算。找准单位 '1', 单位 '1'未知, 用除法计算。找出 3 元对应的分率是解题的关键。\n\n【详解】$3 \\div (1 - 80\\%) \\\\ 3 \\div 0. 2 \\\\ = 1 5 \\text{(元)} $\n\n故答案为：A",
+    "knowledge_points": [
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "21-22 六年级下·山东菏泽·阶段练习",
+    "merged_id": "unit6_301",
+    "original_id": "unit6_choice_301"
+  },
+  {
+    "id": "unit6_choice_302",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一赛车手驾车驶进坡道，上坡用了15秒，下坡速度提高了 $25\\%$，下坡用了（）秒。",
+    "options": [
+      "A. 11",
+      "B. 12",
+      "C. 13"
+    ],
+    "answer": "B",
+    "solution": "【分析】先把这段坡道的距离看成单位'1'，上坡用了15秒，那么上坡的速度就是 $\\frac{1}{15}$，下坡的速度提高了 $25\\%$，那么下坡的速度是上坡速度的（$1 + 25\\%$），用上坡的速度乘（$1 + 25\\%$），即可求出下坡的速度，再用1除以下坡的速度，就是下坡用的时间。解决本题先根据速度、路程、时间三者之间的关系，先求出上坡的速度，再根据分数乘法的意义得出下坡的速度，最后用路程除以速度求出下坡的时间。\n\n【详解】上坡的速度： $1 \\div 15 = \\frac{1}{15}$ 下坡的速度:\n\n$\\frac{1}{15}\\times (1 + 25\\%) \\\\ = \\frac {1}{1 5} \\times \\frac {5}{4} \\\\ =\\frac {1}{1 2} $\n\n下坡用时:$1 \\div \\frac{1}{12} = 12$ (秒) 故答案为: B",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少",
+      "分数与整数的除法",
+      "基础行程问题"
+    ],
+    "source": "18-19六年级下·河北邯郸·期末",
+    "merged_id": "unit6_302",
+    "original_id": "unit6_choice_302"
+  },
+  {
+    "id": "unit6_choice_303",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "把 $30\\%$ 的百分号去掉，原来的数就（）。",
+    "options": [
+      "A. 扩大到原来的 100 倍",
+      "B. 缩小到原来的$\\frac{1}{100}$",
+      "C. 不变"
+    ],
+    "answer": "A",
+    "solution": "【分析】$30\\% = 0.3$，把 $30\\%$ 的百分号去掉，原来的数就由0.3变成30，小数点就向右移动了两位，就表示原来的数扩大了100倍。此题属于考查小数与百分数的互化和小数点的位置移动引起小数的大小变化，掌握小数点的位置移动引起小数的大小变化规律是解题关键。\n\n【详解】$30\\% = 0.3, 30\\% →30$ 即 $0.3 →30$，相当于小数点向右移动了两位，就表示原来的数扩大了 100 倍。故答案为: A",
+    "knowledge_points": [
+      "百分数、分数、小数和比的互化",
+      "小数点位置的移动"
+    ],
+    "source": "18-19六年级上·山东菏泽·阶段练习",
+    "merged_id": "unit6_303",
+    "original_id": "unit6_choice_303"
+  },
+  {
+    "id": "unit6_choice_304",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "女生比男生少 $20 \\%$, 可以说成 ( )。",
+    "options": [
+      "A. 男生比女生多 $20\\%$",
+      "B. 女生是男生的80%",
+      "C. 男生是女生的 $80\\%$"
+    ],
+    "answer": "B",
+    "solution": "【分析】把男生的人数看作单位'1'，女生比男生少 $20\\%$，则女生为 $1 - 20\\% = 80\\%$，据此判断即可。本题考查求一个比另一个多百分之几，明确单位'1'是解题的关键。\n\n【详解】假设男生为1，则女生为 $1 - 20\\% = 80\\%$。\n\n$ (1 - 80\\%) \\div 80\\% \\\\ = 20 \\% \\div 80 \\% \\\\ = 25 \\%, $\n\n则男生比女生多 $25 \\%$ $80 \\% \\div 1 = 80 \\%$ 则女生是男生的 $80\\%$。$1 \\div 80 \\% = 125 \\%$则男生是女生的 $125 \\%$故选：B",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几",
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "19-20 六年级下·湖南怀化·单元测试",
+    "merged_id": "unit6_304",
+    "original_id": "unit6_choice_304"
+  },
+  {
+    "id": "unit6_choice_305",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一种商品的价格比原来降低了 $15\\%$，降低到 180 元，原来多少元？列式是（）。",
+    "options": [
+      "A. $180 \\div 15\\%$",
+      "B. $180 \\times (1 - 15 \\%)$",
+      "C. $180 \\div (1 - 15\\%)$"
+    ],
+    "answer": "C",
+    "solution": "【分析】把原来的价格看作单位'1'，比原来降低了 $15\\%$也就是原来的价格 $1 - 15\\% = 85\\%$, 原来的价格 $85\\%$ 就是 180 元, $180 \\div 85\\%$ 就是原来的价格。注意'降低到'与'降低了'的区别，降低到180元意思是现价是180元。\n\n【详解】$180 \\div (1 - 15\\%)\\\\= 180 \\div 85 \\% \\\\ \\approx 2 1 1. 7 6 (\\text {元}) $\n\n故答案选：C。",
+    "knowledge_points": [
+      "已知比一个数多/少百分之几是多少，求这个数"
+    ],
+    "source": "18-19六年级下·广东汕头·单元测试",
+    "merged_id": "unit6_305",
+    "original_id": "unit6_choice_305"
+  },
+  {
+    "id": "unit6_choice_306",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "生产 101 个零件,合格率最高可达 ( )。",
+    "options": [
+      "A. $101\\%$",
+      "B. $100\\%$",
+      "C. $99\\%$"
+    ],
+    "answer": "B",
+    "solution": "【分析】合格率 = 合格零件个数 ÷ 零件总个数 × 100%, 生产的零件全部合格时, 产品合格率最高, 据此解答。掌握合格率的计算方法是解答题目的关键。\n\n【详解】分析可知，$101 \\div 101 \\times 100\\%\\\\ =1 \\times 100 \\% \\\\ = 100 \\% $\n\n所以，合格率最高可达 $100\\%$。故选: B",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "18-19 六年级下·广东汕头·单元测试",
+    "merged_id": "unit6_306",
+    "original_id": "unit6_choice_306"
+  },
+  {
+    "id": "unit6_choice_307",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一件商品先提价 $15\\%$，后又降价 $15\\%$，原价（）现价。",
+    "options": [
+      "A. 大于",
+      "B. 小于",
+      "C. 等于"
+    ],
+    "answer": "A",
+    "solution": "【分析】一件商品先提价 $15 \\%$, 是以原价为单位 '1', 此时的价格是 1.15, 后又降价 $15 \\%$, 是以此时的价格为单位 '1', 则现价是 0.9775, 再比较大小即可。本题考查百分数，解答本题的关键是找准单位'1'。\n\n【详解】原价：1 现价: $1 \\times (1 + 15\\%) \\times (1 - 15\\%)\\\\=1.1 5 \\times 0. 8 5 \\\\ = 0. 9 7 7 5 $\n\n1 > 0.9775 故答案为: A。",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "16-17 六年级上·全国·课后作业",
+    "merged_id": "unit6_307",
+    "original_id": "unit6_choice_307"
+  },
+  {
+    "id": "unit6_choice_308",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "甲数比乙数多 $25\\%$，乙数是甲数的（）。",
+    "options": [
+      "A. $80\\%$",
+      "B. $\\frac{1}{4}$",
+      "C. $75\\%$",
+      "D. $70\\%$"
+    ],
+    "answer": "A",
+    "solution": "【分析】已知甲数比乙数多 $25 \\%$, 把乙数看作单位 '1',则甲数是 $1 + 25 \\%$, 那么乙数是甲数的 $1 \\div (1 + 25 \\%),$  计算即可。此题解答的关键是把乙数看作单位 '1', 表示出甲数是乙数的百分之几, 列式计算。\n\n【详解】$1 \\div (1 + 25\\%)\\\\= 1 \\div 1. 2 5 \\\\ = 0. 8 \\\\ = 80 \\% $乙数是甲数的 $80 \\%$故选: A。",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "19-20六年级上·全国·单元测试",
+    "merged_id": "unit6_308",
+    "original_id": "unit6_choice_308"
+  },
+  {
+    "id": "unit6_choice_309",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "体育场内跑道总长是 400 米, 其中弯道部分与直道部分长度之比为 $2: 3$, 弯道部分比直道部分短 ( )。",
+    "options": [
+      "A. $66.6\\%$",
+      "B. $50\\%$",
+      "C. $33.3\\%$"
+    ],
+    "answer": "C",
+    "solution": "【分析】用跑道总长 $\\div$ 总份数, 求出一份数, 一份数分别乘弯道和直道对应份数, 求出弯道和直道长度, 弯道和直道长度差 $\\div$ 直道长度即可。关键是理解比的意义, 求一个数比另一个数多百分之几用除法。\n\n【详解】$400 \\div (2 + 3)\\\\ = 4 0 0 \\div 5 \\\\ = 8 0 \\text {(米)} $\n\n$8 0 \\times 2 = 1 6 0 (\\text {米})$\n\n$8 0 \\times 3 = 2 4 0 (\\text {米}) \\\\ (2 4 0 - 1 6 0) \\div 2 4 0 \\\\ = 8 0 \\div 2 4 0 \\\\ \\approx 33.3 \\% $\n\n故答案为: C",
+    "knowledge_points": [
+      "按比分配问题",
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "21-22 六年级上·湖南常德·期末",
+    "merged_id": "unit6_309",
+    "original_id": "unit6_choice_309"
+  },
+  {
+    "id": "unit6_choice_310",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "如果甲数比乙数多 $25 \\%$, 那么乙数比甲数少（）。",
+    "options": [
+      "A. $20\\%$",
+      "B. $25\\%$",
+      "C. 不能确定"
+    ],
+    "answer": "A",
+    "solution": "【分析】根据'甲数比乙数多 $25\\%$'可知，乙数为单位'1'，则甲数为 $1 + 25 \\%$, 再用甲数与乙数的差除以甲数即可解答。明确数学信息和数学问题对应的单位 '1' 不同是解答本题的关键。\n\n【详解】$25\\% \\div (1 + 25\\%)\\\\=0.25 \\div 1. 2 5 \\\\ = 20\\%$; \n\n故答案为: A。",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "2018·山东·小升初真题",
+    "merged_id": "unit6_310",
+    "original_id": "unit6_choice_310"
+  },
+  {
+    "id": "unit6_choice_311",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "如果六（1）班人数的 $60 \\%$ 和六（2）班人数的 $\\frac{2}{3}$ 一样多，那么（）。",
+    "options": [
+      "A. 六（1）班人数多",
+      "B. 六 (2) 班人数多",
+      "C. 无法确定"
+    ],
+    "answer": "A",
+    "solution": "【分析】由题意可知, 六 (1) 班的人数 $\\times 60 \\% =$ 六 (2) 班的人数 $\\times \\frac{2}{3}$, 当乘法算式的乘积一定时, 如果已知因数越小,那么与它相乘的另一个因数越大; 相反地, 已知因数越大,与它相乘的另一个因数就越小; 据此比较六 (1) 班和六(2) 班的人数即可。掌握乘数和积的关系是解答题目的关键。\n\n【详解】六（1）班的人数 $\\times 60\\% =$ 六（2）班的人数 $\\times \\frac{2}{3}$ 则六（1）班的人数 $\\times \\frac{3}{5} =$ 六（2）班的人数 $\\times \\frac{2}{3}$\n\n因为 $\\frac{3}{5}<\\frac{2}{3}$, 所以六 (1) 班的人数 $>6$ (2) 班的人数故答案为: A",
+    "knowledge_points": [
+      "因数和积的大小关系（小数乘法）",
+      "异分母异分子分数的大小比较",
+      "求一个数的几分之几的问题",
+      "求一个数的百分之几是多少"
+    ],
+    "source": "21-22六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_311",
+    "original_id": "unit6_choice_311"
+  },
+  {
+    "id": "unit6_choice_312",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "在53的后面添上'%'，这个数就（）。",
+    "options": [
+      "A. 扩大到原来的 100 倍",
+      "B. 不变",
+      "C. 缩小到原来的 $\\frac{1}{100}$"
+    ],
+    "answer": "C",
+    "solution": "【分析】在 53 的后面添上 '%', 相当于把 53 的小数点向左移动了两位, 缩小到原来的 $\\frac{1}{100}$。掌握百分数和小数互化的方法是解答题目的关键。\n\n【详解】53 的后面添上 '%'是 $53 \\%$, $53 \\% = 0.53$, 53 缩小到原来的 $\\frac{1}{100}$ 即为 0.53。故答案为: C",
+    "knowledge_points": [
+      "小数点位置的移动",
+      "百分数、小数和分数的互化"
+    ],
+    "source": "21-22六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_312",
+    "original_id": "unit6_choice_312"
+  },
+  {
+    "id": "unit6_choice_313",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "把单位 '1'平均分成 100 份, 阴影部分占 35 份, 空白部分占单位 '1' 的 ()。",
+    "options": [
+      "A. $35\\%$",
+      "B. $135\\%$",
+      "C. $65\\%$"
+    ],
+    "answer": "C",
+    "solution": "【分析】由题意可知，阴影部分占'1'的 $35\\%$，空白部分占单位'1'的百分率 $= 1 -$ 阴影部分占单位'1'的百分率。掌握百分数的意义是解答题目的关键。\n\n【详解】$1 - 35\\% = 65\\%$ 故答案为: C",
+    "knowledge_points": [
+      "百分数的意义",
+      "含百分数的运算"
+    ],
+    "source": "21-22 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_313",
+    "original_id": "unit6_choice_313"
+  },
+  {
+    "id": "unit6_choice_314",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "王大伯家上月用电 100 千瓦时, 本月比上月节约了 20 千瓦时, 比上月节约了 ( )。",
+    "options": [
+      "A. $80\\%$",
+      "B. $40\\%$",
+      "C. $20\\%$"
+    ],
+    "answer": "C",
+    "solution": "【分析】B 比 A 少百分之几的计算方法：(A-B) $\\div \\mathrm{A} \\times 100 \\%$, 据此解答。掌握一个数比另一个数少百分之几的计算方法是解答题目的关键。\n\n【详解】$20 \\div 100 \\times 100\\%\\\\ 0.2 \\times 100 \\% \\\\ = 20 \\% $\n\n故答案为: C",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "21-22 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_314",
+    "original_id": "unit6_choice_314"
+  },
+  {
+    "id": "unit6_choice_315",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "在扇形统计图中,其中一个扇形的圆心角是 $72^{\\circ}$, 则这个扇形所表示的数量占总数量的 ( ) $\\%$.",
+    "options": [
+      "A. 72",
+      "B. 60",
+      "C. 40",
+      "D. 20"
+    ],
+    "answer": "D",
+    "solution": "【分析】整个扇形统计图是 $360^{\\circ}$, 用 $72^{\\circ} \\div 360^{\\circ}$ 即可求出这个扇形所表示的数量占总数量的百分之几。本题考查扇形的圆心角, 明确整个扇形统计图是 $360^{\\circ}$ 是解题的关键。\n\n【详解】$72^{\\circ} \\div 360^{\\circ} = \\frac{1}{5} = 20\\%$ 则这个扇形所表示的数量占总数量的 $20\\%$。故答案为: D",
+    "knowledge_points": [
+      "弧、圆心角、扇形的认识",
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "20-21 六年级上·河北秦皇岛·期末",
+    "merged_id": "unit6_315",
+    "original_id": "unit6_choice_315"
+  },
+  {
+    "id": "unit6_choice_316",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "甲数的 $60 \\%$ 等于乙数的 $75 \\%$ （两数均不为 0），那么甲数和乙数比较，（）。",
+    "options": [
+      "A. 甲数大",
+      "B. 乙数大",
+      "C. 一样大"
+    ],
+    "answer": "A",
+    "solution": "【分析】求一个数的百分之几是多少用乘法, 根据积一定,一个数乘的数越大, 其本身越小, 进行分析。关键是理解百分数的意义，整体数量 $\\times$ 部分对应百分率 $=$ 部分数量。\n\n【详解】$60 \\%$ 甲 $= 75 \\%$ 乙, $60 \\%< 75 \\%$, 所以甲 $>$ 乙。故答案为: A",
+    "knowledge_points": [
+      "含百分数的运算",
+      "求一个数的百分之几是多少"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_316",
+    "original_id": "unit6_choice_316"
+  },
+  {
+    "id": "unit6_choice_317",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "某商店按 $5 \\%$ 缴纳营业税 200 元，则商场的营业额是（）元。",
+    "options": [
+      "A. 400",
+      "B. 1000",
+      "C. 4000"
+    ],
+    "answer": "C",
+    "solution": "【分析】把营业额看作单位'1'，根据营业额的 $5 \\%$ 是200元，所以用200除以 $5 \\%$ 就是营业额是多少元。营业税 $=$ 营业额 $\\times$ 税率; 已知一个是的百分之几是多少, 求这个数用除法。\n\n【详解】$200 \\div 5\\% = 4000$ （元） 则该商店的营业额是4000元。故答案为: C",
+    "knowledge_points": [
+      "税率问题"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_317",
+    "original_id": "unit6_choice_317"
+  },
+  {
+    "id": "unit6_choice_318",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "50 克盐溶解在 200 克水中, 盐水的含盐率为 ( )。",
+    "options": [
+      "A. $25\\%$",
+      "B. $20\\%$",
+      "C. $80\\%$"
+    ],
+    "answer": "B",
+    "solution": "【分析】根据含盐率 $=$ 盐的质量 $\\div$ 盐水的质量 $\\times 100 \\%$, 据此解答即可。本题考查含盐率问题, 明确含盐率 $=$ 盐的质量 $\\div$ 盐水的质量 $\\times 100 \\%$ 是解题的关键。\n\n【详解】$50 \\div (50 + 200) \\times 100\\%\\\\50 \\div 250 \\times 100 \\% \\\\ = 0.2 \\times 100 \\% \\\\ = 20 \\% $\n\n则盐水的含盐率为 $20 \\%$。故选: B",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题)"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_318",
+    "original_id": "unit6_choice_318"
+  },
+  {
+    "id": "unit6_choice_319",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "下面各数中可以写成百分数的是 ( )。",
+    "options": [
+      "A. 甲数是乙数的 $\\frac{1}{2}$",
+      "B. 一个菠萝重是 0.5千克",
+      "C. 小明身高 1.5 米"
+    ],
+    "answer": "A",
+    "solution": "【分析】根据百分数的意义，百分数表示两者之间的关系，所以百分数不带单位。本题考查百分数的意义，明确百分数的意义是解题的关键。解答此题注意'分率'和'具体数量'的区别:'具体数量'带有单位名称，而分率不能带单位名称，'具体数量'和'分率'不能互化。\n\n【详解】由分析可知:只有 A 项中的分数没有带单位, 表示甲数与乙数之间的关系, 所以可以写成百分数。故答案为：A",
+    "knowledge_points": [
+      "百分数的意义"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_319",
+    "original_id": "unit6_choice_319"
+  },
+  {
+    "id": "unit6_choice_320",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "'已修公路长度占公路全长的 $68\\%$'，这句话的含义是（）。",
+    "options": [
+      "A. 已修公路 68 千米",
+      "B. 公路全长共 100 千米",
+      "C. 假如公路全长是 100 千米, 已修公路是 68 千米"
+    ],
+    "answer": "C",
+    "solution": "【分析】已修公路长度占公路全长的 $68 \\%$, 是把公路全长看作单位 '1', 由于不知道公路全长具体是多少, 所以选项 A、B 都是错误的, 只有选项 C 是正确的, 据此解答。\n\n【详解】'已修公路长度占公路全长的 $68 \\%$, 是把公路全长看作单位 '1',本题中公路全长具体是多少不知道。故选：C",
+    "knowledge_points": [
+      "百分数的意义"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_320",
+    "original_id": "unit6_choice_320"
+  },
+  {
+    "id": "unit6_fill_321",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "把 5 米长的铁丝平均分成 8 段, 每段占全长的( )%, 每段长( )米。",
+    "options": [],
+    "answer": "$12.5 \\quad \\frac{5}{8}$",
+    "solution": "【分析】根据分数的意义, 把 5 米长的铁丝平均分成 8 段,即把这根铁丝的长度看作单位 '1', 将它平均分成 8 份, 那么每份长度就是全长的 $\\frac{1}{8}$, 将分数化成百分数。每段的长度 = 铁丝总长度 ÷ 所分的段数。求一个数是另一个数的几分之几/百分之几，用除法。\n\n【详解】$1 \\div 8 = \\frac{1}{8} = 12.5\\%$ ，$5 \\div 8 = \\frac{5}{8}$ （米）每段占全长的 $12.5 \\%$, 每段长 $\\frac{5}{8}$ 米。",
+    "knowledge_points": [
+      "分数与除法的关系",
+      "求一个数是另一个数的百分之几 (百分率问题)"
+    ],
+    "source": "17-18 六年级上·吉林·期末",
+    "merged_id": "unit6_321",
+    "original_id": "unit6_fill_321"
+  },
+  {
+    "id": "unit6_fill_322",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一桶油有 210 千克,用去的是剩下的 $40 \\%$, 剩下( )千克。",
+    "options": [],
+    "answer": "150",
+    "solution": "【分析】根据题意, 把剩下的质量看作单位 '1', 则用去的质量是剩下的 $40 \\%$; 总质量 = 用去的质量 + 剩下的质量,所以 210 千克对应的分率是 $(1 + 40 \\%)$ ，单位 '1' 未知,用除法计算, 求出剩下的质量。找准单位 '1', 单位 '1'未知, 用具体的数量 $\\div$ 对应的分率 $=$ 单位 '1'的量。\n\n【详解】$210 \\div (1 + 40\\%)\\\\= 2 1 0 \\div 1. 4 \\\\ = 1 5 0 (\\text {千 克}) $",
+    "knowledge_points": [
+      "单位 '1'的认识与确定",
+      "已知一个数的百分之几, 是多少, 求这个数"
+    ],
+    "source": "20-21 六年级下·湖南衡阳·期中",
+    "merged_id": "unit6_322",
+    "original_id": "unit6_fill_322"
+  },
+  {
+    "id": "unit6_fill_323",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "一件商品原价500元，实际售价400元。实际售价比原价便宜了( )%。",
+    "options": [],
+    "answer": "20",
+    "solution": "【分析】把原价看成单位'1'，先用实际售价除以原价，求出实际售价是原价的百分之几，用原价1减去实际售价占原价的分率，即可求出实际售价比原价便宜百分之几。此题属于求一个数比另一个数少百分之几, 把被比的数量看作单位 '1', 用除法解答。\n\n【详解】$1 - 400 \\div 500\\\\= 1 - 80 \\% \\\\ = 20 \\% $",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "2021·天津河西·小升初真题",
+    "merged_id": "unit6_323",
+    "original_id": "unit6_fill_323"
+  },
+  {
+    "id": "unit6_fill_324",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一件商品若先涨价20%后又下降20%，商品的价格比原来( )了；若先下降20%后又上涨 20%, 商品的价格比原来( )了。(填'上涨'或'下降'）",
+    "options": [],
+    "answer": "下降 下降",
+    "solution": "【分析】将商品的原价看作单位'1'，用单位'1'乘（1+20%）再乘（1-20%），求出一件商品若先涨价20%后又下降20%之后的价格，从而判断出价格是上涨了还是下降了。同理，求出若先下降20%后又上涨20%，商品的价格比原来是上涨了还是下降了。本题考查了含百分数的运算，求比一个数多(少) 百分之几的数是多少, 用乘法。\n\n【详解】$1 \\times (1 + 20\\%) \\times (1 - 20\\%)\\\\=120 \\% \\times 80 \\% \\\\ = 96 \\% $\n\n$96\\% < 1$，所以，一件商品若先涨价 $20\\%$ 后又下降 $20\\%$，商品的价格比原来下降了；\n\n$ 1 \\times (1 - 20 \\%) \\times (1 + 20 \\%) \\\\ = 80 \\% \\times 120 \\% \\\\ = 96 \\% $\n\n$96\\% < 1$，所以，若先下降 $20\\%$ 后又上涨 $20\\%$，商品的价格比原来下降了。",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少",
+      "含百分数的运算"
+    ],
+    "source": "21-22六年级下·全国·单元测试",
+    "merged_id": "unit6_324",
+    "original_id": "unit6_fill_324"
+  },
+  {
+    "id": "unit6_fill_325",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "六年级某班女生人数占全班的 $\\frac{3}{8}$, 那么女生占男生人数的( )%, 女生比男生少()%。",
+    "options": [],
+    "answer": "60 40",
+    "solution": "【分析】六年级某班女生人数占全班的 $\\frac{3}{8}$, 转化成女生人数与全班的人数之比是 $3: 8$, 女生人数占 3 份, 全班人数占 8 份, 所以男生占 5 份, 用女生人数所占的份数除以男生人数所占的份数得到第一问的答案。男生的人数减去女生的人数, 女生少的这部分人数除以男生的人数求出第二问的答案。此题的解题关键是利用分数与比之间的关系，分别找出女生与男生所占的份数，即可求出最终的结果。\n\n【详解】女生人数与全班的人数之比是 $3: 8$, 所以女生人数与男生人数之比是 $3: 5$\n\n$ 3 \\div (8 - 3) \\\\ =3 \\div 5 \\\\ =0. 6 \\\\ = 60 \\%$ \n\n$ \\\\ (5 - 3) \\div 5 \\\\= 2 \\div 5 \\\\ = 0. 4 \\\\ =40 \\% $",
+    "knowledge_points": [
+      "比的应用",
+      "求一个数比另一个数多/少百分之几",
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "21-22 六年级下·全国·单元测试",
+    "merged_id": "unit6_325",
+    "original_id": "unit6_fill_325"
+  },
+  {
+    "id": "unit6_fill_326",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "24 千克是 32 千克的( )%, 24 千米比 32 千米少( )%。",
+    "options": [],
+    "answer": "75 25",
+    "solution": "【分析】求一个数是另一个数的百分之几，用除法计算；求一个数比另一个数少百分之几，用少的部分除以另一个数即可。此题的解题关键是掌握求一个数是另一个数百分之几和求一个数比另一个数少百分之几的计算方法。\n\n【详解】 $24 \\div 32 = 0.75 = 75\\%$ \n\n $\\\\ (32 - 24) \\div 32 \\\\= 8 \\div 32 \\\\= 0.25 \\\\=25 \\% $",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几",
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "21-22 六年级下·全国·单元测试",
+    "merged_id": "unit6_326",
+    "original_id": "unit6_fill_326"
+  },
+  {
+    "id": "unit6_fill_327",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一根长11.2米的绳子剪成5段，每段长( )米，是全长的( )%。",
+    "options": [],
+    "answer": "2.24 20",
+    "solution": "【分析】用全长除以段数，求出每段的长；把全长看作单位'1'，求出每段占单位'1'的百分之几即可。本题考查百分数, 解答本题的关键是掌握求一个数是另一个数的百分之几用除法计算。\n\n【详解】$11.2 \\div 5 = 2.24$ （米）$1 \\div 5 = 20 \\%$",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "21-22六年级下·全国·单元测试",
+    "merged_id": "unit6_327",
+    "original_id": "unit6_fill_327"
+  },
+  {
+    "id": "unit6_fill_328",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "近年来，中国高铁由原来的200千米/时连续进行了两次大提速，第一次提速 $25\\%$，第二次在第一次基础上再提速 $40\\%$，那么，现在的速度是( )千米/时，比原来的速度提高了( )%。",
+    "options": [],
+    "answer": "350 75",
+    "solution": "【分析】（1）把原来高铁的速度看作单位'1'，现在的速度 $=$ 原来的速度 $\\times (1 + 25\\%) \\times (1 + 40\\%)$ ;(2) 现在比原来速度提高的百分率 = (现在的速度 - 原来的速度) ÷ 原来的速度 $\\times 100\\%$ ; 据此解答。A比B多百分之几的计算方法：（A一B）÷B×100%。\n\n【详解】 (1) $200 \\times (1 + 25\\%) \\times (1 + 40\\%)\\\\= 2 0 0 \\times 1. 2 5 \\times 1. 4 \\\\ = 2 5 0 \\times 1. 4 \\\\ = 3 5 0 \\text {(千 米 / 时)} $\n\n(2) $(350 - 200) \\div 200 \\times 100\\%\\\\=150 \\div 200 \\times 100 \\% \\\\ = 0.75 \\times 100 \\% \\\\ = 75 \\% $",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少",
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "21-22六年级上·湖北随州·期末",
+    "merged_id": "unit6_328",
+    "original_id": "unit6_fill_328"
+  },
+  {
+    "id": "unit6_fill_329",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "在一个减法算式中，减数是差的 $50\\%$，已知被减数十减数十差 $= 8$，差是( )。",
+    "options": [],
+    "answer": "$\\frac{8}{3}$ 或 $\\frac{2}{3}$",
+    "solution": "【分析】假设差是未知数 $x$, 求一个数的百分之几是多少,用乘法可以表示出减数是 $50 \\% x$, 利用被减数一减数 = 差,用字母表示出被减数, 代入到被减数十减数十差 = 8 中, 可解出最终的结果。此题的解题关键是弄清题意, 把差设为未知数 $x$ ,找出题中数量间的相等关系, 列出包含 $x$ 的等式, 解方程得到最终的结果。\n\n【详解】假设差是未知数 $x$，减数 $650 \\% x$，被减数 $=$ 减数十 + 差 $650 \\% x + x$，代入到被减数十减数十 + 差 $= 8$ 中，列出方程:$ 50 \\% x + x + 50 \\% x + x = 8 \\\\ 0. 5 x + x + 0. 5 x + x = 8 \\\\ 3 x = 8 $\n\n解得 $x = \\frac{8}{3}$ 或 $x = 2 \\frac{2}{3}$, 即是差的结果。",
+    "knowledge_points": [
+      "列方程解含一个未知数的问题",
+      "求一个数的百分之几是多少"
+    ],
+    "source": "19-20六年级下·湖南怀化·单元测试",
+    "merged_id": "unit6_329",
+    "original_id": "unit6_fill_329"
+  },
+  {
+    "id": "unit6_fill_330",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一种商品原价 80元，现价比原价降低了 $20\\%$，现价( )元；一种商品售价80元，比原价降低了20元，降低了( )%。",
+    "options": [],
+    "answer": "64 20",
+    "solution": "【分析】（1）把商品原价看作单位'1'，现价是原价的$（1-20\\%)$ ，现价 $=$ 原价 $\\times (1 - 20\\%)$(2) 原价 $= \\text{售价} + 20$ 元，售价比原价少的百分率 $= \\text{售价}$ 原价降低的价格 $\\div$ 商品原价 $\\times 100\\%$ ；据此解答。B比A少百分之几的计算方法：$（A-B) \\div A \\times 100 \\%$\n\n【详解】 (1) $80 \\times (1 - 20\\%)= 8 0 \\times 0. 8= 64 (元)$ \n\n(2) $20 \\div (80 + 20) \\times 100\\%\\\\= 20 \\div 100 \\times 100 \\% \\\\ = 20 \\% $",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少",
+      "经济问题",
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "19-20 六年级下·湖南怀化·单元测试",
+    "merged_id": "unit6_330",
+    "original_id": "unit6_fill_330"
+  },
+  {
+    "id": "unit6_fill_331",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "( )是 100 米 $20\\%$ ; 12 千克比( )少 $\\frac{1}{3}$。",
+    "options": [],
+    "answer": "20 18",
+    "solution": "【分析】第一空求一个数的百分之几是多少，用乘法。第二空把要填空的地方设为未知数 $x$ ，根据数量关系： $x$ 减去 12 千克比 $x$ 少的部分等于 12 千克，12 千克比 $x$ 少的部分即 $\\frac{1}{3} x$ 列出方程，求出 $x$ 。此题的关键是掌握求一个数的百分之几是多少和求一个数比另一个数多几分之几的方法, 弄清楚数量关系,列出算式, 求得最终的结果。\n\n【详解】 (1) $100 \\times 20\\% = 20$ (2) 设填空的地方为 $x$, 根据分析列出方程:\n\n$x - \\frac {1}{3} x = 1 2\\\\\\ \\frac {2}{3} x = 1 2\\\\x = 1 2 \\div \\frac {2}{3}$\n\n解得 $x = 18$",
+    "knowledge_points": [
+      "已知比一个数多/少百分之几是多少，求这个数",
+      "求一个数的百分之几是多少"
+    ],
+    "source": "19-20 六年级下·湖南怀化·单元测试",
+    "merged_id": "unit6_331",
+    "original_id": "unit6_fill_331"
+  },
+  {
+    "id": "unit6_fill_332",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "只列式不计算。加工一批零件，完成原计划的 $\\frac{4}{5}$ 后，又加工了 400 个，结果超额完成 $20\\%$，原计划加工零件多少个？",
+    "options": [],
+    "answer": "$400 \\div (120\\% - \\frac{4}{5})$",
+    "solution": "【分析】由题意可知, 把这批零件看作单位 '1', 则超额完成的分率为 $1 + 20 \\%$, 用超额完成的分率减去原计划的 $\\frac{4}{5}$ 即为 400 个所对应的分率, 根据分数除法的意义, 用除法解答即可。本题考查分数除法，明确部分的量÷所对应的分率 $=$ 单位 '1'的量是解题的关键。\n\n【详解】$400 \\div (120\\% - \\frac{4}{5})\\\\= 4 0 0 \\div \\frac {2}{5} \\\\\\ = 1 0 0 0 (\\text {个}) $\n\n答: 原计划加工零件 1000 个。",
+    "knowledge_points": [
+      "分数与整数的除法",
+      "分数除法的应用"
+    ],
+    "source": "18-19 六年级下·广东汕头·单元测试",
+    "merged_id": "unit6_332",
+    "original_id": "unit6_fill_332"
+  },
+  {
+    "id": "unit6_fill_333",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "某班今天有 48名学生到校，请事假、病假各1人，这个班今天的出勤率是（）。",
+    "options": [],
+    "answer": "$96\\%$",
+    "solution": "【分析】出勤率 = 出勤人数 $\\div$ 全班人数 $\\times$ ，据此解答即可。本题考查百分数, 解答本题的关键是掌握出勤率的计算公式。\n\n【详解】$48 \\div (48 + 1 + 1) \\times 100\\%\\\\=48 \\div 50 \\times 100 \\% \\\\ = 96 \\% $",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "18-19六年级下·广东汕头·单元测试",
+    "merged_id": "unit6_333",
+    "original_id": "unit6_fill_333"
+  },{
+    "id": "unit6_fill_334",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一瓶可乐, 喝了它的 $\\frac{3}{5}$, 还剩( )%, 如果剩下 400 毫升, 则喝了( )毫升。",
+    "options": [],
+    "answer": "40 600",
+    "solution": "【分析】把这瓶可乐的总量看作单位 '1', 1—喝的分率 =剩下的分率; 剩下的分率所对应得具体数量是 400 毫升, 用 $400 \\div$ 剩下的分率 = 可乐的总量, 用总量减去剩下 400 毫升即可解答。求单位 '1'的量用除法，具体数量÷分率 = 单位'1'的量。\n\n【详解】$1 - \\frac{3}{5} = \\frac{2}{5} = 40\\%$ \n\n$ 400 \\div 40 \\% - 400 \\\\ = 1 0 0 0 - 4 0 0 \\\\ = 6 0 0 \\text {(毫 升)} $",
+    "knowledge_points": [
+      "百分数、小数和分数的互化",
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "18-19 六年级下·广东汕头·单元测试",
+    "merged_id": "unit6_334",
+    "original_id": "unit6_fill_334"
+  },
+  {
+    "id": "unit6_fill_335",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "若 $\\frac{3}{4} \\mathrm{~A} = \\frac{1}{2} \\mathrm{~B}$, 则 $ A:B = (\\quad): (\\quad) $, A 比 B 少 ()$\\%$。",
+    "options": [],
+    "answer": "2 3 33.3",
+    "solution": "【分析】利用等式的性质, 得到 $A: B = \\frac{1}{2}: \\frac{3}{4} = 2: 3$, 再用少的部分除以 B 的 3 份, 求出 A 比 B 少百分之几即可。本题考查比、百分数, 解答本题的关键是掌握比的化简方法和求一个数比另一个数少百分之几的方法。\n\n【详解】$A:B = \\frac{1}{2}: \\frac{3}{4} = 2: 3\\\\=(2 - 3) \\div 3 \\\\= 1 \\div 3  \\\\ \\approx 33.3 \\%$",
+    "knowledge_points": [
+      "比的化简",
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "18-19 六年级下·广东汕头·单元测试",
+    "merged_id": "unit6_335",
+    "original_id": "unit6_fill_335"
+  },
+  {
+    "id": "unit6_fill_336",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一件卫衣从 150 元降低了 30 元, 现价是原价的( )%。",
+    "options": [],
+    "answer": "80",
+    "solution": "【分析】用原价减去降低的价格求出现价，再用现价除以原价即可。本题考查百分数，解答本题的关键是掌握求一个数是另一个数的百分之几的计算方法。\n\n【详解】$(150 - 30) \\div 150\\\\= 1 2 0 \\div 1 5 0 \\\\ = 80 \\% $",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "18-19 六年级下·广东汕头·单元测试",
+    "merged_id": "unit6_336",
+    "original_id": "unit6_fill_336"
+  },
+  {
+    "id": "unit6_fill_337",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "修一条公路，已修的是未修的 $60\\%$，已经修了840米，这条公路全长是( )米。",
+    "options": [],
+    "answer": "2240",
+    "solution": "【分析】把未修的公路长度看作单位'1'，已修的公路长度占未修公路长度的 $60\\%$，根据'量÷对应的百分率'求出未修部分的长度，最后计算已修的公路长度和未修公路长度之和即可。掌握标准量的计算方法是解答题目的关键。\n\n【详解】$840 \\div 60\\% + 840= 1 4 0 0 + 8 4 0= 2240$ (米)所以，这条公路全长是 2240 米。",
+    "knowledge_points": [
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "21-22六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_337",
+    "original_id": "unit6_fill_337"
+  },
+  {
+    "id": "unit6_fill_338",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "某超市8月的营业额是20万元，如果按营业额的 $5\\%$ 缴纳营业税，某超市8月应缴纳营业税( )万元。",
+    "options": [],
+    "answer": "1",
+    "solution": "【分析】应缴纳营业税 $=$ 营业额 $\\times 5\\%$ ，由题意可知，营业额为20万元，把营业额代入公式计算即可。掌握应纳税额的计算方法是解答题目的关键。\n\n【详解】$20 \\times 5\\% = 1$ （万元）",
+    "knowledge_points": [
+      "税率问题",
+      "求一个数的百分之几是多少"
+    ],
+    "source": "21-22六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_338",
+    "original_id": "unit6_fill_338"
+  },
+  {
+    "id": "unit6_fill_339",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "欣欣服装店9月份卖出上衣 120 件，10 月份卖出的上衣是 9 月份的 $80\\%$，10月份卖出上衣( )件。",
+    "options": [],
+    "answer": "96",
+    "solution": "【分析】把9月份卖出上衣的数量看作单位'1'，10月份卖出上衣的数量=9月份卖出上衣的数量×80%。找准题目中的单位 '1', 求一个数的百分之几是多少, 用乘法计算。\n\n【详解】$120 \\times 80\\% = 96$ （件）",
+    "knowledge_points": [
+      "求一个数的百分之几是多少"
+    ],
+    "source": "21-22六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_339",
+    "original_id": "unit6_fill_339"
+  },
+  {
+    "id": "unit6_fill_340",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "某公司今天到岗人数是190人，缺10人，今天该公司工人的出勤率是( )$\\%$。",
+    "options": [],
+    "answer": "95",
+    "solution": "【分析】出勤率 $=$ 出勤人数 $\\div$ 总人数 $\\times 100\\%$ ，据此解答即可。本题考查百分数, 解答本题的关键是掌握出勤率的计算公式。\n\n【详解】$190 \\div (10 + 190)\\\\= 1 9 0 \\div 2 0 0 \\\\ = 95 \\% $",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "21-22六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_340",
+    "original_id": "unit6_fill_340"
+  },
+  {
+    "id": "unit6_fill_341",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "超市里进了两种牛奶，甲品牌进了40箱，卖出 $70\\%$，乙品牌进了50箱，卖出 $90\\%$，两种牛奶分别卖出( )箱和( )箱。",
+    "options": [],
+    "answer": "28 45",
+    "solution": "【分析】求一个数的百分之几是多少，用乘法计算。本题考查百分数, 解答本题的关键是掌握求一个数的百分之几的计算方法。\n\n【详解】$40 \\times 70\\% = 28$ （箱） $50 \\times 90\\% = 45$ （箱）",
+    "knowledge_points": [
+      "求一个数的百分之几是多少"
+    ],
+    "source": "21-22六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_341",
+    "original_id": "unit6_fill_341"
+  },
+  {
+    "id": "unit6_fill_342",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "六(1)班有男生 25 人, 女生 24 人, 男生人数比女生人数多( )%, 女生人数比男生人数少( )%。",
+    "options": [],
+    "answer": "4.2 4",
+    "solution": "【分析】求一个数比另一个数多（少）百分之几，用多(少) 的部分除以另一个数解答即可。本题考查百分数，解答本题的关键是找准单位'1'。\n\n【详解】$ (25-24) ÷ 24×100\\%\\\\= 1 \\div 24 \\times 100 \\% \\\\ \\approx 4.2 \\%$ \n\n$(25 - 24) \\div 25 \\times 100 \\% \\\\ = 1 \\div 25 \\times 100 \\% \\\\ = 4 \\% $",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "21-22 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_342",
+    "original_id": "unit6_fill_342"
+  },
+  {
+    "id": "unit6_fill_343",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "40 米比 80 米少 $(\\quad)\\% ; 80$ 米比 40 米多 $(\\quad)\\%$ 。",
+    "options": [],
+    "answer": "50 100",
+    "solution": "【分析】40 米比 80 米少 40 米, 把 80 米看作单位 '1', 80米比 40 米多 40 米, 把 40 米看作单位 '1', 先求出一个数比另一个数多 (或少) 的具体量, 再除以单位 '1' 的量, 据此解答。找准单位 '1', 用两数量之差 $\\div$ 单位 '1' 的量是解决此题的关键。\n\n【详解】40米比80米少：$ (8 0 - 4 0) \\div 8 0 \\\\ = 4 0 \\div 8 0 \\\\ = 50 \\% $\n\n80 米比 40 米多:\n\n$ (8 0 - 4 0) \\div 4 0 \\\\ = 4 0 \\div 4 0 \\\\ = 100 \\% $",
+    "knowledge_points": [
+      "百分数、小数和分数的互化",
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "21-22 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_343",
+    "original_id": "unit6_fill_343"
+  },
+  {
+    "id": "unit6_fill_344",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "在括号里填上 '>' '<' 或 '='。 $20 \\% (\\quad) \\frac {1}{2} \\quad \\frac {1}{3} (\\quad) 33 \\% \\quad 0.36(\\quad) 36 \\%$",
+    "options": [],
+    "answer": "< > =",
+    "solution": "【分析】先把百分数和分数都化为小数，再按照小数大小比较的方法即可比较大小。本题主要考查小数、分数、百分数之间的关系及其转化和小数大小的比较\n\n【详解】$20\\% = 0.2, \\frac{1}{2} = 0.5$ \n\n $20 \\% <   \\frac {1}{2} \\\\ \\frac{1}{3} = 0.333\\dots ,33\\% = 0.33 \\\\ \\frac{1}{3} >33\\% \\\\ 36 \\% = 0.36 \\\\ 0.36 = 36 \\% $",
+    "knowledge_points": [
+      "一位小数的大小比较",
+      "多位小数的大小比较",
+      "百分数、小数和分数的互化"
+    ],
+    "source": "21-22 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_344",
+    "original_id": "unit6_fill_344"
+  },
+  {
+    "id": "unit6_fill_345",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "8 比 5 多( )%;5比8少( )%；( )： $8=0.5$ 。",
+    "options": [],
+    "answer": "60 37.5 4",
+    "solution": "【分析】求一个数比另一个数多（少）百分之几，用多（少）的部分除以另一个数即可。本题考查百分数、比, 解答本题的关键是掌握求比的前项是用比的后项乘比值求出。\n\n【详解】$(8 - 5) \\div 5 = 3 \\div 5 = 60\\%$\n\n$(8 - 5) \\div 8 \\\\= 3 \\div 8 = 37.5 \\% \\\\ =8 \\times 0. 5 \\\\= 4 $",
+    "knowledge_points": [
+      "比的意义",
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "21-22 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_345",
+    "original_id": "unit6_fill_345"
+  },
+  {
+    "id": "unit6_fill_346",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "甲乙两数比是 $4: 5$, 则甲数比乙数少 $(\\quad)\\%),$  乙数比甲数多 $(\\quad)\\%$。",
+    "options": [],
+    "answer": "20 25",
+    "solution": "【分析】用甲数比乙数的差除以乙数即可求出甲数比乙数少百分之几；用甲数比乙数的差除以甲数即可求出乙数比甲数多百分之几。求一个数比另一个数多（少）百分之几，用两个数的差除以另一个数即可。\n\n【详解】$(5 - 4) \\div 5 \\\\= 1 \\div 5 \\\\ = 20 \\%$;  \n\n$\\\\ (5 - 4) \\div 4 \\\\ = 1 \\div 4 \\\\ = 25\\%$;",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "2019 六年级下·四川·专题练习",
+    "merged_id": "unit6_346",
+    "original_id": "unit6_fill_346"
+  },
+  {
+    "id": "unit6_fill_347",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "葡萄晒成葡萄干要失去原质量的 $80 \\%$, 现有 320 千克的葡萄可以晒出( )千克的葡萄干; 如果要晒 50 千克的葡萄干需要准备( )千克葡萄。",
+    "options": [],
+    "answer": "64 250",
+    "solution": "【分析】葡萄晒成葡萄干要失去原质量的 $80 \\%$, 可知葡萄干是葡萄质量的 $1 - 80 \\% = 20 \\%$; 已知有 320 千克的葡萄,依据: 求一个数的百分之几是多少, 用乘法解答。已知 50 千克的葡萄干, 它占葡萄质量的 $20 \\%$, 根据: 已知一个数的百分之几是多少, 求这个数用除法计算。已知单位 '1'的量用乘法; 求单位 '1'的量用除法。\n\n【详解】$1 - 80\\% = 20\\%\\\\=320 \\times 20 \\% \\\\ = 3 2 0 \\times 0. 2 \\\\ = 6 4 (\\text {千 克}) $\n\n$50 \\div 20 \\% = 250 (\\text{千克})$",
+    "knowledge_points": [
+      "已知一个数的百分之几是多少，求这个数",
+      "求一个数的百分之几是多少"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_347",
+    "original_id": "unit6_fill_347"
+  },
+  {
+    "id": "unit6_fill_348",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "一次数学测试，满分是100分，小明的正确率是 $95\\%$，小明应得( )分；如果满分是120分，正确率还是 $95\\%$，小明应得( )分。",
+    "options": [],
+    "answer": "95 114",
+    "solution": "【分析】根据：满分 $\\times$ 正确率 $=$ 小明应得的分数，据此解答。求一个数的百分之几是多少，用乘法解答。\n\n【详解】$100 \\times 95\\% = 95$ （分）$120 \\times 95 \\% = 114 (\\text{分})$",
+    "knowledge_points": [
+      "求一个数的百分之几是多少"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_348",
+    "original_id": "unit6_fill_348"
+  },
+  {
+    "id": "unit6_fill_349",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一本书 120 页,已读了 $40 \\%$, 还剩( )页没有读。",
+    "options": [],
+    "answer": "72",
+    "solution": "【分析】由题意可知, 已读了 $40 \\%$, 还剩 $1 - 40 \\%$ 没有读,根据求一个数的百分之几是多少, 用乘法进行计算即可。本题考查求一个数的百分之几是多少, 明确用乘法是解题的关键。\n\n【详解】$120 \\times (1 - 40\\%)= 120 \\times 60 \\%=72 (页)$则还剩 72 页没有读。",
+    "knowledge_points": [
+      "求一个数的百分之几是多少"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_349",
+    "original_id": "unit6_fill_349"
+  },
+  {
+    "id": "unit6_fill_350",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "甲、乙两数的比是 $3: 4$, 甲数是乙数的 $(\\quad)\\% ;$ 甲数比乙数多 $10 \\%$, 乙数比甲数少 $(\\quad)\\%$ (百分号前保留一位小数), 甲数是乙数的 $(\\quad)\\%$ 。",
+    "options": [],
+    "answer": "75 9.1 110",
+    "solution": "【分析】（1）甲、乙两数的比是 $3:4$ ，就是甲数是3份，乙数是 4 份, 甲数是乙数的百分之几用甲数 $\\div$ 乙数列式计算;(2) 甲数比乙数多 $10 \\%$, 把乙数是单位 '1', 那么甲数就是乙数的 $(1 + 10 \\%)$, 求乙数比甲数多百分之几, 那么此时的单位 '1' 就是甲数, 就用两者的差除以甲数即可。找准单位 '1', 找清楚数量关系, 用其中的一个数来表示出另一个数就不难解决。\n\n【详解】$3 \\div 4 = 0.75 = 75\\%$\n\n$ (1 + 10\\% -1)\\div (1 + 10\\%) \\\\ = 0. 1 \\div 1. 1 \\\\ \\approx 0. 0 9 1 \\\\ = 9.1 \\% $\n\n$1 + 10 \\% = 110 \\%$",
+    "knowledge_points": [
+      "比的意义",
+      "求一个数比另一个数多/少百分之几",
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_350",
+    "original_id": "unit6_fill_350"
+  },
+  {
+    "id": "unit6_fill_351",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "王明和李新同学进行抛掷硬币的实验。王明抛100次，正面向上有48次，李新抛50次，正面向上有26次。王明抛硬币正面向上次数是抛掷总次数的( )%, 李新抛硬币正面向上次数是抛掷总次数的( )$\\%$。",
+    "options": [],
+    "answer": "48 52",
+    "solution": "【分析】（1）王明抛硬币正面向上次次数占抛掷总次数的百分率 $=$ 正面向上的次数 $\\div$ 抛掷总次数 $\\times 100\\%$ ;(2) 李新抛硬币正面向上次数占抛掷总次数的百分率 = 正面向上的次数 ÷ 抛掷总次数 $\\times 100\\%$, 据此解答。掌握一个数占另一个数百分之几的计算方法是解答题目的关键。\n\n【详解】 (1) $48 \\div 100 \\times 100\\%\\\\= 0.48 \\times 100 \\% \\\\ = 48 \\% $\n\n(2) $26 \\div 50 \\times 100 \\%\\\\= 0.52 \\times 100 \\%\\\\= 52 \\%$",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_351",
+    "original_id": "unit6_fill_351"
+  },
+  {
+    "id": "unit6_fill_352",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "为了缓解交通拥挤的状况，公路局对云龙大道进行了拓宽改造。路宽由原来的 12 米增加到 21 米，拓宽了( )‰。",
+    "options": [],
+    "answer": "75",
+    "solution": "【分析】云龙大道原来的宽度是单位 '1', (拓宽后的宽度一原来的宽度) ÷原来的宽度 = 扩宽了百分之几, 据此列式计算。此类问题一般用表示单位'1'的量作除数。\n\n【详解】$(21 - 12) \\div 12\\\\= 9 \\div 1 2 \\\\ = 75 \\% $",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_352",
+    "original_id": "unit6_fill_352"
+  },
+  {
+    "id": "unit6_fill_353",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "男性标准体重计算法（单位：身高为厘米，体重为千克）为：（身高一80) $\\times70\\% =$ 标准体重, 张叔叔身高 175 厘米, 据此算法,他的标准体重应是( )千克。",
+    "options": [],
+    "answer": "66.5",
+    "solution": "【分析】由题意可知，将 175 厘米代入公式中进行计算即可。本题考查含百分数的运算, 熟练掌握百分数化小数是解题的关键。\n\n【详解】$(175 - 80) \\times 70\\%\\\\=95 \\times 70 \\% \\\\ = 6 6. 5 \\text {(千 克)} $",
+    "knowledge_points": [
+      "含百分数的运算"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_353",
+    "original_id": "unit6_fill_353"
+  },
+  {
+    "id": "unit6_fill_354",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "刘庄超额完成计划种树任务的 $10\\%$，表示刘庄实际完成种树计划的()。",
+    "options": [],
+    "answer": "$110\\%$",
+    "solution": "【分析】由题意可知, 实际完成的数量占计划的 $1 + 10 \\%$ ,由此解答即可。明确单位'1'是解答本题的关键。\n\n【详解】$1 + 10\\% = 110\\%$",
+    "knowledge_points": [
+      "百分数的其他问题"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_354",
+    "original_id": "unit6_fill_354"
+  },
+  {
+    "id": "unit6_judge_355",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "一次抽奖活动的中奖率是 $1 \\%$, 抽 100 次一定会中奖。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】根据可能性大小的判断方法，结合题干，分析解题即可。本题考查了可能性的大小, 抽奖活动中奖率是 $1 \\%$, 表示每次可能中奖, 也可能不中奖。\n\n【详解】一次抽奖活动的中奖率是 $1 \\%$, 表示中奖率比较低,但是不代表抽 100 次一定会中奖。故答案为: ×",
+    "knowledge_points": [
+      "可能性的大小",
+      "百分数的意义"
+    ],
+    "source": "2021·黑龙江牡丹江·小升初真题",
+    "merged_id": "unit6_355",
+    "original_id": "unit6_judge_355"
+  },
+  {
+    "id": "unit6_judge_356",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "一条公路长 $70 \\%$ 千米。",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】根据百分数的意义，表示一个数是另一个数的百分之几的数叫做百分数。因为百分数只能表示两个数的倍数关系，不能表示具体数量，据此判断。此题考查的目的是理解百分数的意义，明确：百分数只能表示两个数的倍数关系，不能表示具体数量是解题关键。\n\n【详解】因为百分数只能表示两个数的倍数关系，不能表示具体数量。所以，一条公路长 $70 \\%$ 千米，是错误的。故答案为: ×",
+    "knowledge_points": [
+      "百分数的意义"
+    ],
+    "source": "2020·湖南怀化·小升初真题",
+    "merged_id": "unit6_356",
+    "original_id": "unit6_judge_356"
+  },
+  {
+    "id": "unit6_judge_357",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "把 25 千克食盐溶解到 100 克水里, 盐水的含盐率为 $25 \\%$ ( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】用盐的质量加上水的质量，求出盐水的质量。用盐的质量除以盐水的质量，求出盐水的含盐率。本题考查了百分率，求一个数是另一个数的百分之几，用除法。\n\n【详解】$25 \\div (25 + 100)\\\\= 2 5 \\div 1 2 5 \\\\ = 20 \\% $\n\n所以，盐水的含盐率是 $20 \\%$。所以判断错误。",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "21-22 六年级下·全国·单元测试",
+    "merged_id": "unit6_357",
+    "original_id": "unit6_judge_357"
+  },
+  {
+    "id": "unit6_judge_358",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "一道数学题全班有 50 人做, 10 人做错, 这道题的正确率是 $40 \\%$ 。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】正确率 = 做对题目的人数 ÷ 做题目的总人数 × 100%, 把题中数据代入公式求出结果并和题中的正确率比较即可。掌握正确率的计算方法是解答题目的关键。\n\n【详解】 (50-10) $\\div 50 \\times 100\\%$ $= 40 \\div 50 \\times 100\\%$ $= 0.8 \\times 100\\%$ $= 80\\%$ 故答案为：×",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "21-22 六年级上·湖北随州·期末",
+    "merged_id": "unit6_358",
+    "original_id": "unit6_judge_358"
+  },
+  {
+    "id": "unit6_judge_359",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "$\\frac{39}{100}$ 和 $39 \\%$ 都是分母是 100 的分数, 所以它们的意义和大小都完全相同。()",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】 $\\frac{39}{100}$ 表示把单位'1'平均分成100份取其中的39份，它即可以表示一个分率，也可以表示一个数量；而 $39\\%$ 只表示一个数占另一个数的百分之三十九，它们所表示的意义不完全相同，只是大小相等。此题考查的是百分数的意义, 解答此题关键是明确分数与百分数的区别和联系。\n\n【详解】由分析得，39 和 39%都是分母是 100 的分数，它们的大小相同，意义不同。故答案为: ×",
+    "knowledge_points": [
+      "分数的意义",
+      "百分数的意义"
+    ],
+    "source": "21-22 六年级上·河北邯郸·期末",
+    "merged_id": "unit6_359",
+    "original_id": "unit6_judge_359"
+  },
+  {
+    "id": "unit6_judge_360",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "一桶油先用去 $20\\%$ 后，再增加 $20\\%$，油的总重量与原来一样。（ ）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】把这桶油的重量看作单位'1'，第一次用去后的重量=这桶油的重量×（1-20%），第二次增加后的重量=第一次用去后的重量×（1+20%），据此解答。掌握求比一个数多（少）百分之几的数是多少的计算方法是解答题目的关键，在理解情景问题的基础上找到'整体'也就是常说的单位'1'的量。\n\n【详解】把这桶油的重量看作单位'1'第二次增加后的重量:  \n\n$1\\times (1 -20\\%) \\times (1 + 20\\%)\\\\=8 \\times 1.2\\\\=0.96$ $0.96 < 1$,\n\n 即第二次增加后重量比原来的那桶油的重量轻。故答案为: ×",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少",
+      "含百分数的运算"
+    ],
+    "source": "21-22六年级上·广东汕尾·期末",
+    "merged_id": "unit6_360",
+    "original_id": "unit6_judge_360"
+  },
+  {
+    "id": "unit6_judge_361",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "甲数比乙数少 $20\\%$，也就是乙数比甲数多 $20\\%$。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】甲数比乙数少百分之几是把乙数看作单位 '1', 甲数比乙数少的百分率 = (乙数 - 甲数) ÷ 乙数 $\\times 100\\%$ ;\n\n乙数比甲数多百分之几是把甲数看作单位 '1', 乙数比甲数多的百分率 = (乙数 - 甲数) ÷ 甲数 $\\times 100 \\%$; 据此解答。A比B多百分之几的计算方法：（A一B) $\\div \\mathrm{B} \\times 100 \\%$; B 比 A 少百分之几的计算方法: $(A- B) \\div A \\times 100 \\%$\n\n【详解】假设乙数为1 ,甲数: $1 \\times (1 - 20\\%) = 0.8$,乙数比甲数多的百分率： $(1 - 0.8) \\div 0.8 \\times 100\\%$\n\n$ 0.2 \\div 0.8 \\times 100 \\% \\\\ = 0.25 \\times 100 \\% \\\\ = 25 \\% $\n\n所以，甲数比乙数少 $20 \\%$, 乙数比甲数多 $25 \\%$ 。故答案为: ×",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "2022·新疆吐鲁番·小升初真题",
+    "merged_id": "unit6_361",
+    "original_id": "unit6_judge_361"
+  },
+  {
+    "id": "unit6_judge_362",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "在含盐 $25\\%$ 的盐水中，加入同样多的盐和水后，盐水的含盐率不变。（）。",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】含盐率 = 盐的质量 ÷ 盐水的质量 × 100%, 假设出原来盐和盐水的质量, 加入同样多的盐和水后求出现在盐水的含盐率, 最后比较含盐率是否相等, 据此解答。掌握盐水含盐率的计算方法是解答题目的关键。\n\n【详解】在含盐 25 克的 100 克盐水中加入 20 克盐和 20 克水。现在盐的质量: $25 + 20 = 45$ (克)。现在盐水的质量: $100 + 20 + 20 = 140$ (克)现在的含盐率： $45 \\div 140 \\times 100\\%\\\\ \\approx 0.321 \\times 100 \\% \\\\ = 32.1 \\% $\n\n因为 $32.1\\% \\neq 25\\%$, 所以盐水的含盐率改变。所以答案为: ×",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "21-22六年级上·河南驻马店·期末",
+    "merged_id": "unit6_362",
+    "original_id": "unit6_judge_362"
+  },
+  {
+    "id": "unit6_judge_363",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "甲数是乙数的 $80 \\%$, 则乙数比甲数多 $25 \\%$ 。( )",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】把乙数看作单位 '1', 甲数是 $80 \\%$, 求乙数比甲数多百分之几时, 用 $(1 - 80 \\%) \\div 80 \\%$ 即可算出, 进而可以判断对错。求一个数比另一个数多或少百分之几时，用多或少的具体数量÷另一个数×100%。\n\n【详解】$(1 - 80\\%) \\div 80\\%\\\\ 20 \\% \\div 80 \\% \\\\ = 25 \\%$ \n\n所以乙数比甲数多 $25 \\%$ ,故答案为: √。",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "18-19 六年级下·广东汕头·单元测试",
+    "merged_id": "unit6_363",
+    "original_id": "unit6_judge_363"
+  },
+  {
+    "id": "unit6_judge_364",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "甲车间生产零件的合格率比乙车间高，说明甲车间生产的合格数量比乙车间多。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】甲车间合格零件的个数 = 甲车间生产零件总数 $\\times$ 合格率, 乙车间合格零件数 = 乙车间生产零件总数 $\\times$ 合格率,因为甲车间和乙车间零件总数不确定, 所以两个车间生产的合格零件数也就无法比较; 据此解答。明确合格率的含义并能灵活利用是解答本题的关键。\n\n【详解】虽然甲车间生产零件的合格率比乙车间高, 但是甲车间和乙车间零件总数不确定, 所以不能说甲车间生产的合格数量比乙车间多, 原题说法错误;故答案为: ×",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）",
+      "求一个数的百分之几是多少"
+    ],
+    "source": "19-20六年级下·浙江杭州·期末",
+    "merged_id": "unit6_364",
+    "original_id": "unit6_judge_364"
+  },
+  {
+    "id": "unit6_judge_365",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "百分数可以看作分母是 100 的分数。( )",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】百分数是一种特殊的分数，常常不写成分母是 100 的分数形式，而是在原来的分子后面添加上百分号 '%'来表示。表示一个数是另一个数的百分之几的数叫百分数,也叫百分比、百分率。\n\n【详解】百分数可以看作分母是 100 的分数, 说法正确。故答案为: √",
+    "knowledge_points": [
+      "百分数的意义"
+    ],
+    "source": "2022·四川巴中·小升初真题",
+    "merged_id": "unit6_365",
+    "original_id": "unit6_judge_365"
+  },
+  {
+    "id": "unit6_judge_366",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "一种商品的价格先降低 $10\\%$，再提高 $10\\%$，价格比原价低。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】假设原价是 100 元，原价 $\\times$ 降低后对应百分率 $\\times$ 再提高后对应百分率 $=$ 现价，比较即可。关键是确定单位 '1', 整体数量 $\\times$ 部分对应百分率 =部分数量。\n\n【详解】假设原价是100元。$100 \\times (1 - 10 \\%) \\times (1 + 10 \\%) \\\\ = 1 0 0 \\times 0. 9 \\times 1. 1 \\\\ = 9 9 \\text {(元)} $\n\n$99 < 100$, 价格比原价低, 所以原题说法正确。",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "2019·全国·小升初模拟",
+    "merged_id": "unit6_366",
+    "original_id": "unit6_judge_366"
+  }, {
+    "id": "unit6_judge_367",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "$15 m$ 的 $\\frac{2}{5}$ 和 $8m$的 $75\\%$ 同样长。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】求一个数的几分之几（百分之几），用这个数乘几分之几（百分之几）解答即可。本题考查分数乘法、百分数，解答本题的关键是掌握求一个数的几分之几（百分之几）的计算方法。\n\n【详解】$15 \\times \\frac{2}{5} = 6 \\quad (\\mathrm{m})\\\\8 \\times 75 \\% = 6 (m)$\n\n本题说法正确，故答案为：√。",
+    "knowledge_points": [
+      "求一个数的几分之几的问题",
+      "求一个数的百分之几是多少"
+    ],
+    "source": "21-22六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_367",
+    "original_id": "unit6_judge_367"
+  },
+  {
+    "id": "unit6_judge_368",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "某产品的合格率达到 $102\\%$ 。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】根据合格率 $=$ 合格产品数 $\\div$ 产品总数 $\\times 100 \\%$, 即使所有的产品都合格, 合格率最高只能为 $100 \\%$ 进行判断即可。本题主要考查了百分率的问题。\n\n【详解】合格率 $=$ 合格产品数 $\\div$ 产品总数 $\\times 100\\%$; 所以合格率最高为 $100\\%$ ;所以题干说法错误。故答案为: × 。",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "21-22六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_368",
+    "original_id": "unit6_judge_368"
+  },
+  {
+    "id": "unit6_judge_369",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "如果 $A: B = \\frac{3}{7}: \\frac{5}{7}$, 那么 A 比 B 少 $40 \\%$ 。( )",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】先把 $\\mathrm{A}: \\mathrm{B} = \\frac{3}{7}: \\frac{5}{7}$ 根据比的性质进行化简成: $A: B=3:5$, A 是 3 份, B 是 5 份, 然后用它们的差除以 B 即可解答。此题考查的是比的应用，掌握用它们的差÷单位'1'=多（少）的分率是解题关键。\n\n【详解】由 $A: B = \\frac{3}{7}: \\frac{5}{7}$ 得A:B=3:5, A是3份, B是5份,$ (5 - 3) \\div 5 \\\\ = 2 \\div 5 \\\\ = 40 \\% $\n\n故答案: √",
+    "knowledge_points": [
+      "比的化简",
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "21-22 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_369",
+    "original_id": "unit6_judge_369"
+  },
+  {
+    "id": "unit6_judge_370",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "如果女生人数比男生人数少 $20 \\%$, 那么男生人数就比女生人数多 $20 \\%$ ( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】把男生人数看作单位'1'，女生人数占男生的（1）$-20\\%$ ，据此计算男生人数比女生人数多的百分率即可。A比B多百分之几的计算方法：$（A-B) \\div B \\times 100 \\%$\n\n【详解】$[1 - (1 - 20\\%)] \\div (1 - 20\\%) \\times 100\\%\\\\= 0.2 \\div 0.8 \\times 100 \\% \\\\ = 0.25 \\times 100 \\% \\\\ = 25 \\% $\n\n所以，男生人数比女生人数多 $25\\%$。故答案为: ×",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "19-20 五年级下·山东·单元测试",
+    "merged_id": "unit6_370",
+    "original_id": "unit6_judge_370"
+  },
+  {
+    "id": "unit6_judge_371",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "一件商品，先提价 $25\\%$，再降价 $25\\%$ 后，现价就等于原价。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】假设这件商品 100 元，原价 $x$ 提价后对应百分率 $x$ 再降价后对应百分率 $=$ 现价，比较即可。关键是理解百分率的含义，整体数量 $\\times$ 部分对应百分率 $=$ 部分数量。\n\n【详解】假设这件商品100元。\n\n$ 100 \\times (1 + 25 \\%) \\times (1 - 25 \\%) \\\\ = 1 0 0 \\times 1. 2 5 \\times 0. 7 5 \\\\ = 9 3. 7 5 \\text {(元)} $9 3. 7 5 < 1 0 0\n\n故答案为: ×",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "2018 六年级上·全国·专题练习",
+    "merged_id": "unit6_371",
+    "original_id": "unit6_judge_371"
+  },
+  {
+    "id": "unit6_judge_372",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "大牛和小牛的头数比是 4:5, 表示大牛比小牛少 $\\frac{1}{4}$ 。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】B 比 A 少几分之几的计算方法: (A-B) $\\div$ A, 据此解答。本题考查了分数除法的应用，找准标准量是解答题目的关键。\n\n【详解】大牛比小牛少: $(5 - 4) \\div 5 = \\frac{1}{5}$。故判断错误。",
+    "knowledge_points": [
+      "比的应用",
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "2014·浙江宁波·小升初真题",
+    "merged_id": "unit6_372",
+    "original_id": "unit6_judge_372"
+  },
+  {
+    "id": "unit6_judge_373",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "$0.35 \\%$ m和 $0.35^{m}$ 同样长。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】百分数是一种特殊的分数，表示两个数之间的倍比关系，不表示具体的数量，所以后面不带单位名称。表示一个数是另一个数的百分之几的数叫百分数,也叫百分比、百分率。\n\n【详解】$0.35 \\%$ 是百分数，后面不能带单位名称，所以原题说法错误。",
+    "knowledge_points": [
+      "百分数的意义"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_373",
+    "original_id": "unit6_judge_373"
+  },
+  {
+    "id": "unit6_judge_374",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "将 3 克糖放入100克水中，糖的质量占糖水的 $3\\%$ （）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】糖+水=糖水，糖的质量÷糖水质量=糖占糖水的百分之几。此类问题一般用表示单位'1'的量作除数。\n\n【详解】$3 \\div (3 + 100)\\\\= 3 \\div 1 0 3 \\\\ \\approx 2.9 \\% $\n\n故答案为: ×",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_374",
+    "original_id": "unit6_judge_374"
+  },
+  {
+    "id": "unit6_judge_375",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "五年级学生参加劳动的人数占五年级总人数的 $100\\%$，就是五年级的大部分学生都参加了劳动。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】根据题意, 把五年级的总人数看作单位 '1', 五年级学生参加劳动的人数占五年级总人数的 $100 \\%$, 就是五年级的全部学生都参加了劳动; 由此判断即可。此题主要考查了百分数的意义的应用，解答此题的关键是把五年级的总人数看作单位'1'，参加的人数占 $100\\%$ ，就是全部参加。\n\n【详解】五年级学生参加劳动的人数占五年级总人数的 $100\\%$ ，就是五年级的全部学生都参加了劳动，所以本题说法错误；故答案为：×",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "20-21六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_375",
+    "original_id": "unit6_judge_375"
+  },
+  {
+    "id": "unit6_judge_376",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "六年级完成了浇树任务的 $120\\%$，就是超额完成浇树任务的 $20\\%$。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】根据题意, 把六年级的浇树任务看作单位 '1', 用六年级完成的浇树任务的百分率减去 1, 求出超额完成浇树任务的百分之几即可。此题主要考查了百分数的意义的应用, 解答此题的关键是把六年级的浇树任务看作单位 '1', 求出超额完成浇树任务的百分之几即可。\n\n【详解】因为 $120\\% - 1 = 20\\%$ ,所以六年级完成了浇树任务的 $120 \\%$, 就是超额完成浇树任务的 $20 \\%$ 。所以题中说法正确。故答案为: √",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "20-21六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_376",
+    "original_id": "unit6_judge_376"
+  },
+  {
+    "id": "unit6_judge_377",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "$38.62 \\%$ 读作百分之三十八点六十二。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】百分数的读法：与分数的读法相同，先读分母，再读分子；据此判断即可。此题考查百分数的读写法。\n\n【详解】$38.62 \\%$ 读作：百分之三十八点六二，所以原题说法错误；故答案为: ×",
+    "knowledge_points": [
+      "百分数的读法和写法"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_377",
+    "original_id": "unit6_judge_377"
+  },
+  {
+    "id": "unit6_judge_378",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "$79 \\%$ 与 $\\frac{79}{100}$ 读法完全相同，意义也相同。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】根据分数和百分数的意义: $\\frac{79}{100}$ 表示是把单位 '1'平均分成 100 份, 表示其中 79 份的数; $79 \\%$ 表示的是一个数是另一个数的百分之七十九, 百分数不能带单位名称,即不能表示具体的数量; 据此判断即可。此题考查了分数和百分数的意义的区别, 掌握分数和百分数的意义的区别: 百分数表示的是一个数是另一个数的百分之几; 分母是 100 的分数表示是把单位 '1' 平均分成 100 份, 表示其中几份的数是解题关键。\n\n【详解】$79 \\%$ 读作百分之七十九, 它表示的是一个数是另一个数的百分之七十九,79 读作一百分之七十九，表示是把单位 '1'平均分成 100份，表示其中 79 份的数；它们的读法相同，表示的意义也不相同；原题说法错误；故答案为：×",
+    "knowledge_points": [
+      "分数的意义",
+      "百分数的意义"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_378",
+    "original_id": "unit6_judge_378"
+  },
+  {
+    "id": "unit6_judge_379",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "一瓶水重 56.5%千克。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】任何一个百分数都不能表示具体的数量，也不能带单位名称，据此解答。掌握百分数的意义是解答题目的关键。\n\n【详解】百分数表示一个数是另一个数的百分之几，百分数也叫做百分率或百分比，百分数后面不能带单位名称，一瓶水的重量不能用 $56.5\\%$ 千克表示。故答案为: ×",
+    "knowledge_points": [
+      "百分数的意义"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_379",
+    "original_id": "unit6_judge_379"
+  },
+  {
+    "id": "unit6_calc_380",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "解方程。 $40 x + 49 = 89 \\quad \\frac{3}{5} x = 20\\times 25\\% \\quad x - 65\\% x = 70$",
+    "options": [],
+    "answer": "$x = 1$ ； $x = \\frac{25}{3}$ 或 $8\\frac{1}{3}$ ； $x = 200$",
+    "solution": "【分析】（1）根据等式的性质 1 和性质 2，两边先同时减 49，再同时除以 40；(2) 先将右边的计算出来, 再根据等式的性质 2, 两边同时除以 $\\frac{3}{5}$ ; (3) 先将左边进行合并, 再根据等式的性质 2 解方程。\n\n【详解】$40x + 49 = 89$\n\n解: $40 x = 89 - 49\\\\4 0 x = 4 0\\\\x = 1$\n\n$\\frac{3}{5} x = 20\\times 25\\%$\n\n解: $\\frac{3}{5} x = 5\\\\x = 5 \\div \\frac {3}{5}\\\\x = \\frac {2 5}{3} \\text {或} 8 \\frac {1}{3}$\n\n解： $0.35x = 70$\n\n$ x - 65 \\% x = 70 \\\\ x = 7 0 \\div 0. 3 5 \\\\ x = 2 0 0 $",
+    "knowledge_points": [
+      "含百分数的运算",
+      "应用等式的性质 1 和 2 解方程"
+    ],
+    "source": "19-20 六年级下·湖南怀化·单元测试",
+    "merged_id": "unit6_380",
+    "original_id": "unit6_calc_380"
+  },
+  {
+    "id": "unit6_solve_381",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "新华书店搞促销活动，一本《格林童话》降价 $20\\%$ 后，现在售价为24元，《格林童话》原来的售价是多少元？",
+    "options": [],
+    "answer": "30元",
+    "solution": "【分析】把原价看作单位'1'，降价 $20\\%$ 后是24元，实际是表示比原价少 $20\\%$ 的数是24元，按照已知比一个数少百分之几的数是多少，求这个数的方法，可用除法求得结果。此题考查了已知比一个数少百分之几的数是多少,求这个数的问题, 需掌握这类应用题的分析和解答的技巧。\n\n【详解】$24 \\div (1 - 20\\%) \\\\=2 4 \\div 0. 8 \\\\ = 3 0 (\\text {元}) $\n\n答: 《格林童话》原来的售价是 30 元。",
+    "knowledge_points": [
+      "已知比一个数多/少百分之几是多少，求这个数",
+      "含百分数的运算"
+    ],
+    "source": "21-22六年级上·河南安阳·期末",
+    "merged_id": "unit6_381",
+    "original_id": "unit6_solve_381"
+  },
+  {
+    "id": "unit6_solve_382",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "建造一栋楼房，计划投资100万元，实际超用了 $10\\%$，实际投资了多少万元？(只列式不计算)",
+    "options": [],
+    "answer": "$100 \\times (1 + 10\\%)$ 万元",
+    "solution": "【分析】把计划投资的钱数看作单位'1'，实际投资的钱数占计划投资钱数的 $(1 + 10\\%)$，实际投资的钱数 $=$ 计划投资的钱数 $\\times (1 + 10\\%)$，据此解答。掌握求比一个数多百分之几的数是多少的计算方法是解答题目的关键。\n\n【详解】$100 \\times (1 + 10\\%)\\\\= 1 0 0 \\times 1. 1 \\\\ = 1 1 0 (\\text {万 元}) $\n\n答: 实际投资了 110 万元。",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "19-20六年级下·湖南怀化·单元测试",
+    "merged_id": "unit6_382",
+    "original_id": "unit6_solve_382"
+  },
+  {
+    "id": "unit6_solve_383",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一件原价为2000元的服装按'降价 $30\\%$ '的方式出售，降价后便宜了多少钱？",
+    "options": [],
+    "answer": "600元",
+    "solution": "【分析】根据题意，降价 $30\\%$ ，实际是在原价的基础之上，再减去原价的 $30\\%$ ，算出现在的价格。用原价减现价即是便宜的价钱。此题的解题关键是掌握比一个数少百分之几的方法，明确数量关系，列出算式，求出最终的结果。\n\n【详解】现价 $2000 - 2000 \\times 30\\%\\\\=2000-600\\\\=1400 (元)$\n\n便宜的价钱 $20 0 0 - 1 4 0 0=600 (元)$答: 降价后便宜了 600 元。",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "19-20六年级下·湖南怀化·单元测试",
+    "merged_id": "unit6_383",
+    "original_id": "unit6_solve_383"
+  },
+  {
+    "id": "unit6_solve_384",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "建造一栋楼房,计划投资100万元，实际节约了 $10\\%$，节约了多少万元？(只列式不计算)",
+    "options": [],
+    "answer": "10万元",
+    "solution": "【分析】根据求一个数的百分之几是多少，用乘法即可求出节约多少钱。本题考查求一个数的百分之几是多少, 明确用乘法是解题的关键。\n\n【详解】$100 \\times 10\\% = 10$ （万元）答：节约了10万元。",
+    "knowledge_points": [
+      "求一个数的百分之几是多少"
+    ],
+    "source": "19-20六年级下·湖南怀化·单元测试",
+    "merged_id": "unit6_384",
+    "original_id": "unit6_solve_384"
+  },
+  {
+    "id": "unit6_solve_385",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "建造一栋楼房，用了90万元，比计划节约了 $10\\%$，计划投资多少万元？(只列式不计算)",
+    "options": [],
+    "answer": "$90 \\div (1 - 10\\%)$",
+    "solution": "【分析】实际用的比计划节约了 $10\\%$ ，计划投资金额是单位 '1'，实际用的是计划的 $(1 - 10\\%)$ ，根据分数除法的意义，可据此解答。有关百分数的应用题的解法和分数应用题的解法相同，找准单位'1'是解此题的关键。\n\n【详解】$90 \\div (1 - 10\\%)\\\\= 90 \\div 90 \\%\\\\= 9 0 \\div 0. 9 \\\\ = 1 0 0 (\\text {万 元}) $\n\n答：计划投资100万元。",
+    "knowledge_points": [
+      "已知比一个数多/少百分之几是多少，求这个数"
+    ],
+    "source": "19-20六年级下·湖南怀化·单元测试",
+    "merged_id": "unit6_385",
+    "original_id": "unit6_solve_385"
+  },
+  {
+    "id": "unit6_solve_386",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "建造一栋楼房，计划投资100万元，实际用了90万元，节约了百分之几？(只列式不计算)",
+    "options": [],
+    "answer": "$(100 - 90) \\div 100 \\times 100\\%$",
+    "solution": "【分析】分析题意，节约了百分之几，表示的是实际比计划少百分之几，这实际是在求一个数比一个数少百分之几的问题，所以列式是用实际比计划少的投资金额，除以计划的投资金额即可。此题的解题关键是读懂题目的问题是在求什么,把它转化成我们熟悉的一个数比另一个数多百分之几或少百分之几的问题, 这样我们处理问题会容易许多。\n\n【详解】$(100 - 90) \\div 100\\\\=10 \\div 100\\\\=0.1=10%$\n\n答：节约了百分之十。",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "19-20六年级下·湖南怀化·单元测试",
+    "merged_id": "unit6_386",
+    "original_id": "unit6_solve_386"
+  },
+  {
+    "id": "unit6_solve_387",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一个数的 $20\\%$ 比 40 的 $87.5\\%$ 多 5, 这个数是多少? (方程解)",
+    "options": [],
+    "answer": "200",
+    "solution": "【分析】设这个数为 $x$, 根据这个数 $\\times 20 \\% = 40 \\times 87.5 \\% + 5$,列出方程解答即可。本题考查列方程解决问题, 解答本题的关键是掌握题中的等量关系式。\n\n【详解】解：设这个数为 $x$\n\n$20 \\%x = 40\\times 87.5\\% +5\\\\0.2 x = 3 5 + 5\\\\0.2 \\mathrm {x} = 4 0\\\\x = 2 0 0$\n\n答: 这个数是 $200$ 。",
+    "knowledge_points": [
+      "列方程解含一个未知数的问题",
+      "求一个数的百分之几是多少"
+    ],
+    "source": "18-19 六年级下·广东汕头·单元测试",
+    "merged_id": "unit6_387",
+    "original_id": "unit6_solve_387"
+  },
+  {
+    "id": "unit6_solve_388",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "在献爱心捐款活动中，六年级同学捐款是五年级同学捐款的 $80\\%$，两个年级共捐款4500元，五、六年级各捐款多少元？",
+    "options": [],
+    "answer": "五年级 2500 元；六年级 2000 元",
+    "solution": "【分析】把五年级同学的捐款钱数看作单位'1'，六年级同学的捐款钱数=五年级同学的捐款钱数×80%，根据'量÷对应的百分率'求出五年级同学的捐款钱数，最后求出六年级同学的捐款钱数，据此解答。掌握标准量和比较量的计算方法是解答题目的关键。\n\n【详解】五年级： $4500 \\div (1 + 80\\%)\\\\= 4 5 0 0 \\div 1. 8 \\\\ = 2 5 0 0 \\text {(元)} $六年级： $2500 \\times 80\\% = 2000$ （元）\n\n答：五年级同学捐款2500元，六年级同学捐款2000元。",
+    "knowledge_points": [
+      "已知一个数的百分之几是多少，求这个数",
+      "求一个数的百分之几是多少"
+    ],
+    "source": "21-22六年级下·全国·单元测试",
+    "merged_id": "unit6_388",
+    "original_id": "unit6_solve_388"
+  },
+  {
+    "id": "unit6_solve_389",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "挖一条水渠，第一天挖了全长的 $28\\%$，第二天挖了全长的 $32\\%$，两天共挖了 870 米，这条水渠长多少米？",
+    "options": [],
+    "answer": "1450 米",
+    "solution": "【分析】两天挖的长度 $= \\text{全长} \\times (28\\% + 32\\%)$, 据此求出水渠的长度即可。本题考查百分数，解答本题的关键是掌握题中的数量关系式。\n\n【详解】$870 \\div (28\\% + 32\\%)\\\\= 8 7 0 \\div 0. 6 \\\\ = 1 4 5 0 (\\text {米}) $答: 这条水渠长 1450 米。",
+    "knowledge_points": [
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "2022六年级上·全国·专题练习",
+    "merged_id": "unit6_389",
+    "original_id": "unit6_solve_389"
+  },
+  {
+    "id": "unit6_solve_390",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "甲村和乙村合修一条公路，甲修了全长的 $30 \\%$, 乙修了 450 米, 还剩下一半没修, 这条公路全长多少米?",
+    "options": [],
+    "answer": "2250 米",
+    "solution": "【分析】根据题意可得，甲修的与乙修的是全长的一半，则乙修的是全长的 $(1 - 30\\% - \\frac{1}{2})$ ，据此求出全长即可。本题考查百分数，解答本题的关键是找到题中的数量关系式。\n\n【详解】$450 \\div (1 - \\frac{1}{2} - 30\\%)\\\\=45 0 \\div 0. 2\\\\=2250 (米)$\n\n答: 这条公路全长 2250 米。",
+    "knowledge_points": [
+      "分数除法的应用",
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "21-22六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_390",
+    "original_id": "unit6_solve_390"
+  },
+  {
+    "id": "unit6_solve_391",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "为培养学生的创造性思维，新华学校举行科技小制作比赛，六年级上交作品60件，比五年级上交的作品数多 $25\\%$，五年级上交多少件作品？",
+    "options": [],
+    "answer": "48 件",
+    "solution": "【分析】六年级上交作品 60 件 = 五年级上交的作品数 (1 + 25%), 据此解答即可。本题考查百分数，解答本题的关键是掌握题中的数量关系式。\n\n【详解】$60 \\div (1 + 25\\%)\\\\= 6 0 \\div 1. 2 5 \\\\ = 4 8 \\text {(件)} $\n\n答: 五年级上交 48 件作品。",
+    "knowledge_points": [
+      "已知比一个数多/少百分之几是多少，求这个数"
+    ],
+    "source": "21-22六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_391",
+    "original_id": "unit6_solve_391"
+  },
+  {
+    "id": "unit6_solve_392",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "参加学校田径比赛的女选手有 50 人, 男选手人数比女选手多 16, 男选手占参加田径比赛总人数的百分之几?",
+    "options": [],
+    "answer": "$56.9\\%$",
+    "solution": "【分析】计算出参加田径比赛的男选手人数和总人数，根据 A 是 B 的百分之几计算方法：A÷B×100%，即可求得男选手人数占总人数的百分率。掌握求一个数是另一个数百分之几的计算方法是解答题目的关键。\n\n【详解】$(50 + 16) \\div (50 + 16 + 50) \\times 100\\%\\\\= 66 \\div 116 \\times 100 \\% \\\\ \\approx 56.90 \\times 100 \\% \\\\ = 56.9 \\% $\n\n答: 男选手占参加田径比赛总人数的 $56.9 \\%$ .",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "21-22六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_392",
+    "original_id": "unit6_solve_392"
+  },
+  {
+    "id": "unit6_solve_393",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "李师傅生产的500 个零件中, 有 15 个不合格, 这批零件的合格率是多少?",
+    "options": [],
+    "answer": "$97\\%$",
+    "solution": "【分析】合格率 = 合格产品数量 ÷ 零件总数 × 100%, 据此解答即可。本题考查百分数，解答本题的关键是掌握合格率的计算公式。\n\n【详解】$(500 - 15) \\div 500\\\\=4 8 5 \\div 5 0 0 \\\\ = 97 \\% $\n\n答: 这批零件的合格率是 $97 \\%$ .",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "21-22六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_393",
+    "original_id": "unit6_solve_393"
+  },
+  {
+    "id": "unit6_solve_394",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "饲养小组养了36 只白兔, 12 只灰兔, 白兔和灰兔分别占总只数的百分之几?",
+    "options": [],
+    "answer": "白兔 ,$75 \\%$, 灰兔 , $25 \\%$",
+    "solution": "【分析】根据百分数的意义，首先确定单位'1'，一般占'谁'、是'谁'相当于'谁'就把'谁'看作单位'1'，本题把养兔的总数看作单位'1'（作除数），用除法解答即可。此题属于百分数的基本应用题, 求一个数是另一个数的百分之几, 解答关键是确定单位 '1' (作除数), 用除法解答。\n\n【详解】$36 \\div (36 + 12)\\\\= 3 6 \\div 4 8 \\\\ = 0. 7 5 \\\\ = 75 \\% $ \n\n$ 1 2 \\div (3 6 + 1 2) \\\\ = 1 2 \\div 4 8 \\\\ = 0. 2 5 \\\\ = 25 \\% $\n\n答: 白兔占总数的 $75 \\%$, 灰兔占总数的 $25 \\%$ 。",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_394",
+    "original_id": "unit6_solve_394"
+  },
+  {
+    "id": "unit6_solve_395",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "李大伯家的一块菜地，种西红柿的面积占菜地面积的 $38 \\%$, 种黄瓜的面积占菜地面积的 $\\frac{3}{8}$ 。哪种菜占地面积大？",
+    "options": [],
+    "answer": "西红柿占地面积大",
+    "solution": "【分析】首先把两种蔬菜占菜地面积的分率比较大小, 然后根据哪个分数越大, 则这种菜的占地面积最多, 据此判断即可。此题主要考查了分数大小比较的方法, 要熟练掌握。\n\n【详解】$38\\% = 0.38\\\\\\frac {3}{8} = 0. 3 7 5$\n\n$0.38 > 0.375$, 即 $38\\% > \\frac{3}{8}$, 所以西红柿的占地面积大。答：西红柿占地面积大。",
+    "knowledge_points": [
+      "多位小数的大小比较",
+      "分数化小数",
+      "百分数、小数和分数的互化"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_395",
+    "original_id": "unit6_solve_395"
+  },
+  {
+    "id": "unit6_solve_396",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一本故事书有 120 页，小明第一天读了全书的 $30 \\%$, 第二天读了的页数是第一天的 $\\frac{5}{6}$ 。小明第二天读了多少页？",
+    "options": [],
+    "answer": "30页",
+    "solution": "【分析】将全书页数看作单位 '1', 全书页数 $\\times$ 第一天读的对应百分率 $\\times$ 第二天读的对应分率 $=$ 第二天读的页数。注意百分数和分数的单位'1'是不同的， $\\frac{5}{6}$ 的单位'1'是第一天读的页数。\n\n【详解】$120 \\times 30\\% \\times \\frac{5}{6}\\\\= 3 6 \\times \\frac {5}{6} \\\\ = 3 0 \\text {(页)} $\n\n答: 小明第二天读了 30 页。",
+    "knowledge_points": [
+      "连续求一个数的几分之几是多少的问题",
+      "求一个数的几分之几的问题",
+      "求一个数的百分之几是多少"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_396",
+    "original_id": "unit6_solve_396"
+  },
+  {
+    "id": "unit6_solve_397",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "'长沙磁浮快线'是我国首条自主研制的中低速磁悬浮铁路，2015年12月26日开通试运行，车速80千米/小时，明年试运营后，车速将提高 $25\\%$ 达到设计的最高时速，长沙磁悬浮列车设计的最高时速是多少千米/小时？",
+    "options": [],
+    "answer": "100千米/小时",
+    "solution": "【分析】将试运行时的速度看作单位 '1', 试运行的速度 $\\times$ 最高时速的对应百分率 = 最高时速, 据此列式解答。关键是确定单位 '1', 整体数量 $\\times$ 部分对应百分率 =部分数量。\n\n【详解】$80 \\times (1 + 25\\%)= 8 0 \\times 1. 2 5= 100（千米/小时）$答: 长沙磁悬浮列车设计的最高时速是 100 千米/小时。",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_397",
+    "original_id": "unit6_solve_397"
+  },
+  {
+    "id": "unit6_solve_398",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "困难",
+    "question": "阅读下面材料，解答问题。你知道热带雨林吗？它像一条环绕地球的绿色项链，源源不断地供给着所有生物赖以生存的氧气，所以被人们称为'地球之肺'。亚马孙平原是世界上最大的热带雨林区，占地球上热带雨林总面积的 $50\\%$，达650万平方千米；热带雨林贮蓄的淡水约为70万立方千米，约占地表淡水总量的 $23\\%$ ；地球上约1000万个物种中，约有350万种都生存于热带雨林中。亚马逊热带雨林中每平方千米不同种类的植物达 1200 多种, 地球上动植物的 $20 \\%$ 都生长在这里。由于人们的破坏性砍伐, 热带雨林正以每分钟约 50 公顷的速度消失, 科学家估计, 十几年后, 至少将有 50-80 万种动植物种灭绝。如果继续这样砍伐下去, 到 2020 年热带雨林将有 $80 \\%$ 从地球上消失。一摘自联合国《20 世纪环境警示录》(1) 地球上热带雨林总面积是多少万平方千米？ (2) 如果不注意保护, 到 2020 年大约将会有多少万平方千米热带雨林从地球上消失?",
+    "options": [],
+    "answer": "(1) 1300万平方千米(2) 1040 万平方千米",
+    "solution": "【分析】（1）将地球热带雨林总面积看作单位'1'，亚马孙平原的热带雨林面积÷对应百分率=地球热带雨林总面积。 (2) 将地球热带雨林总面积看作单位 '1', 地球热带雨林总面积 $\\times$ 消失的热带雨林对应百分率 $=$ 消失的热带玉林面积。【详解】 (1) $650 \\div 50\\% = 1300$ （万平方千米）答：地球上热带雨林总面积是1300万平方千米。关键是确定单位 '1', 部分数量÷对应百分率 = 整体数量, 整体数量 $\\times$ 部分对应百分率 = 部分数量。\n\n(2) $1300 \\times 80\\% = 1040$ （万平方千米）答: 到 2020 年大约将会有 1040 万平方千米热带雨林从地球上消失。",
+    "knowledge_points": [
+      "已知一个数的百分之几是多少，求这个数",
+      "求一个数的百分之几是多少"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_398",
+    "original_id": "unit6_solve_398"
+  },
+  {
+    "id": "unit6_solve_399",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "光明小学进行校园改造，对学校的围墙进行重修，第一天修了20米，正好是围墙总长度的 $25\\%$，学校要重修的围墙一共长多少米？",
+    "options": [],
+    "answer": "80米",
+    "solution": "【分析】把这段围墙的总长度看成单位 '1', 它的 $25 \\%$ 对应的数量是 20 米, 用除法求出这条围墙的总长度, 解决问题。此题考查的是百分数的应用，解答本题的关键是找出单位'1'，并找出单位'1'的百分之几对应的数量，用除法就可以求出单位'1'的量。\n\n【详解】$20 \\div 25\\%\\\\= 2 0 \\div 0. 2 5 \\\\ = 8 0 \\text {(米)} $\n\n答: 学校要重修的围墙一共长 80 米。",
+    "knowledge_points": [
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "20-21六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_399",
+    "original_id": "unit6_solve_399"
+  },
+  {
+    "id": "unit6_solve_400",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "李师傅对一批机器零件做抽样检查，从抽查的 80 个零件看，有 78 个合格，2 个不合格。 (1) 这批零件的合格率是多少? (2) 不合格率是多少？",
+    "options": [],
+    "answer": "(1) $97.5\\%$ (2) $2.5\\%$",
+    "solution": "【分析】合格率即合格零件个数占生产零件总个数的百分之几，根据'合格率 $=$ 合格零件数 $\\div$ 零件总数 $\\times 100\\%$ '；不合格率是不合格零件个数占零件总个数的百分之几，根据'不合格率 $=$ 不合格零件数 $\\div$ 零件总数 $\\times 100\\%$ '分别列出算式解答即可。此题属于百分率问题, 解答此题关键是用一部分数量 (或全部数量) 除以全部数量乘百分之百即可。\n\n【详解】 (1) 合格率: $78 \\div 80 \\times 100\\%\\\\= 0.975 \\times 100 \\% \\\\ = 97.5 \\% $\n\n答：这批零件的合格率是 $97.5 \\%$。\n\n(2) 不合格率: $2 \\div 80 \\times 100\\%\\\\= 0.025 \\times 100 \\% \\\\ = 2.5\\% $\n\n答：不合格率是 $2.5 \\%$",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_400",
+    "original_id": "unit6_solve_400"
+  },
+    {
+    "id": "unit6_choice_401",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "（21-22六年级上·江西赣州·单元测试）扇形统计图中，已知总数是1500，其中一部分是 $12\\%$ ，这部分是（）。",
+    "options": [
+      "A. 360",
+      "B. 180",
+      "C. 1140"
+    ],
+    "answer": "B",
+    "solution": "【分析】已知一个数, 求这个数的百分之几是多少用乘法计算。在扇形统计图中，已知整体求部分用乘法计算。\n\n【详解】 $1500 \\times 12\\% = 180$ 故答案为: B",
+    "knowledge_points": [
+      "求一个数的百分之几是多少"
+    ],
+    "source": "21-22六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_401",
+    "original_id": "unit6_choice_401"
+  },
+  {
+    "id": "unit6_choice_402",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "(17-18 六年级上·全国·单元测试) 甲射击 40 次, 有 35次命中；乙射击 60 次，有 55 次命中。他们的命中率相比（）。",
+    "options": [
+      "A. 命中率相同",
+      "B. 甲的命中率高",
+      "C. 乙的命中率高"
+    ],
+    "answer": "C",
+    "solution": "【分析】根据'命中率 $= \\frac{\\text { 命中的次数 }}{\\text { 总次数 }} \\times 100\\%$ '，求出两人的命中率，再比较即可。明确命中率的含义是解答本题的关键。\n\n【详解】 $\\frac{35}{40} \\times 100\\% = 87.5\\%$；$\\frac{55}{60}\\times 100\\% = 91.7\\%;$ $91.7\\% > 87.5\\%$ , 所以乙的命中率高;故答案为: C",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "17-18 六年级上·全国·单元测试",
+    "merged_id": "unit6_402",
+    "original_id": "unit6_choice_402"
+  },
+  {
+    "id": "unit6_choice_403",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "(21-22 六年级上·江西赣州·单元测试) 下面几句话中,( ) 中的数可以改成百分数。",
+    "options": [
+      "A. 小明身高 1.41 米",
+      "B. 一盒饼干重 $\\frac{7}{10}$ 千克",
+      "C. 甲的体重是乙的 $\\frac{9}{10}$"
+    ],
+    "answer": "C",
+    "solution": "【分析】百分数表示一个数是另一个数的百分之几的数, 表示一种关系，据此解答即可。明确百分数的含义是解答本题的关键。\n\n【详解】A．小明身高1.41米，1.41米是具体的数，不可以改成百分数； B. 一盒饼干重 $\\frac{7}{10}$ 千克, $\\frac{7}{10}$ 千克是具体的数, 不可以改成百分数; C. 甲的体重是乙的 $\\frac{9}{10}, \\frac{9}{10}$ 表示关系, 可以改成百分数;故答案为: C",
+    "knowledge_points": [
+      "百分数的意义"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_403",
+    "original_id": "unit6_choice_403"
+  },
+  {
+    "id": "unit6_choice_404",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "（21-22六年级上·全国·单元测试）六年级有24名同学没有达到国家体育锻炼标准，全年级体育达标率为 $88\\%$ 体育达标的学生有（）人。",
+    "options": [
+      "A. 200",
+      "B. 88",
+      "C. 176",
+      "D. 信息不够,无法计算"
+    ],
+    "answer": "C",
+    "solution": "【分析】计算出全年级学生的未达标率，根据'量÷对应的分率'计算出六年级学生的总人数，最后计算达标的学生人数即可。分析题意计算出六年级学生的总人数是解答题目的关键。\n\n【详解】 $24 \\div (1 - 88\\%) \\times 88\\%\\\\ 24 \\div 0.12 \\times 88 \\% \\\\ = 200\\times 88\\% \\\\ = 1 7 6 (\\text {人}) $\n\n故答案为: C",
+    "knowledge_points": [
+      "求一个数的百分之几是多少",
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "21-22六年级上·全国·单元测试",
+    "merged_id": "unit6_404",
+    "original_id": "unit6_choice_404"
+  },
+  {
+    "id": "unit6_choice_405",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "（21-22六年级上·全国·单元测试）某商店同时卖出两件商品，每件各卖得36元，其中一件赚了 $25\\%$ ，另一件亏了 $25\\%$ ，则这个商店卖出这两件商品是（）。",
+    "options": [
+      "A. 亏了",
+      "B. 赚了",
+      "C. 持平",
+      "D. 无法判断"
+    ],
+    "answer": "A",
+    "solution": "【分析】把成本价看作单位'1'，分别计算两件商品的成本价，最后成本价和售价比较大小即可。分别计算出两件商品的成本价是解答题目的关键。\n\n【详解】成本价 $1: 36 \\div (1 + 25 \\%)\\\\ =3 6 \\div 1. 2 5 \\\\ = 2 8. 8 \\text {(元)} \\\\ $\n\n成本价 $2: 36 \\div (1 - 25\\%)\\\\=3 6 \\div 0. 7 5 \\\\ = 4 8 \\text {(元)} $\n\n成本价: $28.8 + 48 = 76.8$ (元) 售价： $36 \\times 2 = 72$ （元） 因为 76.8 元 $>72$ 元，成本价大于售价，所以亏本了。故答案为: A",
+    "knowledge_points": [
+      "已知比一个数多/少百分之几是多少，求这个数",
+      "经济问题",
+      "盈亏问题"
+    ],
+    "source": "21-22六年级上·全国·单元测试",
+    "merged_id": "unit6_405",
+    "original_id": "unit6_choice_405"
+  },
+  {
+    "id": "unit6_choice_406",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "(21-22 六年级上·广东中山·期中) 下面几杯糖水中, 最甜的是 ( )。",
+    "options": [
+      "A. 糖 $40 \\mathrm{~g}$ , 水 $160 \\mathrm{~g}$",
+      "B. 糖 $30 \\mathrm{~g}$ , 水 $120 \\mathrm{~g} $",
+      "C . 糖$20 \\mathrm{~g},$ 水 $100 \\mathrm{~g}$",
+      "D. 糖 $10 \\mathrm{~g}$ , 水 $30 \\mathrm{~g}$"
+    ],
+    "answer": "D",
+    "solution": "【分析】根据含糖率 $=$ 糖 $\\div$ 糖水 $\\times 100\\%$ ，求出含糖率，比较即可。 $\\times \\times$ 率 = 要求量（就是 $\\times \\times$ 所代表的信息）/单位 '1'的量 (总量) $\\times 100 \\%$\n\n【详解】\n\nA. $40 \\div (40 + 160) \\times 100\\%\\\\= 40 \\div 200 \\times 100 \\% \\\\ = 20 \\% $\n\nB. $30 \\div (30 + 120) \\times 100\\%\\\\= 30 \\div 150 \\times 100 \\% \\\\ = 20 \\% $\n\nC. $20 \\div (20 + 100) \\times 100 \\%\\\\= 20 \\div 120 \\times 100 \\% \\\\\\approx 16.7 \\%$\n\nD. $10 \\div (10 + 30) \\times 100\\%\\\\=10 \\div 40 \\times 100 \\% \\\\ = 25 \\% $\n\n故答案为: D",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "21-22 六年级上·广东中山·期中",
+    "merged_id": "unit6_406",
+    "original_id": "unit6_choice_406"
+  },
+  {
+    "id": "unit6_choice_407",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "（20-21 六年级上·江西赣州·单元测试）下面的数和 1.5 相等的数有（）个。15% ;150%; $\\frac{3}{2}$ ; 1.50",
+    "options": [
+      "A. 1",
+      "B. 2",
+      "C. 3",
+      "D. 4"
+    ],
+    "answer": "C",
+    "solution": "【分析】把百分数和分数化为小数，观察结果是否等于1.5，据此解答。掌握百分数、分数化小数的方法是解答题目的关键。\n\n【详解】 $15\\% = 0.15; 150\\% = 1.5; \\frac{3}{2} = 1.5; 1.50 = 1.5$所以，与 1.5 相等的数有 3 个。故答案为: C",
+    "knowledge_points": [
+      "小数的性质",
+      "分数化小数",
+      "百分数",
+      "小数和分数的互化"
+    ],
+    "source": "20-21 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_407",
+    "original_id": "unit6_choice_407"
+  },
+  {
+    "id": "unit6_choice_408",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "(20-21 六年级上·江西赣州·单元测试) 下面的分数可以用百分数表示的是 ( )。",
+    "options": [
+      "A. 一根铁丝长 $\\frac{2}{5} \\mathrm{~m}$",
+      "B. 女生人数比男生少 $\\frac{5}{7}$",
+      "C. 一袋大米吃了 $\\frac{5}{8}$ 千克",
+      "D. 商店卖出 $\\frac{5}{2}$ 千克糖果"
+    ],
+    "answer": "B",
+    "solution": "【分析】分数可以表示分率，也可以表示具体的量，表示量时分数后面加上单位名称；而百分数表示两个数之间的倍比关系，后面不能带单位名称。掌握分数和百分数的区别是解答题目的关键。\n\n【详解】A. 一根铁丝长 $\\frac{2}{5} \\mathrm{~m}, \\frac{2}{5} \\mathrm{~m}$ 表示具体的长度, 不可以用百分数表示, 错误;B. 女生人数比男生少 $\\frac{5}{7}, \\frac{5}{7}$ 表示分率, 不表示具体的量, 可以用百分数表示, 正确;C. 一袋大米吃了 $\\frac{5}{8}$ 千克, $\\frac{5}{8}$ 千克表示具体的重量, 不可以用百分数表示，错误；D. 商店卖出 $\\frac{5}{2}$ 千克糖果, $\\frac{5}{2}$ 千克表示具体的重量, 不可以用百分数表示, 错误。故答案为: B",
+    "knowledge_points": [
+      "百分数的意义"
+    ],
+    "source": "20-21 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_408",
+    "original_id": "unit6_choice_408"
+  },
+  {
+    "id": "unit6_choice_409",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "（2014 三年级·全国·课后作业）把 25 克盐放入 100 克水中，盐水的含盐率是（）。",
+    "options": [
+      "A. $20\\%$",
+      "B. $25\\%$",
+      "C. $75\\%$",
+      "D. $80\\%$"
+    ],
+    "answer": "A",
+    "solution": "【分析】含盐率 $=$ 盐的质量 $\\div$ 盐水的质量 $\\times 100\\%$ ，据此解答。掌握百分率的计算方法是解答题目的关键。\n\n【详解】 $25 \\div (25 + 100) \\times 100\\%\\\\= 25 \\div 125 \\times 100 \\% \\\\ = 0.2 \\times 100 \\% \\\\= 20 \\%$\n\n故答案为：A",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）",
+      "含百分数的运算"
+    ],
+    "source": "2014 三年级·全国·课后作业",
+    "merged_id": "unit6_409",
+    "original_id": "unit6_choice_409"
+  },
+  {
+    "id": "unit6_choice_410",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "（21-22六年级上·江西赣州·单元测试）一件商品先涨价 $25\\%$ ，再降价 $25\\%$ ，现价与原价比（）。",
+    "options": [
+      "A. 现价小于原价",
+      "B. 现价与原价相等等",
+      "C. 现价大于原价"
+    ],
+    "answer": "A",
+    "solution": "【分析】假设原价为 100 元，现价 $= \\text{原价} \\times (1 + 25\\%) \\times (1 - 25\\%)$ ，由此计算出现价，再与原价进行比较即可。明确涨价 $25 \\%$ 和降价 $25 \\%$ 对应的单位 '1' 不同是解答本题的关键。\n\n【详解】假设原价为100元。\n\n$100 \\times (1 + 25 \\%) \\times (1 - 25 \\%) \\\\ = 1 0 0 \\times 1. 2 5 \\times 0. 7 5 \\\\ = 9 3. 7 5 \\text {(元)} $\n\n$93.75 < 100$\n\n故答案为: A",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "21-22六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_410",
+    "original_id": "unit6_choice_410"
+  },
+  {
+    "id": "unit6_choice_411",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "(21-22 六年级上·江西赣州·单元测试) 一杯盐水盐与水按 1:4 的比调制而成, 盐水的含盐率是 ( )。",
+    "options": [
+      "A. $25\\%$",
+      "B. $20\\%$",
+      "C. $80\\%$",
+      "D. $75\\%$"
+    ],
+    "answer": "B",
+    "solution": "【分析】根据比的意义, 将盐看成 1 , 水看成 4 , 盐 + 水 = 盐水, 含盐率 = 盐 ÷ 盐水 $\\times 100 \\%$ , 据此分析。 $\\times \\times$ 率 = 要求量（就是 $\\times \\times$ 所代表的信息）/单位 '1' 的量（总量） $\\times 100 \\%$ 。\n\n【详解】 $1 \\div (1 + 4) \\times 100\\%\\\\= 1 \\div 5 \\times 100 \\%\\\\= 20 \\%$\n\n故答案为: B",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）",
+      "比的意义"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_411",
+    "original_id": "unit6_choice_411"
+  },
+  {
+    "id": "unit6_choice_412",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "（21-22 六年级上·江西赣州·单元测试）铁路提速后，从甲地到乙地时间由 16 小时缩短到 10 小时，下列说法错误的是（）。",
+    "options": [
+      "A. 速度比原来提高 $60\\%$",
+      "B. 时间比原来减少 $37.5\\%$",
+      "C. 现在速度是原来的 $62.5\\%$",
+      "D. 现在与原来速度比是8:5"
+    ],
+    "answer": "C",
+    "solution": "【分析】A. 时间差 $\\div$ 较少时间 = 速度提高百分之几; B. 时间差 $\\div$ 较多时间 = 时间比原来减少百分之几; C. 较多时间÷较少时间 = 现在速度是原来的百分之几; D. 将时间比反过来是速度比。关键是理解百分数和比的意义，求一个数占另一个数的百分之几用除法。\n\n【详解】A. $(16 - 10) \\div 10\\\\= 6 \\div 1 0\\\\= 60\\%$\n\n , 选项说法正确;B. $(16-10) \\div 16= 6 \\div 1 6= 37.5\\%$, 选项说法正确; C. $16 \\div 10 = 160 \\%$ , 选项说法错误; D. $16: 10 = 8: 5$ , 选项说法正确。故答案为: C",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）",
+      "比的化简",
+      "求一个数比另一个数多/少百分之几",
+      "比的意义"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_412",
+    "original_id": "unit6_choice_412"
+  },
+  {
+    "id": "unit6_choice_413",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "(21-22 六年级上·江西赣州·单元测试) 一个数添上百分号，这个数就（）。",
+    "options": [
+      "A. 不变",
+      "B. 缩小到原数的 $\\frac{1}{100}$",
+      "C. 扩大 100倍",
+      "D. 增加 100"
+    ],
+    "answer": "B",
+    "solution": "【分析】表示一个数是另一个数的百分之几的数，也叫百分比、百分率。百分数常常不写成分母是 100 的分数形式，而是在原来的分子后面添加上百分号 '%'来表示。关键是理解百分数的意义, 百分数是一种特殊的分数, 表示两个数之间的倍比关系, 不表示具体的数量,所以后面不带单位名称。\n\n【详解】一个数添上百分号, 这个数就缩小到原数的 $\\frac{1}{100}$ 。故答案为: B",
+    "knowledge_points": [
+      "百分数的意义"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_413",
+    "original_id": "unit6_choice_413"
+  },
+  {
+    "id": "unit6_choice_414",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "(21-22 六年级上·江西赣州·单元测试) 一块 2 米长的花布用去 $80 \\%$ , 还剩下 ( )。",
+    "options": [
+      "A. 1.6 米",
+      "B. $20\\%$ 米",
+      "C. $40\\%$ 米",
+      "D. $\\frac{2}{5}$ 米"
+    ],
+    "answer": "D",
+    "solution": "【分析】将画布长度看作单位'1'，画布长度×剩下对应百分率=剩下长度。百分数不能表示具体数量，后面不能带单位。\n\n【详解】 $2 \\times (1 - 80\\%)\\\\= 2 \\times 0. 2 \\\\ = 0. 4 \\\\ = \\frac {2}{5} (\\text {米})$\n\n故答案为: D",
+    "knowledge_points": [
+      "求一个数的百分之几是多少",
+      "百分数的意义"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_414",
+    "original_id": "unit6_choice_414"
+  },
+  {
+    "id": "unit6_choice_415",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "（18-19六年级上·辽宁·期末）一种商品，先降价 $10\\%$ ，后又涨价 $10\\%$ ，现在的价格与原价相比（）。",
+    "options": [
+      "A. 涨了",
+      "B. 不变",
+      "C. 降了"
+    ],
+    "answer": "C",
+    "solution": "【分析】把原价看作单位'1'，降价 $10\\%$ 后相当于原价的 $1 - 10\\% = 90\\%$ ，又涨价 $10\\%$ ，这时把后来的价格看作单位'1'，这时的价格相当于原来的 $90\\% \\times (1 + 10\\%)$ ，计算出结果，与 1 比较即可。这类问题容易认为现在的商品价格与原来相等,此题容易出错。\n\n【详解】 $1 \\times (1 - 10\\%) \\times (1 + 10\\%)\\\\= 1 \\times 0. 9 \\times 1. 1 \\\\ = 0. 9 9 $\n\n因为 $0.99 < 1$ ，所以降低了。则现在的商品价格与原来相比降低了。故选: C",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "18-19六年级上·辽宁·期末",
+    "merged_id": "unit6_415",
+    "original_id": "unit6_choice_415"
+  },
+  {
+    "id": "unit6_choice_416",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "（20-21六年级上·山东聊城·期末）一堆煤用了 $40\\%$ ，还剩这堆煤的（）。",
+    "options": [
+      "A. 60吨",
+      "B. $40\\%$",
+      "C. $60\\%$"
+    ],
+    "answer": "C",
+    "solution": "【分析】这堆煤的实际重量未知，把这堆煤总量看作'1'，用去的占'1'的 $40\\%$ ，剩下的表示为 $(1 - 40\\%)$ ，据此解答。找出题目中的 '1'是表示剩下煤占总量百分率的关键。\n\n【详解】剩下的煤占总量的百分率表示为: $1 - 40\\% = 60\\%$ ; 故答案为: C",
+    "knowledge_points": [
+      "含百分数的运算",
+      "单位'1'的认识与确定"
+    ],
+    "source": "20-21六年级上·山东聊城·期末",
+    "merged_id": "unit6_416",
+    "original_id": "unit6_choice_416"
+  },
+  {
+    "id": "unit6_choice_417",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "（21-22六年级上·全国·单元测试）春蕾小学的一项调查表明，有蛀牙的学生人数占全校学生人数的 $20\\%$ ，没有蛀牙的学生人数占 $80\\%$ 。有蛀牙的学生占全校学生的几分之几呢？（）",
+    "options": [
+      "A. 二分之一",
+      "B. 三分之一",
+      "C. 四分之一",
+      "D. 五分之一"
+    ],
+    "answer": "D",
+    "solution": "【分析】将 $20 \\%$ 化成分数, 求出有蛀牙的学生占全校学生的几分之几。本题考查了分数和百分数的互化，属于简单题，解题时细心即可。\n\n【详解】 $20 \\%$ 等于五分之一，所以，有蛀牙的学生占全校学生的五分之一。故答案为: D",
+    "knowledge_points": [
+      "百分数",
+      "小数和分数的互化"
+    ],
+    "source": "21-22六年级上·全国·单元测试",
+    "merged_id": "unit6_417",
+    "original_id": "unit6_choice_417"
+  },
+  {
+    "id": "unit6_choice_418",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "（21-22六年级上·全国·单元测试）甲、乙两名同学进行投篮比赛，甲投了20次，有1次没有投中，乙投了60次，投中了57次。甲、乙命中率相比，（）。",
+    "options": [
+      "A. 甲高",
+      "B. 乙高",
+      "C. 一样"
+    ],
+    "answer": "C",
+    "solution": "【分析】利用除法分别求出甲乙的命中率，再对比即可。本题考查了百分数的应用，命中率等于命中的次数÷总次数×100%。\n\n【详解】甲:$(2 0 - 1) \\div 2 0 \\\\ = 1 9 \\div 2 0 \\\\ = 95 \\% $\n\n乙: $57 \\div 60 = 95\\%$所以，甲乙的命中率是相等的。故答案为: C",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "21-22六年级上·全国·单元测试",
+    "merged_id": "unit6_418",
+    "original_id": "unit6_choice_418"
+  },
+  {
+    "id": "unit6_fill_419",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "（20-21 六年级上·湖南怀化·单元测试）王庄村计划种树 500 棵，实际种树 600 棵，实际完成计划的( )。",
+    "options": [],
+    "answer": "$120\\%$",
+    "solution": "【分析】把计划种树 500 棵看作单位 '1', 用实际种树棵树 $\\div$ 计划种树棵树即可。根据求一个数是另一个数的百分之几，用除法解答。\n\n【详解】 $600 \\div 500 = 1.2 = 120\\%$实际完成计划的 $120 \\%$ 。",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_419",
+    "original_id": "unit6_fill_419"
+  },
+  {
+    "id": "unit6_fill_420",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·湖南怀化·单元测试) 李庄村种植的果树全部成活，成活率是( )%；张叔果园今年栽了200棵果树，活了198棵，这批果树的成活率是( )%。",
+    "options": [],
+    "answer": "100 99",
+    "solution": "【分析】成活率 $=$ 成活棵数 $\\div$ 总棵数 $\\times 100\\%$ ，据此计算。掌握成活率的计算方法是解答题目的关键。\n\n【详解】（1）李庄村种植的果树全部成活，则成活棵数 $=$ 总棵数，成活率为 $100\\%$ ；\n\n(2) $198 \\div 200 \\times 100 \\%\\\\= 0.99 \\times 100 \\% \\\\ = 99 \\% $",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_420",
+    "original_id": "unit6_fill_420"
+  },
+  {
+    "id": "unit6_fill_421",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·湖南怀化·单元测试) 男生 20 人, 女生 30 人, 男生约占女生人数的( )%, 男生占全班人数的( )%, 女生比男生多( )%。",
+    "options": [],
+    "answer": "66.7 40 50",
+    "solution": "【分析】 (1) A 占 B 的百分之几计算方法: $A \\div B \\times 100\\%$ ;(2) A 比 B 多百分之几的计算方法: (A-B) $\\div \\mathrm{B} \\times 100 \\%$ , 据此解答。掌握 A 占 B 百分之几以及 A 比 B 多百分之几的计算方法是解答题目的关键。\n\n【详解】 (1) $20 \\div 30 \\times 100\\%\\\\ \\approx 0.667 \\times 100 \\%\\\\= 66.7 \\%$\n\n(2) $20 \\div (20 + 30) \\times 100\\%\\\\= 20 \\div 50 \\times 100 \\%\\\\= 0.4 \\times 100 \\%\\\\= 40 \\%$\n\n(3) $(30 - 20) \\div 20 \\times 100\\%\\\\= 10 \\div 20 \\times 100 \\% \\\\ = 0.5 \\times 100 \\% \\\\ = 50 \\% $",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）",
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_421",
+    "original_id": "unit6_fill_421"
+  },
+  {
+    "id": "unit6_fill_422",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "(20-21 六年级上·湖南怀化·单元测试) $14.6 \\%$ 读作( ); 百分之零点零九写作( )。",
+    "options": [],
+    "answer": "百分之十四点六 $0.09\\%$",
+    "solution": "【分析】（1）百分数的读法：先读分母，再读分子，读作'百分之……'；(2) 百分数的写法: 百分数通常不写成分数形式, 而在原来的分子后面加上百分号 '%'来表示, 据此解答。掌握百分数的读写方法是解答题目的关键。\n\n【详解】 $14.6 \\%$ 读作（百分之十四点六）；百分之零点零九写作（0.09%）。",
+    "knowledge_points": [
+      "百分数的读法和写法"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_422",
+    "original_id": "unit6_fill_422"
+  },
+  {
+    "id": "unit6_fill_423",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(21-22 六年级上·江西赣州·单元测试) 甲数是乙数的 $\\frac{7}{8}$ ,甲数比乙数少( )；乙数比甲数多( )。(百分数, 除不尽的保留一位小数)",
+    "options": [],
+    "answer": "$12.5\\%$ ;$14.3\\%$",
+    "solution": "【分析】根据题意可知, 乙数是单位 '1', 则甲数为 $\\frac{7}{8}$ ; 用甲数与乙数的差除以乙数即可求出甲数比乙数少百分之几,用甲数与乙数的差除以甲数即可求出乙数比甲数多百分之几。求一个数比另一个数多（少）百分之几，用两个数的差除以另一个数即可。\n\n【详解】 $(1 - \\frac{7}{8}) \\div 1 = \\frac{1}{8} = 12.5\\%\\\\(1 - \\frac {7}{8}) \\div \\frac {7}{8} \\\\ = \\frac {1}{8} \\div \\frac {7}{8} \\\\ =\\frac {1}{7} \\approx 14.3 \\% $",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_423",
+    "original_id": "unit6_fill_423"
+  },
+  {
+    "id": "unit6_fill_424",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "(21-22 六年级上·江西赣州·单元测试) $\\frac{3}{4}$ 千克是 $\\frac{6}{7}$ 千克的( )%, 30 比 50 少( )%。",
+    "options": [],
+    "answer": "87.5; 40",
+    "solution": "【分析】求一个数是另一数的百分之几，用这个数除以另一个数即可；求一个数比另一个数多（少）百分之几，用两个数的差除以另一个数即可。熟练掌握'求一个数是另一数的百分之几'和'求一个数比另一个数多（少）百分之几'的方法是解答本题的关键。\n\n【详解】 $\\frac{3}{4} \\div \\frac{6}{7} = 87.5\\%$\n\n$ (5 0 - 3 0) \\div 5 0 \\\\ = 2 0 \\div 5 0 \\\\ = 4 0 \\% $",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）",
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_424",
+    "original_id": "unit6_fill_424"
+  },
+  {
+    "id": "unit6_fill_425",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "(21-22 六年级上·江西赣州·单元测试) $59 \\%$ 读作( ),百分之七十六点五写作( )。",
+    "options": [],
+    "answer": "百分之五十九 $76.5\\%$",
+    "solution": "【分析】百分数读法：先读百分号，再读百分号前面的数；百分数写法：先写分子，再写百分号，分子写在百分号的前面。熟练掌握百分数的读写是解答本题的关键。\n\n【详解】 $59 \\%$ 读作百分之五十九;百分之七十六点五写作： $76.5\\%$",
+    "knowledge_points": [
+      "百分数的读法和写法"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_425",
+    "original_id": "unit6_fill_425"
+  },
+  {
+    "id": "unit6_fill_426",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "（21-22六年级上·江西赣州·单元测试）某校开展运动会，四年级参加的人数有40人，五年级参加的人数有60人，六年级参加的人数有 100 人, 如果制成扇形统计图, 那么四年级参加的人数占三个年级总人数的( )%，五年级参加人数占三个年级参加总人数的( )%，六年级参加的人数占三个年级总人数的( )%。",
+    "options": [],
+    "answer": "20 30 50",
+    "solution": "【分析】计算出四、五、六年级的总人数，根据 A 是 B 的百分之几计算方法：A ÷ B × 100%，即可求得。掌握求一个数占另一个数百分之几的计算方法是解答题目的关键。\n\n【详解】总人数： $40 + 60 + 100\\\\=1 0 0 + 1 0 0 \\\\ = 2 0 0 \\text {(人)} $\n\n四年级： $40 \\div 200 \\times 100\\%\\\\=2 \\times 1 0 0 \\\\ = 20 \\% $\n\n五年级： $60 \\div 200 \\times 100\\%\\\\=0.3 \\times 100 \\% \\\\ = 30 \\% $\n\n六年级: $100 \\div 200 \\times 100 \\%\\\\= 0.5 \\times 100 \\% \\\\ = 50 \\%$",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "21-22六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_426",
+    "original_id": "unit6_fill_426"
+  },
+  {
+    "id": "unit6_fill_427",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "（21-22六年级上·江西赣州·单元测试）中国人口约占世界人口的 $21\\%$ ，全世界有60亿人口，中国约有( )亿人口。",
+    "options": [],
+    "answer": "12.6",
+    "solution": "【分析】中国的人口 = 全世界的人口 × 21%，据此计算。已知一个数, 求这个数的百分之几是多少用乘法计算。\n\n【详解】 $60 \\times 21\\% = 12.6$ （亿人）",
+    "knowledge_points": [
+      "求一个数的百分之几是多少"
+    ],
+    "source": "21-22六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_427",
+    "original_id": "unit6_fill_427"
+  },
+  {
+    "id": "unit6_fill_428",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "（21-22六年级上·江西赣州·单元测试）希望小学十月份用电量比九月份节约了 $12\\%$ ，十月份用电量相当于九月份的( )%，九月份用电9600度，十月份的用电量是( )度。",
+    "options": [],
+    "answer": "88; 8448",
+    "solution": "【分析】把九月份用电量看作单位'1'，十月份用电量占九月份的 $(1 - 12\\%)$ , 十月份的用电量 $=$ 九月份的用电量 $\\times$ $(1 - 12\\%)$ 。掌握求比一个数少百分之几的数是多少的计算方法是解答题目的关键。\n\n【详解】 $1 - 12\\% = 88\\%$ \n\n $9600 \\times (1 - 12 \\%) \\\\ = 9600 \\times 88 \\% \\\\ = 8 4 4 8 \\text {(度)} $",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "21-22六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_428",
+    "original_id": "unit6_fill_428"
+  },
+  {
+    "id": "unit6_fill_429",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "(21-22六年级上·江西赣州·单元测试) 修一条长800米的路，第一天修了 $35 \\%$ , 修了 ( )米, 还剩 ( )米没有修。",
+    "options": [],
+    "answer": "280 520",
+    "solution": "【分析】用 $800 \\times 35\\%$ 即可求出修了多少米，用总长度减去修了的米数即可求出剩下的米数。熟练掌握百分数乘法的意义是解答本题的关键。\n\n【详解】 $800 \\times 35\\% = 280$ （米） $800 - 280 = 520$ (米)",
+    "knowledge_points": [
+      "求一个数的百分之几是多少"
+    ],
+    "source": "21-22六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_429",
+    "original_id": "unit6_fill_429"
+  },
+  {
+    "id": "unit6_fill_430",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "(21-22 六年级上·江西赣州·单元测试) 一种树苗的成活率是 $85 \\%$ , 现在要栽 1000 棵这样的树苗, 大约( )棵能成活。",
+    "options": [],
+    "answer": "850",
+    "solution": "【分析】用树苗总棵数乘成活率即可解答。求一个数的百分之几是多少，用乘法解答。\n\n【详解】 $1000 \\times 85\\% = 850$ （棵）",
+    "knowledge_points": [
+      "求一个数的百分之几是多少"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_430",
+    "original_id": "unit6_fill_430"
+  },
+  {
+    "id": "unit6_fill_431",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "(21-22 六年级上·江西赣州·单元测试) 六 (1) 班有男生25人，女生20人，女生人数是男生的( )%，男生占全班人数的( )%。",
+    "options": [],
+    "answer": "80 55.6",
+    "solution": "【分析】用女生人数除以男生人数即可求出女生人数是男生的百分之几；用男生人数除以全班人数即可求出男生占全班人数的百分之几。求一个数是另一数的百分之几，用这个数除以另一个数即可。\n\n【详解】 $20 \\div 25 = 80\\%$ ; $2 5 \\div (2 5 + 2 0) \\\\ = 2 5 \\div 4 5 \\\\ = 55.6 \\% $",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_431",
+    "original_id": "unit6_fill_431"
+  },
+  {
+    "id": "unit6_fill_432",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(21-22 六年级上·全国·单元测试) 周一六 (1) 班有 49人按时上学，1人请假，这天的出勤率是____。",
+    "options": [],
+    "answer": "$98\\%$",
+    "solution": "【分析】出勤率 $=$ 出勤人数 $\\div$ 总人数 $\\times 100\\%$ ，据此计算。掌握生活中百分率的计算方法是解答题目的关键。\n\n【详解】 $49 \\div (49 + 1) \\times 100\\%\\\\49 \\div 50 \\times 100 \\% \\\\ = 0.98 \\times 100 \\% \\\\ = 98 \\% $",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "21-22 六年级上·全国·单元测试",
+    "merged_id": "unit6_432",
+    "original_id": "unit6_fill_432"
+  },
+  {
+    "id": "unit6_fill_433",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(21-22 六年级上·全国·单元测试) 在 $\\frac{1}{3} 、 33.3\\%$ 、 $\\frac{3}{10}$ 、3.3 中, 最大的是 , 最小的是",
+    "options": [],
+    "answer": "3.3 $\\frac{3}{10}$",
+    "solution": "【分析】把题目中的分数和百分数统一化为小数，利用小数比较大小的方法计算即可。掌握多位小数比较大小的方法是解答题目的关键。\n\n【详解】 $\\frac{1}{3} = 0.333\\dots, 33.3\\% = 0.333, \\frac{3}{10} = 0.3, 3.3$因为 $3.3 > 0.333 \\ldots > 0.333 > 0.3$ ，所以 $3.3 > \\frac{1}{3} > 33.3\\% > \\frac{3}{10}$ ，最大的是 3.3，最小的是 $\\frac{3}{10}$ 。",
+    "knowledge_points": [
+      "循环小数的认识与简写",
+      "多位小数的大小比较",
+      "分数化小数",
+      "百分数",
+      "小数和分数的互化"
+    ],
+    "source": "21-22 六年级上·全国·单元测试",
+    "merged_id": "unit6_433",
+    "original_id": "unit6_fill_433"
+  },{
+    "id": "unit6_fill_434",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "(21-22六年级上·全国·单元测试) $\\frac{3}{4} = 12$ : ()=( )%",
+    "options": [],
+    "answer": "16 75",
+    "solution": "【分析】 $\\frac{3}{4} = 3:4 = 0.75 = 75\\% \\frac {3}{4} = \\dot {3}: 4 = (3 \\times 4): (4 \\times 4) = 1 2: 1 6$ 掌握比和分数的关系是解答题目的关键。\n\n【详解】 $\\frac{3}{4} = 12:16 = 75\\%$",
+    "knowledge_points": [
+      "百分数",
+      "分数",
+      "小数和比的互化"
+    ],
+    "source": "21-22六年级上·全国·单元测试",
+    "merged_id": "unit6_434",
+    "original_id": "unit6_fill_434"
+  },
+  {
+    "id": "unit6_fill_435",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·江西赣州·单元测试) $80 \\mathrm{~m}$ 比 $50 \\mathrm{~m}$ 多 $( )\\%$ , ( )比 40 少 $80 \\%$ 。",
+    "options": [],
+    "answer": "60 8",
+    "solution": "【分析】根据一个数比另一个数多/少百分之几列式即可解决问题。本题关键在于对一个数比另一个数多/少百分之几的理解。\n\n【详解】 (1) $(80 \\div 50 - 1) \\times 100\\%\\\\ (1.6 - 1) \\times 100 \\% \\\\ = 0.6 \\times 100 \\% \\\\ = 60 \\% $\n\n(2) $40 \\times (1 - 80\\%)\\\\ =40 \\times 20 \\% \\\\ = 8 $",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少",
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "20-21 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_435",
+    "original_id": "unit6_fill_435"
+  },
+  {
+    "id": "unit6_fill_436",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·江西赣州·单元测试) 跑一段路, 小明用了 40 秒, 小华用了 50 秒, 小明比小华快( )%。",
+    "options": [],
+    "answer": "25",
+    "solution": "【分析】设出这条路的总长度，根据'速度 = 路程 ÷ 时间'计算出小明和小华的速度，小明比小华快的百分率 = （小明的速度 - 小华的速度）÷小华的速度 × 100%，据此计算。A比B多百分之几的计算方法：（A一B) $\\div \\mathrm{B} \\times 100 \\%$ .\n\n【详解】假设这段路的路程为 1\n\n小明速度: $1 \\div 40 = \\frac{1}{40}$\n\n小华速度: $1 \\div 50 = \\frac{1}{50}$\n\n$(\\frac{1}{40} -\\frac{1}{50})\\div \\frac{1}{50}\\times 100\\% \\\\ = \\frac{1}{200}\\div \\frac{1}{50}\\times 100\\% \\\\ =\\frac{1}{200}\\times 50\\times 100\\% \\\\ = \\frac{1}{4}\\times 100\\% \\\\ = 25 \\% $",
+    "knowledge_points": [
+      "基础行程问题",
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "20-21 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_436",
+    "original_id": "unit6_fill_436"
+  },
+  {
+    "id": "unit6_fill_437",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·江西赣州·单元测试) 油菜籽的出油率是 $42 \\%$ , 5t 油菜籽可榨油( )t; 要榨油 1050 千克, 需要油菜籽( )千克。",
+    "options": [],
+    "answer": "2.1 2500",
+    "solution": "【分析】由'出油率 $=$ 油的质量 $\\div$ 油菜籽的质量 $\\times 100\\%$ '可知,油的质量 $=$ 油菜籽的质量 $\\times$ 出油率; 油菜籽的质量 $=$ 油的质量÷出油率，据此解答。灵活运用出油率的计算公式是解答题目的关键。\n\n【详解】 (1) $5 \\times 42\\% = 2.1$ （吨） (2) $1050 \\div 42\\% = 2500$ （千克）",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）",
+      "求一个数的百分之几是多少",
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "20-21 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_437",
+    "original_id": "unit6_fill_437"
+  },
+  {
+    "id": "unit6_fill_438",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·江西赣州·单元测试) $\\frac{3}{8} = $ （）÷24 = 18: ( ) = $\\frac{12}{()}$ = ( ) %。",
+    "options": [],
+    "answer": "9; 48; 32; 37.5",
+    "solution": "【分析】根据分数与除法和比的关系，以及它们通用的基本性质进行填空，先将分数化成小数，再将小数化成百分数，据此分析。分数的分子相当于被除数、比的前项，分母相当于除数、比的后项。\n\n【详解】 $24 \\div 8 \\times 3 = 9; 18 \\div 3 \\times 8 = 48; 12 \\div 3 \\times 8 = 32; 3 \\div 8 =$ $0.375 = 37.5\\%$",
+    "knowledge_points": [
+      "百分数",
+      "分数",
+      "小数和比的互化"
+    ],
+    "source": "20-21 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_438",
+    "original_id": "unit6_fill_438"
+  },
+  {
+    "id": "unit6_fill_439",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "(20-21 六年级上·江西赣州·单元测试) 百分之零点八七写作( )， $21.58\\%$ 读作( )。",
+    "options": [],
+    "answer": "$0.87\\%$ 百分之二十一点五八",
+    "solution": "【分析】写百分数时，先写分子，再写百分号。一个百分数，%前面的数是几，就读作百分之几。百分数常常不写成分母是 100 的分数形式, 而是在原来的分子后面添加上百分号 '%'来表示。\n\n【详解】百分之零点八七写作 $0.87 \\%$ , $21.58 \\%$ 读作百分之二十一点五八。",
+    "knowledge_points": [
+      "百分数的读法和写法"
+    ],
+    "source": "20-21 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_439",
+    "original_id": "unit6_fill_439"
+  },
+  {
+    "id": "unit6_fill_440",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·江西赣州·单元测试) 绿化队修整街心花园, 用去 900 元, 比原计划节省了 300 元, 节省了 $( ) \\%$",
+    "options": [],
+    "answer": "25",
+    "solution": "【分析】先依据原来需要的钱数 = 实际用去钱数十节省钱数，求出原来需要的钱数，再用节省钱数除以原来需要的钱数即可解答。解答本题的关键：依据等量关系式原来需要的钱数 = 实际用去钱数 + 节省钱数，求出原来需要的钱数。\n\n【详解】 $300 \\div (900 + 300)\\\\ 3 0 0 \\div 1 2 0 0 \\\\ = 25 \\% $\n\n则节省了 $25 \\%$",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "20-21 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_440",
+    "original_id": "unit6_fill_440"
+  },
+  {
+    "id": "unit6_fill_441",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·江西赣州·单元测试) 一种树苗经试种成活率在 $75\\% - 80\\%$ 之间，现种植这种树苗，要至少保证成活120棵，需要种植( )棵树苗。",
+    "options": [],
+    "answer": "160",
+    "solution": "【分析】根据植的棵数 = 成活的棵数 ÷ 成活率，因为题中要保证至少成活 129 棵，需除以的成活率是最低的，据此解答即可。此题考查百分率应用题, 解决此题的关键是理解成活率的含义, 并且要注意: '至少保证', 说明成活率为 $75 \\%$ .\n\n【详解】由分析得,$120 \\div 75\\% = 160$ （棵）",
+    "knowledge_points": [
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "20-21 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_441",
+    "original_id": "unit6_fill_441"
+  },
+  {
+    "id": "unit6_fill_442",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "(20-21 六年级上·江西赣州·单元测试) 一次射击练习中, 命中的子弹是 100 发, 没命中的是 25 发, 命中率是( )。",
+    "options": [],
+    "answer": "$80\\%$",
+    "solution": "【分析】命中率 $=$ 命中数 $\\div$ 总数 $\\times 100\\%$ ，据此解答即可。本题主要考查对命中率的理解与运用。\n\n【详解】 $100 \\div (100 + 25) \\times 100\\%\\\\=100 \\div 125 \\times 100 \\% \\\\ = 0.8 \\times 100 \\% \\\\ = 80 \\% $",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）",
+      "分数的四则混合运算"
+    ],
+    "source": "20-21 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_442",
+    "original_id": "unit6_fill_442"
+  },
+  {
+    "id": "unit6_fill_443",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(15-16 六年级上·全国·单元测试) 在 $67\\%$ 、 $\\frac{2}{3}$、0.66、0.667中，最大的数是____，最小的数是_____。",
+    "options": [],
+    "answer": "$67\\%$ 0.66",
+    "solution": "【分析】 $67\\% = 0.67, \\frac{2}{3} = 0.6$ ，在 0.67、0.6、0.66、0.667 中 0.67 的十分位上的数字 7 最大，在 0.6 、0.66、0.667 中 0.667 千分位上面的数字 7 最大，0.66 千分位上面的数字 0 最小，则 $0.66 < 0.6 < 0.667$ ，据此解答。掌握多位小数比较大小的方法是解答题目的关键。\n\n【详解】分析可知, $0.66 < 0.6 < 0.667 < 0.67$ , 则 $0.66 < \\frac{2}{3} < 0.667 < 67\\%$ , 所以最大的数是 $67\\%$ , 最小的数是 $0.66$ 。",
+    "knowledge_points": [
+      "循环小数的认识与简写",
+      "多位小数的大小比较",
+      "分数化小数",
+      "百分数",
+      "小数和分数的互化"
+    ],
+    "source": "15-16 六年级上·全国·单元测试",
+    "merged_id": "unit6_443",
+    "original_id": "unit6_fill_443"
+  },
+  {
+    "id": "unit6_fill_444",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "（20-21 六年级上·福建厦门·单元测试）高速列车提速后，由原来的每小时 250 千米提高到 300 千米，速度提高了（）分之（）。",
+    "options": [],
+    "answer": "百 二十",
+    "solution": "【分析】速度差÷原来速度=提高了百分之几，据此分析。此类问题一般用表示单位'1'的量作除数。\n\n【详解】 $(300 - 250) \\div 250\\\\ 5 0 \\div 2 5 0 \\\\ = 20 \\% $\n\n速度提高了百分之二十。",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "20-21 六年级上·福建厦门·单元测试",
+    "merged_id": "unit6_444",
+    "original_id": "unit6_fill_444"
+  },
+  {
+    "id": "unit6_fill_445",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(21-22 六年级上·江西赣州·单元测试) 某校今年新招收的一年级新生中，男女生人数比是 $3: 2$ ，如果要制作成扇形统计图，得算出女生占总人数的( )%，男生占总人数的( )%。",
+    "options": [],
+    "answer": "40 60",
+    "solution": "【分析】根据男女生人数比可知，男生人数有3份，女生人数有 2 份, 总人数有 5 份。据此利用除法分别求出女生占总人数的百分之几，男生占总人数的百分之几即可。本题考查了百分数, 求一个数是另一个数的百分之几, 用除法。\n\n【详解】 $2 \\div (3 + 2) = 40\\%, 3 \\div (3 + 2) = 60\\%$ , 所以, 女生占总人数的 $40\\%$ , 男生占总人数的 $60\\%$ 。",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）",
+      "比的应用"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_445",
+    "original_id": "unit6_fill_445"
+  },
+  {
+    "id": "unit6_fill_446",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "（21-22六年级上·江西赣州·单元测试）师徒两人计划九月份加工一批零件，结果超额完成 $20\\%$ ，已知师傅完成了 $65\\%$ ，则徒弟完成了( )%。",
+    "options": [],
+    "answer": "55",
+    "solution": "【分析】计划加工的零件数是 $100 \\%$ , 据此利用加法求出超额完成后的零件数是百分之几, 再将其减去师傅完成的 $65 \\%$ , 求出徒弟完成了百分之几即可。本题考查了含百分数的运算，能根据题意正确列式是解题的关键。\n\n【详解】 $100\\% + 20\\% - 65\\% = 55\\%$ ，所以徒弟完成了 $55\\%$ 。",
+    "knowledge_points": [
+      "含百分数的运算"
+    ],
+    "source": "21-22六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_446",
+    "original_id": "unit6_fill_446"
+  },
+  {
+    "id": "unit6_fill_447",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(21-22 六年级上·江西赣州·单元测试) 一种油菜籽的出油率为 $40\\% \\sim 50\\%$ , 200 千克油菜籽最多能榨油( )千克,要保证榨油 120 千克, 至少要准备( )千克油菜籽。",
+    "options": [],
+    "answer": "100 300",
+    "solution": "【分析】第一个空，用油菜籽质量 $\\times$ 最大出油率；第二个空，用油的质量 $\\div$ 最小出油率。 $\\times \\times$ 率 = 要求量（就是 $\\times \\times$ 所代表的信息）/单位 '1' 的量（总量） $\\times 100 \\%$ 。\n\n【详解】 $200 \\times 50\\% = 100$ （千克）$120 \\div 40\\% = 300$ （千克）",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）",
+      "求一个数的百分之几是多少",
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_447",
+    "original_id": "unit6_fill_447"
+  },
+  {
+    "id": "unit6_fill_448",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "(21-22 六年级上·江西赣州·单元测试) 黄豆中蛋白质含量约占 $36 \\%$ , ( )千克黄豆蛋白质含量能达到 7.2 千克。",
+    "options": [],
+    "answer": "20",
+    "solution": "【分析】将黄豆质量看作单位 '1', 蛋白质质量 $\\div$ 对应百分率即可。关键是确定单位 '1', 部分数量: 对应百分率 = 整体数量。\n\n【详解】 $7.2 \\div 36\\% = 20$ （千克）",
+    "knowledge_points": [
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_448",
+    "original_id": "unit6_fill_448"
+  },
+  {
+    "id": "unit6_fill_449",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(21-22 六年级上·江西赣州·单元测试) 把长、宽分别为 $5 \\mathrm{~cm} 、 4 \\mathrm{~cm}$ 的长方形木块锯成最大的正方形, 面积减少( )$\\%$。",
+    "options": [],
+    "answer": "20",
+    "solution": "【分析】长方形里画最大的正方形，正方形边长 $=$ 长方形的宽，用长方形面积一正方形面积 $=$ 减少的面积，减少的面积 $\\div$ 长方形面积即可。长方形面积 $=$ 长 $\\times$ 宽，正方形面积 $=$ 边长 $\\times$ 边长。\n\n【详解】 $5 \\times 4 - 4 \\times 4= 2 0 - 1 6= 4$ （平方厘米）$4 \\div 20 = 20 \\%$",
+    "knowledge_points": [
+      "正方形的面积",
+      "长方形的面积",
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_449",
+    "original_id": "unit6_fill_449"
+  },
+  {
+    "id": "unit6_fill_450",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "（2014 五年级·全国·课后作业）六（1）班今天 48 人到校，请事假有 2 人，今天的出勤率为( )%。",
+    "options": [],
+    "answer": "96",
+    "solution": "【分析】出勤率 $=$ 出勤人数 $\\div$ 总人数 $\\times 100\\%$ ，据此解答。掌握出勤率的计算方法是解答题目的关键。\n\n【详解】 $48 \\div (48 + 2) \\times 100\\%$\n\n$ 48 \\div 50 \\times 100 \\% \\\\ = 0.96 \\times 100 \\% \\\\ = 96 \\% $",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "2014 五年级·全国·课后作业",
+    "merged_id": "unit6_450",
+    "original_id": "unit6_fill_450"
+  },
+  {
+    "id": "unit6_fill_451",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(21-22六年级上·江西赣州·单元测试) 书法兴趣小组有男生15人，女生12人，男生比女生多( )%，女生比男生少( )%。",
+    "options": [],
+    "answer": "25 20",
+    "solution": "【分析】男女生人数差 $\\div$ 女生人数 = 男生比女生多百分之几;男女生人数差 $\\div$ 男生人数 = 女生比男生少百分之几。此类问题一般用表示单位'1'的量作除数。\n\n【详解】 $(15 - 12) \\div 12\\\\=3 \\div 1 2 \\\\ = 25 \\% $\n\n$(1 5 - 1 2) \\div 1 5 \\\\ = 3 \\div 1 5 \\\\ = 20 \\% $",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "21-22六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_451",
+    "original_id": "unit6_fill_451"
+  },
+  {
+    "id": "unit6_fill_452",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "(21-22 六年级上·江西赣州·单元测试) ( )米的 $30 \\%$ 是 1.8 米, 60 千克减少 $20 \\%$ 后是( )千克。",
+    "options": [],
+    "answer": "6 48",
+    "solution": "【分析】（1）求标准量时用'量÷对应的百分率'即可；(2) 求比一个数少百分之几的数是多少: 这个数 $\\times$ (1- 百分率)。掌握百分率的相关计算方法是解答题目的关键。\n\n【详解】 (1) $1.8 \\div 30\\% = 6$ （米）(2) $60 \\times (1 - 20\\%)\\\\= 6 0 \\times 0. 8\\\\=48 (千克)$",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少",
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_452",
+    "original_id": "unit6_fill_452"
+  },
+  {
+    "id": "unit6_fill_453",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "(21-22 六年级上·江西赣州·单元测试) ( ) ÷ 24=18: ( ) $= 6 \\div 16 = \\frac{15}{()} = ()$ %。",
+    "options": [],
+    "answer": "9; 48; 40; 37.5",
+    "solution": "【分析】商不变的性质可知 $6 \\div 16 = 9 \\div 24$ ; 根据除法与分数的关系可知 $6 \\div 16 = \\frac{3}{8}$ , 根据分数的基本性质, 求出 $\\frac{3}{8} = \\frac{15}{40}$ ;根据分数与比的关系可知, $\\frac{3}{8} = 3: 8$ , 根据比的基本性质,求出 $3: 8 = 18: 48; 6 \\div 16 = 0. 3 7 5$ , 小数化成百分数, 小数点向右移动两位，加上百分号，即 $0.375 = 37.5\\%$ 。再根据由此解答即可。熟练掌握分数、除法、比之间的关系以及分数、百分数、小数之间的互化是解答本题的关键。\n\n【详解】 $9 \\div 24 = 18: 48 = 6 \\div 16 = \\frac{15}{40} = 37.5\\%$",
+    "knowledge_points": [
+      "百分数",
+      "分数",
+      "小数和比的互化",
+      "比与分数",
+      "除法的关系"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_453",
+    "original_id": "unit6_fill_453"
+  },
+  {
+    "id": "unit6_fill_454",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "(21-22 六年级上·江西赣州·单元测试) 百分之零点八七写作( ), $120.5\\%$ 读作( )。",
+    "options": [],
+    "answer": "$0.87\\%$ 百分之一百二十点五",
+    "solution": "【分析】百分数常常不写成分母是 100 的分数形式，而是在原来的分子后面添加上 '百分号%'来表示。写百分数时，先写分子，再写百分号。一个百分数，%前面的数是几，就读作百分之几。关键是掌握百分数的读法和写法。\n\n【详解】百分之零点八七写作 $0.87\\%$ ， $120.5\\%$ 读作百分之一百二十点五。",
+    "knowledge_points": [
+      "百分数的读法和写法"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_454",
+    "original_id": "unit6_fill_454"
+  },
+  {
+    "id": "unit6_fill_455",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "(14-15 六年级上·江苏宿迁·阶段练习) 甲: 乙 = 8:5,甲比乙多( )%, 乙比甲少( )%。",
+    "options": [],
+    "answer": "60 37.5",
+    "solution": "【分析】用甲、乙的份数差除以乙的份数即可求出甲比乙多百分之几; 用甲、乙的份数差除以甲的份数即可求出乙比甲少百分之几。求一个数比另一个数多（少）百分之几，用'两个数的差÷另一个数'。\n\n【详解】 $(8 - 5)\\div 5\\\\= 3 \\div 5 \\\\ = 60 \\%$ \n\n$ \\\\ (8 - 5) \\div 8 \\\\ = 3 \\div 8 \\\\ = 37.5 \\% $",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几",
+      "比的意义"
+    ],
+    "source": "14-15 六年级上·江苏宿迁·阶段练习",
+    "merged_id": "unit6_455",
+    "original_id": "unit6_fill_455"
+  },
+  {
+    "id": "unit6_fill_456",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "（21-22六年级上·全国·单元测试）某车间应出勤200人，有一天缺勤4人，这天的出勤率是( )%。",
+    "options": [],
+    "answer": "98",
+    "solution": "【分析】根据'出勤率 $=\\frac{\\text{出勤人数}}{\\text{总人数}} \\times 100\\%$ '，进行解答即可。明确出勤率的含义是解答本题的关键。\n\n【详解】 $\\frac{200 - 4}{200} \\times 100\\% = 98\\%$",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "21-22六年级上·全国·单元测试",
+    "merged_id": "unit6_456",
+    "original_id": "unit6_fill_456"
+  },
+  {
+    "id": "unit6_fill_457",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "(21-22 六年级上·全国·单元测试) 数学书和故事书的数量比是 $5: 6$ , 故事书比数学书多 $(\\quad)\\%$ 。",
+    "options": [],
+    "answer": "20",
+    "solution": "【分析】用故事书比数学书多的份数除以数学书的份数即可。求一个数比另一个数多百分之几，用两个数的差÷另一个数。\n\n【详解】 $(6-5) \\div 5\\\\= 1 \\div 5 \\\\ = 20 \\% $",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几",
+      "比的意义"
+    ],
+    "source": "21-22 六年级上·全国·单元测试",
+    "merged_id": "unit6_457",
+    "original_id": "unit6_fill_457"
+  },
+  {
+    "id": "unit6_sort_458",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "排序题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·江西赣州·单元测试) 把 $0.714, \\frac{5}{7}$ , $72 \\%$ , 0.714 按从小到大的顺序排列起来: ( )。",
+    "options": [],
+    "answer": "$0.714 < \\frac{5}{7} < 0.714 < 72\\%$",
+    "solution": "【分析】根据题目要求，应将题中的分数、百分数化成小数后再比较大小，然后按从小到大的顺序排列起来。本题关键在于都把分数、百分数化为小数再进行比较，从而解决问题。\n\n【详解】 $\\frac{5}{7} = 0.714285\\ldots$ ， $72\\% = 0.72$ ， $0.714 =$ 0.714444....因为 $0.714 < 0.714285\\ldots \\ldots < 0.714444\\ldots \\ldots < 0.72$\n\n所以按从小到大排列为: $0.714 < \\frac{5}{7}< 0.714<72 \\%$ .",
+    "knowledge_points": [
+      "多位小数的大小比较",
+      "百分数",
+      "小数和分数的互化"
+    ],
+    "source": "20-21 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_458",
+    "original_id": "unit6_sort_458"
+  },
+  {
+    "id": "unit6_judge_459",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·湖南怀化·单元测试) 将 2 克盐放入100克水中，盐的质量占盐水的 $2\\%$ 。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】盐水的质量是 $100 + 2 = 102$ 克, 盐的质量占盐水的百分之几, 应用盐的质量除以盐水的质量。本题考查百分数, 解答本题的关键是理解盐水的质量是盐加水的质量。\n\n【详解】 $2 \\div 102 \\approx 0.0196 = 1.96 \\%$ , 所以本题说法错误。故答案为: ×。",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_459",
+    "original_id": "unit6_judge_459"
+  },
+  {
+    "id": "unit6_judge_460",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "（20-21 六年级上·湖南怀化·单元测试）0.8 米 $= 80\\%$ 米 $= \\frac{4}{5}$ 米。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】百分数表示两者之间的关系，百分数不带单位。据此解答即可。本题考查百分数的意义，明确百分数的意义是解题的关键。\n\n【详解】由分析可知:小数带单位不能转化为百分数，原题干说法错误。故答案为: ×",
+    "knowledge_points": [
+      "百分数的意义"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_460",
+    "original_id": "unit6_judge_460"
+  },
+  {
+    "id": "unit6_judge_461",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·湖南怀化·单元测试) 甲数比乙数多 $6 \\%$ , 乙数比甲数就少 $6 \\%$ 。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】由题意可知, 设乙数是 1 , 则甲数为 $1 \\times (1 + 6 \\%) ,$ 然后求出乙数比甲数少多少, 再除以甲数即可。本题考查求一个数比另一个数少百分之几，明确单位 '1'是解题的关键。\n\n【详解】假设乙数是 1 ,\n\n$[ 1 \\times (1 + 6 \\%) - 1 ] \\div [ 1 \\times (1 + 6 \\%) ] \\\\ = 0. 0 6 \\div 1. 0 6 \\\\ \\approx 5.7 \\% $\n\n则原题干说法错误。",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_461",
+    "original_id": "unit6_judge_461"
+  },
+  {
+    "id": "unit6_judge_462",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "(21-22 六年级上·云南·期末) 晓东用 99 颗种子做发芽试验, 99 颗都发芽, 发芽率是 $100 \\%$ 。( )",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】根据发芽率 $=$ 发芽种子数 $\\div$ 总种子数 $\\times 100 \\%$ , 进行分析。 $\\times \\times$ 率 = 要求量（就是 $\\times \\times$ 所代表的信息）/单位 '1' 的量（总量） $\\times 100 \\%$ 。\n\n【详解】 $99 \\div 99 \\times 100\\% = 100\\%$故答案为: √",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "21-22 六年级上·云南·期末",
+    "merged_id": "unit6_462",
+    "original_id": "unit6_judge_462"
+  },
+  {
+    "id": "unit6_judge_463",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "(15-16 六年级上·全国·课后作业) 分母是 100 的分数就是百分数。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【详解】根据百分数的意义可知：表示一个数是另一个数的百分之几的数，叫做百分数，百分数也叫做百分率或百分比。它表示两个数之间的倍、比关系，不表示具体数量，所以分母是 100 的分数不都是百分数。",
+    "knowledge_points": [
+      "百分数的意义"
+    ],
+    "source": "15-16 六年级上·全国·课后作业",
+    "merged_id": "unit6_463",
+    "original_id": "unit6_judge_463"
+  },
+  {
+    "id": "unit6_judge_464",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "（2014 五年级·全国·课后作业）种 120 棵树，成活 120 棵，成活率是 $120\\%$ 。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】可依据 '成活的棵数÷种植的棵数×100%' 来计算种树的成活率。计算百分率，像成活率、及格率、出勤率的结果小于等于 $100\\%$ ；而增长率、降低率则有可能大于 $100\\%$ 。\n\n【详解】 $120 \\div 120 \\times 100\\%\\\\=1 \\times 100 \\% \\\\ = 100 \\% $\n\n原题说法错误。故答案为: ×",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "2014 五年级·全国·课后作业",
+    "merged_id": "unit6_464",
+    "original_id": "unit6_judge_464"
+  },
+  {
+    "id": "unit6_judge_465",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "（21-22六年级上·江西赣州·单元测试）某厂上半月完成计划任务的 $35\\%$ ，下半月完成计划任务的 $80\\%$ ，该厂本月已超额完成计划任务。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】用加法计算上半月和下半月完成的任务占计划任务的百分率，计算结果和 $100\\%$ 比较大小即可。刚好完成计划任务时计算结果等于 $100 \\%$ , 没有完成计划任务时计算结果小于 $100 \\%$ , 超额完成计划任务时计算结果大于 $100 \\%$ 。\n\n【详解】 $35\\% + 80\\% = 115\\%$因为 $115\\% > 100\\%$ ，所以该厂本月已超额完成计划任务。故答案为: √",
+    "knowledge_points": [
+      "含百分数的运算"
+    ],
+    "source": "21-22六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_465",
+    "original_id": "unit6_judge_465"
+  },
+  {
+    "id": "unit6_judge_466",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "(21-22 六年级上·江西赣州·单元测试) $100 \\mathrm{~g}$ 水中放入 $25 \\mathrm{~g}$ 糖, 糖水的含糖率是 $25 \\%$ 。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】先用 '100+25' 求出糖水的质量, 进而根据公式: $\\frac{\\text { 糖 的 质 量 }}{\\text { 糖 水 的 质 量 }} \\times 100 \\%$ , 代入数值进行解答, 进而判断即可。此题属于百分率问题, 都是用一部分数量 (或全部数量) 除以全部数量乘百分之百。\n\n【详解】 $\\frac{25}{100 + 25} \\times 100\\%\\\\=0.2 \\times 100 \\% \\\\ = 20 \\% $糖水的含糖率是 $20\\%$\n\n故答案为: ×",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_466",
+    "original_id": "unit6_judge_466"
+  }, {
+    "id": "unit6_judge_467",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·江西赣州·单元测试) 梨与苹果的重量比是 $5: 8$ , 表示梨的重量比苹果少 $60 \\%$ 。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】已知梨与苹果的重量比是 $5: 8$ , 则梨有 5 份, 苹果有 8 份, 先求出梨比苹果少的重量, 然后再除以苹果的重量即可。本题考查求一个数比另一个数少百分之几，明确单位'1'是解题的关键。\n\n【详解】$ (8-5) \\div 8\\\\= 3 \\div 8 \\\\ = 37.5\\% $\n\n则梨的重量比苹果少 $37.5 \\%$ 。原题干说法错误。故答案为: ×",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "20-21 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_467",
+    "original_id": "unit6_judge_467"
+  },
+  {
+    "id": "unit6_judge_468",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·辽宁·单元测试) $\\frac{3}{4}$ 米 $= 0.75$ 米 $= 75\\%$ 米。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】根据百分数的意义，结合题意，直接判断即可。本题考查了百分数的意义, 百分数表示一个数占另一个数的百分之几, 不能带单位。\n\n【详解】百分数不能带单位，所以 $75 \\%$ 米的说法是错误的。所以判断错误。",
+    "knowledge_points": [
+      "百分数",
+      "小数和分数的互化",
+      "百分数的意义"
+    ],
+    "source": "20-21 六年级上·辽宁·单元测试",
+    "merged_id": "unit6_468",
+    "original_id": "unit6_judge_468"
+  },
+  {
+    "id": "unit6_judge_469",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "(21-22 六年级上·江西赣州·单元测试) 如果甲比乙多 $100\\%$ , 那么乙比甲少一半。( )",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】甲比乙多 $100 \\%$ , 将乙看作 100 , 甲看作 $100+$ 100, 两数差 $\\div$ 甲 = 少百分之几, 据此分析。此类问题一般用表示单位'1'的量作除数。\n\n【详解】 $100 \\div (100 + 100)\\\\= 1 0 0 \\div 2 0 0 \\\\ = 50 \\% \\\\ = 一半 $\n\n故答案为: √",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_469",
+    "original_id": "unit6_judge_469"
+  },
+  {
+    "id": "unit6_judge_470",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "(21-22 六年级上·江西赣州·单元测试) 朱叔叔是个神枪手, 有打靶命中率达 $120\\%$ 的辉煌历史。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】'命中率 $= \\frac{\\text { 命中的次数 }}{\\text { 总次数 }} \\times 100\\%$ '，据此可知命中率最高为 $100\\%$ ，据此判断即可。明确命中率的含义是解答本题的关键。\n\n【详解】打靶命中率最高为 $100 \\%$ , 不可能是 $120 \\%$ , 原题说法错误。故答案为: ×",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_470",
+    "original_id": "unit6_judge_470"
+  },
+  {
+    "id": "unit6_judge_471",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "(21-22 六年级上·江西赣州·单元测试) 某产品合格率 $98\\%$ , 即任抽某产品 100 件, 一定有 2 件不合格。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】合格率 $=$ 合格产品 $\\div$ 抽样产品 $\\times 100\\%$ ，指的是抽样产品的合格率是 $98\\%$ ，据此分析。关键是理解百分率的意义。\n\n【详解】某产品合格率 $98 \\%$ , 即任抽某产品 100 件, 可能抽出的 100 件产品全部合格, 不一定有 2 件不合格, 所以原题说法错误。",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_471",
+    "original_id": "unit6_judge_471"
+  },
+  {
+    "id": "unit6_calc_472",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·湖南怀化·单元测试) 解方程。\n\n$2 x = \\frac {5}{7} \\quad 60 \\% x + 25 = 40 \\quad \\frac {3}{4} x - 50 \\% = 7$",
+    "options": [],
+    "answer": "$x = \\frac{5}{14}; x = 25; x = 28$",
+    "solution": "【分析】根据等式的性质解方程。等式的性质 1: 等式两边同时加上或减去同一个数, 等式仍然成立;等式的性质 2: 等式两边同时乘或除以同一个不是零的数,等式仍然成立;注意写'解'字和'='要对齐。\n\n【详解】 $2x = \\frac{5}{7}$\n\n解: $x = \\frac{5}{7} \\div 2\\\\x = \\frac {5}{7} \\times \\frac {1}{2}\\\\x = \\frac {5}{1 4}$\n\n$60 \\% x + 25 = 40$\n\n解： $0.6x = 40 - 25\\\\0.6 x = 1 5$\n\n$x = 1 5 \\div 0. 6$\n\n$x = 2 5$\n\n$\\frac{3}{4} x - 50\\% x = 7$\n\n解: $\\frac{3}{4} x - \\frac{1}{2} x = 7\\\\ \\frac {1}{4} x = 7\\\\x = 7 \\div \\frac {1}{4}\\\\x = 2 8$",
+    "knowledge_points": [
+      "应用等式的性质 1 和 2 解方程"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试",
+    "merged_id": "unit6_472",
+    "original_id": "unit6_calc_472"
+  },
+  {
+    "id": "unit6_calc_473",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "计算题",
+    "difficulty": "容易",
+    "question": "(21-22 六年级上·江西赣州·单元测试) 求未知数 $x$ 。\n\n$50 \\% \\mathrm {x} - 33 \\% \\mathrm {x} = 34$ $(1 + 25 \\%) x = 35$",
+    "options": [],
+    "answer": "$x = 200; x = 28$",
+    "solution": "【分析】第一题先化简方程为 $0.17 \\mathrm{x} = 34$ ，再左右两边同时除以 0.17 即可；第二题先化简方程为 $1.25 \\mathrm{x} = 35$ , 再左右两边同时除以 1.25 即可。\n\n【详解】 $50\\% \\mathrm{x} - 33\\% \\mathrm{x} = 34$\n\n解： $0.17\\mathrm{x} = 34\\\\1 7 \\mathrm {x} \\div 0. 1 7 = 3 4 \\div 0. 1 7\\\\x = 2 0 0;$\n\n$(1 + 25 \\%) x = 35$\n\n解: $1.25x= 35\\\\1.2 5 x \\div 1. 2 5 = 3 5 \\div 1. 2 5 \\\\ x = 2 8$",
+    "knowledge_points": [
+      "应用等式的性质2解方程",
+      "含百分数的运算"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_473",
+    "original_id": "unit6_calc_473"
+  },
+  {
+    "id": "unit6_calc_474",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "计算题",
+    "difficulty": "容易",
+    "question": "(21-22 六年级上·全国·单元测试) 一个数的 $\\frac{1}{4}$ 与它的 $40 \\%$ 的和是 1300 , 这个数是多少?",
+    "options": [],
+    "answer": "2000",
+    "solution": "【分析】把这个数看成单位 '1', 它的 $\\frac{1}{4}$ 与 $40 \\%$ 的和就是它的 $(\\frac{1}{4} + 40 \\%) ,$ 对应的数量是 1300, 由此用除法求出这个数。\n\n【详解】 $1300 \\div (\\frac{1}{4} + 40\\%)\\\\ =1300 \\div 65 \\% \\\\= 2 0 0 0 $",
+    "knowledge_points": [
+      "分数的四则混合运算",
+      "含百分数的运算"
+    ],
+    "source": "21-22 六年级上·全国·单元测试",
+    "merged_id": "unit6_474",
+    "original_id": "unit6_calc_474"
+  },
+  {
+    "id": "unit6_calc_475",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "计算题",
+    "difficulty": "容易",
+    "question": "(21-22六年级上·全国·单元测试)一个数的 $30\\%$ 是 60, 这个数的 $\\frac{4}{5}$ 是多少?",
+    "options": [],
+    "answer": "160",
+    "solution": "【分析】 $30 \\%$ 的单位'1'是一个数，根据除法的意义，用除法列式求出一个数，再根据分数乘法的意义列式求出答案。\n\n【详解】 $60 \\div 30\\% \\times \\frac{4}{5}\\\\= 2 0 0 \\times \\frac {4}{5} \\\\ = 1 6 0 $",
+    "knowledge_points": [
+      "分数的乘",
+      "除法的混合运算",
+      "含百分数的运算"
+    ],
+    "source": "21-22六年级上·全国·单元测试",
+    "merged_id": "unit6_475",
+    "original_id": "unit6_calc_475"
+  },
+  {
+    "id": "unit6_calc_476",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "计算题",
+    "difficulty": "容易",
+    "question": "(21-22 六年级上·全国·单元测试) 列式计算。\n\n480 的 $20 \\%$ 是多少?",
+    "options": [],
+    "answer": "96",
+    "solution": "【分析】把 480 看成单位 '1', 求它的 $20 \\%$ , 用 480 乘 $20 \\%$ 即可。\n\n【详解】 $480 \\times 20\\%\\\\=4 8 0 \\times 0. 2 \\\\ = 9 6 $",
+    "knowledge_points": [
+      "求一个数的百分之几是多少",
+      "含百分数的运算"
+    ],
+    "source": "21-22 六年级上·全国·单元测试",
+    "merged_id": "unit6_476",
+    "original_id": "unit6_calc_476"
+  },
+  {
+    "id": "unit6_calc_477",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "(21-22 六年级上·全国·单元测试) 12 的 $\\frac{1}{4}$ 比哪一个数多 $20\\%$ ？",
+    "options": [],
+    "answer": "2.5",
+    "solution": "【分析】12 的 $\\frac{1}{4}$ 比这个数多 $20 \\%$ , 那么 12 的 $\\frac{1}{4}$ 是这个数的 $120 \\%$ 。据此利用除法求出这个数即可。\n\n【详解】 $12 \\times \\frac{1}{4} \\div (1 + 20\\%)\\\\=3 \\div 120 \\% \\\\ = 2. 5$\n\n所以，12 的 $\\frac{1}{4}$ 比 2.5 多 $20 \\%$ 。",
+    "knowledge_points": [
+      "分数与整数的乘法",
+      "已知比一个数多/少百分之几是多少，求这个数",
+      "含百分数的运算"
+    ],
+    "source": "21-22 六年级上·全国·单元测试",
+    "merged_id": "unit6_477",
+    "original_id": "unit6_calc_477"
+  },
+  {
+    "id": "unit6_solve_478",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（21-22六年级上·江西赣州·单元测试）便民粮店运来大米3600千克，运来的面粉比大米少 $25\\%$ ，这个粮店运来面粉和大米共多少千克？",
+    "options": [],
+    "answer": "6300千克",
+    "solution": "【分析】运来面粉的质量 = 运来大米的质量 $\\times (1 - 25 \\%)$ ，最后加上大米的质量即可。分析题意求出运来面粉的质量是解答题目的关键。\n\n【详解】 $3600 \\times (1 - 25\\%) + 3600\\\\= 3 6 0 0 \\times 0. 7 5 + 3 6 0 0 \\\\ = 2 7 0 0 + 3 6 0 0 \\\\ = 6 3 0 0 \\text {(千 克)} $\n\n答: 这个粮店运来面粉和大米共 6300 千克。",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少",
+      "含百分数的运算"
+    ],
+    "source": "21-22六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_478",
+    "original_id": "unit6_solve_478"
+  },
+  {
+    "id": "unit6_solve_479",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(21-22 六年级上·全国·单元测试) 只列式, 不计算。某水泥厂三月份原计划生产水泥2700吨，实际生产了2900吨，超产百分之几？",
+    "options": [],
+    "answer": "(2900 - 2700) ÷ 2700 × 100%",
+    "solution": "【分析】计算实际比计划多百分之几，（实际产量一计划产量）÷计划产量×100%。A比B多百分之几的计算方法：$(A-B) \\div B \\times 100 \\%$\n\n【详解】 $(2900- 2700) ÷ 2700×100\\%\\\\= 200 \\div 2700 \\times 100 \\% \\\\ \\approx 0.074 \\times 100 \\% \\\\ = 7.4 \\%$\n\n答: 超产 $7.4 \\%$",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "21-22 六年级上·全国·单元测试",
+    "merged_id": "unit6_479",
+    "original_id": "unit6_solve_479"
+  },
+  {
+    "id": "unit6_solve_480",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(21-22 六年级上·全国·单元测试) 只列式, 不计算。\n\n两辆汽车同时从两地相向开出，出发后2.4小时相遇，已知甲车每小时行 $75 \\mathrm{~km}$ ，乙车每小时行 $65 \\mathrm{~km}$ ，两地相距多少千米？",
+    "options": [],
+    "answer": "$(75 + 65) \\times 2.4$",
+    "solution": "【分析】两地之间的路程 = （甲车的速度 + 乙车的速度）×相遇时间，据此计算。掌握相遇问题中总路程的计算公式是解答题目的关键。\n\n【详解】 $(75 + 65) \\times 2.4\\\\1 4 0 \\times 2. 4 \\\\ = 3 3 6 (\\text {千 米}) $\n\n答: 两地相距 336 千米。",
+    "knowledge_points": [
+      "小数与整数的乘法",
+      "相遇问题"
+    ],
+    "source": "21-22 六年级上·全国·单元测试",
+    "merged_id": "unit6_480",
+    "original_id": "unit6_solve_480"
+  },
+  {
+    "id": "unit6_solve_481",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(21-22 六年级上·全国·单元测试) 只列式, 不计算。会议室地面用面积 $0.09 \\mathrm{~m}^{2}$ 的方砖铺设, 320 块方砖正好铺满。如果改用边长是 $0.4 \\mathrm{~m}$ 的方砖铺设，需要多少块？",
+    "options": [],
+    "answer": "180 块",
+    "solution": "【分析】所需方砖的块数 = 会议室的地面面积 ÷ 每块方砖的面积，据此解答。会议室的面积一定，表示出每块方砖的面积，用小数除法即可解答。\n\n【详解】 $320 \\times 0.09 \\div (0.4 \\times 0.4)\\\\= 3 2 0 \\times 0. 0 9 \\div 0. 1 6 \\\\ = 2 8. 8 \\div 0. 1 6 \\\\ = 1 8 0 \\text {(块)} $\n\n答: 需要 180 块。",
+    "knowledge_points": [
+      "正方形的面积",
+      "小数的乘",
+      "除法混合运算",
+      "归总问题"
+    ],
+    "source": "21-22 六年级上·全国·单元测试",
+    "merged_id": "unit6_481",
+    "original_id": "unit6_solve_481"
+  },
+  {
+    "id": "unit6_solve_482",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（2021六年级上·全国·专题练习）有含盐 $15\\%$ 的盐水20千克，要使盐水的浓度为 $20\\%$ ，需加盐多少千克？",
+    "options": [],
+    "answer": "1.25千克",
+    "solution": "【分析】水的质量没有变, 先求出 20 千克盐水中水的质量, 水的质量 $\\div$ 加盐后的对应百分率, 求出加盐后盐水的质量,再减去原来盐水的质量即可。整体数量 $\\times$ 部分对应百分率 $=$ 部分数量, 部分数量 $\\div$ 对应百分率 $=$ 整体数量。\n\n【详解】 $20 \\times 15\\% = 3$ （千克）\n\n$(20 - 3) \\div (1 - 20 \\%) \\\\ = 1 7 \\div 0. 8 \\\\ = 2 1. 2 5 \\text {(千 克)} $\n\n$21.25 - 20 = 1.25$ （千克）答: 需加盐 1.25 千克。",
+    "knowledge_points": [
+      "求一个数的百分之几是多少",
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "2021六年级上·全国·专题练习",
+    "merged_id": "unit6_482",
+    "original_id": "unit6_solve_482"
+  },
+  {
+    "id": "unit6_solve_483",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（19-20六年级上·全国·单元测试）商场对某品牌的电视机进行促销活动，降价 $10\\%$ ，在此基础上，商场又返还售价的 $5\\%$ 现金，此时买电视机，相当于降价百分之几？",
+    "options": [],
+    "answer": "$14.5\\%$",
+    "solution": "【分析】把原价看作单位'1'，降价 $10\\%$ ，售价相当于原价的 $1 - 10\\%$ ，商场又返还售价 $5\\%$ 的现金，则返还了原价的 $(1 - 10\\%) \\times 5\\%$ ，再加上 $10\\%$ 即为降价的分率。正确运用分数乘法意义解决问题是本题考查知识点，注意单位'1'的变化。\n\n【详解】 $(1 - 10\\%) \\times 5\\% + 10\\%\\\\=90 \\% \\times 5 \\% + 10 \\% \\\\ = 4.5 \\% + 10 \\% \\\\=14.5\\%$\n\n答：相当于降价 $14.5\\%$ 。",
+    "knowledge_points": [
+      "含百分数的运算",
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "19-20六年级上·全国·单元测试",
+    "merged_id": "unit6_483",
+    "original_id": "unit6_solve_483"
+  },
+  {
+    "id": "unit6_solve_484",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（2014·全国·小升初真题）一个长方体木块的长、宽、高分别是 $5 \\mathrm{~cm} 、 4 \\mathrm{~cm} 、 3 \\mathrm{~cm}$ 。如果用它锯成一个最大的正方体，体积要比原来减少百分之几？",
+    "options": [],
+    "answer": "$55\\%$",
+    "solution": "【分析】一个长方体木块的长、宽、高分别是5厘米、4厘米、3厘米，把它锯成最大的正方体，这个正方体的棱长等于3厘米，根据正方体的体积公式： $v = a^3$ ，长方体的体积公式： $v = abh$ ，分别求出它们的体积，然后根据百分数的意义解答。此题主要考查长方体、正方体的体积公式的灵活运用，以及百分数的实际应用，解答此题关键是把它锯成最大的正方体确定这个正方体的棱长等于3厘米。\n\n【详解】 $(5 \\times 4 \\times 3 - 3 \\times 3 \\times 3) \\div (5 \\times 4 \\times 3) \\times 100\\%\\\\=33 \\div 60 \\times 100 \\% \\\\ = 55 \\% $\n\n答：体积要比原来减少 $55 \\%$ 。",
+    "knowledge_points": [
+      "正方体的体积",
+      "求一个数比另一个数多少百分之几",
+      "长方体的体积"
+    ],
+    "source": "2014·全国·小升初真题",
+    "merged_id": "unit6_484",
+    "original_id": "unit6_solve_484"
+  },
+  {
+    "id": "unit6_solve_485",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（20-21 六年级上·江西赣州·单元测试）有盐水 25 千克，含盐 $20\\%$ ，加了一些水后含盐 $8\\%$ ，加了多少水？",
+    "options": [],
+    "answer": "37.5克",
+    "solution": "【分析】加了一些水后含盐 $8 \\%$ , 这一过程中盐的质量不变,先用原来盐水的总质量乘上 $20 \\%$ , 求出盐的质量, 再用盐的质量除以后来的含盐率, 求出后来盐水的总质量, 最后用后来盐水的总质量减去原来盐水的总质量即可求解。本题关键是抓住不变的盐的质量, 用盐的质量作为中间量, 求出后来盐水的质量, 进而求解。\n\n【详解】 $25 \\times 20\\% \\div 8\\% - 25\\\\= 5 \\div 8 \\% - 25 \\\\ = 6 2. 5 - 2 5 \\\\ = 3 7. 5 \\text {(千 克)}$\n\n答: 加了 37.5 千克的水。",
+    "knowledge_points": [
+      "求一个数的百分之几是多少",
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "20-21 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_485",
+    "original_id": "unit6_solve_485"
+  },
+  {
+    "id": "unit6_solve_486",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（20-21 六年级上·江西赣州·单元测试）某装配车间男职工人数的 $40 \\%$ 和女职工人数的 $20 \\%$ 相等, 已知这个车间有女职工 130 名, 男职工人数比女职工人数少多少名?",
+    "options": [],
+    "answer": "65名",
+    "solution": "【分析】根据女职工人数计算出男职工人数的 $40 \\%$ , 再利用 '量 $\\div$ 对应的分率' 求出男职工人数, 最后计算男女工人数之差即可。利用除法计算出男职工人数是解答题目的关键。\n\n【详解】男职工人数： $130 \\times 20\\% \\div 40\\%\\\\= 2 6 \\div 0. 4 \\\\ = 6 5 \\text {(人)} $\n\n$1 3 0 - 6 5 = 6 5 (\\text {人})$ 答: 男职工人数比女职工人数少 65 名。",
+    "knowledge_points": [
+      "求一个数的百分之几是多少",
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "20-21 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_486",
+    "original_id": "unit6_solve_486"
+  },
+  {
+    "id": "unit6_solve_487",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（20-21六年级上·江西赣州·单元测试）一种商品6月的价格比5月涨了 $25\\%$ ，7月的价格比6月降了 $25\\%$ ，7月的价格和5月比是涨了还是降了？变化幅度是多少？",
+    "options": [],
+    "answer": "降了； $6.25\\%$",
+    "solution": "【分析】根据题意，把商品5月份的价格看作单位'1'，可得6月份的价格是 $1 \\times (1 + 25\\%)$ ，然后根据7月份的价格比6月份又降了 $25\\%$ ，可得7月份的价格是： $1 \\times (1 + 25\\%) \\times (1 - 25\\%)$ ；最后把7份的价格和1比较大小，判断出7月份的价格和5月比是涨了还是降了，进而求出变化幅度是多少即可。此题主要考查了百分数的实际应用，解答此题的关键是把5月份的价格看作单位'1'。\n\n【详解】把商品 5 月份的价格看作单位 '1', 可得 6 月份的价格是: $1 \\times (1 + 25 \\%)$所以7月份的价格是：\n\n$1 \\times (1 + 25 \\%) \\times (1 - 25 \\%) \\\\ = 1 \\times 1. 2 5 \\times 0. 7 5 \\\\ = 0. 9 3 7 5 $\n\n$1 - 0.9375 = 0.0625 = 6.25 \\%$答：7月份的价格和5月比是降了，变化幅度是 $6.25\\%$ 。",
+    "knowledge_points": [
+      "比一个数多/少百分之几的数是多少"
+    ],
+    "source": "20-21六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_487",
+    "original_id": "unit6_solve_487"
+  },
+  {
+    "id": "unit6_solve_488",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(20-21 六年级上·江西赣州·单元测试) 一条公路已经修了 $40 \\%$ , 再修 300 米, 就修好这条公路的一半, 这条公路全长多少米?",
+    "options": [],
+    "answer": "3000 米",
+    "solution": "【分析】把公路全长看作单位 '1', 这条公路的一半就是公路全长的 $\\frac{1}{2}$ , 先求出已修长度比全长一半少的距离占总长度的分率, 也就是 300 米占总长度的分率, 依据分数除法意义即可解答。本题关键在于对分数除法意义的理解, 关键是求出 300 米占总长度的分率。\n\n【详解】 $300 \\div (\\frac{1}{2} - 40\\%)\\\\= 300 \\div 10 \\% \\\\ = 3 0 0 0 \\text {(米)} $\n\n答: 这条公路全长 3000 米。",
+    "knowledge_points": [
+      "已知一个数的几分之几是多少，求这个数",
+      "百分数",
+      "小数和分数的互化",
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "20-21 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_488",
+    "original_id": "unit6_solve_488"
+  },
+  {
+    "id": "unit6_solve_489",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（20-21六年级上·江西赣州·单元测试）妈妈买了一件原价340元的上衣，现在便宜了 $20\\%$ ，便宜了多少元？",
+    "options": [],
+    "answer": "68元",
+    "solution": "【分析】把原价看成单位'1'，现价比原价便宜 $20\\%$ ，直接用原价乘 $20\\%$ 即可。本题关键在于理解便宜了 $20\\%$ 的意义。\n\n【详解】 $340 \\times 20\\% = 68$ （元）答: 便宜了 68 元。",
+    "knowledge_points": [
+      "求一个数的百分之几是多少"
+    ],
+    "source": "20-21六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_489",
+    "original_id": "unit6_solve_489"
+  },
+  {
+    "id": "unit6_solve_490",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（20-21六年级上·江西赣州·单元测试）希望小学航模小组有20人，比美术小组少了 $80\\%$ 。希望小学美术小组有多少人？",
+    "options": [],
+    "answer": "100人",
+    "solution": "【分析】希望小学航模小组有 20 人, 比美术小组的人数少了 $80 \\%$ , 即航模小组是美术小组人数的 $1 - 80 \\%$ , 根据分数除法的意义, 用航模小组人数除以其占美术小组人数的分率, 即得美术小组有多少人。首先将美术小组人数当作单位'1'，求出航模小组是美术小组人数的百分之几是完成本题的关键。\n\n【详解】 $20 \\div (1 - 80\\%)\\\\20 \\div 20 \\% \\\\ = 1 0 0 (\\text {人}) $\n\n答: 希望小学美术小组有 100 人。",
+    "knowledge_points": [
+      "已知比一个数多/少百分之几是多少，求这个数"
+    ],
+    "source": "20-21六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_490",
+    "original_id": "unit6_solve_490"
+  },
+  {
+    "id": "unit6_solve_491",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（20-21六年级上·江西赣州·单元测试）一辆汽车从甲地开往乙地，行了全程的 $20\\%$ ，还剩240千米，这辆汽车已经行了多少千米？",
+    "options": [],
+    "answer": "60千米",
+    "solution": "【分析】把从甲地到乙地的路程看作单位 '1', 行了全程的 $20 \\%$ , 还剩全程的 $(1 - 20 \\%)$ , 根据 '量 $\\div$ 对应的分率' 求出总路程, 最后用乘法计算总路程的 $20 \\%$ 是多少即可。掌握标准量和比较量的计算方法是解答题目的关键。\n\n【详解】 $240 \\div (1 - 20\\%) \\times 20\\%\\\\ 2 4 0 \\div 0. 8 \\times 0. 2 \\\\ = 3 0 0 \\times 0. 2 \\\\ = 6 0 (\\text {千 米}) $\n\n答: 这辆汽车已经行了 60 千米。",
+    "knowledge_points": [
+      "求一个数的百分之几是多少",
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "20-21六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_491",
+    "original_id": "unit6_solve_491"
+  },
+  {
+    "id": "unit6_solve_492",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "（21-22六年级上·江西赣州·单元测试）只列式不计算。生产300个零件，有5个不合格，求合格率。",
+    "options": [],
+    "answer": "$\\frac{300 - 5}{300} \\times 100\\%$",
+    "solution": "【分析】根据'合格率 $=\\frac{\\text{合格的数量}}{\\text{总数量}} \\times 100\\%$ '，进行解答即可。明确合格率的含义是解答本题的关键。\n\n【详解】 $\\frac{300 - 5}{300} \\times 100 \\% \\approx 98.3 \\%$",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "21-22六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_492",
+    "original_id": "unit6_solve_492"
+  },
+  {
+    "id": "unit6_solve_493",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（21-22六年级上·江西赣州·单元测试）只列式不计算。一双鞋现价45元，比原来便宜5元，便宜了百分之几？",
+    "options": [],
+    "answer": "$5 \\div (45 + 5) \\times 100\\%$",
+    "solution": "【分析】便宜的钱数÷原价 $=$ 便宜了百分之几，据此列式。关键是先确定原价，此类问题一般用表示单位'1'的量作除数。\n\n【详解】根据分析，列式为：\n\n$5 \\div (45 + 5) \\times 100 \\% \\\\ = 5 \\div 50 \\times 100 \\% \\\\ = 10 \\% $\n\n答：便宜了 $10 \\%$",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "21-22六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_493",
+    "original_id": "unit6_solve_493"
+  },
+  {
+    "id": "unit6_solve_494",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(21-22六年级上·江西赣州·单元测试) 姐姐想买一部苹果手机，分期付款购买要加价 $6\\%$ ，如果改用现金购买可享受原价的 $96 \\%$ 的优惠价。姐姐算了一下，发现分期付款比现金购买要多付498元，这部苹果手机原价是多少元？",
+    "options": [],
+    "answer": "4980元",
+    "solution": "【分析】设这部苹果手机原价是 $x$ 元, 根据原价 $\\times$ 分期付款对应百分率 - 原价 $\\times$ 现金购买对应百分率 = 多付的钱数, 列出方程解答即可。用方程解决问题的关键是找到等量关系。\n\n【详解】解：设这部苹果手机原价是 $x$ 元。\n\n$(1 + 6 \\%) x - 96 \\% x = 498 \\\\ 1. 0 6 x - 0. 9 6 \\mathrm {x} = 4 9 8 \\\\ 0. 1 \\mathrm {x} \\times 1 0 = 4 9 8 \\times 1 0 \\\\ x = 4 9 8 0 $\n\n答: 这部苹果手机原价是 4980 元。",
+    "knowledge_points": [
+      "列方程解含一个未知数的问题"
+    ],
+    "source": "21-22六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_494",
+    "original_id": "unit6_solve_494"
+  },
+  {
+    "id": "unit6_solve_495",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(21-22 六年级上·江西赣州·单元测试) 乘坐飞机时,携带行李超过20千克的部分，每千克按飞机原价的 $1.5\\%$ 购买行李票。李芳从南京乘飞机到北京，票价按 $30\\%$ 优惠仅700元，她携带了30千克行李。(1) 飞机票的原价是多少元？ (2) 她应付行李费多少元?",
+    "options": [],
+    "answer": "(1) 1000 元 (2) 150 元",
+    "solution": "【分析】（1）票价按 $30\\%$ 优惠，说明票价是打七折，根据'原价 $=$ 现价 $\\div$ 折扣'解答即可；(2) 用 $30 - 20$ 求出超过部分的质量, 再乘飞机原价的 $1.5\\%$ 即可。解答本题的关键是明确'按 $30 \\%$ 优惠'是打了七折，再根据现价、原价和折扣之间的关系求出原价，从而进一步解答。\n\n【详解】 (1) $700 \\div (1 - 30\\%)\\\\= 7 0 0 \\div 0. 7 \\\\ = 1 0 0 0 (\\text {元}) $\n\n答: 飞机票的原价是 1000 元;\n\n(2) $(30 - 20) \\times (1000 \\times 1.5\\%)\\\\= 1 0 \\times 1 5 \\\\ = 1 5 0 (\\text {元}) $\n\n答: 她应付行李费 150 元。",
+    "knowledge_points": [
+      "求一个数的百分之几是多少",
+      "折扣问题",
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_495",
+    "original_id": "unit6_solve_495"
+  },
+  {
+    "id": "unit6_solve_496",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（21-22六年级上·江西赣州·单元测试）开零售店的张叔叔以每套60元的价格从制造商那进了若干套玩具，这样制造商除去成本每套赚了 $20\\%$ ，而张叔叔打算以每套盈利 $10\\%$ 的价格销售这批玩具。（请提一个你感兴趣的数学问题并解答？）",
+    "options": [],
+    "answer": "张叔叔应该每套卖多少元；66元（答案不唯一）",
+    "solution": "【分析】可提出张叔叔应该每套卖多少元，求出每套要盈利的钱数再与进价相加即可。（答案不唯一）本题主要考查了百分数乘、除法的意义，也可以提出每套的成本价是多少元。\n\n【详解】张叔叔应该每套卖多少元？\n\n$60 \\times 10 \\% + 60 \\\\ = 6 + 6 0 \\\\ = 6 6 (\\text {元}) $\n\n答: 张叔叔应该每套卖 66 元。(答案不唯一)",
+    "knowledge_points": [
+      "'提问题''填条件'问题",
+      "利润问题"
+    ],
+    "source": "21-22六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_496",
+    "original_id": "unit6_solve_496"
+  },
+  {
+    "id": "unit6_solve_497",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "(21-22 六年级上·江西赣州·单元测试) 进入第三季度,某种商品销售进入淡季，某工厂把这种商品的生产量在上季度基础上下调 $10\\%$ 。预计第四季度销售回暖，生产量将比第三季度增加 $15\\%$ ，第四季度的产量和第二季度比是增加了还是减少了？变化幅度是多少？",
+    "options": [],
+    "answer": "增加了；增加 $3.5\\%$",
+    "solution": "【分析】把第二季度这种商品的生产量看作单位'1'，第三季度这种商品的生产量表示为（ $1 - 10\\%$ ），第四季度这种商品的生产量表示为 $(1 - 10\\%) \\times (1 + 15\\%)$ ，结果和单位'1'比较大小即可。找准标准量计算出第四季度这种商品的生产量是解答题目的关键。\n\n【详解】假设第二季度这种商品的生产量为1第四季度这种商品的生产量:\n\n$1 \\times (1 - 10\\%) \\times (1 + 15\\%)\\\\=0.9 \\times 1. 1 5 \\\\ = 1. 0 3 5 $\n\n因为 $1.035 > 1$ , 所以第四季度的产量和第二季度比产量增加了。\n\n$(1.035 - 1) \\div 1 \\times 100 \\%\\\\=0.035\\times 100\\% \\\\ = 3.5\\% $\n\n第四季度的产量和第二季度比产量增加了 $3.5\\%$答：第四季度的产量和第二季度比产量增加了 $3.5\\%$ 。",
+    "knowledge_points": [
+      "求一个数的百分之几是多少",
+      "含百分数的运算",
+      "求一个数比另一个数多/少百分之几"
+    ],
+    "source": "21-22 六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_497",
+    "original_id": "unit6_solve_497"
+  },
+  {
+    "id": "unit6_solve_498",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "（21-22六年级上·江西赣州·单元测试）只列式不计算。一瓶油吃了 $30\\%$ 正好吃了6千克，剩下多少千克？",
+    "options": [],
+    "answer": "14千克",
+    "solution": "【分析】根据题意可知，'一瓶油的总质量 $\\times 30\\% = 6$ 千克'，据此列除法算式求出总质量，再减去吃了的质量即可。明确百分数除法的意义是解答本题的关键。\n\n【详解】 $6 \\div 30\\% - 6\\\\= 2 0 - 6\\\\= 14（千克$\n\n答：还剩下14千克。",
+    "knowledge_points": [
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "21-22六年级上·江西赣州·单元测试",
+    "merged_id": "unit6_498",
+    "original_id": "unit6_solve_498"
+  },
+  {
+    "id": "unit6_solve_499",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（21-22六年级上·全国·单元测试）一桶汽油，第一次用去总数的 $30\\%$ ，第二次用去总数的 $\\frac{1}{5}$ ，还剩 50L，这桶汽油原来有多少升？",
+    "options": [],
+    "answer": "100 升",
+    "solution": "【分析】利用减法先求出剩下的 $50 \\mathrm{~L}$ 占这桶汽油的几分之几, 再利用除法求出这桶汽油原来有多少升即可。本题考查了分数除法的应用，已知一个数的几分之几是多少，求这个数用除法。\n\n【详解】\n\n$50 \\div (1 - 30\\% - \\frac{1}{5})\\\\=5 0 \\div \\frac {1}{2} \\\\ = 1 0 0 \\text {(升)} $\n\n答: 这桶汽油原来有 $100$ 升。",
+    "knowledge_points": [
+      "分数除法的应用",
+      "分数与整数的除法",
+      "含百分数的运算"
+    ],
+    "source": "21-22六年级上·全国·单元测试",
+    "merged_id": "unit6_499",
+    "original_id": "unit6_solve_499"
+  },
+  {
+    "id": "unit6_solve_500",
+    "unit": "第六单元：百分数",
+    "chapter": "百分数应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "（21-22六年级上·全国·单元测试）新城开发区开发一个项目，实际投资3.6亿元，比原计划多投资0.4亿元，多投资了百分之几？",
+    "options": [],
+    "answer": "$12.5\\%$",
+    "solution": "【分析】用多投资的钱数除以计划投资钱数即可求出多投资了百分之几，用实际投资钱数减去比原计划多投资的钱数即可求出原计划投资钱数。解答此题的关键是先求出原计划投资的钱数。\n\n【详解】 $0.4 \\div (3.6 - 0.4) \\times 100\\%\\\\= 0. 4 \\div 3. 2 \\times 1 0 0\\\\= 12.5\\%$\n\n答：多投资了 $12.5 \\%$",
+    "knowledge_points": [
+      "求一个数是另一个数的百分之几（百分率问题）"
+    ],
+    "source": "21-22六年级上·全国·单元测试",
+    "merged_id": "unit6_500",
+    "original_id": "unit6_solve_500"
+  }
+]
\ No newline at end of file
diff --git "a/\345\205\255\345\271\264\347\272\247\351\242\230\351\233\206/\345\205\255\345\271\264\347\272\247\344\270\212/\345\205\255\345\271\264\347\272\247\344\270\212\347\254\254\345\233\233\345\215\225\345\205\203\346\257\224\346\225\260\346\215\256\351\233\206.json" "b/\345\205\255\345\271\264\347\272\247\351\242\230\351\233\206/\345\205\255\345\271\264\347\272\247\344\270\212/\345\205\255\345\271\264\347\272\247\344\270\212\347\254\254\345\233\233\345\215\225\345\205\203\346\257\224\346\225\260\346\215\256\351\233\206.json"
new file mode 100644
index 0000000000000000000000000000000000000000..13c422717d0ae768caa6959d8811678d97d2f42b
--- /dev/null
+++ "b/\345\205\255\345\271\264\347\272\247\351\242\230\351\233\206/\345\205\255\345\271\264\347\272\247\344\270\212/\345\205\255\345\271\264\347\272\247\344\270\212\347\254\254\345\233\233\345\215\225\345\205\203\346\257\224\346\225\260\346\215\256\351\233\206.json"
@@ -0,0 +1,9117 @@
+[
+    {
+        "id": "unit4_choice_001",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "一根钢管，每锯一次的时间相等，锯成4段用的时间与锯成8段所用的时间的比是（）。",
+        "options": [
+          "A. $1: 2$",
+          "B. $3: 7$",
+          "C. $7: 3$"
+        ],
+        "answer": "B",
+        "solution": "【分析】锯的次数比段数少 1 。锯成 4 段需锯 3 次, 锯成 8 段需锯 7 次。假设锯 1 次所需的时间是 1 分钟, 分别用 1 乘锯的次数可得锯 4 段与锯 8 段所用的时间, 再列比即可。\n\n【详解】假设锯 1 次所需的时间是 1 分钟\n\n$1 \\times (4 - 1)\\\\=1 \\times 3\\\\=3 (分钟)$\n\n$1 \\times (8 - 1)$\n\n$= 1 \\times 7$\n\n=7 (分钟)\n\n一根钢管，每锯一次的时间相等，锯成 4 段用的时间与锯成 8 段所用的时间的比是  $3: 7$  。故答案为: B",
+        "knowledge_points": [
+          "比的意义",
+          "植树问题（两端都不栽）"
+        ],
+        "source": "25-26六年级上·海南海口·单元测试",
+        "merged_id": "unit4_001",
+        "original_id": "unit4_choice_001"
+      },
+      {
+        "id": "unit4_choice_002",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "一个三角形的三个内角的度数的比是  $1:2:3$ , 这个三角形一定是 ( ) 三角形。",
+        "options": [
+          "A. 直角",
+          "B. 钝角",
+          "C. 锐角"
+        ],
+        "answer": "A",
+        "solution": "【分析】根据角的大小, 三角形可分成直角三角形 (有一个角为  $90^{\\circ}$  )、锐角三角形 (三个角都小于  $90^{\\circ}$  ) 和钝角三角形（有一个角大于  $90^{\\circ}$  且小于  $180^{\\circ}$ ）。结合题意，判断三角形是什么三角形，需要确定最大内角的度数；已知三角形的三个内角的度数比是  $1:2:3$ ，可以将三个角的度数分别看作1份、2份、3份，总份数为  $1 + 2 + 3 = 6$ （份）。\n\n【详解】根据分析：三个内角的度数分别为：\n\n$1 8 0 ^ {\\circ} \\times \\frac {1}{6} = 3 0 ^ {\\circ}$$1 8 0 ^ {\\circ} \\times \\frac {2}{6} = 6 0 ^ {\\circ}$$1 8 0 ^ {\\circ} \\times \\frac {3}{6} = 9 0 ^ {\\circ}$因为有一个角是  $90^{\\circ}$ , 所以这个三角形是直角三角形。故答案为: A",
+        "knowledge_points": [
+          "按比分配问题",
+          "三角形的分类"
+        ],
+        "source": "25-26 六年级上·海南海口·单元测试",
+        "merged_id": "unit4_002",
+        "original_id": "unit4_choice_002"
+      },
+      {
+        "id": "unit4_choice_003",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "450 米: 1 千米化成最简整数比是 ( )。",
+        "options": [
+          "A. $450: 1$",
+          "B. $450: 1000$",
+          "C. 9:20"
+        ],
+        "answer": "C",
+        "solution": "【分析】先将单位统一为米, 1 千米 = 1000 米, 再根据比的基本性质化简比。\n\n【详解】450米：1千米  \n=450米:1000米  \n$= 450: 1000$  \n= (450÷50) : (1000÷50)  \n$= 9:20$\n\n450 米 : 1 千米化成最简整数比是  $9: 20$  。故答案为: C",
+        "knowledge_points": [
+          "比的基本性质",
+          "比的化简",
+          "千米和米之间的进率与换算"
+        ],
+        "source": "25-26 六年级上·海南海口·单元测试",
+        "merged_id": "unit4_003",
+        "original_id": "unit4_choice_003"
+      },
+      {
+        "id": "unit4_choice_004",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "一个比, 如果把它的前项扩大到原来的 2 倍, 要使比值不变, 则比的后项 ()。",
+        "options": [
+          "A. 乘2",
+          "B. 除以 2",
+          "C. 加上 2"
+        ],
+        "answer": "A",
+        "solution": "【分析】比的基本性质是：比的前项和后项同时乘或除以相同的数（0 除外），比值不变。据此解答。\n\n【详解】根据比的基本性质可知, 一个比, 如果把它的前项扩大到原来的 2 倍, 要使比值不变, 则比的后项也要乘 2。\n\n故答案为：A",
+        "knowledge_points": [
+          "比的基本性质"
+        ],
+        "source": "25-26 六年级上·海南海口·单元测试",
+        "merged_id": "unit4_004",
+        "original_id": "unit4_choice_004"
+      },
+      {
+        "id": "unit4_choice_005",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "把 4:0.5 化成最简单的整数比是（）。",
+        "options": [
+          "A. 8",
+          "B. $1 : 8$",
+          "C. 8:1"
+        ],
+        "answer": "C",
+        "solution": "【分析】根据比的基本性质，将比的前项和后项同时乘2，消除小数部分，得到最简整数比。\n\n【详解】4:0.5\n\n$= (4 \\times 2): (0. 5 \\times 2) \\\\ = 8: 1 $\n\n把  $4: 0.5$  化成最简单的整数比是  $8: 1$  。故答案为: C",
+        "knowledge_points": [
+          "比的基本性质",
+          "比的化简"
+        ],
+        "source": "25-26 六年级上·海南海口·单元测试",
+        "merged_id": "unit4_005",
+        "original_id": "unit4_choice_005"
+      },
+      {
+        "id": "unit4_choice_006",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "把 12:16 的后项减去  12，要使比值不变，前项就要（）。",
+        "options": [
+          "A. 除以 3",
+          "B. 除以 4",
+          "C. 减去 12",
+          "D. 加上 12"
+        ],
+        "answer": "B",
+        "solution": "【分析】比的前项和后项，同时乘或除以相同的数（0 除外），比值不变，据此分析。\n\n【详解】  $16 \\div (16 - 12)$\n\n$= 1 6 \\div 4 \\\\ = 4 \\\\ 1 2 - 1 2 \\div 4 \\\\ = 1 2 - 3 \\\\ = 9 $\n\n把 12:16 的后项减去 12, 要使比值不变, 前项就要除以 4 或减去 9。故答案为: B",
+        "knowledge_points": [
+          "比的基本性质"
+        ],
+        "source": "24-25 六年级上·重庆忠县·期末",
+        "merged_id": "unit4_006",
+        "original_id": "unit4_choice_006"
+      },
+      {
+        "id": "unit4_choice_007",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "$a:b = \\frac{a}{n}: \\frac{b}{n}$  （b、n均不为 0）运用了（）。",
+        "options": [
+          "A. 比的基本性质",
+          "B. 除法的性质",
+          "C. 商不变规律"
+        ],
+        "answer": "A",
+        "solution": "【分析】比的基本性质：比的前项和后项同时乘或除以相同的数（0 除外），比值不变。\n\n【详解】根据比的基本性质可知, a:b 的前项、后项同时除以 n (b、n 均不为 0 ) , 比值不变; 即 a:b=\n\n$(a \\div n) : (b \\div n) = \\frac {a}{n}: \\frac {b}{n} 。$所以,  $a: b = \\frac{a}{n}: \\frac{b}{n}$  (b、n均不为 0 ) 运用了比的基本性质。故答案为: A",
+        "knowledge_points": [
+          "比的基本性质"
+        ],
+        "source": "2024·山西长治·小升初真题",
+        "merged_id": "unit4_007",
+        "original_id": "unit4_choice_007"
+      },
+      {
+        "id": "unit4_choice_008",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "在美丽乡村建设中,甲、乙两支工程队分别承包两条同样长道路的施工, 当甲工程队修完全长的  $\\frac{3}{5}$  时, 乙工程队已修的与未修的比是9:11, 根据以上信息, 可以知道 ( )。",
+        "options": [
+          "A. 甲工程队完成的多",
+          "B. 乙工程队完成的多",
+          "C. 两个工程队完成的一样长多",
+          "D. 无法判断哪个工程队完成的多"
+        ],
+        "answer": "A",
+        "solution": "【分析】根据题意：甲工程队修完全长的  $\\frac{3}{5}$  时，乙工程队已修的与未修的比是  $9: 11$  ，那么此时乙工程队已修的占全长的  $\\frac{9}{9 + 11} = \\frac{9}{20}$  ，已知甲、乙两支工程队施工的道路同样长，所以只需要比较  $\\frac{3}{5}$  和  $\\frac{9}{20}$  的大小即可。\n\n【详解】由乙工程队已修的与未修的比是  $9: 11$  知: 乙工程队已修的占全长的  $\\frac{9}{9 + 11} = \\frac{9}{20}$ ;又因为此时甲工程队修完全长的  $\\frac{3}{5}, \\frac{3}{5} = \\frac{12}{20}, \\frac{12}{20} > \\frac{9}{20}$ , 所以甲工程队完成的多。故答案为: A",
+        "knowledge_points": [
+          "异分母异分子分数的大小比较",
+          "比的应用"
+        ],
+        "source": "24-25 六年级上·福建福州·期中",
+        "merged_id": "unit4_008",
+        "original_id": "unit4_choice_008"
+      },
+      {
+        "id": "unit4_choice_009",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "把比  $5: 8$  的前项加10，要使比值不变，后项应该（）。",
+        "options": [
+          "A. 加 3",
+          "B. 加 10",
+          "C. 乘3",
+          "D. 扩大为原来的2倍"
+        ],
+        "answer": "C",
+        "solution": "【分析】比的基本性质, 即将比的前项和后项同时乘上或者同时除以相同的数 (0 除外), 比值不变。前项 5 加 10 变成 15，相当于前项 5 乘 3；依据比的基本性质，比的前项乘 3，后项也乘 3，比值不变；后项乘 3 是  $8 \\times 3 = 24$ , 再用  $24 - 8$ , 可得后项加上几, 比值也不变; 据此分析解答。\n\n【详解】  $(5 + 10) \\div 5$\n\n$= 1 5 \\div 5 \\\\ = 3 \\\\ 8 \\times 3 - 8 \\\\ = 2 4 - 8 \\\\ = 1 6 $\n\n把比  $5: 8$  的前项加 10 , 要使比值不变, 后项应该乘 3 或者加上  $16_。$故答案为: C",
+        "knowledge_points": [
+          "比的基本性质"
+        ],
+        "source": "24-25 六年级上·山东济南·期中",
+        "merged_id": "unit4_009",
+        "original_id": "unit4_choice_009"
+      },
+      {
+        "id": "unit4_choice_010",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "甲、乙、丙三辆车共同完成一项运输海洋垃圾的任务，方案 A 是按 3:2:4 分配，方案 B 是按 2:3:5 分配，那么乙车分到的任务数量（）。",
+        "options": [
+          "A. 按 A 方案分到的多",
+          "B. 按 B 方案分到的多",
+          "C. 按两种方案分到的一样多",
+          "D. 无法确定按哪种方案分到的多"
+        ],
+        "answer": "B",
+        "solution": "【分析】根据题意，先结合比的意义分别求出两种方案中乙车分到的任务数量占总任务数量的几分之几; 再比较分数的大小即可解答。\n\n【详解】  $3 + 2 + 4 = 9$  （份）， $2 \\div 9 = \\frac{2}{9}$ ，所以按照方案 A 乙车可分到任务总量的  $\\frac{2}{9}$ ；2+3+5=10（份）， $3 \\div 10 = \\frac{3}{10}$ ，所以按照方案B乙车可分到任务总量的  $\\frac{3}{10}$ ；$\\frac{2}{9} = \\frac{20}{90}, \\frac{3}{10} = \\frac{27}{90}$ , 因为  $\\frac{20}{90}< \\frac{27}{90}$ , 所以  $\\frac{2}{9}< \\frac{3}{10}$ , 所以按照方案 B 乙车分到的任务数量多。故答案为: B",
+        "knowledge_points": [
+          "求一个数占另一个数几分之几",
+          "按比分配问题"
+        ],
+        "source": "24-25 六年级上·贵州六盘水·期中",
+        "merged_id": "unit4_010",
+        "original_id": "unit4_choice_010"
+      },
+      {
+        "id": "unit4_choice_011",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "小明特别爱喝甜的水, 请你帮他选一个最甜的配置方法 ( )。",
+        "options": [
+          "A. $15 \\mathrm{~kg}$  糖和  $6 \\mathrm{~kg}$  水",
+          "B. $10 \\mathrm{~kg}$  糖和  $5 \\mathrm{~kg}$  水",
+          "C. $20 \\mathrm{~kg}$  糖和  $10 \\mathrm{~kg}$  水",
+          "D. $25 \\mathrm{~kg}$  糖和  $15 \\mathrm{~kg}$  水"
+        ],
+        "answer": "A",
+        "solution": "【分析】先根据比的意义写出每种配置方法中糖与水的质量之比, 再求比值, 然后比较大小; 比值越大的, 说明每 kg 水中放的糖越多, 水就越甜。\n\n【详解】A.  $15: 6 = 15 \\div 6 = 2 \\frac{1}{2}$B.  $10: 5 = 10 \\div 5 = 2$  C.  $20: 10 = 20 \\div 10 = 2$  D.  $25: 15 = 25 \\div 15 = 1 \\frac{2}{3}$$2 \\frac {1}{2} > 2 > 1 \\frac {2}{3}$最甜的配置方法是  $15 \\mathrm{~kg}$  糖和  $6 \\mathrm{~kg}$  水。故答案为: A",
+        "knowledge_points": [
+          "比的意义",
+          "求比值"
+        ],
+        "source": "24-25 六年级上·吉林四平·期中",
+        "merged_id": "unit4_011",
+        "original_id": "unit4_choice_011"
+      },
+      {
+        "id": "unit4_choice_012",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "美丽的海洋里住着各种各样的鱼类: 刺鲅 (cì bà) 是活跃在热带海域的一种鱼类, 它每小时可以游 80 千米; 飞鱼能跳出水面, 拍打翼状鳍进行滑翔, 在海里每小时能游 60 千米, 在空中可停留的最长时间约是 48 秒; 因上颌向前延伸呈剑状而得名的剑鱼时速比飞鱼快  $\\frac{7}{6}$  。关于以上三种鱼类, 下面说法正确的是（ ）。",
+        "options": [
+          "A. 刺鲅和飞鱼的时速比是  $3: 4$",
+          "B. 飞鱼的速度比刺鲅的速度慢  $\\frac{1}{3}$",
+          "C. 剑鱼每小时可以游 130 千米",
+          "D. 若一条飞鱼在空中停留 48 秒, 飞行路程是 396 米,则它在空中的速度是 8 米/秒"
+        ],
+        "answer": "C",
+        "solution": "【分析】A. 刺鲅速度 80 千米, 飞鱼速度 60 千米。则刺鲅和飞鱼的时速比为  $80: 40$ , 化简比即可;B. '飞鱼的速度比刺鲅的速度慢几分之几', 将刺鲅的速度（80千米）看作单位 '1', 对应的量  $\\div$  单位 '1' = 对应的分率。即飞鱼的速度比刺鲅的速度少的部分  $\\div$  刺鲅的速度 =飞鱼的速度比刺鲅的速度慢几分之几, 将数据代入计算即可;C. '剑鱼时速比飞鱼快  $\\frac{7}{6}$  ', 飞鱼的速度是单位 '1', 剑鱼是飞鱼的  $(1 + \\frac{7}{6})$ , 单位 '1'已知, 由分数乘法意义知: 飞鱼的速度  $\\times (1 + \\frac{7}{6}) =$  剑鱼的速度, 代入数据计算即可;  D. 速度 = 路程  $\\div$  时间, 将飞鱼的路程 396 米, 时间 48 秒,代入计算即可。从题目中提取数学信息也是数学的重要解题能力之一，从题目中找到自己需要的数学信息，进而结合问题去分析。\n\n【详解】A. 刺鲮和飞鱼的时速比  $= 80: 60 = 4: 3$ , 选项说法不正确;  B. 飞鱼的速度比刺鲅的速度慢:  $(80 - 60) \\div 80 = \\frac{1}{4}$ , 选项说法不正确;  C. 剑鱼每小时可以游:  $60 \\times (1 + \\frac{7}{6}) = 60 \\times \\frac{13}{6} = 130$  (千米),选项说法正确;  D.  $396 \\div 48 = 8.25$  (米/秒), 选项说法不正确。故答案为: C",
+        "knowledge_points": [
+          "求一个数占另一个数几分之几",
+          "基础行程问题",
+          "比的化简",
+          "求比一个数多/少几分之几的数是多少"
+        ],
+        "source": "23-24六年级上·福建厦门·期中",
+        "merged_id": "unit4_012",
+        "original_id": "unit4_choice_012"
+      },
+      {
+        "id": "unit4_choice_013",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "从甲盐库取出  $\\frac{1}{5}$  的盐运到乙盐库, 这时两个盐库所存的盐的质量相等, 原来甲盐库和乙盐库的存盐质量的比是 ( )。",
+        "options": [
+          "A. $5: 3$",
+          "B. 4:5",
+          "C. 6:5",
+          "D. $5: 4$"
+        ],
+        "answer": "A",
+        "solution": "【分析】由 '甲盐库取出  $\\frac{1}{5}$  的盐运到乙盐库后, 两个盐库所存的盐的质量相等' 可知, 把甲的质量看作单位 '1', 把甲平均分成 5 份, 取出其中的 1 份给乙, 则甲还剩下 4 份。此时乙和甲质量相等, 则乙此时也是 4 份, 那么乙原来就是有 3 份。再写出两个盐库的存盐质量的比。\n\n【详解】  $5 - 1 = 4$  （份）$4 - 1 = 3$  (份)原来甲盐库和乙盐库的存盐质量的比是  $5: 3$  。故答案为: A",
+        "knowledge_points": [
+          "比的应用"
+        ],
+        "source": "2024·四川绵阳·小升初真题",
+        "merged_id": "unit4_013",
+        "original_id": "unit4_choice_013"
+      },
+      {
+        "id": "unit4_choice_014",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "把 20 克盐完全溶解在 500 克水中, 盐与盐水的质量比是 ( )。",
+        "options": [
+          "A. $1: 25$",
+          "B. $1: 26$",
+          "C. 26:1"
+        ],
+        "answer": "B",
+        "solution": "【分析】盐水的质量 = 盐的质量 + 水的质量，再根据比的意义求出盐与盐水的质量比。\n\n【详解】20: (20+500)\n\n$= 2 0: 5 2 0 \\\\ = (2 0 \\div 2 0): (5 2 0 \\div 2 0) \\\\ = 1: 2 6 $\n\n盐与盐水的质量比是  $1: 26$  。故答案为: B",
+        "knowledge_points": [
+          "比的化简",
+          "比的意义"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_014",
+        "original_id": "unit4_choice_014"
+      },
+      {
+        "id": "unit4_choice_015",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "一个比，如果把它的前项扩大到原来的3倍，要使比值不变，则比的后项（）。",
+        "options": [
+          "A. 乘3",
+          "B. 除以 3",
+          "C. 加上 3"
+        ],
+        "answer": "A",
+        "solution": "【分析】根据比的基本性质，比的前项和后项同时乘一个数或除以一个数（0 除外）比值不变。据此解答。\n\n【详解】一个比, 如果把它的前项扩大到原来的 3 倍, 要使比值不变, 则比的后项也要扩大到原来的 3 倍, 即乘 3。故答案为: A",
+        "knowledge_points": [
+          "比的基本性质"
+        ],
+        "source": "23-24六年级上·全国·单元测试",
+        "merged_id": "unit4_015",
+        "original_id": "unit4_choice_015"
+      },
+      {
+        "id": "unit4_choice_016",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "六 (1) 班共有学生45人，男、女生人数的比不可能是（）。",
+        "options": [
+          "A. $3: 2$",
+          "B. 4:5",
+          "C. 4:3"
+        ],
+        "answer": "C",
+        "solution": "【分析】把各项选项中的比看作份数，用总人数除以份数和，求出一份数看一份数是否是整数，不是整数的，这个比就不可能是这个班男、女生人数比。\n\n【详解】A.  $3 + 2 = 5, 45 \\div 5 = 9$  ，能整除，所以  $3: 2$  可能是男、女生人数比；B.  $4 + 5 = 9, 45 \\div 9 = 5$ , 能整除, 所以  $4: 5$  可能是男、女生人数比;C.  $4 + 3 = 7, 45 \\div 7 = 6 \\dots 3$ , 不能整除, 所以  $4: 3$  不可能是该班男、女生人数比。故答案为: C",
+        "knowledge_points": [
+          "比的意义",
+          "比的应用"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_016",
+        "original_id": "unit4_choice_016"
+      },
+      {
+        "id": "unit4_choice_017",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "等底等高的三角形和平行四边形, 三角形和平行四边形面积的比是 ( )。",
+        "options": [
+          "A. $2: 1$",
+          "B. $1 : 2$",
+          "C. 4:1"
+        ],
+        "answer": "B",
+        "solution": "【分析】三角形面积  $= \\text{底} \\times \\text{高} \\div 2$ , 平行四边形面积  $= \\text{底} \\times \\text{高}$ ,所以, 等底等高的三角形面积是平行四边形面积的一半,据此解答。\n\n【详解】假设三角形的面积为1，由分析可知, 与它等底等高的平行四边形面积为 2 ,所以，等底等高的三角形和平行四边形面积的比是  $1: 2$ 。故答案为: B",
+        "knowledge_points": [
+          "三角形面积的计算",
+          "平行四边形面积的计算",
+          "比的意义"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_017",
+        "original_id": "unit4_choice_017"
+      },
+      {
+        "id": "unit4_choice_018",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "甲数的  $\\frac{4}{15}$  和乙数相等，甲数和乙数的比是（）。",
+        "options": [
+          "A. 4:15",
+          "B. $15: 4$",
+          "C. $\\frac{4}{15}: 1$"
+        ],
+        "answer": "B",
+        "solution": "【分析】甲数的  $\\frac{4}{15}$  和乙数相等, 说明乙数是甲数的  $\\frac{4}{15}$ , 把甲数看作 '1', 则乙数是  $\\frac{4}{15}$ , 再据此求出甲数和乙数的比, 化简即可。\n\n【详解】  $1: \\frac{4}{15}$\n\n$= (1 \\times 1 5): (\\frac {4}{1 5} \\times 1 5) \\\\ = 1 5: 4 $\n\n甲数和乙数的比是  $15: 4$  。故答案为: B",
+        "knowledge_points": [
+          "比的化简",
+          "比的意义"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_018",
+        "original_id": "unit4_choice_018"
+      },
+      {
+        "id": "unit4_choice_019",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "13:5 的前项乘 2 ,要使比值缩小为原来的  $\\frac{1}{2}$ , 后项应加上 ( )。",
+        "options": [
+          "A. 2",
+          "B. 15",
+          "C. 13",
+          "D. 无法确定"
+        ],
+        "answer": "B",
+        "solution": "【分析】根据比值的求法：比值  $=$  比的前项  $\\div$  比的后项，求出 13:5 的比值；用  $13 \\times 2$  ，求出前项乘 2 后的值；用比值  $\\times \\frac{1}{2}$ ，求出缩小后的比值；再根据比的后项  $=$  比值前项  $\\div$  比值，用前项乘 2 后的值除以缩小后的比值，求出后项，再用求出的后项一原来的后项，即可解答。\n\n【详解】  $13: 5 = \\frac{13}{5}$\n\n$1 3 \\times 2 = 2 6 \\\\ \\frac {1 3}{5} \\times \\frac {1}{2} = \\frac {1 3}{1 0} \\\\ 2 6 \\div \\frac {1 3}{1 0} - 5 \\\\ = 2 6 \\times \\frac {1 0}{1 3} - 5 \\\\ = 2 0 - 5 、、= 1 5$\n\n13:5 的前项乘 2 , 要使比值缩小为原来的  $\\frac{1}{2}$ , 后项应加上 15。故答案为: B",
+        "knowledge_points": [
+          "比与分数、除法的关系",
+          "分数与整数的除法",
+          "求比值",
+          "分数乘分数"
+        ],
+        "source": "24-25 六年级上·全国·单元测试",
+        "merged_id": "unit4_019",
+        "original_id": "unit4_choice_019"
+      },
+      {
+        "id": "unit4_fill_020",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "$12 \\div () =$  ( ) :  $30 = 3: 5 = \\frac{6}{()}$  。",
+        "options": [],
+        "answer": "20; 18; 10",
+        "solution": "【分析】根据题目中的连等式，所有部分的值都等于  $3: 5$ 。分数的分子相当于被除数、比的前项，分母相当于除数、比的后项。利用比的基本性质和分数与除法的关系，分别求出各个括号中的数。\n\n【详解】  $3:5 = (3\\times 4):(5\\times 4) = 12:20 = 12\\div 20$\n\n$3: 5 = (3 \\times 6): (5 \\times 6) = 1 8: 3 0 \\\\ 3: 5 = (3 \\times 2): (5 \\times 2) = 6: 1 0 = \\frac {6}{1 0} \\\\ 1 2 \\div 2 0 = 1 8: 3 0 = 3: 5 = \\frac {6}{1 0} 。 $",
+        "knowledge_points": [
+          "比的基本性质",
+          "比与分数、除法的关系"
+        ],
+        "source": "25-26 六年级上·海南海口·单元测试",
+        "merged_id": "unit4_020",
+        "original_id": "unit4_fill_020"
+      },
+      {
+        "id": "unit4_fill_021",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "被减数、减数、差的和是 70 , 减数和差的比是  $4: 1$ , 减数是( )。",
+        "options": [],
+        "answer": "28",
+        "solution": "【分析】根据'被减数一减数  $=$  差'，则'被减数  $=$  减数十差'。已知'被减数、减数、差的和是 70'，即：被减数十减数十差  $= 70$ ，把'被减数  $=$  减数十差'代入可得：(减数十差) + 减数十差 = 70, 即:  $2 \\times$  (减数十差) = 70。因此，减数十差为  $70 \\div 2 = 35$  。已知 '减数和差的比是  $4: 1$ ', 把减数看作 4 份; 差看作 1 份。则 '减数十差' 的总份数为:  $4 + 1 = 5$  份。那么每份是  $35 \\div 5 = 7$  。减数占 4 份, 所以用 7 乘 4 即可解答。\n\n【详解】被减数一减数  $=$  差    被减数  $=$  减数十差         (减数 + 差) + 减数 + 差 = 70     2 $\\times$ (减数＋差）＝70   减数 + 差  $= 70 \\div 2$   减数十差  $= 35$   $4 + 1 = 5$  （份）   $35 \\div 5 = 7$   $7 \\times 4 = 28$所以减数是  $28 。$",
+        "knowledge_points": [
+          "加、减法的意义和各部分间的关系",
+          "按比分配问题"
+        ],
+        "source": "25-26 六年级上·海南海口·单元测试",
+        "merged_id": "unit4_021",
+        "original_id": "unit4_fill_021"
+      },
+      {
+        "id": "unit4_fill_022",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "从 A 地步行到 B 地，甲需要 4 分钟，乙需要 6 分钟。甲、乙两人所用时间的比是( )，甲、乙两人速度的比是( )。",
+        "options": [],
+        "answer": "$2: 3或\\frac {2}{3};3: 2或\\frac {3}{2} $",
+        "solution": "【分析】已知甲需要 4 分钟, 乙需要 6 分钟, 据题意列比并根据比的基本性质化简;\n\n将路程看作单位 '1', 根据路程  $\\div$  时间 = 速度, 分别求出甲、乙的速度再列比并化简。\n\n【详解】4:6\n\n$= (4 \\div 2): (6 \\div 2) \\\\ = 2: 3 $\n\n$1 \\div 4 = \\frac {1}{4}$\n\n$1 \\div 6 = \\frac {1}{6} \\\\ \\frac {1}{4}: \\frac {1}{6} \\\\ = (\\frac {1}{4} \\times 1 2): (\\frac {1}{6} \\times 1 2) \\\\ = 3: 2$\n\n从 A 地步行到 B 地, 甲需要 4 分钟, 乙需要 6 分钟。甲、乙两人所用时间的比是  $2: 3$ , 甲、乙两人速度的比是  $3: 2$  。",
+        "knowledge_points": [
+          "比的基本性质",
+          "基础行程问题",
+          "比的化简",
+          "比的意义"
+        ],
+        "source": "25-26 六年级上·海南海口·单元测试",
+        "merged_id": "unit4_022",
+        "original_id": "unit4_fill_022"
+      },
+      {
+        "id": "unit4_fill_023",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "甲、乙两数的比是  $5: 3$ , 乙数和甲数的平均数是 3.2, 甲数是( )。",
+        "options": [],
+        "answer": "4",
+        "solution": "【分析】已知甲、乙两数的比是  $5: 3$ , 以及它们的平均数是 3.2, 先用平均数  $\\times 2$ , 求出甲、乙两数的和, 在根据比求出甲数, 据此解答。\n\n【详解】根据分析：甲、乙两数的和:  $3.2 \\times 2 = 6.4$总份数:  $5 + 3 = 8$甲数:  $6.4 \\times \\frac{5}{5 + 3} = 6.4 \\times \\frac{5}{8} = 4$\n\n所以甲数是  $4$  。",
+        "knowledge_points": [
+          "平均数的意义及求法",
+          "按比分配问题"
+        ],
+        "source": "25-26 六年级上·海南海口·单元测试",
+        "merged_id": "unit4_023",
+        "original_id": "unit4_fill_023"
+      },
+      {
+        "id": "unit4_fill_024",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "苹果千克数与梨千克数的比是  $2: 3$ , 单价的比是  $5: 4$ , 苹果与梨总价的比是( )，比值是( )。",
+        "options": [],
+        "answer": "5:6;$\\frac{5}{6}$",
+        "solution": "【分析】已知苹果与梨的千克数比是  $2: 3$ , 单价比是  $5: 4$  。根据总价  $=$  单价  $\\times$  数量, 分别计算苹果和梨的总价, 再求它们的比及比值。总价的比由对应项的乘积之比确定，最后化简比并求比值。\n\n【详解】  $(2\\times 5):(3\\times 4)$    \n\n   $=10:12\\\\= (10\\div 2):(12\\div 2)\\\\=5:6$    \n\n5:6  \n\n$=5÷6  \\\\=\\frac{5}{6}$  \n\n苹果千克数与梨千克数的比是2：3，单价的比是5：4，苹果  与梨总价的比是5：6，比值是  $\\frac{5}{6}$",
+        "knowledge_points": [
+          "比的化简",
+          "比的意义",
+          "求比值"
+        ],
+        "source": "25-26 六年级上·海南海口·单元测试",
+        "merged_id": "unit4_024",
+        "original_id": "unit4_fill_024"
+      },
+      {
+        "id": "unit4_fill_025",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "小红看一本故事书, 已看了  $\\frac{2}{5}$ , 已看的页数与剩下的最简整数比是( )。",
+        "options": [],
+        "answer": "2:3",
+        "solution": "【分析】把这本书的总页数看作单位 '1', 已看了  $\\frac{2}{5}$ , 那么剩下的就是  $1 - \\frac{2}{5}$  。然后求已看页数与剩下页数的比, 再化简成最简整数比, 据此解答。\n\n【详解】剩下的页数占比:  $1 - \\frac{2}{5} = \\frac{3}{5}$已看页数与剩下页数的比:  $\\frac{2}{5}: \\frac{3}{5} = 10:15 = 2:3$已看的页数与剩下的最简整数比是  $2: 3$",
+        "knowledge_points": [
+          "比的化简",
+          "比的意义"
+        ],
+        "source": "25-26 六年级上·海南海口·单元测试",
+        "merged_id": "unit4_025",
+        "original_id": "unit4_fill_025"
+      },
+      {
+        "id": "unit4_fill_026",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "把一根绳子按5:6 剪成甲、乙两段, 已知甲段长 15 米, 则乙段长( )米。",
+        "options": [],
+        "answer": "18",
+        "solution": "【分析】已知绳子按 5:6 剪成甲、乙两段, 甲段对应 5 份,乙段对应 6 份。根据甲段实际长度 15 米, 先求出每份的长度, 再乘乙段的份数即可得到乙段长度。\n\n【详解】  $15 \\div 5 \\times 6$\n\n$= 3 \\times 6 \\\\ = 1 8 (\\text {米}) $\n\n把一根绳子按 5:6 剪成甲、乙两段, 已知甲段长 15 米, 则乙段长 18 米。",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "25-26 六年级上·海南海口·单元测试",
+        "merged_id": "unit4_026",
+        "original_id": "unit4_fill_026"
+      },
+      {
+        "id": "unit4_fill_027",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "A、B 两数的比是  $4: 7$ , 则数 A 是数 B 的( ), 数 B 是数 A 的( )。",
+        "options": [],
+        "answer": "$\\frac{4}{7}$;$\\frac{7}{4}$",
+        "solution": "【分析】已知 A、B 两数的比是  $4: 7$ , 求数 A 是数 B 的几分之几, 用 A 的份数除以 B 的份数; 求数 B 是数 A 的几分之几, 用 B 的份数除以 A 的份数, 据此解答。\n\n【详解】数 A 是数 B 的:  $4 \\div 7 = \\frac{4}{7}$数B是数A的:  $7 \\div 4 = \\frac{7}{4}$则数 A 是数 B 的  $\\frac{4}{7}$ , 数 B 是数 A 的  $\\frac{7}{4}$  。",
+        "knowledge_points": [
+          "比的意义",
+          "比的应用"
+        ],
+        "source": "25-26 六年级上·海南海口·单元测试",
+        "merged_id": "unit4_027",
+        "original_id": "unit4_fill_027"
+      },
+      {
+        "id": "unit4_fill_028",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "$\\frac{9}{5} : (\\quad) =\\frac{3}{5} (): \\frac {3}{4} = 1 2$",
+        "options": [],
+        "answer": "3;9",
+        "solution": "【分析】第一小题：根据比的前项  $\\div$  比值  $=$  比的后项；第二小题：比的前项  $=$  比的后项  $\\times$  比值。\n\n【详解】  $\\frac{9}{5} \\div \\frac{3}{5} = \\frac{9}{5} \\times \\frac{5}{3} = 3$  ，所以  $\\frac{9}{5}: 3: \\frac{3}{5}$$12 \\times \\frac{3}{4} = 9$ , 所以  $9: \\frac{3}{4} = 12$  。",
+        "knowledge_points": [
+          "求比值"
+        ],
+        "source": "25-26 六年级上·全国·单元测试",
+        "merged_id": "unit4_028",
+        "original_id": "unit4_fill_028"
+      },
+      {
+        "id": "unit4_fill_029",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "一个比的后项是  15, 比值是  $\\frac{2}{3}$ , 这个比的前项是 ( )。如果把它的前项除以  $\\frac{1}{5}$ , 后项乘 5 , 比值是 ( )。",
+        "options": [],
+        "answer": "10;$\\frac{2}{3}$",
+        "solution": "【分析】根据比的前项除以后项等于比值，可得比的前项就等于比值乘比的后项，由此得出结果。根据比的基本性质：比的前项和后项同时乘或除以一个相同的数（0 除外），比值不变，据此解答。\n\n【详解】比值是  $\\frac{2}{3}$ , 比的后项是 15 , 则比的前项是  $15 \\times \\frac{2}{3} = 10$  前项除以  $\\frac{1}{5}$  转化为乘 5 , 后项也是乘 5 , 即比的前项、后面同时乘 5 , 根据比的基本性质, 比值不变。",
+        "knowledge_points": [
+          "比的基本性质",
+          "比的意义",
+          "求比值"
+        ],
+        "source": "25-26 六年级上·全国·单元测试",
+        "merged_id": "unit4_029",
+        "original_id": "unit4_fill_029"
+      },
+      {
+        "id": "unit4_fill_030",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "五 (2) 班男生是女生的  $\\frac{2}{5}$ , 则女生与男生的比是____, 男生占全班人数的_____。",
+        "options": [],
+        "answer": "$5:2$;$\\frac{2}{7}$",
+        "solution": "【分析】把五（2）班的女生人数看作单位'1'，设女生人数为a，则男生人数是  $\\frac{2}{5} a$ ，用女生人数比男生人数，再根据比的基本性质：比的前项和后项同时乘或除以相同的数（0除外），比值不变化成最简比；全班人数为  $a + \\frac{2}{5} a$ ，根据求一个数占另一个数的几分之几，用这个数除以另一个数解答，求男生占全班人数的几分之几，用男生人数除以全班人数即可解答。\n\n【详解】设女生人数为  $a$ ，则男生人数是  $\\frac{2}{5} a$ 。\n\n$a: \\frac {2}{5} \\mathrm {a} = 1: \\frac {2}{5} = (1 \\times 5): (\\frac {2}{5} \\times 5) = 5: 2 \\\\ \\frac {2}{5} a \\div (a + \\frac {2}{5} a) \\\\ = \\frac {2}{5} a \\div \\frac {7}{5} a \\\\ = \\frac {2}{5} \\times \\frac {5}{7} \\\\ = \\frac {2}{7} $\n\n所以女生与男生的比是  $5: 2$ , 男生占全班人数的  $\\frac{2}{7}$  。",
+        "knowledge_points": [
+          "求一个数占另一个数几分之几",
+          "比的化简",
+          "比的意义"
+        ],
+        "source": "24-25 六年级下·海南海口·单元测试",
+        "merged_id": "unit4_030",
+        "original_id": "unit4_fill_030"
+      },
+      {
+        "id": "unit4_fill_031",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "甲数除以乙数的商是 0.75 , 乙数和甲数的比是_____。",
+        "options": [],
+        "answer": "4:3",
+        "solution": "【分析】根据除法与比的关系可得：甲数÷乙数 = 甲数 : 乙数  $= 0.75$  ，据此可以假设乙数是 1 , 则甲数  $= 0.75 \\times 1$ , 再根据比的意义写出乙数和甲数的比, 最后根据比的基本性质把结果化成最简整数比即可。\n\n【详解】假设乙数是1，则甲数  $= 0.75 \\times 1 = 0.75$ 。\n\n乙数：甲数$= 1: 0. 7 5 \\\\ = 1: \\frac {3}{4} \\\\ = (1 \\times 4): (\\frac {3}{4} \\times 4) \\\\ = 4: 3 $\n\n甲数除以乙数的商是 0.75，乙数和甲数的比是  $4: 3$ 。",
+        "knowledge_points": [
+          "比与分数、除法的关系",
+          "比的化简"
+        ],
+        "source": "24-25 六年级下·海南海口·单元测试",
+        "merged_id": "unit4_031",
+        "original_id": "unit4_fill_031"
+      },
+      {
+        "id": "unit4_fill_032",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "把  $\\frac{7}{8}: \\frac{7}{16}$  化成最简的整数比是____，比值是_____。",
+        "options": [],
+        "answer": "2:1;2",
+        "solution": "【分析】比的基本性质：比的前项和后项同时乘或除以一个相同的数（0 除外），比值不变，据此把给出的比化成最简整数比；比值  $= \\frac{1}{2}$  的前项  $\\div$  比的后项，据此用化简的最简整数比的前项除以后项即可得到比值。\n\n【详解】  $\\frac{7}{8}:\\frac{7}{16}$\n\n$= (\\frac {7}{8} \\times 1 6): (\\frac {7}{1 6} \\times 1 6) \\\\ = 1 4: 7 \\\\ = (1 4 \\div 7): (7 \\div 7) \\\\ = 2: 1 $\n\n$2: 1 = 2 \\div 1 = 2$\n\n把  $\\frac{7}{8}: \\frac{7}{16}$  化成最简的整数比是  $2: 1$ , 比值是  $2$  。",
+        "knowledge_points": [
+          "比的化简",
+          "求比值"
+        ],
+        "source": "24-25 六年级下·海南海口·单元测试",
+        "merged_id": "unit4_032",
+        "original_id": "unit4_fill_032"
+      },
+      {
+        "id": "unit4_fill_033",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "将  $1: 0.25$  的前项乘 4 , 要使比值不变, 后项应该变成____; 如果它的前后项都除以0.25，比值是____。",
+        "options": [],
+        "answer": "1;4",
+        "solution": "【分析】比的基本性质：比的前项和后项同时乘或除以一个相同的数（0 除外），比值不变；如果  $1:0.25$  的前项乘4，根据比的基本性质，要使比值不变，后项也应该乘4，据此求出后项应该变成多少；如果前后项都除以0.25，比值不变，据此用比的前项除以后项即可得到比值。\n\n【详解】  $0.25 \\times 4 = 1$  $1: 0. 2 5 = 1 \\div 0. 2 5 = 4$将  $1: 0.25$  的前项乘 4 , 要使比值不变, 后项应该变成 1 ; 如果它的前后项都除以 0.25 , 比值是 4 。",
+        "knowledge_points": [
+          "比的基本性质",
+          "求比值"
+        ],
+        "source": "24-25 六年级下·海南海口·单元测试",
+        "merged_id": "unit4_033",
+        "original_id": "unit4_fill_033"
+      },{
+        "id": "unit4_fill_034",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "甲、乙、丙三个数的平均数是 75，甲：乙=2:3，甲：丙=3:5，乙数是( )。",
+        "options": [],
+        "answer": "81",
+        "solution": "【分析】根据比的意义，先统一甲、乙、丙三个数的份数，甲原来平均分成 2 份，后来平均分成 3 份，2 和 3 的最小公倍数是  $2 \\times 3 = 6$  ，根据比的基本性质，甲：乙  $= 2:3 = 6:9$  ，甲：丙  $= 3:5 = 6:10$  ，可知甲、乙、丙三个数的比是  $6:9:10$  ，由题意可知三个数的和是  $75 \\times 3$  ，可用和除以  $(6 + 9 + 10)$  得到每份是多少，再用每份的数量乘乙数的 9 份，即可得解。\n\n【详解】甲：乙  $= 2:3 = 6:9$  甲：丙  $= 3: 5 = 6: 10$ 甲：乙：丙=6:9:10\n\n$(7 5 \\times 3) \\div (6 + 9 + 1 0)$\n\n$= 2 2 5 \\div 2 5$\n\n$=9$\n\n$9 \\times 9 = 81$\n\n甲、乙、丙三个数的平均数是 75 , 甲 : 乙 = 2:3, 甲 : 丙 = 3:5, 乙数是 81 。",
+        "knowledge_points": [
+          "公倍数与最小公倍数",
+          "比的基本性质",
+          "按比分配问题",
+          "比的应用"
+        ],
+        "source": "2025·福建龙岩·小升初真题",
+        "merged_id": "unit4_034",
+        "original_id": "unit4_fill_034"
+      },
+      {
+        "id": "unit4_fill_035",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "《庄子·天下》中有这样一句话：'一尺之棰，日取其半，万世不竭。'意思是说：一尺长的木棒，每天截取一半，永远也截取不完。照这样推算，第三天截取的长度与最初木棒长度的比是( )。",
+        "options": [],
+        "answer": "$1: 8$",
+        "solution": "【分析】假设木棒长度是40米，将木棒长度看作单位'1'，每天截取一半，即每天截取  $\\frac{1}{2}$ ，第一天截取  $\\frac{1}{2}$ ，还剩\n\n$(1 - \\frac{1}{2})$  ；再将第一天截取后剩下的长度看作单位'1'，第二天截取第一天截取后剩下的  $\\frac{1}{2}$ ，还剩第二天截取后剩下长度的  $(1 - \\frac{1}{2})$  ；再将第二天截取后剩下的长度看作单位'1'，第三天截取第二天截取后剩下的  $\\frac{1}{2}$ ，最初木棒长度×第一天截取后剩下的对应分率×第二天截取后剩下的对应分率×第三天截取的对应分率 = 第三天截取的长度。\n\n两数相除又叫两个数的比, 根据比的意义, 写出第三天截取的长度与最初木棒长度的比, 化简即可。\n\n【详解】假设木棒长度是40米。\n\n$4 0 \\times (1 - \\frac {1}{2}) \\times (1 - \\frac {1}{2}) \\times \\frac {1}{2} \\\\ = 4 0 \\times \\frac {1}{2} \\times \\frac {1}{2} \\times \\frac {1}{2} \\\\ = 2 0 \\times \\frac {1}{2} \\times \\frac {1}{2} \\\\ = 1 0 \\times \\frac {1}{2} \\\\ = 5 (\\text {米}) $\n\n$5: 4 0 = (5 \\div 5): (4 0 \\div 5) = 1: 8$\n\n第三天截取的长度与最初木棒长度的比是  $1: 8$  。",
+        "knowledge_points": [
+          "连续求一个数的几分之几是多少的问题",
+          "比的化简",
+          "比的意义"
+        ],
+        "source": "24-25六年级上·湖南岳阳·期中",
+        "merged_id": "unit4_035",
+        "original_id": "unit4_fill_035"
+      },
+      {
+        "id": "unit4_fill_036",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "从甲桶中取出  $\\frac{1}{5}$  的油倒入乙桶, 这时两桶油的质量相等, 原来甲、乙两桶中油的质量之比为( )。",
+        "options": [],
+        "answer": "5:3",
+        "solution": "【分析】将甲桶中原来的油看作单位 '1', 用单位 '1' 减去  $\\frac{1}{5}$ , 求出之后甲桶中的油, 即现在乙桶中的油。再将乙桶现在的减去甲倒进来的, 求出乙桶原来的, 从而求出原来甲、乙两桶中油的质量之比。\n\n【详解】1:  $(1 - \\frac{1}{5} - \\frac{1}{5})$\n\n$= 1: \\frac {3}{5} \\\\ = (1 \\times 5): (\\frac {3}{5} \\times 5) \\\\ = 5: 3 $\n\n所以，原来甲、乙两桶中油的质量之比为  $5: 3$  。",
+        "knowledge_points": [
+          "比的化简",
+          "比的意义"
+        ],
+        "source": "20-21 六年级上·山东·单元测试",
+        "merged_id": "unit4_036",
+        "original_id": "unit4_fill_036"
+      },
+      {
+        "id": "unit4_fill_037",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "$0.8: \\frac{3}{4}$  化成最简单的整数比是( )，比值是( )。",
+        "options": [],
+        "answer": "$16: 15;\\frac{16}{15}或1\\frac{1}{15}$",
+        "solution": "【分析】比的基本性质：比的前项和后项同时乘或除以相同的数（0除外），比值不变。利用'比的基本性质'把比化简成最简单的整数比。用比的前项除以比的后项所得的商，叫做比值。根据比值的意义，用最简比的前项除以比的后项即得比值。\n\n【详解】  $0.8:\\frac{3}{4}$\n\n$= \\frac {4}{5}: \\frac {3}{4} \\\\ = (\\frac {4}{5} \\times 2 0): (\\frac {3}{4} \\times 2 0) \\\\ = 1 6: 1 5 $\n\n16:15\n\n$= 1 6 \\div 1 5 \\\\ = \\frac {1 6}{1 5} $\n\n0.8: $\\frac{3}{4}$ 化成最简单的整数比是  $16: 15$ , 比值是  $\\frac{16}{15}$ 。",
+        "knowledge_points": [
+          "比的化简",
+          "求比值"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_037",
+        "original_id": "unit4_fill_037"
+      },
+      {
+        "id": "unit4_fill_038",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "被减数、减数、差的和是 60 , 减数和差的比是  $3: 2$ , 减数是( )。",
+        "options": [],
+        "answer": "18",
+        "solution": "【分析】根据被减数  $=$  差 + 减数, 则  $60 \\div 2$  就是减数与差的和,又知减数和差的比是  $3: 2$ , 即可知减数占两数之和的  $\\frac{3}{3+2}$ ,根据求一个数的几分之几是多少用乘法计算, 即可得解。\n\n【详解】  $(60 \\div 2) \\times \\frac{3}{3 + 2}$\n\n$ 3 0 \\times \\frac {3}{5}\\\\=18$\n\n减数是 18 。",
+        "knowledge_points": [
+          "加、减法的意义和各部分间的关系",
+          "按比分配问题",
+          "比的应用"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_038",
+        "original_id": "unit4_fill_038"
+      },
+      {
+        "id": "unit4_fill_039",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "鸡的数量比鸭多  $\\frac{1}{4}$ ,鸡和鸭的数量比是 ( ) , 鸭比鸡少  $\\frac{()}{()}$  。",
+        "options": [],
+        "answer": "5:4;$\\frac{1}{5}$",
+        "solution": "【分析】已知鸡的数量比鸭多  $\\frac{1}{4}$ , 把鸭的数量看作 4 份, 则鸡的数量是  $(4 + 1)$  份, 根据比的意义, 写出鸡和鸭的数量比;先用减法求出鸭比鸡少的份数，再除以鸡的数量，即是鸭比鸡少几分之几。\n\n【详解】  $(4 + 1): 4 = 5: 4$\n\n$(5 - 4) \\div (4 + 1) \\\\ = 1 \\div 5 $\n\n$= \\frac {1}{5}$\n\n鸡和鸭的数量比是  $5: 4$ , 鸭比鸡少  $\\frac{1}{5}$  。",
+        "knowledge_points": [
+          "求一个数占另一个数几分之几",
+          "比的意义",
+          "比的应用"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_039",
+        "original_id": "unit4_fill_039"
+      },
+      {
+        "id": "unit4_fill_040",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "从 A 地步行到 B 地,甲需要 2 小时, 乙需要 3 小时。甲、乙两人所用时间的比是 ( ) , 甲、乙两人速度的比是 ( ) 。",
+        "options": [],
+        "answer": "2:3;3:2",
+        "solution": "【分析】已知甲要 2 小时到达, 乙要 3 小时到达, 甲、乙两人所用时间的比是  $2: 3$ , 把两地的距离看作单位 1 , 再根据路程  $\\div$  时间 = 速度, 分别求出甲、乙的速度, 进而求出它们速度的比即可。\n\n【详解】时间比：2:3速度比:  $\\frac{1}{2}: \\frac{1}{3} = (\\frac{1}{2} \\times 6): (\\frac{1}{3} \\times 6) = 3: 2$甲、乙两人所用时间的比是  $2: 3$ , 甲、乙两人速度的比是3:2。",
+        "knowledge_points": [
+          "基础行程问题",
+          "比的化简",
+          "比的意义"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_040",
+        "original_id": "unit4_fill_040"
+      },
+      {
+        "id": "unit4_fill_041",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "甲、乙两数的比是  4:3, 乙数和甲数的平均数是 8.4 , 甲数是( )。",
+        "options": [],
+        "answer": "9.6",
+        "solution": "【分析】根据总数  $=$  平均数  $\\times$  份数, 由此即可求出甲乙两个数的总和, 即  $8.4 \\times 2 = 16.8$ , 根据比的意义, 甲和乙的比是4:3, 则甲相当于 4 份, 乙相当于 3 份, 总共  $4 + 3 = 7$  份,即一份量用总量÷总份数，再用一份量乘甲的份数即可求解。\n\n【详解】  $8.4 \\times 2 \\div (4 + 3) \\times 4$$= 1 6. 8 \\div 7 \\times 4 \\\\ = 2. 4 \\times 4 \\\\ = 9. 6 $\n\n甲数是  $9.6$  。",
+        "knowledge_points": [
+          "平均数的意义及求法",
+          "比的应用"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_041",
+        "original_id": "unit4_fill_041"
+      },
+      {
+        "id": "unit4_fill_042",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "香蕉千克数与芒果千克数的比是  $3: 4$ ，单价的比是  $5: 4$ ，香蕉与芒果总价的比是（），比值是（）。",
+        "options": [],
+        "answer": "15:16;$\\frac{15}{16}$",
+        "solution": "【分析】已知香蕉千克数与芒果千克数的比是  $3: 4$ , 单价的比是  $5: 4$ , 则香蕉的千克数看作 3 , 单价看作 5 , 芒果的千克数看作 4 , 单价看作 4 , 根据单价  $\\times$  数量 = 总价, 据此求出香蕉和芒果的总价, 根据比的意义写出总价的比, 再用香蕉与芒果总价比的前项除以后项, 即可求出比值。\n\n【详解】  $3 \\times 5 = 15$\n\n$4 \\times 4 = 1 6 \\\\ 1 5: 1 6 \\\\ = 1 5 \\div 1 6 = \\frac {1 5}{1 6}$\n\n香蕉千克数与芒果千克数的比是  $3: 4$ , 单价的比是  $5: 4$ , 香蕉与芒果总价的比是  $15: 16$ , 比值是  $\\frac{15}{16}$  。",
+        "knowledge_points": [
+          "比的意义",
+          "求比值"
+        ],
+        "source": "23-24六年级上·全国·单元测试",
+        "merged_id": "unit4_042",
+        "original_id": "unit4_fill_042"
+      },
+      {
+        "id": "unit4_fill_043",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "小红看一本故事书,已看了  $\\frac{3}{7}$ , 已看的页数与剩下的最简整数比是( )。",
+        "options": [],
+        "answer": "3:4",
+        "solution": "【分析】把这本故事书的总页数看作单位 '1', 已看了  $\\frac{3}{7}$ ,则还剩下总页数的  $(1 - \\frac{3}{7})$ ; 根据比的意义, 写出已看的页数与剩下的页数之比, 并化简比。\n\n【详解】  $\\frac{3}{7}: (1 - \\frac{3}{7})$\n\n$= \\frac {3}{7}: \\frac {4}{7} \\\\ = (\\frac {3}{7} \\times 7): (\\frac {4}{7} \\times 7) \\\\ = 3: 4 $\n\n已看的页数与剩下的最简整数比是  $3: 4$  。",
+        "knowledge_points": [
+          "比的化简",
+          "比的意义"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_043",
+        "original_id": "unit4_fill_043"
+      },
+      {
+        "id": "unit4_fill_044",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "把一根绳子按 3:5剪成甲、乙两段，已知甲段长9米，则乙段长( )米。",
+        "options": [],
+        "answer": "15",
+        "solution": "【分析】已知一根绳子按 3:5 剪成甲、乙两段, 可以把甲段看作 3 份, 乙段看作 5 份, 又知甲段长 9 米, 用甲段的长度除以甲段的份数, 求出每份长多少米, 再乘乙段对应的份数即可。\n\n【详解】  $9 \\div 3 \\times 5$\n\n$= 3 \\times 5 \\\\ = 1 5 \\text {(米)} $\n\n则乙段长 15 米。",
+        "knowledge_points": [
+          "比的应用"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_044",
+        "original_id": "unit4_fill_044"
+      },
+      {
+        "id": "unit4_fill_045",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "A、B 两数的比是2:5, 则数 A 是数 B 的( ), 数 B 是数 A 的( )。",
+        "options": [],
+        "answer": "$\\frac{2}{5}$;$\\frac{5}{2}$",
+        "solution": "【分析】根据比的意义，可假设 A 为 2，B 为 5，根据求一个数占另一个数的几分之几，用一个数除以另一个数，分别求出数 A 是数 B 的几分之几以及数 B 是数 A 的几分之几。\n\n【详解】假设A为2，B为5,\n\n$2 \\div 5 = \\frac {2}{5} \\\\ 5 \\div 2 = \\frac {5}{2} $\n\nA、B 两数的比是  $2: 5$ , 则数 A 是数 B 的  $\\frac{2}{5}$ , 数 B 是数 A 的  $\\frac{5}{2}$ 。",
+        "knowledge_points": [
+          "比与分数、除法的关系",
+          "求一个数占另一个数几分之几",
+          "比的意义"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_045",
+        "original_id": "unit4_fill_045"
+      },
+      {
+        "id": "unit4_fill_046",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "5: ( )=35   ( ): 0. 6 = 8",
+        "options": [],
+        "answer": "$\\frac{1}{7}$;4.8",
+        "solution": "【分析】比的前项相当于被除数，比的后项相当于除数，比值相当于商，根据比和除法的关系，可知比的前项  $= \\frac{1}{2}$  的后项  $\\times$  比值，比的后项  $= \\frac{1}{2}$  的前项  $\\div$  比值。据此解答。\n\n【详解】  $5 \\div 35 = \\frac{1}{7}\n\n$0.6 \\times 8 = 4. 8 \\\\ 5: \\frac {1}{7} = 3 5 \\\\ 4. 8: 0. 6 = 8 $",
+        "knowledge_points": [
+          "比与分数、除法的关系",
+          "比的意义"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_046",
+        "original_id": "unit4_fill_046"
+      },
+      {
+        "id": "unit4_fill_047",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "已知  $A: B = 4: 3$ , 且  $B$  比  $A$  少 15 , 则  $A = (\\quad)$  。",
+        "options": [],
+        "answer": "60",
+        "solution": "【分析】由已知，可知A:B=4:3，则把A看作4份，B看\n\n作3份，又知B比A少1份，且B比A少15，由此可求出1份是多少，进一步可以求出A是多少。\n\n【详解】  $15 \\div (4 - 3)$\n\n$= 1 5 \\div 1 \\\\ = 1 5 $\n\n$1 5 \\times 4 = 6 0$所以  $A = 60$ 。",
+        "knowledge_points": [
+          "比的意义",
+          "比的应用"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_047",
+        "original_id": "unit4_fill_047"
+      },
+      {
+        "id": "unit4_fill_048",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "甲数是乙数的  $\\frac{5}{8}$ , 乙数与甲数的比是 ( ) , 乙数占甲、乙两数和的  $\\frac{()}{()}$ 。",
+        "options": [],
+        "answer": "8: 5;$\\frac{8}{13}$",
+        "solution": "【分析】已知甲数是乙数的  $\\frac{5}{8}$ , 则把乙数看作 1 , 甲数看作  $\\frac{5}{8}$ ,根据比的意义, 写出乙数与甲数的比, 再化简即可;再用乙数除以甲、乙两数和，即可求出乙数占甲、乙两数和的几分之几。\n\n【详解】  $1:\\frac{5}{8}$\n\n$= (1 \\times 8): (\\frac {5}{8} \\times 8) \\\\ = 8: 5 $\n\n$1 \\div (1 + \\frac {5}{8}) \\\\ = 1 \\div \\frac {1 3}{8} \\\\ = 1 \\times \\frac {8}{1 3} \\\\ = \\frac {8}{1 3} $\n\n甲数是乙数的  $\\frac{5}{8}$ , 乙数与甲数的比是  $8: 5$ , 乙数占甲、乙两数和的  $\\frac{8}{13}$  。",
+        "knowledge_points": [
+          "求一个数占另一个数几分之几",
+          "比的化简",
+          "比的意义"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_048",
+        "original_id": "unit4_fill_048"
+      },
+      {
+        "id": "unit4_fill_049",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "$20 \\div ()= \\frac{()}{9} = 2: 3 = 8:()$",
+        "options": [],
+        "answer": "30;6;12",
+        "solution": "【分析】比与分数的关系：比的前项相当于分子，比的后项相当于分母，比号相当于分数线；分数的基本性质: 分数的分子和分母同时乘或除以相同的数 (0 除外), 分数的大小不变;分数与除法的关系：分子相当于被除数，分母相当于除数，分数线相当于除号。\n\n【详解】  $2:3 = \\frac{2}{3}$\n\n$\\frac {2}{3} = \\frac {2 \\times 1 0}{3 \\times 1 0} = \\frac {2 0}{3 0}, \\quad \\frac {2 0}{3 0} = 2 0 \\div 3 0$\n\n$\\frac {2}{3} = \\frac {2 \\times 3}{3 \\times 3} = \\frac {6}{9}$\n\n$\\frac {2}{3} = \\frac {2 \\times 4}{3 \\times 4} = \\frac {8}{1 2}, \\frac {8}{1 2} = 8: 1 2$\n\n即  $20 \\div 30 = \\frac{6}{9} = 2: 3 = 8: 12$ 。",
+        "knowledge_points": [
+          "比与分数、除法的关系",
+          "分数的基本性质的应用"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_049",
+        "original_id": "unit4_fill_049"
+      },
+      {
+        "id": "unit4_fill_050",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "甲、乙两数的比是3:8，两数的差是15，两数的和是( )。",
+        "options": [],
+        "answer": "33",
+        "solution": "【分析】将乙数看为 8 份, 甲数看为 3 份, 那么两数的差就是 (8-3) 份, 两数的和就是 (8+3) 份, 用  $15 \\div (8 - 3)$  计算出每份是多少, 再乘总份数  $(8 + 3)$ , 即可求解。\n\n【详解】  $15 \\div (8 - 3) \\times (8 + 3)$\n\n$= 1 5 \\div 5 + 1 1 \\\\ = 3 \\times 1 1 \\\\ = 3 3 $\n\n甲、乙两数的比是  $3: 8$ , 两数的差是 15 , 两数的和是  $33$  。",
+        "knowledge_points": [
+          "比的应用"
+        ],
+        "source": "24-25 六年级上·全国·单元测试",
+        "merged_id": "unit4_050",
+        "original_id": "unit4_fill_050"
+      },
+      {
+        "id": "unit4_fill_051",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "某班有男生 30 人,男生人数和女生人数的比是  $5: 4$ , 女生有( )人, 全班共有( )人。",
+        "options": [],
+        "answer": "24;54",
+        "solution": "【分析】根据题意可知, 男生人数和女生人数的比是  $5: 4$ ,即女生人数是男生人数的  $\\frac{4}{5}$ , 把男生人数看作单位 '1', 用男生人数  $\\times \\frac{4}{5}$ , 求出女生人数; 再把男生人数与女生人数相加, 即可求出全班人数, 据此解答。\n\n【详解】  $30 \\times \\frac{4}{5} = 24$  （人）\n\n$3 0 + 2 4 = 5 4 \\text {(人)}$\n\n某班有男生 30 人, 男生人数和女生人数的比是  $5: 4$ , 女生有 24 人, 全班共有 54 人。",
+        "knowledge_points": [
+          "比的应用"
+        ],
+        "source": "24-25 六年级上·全国·单元测试",
+        "merged_id": "unit4_051",
+        "original_id": "unit4_fill_051"
+      },
+      {
+        "id": "unit4_fill_052",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "$\\frac{3}{5} \\mathrm{~km}: 200 \\mathrm{~m}$  化简后是( )，比值是( )。",
+        "options": [],
+        "answer": "3:1;3",
+        "solution": "【分析】化简比根据比的基本性质，即比的前项和后项，\n\n同时乘或除以相同的数（0 除外），比值不变；求比值直接用比的前项÷后项即可，化简比的结果还是一个比，求比值的结果是一个数。\n\n【详解】  $\\frac{3}{5} \\mathrm{~km}: 200 \\mathrm{~m}=600 \\mathrm{~m}: 200 \\mathrm{~m}=(600 \\div 200): (200 \\div 200)=3: 1=3 \\div 1=3$\n\n$\\frac{3}{5} \\mathrm{~km}: 200 \\mathrm{~m}$  化简后是  $3: 1$ , 比值是  $3$  。",
+        "knowledge_points": [
+          "比的化简",
+          "求比值"
+        ],
+        "source": "24-25 六年级上·全国·单元测试",
+        "merged_id": "unit4_052",
+        "original_id": "unit4_fill_052"
+      },
+      {
+        "id": "unit4_judge_053",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "从学校到邮局，甲用7分钟，乙用8分钟，甲、乙的速度比是  $7:8$ 。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】分析题目, 把学校到邮局的路程看作 '1', 根据速度  $=$  路程  $\\div$  时间求出甲、乙的速度, 再根据比的意义写出甲、乙的速度比, 最后根据比的基本性质把结果化成最简整数比并判断即可。\n\n【详解】  $1 \\div 7 = \\frac{1}{7}$\n\n$1 \\div 8 = \\frac {1}{8}$\n\n甲的速度：乙的速度\n\n$= \\frac {1}{7}: \\frac {1}{8} \\\\ = (\\frac {1}{7} \\times 5 6): (\\frac {1}{8} \\times 5 6) \\\\ = 8: 7 $\n\n即从学校到邮局, 甲用 7 分钟, 乙用 8 分钟, 甲、乙的速度比是  $8: 7$  。故答案为:  ×",
+        "knowledge_points": [
+          "基础行程问题",
+          "比的化简",
+          "比的意义"
+        ],
+        "source": "24-25六年级下·海南海口·单元测试",
+        "merged_id": "unit4_053",
+        "original_id": "unit4_judge_053"
+      },
+      {
+        "id": "unit4_judge_054",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "7:6 的前项和后项同时加上 6 , 比值不变。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】用比的前项除以比的后项所得的商，叫做比值。比的基本性质: 比的前项和后项同时乘或除以相同的数 (0\n\n除外），比值不变。\n\n【详解】  $7: 6 = 7 \\div 6 = \\frac{7}{6}$\n\n$(7 + 6): (6 + 6) = 1 3: 1 2 = 1 3 \\div 1 2 = \\frac {1 3}{1 2} \\\\ \\frac {7}{6} \\neq \\frac {1 3}{1 2} $\n\n所以，7:6的前项和后项同时加上6，比值变化。原题说法错误。故答案为: ×",
+        "knowledge_points": [
+          "比的基本性质",
+          "求比值"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_054",
+        "original_id": "unit4_judge_054"
+      },
+      {
+        "id": "unit4_judge_055",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "正方形的周长与它边长的比值是  $4$ 。（）",
+        "options": [],
+        "answer": "✓",
+        "solution": "【分析】根据正方形的周长  $= \\text{边长} \\times 4$  的逆运算可知，正方形的周长  $\\div$  边长  $= \\frac{1}{2}$  正方形的周长：边长  $= 4$  ，据此解答即可。\n\n【详解】据分析可知，正方形的周长与它边长的比值是  $4$ 。故答案为:  ✓",
+        "knowledge_points": [
+          "正方形的周长",
+          "求比值"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_055",
+        "original_id": "unit4_judge_055"
+      },
+      {
+        "id": "unit4_judge_056",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "$6 \\mathrm{~m}: 7 \\mathrm{~m}$  的比值是  $\\frac{6}{7} \\mathrm{~m}$  。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】根据比的意义可知, 求比值就是用比的前项除以比的后项。\n\n【详解】  $6 \\mathrm{~m}: 7 \\mathrm{~m}$\n\n$= 6 \\div 7 \\\\ = \\frac {6}{7} $\n\n$6 \\mathrm{~m}: 7 \\mathrm{~m}$  的比值是  $\\frac{6}{7}$ , 原题说法错误。故答案为:  ×",
+        "knowledge_points": [
+          "求比值"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_056",
+        "original_id": "unit4_judge_056"
+      },
+      {
+        "id": "unit4_judge_057",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "把  $4: 0.5$  化成最简单的整数比是  $8 。( )$",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】比的基本性质：比的前项和后项同时乘或除以相同的数（0 除外），比值不变。利用'比的基本性质'把比化简成最简单的整数比。\n\n【详解】4:0.5\n\n$= (4 \\div 0. 5): (0. 5 \\div 0. 5) \\\\ = 8: 1 $\n\n把  $4: 0.5$  化成最简单的整数比是  $8: 1$  。原题说法错误。故答案为:  ×",
+        "knowledge_points": [
+          "比的基本性质",
+          "比的化简"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_057",
+        "original_id": "unit4_judge_057"
+      },
+      {
+        "id": "unit4_judge_058",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "牛与羊的头数比是  4:5, 牛的头数比羊少  $\\frac{1}{5}$  。 ( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】牛与羊的头数比是  $4: 5$ , 就是牛是 4 份, 羊是这样的 5 份, 求一个数比另外一个的数多或者少几分之几, 用 (大数一小数)  $\\div$  单位 '1'。\n\n【详解】  $(5 - 4)\\div 5$\n\n$= 1 \\div 5 \\\\ = \\frac {1}{5} $\n\n则牛的头数比羊少  $\\frac{1}{5}$  。故答案为:  √",
+        "knowledge_points": [
+          "用画图法和转化法解决分数问题（比的应用）",
+          "求一个数占另一个数几分之几",
+          "比的意义"
+        ],
+        "source": "24-25 六年级上·全国·单元测试",
+        "merged_id": "unit4_058",
+        "original_id": "unit4_judge_058"
+      },
+      {
+        "id": "unit4_judge_059",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "学校科技小组中男生人数与女生人数的比是  $3: 5$  。如果科技小组有 40 人, 那么女生就有 25 人。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】已知科技小组中男生人数与女生人数的比是  $3: 5$ ,即女生人数占科技小组总人数的  $\\frac{5}{3 + 5}$ , 根据求一个数的几分之几是多少, 用总人数乘  $\\frac{5}{3 + 5}$ , 即可求出女生人数, 据此判断。\n\n【详解】  $40 \\times \\frac{5}{3 + 5}$\n\n$= 4 0 \\times \\frac {5}{8} \\\\ = 2 5 \\text {(人)} $\n\n如果科技小组有 40 人, 那么女生就有 25 人。原题说法正确。故答案为: √",
+        "knowledge_points": [
+          "整数乘分数",
+          "求一个数的几分之几的问题",
+          "比的应用"
+        ],
+        "source": "24-25 六年级上·全国·单元测试",
+        "merged_id": "unit4_059",
+        "original_id": "unit4_judge_059"
+      },
+      {
+        "id": "unit4_judge_060",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "一杯盐水, 盐占盐水的  $\\frac{1}{10}$ , 盐和水的比是  $1: 9$ 。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】盐占盐水的  $\\frac{1}{10}$ , 可知盐水相当于 10 份, 盐是 1 份,水就相当于 (10 - 1) 份, 由此解答即可。\n\n【详解】1：(10-1)\n\n$= 1: 9$\n\n一杯盐水, 盐占盐水的  $\\frac{1}{10}$ , 盐和水的比是  $1: 9$  。原题干说法正确。故答案为: √",
+        "knowledge_points": [
+          "比与分数、除法的关系",
+          "比的意义"
+        ],
+        "source": "2014 六年级·全国·课后作业",
+        "merged_id": "unit4_060",
+        "original_id": "unit4_judge_060"
+      },
+      {
+        "id": "unit4_judge_061",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "$50 \\mathrm{~m}: 50 \\mathrm{dm}$  的比值是  $1 。$  ( )",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】先根据进率 ' $1 \\mathrm{~m} = 10 \\mathrm{dm}$ '统一单位, 再根据求比值的方法, 用比的前项除以比的后项, 据此判断。\n\n【详解】  $50\\mathrm{m}:50\\mathrm{dm}$ $= (50 \\times 10)\\mathrm{dm}:50\\mathrm{dm}$ $= 500:50$ $= 500 \\div 50$ $= 10$  50m: 50dm 的比值是 10。 原题说法错误。 故答案为：×",
+        "knowledge_points": [
+          "毫米、厘米、分米、米之间的进率与换算",
+          "求比值"
+        ],
+        "source": "24-25 六年级上·全国·单元测试",
+        "merged_id": "unit4_061",
+        "original_id": "unit4_judge_061"
+      },
+      {
+        "id": "unit4_calc_062",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "求下面各比的比值。\n\n$\\frac {8}{1 5}: \\frac {4}{5} \\quad 0. 2 5: \\frac {5}{8} \\quad 1 5: \\frac {3}{5} \\quad 4. 5 \\mathrm {d m}: 1 5 \\mathrm {c m}$",
+        "options": [],
+        "answer": "$\\frac{2}{3}; \\frac{2}{5}; 25; 3$",
+        "solution": "【分析】 $\\frac{8}{15}: \\frac{4}{5}$ , 根据比与除法的关系, 用比的前项除以比的后项, 即用  $\\frac{8}{15}$  除以  $\\frac{4}{5}$  。0.25: $\\frac{5}{8}$ , 用比的前项除以比的后项, 即用0.25除以  $\\frac{5}{8}$  。  15:  $\\frac{3}{5}$ , 用比的前项除以比的后项, 即用 15 除以  $\\frac{3}{5}$  。  4.5dm:15cm，因为  $1 \\mathrm{dm} = 10 \\mathrm{~cm}$ ，所以  $4.5 \\mathrm{dm}$  为  $4.5 \\times 10 =$  45cm, 原比变为  $45 \\mathrm{~cm}: 15 \\mathrm{~cm}$  。然后用比的前项除以比的后项, 即用 45 除以 15 。\n\n【详解】  $\\frac{8}{15}:\\frac{4}{5}$\n\n$= \\frac {8}{1 5} \\div \\frac {4}{5} \\\\ = \\frac {8}{1 5} \\times \\frac {5}{4} \\\\ = \\frac {2}{3} \\\\ 0. 2 5: \\frac {5}{8} \\\\ = 0. 2 5 \\div \\frac {5}{8} \\\\ = 0. 2 5 \\times \\frac {8}{5} \\\\ = \\frac {2}{5} $\n\n$1 5: \\frac {3}{5} \\\\ = 1 5 \\div \\frac {3}{5} \\\\ = 1 5 \\times \\frac {5}{3} \\\\ = 2 5 $\n\n$4.5 \\mathrm {d m}: 1 5 \\mathrm {c m} \\\\ 1 \\mathrm {d m} = 1 0 \\mathrm {c m} \\\\ 4. 5 \\times 1 0 = 4 5 (\\mathrm {c m}) \\\\ 4 5 \\mathrm {c m}: 1 5 \\mathrm {c m} \\\\ = 4 5 \\div 1 5 \\\\ = 3 $",
+        "knowledge_points": [
+          "比与分数、除法的关系",
+          "毫米、厘米、分米、米之间的进率与换算",
+          "求比值"
+        ],
+        "source": "25-26 六年级上·海南海口·单元测试",
+        "merged_id": "unit4_062",
+        "original_id": "unit4_calc_062"
+      },
+      {
+        "id": "unit4_calc_063",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "把下面的比化成最简单的整数比并求比值。\n\n$\\frac {3}{5}: \\frac {9}{2 0}$   45%:1.5       0.8千克：600克       $\\frac{5}{12}$ 时：75分",
+        "options": [],
+        "answer": "4:3;$\\frac{4}{3}$;3:10;$\\frac{3}{10}$;4: 3; $\\frac{4}{3}$;1:3,$\\frac{1}{3}$",
+        "solution": "【分析】根据比的基本性质：比的前项和后项同时乘或除以一个相同的数，比值不变，将比化简成最简整数比，再用比的前项除以后项得出比值。先将比的前项和后项同时乘两个分数的分母的最小公倍数为  $12: 9$ , 再同时除以两个数的最大公因数;先将百分数转化为小数, 即  $75\\% = 0.75$  。再将比的前项和后项同时乘 100 转化为两个整数, 最后再同时除以两个数的最大公因数;先根据高级单位转化为低级单位用乘法, 1 千克 = 1000 克,将 0.8 千克乘 1000 转化为以克为单位, 再同时除以两个数的最大公因数;先根据高级单位转化为低级单位用乘法, 1 小时  $= 60$  分, 将  $\\frac{5}{12}$  时乘 60 转化为以分为单位, 再同时除以两个数的最大公因数。\n\n【详解】  $\\frac{3}{5}:\\frac{9}{20}$  12:9\n\n$ (\\frac {3}{5} \\times 2 0): (\\frac {9}{2 0} \\times 2 0) \\\\ \\dot {} (1 2 \\div 3): (9 \\div 3) \\\\ = 4: 3 \\\\ = 4 \\div 3 \\\\ = \\frac {4}{3} \\\\ 45 \\% : 1.5 \\\\ = 0. 4 5: 1. 5 \\\\ = 4 5: 1 5 0 \\\\ = (4 5 \\div 1 5): (1 5 0 \\div 1 5) \\\\ = 3: 1 0 \\\\ = 3 \\div 1 0 \\\\ = \\frac {3}{1 0} $\n\n0.8千克：600克\n\n$= 8 0 0 \\text {克}: 6 0 0 \\text {克} \\\\ = (8 0 0 \\div 2 0 0): (6 0 0 \\div 2 0 0) = 4: 3 \\\\ = 4 \\div 3 \\\\ = \\frac {4}{3} $\n\n$\\frac{5}{12}$ 时：75分\n\n=25分：75分\n\n$= (2 5 \\div 2 5): (7 5 \\div 2 5) \\\\ = 1: 3 \\\\ = 1 \\div 3 \\\\ = \\frac {1}{3} $",
+        "knowledge_points": [
+          "比的基本性质",
+          "比的化简",
+          "求比值"
+        ],
+        "source": "24-25 六年级上·河北张家口·期末",
+        "merged_id": "unit4_063",
+        "original_id": "unit4_calc_063"
+      },
+      {
+        "id": "unit4_calc_064",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "把下面比化成最简单的整数比。\n\n$0.7 5: \\frac {9}{8}$   $\\frac {3}{1 0}: 0. 6$    38千米：200米",
+        "options": [],
+        "answer": "2: 3;1: 2; 190: 1",
+        "solution": "【分析】化简比的方法：根据比的基本性质，比的前项和后项同时乘一个数或除以一个数（0 除外）比值不变。有单位的要先统一单位。据此解答。\n\n【详解】  $0.75:\\frac{9}{8}$\n\n$= (0. 7 5 \\times 8 0 0): (\\frac {9}{8} \\times 8 0 0)\\\\=600:900\\\\=(6 0 0 \\div 3 0 0): (9 0 0 \\div 3 0 0)\\\\=2:3$\n\n$\\frac {3}{1 0}: 0. 6$\n\n$ =(\\frac {3}{1 0} \\times 1 0): (0. 6 \\times 1 0)\\\\=3:6\\\\= (3 \\div 3): (6 \\div 3)\\\\=1:2$\n\n38千米：200米\n\n38000:200\n\n$= (3 8 0 0 0 \\div 2 0 0): (2 0 0 \\div 2 0 0)\\\\=190:1$",
+        "knowledge_points": [
+          "比的基本性质",
+          "比的化简"
+        ],
+        "source": "24-25六年级上·湖南娄底·期中",
+        "merged_id": "unit4_064",
+        "original_id": "unit4_calc_064"
+      },
+      {
+        "id": "unit4_calc_065",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "求下面各比的比值。\n\n(1)  $\\frac{8}{15}: \\frac{4}{5}$  \n(2)  $0.25: \\frac{5}{8}$  \n(3)  $15:\\frac{3}{5}$  \n(4)  $4.5 \\mathrm{dm}: 15 \\mathrm{~cm}$",
+        "options": [],
+        "answer": "(1) $\\frac{2}{3}$;(2)0.4;(3) 25;(4) 3",
+        "solution": "【分析】根据比的意义可知, 求比值就是用比的前项除以比的后项, 单位不统一时, 要统一单位后, 再进行计算。\n\n【详解】 (1)  $\\frac{8}{15}: \\frac{4}{5}$\n\n$= \\frac {8}{1 5} \\div \\frac {4}{5} \\\\ = \\frac {8}{1 5} \\times \\frac {5}{4} \\\\ = \\frac {2}{3} $\n\n(2)  $0.25: \\frac{5}{8}$\n\n$= 0. 2 5 \\div \\frac {5}{8} \\\\ = 0. 2 5 \\times \\frac {8}{5} \\\\ = 0. 4 $\n\n(3)  $15:\\frac{3}{5}$\n\n$= 1 5 \\div \\frac {3}{5} \\\\ = 1 5 \\times \\frac {5}{3} \\\\ = 2 5 $\n\n(4)  $4.5 \\mathrm{dm}: 15 \\mathrm{~cm}$\n\n$= 4 5 \\mathrm {c m}: 1 5 \\mathrm {c m} \\\\ = 4 5 \\div 1 5 \\\\ = 3 $",
+        "knowledge_points": [
+          "求比值"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_065",
+        "original_id": "unit4_calc_065"
+      },
+      {
+        "id": "unit4_calc_066",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "把下面各比化成最简单的整数比。\n\n①  $12: 36$  ②  $3.5: 0.7$  ③  $\\frac{1}{4}: 1.5$  ④  $15 \\text{分}: \\frac{1}{5}$  时",
+        "options": [],
+        "answer": "① $1: 3$ ; ②  $5: 1$ ; ③  $1: 6$ ; ④  $5: 4$",
+        "solution": "【分析】化简比根据比的基本性质, 比的前项和后项同时乘或除以相同的数 (0 除外), 比值不变。①比的前项和后项同时除以 12;②比的前项和后项同时除以 0.7;③比的前项和后项同时乘 4, 再除以公因数;④统一单位再化简。\n\n【详解】①  $12:36=(12\\div 12):(36\\div 12)=1:3$\n\n②  $3.5:0.7=(3.5\\div 0.7):(0.7\\div 0.7)=5:1$\n\n③  $\\frac{1}{4}:1.5=(\\frac{1}{4}\\times 4):(1.5\\times 4)=1:6$\n\n④  $15\\text{分}:\\frac{1}{5}\\text{时}=15\\text{分}:12\\text{分}=(15\\div 3):(12\\div 3)=5:4$",
+        "knowledge_points": [
+          "比的化简",
+          "时、分的认识及换算"
+        ],
+        "source": "23-24六年级上·全国·单元测试",
+        "merged_id": "unit4_066",
+        "original_id": "unit4_calc_066"
+      }, {
+        "id": "unit4_calc_067",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "直接写出得数。\n\n$\\frac {3}{4} \\times 8 =  \\quad \\frac {3}{5} \\times \\frac {5}{6} = \\quad 0 \\times \\frac {3}{8} + \\frac {5}{8}, \\quad 5 - \\frac {3}{7} = $\n\n$2.1 \\div \\frac{3}{8} = \\quad  \\frac{1}{6} + \\frac{5}{6} \\times \\frac{1}{5} = \\quad  \\frac{1}{2} \\times \\frac{1}{3} \\div \\frac{1}{3} = \\quad 3.2: 0.08 = $ （求比值）",
+        "options": [],
+        "answer": "6; $\\frac{1}{2}$; $\\frac{5}{8}$; $4\\frac{4}{7};5.6;$\\frac{1}{3}$;$\\frac{1}{2}$;40",
+        "solution": "【解析】略",
+        "knowledge_points": [
+          "异分母分数加、减法",
+          "求比值",
+          "分数的四则混合运算",
+          "分数乘分数"
+        ],
+        "source": "22-23 六年级上·福建莆田·期中",
+        "merged_id": "unit4_067",
+        "original_id": "unit4_calc_067"
+      },
+      {
+        "id": "unit4_calc_068",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "计算题",
+        "difficulty": "容易",
+        "question": "求下列各比的比值。\n\n$\\frac {1}{5}: 0. 6\\quad  5: 1. 5 \\quad  \\frac {2}{3}: \\frac {4}{5} $\n\n$\\frac {1}{2}: \\frac {5}{8} \\quad  0. 1 6: 2. 4 \\quad 0. 6: 3 $",
+        "options": [],
+        "answer": "$\\frac{1}{3}$;$\\frac{10}{3}$;$\\frac{5}{6}$;$\\frac {4}{5}$; $\\frac {1}{1 5}$; 0.2 ",
+        "solution": "【分析】求比值直接用比的前项÷后项即可，求比值的结果是一个数，据此求出各比的比值。\n\n【详解】  $\\frac{1}{5}:0.6 = \\frac{1}{5}\\div \\frac{3}{5} = \\frac{1}{5}\\times \\frac{5}{3} = \\frac{1}{3}$\n\n$5: 1. 5 = 5 \\div 1. 5 = \\frac {5 0}{1 5} = \\frac {1 0}{3} \\\\ \\frac {2}{3}: \\frac {4}{5} = \\frac {2}{3} \\div \\frac {4}{5} = \\frac {2}{3} \\times \\frac {5}{4} = \\frac {5}{6} \\\\ \\frac {1}{2}: \\frac {5}{8} = \\frac {1}{2} \\div \\frac {5}{8} = \\frac {1}{2} \\times \\frac {8}{5} = \\frac {4}{5} \\\\ 0. 1 6: 2. 4 = 0. 1 6 \\div 2. 4 = \\frac {1 6}{2 4 0} = \\frac {1}{1 5} \\\\ 0. 6: 3 = 0. 6 \\div 3 = 0. 2 $",
+        "knowledge_points": [
+          "求比值"
+        ],
+        "source": "24-25 六年级上·全国·单元测试",
+        "merged_id": "unit4_068",
+        "original_id": "unit4_calc_068"
+      },
+      {
+        "id": "unit4_calc_069",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "化简下列各比。\n\n$ 1 6: 3 0 \\quad 3 5: 1 0   \\quad\\frac {1 1 4}{1 9} $",
+        "options": [],
+        "answer": "8:15;7:2;6:1",
+        "solution": "【分析】根据比的基本性质：比的前项和后项同时乘或以一个不为 0 的数，比值不变，据此解答；根据分数与比的关系：分子做比的前项，分母做比的后项，把分数化成比的形式，再进行化简。\n\n【详解】 16:30\n\n$= (1 6 \\div 2): (3 0 \\div 2) \\\\ = 8: 1 5 $\n\n$3 5: 1 0 \\\\ = (3 5 \\div 5): (1 0 \\div 5) \\\\ = 7: 2 $\n\n$\\frac {1 1 4}{1 9} \\\\ = 1 1 4: 1 9 \\\\ = (1 1 4 \\div 1 9): (1 9 \\div 1 9) \\\\ = 6: 1 $",
+        "knowledge_points": [
+          "比的基本性质",
+          "比的化简"
+        ],
+        "source": "24-25 六年级上·全国·单元测试",
+        "merged_id": "unit4_069",
+        "original_id": "unit4_calc_069"
+      },
+      {
+        "id": "unit4_solve_070",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "一种什锦糖是由水果糖、奶糖和酥糖三种糖配制而成的。其中水果糖、奶糖和酥糖三种糖的比是  $3: 4: 6$ ，如果这三种糖各有 20 千克，当奶糖全部用完时，水果糖还剩下多少千克？酥糖已经增加了多少千克？",
+        "options": [],
+        "answer": "水果糖还剩下5千克，酥糖已经增加了10千克。",
+        "solution": "【分析】根据比的分配，当奶糖用完时，计算出每份的重量，进而求出水果糖和酥糖的使用量，再与原有数量比较得出剩余和增加量。\n\n【详解】  $20 \\div 4 = 5$  （千克）\n\n$2 0 - 5 \\times 3\\\\= 2 0 - 1 5\\\\=5 (千克)$\n\n$5 \\times 6 - 2 0\\\\= 3 0 - 2 0\\\\=10 (千克)$\n\n答: 水果糖还剩下 5 千克; 酥糖已经增加了 10 千克。",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "25-26 六年级上·海南海口·单元测试",
+        "merged_id": "unit4_070",
+        "original_id": "unit4_solve_070"
+      },
+      {
+        "id": "unit4_solve_071",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "一个长方形，宽和长的比是4:5，如果宽再增加6厘米，这时长方形就变成了正方形，求这个长方形的面积是多少平方厘米？",
+        "options": [],
+        "answer": "720平方厘米",
+        "solution": "【分析】已知宽和长的比是  $4: 5$ , 说明长比宽多 1 份。宽增加 6 厘米后变成正方形, 这意味着长比宽多 6 厘米, 所以 1 份对应的长度就是 6 厘米。由此可先求出长和宽, 再根据长方形面积公式计算面积, 据此解答。\n\n【详解】  $5 - 4 = 1$  （份）1 份的长度是 6 厘米 宽为  $4 \\times 6 = 24$  （厘米）长为  $5 \\times 6 = 30$  (厘米)\n\n长方形的面积： $30 \\times 24 = 720$  （平方厘米）答: 这个长方形的面积是 720 平方厘米。",
+        "knowledge_points": [
+          "长方形的面积",
+          "按比分配问题"
+        ],
+        "source": "25-26六年级上·海南海口·单元测试",
+        "merged_id": "unit4_071",
+        "original_id": "unit4_solve_071"
+      },
+      {
+        "id": "unit4_solve_072",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "爷爷家养的鸡比鸭多130只，已知鸡和鸭只数的比是5:3，爷爷家养鸡多少只？",
+        "options": [],
+        "answer": "325只",
+        "solution": "【分析】已知鸡和鸭的只数比是  $5: 3$ , 把鸡的只数看作 5 份;鸭的只数看作 3 份。则鸡比鸭多的份数为:  $5 - 3 = 2$  份。'鸡比鸭多 130 只'，而这 130 只正好对应鸡比鸭多的 2 份。那么每份是  $130 \\div 2 = 65$  （只）。鸡的只数是 5 份，每份是 65 只，根据：鸡的总只数 = 每份只数 × 鸡的份数，把数据代入计算即可。\n\n【详解】鸡和鸭的只数比是  $5: 3$ , 把鸡的只数看作 5 份; 鸭的只数看作 3 份。$5 - 3 = 2$  （份）$130 \\div 2 = 65$  (只)$65 \\times 5 = 325$  (只)答: 爷爷家养鸡 325 只。",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "25-26六年级上·海南海口·单元测试",
+        "merged_id": "unit4_072",
+        "original_id": "unit4_solve_072"
+      },
+      {
+        "id": "unit4_solve_073",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "某校六 (1) 班和六 (2) 班共有 91 人, 六 (1) 班和六 (2) 班人数的比是4:3，六（1）班和六（2）班各有多少人？",
+        "options": [],
+        "answer": "六（1）班有52人；六（2）班有39人",
+        "solution": "【分析】已知两个班的总人数以及人数比，可先求出总份数，再用总人数分别  $\\times$  每个班人数占总份数的分率，从而得到每个班的人数，据此解答。\n\n【详解】总份数： $4 + 3 = 7$六 (1) 班人数:  $91 \\times \\frac{4}{7} = 52$  （人）  六 (2) 班人数:  $91 \\times \\frac{3}{7} = 39$  （人）答: 六 (1) 班有 52 人, 六 (2) 班有 39 人。",
+        "knowledge_points": [
+          "比的应用"
+        ],
+        "source": "25-26 六年级上·海南海口·单元测试",
+        "merged_id": "unit4_073",
+        "original_id": "unit4_solve_073"
+      },
+      {
+        "id": "unit4_solve_074",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "一件上衣的售价是90元，裤子与上衣单价的比是2:3，每条裤子多少元？",
+        "options": [],
+        "answer": "60元",
+        "solution": "【分析】已知裤子与上衣单价的比是  $2: 3$ , 说明上衣的价格对应 3 份, 裤子的价格对应 2 份。上衣售价为 90 元, 用 90 元除以 3 可求出每份的价格, 再乘 2 即可得到裤子的价格。\n\n【详解】  $90 \\div 3 \\times 2= 3 0 \\times 2 = 6 0 (\\text {元}) $答: 每条裤子 60 元。",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "25-26 六年级上·海南海口·单元测试",
+        "merged_id": "unit4_074",
+        "original_id": "unit4_solve_074"
+      },
+      {
+        "id": "unit4_solve_075",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "A、B、C三个箱子里各有一些小球，已知A箱子里的小球数目的1.5倍恰好等于B箱子小球数目的一半；B箱子里小球数目的 $\\frac{5}{4}$ 倍恰好等于C箱子小球数目的 $\\frac{2}{3}$ ，则A、C两个箱子的小球数目之比为多少？",
+        "options": [],
+        "answer": "8:45",
+        "solution": "【分析】设 A 箱子里小球的数目是 1 , A 箱子里的小球数目的 1.5 倍恰好等于 B 箱子小球数目的一半, 即  $1.5 \\times \\mathrm{A}$  箱子里小球的数目 = B 箱子小球的数目 ÷ 2 ;  $1.5 \\times \\mathrm{A}$  箱里小球的数目  $\\times 2 = \\mathrm{B}$  箱里小球的数目; 即 B 箱里小球数目是  $1.5 \\times 1 \\times 2$ ; B 箱子里小球数目的  $\\frac{5}{4}$  倍恰好等于 C 箱子小球数目的  $\\frac{2}{3}$ , 即 B箱子里小球数目  $\\times \\frac{5}{4} = C$  箱子里小球数目的  $\\frac{2}{3}$ , C 箱子里小球数目是 B 箱子里小球数目  $\\times \\frac{5}{4} \\div \\frac{2}{3}$ , 即  $(1.5 \\times 1 \\times 2) \\times \\frac{5}{4} \\div \\frac{2}{3}$ , 据此求出 C 箱子里小球数目;再根据比的意义得出 A、C 两个箱子的小球数目之比，化简比即可。\n\n【详解】设A箱子小球的数目是1。B 箱子小球的数目是:  $1.5 \\times 1 \\times 2 = 3$C 箱子里小球数目是:\n\n$3 \\times \\frac {5}{4} \\div \\frac {2}{3} \\\\ = \\frac {1 5}{4} \\div \\frac {2}{3} \\\\ = \\frac {1 5}{4} \\times \\frac {3}{2} \\\\ = \\frac {4 5}{8} $\n\nA 箱子里小球数目：C 箱子里小球数目  $= 1: \\frac{45}{8}$\n\n$= (1 \\times 8): (\\frac {4 5}{8} \\times 8) \\\\ = 8: 4 5 $\n\n答: A、C 两个箱子的小球数目之比为  $8: 45$  。",
+        "knowledge_points": [
+          "求一个数的几分之几的问题",
+          "比的化简",
+          "比的意义",
+          "分数的乘、除法的混合运算"
+        ],
+        "source": "24-25六年级上·全国·单元测试",
+        "merged_id": "unit4_075",
+        "original_id": "unit4_solve_075"
+      },
+      {
+        "id": "unit4_solve_076",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "甲、乙、丙三个村合修一条水渠，修完后，甲、乙、丙村可灌溉的面积比是8:7:5，原来三个村计划按可灌溉的面积比派出劳力。后来因为丙村抽不出劳力，经协商，丙村应抽出的劳力由甲、乙两村分担，丙村付给甲、乙两村工钱共1350元。结果，甲村共派出60人，乙村共派出40人，问甲、乙两村各应分得工钱多少元？",
+        "options": [],
+        "answer": "甲村 1080 元,乙村 270 元",
+        "solution": "【分析】由甲、乙、丙村可灌溉的面积比为  $8: 7: 5$  可得, 甲村应派出的人数占总人数的  $\\frac{8}{8 + 7 + 5}$ , 乙村派出的人数占总人数的  $\\frac{7}{8 + 7 + 5}$ , 因此, 将甲乙两村实际派出的人数相加计算出总人数, 再用总人数分别乘  $\\frac{8}{8 + 7 + 5}$  和  $\\frac{7}{8 + 7 + 5}$  求出甲、乙两村应派出的人数; 用实际派出的人数减去鹰牌出的人数, 分别求出两村分别多派出的人数, 进而求出多派出的人数比, 最后用丙村付的工钱分别乘多派出人数占两村总多派出人数的分率即可求出两村各应分得的工钱。此题关键理解总人数不变，用原来的比求出甲、乙两村原来的人数，得到丙原来应派出的人数之后再按比例分配。\n\n【详解】  $60 + 40 = 100$  （人）\n\n$1 0 0 \\times \\frac {8}{8 + 7 + 5} = 4 0 (\\text {人}) \\\\ 1 0 0 \\times \\frac {7}{8 + 7 + 5} = 3 5 (\\text {人}) \\\\ 6 0 - 4 0 = 2 0 \\text {(人)} \\\\ 4 0 - 3 5 = 5 \\text {(人)} \\\\ 2 0: 5 = 4: 1 \\\\ 1 3 5 0 \\times \\frac {4}{4 + 1} = 1 0 8 0 (\\text {元}) \\\\ 1 3 5 0 \\times \\frac {1}{4 + 1} = 2 7 0 (\\text {元}) $\n\n答: 甲村应分得 1080 元, 乙村应分得 270 元。",
+        "knowledge_points": [
+          "按比分配问题",
+          "比的应用"
+        ],
+        "source": "24-25六年级上·全国·单元测试",
+        "merged_id": "unit4_076",
+        "original_id": "unit4_solve_076"
+      },
+      {
+        "id": "unit4_solve_077",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "甲、乙、丙三人加工方形和圆形的两种零件，已知甲每加工3个零件中有2个是圆形的；乙每加工4个零件中有3个是圆形的；丙每加工5个零件中有4个是圆形的。这天三人共加工了116个圆形零件，而加工的方形零件个数的比为  $4:3:3$  ，那么这天三人共加工零件多少个？",
+        "options": [],
+        "answer": "156个",
+        "solution": "【分析】根据题意, 甲、乙、丙三人加工的方形零件个数的比为  $4: 3: 3$ , 把比看作份数;因为甲每加工 3 个零件中有 2 个是圆形的，即有 1 个是方形的，所以甲加工圆形零件的个数是方形的 2 倍；那么甲加工方形零件的个数是 4 份，则甲加工圆形零件的个数就是  $4 \\times 2 = 8$  份；因为乙每加工 4 个零件中有 3 个是圆形的，即有 1 个是方形的，所以乙加工圆形零件的个数是方形的 3 倍；那么乙加工方形零件的个数是 3 份，则乙加工圆形零件的个数就是  $3 \\times 3 = 9$  份；因为丙每加工 5 个零件中有 4 个是圆形的，即有 1 个是方形的，所以丙加工圆形零件的个数是方形的 4 倍；那么丙加工方形零件的个数是 3 份，则丙加工圆形零件的个数就是  $3 \\times 4 = 12$  份；甲、乙、丙三人加工的圆形零件个数的比是  $8: 9: 12$ , 即共加工了  $8 + 9 + 12 = 29$  份; 用三人加工圆形零件的总数除以圆形零件的总份数, 求出一份数; 再用一份数乘加工方形零件的总份数即可求出加工方形零件的总数, 然后把方形和圆形零件个数相加求出加工零件的总数。通过每人加工圆形零件个数是方形零件的几倍，得出三人加工圆形零件的个数比，再把比看作份数，求出一份数是解题的关键。\n\n【详解】甲、乙、丙三人加工的圆形零件个数的比是:\n\n$(4 \\times 2): (3 \\times 3): (3 \\times 4) = 8: 9: 1 2$\n\n一份数:$1 1 6 \\div (8 + 9 + 1 2) \\\\ = 1 1 6 \\div 2 9 \\\\ = 4 \\text {(个)} $\n\n方形零件共加工:$4 \\times (4 + 3 + 3) \\\\ = 4 \\times 1 0 \\\\ = 4 0 (\\text {个}) $\n\n一共:  $40 + 116 = 156$  (个)答: 这天三人共加工零件 156 个。",
+        "knowledge_points": [
+          "比的应用"
+        ],
+        "source": "24-25 六年级上·全国·单元测试",
+        "merged_id": "unit4_077",
+        "original_id": "unit4_solve_077"
+      },
+      {
+        "id": "unit4_solve_078",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "甲、乙、丙三个数的和是 1690 , 已知甲数的  $\\frac{1}{3}$  等于乙数的  $\\frac{2}{5}$ , 乙数的  $\\frac{2}{3}$  等于丙数的  $\\frac{3}{7}$  甲、乙、丙三个数分别是多少？",
+        "options": [],
+        "answer": "540;450;700",
+        "solution": "【分析】已知甲数的  $\\frac{1}{3}$  等于乙数的  $\\frac{2}{5}$ , 根据比的基本性质:两内项之积等于两外项之积, 可得甲数: 乙数, 又已知乙数的  $\\frac{2}{3}$  等于丙数的  $\\frac{3}{7}$ , 同理可求出乙数: 丙数, 为了统一乙数在两个比中的份数, 需要求出 5 和 9 的最小公倍数, 即可求出甲、乙、丙三个数之比, 因为甲、乙、丙三个数的和是 1690 , 再按比分配, 即可求出甲、乙、丙三个数分别是多少。\n\n【详解】甲数：乙数  $= \\frac{2}{5}: \\frac{1}{3} = (\\frac{2}{5} \\times 15): (\\frac{1}{3} \\times 15) = 6: 5$  乙数：丙数  $= \\frac{3}{7}: \\frac{2}{3} = (\\frac{3}{7} \\times 21): (\\frac{2}{3} \\times 21) = 9: 14$  因为5和9互质，5和9的最小公倍数： $5 \\times 9 = 45$$6 \\times 9 = 5 4, \\quad 5 \\times 9 = 4 5$甲数：乙数  $= 6: 5 = 54: 45$$5 \\times 9 = 4 5, \\quad 1 4 \\times 5 = 7 0$\n\n乙数：丙数  $= 9: 14 = 45: 70$所以甲数：乙数：丙数  $= 54: 45: 70$总份数:  $54 + 45 + 70 = 169$\n\n甲数占 54 份, 甲数:  $1690 \\times \\frac{54}{169} = 540$乙数占45份，乙数： $1690 \\times \\frac{45}{169} = 450$丙数占 70 份, 丙数:  $1690 \\times \\frac{70}{169} = 700$答: 甲、乙、丙三个数分别是 540、450、700。",
+        "knowledge_points": [
+          "求一个数的几分之几的问题",
+          "分数与整数的除法",
+          "按比分配问题",
+          "分数的四则混合运算"
+        ],
+        "source": "24-25 六年级上·全国·单元测试",
+        "merged_id": "unit4_078",
+        "original_id": "unit4_solve_078"
+      },
+      {
+        "id": "unit4_solve_079",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "甲数是乙数的  $\\frac{3}{10}$ ,乙数与丙数的比是  $8: 9$ , 甲、乙、丙三个数的比是多少?",
+        "options": [],
+        "answer": "12:40:45",
+        "solution": "【分析】已知甲数是乙数的  $\\frac{3}{10}$ , 根据分数与比的关系: 分子\n\n相当于比的前项，分母相当于比的后项，分数线相当于比号，即可求出甲数：乙数；已知乙数与丙数的比是  $8: 9$  ，为了将两个比中乙数的份数统一，需求出 10 和 8 的最小公倍数，再根据比的基本性质：比的前项和后项同时乘一个数 (0 除外），比值不变，即可求解。\n\n【详解】甲数：乙数  $= 3:101 \\quad 0 = 2 \\times 5, 8 = 2 \\times 2 \\times 2$10 和 8 的最小公倍数:  $2 \\times 2 \\times 2 \\times 5 = 40$\n\n$1 0 \\times 4 = 4 0, \\quad 3 \\times 4 = 1 2$甲数：乙数  $= 3: 10 = 12: 40$$8 \\times 5 = 4 0, \\quad 9 \\times 5 = 4 5$乙数：丙数  $= 8: 9 = 40: 45$所以甲：乙：丙=12:40:45答: 甲 : 乙 : 丙  $= 12:40:45$  。",
+        "knowledge_points": [
+          "比的基本性质",
+          "比的化简",
+          "比的应用"
+        ],
+        "source": "24-25 六年级上·全国·单元测试",
+        "merged_id": "unit4_079",
+        "original_id": "unit4_solve_079"
+      },
+      {
+        "id": "unit4_solve_080",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "某商店在奥林匹克官方旗舰店开启预定的第一天，成功向旗舰店预定了'冰墩墩'系列和'雪容融'系列的数量之比为7:8。接着在第二天预定的时候，又成功预定了这两种系列一共41件，此时'冰墩墩'系列和'雪容融'系列数量之比变为15:17。第一天和第二天都只预定不售出，第三天的时候该店开始卖出的同时，继续预定。该店在第三天的售出和预定结束后统计得知，比第二天的吉祥物数量净增14件，此时'冰墩墩'系列和'雪容融'系列的数量之比变为8:9。则在第三天销售和预定结束后'冰墩墩'系列的数量为多少件？",
+        "options": [],
+        "answer": "880 件",
+        "solution": "【分析】设第一天'冰墩墩'系列和'雪容融'数量总和为  $x$ , 因为第一天预定 '冰墩墩' 系列和 '雪容融' 系列的\n\n数量之比为  $7: 8$  。把 '冰墩墩' 看作 7 份, '雪容融' 看作 8 份。所以共有  $7 + 8 = 15$  份, 即  $\\mathrm{x}$  为 15 的倍数, 那么  $\\mathrm{x}$  的个位数是 0 或 5 。第二天预定了这两种系列一共 41 件，那么第二天的数量为  $(x + 41)$ ，此时两种系列数量之比为  $15: 17$ ，把'冰墩墩'看作 15 份，'雪容融'看作 17 份。 $15 + 17 = 32$  份，即  $(x + 41)$  是 32 的倍数，因为 32 是偶数，所以它的倍数一定是偶数，即  $(x + 41)$  是偶数，如果  $x$  的个位是 0 时， $(x + 41)$  是奇数，所以  $x$  个位只能是 5。第三天比第二天的吉祥物数量净增 14 件, 即第三天的数量为:  $x + 41 + 14 = x + 55$ , 此时两种系列数量之比为  $8: 9$ , 把 '冰墩墩' 看作 8 份, '雪容融' 看作 9 份。  $8 + 9 = 17$  份,即  $(x + 55)$  是 17 的倍数, 所以  $x + 55 = 17 \\times k$  。当  $k$  为 110 时, 即  $x$  为:  $17 \\times 110 - 55 = 1870 - 55 = 1815$  件。所以能满足以上条件。那么第三天总数量为  $1815 + 41 + 14 = 1870$  件，'冰墩墩'占  $\\frac{8}{8 + 9}$ , 即 '冰墩墩' 的数量为  $1870 \\times \\frac{8}{8 + 9} = 880$  件。设第一天两种吉祥物总数为  $x$ , 根据各天数量比及倍数关系, 确定  $x$  的值, 再依据第三天总数与比例, 求出\n\n【详解】解：令第一天'冰墩墩'系列和'雪容融'数量总和为  $x$ 。$\\mathrm {x = (7 + 8) k}$ 第二天:  $x + 41 = (15 + 17) \\mathrm{k}, x$  个位数只能为 5 ; 第三天:  $x + 55 = (8 + 9) \\mathrm{k}$ , 为 17 的倍数。$\\mathrm{x} = 17 \\mathrm{k} - 55$ , 并且  $\\mathrm{x}$  为个位数为  $5, \\mathrm{x}$  是 15 的倍数。由此推理可得，x 为 1815 件。$1 8 1 5 + 4 1 + 1 4 = 1 8 7 0 \\text {(件)}$ $1 8 7 0 \\times \\frac {8}{8 + 9} = 8 8 0 \\text {(件)}$\n\n答: 第三天销售和预定结束后 '冰墩墩' 系列的数量为 880 件。'冰墩墩'系列数量。",
+        "knowledge_points": [
+          "列方程解含两个未知数的问题",
+          "按比分配问题",
+          "比的应用"
+        ],
+        "source": "24-25 六年级上·全国·单元测试",
+        "merged_id": "unit4_080",
+        "original_id": "unit4_solve_080"
+      },
+      {
+        "id": "unit4_solve_081",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "重阳节历来就有赏菊花的风俗。如图是某公园的赏菊区，池塘的面积是墨菊区面积的  $\\frac{1}{5}$ , 是玉壶春区面积的  $\\frac{2}{7}$ , 则玉壶春区与墨菊区的面积比是多少？",
+        "options": [],
+        "answer": "7:10",
+        "solution": "【分析】将池塘的面积看作是 '1', 用池塘的面积分别除以池塘占墨菊区的分率和占玉壶春区的分率, 求出墨菊区和玉壶春区的面积, 再把这两个区的面积相比即可。\n\n【详解】  $1 \\div \\frac{1}{5} = 5$    $1 \\div \\frac {2}{7} = \\frac {7}{2}$   $\\frac {7}{2}: 5 = 7: 1 0$答: 玉壶春区与墨菊区的面积比是  $7: 10$  。",
+        "knowledge_points": [
+          "分数与分数的除法",
+          "比的意义",
+          "比的应用"
+        ],
+        "source": "24-25 六年级上·全国·单元测试",
+        "merged_id": "unit4_081",
+        "original_id": "unit4_solve_081"
+      },
+      {
+        "id": "unit4_solve_082",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "低碳出行，从我做起。奇奇、奇妙和聪聪三人同时从图书馆匀速步行去学校，当奇奇到学校时，妙妙离学校还有28米，聪聪距离学校还有60米，当妙妙到学校时，聪聪离学校还有34米，图书馆到学校的距离是多少米？",
+        "options": [],
+        "answer": "476米",
+        "solution": "【分析】奇妙步行最后 28 米的同时, 聪聪步行了  $60 - 34 =$  26 (米), 因为速度比等于路程比, 所以聪聪与奇妙的速度比是  $26: 28 = 13: 14$ , 把图书馆到学校的距离看作单位 '1', 求出当奇妙到达学校时, 求出聪聪距离学校的路程占总路程的分率, 再用聪聪离学校的距离除以对应的分率即可求出图书馆到学校的距离。\n\n【详解】  $60 - 34 = 26$  （米）聪聪的速度：奇妙的速度  $= 26: 28 = 13: 14$\n\n$1 3 \\div 1 4 = \\frac {1 3}{1 4} \\\\ 1 - \\frac {1 3}{1 4} = \\frac {1}{1 4} \\\\ 3 4 \\div \\frac {1}{1 4} \\\\ = 3 4 \\times 1 4 \\\\ = 4 7 6 \\text {(米)} $\n\n答: 图书馆到学校的距离是 476 米。",
+        "knowledge_points": [
+          "分数与整数的除法",
+          "比的应用"
+        ],
+        "source": "24-25六年级上·全国·单元测试",
+        "merged_id": "unit4_082",
+        "original_id": "unit4_solve_082"
+      },
+      {
+        "id": "unit4_solve_083",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "小华对本校2024年春季学期参加足球、篮球和乒乓球三项社团活动的学生进行了统计，参加这3项活动的学生共有360人，其中足球社团人数是篮球社团人数的  $\\frac{3}{5}$ ，篮球社团人数与乒乓球社团人数的比是5:7。参加这3项社团活动的学生各有多少人？(先画线段图分析，再列式解答）\n\n    ![](images\\9335841b03572f1a53a71ef705ba4599.png)",
+        "options": [],
+        "answer": "72人;120人;168人",
+        "solution": "【分析】根据题意, 把乒乓球社团人数看作单位 '1', 篮球社团人数是乒乓球社团人数的  $\\frac{5}{7}$ , 足球社团人数是篮球社团人数的  $\\frac{3}{5}$ , 据此画出线段图。由图可知, 足球社团人数、篮球社团人数、乒乓球社团人数的比是  $3: 5: 7$ , 将 360 人平均分成  $(3 + 5 + 7)$  份, 求出每份是多少人, 再根据每个社团所占的份数, 求出每个社团的人数即可。\n\n【详解】![](images\\9335841b03572f1a53a71ef705ba4599.png)$3 6 0 \\div (3 + 5 + 7) \\\\ = 3 6 0 \\div 1 5 \\\\ = 2 4 \\text {(人)} $\n\n$2 4 \\times 3 = 7 2 (\\text {人}) \\\\ 2 4 \\times 5 = 1 2 0 (\\text {人}) \\\\ 2 4 \\times 7 = 1 6 8 (\\text {人}) $\n\n答: 足球社团有 72 人, 篮球社团有 120 人, 乒乓球社团有 168 人。",
+        "knowledge_points": [
+          "比的基本性质",
+          "比的意义",
+          "按比分配问题",
+          "比的应用"
+        ],
+        "source": "2024·云南昭通·小升初真题",
+        "merged_id": "unit4_083",
+        "original_id": "unit4_solve_083"
+      },
+      {
+        "id": "unit4_solve_084",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "从前有个农民，临死前留下遗言，要把 17 头牛分给三个儿子，其中大儿子分得  $\\frac{1}{2}$ ，二儿子分得  $\\frac{1}{3}$ ，小儿子分得  $\\frac{1}{9}$ ，但不能把牛卖掉或杀掉；三个儿子按照老人的要求怎么也不好分；后来一位邻居顺利地把 17 头牛分完了, 你知道这到底是怎么回事吗?",
+        "options": [],
+        "answer": "大儿子分得9头,二儿子分得6头, 小儿子分得 2 头",
+        "solution": "【分析】把牛的总头数看作单位 '1',  $\\frac{1}{2} + \\frac{1}{3} + \\frac{1}{9} = \\frac{17}{18}, \\frac{17}{18}<$  1, 也就是三兄弟并未将全部牛分完, 那么可以求出三个儿子分牛头数的比,  $\\frac{1}{2}: \\frac{1}{3}: \\frac{1}{9} = 9: 6: 2, 9 + 6 + 2 = 17$ , 则三个儿子分别分得总头数的  $\\frac{9}{17}$ 、 $\\frac{6}{17}$ 、 $\\frac{2}{17}$ , 用总头数分别乘这三个分数即可求出各自分得牛的头数。\n\n【详解】  $\\frac{1}{2}: \\frac{1}{3}: \\frac{1}{9}= (\\frac {1}{2} \\times 1 8): (\\frac {1}{3} \\times 1 8): (\\frac {1}{9} \\times 1 8)  = 9: 6: 2 $\n\n总份数:  $9 + 6 + 2 = 17$大儿子:  $17 \\times \\frac{9}{17} = 9$  (头)   二儿子:  $17 \\times \\frac{6}{17} = 6$  (头)  小儿子:  $17 \\times \\frac{2}{17} = 2$  (头)\n\n答: 大儿子分得 9 头, 二儿子分得 6 头, 小儿子分得 2 头。",
+        "knowledge_points": [
+          "求一个数的几分之几的问题",
+          "按比分配问题"
+        ],
+        "source": "2025 六年级下·全国·专题练习",
+        "merged_id": "unit4_084",
+        "original_id": "unit4_solve_084"
+      },
+      {
+        "id": "unit4_solve_085",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "六（1）班男、女生人数的比是  $5: 3$ , 已知男生比女生多 12 人。\n\n(1) 画图表示数量关系。  \n(2) 男、女生各有多少人?",
+        "options": [],
+        "answer": "(1) 见详解; (2) 男生: 30人; 女生: 18人",
+        "solution": "【分析】（1）分析题目，结合比的意义可知：男生的人数是5份，女生人数是3份，女生比男生少（5- 3）份，即12人，据此画出线段图即可；(2) 根据比的意义可知, 12 人是 (5-3) 份, 据此用除法求出一份是多少，再分别乘男生、女生人数所占的份数即可得到男生、女生人数。\n\n【详解】（1）画线段图如下：\n\n![](images\\0045a9ef66cc1e30618cc8b85a2facc8.png)\n\n(2)  $12 \\div (5 - 3)= 1 2 \\div 2  = 6 (\\text {人}) $$6 \\times 5 = 3 0 (\\text {人}) \\quad 6 \\times 3 = 1 8 (\\text {人}) $答: 男生有 30 人, 女生有 18 人。",
+        "knowledge_points": [
+          "比的意义",
+          "比的应用"
+        ],
+        "source": "2024·安徽宿州·小升初真题",
+        "merged_id": "unit4_085",
+        "original_id": "unit4_solve_085"
+      },
+      {
+        "id": "unit4_solve_086",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "用甲、乙两个粮库存放大米，原来甲粮库存放大米的数量是乙粮库的  $\\frac{5}{7}$ 。如果从乙粮库调 30 吨大米到甲粮库, 那么甲粮库存放大米的数量与乙粮库的比是  $4: 5$  。原来甲、乙两个粮库各存放大米多少吨？",
+        "options": [],
+        "answer": "甲450吨;乙630吨",
+        "solution": "【分析】根据题意, 两个粮库的大米总吨数不变。已知原来甲粮库存放大米的数量是乙粮库的  $\\frac{5}{7}$ , 则原来乙粮库大米的吨数占两个粮库大米总吨数的  $\\frac{7}{5 + 7}$ ;如果从乙粮库调 30 吨大米到甲粮库, 那么甲粮库存放大米的数量与乙粮库的比是  $4: 5$ , 则现在乙粮库大米的吨数占两个粮库大米总吨数的  $\\frac{5}{4 + 5}$ ;那么 30 吨大米占两个粮库大米总吨数的  $(\\frac{7}{5 + 7} - \\frac{5}{4 + 5})$ , 单位 '1' 未知, 根据分数除法的意义求出两个粮库大米的总吨数;再根据求一个数的几分之几是多少，用两个粮库大米的总吨数乘  $\\frac{7}{5 + 7}$ ，求出原来乙粮库大米的吨数；再用两个粮库大米的总吨数减去原来乙粮库大米的吨数，求出原来甲粮库存放大米的吨数。抓住两个粮库的大米总吨数不变，求出30吨占两个粮库大米总吨数的几分之几，然后根据分数除法的意义求出总吨数是解题的关键。\n\n【详解】两个粮库大米的总吨数：\n\n$3 0 \\div (\\frac {7}{5 + 7} - \\frac {5}{4 + 5}) \\\\ = 3 0 \\div (\\frac {7}{1 2} - \\frac {5}{9}) \\\\ = 3 0 \\div (\\frac {2 1}{3 6} - \\frac {2 0}{3 6}) \\\\ = 3 0 \\div \\frac {1}{3 6} \\\\ = 3 0 \\times 3 6 \\\\ = 1 0 8 0 (\\text {吨}) $\n\n原来乙粮库大米的吨数:\n\n$1 0 8 0 \\times \\frac {7}{5 + 7} \\\\ = 1 0 8 0 \\times \\frac {7}{1 2} \\\\ = 6 3 0 (\\text {吨}) $原来甲粮库大米的吨数:\n\n$1080 - 630 = 450$  （吨）\n\n答: 原来甲粮库存放大米 450 吨, 乙粮库存放大米 630 吨。",
+        "knowledge_points": [
+          "求一个数的几分之几的问题",
+          "已知一个数的几分之几是多少, 求这个数",
+          "比与分数、除法的关系",
+          "比的应用"
+        ],
+        "source": "24-25 六年级下·海南海口·单元测试",
+        "merged_id": "unit4_086",
+        "original_id": "unit4_solve_086"
+      },
+      {
+        "id": "unit4_solve_087",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "2024年，林老师参加了双辽乡村（森林）马拉松比赛，她一路坚持，终于跑到了'距离终点还有15千米'的提示牌，此时她已经跑完全程的 $\\frac{9}{14}$ ，真是太了不起了！这场比赛有大量长跑爱好者踊跃参加。参加全程马拉松的有840人，参加半程马拉松的有1000人，参加健康跑的有1200人。已知全程马拉松男女比例为3:1；半程马拉松男女比例为7:3；健康跑男女比例为12:13。\n\n(1) 这次马拉松比赛的全程是多少千米？  \n(2) 参加全程马拉松比赛的男女人数各是多少人？",
+        "options": [],
+        "answer": "(1) 42 千米 (2) 男生 630 人; 女生 210 人",
+        "solution": "【分析】（1）把这次马拉松比赛的全程看作单位'1'，林老师已经跑完全程的  $\\frac{9}{14}$ ，离终点还有 15 千米，即 15 千米占全程的  $(1 - \\frac{9}{14})$ ，单位'1'未知，根据分数除法的意义求出全程。(2) 已知参加全程马拉松的有 840 人, 全程马拉松男女比例为  $3: 1$ , 即参加全程马拉松的男生人数占参加全程马拉松总人数的  $\\frac{3}{3+1}$ , 根据求一个数的几分之几是多少, 用乘法计算, 求出参加全程马拉松的男生人数; 再用参加全程马拉松总人数减去参加全程马拉松的男生人数, 即是参加全程马拉松的女生人数。\n\n【详解】 (1)  $15 \\div (1 - \\frac{9}{14})$\n\n$= 1 5 \\div \\frac {5}{1 4} \\\\ = 1 5 \\times \\frac {1 4}{5} \\\\ = 4 2 \\text {(千 米)} $\n\n答: 这次马拉松比赛的全程是 42 千米。\n\n(2) 男:  $840 \\times \\frac{3}{3 + 1}$\n\n$= 8 4 0 \\times \\frac {3}{4} \\\\ = 6 3 0 (\\text {人}) $\n\n女:  $840 - 630 = 210$  (人)\n\n答: 参加全程马拉松比赛的男生人数是 630 人, 女生人数是 210 人。",
+        "knowledge_points": [
+          "求一个数的几分之几的问题",
+          "已知一个数的几分之几是多少，求这个数",
+          "按比分配问题",
+          "分数的四则混合运算"
+        ],
+        "source": "24-25 六年级上·吉林四平·期中",
+        "merged_id": "unit4_087",
+        "original_id": "unit4_solve_087"
+      },
+      {
+        "id": "unit4_solve_088",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "《周髀算经》中记载: '勾广三, 股修四, 径隅五。'意思是说: 当直角三角形的两条直角边分别为3（勾）和4（股）时，径（弦）则为 5 。后人简单地把这个事实说成 '勾 3 股 4 弦 5 。 '一个直角三角形 3 条边的长度比是  $3: 4: 5$ , 斜边长 10 厘米,另外两条直角边分别长多少厘米?",
+        "options": [],
+        "answer": "6 厘米;8 厘米",
+        "solution": "【分析】已知一个直角三角形3条边的长度比是  $3: 4: 5$ , 直角三角形中斜边最长, 可知这个直角三角形中两条直角边分别占3份和4份, 斜边占5份; 已知斜边长10厘米, 用斜边的长度除以5, 求出一份量, 再用一份量分别乘3、乘4, 求出两条直角边的长度。\n\n【详解】  $10 \\div 5 = 2$  （厘米）\n\n$2 \\times 3 = 6 (\\text {厘 米})$\n\n$2 \\times 4 = 8 (\\text {厘 米})$\n\n答: 另外两条直角边分别长 6 厘米和 8 厘米。",
+        "knowledge_points": [
+          "单归一问题",
+          "比的应用",
+          "三角形的分类"
+        ],
+        "source": "23-24 六年级上·山东临沂·期中",
+        "merged_id": "unit4_088",
+        "original_id": "unit4_solve_088"
+      },
+      {
+        "id": "unit4_solve_089",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "两堆煤，甲堆煤的重量占总重量的  $\\frac{5}{8}$ ，如果从甲堆煤里取出26吨，从乙堆煤里取出10吨，两堆煤剩下重量的比是  $1:1$ ，求甲乙两堆煤共有多少吨？",
+        "options": [],
+        "answer": "64吨",
+        "solution": "【分析】把总重量看作单位 '1', 已知原来甲堆煤的重量占总重量的  $\\frac{5}{8}$ , 原来乙堆煤的重量占总重量的  $(1 - \\frac{5}{8})$ , 假设甲乙两堆煤共有  $x$  吨, 根据分数乘法的意义, 可知原来甲堆煤的重量是  $\\frac{5}{8} x$  吨, 原来乙堆煤的重量是  $(1 - \\frac{5}{8}) x$  吨; 已知两堆煤剩下重量的比是  $1: 1$ , 根据比的意义, 可知两堆煤剩下重量相等, 据此可知原来甲堆煤的重量  $-26$  吨 = 原来乙堆煤的重量  $-10$  吨, 列方程为  $\\frac{5}{8} x - 26 = (1 - \\frac{5}{8}) x - 10$ ,\n\n然后解出方程即可。本题可用列方程解决问题, 找到相应的数量关系是解答本题的关键。\n\n【详解】解：设甲乙两堆煤共有  $\\mathrm{x}$  吨。\n\n$\\frac {5}{8} x - 2 6 = (1 - \\frac {5}{8}) x - 1 0 \\\\ \\frac {5}{8} x - 2 6 = \\frac {3}{8} x - 1 0 \\\\ \\frac {5}{8} x = \\frac {3}{8} x - 1 0 + 2 6 \\\\ \\frac {5}{8} x = \\frac {3}{8} x + 1 6 \\\\ \\frac {5}{8} x - \\frac {3}{8} x = 1 6 \\\\ \\frac {1}{4} x = 1 6 \\\\ x = 1 6 \\div \\frac {1}{4} \\\\ x = 1 6 \\times 4 \\\\ x = 6 4 $\n\n答: 甲乙两堆煤共有 64 吨。",
+        "knowledge_points": [
+          "已知一部分量占总量的几分之几及另一部分量,求总量",
+          "比的应用",
+          "列方程解决稍复杂的实际问题"
+        ],
+        "source": "23-24六年级上·全国·单元测试",
+        "merged_id": "unit4_089",
+        "original_id": "unit4_solve_089"
+      },
+      {
+        "id": "unit4_solve_090",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "加工一批零件，小王加工的个数是小明的  $\\frac{1}{4}$ ，小明加工的个数与小青的比1:3，已知小王加工了40个，小青加工了多少个？",
+        "options": [],
+        "answer": "480个",
+        "solution": "【分析】由题可知, 小王加工了 40 个, 小王加工的个数是小明的  $\\frac{1}{4}$ , 用小王加工的个数除以  $\\frac{1}{4}$ , 求出小明加工的个数,又知小明加工的个数与小青的比  $1: 3$ , 所以小明加工的个数是小青的  $\\frac{1}{3}$ , 所以用小明加工的个数除以  $\\frac{1}{3}$  即可求出小青加工的个数。\n\n【详解】  $40 \\div \\frac{1}{4} \\div \\frac{1}{3}$\n\n$= 4 0 \\times 4 \\times 3 \\\\ = 1 6 0 \\times 3 \\\\ = 4 8 0 \\text {(个)} $\n\n答: 小青加工了 480 个。",
+        "knowledge_points": [
+          "已知一个数的几分之几是多少，求这个数",
+          "比的应用"
+        ],
+        "source": "23-24六年级上·全国·单元测试",
+        "merged_id": "unit4_090",
+        "original_id": "unit4_solve_090"
+      },
+      {
+        "id": "unit4_solve_091",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "爷爷果园里的苹果树比梨树多 320 棵, 已知苹果树和梨树棵数的比是  $7: 5$ , 爷爷的果园里有梨树多少棵?",
+        "options": [],
+        "answer": "800棵",
+        "solution": "【分析】已知苹果树和梨树棵数的比是  $7: 5$ , 所以可以把苹果树棵数看作 7 份, 梨树棵数看作 5 份, 苹果树比梨树多的棵数对应 2 份, 用苹果树比梨树多的棵数  $\\div 2$ , 求出 1 份对应的棵数, 1 份对应的棵数乘梨树的份数求出梨树的棵数。\n\n【详解】  $320 \\div (7 - 5) \\times 5= 3 2 0 \\div 2 \\times 5  = 1 6 0 \\times 5 = 8 0 0 \\text {(棵)} $答: 爷爷的果园里有梨树 800 棵。",
+        "knowledge_points": [
+          "比的应用"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_091",
+        "original_id": "unit4_solve_091"
+      },
+      {
+        "id": "unit4_solve_092",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "某校六（1）和六(2) 班共有 72 人, 六 (1) 班和六 (2) 班人数的比是5:4, 六 (1) 和六 (2) 各有多少人?",
+        "options": [],
+        "answer": "40,32人",
+        "solution": "【分析】已知六（1）班和六（2）班人数的比是  $5: 4$ ，根据比的意义，把六（1）班和六（2）班人数分别看作 5 份和 4 份，然后用总人数除以  $(5 + 4)$  份，即可求出每份是多少，进而用乘法分别求出两个班的人数。\n\n【详解】  $72 \\div (5 + 4)= 7 2 \\div 9 = 8 \\text {(人)} $ 六 (1) 班:  $8 \\times 5 = 40$  (人) 六 (2) 班:  $8 \\times 4 = 32$  (人)\n\n答: 六 (1) 班有 40 人, 六 (2) 班有 32 人。",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "23-24六年级上·全国·单元测试",
+        "merged_id": "unit4_092",
+        "original_id": "unit4_solve_092"
+      },
+      {
+        "id": "unit4_solve_093",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "一件上衣的售价是60元，裤子与上衣单价的比是2:3。每条裤子多少元？",
+        "options": [],
+        "answer": "40元",
+        "solution": "【分析】已知裤子与上衣单价的比是  $2: 3$ , 则说明上衣的价格有 3 份, 裤子的价格有 2 份, 已知上衣售价是 60 元, 用 60 除以 3 可求出每份代表的价格, 再跟裤子价格的份数相乘即可求出结果。\n\n【详解】  $60 \\div 3 \\times 2$ \n\n$= 2 0 \\times 2 \\\\ = 4 0 \\text {(元)} $\n\n答: 每条裤子 40 元。",
+        "knowledge_points": [
+          "比的应用"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_093",
+        "original_id": "unit4_solve_093"
+      },
+      {
+        "id": "unit4_solve_094",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "一个长方形, 宽和长的比是  $\\frac{4}{5}$ , 如果宽再增加 6 厘米, 这时长方形就变成了正方形, 求这个长方形的宽是多少厘米?",
+        "options": [],
+        "answer": "24厘米",
+        "solution": "【分析】已知长方形的宽和长的比是  $\\frac{4}{5}$ , 把宽看作 4 份, 长看作 5 份, 宽比长少 (5-4) 份;如果宽再增加 6 厘米, 这时长方形就变成了正方形, 说明原来宽比长少 6 厘米, 用少的长度除以少的份数, 求出一份数, 再用一份数乘宽的份数, 即可求出宽。\n\n【详解】一份数：\n\n$6 \\div (5 - 4) \\\\ = 6 \\div 1 \\\\ = 6 \\text {(厘 米)} $\n\n宽:  $6 \\times 4 = 24$  (厘米)\n\n答: 这个长方形的宽是 24 厘米。",
+        "knowledge_points": [
+          "正方形的概念及特点",
+          "比的意义",
+          "比的应用"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_094",
+        "original_id": "unit4_solve_094"
+      },
+      {
+        "id": "unit4_solve_095",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "甲、乙、丙三人合作投资兴办服装厂，甲投资30万元，乙投资40万元，丙投资50万元。服装厂去年获得纯利润是48万元。按投资额分配，丙获得利润多少万元？",
+        "options": [],
+        "answer": "20万元",
+        "solution": "【分析】甲投资 30 万元, 乙投资 40 万元, 丙投资 50 万元。根据比的意义和化简, 得出三人投资额的比为  $3: 4: 5$ , 即甲、乙两人各投资多少万元?乙、丙的投资额分别占三人总投资的  $\\frac{3}{3 + 4 + 5} 、 \\frac{4}{3 + 4 + 5} 、 \\frac{5}{3 + 4 + 5}$ ,根据求一个数的几分之几是多少，用乘法计算，即可求出按投资额分配，三人各应获得利润。\n\n【详解】  $30: 40: 50$\n\n$= (3 0 \\div 1 0): (4 0 \\div 1 0): (5 0 \\div 1 0)\\\\=3:4:5$\n\n丙:  $48 \\times \\frac{5}{3 + 4 + 5}$\n\n$= 4 8 \\times \\frac {5}{1 2} \\\\ = 2 0 (\\text {万 元}) $\n\n答：丙获得利润20万元。",
+        "knowledge_points": [
+          "比的化简",
+          "比的意义",
+          "比的应用"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_095",
+        "original_id": "unit4_solve_095"
+      },
+      {
+        "id": "unit4_solve_096",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "A、B两地相距750千米，甲．乙两车同时从 A、B 两地相对开出，相向而行，经过5小时相遇。已知甲、乙两车的速度比是8:7。求甲车每小时行多少千米？",
+        "options": [],
+        "answer": "80千米",
+        "solution": "【分析】总路程÷相遇时间 = 两车速度和，将比的前后项看成份数，速度和 ÷ 总份数，求出一份数，一份数 × 甲车对应份数 = 甲车速度，据此列式解答。\n\n【详解】  $750 \\div 5 = 150$  （千米/时）\n\n$1 5 0 \\div (8 + 7) \\times 8 \\\\ = 1 5 0 \\div 1 5 \\times 8 \\\\ = 1 0 \\times 8 \\\\ = 8 0 \\text {(千 米 / 时)} $\n\n答: 甲车每小时行 80 千米。",
+        "knowledge_points": [
+          "相遇问题",
+          "按比分配问题"
+        ],
+        "source": "23-24六年级上·全国·单元测试",
+        "merged_id": "unit4_096",
+        "original_id": "unit4_solve_096"
+      },
+      {
+        "id": "unit4_solve_097",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "小明、小林、小刚三人的邮票枚数的比为  $7:4:1$  ，小明赠送给小刚24枚邮票，小林也赠送给小刚一些邮票，这时三人的邮票枚数比为8:3:5。小林赠送给小刚多少枚邮票？",
+        "options": [],
+        "answer": "42枚",
+        "solution": "【分析】三人邮票总数不变，将三人邮票总数看作单位'1'，开始小明邮票枚数占邮票总数的  $\\frac{7}{7+4+1}$ ，小明和小林都赠送给小刚邮票后，小明邮票枚数占邮票总数的  $\\frac{8}{8+3+5}$ ，因此24枚邮票占邮票总数的  $(\\frac{7}{7+4+1}-\\frac{8}{8+3+5})$ ，根据部分数量÷对应分率 = 整体数量，求出三人邮票总数，小林邮票数量减少了总数的  $(\\frac{4}{7+4+1}-\\frac{3}{8+3+5})$ ，三人邮票总数×小林邮票数量减少的对应分率 = 小林赠送给小刚的数量，据此列式解答。关键是理解比的意义，明确总数不变，从而确定单位'1'，根据分数乘除法的意义，进行解答。\n\n【详解】  $24 \\div (\\frac{7}{7 + 4 + 1} - \\frac{8}{8 + 3 + 5})$\n\n$= 2 4 \\div (\\frac {7}{1 2} - \\frac {8}{1 6}) \\\\ = 2 4 \\div \\frac {4}{4 8} \\\\ = 2 4 \\times \\frac {4 8}{4}\\\\=288 (枚) $\n\n$2 8 8 \\times (\\frac {4}{7 + 4 + 1} - \\frac {3}{8 + 3 + 5}) \\\\ = 2 8 8 \\times (\\frac {4}{1 2} - \\frac {3}{1 6}) $\n\n$= 2 8 8 \\times \\frac {7}{4 8}\\\\= 42 (枚)$\n\n答: 小林赠送给小刚 42 枚邮票。",
+        "knowledge_points": [
+          "求一个数的几分之几的问题",
+          "已知一个数的几分之几是多少，求这个数",
+          "按比分配问题"
+        ],
+        "source": "23-24 六年级上·全国·课后作业",
+        "merged_id": "unit4_097",
+        "original_id": "unit4_solve_097"
+      },
+      {
+        "id": "unit4_solve_098",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "有甲、乙两筐苹果，甲筐苹果重40千克，乙筐苹果重30千克，要使甲、乙两筐苹果的质量比是3:2，应从乙筐拿出多少千克苹果放入甲筐？",
+        "options": [],
+        "answer": "2千克",
+        "solution": "【分析】把两筐苹果的总质量看作单位 '1', 从乙筐拿出一些苹果放入甲筐后, 甲、乙两筐苹果的质量比是  $3: 2$ , 即现在乙筐苹果的质量占两筐苹果总质量的  $\\frac{2}{3+2}$ , 单位 '1'已知,用两筐苹果的总质量乘  $\\frac{2}{3 + 2}$ , 求出现在乙筐苹果的质量, 再用原来乙筐苹果的质量减去现在乙筐苹果的质量即可。\n\n【详解】  $(40 + 30) \\times \\frac{2}{3 + 2}$\n\n$= 7 0 \\times \\frac {2}{5}\\\\= 28$  （千克）\n\n$3 0 - 2 8 = 2 \\text {(千 克)}$\n\n答: 应从乙筐拿出 2 千克苹果放入甲筐。",
+        "knowledge_points": [
+          "整数乘分数",
+          "求一个数的几分之几的问题",
+          "按比分配问题"
+        ],
+        "source": "24-25六年级上·全国·单元测试",
+        "merged_id": "unit4_098",
+        "original_id": "unit4_solve_098"
+      },
+      {
+        "id": "unit4_solve_099",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "新世纪小学四、五、六年级共有 27 个班, 平均每个班 35 人, 三个年级的人数比是  $2: 3: 4$  。四、五、六年级各有多少人?",
+        "options": [],
+        "answer": "210;15;420",
+        "solution": "【分析】平均每个班人数  $\\times$  班数 = 四五六年级总人数, 将比的各项看成份数, 总人数  $\\div$  总份数, 求出一份数, 一份数分\n\n别乘四、五、六年级的对应份数，即可求出四、五、六年级的人数。\n\n【详解】  $35 \\times 27 \\div (2 + 3 + 4)$\n\n$= 9 4 5 \\div 9 \\\\ = 1 0 5 \\text {(人)} $\n\n$1 0 5 \\times 2 = 2 1 0 (\\text {人}) \\\\ 1 0 5 \\times 3 = 3 1 5 (\\text {人}) \\\\ 1 0 5 \\times 4 = 4 2 0 (\\text {人}) $\n\n答: 四年级有 210 人、五年级有 315 人、六年级有 420 人。",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "24-25 六年级上·全国·单元测试",
+        "merged_id": "unit4_099",
+        "original_id": "unit4_solve_099"
+      },
+      {
+        "id": "unit4_solve_100",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "春节期间，某网站对100万网民的拜年方式进行了调查。结果表明：选择手机短信拜年、打电话拜年及聊天软件拜年方式的人数比是26:9:15。在这调查的 100 万网民中，选择聊天软件拜年的有多少万？",
+        "options": [],
+        "answer": "30万",
+        "solution": "【分析】选择手机短信拜年、打电话拜年及聊天软件拜年\n\n方式的人数比是  $26: 9: 15$ , 即选择聊天软件拜年方式占了总人数的  $\\frac{15}{26 + 9 + 15}$ , 求一个数的几分之几用乘法。\n\n【详解】  $100 \\times \\frac{15}{26 + 9 + 15} = 100 \\times \\frac{15}{50} = 30$  （万）答：选择聊天软件拜年的有 30 万。",
+        "knowledge_points": [
+          "整数乘分数",
+          "按比分配问题"
+        ],
+        "source": "24-25六年级上·全国·单元测试",
+        "merged_id": "unit4_100",
+        "original_id": "unit4_solve_100"
+      },{
+        "id": "unit4_choice_101",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "六(1)班共有35人,男、女生人数的比可能是()",
+        "options": ["A. $5:3$", "B. 4:5", "C. $3:4$", "D. 3:5"],
+        "answer": "C",
+        "solution": "【分析】用总人数分别除以各个选项份数和，求出一份数，看一份数是否是整数；不是整数的，这个比就不可能是这个班男、女生人数比。\n\n【详解】A. $5:3$ $5+3=8$(份)35÷8=4……3; 35不能被8整除, 男、女生人数的比不可能是 $5:3$; 不符合题意;B. $4:5$ $4+5=9$(份)35÷9=3……8; 35不能被9整除, 男、女生人数的比不可能是4:5; 不符合题意;C. $3:4$ $3+4=7$(份)35÷7=5; 35能被7整除, 男、女生人数的比可能是 $3:4$;符合题意;D. $3:5$ $3+5=8$(份)35÷8=4……3; 35不能被8整除, 男、女生人数的比不可能是 $3:5$; 不符合题意。六(1)班共有35人，男、女生人数的比可能是 $3:4$。",
+        "knowledge_points": ["比的应用"],
+        "source": "24-25 六年级上·全国·单元测试",
+        "merged_id": "unit4_101",
+        "original_id": "unit4_choice_101"
+      },
+      {
+        "id": "unit4_choice_102",
+        "unit": "第四单元：比",
+        "chapter": "比的化简",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "$\\frac{1}{6}:\\frac{2}{9}$化成最简整数比是()",
+        "options": ["A. $\\frac{4}{3}$", "B. 3:4", "C. $4:3$", "D. $1:27$"],
+        "answer": "B",
+        "solution": "【分析】最简整数比是比的前项和后项的公因数只有1的整数。将分数化成最简整数比, 根据比的基本性质先将比的前项和后项同时乘两个分数分母的最小公倍数, 转化为整数比, 再同时除以两个整数的最大公因数。\n\n【详解】$\\frac{1}{6}:\\frac{2}{9}\\\\=(\\frac{1}{6}\\times 18):(\\frac{2}{9}\\times 18)\\\\=3:4$",
+        "knowledge_points": ["比的化简", "比的基本性质"],
+        "source": "24-25 六年级上·全国·单元测试",
+        "merged_id": "unit4_102",
+        "original_id": "unit4_choice_102"
+      },
+      {
+        "id": "unit4_choice_103",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "甲、乙、丙三个数的和是300, 甲数是120, 乙数和丙数的比是 $5:4$, 丙数是()",
+        "options": ["A. 80", "B. 100", "C. 180", "D. 120"],
+        "answer": "A",
+        "solution": "【分析】已知甲、乙、丙三个数的和是300, 甲数是120, 则乙、丙数之和是(300-120); 已知乙数和丙数的比是 $5:4$, 则丙数占乙、丙数之和的 $\\frac{4}{5+4}$ 根据求一个数的几分之几是多少, 用乘法计算, 即可求出丙数。\n\n【详解】$(300-120)\\times\\frac{4}{5+4}\\\\=180\\times\\frac{4}{9}\\\\=80\\\\$ 丙数是 $80$。",
+        "knowledge_points": ["整数乘分数", "求一个数的几分之几的问题", "按比分配问题"],
+        "source": "24-25 六年级上·全国·单元测试",
+        "merged_id": "unit4_103",
+        "original_id": "unit4_choice_103"
+      },
+      {
+        "id": "unit4_choice_104",
+        "unit": "第四单元：比",
+        "chapter": "比的意义",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "一堆煤, 用去 $\\frac{3}{5}$, 用去的与剩下的比是()",
+        "options": ["A. 3:5", "B. 2:5", "C. $3:2$", "D. $2:3$"],
+        "answer": "C",
+        "solution": "【分析】将一堆煤看作单位'1', 用去 $\\frac{3}{5}$, 剩下 $(1-\\frac{3}{5})$, 两数相除又叫两个数的比, 根据比的意义, 写出用去的与剩下的对应分率的比, 化简即可。\n\n【详解】$\\frac{3}{5}$: $(1-\\frac{3}{5})\\\\=\\frac{3}{5}:\\frac{2}{5}\\\\=(\\frac{3}{5}\\times 5):(\\frac{2}{5}\\times 5)\\\\=3:2\\\\$ 用去的与剩下的比是 $3:2$。",
+        "knowledge_points": ["比的化简", "比的意义"],
+        "source": "24-25 六年级上·全国·单元测试",
+        "merged_id": "unit4_104",
+        "original_id": "unit4_choice_104"
+      },
+      {
+        "id": "unit4_choice_105",
+        "unit": "第四单元：比",
+        "chapter": "比的意义",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "已知一堆香蕉和橙子的个数之比是 $5:6$, 下列结论中, 正确的是()",
+        "options": ["A. 香蕉有5个, 橙子有6个", "B. 香蕉个数是橙子个数的 $\\frac{5}{11}$", "C. 橙子个数是香蕉个数的 $\\frac{5}{6}$", "D. 香蕉个数是橙子个数的 $\\frac{5}{6}$"],
+        "answer": "D",
+        "solution": "【分析】A. 根据比的意义可知, 香蕉和橙子的个数之比是 $5:6$, 可以把香蕉的个数看作5份, 橙子的个数看作6份,但不能确定香蕉和橙子的个数。B. 用香蕉个数除以橙子个数, 即是香蕉个数是橙子个数的几分之几。C. 用橙子个数除以香蕉个数, 即是橙子个数是香蕉个数的几分之几。D. 用香蕉个数除以橙子个数, 即是香蕉个数是橙子个数的几分之几。\n\n【详解】A. $5:6=10:12=15:18=\\cdots$所以不能确定香蕉和橙子的个数，原题说法错误。B. $5\\div 6=\\frac{5}{6}$, 香蕉个数是橙子个数的 $\\frac{5}{6}$, 原题说法错误。C. $6\\div 5=\\frac{6}{5}$, 橙子个数是香蕉个数的 $\\frac{6}{5}$, 原题说法错误。D. $5\\div 6=\\frac{5}{6}$, 香蕉个数是橙子个数的 $\\frac{5}{6}$, 原题说法正确。",
+        "knowledge_points": ["分数与除法的关系", "比的基本性质", "比的意义", "求一个数占另一个数几分之几"],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_105",
+        "original_id": "unit4_choice_105"
+      },
+      {
+        "id": "unit4_choice_106",
+        "unit": "第四单元：比",
+        "chapter": "比的基本性质",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "现有甲、乙、丙三辆车运送苹果，甲、乙两车运送苹果的质量比是 $7:13$, 乙、丙两车运送苹果的质量比是 $4:3$, 那么()车运送的苹果最少。",
+        "options": ["A. 甲", "B. 乙", "C. 丙", "D. 无法判断"],
+        "answer": "A",
+        "solution": "【分析】根据题意，甲、乙两车运送苹果的质量比是7:13, 乙、丙两车运送苹果的质量比是 $4:3$, 两个比中都有乙，但占的份数不相同，无法组成甲、乙、丙的连比；利用比的基本性质, 让甲:乙中的前项和后项都乘4, 让乙:丙中的前项和后项都乘13, 这样两个比中, 乙数占的份数相同，可以组成甲、乙、丙的连比，再把比看作份数，份数最少的，表示这辆车运送的苹果最少。\n\n【详解】甲:乙$=7:13=(7\\times 4):(13\\times 4)=28:52$ 乙:丙$=4:3=(4\\times 13):(3\\times 13)=52:39$ 甲:乙:丙$=28:52:39$ $28<39<52$那么甲车运送的苹果最少。",
+        "knowledge_points": ["比的基本性质"],
+        "source": "23-24六年级上·全国·单元测试",
+        "merged_id": "unit4_106",
+        "original_id": "unit4_choice_106"
+      },
+      {
+        "id": "unit4_choice_107",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "两个面积相等的正方形拼成一个长方形，如图所示，则这个长方形的周长与两个正方形的周长和之比是()。![](images/f51cda06890aab8f54e51d8f2a06fc770ebbb966fca808ee8bef85ca2c8c810b.jpg)",
+        "options": ["A. $1:3$", "B. 2:1", "C. $3:4$", "D. $1:2$"],
+        "answer": "C",
+        "solution": "【分析】根据题意, 设小正方形的边长为1, 则长方形的长为2、宽为1。根据正方形的周长$=\\text{边长}\\times 4$, 求出一个正方形的周长, 再乘2, 即是两个正方形的周长之和;根据长方形的周长$=(长+宽)\\times 2$, 求出长方形的周长;根据比的意义写出长方形的周长与两个正方形的周长和之比，并化简比。\n\n【详解】设小正方形的边长为 $1$。两个正方形的周长之和: $1\\times 4\\times 2=8$长方形的长: $1\\times 2=2$长方形的周长: $(2+1)\\times 2\\\\=3\\times 2\\\\=6\\\\$ $6:8=(6\\div 2):(8\\div 2)=3:4$所以, 这个长方形的周长与两个正方形的周长和之比是3:4。",
+        "knowledge_points": ["比的化简", "比的意义", "长、正方形周长的应用"],
+        "source": "23-24六年级上·全国·单元测试",
+        "merged_id": "unit4_107",
+        "original_id": "unit4_choice_107"
+      },
+      {
+        "id": "unit4_choice_108",
+        "unit": "第四单元：比",
+        "chapter": "比的基本性质",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "4:13的前项加上8,要使比值不变，后项应()",
+        "options": ["A. 加上26", "B. 乘2", "C. 加上8", "D. 减去8"],
+        "answer": "A",
+        "solution": "【分析】根据比的基本性质：比的前项和后项同时乘或除以一个不为0的数，比值不变；用比的前项+8，再除以4, 求出比的前项扩大到原来的几倍, 后项也扩大到原来的几倍, 求出后项扩大后的数, 再减去原来的后项, 即可求出后项应该加上的数, 据此解答。\n\n【详解】$(4+8)\\div 4\\\\=12\\div 4\\\\=3\\\\$ $13\\times 3-13\\\\=39-13\\\\=26\\\\$ 4:13的前项加上8, 要使比值不变, 后项应加上26或乘3。",
+        "knowledge_points": ["比的基本性质"],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_108",
+        "original_id": "unit4_choice_108"
+      },
+      {
+        "id": "unit4_choice_109",
+        "unit": "第四单元：比",
+        "chapter": "比的意义",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "小张星期天从家出发去书店买书, 已经走了全程的 $\\frac{5}{8}$, 已经走的路程与所剩路程的最简整数比是()",
+        "options": ["A. $\\frac{5}{8}:\\frac{3}{8}$", "B. $5:8$", "C. $5:3$", "D. 3:5"],
+        "answer": "C",
+        "solution": "【分析】把小张家到书店的路程看作单位'1', 已经走了全程的 $\\frac{5}{8}$, 则还剩全程的 $(1-\\frac{5}{8})$;然后根据比的意义写出已经走的路程与所剩路程之比, 并化简比。\n\n【详解】$\\frac{5}{8}$: $(1-\\frac{5}{8})\\\\=\\frac{5}{8}:\\frac{3}{8}\\\\=(\\frac{5}{8}\\times 8):(\\frac{3}{8}\\times 8)\\\\=5:3$ 已经走的路程与所剩路程的最简整数比是 $5:3$。",
+        "knowledge_points": ["比的化简", "比的意义"],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_109",
+        "original_id": "unit4_choice_109"
+      },
+      {
+        "id": "unit4_choice_110",
+        "unit": "第四单元：比",
+        "chapter": "比的化简",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "把 $200\\mathrm{~cm}:4\\mathrm{~m}$化成最简整数比是()",
+        "options": ["A. $1:2$", "B. $50:1$", "C. $2:1$", "D. $1:50$"],
+        "answer": "A",
+        "solution": "【分析】根据比的基本性质：比的前项和后项同时乘或除以一个不为0的数，比值不变，据此解答，注意单位名数的统一。\n\n【详解】$200\\mathrm{~cm}:4\\mathrm{~m}\\\\=200\\mathrm{cm}:400\\mathrm{cm}\\\\=(200\\div 200):(400\\div 200)\\\\=1:2\\\\$ 把 $200\\mathrm{~cm}:4\\mathrm{~m}$化成最简整数比是 $1:2$。",
+        "knowledge_points": ["比的化简", "比的基本性质"],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_110",
+        "original_id": "unit4_choice_110"
+      },
+      {
+        "id": "unit4_choice_111",
+        "unit": "第四单元：比",
+        "chapter": "比的意义",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "a与b的 $\\frac{3}{4}$相等, a与b的比是()",
+        "options": ["A. $1:3$", "B. $4:3$", "C. $3:4$", "D. $4:7$"],
+        "answer": "C",
+        "solution": "【分析】假设 $b=4$, 根据求一个数的几分之几是多少, 用乘法计算, 即可求出 $a$是多少, 再写出 $a$和 $b$的比, 据此解答。\n\n【详解】假设 $b=4, 4\\times\\frac{3}{4}=3$, 则 $a=3$, 所以 $a$与 $b$的比是 $3:4$。",
+        "knowledge_points": ["求一个数的几分之几的问题", "比的意义"],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_111",
+        "original_id": "unit4_choice_111"
+      },
+      {
+        "id": "unit4_choice_112",
+        "unit": "第四单元：比",
+        "chapter": "求比值",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "与 $\\frac{1}{5}:3$比值相等的是()",
+        "options": ["A. $5:3$", "B. 15:1", "C. $1:15$", "D. $3:\\frac{1}{5}$"],
+        "answer": "C",
+        "solution": "【分析】用比的前项除以比的后项即可求出比值, 据此分别求出题干与各项的比值, 再进行对比即可。\n\n【详解】$\\frac{1}{5}:3\\\\=\\frac{1}{5}\\div 3\\\\=\\frac{1}{5}\\times\\frac{1}{3}\\\\=\\frac{1}{15}\\\\$ A. 5:3$=5\\div 3\\\\=\\frac{5}{3}\\\\$ $\\frac{5}{3}\\neq\\frac{1}{15}, 5:3$与 $\\frac{1}{5}:3$比值不相等; B. $15:1\\\\=15\\div 1\\\\=15\\\\$ $15\\neq\\frac{1}{15}, 15:1$与 $\\frac{1}{5}:3$比值不相等; C. $1:15\\\\=1\\div 15\\\\=\\frac{1}{15}\\\\$ $\\frac{1}{15}=\\frac{1}{15}, 1:15$与 $\\frac{1}{5}:3$比值相等; D. $3:\\frac{1}{5}\\\\=3\\times 5\\\\=15\\\\$ $15\\neq\\frac{1}{15}, 3:\\frac{1}{5}$与 $\\frac{1}{5}:3$比值不相等;",
+        "knowledge_points": ["求比值"],
+        "source": "22-23 六年级上·江苏盐城·期中",
+        "merged_id": "unit4_112",
+        "original_id": "unit4_choice_112"
+      },
+      {
+        "id": "unit4_choice_113",
+        "unit": "第四单元：比",
+        "chapter": "比的意义",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "如图是由5个同样的小长方形拼成的, 拼成的图形的长与宽的比是()。![](images/8dc389cfc57881d1b1153da151b9cebc211c08003addf2c08ac91c99d7a200d4.jpg)",
+        "options": ["A. 4:3", "B. 8:5", "C. $3:2$", "D. $6:5$"],
+        "answer": "D",
+        "solution": "【分析】从图中可知, 拼成图形的长等于小长方形的2个长或小长方形的3个宽, 即小长方形的2个长$=3$个宽, 由此可得出小长方形的长与宽的比是 $3:2$;由小长方形的长与宽的比是 $3:2$, 可以设小长方形的长为3份, 宽为2份; 拼成图形的长等于小长方形的2个长, 即6份; 拼成图形的宽等于小长方形的1个长加1个宽, 即5份;根据比的意义写出拼成的图形的长与宽的比。\n\n【详解】小长方形的长与宽的比是 $3:2$;拼成的图形的长与宽的比是:$(3\\times 2):(3+2)=6:5$即拼成的图形的长与宽的比是 $6:5$。",
+        "knowledge_points": ["平面图形的拼接", "比的意义"],
+        "source": "22-23 六年级上·江苏盐城·期末",
+        "merged_id": "unit4_113",
+        "original_id": "unit4_choice_113"
+      },
+      {
+        "id": "unit4_choice_114",
+        "unit": "第四单元：比",
+        "chapter": "比的意义",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "将3克药放入100克水中, 药与药水的比是()",
+        "options": ["A. 3:97", "B. $3:100$", "C. 3:103", "D. 97:3"],
+        "answer": "C",
+        "solution": "【分析】药$+$水$=$药水, 根据比的意义, 写出药与药水的比即可。\n\n【详解】3: $(100+3)=3:103$药与药水的比是 $3:103$。",
+        "knowledge_points": ["比的意义"],
+        "source": "23-24 六年级下·重庆渝北·期末",
+        "merged_id": "unit4_114",
+        "original_id": "unit4_choice_114"
+      },
+      {
+        "id": "unit4_choice_115",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "古希腊时期, 人们认为最美人体的头顶到肚脐的长度与肚脐到足底的长度之比是0.618:1(称为黄金分割比)，著名的'断臂维纳斯'便是如此。此外，最美人体的头顶到咽喉的长度与咽喉到肚脐的长度之比也是0.618:1。若林华满足上述两个黄金分割比，且肚脐到足底的长度为 $108\\mathrm{~cm}$，头的高度约为 $25.5\\mathrm{~cm}$，则林华的身高大约是() $\\mathrm{cm}$。",
+        "options": ["A. 165", "B. 175", "C. 185", "D. 190"],
+        "answer": "B",
+        "solution": "【分析】充分运用黄金分割比例，计算可估计身高。头的高度约为 $25.5\\mathrm{~cm}$，即头顶到咽喉的长度为 $25.5\\mathrm{~cm}$，由头顶至咽喉的长度与咽喉至肚脐的长度之比是 $0.618:1=0.618$，可得咽喉至肚脐的长度为 $25.5\\div 0.618\\approx 41.2\\mathrm{~cm}$，由头顶到咽喉的长度为 $25.5\\mathrm{~cm}$，咽喉至肚脐的长度为 $41.2\\mathrm{~cm}$，肚脐到足底的长度为 $108\\mathrm{~cm}$；由此即可解答。本题考查简单的推理和估算，考查运算能力和推理能力。\n\n【详解】$25.5\\div 0.618\\approx 41.2\\mathrm{~cm}$,由头顶到咽喉的长度为 $25.5\\mathrm{~cm}$, 咽喉至肚脐的长度为41.2cm，肚脐到足底的长度为 $108\\mathrm{~cm}$；可得：林华的身高大约是:$25.5+41.2+108\\\\=66.7+108\\\\=174.7\\\\\\approx 175\\mathrm{(cm)}\\\\$",
+        "knowledge_points": ["比的应用"],
+        "source": "23-24 六年级上·河南周口·期末",
+        "merged_id": "unit4_115",
+        "original_id": "unit4_choice_115"
+      },
+      {
+        "id": "unit4_choice_116",
+        "unit": "第四单元：比",
+        "chapter": "比与分数、除法的关系",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "比的后项相当于除法算式中的()",
+        "options": ["A. 除数", "B. 被除数", "C. 分母"],
+        "answer": "A",
+        "solution": "【分析】根据比与除法的关系，比的前项相当于除法中的被除数，比号相当于除号，比的后项相当于除法中的除数，比值相当于除法中的商，据此选择即可。\n\n【详解】由分析可知：比的后项相当于除法算式中的除数。",
+        "knowledge_points": ["比与分数、除法的关系"],
+        "source": "22-23 六年级上·四川成都·期末",
+        "merged_id": "unit4_116",
+        "original_id": "unit4_choice_116"
+      },
+      {
+        "id": "unit4_choice_117",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "'宫、商、角、徵、羽'是我国古代音乐的基本音阶。基本音阶'商'的发音管比基本音阶'徵'的发音管短 $\\frac{1}{3}$, 则'徵'和'商'的发音管长度比是()",
+        "options": ["A. $3:2$", "B. $2:3$", "C. $4:3$", "D. $3:4$"],
+        "answer": "A",
+        "solution": "【分析】将'徵'的发音管长度看作单位'1'，根据'商'的发音管长度比'徵'的发音管长度短 $\\frac{1}{3}$, 即可理解为当'徵'的发音管长度为3份时, '商'的发音管长度比'徵'的发音管长度少1份, 即'商'的发音管长度是2份,据此得解。\n\n【详解】由分析可知, 将'徵'的发音管长度看作单位'1', 则'商'的发音管长度是'徵'的发音管长度的 $\\frac{2}{3}$,此时'徵'和'商'的发音管长度比为 $3:2$。",
+        "knowledge_points": ["比的应用"],
+        "source": "2024·山西吕梁·小升初真题",
+        "merged_id": "unit4_117",
+        "original_id": "unit4_choice_117"
+      },
+      {
+        "id": "unit4_choice_118",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "据《人民网》报道：教育部在今年4月22日首次向全国中小学生发布《阅读指导目录》，供学生自主选择阅读。《目录》分为小学、初中、高中三个学段，共300种书目，三个学段的书目比是11:10:9，其中小学学段有()种书目。",
+        "options": ["A. 100", "B. 110", "C. 90"],
+        "answer": "B",
+        "solution": "【分析】已知三个学段的书目比是 $11:10:9$, 则把小学、初中、高中三个学段的书目种类看作11份、10份、9份, 用 $300\\div(11+10+9)$即可求出每份是多少, 进而用乘法即可求出11份是多少。\n\n【详解】$300\\div(11+10+9)\\\\=300\\div 30\\\\=10$(种) $10\\times 11=110$(种)小学学段有110种书目。",
+        "knowledge_points": ["比的应用"],
+        "source": "2024六年级下·全国·专题练习",
+        "merged_id": "unit4_118",
+        "original_id": "unit4_choice_118"
+      },{
+        "id": "unit4_choice_119",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "在一个三角形中，三个内角度数的比是 $1:3:5$, 这个三角形是()",
+        "options": ["A. 锐角三角形", "B. 直角三角形", "C. 钝角三角形"],
+        "answer": "C",
+        "solution": "【分析】因为三角形的内角和度数是 $180^{\\circ}$, 已知三个内角度数的比是 $1:3:5$, 则最大的角的度数占内角度数和的 $\\frac{5}{1+3+5}$, 将内角和 $180^{\\circ}$看作单位'1', 根据求一个数的几分之几是多少, 用乘法, 即用 $180^{\\circ}$乘 $\\frac{5}{1+3+5}$可求出最大角的具体度数, 进而判断是什么三角形即可。\n\n【详解】由分析可得:该三角形最大的角的度数为:$180^{\\circ}\\times\\frac{5}{1+3+5}=100^{\\circ}$所以这是一个钝角三角形。",
+        "knowledge_points": ["求一个数的几分之几的问题", "按比分配问题", "三角形的分类", "分数与整数的乘法"],
+        "source": "2021·湖南怀化·小升初真题",
+        "merged_id": "unit4_119",
+        "original_id": "unit4_choice_119"
+      },
+      {
+        "id": "unit4_fill_120",
+        "unit": "第四单元：比",
+        "chapter": "求比值",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "一个比的后项是18，比值是 $\\frac{5}{9}$, 这个比的前项是()",
+        "options": [],
+        "answer": "10",
+        "solution": "【分析】用比的前项除以比的后项所得的商，叫做比值。根据比值的意义可知，比的后项$=\\text{前项}\\div$比值，比的前项$=\\text{比值}\\times$后项；据此解答。\n\n【详解】$\\frac{5}{9}\\times 18=10$,这个比的前项是 $10$。",
+        "knowledge_points": ["分数乘整数", "求比值"],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_120",
+        "original_id": "unit4_fill_120"
+      },
+      {
+        "id": "unit4_fill_121",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "一个三角形, 三个内角度数的比是 $4:6:10$, 最小的内角是()度, 这个三角形是()三角形。",
+        "options": [],
+        "answer": "36 直角",
+        "solution": "【分析】把三个内角度数的比看作份数比，求出总份数，用三角形的内角和180度除以总份数，求出1份是多少度，再乘三个内角各自的份数，分别求出三个内角的度数；三个角都是锐角的三角形是锐角三角形，有一个角是直角的三角形是直角三角形，有一个角是钝角的三角形是钝角三角形，据此判断是什么三角形。\n\n【详解】$180\\div(4+6+10)\\\\=180\\div 20\\\\=9$(度) $9\\times 4=36(\\text{度})$ $9\\times 6=54(\\text{度})$ $9\\times 10=90(\\text{度})$所以最小的内角是36度, 这个三角形是直角三角形。",
+        "knowledge_points": ["三角形的内角和", "按比分配问题", "三角形的分类"],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_121",
+        "original_id": "unit4_fill_121"
+      },
+      {
+        "id": "unit4_fill_122",
+        "unit": "第四单元：比",
+        "chapter": "比的意义",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "读一本故事书, 小明读完需要5天, 小丽读完需要8天。小明和小丽两人读完这本书的时间比是(), 两人读书的效率之比是()。",
+        "options": [],
+        "answer": "5:8 8:5",
+        "solution": "【分析】根据比的意义，用小明读完需要的时间：小丽读完需要的时间；求出小明和小丽两人读完这本书的时间比；把这本书的总页数看作单位'1'用 $1\\div 5$, 求出小明读书的效率；用 $1\\div 8$, 求出小丽读书的效率，再用小明读书的效率：小丽读书的效率，化简，即可解答。\n\n【详解】时间比：5:8 $(1\\div 5):(1\\div 8)\\\\=\\frac{1}{5}:\\frac{1}{8}\\\\=(\\frac{1}{5}\\times 40):(\\frac{1}{8}\\times 40)\\\\=8:5$ 读一本故事书，小明读完需要5天，小丽读完需要8天。小明和小丽两人读完这本书的时间比是 $5:8$, 两人读书的效率之比是 $8:5$。",
+        "knowledge_points": ["比的化简", "比的意义", "有具体量的工程问题"],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_122",
+        "original_id": "unit4_fill_122"
+      },
+      {
+        "id": "unit4_fill_123",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "一盒颜料, 用去的量和剩下的量之比是 $8:3$, 已知这盒颜料共有 $330\\mathrm{~g}$, 那么用去了()g。",
+        "options": [],
+        "answer": "240",
+        "solution": "【分析】已知用去的量和剩下的量之比是 $8:3$, 则用去的量占这盒颜料总质量的 $\\frac{8}{8+3}$, 把这盒颜料的总质量看作单位'1', 单位'1'已知, 用总质量乘 $\\frac{8}{8+3}$, 即是用去的质量。\n\n【详解】$330\\times\\frac{8}{8+3}\\\\=330\\times\\frac{8}{11}\\\\=240\\mathrm{(g)}\\\\$ 那么用去了 $240\\mathrm{~g}$。",
+        "knowledge_points": ["整数乘分数", "求一个数的几分之几的问题", "比与分数、除法的关系"],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_123",
+        "original_id": "unit4_fill_123"
+      },
+      {
+        "id": "unit4_fill_124",
+        "unit": "第四单元：比",
+        "chapter": "比的基本性质",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "给5:6的前项加上25，要使比值不变，后项应该乘()。",
+        "options": [],
+        "answer": "6",
+        "solution": "【分析】比的前项和后项同时乘或除以相同的数(0除外)，比值不变。据此解答。\n\n【详解】给5:6的前项加上25, $5+25=30, 30\\div 5=6$, 相当于前项乘6, 要使比值不变, 后项应该乘6。所以后项应该乘 $6$。",
+        "knowledge_points": ["比的基本性质"],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_124",
+        "original_id": "unit4_fill_124"
+      },
+      {
+        "id": "unit4_fill_125",
+        "unit": "第四单元：比",
+        "chapter": "比的意义",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "小强读一本书, 已读页数与未读页数的比是 $7:9$, 他已经读了这本书的()。",
+        "options": [],
+        "answer": "$\\frac{7}{16}$",
+        "solution": "【分析】根据题意, 已读页数与未读页数的比是 $7:9$, 可以把已读页数看作7份, 未读页数看作9份, 总页数是 $(7+9)$份; 用已读的页数除以总页数, 即是已经读了这本书的几分之几。\n\n【详解】$7\\div(7+9)\\\\=7\\div 16\\\\=\\frac{7}{16}\\\\$ 他已经读了这本书的 $\\frac{7}{16}$。",
+        "knowledge_points": ["分数与除法的关系", "比的意义", "求一个数占另一个数几分之几"],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_125",
+        "original_id": "unit4_fill_125"
+      },
+      {
+        "id": "unit4_fill_126",
+        "unit": "第四单元：比",
+        "chapter": "比与分数、除法的关系",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "$12:()=3:()=0.75=\\frac{()}{12}$。",
+        "options": [],
+        "answer": "16; 4; 9",
+        "solution": "【分析】把0.75化成小数是 $\\frac{3}{4}$, 分数的基本性质: 分数的分子和分母同时乘上或除以相同的数(0除外), 分数的大小不变, 根据分数的基本性质, 把 $\\frac{3}{4}$的分子和分母都乘3就是 $\\frac{9}{12}$; 根据分数与比的关系, $\\frac{3}{4}=3:4$; 比的前项和后项同时乘以或除以同一个不为0的数, 比值不变, 这就是比的基本性质, 根据比的基本性质, 把3:4的前项和后项都乘4就是12:16; 据此解答。\n\n【详解】根据分析：$12:16=3:4=0.75=\\frac{9}{12}$。",
+        "knowledge_points": ["分数的基本性质", "一位或多位小数化分数(约分)", "比的基本性质", "比与分数、除法的关系"],
+        "source": "23-24六年级上·辽宁·单元测试",
+        "merged_id": "unit4_126",
+        "original_id": "unit4_fill_126"
+      },
+      {
+        "id": "unit4_fill_127",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "截至2023年7月20日，南水北调中线工程已向北京输水约90亿立方米，水质始终稳定在地表水环境质量标准二类以上，用于北京自来水厂供水、水源地存蓄和城市河湖补水，它们用水量的比是 $30:11:4$, 照这样计算，城市河湖补水是()亿立方米。",
+        "options": [],
+        "answer": "8",
+        "solution": "【分析】从题意可知: 90亿立方米是向北京输水的总量,它对应的是 $30+11+4=45$份, 先用 $90\\div 45=2$亿立方米,得出1份的量, 再乘4, 可得4份的量, 即城市河湖补水量。\n\n【详解】$90\\div(30+11+4)\\times 4\\\\=90\\div 45\\times 4\\\\=2\\times 4\\\\=8(亿立方米)$ 城市河湖补水是8亿立方米。",
+        "knowledge_points": ["比的应用"],
+        "source": "23-24六年级下·北京丰台·期末",
+        "merged_id": "unit4_127",
+        "original_id": "unit4_fill_127"
+      },
+      {
+        "id": "unit4_fill_128",
+        "unit": "第四单元：比",
+        "chapter": "比的意义",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "某小区总建筑面积$19600\\mathrm{m}^2$，共260户。有地上停车位48个，地下停车位52个，这个小区停车位与住户的比是()。",
+        "options": [],
+        "answer": "5:13",
+        "solution": "【分析】先用 $48+52=100$个求出停车位的总个数，再根据比的意义，写出停车位与住户的比是 $100:260$, 最后根据比的基本性质化简即可。\n\n【详解】$(48+52):260\\\\=100:260\\\\=(100\\div 20):(260\\div 20)\\\\=5:13\\\\$ 这个小区停车位与住户的比是 $5:13$。",
+        "knowledge_points": ["比的化简", "比的意义"],
+        "source": "23-24六年级下·北京丰台·期末",
+        "merged_id": "unit4_128",
+        "original_id": "unit4_fill_128"
+      },
+      {
+        "id": "unit4_fill_129",
+        "unit": "第四单元：比",
+        "chapter": "比的意义",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "走同样一段路程，甲走完需要5小时, 乙走完需要7小时。甲、乙的速度之比是_____。",
+        "options": [],
+        "answer": "7:5",
+        "solution": "【分析】把这段路程看作单位'1', 根据速度$=$路程$\\div$时间,分别求出甲的速度和乙的速度, 再根据比的意义, 用甲的速度:乙的速度, 化简, 即可解答。\n\n【详解】$(1\\div 5):(1\\div 7)\\\\=\\frac{1}{5}:\\frac{1}{7}\\\\=(\\frac{1}{5}\\times 35):(\\frac{1}{7}\\times 35)\\\\=7:5\\\\$ 走同样一段路程，甲走完需要5小时，乙走完需要7小时。甲、乙的速度之比是 $7:5$。",
+        "knowledge_points": ["比的化简", "比的基本性质", "比的意义", "基础行程问题"],
+        "source": "23-24六年级下·湖南岳阳·期末",
+        "merged_id": "unit4_129",
+        "original_id": "unit4_fill_129"
+      },
+      {
+        "id": "unit4_fill_130",
+        "unit": "第四单元：比",
+        "chapter": "比的意义",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "六(2)班男生人数是女生人数的 $\\frac{6}{7}$, 女生与全班人数的比是(), 女生比男生多$\\frac{()}{()}$。",
+        "options": [],
+        "answer": "$7:13; \\frac{1}{6}$",
+        "solution": "【分析】根据分数的意义，可知把女生人数看作7份，男生人数看作6份，全班人数一共有 $(6+7)$份，据此根据比的意义，写出女生与全班人数的比是 $7:(6+7)$；根据求一个数比另一个数多几分之几，用相差数除以另一个数，则用 $(7-6)\\div 6$即可求出女生比男生多几分之几。\n\n【详解】7: $(6+7)=7:13$ $(7-6)\\div 6\\\\=1\\div 6\\\\=\\frac{1}{6}$ 女生与全班人数的比是 $7:13$, 女生比男生多 $\\frac{1}{6}$。",
+        "knowledge_points": ["比的意义", "比与分数、除法的关系", "求一个数占另一个数几分之几"],
+        "source": "22-23 六年级上·浙江杭州·期中",
+        "merged_id": "unit4_130",
+        "original_id": "unit4_fill_130"
+      },
+      {
+        "id": "unit4_fill_131",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如果两个正方形的边长之比是 $4:5$, 那么它们的周长之比是(), 面积之比是()。",
+        "options": [],
+        "answer": "$4:5 \\quad 16:25$",
+        "solution": "【分析】正方形的周长$=\\text{边长}\\times 4$, 正方形的面积$=\\text{边长}\\times$边长, 可以假设小正方形边长为 $4\\mathrm{a}$, 大正方形边长为 $5\\mathrm{a}$, 进而求出周长比和面积比。\n\n【详解】假设小正方形边长为 $4\\mathrm{a}$, 大正方形边长为 $5\\mathrm{a}$, 小正方形周长: $4\\mathrm{a}\\times 4=16\\mathrm{a}$, 面积: $4\\mathrm{a}\\times 4\\mathrm{a}=16\\mathrm{a}^{2}$ 大正方形周长: $5\\mathrm{a}\\times 4=20\\mathrm{a}$, 面积: $5\\mathrm{a}\\times 5\\mathrm{a}=25\\mathrm{a}^{2}$那么它们的周长之比是(16a)：(20a) $=16:20=4:5$面积之比是：$(16a^{2})$：$(25a^{2})=16:25$所以两个正方形的边长之比是 $4:5$, 那么它们的周长之比是 $4:5$, 面积之比是 $16:25$。",
+        "knowledge_points": ["正方形的面积", "正方形的周长", "比的应用"],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_131",
+        "original_id": "unit4_fill_131"
+      },
+      {
+        "id": "unit4_fill_132",
+        "unit": "第四单元：比",
+        "chapter": "比与分数、除法的关系",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "$\\frac{2}{()}=\\frac{6}{21}=\\frac{()}{63}=4\\div()$",
+        "options": [],
+        "answer": "7; 18; 14",
+        "solution": "【分析】根据分数的基本性质, $\\frac{6}{21}$分子和分母都乘3就是 $\\frac{18}{63}$, $\\frac{6}{21}$分子和分母都除以3就是 $\\frac{2}{7}$, $\\frac{2}{7}$分子和分母都乘2就是 $\\frac{4}{14}$, 根据分数与除法的关系, $\\frac{4}{14}=4\\div 14$。\n\n【详解】$\\frac{6}{21}=\\frac{6\\times 3}{21\\times 3}=\\frac{18}{63}\\\\ \\frac{6}{21}=\\frac{6\\div 3}{21\\div 3}=\\frac{2}{7}\\\\ \\frac{2}{7}=\\frac{2\\times 2}{7\\times 2}=\\frac{4}{14}\\\\ \\frac{2}{(7)}=\\frac{6}{21}=\\frac{(18)}{63}=4\\div(14)\\\\$",
+        "knowledge_points": ["分数的基本性质", "比与分数、除法的关系"],
+        "source": "23-24 四年级下·山东泰安·期末",
+        "merged_id": "unit4_132",
+        "original_id": "unit4_fill_132"
+      },
+      {
+        "id": "unit4_fill_133",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "国旗的设计者曾联松是浙江瑞安人，我国《国旗法》规定，国旗的长与宽的比是 $3:2$。天安门广场的国旗是全国升降国旗中最大的，旗长为5米，宽应为()米；若学校选用的国旗宽是1.6米，则这面国旗的面积是()平方米。",
+        "options": [],
+        "answer": "$\\frac{10}{3}$ 3.84",
+        "solution": "【分析】将旗长5米除以对应的份数3份, 求出每份的长度,再乘宽的份数2份, 求出宽应为多少米。将学校用的国旗宽1.6米除以宽对应的份数2份, 求出每份的长度, 再乘长的份数3份, 求出长。根据'长方形面积$=$长$\\times$宽'列式求出这面国旗的面积。\n\n【详解】$5\\div 3\\times 2=\\frac{10}{3}$(米)$1.6\\div 2\\times 3=2.4$(米)$2.4\\times 1.6=3.84$(平方米)天安门广场的国旗是全国升降国旗中最大的，旗长为5米，宽应为 $\\frac{10}{3}$米；若学校选用的国旗宽是1.6米，则这面国旗的面积是3.84平方米。",
+        "knowledge_points": ["小数的乘、除法混合运算", "长方形的面积", "比的应用"],
+        "source": "23-24六年级下·浙江温州·期末",
+        "merged_id": "unit4_133",
+        "original_id": "unit4_fill_133"
+      },{
+        "id": "unit4_fill_134",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(23-24 五年级下·四川·单元测试) 五（1）班有男生 25人，女生 20 人，女生人数是男生人数的  $\\frac{()}{()}$ ，男生人数占全班人数的  $\\frac{()}{()}$ 。",
+        "options": [],
+        "answer": "$\\frac{4}{5}; \\frac{5}{9}$",
+        "solution": "【分析】求女生人数是男生人数的几分之几，用女生人数÷男生人数；求男生占全班人数的几分之几，用男生人数÷全班人数，据此解答。\n\n【详解】$20 \\div 25 = \\frac{4}{5}$\n\n$2 5 \\div (2 5 + 2 0) \\\\ = 2 5 \\div 4 5 \\\\ = \\frac {5}{9} \\\\ $\n\n五（1）班有男生25人，女生20人，女生人数是男生人数的  $\\frac{4}{5}$ ，男生人数占全班人数的  $\\frac{5}{9}$ 。",
+        "knowledge_points": [
+          "分数与除法的关系",
+          "约分的认识及应用",
+          "求一个数占另一个数几分之几"
+        ],
+        "source": "23-24 五年级下·四川·单元测试",
+        "merged_id": "unit4_134",
+        "original_id": "unit4_fill_134"
+      },
+      {
+        "id": "unit4_fill_135",
+        "unit": "第四单元：比",
+        "chapter": "比的意义",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(23-24 六年级上·山西吕梁·期末) 东汉名医张仲景在《金匮要略》中曾写道'茯苓四两，桂枝三两，白术三两，甘草二两'。这就是著名的苓桂术甘汤方。根据这个药方,请你写出两个比：（ ）、（ ）。",
+        "options": [],
+        "answer": "4:3   3:2",
+        "solution": "【分析】根据药方的数值来写比, 把其中一种药材的重量作比的前项，另一种药材的重量作比的后项，中间加上'：'即可。\n\n【详解】茯苓与桂枝的质量比是  $4: 3$白术与甘草的质量比是  $3: 2$因此根据这个药方，可写出两个比是  $4: 3$ 、 $3: 2$ 。",
+        "knowledge_points": [
+          "比的读法、写法及各部分的名称",
+          "比的意义"
+        ],
+        "source": "23-24 六年级上·山西吕梁·期末",
+        "merged_id": "unit4_135",
+        "original_id": "unit4_fill_135"
+      },
+      {
+        "id": "unit4_fill_136",
+        "unit": "第四单元：比",
+        "chapter": "比的意义",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "(23-24 六年级上·福建福州·期中) a 除以 b 的商是  $\\frac{3}{5}$ , a与 b 的比是( )。",
+        "options": [],
+        "answer": "3:5",
+        "solution": "【分析】根据比的意义可知:  $a \\div b = a:b$  。a 除以  $b$  的商是  $\\frac{3}{5}$ ,也就是  $a:b$  的比值是  $\\frac{3}{5}$ , 即  $a:b = \\frac{3}{5}$  。根据分数与比的关系可知:  $\\frac{3}{5} = 3:5$ , 所以  $a:b = 3:5$  。\n\n【详解】$a:b = a \\div b = \\frac{3}{5} = 3:5$所以，a与b的比是  $3: 5$  。",
+        "knowledge_points": [
+          "比的读法、写法及各部分的名称",
+          "比的意义",
+          "比与分数、除法的关系"
+        ],
+        "source": "23-24 六年级上·福建福州·期中",
+        "merged_id": "unit4_136",
+        "original_id": "unit4_fill_136"
+      },
+      {
+        "id": "unit4_fill_137",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（2024 六年级下·全国·专题练习）古代景德镇青花瓷随着海上丝绸之路的传播而为世界所瞩目。制造时，青花瓷外面的釉，所用的材料包括康纳瓦长石、石灰石和高岭土，其中康纳瓦长石和石灰石的比是  $^{3:2}$  。现有 24 吨康纳瓦长石，全部用来制造这种釉，需要( )吨石灰石。",
+        "options": [],
+        "answer": "16",
+        "solution": "【分析】根据题意, 康纳瓦长石和石灰石的比是  $3: 2$ , 则石灰石是康纳瓦长石的  $\\frac{2}{3}$ , 已知康纳瓦长石 24 吨, 求需要石灰石的重量, 把康纳瓦长石的总量看作单位 '1', 根据求一个数的几分之几是多少的计算方法, 用康纳瓦长石的重量  $\\times \\frac{2}{3}$ ,即可求出需要石灰石的重量。\n\n【详解】$24 \\times \\frac{2}{3} = 16$  （吨）古代景德镇青花瓷随着海上丝绸之路的传播而为世界所瞩目。制造时，青花瓷外面的釉，所用的材料包括康纳瓦长石、石灰石和高岭土，其中康纳瓦长石和石灰石的比是  $3: 2$  。现有 24 吨康纳瓦长石，全部用来制造这种釉，需要 16 吨石灰石。",
+        "knowledge_points": [
+          "求一个数的几分之几的问题",
+          "比的应用"
+        ],
+        "source": "2024 六年级下·全国·专题练习",
+        "merged_id": "unit4_137",
+        "original_id": "unit4_fill_137"
+      },
+      {
+        "id": "unit4_fill_138",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(23-24六年级上·山东德州·期中) 《周髀算经》中记载: 勾广三, 股修四, 径隅五。意思是说: 当直角三角形的两条直角边分别为 3 (勾) 和 4 (股) 时, 径隅 (弦) 则为 5 。后人简单地把这个事实说成 '勾三股四弦五'。一个直角三角形三条边的长度是 3:4:5, 斜边长是 25 厘米, 这个三角形的面积则是( )平方厘米。",
+        "options": [],
+        "answer": "150",
+        "solution": "【分析】直角三角形两直角边可以看作底和高，根据比的意义，斜边长  $\\div$  对应份数，求出一份数，一份数分别乘两直角边的对应份数，求出两直角边，根据三角形面积  $=$  底  $\\times$  高  $\\div 2$  ，列式计算即可。\n\n【详解】$25 \\div 5 = 5$  （厘米）$5 \\times 3 = 1 5 (\\text {厘 米})$  $5 \\times 4 = 20$  (厘米)  $15 \\times 20 \\div 2 = 150$  （平方厘米）这个三角形的面积是 150 平方厘米。",
+        "knowledge_points": [
+          "按比分配问题",
+          "三角形的面积"
+        ],
+        "source": "23-24六年级上·山东德州·期中",
+        "merged_id": "unit4_138",
+        "original_id": "unit4_fill_138"
+      },
+      {
+        "id": "unit4_fill_139",
+        "unit": "第四单元：比",
+        "chapter": "比的化简",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(23-24六年级上·河南漯河·期中) 2022年虎年春晚,以北宋画家王希孟的《千里江山图》为灵感创作的舞蹈节目《只此青绿》惊艳了观众，让我们感受到了中国传统文化的魅力。《千里江山图》是中国十大传世名画之一，长约  $12 \\mathrm{~m}$ , 宽约  $0.52 \\mathrm{~m}$ , 这幅画长与宽的比是( ) , 比值是( )。",
+        "options": [],
+        "answer": "$300: 13 \\quad \\frac{300}{13}$",
+        "solution": "【分析】由题意可知, 用这幅画的长比上宽, 再化简即可;用比的前项除以比的后项即可求出比值。\n\n【详解】12:0.52\n\n$= (1 2 \\div 0. 0 4): (0. 5 2 \\div 0. 0 4) \\\\ = 3 0 0: 1 3 \\\\ $\n\n$3 0 0: 1 3 \\\\ = 3 0 0 \\div 1 3 \\\\ = \\frac {3 0 0}{1 3} \\\\ $\n\n则这幅画长与宽的比是  $300: 13$ , 比值是  $\\frac{300}{13}$  。",
+        "knowledge_points": [
+          "比的化简",
+          "求比值",
+          "比的意义"
+        ],
+        "source": "23-24六年级上·河南漯河·期中",
+        "merged_id": "unit4_139",
+        "original_id": "unit4_fill_139"
+      },
+      {
+        "id": "unit4_fill_140",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(18-19 六年级上·全国·期末) 一个等腰三角形, 它的顶角与一个底角的度数比是  $2: 1$ , 这个三角形的顶角是( )°, 底角是( )°。",
+        "options": [],
+        "answer": "90      45",
+        "solution": "【分析】三角形内角和是  $180^{\\circ}$ , 等腰三角形的两个底角相等。根据题中等腰三角形顶角和底角的度数比可知, 顶角度数有 2 份, 底角有 1 份。那么用内角和  $180^{\\circ}$  除以  $(2 + 1 +$  1) 可求出一份的度数, 即底角的度数。将一份的度数再乘 2 , 即可求出顶角的度数。本题考查了比的应用以及等腰三角形的特征, 解题关键是根据比求出一份内角的度数。\n\n【详解】$180^{\\circ} \\div (2 + 1 + 1)$\n\n$= 1 8 0 ^ {\\circ} \\div 4 \\\\ = 4 5 ^ {\\circ} \\\\ $\n\n$4 5 ^ {\\circ} \\times 2 = 9 0 ^ {\\circ}$所以，这个三角形的顶角是  $90^{\\circ}$ ，底角是  $45^{\\circ}$ 。",
+        "knowledge_points": [
+          "三角形的内角和",
+          "等腰三角形和等边三角形的认识及特征",
+          "比的应用"
+        ],
+        "source": "18-19 六年级上·全国·期末",
+        "merged_id": "unit4_140",
+        "original_id": "unit4_fill_140"
+      },
+      {
+        "id": "unit4_fill_141",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(19-20 六年级下·北京·期中) 一个三角形三个角的度数比是  $2:3:5$ , 这是一个( )三角形。",
+        "options": [],
+        "answer": "直角",
+        "solution": "【分析】根据三角形的内角和是  $180^{\\circ}$ , 再由 '一个三角形三个内角度数的比是  $2: 3: 5$ ', 可知三角形的最大内角占内角和的  $\\frac{5}{2 + 3 + 5}$ , 然后根据乘法的意义求出最大角的度数, 再根据三角形的分类确定是什么三角形。\n\n【详解】$180^{\\circ} \\times \\frac{5}{2 + 3 + 5}\\\\= 1 8 0 ^ {\\circ} \\times \\frac {1}{2} \\\\ = 9 0 ^ {\\circ} \\\\ $\n\n有一个角是  $90^{\\circ}$  的三角形就是直角三角形, 所以这是一个直角三角形。\n\n【点睛】本题考查按比分配问题, 明确三角形的内角和是  $180^{\\circ}$  是解题的关键。",
+        "knowledge_points": [
+          "三角形的内角和",
+          "按比分配问题",
+          "三角形的分类"
+        ],
+        "source": "19-20 六年级下·北京·期中",
+        "merged_id": "unit4_141",
+        "original_id": "unit4_fill_141"
+      },
+      {
+        "id": "unit4_fill_142",
+        "unit": "第四单元：比",
+        "chapter": "比的化简",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(2022·重庆·小升初真题) 如果  $\\frac{5}{7} a = \\frac{3}{4} b$ , 则  $a:b=( )：( )$（填最简整数比），如果  $a + b = 150$ ，那么  $a \\div \\frac{3}{8} + b \\div \\frac{3}{8} = (\\quad)$ 。",
+        "options": [],
+        "answer": "21   20  400",
+        "solution": "【分析】假设  $\\frac{5}{7} a = \\frac{3}{4} b = 1$ , 根据互为倒数的两个数的乘积是 1 可知,  $a = \\frac{7}{5}, b = \\frac{4}{3}$ , 然后用  $a$  比上  $b$ , 再根据比的基本性质进行化简即可; 化除法为乘法, 把  $a \\div \\frac{3}{8} + b \\div \\frac{3}{8}$  化为  $a \\times \\frac{8}{3} + b \\times \\frac{8}{3}$ , 然后根据乘法分配律化为  $(a + b) \\times \\frac{8}{3}$ , 再把  $a + b = 150$  代入到式子  $(a + b) \\times \\frac{8}{3}$  中进行计算即可。本题考查比的意义, 熟练运用比的基本性质是解题的关键。\n\n【详解】假设  $\\frac{5}{7} a = \\frac{3}{4} b = 1$则  $a = \\frac{7}{5}, b = \\frac{4}{3}$\n\n$ a: b = \\frac {7}{5}: \\frac {4}{3} \\\\ = (\\frac {7}{5} \\times 1 5): (\\frac {4}{3} \\times 1 5) \\\\ = 2 1: 2 0 \\\\ $因为  $a + b = 150$\n\n则  $a \\div \\frac{3}{8} + b \\div \\frac{3}{8}$\n\n$ = a \\times \\frac {8}{3} + b \\times \\frac {8}{3} \\\\ = (a + b) \\times \\frac {8}{3} \\\\ = 1 5 0 \\times \\frac {8}{3} \\\\ = 4 0 0  $\n\n则如果  $\\frac{5}{7} a = \\frac{3}{4} b$ , 则  $a:b = 21:20$ , 如果  $a + b = 150$ , 那么  $a \\div \\frac{3}{8} + b \\div \\frac{3}{8} = 400$  。",
+        "knowledge_points": [
+          "分数乘法运算律",
+          "比的化简",
+          "比的意义",
+          "分数乘分数"
+        ],
+        "source": "2022·重庆·小升初真题",
+        "merged_id": "unit4_142",
+        "original_id": "unit4_fill_142"
+      },
+      {
+        "id": "unit4_judge_143",
+        "unit": "第四单元：比",
+        "chapter": "比的化简",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "(23-24 六年级上·全国·单元测试) 把 6:10 化简成最简整数比是  $0.6$ 。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】比的基本性质：比的前项和后项同时乘或除以相同的数（0 除外），比值不变。利用'比的基本性质'把比化简成最简整数比。\n\n【详解】$6:10\\\\= (6 \\div 2): (1 0 \\div 2) \\\\ = 3: 5  $\n\n把 6:10 化简成最简整数比是  $3:5$  。原题说法错误。故答案为:  ×",
+        "knowledge_points": [
+          "比的化简",
+          "比的基本性质"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_143",
+        "original_id": "unit4_judge_143"
+      },
+      {
+        "id": "unit4_judge_144",
+        "unit": "第四单元：比",
+        "chapter": "比的化简",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "(17-18 六年级上·全国·课后作业) 前项和后项都是整数的比, 是最简单的整数比。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】比的前项和后项互质，且都是整数，这样的比就是最简整数比。\n\n【详解】前项和后项都是整数的比，不一定是最简单的整数比，如4:6就不是最简整数比。故答案为：×",
+        "knowledge_points": [
+          "比的化简"
+        ],
+        "source": "17-18 六年级上·全国·课后作业",
+        "merged_id": "unit4_144",
+        "original_id": "unit4_judge_144"
+      },
+      {
+        "id": "unit4_judge_145",
+        "unit": "第四单元：比",
+        "chapter": "比的意义",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "(23-24 六年级上·全国·单元测试) 如果甲数与乙数的比是  $1: 2$ , 那么乙数: 甲数  $= 2: 1$  。 ( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】根据比的意义, 如果一个数与另一个数的比为  $a:b$ , 则另一个数与一个数的比为:  $b:a$ , 据此可计算得出答案。\n\n【详解】如果甲数与乙数的比是  $1: 2$ , 则乙数: 甲数  $= 2: 1$  。题干表述正确。故答案为: √",
+        "knowledge_points": [
+          "比的意义"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_145",
+        "original_id": "unit4_judge_145"
+      },
+      {
+        "id": "unit4_judge_146",
+        "unit": "第四单元：比",
+        "chapter": "比的意义",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "(23-24 六年级上·全国·单元测试) 一杯糖水, 糖占糖水的  $\\frac{7}{25}$ , 则糖与水的比是  $7:25$ 。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】已知糖占糖水的  $\\frac{7}{25}$ , 把糖的质量看作 7 份, 糖水的质量看作 25 份, 则水的质量是 (25 - 7) 份, 再根据比的意义得出糖与水的比, 据此判断。\n\n【详解】7: (25-7) = 7:18一杯糖水，糖占糖水的  $\\frac{7}{25}$ ，则糖与水的比是  $7: 18$ 。原题说法错误。故答案为:  ×",
+        "knowledge_points": [
+          "比的意义",
+          "比与分数、除法的关系"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_146",
+        "original_id": "unit4_judge_146"
+      },
+      {
+        "id": "unit4_judge_147",
+        "unit": "第四单元：比",
+        "chapter": "比的化简",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "(23-24 六年级上·全国·单元测试)  $3 \\mathrm{~m}: 21 \\mathrm{dm} = 1: 7$  。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】比的基本性质：比的前项和后项同时乘或除以相同的数（0 除外），比值不变。\n\n比的前项和后项的单位不统一，先根据进率 ' $1\\mathrm{m} = 10\\mathrm{dm}$ '换算单位，再利用'比的基本性质'把比化简成最简单的整数比。\n\n【详解】$3\\mathrm{m}:21\\mathrm{dm}\\\\= (3 \\times 1 0) d m: 2 1 d m \\\\ = 3 0: 2 1 \\\\ = (3 0 \\div 3): (2 1 \\div 3) \\\\ = 1 0: 7  $\n\n$3 m: 2 1 d m = 1 0: 7$原题说法错误。故答案为:  ×",
+        "knowledge_points": [
+          "比的化简",
+          "比的基本性质",
+          "毫米、厘米、分米、米之间的进率与换算"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_147",
+        "original_id": "unit4_judge_147"
+      },
+      {
+        "id": "unit4_judge_148",
+        "unit": "第四单元：比",
+        "chapter": "比的意义",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "（2023 六年级上·全国·专题练习）加工一批零件，如果甲、乙两人的工作效率之比是  $7: 6$  ，那么加工同一批零件他们所用的时间之比是  $7: 6$  。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】把加工这批零件的工作总量看作单位 '1', 已知甲、乙两人的工作效率之比是  $7: 6$ , 即甲的工作效率占 7 份, 乙的工作效率占 6 份; 根据 '工作时间 = 工作总量 ÷ 工作效率', 分别求出甲、乙加工同一批零件所需的时间, 再根据比的意义写出甲、乙的时间之比, 并化简比。\n\n【详解】甲的工作时间:  $1 \\div 7 = \\frac{1}{7}$,乙的工作时间:  $1 \\div 6 = \\frac{1}{6}$\n\n$\\frac {1}{7}: \\frac {1}{6} \\\\ = (\\frac {1}{7} \\times 4 2): (\\frac {1}{6} \\times 4 2) \\\\ = 6:7$\n\n加工同一批零件他们所用的时间之比是  $6: 7$  。原题说法错误。故答案为: ×",
+        "knowledge_points": [
+          "比的化简",
+          "比的意义"
+        ],
+        "source": "2023 六年级上·全国·专题练习",
+        "merged_id": "unit4_148",
+        "original_id": "unit4_judge_148"
+      },
+      {
+        "id": "unit4_judge_149",
+        "unit": "第四单元：比",
+        "chapter": "比的意义",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "(23-24 六年级下·四川凉山·期末) 在 100 克盐水中含有 1 克盐, 盐与盐水的比是  $1: 100$  。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】两数相除又叫两个数的比, 根据比的意义, 即可得出盐与盐水的比, 据此分析。\n\n【详解】在 100 克盐水中含有 1 克盐, 盐与盐水的比是  $1: 100$ , 说法正确。故答案为: √",
+        "knowledge_points": [
+          "比的意义"
+        ],
+        "source": "23-24 六年级下·四川凉山·期末",
+        "merged_id": "unit4_149",
+        "original_id": "unit4_judge_149"
+      },
+      {
+        "id": "unit4_judge_150",
+        "unit": "第四单元：比",
+        "chapter": "比的基本性质",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "(23-24六年级下·内蒙古呼伦贝尔·期末) 5:9 的后项若变成 90，要使比值不变，前项应加 45。（）",
+        "options": [],
+        "answer": "$\\sqrt{}$",
+        "solution": "【分析】根据比的基本性质，比的前项和后项，同时乘或除以相同的数（0 除外），比值不变，进行分析。\n\n【详解】$90 \\div 9 \\times 5 = 50$   $5 0 - 5 = 4 5$   5:9 的后项若变成 90, 要使比值不变, 前项应加 45 , 说法正确。故答案为: √",
+        "knowledge_points": [
+          "比的基本性质"
+        ],
+        "source": "23-24六年级下·内蒙古呼伦贝尔·期末",
+        "merged_id": "unit4_150",
+        "original_id": "unit4_judge_150"
+      },
+      {
+        "id": "unit4_judge_151",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "(23-24 六年级上·海南省直辖县级单位·期末) 六 (6)班男生人数与女生人数的比是  $5: 3$ , 则女生比男生少  $\\frac{2}{5}$  。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】根据男生和女生的人数比可知，男生人数有 5 份，女生人数有 3 份，女生比男生少 2 份。将女生比男生少的份数除以男生的份数，求出女生比男生少几分之几。\n\n【详解】$(5 - 3) \\div 5\\\\= 2 \\div 5 \\\\ = \\frac {2}{5}  $\n\n所以，女生比男生少  $\\frac{2}{5}$  。故答案为:  √",
+        "knowledge_points": [
+          "分数与除法的关系",
+          "比的意义",
+          "求一个数占另一个数几分之几",
+          "比的应用"
+        ],
+        "source": "23-24 六年级上·海南省直辖县级单位·期末",
+        "merged_id": "unit4_151",
+        "original_id": "unit4_judge_151"
+      },
+      {
+        "id": "unit4_judge_152",
+        "unit": "第四单元：比",
+        "chapter": "比的意义",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "(2014 六年级·全国·课后作业) 某场世界杯两球队的比分是  $3: 0$ , 所以比的后项可以是 0 。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】两个数相除又叫做两个数的比。在两个数的比中，比号前面的数叫做比的前项，比号后面的数叫做比的后项。比一般分为两种情况：一种是同类数量的比，表示一个数是另一个数的几倍或几分之几；另一种是两个不同类的量的比。据此解答。\n\n【详解】比的后项相当于除法中的除数，而除数不能为0，比的后项也不能为0。但在足球、排球等体育比赛中，比分的后项可以是0，因为这个比是体现双方得分的多少，所以原题说法错误。故答案为:  $\\times$",
+        "knowledge_points": [
+          "比的意义"
+        ],
+        "source": "2014 六年级·全国·课后作业",
+        "merged_id": "unit4_152",
+        "original_id": "unit4_judge_152"
+      },
+      {
+        "id": "unit4_judge_153",
+        "unit": "第四单元：比",
+        "chapter": "比的意义",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "(2014 六年级·全国·课后作业) 比的后项不能是 0 。()",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】两个数相除又叫做两个数的比。比是一种数量关系，相同于除法、分数，但除法是一种运算，分数是一个数，这就是它们的区别；所以比的后项相当于除法中的除数，零不能作除数，所以比的后项也不能为0。\n\n【详解】由比的意义可知: 比的后项也不能为 0 , 所以说法正确;故答案为:  √",
+        "knowledge_points": [
+          "比的意义"
+        ],
+        "source": "2014 六年级·全国·课后作业",
+        "merged_id": "unit4_153",
+        "original_id": "unit4_judge_153"
+      },
+      {
+        "id": "unit4_judge_154",
+        "unit": "第四单元：比",
+        "chapter": "比的基本性质",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "(21-22 六年级上·河北邯郸·期末) 比的前项扩大到原来的 2 倍, 后项缩小到原来的  $\\frac{1}{3}$ , 比值扩大到原来的 6 倍。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】根据比与除法的关系和商的变化规律可知，比的前项扩大到原来的 2 倍，即前项（被除数）乘 2，则比值(商) 也要乘 2 ; 后项缩小到原来的  $\\frac{1}{3}$ , 即后项 (除数) 除以 3 , 则比值 (商) 反而乘 3 ; 也就是比值扩大到原来的  $(2 \\times 3)$  倍。据此解答。\n\n【详解】由分析可知， $2 \\times 3 = 6$即比的前项扩大到原来的 2 倍, 后项缩小到原来的  $\\frac{1}{3}$ , 比值扩大到原来的 6 倍。说法正确。故答案为:  √",
+        "knowledge_points": [
+          "商的变化规律及应用",
+          "比与分数、除法的关系"
+        ],
+        "source": "21-22 六年级上·河北邯郸·期末",
+        "merged_id": "unit4_154",
+        "original_id": "unit4_judge_154"
+      },
+      {
+        "id": "unit4_judge_155",
+        "unit": "第四单元：比",
+        "chapter": "求比值",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "(22-23 六年级上·河南周口·期末) 4 分米：20 厘米 =40 厘米：20 厘米 = 2 厘米。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】求比值直接用比的前项÷后项，求比值的结果是一个数，不带单位名称，据此分析。\n\n【详解】4分米：20厘米  $= 40$  厘米：20厘米  $= 40 \\div 20 = 2$故答案为:×",
+        "knowledge_points": [
+          "求比值"
+        ],
+        "source": "22-23 六年级上·河南周口·期末",
+        "merged_id": "unit4_155",
+        "original_id": "unit4_judge_155"
+      },
+      {
+        "id": "unit4_judge_156",
+        "unit": "第四单元：比",
+        "chapter": "比的意义",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "(22-23 六年级上·湖北省直辖县级单位·期末) 2022 年卡塔尔世界杯足球比赛中, 阿根廷球队以  $2: 0$  战胜了墨西哥球队, 所以比的后项可以是 0 。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】两个数相除又叫做两个数的比。在两个数的比中，比号前面的数叫做比的前项，比号后面的数叫做比的后项。比一般分为两种情况：一种是同类数量的比，表示一个数是另一个数的几倍或几分之几；另一种是两个不同类的量的比。据此解答。\n\n【详解】比的后项相当于除法中的除数，而除数不能为0，比的后项也不能为0。但在足球、排球等体育比赛中，比分的后项可以是0，因为这个比是体现双方得分的多少。故答案为:  ×",
+        "knowledge_points": [
+          "比的读法、写法及各部分的名称"
+        ],
+        "source": "22-23 六年级上·湖北省直辖县级单位·期末",
+        "merged_id": "unit4_156",
+        "original_id": "unit4_judge_156"
+      },
+      {
+        "id": "unit4_judge_157",
+        "unit": "第四单元：比",
+        "chapter": "比的基本性质",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "(19-20 六年级上·四川·单元测试) 比的前项和后项同时乘或除以非 0 的数, 比值不变。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】根据比的基本性质，比的前项和后项同时乘或除以一个相同的数（0 除外），比值不变，据此举例判断即可。\n\n【详解】题干中并没有说明比的前项和后项同时除以一个相同的数, 如  $2: 3 = \\frac{2}{3}$ , 若比的前项除以 2 , 比的后项除以 3,  $(2 \\div 2): (3 \\div 3) = 1: 1 = 1$ , 此时的比值发生改变, 原题干说法错误。故答案为:  ×",
+        "knowledge_points": [
+          "比的基本性质"
+        ],
+        "source": "19-20 六年级上·四川·单元测试",
+        "merged_id": "unit4_157",
+        "original_id": "unit4_judge_157"
+      },
+      {
+        "id": "unit4_judge_158",
+        "unit": "第四单元：比",
+        "chapter": "比与分数、除法的关系",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "(22-23 六年级上·山东德州·期中) 比和分数都可以用分数的形式表示, 但是分数并不一定表示两个数量的比。()",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】比表示两个量之间的倍比关系，分数是一个数，一个具体的数量，则比可以用分数进行表示，但分数并不一定表示两个数量的比。\n\n【详解】由分析可知:比和分数都可以用分数的形式表示, 但是分数并不一定表示两个数量的比。原说法正确。故答案为:  √",
+        "knowledge_points": [
+          "比与分数、除法的关系"
+        ],
+        "source": "22-23 六年级上·山东德州·期中",
+        "merged_id": "unit4_158",
+        "original_id": "unit4_judge_158"
+      },
+      {
+        "id": "unit4_judge_159",
+        "unit": "第四单元：比",
+        "chapter": "比的基本性质",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "(23-24 六年级上·湖北荆州·期中) 比的前项和后项同时乘 (或除以) 相同的数, 比值不变。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】比的性质需强调，前项和后项同时乘或除以同一个不为 0 的数，比值不变。据此判断。\n\n【详解】比的前项和后项同时乘（或除以）相同的数（0 除外），比值不变。故答案为:  ×",
+        "knowledge_points": [
+          "比的基本性质"
+        ],
+        "source": "23-24 六年级上·湖北荆州·期中",
+        "merged_id": "unit4_159",
+        "original_id": "unit4_judge_159"
+      },
+      {
+        "id": "unit4_judge_160",
+        "unit": "第四单元：比",
+        "chapter": "比的化简",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "(20-21 六年级上·全国·单元测试) 甲数的  $\\frac{2}{3}$  与乙数的  $\\frac{3}{4}$  相等, (甲、乙均不为 0 ), 则甲、乙两数比是 $9: 8$ 。 ()",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】求一个数的几分之几是多少，用这个数乘分率。假设甲数是 3，将 3 乘  $\\frac{2}{3}$  求出甲数的  $\\frac{2}{3}$  是多少，即乙数的  $\\frac{3}{4}$  是多少。将乙数看作单位 '1'，单位 '1'未知，用除法求出乙数。根据比的意义，求出甲数和乙数的比即可。\n\n【详解】令甲数是3，则乙数为：\n\n$ 3 \\times \\frac {2}{3} \\div \\frac {3}{4} \\\\ = 2 \\times \\frac {4}{3} \\\\ = \\frac {8}{3} \\\\ $\n\n$ 3: \\frac {8}{3} \\\\ = (3 \\times 3): (\\frac {8}{3} \\times 3) \\\\ = 9: 8 \\\\ $\n\n√所以，甲、乙两数的比是  $9: 8$  。故答案为: √",
+        "knowledge_points": [
+          "分数与整数的除法",
+          "比的化简",
+          "比的意义",
+          "分数与整数的乘法"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_160",
+        "original_id": "unit4_judge_160"
+      },
+      {
+        "id": "unit4_judge_161",
+        "unit": "第四单元：比",
+        "chapter": "比的基本性质",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "（2020·河北保定·小升初真题）在比中，比的前项和后项同时乘以或者除以一个数，比值不变。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】根据比的基本性质：比的前项和后项同时乘或除以一个不为零的数，比值不变，据此解答。\n\n【详解】根据分析可知，在比中，比的前项和后项同时乘或除以一个不为零的数，比值不变。原题干说法错误。故答案为:  ×",
+        "knowledge_points": [
+          "比的基本性质"
+        ],
+        "source": "2020·河北保定·小升初真题",
+        "merged_id": "unit4_161",
+        "original_id": "unit4_judge_161"
+      },
+      {
+        "id": "unit4_judge_162",
+        "unit": "第四单元：比",
+        "chapter": "比的基本性质",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "(23-24 六年级上·湖北省直辖县级单位·期中) 甲数比乙数少乙数的  $\\frac{3}{7}$ , 乙数与丙数的比是  $5: 6$ , 则甲数与丙数的比是  $10: 21$ . ( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】根据甲数比乙数少  $\\frac{3}{7}$ , 求出甲数与乙数的比, 已知乙数与丙数的比, 根据比的基本性质, 写出三个数的连比,进而确定甲数与丙数的比。此题考查了比的意义和性质, 先求出甲、乙、丙三数的比是解题的关键。\n\n【详解】把乙数看作单位'1'，则甲数是  $(1 - \\frac{3}{7})$  ，甲数与乙数的比是  $(1 - \\frac{3}{7}): 1$ , 化简是  $4: 7$ , 乙数与丙数的比是  $5: 6$ , 甲数: 乙数: 丙数  $= 20: 35: 42$ ; 甲数与丙数的比是  $20: 42$ , 化简得  $10: 21$  。故答案为:√",
+        "knowledge_points": [
+          "比的基本性质",
+          "比的意义"
+        ],
+        "source": "23-24 六年级上·湖北省直辖县级单位·期中",
+        "merged_id": "unit4_162",
+        "original_id": "unit4_judge_162"
+      },
+      {
+        "id": "unit4_judge_163",
+        "unit": "第四单元：比",
+        "chapter": "比的意义",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "(2014 六年级·全国·课后作业) 一场足球比赛的比分是  $2: 0$ , 所以比的后项可以为零。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】比的意义是两个数相除又叫做两个数的比, 比是表示两个数相除。比的前项相当于除法算式中被除数, 比的后项相当于除法算式中的除数, 除数不能为 0 。据此解答。本题要重点理解比的意义, 知道比是表示两个数的关系。\n\n【详解】例如：  $3 \\div 5 = 3: 5$  。3:5表示3和5相除，是表示两个数的关系。一场足球比赛的比分是  $2: 0$ , 比分代表的是具体的数值。比分可以为 0 , 比的后项不能为 0 。故答案为: ×",
+        "knowledge_points": [
+          "比的意义"
+        ],
+        "source": "2014 六年级·全国·课后作业",
+        "merged_id": "unit4_163",
+        "original_id": "unit4_judge_163"
+      },
+      {
+        "id": "unit4_calc_164",
+        "unit": "第四单元：比",
+        "chapter": "求比值",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "(23-24 六年级上·全国·单元测试) 求下面各比的比值。\n\n15:75    5:  $\\frac{15}{4}$      $\\frac{5}{8}:\\frac{15}{24}$    $0.8: 3.2$   $\\frac{4}{5}: 0.2$     $\\frac{8}{5}: 40$",
+        "options": [],
+        "answer": "$\\frac{1}{5}; \\frac{4}{3}; 1$     $\\frac{1}{4}; 4; \\frac{1}{25}$",
+        "solution": "【分析】用比的前项除以比的后项所得的商，叫做比值。\n\n【详解】(1)  $15: 75\\\\ = 1 5 \\div 7 5 \\\\ = \\frac {1}{5} \\\\ $\n\n(2)  $5: \\frac{15}{4}\\\\= 5 \\div \\frac {1 5}{4} \\\\ = 5 \\times \\frac {4}{1 5} \\\\ = \\frac {4}{3} \\\\ $\n\n(3)  $\\frac{5}{8}: \\frac{15}{24}\\\\= \\frac {5}{8} \\div \\frac {1 5}{2 4} \\\\ = \\frac {5}{8} \\times \\frac {2 4}{1 5} \\\\ = 1 \\\\ $\n\n(4)  $0.8: 3.2\\\\= 0. 8 \\div 3. 2 \\\\ = \\frac {1}{4} \\\\ $\t\n\n(5)  $\\frac{4}{5}: 0.2\\\\= 0. 8 \\div 0. 2\\\\= 4$\n\n(6)  $\\frac{8}{5}: 40\\\\= \\frac {8}{5} \\div 4 0 \\\\ = \\frac {8}{5} \\times \\frac {1}{4 0} \\\\ = \\frac {1}{2 5} \\\\ $",
+        "knowledge_points": [
+          "求比值"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_164",
+        "original_id": "unit4_calc_164"
+      },
+      {
+        "id": "unit4_calc_165",
+        "unit": "第四单元：比",
+        "chapter": "比的化简",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "(23-24 六年级上·全国·单元测试) 把下面各比化成最简单的整数比。\n\n$\\frac {2}{9}: \\frac {5}{1 8}$\n\n$7 5: 0. 2 5$\n\n$\\frac {1}{4}: 0. 3 7 5$\n\n$\\frac {2}{5} \\mathrm {k m}: 2 4 0 \\mathrm {m}$\n\n$\\frac {2}{3} \\text {小 时}: 5 0 \\text {分 钟}$\n\n$0.2 5: \\frac {1}{3}$",
+        "options": [],
+        "answer": "4:5; 3:1; 2:3；5: 3; 4: 5; 3: 4",
+        "solution": "【分析】根据比的基本性质可知，比的前项和后项同时乘或除以相同的数（0 除外），比值不变。据此化简比。单位不统一的先统一单位后，再化简比。\n\n【详解】$\\frac{2}{9}:\\frac{5}{18}$\n\n$ = (\\frac {2}{9} \\times 1 8): (\\frac {5}{1 8} \\times 1 8) \\\\ = 4: 5 \\\\ $\n\n$0.7 5: 0. 2 5 \\\\ = (0. 7 5 \\div 0. 2 5): (0. 2 5 \\div 0. 2 5) \\\\ = 3: 1 \\\\ $\n\n$ \\frac {1}{4}: 0. 3 7 5 \\\\ = \\frac {1}{4}: \\frac {3}{8} \\\\ = (\\frac {1}{4} \\times 8): (\\frac {3}{8} \\times 8) \\\\ = 2: 3 \\\\ $\n\n$ \\frac {2}{5} \\mathrm {k m}: 2 4 0 \\mathrm {m} \\\\ = 4 0 0 \\mathrm {m}: 2 4 0 \\mathrm {m} \\\\ = 4 0 0: 2 4 0 \\\\ = (4 0 0 \\div 8 0): (2 4 0 \\div 8 0) \\\\ = 5: 3 \\\\ $\n\n$\\frac{2}{3}  小时：50 分钟=40分钟：50分钟= 4 0: 5 0 \\\\ = (4 0 \\div 1 0): (5 0 \\div 1 0) \\\\ = 4: 5 \\\\ $\n\n$0.2 5: \\frac {1}{3} \\\\ = \\frac {1}{4}: \\frac {1}{3} \\\\ = (\\frac {1}{4} \\times 1 2): (\\frac {1}{3} \\times 1 2) \\\\ = 3: 4 \\\\ $",
+        "knowledge_points": [
+          "比的化简",
+          "比的基本性质"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_165",
+        "original_id": "unit4_calc_165"
+      },
+      {
+        "id": "unit4_calc_166",
+        "unit": "第四单元：比",
+        "chapter": "比的化简和求比值",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "(22-23 六年级上·江苏盐城·期中) 化简下列各比并求比值。\n\n(1)  $81: 9$\n\n(2)  $\\frac{5}{7}: \\frac{1}{14}$\n\n(3)  $0.125: \\frac{1}{4}$\n\n(4) 2 千克:50克",
+        "options": [],
+        "answer": "(1)  $9: 1; 9$(2)  $10: 1; 10$  (3)  $1: 2; 0.5$  (4)  $40: 1; 40$",
+        "solution": "【分析】比的基本性质：比的前项和后项同时乘或除以相同的数（0 除外），比值不变。如果比的前项和后项的单位不统一，先根据进率换算单位，再利用'比的基本性质'把比化简成最简单的整数比。用比的前项除以比的后项所得的商，叫做比值。根据求比值的方法，用最简比的前项除以比的后项即得比值。\n\n【详解】\n\n(1)  $81: 9\\\\= (8 1 \\div 9): (9 \\div 9) \\\\ = 9: 1 $\n\n$9:1\\\\= 9 \\div 1 \\\\ = 9 \\\\ $\n\n(2)  $\\frac{5}{7}: \\frac{1}{14}$\n\n$= (\\frac {5}{7} \\times 1 4): (\\frac {1}{1 4} \\times 1 4)$\n\n$ = 1 0: 1 \\\\ 1 0: 1 \\\\ = 1 0 \\div 1 \\\\ = 1 0 $\n\n(3)  $0.125:\\frac{1}{4}$\n\n$= (0. 1 2 5 \\times 8): (\\frac {1}{4} \\times 8) \\\\ = 1: 2  $\n\n$1:2\\\\= 1 \\div 2 \\\\ = 0. 5 \\\\ $\n\n(4)$ 2千克：50克\\\\= 2 0 0 0 \\text {克}: 5 0 \\text {克} \\\\ = (2 0 0 0 \\div 5 0): (5 0 \\div 5 0) \\\\ = 4 0: 1 \\\\ $\n\n$4 0: 1 \\\\ = 4 0 \\div 1 \\\\ = 4 0$",
+        "knowledge_points": [
+          "比的化简",
+          "求比值"
+        ],
+        "source": "22-23 六年级上·江苏盐城·期中",
+        "merged_id": "unit4_166",
+        "original_id": "unit4_calc_166"
+      },
+      {
+        "id": "unit4_calc_167",
+        "unit": "第四单元：比",
+        "chapter": "比的化简和求比值",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "(22-23 六年级上·江苏盐城·期中) 化简比并求比值。\n\n$\\frac{1}{3}: \\frac{2}{5}$     $0.15: \\frac{3}{4}$     36分：1时",
+        "options": [],
+        "answer": "$5:6;\\frac{5}{6};1:5;\\frac{1}{5};3:5;\\frac{3}{5}$",
+        "solution": "【分析】比的基本性质：比的前项和后项同时乘或除以相同的数（0 除外），比值不变。\n\n如果比的前项和后项的单位不统一，先根据进率换算单位，再利用比的基本性质把比化简成最简单的整数比。\n\n用比的前项除以比的后项所得的商，叫做比值。根据求比值的方法，用最简比的前项除以比的后项即得比值。\n\n【详解】$\\frac{1}{3}:\\frac{2}{5}\\\\= (\\frac {1}{3} \\times 1 5): (\\frac {2}{5} \\times 1 5) \\\\= 5: 6 \\\\ 5: 6 \\\\ = 5 \\div 6 \\\\ = \\frac {5}{6} \\\\ $\n\n$0.1 5: \\frac {3}{4} \\\\ = (0. 1 5 \\times 1 0 0 \\div 1 5): (\\frac {3}{4} \\times 1 0 0 \\div 1 5) \\\\ = 1: 5 $\n\n$ 1: 5 \\\\ = 1 \\div 5 \\\\ = \\frac {1}{5} \\\\ $\n\n$36分:1时\\\\= 3 6 \\text {分}: 6 0 \\text {分} \\\\ = (3 6 \\div 1 2): (6 0 \\div 1 2) \\\\ = 3: 5 \\\\ $\n\n$ 3: 5 \\\\ = 3 \\div 5 \\\\ = \\frac {3}{5}$",
+        "knowledge_points": [
+          "比的化简",
+          "求比值",
+          "比的基本性质",
+          "时、分的认识及换算"
+        ],
+        "source": "22-23 六年级上·江苏盐城·期中",
+        "merged_id": "unit4_167",
+        "original_id": "unit4_calc_167"
+      },
+      {
+        "id": "unit4_calc_168",
+        "unit": "第四单元：比",
+        "chapter": "求比值",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "(23-24 六年级上·辽宁·单元测试) 求下列比的比值。\n\n$\\frac {7}{2 0}: \\frac {1}{4}$   0.25:1    48分：1.2时     $\\frac {1}{5}: \\frac {1}{6}$",
+        "options": [],
+        "answer": "$\\frac{7}{5}; \\frac{1}{4}; \\frac{2}{3}; \\frac{6}{5}$",
+        "solution": "【分析】根据比值  $=$  比的前项  $\\div$  比的后项，代入数据计算，即可解答。\n\n【详解】$\\frac{7}{20}: \\frac{1}{4}$\n\n$ 6 \\frac {7}{2 0} \\div \\frac {1}{4} \\\\ =\\frac {7}{2 0} \\times 4 \\\\ = \\frac {7}{5}  $\n\n$0.2 5: 1 \\\\ = 0. 2 5 \\div 1 \\\\ = \\frac {1}{4} \\\\ $\n\n$48分：1.2时\\\\=48分：72分\\\\= 48 分  \\div 72  分\\\\= \\frac{2}{3}$ \n\n$\\frac{1}{5}: \\frac{1}{6} \\\\ =\\frac{1}{5} \\div \\frac{1}{6} \\\\ = \\frac{1}{5} \\times 6 \\\\=  \\frac{6}{5}$",
+        "knowledge_points": [
+          "求比值"
+        ],
+        "source": "23-24 六年级上·辽宁·单元测试",
+        "merged_id": "unit4_168",
+        "original_id": "unit4_calc_168"
+      },
+      {
+        "id": "unit4_solve_169",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "(23-24六年级上·全国·单元测试) 学校进行大扫除,老师把打扫校园的任务按 2:3:4 的比分配给四、五、六年级，六年级比四年级多分到180平方米。五年级分到多少平方米？",
+        "options": [],
+        "answer": "270 平方米",
+        "solution": "【分析】把校园的总面积看作单位'1'，根据题意，老师把打扫校园的任务按2:3:4的比分配给四、五、六年级；六年级占总任务的  $\\frac{4}{2 + 3 + 4}$ ，五年级占总任务的  $\\frac{3}{2 + 3 + 4}$ ，四年级占总任务的  $\\frac{2}{2 + 3 + 4}$ ，六年级分配的任务比四年级分配的任务多  $\\frac{4}{2 + 3 + 4} - \\frac{2}{2 + 3 + 4}$ ，对应的是180平方米，求单位'1'，用  $180 \\div (\\frac{4}{2 + 3 + 4} - \\frac{2}{2 + 3 + 4})$ ，求出总任务，五年级占总任务的  $\\frac{3}{2 + 3 + 4}$ ，再用总任务  $\\times \\frac{3}{2 + 3 + 4}$ ，即可求出五年级分到多少平方米。\n\n【详解】$180 \\div (\\frac{4}{2 + 3 + 4} - \\frac{2}{2 + 3 + 4}) \\times \\frac{3}{2 + 3 + 4}$\n\n$ = 1 8 0 \\div (\\frac {4}{9} - \\frac {2}{9}) \\times \\frac {3}{9} \\\\ = 1 8 0 \\div \\frac {2}{9} \\times \\frac {3}{9} \\\\ = 1 8 0 \\times \\frac {9}{2} \\times \\frac {3}{9} \\\\ = 2 7 0 \\text {(平 方 米)} \\\\ $\n\n答: 五年级分到 270 平方米。",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "23-24六年级上·全国·单元测试",
+        "merged_id": "unit4_169",
+        "original_id": "unit4_solve_169"
+      },
+      {
+        "id": "unit4_solve_170",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "(23-24六年级上·全国·单元测试) 一个直角三角形,三条边的长度比是  $3: 4: 5$ , 已知最短边和最长边的长度和是48 厘米。这个三角形的面积是多少平方厘米？",
+        "options": [],
+        "answer": "216平方厘米",
+        "solution": "【分析】已知直角三角形三条边的长度比是  $3: 4: 5$ , 那么最短边占 3 份, 最长边占 5 份, 一共占  $(3 + 5)$  份; 用最短边和最长边的长度和除以  $(3 + 5)$  份, 求出一份数, 再用一份数分别乘 3、乘 4, 求出直角三角形的两条直角边长, 也就是这个直角三角形的底和高, 最后根据三角形的面积  $=$  底  $\\times$  高  $\\div 2$ , 求出三角形的面积。\n\n【详解】一份数：\n\n$4 8 \\div (3 + 5) \\\\ = 4 8 \\div 8 \\\\ = 6 \\text {(厘 米)} \\\\ $\n\n两条直角边分别是:$6 \\times 3 = 18$  (厘米) $6 \\times 4 = 24$  (厘米)  三角形的面积：$18 \\times 24 \\div 2  = 432 \\div 2 =216 (平方厘米)$答: 这个三角形的面积是 216 平方厘米。",
+        "knowledge_points": [
+          "三角形面积的应用",
+          "比的应用"
+        ],
+        "source": "23-24六年级上·全国·单元测试",
+        "merged_id": "unit4_170",
+        "original_id": "unit4_solve_170"
+      },
+      {
+        "id": "unit4_solve_171",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "(23-24六年级上·全国·单元测试)六（1）班举办元旦晚会，共有500元活动经费， $\\frac{1}{20}$ 用于购买气球，余下的按7:8:4的比例分别购买苹果、橘子和瓜子。购买橘子花了多少钱？",
+        "options": [],
+        "answer": "200元",
+        "solution": "【分析】将活动经费看作单位 '1', 则购买气球后余下的钱运用分数乘法计算得出; 根据按比分配原则, 的按  $7: 8: 4$  的比例分别购买苹果、橘子和瓜子, 则总的份数为:  $7 + 8 + 4 = 19$ , 根据按比分配原则可计算得出答案。\n\n【详解】活动经费购买气球后余下的经费为:\n\n$5 0 0 \\times (1 - \\frac {1}{2 0}) \\\\ =5 0 0 \\times \\frac {1 9}{2 0} \\\\=  \\text {4 7 5 (元)}  \\\\ $\n\n则购买橘子花的钱为：\n\n$ 4 7 5 \\times \\frac {8}{(7 + 8 + 4)} \\\\ =4 7 5 \\times \\frac {8}{1 9} \\\\ = 2 0 0 \\quad (\\text {元}) \\\\ $\n\n答: 购买橘子花了 200 元。",
+        "knowledge_points": [
+          "求一个数的几分之几的问题",
+          "按比分配问题"
+        ],
+        "source": "23-24六年级上·全国·单元测试",
+        "merged_id": "unit4_171",
+        "original_id": "unit4_solve_171"
+      },
+      {
+        "id": "unit4_solve_172",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "(23-24 六年级上·全国·单元测试) 七彩书店一天共卖出故事书和科普书 260 本, 已知卖出的故事书和科普书的比是8:5，这一天卖出的故事书比科普书多多少本？",
+        "options": [],
+        "answer": "60 本",
+        "solution": "【分析】已知卖出的故事书和科普书的比是  $8: 5$ , 把故事书的本数看作 8 份, 科普书的本数看作 5 份, 一共是  $(8 + 5)$  份, 故事书比科普书多 (8-5) 份;先用故事书和科普书的总本数除以总份数，求出一份数，再用一份数乘故事书比科普书多的份数，即可求出这一天卖出的故事书比科普书多的本数。\n\n【详解】一份数：\n\n$ 2 6 0 \\div (8 + 5) \\\\ = 2 6 0 \\div 1 3 \\\\ = 2 0 \\text {(本)} \\\\ $\n\n故事书比科普书多：\n\n$ 2 0 \\times (8 - 5) \\\\ = 2 0 \\times 3 \\\\ = 6 0 \\text {(本)} \\\\ $\n\n答: 这一天卖出的故事书比科普书多 60 本。",
+        "knowledge_points": [
+          "比的应用"
+        ],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_172",
+        "original_id": "unit4_solve_172"
+      },
+      {
+        "id": "unit4_solve_173",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "(23-24六年级下·湖南常德·期末）我国古代数学巨著《周髀算经》提到'勾三、股四、弦五'，其中的含义是如果一个三角形中三条边的比满足  $3:4:5$ , 则这个三角形是一个直角三角形。已知一个直角三角形的周长是36厘米，则这个直角三角形的面积是多少平方厘米？",
+        "options": [],
+        "answer": "54平方厘米",
+        "solution": "【分析】根据比的意义, 将三角形的周长 36 厘米除以 (3+4+5) , 求出一份边的长度, 从而利用乘法求出其中两条直角边的长度。根据三角形面积  $= \\text{底} \\times \\text{高} \\div 2$ , 列式求出这个三角形的面积即可。\n\n【详解】$36 \\div (3 + 4 + 5)\\\\= 3 6 \\div 1 2\\\\=3 (厘米)  $\n\n$3 \\times 3 = 9$  (厘米)     $4 \\times 3 = 12$  (厘米)      $9 \\times 12 \\div 2 = 108 \\div 2 =54 (平方厘米)$答: 这个三角形的面积是 54 平方厘米。",
+        "knowledge_points": [
+          "三角形面积的计算",
+          "按比分配问题"
+        ],
+        "source": "23-24六年级下·湖南常德·期末",
+        "merged_id": "unit4_173",
+        "original_id": "unit4_solve_173"
+      },
+      {
+        "id": "unit4_solve_174",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（23-24六年级下·四川凉山·期末）某校合唱队原来男、女生人数比是5:3，后来又增加了14名女生，这时男、女生人数比是6:5，原来合唱队一共有多少人？",
+        "options": [],
+        "answer": "96名",
+        "solution": "【分析】根据题意可知, 原来女生人数是男生人数的  $\\frac{3}{5}$ , 增加 14 名女生后, 女生人数是男生人数的  $\\frac{5}{6}$ , 先求出女生人数增加的比例, 即  $\\frac{5}{6} - \\frac{3}{5}$ , 再用增加的女生人数除以求出的女生人数增加的比例, 计算出男生人数, 把男生人数看作单位 '1', 即原来总人数为  $1 + \\frac{3}{5}$ , 用男生人数乘上  $1 + \\frac{3}{5}$  。据此求出答案。\n\n【详解】原来女生人数是男生人数的  $\\frac{3}{5}$ , 增加 14 名女生后,女生人数是男生人数的  $\\frac{5}{6}$  。\n\n$ \\frac {5}{6} - \\frac {3}{5} \\\\ = \\frac {2 5}{3 0} - \\frac {1 8}{3 0} \\\\ = \\frac {7}{3 0} $\n\n男生人数:  $14 \\div \\frac{7}{30}$\n\n$= 1 4 \\times \\frac {3 0}{7} \\\\ = 6 0 \\text {(名)} \\\\ $\n\n原来总人数:  $(1 + \\frac{3}{5}) \\times 60$\n\n$= \\frac {8}{5} \\times 6 0 \\\\ = 9 6 \\text {(名)} \\\\ $\n\n答: 原来合唱队一共有 96 人。",
+        "knowledge_points": [
+          "分数与整数的除法",
+          "分数的四则混合运算",
+          "比的应用"
+        ],
+        "source": "23-24六年级下·四川凉山·期末",
+        "merged_id": "unit4_174",
+        "original_id": "unit4_solve_174"
+      },
+      {
+        "id": "unit4_solve_175",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "(2024 六年级下·全国·专题练习) 一个工厂有三个车间,全厂男、女职工人数之比是  $9: 5$ , 三个车间的人数之比是8:9:11，第一车间男、女职工人数之比是  $3: 1$ ；第二车间男女职工人数之比是  $5: 4$ ; 第三车间男职工比女职工多 150 人。那么这个工厂共有多少人？温馨小提示：利用表格整理信息",
+        "options": [],
+        "answer": "1400人",
+        "solution": "【分析】根据题意，可绘制如下表格：![](images\\3eebecc884ac1bffa28de347b122ada0.png)\n\n由表格信息可知，第三车间男职工比女职工多（7-4）份，用第三车间男职工比女职工多的人数除以多的份数，再乘三个车间男、女职工的总份数即可解答。解答较复杂的按比例分配问题, 可利用表格整理题中的数据, 使数据变得简单、清晰, 从而使解题变得简单一些。\n\n【详解】把题中数据整理成如下表格：![](images\\3eebecc884ac1bffa28de347b122ada0.png)\n\n$ 1 5 0 \\div (7 - 4) \\times 2 8 \\\\ = 1 5 0 \\div 3 \\times 2 8 \\\\ = 5 0 \\times 2 8 \\\\ = 1 4 0 0 (\\text {人}) \\\\ $\n\n答: 这个工厂共有 1400 人。",
+        "knowledge_points": [
+          "按比分配问题",
+          "比的应用"
+        ],
+        "source": "2024 六年级下·全国·专题练习",
+        "merged_id": "unit4_175",
+        "original_id": "unit4_solve_175"
+      },
+      {
+        "id": "unit4_solve_176",
+        "unit": "第四单元：比",
+        "chapter": "工程问题",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "（23-24六年级上·全国·单元测试）有两个同样的仓库 A 和 B，搬运一个仓库里的货物，甲需要 18 小时，乙需要 12 小时，丙需要 9 小时。甲、乙在 A 仓库，丙在 B 仓库，同时开始搬运。中途甲又转向帮助丙搬运。最后两个仓库同时搬完。甲帮助乙、丙各多少小时？",
+        "options": [],
+        "answer": "甲帮乙：6小时；甲帮丙：2小时",
+        "solution": "【分析】把每个仓库的货物看作单位 '1', 根据工作效率  $=$  工作总量  $\\div$  工作时间, 分别求出甲, 乙、丙的工作效率。无论怎么样干活, 最后一起干完, 相当于三个人一块搬完两个仓库的货物, 那么求出搬完一共用的时间。再根据工作总量  $=$  工作效率  $\\times$  工作时间, 用乙的工作效率  $\\times$  搬完一共用的时间, 求出乙的工作总量, 用 1-乙的工作总量, 求出甲在仓库 A 的工作总量, 再除以甲的工作效率, 即可求出甲在仓库 A 的工作时间，即甲帮乙的时间；再用一共用的时间减去甲帮乙时间，即可求出甲在仓库 B 的工作时间，也就是帮丙的时间，据此解答。解答本题的关键是先求出三人同时搬运所需要的时间，再利用工作总量、工作效率和工作时间三者的关系，进行解答。\n\n【详解】把每个仓库的货物看作单位 '1', 则甲的工作效率是  $\\frac{1}{18}$ , 乙的工作效率是  $\\frac{1}{12}$ , 丙的工作效率是  $\\frac{1}{9}$  。\n\n$2 \\div (\\frac {1}{1 8} + \\frac {1}{1 2} + \\frac {1}{9}) \\\\ = 2 \\div (\\frac {2}{3 6} + \\frac {3}{3 6} + \\frac {4}{3 6}) \\\\ = 2 \\div (\\frac {5}{3 6} + \\frac {4}{3 6}) \\\\ = 2 \\div \\frac {1}{4} \\\\ = 2 \\times 4 \\\\ = 8 \\text {(小 时)} \\\\ (1 - \\frac {1}{1 2} \\times 8) \\div \\frac {1}{1 8} \\\\ = (1 - \\frac {2}{3}) \\div \\frac {1}{1 8} \\\\ = \\frac {1}{3} \\times 1 8 \\\\ = 6 \\text {(小 时)} \\\\ 8 - 6 = 2 \\text {(小 时)} $\n\n甲帮乙 6 小时, 甲帮丙 2 小时。",
+        "knowledge_points": [
+          "两人合作的工程问题",
+          "分数的四则混合运算"
+        ],
+        "source": "23-24六年级上·全国·单元测试",
+        "merged_id": "unit4_176",
+        "original_id": "unit4_solve_176"
+      },
+      {
+        "id": "unit4_solve_177",
+        "unit": "第四单元：比",
+        "chapter": "工程问题",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "（23-24六年级上·全国·单元测试）丽丽和小美去参加活动，需要制作礼物。如果丽丽做10小时，小美再做10小时可以完成；如果丽丽做8小时，小美再做13小时也可完成。现在小美单独做，可多少小时完成这项任务？温馨小提示：合作效率是  $\\frac{1}{10}$ ，丽丽8小时小美13小时可以看作合作8小时后小美5小时",
+        "options": [],
+        "answer": "25 小时",
+        "solution": "【分析】如果丽丽做 10 小时, 小美再做 10 小时可以完成,可看成两人合作 10 小时完成, 则两人的合作效率是  $1 \\div 10 = \\frac{1}{10}$ ; 如果丽丽做 8 小时, 小美再做 13 小时也可完成, 可看成两人合作 8 小时, 完成这项任务的  $\\frac{1}{10} \\times 8$ , 剩下的  $(1 - \\frac{1}{10} \\times 8)$ , 由小美单独 5 小时完成, 用 5 除以  $(1 - \\frac{1}{10} \\times 8)$ , 求出小美单独做,可多少小时完成这项任务。本题考查分数乘除法、工程问题, 解答本题的关键是掌握题中的数量关系。\n\n【详解】两人合作效率： $1 \\div 10 = \\frac{1}{10}$ 剩下工作量： $1 - \\frac{1}{10} \\times 8=1 - \\frac {4}{5}   =\\frac {1}{5}  $小美单独完成时间:  $5 \\div \\frac{1}{5} = 5 \\times 5 = 25$  (小时)答: 小美单独做, 可 25 小时完成这项任务。",
+        "knowledge_points": [
+          "两人合作的工程问题",
+          "分数与整数的除法",
+          "分数乘整数"
+        ],
+        "source": "23-24六年级上·全国·单元测试",
+        "merged_id": "unit4_177",
+        "original_id": "unit4_solve_177"
+      },
+      {
+        "id": "unit4_solve_178",
+        "unit": "第四单元：比",
+        "chapter": "工程问题",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "（23-24六年级上·全国·单元测试）一件工作，甲工程队独做12小时可以完成，现在先甲、乙合做4小时，剩下的工作，全部交给乙完成，还需要2小时，乙单独完成这份工作需要多少天？",
+        "options": [],
+        "answer": "9天",
+        "solution": "【分析】把这件工作的总量看作单位'1'，甲工程队独做12小时可以完成，则甲的工作效率是  $\\frac{1}{12}$ ，完成这件工作，甲一共做了4小时，乙做了（4+2）小时，根据'工作量  $=$ 工作效率  $\\times$  工作时间'求出甲4小时完成的工作量，再用总工作量减去甲4小时完成的工作量就是乙（4+2）小时完成的工作量，再根据'工作量  $\\div$  工作时间  $=$  工作效率'求出乙的工作效率，再根据'工作量  $\\div$  工作效率  $=$  工作时间'即可解答。本题考查了工作量、工作效率、工作时间的关系,求出乙的工作效率是解题的关键。\n\n【详解】$(1 - \\frac{1}{12} \\times 4) \\div (4 + 2)$\n\n$= \\frac {2}{3} \\div 6 \\\\ = \\frac {1}{9}  $\n\n$1 \\div \\frac {1}{9} = 9 \\quad (\\text {天})$答: 乙单独完成这份工作需要 9 天。",
+        "knowledge_points": [
+          "两人合作的工程问题",
+          "分数的四则混合运算"
+        ],
+        "source": "23-24六年级上·全国·单元测试",
+        "merged_id": "unit4_178",
+        "original_id": "unit4_solve_178"
+      },
+      {
+        "id": "unit4_solve_179",
+        "unit": "第四单元：比",
+        "chapter": "工程问题",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "（23-24六年级上·全国·单元测试）师徒二人加工一批零件，师傅单独加工需12天完成，先师傅和徒弟一起加工7天，师傅因事离开，徒弟还需3天才能完成全部工程。那么，徒弟单独做完成全工程需多少天？",
+        "options": [],
+        "answer": "24天",
+        "solution": "【分析】根据工作效率  $=$  工作总量  $\\div$  工作时间；把这批零件的总量看作单位'1'，用  $1 \\div 12$  ，求出师傅的工作效率；再根据工作总量  $=$  工作效率  $\\times$  工作时间，用师傅的工作效率  $\\times 7$  ，求出师傅7天的工作量，再用1一师傅7天的工作量，求出徒弟的工作量，再用徒弟的工作量÷徒弟的工作时间，即可求出徒弟的工作效率，再用工作总量÷徒弟的工作效率，即可求出徒弟单独完成全工程需要的时间。求出徒弟的工作效率是解答本题的关键。\n\n【详解】$(1 - \\frac{1}{12} \\times 7) \\div (7 + 3) = (1 - \\frac {7}{1 2}) \\div 1 0 \\\\ = \\frac {5}{1 2} \\div 1 0 \\\\ = \\frac {5}{1 2} \\times \\frac {1}{1 0} \\\\ = \\frac {1}{2 4}  $\n\n$ 1 \\div \\frac {1}{2 4} \\\\ = 1 \\times 2 4 \\\\ = 2 4 \\text {(天)}  $\n\n答: 徒弟单独完成全工程需要 24 天。",
+        "knowledge_points": [
+          "两人合作的工程问题",
+          "分数的四则混合运算"
+        ],
+        "source": "23-24六年级上·全国·单元测试",
+        "merged_id": "unit4_179",
+        "original_id": "unit4_solve_179"
+      },
+      {
+        "id": "unit4_solve_180",
+        "unit": "第四单元：比",
+        "chapter": "工程问题",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "（23-24六年级上·全国·单元测试）加工一批零件，甲单独做要15天，乙单独做要10天完成，甲乙同时工作，全做完时，甲做了2400个，这批零件共有多少个？",
+        "options": [],
+        "answer": "6000 个",
+        "solution": "【分析】甲单独做要 15 天, 乙单独做要 10 天完成, 则甲每天完成全部零件的  $1 \\div 15 = \\frac{1}{15}$ , 乙每天完成全部零件的  $1 \\div 10 = \\frac{1}{10}$ ,两人合作时每天完成全部零件的  $(\\frac{1}{15} + \\frac{1}{10})$ , 据此求出两人合作完成的天数, 再求出两人合作时, 甲完成的占全部零件的分率, 用甲完成的数量除以他完成的占全部零件的分率,求出这批零件共有多少个即可。本题考查工程问题、分数乘除法，解答本题的关键是掌握题中的数量关系式\n\n【详解】合作完成天数:  $1 \\div (\\frac{1}{15} + \\frac{1}{10})\\\\ =1 \\div \\frac {1}{6}\\\\= 1 \\times 6\\\\=6 (天)$\n\n这批零件数量:  $2400 \\div (\\frac{1}{15} \\times 6)$\n\n$ 2 4 0 0 \\div \\frac {2}{5}\\\\=2 4 0 0 \\times \\frac {5}{2}\\\\=6000 (个)$\n\n答: 这批零件共有 6000 个。",
+        "knowledge_points": [
+          "两人合作的工程问题",
+          "求一个数的几分之几的问题",
+          "已知一个数的几分之几是多少，求这个数"
+        ],
+        "source": "23-24六年级上·全国·单元测试",
+        "merged_id": "unit4_180",
+        "original_id": "unit4_solve_180"
+      },
+      {
+        "id": "unit4_solve_181",
+        "unit": "第四单元：比",
+        "chapter": "工程问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（23-24六年级上·全国·单元测试）一批零件，师傅单独加工需要50天完成，徒弟单独加工需要75天完成，现师徒两人共同加工，完成任务时，徒弟比师傅少加工120个，这批零件共有多少个？",
+        "options": [],
+        "answer": "600 个",
+        "solution": "【分析】先算出两人单独的工作效率，用单位'1'除以工作效率之和可以算出合作完成所用的时间；再用两人的工作效率分别乘工作时间算出两人的工作量，120个对应的分率是工作量之差，用120除以工作量之差即可算出零件个数。\n\n【详解】师傅的工作效率:  $1 \\div 50 = \\frac{1}{50}$徒弟的工作效率:  $1 \\div 75 = \\frac{1}{75}$\n\n师徒合作时间:  $1 \\div (\\frac{1}{50} + \\frac{1}{75})\\\\= 1 \\div \\frac {1}{3 0}\\\\=30 (天)$\n\n$1 2 0 \\div (\\frac {1}{5 0} \\times 3 0 - \\frac {1}{7 5} \\times 3 0)\\\\= 1 2 0 \\div (\\frac {3}{5} - \\frac {2}{5})\\\\= 1 2 0 \\div \\frac {1}{5}\\\\=600 (个)$\n\n答：这批零件共有 600 个。",
+        "knowledge_points": [
+          "两人合作的工程问题",
+          "分数的四则混合运算"
+        ],
+        "source": "23-24六年级上·全国·单元测试",
+        "merged_id": "unit4_181",
+        "original_id": "unit4_solve_181"
+      },
+      {
+        "id": "unit4_solve_182",
+        "unit": "第四单元：比",
+        "chapter": "工程问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（23-24六年级上·全国·单元测试）加工一批零件甲单独做6小时完成，乙单独做要10小时完成。甲乙共同做了1小时后再由甲单独做2小时，此时还剩240个没有加工，这批零件共有多少个？",
+        "options": [],
+        "answer": "600 个",
+        "solution": "【分析】把这批零件的个数看作单位 '1', 根据工作效率  $=$工作总量÷工作时间，分别求出甲、乙的工作效率；根据工作量 = 工作效率 × 工作时间，求出甲乙共做了 1 小时的工作量；求出甲单独做2小时的工作量，用1减去甲乙共做了1小时的工作量，减去甲单独做2小时的工作量，求出还剩下几分之几没有完成，对应的是 240 个，再根据分数除法的意义，求出单位'1'，据此解答。\n\n【详解】$240 \\div [1 - (\\frac{1}{6} + \\frac{1}{10}) \\times 1 - \\frac{1}{6} \\times 2]\\\\ = 2 4 0 \\div [ 1 - (\\frac {5}{3 0} + \\frac {3}{3 0}) \\times 1 - \\frac {1}{3} ] \\\\ = 2 4 0 \\div [ 1 - \\frac {4}{1 5} \\times 1 - \\frac {1}{3} ] \\\\ = 2 4 0 \\div [ 1 - \\frac {4}{1 5} - \\frac {5}{1 5} ] \\\\ = 2 4 0 \\div [ \\frac {1 1}{1 5} - \\frac {5}{1 5} ] \\\\ = 2 4 0 \\div \\frac {2}{5} \\\\ = 2 4 0 \\times \\frac {5}{2} \\\\ = 6 0 0 \\text {(个)} \\\\ $\n\n答：这批零件共有 600 个。",
+        "knowledge_points": [
+          "两人合作的工程问题",
+          "分数的四则混合运算"
+        ],
+        "source": "23-24六年级上·全国·单元测试",
+        "merged_id": "unit4_182",
+        "original_id": "unit4_solve_182"
+      },
+      {
+        "id": "unit4_solve_83",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "小华对本校2024年春季学期参加足球、篮球和乒乓球三项社团活动的学生进行了统计，参加这3项活动的学生共有360人，其中足球社团人数是篮球社团人数的 $\\frac{3}{5}$，篮球社团人数与乒乓球社团人数的比是5:7。参加这3项社团活动的学生各有多少人？(先画线段图分析，再列式解答）\n\n![](images\\9335841b03572f1a53a71ef705ba4599.png)",
+        "options": [],
+        "answer": "72人；120人；168人",
+        "solution": "【分析】根据题意, 把乒乓球社团人数看作单位 '1', 篮球社团人数是乒乓球社团人数的 $\\frac{5}{7}$, 足球社团人数是篮球社团人数的 $\\frac{3}{5}$, 据此画出线段图。由图可知, 足球社团人数、篮球社团人数、乒乓球社团人数的比是 $3: 5: 7$, 将 360 人平均分成 $(3 + 5 + 7)$ 份, 求出每份是多少人, 再根据每个社团所占的份数, 求出每个社团的人数即可。\n\n【详解】![](images\\9335841b03572f1a53a71ef705ba4599.png)$3 6 0 \\div (3 + 5 + 7) \\\\ = 3 6 0 \\div 1 5 \\\\ = 2 4 \\text {(人)} $\n\n$2 4 \\times 3 = 7 2 (\\text {人}) \\\\ 2 4 \\times 5 = 1 2 0 (\\text {人}) \\\\ 2 4 \\times 7 = 1 6 8 (\\text {人}) $\n\n答: 足球社团有 72 人, 篮球社团有 120 人, 乒乓球社团有 168 人。",
+        "knowledge_points": ["比的基本性质", "比的意义", "按比分配问题", "比的应用"],
+        "source": "2024·云南昭通·小升初真题",
+        "merged_id": "unit4_83",
+        "original_id": "unit4_solve_83"
+      },
+      {
+        "id": "unit4_solve_184",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "从前有个农民，临死前留下遗言，要把 17 头牛分给三个儿子，其中大儿子分得 $\\frac{1}{2}$，二儿子分得 $\\frac{1}{3}$，小儿子分得 $\\frac{1}{9}$，但不能把牛卖掉或杀掉；三个儿子按照老人的要求怎么也不好分；后来一位邻居顺利地把 17 头牛分完了, 你知道这到底是怎么回事吗?",
+        "options": [],
+        "answer": "大儿子分得 9 头, 二儿子分得 6 头, 小儿子分得 2 头",
+        "solution": "【分析】把牛的总头数看作单位 '1', $\\frac{1}{2} + \\frac{1}{3} + \\frac{1}{9} = \\frac{17}{18}, \\frac{17}{18}<$ 1, 也就是三兄弟并未将全部牛分完, 那么可以求出三个儿子分牛头数的比, $\\frac{1}{2}: \\frac{1}{3}: \\frac{1}{9} = 9: 6: 2, 9 + 6 + 2 = 17$, 则三个儿子分别分得总头数的 $\\frac{9}{17}$、$\\frac{6}{17}$、$\\frac{2}{17}$, 用总头数分别乘这三个分数即可求出各自分得牛的头数。\n\n【详解】$\\frac{1}{2}: \\frac{1}{3}: \\frac{1}{9}= (\\frac {1}{2} \\times 1 8): (\\frac {1}{3} \\times 1 8): (\\frac {1}{9} \\times 1 8) = 9: 6: 2 $\n\n总份数: $9 + 6 + 2 = 17$大儿子: $17 \\times \\frac{9}{17} = 9$ (头) 二儿子: $17 \\times \\frac{6}{17} = 6$ (头) 小儿子: $17 \\times \\frac{2}{17} = 2$ (头)\n\n答: 大儿子分得 9 头, 二儿子分得 6 头, 小儿子分得 2 头。",
+        "knowledge_points": ["求一个数的几分之几的问题", "按比分配问题"],
+        "source": "2025 六年级下·全国·专题练习",
+        "merged_id": "unit4_184",
+        "original_id": "unit4_solve_184"
+      },
+      {
+        "id": "unit4_solve_185",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "六（1）班男、女生人数的比是 $5: 3$, 已知男生比女生多 12 人。\n\n(1) 画图表示数量关系。\n(2) 男、女生各有多少人?",
+        "options": [],
+        "answer": "(1) 见详解; (2) 男生: 30人; 女生: 18人",
+        "solution": "【分析】（1）分析题目，结合比的意义可知：男生的人数是5份，女生人数是3份，女生比男生少（5- 3）份，即12人，据此画出线段图即可；(2) 根据比的意义可知, 12 人是 (5-3) 份, 据此用除法求出一份是多少，再分别乘男生、女生人数所占的份数即可得到男生、女生人数。\n\n【详解】（1）画线段图如下：\n\n![](images\\0045a9ef66cc1e30618cc8b85a2facc8.png)\n\n(2) $12 \\div (5 - 3)= 1 2 \\div 2 = 6 (\\text {人}) $$6 \\times 5 = 3 0 (\\text {人}) \\quad 6 \\times 3 = 1 8 (\\text {人}) $答: 男生有 30 人, 女生有 18 人。",
+        "knowledge_points": ["比的意义", "比的应用"],
+        "source": "2024·安徽宿州·小升初真题",
+        "merged_id": "unit4_185",
+        "original_id": "unit4_solve_185"
+      },
+      {
+        "id": "unit4_solve_186",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "用甲、乙两个粮库存放大米，原来甲粮库存放大米的数量是乙粮库的 $\\frac{5}{7}$。如果从乙粮库调 30 吨大米到甲粮库, 那么甲粮库存放大米的数量与乙粮库的比是 $4: 5$。原来甲、乙两个粮库各存放大米多少吨？",
+        "options": [],
+        "answer": "甲450吨；乙630吨",
+        "solution": "【分析】根据题意, 两个粮库的大米总吨数不变。已知原来甲粮库存放大米的数量是乙粮库的 $\\frac{5}{7}$, 则原来乙粮库大米的吨数占两个粮库大米总吨数的 $\\frac{7}{5 + 7}$;如果从乙粮库调 30 吨大米到甲粮库, 那么甲粮库存放大米的数量与乙粮库的比是 $4: 5$, 则现在乙粮库大米的吨数占两个粮库大米总吨数的 $\\frac{5}{4 + 5}$;那么 30 吨大米占两个粮库大米总吨数的 $(\\frac{7}{5 + 7} - \\frac{5}{4 + 5})$, 单位 '1' 未知, 根据分数除法的意义求出两个粮库大米的总吨数;再根据求一个数的几分之几是多少，用两个粮库大米的总吨数乘 $\\frac{7}{5 + 7}$，求出原来乙粮库大米的吨数；再用两个粮库大米的总吨数减去原来乙粮库大米的吨数，求出原来甲粮库存放大米的吨数。抓住两个粮库的大米总吨数不变，求出30吨占两个粮库大米总吨数的几分之几，然后根据分数除法的意义求出总吨数是解题的关键。\n\n【详解】两个粮库大米的总吨数：\n\n$3 0 \\div (\\frac {7}{5 + 7} - \\frac {5}{4 + 5}) \\\\ = 3 0 \\div (\\frac {7}{1 2} - \\frac {5}{9}) \\\\ = 3 0 \\div (\\frac {2 1}{3 6} - \\frac {2 0}{3 6}) \\\\ = 3 0 \\div \\frac {1}{3 6} \\\\ = 3 0 \\times 3 6 \\\\ = 1 0 8 0 (\\text {吨}) $\n\n原来乙粮库大米的吨数:\n\n$1 0 8 0 \\times \\frac {7}{5 + 7} \\\\ = 1 0 8 0 \\times \\frac {7}{1 2} \\\\ = 6 3 0 (\\text {吨}) $原来甲粮库大米的吨数:\n\n$1080 - 630 = 450$ （吨）\n\n答: 原来甲粮库存放大米 450 吨, 乙粮库存放大米 630 吨。",
+        "knowledge_points": ["求一个数的几分之几的问题", "已知一个数的几分之几是多少, 求这个数", "比与分数、除法的关系", "比的应用"],
+        "source": "24-25 六年级下·海南海口·单元测试",
+        "merged_id": "unit4_186",
+        "original_id": "unit4_solve_186"
+      },
+      {
+        "id": "unit4_solve_187",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "2024年，林老师参加了双辽乡村（森林）马拉松比赛，她一路坚持，终于跑到了'距离终点还有15千米'的提示牌，此时她已经跑完全程的 $\\frac{9}{14}$，真是太了不起了！这场比赛有大量长跑爱好者踊跃参加。参加全程马拉松的有840人，参加半程马拉松的有1000人，参加健康跑的有1200人。已知全程马拉松男女比例为3:1；半程马拉松男女比例为7:3；健康跑男女比例为12:13。\n\n(1) 这次马拉松比赛的全程是多少千米？\n(2) 参加全程马拉松比赛的男女人数各是多少人？",
+        "options": [],
+        "answer": "(1) 42 千米 (2) 男生 630 人; 女生 210 人",
+        "solution": "【分析】（1）把这次马拉松比赛的全程看作单位'1'，林老师已经跑完全程的 $\\frac{9}{14}$，离终点还有 15 千米，即 15 千米占全程的 $(1 - \\frac{9}{14})$，单位'1'未知，根据分数除法的意义求出全程。(2) 已知参加全程马拉松的有 840 人, 全程马拉松男女比例为 $3: 1$, 即参加全程马拉松的男生人数占参加全程马拉松总人数的 $\\frac{3}{3+1}$, 根据求一个数的几分之几是多少, 用乘法计算, 求出参加全程马拉松的男生人数; 再用参加全程马拉松总人数减去参加全程马拉松的男生人数, 即是参加全程马拉松的女生人数。\n\n【详解】 (1) $15 \\div (1 - \\frac{9}{14})$\n\n$= 1 5 \\div \\frac {5}{1 4} \\\\ = 1 5 \\times \\frac {1 4}{5} \\\\ = 4 2 \\text {(千 米)} $\n\n答: 这次马拉松比赛的全程是 42 千米。\n\n(2) 男: $840 \\times \\frac{3}{3 + 1}$\n\n$= 8 4 0 \\times \\frac {3}{4} \\\\ = 6 3 0 (\\text {人}) $\n\n女: $840 - 630 = 210$ (人)\n\n答: 参加全程马拉松比赛的男生人数是 630 人, 女生人数是 210 人。",
+        "knowledge_points": ["求一个数的几分之几的问题", "已知一个数的几分之几是多少，求这个数", "按比分配问题", "分数的四则混合运算"],
+        "source": "24-25六年级上·吉林四平·期中",
+        "merged_id": "unit4_187",
+        "original_id": "unit4_solve_187"
+      },
+      {
+        "id": "unit4_solve_188",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "《周髀算经》中记载: '勾广三, 股修四, 径隅五。'意思是说: 当直角三角形的两条直角边分别为3（勾）和4（股）时，径（弦）则为 5。后人简单地把这个事实说成 '勾 3 股 4 弦 5。' 一个直角三角形 3 条边的长度比是 $3: 4: 5$, 斜边长 10 厘米,另外两条直角边分别长多少厘米?",
+        "options": [],
+        "answer": "6 厘米; 8 厘米",
+        "solution": "【分析】已知一个直角三角形3条边的长度比是 $3: 4: 5$, 直角三角形中斜边最长, 可知这个直角三角形中两条直角边分别占3份和4份, 斜边占5份; 已知斜边长10厘米, 用斜边的长度除以5, 求出一份量, 再用一份量分别乘3、乘4, 求出两条直角边的长度。\n\n【详解】$10 \\div 5 = 2$ （厘米）\n\n$2 \\times 3 = 6 (\\text {厘 米})$\n\n$2 \\times 4 = 8 (\\text {厘 米})$\n\n答: 另外两条直角边分别长 6 厘米和 8 厘米。",
+        "knowledge_points": ["单归一问题", "比的应用", "三角形的分类"],
+        "source": "23-24 六年级上·山东临沂·期中",
+        "merged_id": "unit4_188",
+        "original_id": "unit4_solve_188"
+      },
+      {
+        "id": "unit4_solve_189",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "两堆煤，甲堆煤的重量占总重量的 $\\frac{5}{8}$，如果从甲堆煤里取出26吨，从乙堆煤里取出10吨，两堆煤剩下重量的比是 $1:1$，求甲乙两堆煤共有多少吨？",
+        "options": [],
+        "answer": "64吨",
+        "solution": "【分析】把总重量看作单位 '1', 已知原来甲堆煤的重量占总重量的 $\\frac{5}{8}$, 原来乙堆煤的重量占总重量的 $(1 - \\frac{5}{8})$, 假设甲乙两堆煤共有 $x$ 吨, 根据分数乘法的意义, 可知原来甲堆煤的重量是 $\\frac{5}{8} x$ 吨, 原来乙堆煤的重量是 $(1 - \\frac{5}{8}) x$ 吨; 已知两堆煤剩下重量的比是 $1: 1$, 根据比的意义, 可知两堆煤剩下重量相等, 据此可知原来甲堆煤的重量 $-26$ 吨 = 原来乙堆煤的重量 $-10$ 吨, 列方程为 $\\frac{5}{8} x - 26 = (1 - \\frac{5}{8}) x - 10$,然后解出方程即可。本题可用列方程解决问题, 找到相应的数量关系是解答本题的关键。\n\n【详解】解：设甲乙两堆煤共有 $\\mathrm{x}$ 吨。\n\n$\\frac {5}{8} x - 2 6 = (1 - \\frac {5}{8}) x - 1 0 \\\\ \\frac {5}{8} x - 2 6 = \\frac {3}{8} x - 1 0 \\\\ \\frac {5}{8} x = \\frac {3}{8} x - 1 0 + 2 6 \\\\ \\frac {5}{8} x = \\frac {3}{8} x + 1 6 \\\\ \\frac {5}{8} x - \\frac {3}{8} x = 1 6 \\\\ \\frac {1}{4} x = 1 6 \\\\ x = 1 6 \\div \\frac {1}{4} \\\\ x = 1 6 \\times 4 \\\\ x = 6 4 $\n\n答: 甲乙两堆煤共有 64 吨。",
+        "knowledge_points": ["已知一部分量占总量的几分之几及另一部分量,求总量", "比的应用", "列方程解决稍复杂的实际问题"],
+        "source": "23-24六年级上·全国·单元测试",
+        "merged_id": "unit4_189",
+        "original_id": "unit4_solve_189"
+      },
+      {
+        "id": "unit4_solve_190",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "加工一批零件，小王加工的个数是小明的 $\\frac{1}{4}$，小明加工的个数与小青的比1:3，已知小王加工了40个，小青加工了多少个？",
+        "options": [],
+        "answer": "480 个",
+        "solution": "【分析】由题可知, 小王加工了 40 个, 小王加工的个数是小明的 $\\frac{1}{4}$, 用小王加工的个数除以 $\\frac{1}{4}$, 求出小明加工的个数,又知小明加工的个数与小青的比 $1: 3$, 所以小明加工的个数是小青的 $\\frac{1}{3}$, 所以用小明加工的个数除以 $\\frac{1}{3}$ 即可求出小青加工的个数。\n\n【详解】$40 \\div \\frac{1}{4} \\div \\frac{1}{3}$\n\n$= 4 0 \\times 4 \\times 3 \\\\ = 1 6 0 \\times 3 \\\\ = 4 8 0 \\text {(个)} $\n\n答: 小青加工了 480 个。",
+        "knowledge_points": ["已知一个数的几分之几是多少，求这个数", "比的应用"],
+        "source": "23-24六年级上·全国·单元测试",
+        "merged_id": "unit4_190",
+        "original_id": "unit4_solve_190"
+      },
+      {
+        "id": "unit4_solve_191",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "爷爷果园里的苹果树比梨树多 320 棵, 已知苹果树和梨树棵数的比是 $7: 5$, 爷爷的果园里有梨树多少棵?",
+        "options": [],
+        "answer": "800 棵",
+        "solution": "【分析】已知苹果树和梨树棵数的比是 $7: 5$, 所以可以把苹果树棵数看作 7 份, 梨树棵数看作 5 份, 苹果树比梨树多的棵数对应 2 份, 用苹果树比梨树多的棵数 $\\div 2$, 求出 1 份对应的棵数, 1 份对应的棵数乘梨树的份数求出梨树的棵数。\n\n【详解】$320 \\div (7 - 5) \\times 5= 3 2 0 \\div 2 \\times 5 = 1 6 0 \\times 5 = 8 0 0 \\text {(棵)} $答: 爷爷的果园里有梨树 800 棵。",
+        "knowledge_points": ["比的应用"],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_191",
+        "original_id": "unit4_solve_191"
+      },
+      {
+        "id": "unit4_solve_192",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "某校六（1）和六(2) 班共有 72 人, 六 (1) 班和六 (2) 班人数的比是5:4, 六 (1) 和六 (2) 各有多少人?",
+        "options": [],
+        "answer": "六（1）班有40人，六（2）班有32人",
+        "solution": "【分析】已知六（1）班和六（2）班人数的比是 $5: 4$，根据比的意义，把六（1）班和六（2）班人数分别看作 5 份和 4 份，然后用总人数除以 $(5 + 4)$ 份，即可求出每份是多少，进而用乘法分别求出两个班的人数。\n\n【详解】$72 \\div (5 + 4)= 7 2 \\div 9 = 8 \\text {(人)} $ 六 (1) 班: $8 \\times 5 = 40$ (人) 六 (2) 班: $8 \\times 4 = 32$ (人)\n\n答: 六 (1) 班有 40 人, 六 (2) 班有 32 人。",
+        "knowledge_points": ["按比分配问题"],
+        "source": "23-24六年级上·全国·单元测试",
+        "merged_id": "unit4_192",
+        "original_id": "unit4_solve_192"
+      },
+      {
+        "id": "unit4_solve_193",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "一件上衣的售价是60元，裤子与上衣单价的比是2:3。每条裤子多少元？",
+        "options": [],
+        "answer": "40元",
+        "solution": "【分析】已知裤子与上衣单价的比是 $2: 3$, 则说明上衣的价格有 3 份, 裤子的价格有 2 份, 已知上衣售价是 60 元, 用 60 除以 3 可求出每份代表的价格, 再跟裤子价格的份数相乘即可求出结果。\n\n【详解】$60 \\div 3 \\times 2$\n\n$= 2 0 \\times 2 \\\\ = 4 0 \\text {(元)} $\n\n答: 每条裤子 40 元。",
+        "knowledge_points": ["比的应用"],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_193",
+        "original_id": "unit4_solve_193"
+      },
+      {
+        "id": "unit4_solve_194",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "一个长方形, 宽和长的比是 $\\frac{4}{5}$, 如果宽再增加 6 厘米, 这时长方形就变成了正方形, 求这个长方形的宽是多少厘米?",
+        "options": [],
+        "answer": "24 厘米",
+        "solution": "【分析】已知长方形的宽和长的比是 $\\frac{4}{5}$, 把宽看作 4 份, 长看作 5 份, 宽比长少 (5-4) 份;如果宽再增加 6 厘米, 这时长方形就变成了正方形, 说明原来宽比长少 6 厘米, 用少的长度除以少的份数, 求出一份数, 再用一份数乘宽的份数, 即可求出宽。\n\n【详解】一份数：\n\n$6 \\div (5 - 4) \\\\ = 6 \\div 1 \\\\ = 6 \\text {(厘 米)} $\n\n宽: $6 \\times 4 = 24$ (厘米)\n\n答: 这个长方形的宽是 24 厘米。",
+        "knowledge_points": ["正方形的概念及特点", "比的意义", "比的应用"],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_194",
+        "original_id": "unit4_solve_194"
+      },
+      {
+        "id": "unit4_solve_195",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "甲、乙、丙三人合作投资兴办服装厂，甲投资30万元，乙投资40万元，丙投资50万元。服装厂去年获得纯利润是48万元。按投资额分配，丙获得利润多少万元？",
+        "options": [],
+        "answer": "20万元",
+        "solution": "【分析】甲投资 30 万元, 乙投资 40 万元, 丙投资 50 万元。根据比的意义和化简, 得出三人投资额的比为 $3: 4: 5$, 即甲、乙两人各投资多少万元?乙、丙的投资额分别占三人总投资的 $\\frac{3}{3 + 4 + 5}$、$\\frac{4}{3 + 4 + 5}$、$\\frac{5}{3 + 4 + 5}$,根据求一个数的几分之几是多少，用乘法计算，即可求出按投资额分配，三人各应获得利润。\n\n【详解】$30: 40: 50$\n\n$= (3 0 \\div 1 0): (4 0 \\div 1 0): (5 0 \\div 1 0)\\\\=3:4:5$\n\n丙: $48 \\times \\frac{5}{3 + 4 + 5}$\n\n$= 4 8 \\times \\frac {5}{1 2} \\\\ = 2 0 (\\text {万 元}) $\n\n答：丙获得利润20万元。",
+        "knowledge_points": ["比的化简", "比的意义", "比的应用"],
+        "source": "23-24 六年级上·全国·单元测试",
+        "merged_id": "unit4_195",
+        "original_id": "unit4_solve_195"
+      },
+      {
+        "id": "unit4_solve_196",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "A、B两地相距750千米，甲、乙两车同时从 A、B 两地相对开出，相向而行，经过5小时相遇。已知甲、乙两车的速度比是8:7。求甲车每小时行多少千米？",
+        "options": [],
+        "answer": "80千米",
+        "solution": "【分析】总路程÷相遇时间 = 两车速度和，将比的前后项看成份数，速度和 ÷ 总份数，求出一份数，一份数 × 甲车对应份数 = 甲车速度，据此列式解答。\n\n【详解】$750 \\div 5 = 150$ （千米/时）\n\n$1 5 0 \\div (8 + 7) \\times 8 \\\\ = 1 5 0 \\div 1 5 \\times 8 \\\\ = 1 0 \\times 8 \\\\ = 8 0 \\text {(千 米 / 时)} $\n\n答: 甲车每小时行 80 千米。",
+        "knowledge_points": ["相遇问题", "按比分配问题"],
+        "source": "23-24六年级上·全国·单元测试",
+        "merged_id": "unit4_196",
+        "original_id": "unit4_solve_196"
+      },
+      {
+        "id": "unit4_solve_197",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "小明、小林、小刚三人的邮票枚数的比为 $7:4:1$，小明赠送给小刚24枚邮票，小林也赠送给小刚一些邮票，这时三人的邮票枚数比为8:3:5。小林赠送给小刚多少枚邮票？",
+        "options": [],
+        "answer": "42 枚",
+        "solution": "【分析】三人邮票总数不变，将三人邮票总数看作单位'1'，开始小明邮票枚数占邮票总数的 $\\frac{7}{7+4+1}$，小明和小林都赠送给小刚邮票后，小明邮票枚数占邮票总数的 $\\frac{8}{8+3+5}$，因此24枚邮票占邮票总数的 $(\\frac{7}{7+4+1}-\\frac{8}{8+3+5})$，根据部分数量÷对应分率 = 整体数量，求出三人邮票总数，小林邮票数量减少了总数的 $(\\frac{4}{7+4+1}-\\frac{3}{8+3+5})$，三人邮票总数×小林邮票数量减少的对应分率 = 小林赠送给小刚的数量，据此列式解答。关键是理解比的意义，明确总数不变，从而确定单位'1'，根据分数乘除法的意义，进行解答。\n\n【详解】$24 \\div (\\frac{7}{7 + 4 + 1} - \\frac{8}{8 + 3 + 5})$\n\n$= 2 4 \\div (\\frac {7}{1 2} - \\frac {8}{1 6}) \\\\ = 2 4 \\div \\frac {4}{4 8} \\\\ = 2 4 \\times \\frac {4 8}{4}\\\\=288 (枚) $\n\n$2 8 8 \\times (\\frac {4}{7 + 4 + 1} - \\frac {3}{8 + 3 + 5}) \\\\ = 2 8 8 \\times (\\frac {4}{1 2} - \\frac {3}{1 6}) $\n\n$= 2 8 8 \\times \\frac {7}{4 8}\\\\= 42 (枚)$\n\n答: 小林赠送给小刚 42 枚邮票。",
+        "knowledge_points": ["求一个数的几分之几的问题", "已知一个数的几分之几是多少，求这个数", "按比分配问题"],
+        "source": "23-24六年级上·全国·课后作业",
+        "merged_id": "unit4_197",
+        "original_id": "unit4_solve_197"
+      },
+      {
+        "id": "unit4_solve_198",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "有甲、乙两筐苹果，甲筐苹果重40千克，乙筐苹果重30千克，要使甲、乙两筐苹果的质量比是3:2，应从乙筐拿出多少千克苹果放入甲筐？",
+        "options": [],
+        "answer": "2千克",
+        "solution": "【分析】把两筐苹果的总质量看作单位 '1', 从乙筐拿出一些苹果放入甲筐后, 甲、乙两筐苹果的质量比是 $3: 2$, 即现在乙筐苹果的质量占两筐苹果总质量的 $\\frac{2}{3+2}$, 单位 '1'已知,用两筐苹果的总质量乘 $\\frac{2}{3 + 2}$, 求出现在乙筐苹果的质量, 再用原来乙筐苹果的质量减去现在乙筐苹果的质量即可。\n\n【详解】$(40 + 30) \\times \\frac{2}{3 + 2}$\n\n$= 7 0 \\times \\frac {2}{5}\\\\= 28$ （千克）\n\n$3 0 - 2 8 = 2 \\text {(千 克)}$\n\n答: 应从乙筐拿出 2 千克苹果放入甲筐。",
+        "knowledge_points": ["整数乘分数", "求一个数的几分之几的问题", "按比分配问题"],
+        "source": "24-25六年级上·全国·单元测试",
+        "merged_id": "unit4_198",
+        "original_id": "unit4_solve_198"
+      },
+      {
+        "id": "unit4_solve_199",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "新世纪小学四、五、六年级共有 27 个班, 平均每个班 35 人, 三个年级的人数比是 $2: 3: 4$。四、五、六年级各有多少人?",
+        "options": [],
+        "answer": "四年级 210 人；五年级 315 人；六年级 420 人",
+        "solution": "【分析】平均每个班人数 $\\times$ 班数 = 四五六年级总人数, 将比的各项看成份数, 总人数 $\\div$ 总份数, 求出一份数, 一份数分别乘四、五、六年级的对应份数，即可求出四、五、六年级的人数。\n\n【详解】$35 \\times 27 \\div (2 + 3 + 4)$\n\n$= 9 4 5 \\div 9 \\\\ = 1 0 5 \\text {(人)} $\n\n$1 0 5 \\times 2 = 2 1 0 (\\text {人}) \\\\ 1 0 5 \\times 3 = 3 1 5 (\\text {人}) \\\\ 1 0 5 \\times 4 = 4 2 0 (\\text {人}) $\n\n答: 四年级有 210 人、五年级有 315 人、六年级有 420 人。",
+        "knowledge_points": ["按比分配问题"],
+        "source": "24-25 六年级上·全国·单元测试",
+        "merged_id": "unit4_199",
+        "original_id": "unit4_solve_199"
+      },
+      {
+        "id": "unit4_solve_200",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "春节期间，某网站对100万网民的拜年方式进行了调查。结果表明：选择手机短信拜年、打电话拜年及聊天软件拜年方式的人数比是26:9:15。在这调查的 100 万网民中，选择聊天软件拜年的有多少万？",
+        "options": [],
+        "answer": "30 万",
+        "solution": "【分析】选择手机短信拜年、打电话拜年及聊天软件拜年方式的人数比是 $26: 9: 15$, 即选择聊天软件拜年方式占了总人数的 $\\frac{15}{26 + 9 + 15}$, 求一个数的几分之几用乘法。\n\n【详解】$100 \\times \\frac{15}{26 + 9 + 15} = 100 \\times \\frac{15}{50} = 30$ （万）答：选择聊天软件拜年的有 30 万。",
+        "knowledge_points": ["整数乘分数", "按比分配问题"],
+        "source": "24-25六年级上·全国·单元测试",
+        "merged_id": "unit4_200",
+        "original_id": "unit4_solve_200"
+      },{
+        "id": "unit4_choice_201",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "比的前项扩大到原来的 2 倍, 后项缩小到原来的 $\\frac{1}{4}$, 比值 ( )。",
+        "options": [
+          "A. 缩小到原来的 $\\frac{1}{4}$",
+          "B. 扩大到原来的 2 倍",
+          "C. 扩大到原来的 8 倍",
+          "D. 不变"
+        ],
+        "answer": "C",
+        "solution": "【分析】求比值是用比前项除以后项，得到比值。比的前项扩大到原来的2倍，后项缩小到原来的 $\\frac{1}{4}$，根据比与除法的关系得出答案。本题主要考查的是比与除法的关系，解题的关键是熟练运用比与除法之间的关系，进而得出答案。\n\n【详解】比的前项扩大到原来的 2 倍, 后项缩小到原来的 $\\frac{1}{4}$, 可看作被除数扩大到原来的 2 倍, 除数缩小到原来的 $\\frac{1}{4}$, 则商扩大的为原来的 8 倍, 即比值扩大为原来的 8 倍。故答案为: C。",
+        "knowledge_points": [
+          "求比值、比与分数、除法的关系"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_201",
+        "original_id": "unit4_choice_201"
+      },
+      {
+        "id": "unit4_choice_202",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "把 50 克糖溶解到 500 克水中, 糖和糖水的质量比是 ( )。",
+        "options": [
+          "A. $10: 1$",
+          "B. $1: 10$",
+          "C. 11:1",
+          "D. $1: 11$"
+        ],
+        "answer": "D",
+        "solution": "【分析】糖 + 水 = 糖水, 根据比的意义, 写出糖和糖水的比, 化简即可。两数相除又叫两个数的比。\n\n【详解】$50: (50+500)\\ = 5 0: 5 5 0 \\= 1: 1 1$\n\n故答案为: D",
+        "knowledge_points": [
+          "比的意义、比的化简"
+        ],
+        "source": "21-22 六年级上·云南曲靖·期末",
+        "merged_id": "unit4_202",
+        "original_id": "unit4_choice_202"
+      },
+      {
+        "id": "unit4_choice_203",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "一个比的前项除以 8, 后项乘 $\\frac{1}{8}$, 它的比值 ( )。",
+        "options": [
+          "A. 变大",
+          "B. 不变",
+          "C. 变小"
+        ],
+        "answer": "B",
+        "solution": "【分析】根据比的基本性质，比的前项除以 8，后项乘 $\\frac{1}{8}$ 相当于后项除以 8，比值不变，据此解答。比的前项和后项同时乘或除以相同的数（0 除外），比值不变，这叫做比的基本性质。\n\n【详解】假设比的前项为 24，比的后项为 8，则 $24:8 = 3$。\n\n$(2 4 \\div 8) : (8 \\times \\frac {1}{8}) \\ = 3: 1 \\ = 3$\n\n故答案为: B",
+        "knowledge_points": [
+          "比的基本性质"
+        ],
+        "source": "19-20 六年级下·湖南怀化·单元测试",
+        "merged_id": "unit4_203",
+        "original_id": "unit4_choice_203"
+      },
+      {
+        "id": "unit4_choice_204",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "1 克盐放入 49 克水中，这时盐与水的比是（）。",
+        "options": [
+          "A. 1:48",
+          "B. $1: 49$",
+          "C. $1: 50$"
+        ],
+        "answer": "B",
+        "solution": "【分析】由题意可知, 盐的质量为 1 克, 水的质量为 49 克,根据比的意义求出盐与水的质量比, 据此解答。掌握比的意义是解答题目的关键。\n\n【详解】盐的质量：水的质量 $= 1$ 克：49 克 $= 1: 49$ 故答案为：B",
+        "knowledge_points": [
+          "比的意义"
+        ],
+        "source": "19-20 六年级下·湖南怀化·单元测试",
+        "merged_id": "unit4_204",
+        "original_id": "unit4_choice_204"
+      },
+      {
+        "id": "unit4_choice_205",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "在含盐 $\\frac{1}{6}$ 的盐水中,盐和水的质量比是 ( )。",
+        "options": [
+          "A. $1: 5$",
+          "B. $1: 6$",
+          "C. $1: 7$"
+        ],
+        "answer": "A",
+        "solution": "【分析】含盐 $\\frac{1}{6}$ 的盐水, 将盐水质量看作 6 , 盐的质量看作 1 , 盐水一盐 = 水, 根据比的意义写出盐和水的比即可。两数相除又叫两个数的比。\n\n【详解】1: $(6 - 1) = 1:5$故答案为: A",
+        "knowledge_points": [
+          "比的意义"
+        ],
+        "source": "21-22 六年级上·辽宁抚顺·期末",
+        "merged_id": "unit4_205",
+        "original_id": "unit4_choice_205"
+      },
+      {
+        "id": "unit4_choice_206",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "如果把 $3: 7$ 的前项加上 9，要使它的比值不变，后项应（）。",
+        "options": [
+          "A. 加上 9",
+          "B. 加上 21",
+          "C. 减去 9"
+        ],
+        "answer": "B",
+        "solution": "【分析】比的前项和后项，同时乘或除以相同的数（0 除外），比值不变，据此分析。关键是掌握并灵活运用比的基本性质。\n\n【详解】 $9 \\div 3 \\times 7 = 21$故答案为: B",
+        "knowledge_points": [
+          "比的基本性质"
+        ],
+        "source": "2014 六年级·全国·课后作业",
+        "merged_id": "unit4_206",
+        "original_id": "unit4_choice_206"
+      },
+      {
+        "id": "unit4_choice_207",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "7:10 也可以写成 $\\frac{7}{10}$, 仍读作 ( )。",
+        "options": [
+          "A. 十分之七",
+          "B. 7 比 10",
+          "C. 10 比 7"
+        ],
+        "answer": "B",
+        "solution": "【分析】两个数相除叫做两个数的比，如：3比2记作3:2, 分数的分子相当于除法算式中的被除数, 也相当于比的前项; 分数的分母相当于除法算式中的分母, 也相当于比的后项; 分数线相当于除法算式中的除号, 也相当于比中的比号; 如: 3 比 2 可以记作 $3:2$, 也可以记作 $\\frac{3}{2}$, 据此解答。掌握分数、除法和比的关系，以及比的读写方法是解答题目的关键。\n\n【详解】根据分数与除法的关系，$7:10 = 7 \\div 10 = \\frac{7}{10}$，两个数的比也可以写成分数的形式，所以 $7:10$ 也可以写成 $\\frac{7}{10}$，仍读作'7比10'。故答案为: B",
+        "knowledge_points": [
+          "比的读法、写法及各部分的名称"
+        ],
+        "source": "21-22 六年级上·广西玉林·期末",
+        "merged_id": "unit4_207",
+        "original_id": "unit4_choice_207"
+      },
+      {
+        "id": "unit4_choice_208",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "一个三角形三个内角的度数比是 $2:3:5$, 这个三角形是 ( )。",
+        "options": [
+          "A. 直角三角形",
+          "B. 锐角三角形",
+          "C. 钝角三角形"
+        ],
+        "answer": "A",
+        "solution": "【分析】用 $180^{\\circ}$ 除以总份数求出每份是多少度, 再乘最大角对应的份数求出度数, 进而判断出是什么三角形。本题较易，考查了按比例分配的知识点，关键是先求出最大角的度数。\n\n【详解】 $180^{\\circ} \\div (2 + 3 + 5) \\times 5\\= 1 8 0 ^ {\\circ} \\div 1 0 \\times 5 \\ = 9 0 ^ {\\circ}$\n\n是直角三角形;故答案为: A。",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "2016·全国·小升初真题",
+        "merged_id": "unit4_208",
+        "original_id": "unit4_choice_208"
+      },
+      {
+        "id": "unit4_choice_209",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "两个正方形边长比是 2:3，它们的面积比是（）。",
+        "options": [
+          "A. $2: 3$",
+          "B. $3: 2$",
+          "C. $4: 9$",
+          "D. 9:4"
+        ],
+        "answer": "C",
+        "solution": "【分析】两个正方形的面积比是边长比的平方，据此解答即可。熟记正方形边长比和面积比的关系。\n\n【详解】两个正方形边长比是 $2: 3$, 它们的面积比是 $4: 9$;故答案为: C。",
+        "knowledge_points": [
+          "比的意义"
+        ],
+        "source": "2014 六年级·全国·课后作业",
+        "merged_id": "unit4_209",
+        "original_id": "unit4_choice_209"
+      },
+      {
+        "id": "unit4_choice_210",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "把 8:15 的前项增加16，要使比值不变，后项应该（）。",
+        "options": [
+          "A. 加上 16",
+          "B. 乘 16",
+          "C. 除以 16",
+          "D. 乘3"
+        ],
+        "answer": "D",
+        "solution": "【分析】比的性质：比的前项和比的后项同时乘或除以同一个不为 0 的数，比值不变。本题考查比的性质, 解答本题的关键是掌握比的性质。\n\n【详解】前项增加 16，变成 24，相当于前项乘上 3，要使比值不变，后项也应该乘 3。故选: D。",
+        "knowledge_points": [
+          "比的基本性质"
+        ],
+        "source": "2014 六年级·全国·课后作业",
+        "merged_id": "unit4_210",
+        "original_id": "unit4_choice_210"
+      },
+      {
+        "id": "unit4_choice_211",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "六年级全班人数与女生人数的比是 $5: 2$, 则男生人数与女生人数的比是 ( )。",
+        "options": [
+          "A. $2: 5$",
+          "B. $3: 2$",
+          "C. $2: 3$"
+        ],
+        "answer": "B",
+        "solution": "【分析】由题意可知, 把全班人数平均分成 5 份, 女生人数占其中的 2 份, 则男生人数占总人数的 (5-2) 份, 根据比的意义即可求得男女生的人数比。掌握比的意义并求出男生人数占全班人数的份数是解答题目的关键。\n\n【详解】男生人数：女生人数 $= (5 - 2):2 = 3:2$故答案为: B",
+        "knowledge_points": [
+          "比的意义"
+        ],
+        "source": "21-22 六年级上·湖南怀化·单元测试",
+        "merged_id": "unit4_211",
+        "original_id": "unit4_choice_211"
+      },
+      {
+        "id": "unit4_choice_212",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "两个正方体的棱长比是 $2:5$，它们的体积比是（）。",
+        "options": [
+          "A. $2: 5$",
+          "B. $4: 25$",
+          "C. 8:125"
+        ],
+        "answer": "C",
+        "solution": "【分析】正方体的体积比等于棱长立方的比, 据此解答即可。熟记正方体的体积比与棱长比的关系是解答本题的关键。\n\n【详解】两个正方体的棱长比是 $2: 5$, 它们的体积比是 $2^{3}: 5^{3} = 8: 125$;故答案为: C。",
+        "knowledge_points": [
+          "比的意义、正方体的体积"
+        ],
+        "source": "18-19六年级上·广东广州·期末",
+        "merged_id": "unit4_212",
+        "original_id": "unit4_choice_212"
+      },
+      {
+        "id": "unit4_choice_213",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "一个三角形三个内角的比是2:3:4，这个三角形是（）三角形。",
+        "options": [
+          "A. 锐角",
+          "B. 直角",
+          "C. 钝角"
+        ],
+        "answer": "A",
+        "solution": "【分析】三角形内角和 $180^{\\circ}$, 内角和 $\\div$ 总份数, 求出一份数,一份数 $\\times$ 最大份数, 求出最大的角, 再判断三角形类型即可。关键是理解比的意义，掌握三角形分类标准。\n\n【详解】 $180^{\\circ} \\div (2 + 3 + 4)\\= 1 8 0 ^ {\\circ} \\div 9 \\ = 2 0 ^ {\\circ}$\n\n$2 0 ^ {\\circ} \\times 4 = 8 0 ^ {\\circ}$这个三角形是锐角三角形。故答案为：A",
+        "knowledge_points": [
+          "三角形的分类、按比分配问题"
+        ],
+        "source": "2017·全国·小升初真题",
+        "merged_id": "unit4_213",
+        "original_id": "unit4_choice_213"
+      },
+      {
+        "id": "unit4_choice_214",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "将甲班人数的 $\\frac{1}{8}$ 调入乙班后, 两班人数相等, 原来甲、乙两班的人数比是 ( )。",
+        "options": [
+          "A. $4: 3$",
+          "B. $5: 3$",
+          "C. $5: 4$",
+          "D. $3: 4$"
+        ],
+        "answer": "A",
+        "solution": "【分析】把甲班人数的 $\\frac{1}{8}$ 调入乙班后，两班人数相等，说明甲班人数比乙班人数多甲班人数的 $(\\frac{1}{8} \\times 2) = \\frac{2}{8}$，把甲班人数看作单位 '1', 则乙班人数是甲班人数的 $(1 - \\frac{2}{8})$, 进而根据题意, 进行比即可。解答此题的关键：判断出单位'1'，转化为同一单位'1'下进行比，然后化为最简整数比即可。\n\n【详解】1: $(1 - \\frac{1}{8} \\times 2)\\= 1: \\frac {3}{4} \\ = (1 \\times 4): (\\frac {3}{4} \\times 4), \\ = 4: 3$\n\n故选: A",
+        "knowledge_points": [
+          "比的化简"
+        ],
+        "source": "2014 六年级·全国·课后作业",
+        "merged_id": "unit4_214",
+        "original_id": "unit4_choice_214"
+      },
+      {
+        "id": "unit4_choice_215",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "把 3:5 的后项加上 15，要使比值不变，前项要加上（）。",
+        "options": [
+          "A. 15",
+          "B. 12",
+          "C. 9"
+        ],
+        "answer": "C",
+        "solution": "【分析】3:5的后项加上15，由5变成20，相当于后项乘\n\n4, 根据比的性质, 要使比值不变, 前项也应该乘 4 , 由 3 变成 12, 也可以认为是前项加上 9。此题考查比的性质的运用：比的前项和后项只有同时乘或除以相同的数（0 除外），比值才不变。\n\n【详解】由分析可知:\n\n$(5 + 1 5) \\div 5 \\ = 2 0 \\div 5 \\ = 4 \\ 3 \\times 4 - 3 \\ = 1 2 - 3 \\ = 9$\n\n故选: C",
+        "knowledge_points": [
+          "比的基本性质"
+        ],
+        "source": "20-21 六年级上·湖南怀化·单元测试",
+        "merged_id": "unit4_215",
+        "original_id": "unit4_choice_215"
+      },
+      {
+        "id": "unit4_choice_216",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "23 分钟与 1 小时的比是 ( )。",
+        "options": [
+          "A. 23:1",
+          "B. 23:10",
+          "C. 23:60"
+        ],
+        "answer": "C",
+        "solution": "【分析】根据1小时 $= 60$ 分，根据比的意义列式解答。此题考查的是单位换算, 掌握 1 小时 $= 60$ 分是解答此题的关键。\n\n【详解】$23分：1小时＝23分∶60分＝23∶60$ 故选：C",
+        "knowledge_points": [
+          "比的化简"
+        ],
+        "source": "20-21 六年级上·湖南怀化·单元测试",
+        "merged_id": "unit4_216",
+        "original_id": "unit4_choice_216"
+      },
+      {
+        "id": "unit4_choice_217",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "在 3:5 中，如果前项加 6，要使比值不变，后项应（）。",
+        "options": [
+          "A. 增加 6",
+          "B. 乘6",
+          "C. 乘3"
+        ],
+        "answer": "C",
+        "solution": "【分析】计算出比的前项加 6 之后前项扩大的倍数，根据比的基本性质计算出比的后项即可。掌握比的基本性质是解答题目的关键。\n\n【详解】 $(3 + 6) \\div 3 \\times 5\\= 9 \\div 3 \\times 5 \\ = 3 \\times 5 \\ = 1 5$\n\n$1 5 \\div 5 = 3 \\quad 1 5 - 5 = 1 0$ 则比的后项应乘 3 或加 $10$。故答案为: C",
+        "knowledge_points": [
+          "比的基本性质"
+        ],
+        "source": "20-21 六年级上·湖南怀化·单元测试",
+        "merged_id": "unit4_217",
+        "original_id": "unit4_choice_217"
+      },
+      {
+        "id": "unit4_choice_218",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "故事书的本数是科技书的 $\\frac{2}{5}$, 正确的等量关系式是 ( )。",
+        "options": [
+          "A. 故事书的本数：科技书的本数 $\\frac{2}{5}$",
+          "B. 故事书的本数$\\times \\frac{2}{5} =$ 科技书的本数",
+          "C. 科技书的本数$\\div \\frac{2}{5} = 6$ 故事书的本数"
+        ],
+        "answer": "A",
+        "solution": "【分析】故事书的本数是科技书的 $\\frac{2}{5}$, 运用分数的意义可得: 故事书的本数 $\\div$ 科技书的本书 $= \\frac{2}{5}$, 据此分析各个选项即可得出答案。本题主要考查的是比的应用, 解题的关键是熟练运用比的意义解决问题, 进而得出本题答案。\n\n【详解】A．故事书的本数：科技书的本数 $\\frac{2}{5}$，等量关系正确； B. 故事书的本数 $\\times \\frac{2}{5} = i$ 科技书的本数, 应当运用分数的除法得到科技树木数, 故错误; C. 科技书的本数 $\\div \\frac{2}{5} = i$ 故事书的本数, 应当运用分数乘法得到故事书的本书, 故错误。故答案为: A",
+        "knowledge_points": [
+          "比的应用"
+        ],
+        "source": "21-22 六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_218",
+        "original_id": "unit4_choice_218"
+      },
+      {
+        "id": "unit4_choice_219",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "最简比的前项和后项一定是（）。",
+        "options": [
+          "A. 奇数",
+          "B. 偶数",
+          "C. 互质数"
+        ],
+        "answer": "C",
+        "solution": "【分析】最简比的前项和后项都为整数, 并且是互质数,据此可解答本题。本题主要考查的是最简比的定义, 解题的关键是熟练运用最简比的定义, 进而得出答案。\n\n【详解】最简比的前项和后项都为整数，并且是互质数，如最简比 $2: 3$，前项、后项分别是 2 和 3，且 2 和 3 是互质数。故答案为：C",
+        "knowledge_points": [
+          "比的化简"
+        ],
+        "source": "15-16 六年级·云南曲靖·期中",
+        "merged_id": "unit4_219",
+        "original_id": "unit4_choice_219"
+      },
+      {
+        "id": "unit4_choice_220",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "妈妈调制了 3 杯蜂蜜水，最甜的一杯是（）。",
+        "options": [
+          "A. 蜂蜜与水的质量比是 $1: 7$",
+          "B. $10 \\mathrm{~g}$ 蜂蜜调制了 $75 \\mathrm{~g}$ 蜂蜜水",
+          "C. 蜂蜜与蜂蜜水的质量是 $1: 7$"
+        ],
+        "answer": "C",
+        "solution": "【分析】用蜂蜜的质量除以蜂蜜水的质量, 求出蜂蜜占蜂蜜水的几分之几, 进而选择。此题主要考查了比的意义，认真解答即可。\n\n【详解】A. $1 \\div (1 + 7) = 1 \\div 8 = \\frac{1}{8}$ B. $10 \\div 75 = \\frac{2}{15}$ C. $1 \\div 7 = \\frac{1}{7}$ $\\frac {1}{7} > \\frac {2}{1 5} > \\frac {1}{8}$故答案为: C",
+        "knowledge_points": [
+          "比的意义、求一个数占另一个数几分之几"
+        ],
+        "source": "20-21 六年级上·河北秦皇岛·期中",
+        "merged_id": "unit4_220",
+        "original_id": "unit4_choice_220"
+      },
+      {
+        "id": "unit4_choice_221",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "苹果的重量比梨多 $\\frac{3}{8}$, 则下面描述正确的有 ( ) 个。\n\n① 苹果与梨重量的比是 。\n\n② 梨的重量比苹果少 $\\frac{3}{11}$。\n\n③ 梨与苹果的重量比是 $8: 11$。\n\n④ 苹果的重量是梨的 $\\frac{8}{11}$。",
+        "options": [
+          "A. 1",
+          "B. 2",
+          "C. 3",
+          "D. 4"
+        ],
+        "answer": "B",
+        "solution": "【分析】由题意可知, 把梨的重量看作单位 '1', 苹果的重量占梨重量的 $(1 + \\frac{3}{8})$, 据此计算。掌握比的意义和分数除法的计算方法是解答题目的关键。\n\n【详解】① $(1 + \\frac{3}{8}): 1 = \\frac{11}{8}: 1 = 11: 8$\n\n② $(1 + \\frac{3}{8} - 1) \\div (1 + \\frac{3}{8})\\ = \\frac {3}{8} \\div \\frac {1 1}{8} \\ = \\frac {3}{8} \\times \\frac {8}{1 1} \\ = \\frac {3}{1 1}$\n\n③ $1: (1 + \\frac{3}{8}) = 1: \\frac{11}{8} = 8: 11$④ $(1 + \\frac{3}{8}) \\div 1 = \\frac{11}{8}$由上可知, ②③正确。故答案为: B",
+        "knowledge_points": [
+          "分数除法的应用、比的意义、求一个数占另一个数几分之几"
+        ],
+        "source": "20-21 六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_221",
+        "original_id": "unit4_choice_221"
+      },
+      {
+        "id": "unit4_fill_222",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "阅读材料, 回答问题。根据国家卫健委最新公布的数据，我国儿童青少年总体近视率为 $52.7\\%$。针对此现状，《儿童青少年近视防控公共卫生综合干预技术指南》指出儿童青少年近视普遍易感，必须彩取全人群策略防控，开展近视防控健康教育，增加日间户外活动，减少近距离、长时间用眼。还有一些小学生爱眼护眼小常识：读书写字时眼睛要距离书本一尺，一米的长度和一尺的长度的比是 $3: 1$，看电视时也要保持最佳观看距离，最佳观看距离电视对角线距离 $\\times 4$。注：电视机的寸数就是指电视机的屏幕对角线长度。英寸是一种长度单位，1英寸 $\\text{2.54厘米}$。(1)根据上面的材料可知, 一尺约 $(\\mathrm{~cm})$ 厘米, 60 英寸电视的最佳观看距离是 $(\\mathrm{~cm})$。 (得数保留一位小数) (2)如今高清电视屏幕的长与宽之比由原来的 $4: 3$ 发展为 $16: 9$, 下面四位同学说了自己对 $16: 9$ 的理解, 其中错误的是 ( )。(填名字)\n\n亮亮：如果电视机屏幕长 8 英寸，那么宽应该是 4.5 英寸\n\n乐乐：电视机屏幕的宽是长的 $\\frac{9}{16}$\n\n琪琪：电视机屏幕的长比宽长 $\\frac{7}{9}$\n\n军军：电视机屏幕的长减少7英寸，就和宽一样长了",
+        "options": [],
+        "answer": "(1) 33.3   6.1  (2)军军",
+        "solution": "【分析】（1）根据 1 米 = 100 厘米，一米的长度和一尺的长度的比是 $3: 1$，即一米可平均分成 3 份，用除法计算可得每份是多少，一尺是 1 份，即每份就是一尺的长度。根据 1 英寸 = 2.54 厘米，60 英寸电视的对角线距离就是 60 个 2.54 厘米，可用乘法计算，再根据最佳观看距离电视对角线距离 $\\times 4$，代入数据即可。得数采用'四舍五入法'保留一位小数。(2) 亮亮: 如果电视机屏幕长 8 英寸, 又知长有 16 份, 可用除法求每份的长度, 再乘宽的份数, 看是否是 4.5 英寸即可。乐乐：把长看作 16，宽看作 9，根据求一个数是另一个数的几分之几，用除法计算即可。琪琪：把长看作 16，宽看作 9，根据求一个数比另一个数多几分之几，用多的除以另一个数计算即可。军军：根据比的意义分析，16:9是说长可以平均分为 16 份，宽可平均分为 9 份, 16 和 9 并不是具体长度, 据此解答。\n\n【详解】 (1) 1 米 $= 100$ 厘米 $100 \\div 3 \\approx 33.3$ (厘米)\n\n$2.5 4 \\times 6 0 \\times 4 \\=15 2. 4 \\times 4\\=609.6 (厘米)\\\\approx 6.1 (米)$\n\n一尺约 33.3 厘米, 60 英寸电视的最佳观看距离是 6.1 米。\n\n(2) 亮亮: $8 \\div 16 \\times 9 = 0.5 \\times 9 = 4.5$ (英寸), 如果电视机屏幕长8英寸，那么宽应该是4.5英寸，说法正确。\n\n乐乐: $9 \\div 16 = \\frac{9}{6}$, 电视机屏幕的宽是长的 $\\frac{9}{16}$, 说法正确。琪琪: $(16 - 9) \\div 9 = 7 \\div 9 = \\frac{7}{9}$, 视机屏幕的长比宽长 $\\frac{7}{9}$, 说法正确。军军：根据比的意义分析，16:9是说长可以平均分为 16 份，宽可平均分为 9 份，16 和 9 并不是具体长度。所以电视机屏幕的长减少 7 英寸, 就和宽一样长了, 说法错误。四位同学说了自己对16:9的理解，其中错误的是军军。",
+        "knowledge_points": [
+          "比的应用、厘米和米之间的进率与换算、比的意义、求一个数占另一个数几分之几"
+        ],
+        "source": "24-25 六年级上·贵州六盘水·期中",
+        "merged_id": "unit4_222",
+        "original_id": "unit4_fill_222"
+      },
+      {
+        "id": "unit4_fill_223",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "'二十四节气' 准确反映了自然节律变化，作为农耕文明的产物是我国劳动\n\n人民的智慧结晶。'夏至'是一年中白天最长的日子，今年'夏至'这天某一地区的白昼和黑夜的时间比大约为7:5, 这一天, 该地区黑夜约有( )小时, 白昼约有( )小时。",
+        "options": [],
+        "answer": " 10 14",
+        "solution": "【分析】一天有 24 小时, 将比的前后项看成份数, 一天时间 $\\div$ 总份数 $=$ 一份数, 一份数分别乘黑夜和白昼对应份数,即可求出黑夜和白昼时间。\n\n【详解】 $24 \\div (7 + 5)\\ = 2 4 \\div 1 2 \\ = 2 \\quad (\\text {小 时})$\n\n$2 \\times 5 = 1 0 (\\text {小 时}) \\quad 2 \\times 7 = 1 4 (\\text {小 时})$ 该地区黑夜约有 10 小时, 白昼约有 14 小时。",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "23-24 六年级上·河北邢台·期中",
+        "merged_id": "unit4_223",
+        "original_id": "unit4_fill_223"
+      },
+      {
+        "id": "unit4_fill_224",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "$0.5: \\frac{3}{4}$ 化成最简整数比是( )，比值是( )。",
+        "options": [],
+        "answer": "$2: 3 \\quad \\frac {2}{3}$",
+        "solution": "【分析】（1）先把比的前项化成分数，再根据比的基本性质作答，即比的前项和后项同时乘一个数或除以一个数（0除外）比值不变。(2) 用比的前项除以后项即可。此题主要考查了化简比和求比值的方法, 另外还要注意化简比的结果是一个比, 它的前项和后项都是整数,并且是互质数; 而求比值的结果是一个商, 可以是整数,小数或分数。\n\n【详解】 (1) $0.5:\\frac{3}{4}\\= (0. 5 \\times 4): (\\frac {3}{4} \\times 4)\\= 2: 3$\n\n(2) $0.5:\\frac{3}{4}\\= 0. 5 \\div \\frac {3}{4} \\ = \\frac {2}{3}$",
+        "knowledge_points": [
+          "比的化简、求比值"
+        ],
+        "source": "2021·广东肇庆·小升初真题",
+        "merged_id": "unit4_224",
+        "original_id": "unit4_fill_224"
+      },
+      {
+        "id": "unit4_fill_225",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "一个长方体的棱长总和是 120 厘米，长、宽、高的比是 $5: 3: 2$，这个长方体的体积是( )立方厘米。",
+        "options": [],
+        "answer": "810",
+        "solution": "【分析】棱长总和 $\\div 4$, 先求出一组长宽高的和, 长宽高的和 $\\div$ 总份数, 求出一份数, 一份数分别乘长、宽、高的对应份数, 求出长、宽、高, 根据长方体体积 $=$ 长 $\\times$ 宽 $\\times$ 高, 计算即可。关键是理解比的意义，掌握长方体棱长总和以及体积公式。\n\n【详解】 $120 \\div 4 \\div (5 + 3 + 2)\\=3 0 \\div 1 0\\=3 (厘米)$\n\n$3 \\times 5 = 1 5 (\\text {厘 米}) \\ 3 \\times 3 = 9 \\text {(厘 米)} \\ 3 \\times 2 = 6 (\\text {厘 米}) \\ 1 5 \\times 9 \\times 6 = 8 1 0 (\\text {立 方 厘 米})$",
+        "knowledge_points": [
+          "长方体有关棱长的应用、长方体的体积、按比分配问题"
+        ],
+        "source": "2014 五年级·全国·课后作业",
+        "merged_id": "unit4_225",
+        "original_id": "unit4_fill_225"
+      },
+      {
+        "id": "unit4_fill_226",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "果园里有苹果树、梨树和桃树棵数比是 $7: 2: 1$, 已知梨树和桃树共 870 棵,苹果树有( )棵。",
+        "options": [],
+        "answer": "2030",
+        "solution": "【分析】把果树的总棵数看作单位 '1', 苹果树占果树总棵数的 $\\frac{7}{7+2+1}$, 梨树占果树总棵数的 $\\frac{2}{7+2+1}$, 桃树占果树总棵数的 $\\frac{1}{7+2+1}$, 根据 '量÷对应的分率' 求出果树总棵数, 最后求出苹果树的棵数, 据此解答。掌握按比例分配问题的解题方法是解答题目的关键。\n\n【详解】 $870 \\div (\\frac{2}{7 + 2 + 1} + \\frac{1}{7 + 2 + 1}) \\times \\frac{7}{7 + 2 + 1}$\n\n$= 8 7 0 \\div (\\frac {1}{5} + \\frac {1}{1 0}) \\times \\frac {7}{1 0} \\ = 8 7 0 \\div \\frac {3}{1 0} \\times \\frac {7}{1 0} \\ = 2 9 0 0 \\times \\frac {7}{1 0} \\ = 2 0 3 0 \\text {(棵)}$\n\n所以，苹果树有 2030 棵。",
+        "knowledge_points": [
+          "比的应用、按比分配问题"
+        ],
+        "source": "18-19 六年级下·广东汕头·单元测试",
+        "merged_id": "unit4_226",
+        "original_id": "unit4_fill_226"
+      },
+      {
+        "id": "unit4_fill_227",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "甲数的 $\\frac{2}{5}$ 等于乙数的 $\\frac{3}{4}$, 甲乙两数的最简整数比是( )。",
+        "options": [],
+        "answer": " 15:8",
+        "solution": "【分析】由题意可知, 甲数 $\\times \\frac{2}{5} =$ 乙数 $\\times \\frac{3}{4}$, 假设式子的值为 1 , 求出甲数和乙数的值, 最后根据比的意义求出甲乙两数的最简整数比, 据此解答。掌握比的意义和化简方法是解答题目的关键\n\n【详解】假设甲数 $\\times \\frac{2}{5} =$ 乙数 $\\times \\frac{3}{4} = 1$甲数 $= \\frac{5}{2}$, 乙数 $= \\frac{4}{3}$甲数: 乙数 $= \\frac{5}{2}: \\frac{4}{3} = (\\frac{5}{2} \\times 6): (\\frac{4}{3} \\times 6) = 15: 8$",
+        "knowledge_points": [
+          "比的意义、比的化简"
+        ],
+        "source": "2022·四川绵阳·小升初真题",
+        "merged_id": "unit4_227",
+        "original_id": "unit4_fill_227"
+      },
+      {
+        "id": "unit4_fill_228",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "两个圆直径的比是 $3: 2$，它们周长的比是( )，面积的比是( )。",
+        "options": [],
+        "answer": " 3:2 9:4",
+        "solution": "【分析】圆的周长 $= 2 \\pi r$, 圆的面积 $= \\pi r^{2}$, 直径的比就是半径的比; 设小圆的半径为 2 , 则大圆的半径为 3 , 分别求两个圆的周长和面积, 即可求得其周长比和面积比。此题主要考查圆的周长和面积的计算方法的灵活应用。\n\n【详解】设小圆的半径为2，则大圆的半径为3，周长之比: $(2\\pi \\times 3): (2\\pi \\times 2) = 9\\pi: 6\\pi = 3: 2$面积之比: $(\\pi \\times 3^{2}) : (\\pi \\times 2^{2}) = 9 \\pi: 4 \\pi = 9: 4$",
+        "knowledge_points": [
+          "比的应用、圆的周长及应用、圆的面积及应用"
+        ],
+        "source": "2015·全国·小升初真题",
+        "merged_id": "unit4_228",
+        "original_id": "unit4_fill_228"
+      },
+      {
+        "id": "unit4_fill_229",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "一个直角三角形中,两个锐角度数的比是 $3: 2$, 这两个锐角分别是( )度和( )度。",
+        "options": [],
+        "answer": " 54 36",
+        "solution": "【分析】在一个直角三角形中，两个锐角度数的和是 $90^{\\circ}$,它们的比是 3:2, 根据按比分配的方法求出两个角的度数即可。此题考查了三角形内角和以及利用比的意义求三角形各个角的度数, 解答此题关键是掌握两个锐角度数的和是 $90^{\\circ}$。\n\n【详解】总份数： $3 + 2 = 5$ （份），第一个锐角的度数: $90 \\times \\frac{3}{5} = 54^{\\circ}$ 第二个锐角的度数: $90 \\times \\frac{2}{5} = 36^{\\circ}$",
+        "knowledge_points": [
+          "三角形的内角和、按比分配问题"
+        ],
+        "source": "19-20 六年级下·湖南怀化·期末",
+        "merged_id": "unit4_229",
+        "original_id": "unit4_fill_229"
+      },
+      {
+        "id": "unit4_fill_230",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "一种农药，药和水的比是 $1: 3000$，现有药 0.1 千克，要加水( )千克。",
+        "options": [],
+        "answer": "300",
+        "solution": "【分析】根据比的意义，药÷对应份数×水的对应份数 = 水的质量，据此列式计算。关键是理解比的意义，先确定一份数。\n\n【详解】 $0.1 \\div 1 \\times 3000\\= 0. 1 \\times 3 0 0 0 \\ = 3 0 0 (\\text {千 克})$",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "19-20 六年级下·湖南怀化·单元测试",
+        "merged_id": "unit4_230",
+        "original_id": "unit4_fill_230"
+      },
+      {
+        "id": "unit4_fill_231",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "学校买5个足球花125元，买4个篮球花240元。所买篮球与足球个数的比是( )；买篮球花的总钱数与买足球花的总钱数的比是( )。",
+        "options": [],
+        "answer": " 4:5 48:25",
+        "solution": "【分析】由题意可知，篮球4个，足球5个，根据比的意义求出篮球与足球的个数比；买篮球花的总钱数为 240 元，买足球花的总钱数为 125 元，根据化简整数比的方法求出买篮球花的总钱数与买足球花的总钱数的最简整数比，据此解答。掌握比的意义和化简方法是解答题目的关键。\n\n【详解】（1）篮球个数：足球个数 $= 4:5$(2) $篮球总钱数：足球总钱数= 2 4 0: 1 2 5 \\ = (2 4 0 \\div 5): (1 2 5 \\div 5) \\ = 4 8: 2 5$",
+        "knowledge_points": [
+          "比的意义、比的化简"
+        ],
+        "source": "19-20 六年级下·湖南怀化·单元测试",
+        "merged_id": "unit4_231",
+        "original_id": "unit4_fill_231"
+      },
+      {
+        "id": "unit4_fill_232",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "0.3:0.25 的比值是( )，化成最简整数比是( )。",
+        "options": [],
+        "answer": " 1.2   6:5",
+        "solution": "【分析】先根据比的基本性质移动小数点的位置，把小数比化为整数比，再按照整数比的化简方法计算；比的前项除以后项所得的商即为比值。据此解答。\n\n【详解】 $0.3: 0.25\\= 3 0: 2 5 \\ = (3 0 \\div 5): (2 5 \\div 5) \\ = 6: 5 \\ = 1. 2$",
+        "knowledge_points": [
+          "比的化简、求比值"
+        ],
+        "source": "19-20 六年级下·湖南怀化·单元测试",
+        "merged_id": "unit4_232",
+        "original_id": "unit4_fill_232"
+      },
+      {
+        "id": "unit4_fill_233",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "( ) : $12 = \\frac{1}{3} =$ ()÷9。",
+        "options": [],
+        "answer": " 4 3",
+        "solution": "【分析】分数的分子相当于除法算式中的被除数，也相当于比的前项；分数的分母相当于除法算式中的除数，也相当于比的后项；分数线相当于除法算式中的除号，也相当于比中的比号；根据商不变规律和比的基本性质解答。掌握比、分数、除法之间的关系是解答题目的关键。\n\n【详解】 $\\frac{1}{3} = 1:3 = 1\\div 3$ $1: 3 = (1 \\times 4): (3 \\times 4) = 4: 1 2$ $1 \\div 3 = (1 \\times 3) \\div (3 \\times 3) = 3 \\div 9$ (4) $12 = \\frac{1}{3} = (3) \\div 9$。",
+        "knowledge_points": [
+          "比与分数、除法的关系"
+        ],
+        "source": "19-20 六年级下·湖南怀化·单元测试",
+        "merged_id": "unit4_233",
+        "original_id": "unit4_fill_233"
+      },  {
+        "id": "unit4_fill_234",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "从甲城到乙城，货车要行 5 小时，客车要行 6 小时，货车的速度与客车的速度的最简比是( )。",
+        "options": [],
+        "answer": "6:5",
+        "solution": "【分析】根据题意可知，总路程为单位'1'，货车的速度为$\\frac{1}{5}$, 客车的速度为 $\\frac{1}{6}$, 再写出货车的速度与客车的速度的比即可。明确单位'1'，进而确定货车与客车的速度即可。\n\n【详解】货车的速度与客车的速度的比为 $\\frac{1}{5}: \\frac{1}{6} = 6: 5$。",
+        "knowledge_points": [
+          "比的意义、比的化简"
+        ],
+        "source": "2014 六年级·全国·课后作业",
+        "merged_id": "unit4_234",
+        "original_id": "unit4_fill_234"
+      },
+      {
+        "id": "unit4_fill_235",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "$\\frac{2}{5}: 0.75$ 化成最简整数比是( )，比值是( )。",
+        "options": [],
+        "answer": " $8: 15 \\quad \\frac{8}{15}$",
+        "solution": "【分析】根据比的基本性质，即比的前项和后项同时乘或除以一个相同的数（0 除外）比值不变，进而把比化成最简比；根据求比值的方法，就是用比的前项除以后项即得比值。化简整数比最后的答案是一个比，而求比值最后的答案是一个比值，它可以表示为一个整数、分数或小数。\n\n【 详 解 】$\\frac {2}{5}: 0. 7 5 = \\frac {2}{5}: \\frac {3}{4} = (\\frac {2}{5} \\times 2 0): (\\frac {3}{4} \\times 2 0) = 8: 1 5 \\ \\frac {2}{5}: 0. 7 5 = \\frac {2}{5} \\div \\frac {3}{4} = \\frac {2}{5} \\times \\frac {4}{3} = \\frac {8}{1 5}$",
+        "knowledge_points": [
+          "比的化简、求比值"
+        ],
+        "source": "2017·四川雅安·小升初真题",
+        "merged_id": "unit4_235",
+        "original_id": "unit4_fill_235"
+      },
+      {
+        "id": "unit4_fill_236",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "一个直角三角形的两个锐角的度数比是 $4: 5$, 则这两个锐角分别是( ) 度和() 度。",
+        "options": [],
+        "answer": " 40  50",
+        "solution": "【分析】用两个锐角的度数和除以总份数求出每份多少度,再乘两个锐角各自对应的份数即可。本题较易，熟练掌握按比例分配解决问题的方法是解答本题的关键。\n\n【详解】 $(180^{\\circ} - 90^{\\circ}) \\div (4 + 5)\\= 9 0 ^ {\\circ} \\div 9 \\ = 1 0 ^ {\\circ}$\n\n$1 0 ^ {\\circ} \\times 4 = 4 0 ^ {\\circ}$   $1 0 ^ {\\circ} \\times 5 = 5 0 ^ {\\circ}$",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "17-18 六年级上·全国·课后作业",
+        "merged_id": "unit4_236",
+        "original_id": "unit4_fill_236"
+      },
+      {
+        "id": "unit4_fill_237",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "甲: 乙 $= 3: 4$, 乙: 丙 $= 5: 6$, 甲: 乙: 丙 $=\\_ : \\_ : \\_$。",
+        "options": [],
+        "answer": " 15  20  24",
+        "solution": "【分析】甲乙两数的比是 $3: 4$, 乙丙两数的比是 $5: 6$, 根据比的基本性质, 甲乙两数的比 $3: 4 = (3 \\times 5) : (4 \\times 5) = 15: 20$; 乙丙两数的比 $5: 6 = (5 \\times 4) : (6 \\times 4) = 20: 24$, 所以甲乙丙三数的比为: $15: 20: 24$; 据此解答。掌握比的基本性质是解答题目的关键。\n\n【详解】甲：乙 $= 3:4 = (3 \\times 5)$ ： $(4 \\times 5) = 15:20$ 乙: 丙 $= 5: 6 = (5 \\times 4) : (6 \\times 4) = 20: 24$\n\n甲：乙：丙 $= 15:20:24$",
+        "knowledge_points": [
+          "比的基本性质"
+        ],
+        "source": "2019·浙江宁波·小升初真题",
+        "merged_id": "unit4_237",
+        "original_id": "unit4_fill_237"
+      },
+      {
+        "id": "unit4_fill_238",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "15: ( ) $= \\frac{5}{6} = 35 \\div$ ( )。",
+        "options": [],
+        "answer": "18；20；42",
+        "solution": "【分析】分数的基本性质：分子和分母同时乘或除以同一个不为 0 的数，分数值不变；再利用分数与比、除法的关系进行解答即可。本题考查比与分数、除法的互化，解答本题的关键是掌握比与分数、除法的互化方法。\n\n【详解】 $\\frac{5}{6} = \\frac{20}{24} = 5 \\div 6 = 35 \\div 42 = 5:6 = 15:18$",
+        "knowledge_points": [
+          "比与分数、除法的关系"
+        ],
+        "source": "21-22 六年级上·湖南怀化·单元测试",
+        "merged_id": "unit4_238",
+        "original_id": "unit4_fill_238"
+      },
+      {
+        "id": "unit4_fill_239",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "走一段路, 甲用了 8 小时, 乙用了 10 小时, 甲与乙所行时间的最简整数比是( ), 甲与乙行走的速度比的比值是( )。",
+        "options": [],
+        "answer": " $4: 5 \\quad \\frac{5}{4}$",
+        "solution": "【分析】根据题意可知甲乙走的路程是一样，可以把所走的路程看作单位'1'，由速度 $=$ 路程 $\\div$ 时间，可以求出甲乙的速度，再根据比的意义，就可以求出甲与乙所行时间的最简比和甲与乙行走的速度比的比值。此题考查的目的是理解掌握比的意义, 比的化简方法及应用。\n\n【详解】甲与乙所行时间的最简比是：$8: 1 0 = (8 \\div 2): (1 0 \\div 2) = 4: 5$甲与乙行走的速度比的比值是:\n\n$\\frac {1}{8}: \\frac {1}{1 0} = \\frac {1}{8} \\div \\frac {1}{1 0} = \\frac {1}{8} \\times 1 0 = \\frac {5}{4}$",
+        "knowledge_points": [
+          "比的应用、比的意义、比的化简、求比值"
+        ],
+        "source": "21-22 六年级上·湖南怀化·单元测试",
+        "merged_id": "unit4_239",
+        "original_id": "unit4_fill_239"
+      },
+      {
+        "id": "unit4_fill_240",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "把 $1 \\frac{1}{2}: 0.75$ 化成最简整数比是( )，比值是( )。",
+        "options": [],
+        "answer": " 2:1 2",
+        "solution": "【分析】化简比根据比的基本性质，求比值直接用前项÷后项。化简比的结果还是一个比, 求比值的结果是一个数。\n\n【详解】 $1\\frac{1}{2}:0.75 = 1.5:0.75 = 150:75 = 2:1 = 2$",
+        "knowledge_points": [
+          "比的化简、求比值"
+        ],
+        "source": "2022·广东深圳·小升初真题",
+        "merged_id": "unit4_240",
+        "original_id": "unit4_fill_240"
+      },
+      {
+        "id": "unit4_fill_241",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "已知三个数的平均数是 50, 这三个数的比是 $6: 2: 7$, 那么这三个数分别是( )、( )、( )。",
+        "options": [],
+        "answer": " 60 ;20; 70",
+        "solution": "【分析】根据题意可知, 三个数的和为 $50 \\times 3 = 150$, 用 150 除以总份数求出每份是多少, 再分别乘三个数对应的份数即可。本题考查了按比例分配的知识点, 求出三个数的和是解答本题的关键, 进而求出每份是多少。\n\n【详解】 $50 \\times 3 \\div (6 + 2 + 7)\\= 1 5 0 \\div 1 5 \\ = 1 0; \\ 1 0 \\times 6 = 6 0; \\quad 1 0 \\times 2 = 2 0; \\quad 1 0 \\times 7 = 7 0$",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "21-22 六年级上·河南漯河·期中",
+        "merged_id": "unit4_241",
+        "original_id": "unit4_fill_241"
+      },
+      {
+        "id": "unit4_fill_242",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "把 4:11 的后项加上 33，要使比值不变，那么前项应该乘( )。",
+        "options": [],
+        "answer": "4",
+        "solution": "【分析】把 4:11 的后项加上 33，后项变为 44，扩大到原来的 4 倍，要使比值不变，前项也应扩大到原来的 4 倍，据此解答即可。熟练掌握比的基本性质是解答本题的关键。\n\n【详解】把 4:11 的后项加上 33 , 要使比值不变, 那么前项应该乘 4 。",
+        "knowledge_points": [
+          "比的基本性质"
+        ],
+        "source": "21-22 六年级上·河南漯河·期中",
+        "merged_id": "unit4_242",
+        "original_id": "unit4_fill_242"
+      },
+      {
+        "id": "unit4_fill_243",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "甲、乙两数之比为 8:5, 则乙数是甲数的( ), 甲数是两数之和的( )。",
+        "options": [],
+        "answer": " $\\frac{5}{8}; \\frac{8}{13}$",
+        "solution": "【分析】求一个数是另一个数的几分之几, 用一个数除以另一个数即可。熟练掌握求一个数是另一个数的几分之几的方法是解答本题的关键。\n\n【详解】 $5 \\div 8 = \\frac{5}{8}$, 所以乙数是甲数的 $\\frac{5}{8}$;\n\n$8 \\div (8 + 5) \\ = 8 \\div 1 3 \\ = \\frac {8}{1 3}$\n\n所以甲数是两数之和的 $\\frac{8}{13}$。",
+        "knowledge_points": [
+          "比的意义、求一个数占另一个数几分之几"
+        ],
+        "source": "21-22 六年级上·河南漯河·期中",
+        "merged_id": "unit4_243",
+        "original_id": "unit4_fill_243"
+      },
+      {
+        "id": "unit4_fill_244",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "把 $10 \\mathrm{~g}$ 的糖放入 $100 \\mathrm{~g}$ 的水中，糖占糖水的( )，糖和水的质量比是( )。",
+        "options": [],
+        "answer": " $\\frac{1}{11}$；1:10",
+        "solution": "【分析】糖占糖水的几分之几，先求出糖水的总重量，然后用糖的重量除以糖水的总重量；糖和水的质量比，用糖的重量：水的重量化成最简整数比即可得解。此题考查的是分数除法应用，解答此题关键是掌握求一个数是另一个数的几分之几，关键是看把谁当成了单位'1'，单位'1'的量为除数。\n\n【详解】由分析得:糖占糖水的: $10 \\div (10 + 100)\\= 1 0 \\div 1 1 0 \\ = \\frac {1}{1 1}$\n\n糖和水的质量比是:$10:100  = 1:10$",
+        "knowledge_points": [
+          "比的化简、求一个数占另一个数几分之几"
+        ],
+        "source": "2014 六年级·全国·课后作业",
+        "merged_id": "unit4_244",
+        "original_id": "unit4_fill_244"
+      },
+      {
+        "id": "unit4_fill_245",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如果 $\\frac{3}{5} \\times A = \\frac{3}{2} \\times B$, 且 $A$ 和 $B$ 都不为 0 , 那么 $A: B = (\\quad)$.",
+        "options": [],
+        "answer": "5:2",
+        "solution": "【分析】根据等式的性质 2 , 两边先同时 $\\div$ B, 再同时 $\\times \\frac{5}{3}$,即可得到 A 与 B 的比。两数相除又叫两个数的比。\n\n【详解】 $\\frac{3}{5} \\times A = \\frac{3}{2} \\times B$\n\n解: $\\frac{3}{5} \\times A \\div B = \\frac{3}{2} \\times B \\div B \\\\\\frac {3}{5} \\times A \\div B \\times \\frac {5}{3} = \\frac {3}{2} \\times \\frac {5}{3} \\ A \\div B = \\frac {5}{2} \\ A: B = 5: 2$",
+        "knowledge_points": [
+          "比的意义、比的化简"
+        ],
+        "source": "20-21 六年级上·湖南怀化·单元测试",
+        "merged_id": "unit4_245",
+        "original_id": "unit4_fill_245"
+      },
+      {
+        "id": "unit4_fill_246",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "$\\frac{7}{8}: 0.125$ 的比值是( )，化成最简整数比是( )。",
+        "options": [],
+        "answer": " 7 7:1",
+        "solution": "【分析】（1）用比的前项除以后项即得比值；(2) 根据比的基本性质, 即比的前项和后项同时乘或除以一个相同的数 (0 除外), 比值不变, 进而把比化成最简比。此题考查化简比和求比值的方法，要注意区分：化简比的结果是一个比；而求比值的结果是一个数。\n\n【详解】 $\\frac{7}{8}: 0.125 = \\frac{7}{8} \\div \\frac{1}{8} = 7$\n\n$\\frac {7}{8}: 0. 1 2 5 = \\frac {7}{8}: \\frac {1}{8} = (\\frac {7}{8} \\times 8): (\\frac {1}{8} \\times 8) = 7: 1$",
+        "knowledge_points": [
+          "比的化简、求比值"
+        ],
+        "source": "20-21 六年级上·湖南怀化·单元测试",
+        "merged_id": "unit4_246",
+        "original_id": "unit4_fill_246"
+      },
+      {
+        "id": "unit4_fill_247",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "打同一篇文章,小强用了10分钟，小玲用了12分钟，小强和小玲的打字速度之比是( )。（填最简整数比）",
+        "options": [],
+        "answer": "6:5",
+        "solution": "【分析】把这篇稿件看作单位 '1', 根据工作效率 = 工作总量 ÷ 工作时间, 可求出小强和小玲的工作效率是 $\\frac{1}{10}$ 和 $\\frac{1}{12}$, 然后根据比的基本性质化简比即可。本题考查工作总量、工作时间和工作效率的关系，以及化简比的方法。\n\n【详解】 $\\frac{1}{10}:\\frac{1}{12}\\ = (\\frac {1}{1 0} \\times 6 0): (\\frac {1}{1 2} \\times 6 0) \\ = 6: 5$\n\n则小强和小玲的打字速度之比是 $6: 5$。",
+        "knowledge_points": [
+          "工程问题、比的化简"
+        ],
+        "source": "20-21 六年级上·湖南怀化·单元测试",
+        "merged_id": "unit4_247",
+        "original_id": "unit4_fill_247"
+      },
+      {
+        "id": "unit4_fill_248",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "男生和女生人数比是 $5: 7$, 男生占总人数的 $\\frac{()}{()}$, 女生占总人数 $\\frac{()}{()}$。",
+        "options": [],
+        "answer": " $\\frac{5}{12}; \\frac{7}{12}$",
+        "solution": "【分析】总人数平均分成（5+7）份，男生占其中的5份，女生占其中的7份，据此解答。计算出全班人数的总份数是解答题目的关键。\n\n【详解】男生占总人数： $5 \\div (5 + 7) = \\frac{5}{12}$ 女生占总人数： $7 \\div (5 + 7) = \\frac{7}{12}$",
+        "knowledge_points": [
+          "比的意义、求一个数占另一个数几分之几"
+        ],
+        "source": "20-21 六年级上·湖南怀化·单元测试",
+        "merged_id": "unit4_248",
+        "original_id": "unit4_fill_248"
+      },
+      {
+        "id": "unit4_fill_249",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "在长 5 厘米、宽 4 厘米的长方形中画一个最大的圆, 这个圆的面积是( )。",
+        "options": [],
+        "answer": "12.56 平方厘米",
+        "solution": "【分析】长方形中画一个最大的圆，圆的直径=长方形的宽，根据圆的面积 $= \\pi r^2$，计算即可。关键是理解长方形和圆之间的关系，掌握圆的面积公式。\n\n【详解】 $3.14 \\times (4 \\div 2)^{2}\\= 3. 1 4 \\times 4\\=12.56 (平方厘米)$",
+        "knowledge_points": [
+          "圆的面积及应用"
+        ],
+        "source": "20-21 六年级上·湖南怀化·单元测试",
+        "merged_id": "unit4_249",
+        "original_id": "unit4_fill_249"
+      },
+      {
+        "id": "unit4_fill_250",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "甲圆的半径是 3 厘米，乙圆的直径是 9 厘米，那么，甲、乙两圆直径的比是（：），周长的比是（：），面积的比是（：）。",
+        "options": [],
+        "answer": " 2; 3; 2 ;3; 4; 9",
+        "solution": "【分析】直径 $= \\text{半径} \\times 2$, 根据比的意义, 写出两圆直径比,化简即可, 根据直径比 $= \\text{周长比}$, 平方以后的比是面积比,进行填空。关键是理解比的意义，掌握圆的周长和面积公式。\n\n【详解】 $(3 \\times 2): 9 = 6: 9 = 2: 3; 2^{2}: 3^{2} = 4: 9$甲圆的半径是 3 厘米, 乙圆的直径是 9 厘米, 那么, 甲、乙两圆直径的比是 $2: 3$, 周长的比是 $2: 3$, 面积的比是 $4: 9$。",
+        "knowledge_points": [
+          "圆的周长及应用、比的意义、圆的面积及应用、比的化简"
+        ],
+        "source": "20-21 六年级上·湖南怀化·单元测试",
+        "merged_id": "unit4_250",
+        "original_id": "unit4_fill_250"
+      },
+      {
+        "id": "unit4_fill_251",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "一个圆直径由 5 厘米增加到 10 厘米，周长增加( )厘米，面积增加( )厘米²。",
+        "options": [],
+        "answer": " 15.7; 235.5",
+        "solution": "【分析】（1）圆的周长计算公式是 $C = 2 \\pi r$，如果半径增加 $n$ 厘米，根据周长的计算公式可知周长增加 $2 n \\pi$，列式进行计算即可；(2) 根据题意可知, 增加的部分是环形, 由环形的面积 $=$外圆面积一内圆面积。把数据代入公式解答。本题考查圆的周长、面积的计算, 解答此题应注意在圆中, 如果是圆的半径增加 $\\mathrm{n}$, 则其周长增加 $2 \\mathrm{~n} \\pi$, 周长增加的值与原来圆的半径大小无关。\n\n【详解】 (1) $3.14 \\times 2 \\times (10 - 5)\\= 3. 1 4 \\times 5\\=15.7 (厘米)$\n\n(2) $3.14 \\times (10^{2} - 5^{2})\\ = 3. 1 4 \\times (1 0 0 - 2 5) \\ = 3. 1 4 \\times 7 5 \\=235.5 (平方厘米)$",
+        "knowledge_points": [
+          "圆的周长及应用、圆的面积及应用"
+        ],
+        "source": "20-21 六年级上·湖南怀化·单元测试",
+        "merged_id": "unit4_251",
+        "original_id": "unit4_fill_251"
+      },
+      {
+        "id": "unit4_fill_252",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "一座古钟的分针长 15 厘米, 经过 1 小时扫过的面积是( )平方厘米。",
+        "options": [],
+        "answer": "706.5",
+        "solution": "【分析】根据题干可知，分针 1 小时旋转一周，组成的图形是一个圆形，那么此题就是求这个半径为 18 厘米的圆的面积，利用圆的面积公式即可解答。此题考查圆的面积公式的应用, 关键是根据钟面上分针旋转的特点得出旋转后的图形。\n\n【详解】 $3.14 \\times 15^{2}\\= 3. 1 4 \\times 2 2 5 \\ = 7 0 6. 5 \\text {(平 方 厘 米)}$",
+        "knowledge_points": [
+          "圆的面积及应用"
+        ],
+        "source": "20-21 六年级上·湖南怀化·单元测试",
+        "merged_id": "unit4_252",
+        "original_id": "unit4_fill_252"
+      },
+      {
+        "id": "unit4_fill_253",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "一天当中，一般人生活和睡眠的时间比是 2:1，睡眠时间是( )小时。",
+        "options": [],
+        "answer": "8",
+        "solution": "【分析】一昼夜就是 24 小时, 因为一般人生活和睡眠的时间比是 $2: 1$, 则一天中睡眠的时间的占比为 $\\frac{1}{2 + 1}$。要计算睡眠时间是多少，可按比例分配，列式为： $24 \\times \\frac{1}{2 + 1} = 8$（小时）。解答按比例分配的一类题目, 首先要明确所求数量占总数的分率, 然后用总数与对应的分率相乘, 就是具体的数量。\n\n【详解】 $24 \\times \\frac{1}{2 + 1} = 24 \\times \\frac{1}{3} = 8$ （小时）",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "20-21 六年级上·湖南怀化·单元测试",
+        "merged_id": "unit4_253",
+        "original_id": "unit4_fill_253"
+      },
+      {
+        "id": "unit4_fill_254",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "4.5:0.6 化成最简整数比是( )，比值是( )。",
+        "options": [],
+        "answer": " 15:2 ;7.5",
+        "solution": "【分析】化简比时根据比的基本性质：比的前项、后项乘或除以同一个数（0 除外），比值不变。求比值是将比前项除以后项，可得出比值。本题主要考查的是比的化简及比值，解题的关键是熟练运用比的基本性质及求比值，进而得出答案。\n\n【详解】 $4.5: 0.6 = 15: 2$，即最简比为： $15: 2$；比值为：$4.5: 0. 6 = 4. 5 \\div 0. 6 = 7. 5$",
+        "knowledge_points": [
+          "比的化简、求比值"
+        ],
+        "source": "20-21 六年级上·湖南怀化·单元测试",
+        "merged_id": "unit4_254",
+        "original_id": "unit4_fill_254"
+      },
+      {
+        "id": "unit4_fill_255",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "在 3:5 的中, 比的前项加上 12 , 要使比值不变, 后项应加上( )。",
+        "options": [],
+        "answer": "20",
+        "solution": "【分析】先计算比的前项加上 12 后扩大的倍数，根据比的基本性质，计算出比的后项，最后计算新的后项和原来后项的差即可。掌握比的基本性质是解答题目的关键。\n\n【详解】 $(3 + 12) \\div 3 \\times 5 - 5= 1 5 \\div 3 \\times 5 - 5 \\ = 5 \\times 5 - 5 \\ = 2 5 - 5 \\ = 20$",
+        "knowledge_points": [
+          "比的基本性质"
+        ],
+        "source": "20-21 六年级上·湖南怀化·单元测试",
+        "merged_id": "unit4_255",
+        "original_id": "unit4_fill_255"
+      },
+      {
+        "id": "unit4_fill_256",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "把 3:5 的后项加上 15，要使比值不变，前项要加上( )。",
+        "options": [],
+        "answer": "9",
+        "solution": "【分析】把 3:5 的后项加上 15 , 后项变为 20 , 扩大到原来的 4 倍, 要使比值不变, 前项也要扩大到原来的 4 倍, 变为 12 , 加上 9 , 据此解答即可。熟练掌握比的基本性质是解答本题的关键。\n\n【详解】把 3:5 的后项加上 15 , 要使比值不变, 前项要加上 9 。",
+        "knowledge_points": [
+          "比的基本性质"
+        ],
+        "source": "20-21 六年级上·河南南阳·期中",
+        "merged_id": "unit4_256",
+        "original_id": "unit4_fill_256"
+      },
+      {
+        "id": "unit4_fill_257",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "若两个正方形的边长比是 $4: 3$, 则它们的周长比是( ), 面积比是( )。",
+        "options": [],
+        "answer": " 4:3 16:9",
+        "solution": "【分析】把两个正方形的边长分别看作 4 份、3 份, 再根据正方形的周长公式 $C = 4 a$, 正方形的面积公式 $S = a \\times a$, 分别求出周长及面积, 再写出比即可。此题主要考查了正方形的周长公式 $C = 4a$ 与面积公式 $S = a \\times a$ 的灵活应用，熟记公式是解题关键。\n\n【详解】周长比为： $(4 \\times 4) :(3 \\times 4)= 1 6: 1 2  = 4: 3$面积比为$(4\\times 4):(3\\times 3)= 1 6: 9$",
+        "knowledge_points": [
+          "比的应用、正方形的周长、正方形的面积"
+        ],
+        "source": "18-19 六年级下·全国·单元测试",
+        "merged_id": "unit4_257",
+        "original_id": "unit4_fill_257"
+      },
+      {
+        "id": "unit4_fill_258",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "甲数是 0.4, 乙数是 0.6, 甲数与乙数的比是( ), 比值是( )。",
+        "options": [],
+        "answer": " $2:3\\quad \\frac{2}{3}$",
+        "solution": "【分析】甲数与乙数的比是 $0.4: 0.6$, 根据比的基本性质可化简为最简整数比。比值是用比的前项除以后项即可得到答案。本题主要考查的是比的化简及求比值, 解题的关键是熟练运用比的基本性质化简比, 进而得出答案。\n\n【详解】甲数与乙数的比是 $0.4: 0.6 = 4: 6 = 2: 3$, 比值为: $2: 3 = \\frac{2}{3}$。",
+        "knowledge_points": [
+          "比的化简、求比值"
+        ],
+        "source": "21-22 六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_258",
+        "original_id": "unit4_fill_258"
+      },
+      {
+        "id": "unit4_fill_259",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "$\\frac {()}{4} = 6: 8 = 1 2 \\div () = \\frac {3}{4} = () \\div 4 0$。",
+        "options": [],
+        "answer": "3; 16; 30",
+        "solution": "【分析】根据分数的基本性质：分子、分母同乘或除以一个数（0 除外），大小不变；分数化为除法时，分子为被除数，分母为除数；分数化为比时，分子为前项，分母为后项。据此可得出本题答案。本题主要考查的是比与与除法、分数的转化, 解题的关键是熟练运用转换方法, 进而得出答案。\n\n【详解】 $6:8 = \\frac{6}{8} = \\frac{3}{4}$$6: 8 = 6 \\div 8 = 1 2 \\div 1 6 \\quad 6: 8 = 6 \\div 8 = 3 0 \\div 4 0$",
+        "knowledge_points": [
+          "比的基本性质、比与分数、除法的关系"
+        ],
+        "source": "21-22 六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_259",
+        "original_id": "unit4_fill_259"
+      },
+      {
+        "id": "unit4_fill_260",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "把 $\\frac{2}{9} \\times 4 = \\frac{8}{9}$ 改写成两道除法算式是( )和( )。",
+        "options": [],
+        "answer": " $\\frac{8}{9} \\div 4 = \\frac{2}{9}$    $\\frac{8}{9} \\div \\frac{2}{9} = 4$",
+        "solution": "【分析】 $\\frac{2}{9} \\times 4 = \\frac{8}{9}$, 分数乘法中, 乘数 $=$ 积分被乘数, 被乘数 $=$ 积分乘数, 据此可得出答案。本题主要考查的是分数乘法与除法的关系，解题的关键是合理利用分数乘法、除法之间的关系，进而得出答案。\n\n【详解】 $\\frac{2}{9} \\times 4 = \\frac{8}{9}$, 则可得到除法算式: $\\frac{8}{9} \\div 4 = \\frac{2}{9}, \\frac{8}{9} \\div \\frac{2}{9} = 4$。",
+        "knowledge_points": [
+          "分数与分数的除法、分数乘分数"
+        ],
+        "source": "21-22 六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_260",
+        "original_id": "unit4_fill_260"
+      },
+      {
+        "id": "unit4_fill_261",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "六年级男生与全级人数的比是 $5: 9$，那么全级女生与全级人数的比是( )。",
+        "options": [],
+        "answer": "4:9",
+        "solution": "【分析】根据题意可知, 全级人数是 9 份, 全级男生是 5 份, 则全级女生是 4 份, 据此写出全级女生与全级人数的比即可。明确比的意义是解答本题的关键。\n\n【详解】全级女生与全级人数的比是 $(9 - 5): 9 = 4: 9$",
+        "knowledge_points": [
+          "比的意义"
+        ],
+        "source": "20-21 六年级上·广东江门·期末",
+        "merged_id": "unit4_261",
+        "original_id": "unit4_fill_261"
+      },
+      {
+        "id": "unit4_fill_262",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "一种核桃，500 克核桃仁可以榨油 340 克，榨油量与核桃仁质量的比是( )，比值是( )。",
+        "options": [],
+        "answer": " $17:25;\\frac{17}{25}$",
+        "solution": "【分析】根据比的意义直接写出榨油量与核桃仁质量的比;根据比的性质, 化简成最简整数比, 然后比值用分数的形式表示出来, 即可得解。此题考查的目的是理解掌握比的意义, 比的化简方法及应用。\n\n【详解】 $340: 500 = 17: 25 = \\frac{17}{25}$",
+        "knowledge_points": [
+          "比的应用、比的意义、比的化简、求比值"
+        ],
+        "source": "2021 六年级上·全国·专题练习",
+        "merged_id": "unit4_262",
+        "original_id": "unit4_fill_262"
+      },
+      {
+        "id": "unit4_fill_263",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "$4: 5 = \\frac{20}{()} = (): 20$。",
+        "options": [],
+        "answer": "25;16",
+        "solution": "【分析】由比的基本性质可知, $4: 5 = (4 \\times 4): (5 \\times 4) =$(16) : 20; $4: 5 = \\frac{4}{5} = \\frac{4 \\times 5}{(5 \\times 5)} = \\frac{20}{(25)}$,掌握比和分数的关系并灵活运用比的基本性质是解答题目的关键。\n\n【详解】 $4:5 = \\frac{20}{(25)}=(16):20$",
+        "knowledge_points": [
+          "比的基本性质、比与分数、除法的关系"
+        ],
+        "source": "21-22 六年级上·全国·单元测试",
+        "merged_id": "unit4_263",
+        "original_id": "unit4_fill_263"
+      },
+      {
+        "id": "unit4_fill_264",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "六（1）班有男生 30 人，女生 20 人。男生人数是女生人数的( )倍，女生人数与男生人数的最简单的整数比是( )，女生人数占总人数的( )。",
+        "options": [],
+        "answer": " 1.5 ;2:3  ;$\\frac{2}{5}$",
+        "solution": "【分析】根据求一个数是另一个数的几分之几用除法；根据比的意义写出男女生的人数比再进行化简即可。此题考查的是分数除法的应用, 掌握求一个数是另一个数的几分之几用除法是解题关键。\n\n【详解】男生人数是女生人数的: $30 \\div 20 = 1.5$;女生人数与男生人数的最简单的整数比是:\n\n$2 0: 3 0 \\ = (2 0 \\div 1 0): (3 0 \\div 1 0) \\ = 2: 3$\n\n总人数: $20 + 30 = 50$ (人)    $2 0 \\div 5 0 = \\frac {2}{5}$",
+        "knowledge_points": [
+          "求一个数是另一个数的几倍、比的化简、求一个数占另一个数几分之几"
+        ],
+        "source": "21-22 六年级上·陕西延安·期中",
+        "merged_id": "unit4_264",
+        "original_id": "unit4_fill_264"
+      },
+      {
+        "id": "unit4_fill_265",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "一个正方形的边长是 a 厘米 (a > 0), 它的周长与边长的比是( ), 边长与面积的比是( ), 比值是( )。",
+        "options": [],
+        "answer": " 4:1 ;1:a  ;$\\frac{1}{a}$",
+        "solution": "【分析】根据正方形的周长 $= \\text{边长} \\times 4$, 正方形的面积 = 边长 $\\times$ 边长, 分别求出正方形的周长和面积, 然后求出周长与边长的比, 边长与面积的比即可, 用边长除以面积即可求出比值。本题考查比的意义，明确比的意义是解题的关键。\n\n【详解】正方形的周长： $a \\times 4 = 4a$ （厘米）正方形面积: $a \\times a = a^{2}$周长与边长的比: $4 \\mathrm{a}: \\mathrm{a} = 4: 1$边长与面积的比: $a : a^{2} = 1: a$比值为: $1 \\div a = \\frac{1}{a}$",
+        "knowledge_points": [
+          "正方形的周长、比的化简、求比值、正方形的面积"
+        ],
+        "source": "20-21 六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_265",
+        "original_id": "unit4_fill_265"
+      },
+      {
+        "id": "unit4_fill_266",
+        "unit": "第四单元：比",
+        "chapter": "比",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "$\\frac{4}{5} = (\\quad)\\div 20= 20: () = () : 35 = ()$ (小数)。",
+        "options": [],
+        "answer": " 16 ;25 ;28 ;0.8",
+        "solution": "【分析】根据分数与除法的关系，把 $\\frac{4}{5}$ 变为 $4 \\div 5$，再根据商不变规律：被除数和除数同时乘或除以一个相同的数（0 除外）商不变，然后再根据除法和比的关系化成比的形式，再根据比的性质进行转化即可。此题考查的是分数、除法、比、小数的关系, 掌握它们间的关系是解题关键。\n\n【详解】由分析得，$\\frac {4}{5} = 1 6 \\div 2 0 = 2 0: 2 5 = 2 8: 3 5 = 0. 8$",
+        "knowledge_points": [
+          "百分数、分数、小数和比的互化、比与分数、除法的关系"
+        ],
+        "source": "20-21 六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_266",
+        "original_id": "unit4_fill_266"
+      },  {
+        "id": "unit4_judge_267",
+        "unit": "第四单元：比",
+        "chapter": "比的认识与应用",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "一场足球比赛最后结果是 0:0，说明比的前项、后项都可以为 0。",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】比的意义是两个数相除，又叫做两个数的比，比是表示两个数之间的关系；而一场足球比赛的比分是0：0, 说明本次比赛, 第一队一个球也没有进, 第二队一个球也没有进，这是表示进的球的个数比，与前一个比意义不同，据此判断。此题考查比的意义, 掌握比是表示两个数之间的关系，比的后项不能为 0 ; 要与比赛时进球的比区分开, 那是进球个数的比，它们的意义不同是解题关键。\n\n【详解】比是表示两个数相除，是两个数之间的关系，在比中，比的后项不能为 0 ; 而一场足球比赛的比分是 0:0 ,说明本次比赛，第一队一个球也没有进，第二队一个球也没有进，这是表示进的球的个数比，比号后面的数可以是0，表示一个也没有；所以它们意义不同。故答案为: ×",
+        "knowledge_points": [
+          "比的意义"
+        ],
+        "source": "2020·湖南怀化·小升初真题",
+        "merged_id": "unit4_267",
+        "original_id": "unit4_judge_267"
+      },
+      {
+        "id": "unit4_judge_268",
+        "unit": "第四单元：比",
+        "chapter": "比的认识与应用",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "某班男、女生人数的比是 7:8，男生占全班人数的 $\\frac{7}{15}$。",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】男、女生人数的比是 7:8, 将男生人数看作 7 , 女生人数看作 8 , 男生人数 + 女生人数 = 全班人数, 男生人数 ÷ 全班人数 = 男生占全班人数的几分之几。两数相除又叫两个数的比, 求一个数占另一个数的几分之几用除法。\n\n【详解】$7 \\div (7 + 8)\\\\ = 7 \\div 15 \\\\ = \\frac{7}{15}$\n\n故答案为: √",
+        "knowledge_points": [
+          "比的意义",
+          "求一个数占另一个数几分之几"
+        ],
+        "source": "2022·湖南衡阳·小升初真题",
+        "merged_id": "unit4_268",
+        "original_id": "unit4_judge_268"
+      },
+      {
+        "id": "unit4_judge_269",
+        "unit": "第四单元：比",
+        "chapter": "比的认识与应用",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "如果 $a \\div b = \\frac{5}{6}$，那么 $a:b = 6:5$。",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】根据比和除法、分数的关系，结合题干条件，直接写出 a 和 b 的比。本题考查了比, 掌握比和除法、分数的关系是解题的关键。\n\n【详解】如果 $a \\div b = \\frac{5}{6}$，那么 $a:b = 5:6$。所以判断错误。故答案为: ×",
+        "knowledge_points": [
+          "比的意义",
+          "比与分数、除法的关系"
+        ],
+        "source": "21-22 六年级上·湖北黄石·期末",
+        "merged_id": "unit4_269",
+        "original_id": "unit4_judge_269"
+      },
+      {
+        "id": "unit4_judge_270",
+        "unit": "第四单元：比",
+        "chapter": "比的认识与应用",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "一项工程，甲队40天完工，乙队50天完工。甲乙两队工作效率比是4:5。",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】根据比的意义，时间比反过来就是效率比，据此分析。两数相除又叫两个数的比。\n\n【详解】$50:40 = 5:4$，甲乙两队工作效率比是 $5:4$，所以原题说法错误。",
+        "knowledge_points": [
+          "比的意义",
+          "比的化简"
+        ],
+        "source": "19-20 六年级下·湖南怀化·单元测试",
+        "merged_id": "unit4_270",
+        "original_id": "unit4_judge_270"
+      },
+      {
+        "id": "unit4_judge_271",
+        "unit": "第四单元：比",
+        "chapter": "比的认识与应用",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "大牛和小牛的头数比是 4:5，表示大牛比小牛少 $\\frac{1}{5}$。",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】大牛和小牛的头数比是 4:5，把小牛数量看作 4 份,大牛数量看作 5 份, 再用大牛比小牛少的占小牛数量的几分之几即可。本题考查比、分数除法，解答本题的关键是找准单位'1'。\n\n【详解】$(5 - 4) \\div 5 = 1 \\div 5 = \\frac{1}{5}$，说法正确。故答案为: √。",
+        "knowledge_points": [
+          "分数除法的应用",
+          "比的意义"
+        ],
+        "source": "2022·湖南怀化·小升初真题",
+        "merged_id": "unit4_271",
+        "original_id": "unit4_judge_271"
+      },
+      {
+        "id": "unit4_judge_272",
+        "unit": "第四单元：比",
+        "chapter": "比的认识与应用",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "比的前项乘3，后项除以 $\\frac{1}{3}$，比值的大小不变。",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】后项除以 $\\frac{1}{3}$，相当于后项乘 3，比的前项和后项同时乘 3，比值的大小不变。掌握比的基本性质是解答题目的关键。\n\n【详解】$2:7 = (2 \\times 3):(7 \\div \\frac{1}{3}) = (2 \\times 3):(7 \\times 3) = 6:21 = \\frac{2}{7}$\n\n故答案为: √",
+        "knowledge_points": [
+          "比的基本性质"
+        ],
+        "source": "19-20六年级上·全国·单元测试",
+        "merged_id": "unit4_272",
+        "original_id": "unit4_judge_272"
+      },
+      {
+        "id": "unit4_judge_273",
+        "unit": "第四单元：比",
+        "chapter": "比的认识与应用",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "比与比值是一回事，没有区别。",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】比表示两个数量之间的相除关系；比值是一个具体的数，可以是分数，也可以是小数或整数。掌握比和比值的区别是解答题目的关键。\n\n【详解】根据分析可知，比与比值两者之间有区别。原题说法错误。故答案为: ×",
+        "knowledge_points": [
+          "比的意义",
+          "求比值"
+        ],
+        "source": "20-21 六年级上·湖南怀化·单元测试",
+        "merged_id": "unit4_273",
+        "original_id": "unit4_judge_273"
+      },
+      {
+        "id": "unit4_judge_274",
+        "unit": "第四单元：比",
+        "chapter": "比的认识与应用",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "比的前项相当于被除数，后项相当于除数，比值相当于商。",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】比与分数、除法的关系。不能说比的前项就是被除数（或分子），比与除法、分数之间的对应关系只能用'相当于'来表达。\n\n【详解】\n\n| 名称 | 联系 | 区别 |\n| ---- | ---------------------------- | ---------------------- |\n| 比 | 前项 :（比号） 后项 比值 | 比表示两个量的倍数关系 |\n| 除法 | 被除数 ÷（除号） 除数 商 | 除法是一种运算 |\n| 分数 | 分子 —（分数线） 分母 分数值 | 分数是一种数 |\n\n比的前项相当于被除数，后项相当于除数，比值相当于商，说法正确。故答案为: √",
+        "knowledge_points": [
+          "比与分数、除法的关系"
+        ],
+        "source": "20-21 六年级上·湖南怀化·单元测试",
+        "merged_id": "unit4_274",
+        "original_id": "unit4_judge_274"
+      },
+      {
+        "id": "unit4_judge_275",
+        "unit": "第四单元：比",
+        "chapter": "比的认识与应用",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "一杯糖水有 120克，糖和水的比是 $1:5$，如果在这杯糖水中加入 5 克糖，再加入 25 克水。那么这杯糖水的浓度不变。",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】一杯糖水有 120 克，糖和水的比是 $1:5$，根据按比例分配方法得出糖水中糖和水的重量；保持糖水浓度不变,即糖和水的比还是 $1:5$，根据比的化简即可得出答案。本题主要考查的是按比例分配及比的化简，解题的关键是计算出糖水中糖和水的重量，再进行计算得出答案。\n\n【详解】一杯糖水有 120 克，糖和水的比是 $1:5$，则糖水中糖的重量为: $120 \\times \\frac{1}{6} = 20$ (克)，水的重量为: $120 \\times \\frac{5}{6} = 100$(克)。加入糖和水后糖的重量为 25 克，水的重量为 125 克，此时糖和水的比为: $25:125 = 1:5$，与原糖水浓度相同,即浓度不变。因此本题正确。",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "21-22 六年级上·新疆克孜勒苏·期中",
+        "merged_id": "unit4_275",
+        "original_id": "unit4_judge_275"
+      },
+      {
+        "id": "unit4_calc_276",
+        "unit": "第四单元：比",
+        "chapter": "比的认识与应用",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "把下面各比化成最简单的整数比。\n\n25:75   3.8:$\\frac{19}{20}$     $\\frac{13}{18}:\\frac{5}{6}$",
+        "options": [],
+        "answer": "$1:3;4:1;13:15$",
+        "solution": "【分析】化简比根据比的基本性质，化简比的结果还是一个比。\n\n【详解】$25:75 = 5:15 = 1:3$     $3.8:\\frac{19}{20} = 76:19 = 4:1 \\quad \\frac{13}{18}:\\frac{5}{6} = 13:15$",
+        "knowledge_points": [
+          "比的化简"
+        ],
+        "source": "19-20六年级下·湖南怀化·单元测试",
+        "merged_id": "unit4_276",
+        "original_id": "unit4_calc_276"
+      },
+      {
+        "id": "unit4_calc_277",
+        "unit": "第四单元：比",
+        "chapter": "比的认识与应用",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "求下面各比的比值。\n\n2.1:0.7    20.5:$\\frac{5}{7}$      $\\frac{5}{8}:\\frac{3}{4}$",
+        "options": [],
+        "answer": "3;  $\\frac{287}{10}; \\frac{5}{6}$",
+        "solution": "【分析】比的前项除以比的后项所得的商即为比值，据此计算。\n\n【详解】(1) $2.1:0.7\\\\ = 2.1 \\div 0.7 \\\\ = 3$\n\n(2) $20.5:\\frac{5}{7}\\\\= 20.5 \\div \\frac{5}{7} \\\\ = \\frac{41}{2} \\div \\frac{5}{7} \\\\ = \\frac{41}{2} \\times \\frac{7}{5} \\\\ = \\frac{287}{10}$\n\n(3) $\\frac{5}{8}:\\frac{3}{4}= \\frac{5}{8} \\div \\frac{3}{4} \\\\ = \\frac{5}{8} \\times \\frac{4}{3} \\\\ = \\frac{5}{6}$",
+        "knowledge_points": [
+          "求比值"
+        ],
+        "source": "19-20 六年级下·湖南怀化·单元测试",
+        "merged_id": "unit4_277",
+        "original_id": "unit4_calc_277"
+      },
+      {
+        "id": "unit4_calc_278",
+        "unit": "第四单元：比",
+        "chapter": "比的认识与应用",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "先化简比，再求比值。\n\n25:40       3.6:9            $\\frac{4}{5}:\\frac{2}{3}$\n\n1:0.25      $\\frac{6}{5}:\\frac{2}{5}$         $4:\\frac{4}{3}$",
+        "options": [],
+        "answer": "$5:8; \\frac{5}{8}; 2:5; \\frac{2}{5}; 6:5, \\frac{6}{5}$; 4:1,;4; 3:1; 3; 3:1; 3",
+        "solution": "【分析】'25:40'将比的前项和后项同时除以 5，先求出最简整数比，再将最简整数比的前项除以后项，求出比值;'3.6:9'将比的前项和后项同时除以 1.8，先求出最简整数比，再将最简整数比的前项除以后项，求出比值；$\\frac{4}{5}:\\frac{2}{3}$ 将比的前项和后项同时乘 $\\frac{15}{2}$，先求出最简整数比，再将最简整数比的前项除以后项，求出值；'1:0.25'将比的前项和后项同时乘 4，先求出最简整数比,再将最简整数比的前项除以后项，求出比值; '$\\frac{6}{5}:\\frac{2}{5}$'将比的前项和后项同时乘 $\\frac{5}{2}$，先求出最简整数比，再将最简整数比的前项除以后项，求出比值; '4:$\\frac{4}{3}$'将比的前项和后项同时乘 $\\frac{3}{4}$，先求出最简整数比，再将最简整数比的前项除以后项，求出比值。\n\n【详解】$25:40 = (25 \\div 5):(40 \\div 5) = 5:8, 5 \\div 8 = \\frac{5}{8}$;\n\n$3.6:9 = (3.6 \\div 1.8):(9 \\div 1.8) = 2:5, 2 \\div 5 = \\frac{2}{5}; \\\\ \\frac{4}{5}:\\frac{2}{3} = (\\frac{4}{5} \\times \\frac{15}{2}):(\\frac{2}{3} \\times \\frac{15}{2}) = 6:5, 6 \\div 5 = \\frac{6}{5}; \\\\ 1:0.25 = (1 \\times 4):(0.25 \\times 4) = 4:1, 4 \\div 1 = 4; \\\\ \\frac{6}{5}:\\frac{2}{5} = (\\frac{6}{5} \\times \\frac{5}{2}):(\\frac{2}{5} \\times \\frac{5}{2}) = 3:1, 3 \\div 1 = 3; \\\\ 4:\\frac{4}{3} = (4 \\times \\frac{3}{4}):(\\frac{4}{3} \\times \\frac{3}{4}) = 3:1, 3 \\div 1 = 3$",
+        "knowledge_points": [
+          "比的化简",
+          "求比值"
+        ],
+        "source": "21-22 六年级上·福建福州·单元测试",
+        "merged_id": "unit4_278",
+        "original_id": "unit4_calc_278"
+      },
+      {
+        "id": "unit4_calc_279",
+        "unit": "第四单元：比",
+        "chapter": "比的认识与应用",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "计算下列各题，能用简便方法的用简便方法。\n\n$\\frac{1}{7} \\times \\frac{9}{13} \\div \\frac{5}{14} \\div \\frac{3}{13} \\quad \\frac{5}{8} \\times \\frac{3}{4} + \\frac{1}{4} \\times \\frac{5}{8} + \\frac{5}{8} \\quad 128 \\times \\frac{17}{126} \\\\ 1.5:\\frac{3}{5} \\quad [1 - (\\frac{1}{4} + \\frac{3}{8})] \\div \\frac{1}{4}$",
+        "options": [],
+        "answer": "$\\frac{6}{5}; \\frac{5}{4}; 17\\frac{17}{63}$  ;$2.5; \\frac{3}{2}$",
+        "solution": "【分析】（1）把除法化成乘法，再根据乘法交换律和乘法结合律进行简算；(3) 先约分为 $64 \\times \\frac{17}{63}$，把 64 写成 $(63 + 1)$，再根据乘法分配律进行简算;(2) 根据乘法分配律进行简算; (4) 用前项除以后项求出比值; (5) 分数四则混合运算的顺序与整数一致。\n\n【详解】$\\frac{1}{7} \\times \\frac{9}{13} \\div \\frac{5}{14} \\div \\frac{3}{13}$\n\n$= \\frac{1}{7} \\times \\frac{9}{13} \\times \\frac{14}{5} \\times \\frac{13}{3}$\n\n$= (\\frac{1}{7} \\times \\frac{14}{5}) \\times (\\frac{9}{13} \\times \\frac{13}{3}) \\\\ = \\frac{2}{5} \\times 3 \\\\ = \\frac{6}{5}$\n\n$\\frac{5}{8} \\times \\frac{3}{4} + \\frac{1}{4} \\times \\frac{5}{8} + \\frac{5}{8} \\\\ = (\\frac{3}{4} + \\frac{1}{4} + 1) \\times \\frac{5}{8} \\\\ = 2 \\times \\frac{5}{8} \\\\ = \\frac{5}{4}$\n\n$128 \\times \\frac{17}{126} \\\\ = 64 \\times \\frac{17}{63} \\\\ = (63 + 1) \\times \\frac{17}{63} \\\\ = 63 \\times \\frac{17}{63} + 1 \\times \\frac{17}{63} \\\\ = 17\\frac{17}{63}$\n\n$1.5:\\frac{3}{5} \\\\ = 1.5 \\div 0.6 \\\\ = 2.5$\n\n$[1 - (\\frac{1}{4} + \\frac{3}{8})] \\div \\frac{1}{4}\\\\ = [1 - (\\frac{2}{8} + \\frac{3}{8})] \\div \\frac{1}{4} \\\\ = [1 - \\frac{5}{8}] \\div \\frac{1}{4} \\\\ = \\frac{3}{8} \\div \\frac{1}{4} \\\\ = \\frac{3}{8} \\times 4 \\\\ = \\frac{3}{2}$",
+        "knowledge_points": [
+          "分数的四则混合运算",
+          "求比值",
+          "分数乘法运算律"
+        ],
+        "source": "21-22 六年级上·湖南益阳·期末",
+        "merged_id": "unit4_279",
+        "original_id": "unit4_calc_279"
+      },
+      {
+        "id": "unit4_calc_280",
+        "unit": "第四单元：比",
+        "chapter": "比的认识与应用",
+        "question_type": "计算题",
+        "difficulty": "容易",
+        "question": "求下面各比的比值。\n\n$24:60=$     $\\frac{1}{12}:\\frac{5}{24}=$     $0.5:0.75=$   $0.6:\\frac{2}{5}=$",
+        "options": [],
+        "answer": "$0.4; \\frac{2}{5}; \\frac{2}{3}; \\frac{3}{2}$",
+        "solution": "【分析】①可用 24 除以 60，得到比值 0.4；② 可按分数除法计算出商，得到比值 $\\frac{2}{5}$; ③ 可按小数除法计算出商，除不尽的就写成分数的形式，再按分数约分的方法进行化简，得到比值 $\\frac{2}{3}$; ④ 可先把小数 0.6 化为分数 $\\frac{3}{5}$，再按分数除法计算出商，得到比值 $\\frac{3}{2}$。\n\n【详解】$24:60 = \\frac{1}{2} 24 \\div 60 = 0.4$\n\n$\\frac{1}{12}:\\frac{5}{24} = \\frac{1}{12} \\times \\frac{24}{5} = \\frac{2}{5} \\\\ 0.5:0.75 = 0.5 \\div 0.75 = \\frac{50}{75} = \\frac{2}{3} \\\\ 0.6:\\frac{2}{5} = \\frac{3}{5} \\div \\frac{2}{5} = \\frac{3}{5} \\times \\frac{5}{2} = \\frac{3}{2}$",
+        "knowledge_points": [
+          "求比值"
+        ],
+        "source": "21-22 六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_280",
+        "original_id": "unit4_calc_280"
+      },
+      {
+        "id": "unit4_calc_281",
+        "unit": "第四单元：比",
+        "chapter": "比的认识与应用",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "先化简比，再求比值。\n\n$\\frac{3}{8}:1.2 \\quad 0.56:40 \\quad 6.5 \\mathrm{t}:1500 \\mathrm{~kg}$",
+        "options": [],
+        "answer": "$5:16$ / $\\frac{5}{16}$  ;  $7:500$ / $\\frac{7}{500}$ ;  $13:3$ / $\\frac{13}{3}$",
+        "solution": "【分析】化简比根据比的基本性质，求比值直接用前项÷后项，化简比的结果还是一个比，求比值的结果是一个数。\n\n【详解】$\\frac{3}{8}:1.2 = 3:9.6 = 30:96 = 5:16 = \\frac{5}{16}\\quad 0.56:40 = 56:4000 = 7:500 = \\frac{7}{500} \\\\ 6.5 \\mathrm{t}:1500 \\mathrm{kg} = 6500:1500 = 65:15 = 13:3 = \\frac{13}{3}$",
+        "knowledge_points": [
+          "比的化简",
+          "求比值"
+        ],
+        "source": "20-21 六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_281",
+        "original_id": "unit4_calc_281"
+      },
+      {
+        "id": "unit4_calc_282",
+        "unit": "第四单元：比",
+        "chapter": "比的认识与应用",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "能简算的要简算。\n\n$\\frac{7}{15} \\times 23 - 8 \\div \\frac{15}{7} \\quad (\\frac{4}{5} + \\frac{1}{4} - \\frac{1}{3}) \\div \\frac{1}{60}$",
+        "options": [],
+        "answer": "7; 43",
+        "solution": "【分析】（1）把除法化为乘法，然后运用乘法分配律进行计算即可。(2) 运用乘法分配律进行计算即可。\n\n【详解】$\\frac{7}{15} \\times 23 - 8 \\div \\frac{15}{7}$\n\n$= \\frac{7}{15} \\times 23 - 8 \\times \\frac{7}{15} \\\\ = \\frac{7}{15} \\times (23 - 8) \\\\ = \\frac{7}{15} \\times 15 \\\\ = 7$\n\n$(\\frac{4}{5} + \\frac{1}{4} - \\frac{1}{3}) \\div \\frac{1}{60} \\\\ = (\\frac{4}{5} + \\frac{1}{4} - \\frac{1}{3}) \\times 60 = \\frac{4}{5} \\times 60 + \\frac{1}{4} \\times 60 - \\frac{1}{3} \\times 60 \\\\ = 48 + 15 - 20 \\\\ = 63 - 20 \\\\ = 43$",
+        "knowledge_points": [
+          "分数与整数的除法",
+          "分数乘法运算律"
+        ],
+        "source": "20-21 六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_282",
+        "original_id": "unit4_calc_282"
+      },
+      {
+        "id": "unit4_calc_283",
+        "unit": "第四单元：比",
+        "chapter": "比的认识与应用",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "直接写得数。\n\n$\\frac{2}{3} \\times \\frac{5}{3}=$ $\\frac{2}{5} - \\frac{1}{6}=$ $\\frac{5}{9} \\div \\frac{1}{3}=$ $():$$\\frac{1}{4}=1.5$",
+        "options": [],
+        "answer": "$\\frac{10}{9}; \\frac{7}{30}; \\frac{5}{3}; \\frac{3}{8}$",
+        "solution": "【详解】略",
+        "knowledge_points": [
+          "分数与分数的除法",
+          "分数乘分数",
+          "求比值"
+        ],
+        "source": "20-21 六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_283",
+        "original_id": "unit4_calc_283"
+      },
+      {
+        "id": "unit4_solve_284",
+        "unit": "第四单元：比",
+        "chapter": "比的认识与应用",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "学校运来30捆树苗，每捆5棵，按4:6分给五、六年级学生种植，每个班级各分得多少棵树苗？",
+        "options": [],
+        "answer": "60棵；90棵",
+        "solution": "【分析】先利用乘法求出总共有多少棵树苗，再根据比将五、六年级分得的树苗占总树苗的几分之几表示出来，最后利用乘法求出每个年级各分得多少棵树苗。本题考查了比的应用，能根据比求出五、六年级分得树苗占总树苗的几分之几是解题的关键。\n\n【详解】$30 \\times 5 = 150$ （棵）五年级: $150 \\times \\frac{4}{4 + 6} = 60$ (棵)  六年级: $150 \\times \\frac{6}{4 + 6} = 90$ (棵)  答: 五年级分得树苗 60 棵，六年级分得树苗 90 棵。",
+        "knowledge_points": [
+          "比的应用",
+          "按比分配问题"
+        ],
+        "source": "2021·黑龙江牡丹江·小升初真题",
+        "merged_id": "unit4_284",
+        "original_id": "unit4_solve_284"
+      },
+      {
+        "id": "unit4_solve_285",
+        "unit": "第四单元：比",
+        "chapter": "比的认识与应用",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "六年级学生报名参加数学兴趣小组，参加的同学是六年级总人数的 $\\frac{1}{3}$，后来又有 20 人参加，这时参加的同学与未参加同学的人数比是 $3:4$。六年级一共有多少人？",
+        "options": [],
+        "answer": "210人",
+        "solution": "【分析】把六年级的学生总数看作单位 '1'，原来参加兴趣小组的人数占总人数的 $\\frac{1}{3}$，现在参加兴趣小组的人数占总人数的 $(\\frac{3}{3 + 4})$，后来又参加的 20 人对应的分率为两个分数的分率之差，利用 '量÷对应的分率' 即可求得六年级的总人数,据此解答。题中六年级学生的总人数不变，找出后来又参加人数对应的分率是解答题目的关键。\n\n【详解】$20 \\div (\\frac{3}{3 + 4} - \\frac{1}{3})= 20 \\div (\\frac{3}{7} - \\frac{1}{3}) \\\\ = 20 \\div \\frac{2}{21} \\\\ = 210 \\text{(人)}$答: 六年级一共有 210 人。",
+        "knowledge_points": [
+          "分数除法的应用",
+          "比的应用",
+          "按比分配问题"
+        ],
+        "source": "2019·湖北黄冈·小升初真题",
+        "merged_id": "unit4_285",
+        "original_id": "unit4_solve_285"
+      },
+      {
+        "id": "unit4_solve_286",
+        "unit": "第四单元：比",
+        "chapter": "比的认识与应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "甲乙两地相距360千米，一辆客车和一辆货车分别从甲、乙两地相对开出，经过3小时相遇。已知货车和客车的速度比是5:7，客车每小时行驶多少千米？",
+        "options": [],
+        "answer": "70千米",
+        "solution": "【分析】根据'速度 = 路程 ÷ 时间'，用甲、乙两地的距离除以两车相遇时间，就是客车与货车的速度之和，再把客车、货车的速度之和平均分成（5+7）份，先用除法求出1份是多少千米，再用乘法求出7份（客车的速度）是多少千米。此题是考查按比例分配，解答本题的关键是按比例分配解题的计算方法。\n\n【详解】$360 \\div 3 \\div (5 + 7) \\times 7\\\\= 120 \\div 12 \\times 7 \\\\ = 10 \\times 7\\\\= 70 (千米)$\n\n答: 客车每小时行驶 70 千米。",
+        "knowledge_points": [
+          "按比分配问题",
+          "基础行程问题"
+        ],
+        "source": "2021·河北保定·小升初真题",
+        "merged_id": "unit4_286",
+        "original_id": "unit4_solve_286"
+      },
+      {
+        "id": "unit4_solve_287",
+        "unit": "第四单元：比",
+        "chapter": "比的认识与应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "加工一批零件，第一天加工了总数的 $\\frac{2}{7}$，第二天比第一天多加工 5 个，这时已加工的与未加工的零件个数比是 $7:5$，这批零件共多少个？",
+        "options": [],
+        "answer": "420 个",
+        "solution": "【分析】把这批零件的总个数看作单位 '1'，第一天和第二天加工的零件总个数占这批零件总个数的 $(\\frac{7}{5 + 7})$，计算出第二天加工的零件个数占零件总个数的分率，根据 '量÷对应的分率' 求出这批零件的总个数，据此解答。本题考查分数除法的应用，找出量和对应的分率是解答题目的关键。\n\n【详解】$5 \\div (\\frac{7}{5 + 7} - \\frac{2}{7} - \\frac{2}{7})\\\\ = 5 \\div [\\frac{7}{12} - (\\frac{2}{7} + \\frac{2}{7})] \\\\ = 5 \\div [\\frac{7}{12} - \\frac{4}{7}] \\\\ = 5 \\div \\frac{1}{84} \\\\ = 420 \\text{(个)}$\n\n答: 这批零件共 420 个。",
+        "knowledge_points": [
+          "分数除法的应用",
+          "分数的四则混合运算",
+          "比的应用"
+        ],
+        "source": "19-20六年级下·湖南怀化·单元测试",
+        "merged_id": "unit4_287",
+        "original_id": "unit4_solve_287"
+      },
+      {
+        "id": "unit4_solve_288",
+        "unit": "第四单元：比",
+        "chapter": "比的认识与应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "一个长方形周长为 98 厘米，长与宽的比是 $2:5$，它的面积是多少平方厘米？",
+        "options": [],
+        "answer": "490平方厘米",
+        "solution": "【分析】先求出长宽和，长宽和÷总份数，求出一份数，一份数分别乘长和宽的对应份数即可求出长和宽，根据长方形面积 = 长 × 宽，列出方程解答即可。关键是理解比的意义，掌握长方形周长和面积公式。\n\n【详解】$98 \\div 2 \\div (2 + 5)\\\\= 49 \\div 7\\\\=7 (厘米)$\n\n$7 \\times 2 = 14$ (厘米)   $7 \\times 5 = 35$ (厘米)    $35 \\times 14 = 490$ （平方厘米）答: 它的面积是 490 平方厘米。",
+        "knowledge_points": [
+          "长方形的周长",
+          "长方形的面积",
+          "按比分配问题"
+        ],
+        "source": "19-20 六年级下·湖南怀化·单元测试",
+        "merged_id": "unit4_288",
+        "original_id": "unit4_solve_288"
+      },
+      {
+        "id": "unit4_solve_289",
+        "unit": "第四单元：比",
+        "chapter": "比的认识与应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "六年级 3 个班共植树 180 棵，一、二、三班种植棵数的比是 $4:5:6$。三个班各植树多少棵？",
+        "options": [],
+        "answer": "48棵；60棵；72棵",
+        "solution": "【分析】总棵数÷总份数，求出一份数，一份数分别乘三个班的对应份数即可。关键是理解比的意义，将比的前后项看成份数。\n\n【详解】$180 \\div (4 + 5 + 6)\\\\= 180 \\div 15 \\\\=12 (棵)$\n\n$= 12$ (棵)\n\n$12 \\times 4 = 48$ (棵)$12 \\times 5 = 60$ (棵)   $12 \\times 6 = 72$ (棵)\n\n答: 三个班各植树 48 棵、60 棵、72 棵。",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "19-20 六年级下·湖南怀化·单元测试",
+        "merged_id": "unit4_289",
+        "original_id": "unit4_solve_289"
+      },
+      {
+        "id": "unit4_solve_290",
+        "unit": "第四单元：比",
+        "chapter": "比的认识与应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "学校买来 270本儿童图书，按4:5借给五、六年级。每个年级各借了多少本？",
+        "options": [],
+        "answer": "五年级 120 本; 六年级 150 本",
+        "solution": "【分析】五年级借的图书数量占图书总数量的 $\\frac{4}{4 + 5}$，六年级借的图书数量占图书总数量的 $\\frac{5}{4 + 5}$，根据比的应用求出五六年级借的图书数量，据此解答。掌握按比例分配问题的解题方法是解答题目的关键。\n\n【详解】五年级：$270 \\times \\frac{4}{4 + 5} = 120$ （本）六年级: $270 \\times \\frac{5}{4 + 5} = 150$ (本)答: 五年级借了 120 本书，六年级借了 150 本书。",
+        "knowledge_points": [
+          "比的应用",
+          "按比分配问题"
+        ],
+        "source": "19-20 六年级下·湖南怀化·单元测试",
+        "merged_id": "unit4_290",
+        "original_id": "unit4_solve_290"
+      },
+      {
+        "id": "unit4_solve_291",
+        "unit": "第四单元：比",
+        "chapter": "比的认识与应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "圆圆看一本故事书，看了224页，看的页数与没看的页数的比是 $2:3$，这本书有多少页？",
+        "options": [],
+        "answer": "560页",
+        "solution": "【分析】把这本书的总页数看作单位 '1'，则已看的 225 页所对应的分率是 $\\frac{2}{2 + 3}$，用对应量除以对应分率，就是这本书的总页数。此题考查的是比的应用，解答此题的关键是：求出225页所对应的分率，利用对应量除以对应分率，即可求出单位'1'对应的量。\n\n【详解】$224 \\div \\frac{2}{2 + 3}\\\\ = 224 \\div \\frac{2}{5} \\\\ = 560 \\text{(页)}$\n\n答：这本书有560页。",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "20-21六年级上·湖南怀化·单元测试",
+        "merged_id": "unit4_291",
+        "original_id": "unit4_solve_291"
+      },
+      {
+        "id": "unit4_solve_292",
+        "unit": "第四单元：比",
+        "chapter": "比的认识与应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "一个圆形花坛的周长是50.24米，在里面种两种花，种菊花的面积与茶花的面积比是 $2:3$，这两种花的面积分别是多少？",
+        "options": [],
+        "answer": "80.384 平方米；120.576 平方米",
+        "solution": "【分析】将花坛面积看作单位 '1'，平均分成 5 份，种菊花占 2 份，种茶花占 3 份，先算出圆形花坛的面积再计算分别占多少即可，圆形的面积公式为: $\\pi r^{2}$。本题关键在于对比的理解，并且要会通过圆的周长找到圆的半径从而求出圆的面积。\n\n【详解】$50.24 \\div 3.14 \\div 2 = 8$ （米） $3.14 \\times 8^{2} = 200.96$ （平方米）\n\n$200.96 \\div (2 + 3) \\\\ = 200.96 \\div 5 \\\\ = 40.192 \\text{(平方米)}$\n\n$40.192 \\times 2 = 80.384$ （平方米）  $40.192 \\times 3 = 120.576$ （平方米）\n\n答: 种植菊花的面积是 80.384 平方米，种植茶花的面积是 120.576 平方米。",
+        "knowledge_points": [
+          "圆的周长及应用",
+          "圆的面积及应用",
+          "按比分配问题"
+        ],
+        "source": "20-21六年级上·湖南怀化·单元测试",
+        "merged_id": "unit4_292",
+        "original_id": "unit4_solve_292"
+      },
+      {
+        "id": "unit4_solve_293",
+        "unit": "第四单元：比",
+        "chapter": "比的认识与应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "两个大小不等的圆形粮仓，小粮仓的底面周长是 12.56 米，它的占地面积是大粮仓的 $\\frac{1}{3}$，大粮仓占地面积是多少平方米？",
+        "options": [],
+        "answer": "37.68平方米",
+        "solution": "【分析】根据小粮仓的底面周长计算出小粮仓的底面半径,再利用圆的面积公式计算出小粮仓的占地面积，大粮仓的占地面积 = 小粮仓的占地面积 $\\div \\frac{1}{3}$，据此解答。灵活运用圆的周长和面积计算公式是解答题目的关键。\n\n【详解】小粮仓的底面半径：$12.56 \\div 3.14 \\div 2\\\\= 4 \\div 2 \\\\ = 2 \\text{(米)}$\n\n小粮仓的占地面积：$3.14 \\times 2^{2}\\\\= 3.14 \\times 4\\\\= 12.56$ （平方米）\n\n大粮仓的占地面积: $12.56 \\div \\frac{1}{3} = 37.68$ (平方米)答: 大粮仓占地面积是 37.68 平方米。",
+        "knowledge_points": [
+          "分数除法的应用",
+          "圆的周长及应用",
+          "圆的面积及应用"
+        ],
+        "source": "20-21 六年级上·湖南怀化·单元测试",
+        "merged_id": "unit4_293",
+        "original_id": "unit4_solve_293"
+      },
+      {
+        "id": "unit4_solve_294",
+        "unit": "第四单元：比",
+        "chapter": "比的认识与应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "张师傅加工一批零件，第一天完成的个数与零件总个数的比是 $1:3$。如果再加工 15 个，就可以完成这批零件的 $\\frac{1}{2}$。这批零件共有多少个？",
+        "options": [],
+        "answer": "90个",
+        "solution": "【分析】根据题意可知，第一天完成的个数占总个数的 $\\frac{1}{3}$，用 $\\frac{1}{2} - \\frac{1}{3}$ 求出再加工的 15 个占总个数的百分比，再根据分数除法的意义解答即可。明确再加工的 15 个占总个数的几分之几是解答本题的关键。\n\n【详解】$15 \\div (\\frac{1}{2} - \\frac{1}{3})\\\\ = 15 \\div \\frac{1}{6} \\\\ = 90 (\\text{个})$\n\n答：这批零件共有 90 个。",
+        "knowledge_points": [
+          "比的应用"
+        ],
+        "source": "2015·湖北·小升初真题",
+        "merged_id": "unit4_294",
+        "original_id": "unit4_solve_294"
+      },
+      {
+        "id": "unit4_solve_295",
+        "unit": "第四单元：比",
+        "chapter": "比的认识与应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "明明按如下标准配制了两杯糖水：第一杯，糖和水的比是 $1:10$；第二杯，用 10 克糖配制 125 克糖水。哪杯糖水更甜？你是怎么想的？",
+        "options": [],
+        "answer": "第一杯",
+        "solution": "【分析】根据'含糖率 = 糖的质量 ÷ 糖水的质量 × 100%'计算出两杯糖水的含糖率，含糖率高的糖水更甜。掌握生活中百分率的计算方法是解答题目的关键。\n\n【详解】第一杯：$1 \\div (1 + 10) \\times 100\\%\\\\ = 1 \\div 11 \\times 100\\% \\\\ \\approx 0.091 \\times 100\\% \\\\ = 9.1\\%$\n\n第二杯：$10 \\div 125 \\times 100\\%\\\\= 0.08 \\times 100\\% \\\\ = 8\\%$\n\n因为 $9.1\\% > 8\\%$，所以第一杯糖水更甜。答: 第一杯糖水更甜。",
+        "knowledge_points": [
+          "求一个数是另一个数的百分之几（百分率问题）"
+        ],
+        "source": "21-22 六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_295",
+        "original_id": "unit4_solve_295"
+      },
+      {
+        "id": "unit4_solve_296",
+        "unit": "第四单元：比",
+        "chapter": "比的认识与应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "水果店有桃子、芒果、香蕉共 330 千克，其中桃子与芒果的质量比是 $5:4$，桃子与香蕉的质量比是 $2:3$，水果店有三种水果各多少千克",
+        "options": [],
+        "answer": "芒果: 80 千克; 桃子: 100 千克; 香蕉: 150 千克",
+        "solution": "【分析】由题意可知，桃子与芒果的质量比是 $5:4$，桃子与香蕉的质量比是 $2:3$，先它们按照比的基本性质化成三联比,然后再按比分配进行计算即可。本题考查按比分配，明确芒果、桃子和香蕉的份数是解题的关键。\n\n【详解】芒果：桃子 = 4:5     桃子：香蕉 = 2: 3    芒果：桃子：香蕉=8:10:15\n\n$8 + 10 + 15 = 33$ （份）  $330 \\times \\frac{8}{33} = 80$ （千克） $330 \\times \\frac{10}{33} = 100$ (千克)$330 \\times \\frac{15}{33} = 150$ (千克)\n\n答: 芒果有 80 千克，桃子有 100 千克，香蕉有 150 千克。",
+        "knowledge_points": [
+          "比的化简",
+          "按比分配问题"
+        ],
+        "source": "20-21 六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_296",
+        "original_id": "unit4_solve_296"
+      },
+      {
+        "id": "unit4_solve_297",
+        "unit": "第四单元：比",
+        "chapter": "比的认识与应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "等腰三角形的周长是 $70 \\mathrm{~cm}$，一条腰与底边的长度比是 $3:4$，这个三角形的底边长是多少厘米?",
+        "options": [],
+        "answer": "28 厘米",
+        "solution": "【分析】等腰三角形两腰相等，用周长 ÷ 总份数 × 底边对应份数 = 底边长，据此列式解答。关键是理解比的意义，熟悉等腰直角三角形的特征。\n\n【详解】$70 \\div (3 + 3 + 4) \\times 4\\\\= 70 \\div 10 \\times 4\\\\=28 (厘米)$\n\n答: 这个三角形的底边长是 28 厘米。",
+        "knowledge_points": [
+          "按比分配问题",
+          "等腰三角形和等边三角形的认识及特征"
+        ],
+        "source": "20-21 六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_297",
+        "original_id": "unit4_solve_297"
+      },
+      {
+        "id": "unit4_solve_298",
+        "unit": "第四单元：比",
+        "chapter": "比的认识与应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "水是由氢和氧按 1:8 的质量比化合而成的，6.3 千克的水中含氧多少千克?",
+        "options": [],
+        "answer": "0.7千克",
+        "solution": "【分析】因为氢和氧按 $1:8$ 化合成水，氢占水的 $\\frac{1}{1+8}$，然后用乘法解答即可。此题考查的是按比分配的方法，关键是求出氧占水的几分之几，然后再根据一个数乘分数的意义，用乘法列式解答。\n\n【详解】$6.3 \\times \\frac{1}{1 + 8} = 0.7$ （千克）答: 6.3 千克的水中含氧 0.7 千克。",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "20-21 六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_298",
+        "original_id": "unit4_solve_298"
+      },
+      {
+        "id": "unit4_solve_299",
+        "unit": "第四单元：比",
+        "chapter": "比的认识与应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "学校新到一批图书，按六年级三个班的人数分配到各班，一班有58人，二班有62人，三班有60人。结果一班分得29本，二、三班各分得多少本图书？",
+        "options": [],
+        "answer": "二班：31本；三班：30本",
+        "solution": "【分析】一班有 58 人，已知一班分到 29 本，可知一班每人分到 $29 \\div 58 = \\frac{1}{2}$ 本，再分别乘二班、三班的人数即可得二班、三班各应分得多少本图书。本题考查了按比例分配应用题，关键是得出每人分到 $29 \\div 58 = \\frac{1}{2}$ 本。\n\n【详解】$29 \\div 58 = \\frac{1}{2}$ (本)  $62 \\times \\frac{1}{2} = 31 \\text{(本)} \\quad60 \\times \\frac{1}{2} = 30 (\\text{本})$答: 二、三班各分得 31 本和 30 本图书。",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "20-21六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_299",
+        "original_id": "unit4_solve_299"
+      },
+      {
+        "id": "unit4_solve_300",
+        "unit": "第四单元：比",
+        "chapter": "比的认识与应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "两地相距 600 千米，甲、乙两车同时从两地相对开出，5 小时后相遇，已知甲、乙两车速度比是 $5:3$，两车每小时各行多少千米?",
+        "options": [],
+        "answer": "75千米、45千米",
+        "solution": "【分析】路程÷相遇时间 = 速度和，速度和 ÷ 总份数 = 一份数，一份数分别乘两车速度对应份数即可。关键是理解比的意义，理解速度、时间、路程之间的关系。\n\n【详解】$600 \\div 5 = 120$ （千米）\n\n$120 \\div (5 + 3) \\\\ = 120 \\div 8 \\\\ = 15 \\text{(千米)}$\n\n$15 \\times 5 = 75 (\\text{千米})$  $15 \\times 3 = 45 (\\text{千米})$答: 甲车每小时行 75 千米、乙车每小时行 45 千米。",
+        "knowledge_points": [
+          "按比分配问题",
+          "相遇问题"
+        ],
+        "source": "20-21 六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_300",
+        "original_id": "unit4_solve_300"
+      },{
+        "id": "unit4_choice_301",
+        "unit": "第四单元：比",
+        "chapter": "比的意义、比的应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "(21-22 六年级上·江西赣州·单元测试) 把一张长 $6 \\mathrm{~cm}$ 、宽 $2 \\mathrm{~cm}$ 的长方形纸对折一次后, 长与宽的比可能是 ( ) 和 ( )。",
+        "options": [
+          "A. $6: 1; 3: 1$",
+          "B. $3: 1; 3: 2$",
+          "C. $3: 2; 6: 1$",
+          "D. $3: 1; 2: 3$"
+        ],
+        "answer": "C",
+        "solution": "【分析】可以沿着长方形的长对折, 也可以沿着长方形的宽对折, 画出图形, 先计算对折之后长与宽的值, 再求长与宽的比即可。分析图形根据比的意义求出长和宽的比。\n【详解】![](images\\330eec824a5c90fdde85020a8bbad15c.png)\n图一：6cm：（2÷2）$\\mathrm{cm} = 6:1$ 图二：$(6 \\div 2) \\mathrm{cm}: 2 \\mathrm{~cm} = 3: 2$ 故答案为: C",
+        "knowledge_points": [
+          "比的意义",
+          "比的应用"
+        ],
+        "source": "21-22 六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_301",
+        "original_id": "unit4_choice_301"
+      },
+      {
+        "id": "unit4_choice_302",
+        "unit": "第四单元：比",
+        "chapter": "比的意义",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "(21-22 六年级上·江西赣州·单元测试) 已知甲比乙多 $\\frac{1}{4}$ ,下面理解错误的是 ( )。",
+        "options": [
+          "A. 甲是乙的 $\\frac{5}{4}$",
+          "B. 乙与甲的比是 $4: 5$",
+          "C. 乙比甲少 $\\frac{1}{4}$"
+        ],
+        "answer": "C",
+        "solution": "【分析】甲比乙多 $\\frac{1}{4}$ , 把乙看作单位 '1', 假设乙是 1, 则甲是 $(1 + \\frac{1}{4})$ , 据此计算。A比B多几分之几表示为：(A-B)÷B；B比A少几分之几表示为：(A-B)÷A。\n【详解】A．甲÷乙= $(1 + \\frac{1}{4}) \\div 1 = \\frac{5}{4}$ ，正确；B. 乙: 甲 $= 1: (1 + \\frac{1}{4}) = 1: \\frac{5}{4} = 4: 5$ , 正确; C. $(1 + \\frac{1}{4} - 1) \\div (1 + \\frac{1}{4}) = \\frac{1}{4} \\div \\frac{5}{4} = \\frac{1}{5}$ , 错误。故答案为: C",
+        "knowledge_points": [
+          "分数与分数的除法",
+          "求一个数占另一个数几分之几",
+          "比的意义"
+        ],
+        "source": "21-22 六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_302",
+        "original_id": "unit4_choice_302"
+      },
+      {
+        "id": "unit4_choice_303",
+        "unit": "第四单元：比",
+        "chapter": "比的基本性质",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "(21-22 六年级上·江西赣州·单元测试) 3:2 的前项增加3, 以下对于后项的处理可使比值不变的是 ( ) 和 ( )。",
+        "options": [
+          "A. 加 3 ; 加 6",
+          "B. 乘2；加2",
+          "C. 加2；加6",
+          "D. 加 6 ; 加 3"
+        ],
+        "answer": "B",
+        "solution": "【分析】可分别按 4 个选项中的叙述, 依次求得变化后的比,再做判断。通过依次计算几种方法变化后的比, 能够加深学生们对于比的性质的理解: 前项和后项同时乘或除以一个不为 0 的数, 比值不变。\n【详解】A. $(3 + 3)$ ： $(2 + 3) = 6:5$ ，不合题意；(3+3) : (2+6) $= 6:8 = 3:4$ , 不合题意;B. $(3 + 3) : (2 \\times 2) = 6: 4 = 3: 2$ , 符合题意; (3+3) : (2+2) =6:4=3:2, 符合题意;C. $(3 + 3): (2 + 2) = 6: 4 = 3: 2$ , 符合题意;(3+3) : (2+6) = 6:8=3:4, 不合题意;D. $(3 + 3): (2 + 6) = 6: 8 = 3: 4$ , 不合题意;(3+3) : (2+3) =6:5, 不合题意。故答案为: B",
+        "knowledge_points": [
+          "比的基本性质"
+        ],
+        "source": "21-22 六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_303",
+        "original_id": "unit4_choice_303"
+      },
+      {
+        "id": "unit4_choice_304",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "（2020·全国·小升初模拟）一辆客车从甲地到乙地，第一天行驶了全程的 $\\frac{1}{5}$ ，第二天行驶了 450 千米，这时已行路程和剩下路程的比是 $3:7$ 。甲乙两地相距（ ）千米。",
+        "options": [
+          "A. 750",
+          "B. 4500",
+          "C. 2250"
+        ],
+        "answer": "B",
+        "solution": "【分析】根据题意可知, 两天行驶的总路程占全程的 $\\frac{3}{3 + 7}$ ,第二天行驶的路程占全程的 $\\frac{1}{5}$ , 正好是 450 千米, 再根据分数除法的意义解答即可。求出第二天行驶的路程占全程的几分之几是解答本题的关键, 再根据分数除法的意义解答。\n【详解】$450 \\div (\\frac{3}{3 + 7} - \\frac{1}{5})\\\\= 4 5 0 \\div \\frac {1}{1 0}\\\\= 4500$ （千米）\n故正确答案为: B",
+        "knowledge_points": [
+          "求一个数的几分之几的问题",
+          "分数除法的应用",
+          "比的应用"
+        ],
+        "source": "2020·全国·小升初模拟",
+        "merged_id": "unit4_304",
+        "original_id": "unit4_choice_304"
+      },
+      {
+        "id": "unit4_choice_305",
+        "unit": "第四单元：比",
+        "chapter": "比的意义",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "(21-22 六年级上·全国·单元测试) 在蜂蜜水中, 蜂蜜占蜂蜜水的 $\\frac{1}{10}$ , 蜂蜜和水的比是 ( )。",
+        "options": [
+          "A. $1: 10$",
+          "B. $1 : 9$",
+          "C. 45",
+          "D. 5"
+        ],
+        "answer": "B",
+        "solution": "【分析】蜂蜜占蜂蜜水的 $\\frac{1}{10}$ , 据此可知蜂蜜的量为 1 份, 蜂蜜水的总量为 10 份, 水的量为 $(10 - 1)$ 份, 再进一步求出蜂蜜和水的比即可。明确蜂蜜和水各占的份数是解答本题的关键。\n【详解】蜂蜜占蜂蜜水的 $\\frac{1}{10}$ , 蜂蜜和水的比是 $1: (10 - 1) = 1: 9$ 故答案为: B",
+        "knowledge_points": [
+          "比的意义"
+        ],
+        "source": "21-22 六年级上·全国·单元测试",
+        "merged_id": "unit4_305",
+        "original_id": "unit4_choice_305"
+      },
+      {
+        "id": "unit4_choice_306",
+        "unit": "第四单元：比",
+        "chapter": "求比值",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "(21-22 六年级上·全国·单元测试) 在下面各比中, 与 $0.5:0.6$ 的比值相等的比是 ( )。",
+        "options": [
+          "A. $\\frac{1}{5}: \\frac{1}{6}$",
+          "B. $\\frac{1}{2}: \\frac{3}{5}$",
+          "C. $25: 26$",
+          "D. $32: 26$"
+        ],
+        "answer": "B",
+        "solution": "【分析】用比的前项除以比的后项求比值即可；分别求出每个选项的比值，再进行判断即可。熟记求比值的方法是解答本题的关键。\n【详解】$0.5: 0.6 = \\frac{5}{6}$ ; A. $\\frac{1}{5}: \\frac{1}{6} = \\frac{6}{5}$ ; B. $\\frac{1}{2}: \\frac{3}{5} = \\frac{5}{6}$ ; C. $25: 26 = \\frac{25}{26}$ ; D. $32: 26 = \\frac{16}{13}$ ;故答案为: B。",
+        "knowledge_points": [
+          "求比值"
+        ],
+        "source": "21-22 六年级上·全国·单元测试",
+        "merged_id": "unit4_306",
+        "original_id": "unit4_choice_306"
+      },
+      {
+        "id": "unit4_choice_307",
+        "unit": "第四单元：比",
+        "chapter": "比的意义、比的化简",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "(18-19 六年级上·全国·期中) 从武汉到北京, 甲车用了 8 小时, 乙车用了 10 小时。甲、乙两车速度的最简整数比是 ( )。",
+        "options": [
+          "A. $8: 10$",
+          "B. $5: 4$",
+          "C. $\\frac{1}{8}: \\frac{1}{10}$",
+          "D. 4:5"
+        ],
+        "answer": "B",
+        "solution": "【分析】根据题意可知，从武汉到北京的总路程为单位'1'，则甲车的速度为 $1 \\div 8 = \\frac{1}{8}$ ，乙车的速度为 $1 \\div 10 = \\frac{1}{10}$再写出甲、乙两车速度的最简整数比即可。明确题目中的单位 '1', 分别求出甲车和乙车的速度, 再进一步解答即可。\n【详解】$1 \\div 8 = \\frac{1}{8}$ ;$1 \\div 1 0 = \\frac {1}{1 0};$甲、乙两车速度的比为 $\\frac{1}{8}: \\frac{1}{10} = 5: 4$ ;故答案为: B。",
+        "knowledge_points": [
+          "比的意义",
+          "比的化简"
+        ],
+        "source": "18-19 六年级上·全国·期中",
+        "merged_id": "unit4_307",
+        "original_id": "unit4_choice_307"
+      },
+      {
+        "id": "unit4_choice_308",
+        "unit": "第四单元：比",
+        "chapter": "比的意义、比的化简",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "(21-22 六年级上·全国·单元测试) 当 $x \\times \\frac{2}{3} = y \\times \\frac{3}{4} (x \\cdot y \\neq 0)$ 时, $x: y =$ ( )。",
+        "options": [
+          "A. 8:9",
+          "B. 9:8",
+          "C. $1: 2$",
+          "D. 2:1"
+        ],
+        "answer": "B",
+        "solution": "【分析】假设 $x \\times \\frac{2}{3} = y \\times \\frac{3}{4} = 1$ , 分别求出 $x$ 和 $y$ , 再写出它们之间的比即可。本题采用了假设法，使题目具体化，分别求出 $x$ 和 $y$ ，再进一步解答。\n【详解】假设 $x \\times \\frac{2}{3} = y \\times \\frac{3}{4} = 1$则 $x = \\frac{3}{2}, y = \\frac{4}{3}$$x: y = \\frac {3}{2}: \\frac {4}{3} = 9: 8$故答案为: B",
+        "knowledge_points": [
+          "比的意义",
+          "倒数的认识及应用",
+          "比的化简"
+        ],
+        "source": "21-22 六年级上·全国·单元测试",
+        "merged_id": "unit4_308",
+        "original_id": "unit4_choice_308"
+      },
+      {
+        "id": "unit4_choice_309",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "(21-22 六年级上·全国·单元测试) 甲与乙的比是 $3: 4$ , 甲、乙两数的和是 210 , 求甲、乙两数各是几? 正确的是 ( )。",
+        "options": [
+          "A. 甲: 75, 乙: 135",
+          "B. 甲: 80, 乙: 130",
+          "C. 甲: 100, 乙: 110",
+          "D. 甲: 90, 乙: 120"
+        ],
+        "answer": "D",
+        "solution": "【分析】甲与乙的比是 $3: 4$ , 甲占甲乙两数和的 $\\frac{3}{3 + 4}$ , 乙占甲乙两数和的 $\\frac{4}{3 + 4}$ , 根据求一个数的几分之几是多少用乘法计算甲乙各是多少。据此解答再选择答案。此题考查的是按比例分配问题，按比例分配的方法是，将已知整数比或分率比变为按份数分配，把比的各项相加得到总份数，各项和总份数的比就是各分量在总量中占的份数，由此求的各个分量。\n【详解】甲数: $210 \\times \\frac{3}{3 + 4} = 90$ ,乙数: $210 \\times \\frac{4}{3 + 4} = 120$ ;故答案为: D。",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "21-22 六年级上·全国·单元测试",
+        "merged_id": "unit4_309",
+        "original_id": "unit4_choice_309"
+      },
+      {
+        "id": "unit4_choice_310",
+        "unit": "第四单元：比",
+        "chapter": "比的意义",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "(21-22 六年级上·全国·期中) 下列叙述中, 与众不同的 '比' 是 ( )。",
+        "options": [
+          "A. 国旗法规定：国旗的长与宽的比必须是 $3: 2$ 。",
+          "B. 在这场足球比赛中, 601 班 0:5 惨败给 602 班。",
+          "C. 603 班男女生人数的比是 $8: 7$ 。",
+          "D. 等腰直角三角形的顶角与底角度数比是 $2: 1$ 。"
+        ],
+        "answer": "B",
+        "solution": "【分析】根据比的意义，结合选项，选出最与众不同的即可。本题考查了比的意义，比表示两个量之间的倍数关系。\n【详解】国旗的长和宽的比、男生女生的人数比以及等腰直角三角形的顶角与底角度数比，都表示出了两个量之间的倍数关系，但是601班0:5惨败给602班中的这个比不存在倍数关系，只是一种计分方式。所以，与众不同的'比'是B选项。故答案为: B",
+        "knowledge_points": [
+          "比的意义"
+        ],
+        "source": "21-22 六年级上·全国·期中",
+        "merged_id": "unit4_310",
+        "original_id": "unit4_choice_310"
+      },
+      {
+        "id": "unit4_choice_311",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题、长方体的体积",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "(18-19 六年级下·河南洛阳·期末) 一个长方体, 相交于同一个顶点的三条棱长度的比是 $1: 1: 2$ , 这三条棱的和是 80 厘米, 这个长方体的体积是 ( ) 立方厘米。",
+        "options": [
+          "A. 3200",
+          "B. 4000",
+          "C. 16000"
+        ],
+        "answer": "C",
+        "solution": "【分析】三条棱的和÷总份数，求出一份数，一份数分别乘长宽高对应份数，求出长、宽、高，根据长方体体积=长×宽×高，列式计算即可。关键是理解比的意义，掌握长方体体积公式。\n【详解】$80 \\div (1 + 1 + 2)\\\\= 8 0 \\div 4 \\\\ = 2 0 \\text {(厘 米)} $\n$2 0 \\times 2 = 4 0 (\\text {厘 米})$ $20 \\times 20 \\times 40 = 16000$ （立方厘米）故答案为: C",
+        "knowledge_points": [
+          "按比分配问题",
+          "长方体的体积"
+        ],
+        "source": "18-19 六年级下·河南洛阳·期末",
+        "merged_id": "unit4_311",
+        "original_id": "unit4_choice_311"
+      },
+      {
+        "id": "unit4_choice_312",
+        "unit": "第四单元：比",
+        "chapter": "比的意义、比的化简",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "(21-22 六年级上·全国·单元测试) 甲是乙的 2.5 倍, 那么甲与乙的最简比是 ( )。",
+        "options": [
+          "A. 25:10",
+          "B. $10: 25$",
+          "C. $2: 5$",
+          "D. $5: 2$"
+        ],
+        "answer": "D",
+        "solution": "【分析】由 '甲是乙的 2.5 倍' 可知, 把乙数看作 1 , 则甲数是 2.5 , 写出相应的比, 再根据比的基本性质化简即可。本题较易，考查了有关比的意义和化简比的知识。\n【详解】甲与乙的最简比是：$2.5:1= (2. 5 \\times 1 0): (1 \\times 1 0) = 2 5: 1 0 = (2 5 \\div 5): (1 0 \\div 5) = 5: 2; $故答案为: D",
+        "knowledge_points": [
+          "比的意义",
+          "比的化简"
+        ],
+        "source": "21-22 六年级上·全国·单元测试",
+        "merged_id": "unit4_312",
+        "original_id": "unit4_choice_312"
+      },
+      {
+        "id": "unit4_choice_313",
+        "unit": "第四单元：比",
+        "chapter": "比的意义、比的化简",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "(21-22 六年级上·全国·单元测试) 一种长方形屏幕长与宽的比是 $16: 9$ , 下面几种规格屏幕合格的 ( )。",
+        "options": [
+          "A. 长 1.6 米, 宽 1 米",
+          "B. 长 $\\frac{4}{5}$ 米, 宽 $\\frac{9}{20}$ 米",
+          "C. 长 1.2 米, 宽 80 厘米",
+          "D. 以上都不对"
+        ],
+        "answer": "B",
+        "solution": "【分析】每个选项中长方形的屏幕的长和宽的比只要等于 16:9, 就符合标准, 屏幕合格, 否则不合格, 据此解答即可。本题较易，考查了有关比的意义和化简比的知识，掌握基础知识是关键。\n【详解】A. 1.6 米: 1 米 $= 16: 10 \\neq 16: 9$ , 所以不属于规格标准;B. $\\frac{4}{5}$ 米: $\\frac{9}{20}$ 米 $= 16: 9$ , 所以这个规格屏幕合格;C. $1.2 \\text{ 米}: 80$ 厘米 $= 120: 80 = 12: 8 \\neq 16: 9$ , 所以不属于规格标准;故答案为: B",
+        "knowledge_points": [
+          "比的意义",
+          "比的化简"
+        ],
+        "source": "21-22 六年级上·全国·单元测试",
+        "merged_id": "unit4_313",
+        "original_id": "unit4_choice_313"
+      },
+      {
+        "id": "unit4_choice_314",
+        "unit": "第四单元：比",
+        "chapter": "比的意义、比的化简",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "（20-21六年级下·广东佛山·期末）一种自行车，如果前齿轮转3圈时，后齿轮要转8圈。这种自行车前、后齿轮的齿数可能是（）。",
+        "options": [
+          "A. 46 和 20",
+          "B. 48 和 18",
+          "C. 40和32",
+          "D. 38 和 16"
+        ],
+        "answer": "B",
+        "solution": "【分析】根据前后轮齿数比反过来是转动圈数比，将各选项写成比的形式，化简，结果是 $8: 3$ 即可。关键是理解比的意义，两数相除又叫两个数的比。\n【详解】A．$46:20 = 23:10$ B. $48: 18 = 8: 3$ C. $40: 32 = 5: 4$ D. $38: 16 = 19: 8$ 故答案为: B",
+        "knowledge_points": [
+          "比的意义",
+          "比的化简"
+        ],
+        "source": "20-21六年级下·广东佛山·期末",
+        "merged_id": "unit4_314",
+        "original_id": "unit4_choice_314"
+      },
+      {
+        "id": "unit4_choice_315",
+        "unit": "第四单元：比",
+        "chapter": "求一个数占另一个数几分之几、比的意义",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "(20-21 六年级下·广东广州·期末) 东东、青青、芳芳三人分一包糖果，准备按 $3: 2: 5$ 或 $1: 2: 3$ 来分配，两种分法中（）分得的一样多。",
+        "options": [
+          "A. 东东",
+          "B. 青青",
+          "C. 芳芳",
+          "D. 无法确定"
+        ],
+        "answer": "C",
+        "solution": "【分析】分别用三人对应份数÷总份数，两种分法中，得到分率相同的说明分得一样多。关键是理解比的意义, 分数的分子相当于被除数,分母相当于除数。\n【详解】$3 + 2 + 5 = 10$ 、$1 + 2 + 3 = 6$ 京东: $3 \\div 10 = \\frac{3}{10}, 1 \\div 6 = \\frac{1}{6}$ 青青: $2 \\div 10 = \\frac{1}{5},$ $2 \\div 6 = \\frac{1}{3}$\n芳芳: $5 \\div 10 = \\frac{1}{2}, 3 \\div 6 = \\frac{1}{2}$ 故答案为: C",
+        "knowledge_points": [
+          "求一个数占另一个数几分之几",
+          "比的意义"
+        ],
+        "source": "20-21 六年级下·广东广州·期末",
+        "merged_id": "unit4_315",
+        "original_id": "unit4_choice_315"
+      },
+      {
+        "id": "unit4_choice_316",
+        "unit": "第四单元：比",
+        "chapter": "比的基本性质",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "(20-21 六年级下·广东佛山·期末) 在 2:5 中, 如果比的前项加上 6 , 要使比值不变, 后项应该 ( )。",
+        "options": [
+          "A. 加 6",
+          "B. 乘3",
+          "C. 乘4",
+          "D. 乘6"
+        ],
+        "answer": "C",
+        "solution": "【分析】在 2:5 中, 如果比的前项加上 6 , 则前项变为 8 ,扩大到原来的 4 倍, 要使比值不变, 后项也应扩大到原来的 4 倍, 据此解答即可。熟练掌握比的性质是解答本题的关键。\n【详解】在 2:5 中, 如果比的前项加上 6 , 要使比值不变,后项应该乘 $4 \\text{。}$故答案为: C",
+        "knowledge_points": [
+          "比的基本性质"
+        ],
+        "source": "20-21 六年级下·广东佛山·期末",
+        "merged_id": "unit4_316",
+        "original_id": "unit4_choice_316"
+      },
+      {
+        "id": "unit4_choice_317",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "(20-21 六年级下·辽宁大连·期末) 某校六年一班有学生 48 人, 这个班男、女生人数的比可能是 ( )。",
+        "options": [
+          "A. $5: 2$",
+          "B. $7: 8$",
+          "C. 6:11",
+          "D. 9:7"
+        ],
+        "answer": "D",
+        "solution": "【分析】根据实际可知, 男女生的人数比的前项与后项之和是总人数的因数, 得出的男、女生人数才是整数, 据此解答。此题考查了比的应用, 明确求出的一份量是整数,是解题关键。\n【详解】A．$5 + 2 = 7$ ，7不是48的因数。B. $7 + 8 = 15, 15$ 不是 48 的因数。C. $6 + 11 = 17, 17$ 不是 48 的因数。D. $9 + 7 = 16, 16$ 是 48 的因数。故选择：D",
+        "knowledge_points": [
+          "比的应用"
+        ],
+        "source": "20-21 六年级下·辽宁大连·期末",
+        "merged_id": "unit4_317",
+        "original_id": "unit4_choice_317"
+      },
+      {
+        "id": "unit4_choice_318",
+        "unit": "第四单元：比",
+        "chapter": "比的意义、比的化简",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "(20-21 六年级下·湖南岳阳·期末) 甲数的 $\\frac{2}{5}$ 等于乙数的 $\\frac{4}{9}$ (甲、乙均不为 0), 甲数与乙数的最简整数比是 ( )。",
+        "options": [
+          "A. 9:10",
+          "B. 10: 9",
+          "C. 8: 35",
+          "D. 35: 8"
+        ],
+        "answer": "B",
+        "solution": "【分析】假设甲数 $\\times \\frac{2}{5} =$ 乙数 $\\times \\frac{4}{9} = 1$ , 分别计算出甲数和乙数,再写出它们之间的比即可。本题采用了假设法，假设法使题目具体化、简单化。\n【详解】假设甲数 $\\times \\frac{2}{5} =$ 乙数 $\\times \\frac{4}{9} = 1$ 则甲数 $= \\frac{5}{2}$ , $乙数= \\frac{9}{4}$ 甲数与乙数的比为: $\\frac{5}{2}: \\frac{9}{4} = 10: 9$故答案为: B",
+        "knowledge_points": [
+          "比的意义",
+          "比的化简"
+        ],
+        "source": "20-21 六年级下·湖南岳阳·期末",
+        "merged_id": "unit4_318",
+        "original_id": "unit4_choice_318"
+      },
+      {
+        "id": "unit4_choice_319",
+        "unit": "第四单元：比",
+        "chapter": "比的意义、比的化简",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "(2020·陕西宝鸡·小升初真题) 在 100 克盐水中, 含盐 20 克, 盐与水的比是 ( )。",
+        "options": [
+          "A. $1: 4$",
+          "B. 1:5",
+          "C. $1: 6$"
+        ],
+        "answer": "A",
+        "solution": "【分析】根据题意可知, 水的质量为 $100 - 20 = 80$ 克, 再用盐的质量比水的质量即可。解答本题的关键是明确水的质量, 切勿将盐水的质量看作水的质量。\n【详解】盐与水的比为 $20: (100 - 20) = 1: 4$ ;故答案为: A",
+        "knowledge_points": [
+          "比的意义",
+          "比的化简"
+        ],
+        "source": "2020·陕西宝鸡·小升初真题",
+        "merged_id": "unit4_319",
+        "original_id": "unit4_choice_319"
+      },
+      {
+        "id": "unit4_choice_320",
+        "unit": "第四单元：比",
+        "chapter": "比的化简、比的应用、比的基本性质",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "（19-20 六年级下·贵州铜仁·期末）把 10 克糖溶入 100 克水中，糖水与糖的重量比是（）。",
+        "options": [
+          "A. $1: 9$",
+          "B. 11:1",
+          "C. $1: 11$"
+        ],
+        "answer": "B",
+        "solution": "【分析】糖加水是糖水的重量，用糖水的重量比上糖的重量，然后化简即可。化成最简整数比应用到比的基本性质, 前项和后项同时乘上或者除以同一个不为零的数, 比值不变。\n【详解】$10 + 100: 10 = 110: 10 = 11: 1$ 。故答案为: B。",
+        "knowledge_points": [
+          "比的化简",
+          "比的应用",
+          "比的基本性质"
+        ],
+        "source": "19-20 六年级下·贵州铜仁·期末",
+        "merged_id": "unit4_320",
+        "original_id": "unit4_choice_320"
+      },
+      {
+        "id": "unit4_choice_321",
+        "unit": "第四单元：比",
+        "chapter": "求比值、正方形的面积、圆的面积及应用",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "（2014 三年级·全国·课后作业）周长相等的正方形和圆，它们面积的比是（）",
+        "options": [
+          "A. $1: 1$",
+          "B. $2: 3$",
+          "C. $4: \\pi$",
+          "D. $\\pi : 4$"
+        ],
+        "answer": "D",
+        "solution": "【分析】设周长为 $a$ , 则正方形的边长为 $\\frac{a}{4}$ , 圆的半径 $\\frac{a}{2 \\pi}$ ,为三, 利用长方形和圆的面积公式求得出二者的面积, 再用正方形的面积除以圆的面积, 即可求得它们的面积比。此题主要考查正方形和圆的面积公式的灵活应用。\n【详解】假定正方形和圆的周长都为 $a$ , 则:正方形的边长: $a \\div 4 = \\frac{a}{4}$ 正方形的面积: $\\frac{a}{4} \\times \\frac{a}{4} = \\frac{a^{2}}{16}$ 圆的半径为: $a \\div \\pi \\div 2 = \\frac{a}{2 \\pi}$ 圆的面积: $\\pi \\times \\frac{a}{2 \\pi} \\times \\frac{a}{2 \\pi} = \\frac{a^{2}}{4 \\pi}$ 正方形和圆的面积比为: $\\frac{a^{2}}{16}: \\frac{a^{2}}{4 \\pi} = \\frac{a^{2}}{16} \\times \\frac{4 \\pi}{a^{2}} = \\frac{\\pi}{4} = \\pi: 4$\n故答案为: D",
+        "knowledge_points": [
+          "求比值",
+          "正方形的面积",
+          "圆的面积及应用"
+        ],
+        "source": "2014 三年级·全国·课后作业",
+        "merged_id": "unit4_321",
+        "original_id": "unit4_choice_321"
+      },
+      {
+        "id": "unit4_choice_322",
+        "unit": "第四单元：比",
+        "chapter": "长方形的面积、比的应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "（18-19六年级下·贵州·期末）一个长方形的周长是50米，长与宽的比是3:2，它的面积是（）平方米。",
+        "options": [
+          "A. 600",
+          "B. 300",
+          "C. 200",
+          "D. 150"
+        ],
+        "answer": "D",
+        "solution": "【分析】因为长方形的周长是 50 厘米, 那么长方形的条长与宽的和是: $50 \\div 2 = 25$ (厘米), 又因为长与宽的比是:\n3:2 可知：长与宽的和的总份数是： $3 + 2 = 5$ （份），用长与宽的和除以总份数求出每份的长度，再乘长、宽分别占的份数，即可求出长与宽的长度，再根据长方形面积 $=$ 长 $\\times$ 宽，计算即可。此题主要考查比的运用, 关键是根据比求出每份长度, 进而求出长与宽的长度。\n【详解】$50 \\div 2 = 25$ （厘米）\n长为: $25 \\div (3 + 2) \\times 3\\\\= 2 5 \\div 5 \\times 3\\\\= 5 \\times 3 \\\\=15 (厘米)$\n宽为: $25 \\div (3 + 2) \\times 2\\\\= 2 5 \\div 5 \\times 2\\\\ = 5 \\times 2 \\\\ = 1 0 (\\text {厘 米}) $\n面积为: $15 \\times 10 = 150$ (平方厘米)所以，它的面积是 150 平方厘米。故答案为: D",
+        "knowledge_points": [
+          "长方形的面积",
+          "比的应用"
+        ],
+        "source": "18-19六年级下·贵州·期末",
+        "merged_id": "unit4_322",
+        "original_id": "unit4_choice_322"
+      },
+      {
+        "id": "unit4_fill_323",
+        "unit": "第四单元：比",
+        "chapter": "比的化简",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(20-21 六年级上·江西赣州·单元测试) 一本书, 已读的页数是总页数的 $\\frac{3}{4}$ , 未读的与已读的页数比是( )。",
+        "options": [],
+        "answer": "1:3",
+        "solution": "【分析】把这本书总页数看作单位 '1', 已读的页数是总页数的 $\\frac{3}{4}$ , 未读页数占总页数的 $(1 - \\frac{3}{4})$ , 据此计算。掌握分数比的化简方法是解答题目的关键。\n【详解】$(1 - \\frac{3}{4}): \\frac{3}{4}\\\\ = \\frac {1}{4}: \\frac {3}{4} \\\\ = (\\frac {1}{4} \\times 4): (\\frac {3}{4} \\times 4) \\\\ = 1: 3 $",
+        "knowledge_points": [
+          "比的化简"
+        ],
+        "source": "20-21 六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_323",
+        "original_id": "unit4_fill_323"
+      },
+      {
+        "id": "unit4_fill_324",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题、平均数的意义及求法",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(20-21 六年级上·江西赣州·单元测试) 甲、乙、丙三个数的平均数是 60 , 甲、乙、丙三个数的比是 $2: 3: 1$ , 则甲、乙、丙三个数分别是( )、( )、( )。",
+        "options": [],
+        "answer": "60; 90; 30",
+        "solution": "【分析】根据题意, 三个数的和为 $60 \\times 3 = 180$ , 甲乙丙的比是 $2: 3: 1$ , 然后根据一个数乘分数的意义, 分别求出甲乙丙三个数。此题考查的是按比分配的方法, 解题关键是先求出三个数的和, 根据按比例分配的方法解答即可。\n【详解】$60 \\times 3 = 180, 2 + 3 + 1 = 6$ 甲: $180 \\times \\frac{2}{6} = 60$ 乙: $180 \\times \\frac{3}{6} = 90$ 丙: $180 \\times \\frac{1}{6} = 30$",
+        "knowledge_points": [
+          "按比分配问题",
+          "平均数的意义及求法"
+        ],
+        "source": "20-21 六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_324",
+        "original_id": "unit4_fill_324"
+      },
+      {
+        "id": "unit4_fill_325",
+        "unit": "第四单元：比",
+        "chapter": "比的意义、比的化简",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（20-21六年级上·江西赣州·单元测试）六（1）班有65个同学，其中女同学有30个，男同学人数与全班人数的比是( )，女同学人数与男同学人数的比是( )。",
+        "options": [],
+        "answer": "7:13 6:7",
+        "solution": "【分析】已知六（1）班有 65 个同学，其中女同学有 30 个，则男同学有 $65 - 30 = 35$ 人，则用男同学人数与全班人数比即可；然后用女同学人数与男同学人数比即可，然后用比的基本性质化简比即可解答。本题考查比的意义，明确比的基本性质是解题的关键。\n【详解】$(65-30) : 65 = 35:65 = (35 \\div 5): (65 \\div 5)=7:13$ $30: (65-30) = 30:35= (30 \\div 5): (35 \\div 5) = 6:7$",
+        "knowledge_points": [
+          "比的意义",
+          "比的化简"
+        ],
+        "source": "20-21六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_325",
+        "original_id": "unit4_fill_325"
+      },
+      {
+        "id": "unit4_fill_326",
+        "unit": "第四单元：比",
+        "chapter": "比的化简",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(19-20 六年级上·江苏·期末) 把 25 克糖溶解在 100 克水中, 糖与糖水的比是( )。",
+        "options": [],
+        "answer": "1:5",
+        "solution": "【分析】25克糖完全溶解在100克水里，糖水为（25+）100) 克, 进而根据题意, 求出糖与糖水的比, 进行选择即可。此题考查了比的意义，应明确：糖 + 水 = 糖水。\n【详解】$25: (25+100)= 2 5: 1 2 5 \\\\ = (2 5 \\div 2 5): (1 2 5 \\div 2 5) \\\\ = 1: 5 $",
+        "knowledge_points": [
+          "比的化简"
+        ],
+        "source": "19-20 六年级上·江苏·期末",
+        "merged_id": "unit4_326",
+        "original_id": "unit4_fill_326"
+      },
+      {
+        "id": "unit4_fill_327",
+        "unit": "第四单元：比",
+        "chapter": "比与分数、除法的关系",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(20-21 六年级上·江西赣州·单元测试) 甲数除以乙数的商是 4 , 则乙数与甲数的比是( )。",
+        "options": [],
+        "answer": "1:4",
+        "solution": "【分析】甲数除以乙数的商是 4 , 也就是甲数是乙数的 4 倍,把乙数看作单位 '1', 因此乙数与甲数的比是 $1: 4$ , 据此解答即可。此题主要利用比与分数、除法的关系等知识解答。\n【详解】由分析知，甲数除以乙数的商是 4 , 把乙数看作单位 '1', 因此乙数与甲数的比是 $1: 4$ 。",
+        "knowledge_points": [
+          "比与分数、除法的关系"
+        ],
+        "source": "20-21 六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_327",
+        "original_id": "unit4_fill_327"
+      },
+      {
+        "id": "unit4_fill_328",
+        "unit": "第四单元：比",
+        "chapter": "求一个数占另一个数几分之几、比的应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(2014 六年级·全国·课后作业) 甲、乙两数的比是4:5, 甲数是乙数的( ), 乙数是甲、乙两数和的( )。",
+        "options": [],
+        "answer": "$\\frac{4}{5}$ $\\frac{5}{9}$",
+        "solution": "【分析】把乙数看作 5 份, 甲为 4 份, 根据求一个数是另一个数的几分之几用除法。此题属于求一个数是另一个数的几分之几, 掌握求一个数是另一个数的几分之几用除法是解题关键。\n【详解】把乙数看作5份，甲为4份，甲数是乙数的: $4 \\div 5 = \\frac{4}{5}$乙数是甲、乙两数和的:\n$ 5 \\div (4 + 5) \\\\ = 5 \\div 9 \\\\= \\frac {5}{9}$",
+        "knowledge_points": [
+          "求一个数占另一个数几分之几",
+          "比的应用"
+        ],
+        "source": "2014 六年级·全国·课后作业",
+        "merged_id": "unit4_328",
+        "original_id": "unit4_fill_328"
+      },
+      {
+        "id": "unit4_fill_329",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题、长方形的周长、长方形的面积",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(19-20 六年级上·江苏淮安·期中) 一个长方形的周长是 32 厘米, 长与宽的比是 $5: 3$ , 它的面积是( )平方厘米。",
+        "options": [],
+        "answer": "60",
+        "solution": "【分析】先求出一组长和宽的和，长宽和 $\\div$ 总份数，求出一份数，一份数分别乘长和宽的对应份数，求出长和宽，再根据长方形面积公式进行计算。关键是理解比的意义, 掌握长方形周长和面积公式。\n【详解】$32 \\div 2 \\div (5 + 3)\\\\ = 1 6 \\div 8 \\\\ = 2 \\text {(厘 米)} $\n$2 \\times 5 = 1 0 (\\text {厘 米})$ $2 \\times 3 = 6$ (厘米) $10 \\times 6 = 60$ （平方厘米）",
+        "knowledge_points": [
+          "按比分配问题",
+          "长方形的周长",
+          "长方形的面积"
+        ],
+        "source": "19-20 六年级上·江苏淮安·期中",
+        "merged_id": "unit4_329",
+        "original_id": "unit4_fill_329"
+      },
+      {
+        "id": "unit4_fill_330",
+        "unit": "第四单元：比",
+        "chapter": "比的意义、三角形的内角和",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(16-17 六年级上·广西玉林·期末) 在一个直角三角形中, 两个锐角的度数比是 $1: 2$ , 这两个锐角的度数分别是 $( )^{\\circ} \\text{和} ()^{\\circ}$ 。",
+        "options": [],
+        "answer": "30 60",
+        "solution": "【分析】直角三角形两锐角和是 90 度, 两个锐角共 $1 + 2$ 份,先求出一份对应的度数, 再分别乘两个锐角的份数即可。关键是知道三角形的内角和，理解比的意义。\n【详解】$90 \\div (1 + 2)= 9 0 \\div 3 = 3 0 \\text {(度)} $ $30 \\times 2 = 60$ (度)",
+        "knowledge_points": [
+          "比的意义",
+          "三角形的内角和"
+        ],
+        "source": "16-17 六年级上·广西玉林·期末",
+        "merged_id": "unit4_330",
+        "original_id": "unit4_fill_330"
+      },
+      {
+        "id": "unit4_fill_331",
+        "unit": "第四单元：比",
+        "chapter": "分母在 10 以内的同分母分数加、减法、分数与整数的除法",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "(21-22 六年级上·江西赣州·单元测试) ( ) $\\times \\frac{1}{3} = \\frac{7}{5} -$ ( ) = $\\frac {5}{6} \\div (\\quad) = \\frac {3}{4} + (\\quad)$ 。",
+        "options": [],
+        "answer": "3 $\\frac{2}{5}$ $\\frac{5}{6}$ $\\frac{1}{4}$",
+        "solution": "【分析】这是一道开放题，且答案不唯一。从最简便的角度去思考，可假设四个算式的结果均等于 1，再根据四则运算的性质依次求出每个算式的空缺部分。本题通过运用分数的加减乘除运算，依次求出每个式子的未知部分；前提是能够适当的定下几个算式的统一结果。\n【详解】$1 \\div \\frac{1}{3} = 3=\\frac {7}{5} - 1 = \\frac {2}{5}$ $\\frac {5}{6} \\div 1 = \\frac {5}{6}$ $1 - \\frac {3}{4} = \\frac {1}{4}$(答案不唯一)",
+        "knowledge_points": [
+          "分母在 10 以内的同分母分数加、减法",
+          "分数与整数的除法"
+        ],
+        "source": "21-22 六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_331",
+        "original_id": "unit4_fill_331"
+      },
+      {
+        "id": "unit4_fill_332",
+        "unit": "第四单元：比",
+        "chapter": "分数与整数的除法",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（21-22六年级上·江西赣州·单元测试）把6米长的钢筋截成 $\\frac{3}{2}$ 米长的小段，（）次截完，截的每段占全长的 $\\frac{()}{( )}$ 。",
+        "options": [],
+        "answer": "$3; \\frac{1}{4}$",
+        "solution": "【分析】已知要把 6 米长的钢筋截成 $\\frac{3}{2}$ 米长的小段, 求几次截完, 则用 $6 \\div \\frac{3}{2}$ 求出能截成几段, 用段数减 1 , 就是截的次数; 根据分数的意义可知, 每段占全长的几分之几, 一共有几段就是几分之一。据此解答即可。本题考查分数除法，完成本题要注意截得的段数 $=$ 截的次数 +1。\n【详解】$6 \\div \\frac{3}{2} - 1= 4 - 1= 3 (次)$$1 \\div 4 = \\frac {1}{4}$则 3 次截完, 截的每段占全长的 $\\frac{1}{4}$ 。",
+        "knowledge_points": [
+          "分数与整数的除法"
+        ],
+        "source": "21-22六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_332",
+        "original_id": "unit4_fill_332"
+      },
+      {
+        "id": "unit4_fill_333",
+        "unit": "第四单元：比",
+        "chapter": "分数与分数的除法、用字母表示数",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(21-22 六年级上·江西赣州·单元测试) $\\frac{2}{9} \\times 15 = \\frac{10}{3}$ , 则 $\\frac{10}{3} \\div (\\quad) = 15$ ; 若 $S = a b h$ , 则 $s \\div a \\div (\\quad) = (\\quad)$ 。",
+        "options": [],
+        "answer": "$\\frac{2}{9}$ $\\mathrm{b}$ $\\mathrm{h}$",
+        "solution": "【分析】根据乘法各部分之间的关系：一个因数 $=$ 积 $\\div$ 另一个因数，据此解答。此题考查的是乘法各部分之间的关系的应用，掌握乘法各部分之间的关系是关键。\n【详解】由分析可得,$\\frac {2}{9} \\times 1 5 = \\frac {1 0}{3}, \\text {则} \\frac {1 0}{3} \\div \\frac {2}{9} = 1 5;$若 $S = a b h$ , 则 $s \\div a \\div b = h$",
+        "knowledge_points": [
+          "分数与分数的除法",
+          "用字母表示数"
+        ],
+        "source": "21-22 六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_333",
+        "original_id": "unit4_fill_333"
+      },{
+        "id": "unit4_fill_334",
+        "unit": "第四单元：比",
+        "chapter": "比的化简、比的应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(21-22 六年级上·江西赣州·单元测试) 打一份稿件,甲、乙工效比是 $3: 2$ , 若甲 6 分钟完成, 则乙( )分钟完成。",
+        "options": [],
+        "answer": "9",
+        "solution": "【分析】已知打这份稿件时甲乙的功效之比, 可把打这份稿件的任务量看作单位 '1', 求出甲乙打这份稿件的时间之比; 再结合题意: 甲在 6 分钟完成, 可通过计算求得乙完成的时间。本题运用了比的性质求得乙的完成时间, 其中还包含了比的化简, 此外还需要我们熟悉数量关系式: 工作时间 = 工作总量 ÷ 功效。\n【详解】甲乙完成时间之比如下表:\n$\\frac {1}{3}: \\frac {1}{2} = 2: 3 \\\\ 2: 3 = (2 \\times 3): (3 \\times 3) = 6: 9 $",
+        "knowledge_points": [
+          "比的化简",
+          "比的应用"
+        ],
+        "source": "21-22 六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_334",
+        "original_id": "unit4_fill_334"
+      },
+      {
+        "id": "unit4_fill_335",
+        "unit": "第四单元：比",
+        "chapter": "分数与分数的除法、分数与整数的除法、被除数和商的大小关系（小数除法）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(21-22 六年级上·江西赣州·单元测试) 在括号里填上 '>' '<' 或 '='。$\\frac {9}{1 0} \\div \\frac {1}{6} (\\quad) \\frac {9}{1 0} \\quad \\frac {4}{5} \\div 5 (\\quad) \\frac {4}{5} \\times 1 \\quad 1 \\div \\frac {1 2}{1 1} (\\quad) 1 \\div \\frac {1 3}{1 2}$",
+        "options": [],
+        "answer": "> < <",
+        "solution": "【分析】（1）当被除数大于 0 时，被除数除以小于 1 的数，所得结果一定大于原来这个数， $\\frac{1}{6}<1, \\frac{9}{10} \\div \\frac{1}{6}>\\frac{9}{10}$ ;(2) 被除数大于 0 时, 被除数除以大于 1 的数, 所得结果一定小于原来这个数 $\\frac{4}{5} \\div 5 < \\frac{4}{5}, \\frac{4}{5} \\times 1 = \\frac{4}{5}$ , 则 $\\frac{4}{5} \\div 5 < \\frac{4}{5} \\times 1$ ;(3) $1 \\div \\frac{12}{11} = 1 \\div 1 \\frac{1}{11}, 1 \\div \\frac{13}{12} = 1 \\div 1 \\frac{1}{12}, 1 \\frac{1}{11} > 1 \\frac{1}{12}$ , 被除数相同时,除数越小商越大, 除数越大商越小, $1 \\div 1 \\frac{1}{11} < 1 \\div 1 \\frac{1}{12}$ , 则 $1 \\div \\frac{12}{11} < 1 \\div \\frac{13}{12}$ 。本题也可根据分数乘除法的计算方法直接计算出式子结果比较大小。\n【详解】$\\frac{9}{10} \\div \\frac{1}{6}$ （ $>$ ） $\\frac{9}{10} \\quad \\frac{4}{5} \\div 5$ （ $<$ ）$\\frac{4}{5} \\times 1 $ $ 1 \\div \\frac{12}{11}(< ) 1 \\div \\frac {1 3}{1 2}$",
+        "knowledge_points": [
+          "分数与分数的除法",
+          "分数与整数的除法",
+          "被除数和商的大小关系（小数除法）"
+        ],
+        "source": "21-22 六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_335",
+        "original_id": "unit4_fill_335"
+      },
+      {
+        "id": "unit4_fill_336",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题、分数除法的应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（19-20六年级上·辽宁大连·期末）要配制一种盐与盐水的比是 $1: 100$ 的盐水溶液，现有5克盐，应加水( )克。",
+        "options": [],
+        "answer": "495",
+        "solution": "【分析】根据题意可知, 盐占盐水的 $\\frac{1}{100}$ , 则 '盐水的质量 $\\times$ $\\frac{1}{100} =$ 盐的质量', 据此列除法算式求出盐水的质量, 进而求出水的质量即可。本题考查了按比例分配的知识点, 先明确盐占盐水的几分之几是解答本题的关键, 也可以先用 $5 \\div 1$ 求出每份是多少克, 再乘水对应的份数。\n【详解】盐占盐水的 $\\frac{1}{100}$ ;$5 \\div \\frac {1}{1 0 0} - 5 \\\\ = 5 0 0 - 5 \\\\ = 4 9 5 \\text {(克)} $",
+        "knowledge_points": [
+          "按比分配问题",
+          "分数除法的应用"
+        ],
+        "source": "19-20六年级上·辽宁大连·期末",
+        "merged_id": "unit4_336",
+        "original_id": "unit4_fill_336"
+      },
+      {
+        "id": "unit4_fill_337",
+        "unit": "第四单元：比",
+        "chapter": "比的意义",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(19-20 六年级上·河南安阳·期中) 走一段路, 甲用了 3 小时, 乙用了 2 小时, 甲与乙行走的速度比是( )。",
+        "options": [],
+        "answer": "2:3",
+        "solution": "【分析】根据比的意义，两数相除又叫两个数的比，将时间比反过来就是速度比，据此分析。同样的路程，时间越少，速度越快。\n【详解】走一段路, 甲用了 3 小时, 乙用了 2 小时, 甲与乙行走的速度比是 $2: 3$ 。",
+        "knowledge_points": [
+          "比的意义"
+        ],
+        "source": "19-20 六年级上·河南安阳·期中",
+        "merged_id": "unit4_337",
+        "original_id": "unit4_fill_337"
+      },
+      {
+        "id": "unit4_fill_338",
+        "unit": "第四单元：比",
+        "chapter": "三角形的内角和、比的应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（20-21 六年级上·山东临沂·期末）在一个直角三角形中，最大角与最小角的度数的比是 $5: 2$ ，那么这个直角三角形中的两个锐角度数分别为( )和( )。",
+        "options": [],
+        "answer": "$36^{\\circ} \\quad 54^{\\circ}$",
+        "solution": "【分析】在直角三角形中最大角为直角 $90^{\\circ}$ , 根据最大角与最小角的度数的比是 $5: 2$ , 计算出最小角, 另一个锐角 $= 90^{\\circ} -$ 最小角。根据比的应用由最大角计算出最小角是解答本题的关键。\n【详解】最小角: $90^{\\circ} \\div 5 \\times 2 = 36^{\\circ}$ 另一个锐角: $90^{\\circ} - 36^{\\circ} = 54^{\\circ}$",
+        "knowledge_points": [
+          "三角形的内角和",
+          "比的应用"
+        ],
+        "source": "20-21 六年级上·山东临沂·期末",
+        "merged_id": "unit4_338",
+        "original_id": "unit4_fill_338"
+      },
+      {
+        "id": "unit4_fill_339",
+        "unit": "第四单元：比",
+        "chapter": "求比值、比的意义",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "(21-22 六年级上·全国·单元测试) A 是 7, B 是 4, A:B=( ), 比值是( )。",
+        "options": [],
+        "answer": "$7 4 \\frac{7}{4}$",
+        "solution": "【分析】A:B=7:4，用比的前项除以比的后项求比值即可。熟练掌握比的意义以及求比值的方法是解答本题的关键。\n【详解】A:B=7:4; $7: 4 = \\frac {7}{4}$",
+        "knowledge_points": [
+          "求比值",
+          "比的意义"
+        ],
+        "source": "21-22 六年级上·全国·单元测试",
+        "merged_id": "unit4_339",
+        "original_id": "unit4_fill_339"
+      },
+      {
+        "id": "unit4_fill_340",
+        "unit": "第四单元：比",
+        "chapter": "比的意义、比的化简",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "(21-22 六年级上·全国·单元测试) 甲数除以乙数的商是 2 , 那么甲数与乙数的最简整数比是( ) : ( )。",
+        "options": [],
+        "answer": "2 1",
+        "solution": "【分析】甲数÷乙数 = 2，假设乙数是 1，则甲数是 2，据此写出甲数与乙数的最简整数比即可。本题采用了假设法，写出具体的甲数和乙数的值，再进一步解答即可。\n【详解】甲数除以乙数的商是 2 , 假设乙数是 1 , 则甲数是 2 ;甲数与乙数的最简整数比是 $2: 1$ 。",
+        "knowledge_points": [
+          "比的意义",
+          "比的化简"
+        ],
+        "source": "21-22 六年级上·全国·单元测试",
+        "merged_id": "unit4_340",
+        "original_id": "unit4_fill_340"
+      },
+      {
+        "id": "unit4_fill_341",
+        "unit": "第四单元：比",
+        "chapter": "平角、周角的认识及特征、比的意义、比的化简、一般时间与钟面指针的位置",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(21-22 六年级上·全国·单元测试) 钟面上, 6 时整, 时针和分针成( )角, 钟表上的分针与时针速度的比是( )。",
+        "options": [],
+        "answer": "平 12:1",
+        "solution": "【分析】（1）在钟面上，一共有 12 个大空格，时针与分针所夹的每一个空格是 $30^{\\circ}$ ，6 时整，时针指向 6，分针指向12, 时针分针相差 6 个大格, 相差 $30^{\\circ} \\times 6 = 180^{\\circ}$ , 为平角; (2) 表上分针每转动一周，时针转动 1 小时，可知分针转动 360 度，时针转动 30 度；据此解答。本题依据角的定义进行解答, 应明确: 钟面上,一共有 12 个大空格, 时针与分针所夹的每一个空格是 $30^{\\circ}$ ; 表上分针每转动一周, 时针转动 1 小时。\n【详解】根据分析知, 当钟面上 6 时整, 时针与分针之间有 6 个大格是 $180^{\\circ}$ , 是平角;钟表上的分针与时针速度的比是 $360: 30 = 12: 1$ 。",
+        "knowledge_points": [
+          "平角、周角的认识及特征",
+          "比的意义",
+          "比的化简",
+          "一般时间与钟面指针的位置"
+        ],
+        "source": "21-22 六年级上·全国·单元测试",
+        "merged_id": "unit4_341",
+        "original_id": "unit4_fill_341"
+      },
+      {
+        "id": "unit4_fill_342",
+        "unit": "第四单元：比",
+        "chapter": "求比值、比的意义、比的化简",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（21-22六年级上·全国·期中）从甲地到乙地，一辆客车2小时行驶了120千米。这辆客车所行驶的路程和时间的比是( )，比值是( )。",
+        "options": [],
+        "answer": "60:1 60",
+        "solution": "【分析】根据比的意义写出路程和时间的比，化简即可，求比值直接用前项÷后项。两数相除又叫两个数的比。\n【详解】$120: 2 = 60: 1 = 60$",
+        "knowledge_points": [
+          "求比值",
+          "比的意义",
+          "比的化简"
+        ],
+        "source": "21-22六年级上·全国·期中",
+        "merged_id": "unit4_342",
+        "original_id": "unit4_fill_342"
+      },
+      {
+        "id": "unit4_fill_343",
+        "unit": "第四单元：比",
+        "chapter": "求一个数占另一个数几分之几、比的应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（19-20六年级下·山东济南·期末）五（1）班女生人数与男生人数的比是 $10:9$ ，那么男生人数比女生人数少( )，男生人数占全班人数的( )。",
+        "options": [],
+        "answer": "$\\frac{1}{10} ;\\frac{9}{19}$",
+        "solution": "【分析】根据女生和男生人数比，将女生人数看作10，男 生人数看作9，男女生人数差÷女生人数=男生比女生少几分之几；男生人数+女生人数=全班人数，男生人数÷全班人数=男生人数占全班人数的几分之几。关键是理解比的意义，求一个数占另一个数的几分之几用除法，此类问题一般用表示单位'1'的量作除数。\n【详解】$(10 - 9) \\div 10\\\\= 1 \\div 1 0 \\\\ = \\frac {1}{1 0} $\n$ 9 \\div (1 0 + 9) \\\\ = 9 \\div 1 9 \\\\ = \\frac {9}{1 9} $",
+        "knowledge_points": [
+          "求一个数占另一个数几分之几",
+          "比的应用"
+        ],
+        "source": "19-20六年级下·山东济南·期末",
+        "merged_id": "unit4_343",
+        "original_id": "unit4_fill_343"
+      },
+      {
+        "id": "unit4_fill_344",
+        "unit": "第四单元：比",
+        "chapter": "求比值、比的化简",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(19-20 六年级下·河南·期末) $\\frac{3}{8}: \\frac{1}{4}$ 的比值是( )，把 $4: 0.8$ 化成最简单的整数比是( )。",
+        "options": [],
+        "answer": "$\\frac{3}{2} \\quad5:1$",
+        "solution": "【分析】求比值直接用前项÷后项, 化简比根据比的基本性质。求比值的结果是一个数, 化简比的结果还是一个比。\n【详解】$\\frac{3}{8}: \\frac{1}{4} = \\frac{3}{8} \\times 4 = \\frac{3}{2}$ $4: 0. 8 = 4 0: 8 = 5: 1$",
+        "knowledge_points": [
+          "求比值",
+          "比的化简"
+        ],
+        "source": "19-20 六年级下·河南·期末",
+        "merged_id": "unit4_344",
+        "original_id": "unit4_fill_344"
+      },
+      {
+        "id": "unit4_fill_345",
+        "unit": "第四单元：比",
+        "chapter": "比的化简、比的基本性质",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(21-22 六年级上·全国·单元测试) 甲: 乙 $= 4: 5$ , 乙:丙 $= 3: 7$ , 那么甲: 乙: 丙 $=$ ( )。",
+        "options": [],
+        "answer": "12:15:35",
+        "solution": "【分析】甲：乙 $= 4:5 = \\frac{4}{5}$ ，则甲 $= 2 \\times \\frac{4}{5}$ 乙：丙 $= 3: 7 = \\frac{3}{7}$ ，则丙 $= 乙 \\div \\frac{3}{7} = 乙 \\times \\frac{7}{3}$ 求甲：乙：丙是多少，就把甲换成乙 $\\times \\frac{4}{5}$ , 丙换成乙 $\\times \\frac{7}{3}$ , 所以甲: 乙: 丙 $= (\\text {乙} \\times \\frac{4}{5})$ : 乙: (乙 $\\times \\frac{7}{3}$ ), 然后化简即可解答。比的前项和后项, 同时乘或除以相同的数 (0 除外), 比值不变。\n【详解】甲：乙 $= 4:5 = \\frac{4}{5}$ ，则甲 $= 2 \\times \\frac{4}{5}$ 乙：丙 $= 3: 7 = \\frac{3}{7}$ ，则丙 $= \\text{乙} \\div \\frac{3}{7} = \\text{乙} \\times \\frac{7}{3}$ , 甲：乙：丙= $(\\text{乙} \\times \\frac{4}{5}) ：乙： (\\text{乙} \\times \\frac{7}{3})\\\\= \\frac {4}{5}: 1: \\frac {7}{3} \\\\ = (1 5 \\times \\frac {4}{5}): (1 5 \\times 1): (1 5 \\times \\frac {7}{3}) \\\\ = 1 2: 1 5: 3 5 $",
+        "knowledge_points": [
+          "比的化简",
+          "比的基本性质"
+        ],
+        "source": "21-22 六年级上·全国·单元测试",
+        "merged_id": "unit4_345",
+        "original_id": "unit4_fill_345"
+      },
+      {
+        "id": "unit4_fill_346",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题、三角形的周长、三角形的面积",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（19-20六年级上·辽宁·单元测试）一个直角三角形的周长是24厘米，三条边的比是 $3:4:5$ ，这个三角形的面积是( )平方厘米。",
+        "options": [],
+        "answer": "24",
+        "solution": "【分析】把三角形周长按照 $3:4:5$ 的比分配, 分别求出三角形三条边的长度。直角三角形中斜边是最长的, 所以三角形两条直角边就是两条较短的边。根据三角形面积 $=$ 底 $\\times$ 高 $\\div 2$ 计算面积即可。本题考查比的应用、三角形的周长和面积等知识的综合应用。用按比例分配的方法求出三角形的底和高是解题的关键。\n【详解】$3 + 4 + 5 = 12$ $24 \\div 12 = 2$ （厘米） $3 \\times 2 = 6$ (厘米) $4 \\times 2 = 8$ (厘米) $5 \\times 2 = 10$ (厘米) 三角形面积: $6 \\times 8 \\div 2 = 24$ (平方厘米)",
+        "knowledge_points": [
+          "按比分配问题",
+          "三角形的周长",
+          "三角形的面积"
+        ],
+        "source": "19-20六年级上·辽宁·单元测试",
+        "merged_id": "unit4_346",
+        "original_id": "unit4_fill_346"
+      },
+      {
+        "id": "unit4_fill_347",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题、比的意义、比的化简",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "(19-20 六年级下·四川南充·期末) 甲数比乙数少 $\\frac{1}{4}$ , 乙数与甲数的比是( )；如果乙数为 20 , 则甲数是( )。",
+        "options": [],
+        "answer": "4:3 15",
+        "solution": "【分析】根据题意可知, 乙数为 1 , 则甲数为 $1 - \\frac{1}{4} = \\frac{3}{4}$ , 进而写出乙数和甲数的比即可；用 20 除以乙数所占的份数求出每份是多少，再乘甲数占的份数即可。\n【详解】1: $(1 - \\frac{1}{4}) = 1:\\frac{3}{4} = 4:3;$根据题意, 明确乙数为 1 , 甲数为 $\\frac{3}{4}$ 是解答本题的关键, 进而写出它们之间的比, 再根据按比例分配的知识点求出甲数。则乙数与甲数的比为 $4: 3$ ;$2 0 \\div 4 \\times 3 = 1 5$",
+        "knowledge_points": [
+          "按比分配问题",
+          "比的意义",
+          "比的化简"
+        ],
+        "source": "19-20 六年级下·四川南充·期末",
+        "merged_id": "unit4_347",
+        "original_id": "unit4_fill_347"
+      },
+      {
+        "id": "unit4_fill_348",
+        "unit": "第四单元：比",
+        "chapter": "比的应用、分数与整数的乘法",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(20-21 六年级下·江西吉安·期末) 甲乙两书架原有图书共 240 本, 其中乙书架占 $\\frac{4}{5}$ 。现在又买一些书都放到了甲书架, 这时甲与乙书架的本数比是 $2:3$ 。现在甲乙两书架共有( )本书。",
+        "options": [],
+        "answer": "320",
+        "solution": "【分析】先利用乘法求出乙书架的数量，再根据增加书后的比求出增加后的甲书架的书量，最后利用加法求出现在甲乙两个书架的总数量即可。本题考查了比的应用, 能根据比求出甲书架现在的数量是解题的关键。\n【详解】$240 \\times \\frac{4}{5} = 192$ （本）， $192 \\times \\frac{2}{3} = 128$ （本）， $128 + 192 = 320$ （本），所以现在甲乙两书架共有 320 本。",
+        "knowledge_points": [
+          "比的应用",
+          "分数与整数的乘法"
+        ],
+        "source": "20-21 六年级下·江西吉安·期末",
+        "merged_id": "unit4_348",
+        "original_id": "unit4_fill_348"
+      },
+      {
+        "id": "unit4_fill_349",
+        "unit": "第四单元：比",
+        "chapter": "比的化简、比的基本性质",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "(20-21 六年级下·江西吉安·期末) 把 $50 \\mathrm{~kg}: 150 \\mathrm{~kg}$ 化成最简整数比是( )。",
+        "options": [],
+        "answer": "1:3",
+        "solution": "【分析】根据比的基本性质：比的前项和后项同时乘或除以一个数（0 除外），比的大小不变， $50\\mathrm{kg}:150\\mathrm{kg}$ 的前项和后项同时除以 50，即可得出答案。本题主要考查的是比的基本性质, 解题的关键是熟记比的基本性质并加以运用。\n【详解】最简整数比为:$5 0 \\mathrm {k g}: 1 5 0 \\mathrm {k g} = (5 0 \\div 5 0) \\mathrm {k g}: (1 5 0 \\div 5 0) \\mathrm {k g} = 1: 3 。$",
+        "knowledge_points": [
+          "比的化简",
+          "比的基本性质"
+        ],
+        "source": "20-21 六年级下·江西吉安·期末",
+        "merged_id": "unit4_349",
+        "original_id": "unit4_fill_349"
+      },
+      {
+        "id": "unit4_fill_350",
+        "unit": "第四单元：比",
+        "chapter": "比的意义、因数和倍数的求法",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（2021·辽宁沈阳·小升初真题）围棋组人数在 $30 \\sim 40$ 之间，男生与女生的人数比是 $5: 7$ ，围棋组有( )人。",
+        "options": [],
+        "answer": "36",
+        "solution": "【分析】由于男生与女生人数的比是 $5: 7$ , 则男生相当于 5 份, 女生是 7 份, 总人数: $5 + 7 = 12$ 份, 由此即可知道围棋组人数是 12 的倍数, 由此即可找出 12 的倍数, 并且在 30 到 40 之间即可。本题主要考查比的意义以及找公倍数的方法, 要注意总人数的范围。\n【详解】由分析可知, 围棋组人数是 12 的倍数, 12 的倍数有: 12、24、36……$30 < 36 < 40$由此即可知道围棋组有 36 人。",
+        "knowledge_points": [
+          "比的意义",
+          "因数和倍数的求法"
+        ],
+        "source": "2021·辽宁沈阳·小升初真题",
+        "merged_id": "unit4_350",
+        "original_id": "unit4_fill_350"
+      },
+      {
+        "id": "unit4_fill_351",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(20-21 六年级下·吉林通化·期末) 有两杯盐水, 甲杯中盐与水的比是 $1:7$ , 乙杯中盐与水的比是 $2:8$ , ( )杯中的盐水咸。",
+        "options": [],
+        "answer": "乙",
+        "solution": "【分析】已知甲杯中盐与水的比是 $1: 7$ , 乙杯中盐与水的比是 $2: 8$ , 则可依据公式: $\\frac{\\text { 盐 }}{\\text { 盐 } + \\text { 水 }}$ 来求得每杯盐水的浓度, 经过比较后做判断。通过比与分数的联系, 将盐与水的比转化为盐水的浓度, 这样比较更直接; 应用公式时, 注意盐、水及盐水几个量的区别。\n【详解】甲杯盐水的浓度: $\\frac{1}{1 + 7} = \\frac{1}{8}$ ;乙杯盐水的浓度: $\\frac{2}{2 + 8} = \\frac{1}{5}$ ;因为 $\\frac{1}{8}<\\frac{1}{5}$ , 所以乙杯盐水的浓度更大, 即乙杯中的盐水更咸。",
+        "knowledge_points": [
+          "比的应用"
+        ],
+        "source": "20-21 六年级下·吉林通化·期末",
+        "merged_id": "unit4_351",
+        "original_id": "unit4_fill_351"
+      },
+      {
+        "id": "unit4_fill_352",
+        "unit": "第四单元：比",
+        "chapter": "求比值、倒数的认识及应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(2021·贵州遵义·小升初模拟) 0.25:1.5 的比值是( ),和( )互为倒数。",
+        "options": [],
+        "answer": "$\\frac{1}{6}; 6$",
+        "solution": "【分析】$0.25: 1.5$ 的比可化为除法 $0.25 \\div 1.5 = \\frac{1}{6}$ , 即比值为 $\\frac{1}{6}$ ;两个数的乘积为 1 , 则这两个数互为倒数, 据此可得出答案。半屉主要考查的是比值和倒数的求法，解题的关键是将比化为除法求出比值，进而解出答案。\n【详解】0.25:1.5的比可化为除法 $0.25 \\div 1.5 = \\frac{1}{6}$ , 即比值为 $\\frac{1}{6}$ ; $\\frac{1}{6}$ 与6相乘得到1, 即 $1 \\div \\frac{1}{6} = 6$ , 即它和6互为倒数。",
+        "knowledge_points": [
+          "求比值",
+          "倒数的认识及应用"
+        ],
+        "source": "2021·贵州遵义·小升初模拟",
+        "merged_id": "unit4_352",
+        "original_id": "unit4_fill_352"
+      },
+      {
+        "id": "unit4_fill_353",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题、等腰三角形和等边三角形的认识及特征、三角形的内角和、比的应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(20-21 六年级下·广西南宁·期末) 一个三角形内角的度数比是 $1:4:1$ , 它是( )三角形, 它的最大内角的度数是()。",
+        "options": [],
+        "answer": "等腰 $120^{\\circ}$",
+        "solution": "【分析】根据内角度数比，结合三角形的分类，填出第一空；根据三角形的内角度数比，将这个三角形最大的内角占内角和 $180^{\\circ}$ 的几分之几表示出来，再利用乘法求出它的度数即可。本题考查了比的应用, 解题关键在于, 根据内角的度数比将最大内角占内角和的几分之几表示出来。\n【详解】由于这个三角形内角的度数比是 $1: 4: 1$ , 所以这个三角形有两个内角是相等的, 所以它是等腰三角形;\n180°× $\\frac{4}{1+4+1}=120^{\\circ}$ , 所以, 它的最大内角度数是 $120^{\\circ}$ 。",
+        "knowledge_points": [
+          "按比分配问题",
+          "等腰三角形和等边三角形的认识及特征",
+          "三角形的内角和",
+          "比的应用"
+        ],
+        "source": "20-21 六年级下·广西南宁·期末",
+        "merged_id": "unit4_353",
+        "original_id": "unit4_fill_353"
+      },
+      {
+        "id": "unit4_fill_354",
+        "unit": "第四单元：比",
+        "chapter": "求比值、时、分的认识及换算、比的化简",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(19-20 六年级下·湖南郴州·期末) $\\frac{5}{8}: \\frac{1}{4}$ 的比值是( )；把 $\\frac{2}{5}$ 小时：30 分钟化成最简单的整数比是( )。",
+        "options": [],
+        "answer": "$\\frac{5}{2}$ 4:5",
+        "solution": "【分析】运用比与除法的关系，可得出比值；时间单位换算中，1小时=60分钟，运用比的基本性质可得出答案。本题主要考查的是比的基本性质, 解题的关键是运用比的基本性质化简为整数比。\n【详解】$\\frac{5}{8}: \\frac{1}{4} = \\frac{5}{8} \\div \\frac{1}{4} = \\frac{5}{2}$ , 它的比值就为 $\\frac{5}{2}$ ; 30 分钟 $= \\frac{1}{2}$ 小时,即 $\\frac{2}{5}: \\frac{1}{2} = (\\frac{2}{5} \\times 10): (\\frac{1}{2} \\times 10) = 4: 5$ 。",
+        "knowledge_points": [
+          "求比值",
+          "时、分的认识及换算",
+          "比的化简"
+        ],
+        "source": "19-20 六年级下·湖南郴州·期末",
+        "merged_id": "unit4_354",
+        "original_id": "unit4_fill_354"
+      },
+      {
+        "id": "unit4_fill_355",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(20-21 六年级下·河南平顶山·期末) 我国四大发明之一黑火药是用木炭、硫磺、火硝按 $3: 2: 1$ 配制而成。其中木炭占黑火药的____；如果有 12 千克的木炭，则需要____千克的硫磺来配制黑火药。",
+        "options": [],
+        "answer": "$\\frac{1}{2}$ 8",
+        "solution": "【分析】首先算出黑火药的总份数，然后找到木炭的份数即可求出木炭占黑火药的比重；按比例分配问题常用解题方法，已知部分量，可知转化为已知一个数的几分之几是多少，求这个数用除法，求一个数的几分之几是多少用乘法即可。本题考查按比例分配解决问题, 注意已知几个量的比和几个量的和, 求部分量, 可以转化成求单位 '1'的几分之几是多少的问题。\n【详解】$3 \\div (3 + 2 + 1)\\\\= 3 \\div 6 \\\\ = \\frac {1}{2}$\n$12 \\div \\frac{1}{2} = 24$ (千克)\n$2 4 \\times \\frac {2}{3 + 2 + 1} \\\\ = 2 4 \\times \\frac {2}{6} \\\\ = 8 \\text {(千 克)} $",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "20-21 六年级下·河南平顶山·期末",
+        "merged_id": "unit4_355",
+        "original_id": "unit4_fill_355"
+      },
+      {
+        "id": "unit4_judge_356",
+        "unit": "第四单元：比",
+        "chapter": "比的意义、比的基本性质",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "(20-21 六年级上·江西赣州·单元测试) 若 $A: B = 5: 3$ , 则 $A = 5; B = 3$ 。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】A:B=5:3, 此关系式只是确定两数的份数比, 而不是具体的数量比, 所以此说法错误。此题考查的是区别数量和份数，明确已知的是份数比是解题关键。\n【详解】由分析得,A:B=5:3, 是两数的份数关系, 而不表示具体的数量,故答案为: ×",
+        "knowledge_points": [
+          "比的意义",
+          "比的基本性质"
+        ],
+        "source": "20-21 六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_356",
+        "original_id": "unit4_judge_356"
+      },
+      {
+        "id": "unit4_judge_357",
+        "unit": "第四单元：比",
+        "chapter": "比的基本性质",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "(19-20 六年级上·山东·单元测试) 若 $a:b = 3:4$ 。则 $a = 3$ , $b = 4$ 。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】根据比的意义和比的基本性质进行分析。比的前项和后项, 同时乘或除以相同的数 (0 除外), 比值不变。\n【详解】$a:b = 3:4$ , a 也可能是 6, b 也可能是 8, 所以原题说法错误。",
+        "knowledge_points": [
+          "比的基本性质"
+        ],
+        "source": "19-20 六年级上·山东·单元测试",
+        "merged_id": "unit4_357",
+        "original_id": "unit4_judge_357"
+      },
+      {
+        "id": "unit4_judge_358",
+        "unit": "第四单元：比",
+        "chapter": "比的化简",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "(21-22 六年级上·江西赣州·单元测试) 身高 152 厘米的我站在9米高的树旁，我与树高的比是 $152: 9$ 。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】先转化单位，再化简 152 厘米：9 米即可。比的前项和后项单位不相同时应先转化为相同单位, 再化简比计算。\n【详解】152 厘米：9 米\n$ = 1 5 2 \\text {厘 米}: (9 \\times 1 0 0) \\text {厘 米} \\\\ = 1 5 2: 9 0 0 \\\\ = (1 5 2 \\div 4): (9 0 0 \\div 4) \\\\ = 3 8: 2 2 5 $\n故答案为: ×",
+        "knowledge_points": [
+          "比的化简"
+        ],
+        "source": "21-22 六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_358",
+        "original_id": "unit4_judge_358"
+      },
+      {
+        "id": "unit4_judge_359",
+        "unit": "第四单元：比",
+        "chapter": "比的意义",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "(21-22 六年级上·全国·期中) 如果 $a:b = 8:5$ , 那么 $a = 8, b = 5$ 。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】根据比的意义，结合题意，直接判断即可。本题考查了比，明确比的意义是解题的关键。\n【详解】如果 $a:b = 8:5$ , 那么 $a$ 和 $b$ 的最简整数比是 $8:5$ , 并不代表 $a = 8, b = 5$ 。所以判断错误。",
+        "knowledge_points": [
+          "比的意义"
+        ],
+        "source": "21-22 六年级上·全国·期中",
+        "merged_id": "unit4_359",
+        "original_id": "unit4_judge_359"
+      },
+      {
+        "id": "unit4_judge_360",
+        "unit": "第四单元：比",
+        "chapter": "互质数的认识、比的化简",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "（2018 六年级上·全国·专题练习）最简整数比的前项和后项一定互质。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】比的前项和后项是互质数的比叫做最简单的整数比, 据此解答。掌握最简整数比的意义是解答本题的关键。\n【详解】最简整数比的前项和后项只有公因数 1, 是互质数,如 5:3故答案为: √",
+        "knowledge_points": [
+          "互质数的认识",
+          "比的化简"
+        ],
+        "source": "2018 六年级上·全国·专题练习",
+        "merged_id": "unit4_360",
+        "original_id": "unit4_judge_360"
+      },
+      {
+        "id": "unit4_judge_361",
+        "unit": "第四单元：比",
+        "chapter": "比的化简",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "(21-22 六年级上·全国·单元测试) 一个比的比值是 3.2，这个比化成最简比是 $16:5$ 。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】一个比的比值是 3.2，假设这个比的后项为 1，则前项为 3.2，即 $3.2: 1$ ，再将其化成最简比即可。本题采用了假设法，写出具体的前项和后项，再化成最简比即可。\n【详解】假设这个比的后项为1，则前项为3.2；3.2:1 = (3.2×10) : (1×10) =32:10=16:5，原题正确;故答案为：√。",
+        "knowledge_points": [
+          "比的化简"
+        ],
+        "source": "21-22 六年级上·全国·单元测试",
+        "merged_id": "unit4_361",
+        "original_id": "unit4_judge_361"
+      },
+      {
+        "id": "unit4_judge_362",
+        "unit": "第四单元：比",
+        "chapter": "比的意义",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "(2014 六年级·全国·课后作业) 一场球赛的比分是4:0, 因此比的后项可以是 0 。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】数学中的比表示两个数相除，是两个数之间的关系；本题中 4:0 表示两个队的得分情况，没有进球就是 0 分，两者意义不同，据此解答即可。理解熟记比的意义是解答本题的关键。\n【详解】一场球赛的比分是 $4: 0$ , 与数学中的比意义不同,比的后项不可以是 0 , 原题说法错误;故答案为: $\\times$ 。",
+        "knowledge_points": [
+          "比的意义"
+        ],
+        "source": "2014 六年级·全国·课后作业",
+        "merged_id": "unit4_362",
+        "original_id": "unit4_judge_362"
+      },
+      {
+        "id": "unit4_judge_363",
+        "unit": "第四单元：比",
+        "chapter": "比的意义、比的化简",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "(21-22 六年级上·全国·单元测试) 小明身高是125cm，爸爸身高是 $1.8 \\mathrm{~m}$ ，小明和爸爸身高比是 125: 1.8（）。",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】小明和爸爸身高比是 125 厘米：1.8 米，再化成最简整数比即可。熟记比的意义和化简比的方法是解答本题的关键。\n【详解】小明和爸爸身高比:\n$125 厘米：1.8 米 \\\\=125 厘米: 180 厘米\\\\ = 25:36 $\n故答案为: ×",
+        "knowledge_points": [
+          "比的意义",
+          "比的化简"
+        ],
+        "source": "21-22 六年级上·全国·单元测试",
+        "merged_id": "unit4_363",
+        "original_id": "unit4_judge_363"
+      },
+      {
+        "id": "unit4_judge_364",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "(21-22 六年级上·全国·单元测试) 甲数是乙数的 7 倍,那么甲: 乙 $= 7: 1$ 。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】把乙数看作单位'1'，甲数就是'7'，甲乙两数的比即7:1；据此判断。此题考查的是比的应用。\n【详解】假设乙数是 1 , 甲数就是 7 , 甲 : 乙 = 7 : 1 。故答案为: √ 。",
+        "knowledge_points": [
+          "比的应用"
+        ],
+        "source": "21-22 六年级上·全国·单元测试",
+        "merged_id": "unit4_364",
+        "original_id": "unit4_judge_364"
+      },
+      {
+        "id": "unit4_judge_365",
+        "unit": "第四单元：比",
+        "chapter": "比的意义",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "(21-22 六年级上·全国·期中) 4:5 可以写成 $\\frac{4}{5}$ , 仍读作 '4 比 5'。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】两数相除又叫两个数的比, 比有两种形式, 可以写成算式形式, 也可以写成分数形式。关键是理解比的意义，掌握比的两种书写形式。\n【详解】4:5 可以写成 $\\frac{4}{5}$ , 仍读作 '4 比 5', 说法正确。故答案为: √",
+        "knowledge_points": [
+          "比的意义"
+        ],
+        "source": "21-22 六年级上·全国·期中",
+        "merged_id": "unit4_365",
+        "original_id": "unit4_judge_365"
+      },
+      {
+        "id": "unit4_judge_366",
+        "unit": "第四单元：比",
+        "chapter": "比的意义",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "(21-22 六年级上·全国·单元测试) 从学校走到电影院,甲用 4 分钟, 乙用 5 分钟, 甲和乙的速度比是 $5: 4$ 。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】把学校到电影院的总路程看作单位 '1', 根据甲、乙所用的时间分别求出他们的速度, 进而写出对应的速度比并化简比。关键是理解比的意义，路程一样，时间越少速度越快。\n【详解】甲的速度：乙的速度\n$= \\frac {1}{4}: \\frac {1}{5} \\\\ = 5: 4 $\n甲和乙的速度比是 $5: 4$故答案为: √",
+        "knowledge_points": [
+          "比的意义"
+        ],
+        "source": "21-22 六年级上·全国·单元测试",
+        "merged_id": "unit4_366",
+        "original_id": "unit4_judge_366"
+      },
+      {
+        "id": "unit4_judge_367",
+        "unit": "第四单元：比",
+        "chapter": "比的意义",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "(21-22 六年级上·全国·单元测试) 甲数是乙数的 3 倍,乙数和甲数的比是 $1: 3$ 。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】甲数是乙数的 3 倍, 乙是 1 , 则甲就是 3 , 由此即可求出乙和甲数的比; 进而判断即可。关键是理解'倍'和比的意义，两数相除又叫两个数的比。\n【详解】甲数是乙数的 3 倍, 乙是 1 , 则甲就是 3 , 则乙数和甲数的比是 $1: 3$ , 所以本题说法正确。故答案为: √",
+        "knowledge_points": [
+          "比的意义"
+        ],
+        "source": "21-22 六年级上·全国·单元测试",
+        "merged_id": "unit4_367",
+        "original_id": "unit4_judge_367"
+      },
+      {
+        "id": "unit4_judge_368",
+        "unit": "第四单元：比",
+        "chapter": "求一个数占另一个数几分之几、比的意义",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "(21-22 六年级上·全国·单元测试) 若小红与小花的写字数量的比是 $7: 4$ , 则小红比小花多写 $\\frac{3}{4}$ 。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】把小红与小花的写字数量分别看作 7 和 4 , 求小红比小花多写几分之几就用小红比小花多写的数量除以小花写的数量即可解答。明确 '求小红比小花多写几分之几就用小红比小花多写的数量除以小花写的数量'是解答本题的关键。\n【详解】$(7 - 4)\\div 4\\\\= 3 \\div 4 \\\\ = \\frac {3}{4}$\n故答案为:√。",
+        "knowledge_points": [
+          "求一个数占另一个数几分之几",
+          "比的意义"
+        ],
+        "source": "21-22 六年级上·全国·单元测试",
+        "merged_id": "unit4_368",
+        "original_id": "unit4_judge_368"
+      },
+      {
+        "id": "unit4_judge_369",
+        "unit": "第四单元：比",
+        "chapter": "分数的意义、比与分数、除法的关系",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "(20-21 六年级下·山东菏泽·期末) 红旗面数是黄旗的 $\\frac{2}{7}$ ,红旗面数是这两种彩旗总数的 $\\frac{7}{9}$ 。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】红旗面数是黄旗的七分之二，说明红旗面数与黄旗的面数的比是 $2: 7$ , 把红旗面数看作 2 份, 黄旗面数看作 7 份, 则红旗面数是这两种彩旗总数的九分之二。本题考查分数与比的关系、分数的意义, 解答本题的关键是掌握分数的意义。\n【详解】根据分析可得，本题说法错误。故答案为: ×",
+        "knowledge_points": [
+          "分数的意义",
+          "比与分数、除法的关系"
+        ],
+        "source": "20-21 六年级下·山东菏泽·期末",
+        "merged_id": "unit4_369",
+        "original_id": "unit4_judge_369"
+      },
+      {
+        "id": "unit4_judge_370",
+        "unit": "第四单元：比",
+        "chapter": "比与分数、除法的关系",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "(2021·江苏苏州·小升初真题) 甲数和乙数的比是2:3, 乙数是丙数的 $\\frac{6}{5}$ , 则甲、丙两数的比是 $4: 5$ 。 ( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】由题意可知, 甲数和乙数的比是 $2: 3$ , 乙数是丙数的 $\\frac{6}{5}$ , 可得乙数: 丙数 $= 6: 5$ , 两个比中乙数的份数是 3 和6, 3 和 6 的最小公倍数是 6 , 所以 $2: 3 = 4: 6$ , 进而得出甲、乙、丙的比是 $4: 6: 7$ ; 据此解答即可。本题主要考查了分数与比的相互转化，把分数化成两个数的比，再把两个比中的乙数根据比的基本性质进行转化，都转化为同一个数是解答此题的关键。\n【详解】乙数：丙数 $= 6:5$ 甲数：乙数 $= 2: 3 = 4: 6$ 甲：乙：丙=4:6:5 所以甲、丙两数的比是 $4: 5$ 。故答案为: √",
+        "knowledge_points": [
+          "比与分数、除法的关系"
+        ],
+        "source": "2021·江苏苏州·小升初真题",
+        "merged_id": "unit4_370",
+        "original_id": "unit4_judge_370"
+      },
+      {
+        "id": "unit4_judge_371",
+        "unit": "第四单元：比",
+        "chapter": "比的基本性质",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "(19-20 六年级下·湖南郴州·期末) 若 $a:b = 11:7$ , 则 $a$ $= 11, b = 7$ 。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】根据比的基本性质化简整数比, $a:b = 11:7$ , 只能说明 $a:b$ 的最简整数比是 $11:7$ , 并不代表 $a$ 和 $b$ 原来的是就是 11 和 7 , 据此可解答。本题考查比的基本性质, 明确两个数的比并不代表就是原来的数是解题的关键。\n【详解】a:b=11:7，只能说明 a:b 的最简整数比是 11:7，并不代表 a 和 b 原来的是就是 11 和 7，故说法错误。",
+        "knowledge_points": [
+          "比的基本性质"
+        ],
+        "source": "19-20 六年级下·湖南郴州·期末",
+        "merged_id": "unit4_371",
+        "original_id": "unit4_judge_371"
+      },
+      {
+        "id": "unit4_judge_372",
+        "unit": "第四单元：比",
+        "chapter": "基础行程问题、比的意义、比的化简",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "（2014 六年级·全国·课后作业）从甲地到乙地，A 车用了 6 小时，B 车用了 8 小时，A、B 两车的速度比是 $3:4$ 。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】将两地的距离看成单位 '1', 根据 A 车和 B 车分别用的时间, 先分别求出两车的速度, 写出比并化简即可。此题关键是先求出两车的速度，进而写比并化简比。\n【详解】$1 \\div 6 = \\frac{1}{6}$$1 \\div 8 = \\frac {1}{8}$ $\\frac{1}{6}: \\frac{1}{8} = 4: 3$ , 即 A、B 两车的速度比是 $4: 3$ 。故答案为: ×",
+        "knowledge_points": [
+          "基础行程问题",
+          "比的意义",
+          "比的化简"
+        ],
+        "source": "2014 六年级·全国·课后作业",
+        "merged_id": "unit4_372",
+        "original_id": "unit4_judge_372"
+      },
+      {
+        "id": "unit4_judge_373",
+        "unit": "第四单元：比",
+        "chapter": "比的基本性质",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "(19-20 六年级下·江苏连云港·期末) 比的前项乘 2 , 比的后项除以 2 , 比值扩大 4 倍。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】根据比的性质'比的前项和后项同时乘或除以一个相同的数（0 除外），比值不变'，可知比的前项乘2，后项除以2，比值变了，扩大了4倍；此题也可以举例子进行验证。此题考查比的性质的运用：只有比的前项和后项同时乘或除以一个相同的数（0 除外），比值才不变；否则比值会改变。\n【详解】如比： $6: 2 = 3$比的前项乘2，由6变成12，后项除以2，由2变成1，则比变为: $12: 1 = 12$ , 比值扩大了: $12 \\div 3 = 4$ 倍 所以比的前项乘 2 , 后项除以 2 , 比值扩大 4 倍, 原题说法正确 故答案为: √",
+        "knowledge_points": [
+          "比的基本性质"
+        ],
+        "source": "19-20 六年级下·江苏连云港·期末",
+        "merged_id": "unit4_373",
+        "original_id": "unit4_judge_373"
+      },
+      {
+        "id": "unit4_calc_374",
+        "unit": "第四单元：比",
+        "chapter": "比与分数、除法的关系、分数乘小数、分数与整数的乘法、应用等式的性质 2 解方程",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "(20-21 六年级上·江西赣州·单元测试) 解方程。$2 x - \\frac {4}{5} x = 1. 2 \\quad x: 8 = \\frac {3}{1 6}$",
+        "options": [],
+        "answer": "$x = 1; x = \\frac{3}{2}$",
+        "solution": "【分析】根据等式的性质：1.等式两边同时加或减去同一个数，等式仍然成立； 2.等式两边同时乘或除以一个不为 0 的数, 等式仍然成立,据此解答。\n【详解】$2\\mathrm{x} - \\frac{4}{5}\\mathrm{x} = 1.2$\n解: $\\frac{6}{5} x = 1.2\\\\ x = 1. 2 \\div \\frac {6}{5} \\\\ x = 1 \\\\ x: 8 = \\frac {3}{1 6} $\n解: $x \\div 8 = \\frac{3}{16}\\\\x = \\frac {3}{1 6} \\times 8 \\\\ x = \\frac {3}{2} $",
+        "knowledge_points": [
+          "比与分数、除法的关系",
+          "分数乘小数",
+          "分数与整数的乘法",
+          "应用等式的性质 2 解方程"
+        ],
+        "source": "20-21 六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_374",
+        "original_id": "unit4_calc_374"
+      },
+      {
+        "id": "unit4_calc_375",
+        "unit": "第四单元：比",
+        "chapter": "比的化简",
+        "question_type": "计算题",
+        "difficulty": "容易",
+        "question": "(21-22 六年级上·全国·单元测试) 把下面各比化成最简单的整数比。(写出化简过程)25:20 20:0.2 $\\frac {5}{6}: \\frac {2}{9}$",
+        "options": [],
+        "answer": "$5: 4; 100: 1; 15: 4$",
+        "solution": "【分析】化简比的方法:\n整数比：前、后项都除以它们的最大公因数；\n小数比：前、后项扩大相同倍数，化成整数比，再按整数比化简；\n分数比：前、后项都乘它们分母的最小公倍数，化成整数比，再化简。\n【详解】$25: 20 = (25 \\div 5): (20 \\div 5) = 5: 4$ $2 0: 0. 2 = 2 0 0: 2 = (2 0 0 \\div 2): (2 \\div 2) = 1 0 0: 1$ $\\frac {5}{6}: \\frac {2}{9} = (\\frac {5}{6} \\times 1 8): (\\frac {2}{9} \\times 1 8) = 1 5: 4$",
+        "knowledge_points": [
+          "比的化简"
+        ],
+        "source": "21-22 六年级上·全国·单元测试",
+        "merged_id": "unit4_375",
+        "original_id": "unit4_calc_375"
+      },
+      {
+        "id": "unit4_calc_376",
+        "unit": "第四单元：比",
+        "chapter": "分数的四则混合运算、应用等式的性质 2 解方程",
+        "question_type": "计算题",
+        "difficulty": "容易",
+        "question": "(21-22 六年级上·江西赣州·单元测试) 求未知数 $x$ 。$\\frac {5}{8} x = 2 0 \\quad x + \\frac {2}{5} x = 2 1$",
+        "options": [],
+        "answer": "$x = 32; x = 15$",
+        "solution": "【分析】①根据等式性质2，将方程左右两边同时除以 $\\frac{5}{8}$\n求得方程的解;\n(2) 先逆用乘法分配律, 合并未知数前面的系数, 再应用等式性质 2 , 将方程左右两边同时除以 $\\frac{7}{5}$ , 求得方程的解。\n【详解】$\\frac{5}{8} x = 20$\n解: $x = 20 \\div \\frac{5}{8}\\\\x = 2 0 \\times \\frac {8}{5}\\\\x = 3 2$\n$x + \\frac {2}{5} x = 2 1$\n解: $x(1 + \\frac{2}{5}) = 21\\\\ \\frac {7}{5} x = 2 1\\\\x = 2 1 \\div \\frac {7}{5}\\\\x = 1 5$",
+        "knowledge_points": [
+          "分数的四则混合运算",
+          "应用等式的性质 2 解方程"
+        ],
+        "source": "21-22 六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_376",
+        "original_id": "unit4_calc_376"
+      },
+      {
+        "id": "unit4_calc_377",
+        "unit": "第四单元：比",
+        "chapter": "求比值、比的化简",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "(21-22 六年级上·江西赣州·单元测试) 化简或求比值。$\\frac{1}{4}: \\frac{1}{8}$ (求比值) 200千克: $\\frac{2}{5}$ 吨 (化简)",
+        "options": [],
+        "answer": "2; 1:2",
+        "solution": "【分析】①可把比号换成除号，运用分数除法求得比值；② 可先把 $\\frac{2}{5}$ 吨化成以千克为单位的数，再按整数比化简的方法，比的前项后项同时除以它们的最大公因数，来化简这个比。\n【详解】$\\frac{1}{4}:\\frac{1}{8} = \\frac{1}{4}\\div \\frac{1}{8} = 2$\n200千克: $\\frac{2}{5}$ 吨 = 200 千克 : $(\\frac{2}{5} \\times 1000$ 千克 $) = 200$ 千克\n: 400 千克 = (200 ÷ 200) : (400 ÷ 200) = 1:2",
+        "knowledge_points": [
+          "求比值",
+          "比的化简"
+        ],
+        "source": "21-22 六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_377",
+        "original_id": "unit4_calc_377"
+      },
+      {
+        "id": "unit4_calc_378",
+        "unit": "第四单元：比",
+        "chapter": "求比值",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "(21-22 六年级上·全国·单元测试) 求比值。$\\frac {3}{5}: \\frac {7}{8} \\quad 2. 4: 0. 8$",
+        "options": [],
+        "answer": "$\\frac{24}{35}$ ; 3",
+        "solution": "【分析】用比的前项除以比的后项，所得的商就是比值。据此解答即可。\n【详解】(1) $\\frac{3}{5}: \\frac{7}{8}$\n$ = \\frac {3}{5} \\div \\frac {7}{8} \\\\ = \\frac {2 4}{3 5} $\n(2) $2.4: 0.8$\n$= 2. 4 \\div 0. 8 \\\\ = 3 $",
+        "knowledge_points": [
+          "求比值"
+        ],
+        "source": "21-22 六年级上·全国·单元测试",
+        "merged_id": "unit4_378",
+        "original_id": "unit4_calc_378"
+      },
+      {
+        "id": "unit4_calc_379",
+        "unit": "第四单元：比",
+        "chapter": "比的化简",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "(21-22 六年级上·全国·单元测试) 化简比。$0.3: 0. 0 9 \\quad \\frac {3}{4}: 0. 2 5$",
+        "options": [],
+        "answer": "(1) $10: 3$ (2) $3: 1$",
+        "solution": "【分析】（1）比的前项和后项同时扩大 100 倍化作整数比,再把前项和后项除以它们的最大公因数;\n(2) 比的前项和后项同时扩大 4 倍。\n【详解】(1) $0.3: 0.09 = (0.3 \\times 100): (0.09 \\times 100) = 30: 9 = (30 \\div 3): (9 \\div 3) = 10: 3$\n(2) $\\frac{3}{4}: 0.25 = (\\frac{3}{4} \\times 4): (0.25 \\times 4) = 3: 1$",
+        "knowledge_points": [
+          "比的化简"
+        ],
+        "source": "21-22 六年级上·全国·单元测试",
+        "merged_id": "unit4_379",
+        "original_id": "unit4_calc_379"
+      },
+      {
+        "id": "unit4_calc_380",
+        "unit": "第四单元：比",
+        "chapter": "除数是小数的小数除法、异分母分数加、减法、分数与整数的除法、分数乘分数",
+        "question_type": "计算题",
+        "difficulty": "容易",
+        "question": "(21-22 六年级上·全国·单元测试) 直接写出得数。$ \\frac {1}{2} \\times \\frac {3}{4} = \\quad \\frac {2}{5} + \\frac {1}{3} = \\quad 0. 1 2 5 \\times 1. 7 \\times 8 = \\quad 2 - 0. 5 3 = \\quad \\frac {5}{8} \\div 1 0 = \\\\ \\frac {1}{5} - \\frac {1}{6} = \\quad 1 \\div \\frac {5}{6} = \\quad 2. 9 \\times 3 \\div 2. 9 \\times 3 = \\quad 1 2: \\frac {4}{5} = \\quad 7 \\div 1. 4 = \\\\ $",
+        "options": [],
+        "answer": "$\\frac{3}{8}; \\frac{11}{15}; 1.7; 1.47; \\frac{1}{16}$ ; $\\frac {1}{3 0}; 1 \\frac {1}{5}; 9; 1 5; 5$",
+        "solution": "【详解】略",
+        "knowledge_points": [
+          "除数是小数的小数除法",
+          "异分母分数加、减法",
+          "分数与整数的除法",
+          "分数乘分数"
+        ],
+        "source": "21-22 六年级上·全国·单元测试",
+        "merged_id": "unit4_380",
+        "original_id": "unit4_calc_380"
+      },
+      {
+        "id": "unit4_calc_381",
+        "unit": "第四单元：比",
+        "chapter": "求比值、比的化简",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "(21-22 六年级上·全国·单元测试) 化简下列各比, 并求比值。$ 4 8: 4 0 \\quad \\frac {3}{4}: \\frac {7}{2 0} \\quad 0. 7 5: 2 \\quad \\frac {1}{5}: 0. 6 \\quad \\frac {1}{5}t : \\frac {3}{4} k g $",
+        "options": [],
+        "answer": "6:5; 1.2; 15:7; $2 \\frac{1}{7}$ ; 3:8; $\\frac{3}{8}$ ;$1: 3; \\frac {1}{3}; 8 0 0: 3; 2 6 6 \\frac {2}{3}$",
+        "solution": "【分析】（1）根据比的基本性质，即比的前项和后项同时乘或除以一个相同的数（0 除外）比值不变，进而把比化成最简比；\n(2) 用比的前项除以后项, 所得的商即为比值。\n【详解】$48: 40 = (48 \\div 8): (40 \\div 8) = 6: 5$\n$4 8: 4 0 = 4 8 \\div 4 0 = 1. 2\\\\ \\frac {3}{4}: \\frac {7}{2 0} = (\\frac {3}{4} \\times 2 0): (\\frac {7}{2 0} \\times 2 0) = 1 5: 7,$\n$\\frac {3}{4} \\div \\frac {7}{2 0} = \\frac {3}{4} \\div \\frac {7}{2 0} = 2 \\frac {1}{7}$\n$0.7 5: 2 = 7 5: 2 0 0 = 3: 8,$\n$0.7 5: 2 = 0. 7 5 \\div 2 = \\frac {3}{8}$\n$\\frac {1}{5}: 0. 6 = 2: 6 = 1: 3$\n$\\frac {1}{5}: 0. 6 = 0. 2 \\div 0. 6 = \\frac {1}{3}$\n$\\frac {1}{5} t: \\frac {3}{4} k g = 2 0 0 k g: 0. 7 5 k g = 8 0 0: 3$\n$\\frac {1}{5} t: \\frac {3}{4} k g = 2 0 0 k g \\div 0. 7 5 k g = 2 6 6 \\frac {2}{3}$",
+        "knowledge_points": [
+          "求比值",
+          "比的化简"
+        ],
+        "source": "21-22 六年级上·全国·单元测试",
+        "merged_id": "unit4_381",
+        "original_id": "unit4_calc_381"
+      },
+      {
+        "id": "unit4_calc_382",
+        "unit": "第四单元：比",
+        "chapter": "比的化简",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "(20-21 六年级下·湖北随州·期末) 求最简整数比。$\\frac{3}{4}: 0.25$ $\\frac{2}{3}$ 小时：30分 2500米：5千米",
+        "options": [],
+        "answer": "$3: 1; 4: 3; 1: 2$",
+        "solution": "【分析】比的性质：比的前项和比的后项同时乘或除以同一个不为零的数，比值不变。在化简比的同时注意单位换算。本题考查化简比, 解答本题的关键是掌握比的性质。\n【详解】$\\frac{3}{4}: 0.25 = (\\frac{3}{4} \\times 4): (0.25 \\times 4) = 3: 1$ $\\frac{2}{3}$ 小时: 30 分 $(\\frac{2}{3} \\times 60): 30 \\div 4: 3$\n$2500 米 : 5 千米 =(2500 \\div 2500): (5000 \\div 2500)= 1: 2$",
+        "knowledge_points": [
+          "比的化简"
+        ],
+        "source": "20-21 六年级下·湖北随州·期末",
+        "merged_id": "unit4_382",
+        "original_id": "unit4_calc_382"
+      },
+      {
+        "id": "unit4_solve_383",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "(20-21 六年级上·湖北荆门·期末) 甲、乙两支清洁队共有 100 人, 已知甲队人数的 $\\frac{1}{3}$ 与乙人数的 $\\frac{1}{2}$ 相等, 甲、乙两队各有多少人?",
+        "options": [],
+        "answer": "甲队：60人，乙队：40人",
+        "solution": "【分析】根据'甲队人数的 $\\frac{1}{3}$ 与乙人数的 $\\frac{1}{2}$ 相等'可知，甲队与乙队的人数比为 $3: 2$ ，再用总人数除以总份数求出每份是多少人，再乘甲队与乙队各自对应的份数即可。解答本题的关键是写出甲队与乙队的人数比, 再根据按比例分配的知识点解答。\n【详解】甲队与乙队的人数比为 $3:2$ ; $1 0 0 \\div (3 + 2) = 2 0 (\\text {人});$ $2 0 \\times 3 = 6 0 (\\text {人});$ $2 0 \\times 2 = 4 0 (\\text {人});$答: 甲队有 60 人, 乙队有 40 人。",
+        "knowledge_points": [
+          "比的应用"
+        ],
+        "source": "20-21 六年级上·湖北荆门·期末",
+        "merged_id": "unit4_383",
+        "original_id": "unit4_solve_383"
+      },
+      {
+        "id": "unit4_solve_384",
+        "unit": "第四单元：比",
+        "chapter": "求一个数的几分之几的问题、比的应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（20-21 六年级上·福建厦门·单元测试）学校一张课桌比一把椅子贵 12 元，椅子的单价是课桌单价的 $\\frac{3}{5}$ ，课桌每张多少元？",
+        "options": [],
+        "answer": "30元",
+        "solution": "【分析】把课桌的价格看作5份的量，椅子的价格看作3份的量，则课桌的价格就比椅子的价格多两份的量，而'一张课桌比一把椅子贵12元'，多的12元所对应的份数是2份，从而可以求出一份的量是多少，进而可以求出课桌的价格。解决此题的关键是找出对应量和对应分率，从而轻松求解。\n【详解】$12 \\div (5 - 3)\\\\ = 1 2 \\div 2 \\\\ = 6 (\\text {元}) $\n$6 \\times 5 = 3 0 (\\text {元})$ 答: 课桌每张30元。",
+        "knowledge_points": [
+          "求一个数的几分之几的问题",
+          "比的应用"
+        ],
+        "source": "20-21 六年级上·福建厦门·单元测试",
+        "merged_id": "unit4_384",
+        "original_id": "unit4_solve_384"
+      },
+      {
+        "id": "unit4_solve_385",
+        "unit": "第四单元：比",
+        "chapter": "分数与整数的乘法",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "(21-22 六年级上·江西赣州·单元测试) 只列式, 不计算。一长方体长 8 米, 宽是长的 $\\frac{1}{4}$ , 同时又是高的一半, 求长方体的高。",
+        "options": [],
+        "answer": "$8 \\times \\frac{1}{4} \\times 2$",
+        "solution": "【分析】根据题意可知, 把长看作单位 '1', 根据求一个数的几分之几是多少用乘法求出宽, 宽同时又是高的一半,用宽乘 2 得解。此题关键找准单位'1'，单位'1'已知用乘法。\n【详解】$8 \\times \\frac{1}{4} \\times 2\\\\= 2 \\times 2\\\\= 4$ (米)\n答: 长方体的高为 4 米。",
+        "knowledge_points": [
+          "分数与整数的乘法"
+        ],
+        "source": "21-22 六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_385",
+        "original_id": "unit4_solve_385"
+      },
+      {
+        "id": "unit4_solve_386",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "(21-22 六年级上·江西赣州·单元测试) 只列式, 不计算。药液和水按 1:500 配制成一种药水, 5 千克药液可配药水多少千克?",
+        "options": [],
+        "answer": "2505千克",
+        "solution": "【分析】由题意可知, 药液占药水质量的 $(\\frac{1}{1 + 500})$ , 药水质量 = 药液质量 ÷ 药液占药水质量的分率。本题考查了比的相关知识点在生活实际中的应用。\n【详解】$5 \\div \\frac{1}{1 + 500}\\\\= 5 \\div \\frac {1}{5 0 1} \\\\ = 5 \\times 5 0 1 \\\\=2505 (千克)$\n答: 5 千克药液可配药水 2505 千克。",
+        "knowledge_points": [
+          "比的应用"
+        ],
+        "source": "21-22 六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_386",
+        "original_id": "unit4_solve_386"
+      },
+      {
+        "id": "unit4_solve_387",
+        "unit": "第四单元：比",
+        "chapter": "分数除法的应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "(21-22 六年级上·江西赣州·单元测试) 只列式, 不计算。一桶油吃了 $\\frac{3}{8}$ 正好是 9 千克, 这桶油原来有多少千克?",
+        "options": [],
+        "answer": "$9 \\div \\frac{3}{8}$",
+        "solution": "【分析】把一桶油看作单位 '1', 已知单位 '1'的 $\\frac{3}{8}$ 是 9 千克, 求单位 '1'用量除以对应的分率; 据此解答。已知一个数的几分之几是多少, 求这个数, 用除法计算。\n【详解】$9 \\div \\frac{3}{8} = 9 \\times \\frac{8}{3} = 24$ （千克）答: 这桶油原来有 24 千克。",
+        "knowledge_points": [
+          "分数除法的应用"
+        ],
+        "source": "21-22 六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_387",
+        "original_id": "unit4_solve_387"
+      },
+      {
+        "id": "unit4_solve_388",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题、比的化简、比的应用",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "(21-22 六年级上·江西赣州·单元测试) 六 (1) 班和六(2) 班共有 100 人，在六（1）班男生人数占全班人数的 $\\frac{2}{5}$ ，在六（2）班男生人数占全班人数的 $\\frac{3}{5}$ ，两个班的女生人数一样多，两个班各有多少名学生？",
+        "options": [],
+        "answer": "六（1）40人；六（2）60人",
+        "solution": "【分析】由题意可知, 六 (1) 班女生人数占全班人数的 $(1 - \\frac{2}{5})$ , 六 (2) 班女生人数占全班人数的 $(1 - \\frac{3}{5})$ , 根据两班女生人数一样多计算出两班总人数的比, 最后根据按比例分配计算出两班各有多少人。根据两班的人数关系计算出两班总人数的比是解答本题的关键。\n【详解】六（1）班总数 $\\times (1 - \\frac{2}{5}) =$ 六（2）班总数 $\\times (1 - \\frac{3}{5})$六 (1) 班总数 $\\times \\frac{3}{5} =$ 六 (2) 班总数 $\\times \\frac{2}{5}$六 (1) 班总数 $\\div$ 六 (2) 班总数 $= \\frac{2}{5} \\div \\frac{3}{5}$六 (1) 班总数 $\\div$ 六 (2) 班总数 $= \\frac{2}{3}$六（1）班总数：六（2）班总数 $= 2:3$ 六 (1) 班总数: $100 \\times \\frac{2}{2 + 3} = 40$ （人） 六 (2) 班总数: $100 \\times \\frac{3}{2 + 3} = 60$ （人）答: 六 (1) 班有 40 人, 六 (2) 班有 60 人。",
+        "knowledge_points": [
+          "按比分配问题",
+          "比的化简",
+          "比的应用"
+        ],
+        "source": "21-22 六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_388",
+        "original_id": "unit4_solve_388"
+      },
+      {
+        "id": "unit4_solve_389",
+        "unit": "第四单元：比",
+        "chapter": "分数除法的应用、分数的四则混合运算、除法求一个数的几分之一/几是多少",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "(21-22 六年级上·江西赣州·单元测试) 六 (1) 班班主任安排 $\\frac{1}{5}$ 的学生去打扫整洁区, 这时教室里还剩下 36 人, 六(1) 班有学生多少人?",
+        "options": [],
+        "answer": "45人",
+        "solution": "【分析】把六（1）班学生人数看作单位'1'，根据'量÷对应分率'即可求得。掌握标准量的计算方法是解答本题的关键。\n【详解】$36 \\div (1 - \\frac{1}{5})\\\\= 3 6 \\div \\frac {4}{5} \\\\ = 3 6 \\times \\frac {5}{4} \\\\ = 4 5 \\text {(人)} $\n答: 六 (1) 班有学生 45 人。",
+        "knowledge_points": [
+          "分数除法的应用",
+          "分数的四则混合运算",
+          "除法求一个数的几分之一/几是多少"
+        ],
+        "source": "21-22 六年级上·江西赣州·单元测试",
+        "merged_id": "unit4_389",
+        "original_id": "unit4_solve_389"
+      },
+      {
+        "id": "unit4_solve_390",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题、三角形的内角和、三角形的分类",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（19-20六年级上·辽宁大连·期末）三角形的三个角的度数比是 $1:2:3$ ，求：(1) 这三个内角的度数分别是多少度? (2) 这个三角形是什么形状的三角形?",
+        "options": [],
+        "answer": "(1) $30^{\\circ} 、 60^{\\circ}$ 和 $90^{\\circ}$ ; (2) 直角三角形",
+        "solution": "【分析】（1）三角形的内角和为 $180^{\\circ}$ ，根据三角形的三个角的度数比是 $1: 2: 3$ , 先用 $180^{\\circ}$ 除以 6 求出一份的度数, 再利用乘法分别求出三个角的度数即可;(2) 根据第一问的结果, 结合三角形的分类, 直接判断出这个三角形的形状即可。本题考查了比和三角形, 明确三角形的内角和是 $180^{\\circ}$ 是解题的关键。\n【详解】(1) $180^{\\circ} \\div (1 + 2 + 3)\\\\ = 1 8 0 ^ {\\circ} \\div 6 \\\\ = 3 0 ^ {\\circ} $\n$3 0 ^ {\\circ} \\times 2 = 6 0 ^ {\\circ}, 3 0 ^ {\\circ} \\times 3 = 9 0 ^ {\\circ}$答: 这三个内角的度数分别是 $30^{\\circ} 、 60^{\\circ}$ 和 $90^{\\circ}$ 。\n(2) 由于这个三角形有个内角是 $90^{\\circ}$ , 所以这是一个直角三角形。",
+        "knowledge_points": [
+          "按比分配问题",
+          "三角形的内角和",
+          "三角形的分类"
+        ],
+        "source": "19-20六年级上·辽宁大连·期末",
+        "merged_id": "unit4_390",
+        "original_id": "unit4_solve_390"
+      },
+      {
+        "id": "unit4_solve_391",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "(21-22 六年级上·全国·单元测试) 苹果饮料是由 3 份纯苹果汁和 7 份水冲兑的。一瓶苹果饮料的体积是$280 \\mathrm{~mL}$ , 它用了纯苹果汁和水各多少毫升?",
+        "options": [],
+        "answer": "纯苹果汁 84 毫升, 水 196 毫升",
+        "solution": "【分析】根据题意可知, 苹果饮料中纯苹果汁量与水的量的比为 $3: 7$ ; 用饮料的总量除以总份数求出每份是多少毫升,再乘纯苹果汁与水各占的份数即可。求出每份是多少毫升是解答本题的关键，再乘纯苹果汁与水各占的份数。\n【详解】苹果饮料中纯苹果汁量与水量的比为 $3: 7$ ;\n$ 2 8 0 \\div (3 + 7) \\\\ = 2 8 0 \\div 1 0 \\\\ = 2 8 \\text {(毫 升)} ; $\n$2 8 \\times 3 = 8 4 (\\text {毫 升}); 2 8 \\times 7 = 1 9 6 (\\text {毫 升}); $答: 纯苹果汁 84 毫升, 水 196 毫升。",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "21-22 六年级上·全国·单元测试",
+        "merged_id": "unit4_391",
+        "original_id": "unit4_solve_391"
+      },
+      {
+        "id": "unit4_solve_392",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（20-21六年级上·福建福州·期中）乐乐一家三口和陶陶一家四口到餐馆用餐，餐费共840元，两家决定按人数分摊餐费，陶陶家应付餐费多少元？",
+        "options": [],
+        "answer": "480元",
+        "solution": "【分析】根据题意可知, 乐乐家与陶陶家的人数比为 $3: 4$ ,用总费用除以总份数即可求出每份是多少元, 再乘陶陶家对应的份数即可。本题考查了按比例分配的知识点, 明确两家的人数比是解答本题的关键。\n【详解】乐乐家与陶陶家的人数比为 $3:4$ ;\n$ 8 4 0 \\div (3 + 4) \\times 4 \\\\ = 8 4 0 \\div 7 \\times 4 \\\\ = 1 2 0 \\times 4 \\\\ = 4 8 0 \\text {(元)}; $答: 陶陶家应付餐费 480 元。",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "20-21六年级上·福建福州·期中",
+        "merged_id": "unit4_392",
+        "original_id": "unit4_solve_392"
+      },
+      {
+        "id": "unit4_solve_393",
+        "unit": "第四单元：比",
+        "chapter": "长方形的周长、长方形的面积、比的应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（19-20 六年级下·山东济南·期末）一个长方形操场的周长是 200 米，长与宽的比是 $3: 2$ ，操场的面积是多少平方米？",
+        "options": [],
+        "answer": "2400 平方米",
+        "solution": "【分析】我们运用和比问题的解答方法分别求出长方形的长与宽，然后再运用长方形的面积公式进行解答。本题运用'和÷两个数的份数的和=每份的数量'\n【详解】长方形的长是:\n$2 0 0 \\div 2 \\div (3 + 2) \\times 3 \\\\ = 1 0 0 \\div 5 \\times 3 \\\\ = 6 0 \\text {(米)} $\n长方形的宽:\n$ 2 0 0 \\div 2 - 6 0, \\\\ = 1 0 0 - 6 0, \\\\ = 4 0 \\text {(米)} $\n长方形的面积:$6 0 \\times 4 0 = 2 4 0 0 (\\text {平 方 米})$答: 操场的面积是 2400 平方米。进一步求出长与宽，然后求操场的面积。",
+        "knowledge_points": [
+          "长方形的周长",
+          "长方形的面积",
+          "比的应用"
+        ],
+        "source": "19-20 六年级下·山东济南·期末",
+        "merged_id": "unit4_393",
+        "original_id": "unit4_solve_393"
+      },
+      {
+        "id": "unit4_solve_394",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题、求一个数的几分之几的问题、比的应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "(18-19六年级下·河南洛阳·期末) 三、四、五年级学生参加学校乐团的一共有63人，其中三年级人数占乐团总人数的 $\\frac{4}{9}$ , 四年级和五年级参加乐团人数的比为 $4: 3$ 。三、四、年级报名参加乐团的分别有多少人？",
+        "options": [],
+        "answer": "三年级28人、四年级20人、五年级15人",
+        "solution": "【分析】用总人数乘三年级人数占乐团总人数的 $\\frac{4}{9}$ , 求出三年级参加乐团的人数;根据四年级和五年级参加乐团人数的比，将四、五年级参加乐团的人数占这两个年级参加乐团总人数的几分之几分别表示出来，再利用乘法分别求出四、五年级参加乐团的人数即可。本题考查了比的应用, 能根据比求出四、五年级参加乐团的人数占这两个年级的几分之几是解题的关键。\n【详解】三年级: $63 \\times \\frac{4}{9} = 28$ (人)\n四年级：\n$(6 3 - 2 8) \\times \\frac {4}{4 + 3} \\\\ = 3 5 \\times \\frac {4}{7} \\\\ = 2 0 \\text {(人)} $\n五年级：\n$(6 3 - 2 8) \\times \\frac {3}{4 + 3} \\\\ = 3 5 \\times \\frac {3}{7} \\\\ = 15$ （人）\n答: 三、四、年级报名参加乐团的分别有 28 人、20 人、15 人。",
+        "knowledge_points": [
+          "按比分配问题",
+          "求一个数的几分之几的问题",
+          "比的应用"
+        ],
+        "source": "18-19六年级下·河南洛阳·期末",
+        "merged_id": "unit4_394",
+        "original_id": "unit4_solve_394"
+      },
+      {
+        "id": "unit4_solve_395",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（20-21 六年级下·全国·期末）朝阳小学把 250 本图书分给'学法'中队和'明史'中队的少先队员，两个中队分得图书的比是 $2:3$ ，每个中队各分得图书多少本？",
+        "options": [],
+        "answer": "学法'中队分得图书 100 本，'明史'中队分得图书 150 本",
+        "solution": "【分析】用总本数除以总份数求出每份多少本, 再乘每个中队各自对应的份数即可。本题考查了按比例分配的知识点, 求出每份多少本是解答本题的关键。\n【详解】$250 \\div (2 + 3)\\\\= 2 5 0 \\div 5 \\\\ = 5 0 \\text {(本)} $\n$5 0 \\times 2 = 1 0 0 (\\text {本}) \\quad 5 0 \\times 3 = 1 5 0 (\\text {本}) $答: '学法' 中队分得图书 100 本, '明史' 中队分得图书 150 本。",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "20-21 六年级下·全国·期末",
+        "merged_id": "unit4_395",
+        "original_id": "unit4_solve_395"
+      },
+      {
+        "id": "unit4_solve_396",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题、基础行程问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "(20-21 六年级下·江西吉安·期末) 甲乙两车同时从A、B两城的中点背向而行，5小时后，甲到达了A地，乙车离B地还有120千米，已知甲乙两车的速度比是7:5，求A、B 两地相距多少千米？",
+        "options": [],
+        "answer": "840千米",
+        "solution": "【分析】![](images\\1b0f0b46ace2a84d37c4a2b54005e88d.png), 根据路程比 $=$ 速度比,甲乙两车路程比也是 $7: 5$ , 甲到达 A 地, 行了全程的一半,一半是 7 份, 120 千米对应 7-5 份, 求出一份数, 用一份数 $\\times$ 全程总份数即可。关键是理解比的意义，确定份数，先求出一份数。\n【详解】$120 \\div (7 - 5) \\times 7 \\times 2\\\\= 1 2 0 \\div 2 \\times 7 \\times 2 \\\\ = 8 4 0 (\\text {千 米}) $\n答: A、B 两地相距 840 千米。",
+        "knowledge_points": [
+          "按比分配问题",
+          "基础行程问题"
+        ],
+        "source": "20-21 六年级下·江西吉安·期末",
+        "merged_id": "unit4_396",
+        "original_id": "unit4_solve_396"
+      },
+      {
+        "id": "unit4_solve_397",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题、分数的四则混合运算",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "(19-20 六年级下·河南南阳·期末) 李华家的菜地有$800 \\mathrm{~m}^{2}$ , 其中 $\\frac{3}{5}$ 已经种了西红柿, 剩下的按 3:1 的面积比种黄瓜和茄子。黄瓜的面积是多少平方米？",
+        "options": [],
+        "answer": "240 平方米",
+        "solution": "【分析】把菜地总面积看作单位 '1', 则剩下的面积是菜地总面积的 $(1 - \\frac{3}{5})$ , 乘总面积求出剩下的面积, 根据黄瓜和茄子的面积之比, 按比例分配求出黄瓜的面积即可。此题考查了分数与比的综合应用，先求出剩下的面积是解题关键。\n【详解】$800 \\times (1 - \\frac{3}{5}) \\div (3 + 1) \\times 3\\\\= 3 2 0 \\div 4 \\times 3 \\\\ = 2 4 0 \\text {(平 方 米)} $\n答: 黄瓜的面积是 240 平方米。",
+        "knowledge_points": [
+          "按比分配问题",
+          "分数的四则混合运算"
+        ],
+        "source": "19-20 六年级下·河南南阳·期末",
+        "merged_id": "unit4_397",
+        "original_id": "unit4_solve_397"
+      },
+      {
+        "id": "unit4_solve_398",
+        "unit": "第四单元：比",
+        "chapter": "分数的四则混合运算、比的应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（19-20六年级下·河南南阳·期末）李亮打印一份稿件，第一天打印的页数与总页数的比是 $1:3$ 。如果第二天打印 15 页，那么就可以打印全部稿件的一半。这份稿件一共有多少页？",
+        "options": [],
+        "answer": "90页",
+        "solution": "【分析】由第一天打印的页数与总页数的比是 $1: 3$ , 可知第一天打印了总页数的 $\\frac{1}{3}$ , 两天一共打印了总页数的 $\\frac{1}{2}$ , 所以第二天打印了总页数的 $(\\frac{1}{2} - \\frac{1}{3})$ , 已知第二天打印了 15 天,根据分数除法的意义, 用除法即可求出总页数。此题考查了分数除法与比的综合应用，找出第二天打印的页数对应的分率是解题关键。\n【详解】$15 \\div (\\frac{1}{2} - \\frac{1}{3})\\\\= 1 5 \\div \\frac {1}{6} \\\\ = 9 0 \\text {(页)} $\n答: 这份稿件一共有 90 页。",
+        "knowledge_points": [
+          "分数的四则混合运算",
+          "比的应用"
+        ],
+        "source": "19-20六年级下·河南南阳·期末",
+        "merged_id": "unit4_398",
+        "original_id": "unit4_solve_398"
+      },
+      {
+        "id": "unit4_solve_399",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（20-21六年级上·浙江·期末）某校合唱队男生与女生的人数比是5:8，已知男生比女生少12人，这个合唱队一共有多少人？",
+        "options": [],
+        "answer": "52人",
+        "solution": "【分析】男生与女生的人数比是 $5: 8$ , 则总人数所占份数就为 $5 + 8 = 13$ , 男生所占份数比女生所占份数少 $8 - 5 = 3$ 份, 且已知男生比女生少 12 人, 可求出一份是多少人, 再乘总份数得到最后的答案。本题主要考查的是比的应用, 解题的关键是熟悉比中份数的多少, 并运用多少关系进行解答。\n【详解】男生所占份数比女生所占份数少 $8 - 5 = 3$ , 已知男生比女生少 12 人, 则占一份的人数为: $12 \\div 3 = 4$ (人); 总人数所占份数就为 $5 + 8 = 13$ , 故合唱队总人数为: $13 \\times 4 = 52$ (人)。\n答: 这个合唱队一共有 52 人。",
+        "knowledge_points": [
+          "比的应用"
+        ],
+        "source": "20-21六年级上·浙江·期末",
+        "merged_id": "unit4_399",
+        "original_id": "unit4_solve_399"
+      },
+      {
+        "id": "unit4_solve_400",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "(17-18 六年级下·湖南长沙·期末) 一种氯化钠注射液是由氯化钠和水按 1:9 的比配制而成的。一瓶 $50 \\mathrm{ml}$ 氯化钠注射液含氯化钠和水各多少毫升？",
+        "options": [],
+        "answer": "5毫升；45毫升",
+        "solution": "【分析】把氯化钠注射液看作单位 '1', 氯化钠和水按 $1:9$ 的比配制而成的, 就是氯化钠占氯化钠注射液的 $\\frac{1}{10}$ , 水占氯化钠注射液的 $\\frac{9}{10}$ , 代入数据计算即可。把氯化钠注射液看作单位 '1', 求一个数的几分之几是多少要用乘法, 这是解决此题的关键。\n【详解】$1 + 9 = 10$ 氯化钠: $50 \\times \\frac{1}{10} = 5$ (毫升) 水: $50 \\times \\frac{9}{10} = 45$ (毫升) 答: 一瓶 $50 \\mathrm{ml}$ 氯化钠注射液含氯化钠 5 毫升, 含水 45 毫升。",
+        "knowledge_points": [
+          "比的应用"
+        ],
+        "source": "17-18 六年级下·湖南长沙·期末",
+        "merged_id": "unit4_400",
+        "original_id": "unit4_solve_400"
+      }, {
+        "id": "unit4_choice_401",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "参加数学测试的男生与女生人数的比是 $2: 1$, 平均分是 86, 其中男生的平均分是 84, 则女生的平均分是 ( )。",
+        "options": [
+          "A. 87",
+          "B. 88",
+          "C. 89",
+          "D. 90"
+        ],
+        "answer": "D",
+        "solution": "【分析】根据题意, 可找出等量关系: 女生的平均分 $\\times 1+$ 男生的平均分 $\\times 2=$ 全班平均分 $\\times 3$, 设女生的平均分是 $x$, 列解方程即可解答。本题的关键是找出等量关系：女生的平均分 $\\times$ 女生占的份数 + 男生的平均分 $\\times$ 男生占的份数 = 全班平均分 $\\times$ 总份数，然后列方程解答。\n【详解】解：设女生的平均分是 $x$，由题意得：\n$x + 8 4 \\times 2 = 8 6 \\times (1 + 2)\\\\x + 1 6 8 = 2 5 8\\\\x = 90$\n故答案为: D",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "18-19六年级下·浙江·期末",
+        "merged_id": "unit4_401",
+        "original_id": "unit4_choice_401"
+      },
+      {
+        "id": "unit4_choice_402",
+        "unit": "第四单元：比",
+        "chapter": "求比值、基础行程问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "亮亮每天从家到学校上学,步行需要 15 分钟, 骑车需要 9 分钟, 他步行与骑车速度的最简整数比是 ( )。",
+        "options": [
+          "A. 15:9",
+          "B. 3:5",
+          "C. $\\frac{1}{15}: \\frac{1}{9}$",
+          "D. 5:3"
+        ],
+        "answer": "B",
+        "solution": "【分析】把从家到学校的路程看做单位 '1', 先分别求出他步行和骑车的速度, 再进一步根据比的性质化成最简比。解决此题关键是将从家到学校的路程看作单位'1'，求出各自的速度，进一步算出比并化简。\n【详解】解：设从家到学校的路程是'1'，步行的速度: ,$1 \\div 15 = \\frac{1}{15}$骑车的速度: $1 \\div 9 = \\frac{1}{9}$步行与骑车速度的最简整数比: $\\frac{1}{15}: \\frac{1}{9} = \\frac{3}{5} = 3: 5$,故答案为: B",
+        "knowledge_points": [
+          "求比值",
+          "基础行程问题"
+        ],
+        "source": "18-19 六年级下·浙江·期末",
+        "merged_id": "unit4_402",
+        "original_id": "unit4_choice_402"
+      },
+      {
+        "id": "unit4_choice_403",
+        "unit": "第四单元：比",
+        "chapter": "比的化简、基础行程问题、比的应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "从甲地到乙地, A 要 1 小时, B 要 1.5 小时。A、B 的速度比是 ( )。",
+        "options": [
+          "A. $1: 1.5$",
+          "B. $2: 3$",
+          "C. $3: 2$"
+        ],
+        "answer": "C",
+        "solution": "【分析】将甲乙两地的距离看作单位 1 , 据此将 A、B 的速度表示出来, 再做比即可。本题考查了比的应用, 明确速度等于路程除以时间, 会化简比是解题的关键。\n【详解】1: $(1 \\div 1.5) = 3: 2$, 所以速度比是 $3: 2$ 。故答案为: C",
+        "knowledge_points": [
+          "比的化简",
+          "基础行程问题",
+          "比的应用"
+        ],
+        "source": "18-19 六年级下·云南红河·期中",
+        "merged_id": "unit4_403",
+        "original_id": "unit4_choice_403"
+      },
+      {
+        "id": "unit4_choice_404",
+        "unit": "第四单元：比",
+        "chapter": "相遇问题、比的应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "货车和客车分别从甲、乙两地同时出发相向而行，经过2.4小时相遇，货车的速度是客车速度的 $75\\%$ 。货车与客车所行驶的路程的比是（）。",
+        "options": [
+          "A. 3:4",
+          "B. $4: 3$",
+          "C. $7: 4$",
+          "D. $4: 7$"
+        ],
+        "answer": "A",
+        "solution": "【分析】将客车的速度看成 100 , 货车的速度就是 $100 \\times 0.75$ =75, 然后根据路程 = 速度 × 时间分别求出货车和客车的路程, 再写成比的形式化简即可。此题主要考查学生对比的理解与实际应用。\n【详解】假设客车的速度是 $100$ 。货车速度： $100 \\times 0.75 = 75$客车路程: $100 \\times 2.4 = 240$\n货车路程： $75 \\times 2.4 = 180$ 货车路程：客车路程 $= 180: 240 = 3: 4$ 故答案为: A",
+        "knowledge_points": [
+          "相遇问题",
+          "比的应用"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit4_404",
+        "original_id": "unit4_choice_404"
+      },
+      {
+        "id": "unit4_choice_405",
+        "unit": "第四单元：比",
+        "chapter": "比的意义、比的化简",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "把 $25 \\mathrm{~g}$ 糖放入 $200 \\mathrm{~g}$ 水中溶解成糖水, 那么糖与糖水的比是 ( )。",
+        "options": [
+          "A. $1: 8$",
+          "B. $1: 9$",
+          "C. $1: 7$",
+          "D. $1: 10$"
+        ],
+        "answer": "B",
+        "solution": "【分析】根据比的意义, 比的前项是糖的质量, 比的后项是糖水的质量, 据此写出糖与糖水的比即可。此题考查了比的意义，注意化到最简。\n【详解】25: (200+25), 化简得 $1: 9$ 。故选择：B",
+        "knowledge_points": [
+          "比的意义",
+          "比的化简"
+        ],
+        "source": "20-21 六年级下·全国·单元测试",
+        "merged_id": "unit4_405",
+        "original_id": "unit4_choice_405"
+      },
+      {
+        "id": "unit4_choice_406",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "疫情期间医生们夜以继日、争分夺秒的工作着，他们是最美的逆行者。张医生在某日的工作和休息时间的比是7:5。他这一天工作的时间是（）。",
+        "options": [
+          "A. 14 小时",
+          "B. 10 小时",
+          "C. 7 小时",
+          "D. 5 小时"
+        ],
+        "answer": "A",
+        "solution": "【分析】由比的意义知: 张医生在某日的工作和休息时间的比是 $7: 5$, 则是把张医生一天的时间分成 $7 + 5 = 12$ 等份,工作时间是其中的 7 等份, 据此解答。\n【详解】 $7 \\div (7 + 5) = \\frac{7}{12}$ $2 4 \\times \\frac {7}{1 2} = 1 4 \\text {(时)}$故答案为: A",
+        "knowledge_points": [
+          "比的应用"
+        ],
+        "source": "20-21 六年级上·江苏·单元测试",
+        "merged_id": "unit4_406",
+        "original_id": "unit4_choice_406"
+      },
+      {
+        "id": "unit4_choice_407",
+        "unit": "第四单元：比",
+        "chapter": "比的意义、基础行程问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "从甲地到乙地, 小青要走 20 分钟, 小蓝要走 16 分钟。小青与小蓝的速度之比是 ( )。",
+        "options": [
+          "A. $4: 5$",
+          "B. 5:4",
+          "C. $4: 9$",
+          "D. 9:5"
+        ],
+        "answer": "A",
+        "solution": "【分析】把甲地到乙地的路程看作单位 '1', 根据路程÷时间 = 速度分别求出小青与小蓝的速度, 两者相比即可。考查了比, 解题的关键是明确路程、时间、速度之间的关系。\n【详解】 $\\frac{1}{20}: \\frac{1}{16} = (\\frac{1}{20} \\times 80): (\\frac{1}{16} \\times 80) = 4: 5$ 故答案为：A",
+        "knowledge_points": [
+          "比的意义",
+          "基础行程问题"
+        ],
+        "source": "18-19 六年级下·浙江·期末",
+        "merged_id": "unit4_407",
+        "original_id": "unit4_choice_407"
+      },
+      {
+        "id": "unit4_choice_408",
+        "unit": "第四单元：比",
+        "chapter": "分数除法的应用、按比分配问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "学校把购进的图书的 $\\frac{3}{10}$ 按照 3:2:7 分配给四、五、六年级，已知六年级分到 42本，学校共购进（）本图书。",
+        "options": [
+          "A. 240",
+          "B. 260",
+          "C. 300"
+        ],
+        "answer": "A",
+        "solution": "【分析】根据题意可知, 已知量 42 本对应的份数是 7 份,据此求出每份是多少, 再乘总份数即可求出购进图书的 $\\frac{3}{10}$ 是多少本, 再根据 '购进图书的总本数 $\\times \\frac{3}{10} =$ 总共分的本数' 求出总本数即可。解答本题的关键是求出购进图书的 $\\frac{3}{10}$ 是多少本, 再根据分数除以的意义求出总本数。\n【详解】 $42 \\div 7 \\times (3 + 2 + 7) \\div \\frac{3}{10}\\\\= 7 2 \\div \\frac {3}{1 0} \\\\ = 2 4 0 \\text {(本)} $\n故答案为: A。",
+        "knowledge_points": [
+          "分数除法的应用",
+          "按比分配问题"
+        ],
+        "source": "2021 六年级下·全国·专题练习",
+        "merged_id": "unit4_408",
+        "original_id": "unit4_choice_408"
+      },
+      {
+        "id": "unit4_choice_409",
+        "unit": "第四单元：比",
+        "chapter": "比的意义、比的化简",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "甲数除以乙数商是 0.4 , 甲数与乙数的最简单整数比是 ( )。",
+        "options": [
+          "A. $0.4: 1$",
+          "B. $5: 2$",
+          "C. $4: 10$",
+          "D. $2: 5$"
+        ],
+        "answer": "D",
+        "solution": "【分析】根据甲数除以乙数商是 0.4，将甲数看成 0.4，乙数看成 1，根据比的意义，写出两数比，化简即可。关键是理解比的意义，两数相除又叫两个数的比，化简比要根据比的基本性质化简。\n【详解】 $0.4: 1 = 4: 10 = 2: 5$ 故答案为: D",
+        "knowledge_points": [
+          "比的意义",
+          "比的化简"
+        ],
+        "source": "18-19 六年级上·全国·课后作业",
+        "merged_id": "unit4_409",
+        "original_id": "unit4_choice_409"
+      },
+      {
+        "id": "unit4_choice_410",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "甲、乙两种洗发水的售价比是 $4: 5$, 乙种洗发水售价是 27 元, 甲种洗发水的售价是多少元? ( )",
+        "options": [
+          "A. 33.75",
+          "B. 15",
+          "C. 21.6"
+        ],
+        "answer": "C",
+        "solution": "【分析】根据题意可知，已知量 27 元对应的份数是 5 份，据此求出每份是多少元，再乘甲种洗发水对应的份数即可。解答本题的关键是要明确已知量对应的份数是多少。\n【详解】 $27 \\div 5 \\times 4\\\\= 5. 4 \\times 4 \\\\ = 2 1. 6 \\text {(元)}; $\n故答案为: C",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "20-21 六年级上·辽宁沈阳·期末",
+        "merged_id": "unit4_410",
+        "original_id": "unit4_choice_410"
+      },
+      {
+        "id": "unit4_choice_411",
+        "unit": "第四单元：比",
+        "chapter": "比的意义、比的化简",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "把 $20 \\mathrm{~g}$ 盐放入 $200 \\mathrm{~g}$ 水中，盐和水的质量比是（）。",
+        "options": [
+          "A. $1: 10$",
+          "B. $10: 1$",
+          "C. $1 : {11}$"
+        ],
+        "answer": "A",
+        "solution": "【分析】根据比的意义，写出盐和水的质量比，化简即可。两数相除又叫两个数的比, 化简比根据比的基本性质。\n【详解】 $20: 200 = 1: 10$故答案为: A",
+        "knowledge_points": [
+          "比的意义",
+          "比的化简"
+        ],
+        "source": "20-21 六年级上·吉林松原·期末",
+        "merged_id": "unit4_411",
+        "original_id": "unit4_choice_411"
+      },
+      {
+        "id": "unit4_choice_412",
+        "unit": "第四单元：比",
+        "chapter": "求比值",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "3:5 的前项和后项同时加 5 , 比值 ( )。",
+        "options": [
+          "A. 不变",
+          "B. 变大了",
+          "C. 变小了"
+        ],
+        "answer": "B",
+        "solution": "【分析】求出 $3: 5$ 的比值, 根据题干描述, 将前项和后项同时加 5 , 求出比值, 与 $3: 5$ 的比值比较即可。\n【详解】 3: 5 = 0.6   (3+5) : (5+5)  = 8:10  = 0.8  0.8 > 0.6 故答案为: B 比的前项和后项同时乘或除以相同的数（0 除外），比值不变。",
+        "knowledge_points": [
+          "求比值"
+        ],
+        "source": "20-21 六年级上·吉林松原·期末",
+        "merged_id": "unit4_412",
+        "original_id": "unit4_choice_412"
+      },
+      {
+        "id": "unit4_choice_413",
+        "unit": "第四单元：比",
+        "chapter": "三角形的内角和、按比分配问题、三角形的分类",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "如果一个三角形三个内角度数的比是 $1: 2: 3$，这个三角形是（）三角形。",
+        "options": [
+          "A. 不确定",
+          "B. 钝角",
+          "C. 锐角",
+          "D. 直角"
+        ],
+        "answer": "D",
+        "solution": "【分析】三角形三个内角度数的比是 $1:2:3$, 把三角形的三个角度数分别看作 1 份、2 份、3 份, 则最大角占内角和的 $\\frac{3}{1 + 2 + 3}$, 再根据最大角来判断三角形的类型即可。本题考查按比例分配、以及三角形的类型, 解答本题的关键是掌握按比例分配解决问题的方法。\n【详解】 $180 \\times \\frac{3}{1 + 2 + 3}\\\\= 1 8 0 \\times \\frac {3}{6}\\\\= 90$ (度)\n三角形是直角三角形。故答案为: D",
+        "knowledge_points": [
+          "三角形的内角和",
+          "按比分配问题",
+          "三角形的分类"
+        ],
+        "source": "2014·全国·小升初真题",
+        "merged_id": "unit4_413",
+        "original_id": "unit4_choice_413"
+      },
+      {
+        "id": "unit4_choice_414",
+        "unit": "第四单元：比",
+        "chapter": "比的基本性质",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "把 $4: 5$ 的前项加上8, 要使比值不变, 后项要 ( )。",
+        "options": [
+          "A. 加 8",
+          "B. 加 15",
+          "C. 乘2",
+          "D. 乘3"
+        ],
+        "answer": "D",
+        "solution": "【分析】根据比的基本性质，用（前项+8）÷前项，求出前项扩大的倍数,后项乘相同的倍数即可。比的前项和后项同时乘或除以相同的数（0 除外）,比值不变。\n【详解】$(4 + 8) \\div 4 = 1 2 \\div 4 = 3$后项要乘3。故答案为: D",
+        "knowledge_points": [
+          "比的基本性质"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_414",
+        "original_id": "unit4_choice_414"
+      },
+      {
+        "id": "unit4_choice_415",
+        "unit": "第四单元：比",
+        "chapter": "比的意义、比的化简",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "把20克糖放入100克水中，糖与水的比是（）。",
+        "options": [
+          "A. $1: 5$",
+          "B. 5:1",
+          "C. 1:6",
+          "D. 6:1"
+        ],
+        "answer": "A",
+        "solution": "【分析】写出糖与水的比，化到最简即可。此题考查了比的意义与化简，属于基础类题目。\n【详解】糖与水的比是 $20: 100$ ，化简得 $1: 5$ 。故选择：A",
+        "knowledge_points": [
+          "比的意义",
+          "比的化简"
+        ],
+        "source": "19-20六年级上·湖北武汉·期末",
+        "merged_id": "unit4_415",
+        "original_id": "unit4_choice_415"
+      },
+      {
+        "id": "unit4_choice_416",
+        "unit": "第四单元：比",
+        "chapter": "工程问题、比的化简",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "同修一条路，甲队2小时修7千米，乙队3小时修10千米，甲、乙两队的工作效率的比是（）。",
+        "options": [
+          "A. 21:20",
+          "B. $20: 21$",
+          "C. $7: 10$",
+          "D. $10: 7$"
+        ],
+        "answer": "A",
+        "solution": "【分析】可依据工作效率 $=$ 总做总量 $\\div$ 工作时间, 分别计算出甲乙两队的工作效率, 再相比即可。因为是同修一条路，所以可把修路的米数看作工作总量，结合相关公式来解答。注意可应用比与除法的关系来计算。\n【详解】\n|      | 工作总量 | 工作时间 | 工作效率 |\n| ---- | -------- | -------- | -------- |\n| 甲   | 7千米    | 2小时    | 7/2      |\n| 乙   | 10千米   | 3小时    | 10/3     |\n$\\frac {7}{2}: \\frac {1 0}{3} = \\frac {7}{2} \\times \\frac {3}{1 0} = 2 1: 2 0$故答案为: A。",
+        "knowledge_points": [
+          "工程问题",
+          "比的化简"
+        ],
+        "source": "18-19六年级上·辽宁·单元测试",
+        "merged_id": "unit4_416",
+        "original_id": "unit4_choice_416"
+      },
+      {
+        "id": "unit4_choice_417",
+        "unit": "第四单元：比",
+        "chapter": "比的意义",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "下面的描述错误的是（）。",
+        "options": [
+          "A. 同一种树叶, 长与宽的比值都比较接近。",
+          "B. 树叶长与宽的比值越大, 树叶就越狭长。",
+          "C. 树叶的长一般要沿树叶主叶脉的方向测量, 量出树叶最长部分的长度（含叶柄的长）。"
+        ],
+        "answer": "C",
+        "solution": "【分析】两个同类量中一个量是另一个量的几倍或者几分之几，叫做这两个量的比。也可以把两个数相除，叫做两个数的比。据此解答。首先要理解比值的含义，其次要把树叶近似地看成长方形，去研究树叶的形状与其长和宽的比值的关系。\n【详解】A．同一种树叶，因为形状相近，所以长与宽的比值都比较接近，A正确；  B. 当长与宽的比值越大时, 意味着树叶的长是宽的更大的倍数, 故树叶就越狭长, B 正确;  C. 在测量树叶的长度时, 不应该包含叶柄的长, C 错误。故答案为: C。",
+        "knowledge_points": [
+          "比的意义"
+        ],
+        "source": "19-20六年级上·贵州贵阳·期末",
+        "merged_id": "unit4_417",
+        "original_id": "unit4_choice_417"
+      },
+      {
+        "id": "unit4_choice_418",
+        "unit": "第四单元：比",
+        "chapter": "比的基本性质",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "如果7:9的前项增加21，要使它的比值不变，后项应增加（）。",
+        "options": [
+          "A. 21",
+          "B. 18",
+          "C. 9",
+          "D. 27"
+        ],
+        "answer": "D",
+        "solution": "【分析】比的基本性质，比的前项增加 21，扩大 4 倍，保证比值不变，则比的后项也应扩大 4 倍，变成 36，即可求出增加的值。本题考查比的基本性质：比的前项和后项同时乘或除以同一个不为 0 的数，比值不变。\n【详解】 $7:9 = (7 + 21):36\\\\3 6 - 9 = 2 7$",
+        "knowledge_points": [
+          "比的基本性质"
+        ],
+        "source": "19-20六年级下·江苏·单元测试",
+        "merged_id": "unit4_418",
+        "original_id": "unit4_choice_418"
+      },
+      {
+        "id": "unit4_choice_419",
+        "unit": "第四单元：比",
+        "chapter": "三角形的内角和、按比分配问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "一个等腰三角形，底角和顶角的度数比是 $1: 4$，底角是( )度。",
+        "options": [],
+        "answer": "$30^{\\circ}$",
+        "solution": "【分析】因为等腰三角形两个底角相等，所以这个等腰三角形的 3 个内角的度数比为 $1: 1: 4$, 三角形的内角和是 $180^{\\circ}$, 利用按比例分配的方法求出底角的度数。按比例分配应用题解答方法：把比转化为分数，用分数方法解答。\n【详解】 $180^{\\circ} \\times \\frac{1}{1 + 1 + 4} = 180^{\\circ} \\times \\frac{1}{6} = 30^{\\circ}$",
+        "knowledge_points": [
+          "三角形的内角和",
+          "按比分配问题"
+        ],
+        "source": "2021 六年级下·甘肃兰州·学业考试",
+        "merged_id": "unit4_419",
+        "original_id": "unit4_choice_419"
+      },
+      {
+        "id": "unit4_choice_420",
+        "unit": "第四单元：比",
+        "chapter": "比的意义、求比值、比的化简",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "张师傅8小时做400个零件，工作总量与工作时间的比是( )，比值是( )。",
+        "options": [],
+        "answer": "50:1 50",
+        "solution": "【分析】根据题意写出工作总量与工作时间的比即可，要化成最简整数比；用比的前项除以比的后项求比值即可。明确比的意义以及求比值和化简比的方法是解答本题的关键。\n【详解】工作总量与工作时间的比是 $400: 8 = 50: 1$; $50: 1 = 50 \\div 1 = 50$",
+        "knowledge_points": [
+          "比的意义",
+          "求比值",
+          "比的化简"
+        ],
+        "source": "19-20六年级下·广西南宁·期末",
+        "merged_id": "unit4_420",
+        "original_id": "unit4_choice_420"
+      },
+      {
+        "id": "unit4_choice_421",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "商店运来一批电冰箱，卖了18台，卖出的台数与剩下的台数比是3:2，运来电冰箱_台。",
+        "options": [],
+        "answer": "30",
+        "solution": "【分析】用卖出的台数除以卖出的份数求出每份是多少台,用每份的台数乘总份数即可求出运来电冰箱的总台数。本题关键是根据卖出的台数和剩下的台数比求出每份数，继而求总量。\n【详解】 $18 \\div 3 \\times (3 + 2)\\\\= 6 \\times 5 \\\\ = 3 0 \\text {(台)} $",
+        "knowledge_points": [
+          "比的应用"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit4_421",
+        "original_id": "unit4_choice_421"
+      },
+      {
+        "id": "unit4_choice_422",
+        "unit": "第四单元：比",
+        "chapter": "求比值、比的化简",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "已知 $a \\times 4 = b \\times 6$ (a和 b 均不为零), 则 b : a 写成最简整数比是( ), 比值()。",
+        "options": [],
+        "answer": "$2:3 \\quad \\frac{2}{3}$",
+        "solution": "【分析】根据题意, 可以假设 $b$ 是 1 , 根据 $a, b$ 的等量关系,可以求出 $a$, 再根据比的意义, 求出 $a 、 b$ 的比, 再化简和求比值即可。根据题意，假设一个数的值是多少，再根据题目给出的等量关系式，求出另一个数，就可求出结果。用假设法，可以把复杂问题简单化。\n【详解】假设 $b$ 是 1 , 则 $a \\times 4 = 6$ , $ a = \\frac {6}{4} = \\frac {3}{2}, b: a = 1: \\frac {3}{2} = 2: 3 = \\frac {2}{3} $",
+        "knowledge_points": [
+          "求比值",
+          "比的化简"
+        ],
+        "source": "20-21 六年级下·山东菏泽·期中",
+        "merged_id": "unit4_422",
+        "original_id": "unit4_choice_422"
+      },
+      {
+        "id": "unit4_choice_423",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "一种药水是药粉和水按 1:30 的比配成的，现在有药粉 30 千克，可以配制( )千克的药水。",
+        "options": [],
+        "answer": "930",
+        "solution": "【分析】根据题意得：药粉占 1 份，是 30 千克，水占 30 份，药水则是 $1 + 30 = 31$ 份，用一份的重量乘药水总份数就是药水的重量；据此解答即可。此题主要考查比的意义的灵活运用。关键是明确每份的重量和药水的总份数是多少。\n【详解】 $30 \\times (1 + 30)\\\\= 3 0 \\times 3 1\\\\=930 (千克)$",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "20-21 六年级下·全国·单元测试",
+        "merged_id": "unit4_423",
+        "original_id": "unit4_choice_423"
+      },
+      {
+        "id": "unit4_choice_424",
+        "unit": "第四单元：比",
+        "chapter": "比的意义、因数和倍数的求法、比的应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "学校举行运动会，参加比赛的运动员在95—100人之间，男运动员的人数是女运动员的 $\\frac{3}{4}$ 。男运动员有( )人, 女运动员有( )人。",
+        "options": [],
+        "answer": "42 56",
+        "solution": "【分析】根据'男运动员的人数是女运动员的 $\\frac{3}{4}$'可知，男运动员的人数与女运动员的人数比为 $3: 4$，则总人数必须是总份数 7 的倍数，据此解答即可。\n【详解】 $3 + 4 = 7$，则总人数必须是总份数7的倍数。7 的倍数有:7、14、21、28、35、42、49、56、63、70、77、84、91、98、105…因为总人数在 95—100 人之间，所以总人数为 98 人。解答本题的关键是明确男运动员与女运动员的人数比，进而明确总人数必须是总份数 7 的倍数，求出总人数，再进一步解答。\n$9 8 \\div (3 + 4) \\\\ = 9 8 \\div 7 \\\\ = 1 4 \\text {(人)} $\n男： $14 \\times 3 = 42$ （人）  女: $14 \\times 4 = 56$ (人)",
+        "knowledge_points": [
+          "比的意义",
+          "因数和倍数的求法",
+          "比的应用"
+        ],
+        "source": "20-21六年级下·全国·期中",
+        "merged_id": "unit4_424",
+        "original_id": "unit4_choice_424"
+      },
+      {
+        "id": "unit4_choice_425",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题、长方形的周长、长方形的面积",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "用一根长 $20 \\mathrm{~cm}$ 的铁丝围成一个长与宽的比是 $7:3$ 的长方形, 这个长方形的面积是( ) $\\mathrm{cm}^{2}$ 。",
+        "options": [],
+        "answer": "24",
+        "solution": "【分析】铁丝的长度就是长方形的周长，依据长方形的周长公式求出长和宽的和是 $20 \\div 2 = 10$ （厘米），长与宽分别是7份和3份，再利用按比例分配的方法，求出1份的长度即可求出长和宽的值，进而利用长方形的面积公式 $S = ab$ 即可求解。此题主要考查按比例分配的知识以及长方形的周长和面积的计算方法的灵活应用。\n【详解】 $20 \\div 2 = 10$ （厘米）\n$1 0 \\div (7 + 3)\\\\= 1 0 \\div 1 0\\\\= 1$ (厘米)\n$1 \\times 7 = 7$ (厘米)   $1 \\times 3 = 3$ (厘米)   $3 \\times 7 = 21$ (平方厘米)  这个长方形的面积是 24 平方厘米。",
+        "knowledge_points": [
+          "按比分配问题",
+          "长方形的周长",
+          "长方形的面积"
+        ],
+        "source": "18-19 六年级下·云南曲靖·期末",
+        "merged_id": "unit4_425",
+        "original_id": "unit4_choice_425"
+      },
+      {
+        "id": "unit4_choice_426",
+        "unit": "第四单元：比",
+        "chapter": "求比值、比的化简",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "把 420 米：6 千米,化成最简单的整数比是( )，比值是( )。",
+        "options": [],
+        "answer": "$7: 100 \\quad 0.07$",
+        "solution": "【分析】根据比的基本性质化简整数比即可；用比的前项除以比的后项求比值即可。熟记求比值和化简比的方法，切勿混淆。\n【详解】420米：6千米 $= 420$ 米：6000米 $=(4 2 0 \\div 6 0): (6 0 0 0 \\div 6 0) = 7: 1 0 0;$420 米 : 6 千米 = 420 米 : 6000 米 = 420 ÷ 6000 = 0.07",
+        "knowledge_points": [
+          "求比值",
+          "比的化简"
+        ],
+        "source": "19-20 六年级下·陕西宝鸡·期末",
+        "merged_id": "unit4_426",
+        "original_id": "unit4_choice_426"
+      },
+      {
+        "id": "unit4_choice_427",
+        "unit": "第四单元：比",
+        "chapter": "比的意义、比的化简",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "一份文件，甲要40分钟完成，乙要30分钟完成，甲和乙所用的时间比是( ), 工作效率比是( )。",
+        "options": [],
+        "answer": "4:3 3:4",
+        "solution": "【分析】根据比的意义，写出两人时间比，化简即可，将时间比反过来就是效率比。两数相除又叫两个数的比, 同一份文件, 所用时间越长, 效率越低。\n【详解】 $40:30 = 4:3$一份文件，甲要40分钟完成，乙要30分钟完成，甲和乙所用的时间比是 $4: 3$ ，工作效率比是 $3: 4$ 。",
+        "knowledge_points": [
+          "比的意义",
+          "比的化简"
+        ],
+        "source": "20-21六年级上·广东河源·期末",
+        "merged_id": "unit4_427",
+        "original_id": "unit4_choice_427"
+      },
+      {
+        "id": "unit4_choice_428",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题、最简分数",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "一个最简分数，分子与分母的和是 58，如果分子、分母都加上 7，所得到的新分数化简后是 $\\frac{1}{3}$，原来这个分数是____。",
+        "options": [],
+        "answer": "$\\frac{11}{47}$",
+        "solution": "【分析】根据题意，先求出新分数分子与分母的和，然后求出新分数的分子与分母的总份数以及分子、分母各占总份数的几分之几，进而求出新分数的分子与分母，再分别求出原来分数的分子与分母即可。解答本题的关键是先根据按比例分配的方法求出新分数的分子和分母，进而求出原来的分数分子和分母。\n【详解】 $58 + 7 \\times 2\\\\ = 5 8 + 1 4 \\\\ = 7 2$ \n$ \\\\ 1 + 3 = 4; \\\\ 7 2 \\times \\frac {1}{4} = 1 8; \\\\ 7 2 \\times \\frac {3}{4} = 5 4; \\\\ 1 8 - 7 = 1 1; \\\\ 5 4 - 7 = 4 7; $\n原来这个分数是 $\\frac{11}{47}$ 。",
+        "knowledge_points": [
+          "按比分配问题",
+          "最简分数"
+        ],
+        "source": "20-21 五年级下·全国·单元测试",
+        "merged_id": "unit4_428",
+        "original_id": "unit4_choice_428"
+      },
+      {
+        "id": "unit4_choice_429",
+        "unit": "第四单元：比",
+        "chapter": "比的意义、工程问题",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "完成同一项工作,甲用 8 小时, 乙用 9 小时, 甲、乙工作效率的比是( )。",
+        "options": [],
+        "answer": "9:8",
+        "solution": "【分析】根据比的意义，将时间比反过来就是效率比，据此写出甲乙效率比即可。两数相除又叫两个数的比, 用的时间越少, 工作效率越高。\n【详解】完成同一项工作, 甲用 8 小时, 乙用 9 小时, 甲、乙工作效率的比是 $9: 8$ 。",
+        "knowledge_points": [
+          "比的意义",
+          "工程问题"
+        ],
+        "source": "20-21 六年级上·山东济宁·期末",
+        "merged_id": "unit4_429",
+        "original_id": "unit4_choice_429"
+      },
+      {
+        "id": "unit4_choice_430",
+        "unit": "第四单元：比",
+        "chapter": "比的基本性质、求比值、比的化简",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "$\\frac{3}{4}$ 时：30 分，化成最简整数比是( )，比值是( )。",
+        "options": [],
+        "answer": "$3: 2 \\quad \\frac{3}{2}$",
+        "solution": "【分析】比的性质：比的前项和比的后项乘或除以同一个不为 0 的数，比值不变；比的前后项单位不统一时先转化成相同单位。本题考查比的化简与求值, 解答本题的关键是掌握化简比的方法。\n【详解】 $\\frac{3}{4}$ 时 $= 45$ 分    $\\frac{3}{4} 时：30分= 45  分：30分  = 3:2 = \\frac{3}{2}$",
+        "knowledge_points": [
+          "比的基本性质",
+          "求比值",
+          "比的化简"
+        ],
+        "source": "20-21 六年级上·贵州铜仁·期末",
+        "merged_id": "unit4_430",
+        "original_id": "unit4_choice_430"
+      },
+      {
+        "id": "unit4_choice_431",
+        "unit": "第四单元：比",
+        "chapter": "比的基本性质、分数化小数、比与分数、除法的关系",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "$9 \\div (\\quad) = (\\quad): 16 = \\frac{3}{8} =$ （ ）（填小数）。",
+        "options": [],
+        "answer": "24   6   0.375",
+        "solution": "【分析】把分数转化成小数，用分子除以分母求出；分数的分子、除法的被除数相当于比的前项，分数的分母、除法的除数相当于比的后项，分数值、除法的商相当于比值，据此解答即可。本题考查分数、小数、除法、比的互化，解答本题的关键是掌握比的性质。\n【详解】 $\\frac{3}{8} = 3 \\div 8 = 0.375$$ \\frac {3}{8} = 3 \\div 8 = (3 \\times 3) \\div (8 \\times 3) = 9 \\div 2 4 \\quad \\frac {3}{8} = 3: 8 = (3 \\times 2): (8 \\times 2) = 6: 1 6 $",
+        "knowledge_points": [
+          "比的基本性质",
+          "分数化小数",
+          "比与分数",
+          "除法的关系"
+        ],
+        "source": "20-21 六年级上·全国·期末",
+        "merged_id": "unit4_431",
+        "original_id": "unit4_choice_431"
+      },
+      {
+        "id": "unit4_choice_432",
+        "unit": "第四单元：比",
+        "chapter": "比的意义、求比值、比的化简",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "把 5 克盐溶于 95 克水中, 盐与盐水的比是( ), 比值是( )。",
+        "options": [],
+        "answer": "$1: 20 \\quad \\frac{1}{20}$",
+        "solution": "【分析】由题意可知, 盐水的质量为 $95 + 5 = 100$ (克),进而写出盐与盐水的比和比值即可。解答本题的关键是明确盐水的质量, 切勿将水的质量看出盐水的质量。\n【详解】盐与盐水的比为 $5: (95 + 5) = 5: 100 = 1: 20$; $1: 20 = 1 \\div 20 = \\frac{1}{20}$ 。",
+        "knowledge_points": [
+          "比的意义",
+          "求比值",
+          "比的化简"
+        ],
+        "source": "20-21 六年级上·甘肃金昌·期末",
+        "merged_id": "unit4_432",
+        "original_id": "unit4_choice_432"
+      },
+      {
+        "id": "unit4_choice_433",
+        "unit": "第四单元：比",
+        "chapter": "求比值、比的化简",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "$\\frac{3}{4} \\mathrm{~L}: 250 \\mathrm{~mL}$ 的最简单的整数比是( )，比值是( )。",
+        "options": [],
+        "answer": "3:1 3",
+        "solution": "【分析】根据比的基本性质化简整数比即可，用比的前项除以比的后项求比值即可。熟记求比值和化简比的方法，切勿混淆。\n【详解】 $\\frac{3}{4} \\mathrm{~L}: 250 \\mathrm{~mL} = 750 \\mathrm{~mL}: 250 \\mathrm{~mL} =$$(7 5 0 \\div 2 5 0): (2 5 0 \\div 2 5 0) = 3: 1;$$\\frac {3}{4} \\mathrm {L}: 2 5 0 \\mathrm {m L} = 7 5 0 \\mathrm {m L}: 2 5 0 \\mathrm {m L} = 7 5 0 \\div 2 5 0 = 3$",
+        "knowledge_points": [
+          "求比值",
+          "比的化简"
+        ],
+        "source": "20-21 六年级上·河北张家口·期末",
+        "merged_id": "unit4_433",
+        "original_id": "unit4_choice_433"
+      },{
+        "id": "unit4_choice_434",
+        "unit": "第四单元：比",
+        "chapter": "比的基本性质",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "在 2:3 中, 如果前项加上 4 , 要使比值不变, 后项应加上( )。",
+        "options": [],
+        "answer": "6",
+        "solution": "【分析】根据比的性质，前项和后项同时乘或除以相同的数（0 除外），比值的大小不变。据此即可求解。此题考查比的性质的运用, 掌握比的性质并灵活运用是解题的关键。\n【详解】2:3 的前项加 4 , 可知比的前项由 2 变成了 6 , 相当于前项乘 3 , 要使比值不变, 后项也应该乘 3 , 由 3 变成 9 , 也可以认为是后项加上 $9 - 3 = 6$ 。",
+        "knowledge_points": [
+          "比的基本性质"
+        ],
+        "source": "19-20 六年级上·辽宁·单元测试",
+        "merged_id": "unit4_434",
+        "original_id": "unit4_choice_434"
+      },
+      {
+        "id": "unit4_choice_435",
+        "unit": "第四单元：比",
+        "chapter": "比的意义、因数和倍数的求法、比的应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "四川动物园有熊猫数在 40 只与 50 只之间，雄与雌的数量比是 $5: 6$，这个动物园雄熊猫有( )只，雌熊猫有( )只。",
+        "options": [],
+        "answer": "20 24",
+        "solution": "【分析】根据雌雄数量比，求出熊猫数的总份数。又由于熊猫数是40只与50只之间，由此确定下总的熊猫数，从而求出雌雄熊猫各有多少只。本题考查了比的应用, 能够根据雄雌比以及熊猫总数的范围确定熊猫总数是解题的关键。\n【详解】 $5 + 6 = 11$, 由于 40 到 50 之间只有 44 是 11 的倍数,所以总的熊猫数是 44 只, 平均分成 11 份, 每一份是 4 只。所以, 雄熊猫: $4 \\times 5 = 20$ (只)雌熊猫: $4 \\times 6 = 24$ (只)",
+        "knowledge_points": [
+          "比的意义",
+          "因数和倍数的求法",
+          "比的应用"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_435",
+        "original_id": "unit4_choice_435"
+      },
+      {
+        "id": "unit4_choice_436",
+        "unit": "第四单元：比",
+        "chapter": "比的意义、工程问题",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "一项工程，甲要 12 天完成，乙要 20 天完成，甲和乙所用工作时间的比是()，甲和乙的工作效率的比是( )。",
+        "options": [],
+        "answer": "3:5 5:3",
+        "solution": "【分析】根据题意直接写出甲和乙所用工作时间的比即可;甲的工作效率为 $\\frac{1}{12}$, 乙的工作效率为 $\\frac{1}{20}$, 据此写出甲和乙工作效率比即可。明确甲和乙的工作效率是解答本题的关键。\n【详解】甲和乙所用工作时间的比是 $12: 20 = 3: 5$;甲和乙的工作效率的比是 $\\frac{1}{12}: \\frac{1}{20} = 5: 3$ 。",
+        "knowledge_points": [
+          "比的意义",
+          "工程问题"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_436",
+        "original_id": "unit4_choice_436"
+      },
+      {
+        "id": "unit4_choice_437",
+        "unit": "第四单元：比",
+        "chapter": "比的意义、比的应用",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "动物园里老虎数量是狮子数量的 $\\frac{4}{5}$ 。狮子与老虎数量的比是( )，老虎数量与老虎狮子总数的比是( )。",
+        "options": [],
+        "answer": "5:4 4:9",
+        "solution": "【分析】由题意可知, 狮子数量为 1 , 则老虎数量是 $\\frac{4}{5}$, 由此写出狮子与老虎数量的比和老虎数量与老虎狮子总数的比即可。明确狮子数量和老虎数量是解答本题的关键。\n【详解】狮子与老虎数量的比是 $1: \\frac{4}{5} = 5: 4$;老虎数量与老虎狮子总数的比是 $\\frac{4}{5}: (1 + \\frac{4}{5}) = 4: 9$ 。",
+        "knowledge_points": [
+          "比的意义",
+          "比的应用"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_437",
+        "original_id": "unit4_choice_437"
+      },
+      {
+        "id": "unit4_choice_438",
+        "unit": "第四单元：比",
+        "chapter": "求比值、比的化简",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "跑车俱乐部一辆跑车2小时行驶了300千米，那么这辆跑车行驶的路程和时间的最简单的整数比是( )，比值是( )。",
+        "options": [],
+        "answer": "150:1 150",
+        "solution": "【分析】根据题意，直接做比即可。求比值时，用比的前项除以后项即可。本题考查了比的化简和求值, 求比时, 最终结果要化成最简整数比, 求比值时, 用比的前项除以后项即可。\n【详解】 $300:2 = 1 5 0: 1 = 1 5 0 \\div 1 = 1 5 0$\n所以，路程和时间的最简单的整数比是 $150: 1$ ，比值是 $150$ 。",
+        "knowledge_points": [
+          "求比值",
+          "比的化简"
+        ],
+        "source": "20-21六年级上·全国·单元测试",
+        "merged_id": "unit4_438",
+        "original_id": "unit4_choice_438"
+      },
+      {
+        "id": "unit4_choice_439",
+        "unit": "第四单元：比",
+        "chapter": "求比值、比的化简",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "把 $2.4: 1.8$ 化成最简单的整数比是( )，它的比值是( )。",
+        "options": [],
+        "answer": "$4: 3 \\quad \\frac{4}{3}$",
+        "solution": "【分析】先将 2.4:1.8 的前项和后项同时扩大十倍，变成整数比，之后再化简求值即可。本题考查了比的化简和求比值，比的前项和后项同时是小数时，先将比的前项和后项同时扩大相同的倍数，变成整数比，之后再化简成最简整数比。求比值时，用比的前项除以后项。\n【详解】$2.4:1.8 = (2. 4 \\times 1 0): (1. 8 \\times 1 0) = 2 4: 1 8 = (2 4 \\div 6): (1 8 \\div 6) = 4: 3$\n$4: 3 = 4 \\div 3 = \\frac {4}{3}$\n所以, 2.4:1.8 化成最简单的整数比是 $4:3$, 它的比值是 $\\frac{4}{3}$ 。",
+        "knowledge_points": [
+          "求比值",
+          "比的化简"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_439",
+        "original_id": "unit4_choice_439"
+      },
+      {
+        "id": "unit4_choice_440",
+        "unit": "第四单元：比",
+        "chapter": "求比值、比的化简",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "一个比的前项是 21, 后项是 15 , 最简比写作( )，读作( )，比值是( )。",
+        "options": [],
+        "answer": "$7: 5 \\quad 7$ 比 $5 \\quad \\frac{7}{5}$",
+        "solution": "【分析】比的前项和后项都是整数时，同时除以两项的最大公因数，得到最简整数比。求比值时，用比的前项除以后项即可。本题考查了比的化简和求比值，属于基础题，计算时细心即可。\n【详解】$21:15 = (2 1 \\div 3): (1 5 \\div 3) = 7: 5$\n$7: 5 = 7 \\div 5 = \\frac {7}{5}$所以, 最简整数比是 $7: 5$, 读作 $7$ 比 $5$, 比值为 $\\frac{7}{5}$ 。",
+        "knowledge_points": [
+          "求比值",
+          "比的化简"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_440",
+        "original_id": "unit4_choice_440"
+      },
+      {
+        "id": "unit4_choice_441",
+        "unit": "第四单元：比",
+        "chapter": "比的意义、比的应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "小鹏拥有的蓝卡片与红卡片的比是 $9: 8$, 红卡片比蓝卡片少 $\\frac{6}{6}$; 蓝卡片比红卡片多 $\\frac{6}{6}$ 。",
+        "options": [],
+        "answer": "$\\frac{1}{9}; \\frac{1}{8}$",
+        "solution": "【分析】根据两种卡片的数量比，将蓝卡片看作9，红卡片看作8，求红卡片比蓝卡片少几分之几，用差÷蓝卡片；求蓝卡片比红卡片多几分之几，用差÷红卡片。关键是理解比的意义，将比的前后项看成份数。\n【详解】 $9 - 8 = 1$$1 \\div 9 = \\frac {1}{9}  1 \\div 8 = \\frac {1}{8}$",
+        "knowledge_points": [
+          "比的意义",
+          "比的应用"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_441",
+        "original_id": "unit4_choice_441"
+      },
+      {
+        "id": "unit4_choice_442",
+        "unit": "第四单元：比",
+        "chapter": "比的意义、比的应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "天鸿农场鸡与鸭的比是 $5: 4$ 。表示鸡是 ( ) 份, 鸭是 ( ) 份, 鸡鸭总数是 ( ) 份。鸡是鸡鸭总数的 $\\frac{6}{6}$, 鸭是鸡鸭总数的 $\\frac{6}{6}$ 。",
+        "options": [],
+        "answer": "5; 4; 9; $\\frac{5}{9}$; $\\frac{4}{9}$",
+        "solution": "【分析】根据比的意义 5:4 表示鸡是 5 份, 鸭是 4 份, 鸡鸭总数是 9 份。鸡÷鸡鸭总数 = 鸡是鸡鸭总数的几分之几;鸭÷鸡鸭总数 = 鸭是鸡鸭总数的几分之几。本题是求一个数是另一个数的几分之几, 关键是看把谁当成了单位 '1', 单位 '1'的量为除数。\n【详解】 $5 + 4 = 9$ （份）鸡是 5 份，鸭是 4 份，鸡鸭总数是 9 份。$5 \\div (5 + 4) = 5 \\div 9 = \\frac {5}{9}$$4 \\div (5 + 4) = 4 \\div 9 = \\frac {4}{9}$鸡是鸡鸭总数的 $\\frac{5}{9}$, 鸭是鸡鸭总数的 $\\frac{4}{9}$ 。",
+        "knowledge_points": [
+          "比的意义",
+          "比的应用"
+        ],
+        "source": "20-21六年级上·全国·单元测试",
+        "merged_id": "unit4_442",
+        "original_id": "unit4_choice_442"
+      },
+      {
+        "id": "unit4_choice_443",
+        "unit": "第四单元：比",
+        "chapter": "比的意义、比的化简",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "编程公司今日请假 1人, 出勤 24 人。缺勤人数与总人数的最简整数比是( )。",
+        "options": [],
+        "answer": "1:25",
+        "solution": "【分析】请假 1 人, 出勤 24 人, 总人数是 $1 + 24$, 根据比的意义, 写出缺勤人数与总人数的比即可。关键是理解比的意义，两数相除又叫两个数的比。\n【详解】1: $(1 + 24) = 1:25$",
+        "knowledge_points": [
+          "比的意义",
+          "比的化简"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_443",
+        "original_id": "unit4_choice_443"
+      },
+      {
+        "id": "unit4_choice_444",
+        "unit": "第四单元：比",
+        "chapter": "比的基本性质、比与分数、除法的关系",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "28: ( ) $= 1.4 =$ ( ) $\\div 50 = \\frac{56}{()}$ 。",
+        "options": [],
+        "answer": "20; 70; 40",
+        "solution": "【分析】先将 1.4 化成分数, 再根据分数与除法和比的关系,以及它们通用的基本性质进行填空。比的前项和后项同时乘或除以相同的数（0 除外），比值不变，分数和除法也有相同的性质。\n【详解】 $1.4 = \\frac{7}{5}$ ， $28 \\div 7 \\times 5 = 20$ ； $50 \\div 5 \\times 7 = 70$ ； $56 \\div 7 \\times 5 = 40$",
+        "knowledge_points": [
+          "比的基本性质",
+          "比与分数",
+          "除法的关系"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_444",
+        "original_id": "unit4_choice_444"
+      },
+      {
+        "id": "unit4_choice_445",
+        "unit": "第四单元：比",
+        "chapter": "比的基本性质",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "5:8 的前项扩大到原来的 4 倍，要使比值不变，后项应该是( )；如果前项加 15，要使比值不变，后项应加( )。",
+        "options": [],
+        "answer": "32 24",
+        "solution": "【分析】比的性质：比的前项和比的后项乘或除以同一个不为 0 的数，比值不变，据此解答即可。本题考查比的性质, 解答本题的关键是掌握比的性质。\n【详解】5:8 的前项扩大到原来的 4 倍, 要使比值不变, 后项应该是 32 ; 如果前项加 15 , 要使比值不变, 后项应加 24。",
+        "knowledge_points": [
+          "比的基本性质"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_445",
+        "original_id": "unit4_choice_445"
+      },
+      {
+        "id": "unit4_choice_446",
+        "unit": "第四单元：比",
+        "chapter": "比的意义、求比值、比的化简",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "把 10 克盐溶入 100 克水中，盐和水的比是( )，比值是( )；盐和盐水的比是( )，比值是( )。",
+        "options": [],
+        "answer": "$1: 10 \\quad \\frac{1}{10} \\quad 1: 11 \\quad \\frac{1}{11}$",
+        "solution": "【分析】根据题意, 盐水的质量为 $(100 + 10)$ 克, 由此写出盐和水、盐和盐水的质量比即可, 一定要化成最简整数比; 用比的前项除以比的后项求比值即可;本题属于基础性题目, 一定要熟练掌握比的有关知识。\n【详解】盐和水的比是 $10: 100 = 1: 10$, 比值是 $\\frac{1}{10}$;盐和盐水的比是 $10: (10 + 100) = 1: 11$, 比值是 $\\frac{1}{11}$ 。",
+        "knowledge_points": [
+          "比的意义",
+          "求比值",
+          "比的化简"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_446",
+        "original_id": "unit4_choice_446"
+      },
+      {
+        "id": "unit4_choice_447",
+        "unit": "第四单元：比",
+        "chapter": "比的意义",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "6:13 的前项是( ),后项是( ), 比值是( )。",
+        "options": [],
+        "answer": "6 13 $\\frac{6}{13}$",
+        "solution": "【分析】在两个数的比中，比号前面的数叫做比的前项，比号后面的数叫做比的后项，比的前项除以比的后项所得的商，叫做比值。本题考查比的意义, 解答本题的关键是掌握比的意义。\n【详解】6:13 的前项是 6 , 后项是 13 , 比值是 $\\frac{6}{13}$ 。",
+        "knowledge_points": [
+          "比的意义"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_447",
+        "original_id": "unit4_choice_447"
+      },
+      {
+        "id": "unit4_choice_448",
+        "unit": "第四单元：比",
+        "chapter": "比的意义、基础行程问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "走完同一条路, 小明用 4 小时, 小红用 3 小时, 小明和小红两人速度比是( )。",
+        "options": [],
+        "answer": "3:4",
+        "solution": "【分析】假设路程为单位 1 , 根据 '速度 = 路程 ÷ 时间' 分别求出小明和小红的速度, 之后作比即可。本题考查了比的应用, 解题时, 除了要熟练使用 '速度 = 路程 ÷ 时间' 外, 还应注意比的结果应化简成最简整数比。\n【详解】小明速度： $1 \\div 4 = \\frac{1}{4}$   小红速度: $1 \\div 3 = \\frac{1}{3}$  速度比: $\\frac{1}{4}: \\frac{1}{3} = 3: 4$   所以，小明和小红的速度比为 $3: 4$ 。",
+        "knowledge_points": [
+          "比的意义",
+          "基础行程问题"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_448",
+        "original_id": "unit4_choice_448"
+      },
+      {
+        "id": "unit4_choice_449",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "一道减法算式中差与减数的比是 $4: 5$, 那么差是被减数的( )。",
+        "options": [],
+        "answer": "$\\frac{4}{9}$",
+        "solution": "【分析】要求差是被减数的几分之几，根据加法和减法的关系表示出各自占的份数，进一步列式解答。本题考查被减数、减数及差的关系即：被减数 $=$ 差 + 减数，同时考查：求一个数是另一个数的几分之几。\n【详解】 $4 \\div (4 + 5) = 4 \\div 9 = \\frac {4}{9}$",
+        "knowledge_points": [
+          "比的应用"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_449",
+        "original_id": "unit4_choice_449"
+      },
+      {
+        "id": "unit4_choice_450",
+        "unit": "第四单元：比",
+        "chapter": "求比值、比的化简",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "A÷B=15:18, A 与B 的最简整数比是( )，比值是( )。",
+        "options": [],
+        "answer": "$5: 6 \\quad \\frac{5}{6}$",
+        "solution": "【分析】根据比的基本性质化简比, 比的前项和后项同时乘或除以一个相同的数 (零除外) 。比值不变; 求比值就用比的前项除以后项即可。此题考查的是根据比的基本性质化简比以及求比值的方法，要注意化简比和求比值的区别。\n【详解】 $\\mathrm{A} \\div \\mathrm{B} = 15:18 = (15 \\div 3): (18 \\div 3) = 5:6$  $A \\div B = 1 5: 1 8 = 1 5 \\div 1 8 = \\frac {5}{6}$",
+        "knowledge_points": [
+          "求比值",
+          "比的化简"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_450",
+        "original_id": "unit4_choice_450"
+      },
+      {
+        "id": "unit4_choice_451",
+        "unit": "第四单元：比",
+        "chapter": "求比值、比的化简",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "25 分: 0.75 时化成最简比是( )，比值是( )。",
+        "options": [],
+        "answer": "$5: 9 \\quad \\frac{5}{9}$",
+        "solution": "【分析】根据时分的进率，先将 0.75 时化成分，之后再根据比的性质化简成最简整数比即可；求比值就用比的前项除以后项即可。本题考查了比的性质, 比的前项和后项同时缩小到原来的几分之几, 比的比值不变。\n【详解】0.75时 $= 45$ 分，所以25分：0.15时=25分：45分=5:9  25分：0.15时=25分：45分=5:9= $\\frac{5}{9}$",
+        "knowledge_points": [
+          "求比值",
+          "比的化简"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_451",
+        "original_id": "unit4_choice_451"
+      },
+      {
+        "id": "unit4_choice_452",
+        "unit": "第四单元：比",
+        "chapter": "比的基本性质、比与分数、除法的关系、分数化小数",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "$3: 4 = \\frac{6}{6} = 9 \\div$ ( ) $= (\\quad) \\div 60 = (\\quad) \\quad$ (小数)。",
+        "options": [],
+        "answer": "3；4；12；45；0.75",
+        "solution": "【分析】解答此题的突破口是 $3: 4$, 根据比与分数的关系, $3: 4 = \\frac{3}{4}$; 根据比与除法的关系, $3: 4 = 3 \\div 4$, 再根据商不变的性质, 被除数、除数都乘 3 就是 $9 \\div 12$; 被除数、除数都乘 15 就是 $45 \\div 60$, 结果化成小数即是 0.75。本题主要是考查比、除式、分数、小数之间的关系及转化，利用它们之间的关系和性质进行转化即可。\n【详解】 $3:4 = \\frac{3}{4} = 9\\div (12) = (45)\\div 60 = (0.75)$ 小数",
+        "knowledge_points": [
+          "比的基本性质",
+          "比与分数",
+          "除法的关系",
+          "分数化小数"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_452",
+        "original_id": "unit4_choice_452"
+      },
+      {
+        "id": "unit4_choice_453",
+        "unit": "第四单元：比",
+        "chapter": "比的意义、基础行程问题",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "甲、乙二人匀速行走，如果二人所走的时间相等，那么甲、乙所走的路程比就等于他们的速度比。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】根据：路程 = 速度 × 时间，甲、乙的路程比 = 速度 × 时间的比，如果时间相等，甲、乙的路程比 = 速度比，据此解答。本题考查时间、速度、路程三者的关系，以及比的意义。\n【详解】有分析可知，甲、乙二人均速行走，如果二人所走的时间相等，那么甲、乙所走的路程比就等于他们的速度比，是正确的。故正确答案为:√",
+        "knowledge_points": [
+          "比的意义",
+          "基础行程问题"
+        ],
+        "source": "19-20六年级下·河南省直辖县级单位·期末",
+        "merged_id": "unit4_453",
+        "original_id": "unit4_choice_453"
+      },
+      {
+        "id": "unit4_choice_454",
+        "unit": "第四单元：比",
+        "chapter": "比的意义",
+        "question_type": "判断题",
+        "difficulty": "困难",
+        "question": "两个数相除又叫做两个数的比。（ ）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】根据比的意义，两个数相除又叫做两个数的比。\n【详解】根据比的意义，两个数相除又叫做两个数的比，说法正确。故答案为:√",
+        "knowledge_points": [
+          "比的意义"
+        ],
+        "source": "14-15六年级上·浙江温州·期末",
+        "merged_id": "unit4_454",
+        "original_id": "unit4_choice_454"
+      },
+      {
+        "id": "unit4_choice_455",
+        "unit": "第四单元：比",
+        "chapter": "分数与除法的关系、比的读法、写法及各部分的名称",
+        "question_type": "判断题",
+        "difficulty": "困难",
+        "question": "'$\\frac{1}{2}$'既可以看作分数'二分之一'，也可以看作是比'1:2'。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】两个数的比也可以写出分数的形式, 但是在读时,要读成几比几, 不能读成几分之几, 据此解答即可。本题主要考查了分数与除法的关系, 根据这个关系比也可以写成分数的形式。\n【详解】'$\\frac{1}{2}$'既可以看作分数'二分之一'，也可以看作是比'$1:2$'，原题说法正确；故答案为: √ 。",
+        "knowledge_points": [
+          "分数与除法的关系",
+          "比的读法",
+          "写法及各部分的名称"
+        ],
+        "source": "20-21 六年级上·吉林长春·期末",
+        "merged_id": "unit4_455",
+        "original_id": "unit4_choice_455"
+      },
+      {
+        "id": "unit4_choice_456",
+        "unit": "第四单元：比",
+        "chapter": "三角形的内角和、按比分配问题、三角形的分类",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "一个等腰三角形的一个底角的度数是顶角度数的 $\\frac{1}{2}$, 则它一定是直角三角形。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】一个底角的度数是顶角度数的 $\\frac{1}{2}$, 把底角的度数看作 1 份, 则顶角度数是 2 份, 那么顶角度数占三角形内角和的 $\\frac{2}{1 + 1 + 2}$, 求出顶角度数判断即可。本题考查三角形、按比例分配, 解答本题的关键是掌握按比例分配解决问题的方法。\n【详解】顶角为：$1 8 0 ^ {\\circ} \\times \\frac {2}{1 + 1 + 2} = 9 0 ^ {\\circ}$所以本题说法正确。故答案为: √ 。",
+        "knowledge_points": [
+          "三角形的内角和",
+          "按比分配问题",
+          "三角形的分类"
+        ],
+        "source": "20-21 六年级上·全国·期末",
+        "merged_id": "unit4_456",
+        "original_id": "unit4_choice_456"
+      },
+      {
+        "id": "unit4_choice_457",
+        "unit": "第四单元：比",
+        "chapter": "求比值",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "求下列比的比值。\n$\\frac {4}{3}: \\frac {8}{9} \\quad 0. 6: 0. 0 4 \\quad 0. 2 5: \\frac {1}{5}$\n$ \\frac {7}{2 2}: \\frac {3}{2 2}\\quad0. 4 2: 1 2. 6\\quad \\frac {3}{1 1}: 6$",
+        "options": [],
+        "answer": "$\\frac{3}{2}$; 15; 1.25  $\\frac {7}{3}; \\frac {1}{3 0}; \\frac {1}{2 2}$",
+        "solution": "【分析】求比值时，用比的前项除以后项即可。比值一般用分数表示，也可以用小数或整数表示。\n【详解】 $\\frac{4}{3}:\\frac{8}{9}= \\frac {4}{3} \\div \\frac {8}{9} = \\frac {4}{3} \\times \\frac {9}{8} = \\frac {3}{2};$\n$0.6: 0. 0 4 = 0. 6 \\div 0. 0 4 = 1 5;$\n$0.2 5: \\frac {1}{5} = 0. 2 5 \\div \\frac {1}{5} = 0. 2 5 \\times 5 = 1. 2 5$\n$ \\frac {7}{2 2}: \\frac {3}{2 2} = \\frac {7}{2 2} \\div \\frac {3}{2 2} = \\frac {7}{2 2} \\times \\frac {2 2}{3} = \\frac {7}{3};$\n$0.4 2: 1 2. 6 = 0. 4 2 \\div 1 2. 6 = \\frac {1}{3 0};$\n$\\frac {3}{1 1}: 6 = \\frac {3}{1 1} \\times \\frac {1}{6} = \\frac {1}{2 2}$",
+        "knowledge_points": [
+          "求比值"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_457",
+        "original_id": "unit4_choice_457"
+      },
+      {
+        "id": "unit4_choice_458",
+        "unit": "第四单元：比",
+        "chapter": "比的化简",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "化简下面各比。\n1.5:3.75    $\\frac{3}{2}$ 小时：45分钟    242:22     $\\frac{6}{5}$:24.4",
+        "options": [],
+        "answer": "2:5; 2:1; 11:1; 3:61",
+        "solution": "【分析】化简比的结果应为最简整数比。比的前项和后项同为整数时，前项和后项同时除以最大公因数得到最简整数比；比的前项和后项同为分数时，前项和后项同时乘分母的最小公倍数得到整数比，之后再化成最简整数比。比中含有单位的，先将单位化统一，之后再化简。\n【详解】$1.5:3.75 = (1. 5 \\times 1 0 0): (3. 7 5 \\times 1 0 0) = 1 5 0: 3 7 5 = (1 5 0 \\div 7 5): (3 7 5 \\div 7 5) = 2: 5$\n$\\frac{3}{2}$ 小时：45 分钟$= (\\frac {3}{2} \\times 6 0) \\text {分 钟}: 4 5 \\text {分 钟} = 9 0: 4 5 = (9 0 \\div 4 5): (4 5 \\div 4 5) = 2: 1$\n$2 4 2: 2 2 = (2 4 2 \\div 2 2): (2 2 \\div 2 2) = 1 1: 1$\n$ \\frac {6}{5}: 2 4. 4 = (\\frac {6}{5} \\times 5): (2 4. 4 \\times 5) = 6: 1 2 2 = (6 \\div 2): (1 2 2 \\div 2) = 3: 6 1$",
+        "knowledge_points": [
+          "比的化简"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_458",
+        "original_id": "unit4_choice_458"
+      },
+      {
+        "id": "unit4_choice_459",
+        "unit": "第四单元：比",
+        "chapter": "比的化简",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "化简下列各比。\n$\\frac{8}{7}: 24 \\frac{5}{3}: 1.5$     0.6  平方米：50平方分米      $\\frac{3}{2}: \\frac{15}{7}$",
+        "options": [],
+        "answer": "$1: 21; 10: 9; 6: 5; 7: 10$",
+        "solution": "【分析】比的前项和后项同时乘或除以相同的数（0 除外），比值不变。根据比的基本性质进行化解。\n【详解】 $\\frac{8}{7}: 24 = (\\frac{8}{7} \\times \\frac{7}{8}): (24 \\times \\frac{7}{8}) = 1: 21$$\\frac {5}{3}: 1. 5 = (\\frac {5}{3} \\times 3 0): (1. 5 \\times 3 0) = 5 0: 4 5 = 1 0: 9$$0.6 平方米 : 50 平方分米 = 60 平方分米 : 50 平方分米 =6 0: 5 0 = 6: 5$$\\frac {3}{2}: \\frac {1 5}{7} = (\\frac {3}{2} \\times 1 4): (\\frac {1 5}{7} \\times 1 4) = 2 1: 3 0 = 7: 1 0$",
+        "knowledge_points": [
+          "比的化简"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_459",
+        "original_id": "unit4_choice_459"
+      },
+      {
+        "id": "unit4_choice_460",
+        "unit": "第四单元：比",
+        "chapter": "求比值",
+        "question_type": "计算题",
+        "difficulty": "困难",
+        "question": "求下列各比的比值。\n0.25:1.25        $\\frac{3}{2}: \\frac{5}{6} \\quad$$\\frac{45}{108}$      0.4平方米：5平方分米\n$\\frac {7}{1 0}: \\frac {3}{5}$             $0.1 2 5: 4$        $5: \\frac {2}{5}$       0.1 5: 3. 5",
+        "options": [],
+        "answer": "$\\frac{1}{5}; 1.8; \\frac{5}{12}; 8$  $\\frac {7}{6}, \\frac {1}{3 2}, \\frac {2 5}{2}, \\frac {3}{7 0}$",
+        "solution": "【分析】用比的前项除以后项即可求出比值。\n【详解】 $0.25: 1.25 = 0.25 \\div 1.25 = \\frac{1}{5}$;$ \\frac {3}{2}: \\frac {5}{6} = \\frac {3}{2} \\div \\frac {5}{6} = 1. 8; \\frac {4 5}{1 0 8} = 4 5 \\div 1 0 8 = \\frac {5}{2};$0.4平方米：5平方分米 = 40 平方分米；5平方分米 = 40 ÷ 5= 8;$\\frac {7}{1 0}: \\frac {3}{5} = \\frac {7}{1 0} \\div \\frac {3}{5} = \\frac {7}{6};$$0.1 2 5: 4 = 0. 1 2 5 \\div 4 = \\frac {1}{3 2}; 5: \\frac {2}{5} = 5 \\div \\frac {2}{5} = \\frac {2 5}{2}; \\quad 0. 1 5: 3. 5 = 0. 1 5 \\div 3. 5 = \\frac {3}{7 0}$",
+        "knowledge_points": [
+          "求比值"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_460",
+        "original_id": "unit4_choice_460"
+      },
+      {
+        "id": "unit4_choice_461",
+        "unit": "第四单元：比",
+        "chapter": "等式的性质、解方程",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "解方程。\n$\\frac {4}{5} \\div x = 4 \\quad x: 1. 2 5 = 8 \\quad x - \\frac {4}{5} x = 6 0$",
+        "options": [],
+        "answer": "$x = \\frac{1}{5}; x = 10; x = 300$",
+        "solution": "【分析】根据等式的性质解方程，等式的左右两边加减乘除同一个数（零除外），等式仍然成立。解方程时注意写上'解'字。\n【详解】 $\\frac{4}{5} \\div x = 4$\n解: $x = \\frac{4}{5} \\div 4$\n$x = \\frac {1}{5}$\n$x: 1. 2 5 = 8$\n解: $x = 1.25 \\times 8$\n$x = 1 0$\n$x - \\frac {4}{5} x = 6 0$\n解: $\\frac{1}{5} x = 60$\n$ \\frac {1}{5} x \\div \\frac {1}{5} = 6 0 \\div \\frac {1}{5} x = 3 0 0$",
+        "knowledge_points": [
+          "等式的性质",
+          "解方程"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_461",
+        "original_id": "unit4_choice_461"
+      },
+      {
+        "id": "unit4_choice_462",
+        "unit": "第四单元：比",
+        "chapter": "求比值",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "求下列各比的比值。\n3.2:8      $\\frac{1}{6}: \\frac{2}{9}$      0.125:  $\\frac{5}{8}$       $\\frac{2}{5}$ 时：10分",
+        "options": [],
+        "answer": "$0.4; \\frac{3}{4}; 0.2; \\frac{12}{5}$",
+        "solution": "【分析】求比值时，利用比的前项除以后项即可。比中含有单位时，先统一单位之后再求比值。比值通常用分数表示，也可以用小数或整数表示。\n【详解】 $3.2: 8 = 3.2 \\div 8 = 0.4$;$\\frac {1}{6}: \\frac {2}{9} = \\frac {1}{6} \\div \\frac {2}{9} = \\frac {1}{6} \\times \\frac {9}{2} = \\frac {3}{4};$$0.1 2 5: \\frac {5}{8} = 0. 1 2 5 \\div \\frac {5}{8} = 0. 1 2 5 \\times \\frac {8}{5} = 0. 2;$\n$\\frac{2}{5}$ 时 : $10 \\text{分} = (\\frac{2}{5} \\times 60)$ 分 : $10 \\text{分} = \\frac{2}{5} \\times 60 \\div 10 = \\frac{2}{5} \\times 6 = \\frac{12}{5}$",
+        "knowledge_points": [
+          "求比值"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_462",
+        "original_id": "unit4_choice_462"
+      },
+      {
+        "id": "unit4_choice_463",
+        "unit": "第四单元：比",
+        "chapter": "相遇问题、比的应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "甲车每小时行80千米，甲、乙两车的速度比是4:5，现在两车同时从A、B两地相对开出，3小时相遇，A、B两地相距多少千米？",
+        "options": [],
+        "answer": "540千米",
+        "solution": "【分析】由甲、乙两车的速度比是 $4: 5$, 则乙车的速度是甲车的速度的 $\\frac{5}{4}$, 用甲车速度 $\\times \\frac{5}{4} =$ 乙车速度, 根据速度和 $\\times$ 相遇时间 = 路程。据此列式解答。此题关键是先求出乙车的速度，再根据速度、时间、路程的关系解答。\n【详解】 $(80 \\times \\frac{5}{4} + 80) \\times 3 = (1 0 0 + 8 0) \\times 3 = 1 8 0 \\times 3 = 5 4 0 \\text {(千 米)}$\n答: A、B 两地相距 540 千米。",
+        "knowledge_points": [
+          "相遇问题",
+          "比的应用"
+        ],
+        "source": "19-20六年级下·江西吉安·期末",
+        "merged_id": "unit4_463",
+        "original_id": "unit4_choice_463"
+      },
+      {
+        "id": "unit4_choice_464",
+        "unit": "第四单元：比",
+        "chapter": "分数除法的应用、按比分配问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "小明看一本故事书,第一天看了 36 页, 第二天看了全书的 $\\frac{2}{5}$, 此时已看的页数和未看的页数比是 $7: 3$ 。这本故事书一共有多少页?",
+        "options": [],
+        "answer": "120页",
+        "solution": "【分析】根据已看的页数与未看的页数比是 $7: 3$ 可知：已看的页数是全书的 $\\frac{7}{10}$ ，而第二天看了全书的 $\\frac{2}{5}$ ，可以求出第一天看了全书的 $\\frac{3}{10}$ ，而已知第一天看了 36 页，单位 '1' 未知，用除法，从而计算出全书的页数。\n【详解】根据已看的页数与未看的页数比是 $7: 3$, 可知已看的页数是全书的 $\\frac{7}{10}$;$3 6 \\div (\\frac {7}{1 0} - \\frac {2}{5})= 3 6 \\div \\frac {3}{1 0}=120（页）$；\n答: 这本故事书一共有 120 页。求出两天共看的页数占全书的几分之几是解答本题的关键, 再根据分数除法的意义解答。",
+        "knowledge_points": [
+          "分数除法的应用",
+          "按比分配问题"
+        ],
+        "source": "19-20 六年级下·江西赣州·期末",
+        "merged_id": "unit4_464",
+        "original_id": "unit4_choice_464"
+      },
+      {
+        "id": "unit4_choice_465",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "去年一月，爸爸、李叔叔和陈叔叔三人分别投资了5万、20万、10万元合资办了一个股份公司。今年一月经过核算，公司可用于分配的赢利共有6.3万元。他们三个人合资时约定：'公司每年可用于分配的赢利按个人出资的比例分配。'这样爸爸可以分得赢利多少万元？",
+        "options": [],
+        "answer": "0.9万元",
+        "solution": "【分析】先想爸爸的投资金额占三人投资总金额的分率，得到的分率再乘总的可分配盈利，即可解出答案。本题主要考查的是按照比例分配的应用, 需要牢记解决此类问题先算出分配的分率, 再乘分配的总数, 得出最后的答案。\n【详解】 $6.3 \\div (5 + 20 + 10) \\times 5= 6. 3 \\div 3 5 \\times 5 = 0. 1 8 \\times 5=0.9（万元）$或 $6.3 \\times \\frac{5}{(5 + 20 + 10)}= 6. 3 \\times \\frac {5}{3 5}= 0.9$ （万元）\n答: 这样爸爸可以分得赢利 0.9 万元。",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "19-20六年级下·江西赣州·期末",
+        "merged_id": "unit4_465",
+        "original_id": "unit4_choice_465"
+      },
+      {
+        "id": "unit4_choice_466",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "四年级共有运动员36人，女运动员与男运动员人数的比是1:3，女运动员有多少人？",
+        "options": [],
+        "answer": "9人",
+        "solution": "【分析】用运动员总人数除以总份数即可求出每份多少人,再乘女运动员对应的份数即可。本题考查了按比例分配的知识点, 求出每份多少人是解答本题的关键。\n【详解】 $36 \\div (1 + 3) \\times 1 = 9 \\times 1 = 9 (\\text {人})$\n答：女运动员有9人。",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "20-21 六年级上·山东德州·期末",
+        "merged_id": "unit4_466",
+        "original_id": "unit4_choice_466"
+      }, {
+        "id": "unit4_choice_467",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "农业基地里的樱桃树比苹果树少350棵，樱桃树与苹果树的棵数比是 $3: 5$，基地里的樱桃树和苹果树各有多少棵？",
+        "options": [],
+        "answer": "樱桃树 525 棵，苹果树 875 棵",
+        "solution": "【分析】由题意可知，樱桃树比苹果树的棵数少（5-3）份，已知樱桃树比苹果树少350棵，把350棵平均分成（5-3）份，先用除法求出1份的棵数，再用乘法分别求出3份（樱桃树）、5份（苹果树）棵数即可。此题属于按比例分配问题。由题意可知，樱桃树比苹果树的棵数少（5-3）份，又知樱桃树比苹果树少350棵，关键是求出1份的棵数，进而求出3份、5份的棵数。\n【详解】 $350 \\div (5 - 3)= 3 5 0 \\div 2= 175$ （棵）；$175 \\times 3 = 525$ （棵）；$175 \\times 5 = 875$ （棵）；\n答：基地里的樱桃树 525 棵，苹果树 875 棵。",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit4_467",
+        "original_id": "unit4_choice_467"
+      },
+      {
+        "id": "unit4_choice_468",
+        "unit": "第四单元：比",
+        "chapter": "列方程解含一个未知数的问题、分数与整数的除法、相遇问题、比的应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "甲乙两地相距 800千米，一列客车和一列货车同时从两地出发，相向而行，4小时后两车途中相遇，已知客车和货车的速度比是3:2，两车速度各是多少？",
+        "options": [],
+        "answer": "客车速度每小时 120 千米，货车速度每小时 80 千米",
+        "solution": "【分析】两个车的速度都是未知数，将客车的速度设为 $x$ 千米每小时，根据两车的速度比将货车的速度表示出来。两车相遇时，两车行驶的路程和恰好等于甲乙两地的距离，据此列方程解方程即可。本题考查了比的应用和相遇问题, 两车相遇时,两车行驶的路程和等于甲乙两地的距离。\n【详解】解：设客车的速度为 $x$ 千米每小时，则货车的速度 $\\frac{2}{3} x$ 千米每小时。\n$4 \\times (x + \\frac {2}{3} x) = 8 0 0 \\frac {5}{3} x = 8 0 0 \\div 4 \\frac {5}{3} x = 2 0 0 x = 2 0 0 \\div \\frac {5}{3} x = 1 2 0 \\frac {2}{3} \\times 1 2 0 = 8 0 (\\text {千 米 每 小 时})$\n答: 客车的速度是每小时 120 千米, 货车的速度是每小时 80 千米。",
+        "knowledge_points": [
+          "列方程解含一个未知数的问题",
+          "分数与整数的除法",
+          "相遇问题",
+          "比的应用"
+        ],
+        "source": "20-21 六年级上·贵州黔东南·期末",
+        "merged_id": "unit4_468",
+        "original_id": "unit4_choice_468"
+      },
+      {
+        "id": "unit4_choice_469",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题、工程问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "加工一批零件，王师傅每小时加工36个，与李师傅每小时加工个数的比是 $4:5$，两人共同加工5小时，可以加工多少个零件？",
+        "options": [],
+        "answer": "405 个",
+        "solution": "【分析】根据两位师傅的工作效率比，用王师傅每小时加工个数 $\\div$ 对应份数 $\\times$ 李师傅每小时加工个数的对应份数 $=$ 李师傅每小时加工个数，两位师傅每小时加工个数和 $\\times$ 工作时间 $=$ 加工的总个数。关键是理解比的意义，掌握工作效率、工作时间、工作总量之间的关系。\n【详解】 $36 \\div 4 \\times 5 = 45$ （个）\n$(4 5 + 3 6) \\times 5 = 8 1 \\times 5 = 4 0 5 \\text {(个)}$\n答: 可以加工 405 个零件。",
+        "knowledge_points": [
+          "按比分配问题",
+          "工程问题"
+        ],
+        "source": "2022·安徽合肥·小升初真题",
+        "merged_id": "unit4_469",
+        "original_id": "unit4_choice_469"
+      },
+      {
+        "id": "unit4_choice_470",
+        "unit": "第四单元：比",
+        "chapter": "工程问题、分数的四则混合运算、比的应用",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "修一条路，第一天修了总长的 $\\frac{1}{6}$，第二天修24米，第三天修的与前两天修的比是3:2，还剩 $\\frac{1}{4}$ 没修，这条路全长是多少米？",
+        "options": [],
+        "answer": "180 米",
+        "solution": "【分析】根据题意先将第三天修的占全长的几分之几求出来，再利用减法求出第二天修的占全长的几分之几，最后利用除法求出这条路全长是多少米。本题考查了分数除法的应用, 能够依据题意求出第二天修的占全长的几分之几是解题的关键。\n【详解】 $24 \\div [1 - \\frac{1}{4} - \\frac{1}{6} - (1 - \\frac{1}{4}) \\times \\frac{3}{3 + 2}]$\n$= 2 4 \\div [ 1 - \\frac {1}{4} - \\frac {1}{6} - \\frac {3}{4} \\times \\frac {3}{5} ] = 2 4 \\div [ 1 - \\frac {1}{4} - \\frac {1}{6} - \\frac {9}{2 0} ] = 2 4 \\div \\frac {2}{1 5} = 1 8 0 \\text {(米)}$\n答: 这条路全长是 180 米。",
+        "knowledge_points": [
+          "工程问题",
+          "分数的四则混合运算",
+          "比的应用"
+        ],
+        "source": "19-20六年级上·天津·期中",
+        "merged_id": "unit4_470",
+        "original_id": "unit4_choice_470"
+      },
+      {
+        "id": "unit4_choice_471",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "一种杀虫剂是把药粉和水按照 $1: 100$ 的比配制成的, 如果要配制 5050 克药水,需要药粉和水各多少克？",
+        "options": [],
+        "answer": "药粉 50 克, 水 5000 克",
+        "solution": "【分析】首先求药粉和水的总份数, 再求药粉占药水的几分之几, 最后求出药粉和水的克数, 列式解答即可。此题主要考查按比例分配应用题的特点：已知两个数的比（三个数的比），两个数的和（三个数的和），求这两个数（三个数），用按比例分配解答。\n【详解】总份数： $1 + 100 = 101$ （份），药粉的克数: $5050 \\times \\frac{1}{101} = 50$ (克),水的克数: $5050 \\times \\frac{100}{101} = 5000$ (克),答: 需要药粉 50 克, 水 5000 克。",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "18-19 六年级下·陕西宝鸡·期末",
+        "merged_id": "unit4_471",
+        "original_id": "unit4_choice_471"
+      },
+      {
+        "id": "unit4_choice_472",
+        "unit": "第四单元：比",
+        "chapter": "长方体的体积、按比分配问题、长方体有关棱长的应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "一种长方体箱子的棱长总和是 $168 \\mathrm{~cm}$, 长、宽、高的比是 $3: 2: 1$ 。这个长方体的体积是多少立方厘米？",
+        "options": [],
+        "answer": "2058 立方厘米",
+        "solution": "【分析】先根据棱长总和 $\\div 4$ 求出 1 条长、宽、高的长度和,然后根据长占长宽高之和的 $\\frac{3}{3 + 2 + 1}$, 宽占长宽高之和的 $\\frac{2}{3 + 2 + 1}$,高占长宽高之和的 $\\frac{1}{3 + 2 + 1}$, 求出长宽高的长度, 再利用长方体的体积公式求出长方体的体积即可。本题考查长方体的体积、按比例分配问题, 掌握按比例分配解决问题的方法是解答本题的关键。\n【详解】 $168 \\div 4 = 42$ （厘米）$42 \\times \\frac{3}{3 + 2 + 1} = 21$ (厘米)$42 \\times \\frac{2}{3 + 2 + 1} = 14$ (厘米)$42 \\times \\frac{1}{3 + 2 + 1} = 7$ (厘米)\n$21 \\times 14 \\times 7 = 2058$ (厘米)答: 这个长方体的体积是 2058 立方厘米。",
+        "knowledge_points": [
+          "长方体的体积",
+          "按比分配问题",
+          "长方体有关棱长的应用"
+        ],
+        "source": "20-21 六年级下·浙江·期末",
+        "merged_id": "unit4_472",
+        "original_id": "unit4_choice_472"
+      },
+      {
+        "id": "unit4_choice_473",
+        "unit": "第四单元：比",
+        "chapter": "比的应用、流水行船问题",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "两船分别从上游 A 地和下游 B 地同时相向前进，水的流速是每分 30 米。两船在静水中的速度都是每分行 600 米。两船分别从 AB 两地同时出发，这时水流速度为平时的 3 倍。所以两船相遇的地点比平时相遇点相差 1200 米。求 $A$、$B$ 两地间的水路的长度。",
+        "options": [],
+        "answer": "24000 米",
+        "solution": "【分析】通过题目分析，在水的流速是每分30米的时候，速度比是 $(600 + 30)$ : $(600 - 30) = 21:19$，由于相遇问题，走的时间相同，即可以知道速度比 $=$ 路程比 $= 21:19$ 可以知道从上游到下游的船走了全程的 $\\frac{21}{40}$；当水流速度是平时的3倍时候，即这一天的速度比是 $(600+$30×3）： $(600 - 30 \\times 3) = 23:17$，即路程比也是 $23:17$，此时从上游到下游的船走了全程的 $\\frac{23}{40}$，从上游到下游的船在水速快的时候比水速慢的时候多走了 1200 米，即 1200 米对应分率是 $(\\frac{23}{40} - \\frac{21}{40})$，用 $1200 \\div (\\frac{23}{40} - \\frac{21}{40})$ 即可求出单位'1'，即 A、B 两地间的水路的长度。本题主要考查比的应用, 同时要注意相遇问题同时出发到相遇, 两个船走的时间是相同的, 此时速度比 =路程比。\n【详解】平时速度比： $(600 + 30) : (600 - 30) =6 3 0: 5 7 0 = 2 1: 1 9$由于路程比 $=$ 速度比 $= 21: 19$\n水流速度变快的速度比: $(600 + 30 \\times 3): (600 - 30 \\times 3)= 6 9 0: 5 1 0 = 2 3: 1 7$即此时的路程比 = 速度比 = 23:17A、B两地间的水路的长度:$ 1 2 0 0 \\div (\\frac {2 3}{4 0} - \\frac {2 1}{4 0}) = 1 2 0 0 \\div \\frac {1}{2 0} = 2 4 0 0 0 \\text {(米)}$\n答: A、B 间的水路的长度是 24000 米。",
+        "knowledge_points": [
+          "比的应用",
+          "流水行船问题"
+        ],
+        "source": "20-21 六年级下·浙江·期中",
+        "merged_id": "unit4_473",
+        "original_id": "unit4_choice_473"
+      },
+      {
+        "id": "unit4_choice_474",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "某奥运会一块金牌的总重量是 412 克，其金牌的含金量与金牌的总重比是 6:412，这次奥运会一共制作了 302 块金牌，共消耗黄金多少克？",
+        "options": [],
+        "answer": "1812克",
+        "solution": "【分析】含金量占总重量的 $\\frac{6}{412}$, 先求出 302 块金牌总重量,再求 302 块总重量的 $\\frac{6}{412}$ 是多少即可。此题考查了比的应用, 比可以转化为分数。比号相当于分号。\n【详解】 $6:412 = \\frac{6}{412}$ $3 0 2 \\times 4 1 2 = 1 2 4 4 2 4 \\text {(克)}$ $1 2 4 4 2 4 \\times \\frac {6}{4 1 2} = 1 8 1 2 \\text {(克)}$ 答: 共消耗黄金 1812 克。",
+        "knowledge_points": [
+          "比的应用"
+        ],
+        "source": "2020·浙江·小升初真题",
+        "merged_id": "unit4_474",
+        "original_id": "unit4_choice_474"
+      },
+      {
+        "id": "unit4_choice_475",
+        "unit": "第四单元：比",
+        "chapter": "分数除法的应用、按比分配问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "一个书架有三层，第一层书的本数是总数的 $\\frac{10}{33}$，第二层的书与第三层的比是 $12:11$，其中第二层有 48 本。这个书架一共有多少本书？",
+        "options": [],
+        "answer": "132 本",
+        "solution": "【分析】根据第二层的书与第三层的比，用第二层本数÷对应份数，求出一份数，一份数×第二、三层的总份数，求出第二、三层的总本数，将书架三层总本数看作单位'1'，用第二、三层的总本数÷对应分率即可。关键是理解比和分数除法的意义，按比例分配问题的关键是先求出一份数，部分数量÷对应分率=整体数量。\n【详解】 $48 \\div 12 \\times (12 + 11) \\div (1 - \\frac{10}{33})$\n$= 4 \\times 2 3 \\div \\frac {2 3}{3 3} = 4 \\times 2 3 \\times \\frac {3 3}{2 3} = 1 3 2 \\text {(本)}$答: 这个书架一共有 132 本书。",
+        "knowledge_points": [
+          "分数除法的应用",
+          "按比分配问题"
+        ],
+        "source": "19-20六年级上·内蒙古呼和浩特·期末",
+        "merged_id": "unit4_475",
+        "original_id": "unit4_choice_475"
+      },
+      {
+        "id": "unit4_choice_476",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题、基础行程问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "甲、乙两列火车同时从A、B两城相向开出，4小时相遇。相遇时，两车所行路程的比是3:4，已知甲车每时行60千米，求A、B两城相距多少千米？",
+        "options": [],
+        "answer": "560千米",
+        "solution": "【分析】根据速度 $\\times$ 时间 = 路程, 求出甲车行驶路程, 甲车行驶路程 $\\div$ 对应份数, 求出一份数, 一份数 $\\times$ 总份数即可。关键是理解速度、时间、路程之间的关系，以及比的意义。\n【详解】 $60 \\times 4 \\div 3 \\times (3 + 4)= 8 0 \\times 7=560 (千米)$答: A、B 两城相距 560 千米。",
+        "knowledge_points": [
+          "按比分配问题",
+          "基础行程问题"
+        ],
+        "source": "15-16 六年级·甘肃·期末",
+        "merged_id": "unit4_476",
+        "original_id": "unit4_choice_476"
+      },
+      {
+        "id": "unit4_choice_477",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "84 消毒液是一种以次氯酸钠为主的高效消毒剂，在用于餐具消毒时，原液与水的配比是 $1: 9$, 现在食堂工作人员要配制这种消毒液 $360 \\mathrm{~L}$, 需要84消毒原液多少L?",
+        "options": [],
+        "answer": "$36 \\mathrm{~L}$",
+        "solution": "【分析】原液与水的配比是 $1: 9$, 即一份原液配 9 份水, 原液与消毒液的比为 $1: 10$, 现在要配制消毒液 $360 \\mathrm{~L}$, 根据比的应用，可求出需要消毒原液的量。明确原液与消毒液的比为 $1: 10$ 是解答此题的关键。\n【详解】 $9 + 1 = 10$ $3 6 0 \\div 1 0 \\times 1 = 3 6 (\\mathrm {L})$答: 需要 84 消毒原液 $36 \\mathrm{~L}$ 。",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "20-21 六年级上·浙江·期末",
+        "merged_id": "unit4_477",
+        "original_id": "unit4_choice_477"
+      },
+      {
+        "id": "unit4_choice_478",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "果果和乐乐在'慈善一日捐'活动中为灾区共捐款98元，果果和乐乐捐款的钱数比是 3:4，果果和乐乐各捐款多少元？",
+        "options": [],
+        "answer": "果果捐款42元，乐乐捐款56元",
+        "solution": "【分析】已知两人两人捐款的钱数比是 $3: 4$, 可把两人捐款总钱数平均分成 7 份, 果果占 3 份, 乐乐占 4 份, 按比例分配计算即可。此题考查了按比例分配问题，也可分别求出果果和乐乐的捐款占捐款总数的几分之几，再计算。\n【详解】 $98 \\div (3 + 4) = 9 8 \\div 7 = 1 4 \\text {(元)}$\n如果: $14 \\times 3 = 42$ (元)  乐乐: $14 \\times 4 = 56$ (元)  答: 果果捐款 42 元, 乐乐捐款 56 元。",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "19-20六年级上·河南濮阳·期末",
+        "merged_id": "unit4_478",
+        "original_id": "unit4_choice_478"
+      },
+      {
+        "id": "unit4_choice_479",
+        "unit": "第四单元：比",
+        "chapter": "分数除法的应用、比的意义、按比分配问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "一项工程由甲队单独完成, 需要 60 天; 由乙队单独完成, 需要 40 天。(1) 如果两队全程合作，共同完成这项工程，需要多少天？ (2) 两队合作完工后，按完成的工作量，总共领到15万元工程款，那么甲队、乙队分别能领到多少万元工程款？",
+        "options": [],
+        "answer": "(1) 24 天(2) 甲6万元; 乙9万元",
+        "solution": "【分析】（1）由题可知，是将这项工程看作单位'1'，则甲队的工作效率为 $\\frac{1}{60}$，乙队的工作效率为 $\\frac{1}{40}$，根据'工作总量÷工作效率 = 工作时间'解答即可；(2) 求出甲队和乙队各自完成工作总量的几分之几, 再写出它们之间的比, 按照完成工作总量的比进行分配即可。解答本题的关键是明确单位 '1', 进而求出他们的工作效率和需要的时间; 第二问中, 根据各自完成工作总量的多少来分配工程款最合理。\n【详解】 (1) $1 \\div ( \\frac{1}{60} + \\frac{1}{40} ) = 1 \\div \\frac {1}{2 4} = 2 4 \\text {(天)};$\n答：共同完成这项工程，需要24天。\n(2) 甲队完成: $24 \\times \\frac{1}{60} = \\frac{2}{5}$;乙队完成: $24 \\times \\frac{1}{40} = \\frac{3}{5}$;甲队和乙队完成工作总量的比为 $\\frac{2}{5}: \\frac{3}{5} = 2: 3$;\n$ 1 5 \\div (2 + 3) = 1 5 \\div 5 = 3 (\\text {万 元});$\n$2 \\times 3 = 6 (\\text {万 元});$$3 \\times 3 = 9$ （万元）；答: 甲队领到 6 万元, 乙队领到 9 万元。",
+        "knowledge_points": [
+          "分数除法的应用",
+          "比的意义",
+          "按比分配问题"
+        ],
+        "source": "20-21 六年级上·浙江温州·期末",
+        "merged_id": "unit4_479",
+        "original_id": "unit4_choice_479"
+      },
+      {
+        "id": "unit4_choice_480",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题、三角形的面积",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "用96厘米长的铁丝围成一个直角三角形，这个直角三角形三条边的长度比是3:4:5，这个三角形的面积是多少？",
+        "options": [],
+        "answer": "384平方厘米",
+        "solution": "【分析】直角三角形三条边最长的是斜边，求出两条直角边的长度，再根据三角形的面积公式求出三角形的面积；一条直角边的长度占铁丝总长度的 $\\frac{3}{3 + 4 + 5}$，另一条直角边的长度占铁丝总长度的 $\\frac{4}{3 + 4 + 5}$，据此求出两条直角边的长度，再求出三角形的面积即可。本题考查按比例分配问题、三角形面积，解答本题的关键是掌握按比例分配解题的方法。\n【详解】 $96 \\times \\frac{3}{3 + 4 + 5} = 24$ （厘米）$96 \\times \\frac{4}{3 + 4 + 5} = 32$ （厘米）\n$3 2 \\times 2 4 \\div 2 = 7 6 8 \\div 2 =384 (平方厘米)$\n答: 这个三角形面积是 384 平方厘米。",
+        "knowledge_points": [
+          "按比分配问题",
+          "三角形的面积"
+        ],
+        "source": "18-19六年级上·湖北宜昌·期中",
+        "merged_id": "unit4_480",
+        "original_id": "unit4_choice_480"
+      },
+      {
+        "id": "unit4_choice_481",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题、相遇问题、比的应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "两地相距 450 千米, 客车和货车同时从两地相向而行，经过3小时相遇，客车和货车速度比是 7:5，两车的速度各是多少？",
+        "options": [],
+        "answer": "客车的速度为 87.5 千米/时，货车的速度为 62.5 千米/时",
+        "solution": "【分析】用 $450 \\div 3$ 求出两车的速度和, 再除以总份数求出每份是多少, 再乘两车速度各自对应的份数即可。求出两车的速度和是解答本题的关键。\n【详解】 $450 \\div3 \\div(7+5) = 1 5 0 \\div 1 2 =12.5$\n$12.5 \\times 7 = 87.5$ （千米/时） $12.5 \\times 5 = 62.5$ （千米/时）答: 客车的速度为 87.5 千米/时, 货车的速度为 62.5 千米/时。",
+        "knowledge_points": [
+          "按比分配问题",
+          "相遇问题",
+          "比的应用"
+        ],
+        "source": "20-21 六年级上·全国·期末",
+        "merged_id": "unit4_481",
+        "original_id": "unit4_choice_481"
+      },
+      {
+        "id": "unit4_choice_482",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "某班有学生45 人, 男生人数和女生人数的比是 $5: 4$, 男生女生各有多少人?",
+        "options": [],
+        "answer": "男生有 25 人，女生有 20 人",
+        "solution": "【分析】把总人数看作单位'1'，则男生人数就是5份，女生人数就是 4 份, 则总人数就是 $5 + 4 = 9$ 份, 先求出一份是多少即可解答。此题考查按比例分配的知识, 关键是求出一份是多少。\n【详解】 $45 \\div (5 + 4) = 4 5 \\div 9 =5$\n$= 5$ （人）\n$ 5 \\times 5 = 2 5 (\\text {人}) \\quad 5 \\times 4 = 2 0 (\\text {人})$ 答: 男生有 25 人, 女生有 20 人。",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "20-21 六年级上·内蒙古呼伦贝尔·期末",
+        "merged_id": "unit4_482",
+        "original_id": "unit4_choice_482"
+      },
+      {
+        "id": "unit4_choice_483",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "学校计划绿化一块 240 平方米的空地，先划出总面积的 $\\frac{5}{12}$ 种树，剩下的地按 4:3 的比种草和种花。种草和种花的面积各是多少平方米？",
+        "options": [],
+        "answer": "种草: 80 平方米, 种花: 60 平方米",
+        "solution": "【分析】把总面积看作单位 '1', 剩余部分的面积为总面积的 $(1 - \\frac{5}{12})$, 根据一个数乘分数的意义, 用乘法求出剩余部分的面积; 进而把 '剩下的地按 4:3 的比种草和种花' 理解为种草的面积占剩下的地 $\\frac{4}{4 + 3}$, 种花的面积占剩下的地 $\\frac{3}{4 + 3}$, 用乘法解答即可。此题先求出剩下的面积是多少平方米, 再根据按比例分配的知识解答。\n【详解】 $240 \\times (1 - \\frac{5}{12}) = 2 4 0 \\times \\frac {7}{1 2} = 140（平方米）$\n$140 \\times \\frac{4}{4 + 3} = 80$    $140 \\times \\frac{3}{4 + 3} = 60$ （平方米）\n答: 种草的面积是 80 平方米, 种花的面积 60 平方米。",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "20-21 六年级上·甘肃武威·期末",
+        "merged_id": "unit4_483",
+        "original_id": "unit4_choice_483"
+      },
+      {
+        "id": "unit4_choice_484",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "由于新冠疫情, 天鸿小学开学将进行消毒。用消毒液与水按 1:200 的比例，如果配制 402 千克消毒水，需要消毒液多少千克？",
+        "options": [],
+        "answer": "2千克",
+        "solution": "【分析】消毒水是消毒液和水的总和，根据消毒液和水的比求出消毒液占消毒水的几分之几，之后利用乘法求出消毒液的重量即可。本题考查了比的应用, 能够根据题意求出消毒液是消毒水的几分之几是解题的关键。\n【详解】 $402 \\times \\frac{1}{1 + 200} = 4 0 2 \\times \\frac {1}{2 0 1} = 2$ （千克）\n答: 需要消毒液 2 千克。",
+        "knowledge_points": [
+          "比的应用"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_484",
+        "original_id": "unit4_choice_484"
+      },
+      {
+        "id": "unit4_choice_485",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "百果园运来苹果和梨子两种水果，苹果和梨子的质量比是4:5，苹果是200千克，梨子是多少千克？",
+        "options": [],
+        "answer": "250千克",
+        "solution": "【分析】根据苹果和梨子的比，以及苹果的重量，直接求出梨子的重量。本题考查了比的应用, 能够根据苹果和梨子的比求出梨子的重量是解题的关键。\n【详解】 $200 \\div 4 \\times 5 = 5 0 \\times 5 =250 (千克)$\n答: 梨子是 250 千克。",
+        "knowledge_points": [
+          "比的应用"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_485",
+        "original_id": "unit4_choice_485"
+      },
+      {
+        "id": "unit4_choice_486",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "两班共捐出720本书，六（1）班有学生46名，六（2）班有学生44名。如果每位学生捐出本数相同，两个班各应捐出多少本书？",
+        "options": [],
+        "answer": "六（1）班368本；六（2）班352本",
+        "solution": "【分析】由于每个学生捐出的本数相同，所以六（1）班和六（2）班的学生人数比就等于捐书的数量比。据此求出两个班级捐书的数量分别占总数的几分之几，最后利用乘法分别求出两个班的捐书数量。本题考查了比的应用, 能够根据题意求出两个班级的捐书数量比是解题的关键。\n【详解】数量比： $46: 44 = 23: 22$\n六（1）班：$ 7 2 0 \\times \\frac {2 3}{2 3 + 2 2} = 7 2 0 \\times \\frac {2 3}{4 5} = 3 6 8 \\text {(本)}$\n六 (2) 班: $720 - 368 = 352$ (本)答: 六 (1) 班和六 (2) 班各应捐出 368 本和 352 本。",
+        "knowledge_points": [
+          "比的应用"
+        ],
+        "source": "20-21六年级上·全国·单元测试",
+        "merged_id": "unit4_486",
+        "original_id": "unit4_choice_486"
+      },
+      {
+        "id": "unit4_choice_487",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "近年来网球运动越来越受欢迎，甲、乙两校网球俱乐部原来的球迷人数之比是2:5。后来甲校球迷人数增加36人，乙校球迷人数不变，现在甲、乙两校球迷的人数比为5:8。现在甲、乙两校各有网球球迷多少人？",
+        "options": [],
+        "answer": "甲校100人；乙校160人",
+        "solution": "【分析】因为乙校球迷人数不变，将前后两个比统一成后项相同的比，即每一份的多少相同，甲校增加36人，增加了25-16份，先求出一份数，再用份数分别乘现在两校的份数即可。关键是将两个比进行统一，再按比例进行分配。\n【详解】 $2:5 = 16:40;5:8 = 25:40$\n$ 3 6 \\div (2 5 - 1 6) = 3 6 \\div 9 = 4 \\text {(人)}$\n$4 \\times 2 5 = 1 0 0 (\\text {人})$ $4 \\times 4 0 = 1 6 0 (\\text {人})$ 答: 甲校有网球球迷 100 人, 乙校有网球球迷 160 人。",
+        "knowledge_points": [
+          "比的应用"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_487",
+        "original_id": "unit4_choice_487"
+      },
+      {
+        "id": "unit4_choice_488",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "小新家里有一袋总质量是 420 克的铜和锌，现在按锌、铜比是 $1:2$ 分类，但是需要添加 60 克铜。这袋中原有锌、铜各多少克？",
+        "options": [],
+        "answer": "锌 160 克; 铜 260 克",
+        "solution": "【分析】先求出后来锌和铜的总克数，再根据'锌和铜按1:2 分类, '把锌看作 1 份, 铜看作 2 份, 由此即可求出 1份，即锌的克数，进而求出现有铜的克数，再减60克就是原有铜的质量。关键是先求出后来两种金属的总量，再用按比例分配的方法，求出锌的含量，进而求出铜的含量。\n【详解】 $(420 + 60) \\div (1 + 2) = 4 8 0 \\div 3 = 1 6 0 \\text {(克)}$\n$1 6 0 \\times 2 - 6 0 = 3 2 0 - 6 0 = 2 6 0 \\text {(克)}$\n答: 这袋中原有锌 160 克、铜 260 克。",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_488",
+        "original_id": "unit4_choice_488"
+      },
+      {
+        "id": "unit4_choice_489",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "老张和小张两人月工资之比是 $5: 7$，小张每月比老张多 480 元。小张月工资是多少元？",
+        "options": [],
+        "answer": "1680元",
+        "solution": "【分析】把老张的工资看作 5 份, 小张的工资就是 7 份, 相差 2 份, 2 份对应的是 480 元, 用 $480 \\div 2$ 求出 1 份的数量,再乘 7 即可。此题关键是根据具体的数量差除以份数差，得出1份的数量是多少。\n【详解】 $480 \\div (7 - 5) \\times 7 = 4 8 0 \\div 2 \\times 7 = 1 6 8 0 (\\text {元})$\n答: 小张月工资是 1680 元。",
+        "knowledge_points": [
+          "比的应用"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_489",
+        "original_id": "unit4_choice_489"
+      },
+      {
+        "id": "unit4_choice_490",
+        "unit": "第四单元：比",
+        "chapter": "求一个数的几分之几的问题、按比分配问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "实验小学图书馆共有 540 儿童读物，其中儿童作文书占 $\\frac{1}{3}$，其余是儿童绘画书和儿童音乐书，儿童绘画书和儿童音乐书的比是 $3:2$，儿童绘画书有多少本？",
+        "options": [],
+        "answer": "216 本",
+        "solution": "【分析】将总本数看作单位 '1', 用总本数 $\\times$ 儿童作文书的对应分率, 求出儿童作文书的本数, 总本数 - 儿童作文书的本数 = 儿童绘画书和儿童音乐书的本数, 根据, 儿童绘画书和儿童音乐书的比，两种书共 $3 + 2$ 份，先求出一份数，再用一份数 $\\times$ 儿童绘画书的对应份数即可。关键是理解分数乘法和比的意义，先求出儿童绘画书和儿童音乐书的数量。\n【详解】 $540 - 540 \\times \\frac{1}{3}$\n$ = 5 4 0 - 1 8 0 = 3 6 0 \\text {(本)}$\n$ 3 6 0 \\div (3 + 2) \\times 3 = 3 6 0 \\div 5 \\times 3 = 2 1 6 \\text {(本)}$\n答: 儿童绘画书有 216 本。",
+        "knowledge_points": [
+          "求一个数的几分之几的问题",
+          "按比分配问题"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_490",
+        "original_id": "unit4_choice_490"
+      },
+      {
+        "id": "unit4_choice_491",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "小王家要用混凝土盖房子，一种混凝土要把水泥、沙子、石子按2:4:5的比例配制。要配制132吨这样的混凝土，需要水泥、沙子、石子各多少吨？",
+        "options": [],
+        "answer": "水泥24吨；沙子48吨；石子60吨",
+        "solution": "【分析】'水泥、沙子和石子的比是 $2: 4: 5$', 水泥就占混凝土总数的 $\\frac{2}{2 + 4 + 5}$, 沙子就占混凝土总数的 $\\frac{4}{2 + 4 + 5}$, 石子就占混凝土总数的 $\\frac{5}{2 + 4 + 5}$; 混凝土 132 吨, 分别用乘法解答即可。本题的关键是根据比与分数的关系求出水泥、沙子、石子各占总数的几分之几, 再根据分数乘法的意义列式解答。\n【详解】 $132 \\times \\frac{2}{2 + 4 + 5} = 24$ （吨） $132 \\times \\frac{4}{2 + 4 + 5} = 48$ (吨)   $132 \\times \\frac{5}{2 + 4 + 5} = 60$ (吨)\n答: 需要水泥 24 吨、沙子 48 吨、石子 60 吨。",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "20-21六年级上·全国·单元测试",
+        "merged_id": "unit4_491",
+        "original_id": "unit4_choice_491"
+      },
+      {
+        "id": "unit4_choice_492",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "超市配制消毒水消灭新冠病毒，这种消毒水是用药粉和水按照 $1: 100$ 配制成的。要配制这种消毒水 3030 克，需要药粉和水各多少克？",
+        "options": [],
+        "answer": "药粉30克；水3000克",
+        "solution": "【分析】根据比与分数的关系知: 药粉就占了这种消毒水的 $\\frac{1}{1 + 100}$, 水就占了这种消毒水的 $\\frac{100}{1 + 100}$, 已知要配制这种消毒水 3030 克, 用乘法分别计算即可求出需要药粉和水各多少千克。本题的关键是根据比与分数的关系, 求出水和药粉各占了消毒水的几分之几, 再根据分数乘法的意义列式解答。\n【详解】 $3030 \\times \\frac{1}{1 + 100} = 30$ （克） $3030 \\times \\frac{100}{1 + 100} = 3000$ (克)  答: 需要药粉 30 克, 水 3000 克。",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_492",
+        "original_id": "unit4_choice_492"
+      },
+      {
+        "id": "unit4_choice_493",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "学校买回 1000 本新图书，准备把其中的 $\\frac{2}{5}$ 借给高年级同学。剩下的书按 3:2 借给中年级和低年级同学。高年级、中年级和低年级同学各借了多少本书？",
+        "options": [],
+        "answer": "高年级 400 本；中年级 360 本；低年级 240 本",
+        "solution": "【分析】用 $1000 \\times \\frac{2}{5}$ 求出高年级借的本数；用总本数减去高年级借的本数，再除以总份数即可求出每份多少本，再乘中年级和低年级各自对应的份数，即可求出中年级和低年级同学借的本数，由此解答即可。求出中年级和低年级共借多少本，进而求出每份是多少本是解答本题的关键。\n【详解】 $1000 \\times \\frac{2}{5} = 400$ （本）；$(1 0 0 0 - 4 0 0) \\div (3 + 2) = 6 0 0 \\div 5 = 1 2 0 \\text {(本)};$\n$ 1 2 0 \\times 3 = 3 6 0 (\\text {本}); \\quad 1 2 0 \\times 2 = 2 4 0 (\\text {本});$答: 高年级 400 本, 中年级 360 本, 低年级 240 本.",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_493",
+        "original_id": "unit4_choice_493"
+      },
+      {
+        "id": "unit4_choice_494",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题、长方形的面积",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "一个长方形的周长是 60 厘米，长和宽的比是 $2: 1$，这个长方形的面积是多少平方厘米？",
+        "options": [],
+        "answer": "200平方厘米",
+        "solution": "【分析】先用周长 $\\div 2$, 求出 1 组长和宽的和, 根据长和宽的比可知, 长和宽共 $2 + 1$ 份, 先求出一份数, 用一份数分别乘长和宽的分数, 求出长和宽, 再根据长方形面积公式计算即可。关键是理解比的意义，掌握长方形的周长和面积公式。\n【详解】 $60 \\div 2 \\div (2 + 1) = 3 0 \\div 3 = 10$ (厘米)\n$ 1 0 \\times 2 = 2 0 (\\text {厘 米}) \\quad 1 0 \\times 1 = 1 0 (\\text {厘 米}) \\quad20 \\times 10 = 200$ （平方厘米）答: 这个长方形的面积是 200 平方厘米。",
+        "knowledge_points": [
+          "按比分配问题",
+          "长方形的面积"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_494",
+        "original_id": "unit4_choice_494"
+      },
+      {
+        "id": "unit4_choice_495",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题、相遇问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "甲乙两车同时从两地出发，相同而行，两地相距480千米，经过3小时相遇，甲乙两车的速度比为 $3: 5$ 。甲乙两车的速度分别是多少千米/小时?",
+        "options": [],
+        "answer": "甲车每小时行 60 千米；乙车每小时行 100 千米",
+        "solution": "【分析】甲乙两地相距 480 千米, 3 小时后相遇, 则两车的速度和为每小时: $480 \\div 3 = 160$ 千米, 又甲、乙两车的速度比是 $3: 5$, 则甲车的速度是两车速度和的 $\\frac{3}{3 + 5}$, 由此可知,甲车的速度是每小时 $160 \\times \\frac{3}{3 + 5}$ 千米, 进而即能求出乙车的速度。首先根据路程 $\\div$ 相遇时间 $=$ 速度和, 求出两车的速度和是完成本题的关键。\n【详解】 $(480 \\div 3) \\times \\frac{3}{3 + 5}=60 (千米/小时)$\n$ 4 8 0 \\div 3 - 6 0 = 1 6 0 - 6 0 =100（千米/小时）$\n答: 甲车每小时行 60 千米, 乙车每小时行 100 千米。",
+        "knowledge_points": [
+          "按比分配问题",
+          "相遇问题"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_495",
+        "original_id": "unit4_choice_495"
+      },
+      {
+        "id": "unit4_choice_496",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "小明有邮票 46 张，小军有邮票 38 张，小军给小明多少张邮票后，小明和小军的邮票数之比是 $4: 3$?",
+        "options": [],
+        "answer": "2 张",
+        "solution": "【分析】根据题意先求出小明和小军的邮票总数，之后根据邮票比，计算出小军给了小明邮票后的剩下的邮票数。将小明原有邮票数和剩下的邮票数做差求出小军给小明的邮票数。本题考查了比的应用, 能够根据邮票变化后的比求出小军余下的邮票是解题的关键。\n【详解】小军剩下的邮票数：\n$(4 6 + 3 8) \\times \\frac {3}{3 + 4} = 8 4 \\times \\frac {3}{7} = 3 6 \\text {(张)}$\n小军给小明: $38 - 36 = 2$ (张)答: 小军给小明 2 张邮票后, 小明和小军的邮票数之比是 $4: 3$ 。",
+        "knowledge_points": [
+          "比的应用"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_496",
+        "original_id": "unit4_choice_496"
+      },
+      {
+        "id": "unit4_choice_497",
+        "unit": "第四单元：比",
+        "chapter": "比的应用、长方体的体积",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "商场要给长方体玻璃柜台各边安上铝合金条，共用铝合金 $128 \\mathrm{~m}$，已知柜台的长宽高的比为 $4:3:1$。这个柜台的体积是多少？",
+        "options": [],
+        "answer": "768立方米",
+        "solution": "【分析】长方体共有长宽高各 4 条, 根据长宽高的比以及共用铝合金的总长, 分别将这个柜台的长宽高求出来, 之后求出长方体的体积即可。本题考查了比的应用和长方体的体积, 熟练使用长方体的体积公式是解题的关键。\n【详解】一组长宽高的长度: $128 \\div 4 = 32$ (米)\n长: $32 \\times \\frac{4}{4 + 3 + 1} = 3 2 \\times \\frac {4}{8} = 1 6 \\text {(米)}$\n宽: $32 \\times \\frac{3}{4 + 3 + 1} = 3 2 \\times \\frac {3}{8} = 1 2 \\text {(米)}$\n高: $32 - 16 - 12 = 1 6 - 1 2 = 4 \\text {(米)}$\n柜台体积： $16 \\times 12 \\times 4 = 1 9 2 \\times 4 = 7 6 8 \\text {(立 方 米)}$\n答: 这个柜台的体积是 768 立方米。",
+        "knowledge_points": [
+          "比的应用",
+          "长方体的体积"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_497",
+        "original_id": "unit4_choice_497"
+      },
+      {
+        "id": "unit4_choice_498",
+        "unit": "第四单元：比",
+        "chapter": "按比分配问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "刘阿姨这个月加班20天，王阿姨这个月加班25天，两人一共可得加班费2700元，请你想想怎样分配这笔加班费呢？",
+        "options": [],
+        "answer": "刘阿姨 1200 元; 王阿姨 1500 元",
+        "solution": "【分析】刘阿姨与王阿姨的加班天数比为 $4: 5$, 可以按加班天数的比来分配 2700 元的加班费, 据此解答即可。本题考查按比例分配问题, 解答本题的关键是掌握按比例分配解答问题的方法。\n【详解】 $20: 25 = 4: 5$ 刘阿姨与王阿姨的加班天数比为 $4: 5$, 可以按加班天数的比来分配 2700 元的加班费。\n刘阿姨: $2700 \\times \\frac{4}{4 + 5} = 1200$ (元)王阿姨: $2700 \\times \\frac{5}{4 + 5} = 1500$ (元)答: 刘阿姨分得 1200 元加班费, 王阿姨分得 1500 元加班费。",
+        "knowledge_points": [
+          "按比分配问题"
+        ],
+        "source": "20-21六年级上·全国·单元测试",
+        "merged_id": "unit4_498",
+        "original_id": "unit4_choice_498"
+      },
+      {
+        "id": "unit4_choice_499",
+        "unit": "第四单元：比",
+        "chapter": "比的应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "某什锦糖果要将话梅糖和软糖按5:3的比混合，超市有20千克话梅糖和10.5千克软糖，软糖正好用完，话梅糖还剩多少千克？",
+        "options": [],
+        "answer": "2.5千克",
+        "solution": "【分析】由于软糖正好用完时 3 份对应 10.5 千克, 其中的一份是 3.5 , 则话梅糖的 5 份为 17.5 , 再用话梅糖的总量减去用掉的量得到剩下的话梅糖数量。本题考查了比的应用, 能够根据题意列式求出用掉了的话梅数是解题的关键。\n【详解】用掉的话梅糖：$ 5 \\times (1 0. 5 \\div 3) = 5 \\times 3. 5 = 1 7. 5 (\\text {千 克})$还剩的话梅糖: $20 - 17.5 = 2.5$ (千克)\n答: 话梅糖还剩 2.5 千克。",
+        "knowledge_points": [
+          "比的应用"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_499",
+        "original_id": "unit4_choice_499"
+      },
+      {
+        "id": "unit4_choice_500",
+        "unit": "第四单元：比",
+        "chapter": "比的应用、三角形的内角和",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "在等腰三角形ABC中，顶角与一个底角的比是 $2:7$，则顶角和底角分别是多少度？",
+        "options": [],
+        "answer": "顶角 $22.5^{\\circ}$; 底角 $78.75^{\\circ}$",
+        "solution": "【分析】由于等腰三角形的两个底角是相等的，所以顶角和两个底角的比为：2:7:7。又因为三角形的内角和是$180^{\\circ}$, 平均分成 16 份, 每份是 $180 \\div 16 = 11.25^{\\circ}$, 据此求出顶角和底角分别是多少度即可。本题考查了比的应用。在做这类题目时，除了要掌握比的意义以外，还要熟练使用三角形的相关结论，比如，等腰三角形的底角是相等的，三角形的内角和是 $180^{\\circ}$ 。\n【详解】据题易知，顶角和两个底角的比为：2:7:7,每一份的度数: $180 \\div 16 = 11.25$ (度)顶角度数: $11.25 \\times 2 = 22.5$ (度)底角度数: $11.25 \\times 7 = 78.75$ (度)答: 顶角和底角分别是 $22.5^{\\circ}$ 和 $78.75^{\\circ}$ 。",
+        "knowledge_points": [
+          "比的应用",
+          "三角形的内角和"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit4_500",
+        "original_id": "unit4_choice_500"
+      }
+]
\ No newline at end of file
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@@ -0,0 +1,8978 @@
+[
+    {
+      "id": "unit1_choice_001",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "下列哪个数在数轴上最接近 0 。( )",
+      "options": [
+        "A. 1",
+        "B. $-\\frac{1}{2}$",
+        "C. $\\frac{1}{3}$",
+        "D. -1"
+      ],
+      "answer": "C",
+      "solution": "【分析】0 是正数、负数的分界点,比 0 大的是正数,正数的数字前面的 '+' 可以省略不写;比 0 小的是负数,负数的数字前面的 '-' 不能省略;先分别找出各数与 0 的距离, 再根据分数大小比较的方法进行比较, 找出与 0 距离最小的数, 即是最接近 0 的数。分数大小的比较:分子相同时,分母越大,分数值反而越小。\n【详解】A．1与0相距1;B. $-\\frac{1}{2}$ 与 0 相距 $\\frac{1}{2}$; C. $\\frac{1}{3}$ 与 0 相距 $\\frac{1}{3}$; D. -1 与 0 相距 1 ;$\\frac {1}{3} < \\frac {1}{2} < 1$所以,这些数中,$\\frac{1}{3}$ 在数轴上最接近 0。故答案为: C",
+      "knowledge_points": [
+        "正负数在数轴上的表示",
+        "同分子分数的大小比较"
+      ],
+      "source": "2022·山东潍坊·小升初真题",
+      "merged_id": "unit1_001",
+      "original_id": "unit1_choice_001"
+    },
+    {
+      "id": "unit1_choice_002",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "牛奶盒上标有净含量: $220 \\mathrm{~mL} \\pm 10 \\mathrm{~mL}$, 实际这种牛奶每盒最多与最少相差不多于 ( )。",
+      "options": [
+        "A. $10 \\mathrm{~mL}$",
+        "B. $20 \\mathrm{~mL}$",
+        "C. $15 \\mathrm{~mL}$"
+      ],
+      "answer": "B",
+      "solution": "【分析】可先根据'净含量: $220\\mathrm{mL} \\pm 10\\mathrm{mL}$'求出这种牛奶每盒最多和最少的含量,再计算它们的差值。'$220\\mathrm{mL} \\pm 10\\mathrm{mL}$'表示这种牛奶的净含量在 $220\\mathrm{mL}$ 的基础上可以有 $10\\mathrm{mL}$ 的误差范围。\n【详解】每盒最多: $220 + 10 = 230(\\mathrm{mL})$每盒最少: $220 - 10 = 210 \\text{ (mL)}$相差: $230 - 210 = 20 \\, (\\mathrm{mL})$实际这种牛奶每盒最多与最少相差不多于 $20 \\mathrm{~mL}$ 。故答案为: B",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "24-25 六年级下·海南海口·单元测试",
+      "merged_id": "unit1_002",
+      "original_id": "unit1_choice_002"
+    },
+    {
+      "id": "unit1_choice_003",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "下列说法中, 正确的是 ( )。",
+      "options": [
+        "A. 某城市一天的气温是 $-3^{\\circ} \\mathrm{C} - 6^{\\circ} \\mathrm{C}$, 最高气温和最低气温相差 $3^{\\circ} \\mathrm{C}$ 。",
+        "B. 305000 是由 3 个十万和 5 个千组成的。",
+        "C. 比 10 小的整数一共有 10 个。"
+      ],
+      "answer": "B",
+      "solution": "【分析】A. 计算一正一负的两个数的差, 直接用两数字相加即可得解。 B. 观察可知, 在 305000 中, 3 在十万位上,表示 3 个十万, 5 在千位上, 表示 5 个千。 C. 比 10 小的正整数有 9 个, 0 也比 10 小, 还有无数的负数比 10 小。\n【详解】A. $3^{\\circ} \\mathrm{C} + 6^{\\circ} \\mathrm{C} = 9^{\\circ} \\mathrm{C}$, 某城市一天的气温是 $-3^{\\circ} \\mathrm{C}-$ $6^{\\circ} \\mathrm{C}$, 最高气温和最低气温相差 $9^{\\circ} \\mathrm{C}$, 原题说法错误。 B. 据分析可知, 3 在十万位上, 5 在千位上, 305000 是由 3 个十万和 5 个千组成的, 原题说法正确。 C. 据分析可知, 比 10 小的整数有无数个, 原题说法错误。故答案为: B",
+      "knowledge_points": [
+        "整数的数级、数位和计数单位的认识",
+        "正负数的意义及应用",
+        "正负数的大小比较"
+      ],
+      "source": "24-25 六年级下·海南海口·单元测试",
+      "merged_id": "unit1_003",
+      "original_id": "unit1_choice_003"
+    },
+    {
+      "id": "unit1_choice_004",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "一袋薯片外包装上标注 '净含量 (120±5) 克', 工商部门抽查了 4 袋薯片,下面 ( ) 是不合格的。",
+      "options": [
+        "A. 121 克",
+        "B. 124克",
+        "C. 126 克"
+      ],
+      "answer": "C",
+      "solution": "【分析】用正负数来表示具有意义相反的两种量: '净含量 $(120 \\pm 5)$ 克', 以 120 克为标准, 即净含量在 $(120-$ 5) 克到 $(120 + 5)$ 克之间, 在这个区间范围内的重量都是合格的, 据此可得出答案。\n【详解】 $120 - 5 = 115$ (克),$120 + 5 = 125$ (克)A. 121 克在 115 克到 125 克之间, 即符合标准; B. 124 克在 115 克到 125 克之间, 即符合标准; C. 126 克不在 115 克到 125 克之间, 即不符合标准。故答案为: C",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "24-25 六年级下·海南海口·单元测试",
+      "merged_id": "unit1_004",
+      "original_id": "unit1_choice_004"
+    },
+    {
+      "id": "unit1_choice_005",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "在 -3、-0.5、0、-0.1这四个数中,最大的数是( )。",
+      "options": [
+        "A. -3",
+        "B. -0.1",
+        "C. 0"
+      ],
+      "answer": "C",
+      "solution": "【分析】正数大于 0 和一切负数, 0 大于一切负数; 负数的大小比较方法是去掉 '-' 号后, 大的数反而小。\n【详解】0 大于所有的负数,所以 0 最大。故答案为: C",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "24-25 六年级下·海南海口·单元测试",
+      "merged_id": "unit1_005",
+      "original_id": "unit1_choice_005"
+    },
+    {
+      "id": "unit1_choice_006",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "我们把高出海平面记作正,低于海平面记作负,甲地海拔高度是40米,乙地海拔高度是 25 米,丙地海拔高度是 -20 米。甲、乙、丙三地最高的地方比最低的地方高( )米。",
+      "options": [
+        "A. 15",
+        "B. 60",
+        "C. 45",
+        "D. 5"
+      ],
+      "answer": "B",
+      "solution": "【分析】甲地海拔高度是 40 米记作: +40 米; 乙地海拔高度是 25 米记作: +25 米; 丙地海拔高度记作: -20 米; 然后进行比较即可, 再求得最高的地方比最低的地方高出的米数, 据此解答。\n【详解】 $+40 > +20 > -20$ $40 + 20 = 60$ (米)我们把高出海平面记作正,低于海平面记作负,甲地海拔高度是40米,乙地海拔高度是25米,丙地海拔高度是-20米。甲、乙、丙三地最高的地方比最低的地方高60米。故答案为: B",
+      "knowledge_points": [
+        "正负数的意义及应用",
+        "正负数的大小比较"
+      ],
+      "source": "24-25 六年级下·全国·单元测试",
+      "merged_id": "unit1_006",
+      "original_id": "unit1_choice_006"
+    },
+    {
+      "id": "unit1_choice_007",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "《九章算术》中注'今两算得失相反,要令正负以名之'意思是:有两数若其意义相反,则分别叫作正数和负数。若气温为零上 $10^{\\circ} \\mathrm{C}$ 记作 $+10^{\\circ} \\mathrm{C}$,则 $-8^{\\circ} \\mathrm{C}$ 表示气温为( )。",
+      "options": [
+        "A. 零上 $8^{\\circ} \\mathrm{C}$",
+        "B. 零下 $8^{\\circ} \\mathrm{C}$",
+        "C. 零上 $2^{\\circ} \\mathrm{C}$"
+      ],
+      "answer": "B",
+      "solution": "【分析】正数与负数表示意义相反的两种量,规定其中一个为正,则和它意义相反的就为负。规定零上为正,则零下就为负。负数的写法是:先写'-'号,然后再写后面的数字,数字要用阿拉伯数字进行书写。正数的写法是:在写正数时,数字前写 $+$ 号或省略'+'号两种形式都可以。\n【详解】根据分析可得:若气温为零上 $10^{\\circ}\\mathrm{C}$ 记作 $+10^{\\circ}\\mathrm{C}$,则 $-8^{\\circ} \\mathrm{C}$ 表示气温为零下 $8^{\\circ} \\mathrm{C}$ 。故答案为: B",
+      "knowledge_points": [
+        "正负数的读法和写法",
+        "温度的应用",
+        "正负数的意义及应用"
+      ],
+      "source": "24-25 六年级下·全国·单元测试",
+      "merged_id": "unit1_007",
+      "original_id": "unit1_choice_007"
+    },
+    {
+      "id": "unit1_choice_008",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "某品牌SUV汽车三月份的产量比二月份增长了 $-2.5\\%$ ,'$-2.5\\%$'表示( )。",
+      "options": [
+        "A. 增长了 $2.5\\%$",
+        "B. 下降了 $2.5\\%$",
+        "C. 不增不减降",
+        "D. 无法确定"
+      ],
+      "answer": "B",
+      "solution": "【分析】分析题目,正数与负数表示意义相反的两种量,把二月份的汽车产量看作分界线,则三月份比二月份增长记为正,三月份比二月份减少记为负,据此结合百分数的意义解答。\n【详解】'-2.5%'表示下降了 $2.5\\%$ 。某品牌SUV汽车三月份的产量比二月份增长了-2.5%, '- 2.5%'表示下降了 2.5%。故答案为: B",
+      "knowledge_points": [
+        "百分数的意义",
+        "正负数的意义及应用"
+      ],
+      "source": "23-24六年级下·四川乐山·期中",
+      "merged_id": "unit1_008",
+      "original_id": "unit1_choice_008"
+    },
+    {
+      "id": "unit1_choice_009",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "规定 10t 记为 0t、11t 记为 +1t、则下列说法错误的是 ( )。",
+      "options": [
+        "A. 8t 记为 -8t",
+        "B. $15 \\mathrm{t}$ 记为 $+5 \\mathrm{t}$",
+        "C. 6t 记为 $-4 \\mathrm{t}$",
+        "D. +3t 表示质量为 $13 \\mathrm{t}$"
+      ],
+      "answer": "A",
+      "solution": "【分析】规定 $10 \\mathrm{t}$ 记为 $0 \\mathrm{t} 、 11 \\mathrm{t}$ 记为 $+1 \\mathrm{t}$, 说明 10 吨以上记为正, 10 吨以下记为负, 据此解答即可。\n【详解】A.8t记为-2t,原说法错误;B. $15 \\mathrm{t}$ 记为 $+5 \\mathrm{t}$, 原说法正确; C. 6t 记为 -4t,原说法正确; D. +3t 表示质量为 $13 \\mathrm{t}$, 原说法正确;故答案为: A",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "23-24 六年级上·河北邯郸·期中",
+      "merged_id": "unit1_009",
+      "original_id": "unit1_choice_009"
+    },
+    {
+      "id": "unit1_choice_010",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "$\\frac{7}{10}$ 和 $+\\frac{7}{10}$ 的读法( )。",
+      "options": [
+        "A. 相同",
+        "B. 不同",
+        "C. 无法确定"
+      ],
+      "answer": "B",
+      "solution": "【分析】$\\frac{7}{10}$ 是一个一般的分数,先读分母,再读分数线,分数线读作分之,再读分子; $+\\frac{7}{10}$ 表示正数,先读正号,再读分数。\n【详解】$\\frac{7}{10}$ 读作: 十分之七;$\\frac{7}{10}$ 读作: 正十分之七。所以 $\\frac{7}{10}$ 和 $+\\frac{7}{10}$ 的读法不同。故答案为: B",
+      "knowledge_points": [
+        "分数的初步认识",
+        "正负数的读法和写法"
+      ],
+      "source": "23-24 六年级上·河北邯郸·期中",
+      "merged_id": "unit1_010",
+      "original_id": "unit1_choice_010"
+    },
+    {
+      "id": "unit1_choice_011",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "一种食品的包装袋上有 '净重 $(250 \\pm 5) \\mathrm{~g}$'的标记, 则下列食品中质量合格的是( )。",
+      "options": [
+        "A. $243 \\mathrm{~g}$",
+        "B. $244 \\mathrm{~g}$",
+        "C. $252 \\mathrm{~g}$"
+      ],
+      "answer": "C",
+      "solution": "【分析】正数、负数表示两种相反意义的量。'净重(250±5) g 的含义, 即 $250 \\mathrm{~g}$ 是这种食品的标准净重, 实际每袋最多不超过 $(250 + 5) \\mathrm{~g}$, 最少不低于 $(250 - 5) \\mathrm{~g}$,据此解答\n【详解】 $250 - 5 = 245$ (g) $250 + 5 = 255$ (g)这种食品的净重在 $245 \\mathrm{~g}$ 至 $255 \\mathrm{~g}$ 之间都是合格的。A. $243 < 245$, 该选项质量不合格。 B. $244 < 245$, 该选项质量不合格。 C. $245 < 251 < 250$, 该选项质量合格。故答案为: C",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "22-23 六年级下·湖南永州·期末",
+      "merged_id": "unit1_011",
+      "original_id": "unit1_choice_011"
+    },
+    {
+      "id": "unit1_choice_012",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "哈尔滨某一天的夜间温度为零下 $20^{\\circ} \\mathrm{C}$, 记作 $-20^{\\circ} \\mathrm{C}$, 第二天中午温度上升了 $6^{\\circ} \\mathrm{C}$, 这时的温度记作 ( )。",
+      "options": [
+        "A. $6^{\\circ} \\mathrm{C}$",
+        "B. $26^{\\circ} \\mathrm{C}$",
+        "C. $14^{\\circ} \\mathrm{C}$",
+        "D. $-14^{\\circ} \\mathrm{C}$"
+      ],
+      "answer": "D",
+      "solution": "【分析】正负数可以表示相反意义的量, 以 $0^{\\circ} \\mathrm{C}$ 为标准, 低于 $0^{\\circ} \\mathrm{C}$ 记为负, 夜间温度 + 上升的温度 = 中午温度, 计算时,不管负号, 直接用 $20^{\\circ} \\mathrm{C} - 6^{\\circ} \\mathrm{C}$, 计算出的温度前边添上负号即可。\n【详解】 $20^{\\circ} \\mathrm{C} - 6^{\\circ} \\mathrm{C} = 14^{\\circ} \\mathrm{C}$ 这时的温度记作 $-14^{\\circ} \\mathrm{C}$ 。故答案为: D",
+      "knowledge_points": [
+        "温度的应用"
+      ],
+      "source": "23-24 六年级下·四川凉山·期末",
+      "merged_id": "unit1_012",
+      "original_id": "unit1_choice_012"
+    },
+    {
+      "id": "unit1_choice_013",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "- 2.5、+6、0、-$\\frac{1}{7}$ 、 18 、 + 3.6, 其中正数有 ( ) 个。",
+      "options": [
+        "A. 2",
+        "B. 3",
+        "C. 4"
+      ],
+      "answer": "B",
+      "solution": "【分析】0 既不是正数也不是负数。比 0 大的是正数,正数的数前面的 '+' 可以省略不写;比 0 小的是负数,负数的数前面的 '-' 不能省略。\n【详解】-2.5、+6、0、- $\\frac{1}{7}$ 、18、+3.6,其中正数是:+6、18、+3.6,所以正数有3个。故答案为: B",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "23-24 六年级下·河北邯郸·期末",
+      "merged_id": "unit1_013",
+      "original_id": "unit1_choice_013"
+    },
+    {
+      "id": "unit1_choice_014",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "中国从很早就开始使用负数。其中给出了用红色算筹表示正数,黑色算筹表示负数的是我国古代数学家( )。",
+      "options": [
+        "A. 祖冲之",
+        "B. 杨辉",
+        "C. 刘徽"
+      ],
+      "answer": "C",
+      "solution": "【分析】中国是最早认识和使用负数的国家。1700多年前,我国数学家刘徽在注解《九章算术》时, 明确提出了正数和负数的概念。他在算筹中规定 '正算赤, 负算黑', 即用红色算筹表示正数, 黑色算筹表示负数。这个记载, 比国外早了七八百年。\n【详解】通过分析可得:给出了用红色算筹表示正数,黑色算筹表示负数的是我国古代数学家刘徽。故答案为: C",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "23-24六年级下·重庆万州·期末",
+      "merged_id": "unit1_014",
+      "original_id": "unit1_choice_014"
+    },
+    {
+      "id": "unit1_choice_015",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "下面是我国四座城市某天的最低气温: 北京 - 6°C, 沈阳 - 15°C, 深圳 12°C,南昌 2°C。气温最低的城市是 ( )。",
+      "options": [
+        "A. 沈阳",
+        "B. 北京",
+        "C. 南昌"
+      ],
+      "answer": "A",
+      "solution": "【分析】负数 $< 0<$ 正数,比较两个负数的大小,看负号后面的数,负号后面的数越大,这个负数反而越小,据此比较4个城市的最低气温。\n【详解】 $-15^{\\circ} \\mathrm{C} < -6^{\\circ} \\mathrm{C} < 2^{\\circ} \\mathrm{C} < 12^{\\circ} \\mathrm{C}$, 气温最低是沈阳。下面是美国四座城市某天的最低气温: 北京 $-6^{\\circ} \\mathrm{C}$, 沈阳 $-15^{\\circ} \\mathrm{C}$, 深圳 $12^{\\circ} \\mathrm{C}$, 南昌 $2^{\\circ} \\mathrm{C}$ 。气温最低的城市是沈阳。故答案为: A",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "23-24 六年级下·江西九江·期末",
+      "merged_id": "unit1_015",
+      "original_id": "unit1_choice_015"
+    },
+    {
+      "id": "unit1_choice_016",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "- 1、0、 $\\frac{1}{2}$ 、-2 的大小顺序是( )。",
+      "options": [
+        "A. $\\frac{1}{2} > 0 > -1 > -2$",
+        "B. $-1 < -2 < \\frac{1}{2} < 0$",
+        "C. $\\frac{1}{2} > 0 > -2 > -1$",
+        "D. $-2 > -1 > \\frac{1}{2} > 0$"
+      ],
+      "answer": "A",
+      "solution": "【分析】负数 $< 0<$ 正数;负数比较大小,去掉负号的数值越大,则这个负数越小。据此解答。\n【详解】因为, $0 < \\frac{1}{2} < 1 < 2$所以, $\\frac{1}{2} > 0 > -1 > -2$1、0、$\\frac{1}{2}$ 、-2的大小顺序是 $\\frac{1}{2} > 0 > -1 > -2$ 。\n故答案为: A",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "23-24 六年级下·河北邢台·期中",
+      "merged_id": "unit1_016",
+      "original_id": "unit1_choice_016"
+    },
+    {
+      "id": "unit1_choice_017",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "$-\\frac{5}{9}$ 读作 ( )。",
+      "options": [
+        "A. 九分之五",
+        "B. 负九分之五",
+        "C. 负五分之九"
+      ],
+      "answer": "B",
+      "solution": "【分析】负数的读法是: 先读 '负', 再读数, 读分数时,先读分母, 再读分子, 据此分析。\n【详解】 $-\\frac{5}{9}$ 读作负九分之五。故答案为: B",
+      "knowledge_points": [
+        "正负数的读法和写法"
+      ],
+      "source": "23-24 六年级下·云南楚雄·期中",
+      "merged_id": "unit1_017",
+      "original_id": "unit1_choice_017"
+    },
+    {
+      "id": "unit1_choice_018",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "一月份,'小砂糖桔'从南宁去哈尔滨游学。南宁当日的气温为 $8^{\\circ} \\mathrm{C} \\sim 15^{\\circ} \\mathrm{C}$, 哈尔滨的气温为 $-13^{\\circ} \\mathrm{C} \\sim -24^{\\circ} \\mathrm{C}$, 途中'小砂糖桔'会感到气温逐渐 ( )。",
+      "options": [
+        "A. 升高",
+        "B. 降低",
+        "C. 变暖"
+      ],
+      "answer": "B",
+      "solution": "【分析】用正数和负数表示相反意义的两种量。表示气温时, 零上气温用正数表示, 零下气温用负数表示。在数轴上, 负数在 0 的左边, 正数在 0 的右边, 正数大于负数, 零下气温比较大小时, 数字越大气温反而越小。据此解答。\n【详解】根据分析,可得:因为正数大于负数,哈尔滨的气温比南宁的气温低,所以从南宁去哈尔滨游学,途中'小砂糖桔'会感到气温逐渐降低。故答案为: B",
+      "knowledge_points": [
+        "温度的应用",
+        "正负数的意义及应用",
+        "温度的认识及比较"
+      ],
+      "source": "23-24六年级下·广西贵港·期中",
+      "merged_id": "unit1_018",
+      "original_id": "unit1_choice_018"
+    },
+    {
+      "id": "unit1_choice_019",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "一种饼干包装袋上标着: 净重 $(150 \\pm 5) \\mathrm{~g}$, 表示这种饼干标准的质量是150g,实际每袋最少不少于( )g 才符合标准。",
+      "options": [
+        "A. 155",
+        "B. 150",
+        "C. 145"
+      ],
+      "answer": "C",
+      "solution": "【分析】正负数可以表示相反意义的量,以标准质量为标准,高于标准质量记为正,低于标准质量记为负,净重(150±5) g, 表示这种饼干标准的质量是 $150 \\mathrm{~g}$, 实际每袋最少不少于 $(150 - 5) \\mathrm{g}$, 最多不多于 $(150 + 5) \\mathrm{g}$, 据此分析。\n【详解】 $150 - 5 = 145$ (g) 实际每袋最少不少于 $145 \\mathrm{~g}$ 才符合标准。故答案为: C",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "23-24 六年级下·广东江门·期中",
+      "merged_id": "unit1_019",
+      "original_id": "unit1_choice_019"
+    },
+    {
+      "id": "unit1_choice_020",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "某饼干包装袋上标着:净重( $150 \\pm 5$ 克),表示这种饼干的标准质量是150克,实际每袋不多于( )克。",
+      "options": [
+        "A. 145",
+        "B. 150",
+        "C. 155"
+      ],
+      "answer": "C",
+      "solution": "【分析】正负数是用来表示具有相反意义的两种量,'净重( $150 \\pm 5$ )克',表示以150克为标准,超过部分为正,不足的部分为负,也就是这种饼干最重为 $(150 + 5)$ 克,最轻为 (150- 5) 克, 据此解答即可。\n【详解】由分析可得:某饼干标有净重( $150 \\pm 5$ )克,这种饼干的标准重是 $150$ 克;其最重不超过: $150 + 5 = 155$ (克)其最轻不低于: $150 - 5 = 145$ (克)所以实际不多于 155 克。故答案为: C",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "22-23五年级下·上海·单元测试",
+      "merged_id": "unit1_020",
+      "original_id": "unit1_choice_020"
+    },
+    {
+      "id": "unit1_choice_021",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "困难",
+      "question": "$(3, -1)$ 、 $(-5, \\frac{1}{2})$ , $(7, -\\frac{1}{3})$ , $(-9, \\frac{1}{4})$ …根据这组有序数对的排列规律,可确定第 10 个数对是( )。",
+      "options": [
+        "A. $(-21, \\frac{1}{10})$",
+        "B. (21, - $\\frac{1}{10}$ )",
+        "C. (-15, $\\frac{1}{10}$",
+        "D. (15, - $\\frac{1}{10}$ )"
+      ],
+      "answer": "A",
+      "solution": "【分析】先看每组数对中的第一个数,不看正负号,每个数都比前一个数多 2,奇数项的数是正数,偶数项的数是负数; 再看每组数对中的第二个数, 每一项的分母比前一个数的分母多 1 , 分子都是 1 , 奇数项的数是负数, 偶数项的数是正数, 据此解答。解答本题的关键是根据已知的数得出前后数之间的变化关系的规律,再利用规律解决问题。\n【详解】第10个数对是偶数项:\n$3 + (1 0 - 1) \\times 2 \\\\ = 3 + 9 \\times 2 \\\\ = 2 1 $\n第一个数是:-21,第二个数是: $\\frac{1}{10}$,所以这个数对是(-21, $\\frac{1}{10}$ )。故答案为: A",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "23-24 六年级上·北京·单元测试",
+      "merged_id": "unit1_021",
+      "original_id": "unit1_choice_021"
+    },
+    {
+      "id": "unit1_choice_022",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "在 -3、0、5、 $-5 \\frac{1}{2}$ 、 +4.9、15%、+2019 中, 是负数的有 ( ) 个。",
+      "options": [
+        "A. 2",
+        "B. 3",
+        "C. 4",
+        "D. 5"
+      ],
+      "answer": "A",
+      "solution": "【分析】大于 0 的数叫做正数, 正数用 '+' 表示, 正号可以省略不写, 小于 0 的数叫做负数, 负数用 '- ' 表示, 负号不可以省略, 0 既不是正数也不是负数, 据此解答。本题主要考查正负数的辨认, 掌握正负数的概念是解答题目的关键。\n【详解】分析可知,在 -3、0、5、-5 $\\frac{1}{2}$ 、+4.9、15%、+2019中,负数有 -3、-5 $\\frac{1}{2}$ ,一共2个。故答案为: A",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "22-23 六年级下·云南楚雄·期中",
+      "merged_id": "unit1_022",
+      "original_id": "unit1_choice_022"
+    },
+    {
+      "id": "unit1_choice_023",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "( ) 不是 -4 与 -2 之间的数。",
+      "options": [
+        "A. -3",
+        "B. -3.5",
+        "C. -1"
+      ],
+      "answer": "C",
+      "solution": "【分析】根据在数轴上,0 的右面是正数,依次为1、2、3、…,0的左边是负数,依次为-1、-2、-3…,所以 -4 与 -2 之间的数有 -3、-2.5、-3.5 等,由此解答即可。明确 -4 与 -2 之间的数有哪些, 是解答此题的关键。\n【详解】-1不是-4与-2之间的数,-1比-2大。故答案为: C",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "21-22 六年级下·内蒙古通辽·期中",
+      "merged_id": "unit1_023",
+      "original_id": "unit1_choice_023"
+    },
+    {
+      "id": "unit1_choice_024",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "下面所给的温度中,最高的是 ( )。",
+      "options": [
+        "A. $0^{\\circ} \\mathrm{C}$",
+        "B. $-9^{\\circ} \\mathrm{C}$",
+        "C. $-5^{\\circ} \\mathrm{C}$",
+        "D. $-10^{\\circ} \\mathrm{C}$"
+      ],
+      "answer": "A",
+      "solution": "【分析】0 既不是正数也不是负数。比 0 大的是正数,比 0 小的是负数,负号后面的数越大,这个负数就越小。本题考查用正负数表示温度以及大小的比较方法。\n【详解】 $0^{\\circ} \\mathrm{C} > -5^{\\circ} \\mathrm{C} > -9^{\\circ} \\mathrm{C} > -10^{\\circ} \\mathrm{C}$最高的是 $0^{\\circ} \\mathrm{C}$ 。故答案为: A",
+      "knowledge_points": [
+        "正负数的大小比较",
+        "温度的认识及比较"
+      ],
+      "source": "21-22 六年级下·全国·单元测试",
+      "merged_id": "unit1_024",
+      "original_id": "unit1_choice_024"
+    },
+    {
+      "id": "unit1_choice_025",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "天气预报说安徽今天的最低气温零下 $8^{\\circ} \\mathrm{C}$,可以记作( )。",
+      "options": [
+        "A. $+8^{\\circ} \\mathrm{C}$",
+        "B. $8^{\\circ} \\mathrm{C}$",
+        "C. $-8^{\\circ} \\mathrm{C}$"
+      ],
+      "answer": "C",
+      "solution": "【分析】负数与正数表示意义相反的量, 因此 $0^{\\circ} \\mathrm{C}$ 以上为正, 则 $0^{\\circ} \\mathrm{C}$ 以下为负, 依此选择。熟练掌握负数的意义及应用是解答此题的关键。\n【详解】根据分析可知,天气预报说安徽今天的最低气温零下 $8^{\\circ} \\mathrm{C}$, 可以记作 $-8^{\\circ} \\mathrm{C}$ 。故答案为: C",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "22-23四年级上·湖南湘西·期末",
+      "merged_id": "unit1_025",
+      "original_id": "unit1_choice_025"
+    },
+    {
+      "id": "unit1_choice_026",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "规定10吨记为0吨,11吨记为 $+1$ 吨,则下列说法错误的是( )。",
+      "options": [
+        "A. 8吨记为 +8 吨",
+        "B. 15 吨记为 +5 吨",
+        "C. 6 吨记为 -4 吨"
+      ],
+      "answer": "A",
+      "solution": "【分析】根据'规定10吨记为0吨,11吨记为+1吨。'可知: 以 10 吨为基准, 11 吨比 10 吨多 1 吨, 记作 +1 吨。即低于 10 吨的吨数用负数表示, 高于 10 吨的吨数用正数表示, 据此解答。用正数和负数表示具有相反意义的量时,要明确'基准'。\n【详解】A.8吨比10吨少2吨,所以8吨记为-2吨;B. 15 吨比 10 吨多 5 吨, 所以 15 吨记为 +5 吨; C. 6 吨比 10 吨少 4 吨, 所以 6 吨记为 -4 吨。故答案为: A",
+      "knowledge_points": [
+        "正负数的意义及应用",
+        "正负数的概念及辨认"
+      ],
+      "source": "22-23六年级下·全国·单元测试",
+      "merged_id": "unit1_026",
+      "original_id": "unit1_choice_026"
+    },
+    {
+      "id": "unit1_choice_027",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "乒乓球被誉为我国的'国球',在正规比赛中,乒乓球的标准质量为2.7克。一位质检员检验乒乓球质量时,把一个超出标准质量 0.15克的乒乓球记作 +0.15,那么另一个低于标准质量 0.03 克的乒乓球记作( )。",
+      "options": [
+        "A. -0.12",
+        "B. -0.03",
+        "C. $+0.12$"
+      ],
+      "answer": "B",
+      "solution": "【分析】正数、负数表示两种相反意义的量。乒乓球的标准质量为 2.7 克, 超出标准质量的记作正, 那么低于标准质量的就记作负, 据此解答即可。掌握正负数的意义,超出基准的量用正数表示,低于基准的量用负数表示。\n【详解】把一个超出标准质量 0.15 克的乒乓球记作 +0.15,那么另一个低于标准质量 0.03 克的乒乓球记作 -0.03。故答案为: B",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "22-23六年级下·全国·单元测试",
+      "merged_id": "unit1_027",
+      "original_id": "unit1_choice_027"
+    },
+    {
+      "id": "unit1_choice_028",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "下列说法正确的是 ( )。",
+      "options": [
+        "A. 0 是最小的数",
+        "B. 0 既是正数也是负数",
+        "C. 所有的正数都比负数大"
+      ],
+      "answer": "C",
+      "solution": "【分析】比 0 大的数是正数, 如 3、500、4.7、 $\\frac{7}{8}$, 这些数都是正数。正数可以在数字前加'+'(正号),一般情况下可省略不写。比0小的数是负数,也可以说在正数的前面添上负号'-'的数,如-3、-500、-4.7,这些数都是负数。特别注意:0既不是正数,也不是负数。原点左边的为负数,原点右边的为正数,在数轴上的数从左到右依次变大。正负数的大小比较:两个负数比较大小,距离原点近的数大;正数大于负数;两个正数比较大小,距离原点远的数大。据此解答。此题主要考查正负数的定义以及正负数的大小比较。\n【详解】A.负数比0小,所以0并不是最小的数,原题说法错误; B. 0 既不是正数, 也不是负数。原题说法错误; \nC. 所有的正数都比负数大, 原题说法正确。故答案为: C",
+      "knowledge_points": [
+        "正负数的大小比较",
+        "正负数的概念及辨认"
+      ],
+      "source": "22-23 六年级下·全国·单元测试",
+      "merged_id": "unit1_028",
+      "original_id": "unit1_choice_028"
+    },
+    {
+      "id": "unit1_choice_029",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "两个冷库, 甲的温度是 $-12^{\\circ} \\mathrm{C}$, 乙的温度是 $-10^{\\circ} \\mathrm{C}$, ( ) 的温度高一些。",
+      "options": [
+        "A. 一样",
+        "B. 甲",
+        "C. 乙"
+      ],
+      "answer": "C",
+      "solution": "【分析】比 $0^{\\circ} \\mathrm{C}$ 低的温度叫零下温度, 通常在数字前加 '- ' 表示。例如, -3°C表示零下3摄氏度。温度是零下的, 数字越大, 温度越低, $0^{\\circ} \\mathrm{C}$ 比任何一个零下几摄氏度都高。据此分析。关键是掌握负数比大小的方法, 两个负数离原点越近的数值越大。\n【详解】 $12 > 10$ 所以乙的温度高一些。 故答案为: C",
+      "knowledge_points": [
+        "温度的认识及比较"
+      ],
+      "source": "21-22 五年级上·江苏淮安·期中",
+      "merged_id": "unit1_029",
+      "original_id": "unit1_choice_029"
+    },
+    {
+      "id": "unit1_choice_030",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "铭铭上次考试得了 88 分, 可看到在计算平均分时老师却用了这样 ' - 2' 的记录, 铭铭班上次考试他们班的平均分是 ( )。",
+      "options": [
+        "A. 86 分",
+        "B. 88 分",
+        "C. 90分",
+        "D. 92 分"
+      ],
+      "answer": "C",
+      "solution": "【分析】由题意可知, 高于平均分的用正数表示, 则低于平均分的用负数表示, 用 88 加上 2 即可求出他们班的平均分。本题考查正负数的意义及应用, 明确高于平均分的用正数表示, 则低于平均分的用负数表示是解题的关键。\n【详解】 $88 + 2 = 90$ (分)则铭铭班上次考试他们班的平均分是90分。故答案为: C",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "22-23 五年级上·河南洛阳·期中",
+      "merged_id": "unit1_030",
+      "original_id": "unit1_choice_030"
+    },
+    {
+      "id": "unit1_choice_031",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "负数与正数比较,( )。",
+      "options": [
+        "A. 负数比正数大",
+        "B. 负数比正数小",
+        "C. 正数和负数一样大"
+      ],
+      "answer": "B",
+      "solution": "【分析】画 出 数轴,如图:![](images\\5ed4eb660abd94ca532e22ce61131954.png),比较正数负数、0之间的关系即可得解。本题的解题关键是通过数轴, 掌握正负数之间的大小关系。\n【详解】通过数轴可以看出:负数都在 0 的左边,它们比 0 小,而正数都在 0 的右边,它们比 0 大,正数也比负数大;即负数 $< 0<$ 正数。故答案为: B",
+      "knowledge_points": [
+        "正负数在数轴上的表示",
+        "正负数的大小比较"
+      ],
+      "source": "20-21五年级下·湖南邵阳·期末",
+      "merged_id": "unit1_031",
+      "original_id": "unit1_choice_031"
+    },
+    {
+      "id": "unit1_choice_032",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "如果规定向东为正,那么小明向东走 $10 \\mathrm{~m}$,记作 $+10 \\mathrm{~m}$,向西走 $5 \\mathrm{~m}$,记作( )。",
+      "options": [
+        "A. $5 \\mathrm{~m}$",
+        "B. -5",
+        "C. -5m",
+        "D. -15m"
+      ],
+      "answer": "C",
+      "solution": "【分析】东和西是具有相反意义的两种量, 如果规定向东为正, 用 '+' 表示, 那么向西为负, 用 '- ' 表示, 向西的路程前面加上负号即可。本题主要考查正负数的意义及应用, 掌握正负数的表示方法是解答题目的关键。\n【详解】分析可知, 如果规定向东为正, 那么小明向东走 $10 \\mathrm{~m}$, 记作 $+10 \\mathrm{~m}$, 向西走 $5 \\mathrm{~m}$, 记作 $-5 \\mathrm{~m}$ 。\n故答案为: C",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·广东江门·期末",
+      "merged_id": "unit1_032",
+      "original_id": "unit1_choice_032"
+    },
+    {
+      "id": "unit1_choice_033",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "期中学业检测,东东数学成绩比全班的平均分高 3 分,记作 +3 分。若兰兰的数学成绩记作 -4 分,则兰兰比东东少( )。",
+      "options": [
+        "A. -7分",
+        "B. 7 分",
+        "C. -3分",
+        "D. -4 分"
+      ],
+      "answer": "B",
+      "solution": "【分析】用正负数来表示具有意义相反的两种量:以平均分为标准记为 0,超过部分为正,不足的部分为负,由此进行解答即可。此题主要考查正负数的意义, 正数与负数表示意义相反的两种量, 看清规定哪一个为正, 则和它意义相反的就为负。\n【详解】东东比平均分高3分,记作 $+3$ 分。兰兰记作 $-4$ 分,说明兰兰比平均分低4分。 $4 + 3 = 7$ (分)\n所以兰兰比东东少 7 分。故答案为: B",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·江苏盐城·期末",
+      "merged_id": "unit1_033",
+      "original_id": "unit1_choice_033"
+    },{
+      "id": "unit1_choice_034",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "如果顺时针旋转 $40^{\\circ}$ 记作 $+40^{\\circ}$, 那么逆时针旋转 $55^{\\circ}$ 记作 ( )",
+      "options": [
+        "A. $+55^{\\circ}$",
+        "B. -55°",
+        "C. 无法表示"
+      ],
+      "answer": "B",
+      "solution": "【分析】用正负数来表示具有意义相反的两种量,如果规定其中一个为正,则相对的就用负表示。顺时针旋转记作'+',那么逆时针旋转就记作'-',据此解答。此题主要考查正负数的意义, 正数与负数表示意义相反的两种量, 看清规定哪一个为正, 则和它意义相反的就为负。\n【详解】根据分析得,如果顺时针旋转 $40^{\\circ}$ 记作 $+40^{\\circ}$,那么逆时针旋转 $55^{\\circ}$ 记作 $-55^{\\circ}$ 。故答案为: B",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "21-22 六年级下·河北保定·期末",
+      "merged_id": "unit1_034",
+      "original_id": "unit1_choice_034"
+    },
+    {
+      "id": "unit1_fill_035",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "在 -5, +3, 1, -2, -9和4中,最小的数是____,最靠近0的数是____。",
+      "options": [],
+      "answer": "-9;1",
+      "solution": "【分析】所有的负数都小于正数;负数比较大小时,去掉负号的数字越大,对应的负数反而越小;也可以利用数轴比较数的大小,数轴上的数字,从左向右,越来越大。看哪一个数最靠近 0 , 不看正负, 只看数字的大小, 1、2、3、4、5、9里面1最小,所以最靠近0的数是1。\n【详解】由分析可知,在 -5, +3, 1, -2, -9 和 4 中,最小的数是 -9,最靠近 0 的数是 1。",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "24-25 六年级下·海南海口·单元测试",
+      "merged_id": "unit1_035",
+      "original_id": "unit1_fill_035"
+    },
+    {
+      "id": "unit1_fill_036",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "某地区当天的气温是 $-7 \\sim 8^{\\circ} \\mathrm{C}$, 该地区当天的温度差是( )°C。",
+      "options": [],
+      "answer": "15",
+      "solution": "【分析】比 $0^{\\circ} \\mathrm{C}$ 低的温度叫零下温度, 通常在数字前面加 '- ' (负号); 比 $0^{\\circ} \\mathrm{C}$ 高的温度叫零上温度, 通常在数字前面加 '+' (正号), 也可以省略不写。根据题意, 某地区当天的气温是 $-7 \\sim 8^{\\circ} \\mathrm{C}, -7^{\\circ} \\mathrm{C}$ 与 $0^{\\circ} \\mathrm{C}$ 相差 $7^{\\circ} \\mathrm{C}, 8^{\\circ} \\mathrm{C}$ 与 $0^{\\circ} \\mathrm{C}$ 相差 $8^{\\circ} \\mathrm{C}$, 所以 $-7 \\sim 8^{\\circ} \\mathrm{C}$ 相差 $(7+$ 8) $^\\circ \\mathrm{C}$, 据此解答。\n【详解】 $7 + 8 = 15^{\\circ} \\mathrm{C}$某地区当天的气温是 $-7 \\sim 8^{\\circ} \\mathrm{C}$, 该地区当天的温度差是 $15^{\\circ} \\mathrm{C}$ 。",
+      "knowledge_points": [
+        "温度的应用",
+        "正负数的意义及应用"
+      ],
+      "source": "24-25 六年级下·海南海口·单元测试",
+      "merged_id": "unit1_036",
+      "original_id": "unit1_fill_036"
+    },
+    {
+      "id": "unit1_fill_037",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "如果将一袋装食品净重 $195 \\mathrm{~g}$, 记为 ' $-5 \\mathrm{~g}$ ', 另一袋食品记为 ' $+3 \\mathrm{~g}$ ', 净重 $\\_ \\mathrm{g}$ 。",
+      "options": [],
+      "answer": "203",
+      "solution": "【分析】根据题意可知, 记为 ' - 5 ' , 表示比标准重量少 $5 \\mathrm{~g}$, 用 $195 + 5$, 求出袋装食品的标准重量; '+3' 表示比标准重量多 $3 \\mathrm{~g}$, 用标准重量 +3 , 即可解答。\n【详解】 $195 + 5 + 3\\\\= 2 0 0 + 3 \\\\ = 2 0 3 \\mathrm {(g)} $\n如果将一袋装食品净重 $195 \\mathrm{~g}$, 记为 ' $-5 \\mathrm{~g}$ ' , 另一袋食品记为 ' $+3 \\mathrm{~g}$ ', 净重 $203 \\mathrm{~g}$ 。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "24-25 六年级下·海南海口·单元测试",
+      "merged_id": "unit1_037",
+      "original_id": "unit1_fill_037"
+    },
+    {
+      "id": "unit1_fill_038",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "在-7、6、0、$+10$、 $+\\frac{3}{5}$ 、-96、-20中,正数有( )个,负数有( )个,最大的数是( ),最小的数是( )。",
+      "options": [],
+      "answer": "3;3;+10或10;-96",
+      "solution": "【分析】0 既不是正数也不是负数; 有 '-' 号的都是负数; 正数可以在数字前加 '+' (正号), 一般情况下可省略不写;\n正负数的大小比较: 两个负数比较大小, 距离原点近的数大; 正数大于负数; 两个正数比较大小, 距离原点远的数大。\n【详解】正数有:6、+10、 $\\frac{+3}{5}$ ,三个;负数有:-7、-96、-20,三个;7、6、0、+10、 $\\frac{+3}{5}$ 、-96、-20中,正数最大的是 +10,负数里面最小的是 -96。即在 -7、6、0、+10、 $\\frac{+3}{5}$ 、-96、-20 中,正数有 3 个,负数有 3 个,最大的数是 +10(或 10),最小的数是 -96。",
+      "knowledge_points": [
+        "正负数的大小比较",
+        "正负数的概念及辨认"
+      ],
+      "source": "24-25 六年级下·海南海口·单元测试",
+      "merged_id": "unit1_038",
+      "original_id": "unit1_fill_038"
+    },
+    {
+      "id": "unit1_fill_039",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "如果把潜水艇在水下 20 米处记作 -20 米, 那么它上浮 8 米后, 这时它的位置可以记作( )米。",
+      "options": [],
+      "answer": "-12",
+      "solution": "【分析】正负数表示一组意义相反的数量,水上记为正,则水下就记为负,由此解答即可。\n【详解】如果把潜水艇在水下 20 米处记作 -20 米, 那么它上浮 8 米后, $20 - 8 = 12$ (米)这时它的位置可以记作 -12 米。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "24-25 六年级下·海南海口·单元测试",
+      "merged_id": "unit1_039",
+      "original_id": "unit1_fill_039"
+    },
+    {
+      "id": "unit1_fill_040",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "去年冬天某地某天凌晨气温是 -3 摄氏度,中午的气温是 9 摄氏度,中午的气温比凌晨高( )摄氏度,傍晚的气温比中午下降了5摄氏度,傍晚的气温是( )摄氏度。",
+      "options": [],
+      "answer": "12;4",
+      "solution": "【分析】凌晨气温是 -3 摄氏度,说明凌晨气温在零下 3 摄氏度,中午的气温是 9 摄氏度,说明中午气温是零上 9 摄氏度,则中午的气温比凌晨高( $3 + 9$ )摄氏度;用中午气温减去下降的温度,即 $9 - 5 = 4$ (摄氏度)得到傍晚的温度即可。\n【详解】去年冬天某地某天凌晨气温是 -3 摄氏度, 中午的气温是 9 摄氏度, 中午的气温比凌晨高 12 摄氏度, 傍晚的气温比中午下降了 5 摄氏度, 傍晚的气温是 4 摄氏度。",
+      "knowledge_points": [
+        "利用正负数解决实际问题",
+        "正负数的意义及应用"
+      ],
+      "source": "24-25 六年级下·海南海口·单元测试",
+      "merged_id": "unit1_040",
+      "original_id": "unit1_fill_040"
+    },
+    {
+      "id": "unit1_fill_041",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "小芳在超市买了一袋味精,发现包装袋上印有'净含量 $100 \\pm 3$ 克'的字样,意思指的是这种袋装味精每袋质量在( )克至( )克之间是符合产品要求的。",
+      "options": [],
+      "answer": "97;103",
+      "solution": "【分析】首先应弄清'净含量100克±3克'的含义,也就是说这袋味精的标准重量是100克,实际每袋重量最多不超过 $(100 + 3)$ 克,最少不能少于 $(100 - 3)$ 克,据此解答。\n【详解】 $100 - 3 = 97$ (克) $100 + 3 = 103$ (克)小芳在超市买了一袋味精,发现包装袋上印有'净含量$100 \\pm 3$ 克'的字样,意思指的是这种袋装味精每袋质量在97克至103克之间是符合产品要求的。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "24-25 六年级下·海南海口·单元测试",
+      "merged_id": "unit1_041",
+      "original_id": "unit1_fill_041"
+    },
+    {
+      "id": "unit1_fill_042",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "'-23'读作( ); 正九分之二写作( ), - $\\frac{3}{4}$ 读作( )。",
+      "options": [],
+      "answer": "负二十三 ;+$\\frac{2}{9}$或$\\frac{2}{9}$;负四分之三",
+      "solution": "【分析】正数的读法:先读'正'(如果'+'没有写,不需要读正字),数字部分按数的读法去读;正数的写法:先写'+',再写数;负数的读法:先读'负',数字部分按数的读法去读;负数的写法:先写'-',再写数。\n【详解】 '-23'读作:负二十三正九分之二写作: $+ \\frac{2}{9}$ $-\\frac{3}{4}$ 读作: 负四分之三",
+      "knowledge_points": [
+        "正负数的读法和写法"
+      ],
+      "source": "24-25 六年级下·海南海口·单元测试",
+      "merged_id": "unit1_042",
+      "original_id": "unit1_fill_042"
+    },
+    {
+      "id": "unit1_fill_043",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "体育锻炼标准规定:六年级男生每分钟跳绳65个为合格。超过标准的个数用正数表示,不足的个数用负数表示。有6位男同学的成绩分别记为 $+5$ 、-4、0、 $+2$ 、-3、-6,这6位男同学中合格的有( )人,6位男同学实际平均每人每分钟跳( )个。",
+      "options": [],
+      "answer": "3;64",
+      "solution": "【分析】正数与负数表示意义相反的两种量,规定其中一个为正,则和它意义相反的就为负。规定超过标准的个数用正数表示,不足的个数用负数表示。据此判断合格人数。求出 6 位男同学的跳绳总个数, 再除以 6 , 即可求出平均成绩,即平均每人每分钟跳的个数。\n【详解】成绩分别记为 $+5 、 0 、 +2$ 是合格成绩。\n$(6 5 \\times 6 + 5 - 4 + 0 + 2 - 3 - 6) \\div 6 \\\\ = (3 9 0 + 5 - 4 + 0 + 2 - 3 - 6) \\div 6 \\\\ = 3 8 4 \\div 6 \\\\ = 6 4 (\\text {个}) $\n这 6 位男同学中合格的有 3 人, 6 位男同学实际平均每人每分钟跳 64 个。",
+      "knowledge_points": [
+        "正负数的意义及应用",
+        "平均数的意义及求法"
+      ],
+      "source": "24-25 六年级下·全国·单元测试",
+      "merged_id": "unit1_043",
+      "original_id": "unit1_fill_043"
+    },
+    {
+      "id": "unit1_fill_044",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "如果以公元元年为界,'诗圣'杜甫出生于公元后712年,记作 $+712$ 年。著名的爱国诗人屈原约出生于-340年,那么-340年表示( )。",
+      "options": [],
+      "answer": "公元前 340 年",
+      "solution": "【分析】以公元元年为界,公元后记为正,则公元前记为负,据此填空。\n【详解】著名的爱国诗人屈原约出生于 -340 年, 那么 -340 年表示公元前 340 年。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "24-25 六年级下·全国·单元测试",
+      "merged_id": "unit1_044",
+      "original_id": "unit1_fill_044"
+    },
+    {
+      "id": "unit1_fill_045",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "从站台向西走 $60 \\mathrm{~m}$ 记作: $-60 \\mathrm{~m}$, 那么先从站台向东走 $100 \\mathrm{~m}$, 再往西走 $20 \\mathrm{~m}$ 记作( ) $\\mathrm{m}$ 。",
+      "options": [],
+      "answer": "$+80$",
+      "solution": "【分析】正、负数表示相反意义的量,从站台向西走 $60 \\mathrm{~m}$ 记作: $-60 \\mathrm{~m}$,那么先从站台向东走 $100 \\mathrm{~m}$,记作: $+100 \\mathrm{~m}$,再往西走 $20 \\mathrm{~m}$ 记作: $-20 \\mathrm{~m}$,据此用向东走的米数减去向西走的米数,就是最后向东走的米数。\n【详解】先从站台向东走 $100 \\mathrm{~m}$, 记作: $+100 \\mathrm{~m}$, 再往西走 $20 \\mathrm{~m}$ 记作: $-20 \\mathrm{~m}$; $100 - 20 = 80 \\text{(m)}$最后向东走了 $80 \\mathrm{~m}$, 所以先从站台向东走 $100 \\mathrm{~m}$, 再往西走 $20 \\mathrm{~m}$ 记作 $+80 \\mathrm{~m}$ 。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "24-25 五年级上·湖南邵阳·期中",
+      "merged_id": "unit1_045",
+      "original_id": "unit1_fill_045"
+    },
+    {
+      "id": "unit1_fill_046",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "一种袋装食品, 超过标准净重的部分记为正数,不足部分记为负数。一袋食品净重 104 克,记为 +4 克,那么这种食品的标准净重是( )克:记为-3克的食品净重为( )克。",
+      "options": [],
+      "answer": "100;97",
+      "solution": "【分析】根据正负数表示一组相反意义的量,超过标准净重的部分记为正数,不足部分记为负数。已知净重 104 克,记为 +4 克,则超过标准量 4 克,因此用 104 减 4 可得标准净重。记为 -3 克即为负数,表示比标准净重少 3 克,所以用标准净重减 3 即可得解。\n【详解】 $104 - 4 = 100$ (克)$100 - 3 = 97$ (克)一种袋装食品,超过标准净重的部分记为正数,不足部分记为负数。一袋食品净重 104 克,记为 +4 克,那么这种食品的标准净重是 100 克:记为 -3 克的食品净重 97 克。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "23-24 六年级下·四川内江·期末",
+      "merged_id": "unit1_046",
+      "original_id": "unit1_fill_046"
+    },
+    {
+      "id": "unit1_fill_047",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "商丘最高气温 + 14 摄氏度表示( ), 零下 10 摄氏度可以表示为( )。",
+      "options": [],
+      "answer": "零上14摄氏度或14°C或+14°C;-10 摄氏度或-10°C",
+      "solution": "【分析】零上温度与零下温度是一对具有相反意义的量,如果一种用'+'表示,则另一种就用'-'表示。通常零上温度摄氏度数前加记'+'(或省略'+'),零下温度摄氏度数前加'-'。据此解答。\n【详解】最高气温 + 14 摄氏度表示零上 14 摄氏度( $14^{\\circ} \\mathrm{C}$ 或者 + 14°C), 零下 10 摄氏度可以表示为 - 10 摄氏度 $(- 10^{\\circ} \\mathrm{C})$ 。",
+      "knowledge_points": [
+        "温度的认识及比较"
+      ],
+      "source": "23-24 四年级上·河南商丘·期末",
+      "merged_id": "unit1_047",
+      "original_id": "unit1_fill_047"
+    },
+    {
+      "id": "unit1_fill_048",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "在 0, 2.6, +2, -8, 100, $-\\frac{4}{7}$ 这 6 个数中, 整数有( )个, 负数有( )个。",
+      "options": [],
+      "answer": "4;2",
+      "solution": "【分析】整数包含正整数、0、负整数,负数前面需要'-'。\n【详解】0 是整数,+2 是正整数,-8 是负整数,100 是整数;-8, - $\\frac{4}{7}$ 是负数;在 $0, 2.6, +2, -8, 100, -\\frac{4}{7}$ 这 6 个数中,整数有 4 个,负数有 2 个。",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "23-24 六年级下·广西桂林·期末",
+      "merged_id": "unit1_048",
+      "original_id": "unit1_fill_048"
+    },
+    {
+      "id": "unit1_fill_049",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "在-3,-0.5,0,$-3,-0.5,0,\\frac{3}{2}, 96, 12.6, \\frac{15}{26}$ 这些数中,正数有( ),负数有( ),( )既不是正数,也不是负数。",
+      "options": [],
+      "answer": "$\\frac{3}{2}$、96、12.6、$\\frac{15}{26}$;-3、-0.5;0",
+      "solution": "【分析】比 0 大的数是正数, 正数可以在数字前加 '+'(正号), 一般情况下可省略不写。比 0 小的数是负数, 也可以说在正数的前面添上负号 '-' 的数。0 既不是正数, 也不是负数。\n【详解】根据分析可得:在 -3, -0.5, 0, $\\frac{3}{2}$, 96, 12.6, $\\frac{15}{26}$ 这些数中, 正数有 $\\frac{3}{2}$ 、96、12.6、 $\\frac{15}{26}$, 负数有 -3、-0.5, 0 既不是正数, 也不是负数。",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "23-24六年级下·湖南常德·期末",
+      "merged_id": "unit1_049",
+      "original_id": "unit1_fill_049"
+    },
+    {
+      "id": "unit1_fill_050",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "湘潭市某天最低气压温 -1 摄氏度,最高气温3摄氏度,这一天的温差是( )摄氏度。",
+      "options": [],
+      "answer": "4",
+      "solution": "【分析】负数表示和正数意义相反的量,当正数表示零上温度时,负数表示零下温度。这一天的最低温度是负数,最高温度是正数,那么把零下温度和零上温度的数值相加,即可求出这天的温差。\n【详解】 $3 + 1 = 4$ (摄氏度)所以,这一天的温差是 4 摄氏度。",
+      "knowledge_points": [
+        "温度的应用",
+        "正负数的意义及应用"
+      ],
+      "source": "23-24六年级下·湖南湘潭·期末",
+      "merged_id": "unit1_050",
+      "original_id": "unit1_fill_050"
+    },
+    {
+      "id": "unit1_fill_051",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "通常人体腋窝温度范围是 $36^{\\circ} \\mathrm{C} \\sim 37^{\\circ} \\mathrm{C}​$, 如果把人体腋窝温度标准定在36.5°C, 37.2°C记作 +0.7°C, 那么 35.6°C可以记作( )°C。",
+      "options": [],
+      "answer": "-0.9",
+      "solution": "【分析】正数和负数是一对表示相反意义的两个量。如果把人体腋窝温度标准定在 $36.5^{\\circ} \\mathrm{C}$, 超过标准温度的记作正数,则低于标准温度记作负数, 据此解答。\n【详解】 $36.5 - 35.6 = 0.9^{\\circ} \\mathrm{C}$ 那么 $35.6^{\\circ} \\mathrm{C}$ 可以记作 $-0.9^{\\circ} \\mathrm{C}$ 。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "23-24 六年级下·湖北·期末",
+      "merged_id": "unit1_051",
+      "original_id": "unit1_fill_051"
+    },
+    {
+      "id": "unit1_fill_052",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "2024年5月3日17时27分,'嫦娥六号'成功发射,开启了世界首次月球背面采样返回之旅。从地球到月球最远的远地点平均距离为405500千米,横线上的数省略万位后面的尾数求它的近似数约为( )万千米。在之前的'嫦娥探月'中,实测月球背面最低气温为零下一百九十六度,可记作( )°C。",
+      "options": [],
+      "answer": "41;-196",
+      "solution": "【分析】省略'万'以后的尾数求近似数,根据千位上数字数字是4或者比4小,就把尾数去掉。如果千位上的数是5或者比5大,就把尾数舍去并且在它的前一位进'1';通常用正负数表示具有相反意义的两种量,零下温度用负数表示,零上温度用正数表示,据此写出气温即可。\n【详解】由分析可得:405500千米≈41万千米零下一百九十六度可记作 $-196^{\\circ} \\mathrm{C}$ 。综上所述: 从地球到月球最远的远地点平均距离为 $405500$ 千米, 横线上的数省略万位后面的尾数求它的近似数约为 41 万千米。在之前的 '嫦娥探月' 中, 实测月球背面最低气温为零下一百九十六度, 可记作 $-196^{\\circ} \\mathrm{C}$ 。",
+      "knowledge_points": [
+        "正负数的读法和写法",
+        "整数的近似数"
+      ],
+      "source": "23-24六年级下·浙江杭州·期末",
+      "merged_id": "unit1_052",
+      "original_id": "unit1_fill_052"
+    },
+    {
+      "id": "unit1_fill_053",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "2022年冬奥会在北京开幕, 其中花样滑冰比赛场地的冰面温度达到 $-5^{\\circ} \\mathrm{C} \\sim -3^{\\circ} \\mathrm{C}​$, 冰面的最低温度是 $\\_{}^{\\circ} \\mathrm{C}​$, 最高温度是 $\\_{}^{\\circ} \\mathrm{C}​$, 相差$\\_ \\mathrm{^\\circ C}​$",
+      "options": [],
+      "answer": "-5;-3;2",
+      "solution": "【分析】正数与负数表示意义相反的两种量, 气温在 $0^{\\circ} \\mathrm{C}$ 以上记为正, 气温在 $0^{\\circ} \\mathrm{C}$ 以下记为负; 正数大于 0 , 负数小于 0 , 正数大于负数。两个负数比较大小, 负号后面的数越大,负数越小; 花样滑冰比赛场地的冰面温度均在 $0^{\\circ} \\mathrm{C}$ 以下, 用 $5^{\\circ} \\mathrm{C}$ 减去 $3^{\\circ} \\mathrm{C}$, 所得差即为温差\n【详解】根据分析可得:$5 ^ {\\circ} \\mathrm {C} < - 3 ^ {\\circ} \\mathrm {C}$$5 ^ {\\circ} \\mathrm {C} - 3 ^ {\\circ} \\mathrm {C} = 2 ^ {\\circ} \\mathrm {C}$冰面的最低温度是 $-5^{\\circ} \\mathrm{C}$, 最高温度是 $-3^{\\circ} \\mathrm{C}$, 相差 $2^{\\circ} \\mathrm{C}$ 。",
+      "knowledge_points": [
+        "正负数的意义及应用",
+        "温度的认识及比较"
+      ],
+      "source": "23-24 六年级下·河北邯郸·期末",
+      "merged_id": "unit1_053",
+      "original_id": "unit1_fill_053"
+    },
+    {
+      "id": "unit1_fill_054",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "在 ( ) 里填上'>' '<'或'='。\n$4(\\quad) - 4 \\quad -\\frac{2}{3} (\\quad) - \\frac{1}{3} \\quad 20\\% (\\quad) \\frac{1}{5} \\quad -0.75(\\quad)75\\%$",
+      "options": [],
+      "answer": ">; <; =; <",
+      "solution": "【分析】两个负数比较大小,距离原点近的数大;正数大于负数;第一、二、四小题据此解答;把百分数和分数化成小数,再根据小数比较大小的方法进行比较,第三小题据此解答。\n【详解】4和-4\n$4 > - 4 \\\\ - \\frac {2}{3} \\text {和} - \\frac {1}{3} \\\\ - \\frac {2}{3} < - \\frac {1}{3} $\n20%和 $\\frac{1}{5}$\n$20 \\% = 0.2 \\\\ 1 \\div 5 = 0. 2 $\n因为 $0.2 = 0.2$, 所以 $20\\% = \\frac{1}{5}$ \n-0.75和 $75\\%$ \n $-0.75 < 75\\%$",
+      "knowledge_points": [
+        "分数化小数、百分数、小数和分数的互化",
+        "一位小数的大小比较",
+        "正负数的大小比较"
+      ],
+      "source": "23-24 六年级下·云南楚雄·期中",
+      "merged_id": "unit1_054",
+      "original_id": "unit1_fill_054"
+    },
+    {
+      "id": "unit1_fill_055",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "比较大小, 在括号里填上 '>' '<' 或 '='。\n$- 7 (\\quad) - 5 \\quad 1. 5 (\\quad) - 5. 2 \\quad 0 (\\quad) - 2. 4 \\quad - 3. 1 (\\quad) 3. 1$",
+      "options": [],
+      "answer": "<; >; >; <",
+      "solution": "【分析】负数都比 0 小, 而正数都比 0 大, 负数都比正数小。负号后面的数越大, 这个数就越小。据此解答。\n【详解】-7和-5\n$- 7 < - 5$\n1.5和 - 5.2\n$1.5 > - 5.2$\n0和-2.4\n-3.1和3.1\n$0 > - 2. 4 \\\\ - 3. 1 < 3. 1 $",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "23-24 六年级下·广西贵港·期中",
+      "merged_id": "unit1_055",
+      "original_id": "unit1_fill_055"
+    },
+    {
+      "id": "unit1_fill_056",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "$+ \\frac{3}{8}$ 读作: ( ), 负三点零二写作: ( )。",
+      "options": [],
+      "answer": "正八分之三;-3.02",
+      "solution": "【分析】根据正负数知识,先读正负号,再读几分之几;写正负数的时候,前面要加'+'、'-'(也可以省略'+'),据此解答即可。\n【详解】 $+ \\frac{3}{8}$ 读作:正八分之三,负三点零二写作:-3.02。",
+      "knowledge_points": [
+        "正负数的读法和写法"
+      ],
+      "source": "23-24 六年级下·广东江门·期中",
+      "merged_id": "unit1_056",
+      "original_id": "unit1_fill_056"
+    },
+    {
+      "id": "unit1_fill_057",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "在 6、3、0、-18、+7 中____是正数,____是负数,____既不是正数,也不是负数。",
+      "options": [],
+      "answer": "6、3、+7; -18; 0",
+      "solution": "【分析】根据正负数的意义,负数是小于0的数,在6、3、0、-18、+7中,6、3、+7是正数,-18是负数,0 既不是正数, 也不是负数, 据此解答即可。\n【详解】在 6、3、0、-18、+7 中,6、3、+7 是正数,-18 是负数,0 既不是正数,也不是负数。",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "22-23 六年级下·吉林白城·期中",
+      "merged_id": "unit1_057",
+      "original_id": "unit1_fill_057"
+    },
+    {
+      "id": "unit1_fill_058",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "一艘潜水艇所在的位置是海拔 - 150 米,如果它上浮 20 米,那么它现在所在的位置是海拔( )米。",
+      "options": [],
+      "answer": "-130",
+      "solution": "【分析】正负数是表示意义相反的两种量,根据潜水艇所在海拔是 -150 米,如果它上浮 20 米,直接用负数的数值减去正数,加上负号即可。\n【详解】由分析可得:它现在所在的位置: $150 - 20 = 130$ (米)综上所述:一艘潜水艇所在的位置是海拔 -150 米,如果它上浮 20 米,那么它现在所在的位置是海拔 -130 米。",
+      "knowledge_points": [
+        "利用正负数解决实际问题",
+        "正负数的意义及应用"
+      ],
+      "source": "23-24 五年级上·安徽马鞍山·期末",
+      "merged_id": "unit1_058",
+      "original_id": "unit1_fill_058"
+    },
+    {
+      "id": "unit1_fill_059",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "请用正负数表示:我国最大的湖泊是青海湖,水面高出海平面 $3193 \\mathrm{~m}$,记为( )m;位于阿拉伯半岛的死海,水面低于海平面约 $400 \\mathrm{~m}$,记为( )m。",
+      "options": [],
+      "answer": "+3193;-400",
+      "solution": "【分析】通常把海平面的高度记作0,高出海平面和低于海平面是两个具有相反意义的量,高出海平面的记作'+',低于海平面的记作'-'。据此解答。\n【详解】我国最大的湖泊是青海湖,水面高出海平面 $3193 \\mathrm{~m}$,记为 $+3193 \\mathrm{~m}$ ;位于阿拉伯半岛的死海,水面低于海平面约 $400 \\mathrm{~m}$,记为 $-400 \\mathrm{~m}$ 。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "2024 六年级下·全国·专题练习",
+      "merged_id": "unit1_059",
+      "original_id": "unit1_fill_059"
+    },
+    {
+      "id": "unit1_fill_060",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "某市在 2023 年 1 月 1 日最高气温为 $11^{\\circ} \\mathrm{C}$, 记作( )°C, 最低气温为零下 $2^{\\circ} \\mathrm{C}$, 记作( )°C。",
+      "options": [],
+      "answer": "+11或11;-2",
+      "solution": "【分析】比 0 大的数叫正数, 比 0 小的数叫负数, 在写正数时, 数字前写 + ' 号或省略 '+' 号两种形式都可以。写负数时先写 '- ' 号, 然后再写后面的数字, 数字要用阿拉伯数字进行书写。关键是理解正负数的意义,掌握正负数的写法。\n【详解】某市在 2023 年 1 月 1 日最高气温为 $11^{\\circ} \\mathrm{C}$, 记作 $+11^{\\circ} \\mathrm{C}$, 最低气温为零下 $2^{\\circ} \\mathrm{C}$, 记作 $-2^{\\circ} \\mathrm{C}$ 。",
+      "knowledge_points": [
+        "正负数的读法和写法",
+        "正负数的概念及辨认"
+      ],
+      "source": "18-19 六年级下·云南昆明·阶段练习",
+      "merged_id": "unit1_060",
+      "original_id": "unit1_fill_060"
+    },
+    {
+      "id": "unit1_fill_061",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "向东走 4 米记作 +4 米, -3 米表示( )。",
+      "options": [],
+      "answer": "向西走3米",
+      "solution": "【分析】用正负数来表示具有意义相反的两种量:把向东走的米数记为正,则把向西走的米数记为负;据此解答。此题主要考查正负数的意义, 正数与负数表示意义相反的两种量, 看清规定哪一个为正, 则和它意义相反的就为负。\n【详解】由分析可得: 向东走 4 米记作 +4 米, -3 米表示向西走 3 米。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "22-23 六年级下·全国·单元测试",
+      "merged_id": "unit1_061",
+      "original_id": "unit1_fill_061"
+    },
+    {
+      "id": "unit1_fill_062",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "在 -7, +36, 9, -20, -140, +175, 0, -102, 1, -0.3 中, 正数有( ), 负数有( )。",
+      "options": [],
+      "answer": "+36、9、+175、1; -7、-20、-140、-102、-0.3",
+      "solution": "【分析】0 既不是正数也不是负数,比 0 大的是正数,正数的数前面的 '+' 可以省略不写;比 0 小的是负数,负数的数前面的 '-' 不能省略。本题主要是考查正、分数的意义,属于基础知识。注意,0 既不是正数也不是负数;负数中的 '-' 不能省略。\n【详解】由分析可得:正数是:+36、9、+175、1 负数是:-7、-20、-140、-102、-0.3",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "22-23 六年级下·全国·单元测试",
+      "merged_id": "unit1_062",
+      "original_id": "unit1_fill_062"
+    },
+    {
+      "id": "unit1_fill_063",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "学校举行科学知识竞赛,抢答题的评分规则是答对一题得 20 分,记作 +20 分;答错一题扣 10 分,应记作( )分。",
+      "options": [],
+      "answer": "-10",
+      "solution": "【分析】正负数主要用来表示具有相反意义的两种量,如果规定其中一个为正,那么相反的量就用负来表示,答为'+',那么答错为'-',数字前面加上负号,据此解答。本题主要考查正负数的意义及应用, 正数与负数表示意义相反的两种量, 分清哪一个为正, 则意义相反的量就为负。\n【详解】分析可知,学校举行科学知识竞赛,抢答题的评分规则是答对一题得20分,记作 $+20$ 分;答错一题扣10分,应记作-10分。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 五年级上·河南洛阳·期末",
+      "merged_id": "unit1_063",
+      "original_id": "unit1_fill_063"
+    },
+    {
+      "id": "unit1_fill_064",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "如果小军向东走 30米,记作 +30 米,那么李刚向西走 50 米,记作( )米。\n如果小军向北走 40 米,记作 +40 米,那么李刚走'-40'米', 表示他( )。",
+      "options": [],
+      "answer": "-50;向南走40米",
+      "solution": "【分析】此题主要用正负数来表示具有意义相反的两种量:向东走记为正, 则向西走就记为负; 向北走记为正, 则向南走就记为负: 直接得出结论即可。此题主要考查正负数的意义,正数与负数表示意义相反的两种量,看清规定哪一个为正,则和它意义相反的就为负。\n【详解】如果小军向东走 30 米,记作 +30 米,那么李刚向西走 50 米,记作 -50 米。如果小军向北走 40 米,记作 +40 米,那么李刚走 '-40 米',表示他向南走 40 米。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "21-22 六年级上·重庆潼南·期末",
+      "merged_id": "unit1_064",
+      "original_id": "unit1_fill_064"
+    },
+    {
+      "id": "unit1_fill_065",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "某商店一月份平均每天的销售额为15000元,如果把它记作'0'元,那么1月18日的销售额记作'+400'元,表示当日的实际销售额为( );1月25日的实际销售额为12000元,当日的销售额可以记作( )。",
+      "options": [],
+      "answer": "15400元;-3000元",
+      "solution": "【分析】正负数主要用来表示具有相反意义的两种量,如果规定其中一个为正,那么相反的量就用负来表示,以15000元为标准,超过15000元用'+'表示,那么不足15000元用'-'表示,'+400'元表示超过15000元了400元,12000元不足15000元用'-'表示,求出两数之差,在数字前面加上'-',据此解答。本题主要考查正负数的意义及应用,正数与负数表示意义相反的两种量,分清哪一个为正,则意义相反的量就为负。\n【详解】 $15000 + 400 = 15400$ (元)$15000 - 12000 = 3000$ (元)分析可知,'+400'元表示当日的实际销售额为15400元,1月25日的实际销售额为12000元,当日的销售额可以记作 - 3000 元。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级上·重庆黔江·期末",
+      "merged_id": "unit1_065",
+      "original_id": "unit1_fill_065"
+    },
+    {
+      "id": "unit1_fill_066",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "比0大3的数记作+3,比0小5的数记作( ),读作( )。",
+      "options": [],
+      "answer": "-5;负五",
+      "solution": "【分析】根据正负数的意义,以'0'为分界线,规定大于0的为正数, 小于 0 的为负数, 所以, 比 0 大 3 的数记作 +3 , 比 0 小 5 的数记作: -5 ; 再根据 '负数的读法: 先读'负', 再读数', 读出这个数即可。解答此题首先要知道以 0 为标准, 规定超出标准的为正, 低于标准的为负, 由此用正负数解答问题。\n【详解】-5读作:负五所以,比 0 大 3 的数记作 +3,比 0 小 5 的数记作:-5,读作:负五。",
+      "knowledge_points": [
+        "正负数的读法和写法",
+        "正负数的意义及应用"
+      ],
+      "source": "22-23 五年级上·福建宁德·期中",
+      "merged_id": "unit1_066",
+      "original_id": "unit1_fill_066"
+    },
+    {
+      "id": "unit1_fill_067",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "红星面粉厂生产的一种袋装面粉, 注明 '净含量 $10 \\mathrm{~kg} \\pm 0.5 \\mathrm{~kg}$', 李叔叔买了一袋这样的面粉, 质量最少是( ) kg。",
+      "options": [],
+      "answer": "9.5",
+      "solution": "【分析】正数、负数表示两种相反意义的量。'净含量10kg±0.5kg'的含义,即10kg是这种面粉的标准净重,实际每袋最多不超过(10+0.5)kg,最少不低于(10-0.5)kg,据此解答。掌握正负数的意义,知道以哪个数为标准,规定超出标准的为正,低于标准的为负。\n【详解】 $10 - 0.5 = 9.5(\\mathrm{kg})$李叔叔买了一袋这样的面粉,质量最少是 $9.5 \\mathrm{~kg}$ 。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "21-22 六年级下·黑龙江大兴安岭地·期末",
+      "merged_id": "unit1_067",
+      "original_id": "unit1_fill_067"
+    },
+    {
+      "id": "unit1_fill_068",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "淘气从家向东走 20 米,所在位置记 +20 米,如果他从家先向西走 300 米,再向东走 120 米,所在的位置记作( )米。",
+      "options": [],
+      "answer": "-180",
+      "solution": "【分析】用正负数表示意义相反的两种量:规定向东走记为正,那么向西走记为负,据此解答。此题主要考查正负数的意义, 正数与负数表示意义相反的两种量, 看清规定哪一个为正, 则和它意义相反的就为负。\n【详解】以淘气家为起点,向东走为正,向西走为负。如果他从家先向西走 300 米,再向东走 120 米,所在的位置记作:-180 米。",
+      "knowledge_points": [
+        "正负数的意义及应用",
+        "正负数的概念及辨认"
+      ],
+      "source": "2022·江西景德镇·小升初真题",
+      "merged_id": "unit1_068",
+      "original_id": "unit1_fill_068"
+    },
+    {
+      "id": "unit1_fill_069",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "如果向东走 40 米,记作: -40 米, 那么向东走 20 米应记作( )米; 向西走 30 米记作( )。",
+      "options": [],
+      "answer": "-20;+30米",
+      "solution": "【分析】正数、负数表示两种相反意义的量。如果规定向东走记为负,那么向西走就记为正,据此解答。掌握正负数的意义,看清规定哪一个为正,则和它意义相反的就为负。\n【详解】如果向东走 40 米,记作:-40 米,那么向东走 20米应记作 -20 米; 向西走 30 米记作 +30 米。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "21-22 六年级下·江西景德镇·期末",
+      "merged_id": "unit1_069",
+      "original_id": "unit1_fill_069"
+    },
+    {
+      "id": "unit1_fill_070",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "网络上某个买菜的小程序中 '尖椒 $500 \\mathrm{~g} \\pm 50 \\mathrm{~g}$, 价钱 1.89 元', 意思是 1.89 元最多可以买到尖椒( )g, 最少可以买到尖椒( )g。",
+      "options": [],
+      "answer": "550;450",
+      "solution": "【分析】尖椒 $500 \\mathrm{~g} \\pm 50 \\mathrm{~g}$, 表示最多不超过 $500 + 50 \\mathrm{~g}$, 最少不低于 $500 - 50 \\mathrm{~g}$, 据此解答即可。此题首先要知道以谁为标准,规定超出标准的为正,低于标准的为负,由此用正负数解答问题。\n【详解】 $500 + 50 = 550$ (g) $5 0 0 - 5 0 = 4 5 0 \\quad (\\mathrm {g})$意思是 1.89 元最多可以买到尖椒 $550 \\mathrm{~g}$, 最少可以买到尖椒 $450 \\mathrm{~g}$ 。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "21-22 六年级下·河北保定·期末",
+      "merged_id": "unit1_070",
+      "original_id": "unit1_fill_070"
+    },
+    {
+      "id": "unit1_fill_071",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "人体正常腋下温度平均为36-37摄氏度,如果我们把人体体温标准定在36.5摄氏度,37 摄氏度可记作 +0.5 摄氏度,那么 35.6 摄氏度可以记作( )摄氏度,37.4摄氏度可以记作( )摄氏度。",
+      "options": [],
+      "answer": "-0.9;+ 0.9",
+      "solution": "【分析】以 36.5 摄氏度为标准, 高于 36.5 摄氏度记为正,低于 36.5 摄氏度记为负, 据此分析。关键是理解正负数的意义, 正负数可以表示相反意义的量。\n【详解】 $36.5 - 35.6 = 0.9$ (摄氏度)$37.4 - 36.5 = 0.9$ (摄氏度)35.6 摄氏度可以记作 -0.9 摄氏度, 37.4 摄氏度可以记作 +0.9 摄氏度。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "21-22 六年级下·山西吕梁·期末",
+      "merged_id": "unit1_071",
+      "original_id": "unit1_fill_071"
+    },
+    {
+      "id": "unit1_fill_072",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "51002003 读作: ( ) 负六分之五写作: ( )。",
+      "options": [],
+      "answer": "五千一百万二千零三;$-\\frac{5}{6}$",
+      "solution": "【分析】根据大数的读法、负数的写法,直接填空即可。本题考查了大数的读法以及负数的写法, 属于基础题, 读数时先分级, 写负数时先写负号。\n【详解】51002003 读作:五千一百万二千零三,负六分之五写作: $-\\frac{5}{6}$ 。",
+      "knowledge_points": [
+        "亿以内数的读、写法",
+        "正负数的读法和写法"
+      ],
+      "source": "2022·湖南常德·小升初真题",
+      "merged_id": "unit1_072",
+      "original_id": "unit1_fill_072"
+    },
+    {
+      "id": "unit1_sort_073",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "排序题",
+      "difficulty": "容易",
+      "question": "把 $-27$ 、7.5、0、 $-3$ 、 $-6$ 这五个数按从小到大的顺序排列。\n( ) $<$ () $<$() $<$ ()$<$()",
+      "options": [],
+      "answer": "-27; -6; -3; 0; 7.5",
+      "solution": "【分析】负数 $< 0<$ 正数;负数比大小,不管负号,数值越大的负数越小,据此分析。\n【详解】 $-27 < -6 < -3 < 0 < 7.5$",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "22-23六年级下·广西贵港·期中",
+      "merged_id": "unit1_073",
+      "original_id": "unit1_sort_073"
+    },
+    {
+      "id": "unit1_judge_074",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "在 - 23、6、0 和 $+\\frac{3}{10}$ 中, 最小的数是 - 23。( )",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】在数轴上,0 的右边是正数,数字越大,离 0 越远,数值就越大;0 的左边是负数,数字越大,离 0 越远,数值反而就越小;也就是负数都比 0 小,正数都比 0 大,正数都比负数大。\n【详解】 $6 > + \\frac{3}{10} >0 > - 23$在 -23、6、0 和 $+\\frac{3}{10}$ 中, 最小的数是 -23。原题说法正确。故答案为: √",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "24-25 六年级下·海南海口·单元测试",
+      "merged_id": "unit1_074",
+      "original_id": "unit1_judge_074"
+    },
+    {
+      "id": "unit1_judge_075",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "在数轴上表示数时, 所有的负数都在 0 的左边, 所有的正数都在 0 的右边。 ()",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】正数指的是比 0 大的数, 负数指的是比 0 小的数,根据数轴的特征, 数轴上 0 左边的数都比 0 小, 0 右边的数都比 0 大, 据此解答。\n【详解】用数轴上的点表示正数和负数的时候, 正数都在 0 的右边, 负数都在 0 的左边。故答案为: √",
+      "knowledge_points": [
+        "正负数在数轴上的表示",
+        "正负数的概念及辨认"
+      ],
+      "source": "24-25 六年级下·海南海口·单元测试",
+      "merged_id": "unit1_075",
+      "original_id": "unit1_judge_075"
+    },
+    {
+      "id": "unit1_judge_076",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "数轴上 '-5'在 '-8'的左边。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】在数轴上,0 的右边是正数,数字越大,离 0 越远,数值就越大;0 的左边是负数,数字越大,离 0 越远,数值反而就越小。\n【详解】 $-5 > -8$数轴上'-5'在'-8'的右边。原题说法错误。故答案为: ×",
+      "knowledge_points": [
+        "正负数在数轴上的表示",
+        "正负数的大小比较"
+      ],
+      "source": "24-25 六年级下·海南海口·单元测试",
+      "merged_id": "unit1_076",
+      "original_id": "unit1_judge_076"
+    },
+    {
+      "id": "unit1_judge_077",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "任意一个正数都比负数大。( )",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】在数轴上,负数都在 0 的左边,它们比 0 小,而正数都在 0 的右边,它们比 0 大,正数都比负数大。\n【详解】由分析可知: 任意一个正数都比负数大的说法是正确的。故答案为: √",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "23-24 六年级下·四川遂宁·期末",
+      "merged_id": "unit1_077",
+      "original_id": "unit1_judge_077"
+    },
+    {
+      "id": "unit1_judge_078",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "+103 读作: 一百零三。()",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】正数的读法是:在读正数时,数的前面有'+'号时,一定要读出'正'字;省略'+'号的,这个'正'字也要省略不读。据此解答。\n【详解】+103读作:正一百零三。原题干说法错误。故答案为:×",
+      "knowledge_points": [
+        "正负数的读法和写法"
+      ],
+      "source": "23-24 六年级上·河北邯郸·期中",
+      "merged_id": "unit1_078",
+      "original_id": "unit1_judge_078"
+    },
+    {
+      "id": "unit1_judge_079",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "零上 3 摄氏度与零下 1 摄氏度相差 2 摄氏度。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】以0摄氏度为界,零上3摄氏度比0摄氏度高3摄氏度,零下1摄氏度比0摄氏度低1摄氏度,将两段温度差相加即可。\n【详解】 $3 + 1 = 4$ (摄氏度)零上 3 摄氏度与零下 1 摄氏度相差 4 摄氏度, 所以原题说法错误。故答案为:×",
+      "knowledge_points": [
+        "利用正负数解决实际问题",
+        "正负数的意义及应用"
+      ],
+      "source": "23-24 六年级下·湖南衡阳·期中",
+      "merged_id": "unit1_079",
+      "original_id": "unit1_judge_079"
+    },
+    {
+      "id": "unit1_judge_080",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "甲、乙两个冷库,甲冷库的温度是 $-12^{\\circ} \\mathrm{C}$,乙冷库的温度是 $-11^{\\circ} \\mathrm{C}$,甲冷库的温度高一些。()",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】比较两个负数的大小, 负号后面的数越大, 负数越小。\n【详解】通过分析可得: $-12^{\\circ} \\mathrm{C}< -11^{\\circ} \\mathrm{C}$, 则乙冷库的温度高一些。原题说法错误。故答案为: $x$",
+      "knowledge_points": [
+        "温度的认识及比较"
+      ],
+      "source": "23-24六年级下·河南信阳·期末",
+      "merged_id": "unit1_080",
+      "original_id": "unit1_judge_080"
+    },
+    {
+      "id": "unit1_judge_081",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "比海拔 -95m 还要低 5m 是 -100m。( )",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】正数与负数表示意义相反的两种量,规定其中一个为正,则和它意义相反的就为负。比海拔 $-95 \\mathrm{~m}$ 还要低 $5 \\mathrm{~m}$ 表示的是离海平面的距离比 $95 \\mathrm{~m}$ 还要靠下 $5 \\mathrm{~m}, 95 + 5 = 100(\\mathrm{m})$ ,据此判断。\n【详解】 $95 + 5 = 100(\\mathrm{m})$所以比海拔 $-95 \\mathrm{~m}$ 还要低 $5 \\mathrm{~m}$ 是 $-100 \\mathrm{~m}$, 原题说法正确。故答案为: √",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "23-24 六年级下·河北沧州·期末",
+      "merged_id": "unit1_081",
+      "original_id": "unit1_judge_081"
+    },
+    {
+      "id": "unit1_judge_082",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "在 $7 、 -6 、 0 、 \\frac{+2}{5} 、 -6.3 、 +13$ 中, 正数有 4 个。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】除 0 以外,带正号 '+' 和不带任何符号的都是正数;带负号 '- ' 的都是负数;0 不是正数也不是负数,据此分析。\n【详解】由分析可得:在 $7 、 - 6 、 0 、 \\frac {+ 2}{5} 、 - 6 . 3 、 + 1 3$ 中,正数有 $7 、 \\frac {+ 2}{5} 、 + 1 3$ ,一共是3个,所以原题说法错误。故答案为: ×",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "23-24 六年级下·云南楚雄·期中",
+      "merged_id": "unit1_082",
+      "original_id": "unit1_judge_082"
+    },
+    {
+      "id": "unit1_judge_083",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "0 大于所有的负数,小于所有的正数。( )",
+      "options": [],
+      "answer": "√",
+      "solution": "【详解】正数大于 0 ; 负数小于 0 ; 0 既不是正数也不是负数。如: $-7 < 0 < +9$ 。故答案为: √",
+      "knowledge_points": [
+        "正负数的大小比较",
+        "正负数的概念及辨认"
+      ],
+      "source": "23-24 六年级下·山东济宁·期中",
+      "merged_id": "unit1_083",
+      "original_id": "unit1_judge_083"
+    },
+    {
+      "id": "unit1_judge_084",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "在 0 和 -5 之间只有 4 个负数。()",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】比 0 大的数是正数, 比 0 小的数是负数。负数除了负整数, 还有负小数、负分数, 据此判断。\n【详解】在 0 和 -5 之间的负整数有:-1、-2、-3、-4,只有 4 个负整数;但在 0 和 -5 之间的负小数、负分数有无数个;所以,0 和 -5 之间有无数个负数。原题说法错误。故答案为: ×",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "2014 六年级·全国·课后作业",
+      "merged_id": "unit1_084",
+      "original_id": "unit1_judge_084"
+    },
+    {
+      "id": "unit1_judge_085",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "公共汽车记录乘客变化情况下车3人记作-3,上车5人应该记作 $+5$ 。()",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】正负数可以表示相反意义的量,如果下车人数记为负,则上车人数记为正,据此分析。\n【详解】公共汽车记录乘客变化情况下车 3 人记作 -3, 上车 5 人应该记作 +5 , 说法正确。故答案为: √",
+      "knowledge_points": [
+        "利用正负数解决实际问题",
+        "正负数的意义及应用"
+      ],
+      "source": "22-23 五年级上·江苏连云港·期末",
+      "merged_id": "unit1_085",
+      "original_id": "unit1_judge_085"
+    },
+    {
+      "id": "unit1_judge_086",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "所有自然数都是正数。()",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】自然数:用来表示物体个数的 0,1,2,3,4……都叫自然数;比 0 大的数是正数,比 0 小的数是负数,0 既不是正数也不是负数。\n【详解】自然数包括 0 , 但 0 不是正数。故答案为: ×",
+      "knowledge_points": [
+        "自然数的认识",
+        "正负数的概念及辨认"
+      ],
+      "source": "2014 六年级·全国·课后作业",
+      "merged_id": "unit1_086",
+      "original_id": "unit1_judge_086"
+    },
+    {
+      "id": "unit1_judge_087",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "数轴上-7在-8的左边。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】在数轴上,0 的右边是正数,数字越大,离 0 越远,数值就越大;0 的左边是负数,数字越大,离 0 越远,数值反而就越小。本题考查正负数在数轴上的表示以及负数大小的比较。\n【详解】因为 $-7 > -8$ ,所以数轴上-7在-8的右边。原题说法错误。故答案为: ×",
+      "knowledge_points": [
+        "正负数在数轴上的表示",
+        "正负数的大小比较"
+      ],
+      "source": "22-23六年级下·广东汕尾·期末",
+      "merged_id": "unit1_087",
+      "original_id": "unit1_judge_087"
+    },
+    {
+      "id": "unit1_judge_088",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "所有正数都大于负数。",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】在数轴上,数值大的在右边,数值小的在左边,根据对数轴的认识可知,在数轴上负数在 0 的左边,正数在 0 的右边,所以正数大于负数;据此解答。\n【详解】根据分析可知, 所有正数都大于负数; 原题干说法正确。故答案为: √",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "2014 六年级·全国·课后作业",
+      "merged_id": "unit1_088",
+      "original_id": "unit1_judge_088"
+    },
+    {
+      "id": "unit1_judge_089",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "温度 $0^{\\circ} \\mathrm{C}$ 就是没有温度。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】0 可以表示没有, 可以用来占位, 还可以表示分界点。比如在此题中, $0^{\\circ} \\mathrm{C}$ 就表示零上温度和零下温度的分界点, 把冰水混合物的温度规定为 $0^{\\circ} \\mathrm{C}$, 比这个温度高的为零上温度, 比这个温度低的为零下温度, 并不是没有温度。据此解答。\n【详解】根据分析得, 温度 $0^{\\circ} \\mathrm{C}$ 是水结成冰时的温度, 同时也是零上温度和零下温度的分界点, 据此可知温度 $0^{\\circ} \\mathrm{C}$ 不是没有温度, 也是温度中的一个具体的值。故答案为: ×",
+      "knowledge_points": [
+        "温度的认识及比较"
+      ],
+      "source": "2014·全国·小升初真题",
+      "merged_id": "unit1_089",
+      "original_id": "unit1_judge_089"
+    },
+    {
+      "id": "unit1_judge_090",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "甲市 15 摄氏度, 乙市- 20 摄氏度, 丙市 -6 摄氏度, 这三个城市中乙市的气温最低。( )",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】0既不是正数也不是负数。比0大的是正数,正数的数字越大,数值就越大;比0小的是负数,负数的数字越大,数值反而就越小;也就是负数都比0小,正数都比0大,正数都比负数大。正数的数字前面的'+'可以省略不写,负数的数字前面的'-'不能省略。掌握正、负数大小比较的方法是解题的关键。\n【详解】 $-20 < -6 < 15$ 乙市的气温 $<$ 丙市的气温 $<$ 甲市的气温这三个城市中乙市的气温最低。原题说法正确。故答案为: √",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "2022·河南漯河·小升初真题",
+      "merged_id": "unit1_090",
+      "original_id": "unit1_judge_090"
+    },
+    {
+      "id": "unit1_judge_091",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "甲市 $9^{\\circ} \\mathrm{C}$, 乙市 $-12^{\\circ} \\mathrm{C}$, 丙市 $-3^{\\circ} \\mathrm{C}$, 这三个城市中, 乙市的温度最低。( )",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】比 $0^{\\circ} \\mathrm{C}$ 低的温度叫零下温度, 通常在数字前加 '- ' 表示。例如, -3°C表示零下 3 摄氏度。比 $0^{\\circ} \\mathrm{C}$ 高的温度叫零上温度, 在数字前加 '+' 表示, 一般情况下 '+' 可省略不写。例如, +4°C表示零上 4 摄氏度, 也可以写成 $4^{\\circ} \\mathrm{C}$ 。温度是零下的,数字越大,温度越低, $0^{\\circ} \\mathrm{C}$ 比任何一个零下几摄氏度都高; 温度是零上的, 数字越大, 温度越高, $0^{\\circ} \\mathrm{C}$ 比任何一个零上几摄氏度都低。据此比较即可。\n【详解】所以甲市 $9^{\\circ} \\mathrm{C}$, 乙市 $-12^{\\circ} \\mathrm{C}$, 丙市 $-3^{\\circ} \\mathrm{C}$, 这三个城市中, 乙市的温度最低, 说法正确。故答案为: √",
+      "knowledge_points": [
+        "温度的认识及比较"
+      ],
+      "source": "21-22 六年级下·云南保山·期中",
+      "merged_id": "unit1_091",
+      "original_id": "unit1_judge_091"
+    },
+    {
+      "id": "unit1_judge_092",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "存入银行1000元,记作 $+1000$ 元,那么取出600元,记作-600元。()",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】存入和取出是具有相反意义的量,如果存入用'+'表示,那么取出就用'-'表示,据此解答即可。本题主要考查正负数的意义及应用,找出相反意义的两种量是解答题目的关键。\n【详解】由分析可知:存入银行 1000 元,记作 +1000 元,那么取出 600 元,记作 -600 元。原题干说法正确。故答案为: √",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "21-22五年级上·海南海口·期末",
+      "merged_id": "unit1_092",
+      "original_id": "unit1_judge_092"
+    },
+    {
+      "id": "unit1_judge_093",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "在'-31,9.5,-5,156,0'这几个数中,正数有3个。()",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】大于 0 的数叫做正数, 正数用 '+' 表示, 正号可以省略不写, 小于 0 的数叫做负数, 负数用 '- ' 表示, 负号不可以省略, 0 既不是正数也不是负数, 据此解答。本题主要考查正负数的认识, 掌握正数、负数的概念是解答题目的关键。\n【详解】分析可知,在'-31,9.5,-5,156,0'这几个数中,9.5和156是正数,-31和-5是负数,0既不属于正数,也不属于负数,所以正数有2个。故答案为: ×",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "20-21 六年级上·重庆黔江·期末",
+      "merged_id": "unit1_093",
+      "original_id": "unit1_judge_093"
+    },
+    {
+      "id": "unit1_judge_094",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "整数都大于 0 , 负数都小于 0。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】整数包括负整数、0、正整数,小于0的数就是负数。据此判断即可。本题考查整数和负数, 明确负数的定义是解题的关键。\n【详解】由分析可知:负整数小于 0 , 所以原题干说法错误。故答案为: ×",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "21-22 六年级下·云南保山·期末",
+      "merged_id": "unit1_094",
+      "original_id": "unit1_judge_094"
+    },
+    {
+      "id": "unit1_judge_095",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "包装袋上的净重(150±5 克) 的意思是实际每袋最少不少于 145 克。( )",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】由题意可知, 包装袋上的净重 $(150 \\pm 5$ 克) 的意思是实际每袋最多不超过 $150 + 5 = 155$ 克, 每袋最少不少于 $150 - 5 = 145$ 克。据此判断即可。本题考查正负数的意义, 明确 $150 \\pm 5$ 克的含义是解题的关键。\n【详解】由分析可知:包装袋上的净重( $150 \\pm 5$ 克)的意思是实际每袋最少不少于 145 克。原题干说法正确。故答案为: √",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "2021·湖南郴州·小升初真题",
+      "merged_id": "unit1_095",
+      "original_id": "unit1_judge_095"
+    },
+    {
+      "id": "unit1_judge_096",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "如果地上三层记为+3层,那么地下2层记为-2层。()",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】此题主要用正负数来表示具有意义相反的两种量:地上记为正,则地下就记为负,直接得出结论即可。此题主要考查正负数的意义, 正数与负数表示意义相反的两种量, 看清规定哪一个为正, 则和它意义相反的就为负。\n【详解】由分析可知:如果地上三层记为 +3 层, 那么地下 2 层记为 -2 层。原题干说法正确。故答案为: √",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21六年级下·浙江金华·期末",
+      "merged_id": "unit1_096",
+      "original_id": "unit1_judge_096"
+    },
+    {
+      "id": "unit1_solve_097",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "六 (1) 班平均体重为 33.5 千克, 以超出平均体重部分为正, 低于平均体重部分为负,小红的体重记为 +4.4 千克,小丽的体重记为 -2.6 千克。两人的实际体重分别是多少?",
+      "options": [],
+      "answer": "小红37.9千克;小丽30.9千克",
+      "solution": "【分析】正负数主要用来表示具有相反意义的两种量,以超出平均体重部分为正,低于平均体重部分为负,+4.4千克表示超出平均体重4.4千克,-2.6千克表示低于平均体重2.6千克,据此解答。\n【详解】 $33.5 + 4.4 = 37.9$ (千克) $33.5 - 2.6 = 30.9$ (千克)答: 小红的实际体重是 37.9 千克, 小丽的实际体重是 30.9 千克。",
+      "knowledge_points": [
+        "一位小数的进位加法、退位减法",
+        "一位小数的不进位加法、不退位减法",
+        "正负数的意义及应用"
+      ],
+      "source": "24-25 六年级下·海南海口·单元测试",
+      "merged_id": "unit1_097",
+      "original_id": "unit1_solve_097"
+    },
+    {
+      "id": "unit1_solve_098",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "豆豆的学校在公园的东边 800 米处,记作 +800 米。现在他以每分钟 70 米的速度从学校往西走了 15 分钟,他一共走了多少米?这时他在公园的哪个方向?他所在的位置可以记作多少米?",
+      "options": [],
+      "answer": "1050米;西边;-250米",
+      "solution": "【分析】正数、负数表示两种相反意义的量。如果规定公园的东边记作正,那么公园的西边就记作负。已知豆豆以每分钟 70 米的速度从学校往西走了 15 分钟, 根据 '速度 $\\times$ 时间 $=$ 路程' 求出他一共走的路程。用豆豆向西走的这段距离与 800 米进行比较, 如果超过 800 米, 则他在公园的西边, 用走的这段距离减去 800 , 即是他与公园的距离, 并用负数表示。\n【详解】 $70 \\times 15 = 1050$ (米) $1050 - 800 = 250$ (米)答: 他一共走了 1050 米, 这时他在公园的西边, 他所在的位置可以记作 -250 米。",
+      "knowledge_points": [
+        "正负数的意义及应用",
+        "基础行程问题"
+      ],
+      "source": "23-24 六年级下·甘肃武威·期中",
+      "merged_id": "unit1_098",
+      "original_id": "unit1_solve_098"
+    },
+    {
+      "id": "unit1_solve_099",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "冬奥会的冰面制作,过程复杂、标准严苛。制作一个1800平方米的标准冰场,任意3平方米内的最大高度差不超过2毫米。如果冰球赛场的冰层标准厚度是4厘米,某3平方米内的最大厚度为4.1厘米,记作 $+1$ 毫米,那么该区域最小厚度为3.9厘米,记作多少毫米?符合要求吗?",
+      "options": [],
+      "answer": "-1毫米;符合要求",
+      "solution": "【分析】正数和负数是一对表示相反意义的两个量,高于标准厚度记作 $+1$ 毫米,那么低于标准厚度应记作 $-1$ 毫米;标准冰场是任意 3 平方米内的最大高度差不超过 2 毫米,那么只要计算出该区域最大厚度和最小厚度差,与标准厚度差相比,不超过2毫米则符合要求,超过2毫米则不符合要求,据此解答。\n【详解】把 3 平方米内的最大厚度记为 +1 毫米, 则该区域最小厚度应记作 -1 毫米。4.1-3.9=0.2(厘米)0.2 厘米=2 毫米任意 3 平方米内的最大高度差不超过 2 毫米, 符合要求。答:该区域最小厚度为 3.9 厘米,记作 -1 毫米,符合要求。\n$4.1 - 3.9 = 0.2$ (厘米)",
+      "knowledge_points": [
+        "正负数的意义及应用",
+        "毫米、厘米、分米、米之间的进率与换算"
+      ],
+      "source": "23-24六年级下·全国·课后作业",
+      "merged_id": "unit1_099",
+      "original_id": "unit1_solve_099"
+    },
+    {
+      "id": "unit1_solve_100",
+      "unit": "第一单元:负数",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "困难",
+      "question": "学校为了普及低碳环保知识,举行了知识竞赛,共10道抢答题。评分规则是答对一道题加20分,答错或不答一道题扣10分。如果把加20分记作 $+20$ 分,那么扣10分应记作多少分呢?蓝蓝在本次竞赛中的得分是110分,她答对了几道题?",
+      "options": [],
+      "answer": "-10分;7道",
+      "solution": "【分析】(1)在用正、负数表示两种具有相反意义的量时,要先规定哪种量为正(或负)。如果一种量用正数表示,那么另一种与它相反的量就用负数表示。据此解答即可。(2) 设她答对了 $x$ 道题, 则答错或不答 $(10 - x)$ 道题。答对一道题加 20 分, 答对 $x$ 道题加 $20x$ 分; 答错或不答一道题扣 10 分, 答错或不答 $(10 - x)$ 道题扣 $10 \\times (10 - x)$ 分。根据等量关系 '加的总分一扣的总分 $= 110$' 列出方程。在用正、负数表示两种具有相反意义的量时,一般用正数表示增加、上升、超出……用负数表示减少、下降、不足……\n【详解】把加 20 分记作 +20 分, 即规定加分为正, 那么扣分为负。所以扣 10 分应记作 -10 分。\n解:设她答对了 $\\mathbf{x}$ 道题。\n$2 0 x - 1 0 \\times (1 0 - x) = 1 1 0\\\\2 0 x - 1 0 0 + 1 0 x = 1 1 0\\\\3 0 x - 1 0 0 = 1 1 0 \\\\ 3 0 x = 1 1 0 + 1 0 0 \\\\ 3 0 \\mathrm {x} = 2 1 0 \\\\ x = 2 1 0 \\div 3 0 \\\\ x = 7 $\n答: 扣 10 分应记作 -10 分。她答对了 7 道题。",
+      "knowledge_points": [
+        "利用正负数解决实际问题",
+        "列方程解含两个未知数的问题",
+        "鸡兔同笼",
+        "正负数的意义及应用"
+      ],
+      "source": "21-22五年级下·山东·单元测试",
+      "merged_id": "unit1_100",
+      "original_id": "unit1_solve_100"
+    },{
+      "id": "unit1_choice_101",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "在 ' - 5, - 100, 2'这三个数中，最大的数是 '（）'。",
+      "options": [
+        "A. -5",
+        "B. - 100",
+        "C. 2"
+      ],
+      "answer": "C",
+      "solution": "【分析】大于 0 的数是正数; 小于 0 的数是负数; 则正数大于负数。据此解答即可。本题考查正负数的大小比较，明确正数大于负数是解题的关键。\n【详解】由分析可知:-5, -100, 2 这三个数中, 2 是正数, 所以最大的数是 2。故答案为: C",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "21-22 六年级下·浙江绍兴·期末",
+      "merged_id": "unit1_choice_101",
+      "original_id": "unit1_choice_101"
+    },
+    {
+      "id": "unit1_choice_102",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "$4^{\\circ} \\mathrm{C}$  比  $-8^{\\circ} \\mathrm{C}$  高（） $^{\\circ} \\mathrm{C}$ 。",
+      "options": [
+        "A. 10",
+        "B. 12",
+        "C. 4",
+        "D. 6"
+      ],
+      "answer": "B",
+      "solution": "【分析】求  $4^{\\circ} \\mathrm{C}$  比  $-8^{\\circ} \\mathrm{C}$  高多少度, 先求出  $4^{\\circ} \\mathrm{C}$  和  $0^{\\circ} \\mathrm{C}$  相差多少、 $0^{\\circ} \\mathrm{C}$  和  $-8^{\\circ} \\mathrm{C}$  相差多少, 再将求得的相差的两数相加即可。本题主要考查正负数的意义和应用。\n【详解】$4^{\\circ} \\mathrm{C}$  和  $0^{\\circ} \\mathrm{C}$  相差  $4^{\\circ} \\mathrm{C}, 0^{\\circ} \\mathrm{C}$  和  $-8^{\\circ} \\mathrm{C}$  相差  $8^{\\circ} \\mathrm{C}$ ;$4 ^ {\\circ} \\mathrm {C} + 8 ^ {\\circ} \\mathrm {C} = 1 2 ^ {\\circ} \\mathrm {C}$所以  $4^{\\circ} \\mathrm{C}$  比  $-8^{\\circ} \\mathrm{C}$  高  $12^{\\circ} \\mathrm{C}$  。故答案为: B",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "2022·四川凉山·小升初真题",
+      "merged_id": "unit1_choice_102",
+      "original_id": "unit1_choice_102"
+    },
+    {
+      "id": "unit1_choice_103",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "在 +8.6、-2.03、0、-100、17、 $-\\frac{1}{2}$ 、 $-\\frac{3}{4}$ 、90 这些数中，下列说法正确的是（）。",
+      "options": [
+        "A. 有 3 个负数",
+        "B. 0 是正数",
+        "C. 有 4 个正数",
+        "D.  $-\\frac{3}{4}$  读作负四分之三"
+      ],
+      "answer": "D",
+      "solution": "【分析】根据正、负数的意义，数的前面加有'+'号的数，就是正数（'+'可省略不写）；数的前面加有'-'号的数，就是负数，0既不是正数也不是负数，据此判断即可。本题考查正负数的认识，根据正负数的概念进行解答。\n【详解】在  $+8.6 、 -2.03 、 0 、 -100 、 17 、 -\\frac{1}{2} 、 -\\frac{3}{4} 、 90$  这些数中，负数有：-2.03、-100、  $-\\frac{1}{2}$  、  $-\\frac{3}{4}$ ，有4个；正数有：+8.6、90，有2个；0既不是正数也不是负数； $-\\frac{3}{4}$  读作负四分之三。故答案为: D",
+      "knowledge_points": [
+        "正负数的读法和写法",
+        "正负数的概念及辨认"
+      ],
+      "source": "2022·陕西安康·小升初真题",
+      "merged_id": "unit1_choice_103",
+      "original_id": "unit1_choice_103"
+    },
+    {
+      "id": "unit1_choice_104",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "一种饼干包装袋上标着：净重（ $150 \\pm 5$  克），表示这种饼干标准的质量是150克，实际每袋最多不多于（）克。",
+      "options": [
+        "A. 155",
+        "B. 150",
+        "C. 145",
+        "D. 153"
+      ],
+      "answer": "A",
+      "solution": "【分析】根据用正负数来表示具有意义相反的两种量：净重量（ $150 \\pm 5$  克），表示这种饼干的标准的质量是 150 克，实际每袋最多不超过  $150 + 5 = 155$  克，据此解答。此题主要考查正负数的意义, 正数与负数表示意义相反的两种量, 看清规定哪一个为正, 和它意义相反的就为负。\n【详解】$150 + 5 = 155$  （克）故答案为：A",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "2019·福建厦门·小升初真题",
+      "merged_id": "unit1_choice_104",
+      "original_id": "unit1_choice_104"
+    },
+    {
+      "id": "unit1_choice_105",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "下列说法错误的是 ( )。",
+      "options": [
+        "A. 不带 '-' 的数一定是正数",
+        "B. 在直线上表示数, 0是正数和负数的分界点",
+        "C. 0 既不是正数也不是负数",
+        "D. 正数一定大于负数"
+      ],
+      "answer": "A",
+      "solution": "【分析】数轴上的数以 0 为分界点, 0 右边的数大于 0 , 大于 0 的数为正数, '+' 可以省略, 0 左边的数小于 0 , 小于 0 的数为负数, '- ' 不可以省略, 0 既不是正数也不是负数, 数轴上越往右边数越大, 越往左边数越小, 正数一定大于负数, 据此解答。掌握正负数的意义和表示方法是解答题目的关键。\n【详解】分析可知, 不带 '- ' 的数可能是正数, 也可能是 0, 0 既不是正数也不是负数。故答案为: A",
+      "knowledge_points": [
+        "正负数在数轴上的表示",
+        "正负数的意义及应用"
+      ],
+      "source": "21-22 六年级下·全国·期末",
+      "merged_id": "unit1_choice_105",
+      "original_id": "unit1_choice_105"
+    },
+    {
+      "id": "unit1_choice_106",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "以小民家为起点,向东走为正, 向西走为负, 如果小民从家走了 +30 米, 又走了 -30 米, 这时小民离家的距离是 ( ) 米。",
+      "options": [
+        "A. 0",
+        "B. -30",
+        "C. 60"
+      ],
+      "answer": "A",
+      "solution": "【分析】正数、负数表示两种相反意义的量；规定向东走为正，向西走为负，小民从家走了 +30 米，即小民向东走了 30 米；又走了 -30 米，即小民又向西走了 30 米，那么此时小民回到了起点；据此解答。掌握正负数的意义，看清规定哪一个为正，则和它意义相反的就为负。\n【详解】小民从家走了 +30 米, 又走了 -30 米, 刚好回到起点, 这时小民离家的距离是 0 米。故答案为: A",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "14-15 五年级下·山东青岛·单元测试",
+      "merged_id": "unit1_choice_106",
+      "original_id": "unit1_choice_106"
+    },
+    {
+      "id": "unit1_choice_107",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "小敏从家向东走了$300 \\mathrm{~m}$ , 又向西走了  $600 \\mathrm{~m}$ , 这时小敏离家距离是 ( )。",
+      "options": [
+        "A. -300m",
+        "B.  $300 \\mathrm{~m}$",
+        "C.  $600 \\mathrm{~m}$",
+        "D.  $900 \\mathrm{~m}$"
+      ],
+      "answer": "B",
+      "solution": "【分析】小敏从家向东走了  $300 \\mathrm{~m}$ , 这时小敏在家的正东方向  $300 \\mathrm{~m}$  处, 这时又向西走了  $600 \\mathrm{~m}$ , 则相当于小敏回到家,又向西走了  $300 \\mathrm{~m}$ , 这时小敏在家的正西方向  $300 \\mathrm{~m}$  处; 由此解答即可。此题主要考查正负数的意义以及方向, 可以动手画图来解题。\n【详解】$600 - 300 = 300 \\text{ (m)}$故答案为: B",
+      "knowledge_points": [
+        "正负数的意义及应用",
+        "东、南、西、北方向"
+      ],
+      "source": "19-20 六年级下·全国·期中",
+      "merged_id": "unit1_choice_107",
+      "original_id": "unit1_choice_107"
+    },
+    {
+      "id": "unit1_choice_108",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "冬季的一天，室外温度为  $-9^{\\circ} \\mathrm{C}$ ，室内的温度是  $20^{\\circ} \\mathrm{C}$ ，则室内外温度相差（）。",
+      "options": [
+        "A.  $11^{\\circ} \\mathrm{C}$",
+        "B.  $29^{\\circ} \\mathrm{C}$",
+        "C.  $-29^{\\circ} \\mathrm{C}$",
+        "D.  $-11^{\\circ} \\mathrm{C}$"
+      ],
+      "answer": "B",
+      "solution": "【分析】以  $0^{\\circ} \\mathrm{C}$  为分界点, 高于  $0^{\\circ} \\mathrm{C}$  用 '+' 表示, 其中正号可以省略, 低于  $0^{\\circ} \\mathrm{C}$  用 '- ' 表示, -9°C到  $0^{\\circ} \\mathrm{C}$  温差为  $9^{\\circ} \\mathrm{C}, 0^{\\circ} \\mathrm{C}$  到  $20^{\\circ} \\mathrm{C}$  温差为  $20^{\\circ} \\mathrm{C}$ , 最后用加法求出室内外的温差即可。\n【详解】分析可知，-9°C到0°C温差为9°C，0°C到20°C温差为20°C；则  $-9^{\\circ} \\mathrm{C}$  到  $20^{\\circ} \\mathrm{C}$  温差为:$9 ^ {\\circ} \\mathrm {C} + 2 0 ^ {\\circ} \\mathrm {C} = 2 9 ^ {\\circ} \\mathrm {C}$故答案为: B",
+      "knowledge_points": [
+        " 温度的应用"
+      ],
+      "source": "19-20六年级上·全国·期末",
+      "merged_id": "unit1_choice_108",
+      "original_id": "unit1_choice_108"
+    },
+    {
+      "id": "unit1_choice_109",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "大于 -3 且小于 3 的数有（）个。",
+      "options": [
+        "A. 4",
+        "B. 5",
+        "C. 无数"
+      ],
+      "answer": "C",
+      "solution": "【分析】大于 -3 且小于 3 的整数有：-2、-1、0、1、2。大于 -3 且小于 3 的一位小数有：-2.1、-1.1、0.1、1.1、2.1……大于 -3 且小于 3 的两位小数有：-2.11、-1.11、0.11、1.11、2.11……大于 -3 且小于 3 还有三位小数、四位小数、五位小数……据此判断即可。此题主要考查了正、负数大小的比较, 解答此题的关键是根据小数的位数分类讨论, 求出大于 -3 而小于 3 的数有多少个。\n【详解】大于 -3 且小于 3 的整数的个数是有限的, 而小数的个数是无限的。故答案为: C",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "18-19 六年级下·全国·期末",
+      "merged_id": "unit1_choice_109",
+      "original_id": "unit1_choice_109"
+    },
+    {
+      "id": "unit1_choice_110",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "如果 140 厘米为六一班女生平均身高, 低于平均身高为负。小红的身高记作  $+2 \\mathrm{~cm}$ , 那么小红的身高是 ( ) cm。",
+      "options": [
+        "A. 138",
+        "B.  $+2$",
+        "C. 142",
+        "D. 无法确定"
+      ],
+      "answer": "C",
+      "solution": "【分析】以六一班女生平均身高为标准，已知平均身高，超出的记为正，不足的记为负，小红的身高记作  $+2 \\mathrm{~cm}$ ，说明她的身高是超出平均身高的，考虑正负数加减法即可解决问题。此题的解题关键是理解负数的概念和意义，弄清题目中是以平均身高为标准，再利用正负数加减法解答问题。\n【详解】$+2 \\mathrm{~cm}$  表示超出平均身高  $2 \\mathrm{~cm}$ ;$1 4 0 + 2 = 1 4 2 \\quad (\\mathrm {c m})$故答案为: C",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "21-22 六年级下·河南南阳·期中",
+      "merged_id": "unit1_choice_110",
+      "original_id": "unit1_choice_110"
+    },
+    {
+      "id": "unit1_choice_111",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "某天菏泽气温 -8°C, 北京 -10°C, 济南 -5°C, 则三个城市中气温最高的是 ()。",
+      "options": [
+        "A. 济南",
+        "B. 北京",
+        "C. 荷泽"
+      ],
+      "answer": "A",
+      "solution": "【分析】负数比较大小，先不考虑负号，数字部分大的数反而小。熟练掌握正负数大小的比较方法是解答本题的关键。\n【详解】-  $5^{\\circ} \\mathrm{C} > -8^{\\circ} \\mathrm{C} > -10^{\\circ} \\mathrm{C}$ , 济南的气温最高。故答案为: A",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "21-22 六年级下·山东菏泽·阶段练习",
+      "merged_id": "unit1_choice_111",
+      "original_id": "unit1_choice_111"
+    },
+    {
+      "id": "unit1_choice_112",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "一种葡萄干包装袋上标着: 净重  $(500 \\pm 10 \\mathrm{~g})$ , 表示这种葡萄干标准的质量是  $500 \\mathrm{~g}$ , 实际每袋最少不少于 ( )。",
+      "options": [
+        "A.  $490 \\mathrm{~g}$",
+        "B.  $500 \\mathrm{~g}$",
+        "C.  $510 \\mathrm{~g}$",
+        "D.  $495 \\mathrm{~g}$"
+      ],
+      "answer": "A",
+      "solution": "【分析】因为把这种葡萄干每袋的标准质量 500 克记为 0 ,即 500 克为标准，超出的记为正，不足的记为负，由此解决问题。此题首先要知道以谁为标准，规定超出标准的为正，低于标准的为负，由此用正负数解答问题。\n【详解】$500 - 10 = 490$  (g)故答案为: A",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "21-22 六年级下·全国·单元测试",
+      "merged_id": "unit1_choice_112",
+      "original_id": "unit1_choice_112"
+    },
+    {
+      "id": "unit1_choice_113",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "在直线上，规定向右为正，点A先向右移动5个单位长度，再向左移动3个单位长度，点A现在的位置是-5，点A原来的位置是（）。",
+      "options": [
+        "A. 3",
+        "B. -3",
+        "C. -7",
+        "D. 7"
+      ],
+      "answer": "C",
+      "solution": "【分析】向右和向左是两个具有相反意义的量，向右记作'+'，向左记作'-'。从现在的位置移到原来的位置，应该是向右移动 3 个单位长度, 再向左移动 5 个单位长度。利用正负数加减法求出结果。本题主要考查正、负数的意义及其应用，利用正负数加减法求出结果。\n【详解】$(-5) + 3 + (-5) = -7$故答案为: C",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20六年级下·全国·阶段练习",
+      "merged_id": "unit1_choice_113",
+      "original_id": "unit1_choice_113"
+    },
+    {
+      "id": "unit1_choice_114",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "体育锻炼标准规定:13岁男生每分钟做22个仰卧起坐为达标，超过标准的个数用正数表示，不足的个数用负数表示。八位同学的成绩分别记录为：+3，-1，+1，0，-2，+2，+4，-3。这八位同学中达标的有（）人。",
+      "options": [
+        "A. 4",
+        "B. 5",
+        "C. 6",
+        "D. 8"
+      ],
+      "answer": "B",
+      "solution": "【分析】此题主要用正负数来表示具有意义相反的两种量:超过标准的个数用正数表示, 不足的个数用负数表示, 直接得出结论即可。此题主要考查正负数的意义, 正数与负数表示意义相反的两种量, 看清规定哪一个为正, 则和它意义相反的就为负。\n【详解】超过标准的个数用正数表示，不足的个数用负数表示，成绩分别记录为：+3，+1，0，+2，+4达标，所以这八位同学中有5位达标。故答案为: B。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "21-22 六年级下·全国·单元测试",
+      "merged_id": "unit1_choice_114",
+      "original_id": "unit1_choice_114"
+    },
+    {
+      "id": "unit1_choice_115",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "下面每组中的两个量不具有相反意义的是 ( )。",
+      "options": [
+        "A. '盈利600元'与'亏损900元'",
+        "B. 浪费10千瓦时电与节约10千瓦时电",
+        "C. 下降  $10 \\mathrm{~m}$  与降低  $9 \\mathrm{~m}$",
+        "D. 收入 1000 元与支出 1000元"
+      ],
+      "answer": "C",
+      "solution": "【分析】根据负数的意义，可得：盈利记为'+'，则亏损记为'-'；节约记为'+'，则浪费记为'-'；收入记为'+'，则支出记为'-'；上升记为'+'，则下降记为'-'，据此判断即可。此题主要考查了负数的意义及其应用, 要熟练掌握。\n【详解】A. '盈利 600 元'与'亏损 900 元'，盈利和亏损是具有相反意义的量；  B. 浪费 10 千瓦时电与节约 10 千瓦时电, 浪费和节约是具有相反意义的量;  C. 下降  $10 \\mathrm{~m}$  与降低  $9 \\mathrm{~m}$ , 下降和降低不是具有相反意义的量;  D. 收入 1000 元与支出 1000 元, 收入和支出是具有相反意义的量;故答案为: C",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "21-22 六年级下·全国·单元测试",
+      "merged_id": "unit1_choice_115",
+      "original_id": "unit1_choice_115"
+    },
+    {
+      "id": "unit1_choice_116",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "下面比较大小正确的是（）。",
+      "options": [
+        "A.  $0 < -1$",
+        "B.  $-3 > -2$",
+        "C.  $-\\pi < -3.1$"
+      ],
+      "answer": "C",
+      "solution": "【分析】根据负数的大小比较方法，一一分析各个选项的正误即可。本题考查了负数的大小比较, 负数比较大小时,负号后的值越大, 对应的负数越小。\n【详解】A.  $0 > -1$ , 所以原比较错误;  B.  $-3 < -2$ , 所以原比较错误;  C.  $-\\pi < -3.1$ , 所以原比较正确。故答案为: C",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "19-20 六年级下·湖南怀化·单元测试",
+      "merged_id": "unit1_choice_116",
+      "original_id": "unit1_choice_116"
+    },
+    {
+      "id": "unit1_choice_117",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "在数轴上想从原点到 -8, 应 ( ) 运动才行。",
+      "options": [
+        "A. 向左",
+        "B. 向右",
+        "C. 无法确定"
+      ],
+      "answer": "A",
+      "solution": "【分析】- 8 是负数，在数轴原点的左边。据此解题。本题考查了负数在数轴上的位置, 明确负数和原点的相对位置是解题的关键。\n【详解】在数轴上想从原点到 -8，应向左运动才行。故答案为：A",
+      "knowledge_points": [
+        "正负数在数轴上的表示"
+      ],
+      "source": "19-20 六年级下·湖南怀化·单元测试",
+      "merged_id": "unit1_choice_117",
+      "original_id": "unit1_choice_117"
+    },
+    {
+      "id": "unit1_choice_118",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "全班同学的平均身高是  $145 \\mathrm{~cm}$ , 记作 '0', 那么小华身高  $144 \\mathrm{~cm}$ , 应记作( )。",
+      "options": [
+        "A.  $+144 \\mathrm{~cm}$",
+        "B.  $144 \\mathrm{~cm}$",
+        "C. -1cm"
+      ],
+      "answer": "C",
+      "solution": "【分析】用 145 厘米减去 144 厘米, 求出小华身高和 '0' 的差距。又由于  $144 < 145$ , 所以小华身高  $144 \\mathrm{~cm}$ , 应记作负数。据此解题。本题考查了负数的应用, 表示比 '0'还少的量记为负数。\n【详解】$145 - 144 = 1$  （厘米），所以小华身高  $144 \\mathrm{~cm}$ ，应记作  $-1 \\mathrm{~cm}$ 。故答案为: C",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·湖南怀化·单元测试",
+      "merged_id": "unit1_choice_118",
+      "original_id": "unit1_choice_118"
+    },
+    {
+      "id": "unit1_choice_119",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "对'0'的描述错误的是（）。",
+      "options": [
+        "A. 0 是自然数",
+        "B. 0 不是正数也不是负数",
+        "C. 既可以是正数也可以是负数"
+      ],
+      "answer": "C",
+      "solution": "【分析】根据 0 的概念和特征，——判断各个选项的正误，从而选出正确选项。本题考查了 0 的分类, 明确 '0 既不是正数也不是负数' 是解题的关键。\n【详解】A．0是自然数，原说法正确；  B. 0 不是正数也不是负数, 原说法正确;  C. 0 不是正数也不是负数, 原说法错误。故答案为: C",
+      "knowledge_points": [
+        "自然数的认识",
+        "正负数的概念及辨认"
+      ],
+      "source": "19-20六年级下·湖南怀化·单元测试",
+      "merged_id": "unit1_choice_119",
+      "original_id": "unit1_choice_119"
+    },
+    {
+      "id": "unit1_choice_120",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "如果规定向南走为正，那么 -50 米表示的意义是（）。",
+      "options": [
+        "A. 向东走 50 米",
+        "B. 向西走 50 米",
+        "C. 向北走 50 米"
+      ],
+      "answer": "C",
+      "solution": "【分析】和南为反方向的是北，和正数相反的是负数。所以，如果规定向南走为正，那么 -50 米表示向北走。据此解题。本题考查了正负数的意义和应用, 负数表示和正数意义相反的量。\n【详解】如果规定向南走为正，那么-50米表示的意义是向北走50米。故答案为: C",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·湖南怀化·单元测试",
+      "merged_id": "unit1_choice_120",
+      "original_id": "unit1_choice_120"
+    },
+    {
+      "id": "unit1_choice_121",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "-5, -45, +7, +1.3, 0, -1, 负数有（）。",
+      "options": [
+        "A. 2 个",
+        "B. 3 个",
+        "C. 4 个"
+      ],
+      "answer": "B",
+      "solution": "【分析】带有 '- ' 的数是负数, 由此进行分类, 再数出负数的个数。此题考查负数的辨识：带有'-'的数就是负数。\n【详解】-5，-45，+7，+1.3，0，-1，负数有-5，-45，-1，共有3个。故答案为: B。",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "21-22 六年级下·全国·单元测试",
+      "merged_id": "unit1_choice_121",
+      "original_id": "unit1_choice_121"
+    },
+    {
+      "id": "unit1_choice_122",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "下面各数中, ( ) 最接近 0 。",
+      "options": [
+        "A. -3",
+        "B. -1",
+        "C.  $+2$",
+        "D. 5"
+      ],
+      "answer": "B",
+      "solution": "【分析】和 0 的差最小的数, 最接近 0 。据此先求出各个选项中的数和 0 的差，再选出正确选项即可。本题考查了正负数, 明确正负数的意义是解题的关键。\n【详解】A.  $0 - (-3) = 3;$B.  $0 - (-1) = 1$ ;  C.  $2 - 0 = 2$ ;  D.  $5 - 0 = 5$ ;3、1、2、5这四个数中1最小，所以-1最接近0。故答案为: B",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "2014 六年级·全国·课后作业",
+      "merged_id": "unit1_choice_122",
+      "original_id": "unit1_choice_122"
+    },
+    {
+      "id": "unit1_choice_123",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "下雪了! 同学们到操场堆雪人。此时的室外温度可能是 ( )。",
+      "options": [
+        "A.  $-12^{\\circ} \\mathrm{C}$",
+        "B.  $10^{\\circ} \\mathrm{C}$",
+        "C.  $2^{\\circ} \\mathrm{C}$"
+      ],
+      "answer": "A",
+      "solution": "【分析】根据生活常识可知，下雪时室外温度在零度以下，据此即可选择。本题主要考查学生正负数知识的掌握。\n【详解】根据分析可知，下雪时室外温度在零度以下，三个选项只有 A 是零度以下。故答案为：A。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "21-22 四年级上·辽宁营口·期末",
+      "merged_id": "unit1_choice_123",
+      "original_id": "unit1_choice_123"
+    },
+    {
+      "id": "unit1_choice_124",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "下列数中，最接近 -2 的数是（）。",
+      "options": [
+        "A. 1",
+        "B. -3",
+        "C. 0",
+        "D. 2.1"
+      ],
+      "answer": "B",
+      "solution": "【分析】求与 -2 最接近的数, 即求到 -2 的距离最近的点,据此解答即可。先求出每个数到 -2 的距离是解答本题的关键。\n【详解】A. 1 到 -2 的距离为 3 ;B. -3 到 -2 的距离为 1 ;C. 0 到 -2 的距离为 2 ;D. 2.1 到 -2 的距离为 4.1;$1 <   2 <   3 <   4.1;$故答案为: B。",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "19-20五年级上·安徽芜湖·期末",
+      "merged_id": "unit1_choice_124",
+      "original_id": "unit1_choice_124"
+    },
+    {
+      "id": "unit1_choice_125",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "如果规定前进、收入、增加为正，那么下面错误的语句是（）。",
+      "options": [
+        "A. -18 米表示后退 18 米",
+        "B. -42人表示增加42人",
+        "C. -4万元表示支出4万元"
+      ],
+      "answer": "B",
+      "solution": "【分析】生活中，通常用正负数表示具有相反意义的两种量，如果规定前进、收入、增加为正，那么后退、支出、减少为负，据此解答。此题主要考查正负数的意义, 正数与负数表示意义相反的两种量, 看清规定哪一个为正, 则和它意义相反的就为负。\n【详解】如果规定前进、收入、增加为正，那么-42人表示减少 42 人, 其它选项表述正确。故答案为: B",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "18-19六年级下·全国·单元测试",
+      "merged_id": "unit1_choice_125",
+      "original_id": "unit1_choice_125"
+    },
+    {
+      "id": "unit1_choice_126",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "向前3步记作  $+3$  ，原地不动记作（）。",
+      "options": [
+        "A. 1",
+        "B.  $+1$",
+        "C. 0",
+        "D. -1"
+      ],
+      "answer": "C",
+      "solution": "【分析】在一对具有相反意义的量中，规定向前为正，则向后就为负，原地不动记作 0；由此解答即可。解题关键是理解'正'和'负'的相对性，确定一对具有相反意义的量。\n【详解】向前3步记作  $+3$ , 原地不动记作  $0$  。故答案为: C。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 五年级上·山西临汾·期中",
+      "merged_id": "unit1_choice_126",
+      "original_id": "unit1_choice_126"
+    },
+    {
+      "id": "unit1_choice_127",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "在－4、24、-8、4.3、0、54中，正数有（）个。",
+      "options": [
+        "A. 2",
+        "B. 3",
+        "C. 4",
+        "D. 5"
+      ],
+      "answer": "B",
+      "solution": "【分析】先找出 -4、24、-8、4.3、0、54 中的正数，再统计出正数的数量即可。本题考查了正负数的认识, 0 既不是正数也不是负数。\n【详解】- 4、24、-8、4.3、0、54中的正数有：24、4.3、54，一共有 3 个正数。故答案为: B",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "19-20六年级下·山东济南·期末",
+      "merged_id": "unit1_choice_127",
+      "original_id": "unit1_choice_127"
+    },
+    {
+      "id": "unit1_choice_128",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "体育老师对六年级女生进行跳绳测试。如果以每分钟70个为合格，王红每分钟跳80个，记作  $+10$  个。李晴的成绩记作-5个，则她跳了（）个。",
+      "options": [
+        "A. 75",
+        "B. 65",
+        "C. 5",
+        "D. -5"
+      ],
+      "answer": "B",
+      "solution": "【分析】以 70 个为标准, 多于 70 个的数量记为正, 少于 70 个的数量记为负, 据此分析。关键是理解正负数的意义, 正负数可以表示相反意义的量。\n【详解】$70 - 5 = 65$  （个）故答案为: B",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "2021·北京朝阳·小升初真题",
+      "merged_id": "unit1_choice_128",
+      "original_id": "unit1_choice_128"
+    },
+    {
+      "id": "unit1_choice_129",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "在下面各数中, 最接近0的是（）。",
+      "options": [
+        "A. -2",
+        "B. 1",
+        "C. 3",
+        "D. 2"
+      ],
+      "answer": "B",
+      "solution": "【分析】根据题意，分别求出每个数与0的差分别是多少，差越小, 则越接近 0 , 据此解答即可。借助数轴可以更加直观的进行解答。\n【详解】$0 - (-2) = 2$ $1 - 0 = 1$ $3 - 0 = 3$ $2 - 0 = 2$$1 <   2 = 2 <   3$所以最接近 0 的是 1。故答案为: B",
+      "knowledge_points": [
+        "正负数在数轴上的表示",
+        "正负数的意义及应用"
+      ],
+      "source": "18-19 六年级下·山西大同·期末",
+      "merged_id": "unit1_choice_129",
+      "original_id": "unit1_choice_129"
+    },
+    {
+      "id": "unit1_choice_130",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "一袋盐的外包装袋上标着'净重  $(200 \\pm 5)$  克'，表示这装盐的标准净重是 200 克，实际这袋盐的净重最少是（）克。",
+      "options": [
+        "A. 205",
+        "B. 200",
+        "C. 195"
+      ],
+      "answer": "C",
+      "solution": "【分析】净重  $(200 \\pm 5)$  克, 表示一袋盐的重量在  $(200-$  5 ) 克和  $(200 + 5)$  克之间, 据此解答。此题考查了正负数的意义及应用, 明确正负数表示意义相反的两个量。\n【详解】$200 - 5 = 195$  （克），实际这袋盐的净重最少是 195 克。故选择：C。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "2021·新疆乌鲁木齐·小升初真题",
+      "merged_id": "unit1_choice_130",
+      "original_id": "unit1_choice_130"
+    },
+    {
+      "id": "unit1_choice_131",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "希望小学六年级开展'学党史'知识竞赛，共设10道题；每道题10分，答对一题得 10 分, 用 +10 分表示, 答错一题扣 10 分, 用 -10 分表示, 亮亮最后答对了 6 题, 答错了 4 题, 他得了 ( ) 分。",
+      "options": [
+        "A. 20",
+        "B. 40",
+        "C. 60",
+        "D. 80"
+      ],
+      "answer": "A",
+      "solution": "【分析】利用乘法分别计算出亮亮答对的得分，以及答错的扣分，再做差计算出他的最终得分即可。本题考查了正负数的意义和应用, 负数是表示与正数意义相反的量。\n【详解】$6 \\times 10 - 4 \\times 10\\\\= 6 0 - 4 0 \\\\ = 2 0 (分) $所以，亮亮他得了 20 分。故答案为: A",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·河南濮阳·期末",
+      "merged_id": "unit1_choice_131",
+      "original_id": "unit1_choice_131"
+    },
+    {
+      "id": "unit1_choice_132",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "下面每组中的两个量不是具有相反意义的量是（）。",
+      "options": [
+        "A. 收入 100 元与支出 100 元",
+        "B. 盈利 500 元与亏损 500 元",
+        "C. 增产 2 吨与减产 2 吨",
+        "D. 向东走 8 米与向南走 8 米"
+      ],
+      "answer": "D",
+      "solution": "【分析】属性相同，但表示的意义相反的量叫做相反意义的量。必须满足两个条件：1、它们是同一属性的量;  2、它们的意义相反。据此解答。本题考查负数的意义，明确相反意义的量的概念是解题的关键。\n【详解】由分析可知，向东和向南不具有相反的意义。故选：D",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·河南濮阳·期末",
+      "merged_id": "unit1_choice_132",
+      "original_id": "unit1_choice_132"
+    },
+    {
+      "id": "unit1_choice_133",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "在 +6、-4、- 1.8、0、 $\\frac{3}{4}$ 、 $-\\frac{2}{11}$ 、-0.303中，负数有（）个。",
+      "options": [
+        "A. 2",
+        "B. 3",
+        "C. 4",
+        "D. 5"
+      ],
+      "answer": "C",
+      "solution": "【分析】小于 0 的数是负数, 0 既不是负数也不是正数, 据此解答即可。本题考查负数的意义, 明确负数的意义是解题的关键。\n【详解】由分析可知, 负数有: -4、-1.8、 $-\\frac{2}{11}$ 、-0.303 共有 4 个。故选: C",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "20-21 六年级下·河南濮阳·期末",
+      "merged_id": "unit1_choice_133",
+      "original_id": "unit1_choice_133"
+    },{
+      "id": "unit1_choice_134",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "在 -6, -0.6, 0,0.06 这四个数中，最大的负数是（）。",
+      "options": [
+        "A. -6",
+        "B. -0.6",
+        "C. 0",
+        "D. -0.06"
+      ],
+      "answer": "D",
+      "solution": "【分析】负数大小比较的方法：数字越大，这个数越小，反之则越大，据此解答即可。解答本题的关键是注意问题, 求最大的负数, 要把 0 排除在外。\n【详解】$-0.06 > -0.6 > -6$ ;故答案为: D。",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "19-20 六年级下·江西赣州·期末",
+      "merged_id": "unit1_choice_134",
+      "original_id": "unit1_choice_134"
+    },
+    {
+      "id": "unit1_fill_135",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "一种食品包装袋上标着：净重（ $275 \\pm 5$  克），表示这种食品每袋最多不超过( )克。",
+      "options": [],
+      "answer": "280",
+      "solution": "【分析】正数、负数表示两种相反意义的量。' $275 \\pm 5$  克'的含义，即275克是这种食品的标准净重，实际每袋最多不超过  $(275 + 5)$  克，最少不低于  $(275 - 5)$  克，据此解答。掌握正负数的意义，知道以哪个数为标准，规定超出标准的为正，低于标准的为负。\n【详解】$275 + 5 = 280$  （克）",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "2022·广东佛山·小升初真题",
+      "merged_id": "unit1_fill_135",
+      "original_id": "unit1_fill_135"
+    },
+    {
+      "id": "unit1_fill_136",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "在 24、-3.5、+6.2、 $-\\frac{5}{6}$ 、 $20\\%$ 、-0.8、0、 $\\frac{9}{2}$ 、-2022 中, 正数有( ),负数有( )。",
+      "options": [],
+      "answer": "24、+6.2、20%、 $\\frac{9}{2}$ #### -3.5、- $\\frac{5}{6}$ 、-0.8、-2022",
+      "solution": "【分析】正负数：像 +30，+3.14，+15……这样的数，都是正数，前面的 + 可以省略；像 -10，-0.8……这样的数，是负数，负号不能省略。0 既不是正数也不是负数。本题主要考查学生对正负数概念的掌握和灵活运用。\n【详解】根据分析得，正数有24、+6.2、20%、 $\\frac{9}{2}$ ；负数有-3.5、- $\\frac{5}{6}$ 、-0.8、-2022。",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "21-22 六年级下·山东济宁·期末",
+      "merged_id": "unit1_fill_136",
+      "original_id": "unit1_fill_136"
+    },
+    {
+      "id": "unit1_fill_137",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "一天，沈阳市的最低气温是零下7摄氏度，记作  $-7^{\\circ} \\mathrm{C}$ ，当天，杭州市的最低气温比沈阳市高9摄氏度。则杭州市的最低气温记作（ ） $^{\\circ} \\mathrm{C}$ 。",
+      "options": [],
+      "answer": "2",
+      "solution": "【分析】通常用正负数表示具有相反意义的两种量，零下温度通常用负数表示，零上温度用正数表示，据此解答。只有具有相反意义的量才可以用正负数进行表示,南和北、东和西、海平面以上和海平面以下等都是具有相反意义的量。\n【详解】由分析可得：杭州市的最低气温记作  $2^{\\circ} \\mathrm{C}$  。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "21-22六年级下·福建泉州·期末",
+      "merged_id": "unit1_fill_137",
+      "original_id": "unit1_fill_137"
+    },
+    {
+      "id": "unit1_fill_138",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "稻花香大米的包装袋上注明：净重10千克  $\\pm 0.25$  千克。任意取两袋大米质量最多相差（ )千克。",
+      "options": [],
+      "answer": "0.5",
+      "solution": "【分析】根据'正'和'负'所表示的意义得出每袋大米的最多含量和最小含量，再两者相减即可得出答案。此题考查了'正'和'负'所表示的意义, 及其如何理解每袋大米的最多含量和最小含量是解题关键。\n【详解】$10 + 0.25 = 10.25$  （千克）$10 - 0.25 = 9.75$  （千克）$10.25 - 9.75 = 0.5$  （千克）",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "2022·河南开封·小升初真题",
+      "merged_id": "unit1_fill_138",
+      "original_id": "unit1_fill_138"
+    },
+    {
+      "id": "unit1_fill_139",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "如果下降 50 米记作 -50 米, 那么上升 500 米记作( )米。",
+      "options": [],
+      "answer": "+500 / 500",
+      "solution": "【分析】如果下降记为负，那么上升记为正，据此填空。关键是理解正负数的意义, 正负数可以表示相反意义的量。\n【详解】如果下降 50 米记作 -50 米, 那么上升 500 米记作 +500 米。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "2022·广东汕头·小升初真题",
+      "merged_id": "unit1_fill_139",
+      "original_id": "unit1_fill_139"
+    },
+    {
+      "id": "unit1_fill_140",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "如果把一组同学的平均身高  $138 \\mathrm{~cm}$  记作  $0 \\mathrm{~cm}$  。那么小飞身高  $136 \\mathrm{~cm}$  应记作( )cm，小刚身高143cm应记作( )cm。",
+      "options": [],
+      "answer": "-2#### +5",
+      "solution": "【分析】此题主要用正负数来表示具有意义相反的两种量:以平均身高  $138 \\mathrm{~cm}$  为标准记为 0 , 超过部分为正, 不足的部分为负, 直接得出结论即可。此题首先要知道以谁为标准，规定超出标准的为正，低于标准的为负，由此用正负数解答问题。\n【详解】由分析可得: 小飞身高  $136 \\mathrm{~cm}$  应记作  $-2 \\mathrm{~cm}$ , 小刚身高  $143 \\mathrm{~cm}$  应记作  $+5 \\mathrm{~cm}$  。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "18-19 六年级下·云南昆明·期末",
+      "merged_id": "unit1_fill_140",
+      "original_id": "unit1_fill_140"
+    },
+    {
+      "id": "unit1_fill_141",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "小云刚开始的位置在 0 处，如果小云从 0 点向东行 5 米，表示为 +5 米，那么小云现在 -8 米处，说明他是向( )行( )米，如果小华从 0 点先向东行 5 米，又向西行 8 米，这时小云的位置在( )米处。",
+      "options": [],
+      "answer": "西#### 8#### -3",
+      "solution": "【分析】向东行 5 米, 表示为 +5 米, 也就是向东为 '正',那么向西为 '负', 所以 -8 米应该是向西走 8 米, 而先向东行 5 米, 又向西行 8 米, 相当于是向西走 3 米。本题考查的是正负数的实际应用, 可以用正负数表示一组具有相反意义的量。\n【详解】小云现在 -8 米处，说明他是向西行 8 米；  $8 - 5 = 3$  （米）如果小华从 0 点先向东行 5 米, 又向西行 8 米, 这时小云的位置在 -3 米处。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·山西忻州·期中",
+      "merged_id": "unit1_fill_141",
+      "original_id": "unit1_fill_141"
+    },
+    {
+      "id": "unit1_fill_142",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "如果向东走$8\\mathrm{m}$  ，记为  $+8\\mathrm{m}$  ，那么向西走  $5\\mathrm{m}$  ，记为( )m，( )m表示起点。",
+      "options": [],
+      "answer": "-5#### 0",
+      "solution": "【分析】此题主要用正负数来表示具有意义相反的两种量:向向东走记为正, 则向西走就记为负, 直接得出结论即可。此题主要考查正负数的意义, 正数与负数表示意义相反的两种量, 看清规定哪一个为正, 则和它意义相反的就为负。\n【详解】如果向东走  $8 \\mathrm{~m}$  记作  $+8 \\mathrm{~m}$ , 那么向西走  $5 \\mathrm{~m}$  记作  $- 5 \\mathrm{~m}, 0 \\mathrm{~m}$  表示起点。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "21-22 六年级下·广东江门·阶段练习",
+      "merged_id": "unit1_fill_142",
+      "original_id": "unit1_fill_142"
+    },
+    {
+      "id": "unit1_fill_143",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "温度越低就越冷, -2°C与-12°C相比, ( )温度低。",
+      "options": [],
+      "answer": "-$12^{\\circ} \\mathrm{C}$",
+      "solution": "【分析】比  $0^{\\circ} \\mathrm{C}$  低的温度叫零下温度, 通常在数字前加 '- ' 表示。例如, -3°C表示零下3摄氏度。温度是零下的, 数字越大, 温度越低,  $0^{\\circ} \\mathrm{C}$  比任何一个零下几摄氏度都高; 据此解答。\n【详解】温度越低就越冷，-2°C与-12°C相比，-12°C温度低。",
+      "knowledge_points": [
+        "温度的认识及比较"
+      ],
+      "source": "21-22 六年级下·广东江门·阶段练习",
+      "merged_id": "unit1_fill_143",
+      "original_id": "unit1_fill_143"
+    },
+    {
+      "id": "unit1_fill_144",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "+ 6.8 读作( )，负八分之七写作( )。",
+      "options": [],
+      "answer": "正六点八#### $-\\frac{7}{8}$",
+      "solution": "【分析】正负数的读法：先读正负号，再读数即可；正负数的写法：先写正负号，再写数即可。此题考查正负数的读写法。\n【详解】由分析可知:+6.8 读作 (正六点八), 负八分之七写作  $(-\\frac{7}{8})$  。",
+      "knowledge_points": [
+        "正负数的读法和写法"
+      ],
+      "source": "21-22 六年级下·广东江门·阶段练习",
+      "merged_id": "unit1_fill_144",
+      "original_id": "unit1_fill_144"
+    },
+    {
+      "id": "unit1_fill_145",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "某地二月份某天上午气温是  $-2^{\\circ} \\mathrm{C}$ , 中午时气温上升了  $7^{\\circ} \\mathrm{C}$ , 此时的气温是  $( )^{\\circ} \\mathrm{C} 。$",
+      "options": [],
+      "answer": "5",
+      "solution": "【分析】要求中午的气温就用早上的气温加上上升的气温就可以得到中午的气温。本题是一道有理数的加法计算题, 考查了加法的意义和有理数加法的法则。\n【详解】由题意，得  $-2 + 7 = 5$",
+      "knowledge_points": [
+        "利用正负数解决实际问题"
+      ],
+      "source": "20-21 六年级下·福建莆田·期中",
+      "merged_id": "unit1_fill_145",
+      "original_id": "unit1_fill_145"
+    },
+    {
+      "id": "unit1_fill_146",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "所有的负数都比 0( ), 所有的正数都比 0( )。所有负数都比正数( )。既不是正数也不是负数的是( )。",
+      "options": [],
+      "answer": "小#### 大#### 小#### 0",
+      "solution": "【分析】比 0 大的数叫正数, 比 0 小的数叫负数, 负数  $<$  正数, 0 既不是正数也不是负数, 据此分析。关键是掌握正负数的意义，能够正确辨认正数和负数。\n【详解】所有的负数都比 0 小, 所有的正数都比 0 大。所有负数都比正数小。既不是正数也不是负数的是 0 。",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "21-22 六年级下·湖南永州·阶段练习",
+      "merged_id": "unit1_fill_146",
+      "original_id": "unit1_fill_146"
+    },
+    {
+      "id": "unit1_fill_147",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "六一班的数学平均分记作0分，那么高于平均分5分记作( )，-12分表示( )。",
+      "options": [],
+      "answer": "+5分#### 低于平均分12分",
+      "solution": "【分析】以平均分为标准，超出的记为正，不足的记为负，由此解决问题。此题首先以平均分为标准，规定高出标准的为正，低于标准的为负，由此用正负数解答问题。\n【详解】六一班的数学平均分记作 0 分, 那么高于平均分 5 分记作 +5 分, -12 分表示低于平均分 12 分。",
+      "knowledge_points": [
+        "正负数的意义及应用",
+        "正负数的读法和写法"
+      ],
+      "source": "21-22六年级下·山东菏泽·阶段练习",
+      "merged_id": "unit1_fill_147",
+      "original_id": "unit1_fill_147"
+    },
+    {
+      "id": "unit1_fill_148",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "在数轴上, 0 的左边是( )数, 右边是( )数。",
+      "options": [],
+      "answer": "负#### 正",
+      "solution": "【分析】在数轴上表示正负数，所有的负数都在 0 的左边，正数都在 0 的右边；0 是正、负数的分界点，它既不是正数也不是负数。此题的解题关键是充分认识数轴, 掌握正负数在数轴上的表示。\n【详解】数轴是规定了原点（0 点）、方向和单位长度的直线，在原点（0 点）左边的点所表示的数都是负数，右边的点表示的数都是正数。",
+      "knowledge_points": [
+        "正负数在数轴上的表示"
+      ],
+      "source": "21-22 六年级下·山东菏泽·阶段练习",
+      "merged_id": "unit1_fill_148",
+      "original_id": "unit1_fill_148"
+    },
+    {
+      "id": "unit1_fill_149",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "某日, 庐山的气温正中午时为零上  $9^{\\circ} \\mathrm{C}$ , 记作( ), 夜间 12 时的气温为零下  $3^{\\circ} \\mathrm{C}$ , 记作( ), 这两个时间的气温相差( )。",
+      "options": [],
+      "answer": "$+9^{\\circ} \\mathrm{C}$#### $ -3^{\\circ} \\mathrm{C} $#### $12^{\\circ} \\mathrm{C}$",
+      "solution": "【分析】以  $0^{\\circ} \\mathrm{C}$  为标准, 超出标准的为正, 读作零上多少  ${ }^{\\circ} \\mathrm{C}$ ,写的时候前面加正号; 低于标准的为负, 读作零下多少  ${ }^{\\circ} \\mathrm{C}$ ,写的时候前面加负号。根据正负数加减法则求出温度差。此题首先清楚以什么为标准，规定超出标准的为正，低于标准的为负，再根据正负数加减法则求出温度差。\n【详解】庐山的气温正中午时为零上  $9^{\\circ} \\mathrm{C}$ , 记作  $+9^{\\circ} \\mathrm{C}$ ; 夜间 12 时的气温为零下  $3^{\\circ} \\mathrm{C}$ , 记作  $-3^{\\circ} \\mathrm{C}$  。 $9^{\\circ} \\mathrm{C} + 3^{\\circ} \\mathrm{C} = 12^{\\circ} \\mathrm{C}$ , 即这两个时间的气温相差  $12^{\\circ} \\mathrm{C}$  。",
+      "knowledge_points": [
+        "利用正负数解决实际问题",
+        "正负数的意义及应用"
+      ],
+      "source": "21-22 六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_149",
+      "original_id": "unit1_fill_149"
+    },
+    {
+      "id": "unit1_fill_150",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "一艘潜水艇所处的位置是 - 130m, 如果它先下潜 30m, 再上升 50m, 所处的位置是( )m。",
+      "options": [],
+      "answer": "- 110",
+      "solution": "【分析】下潜和上升是两个具有相反意义的量，上升记作'+'，下潜记作'-'。利用正负数加减法求出结果。本题主要考查正、负数的意义及其应用，利用正负数加减法求出结果。\n【详解】$(-130) + (-30) + 50 = -110(\\mathrm{m})$",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "21-22 六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_150",
+      "original_id": "unit1_fill_150"
+    },
+    {
+      "id": "unit1_fill_151",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "一座楼高 28 层, 如果把王阿姨所在的第 16 层记作 0 层, 那么她下到第 6 层时可以记作( )层。",
+      "options": [],
+      "answer": "- 10",
+      "solution": "【分析】上和下是两个具有相反意义的量，通常上记作'+'，下记作'-'，把16层记作0层，向上几层记作'+几层'，向下几层读作记作'-几层'，她下到第6层，向下下了（16-6）层，即10层，记作'-10层'。此题考查正、负数的意义及应用, 关键是确定具有相反意义的量。\n【详解】一座楼高 28 层, 如果把王阿姨所在的第 16 层记作 0 层, 那么她下到第 6 层时可以记作 -10 层。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "21-22 六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_151",
+      "original_id": "unit1_fill_151"
+    },
+    {
+      "id": "unit1_fill_152",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "在存折的'存入（+）或支出（-）'栏目中，'+2000'表示( )，'-800'表示( )。",
+      "options": [],
+      "answer": "存入2000元#### 支出800元",
+      "solution": "【分析】存入和支出是两个具有相反意义的量，通常存入记作'+'，支出记作'-'。本题是考查正、负数的意义及其应用，属于基础题。\n【详解】在存折的'存入（+）或支出（-）'栏目中，'+2000'表示存入2000元，'-800'表示支出800元。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "21-22六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_152",
+      "original_id": "unit1_fill_152"
+    },
+    {
+      "id": "unit1_fill_153",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "海平面的高度一般用( )m 表示。马里亚纳海沟海拔 -11034m，表示比海平面( )11034m。",
+      "options": [],
+      "answer": "0#### 低",
+      "solution": "【分析】此题主要用正负数来表示具有意义相反的两种量:高于海平面记为正, 低于海平面记为负, 海平面的高度一般用  $0 \\mathrm{~m}$  表示, 直接写出结论即可。此题主要考查正负数的意义, 正数负数表示意义相反的两种量, 看清规定哪一个为正, 则和它意义相反的就为负。\n【详解】海平面的高度一般用  $0 \\mathrm{~m}$  表示。马里亚纳海沟海拔  $-11034 \\mathrm{~m}$ , 表示比海平面低  $11034 \\mathrm{~m}$  。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "21-22 六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_153",
+      "original_id": "unit1_fill_153"
+    },
+    {
+      "id": "unit1_fill_154",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "在 - 5.08, -  $\\frac{1}{4}$ , -0.13, 0, +3.5 中, 负数有( )个, ( )既不是正数也不是负数。",
+      "options": [],
+      "answer": "3#### 0",
+      "solution": "【分析】像 10、20、100 这样的数，是正数，正数前面可以添加 '+' 号，例 +10、+20、+100 等；像 -10、-20 这样的数，是负数，负号不能省略；0 既不正数，也不是负数。本题主要考查学生对正负数辨认知识的掌握和灵活运用。\n【详解】在  $-5.08$  、  $-\\frac{1}{4}$ ,  $-0.13, 0, +3.5$  中, 正数有:+3.5; 负数有: -5.08、 $-\\frac{1}{4}$  、 -0.13, 负数有 3 个; 0 既不是正数, 也不是负数。",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "21-22 六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_154",
+      "original_id": "unit1_fill_154"
+    },
+    {
+      "id": "unit1_fill_155",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "在 -9,  $-\\frac{1}{5}$,  $0, + 10, + \\frac{7}{2}, 13, 0.8$  这些数中, 正数的有( ), 负数的有( )，自然数有( )，整数有( )。",
+      "options": [],
+      "answer": "$+10$; $+\\frac{7}{2}$; 13; 0.8#### -9;$ -\\frac{1}{5}$#### 0; +10; 13#### 0, + 10; 1 3; - 9",
+      "solution": "【分析】正数大于 0 , 负数前面带负号, 0 既不是正数也不是负数, 再根据自然数和整数的概念即可。本题考查负数, 解答本题的关键是掌握负数的概念。\n【详解】正数的有：  $+10; +\\frac{7}{2}; 13; 0.8$ 负数的有：-9 $-\\frac{1}{5}$  自然数: 0; +10; 13 整数: 0, +10; 13; -9",
+      "knowledge_points": [
+        "自然数的认识",
+        "正负数的概念及辨认"
+      ],
+      "source": "18-19 六年级下·广东汕头·单元测试",
+      "merged_id": "unit1_fill_155",
+      "original_id": "unit1_fill_155"
+    },
+    {
+      "id": "unit1_fill_156",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "小英从 0 点出发, 先向东走 2 米, 表示为  $+2 \\mathrm{~m}$ ; 接着再向西走  $5 \\mathrm{~m}$  到达 B 点, B 点表示为( ); 最后向西再走  $7 \\mathrm{~m}$  到达 C 点, C 点离 O 点的距离为( )。",
+      "options": [],
+      "answer": "-3#### 10米",
+      "solution": "【分析】小英从 0 点出发，先向东走 2 米，表示为  $+2 \\mathrm{~m}$ ，此时以  $+2 \\mathrm{~m}$  为起点，接着再向西走  $5 \\mathrm{~m}$  到达 B 点，B 点表示为 -3 米；以 -3 米为起点最后向西再走  $7 \\mathrm{~m}$  到达 C 点，C点在 -10 米，则 C 点离 O 点的距离为 10 米。本题考查负数, 解答本题的关键是掌握在直线上表示数的方法。\n【详解】B点表示为-3米；C点离O点的距离为10米。",
+      "knowledge_points": [
+        "正负数在数轴上的表示"
+      ],
+      "source": "18-19 六年级下·广东汕头·单元测试",
+      "merged_id": "unit1_fill_156",
+      "original_id": "unit1_fill_156"
+    },
+    {
+      "id": "unit1_fill_157",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "下面是育新小学去年下期的收支情况，请你用正负数记录一下学校的收入和支出情况。9月份收入7710元，表示为( )元；10月份支出6180元，表示为( )元；11月份收入1100元，表示为( )元；12月份支出600元，表示为( )元；一月份支出1300元，表示为( )元；这个学期学校有剩余( )元。",
+      "options": [],
+      "answer": "+7710#### -6180#### +1100#### -600#### -1300#### 730",
+      "solution": "【分析】正负数表示两种相反意义的量，收入记为正，支出记为负，据此解答即可。本题考查负数, 解答本题的关键是掌握正负数的意义。\n【详解】收入：  $7710 + 1100 = 8810$  （元）支出:  $6180 + 600 + 1300\\\\ 6 7 8 0 + 1 3 0 0 \\\\ = 8 0 8 0 (\\text {元}) $剩余： $8810 - 8080 = 730$  （元）9月份收入7710元，表示为（+7710）元；10月份支出6180 元，表示为（- 6180）元；11月份收入1100元，表示为  $(+1100)$  元; 12 月份支出 600 元, 表示为  $(-600)$  元;一月份支出 1300 元，表示为（-1300）元；这个学期学校有剩余（730）元。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20六年级下·湖南怀化·单元测试",
+      "merged_id": "unit1_fill_157",
+      "original_id": "unit1_fill_157"
+    },
+    {
+      "id": "unit1_fill_158",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "在括号里填上'>''<'或'='。$- 3 (\\quad) - 4 \\quad 0 (\\quad) 2. 3 \\quad - 0. 3 5 (\\quad) 0 \\\\ - 3 0 (\\quad) - 3 0. 3 \\quad \\frac {4}{5} (\\quad) - \\frac {1}{4} \\quad \\frac {4}{5} (\\quad) 1. 2 5  $",
+      "options": [],
+      "answer": "$>$#### $ <$#### $ <$#### $>$#### $>$#### $ <$",
+      "solution": "【分析】根据正负数的大小比较方法，直接填空即可。本题考查了正负数的大小比较, 正数永远大于负数, 0 永远大于负数。\n【详解】$-3 > -4; 0 < 2.3; -0.35 < 0$ ;$3 0 > - 3 0. 3; \\frac {4}{5} > - \\frac {1}{4}; \\frac {4}{5} <   1. 2 5$",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "19-20六年级下·湖南怀化·单元测试",
+      "merged_id": "unit1_fill_158",
+      "original_id": "unit1_fill_158"
+    },
+    {
+      "id": "unit1_fill_159",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "某气象局预报明天温度为  $-5^{\\circ} \\mathrm{C} - 12^{\\circ} \\mathrm{C}$ , 则这天的最低气温是( )。",
+      "options": [],
+      "answer": "-$12^{\\circ} \\mathrm{C}$",
+      "solution": "【分析】零下温度记为负, 符号后面的数越大, 温度越低,据此解答即可。本题考查负数, 解答本题的关键是掌握负数的意义。\n【详解】这天的最低气温是  $-12^{\\circ} \\mathrm{C}$  。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·湖南怀化·单元测试",
+      "merged_id": "unit1_fill_159",
+      "original_id": "unit1_fill_159"
+    },
+    {
+      "id": "unit1_fill_160",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "数轴上所有的负数都在 0 的( )边, 所有正数都在 0 的( )边。",
+      "options": [],
+      "answer": "左#### 右",
+      "solution": "【分析】根据正负数在数轴上的位置，结合题干，直接填空即可。 本题考查了数轴, 明确正负数在数轴上的位置是解题的关键。\n【详解】数轴上所有的负数都在 0 的左边，所有正数都在 0 的右边。",
+      "knowledge_points": [
+        "正负数在数轴上的表示"
+      ],
+      "source": "19-20 六年级下·湖南怀化·单元测试",
+      "merged_id": "unit1_fill_160",
+      "original_id": "unit1_fill_160"
+    },
+    {
+      "id": "unit1_fill_161",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "+ 9.06 读作( ), -  $\\frac{2}{5}$  读作( )。",
+      "options": [],
+      "answer": "正九点零六#### 负五分之二",
+      "solution": "【分析】根据正负数的读法，结合题干，直接读数即可。本题考查了正负数的读法，属于基础题，读数时需细心。\n【详解】+9.06读作正九点零六，$-\\frac{2}{5}$  读作负五分之二。",
+      "knowledge_points": [
+        "正负数的读法和写法"
+      ],
+      "source": "19-20 六年级下·湖南怀化·单元测试",
+      "merged_id": "unit1_fill_161",
+      "original_id": "unit1_fill_161"
+    },
+    {
+      "id": "unit1_fill_162",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "比海平面高  $600 \\mathrm{~m}$  的地方，它的高度记作( )，比海平面低  $400 \\mathrm{~m}$  的地方，它的高度记作( )。",
+      "options": [],
+      "answer": "$+600\\mathrm{m}$#### $ - 400\\mathrm{m}$",
+      "solution": "【分析】正负数表示两种相反意义的量，比海平面高记为正，比海平面低记为负，据此解答即可。本题考查负数, 解答本题的关键是掌握正负数的意义。\n【详解】比海平面高  $600 \\mathrm{~m}$  的地方, 它的高度记作  $+600 \\mathrm{~m}$ ,比海平面低  $400 \\mathrm{~m}$  的地方, 它的高度记作  $-400 \\mathrm{~m}$  。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20六年级下·湖南怀化·单元测试",
+      "merged_id": "unit1_fill_162",
+      "original_id": "unit1_fill_162"
+    },
+    {
+      "id": "unit1_fill_163",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "在 1.8, -2, +2, 0, -28, -1.5 这些数中, 正数有( ), 负数有( ), 最小的数是( ), 最大的数是( )。",
+      "options": [],
+      "answer": "1.8，+2#### -2，-28，-1.5#### -28#### +2",
+      "solution": "【分析】大于 0 的数是正数, 小于 0 的数是负数, 正数比负数大, 0 既不是正数也不是负数。据此解答即可。本题考查正数和负数的定义及大小比较，明确正数大于 0，负数小于 0 是解题的关键。\n【详解】由分析可知：正数有：1.8，+2，负数有：-2，-28，-1.5。2 > 1.8 > 0 > -1.5 > -2 > -28，最小的数是 -28，最大的数是 +2。",
+      "knowledge_points": [
+        "正负数的大小比较",
+        "正负数的概念及辨认"
+      ],
+      "source": "19-20 六年级下·湖南怀化·单元测试",
+      "merged_id": "unit1_fill_163",
+      "original_id": "unit1_fill_163"
+    },
+    {
+      "id": "unit1_fill_164",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "在数轴上, 从左往右的顺序就是数从____到____的顺序。",
+      "options": [],
+      "answer": "小#### 大",
+      "solution": "【详解】数轴是规定了原点（0 点）、方向和单位长度的直线，在数轴上，所有负数都在原点（0 点）的左边，越往左表示的数越小，所有正数在右边，越往右表示的数越大，因此，数轴上数值从左到右的顺序就是从小到大的顺序。",
+      "knowledge_points": [
+        "正负数在数轴上的表示"
+      ],
+      "source": "21-22 六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_164",
+      "original_id": "unit1_fill_164"
+    },
+    {
+      "id": "unit1_fill_165",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "今年入冬以来, 洛阳市偃师区的最低气温是零下  $6^{\\circ} \\mathrm{C}$ , 记作  $(\\quad) ^{\\circ} \\mathrm{C}$ ; 假师区南北高中间低, 最低处海拔约 +115 米, 表示比海平面( )115 米 (填 '高' 或 '低')。",
+      "options": [],
+      "answer": "-6#### 高",
+      "solution": "【分析】正负数表示一组相反意义的量, 此题中零上温度和零下温度就是一组相反意义的量, 零上温度记作正, 零下温度记作负; 海平面以上和以下是一组相反意义的量,海平面以上记作正，海平面以下记作负，据此解答即可。明确正负数表示的意义是解答本题的关键。\n【详解】洛阳市偃师区的最低气温是零下  $6^{\\circ} \\mathrm{C}$ , 记作  $- 6^{\\circ} \\mathrm{C}$ ; 最低处海拔约 +115 米, 表示比海平面高 115 米。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "21-22 五年级上·河南洛阳·期末",
+      "merged_id": "unit1_fill_165",
+      "original_id": "unit1_fill_165"
+    },
+    {
+      "id": "unit1_fill_166",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "在  $75.2 \\%$ , 0.8,  $\\frac{3}{5}$ , 和 -0.9 这四个数中, 最大的是( ), 最小的是( )。",
+      "options": [],
+      "answer": "$0.8 \\quad -0.9$",
+      "solution": "【分析】正数大于负数，再把百分数和分数化成小数，再按小数比较大小的方法进行比较即可。本题考查小数、分数、百分数的互化、负数, 解答本题的关键是掌握小数、分数、百分数的互化的方法。\n【详解】$75.2\\% = 0.752$  $\\frac {3}{5} = 0. 6$   $0.8 > 75.2 \\% > \\frac{3}{5} > - 0.9$",
+      "knowledge_points": [
+        "百分数、小数和分数的互化",
+        "正负数的大小比较"
+      ],
+      "source": "21-22 六年级上·吉林四平·期末",
+      "merged_id": "unit1_fill_166",
+      "original_id": "unit1_fill_166"
+    },
+    
+    {
+      "id": "unit1_fill_167",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "3.2,  $-\\frac{4}{9}$, 8, 0, 23, -7.8, +5.6, -4, -95这些数中, 正数有( )个, 负数有( )个。",
+      "options": [],
+      "answer": "4#### 4",
+      "solution": "【分析】通常情况下，把数分为正数和负数，正数的前面加'+'号或不写任何符号，负数前面一定加'-'号。此题考查了数的认识, 掌握正数和负数的意义是解题关键。\n【详解】由分析得,正数有：3.2，8，23，+5.6共4个；负数有:  $-\\frac{4}{9}, -7.8, -4, -95$  共 4 个。",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "21-22 五年级上·山西大同·期中",
+      "merged_id": "unit1_fill_167",
+      "original_id": "unit1_fill_167"
+    },
+    {
+      "id": "unit1_fill_168",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "如果收入 300 元记作 +300 元, 那么支出 70 元, 记作( )。",
+      "options": [],
+      "answer": "-70",
+      "solution": "【分析】首先审清题意，明确'正'和'负'所表示的意义；再根据题意作答。此题主要考查正负数的意义, 正数与负数表示意义相反的两种量, 看清规定哪一个为正, 则和它意义相反的就为负。\n【详解】收入为'+'，则支出为'-'，那么支出70元，记作-70元。",
+      "knowledge_points": [
+        "正负数的读法和写法",
+        "正负数的概念及辨认"
+      ],
+      "source": "20-21 四年级上·河南商丘·期末",
+      "merged_id": "unit1_fill_168",
+      "original_id": "unit1_fill_168"
+    },
+    {
+      "id": "unit1_fill_169",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "某品牌的大米包装袋上的质量标识为：' $50 \\pm 0.5 \\mathrm{~kg}$ '。质检人员随机抽测了四袋该品牌大米的质量，依次记录为：50.4kg，50.1kg，49.7kg，49.4kg，则所抽测的四袋大米中，符合该品牌大米包装袋上的质量标识要求的有_袋。",
+      "options": [],
+      "answer": "3",
+      "solution": "【分析】先求出大米的合格重量的范围即可判断。此题考查的是正负数的意义，解答此题关键是理解'正'和'负'的相对性。\n【详解】质量标识为: ' $50 \\pm 0.5 \\mathrm{~kg}$ '表示 50 上下 0.5 即 49.5 到 50.5 之间为合格, 分析选项可得  $49.4 \\mathrm{~kg}$  不在范围内属于不合格, 其余 3 袋在范围内, 合格。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "21-22六年级上·山东东营·期中",
+      "merged_id": "unit1_fill_169",
+      "original_id": "unit1_fill_169"
+    },
+    {
+      "id": "unit1_fill_170",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "在 5、-12、+ 7、0、-8、1.2这些数中，最大的是( )，最小的是( )。",
+      "options": [],
+      "answer": "+7#### -12",
+      "solution": "【分析】根据正数及负数的意义，正数都比 0 大，负数都比 0 小，然后根据数的大小比较的方法进行比较即可。本题是考查正、负数的意义。注意，正数都大于0，负数都小于0，0既不是正数，也不是负数。\n【详解】在 5、-12、+7、0、-8、1.2 这些数中，正数有 5、+7、1.2；负数有 -12、-8。这些数中，最大的数是 +7，最小的数是 -12。",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "21-22 五年级上·河南洛阳·期中",
+      "merged_id": "unit1_fill_170",
+      "original_id": "unit1_fill_170"
+    },
+    {
+      "id": "unit1_fill_171",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "一袋面粉的包装袋上标有 ' $25 \\pm 1 \\mathrm{~kg}$ '的字样, 这表示这袋化肥的重量在( )-()之间。",
+      "options": [],
+      "answer": "24千克#### 26千克",
+      "solution": "【分析】一袋面粉的包装袋上标有 '  $25 \\pm 1 \\mathrm{~kg}$  ' 的字样, 这表示这袋化肥的重量 (25 - 1) 千克和 (25 + 1) 千克之间,据此解答即可。明确 ' $25 \\pm 1 \\mathrm{~kg}$ '的含义是解答本题的关键。\n【详解】一袋面粉的包装袋上标有 '  $25 \\pm 1 \\mathrm{~kg}$  ' 的字样, 这表示这袋化肥的重量在 24 千克-26 千克之间。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "21-22 五年级上·河南洛阳·阶段练习",
+      "merged_id": "unit1_fill_171",
+      "original_id": "unit1_fill_171"
+    },
+    {
+      "id": "unit1_fill_172",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "某地某一天的最低气温是  $-7^{\\circ} \\mathrm{C}$ , 最高气温是  $10^{\\circ} \\mathrm{C}$ , 这一天的最高气温与最低气温相差( )°C。",
+      "options": [],
+      "answer": "17",
+      "solution": "【分析】-7°C比0°C低7°C，10°C比0°C高10°C，将与0°C相差的两个温度加起来即可。比 0 小的数叫做负数, 负数与正数表示意义相反的量。\n【详解】$7 + 10 = 17 \\quad (^{\\circ} \\mathrm{C})$",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·河南许昌·期末",
+      "merged_id": "unit1_fill_172",
+      "original_id": "unit1_fill_172"
+    },
+    {
+      "id": "unit1_fill_173",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "电梯从 5 层下降到 -2 层, 下降了( )层。",
+      "options": [],
+      "answer": "6",
+      "solution": "【分析】电梯没有 0 层, 从 5 层下降到 1 层, 下降了  $5 - 1$  层, 从 1 层下降到 -2 层, 下降了 2 层, 据此分析。关键是理解正负数的意义及实际应用，注意电梯没有0层。\n【详解】$5 - 1 + 2 = 6$  （层）",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·山东聊城·期末",
+      "merged_id": "unit1_fill_173",
+      "original_id": "unit1_fill_173"
+    },
+    {
+      "id": "unit1_fill_174",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "从站台向西走  $60 \\mathrm{~m}$  记作  $-60 \\mathrm{~m}$ , 那么从站台向东走  $100 \\mathrm{~m}$  记作( )  $\\mathrm{m}$  。",
+      "options": [],
+      "answer": "$+100$",
+      "solution": "【分析】如果向西走记为负，那么向东走记为正，据此分析。正负数可以表示相反意义的量。\n【详解】从站台向西走  $60 \\mathrm{~m}$  记作  $-60 \\mathrm{~m}$ , 那么从站台向东走  $100 \\mathrm{~m}$  记作  $+100 \\mathrm{~m}$  。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·江西吉安·期末",
+      "merged_id": "unit1_fill_174",
+      "original_id": "unit1_fill_174"
+    },
+    {
+      "id": "unit1_fill_175",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "月球表面最高温度约是零上  $127^{\\circ} \\mathrm{C}$ ，可记作( )°C。最低温度是零下  $184^{\\circ} \\mathrm{C}$ ，记作( )°C。",
+      "options": [],
+      "answer": "+127#### -184",
+      "solution": "【分析】零上温度记为正，零下温度记为负，据此分析。正负数可以表示相反意义的量。\n【详解】月球表面最高温度约是零上  $127^{\\circ} \\mathrm{C}$ , 可记作 +$127^{\\circ} \\mathrm{C}$  。最低温度是零下  $184^{\\circ} \\mathrm{C}$ , 记作  $-184^{\\circ} \\mathrm{C}$  。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21六年级下·云南玉溪·期末",
+      "merged_id": "unit1_fill_175",
+      "original_id": "unit1_fill_175"
+    },
+    {
+      "id": "unit1_fill_176",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "以林林家为起点，向东走为正，向西走为负，林林从家出发，向东走了50米，记作( )米；那么从家出发向西走了80米，记作( )米。",
+      "options": [],
+      "answer": "+50#### -80",
+      "solution": "【分析】根据题意可知，向东走记作正，向西走记作负，则向东走50米，记作  $+50$  米，向西走80米，记作-80米，据此解答即可。本题考查的是正数和负数意义的运用，解题的关键是理解'正'和'负'的相对性。\n【详解】向东走 50 米，记作 +50 米，向西走了 80 米，记作 -80 米。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21六年级下·湖南娄底·期末",
+      "merged_id": "unit1_fill_176",
+      "original_id": "unit1_fill_176"
+    },
+    {
+      "id": "unit1_fill_177",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "温度越低就越冷， $-4^{\\circ} \\mathrm{C}$  和  $-16^{\\circ} \\mathrm{C}$  相比，（ ）温度低。",
+      "options": [],
+      "answer": "$-16^{\\circ} \\mathrm{C}$",
+      "solution": "【分析】负数大小比较的方法：数字越大，这个数越小，反之则越大，据此解答即可。熟练掌握负数大小比较的方法是解答本题的关键。\n【详解】$-4^{\\circ} \\mathrm{C} > -16^{\\circ} \\mathrm{C}$ , 所以  $-16^{\\circ} \\mathrm{C}$  温度低。",
+      "knowledge_points": [
+        "正负数的大小比较",
+        "温度的认识及比较"
+      ],
+      "source": "19-20六年级下·河南南阳·期末",
+      "merged_id": "unit1_fill_177",
+      "original_id": "unit1_fill_177"
+    },
+    {
+      "id": "unit1_fill_178",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "如果某股票上涨1.50元记作  $\\neq 1.50$  元，那么下降0.66元记为( )元。",
+      "options": [],
+      "answer": "-0.66",
+      "solution": "【分析】如果规定股票上涨为正，则下降就是负数，据此判断。此题主要考查正负数的意义, 正数与负数表示意义相反的两种量, 看清规定哪一个为正, 则和它意义相反的就为负。\n【详解】如果某股票上涨 1.50 元记作  $+1.50$  元, 那么下降 0.66 元记为  $-0.66$  。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·重庆渝中·期末",
+      "merged_id": "unit1_fill_178",
+      "original_id": "unit1_fill_178"
+    },
+    {
+      "id": "unit1_fill_179",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "在  $3.14, -3.144, \\pi, 3.14, -3.14, 314 \\%$  中, 最大的是( ), 最小的是( )。",
+      "options": [],
+      "answer": "$\\pi$#### - 3.144",
+      "solution": "【分析】正数  $>$  负数, 两负数比大小, 不管负号, 数值大的负数小。关键是掌握正负数大小比较方法，含有百分数将百分数化成小数再比较。\n【详解】-3.144<-3.14<3.14=314%<3.14<π，最大的是π，最小的是-3.144。",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "19-20 六年级下·云南曲靖·期末",
+      "merged_id": "unit1_fill_179",
+      "original_id": "unit1_fill_179"
+    },
+    {
+      "id": "unit1_fill_180",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "一幢住宅楼共 16 层，如果把第 10 层记作 0 层，那么第六层记作( )层，第 15 层记作( )层。",
+      "options": [],
+      "answer": "-4#### +5",
+      "solution": "【分析】上和下是两个具有相反意义的量，通常上记作'+'，下记作'-'，把10层记作0层，向上几层记作'+几层'，向下几层读作记作'-几层'，依此填空即可。此题是考查正、负数的意义及应用。首先确定具有相反意义的量，把一个记作'+'，另一个记作'-'。\n【详解】一幢住宅楼共 16 层, 如果把第 10 层记作 0 层, 那么第六层记作 -4 层, 第 15 层记作 +5 层。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 四年级下·北京·期末",
+      "merged_id": "unit1_fill_180",
+      "original_id": "unit1_fill_180"
+    },
+    {
+      "id": "unit1_judge_181",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "零上 2 摄氏度与零下 5 摄氏度, 相差 3 摄氏度。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】零上 2 摄氏度可以用 2 表示, 零下 5 摄氏度可以用 -5 表示, 在数轴上表示出这两个数, 看一下中间差了几。此题首先要知道以谁为标准，规定高于0度的为正，低于 0 度的为负, 由此用正负数解答问题。\n【详解】![](images/edacc8c2f5e71f0c57b5e0f2dfa28341a239b17f992508e90aba995648d74d48.jpg)由图可知, 零上 2 摄氏度与零下 5 摄氏度, 相差 7 摄氏度, 原题说法错误;故答案为:  ×",
+      "knowledge_points": [
+        "正负数在数轴上的表示",
+        "正负数的意义及应用"
+      ],
+      "source": "21-22 六年级下·山东济宁·期末",
+      "merged_id": "unit1_judge_181",
+      "original_id": "unit1_judge_181"
+    },
+    {
+      "id": "unit1_judge_182",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "0 和 -6 之间有 5 个负数。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】根据负数的意义，0 和 -6 之间不止是 -1、-2、-3、-4、-5，它们之间还有 -1.1，-0.01、-5.467 等等负数。据此判断。本题考查了负数，掌握负数的意义是解题的关键。\n【详解】0和-6之间有无数个负数。故答案为:×",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "2022·山东菏泽·小升初真题",
+      "merged_id": "unit1_judge_182",
+      "original_id": "unit1_judge_182"
+    },
+    {
+      "id": "unit1_judge_183",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "零下  $12^{\\circ} \\mathrm{C}$  可以用'-12°C'表示。（）",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】$0^{\\circ} \\mathrm{C}$  以上用正数表示, 则  $0^{\\circ} \\mathrm{C}$  以下用负数表示。据此解答即可。本题考查正负数的应用，明确正负数表示的意义是解题的关键。\n【详解】由分析可知：零下  $12^{\\circ} \\mathrm{C}$  低于  $0^{\\circ} \\mathrm{C}$ , 可以用 '-  $12^{\\circ} \\mathrm{C}$ ' 表示, 原题说法正确。故答案为:  √",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20五年级上·全国·课后作业",
+      "merged_id": "unit1_judge_183",
+      "original_id": "unit1_judge_183"
+    },
+    {
+      "id": "unit1_judge_184",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "数轴上左边的数一定比右边的数大。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】数轴上越往右边数字越大，越往左边数字越小。所以数轴上左边的数一定比右边的数小, 不是大。本题考查了数轴的认识, 解答本题要明确数轴上的点表示的数, 右边的数比左边的数大。\n【详解】数轴上右边的数一定比左边的数大。故答案为: ×",
+      "knowledge_points": [
+        "正负数在数轴上的表示",
+        "正负数的大小比较"
+      ],
+      "source": "2021·贵州黔东南·小升初真题",
+      "merged_id": "unit1_judge_184",
+      "original_id": "unit1_judge_184"
+    },
+    {
+      "id": "unit1_judge_185",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "负数都比正数小。( )",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】以 0 为分界点, 大于的数叫做正数, 用 '+' 表示,小于 0 的数叫做负数, 用 '- ' 表示, 0 既不是正数也不是负数, 据此解答。掌握正负数与 0 的大小关系是解答题目的关键。\n【详解】分析可知, 负数小于 0 , 正数大于 0 , 如:  $-2<$  2, 所以负数都比正数小。所以原题说法正确。",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_judge_185",
+      "original_id": "unit1_judge_185"
+    },
+    {
+      "id": "unit1_judge_186",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "带有负号的就是负数, 其余的就是正数。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】我们知道整数包括正整数、负整数和 0, 0 既不是正数, 也不是负数。据此判断答案错误。本题是考查整数的意义及正、负数的意义。注意 0 是整数, 它既不是正数, 也不是负数。\n【详解】带有负号的就是负数, 其余的就是正数或 0 。故答案为:  ×",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "20-21 六年级下·湖北十堰·期中",
+      "merged_id": "unit1_judge_186",
+      "original_id": "unit1_judge_186"
+    },
+    {
+      "id": "unit1_judge_187",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "气温升高  $3^{\\circ} \\mathrm{C}$ , 记作  $+3^{\\circ} \\mathrm{C}$ ; 降低  $2^{\\circ} \\mathrm{C}$ , 则记作  $-2^{\\circ} \\mathrm{C}$  。( )",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】此题主要用正负数来表示具有意义相反的两种量:温度上升记为正, 则温度下降就记为负, 直接得出结论即可。此题主要考查正负数的意义, 正数与负数表示意义相反的两种量, 看清规定哪一个为正, 则和它意义相反的就为负。\n【详解】由分析可知：气温升高  $3^{\\circ} \\mathrm{C}$ , 记作  $+3^{\\circ} \\mathrm{C}$ ; 降低  $2^{\\circ} \\mathrm{C}$ , 则记作  $-2^{\\circ} \\mathrm{C}$  。说法正确。故答案为:  √",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·湖南怀化·期中",
+      "merged_id": "unit1_judge_187",
+      "original_id": "unit1_judge_187"
+    },
+    {
+      "id": "unit1_judge_188",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "正数都比负数大。( )",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】在数轴上，从左到右的顺序就是数从小到大的顺序，所有的正数都在负数的右边，所以正数都比负数大。此题考查正负数的大小比较, 解答本题的关键是掌握正数大于所有的负数, 正数大于 0, 0 大于所有的负数。\n【详解】在数轴上，负数都在 0 的左边，它们比 0 小，而正数都在 0 的右边，它们比 0 大，正数都比负数大。故答案为：√ 。",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "2021·河北邯郸·小升初真题",
+      "merged_id": "unit1_judge_188",
+      "original_id": "unit1_judge_188"
+    },
+    {
+      "id": "unit1_judge_189",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "甲地海拔高度为  $-180 \\mathrm{~m}$ , 乙地海拔高度为  $-200 \\mathrm{~m}$ , 甲地比乙地高。( )",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】根据数轴上原点的左边为负数，数轴上的数从左到右依次增加，据此解答即可。本题考查负数的大小比较, 明确负数的大小比较的方法是解题的关键。\n【详解】由分析可知：180> - 200所以, 甲地海拔高度为  $-180 \\mathrm{~m}$ , 乙地海拔高度为  $-200 \\mathrm{~m}$ ,甲地比乙地高。故原题干说法正确。",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "19-20 六年级下·湖南怀化·单元测试",
+      "merged_id": "unit1_judge_189",
+      "original_id": "unit1_judge_189"
+    },
+    {
+      "id": "unit1_judge_190",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "$-3^{\\circ} \\mathrm{C} = 3^{\\circ} \\mathrm{C}$  。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】根据正负数的意义，结合题干，直接分析判断即可。本题考查了正负数的意义, 负数表示和正数意义相反的量。\n【详解】-  $3^{\\circ} \\mathrm{C}$  表示零下  $3^{\\circ} \\mathrm{C}, 3^{\\circ} \\mathrm{C}$  表示零上  $3^{\\circ} \\mathrm{C}$ , 所以 -  $3^{\\circ} \\mathrm{C}$ 和  $3^{\\circ} \\mathrm{C}$  不相等。所以判断错误。",
+      "knowledge_points": [
+        "正负数的意义及应用",
+        "正负数的概念及辨认"
+      ],
+      "source": "19-20 六年级下·湖南怀化·单元测试",
+      "merged_id": "unit1_judge_190",
+      "original_id": "unit1_judge_190"
+    },
+    {
+      "id": "unit1_judge_191",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "困难",
+      "question": "+18读作加18，-18读作减18。（ ）",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】负数的读法：先读'负'，再读数。正数的读法：先读'正'，再读数。本题考查了正负数的读法，属于基础题，读数时细心是解题的关键。\n【详解】+18读作正十八，-18读作负十八。所以判断错误。故答案为:  ×",
+      "knowledge_points": [
+        "正负数的读法和写法"
+      ],
+      "source": "19-20六年级下·湖南怀化·单元测试",
+      "merged_id": "unit1_judge_191",
+      "original_id": "unit1_judge_191"
+    },
+    {
+      "id": "unit1_judge_192",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "1 比 -4 大,  $-\\frac{1}{2}<\\frac{1}{2}$  。( )",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】根据'正数都大于零，负数都小于零，正数大于负数'判断题目中两组数值的大小。本题考查正负数大小的比较，牢记'正数都大于零，负数都小于零，正数大于负数'的法则。\n【详解】1是正数，-4是负数，1比-4大 $-\\frac{1}{2}$  是负数，  $\\frac{1}{2}$  是正数，  $-\\frac{1}{2} < \\frac{1}{2}$故答案为: √",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "18-19 六年级下·山西临汾·期末",
+      "merged_id": "unit1_judge_192",
+      "original_id": "unit1_judge_192"
+    },
+    {
+      "id": "unit1_judge_193",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "最大的负整数大于最小的正整数。（ ）",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】数轴是规定了原点、方向和单位长度的直线，在零点左边所表示的数都是负数，在0点右边所表示的数都是正数，数轴上，从左到右的顺序就是数从小到大的顺序。0是正数和负数的分界点，负数  $< 0<$  正数，负号后面的数越大，这个数就越小。掌握正负数大小的比较方法是解题的关键。\n【详解】负数  $< 0<$  正数, 所以最大的负整数不可能大于最小的正整数, 原题说法错误。故答案为: ×",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "2020·湖南长沙·小升初真题",
+      "merged_id": "unit1_judge_193",
+      "original_id": "unit1_judge_193"
+    },
+    {
+      "id": "unit1_judge_194",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "随州某日气温$-5{ }^{\\circ} \\mathrm{C} \\sim 8^{\\circ} \\mathrm{C}$ , 这天的最大温差是  $3^{\\circ} \\mathrm{C}$  。 ( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】-5与0相差5，0与8相差8，将两个差加起来是最大温差。关键是理解正负数的意义，比0小的数叫负数。\n【详解】$5 + 8 = 13 (^{\\circ} \\mathrm{C})$ , 这天的最大温差是  $13^{\\circ} \\mathrm{C}$ , 所以原题说法错误, 故答案为: ×。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·湖北随州·期末",
+      "merged_id": "unit1_judge_194",
+      "original_id": "unit1_judge_194"
+    },
+    {
+      "id": "unit1_solve_195",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "小红到超市买了一包薯片，外包装上印有净重'（ $100 \\pm 3\\mathrm{g}$ ）'的字样，小红将薯片倒出来称了一下发现只有  $98\\mathrm{g}$ ，请问这包薯片的重量合格吗，你的判断依据是什么？",
+      "options": [],
+      "answer": "合格，理由见详解",
+      "solution": "【分析】因为外包装上印有净重 '  $(100 \\pm 3 \\mathrm{~g})$  ' 的字样,即这包薯片的重量在  $(100 - 3) \\mathrm{~g} \\sim (100 + 3) \\mathrm{~g}$  之间, 只要在这个范围内, 就合格; 由此解答即可。此题首先要知道以谁为标准，规定超出标准的为正，低于标准的为负，由此用正负数解答问题。\n【详解】$100 - 3 = 97$  （克） $100 + 3 = 103$  （克）答: 所以范围在 97 克 -103 克之间, 所以小红将薯片倒出来称了一下发现只有  $98 \\mathrm{~g}$ , 这包薯片的重量合格。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "21-22六年级下·全国·假期作业",
+      "merged_id": "unit1_solve_195",
+      "original_id": "unit1_solve_195"
+    },
+    {
+      "id": "unit1_solve_196",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "如果海平面的高度为0米，一潜水艇在海水下40米处航行，一条鲨鱼在潜水艇上方10米处游动，试用正负数分别表示潜水艇和鲨鱼的高度。",
+      "options": [],
+      "answer": "潜水艇：-40米；鲨鱼：-30米",
+      "solution": "【分析】鲨鱼和潜水艇都在水平面以下, 即 0 米以下, 即用负数表示, 即潜水艇: -40 米, 鲨鱼: -30 米; 由此解答即可。此题首先要知道以谁为标准，规定超出标准的为正，低于标准的为负，由此用正负数解答问题。\n【详解】答：潜水艇的高度是：-40米，鲨鱼的高度是： $-40 + 10 = -30$  米",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "21-22六年级下·全国·假期作业",
+      "merged_id": "unit1_solve_196",
+      "original_id": "unit1_solve_196"
+    },
+    {
+      "id": "unit1_solve_197",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "如果向东走为正,向西走为负。小勇从家先向东走 300 米, 记作 +300 米, 再返回走了 500 米。(1) 这时小勇离家多少米？记作多少米？(2) 画一条直线表示小勇在东西两个端点的位置和走的米数。",
+      "options": [],
+      "answer": " (1) 200 米; -200 米; (2）见详解",
+      "solution": "【分析】（1）以小勇家为分界点，从小勇家向东走用'+'表示，从小勇家向西走用'-'表示，先向正东走300 米, 再向正西走 500 米, 此时小勇在家的西边 200 米处;(2) 画出数轴, 原点为小勇家, 正方向为东, 单位长度为100 米，在数轴上 300 处标注 +300 米，-200 处标注 -200米。掌握数轴上数的表示方法是解答题目的关键。\n【详解】(1)  $500 - 300 = 200$  （米），此时小勇的位置在家的正西方向 200 米，记作 -200 米。答: 这时小勇离家 200 米, 记作 -200 米。![](images/c2200120c44cc860aa5b786018c6b362b3fc56ed67501913cf203c3204f10708.jpg)(2)",
+      "knowledge_points": [
+        "正负数在数轴上的表示",
+        "正负数的意义及应用"
+      ],
+      "source": "17-18 六年级下·湖南湘西·期中",
+      "merged_id": "unit1_solve_197",
+      "original_id": "unit1_solve_197"
+    },
+    {
+      "id": "unit1_solve_198",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "在一次数学速算比赛中，大（1）班的平均分为85分，把高于平均分的记作正数，低于平均分的记作负数。（如86分记作  $+1$  分，84分记作-1分)(1) 英英得了 89 分。应记作 ( ) 分; 花花得了 85 分,应记作 ( ) 分。(2) 老师将第一排的三名同学的成绩简记为 +8 分、-5 分和 +3 分, 这三名同学的平均分是多少?",
+      "options": [],
+      "answer": " (1)  $+4; 0$  (2) 87 分",
+      "solution": "【分析】（1）大于平均分的分数，就用得分减去平均分，再加上正号，小于平均分的分数，就用平均分减去得分，再加上负号即可；(2) 应用上面的方法, 计算出三名同学的得分, 再除以 3 即可。此题首先以平均分为标准，规定超出标准的为正，低于标准的为负，由此用正负数解答问题。\n【详解】(1)  $89 - 85 = 4$ , 记作 +4 分,  $85 - 85 = 0$ , 记作 0 分;(2)  $85 + 8 = 93$  (分)$8 5 - 5 = 8 0 \\text {(分)} \\\\ 8 5 + 3 = 8 8 \\text {(分)} \\\\ (9 3 + 8 0 + 8 8) \\div 3 \\\\ = 2 6 1 \\div 3 \\\\ = 8 7 \\text {(分)}  $答: 这三名同学的平均分是 87 分。",
+      "knowledge_points": [
+        "正负数的意义及应用",
+        "平均数的意义及求法"
+      ],
+      "source": "21-22六年级下·全国·单元测试",
+      "merged_id": "unit1_solve_198",
+      "original_id": "unit1_solve_198"
+    },
+    {
+      "id": "unit1_solve_199",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "下表记录了一辆公共汽车从起点站到终点站途中乘客数量的变化情况。起点站：+22人；中间第1站：+6人、-3人；中间第2站：+4人；中间第3站：-14人，终点站：？人(1) 说说中间3个站的上下人数各是多少？(2) 中间的 3 个站, 哪个站没有人上车? 哪个站没有人下车?(3) 根据数学信息, 请你提出一个数学问题?",
+      "options": [],
+      "answer": "见详解",
+      "solution": "【分析】（1）正数表示上车人数，负数表示下车人数，据此结合中间3个站的乘客数量的变化情况，直接描述出中间3个站的上下人数各是多少；(2) 根据 (1) 的答案, 结合 (2) 的问题, 直接做答;  (3) 问题: 终点站有多少人下车? 解答: 用上车总人数减去已经下车总人数, 求出终点站的下车人数。本题考查了正负数的意义及应用, 负数表示和正数意义相反的量。\n【详解】（1）答：中间第1站：上车6人，下车3人；中间第 2 站：上车 4 人，下车 0 人；中间第 3 站：上车 0 人，下车 14 人。(2) 答: 中间第 3 站没有人上车, 中间第 2 站没有人下车。(3) 问题: 终点站有多少人下车?$2 2 + 6 + 4 - 3 - 1 4 = 1 5 \\text {(人)}$答: 终点站一共有 15 人下车。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20六年级下·湖南怀化·单元测试",
+      "merged_id": "unit1_solve_199",
+      "original_id": "unit1_solve_199"
+    },
+    {
+      "id": "unit1_solve_200",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "体育课上，初一某班对女生进行仰卧起坐测试，以每分钟30个为标准，超过的部分用正数表示，不足的部分用负数表示，其中10名女生的成绩如下（单位：个）$5, - 3, 0, + 1 0, + 7, - 2, - 5, 0, + 1, + 3$求这10名女生各做了多少仰卧起坐？",
+      "options": [],
+      "answer": "这10名女生分别做了35个，27个，30个，40个,37 个, 28 个, 25 个, 30 个, 31 个, 33 个。",
+      "solution": "【分析】负数表示比标准数量少，正数表示比标准数量多。本题考查正负数的应用，明确正负数的意义是解答此题的关键。\n【详解】$30 + 5 = 35$  （个）$3 0 - 3 = 2 7 \\text {(个)}$ $3 0 + 0 = 3 0 \\text {(个)}$  $3 0 + 1 0 = 4 0 \\text {(个)}$ $3 0 + 7 = 3 7 \\text {(个)}$$3 0 - 2 = 2 8 \\text {(个)}$ $3 0 - 5 = 2 5 \\text {(个)}$ $3 0 + 0 = 3 0 (\\text {个})$ $3 0 + 1 = 3 1 \\text {(个)}$$3 0 + 3 = 3 3 (\\text {个})$答: 这 10 名女生仰卧起坐的成绩分别为: 35 个, 27 个, 30 个, 40 个, 37 个, 28 个, 25 个, 30 个, 31 个, 33 个。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "21-22六年级下·全国·单元测试",
+      "merged_id": "unit1_solve_200",
+      "original_id": "unit1_solve_200"
+    },{
+      "id": "unit1_choice_201",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "人体正常体温平均为$36^{\\circ} \\mathrm{C} \\sim 37^{\\circ} \\mathrm{C}$, 如果我们把人体体温标准定在 $36.5^{\\circ} \\mathrm{C}, 37^{\\circ} \\mathrm{C}$ 可以记作 $+0.5^{\\circ} \\mathrm{C}$, 那么 $35.8^{\\circ} \\mathrm{C}$ 可以记作 ( )。",
+      "options": [
+        "A. $-0.7^{\\circ} \\mathrm{C}$",
+        "B. $+35.8^{\\circ} \\mathrm{C}$",
+        "C. $-0.2^{\\circ} \\mathrm{C}$"
+      ],
+      "answer": "A",
+      "solution": "【分析】正负数表示相反意义的量, 以 $36.5^{\\circ} \\mathrm{C}$ 为标准, 高于 $36.5^{\\circ} \\mathrm{C}$ 的部分记作正, 则低于 $36.5^{\\circ} \\mathrm{C}$ 的温度就记作负。本题主要考查的是正负数的意义，解题的关键是确定好标准点之后，高于标准点的记作正数，低于标准点的记作负数。\n\n【详解】$36.5^{\\circ} \\mathrm{C} - 35.8^{\\circ} \\mathrm{C} = 0.7^{\\circ} \\mathrm{C}$, 所以 $35.8^{\\circ} \\mathrm{C}$ 可以记作 $-0.7^{\\circ} \\mathrm{C}$。故答案为: A",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "2021·广西钦州·小升初真题",
+      "merged_id": "unit1_choice_201",
+      "original_id": "unit1_choice_201"
+    },
+    {
+      "id": "unit1_choice_202",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "在 -4, -9, -$\\frac{3}{8}$, -0.1 这些数中, 最大的数是 ( )。",
+      "options": [
+        "A. -4",
+        "B. -9",
+        "C. $-\\frac{3}{8}$",
+        "D. -0.1"
+      ],
+      "answer": "D",
+      "solution": "【分析】比较两个负数大小时，负号后面的数越大，这个负数越小，据此解答。考查了负数大小比较的方法，学生应掌握。\n\n【详解】因为 $0.1 < \\frac{3}{8} < 4 < 9$, 所以 $-0.1 > -\\frac{3}{8} > -4 > -9$, 最大的数是 $-0.1$。故答案为: D。",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_choice_202",
+      "original_id": "unit1_choice_202"
+    },
+    {
+      "id": "unit1_choice_203",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "如果规定从原点出发，向南走为正，那么 $-100\\mathrm{m}$ 表示的意义是（）。",
+      "options": [
+        "A. 向东走 $100 \\mathrm{~m}$",
+        "B. 向西走 $100 \\mathrm{~m}$",
+        "C. 向北走100m",
+        "D. 向南走 $100 \\mathrm{~m}$"
+      ],
+      "answer": "C",
+      "solution": "【分析】如果向南走记为正，那么向北走记为负，据此分析。关键是理解正负数的意义, 正负数可以表示相反意义的量。\n\n【详解】如果规定从原点出发，向南走为正，那么 $-100\\mathrm{m}$ 表示的意义是向北走 $100\\mathrm{m}$。故答案为: C",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "14-15 六年级下·湖北黄冈·阶段练习",
+      "merged_id": "unit1_choice_203",
+      "original_id": "unit1_choice_203"
+    },
+    {
+      "id": "unit1_choice_204",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "某天中午 12 时的气温为$3^{\\circ} \\mathrm{C}$, 若下午 6 时的气温比 12 时低 $7^{\\circ} \\mathrm{C}$, 则下午 6 时的气温是（）。",
+      "options": [
+        "A. $-3^{\\circ} \\mathrm{C}$",
+        "B. $-4^{\\circ} \\mathrm{C}$",
+        "C. $+4^{\\circ} \\mathrm{C}$",
+        "D. $+ 3^{\\circ} \\mathrm{C}$"
+      ],
+      "answer": "B",
+      "solution": "【分析】先求出比 $3^{\\circ} \\mathrm{C}$ 低 $7^{\\circ} \\mathrm{C}$, 是零下 $4^{\\circ} \\mathrm{C}$, 再根据温度表示方法, 零上温度都用正数表示, 零下温度都用负数表示, 正数前面的 '+' 可以省略不写, 但负数前面的 '-' 不可以省略不写, 即可解答此题。本题主要考查正负数的应用, 解题时注意负号不能省略。\n\n【详解】因为比 $3^{\\circ} \\mathrm{C}$ 低 $7^{\\circ} \\mathrm{C}$, 是零下 $4^{\\circ} \\mathrm{C}$, 表示为: $-4^{\\circ} \\mathrm{C}$; 所以下午 6 时的气温是 $-4^{\\circ} \\mathrm{C}$。故答案为: B",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "2014 六年级·全国·课后作业",
+      "merged_id": "unit1_choice_204",
+      "original_id": "unit1_choice_204"
+    },
+    {
+      "id": "unit1_choice_205",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "把学校的位置记作0m，从学校出发向东走 $100\\mathrm{m}$ 到书店，书店的位置记作 $+$ $100\\mathrm{m}$, 再从书店出发向西走 $600\\mathrm{m}$ 是报亭。报亭的位置应记作（） $\\mathrm{m}_{\\circ}$",
+      "options": [
+        "A. -500",
+        "B. -600",
+        "C. -700",
+        "D. +100"
+      ],
+      "answer": "A",
+      "solution": "【分析】已知先从学校出发向东走 100 米到书店, 再从书店出发向西走 600 米是报亭。则这 600 米可看作两份, 一份是学校以东 100 米, 另一份是学校以西 500 米; 又因为书店的位置记作 +100 米, 则学校以西就记作负数。结合具体数据可知报亭的位置记作 -500 米。先确定报亭相对于学校的实际距离，再判断报亭位于学校的方向；两者相结合可将报亭的位置准确写出。\n\n【详解】报亭位于学校以西 500 米, 就记作 - 500 米。故答案为: A。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·河南南阳·期中",
+      "merged_id": "unit1_choice_205",
+      "original_id": "unit1_choice_205"
+    },
+    {
+      "id": "unit1_choice_206",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "在0和-6之间有（）个负数。",
+      "options": [
+        "A. 5",
+        "B. 6",
+        "C. 无数"
+      ],
+      "answer": "C",
+      "solution": "【分析】在 0 和 -6 之间的负数包括负整数、负分数, 有无数个负数, 据此解答即可。此题的关键是明确本题没有说明只有整数，然后再进一步解答。\n\n【详解】在0和-6之间有无数个负数。故答案为: C",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "20-21六年级下·全国·单元测试",
+      "merged_id": "unit1_choice_206",
+      "original_id": "unit1_choice_206"
+    },
+    {
+      "id": "unit1_choice_207",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "天气预报报道某地区今天气温是 $0^{\\circ} \\mathrm{C} \\sim 4^{\\circ} \\mathrm{C}$，明天气温是 $4^{\\circ} \\mathrm{C} \\sim -1^{\\circ} \\mathrm{C}$。该地区明天的最低气温和今天最低气温相比，（）。",
+      "options": [
+        "A. 上升了",
+        "B. 下降了",
+        "C. 不变",
+        "D. 无法判断"
+      ],
+      "answer": "B",
+      "solution": "【分析】以 0 为临界点, 正数都大于 0, 负数都小于 0, 0 不是正数也不是负数。据此比较明天和今天的最低气温即可。考查了正负数大小的比较，明确正数 $>0>$ 负数是解题关键。\n\n【详解】由题意可知, 今天的最低气温是 $0^{\\circ} \\mathrm{C}$, 明天的最低气温是 $-1^{\\circ} \\mathrm{C}$; $0^{\\circ} \\mathrm{C} > -1^{\\circ} \\mathrm{C}$; 即明天的最低气温比今天降低了;故答案为: B",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "20-21六年级下·全国·单元测试",
+      "merged_id": "unit1_choice_207",
+      "original_id": "unit1_choice_207"
+    },
+    {
+      "id": "unit1_choice_208",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "规定向东为正。小蚂蚁从家出发向东爬了 $6 \\mathrm{~m}$ 后，没发现食物，于是继续向东爬了 $2 \\mathrm{~m}$, 还是没有收获，最后它爬了 $-10 \\mathrm{~m}$, 终于找到了食物，食物在小蚂蚁家的（）处。",
+      "options": [
+        "A. 东边 $18 \\mathrm{~m}$",
+        "B. 西边 $2 m$",
+        "C. 东边 $2 m$",
+        "D. 西边 $18 \\mathrm{~m}$"
+      ],
+      "answer": "B",
+      "solution": "【分析】规定了向东为正，则向西即为负。根据小蚂蚁爬行的方向和数据，解答即可。明确正负数的意义是解题关键：一般地，对于具有相反意义的量, 我们可把其中一种意义的量规定为正数,与它意义相反的量即为负数。\n\n【详解】由题意，向东一共爬行了 $6 + 2 = 8$（m）；爬了 $-10 \\mathrm{~m}$ 即向西爬了 $10 \\mathrm{~m}$; 先向东爬 $8 \\mathrm{~m}$, 再向西爬 $10 \\mathrm{~m}$, 则小蚂蚁在出发点的西边 $2 \\mathrm{~m}$; 故答案为: B",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_choice_208",
+      "original_id": "unit1_choice_208"
+    },
+    {
+      "id": "unit1_choice_209",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "在投篮比赛中，投进一球记作 $+2$ 分，投失一球记作 $-1$ 分。张华共投进 8 球，投失 2 球，他的得分是（）分。",
+      "options": [
+        "A. 8",
+        "B. -2",
+        "C. 14",
+        "D. $+16$"
+      ],
+      "answer": "C",
+      "solution": "【分析】根据题意，投进一球记作 $+2$ 分，投失一球记作 $-1$ 分；据此可计算出张华投进球的总分数和失球的分数，进而计算出他的总得分。明确正负数的意义是解题关键。\n\n【详解】$8 \\times 2 - 2 \\times 1\\\\= 1 6 - 2 \\\\ = 1 4 \\text {(分)} $\n\n故答案为: C",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_choice_209",
+      "original_id": "unit1_choice_209"
+    },
+    {
+      "id": "unit1_choice_210",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "下面的量中, 适合用 $-150 \\mathrm{~kg}$ 表示的是 ( )。",
+      "options": [
+        "A. 商店购进 $150 \\mathrm{~kg}$ 梨",
+        "B. 三袋大米重 $150 \\mathrm{~kg}$",
+        "C. 工人运走货物 $150 \\mathrm{~kg}$",
+        "D. 李叔叔家的小麦比去年增产 $150 \\mathrm{~kg}$"
+      ],
+      "answer": "C",
+      "solution": "【分析】一般的，正负数可以表示相反意义的两个量，如买进用 +，则卖出用 -。一般地，对于具有相反意义的量，我们可把其中一种意义的量规定为正数，把与它意义相反的量规定为负数。\n\n【详解】根据正负数的意义可知:A. 商店购进 $150 \\mathrm{~kg}$ 梨, 一般用 $+150 \\mathrm{~kg}$ 表示;B. 三袋大米重 $150 \\mathrm{~kg}$, 表示重量直接用 $150 \\mathrm{~kg}$ 即可;C. 工人运走货物 $150 \\mathrm{~kg}$, 可以用 $-150 \\mathrm{~kg}$ 表示;D. 李叔叔家的小麦比去年增产 $150 \\mathrm{~kg}$, 一般用 $+150 \\mathrm{~kg}$ 表示。故答案为: C",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_choice_210",
+      "original_id": "unit1_choice_210"
+    },
+    {
+      "id": "unit1_choice_211",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "'蛟龙号'是我国第一艘自行设计、自主集成研制、目前世界上下潜能力最深的深海载人潜水器。如果'蛟龙号'从海拔 $-3000 \\mathrm{~m}$ 处上升 $200 \\mathrm{~m}$, 那么它上升后的海拔高度是（）。",
+      "options": [
+        "A. $-200 \\mathrm{~m}$",
+        "B. $-2800 m$",
+        "C. $-3200 m$",
+        "D. $200 \\mathrm{~m}$"
+      ],
+      "answer": "B",
+      "solution": "【分析】要求 $-3000 \\mathrm{~m}$ 处上升 $200 \\mathrm{~m}$ 后的海拔高度, 只要用 $3000 - 200$ 求出其结果, 在结果前面加上负号就可以了。要读懂题意, 根据题意列出式子, 在计算中注意确定结果的符号。\n\n【详解】$3000 - 200 = 2800$ 即上升后的海拔高度为 $-2800 \\mathrm{~m}$; 故答案为: B",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_choice_211",
+      "original_id": "unit1_choice_211"
+    },
+    {
+      "id": "unit1_choice_212",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "一包饼干的标签上标有'净重： $(150g \\pm 5g)$'的字样，表示这种饼干每包的质量范围是（）。",
+      "options": [
+        "A. $145 \\mathrm{~g}-150 \\mathrm{~g}$",
+        "B. $150 g- 160 g$",
+        "C. $145 g-155 g$",
+        "D. $140 \\mathrm{~g}-150 \\mathrm{~g}$"
+      ],
+      "answer": "C",
+      "solution": "【分析】由题干可知, 饼干的质量最高是 $150 \\mathrm{~g} + 5 \\mathrm{~g}$; 最少是 $150 \\mathrm{~g} - 5 \\mathrm{~g}$; 据此解答即可。明确'净重： $(150g \\pm 5g)$'的含义是解题关键。\n\n【详解】$150 + 5 = 155(\\mathrm{g}) $ $1 5 0 - 5 = 1 4 5 \\quad (\\mathrm {g})$ 即这种饼干每包的质量范围在 $145 \\mathrm{~g} - 155 \\mathrm{~g}$ 之间;故答案为: C",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_choice_212",
+      "original_id": "unit1_choice_212"
+    },
+    {
+      "id": "unit1_choice_213",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "某河流的警戒水位是 $3m$，高于警戒水位记作正数，低于警戒水位记作负数，下列说法错误的是（）。",
+      "options": [
+        "A. 水位为 $3.12 \\mathrm{~m}$ 时, 记作 $3.12 \\mathrm{~m} \\quad$",
+        "B. 水位为 $2.8 \\mathrm{~m}$ 时, 记作 $-0.2 \\mathrm{~m}$",
+        "C. 水位为 $3.08 \\mathrm{~m}$ 时, 记作 $+0.08 \\mathrm{~m}$",
+        "D. $-2 \\mathrm{~m}$ 表示比警戒水位低 $2 \\mathrm{~m}$"
+      ],
+      "answer": "A",
+      "solution": "【分析】根据负数的意义，高于警戒水位记为'+'，则低于警戒水位记为'-'，据此判断即可。此题主要考查了负数的意义及其应用，要熟练掌握，解答此题的关键是要明确：高于警戒水位记为'+'，则低于警戒水位记为'-'。\n\n【详解】A. 水位为 $3.12 \\mathrm{~m}$ 时, 高于警戒水位, $3.12 - 3 =$ 0.12，记作 $+0.12 \\mathrm{~m}$，本选项错误；B. 水位为 $2.8 \\mathrm{~m}$ 时, 低于警戒水位, $3 - 2.8 = 0.2$, 记作 $-0.2 \\mathrm{~m}$, 本选项正确;C. 水位为 $3.08 \\mathrm{~m}$ 时, 高于警戒水位, $3.08 - 3 = 0.08$, 记作+0.08 m, 本选项正确;D. $-2m$ 表示比警戒水位低 $2m$, 说法正确。故答案为: A",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_choice_213",
+      "original_id": "unit1_choice_213"
+    },
+    {
+      "id": "unit1_choice_214",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "下列说法中错误的是（）。",
+      "options": [
+        "A. 最大的负数是 $-1$",
+        "B. 0 既不是正数, 也不是负数",
+        "C. 正数一定大于负数",
+        "D. $-3^{\\circ} \\mathrm{C}$ 高于 $-5^{\\circ} \\mathrm{C}$"
+      ],
+      "answer": "A",
+      "solution": "【分析】根据对正数和负数的认识，逐项分析即可。熟练掌握正负数的含义及负数大小的比较方法是。\n\n【详解】A．小于0的数叫负数，-0.1> -1，最大的负数是-1说法错误;B. 0 既不是正数, 也不是负数, 说法正确;C. 正数都大于 0, 负数都小于 0, 所以正数一定大于负数说法正确;D. 负号后面的数字越小, 这个数越大; $-3^{\\circ} \\mathrm{C}$ 高于 $-5^{\\circ} \\mathrm{C}$ 说法正确;故答案为：A。",
+      "knowledge_points": [
+        "正负数的概念及辨认",
+        "正负数的大小比较"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_choice_214",
+      "original_id": "unit1_choice_214"
+    },
+    {
+      "id": "unit1_choice_215",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "在 -34, 0, $\\frac{1}{4}, -\\frac{1}{2}$, $\\frac{+7}{10}, -2, 1, 0.833, +4 \\frac{3}{4}, -10.5$ 中, 正数有（）个。",
+      "options": [
+        "A. 3",
+        "B. 4",
+        "C. 5",
+        "D. 6"
+      ],
+      "answer": "C",
+      "solution": "【分析】根据正、负数的意义，数的前面加有'-'号的数，就是负数；数的前面加有'+'号的数就是正数，正数前面的'+'可以省略不写；0既不是正数也不是负数，据此判断即可。此题考查正、负数的意义和分类。基础知识, 要熟练掌握。\n\n【详解】由分析可得，在 $-34, 0, \\frac{1}{4}, -\\frac{1}{2}, \\frac{+7}{10}, -2, 1, 0.833, +4 \\frac{3}{4}, -10.5$ 中，整数有： $\\frac{1}{4}, \\frac{+7}{10}, 1, 0.833, +4 \\frac{3}{4}$，共5个。故答案为: C",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_choice_215",
+      "original_id": "unit1_choice_215"
+    },
+    {
+      "id": "unit1_choice_216",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "小明家的冰箱，冷藏室的温度是'4℃'，冷冻室的温度是'﹣5℃'，这个冰箱的冷藏室与冷冻重的温度相差（    ）℃。",
+      "options": [
+        "A. -1",
+        "B. -9",
+        "C. 1",
+        "D. 9"
+      ],
+      "answer": "D",
+      "solution": "【分析】这个冰箱的冷藏室与冷冻重的温度相差的度数 = 冷藏室的温度 - 冷冻室的温度。解决本题需要注意的是：两个温度的相差值不能是负数。\n\n【详解】$4 - (-5) = 9^{\\circ} \\mathrm{C}$, 所以这个冰箱的冷藏室与冷冻室的温度相差 $9^{\\circ} \\mathrm{C}$。故答案为: D。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21六年级下·全国·课后作业",
+      "merged_id": "unit1_choice_216",
+      "original_id": "unit1_choice_216"
+    },
+    {
+      "id": "unit1_choice_217",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "甲地海拔 -50 米,乙地海拔 12 米, 甲、乙两地海拔高度相差 ( ) 米。",
+      "options": [
+        "A. 38",
+        "B. 62",
+        "C. -38",
+        "D. 48"
+      ],
+      "answer": "B",
+      "solution": "【分析】甲乙两地海拔高度差 = 甲地海拔高度的绝对值 + 乙地海拔高度的绝对值。正、负数的运算，两个高度的相差值不能是负数。\n\n【详解】$50 + 12 = 62$ （米）；故答案为: B。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·全国·课后作业",
+      "merged_id": "unit1_choice_217",
+      "original_id": "unit1_choice_217"
+    },
+    {
+      "id": "unit1_choice_218",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "在一次体检中，李丽、王强、张军、沈艳、孙悦的体重分别是34 kg、40 kg、38 kg、36 kg、32 kg。如果把他们 5 人的平均体重记作 $0 \\mathrm{~kg}$, 超过平均体重的记为正, 那么张军的体重应记作 ( )。",
+      "options": [
+        "A. $-4 \\mathrm{~kg}$",
+        "B. $-2 \\mathrm{~kg}$",
+        "C. $+2 \\mathrm{~kg}$",
+        "D. $+4 \\mathrm{~kg}$"
+      ],
+      "answer": "C",
+      "solution": "【分析】用 5 人的体重和除以总人数求出平均体重是$(34 + 40 + 38 + 36 + 32) \\div 5 = 36(\\mathrm{~kg})$ ，张军的体重是 $38 \\mathrm{~kg}, 38 > 36$，$38 - 36 = 2(\\mathrm{~kg})$ 根据超过平均体重的记为正，可知张军的体重应记作 $+2 \\mathrm{~kg}$。以 5 个人的平均体重为基准, 就要先计算出平均体重; 再把张军的体重与平均体重作差即可。\n\n【详解】由分析得:\n\n$(3 4 + 4 0 + 3 8 + 3 6 + 3 2) \\div 5\\\\= 1 8 0 \\div 5\\\\= 3 6 (\\mathrm {k g})\\\\3 8 - 3 6 = 2 (\\mathrm {k g})$\n\n所以张军的体重应记作 $+2 \\mathrm{~kg}$。故答案为: C。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_choice_218",
+      "original_id": "unit1_choice_218"
+    },
+    {
+      "id": "unit1_choice_219",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "王叔叔住7楼，某天他将轿车停在地下车库（-1层），然后乘电梯上楼。电梯需上升（）层他才能到家。",
+      "options": [
+        "A. 6",
+        "B. 7",
+        "C. 8",
+        "D. 9"
+      ],
+      "answer": "B",
+      "solution": "【分析】地下车库在地下 1 层, 王叔叔家在地上 7 层, 所以电梯要上升 $7 - 1 + 1 = 7$ (层)。其中 7-1 表示 7 楼至 1 楼存在 6 层楼梯, 再加上 -1 层与地上 1 层之间的 1 层楼梯, 就得到电梯需上升几层。需要注意的是走进楼房直接就是 1 楼, 所以两层楼之间存在 1 层楼梯, 其次理解 -1 层的含义就是地下一层。\n\n【详解】由分析得：$7-1+1=6+1=7 (层) $故答案为: B",
+      "knowledge_points": [
+        "利用正负数解决实际问题"
+      ],
+      "source": "20-21六年级下·全国·单元测试",
+      "merged_id": "unit1_choice_219",
+      "original_id": "unit1_choice_219"
+    },
+    {
+      "id": "unit1_choice_220",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "在 23、-4、0、+108、-6 中, 正数有 ( ) 个。",
+      "options": [
+        "A. 1",
+        "B. 2",
+        "C. 3"
+      ],
+      "answer": "B",
+      "solution": "【分析】大于 0 的数叫做正数，据此解答。此题考查正、负数的辨认，明确 0 既不是正数也不是负数。\n\n【详解】在 23、-4、0、+108、-6 中，正数有 23、+108，共 2 个。故选择：B。",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "20-21 五年级上·江苏盐城·期中",
+      "merged_id": "unit1_choice_220",
+      "original_id": "unit1_choice_220"
+    },
+    {
+      "id": "unit1_choice_221",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "实数-17的相反数是（）。",
+      "options": [
+        "A. 17",
+        "B. 117",
+        "C. -17",
+        "D. - 117"
+      ],
+      "answer": "A",
+      "solution": "【分析】相反数是只有符号不同的数字, 负数的相反数是正数, 由此解答即可。负数的相反数是正数。\n\n【详解】实数 - 17 的相反数是 17。故答案为: A。",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "20-21四年级上·四川·单元测试",
+      "merged_id": "unit1_choice_221",
+      "original_id": "unit1_choice_221"
+    },
+    {
+      "id": "unit1_choice_222",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "一种食品包装袋上标着：净重（ $275 \\pm 5$ 克），表示这种食品每袋最多不超过（）克。",
+      "options": [
+        "A. 270",
+        "B. 280",
+        "C. 290",
+        "D. 300"
+      ],
+      "answer": "B",
+      "solution": "【分析】净重（ $275 \\pm 5$ 克），表示以 $275$ 克为标准量，食品的质量在比标准量多 5 克和比标准量少 5 克之间，据此解答。此题主要考查正负数的意义及应用，在日常生活中它们通常表示两种相反意义的量，正数表示多出的质量，那负数表示减少的量。\n\n【详解】$275 + 5 = 280$ （克）这种食品每袋最多不超过 280 克。故选择：B。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 五年级上·江苏·单元测试",
+      "merged_id": "unit1_choice_222",
+      "original_id": "unit1_choice_222"
+    },
+    {
+      "id": "unit1_choice_223",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "在下列各数中, 最接近 0 的是 ( )。",
+      "options": [
+        "A. 1",
+        "B. 0.1",
+        "C. -1"
+      ],
+      "answer": "B",
+      "solution": "【分析】分别与 0 求差，差越小越接近 0，据此分析。本题考查了正负数的认识，比0小的数是负数。\n\n【详解】A. $1 - 0 = 1$; B. $0.1 - 0 = 0.1$; C. $0 - (-1) = 1$ $0.1 < 1$, 最接近 0 的是 $0.1$。故答案为: B",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·山东菏泽·期末",
+      "merged_id": "unit1_choice_223",
+      "original_id": "unit1_choice_223"
+    },
+    {
+      "id": "unit1_choice_224",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "下列每组中两个量不是具有相反意义的量是 ( )。",
+      "options": [
+        "A. 收入 100 元与支出 70 元",
+        "B. 浪费 1 吨煤与节约 1 吨煤",
+        "C. 增产 45 吨与减产 2 吨",
+        "D. 向东走 5 米与向南走 5 米"
+      ],
+      "answer": "D",
+      "solution": "【分析】根据正负数的意义，可得：收入记为'+'，则支出记为'-'；节约记为'+'，则浪费记为'-'；增产记为'+'，则减产记为'-'；向东记为'+'，则向西记为'-'，据此判断即可。此题主要用正负数来表示具有意义相反的两种量。\n\n【详解】A：收入100元与支出70元是具有相反意义的量。  B: 浪费 1 吨煤与节约 1 吨煤是具有相反意义的量。  C: 增产 45 吨与减产 2 吨是具有相反意义的量。  D: 向东与向西是具有相反意义的量, 向东与向南不是具有相反意义的量。故答案为: D",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_choice_224",
+      "original_id": "unit1_choice_224"
+    },
+    {
+      "id": "unit1_choice_225",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "下列每组中的两个量不具有相反意义的一组量是（）。",
+      "options": [
+        "A. 收入 40 元与支出 40 元",
+        "B. 浪费 1 吨水与节约 1 吨水",
+        "C. 向东走 4 米与向北走 4 米",
+        "D. 增产 12 吨与减产 12吨"
+      ],
+      "answer": "C",
+      "solution": "【分析】可以根据词语上的反义词意义，进行分析。本题考查了正负数的意义, 正负数可以表示相反意义的量。\n\n【详解】A．收入40元与支出40元，收入与支出相反，选项具有相反意义；  B. 浪费 1 吨水与节约 1 吨水, 浪费与节约相反, 选项具有相反意义;  C. 向东走 4 米与向北走 4 米, 东与西相反, 选项不具有相反意义;  D. 增产 12 吨与减产 12 吨, 增产与减产相反, 具有相反意义。故答案为: C",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_choice_225",
+      "original_id": "unit1_choice_225"
+    },
+    {
+      "id": "unit1_choice_226",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "如果 +8 分表示比平均分高 8 分, 那么 -10 分表示 ( )。",
+      "options": [
+        "A. 比平均分低 10 分",
+        "B. 比平均分高 10 分",
+        "C. 和平均分相等",
+        "D. 比平均分低 18 分"
+      ],
+      "answer": "A",
+      "solution": "【分析】以平均分为标准，高于平均分记为正，低于平均分记为负，据此选择。本题考查了正负数的意义，正负数可以表示相反意义的量。\n\n【详解】如果 +8 分表示比平均分高 8 分, 那么 -10 分表示比平均分低 10 分。故答案为: A",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_choice_226",
+      "original_id": "unit1_choice_226"
+    },
+    {
+      "id": "unit1_choice_227",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "在 $\\frac{1}{3}$ 、 -1.5、0、+ 3.9、12、- $\\frac{2}{5}$ 中，负数的个数有（）个。",
+      "options": [
+        "A. 2",
+        "B. 3",
+        "C. 4",
+        "D. 5"
+      ],
+      "answer": "A",
+      "solution": "【分析】正数都大于 0, 正数都带有 '+' 号或不带任何符号; 负数都小于 0, 负数都带有 '- ' 号; 0 不是正数也不是负数。0 既不是正数, 也不是负数; '+' 可以省略不写。\n\n【详解】在 $\\frac{1}{3} 、 -1.5 、 0 、 +3.9 、 12 、 -\\frac{2}{5}$ 中，负数的个数有 -1.5、- $\\frac{2}{5}$ 这2个。故答案为: A",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_choice_227",
+      "original_id": "unit1_choice_227"
+    },
+    {
+      "id": "unit1_fill_228",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "海平面的海拔高度记为0米，高于海平面记为正，寻乌县的最高峰项山甑的主峰海拔1529.8米，可记为( )米；-150米表示( )。",
+      "options": [],
+      "answer": "+1529.8  低于海平面150米",
+      "solution": "【分析】根据正负数的意义，高于海平面的记为正，低于海平面的记为负，进行解答。本题考查正负数的意义, 正负数可以表示相反意义的量。\n\n【详解】海平面的海拔高度记为 0 米, 高于海平面记为正,寻乌县的最高峰项山甑的主峰海拔 1529.8 米, 可记为 +1529.8 米, -150 米表示低于海平面 150 米。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20六年级下·江西赣州·期末",
+      "merged_id": "unit1_fill_228",
+      "original_id": "unit1_fill_228"
+    },
+    {
+      "id": "unit1_fill_229",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "在括号里填上'<' '>'或者'='。\n\n$8 (\\quad) - 6 - 0. 5 (\\quad) 0 \\frac {3}{4} (\\quad) \\frac {4}{7}$",
+      "options": [],
+      "answer": "<   <    >",
+      "solution": "【分析】①不管负号，数值越大的负数越小。② 利用负数小于 0, 比较大小即可。  ③ 先通分, 化成同分母分数后, 利用分子大的分数大, 比较大小即可。此题考查了正数, 负数比较大小的方法, 知道正数大于 0, 负数小于 0, 正数大于一切负数; 两个负数, 数值大的反而小是解题的关键。\n\n【详解】①因为 $8 > 6$, 所以 $-8 < -6$;②  $-0.5 < 0$, (负数小于 0);  ③因为 $\\frac{3}{4} = \\frac{21}{28}, \\frac{4}{7} = \\frac{16}{28}, \\frac{21}{28} > \\frac{16}{28}$, 所以 $\\frac{3}{4} > \\frac{4}{7}$。",
+      "knowledge_points": [
+        "异分母异分子分数的大小比较",
+        "正负数的大小比较"
+      ],
+      "source": "20-21六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_229",
+      "original_id": "unit1_fill_229"
+    },
+    {
+      "id": "unit1_fill_230",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "4，-8，79，3.6,$\\frac{2}{5}, 0.91, 29$, 这些数中( )是自然数, ( )是整数, ( )是质数, ( )是合数。",
+      "options": [],
+      "answer": "4，79，29 4，-8，79，29 79，29 4",
+      "solution": "【分析】首先判断出自然数、整数有哪些；然后根据质数、合数的意义判断出质数、合数分别有哪些即可。此题主要考查了自然数、整数以及质数、合数的判断, 解答此题的关键是要熟练掌握这些数的意义。\n\n【详解】4，79，29是自然数；4, -8, 79, 29 是整数;  79, 29 是质数;  4 是合数。",
+      "knowledge_points": [
+        "自然数的认识",
+        "质数与合数",
+        "正负数的概念及辨认"
+      ],
+      "source": "20-21六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_230",
+      "original_id": "unit1_fill_230"
+    },
+    {
+      "id": "unit1_fill_231",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "在 $1 、 5 、 1 \\frac {1}{4} 、 - 3 、 $-0.15、+4、- $\\frac{1}{4}$ 中，整数有( )个，负数有( )个，如果把这些数标在数轴上，离0最近的数是( )。",
+      "options": [],
+      "answer": "4   3   -0.15",
+      "solution": "【分析】整数包括正整数、负整数、0；整数有：1、5、-3、+4。负数前面有 '-' 号; 负数有: -3、-0.15、 $-\\frac{1}{4}$。根据数轴可得出：离0最近的数是 $0.15$。这道题目考查的是数的认识以及对数轴的认识。\n\n【详解】在 $1 、 5 、 1 \\frac {1}{4} 、 - 3 、 - 0. 1 5 、 + 4 、 - \\frac {1}{4}$ 中, 整数有 4 个, 负数有 3 个, 如果把这些数标在数轴上, 离 0 最近的数是 0.15。",
+      "knowledge_points": [
+        "正负数在数轴上的表示",
+        "正负数的概念及辨认"
+      ],
+      "source": "2020·湖北孝感·小升初真题",
+      "merged_id": "unit1_fill_231",
+      "original_id": "unit1_fill_231"
+    },
+    {
+      "id": "unit1_fill_232",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "通常情况下，我们可以把零下 $6^{\\circ} \\mathrm{C}$ 记作 $(\\quad) ^{\\circ} \\mathrm{C}$。",
+      "options": [],
+      "answer": "-6",
+      "solution": "【分析】用正负数表示意义相反的两种量: 温度零上记作正, 则温度零下就记作负; 由此得解。此题主要考查正负数的意义, 正数与负数表示意义相反的两种量, 看清规定哪一个为正, 则和它意义相反的就为负。\n\n【详解】通常情况下，我们可以把零下 $6^{\\circ} \\mathrm{C}$ 记作 $-6^{\\circ} \\mathrm{C}$。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_232",
+      "original_id": "unit1_fill_232"
+    },
+    {
+      "id": "unit1_fill_233",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "比大小。\n\n$-1.5 (\\quad) - \\frac {1}{2} 0. 7 (\\quad) 0 - 8 (\\quad) \\frac {1}{3}$",
+      "options": [],
+      "answer": "<   >   <",
+      "solution": "【分析】几个正、负数比较大小, 可以借助数轴比较它们的大小, 在数轴上, 从左到右的顺序就是数从小到大的顺序; 正数的大小比较简单, 数字越大, 数值也就越大; 负数大小比较是看负号后面的数字, 数字越大的反而越小。此题考查正负数的大小比较。\n\n【详解】负号后面的数字越大, 数越小。所以 $-1.5 < -\\frac{1}{2}$; 正数 $>0$, 所以 $0.7 > 0$; 正数 $>$ 负数, -8 为负数, $\\frac{1}{3}$ 为正数, 所以 $-8 < \\frac{1}{3}$。",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_233",
+      "original_id": "unit1_fill_233"
+    },{
+      "id": "unit1_fill_234",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "武汉某一天的最高气温是零上 $6^{\\circ} \\mathrm{C}$, 记作 $+6^{\\circ} \\mathrm{C}$, 最低气温是零下 $3^{\\circ} \\mathrm{C}$, 记作( ),这一天的温差是( )°C。",
+      "options": [],
+      "answer": "-3°C   9",
+      "solution": "【分析】正负数表示一组相反意义的量，此题中零上和零下就是一组相反意义的量，零上记作正，零下就记作负，据此解答即可；用最高气温减去最低气温即可求出相差的温度。明确正负数的意义是解答本题的关键。\n\n【详解】零上 $6^{\\circ} \\mathrm{C}$, 记作 $+6^{\\circ} \\mathrm{C}$, 最低气温是零下 $3^{\\circ} \\mathrm{C}$, 记作 $-3^{\\circ} \\mathrm{C}$; $6 - (-3) = 9^{\\circ} \\mathrm{C}$",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "2014 六年级·全国·课后作业",
+      "merged_id": "unit1_fill_234",
+      "original_id": "unit1_fill_234"
+    },
+    {
+      "id": "unit1_fill_235",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "一次数学测试, 六(1) 班的平均成绩是 85 分。如果把平均分记作 0 分, 高于平均分的记作正, 那么小军的成绩是 90 分, 记作 ( ) 分, 小文的成绩是 80 分, 应记作 ( ) 分。",
+      "options": [],
+      "answer": "$+5$  $ - 5$",
+      "solution": "【分析】此题主要用正负数来表示具有意义相反的两种量: 选85分为标准记为0，超过部分为正，不足的部分为负, 直接得出结论即可。此题主要考查正负数的意义, 正数与负数表示意义相反的两种量, 看清规定哪一个为正, 则和它意义相反的就为负。\n\n【详解】$90 - 85 = 5; 90$ 分应该记作 $+5$ 分；$85 - 80 = 5$; 80 分应该记作 -5 分。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_235",
+      "original_id": "unit1_fill_235"
+    },
+    {
+      "id": "unit1_fill_236",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "与北京时间相比，早1小时记作 $+1$ 时，晚2小时记作 $-2$ 时……同一天内，现在的北京时间是8时，东京时间是9时，记作( )；莫斯科时间是3时，记作( )；罗马时间记作一7时，现在罗马时间是( )。",
+      "options": [],
+      "answer": "+1时    -5时    1时",
+      "solution": "【分析】根据题意，以北京时间为标准，比北京时间早记为正，比北京时间晚记为负，据此计算即可。考查了负数的意义和应用，基础题。\n\n【详解】东京： $9 - 8 = 1$；记作 $+1$ 时；莫斯科: $8 - 3 = 5$; 记作 $-5$ 时;罗马： $8 - 7 = 1$，所以现在罗马是1时。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_236",
+      "original_id": "unit1_fill_236"
+    },
+    {
+      "id": "unit1_fill_237",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "如果规定标准体温为 $36.5^{\\circ} \\mathrm{C}, 37^{\\circ} \\mathrm{C}$ 记作 $+0.5^{\\circ} \\mathrm{C}$, 那么 $36.1^{\\circ} \\mathrm{C}$ 记作()。",
+      "options": [],
+      "answer": "$-0.4^{\\circ} \\mathrm{C}$",
+      "solution": "【分析】此题主要用正负数来表示具有意义相反的两种量:把人体的体温标准定在 $36.5^{\\circ} \\mathrm{C}$ 记为 $0^{\\circ} \\mathrm{C}$, 超过正常体温 $36.5^{\\circ} \\mathrm{C}$ 的记为正, 则低于正常体温 $36.5^{\\circ} \\mathrm{C}$ 的就记为负, 直接得出结论即可。此题主要考查正负数的意义, 正数与负数表示意义相反的两种量, 看清规定哪一个为正, 则和它意义相反的就为负。\n\n【详解】$36.5 - 36.1 = 0.4$ 因为 $36.5 > 36.1$ 故 $36.1^{\\circ} \\mathrm{C}$ 记作 $-0.4^{\\circ} \\mathrm{C}$",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_237",
+      "original_id": "unit1_fill_237"
+    },
+    {
+      "id": "unit1_fill_238",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "困难",
+      "question": "通常，我们规定海平面的海拔高度为 $0 \\mathrm{~m}$, 高于海平面的为正。'华山南峰的海拔为 $+2154.9 \\mathrm{~m}$'表示（ ）；'太平洋最深处的海拔为 $-11034 \\mathrm{~m}$'表示（ ）。",
+      "options": [],
+      "answer": "华山南峰比海平面高 $2154.9 \\mathrm{~m}$  太平洋最深处比海平面低 $11034 \\mathrm{~m}$",
+      "solution": "【分析】根据正负数的意义，高于海平面的为正，则低于海平面的就为负。据此解答即可。相反意义的两个量，如果其中一个用正数表示，那么另一个就用负数表示。\n\n【详解】由分析可得：'华山南峰的海拔为 $+2154.9 \\mathrm{~m}$' 表示华山南峰比海平面高2154.9 m;'太平洋最深处的海拔为 $-11034 \\mathrm{~m}$' 表示太平洋最深处比海平面低 $11034 \\mathrm{~m}$。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_238",
+      "original_id": "unit1_fill_238"
+    },
+    {
+      "id": "unit1_fill_239",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "仓库原有 50 吨货物,先运进60吨，记作 $+60$ 吨；又运出40吨，记作( )吨，此时，仓库里还有货物( )吨。",
+      "options": [],
+      "answer": "$-40$    70",
+      "solution": "【分析】运进和运出是一对具有相反意义的量，规定运进记为 +，则运出记为 -，据此解答即可。在一对具有相反意义的量中, 先规定其中一个为正, 则另一个就用负表示。\n\n【详解】运进 60 吨, 记作 +60 吨; 则运出 40 吨, 记作 -40吨; $50 + 60 - 40 = 70$ （吨）",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_239",
+      "original_id": "unit1_fill_239"
+    },
+    {
+      "id": "unit1_fill_240",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "困难",
+      "question": "$-\\frac{5}{8}$ 读作( )；+0.9 读作( )。",
+      "options": [],
+      "answer": "负八分之五   正零点九",
+      "solution": "【分析】这里的'+'、'-'不是运算符号，而是正负号，+读作正，-读作负，其余按照分数或者小数读法正常读出即可。此题考查正负数的读法，要注意不能读成'加几'或减几'，要读成'正几或负几'。\n\n【详解】据分析可知：$-\\frac{5}{8}$ 读作负八分之五； $+0.9$ 读作正零点九。",
+      "knowledge_points": [
+        "正负数的读法和写法"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_240",
+      "original_id": "unit1_fill_240"
+    },
+    {
+      "id": "unit1_fill_241",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "某地区 2020 年冬季的平均温度是零下 20 摄氏度，记作( )°C。正常情况下，水沸腾时的温度为 100 摄氏度，记作( )°C。",
+      "options": [],
+      "answer": "-20  100",
+      "solution": "【分析】通常，规定 $0^{\\circ} \\mathrm{C}$ 为标准， $0^{\\circ} \\mathrm{C}$ 以上记为正， $0^{\\circ} \\mathrm{C}$ 以下记为负，据此解答即可。理解正负数的意义、明确温度的表示方法是解题关键。\n\n【详解】由分析可得:零下 20 摄氏度, 记作 $-20^{\\circ} \\mathrm{C}$;100 摄氏度，记作 $100^{\\circ} \\mathrm{C}$。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_241",
+      "original_id": "unit1_fill_241"
+    },
+    {
+      "id": "unit1_fill_242",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "在 -5, 6, +3, 1, 0, -2, -7中，正数有( )个；如果向东走200米记作+200米，那么向西走500米记作( )米。",
+      "options": [],
+      "answer": "3   -500",
+      "solution": "【分析】除 0 以外, 正数的前面带正号 '+' 或什么符号也不带, 据此找到正数, 数出个数; 向东走记为正, 那么向西走记为负, 据此分析。关键是理解正负数的意义, 正负数可以表示相反意义的量, 0 不是正数也不是负数。\n\n【详解】在 -5, 6, +3, 1, 0, -2, -7 中，正数有 6、+3、1，共 3 个；如果向东走 200 米记作 +200 米，那么向西走 500 米记作 -500 米。",
+      "knowledge_points": [
+        "正负数的概念及辨认",
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_242",
+      "original_id": "unit1_fill_242"
+    },
+    {
+      "id": "unit1_fill_243",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "在 $1 、 5 、 { }^{1} \\frac{1}{4} 、 - 3 、 - 0.15 、 + 4 、 - \\frac{1}{4}$ 中, 正数有( )个, 负数有( )个,\n\n如果把这些数标在数轴上，离0最近的数是( )。",
+      "options": [],
+      "answer": "4  3    -0.15",
+      "solution": "【分析】本题中，数字前面有正号或没有符号的是正数，数字前面带负号的是负数；不管正负号，哪个数的数值最小，哪个数离0最近。关键是辨认正负数，比0大的数叫正数，比0小的数叫负数，数轴上，正数在0的右边，负数在0的左边。\n\n【详解】正数有：1、5、 $1\\frac{1}{4}$ 、+4，共4个；负数有：-3、-0.15、 $-\\frac{1}{4}$，共3个；不看正、负号，只比较数字的大小：\n\n$1 \\frac {1}{4} = 1. 2 5\\\\ \\frac {1}{4} = 0. 2 5 \\\\0.1 5 <   0. 2 5 <   1 <   1. 2 5 <   3 <   4 <   5$\n\n即 $0.15 < \\frac{1}{4} < 1 < 1 \\frac{1}{4} < 3 < 4 < 5$ 所以离 0 最近的数是 -0.15。",
+      "knowledge_points": [
+        "分数化小数",
+        "正负数在数轴上的表示",
+        "正负数的概念及辨认"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_243",
+      "original_id": "unit1_fill_243"
+    },
+    {
+      "id": "unit1_fill_244",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "2.4, - $\\frac{3}{7}$, 8, 0, 46, -8.3, +4.5, -2, -86这些数中, 正数有( )个, 负数有( )个。",
+      "options": [],
+      "answer": "4  4",
+      "solution": "【分析】前边带正号或不带负号的是正数，前边带负号的是负数，0既不是正数也不是负数。据此数出正负数的个数即可。关键是正确辨认正负数，0不是正数也不是负数。\n\n【详解】2.4，- $\\frac{3}{7}$，8，0，46，-8.3，+4.5，-2，-86这些数中，正数有2.4，8，46，+4.5，共4个；负数有- $\\frac{3}{7}$，-8.3，-2，-86，共4个。",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_244",
+      "original_id": "unit1_fill_244"
+    },
+    {
+      "id": "unit1_fill_245",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "在32、-4、+16、-58、20、-9、0、+43中，正数有( )，负数有( )，( )既不是正数，也不是负数。",
+      "options": [],
+      "answer": "32、+16、20、+43     -4、-58、-9 、 0",
+      "solution": "【分析】根据比 0 小的数叫做负数, 比 0 大的数叫正数, 进行填空。正数 $>0>$ 负数，0既不是正数也不是负数。\n\n【详解】在 32、-4、+16、-58、20、-9、0、+43 中，正数有 32、+16、20、+43，负数有 -4、-58、-9，0 既不是正数，也不是负数。",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "20-21六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_245",
+      "original_id": "unit1_fill_245"
+    },
+    {
+      "id": "unit1_fill_246",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "如果从学校出发向南行600米，表示为'-600'米。那么'+150'米，表示在学校的( )方( )米的位置。",
+      "options": [],
+      "answer": "北  150",
+      "solution": "【分析】根据正负数的意义，向南行记为负，那么向北行记为正，据此填空。正负数可以表示相反意义的量。\n\n【详解】如果从学校出发向南行 600 米, 表示为 '- 600' 米。那么 '+150' 米, 表示在学校的北方 150 米的位置。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_246",
+      "original_id": "unit1_fill_246"
+    },
+    {
+      "id": "unit1_fill_247",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "超市盈利 20 万元记作 +20 万元，亏损 15 万元应记作( )万元。",
+      "options": [],
+      "answer": "- 15",
+      "solution": "【分析】超市盈利20万元记作+20万元，意思是赚的钱记作正数；相对应的，赔了钱就记作负数，则题目中亏损15万元应记作 -15 万元。用正负数表示具有相反意义的量时，通常把盈利、上升、增长这一类量记作正数，而把亏损、下降、减少这一类量记作负数。\n\n【详解】超市盈利 20 万元记作 +20 万元，亏损 15 万元应记作 $(-15)$ 万元。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_247",
+      "original_id": "unit1_fill_247"
+    },
+    {
+      "id": "unit1_fill_248",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "如果把平均成绩记为 0 分, +8 分表示比平均成绩____, -15 分表示____, 比平均成绩少 3 分, 记作________。",
+      "options": [],
+      "answer": "高 8 分 (或多 8 分)       比平均成绩少 (或低）15分     -3分",
+      "solution": "【分析】正、负数表示一组相反意义的量，比平均成绩高。用正数表示，那么比平均成绩低就用负数表示。明确正负数的意义是解答本题的关键。\n\n【详解】如果把平均成绩记为 0 分, +8 分表示比平均成绩高 8 分, -15 分表示比平均成绩低 15 分, 比平均成绩少 3 分, 记作 -3 分。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_248",
+      "original_id": "unit1_fill_248"
+    },
+    {
+      "id": "unit1_fill_249",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "如果学校东面 12 米记为 (+12) 米, 那么, (-9) 米表示____。",
+      "options": [],
+      "answer": "学校西面 9 米",
+      "solution": "【分析】正负数表示一组相反意义的量，学校东面记作正，那么学校西面就记作负。明确正负数的意义是解答本题的关键。\n\n【详解】（-9）米表示学校西面9米。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_249",
+      "original_id": "unit1_fill_249"
+    },
+    {
+      "id": "unit1_fill_250",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "如果规定向东为正，那么向西走 $2 \\mathrm{~m}$ 记作 $\\_ \\mathrm{m}$。",
+      "options": [],
+      "answer": "-2",
+      "solution": "【分析】向东记为正，那么向西记为负，所以向西几米，就在距离前面加上'-'。互为相反意义的量的应用，一般把向东走、盈利、收入、增长的量规定为'+'，则与之相反的量就规定为'-'。\n\n【详解】如果规定向东为正，那么向西走 $2 \\mathrm{~m}$ 记作 $- 2 \\mathrm{~m}$。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "2020·北京东城·小升初真题",
+      "merged_id": "unit1_fill_250",
+      "original_id": "unit1_fill_250"
+    },
+    {
+      "id": "unit1_fill_251",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "小乐同学体重为48.5 千克，超出标准体重 3.5 千克，若标准体重记为 0 千克，小乐同学的体重记为 +3.5 千克，小刚体重为 42 千克，应记为____千克。",
+      "options": [],
+      "answer": "-3",
+      "solution": "【分析】标准体重 $=$ 小乐同学的体重一小乐同学超出标准体重的千克数；小刚同学的体重小于标准体重，应记为负，所以小刚低于标准体重的千克数 $=$ 标准体重一小刚同学的体重，据此作答即可。正负数表示一组相反意义的量, 此题中超出标准体重和低于标准体重就是一组相反意义的量, 一定要先求出标准体重。\n\n【详解】$48.5 - 3.5 = 45$ （千克） $45 - 42 = 3$ (千克)小刚的体重记为 -3 千克。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_251",
+      "original_id": "unit1_fill_251"
+    },
+    {
+      "id": "unit1_fill_252",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "以亮亮家为起点，向东走为正，向西走为负。如果亮亮从家走了 +50 米，又走了 -60 米。这时，亮亮在家的____，一共走了____米。",
+      "options": [],
+      "answer": "西边 110",
+      "solution": "【分析】由题意可知，走了 +50 米，表示向东走了 50 米，走了 -60 米，表示向西走了 60 米，所以，亮亮在家的西边 10 米处，据此解答即可。正负数表示一组相反意义的量, 此题中向东向西就是一组相反意义的量, 向东记作正, 向西就记作负。\n\n【详解】$60 + 50 = 110$ （米）亮亮在家的西边，一共走了110米。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_252",
+      "original_id": "unit1_fill_252"
+    },
+    {
+      "id": "unit1_fill_253",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "一筐苹果, 如果把 52 千克记作 +2 千克, 那么 48 千克应记作____千克。",
+      "options": [],
+      "answer": "-2",
+      "solution": "【分析】根据题意可知，50千克以上，用正数表示，50千克以下，用负数表示。数值等于苹果的重量与 50 的差，据此解答即可。解答本题的关键是要明确是以 50 千克作为标准量。\n\n【详解】一筐苹果, 如果把 52 千克记作 +2 千克, 那么 48 千克应记作 -2 千克。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_253",
+      "original_id": "unit1_fill_253"
+    },
+    {
+      "id": "unit1_fill_254",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "困难",
+      "question": "笑笑班同学拍球成绩平均每分钟拍 75 下, 如果把笑笑拍了 74 下, 记作 '-1'下, 奇思拍了 81 下, 记作( )下, 妙想拍了 70 下, 记作( )下。",
+      "options": [],
+      "answer": "$+6$    $ - 5$",
+      "solution": "【分析】拍了 74 下, 记作 '-1'下, 也就是把 75 下记作 '0', 计算出拍的次数与 75 之间的相差, 大于 75 的记作正数, 小于 75 的记作负数。本题主要考查学生对正负数知识的理解和灵活运用。\n\n【详解】笑笑班同学拍球成绩平均每分钟拍 75 下, 如果把笑笑拍了 74 下, 记作 '-1' 下, 奇思拍了 81 下, 记作 +6 下 (或 6 下), 妙想拍了 70 下, 记作 -5 下。",
+      "knowledge_points": [
+        "正负数的读法和写法",
+        "正负数的概念及辨认"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_254",
+      "original_id": "unit1_fill_254"
+    },
+    {
+      "id": "unit1_fill_255",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "某校六年级数学竞赛共有 10 道题, 规定: 答对 1 道题记作 $+10$ 分, 答错 1 道题记作 $-10$ 分, 不答记作 0 分。小明答错 3 道题, 2 道题未答,其余题都答对了。小明最后得( )分。",
+      "options": [],
+      "answer": "20",
+      "solution": "【分析】总共 10 道题, 小明答错 3 道题得 -30 分, 有 2 道题未答得 0 分, 剩下的 5 道题都答对得 50 分, $50 - 30 = 20$ (分), 所以本次数学竞赛小明最后得 20 分。仔细审题后弄清楚本次竞赛计分规则，对于答错的题目计分时要在答对题目得分基础上扣下去相应的分数，这与平常考试计分规则是不同的。\n\n【详解】由分析得:某校六年级数学竞赛共有 10 道题, 规定: 答对 1 道题记作 +10 分, 答错 1 道题记作 -10 分, 不答记作 0 分。小明答错 3 道题, 2 道题未答, 其余题都答对了。小明最后得 (20) 分。",
+      "knowledge_points": [
+        "利用正负数解决实际问题"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_255",
+      "original_id": "unit1_fill_255"
+    },
+    {
+      "id": "unit1_fill_256",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "某种袋装食盐的标准净重为 $500 \\mathrm{~g}$。如果把它的标准净重记为 $0 \\mathrm{~g}$, 超过标准净重记为正，那么一袋食盐净重 $504 \\mathrm{~g}$，记作（ ）g；另一袋食盐净重 $498 \\mathrm{~g}$，记作（ ）g。",
+      "options": [],
+      "answer": "$+4 $   $- 2$",
+      "solution": "【分析】根据正、负数的意义，超过标准净重记为正，则少于标准净重记为负，由此解答即可。本题考查的是正数和负数意义的运用, 解题的关键是理解 '正' 和 '负' 的相对性。\n\n【详解】$504 \\mathrm{~g}$ 超过标准净重 $4 \\mathrm{~g}$, 所以应记为 $+4 \\mathrm{~g}; 498 \\mathrm{~g}$ 少于标准净重 $2 \\mathrm{~g}$, 所以应记为 $- 2 \\mathrm{~g}$。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_256",
+      "original_id": "unit1_fill_256"
+    },
+    {
+      "id": "unit1_fill_257",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "电子支付钱包的零钱明细中显示 +145.00 元表示收入 145.00 元。若支出 88.00 元, 则零钱明细中会显示( )元。",
+      "options": [],
+      "answer": "- 88.00",
+      "solution": "【分析】+145.00由题意可知，收入记作正，则支出记作负，由此解答即可。本题考查的是正数和负数意义的运用, 解题的关键是理解 '正' 和 '负' 的相对性。\n\n【详解】元表示收入 145.00 元。若支出 88.00 元, 则零钱明细中会显示 - 88.00 元",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_257",
+      "original_id": "unit1_fill_257"
+    },
+    {
+      "id": "unit1_fill_258",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "以强强家为起点,向东走为正，向西走为负。强强从家出发，先向东走 $30 \\mathrm{~m}$，记作（ ），又向西走 $30 \\mathrm{~m}$，记作（ ），这时强强与家的距离是( ） $\\mathrm{m}$。",
+      "options": [],
+      "answer": "$+30\\mathrm{m}$   $ - 30\\mathrm{m}$  0",
+      "solution": "【分析】向东走为正，向西走为负，所以向东走 $30 \\mathrm{~m}$ 记作 $+ 30 \\mathrm{~m}$，向西走 $30 \\mathrm{~m}$ 记作 $- 30 \\mathrm{~m}$。强强从家出发，先向东走 $30 \\mathrm{~m}$，再向西走 $30 \\mathrm{~m}$，此时回到了原处，故强强与家的距离是 $0 \\mathrm{~m}$。正负数意义的实际应用，用这种记数的方法不仅能够表示行走的距离，同时也能够表示出行走的方向。\n\n【详解】由分析得：以强强家为起点，向东走为正，向西走为负。强强从家出发，先向东走 $30 \\mathrm{~m}$，记作 $(+30 \\mathrm{~m})$，又向西走 $30 \\mathrm{~m}$，记作 $(-30 \\mathrm{~m})$，这时强强与家的距离是 $(0) \\mathrm{m}$。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_258",
+      "original_id": "unit1_fill_258"
+    },
+    {
+      "id": "unit1_fill_259",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "卢特沙漠位于伊朗境内，是世界上最热的地区之一，被称为'烤熟的小麦'，这里有记录的沙漠表面温度最高达 $71^{\\circ} \\mathrm{C}$，记作 $(\\quad) ^{\\circ} \\mathrm{C}$。南极洲位于地球最南端，是'七大洲'中平均气温最低的一个洲，被称为'冰雪高原'，这里的温度曾达到零下 $94.7^{\\circ} \\mathrm{C}$，记作 $(\\quad) ^{\\circ} \\mathrm{C}$。",
+      "options": [],
+      "answer": "71   -94.7",
+      "solution": "【分析】比 $0^{\\circ} \\mathrm{C}$ 高的温度，在数前加'+'，一般情况下可省略不写；比 $0^{\\circ} \\mathrm{C}$ 低的温度，在数前加'-'，不可以省略，由此解答即可。高于 $0^{\\circ} \\mathrm{C}$ 的温度记作正数, 低于 $0^{\\circ} \\mathrm{C}$ 的温度记作负数。\n\n【详解】温度最高达 $71^{\\circ} \\mathrm{C}$, 记作: $71^{\\circ} \\mathrm{C}$;零下 $94.7^{\\circ} \\mathrm{C}$, 记作: $-94.7^{\\circ} \\mathrm{C}$",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·全国·课后作业",
+      "merged_id": "unit1_fill_259",
+      "original_id": "unit1_fill_259"
+    },
+    {
+      "id": "unit1_fill_260",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "新冠肺炎疫情期间，某市统计现有确诊人数，增加记作正，减少记作负。2020年2月某日，该市现有确诊人数增加2例，记作( )例。 -5例表示( )。",
+      "options": [],
+      "answer": "+2    现有确诊人数减少5例",
+      "solution": "【分析】因为在现有确诊人数中，增加记作正，减少记作负，故现有确诊人数增加 2 例就记作 +2 例；-5 例就表示现有确诊人数减少 5 例。本题考查了正负数实际意义的应用，在把其中一种量记作正数时，与之相反的量就记作负数。\n\n【详解】由分析得：新冠肺炎疫情期间，某市统计现有确诊人数，增加记作正，减少记作负。2020年2月某日，该市现有确诊人数增加2例，记作（+2）例。-5例表示（现有确诊人数减少5例）。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_260",
+      "original_id": "unit1_fill_260"
+    },
+    {
+      "id": "unit1_fill_261",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "在赵明的理财账户上，如果存入 6000 元用 +6000 元表示，那么取出 600 元用( )元表示。",
+      "options": [],
+      "answer": "- 600",
+      "solution": "【分析】此题主要用正负数来表示具有相反意义的两种量:存入银行记为正, 则从银行取出就记为负; 直接得出结论即可。此题主要考查正负数的意义, 正数与负数表示意义相反的两种量, 看清规定哪一个记为正, 则和它意义相反的就记为负。\n\n【详解】由分析可得，在赵明的理财账户上，如果存入6000元用 + 6000 元表示，那么取出 600 元用 - 600 元表示。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20五年级上·江苏盐城·期末",
+      "merged_id": "unit1_fill_261",
+      "original_id": "unit1_fill_261"
+    },
+    {
+      "id": "unit1_fill_262",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "如果把一个人的正常体温 $37^{\\circ} \\mathrm{C}$ 记为 $0^{\\circ} \\mathrm{C}$，则体温达到 $38.5^{\\circ} \\mathrm{C}$ 应记为（ ） $^{\\circ} \\mathrm{C}$，体温为 $36.9^{\\circ} \\mathrm{C}$ 应记为（ ） $^{\\circ} \\mathrm{C}$.",
+      "options": [],
+      "answer": "$+1.5$    $ - 0.1$",
+      "solution": "【分析】此题主要用正负数来表示具有意义相反的两种量，超过标准体温记为正，低于标准体温记为负。\n\n【详解】$38.5 - 37 = 1.5$，记作 $+1.5^{\\circ} \\mathrm{C}$；$37 - 36.9 = 0.1$，记作 $-0.1^{\\circ} \\mathrm{C}$。",
+      "knowledge_points": [
+        "负数的认识"
+      ],
+      "source": "18-19六年级下·全国·课后作业",
+      "merged_id": "unit1_fill_262",
+      "original_id": "unit1_fill_262"
+    },
+    {
+      "id": "unit1_fill_263",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "负数都比 0( ),所有的正数都在 0 的( )边.",
+      "options": [],
+      "answer": "小   右",
+      "solution": "【分析】负数都小于0，在数轴上，正数在0的右边，负数在0的左边。\n\n【详解】负数都比0小，所有的正数都在0的右边。",
+      "knowledge_points": [
+        "负数的认识"
+      ],
+      "source": "18-19 六年级下·全国·课后作业",
+      "merged_id": "unit1_fill_263",
+      "original_id": "unit1_fill_263"
+    },
+    {
+      "id": "unit1_fill_264",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "如果水位下降 $2 \\mathrm{~cm}$ 时水位变化记作 $- 2 \\mathrm{~cm}$，那么水位上升 $1 \\mathrm{~cm}$ 时水位变化记作( ), 水位不升不降时水位变化记作( ).",
+      "options": [],
+      "answer": "$+1 \\mathrm{~cm} \\quad 0 \\mathrm{~cm}$",
+      "solution": "【分析】水位下降记作负，则水位上升记作正；水位不升不降即没有变化，记作0。\n\n【详解】如果水位下降 $2 \\mathrm{~cm}$ 时水位变化记作 $- 2 \\mathrm{~cm}$，那么水位上升 $1 \\mathrm{~cm}$ 时水位变化记作 $+1 \\mathrm{~cm}$, 水位不升不降时水位变化记作 $0 \\mathrm{~cm}$。",
+      "knowledge_points": [
+        "负数的认识"
+      ],
+      "source": "2014 六年级·全国·课后作业",
+      "merged_id": "unit1_fill_264",
+      "original_id": "unit1_fill_264"
+    },
+    {
+      "id": "unit1_fill_265",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "如果比测试平均分高 4 分记作 +4 分, 那么 -3 分表示( ).",
+      "options": [],
+      "answer": "比测试平均分低3分",
+      "solution": "【分析】高于平均分记为正，则低于平均分记为负。\n\n【详解】如果比测试平均分高 4 分记作 +4 分, 那么 -3 分表示比测试平均分低3分。",
+      "knowledge_points": [
+        "负数的认识"
+      ],
+      "source": "2014 六年级·全国·课后作业",
+      "merged_id": "unit1_fill_265",
+      "original_id": "unit1_fill_265"
+    },
+    {
+      "id": "unit1_fill_266",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "在一次智力竞赛中六(1) 班的平均分为 79 分, 把高于平均分的记作正数, 低于平均分的记作负数, 张媛得了 82 分记作_分, 黄涛得了 79 分记作_分。老师将一个小组四名同学的成绩记作+8分, -5分, +6分, -1分。这四名同学的平均分是_。",
+      "options": [],
+      "answer": "+3    0   81分",
+      "solution": "【分析】此题主要用正负数来表示具有意义相反的两种量:平均分 79 分记作 0, 高于 79 分记为正, 则低于 79 分就记为负, 直接得出结论即可. 要求老师将一个小组四名同学的成绩记作 +8 分, -5 分, +6 分, -1 分. 这四名同学的平均分是多少分? 先求出四名同学的分数和, 除以 4, 用 79 加这个数, 即可得解.此题主要考查正负数的意义, 正数与负数表示意义相反的两种量, 看清规定哪一个为正, 则和它意义相反的就为负.\n\n【详解】$82 - 79 = 3$ 在一次智力竞赛中六（1）班的平均分为79分，把高于平均分的记作正数，低于平均分的记作负数，张媛得了82分记作+3分，黄涛得了79分记作0分.\n\n故答案为+3，0.$8 + (- 5) + 6 + (- 1) = 8,$ $8 \\div 4 = 2$, $79 + 2 = 81$ （分）；答: 这四名同学的平均分是 81 分.",
+      "knowledge_points": [
+        "负数的认识"
+      ],
+      "source": "2014 六年级·全国·课后作业",
+      "merged_id": "unit1_fill_266",
+      "original_id": "unit1_fill_266"
+    },
+    {
+      "id": "unit1_fill_267",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "一个物体可以上下平移,若设向下平移为正，那么向上平移 $30 \\mathrm{~cm}$ 应记作 \\_,\n\n'+45cm'表示_。",
+      "options": [],
+      "answer": "- 30cm  向下平移 $45 \\mathrm{~cm}$",
+      "solution": "【分析】此题考查了用正负数表示意义相反的两种量：把向下平移记作正数，则向上平移就记作负数；由此得解。此题主要考查正负数的意义, 正数与负数表示意义相反的两种量, 看清规定哪一个为正, 则和它意义相反的就为负。\n\n【详解】一个物体可以上下平移，若设向下平移为正，那么向上平移 $30 \\mathrm{~cm}$ 应记作 $-30 \\mathrm{~cm}$，'+45cm'表示向下平移 $45 \\mathrm{~cm}$。故答案为 $-30 \\mathrm{~cm}$, 向下平移 $45 \\mathrm{~cm}$。",
+      "knowledge_points": [
+        "负数的认识"
+      ],
+      "source": "2014 六年级·全国·课后作业",
+      "merged_id": "unit1_fill_267",
+      "original_id": "unit1_fill_267"
+    },
+    {
+      "id": "unit1_fill_268",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "大于-3 小于+4 的整数有( )个.",
+      "options": [],
+      "answer": "6",
+      "solution": "【分析】大于-3小于+4的整数包括：-2、-1、0、1、2、3，共6个。\n\n【详解】大于-3 小于+4 的整数有6个。",
+      "knowledge_points": [
+        "整数大小的比较",
+        "负数的认识"
+      ],
+      "source": "2014 六年级·全国·课后作业",
+      "merged_id": "unit1_fill_268",
+      "original_id": "unit1_fill_268"
+    },
+    {
+      "id": "unit1_fill_269",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "一个物体可以上下移动,若设向下移动为正，那么向上移动 $30 \\mathrm{~cm}$ 应记作____，'+45dm'表示____.",
+      "options": [],
+      "answer": "-30cm    向下移动 $45 \\mathrm{~dm}$",
+      "solution": "【分析】此题考查了用正负数表示意义相反的两种量：把向下平移记作正数，则向上平移就记作负数；由此得解。\n\n【详解】一个物体可以上下移动,若设向下移动为正，那么向上移动 $30 \\mathrm{~cm}$ 应记作 $-30\\mathrm{~cm}$，'+45dm'表示向下移动 $45 \\mathrm{~dm}$。",
+      "knowledge_points": [
+        "负数的认识"
+      ],
+      "source": "2014 六年级·全国·课后作业",
+      "merged_id": "unit1_fill_269",
+      "original_id": "unit1_fill_269"
+    },
+    {
+      "id": "unit1_fill_270",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "如果把零上 $15^{\\circ} \\mathrm{C}$ 记作 $+15^{\\circ} \\mathrm{C}$, 那么零下 $8^{\\circ} \\mathrm{C}$ 记作( )。",
+      "options": [],
+      "answer": "$-8^{\\circ} \\mathrm{C}$",
+      "solution": "【分析】此题主要用正负数来表示具有意义相反的两种量:零上气温记为正, 则零下气温就记为负, 直接得出结论即可。此题主要考查正负数的意义, 正数与负数表示意义相反的两种量, 看清规定哪一个为正, 则和它意义相反的就为负。\n\n【详解】如果把零上 $15^{\\circ} \\mathrm{C}$ 记作 $+15^{\\circ} \\mathrm{C}$, 那么零下 $8^{\\circ} \\mathrm{C}$ 记作 $-8^{\\circ} \\mathrm{C}$。故答案为: $-8^{\\circ} \\mathrm{C}$",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "2020 六年级下·全国·专题练习",
+      "merged_id": "unit1_fill_270",
+      "original_id": "unit1_fill_270"
+    },
+    {
+      "id": "unit1_fill_271",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "在 -3, 0, 3, 5, -2, 9, 120, $\\frac{2}{3}$, -1, 7, $-\\frac{2}{5}$ 中, 正数有( )个, 负数有( )个, 既不是正数也不是负数的有( )个。",
+      "options": [],
+      "answer": "6 4 1",
+      "solution": "【分析】根据正数的意义，以前学过的 1、2、3、…这样的数叫做正数，正数前面也可以加'+'号；根据负数的意义，为了表示两种相反意义的量，这里出现了一种新的数，像 -1、-2、-3、…这样的数叫做负数；0 即不是正数也不是负数。本题是考查正、负数的意义，明确正数、负数的含义，是解答此题的关键。\n\n【详解】在 $-3, 0, 3, 5, -2, 9, 120, \\frac{2}{3}, -1, 7, -\\frac{2}{5}$中,正数有: 3、5、9、120、 $\\frac{2}{3}$、7, 共6个,负数有：-3、-2、-1、- $\\frac{2}{5}$，共4个，既不是正数也不是负数的有 0, 只有 1 个;故答案为: 6, 4, 1",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "2020 六年级下·全国·专题练习",
+      "merged_id": "unit1_fill_271",
+      "original_id": "unit1_fill_271"
+    },
+    {
+      "id": "unit1_fill_272",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "如果小华家月收入7500元记作+7500元，那么他家这个月水、电、煤气支出\n\n600元应记作( )元。",
+      "options": [],
+      "answer": "- 600",
+      "solution": "【分析】正负数表示相反意义的量，收入记为正，则支出记为负，据此解答即可。本题考查正负数的意义, 解答本题的关键是掌握正负数的意义。\n\n【详解】如果小华家月收入 7500 元记作 +7500 元, 那么他家这个月水、电、煤气支出 600 元应记作 -600 元。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "2020·黑龙江鸡西·小升初真题",
+      "merged_id": "unit1_fill_272",
+      "original_id": "unit1_fill_272"
+    },
+    {
+      "id": "unit1_fill_273",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "甲、乙两人各买了一袋同样的大米，包装袋上标注'净质量： $50\\mathrm{kg} \\pm 5\\mathrm{g}$'，那么，他俩购买的大米质量最多相差( )g。",
+      "options": [],
+      "answer": "10",
+      "solution": "【分析】利用 '净质量: $50 \\mathrm{~kg} \\pm 5 \\mathrm{~g}$' 即可求出在合格范围内,大米最重为 $50 \\mathrm{~kg} + 5 \\mathrm{~g}$; 最轻为 $50 \\mathrm{~kg} - 5 \\mathrm{~g}$, 最轻最重做差即可求解。此题考查合格范围的实际应用，不低于最低值，不高于最高值都是合格的。\n\n【详解】大米最重为 $50 \\mathrm{~kg} + 5 \\mathrm{~g} = 50005 \\mathrm{~g}$ 最轻为 $50 \\mathrm{~kg} - 5 \\mathrm{~g} = 49995 \\mathrm{~g}$ $50005 - 49995 = 10 \\mathrm{~g}$ 故答案为: 10",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20六年级下·河南郑州·期末",
+      "merged_id": "unit1_fill_273",
+      "original_id": "unit1_fill_273"
+    },
+    {
+      "id": "unit1_fill_274",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "如果自行车车条的长度比标准长度长 $2 \\mathrm{~mm}$, 记作 $+2 \\mathrm{~mm}$, 那么比标准长度短 $1.5 \\mathrm{~mm}$, 记作( )。",
+      "options": [],
+      "answer": "- 1.5mm",
+      "solution": "【分析】根据正数和负数的意义，比标准长度长记作正数，则比标准长度短记作负数。此题主要考查正负数的意义，正数与负数表示意义相反的两种量，看清规定哪一个为正，则和它意义相反的就为负。\n\n【详解】由已知比标准长度长 $2 \\mathrm{~mm}$, 记作 $+2 \\mathrm{~mm}$;则比标准长度短 $1.5 \\mathrm{~mm}$, 应记为 $-1.5 \\mathrm{~mm}$。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_274",
+      "original_id": "unit1_fill_274"
+    },
+    {
+      "id": "unit1_fill_275",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "0.5、-3、+90%、12、0、-73.2、+6.1、+ $\\frac{3}{2}$ 这几个数中，正数有( )\n\n), 负数有( ), ( )既不是正数, 也不是负数。",
+      "options": [],
+      "answer": "$0.5 、 + 90 \\%  、12、+6.1 + \\frac{3}{2}$     -3、-73.2   0",
+      "solution": "【分析】大于 0 的是正数, 小于 0 的是负数, 0 既不是正数,也不是负数。本题考查了正负数的认识，正数 $>0>$ 负数。\n\n【详解】0.5、-3、+90%、12、0、-73.2、+6.1、+ $\\frac{3}{2}$ 这几个数中，正数有0.5、+90%、12、+6.1、+ $\\frac{3}{2}$，负数有 -3、-73.2，0既不是正数，也不是负数。",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_275",
+      "original_id": "unit1_fill_275"
+    },
+    {
+      "id": "unit1_fill_276",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "如果电梯上升到9层，记作 $+9$ 层，那么下降到地下1层，记作( )层；停车场在地下3层，记作( )层。",
+      "options": [],
+      "answer": "-1  -3",
+      "solution": "【分析】从第一层开始，往上记为正，低于第一层记为负，据此填空。本题考查了正负数的意义, 正负数可以表示相反意义的量。\n\n【详解】如果电梯上升到 9 层, 记作 +9 层, 那么下降到地下 1 层, 记作 -1 层; 停车场在地下 3 层, 记作 -3 层。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_276",
+      "original_id": "unit1_fill_276"
+    },
+    {
+      "id": "unit1_fill_277",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "一种袋装食盐标准质量为 $350 \\mathrm{~g}$, 质检工作人员为了解该种食盐每袋的净重与标准的误差, 把食盐净重 $352 \\mathrm{~g}$ 记为 $+2 \\mathrm{~g}$, 那么食盐净重\n\n$349 \\mathrm{~g}$ 就记为( ) $\\mathrm{g}$。",
+      "options": [],
+      "answer": "-1",
+      "solution": "【分析】本题是用正负数来表示具有相反意义的量, $350 \\mathrm{~g}$是分界线, 比 $350 \\mathrm{~g}$ 多记为正, 比 $350 \\mathrm{~g}$ 少记为负, 据此来解题。本题考查正负数的意义, 正数与负数表示相反意义的量, 看清规定题目是规定哪个量为正, 和它意义相反的量就记为负。\n\n【详解】352 克比标准量 $350 \\mathrm{~g}$ 多 $2 \\mathrm{~g}$, 记为 +2 克, 349 克比标准量 $350 \\mathrm{~g}$ 少 $1 \\mathrm{~g}$, 记为 -1 g。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_277",
+      "original_id": "unit1_fill_277"
+    },
+    {
+      "id": "unit1_fill_278",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "如果某地防汛的警戒水位记作 $0 \\mathrm{~m}$, 正数表示高于警戒水位的水面高度, 那么高于警戒水位 $1.2 \\mathrm{~m}$, 记为( ) $\\mathrm{m}$, 低于警戒水位 $0.5 \\mathrm{~m}$,\n\n记为( )m。",
+      "options": [],
+      "answer": "$+1.2 - 0.5$",
+      "solution": "【分析】以警戒水位为标准，高于警戒水位记为正，低于警戒水位记为负，据此填空。本题考查了正负数的意义, 正负数可以表示相反意义的量。\n\n【详解】如果某地防汛的警戒水位记作 $0 \\mathrm{~m}$, 正数表示高于警戒水位的水面高度, 那么高于警戒水位 $1.2 \\mathrm{~m}$, 记为 +1.2m, 低于警戒水位 $0.5 \\mathrm{~m}$, 记为 -0.5m。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_278",
+      "original_id": "unit1_fill_278"
+    },
+    {
+      "id": "unit1_fill_279",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "温度越低就越冷,-4°C与 -15°C相比, ( )温度低。",
+      "options": [],
+      "answer": "$-15^{\\circ} \\mathrm{C}$",
+      "solution": "【分析】比 $0^{\\circ} \\mathrm{C}$ 低的温度叫零下温度, 通常在数字前加 '- ' 表示。例如, -3°C表示零下3摄氏度。温度是零下的, 数字越大, 温度越低, $0^{\\circ} \\mathrm{C}$ 比任何一个零下几摄氏度都高。据此分析。\n\n【详解】由分析得：所以 $-4^{\\circ} \\mathrm{C}$ 与 $-15^{\\circ} \\mathrm{C}$ 相比, $-15^{\\circ} \\mathrm{C}$ 温度低。",
+      "knowledge_points": [
+        "温度的认识及比较"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_fill_279",
+      "original_id": "unit1_fill_279"
+    },
+    {
+      "id": "unit1_judge_280",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "0 是正数和负数的分界点。( )",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】0 是非负、非正的中性数，0 大于一切负数，小于一切正数，是正负数的分界点。结合数轴可知, 正数在 0 的右边, 负数在 0 的左边, 0 可以看作是正数和负数的分界点。\n\n【详解】由分析得：数字0是正数和负数的分界点。原题说法正确。故答案为: √。",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "2023·陕西西安·小升初真题",
+      "merged_id": "unit1_judge_280",
+      "original_id": "unit1_judge_280"
+    },
+    {
+      "id": "unit1_judge_281",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "如果想从起点到－1.5 处，那么应从起点开始向右运动 1.5 个单位长度。（）",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】一般在直线上用 0 表示起点, 表示正数应从 0 处起向右运动, 表示负数应从 0 处起向左运动。关键是理解正负数在数轴上的表示方法，本题错在误认为 0 的右边是负数。\n\n【详解】从起点到 -1.5 处，应向左运动，而不是向右运动。故答案为：×",
+      "knowledge_points": [
+        "正负数在数轴上的表示"
+      ],
+      "source": "20-21六年级下·全国·单元测试",
+      "merged_id": "unit1_judge_281",
+      "original_id": "unit1_judge_281"
+    },
+    {
+      "id": "unit1_judge_282",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "存折中'2000'表示取数，'-500'为取出的钱数。（）",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】存入的钱数用正数表示，取出的钱数用负数表示，本题错在把正、负数的意义弄混淆了。关键是理解正负数的意义, 正负数可以表示相反意义的量。\n\n【详解】'2000'为存入的钱，'- 500'为取出的钱数。存入为正，取出为负。故答案为:×",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_judge_282",
+      "original_id": "unit1_judge_282"
+    },
+    {
+      "id": "unit1_judge_283",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "- 5°C前面的 '- '可以省略不写。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】$0^{\\circ} \\mathrm{C}$是零上温度和零下温度的分界点。零下温度用 '-'（符号）表示，零上温度用'+'（正号）表示或省略不写。零上温度和零下温度是两种相反意义的量。本题错在误认为 '- ' 和 '+' 一样, 可以省略。没有正确理解正负数的意义。\n\n【详解】-5°C前面的'-'不可以省略不写。故答案为: ×",
+      "knowledge_points": [
+        "正负数的读法和写法",
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_judge_283",
+      "original_id": "unit1_judge_283"
+    },
+    {
+      "id": "unit1_judge_284",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "- $3^{\\circ} \\mathrm{C}$ 比 $-10^{\\circ} \\mathrm{C}$ 低 $7^{\\circ} \\mathrm{C}$。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】两个数都是负数，去掉负号后比较出大小，比较大的数实际小，比较小的数实际大，据此解答即可熟练掌握负数大小比较的方法是解答本题的关键。。\n\n【详解】$3 < 10$, 所以 $-3^{\\circ} \\mathrm{C} > -10^{\\circ} \\mathrm{C}$, 则 $-3^{\\circ} \\mathrm{C}$ 高于 $-10^{\\circ} \\mathrm{C}$, 原题说法错误;故答案为: ×。",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_judge_284",
+      "original_id": "unit1_judge_284"
+    },
+    {
+      "id": "unit1_judge_285",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "生活中也有像' $-2.5\\%$ '这样的百分数。（）",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】百分比的意义：表示一个数是另一个数的百分之几的数，叫百分数。本题考查了百分数的意义，属于基础题。\n\n【详解】根据百分数的意义，百分数只能表示一个数占另一个数的百分之几，在生活中，表示负增长时百分数可表示为负，如人口负增长，就是新增的人口比死亡的人数要少，如意大利的人口增长率为 $-0.05\\%$ ，比如三亚'双节'客房报价同比下调 $2.5\\%$ ，也就是同比增长为 $-2.5\\%$ 。所以，题干叙述正确。故答案为: √",
+      "knowledge_points": [
+        "正负数的意义及应用",
+        "百分数的意义"
+      ],
+      "source": "20-21六年级上·全国·单元测试",
+      "merged_id": "unit1_judge_285",
+      "original_id": "unit1_judge_285"
+    },
+    {
+      "id": "unit1_judge_286",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "大于-3而小于+1的自然数只有3个。（）",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】大于 -3 而小于 +1 的整数有 3 个, 即 -2、-1、0, 其中自然数只有一个 0。本题主要考查学生对负数、自然数知识的掌握和灵活运用。\n\n【详解】大于 $-3$ 而小于 $+1$ 的自然数只有一个 0, 所以判断错误。故答案为: ×",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "20-21四年级上·全国·单元测试",
+      "merged_id": "unit1_judge_286",
+      "original_id": "unit1_judge_286"
+    },
+    {
+      "id": "unit1_judge_287",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "用有正数和负数的直线可以表示距离和相反的方向。( )",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】用有正数和负数的直线可以表示距离和相反的方向，像这样在直线上表示出 0、正数和负数的数的线叫数轴，在数轴上，从左到右的顺序就是数从小到大的顺序。所有的正负数都可以在数轴上找到它的位置，数轴上右边的数比左边的数大。\n\n【详解】用有正数和负数的直线可以表示距离和相反的方向, 此说法正确。故答案为: √",
+      "knowledge_points": [
+        "正负数在数轴上的表示"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_judge_287",
+      "original_id": "unit1_judge_287"
+    },
+    {
+      "id": "unit1_judge_288",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "欢欢向东走 $30\\mathrm{m}$ 记作 $+30\\mathrm{m}$，那么她向北走 $90\\mathrm{m}$，就记作 $-90\\mathrm{m}$。（）",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】主要用正负数来表示具有意义相反的两种量：向东走 $30 \\mathrm{~m}$，记作 $+30 \\mathrm{~m}$，和东相反的方向是西，她向西走记作负数。据此解答。此题主要考查正负数的意义，看清规定哪一个为正，则和它意义相反的就为负。\n\n【详解】欢欢向东走 $30 \\mathrm{~m}$, 记作 $+30 \\mathrm{~m}$, 那么她向西走 $90 \\mathrm{~m}$, 就记作 $-90 \\mathrm{~m}$, 而不是向北走。原题说法错误。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_judge_288",
+      "original_id": "unit1_judge_288"
+    },
+    {
+      "id": "unit1_judge_289",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "'+' 可以省略不写, '- ' 也可以省略不写。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】正数都大于 0, 正数都带有 '+' 号或不带任何符号, '+' 可以省略不写; 负数都小于 0, 负数都带有 '- ' 号; 据此判断。只有'+'可以省略不写。\n\n【详解】'+'可以省略不写，'-'不可以省略不写。原说法错误。故答案为: ×",
+      "knowledge_points": [
+        "正负数的读法和写法"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_judge_289",
+      "original_id": "unit1_judge_289"
+    },
+    {
+      "id": "unit1_judge_290",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "汽油凝固的温度是零下 $17^{\\circ} \\mathrm{C}$, 记作 $+17^{\\circ} \\mathrm{C}$。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】正数与负数表示两种具有相反意义的量, 0 是正数与负数的分界点。所以一般用正数表示零上温度，用负数表示零下温度；据此解答。本题考查了正负数的意义与应用, 关键是要掌握正负数的意义。\n\n【详解】由分析可知: 汽油凝固的温度是零下 $17^{\\circ} \\mathrm{C}$, 应该记作 $-17^{\\circ} \\mathrm{C}$; 所以原题说法错误。故答案为: ×",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_judge_290",
+      "original_id": "unit1_judge_290"
+    },
+    {
+      "id": "unit1_solve_291",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "解答题",
+      "difficulty": "困难",
+      "question": "把几个数用大括号括起来，中间用逗号断开，如：{1，2，8}、{-2，7， $\\frac{3}{4}$，19}，我们称之为集合，其中的数称为集合的元素。如果一个集合满足：当有理数 a 是集合的元素时，有理数 10-a 也必是这个集合的元素，这样的集合我们称为'好的集合'。例如集合 \\{10, 0\\} 就是一个'好的集合'。(1) 集合 $\\{-2, 7, \\frac{3}{4}, 19\\}$ （）（填'是'或'不是'）'好的集合'。  (2) 请你再写出两个'好的集合'（不得与上面出现过的集合重复）。  (3) 在所有 '好的集合' 中, 元素个数最少的集合是 ( )。",
+      "options": [],
+      "answer": "（1）不是；（2）{3，7}，{1，9}；（3）{5}",
+      "solution": "【分析】'好的集合'是指：当有理数 $a$ 是集合的元素时，有理数 $10 - a$ 也必是这个集合的元素。可理解为集合里有两个数相加等于 $10$。(1) 任意找一个数验证, 例如 $7, 10 - 7 = 3, 3$ 不是集合中的元素, 所以这个集合不是 '好的集合';(2) $3 + 7 = 10, 1 + 9 = 10$ ，可写出 $\\{3, 7\\}, \\{1, 9\\}$; (3) 元素个数最少的集合应是 $10 - a = a$ 时, 解得 $a = 5$, 即 $\\{5\\}$。解答此题的关键是明确 '好的集合' 的定义, 然后再进一步解答。\n\n【详解】（1）集合 $\\{-2, 7, \\frac{3}{4}, 19\\}$ 不是'好的集合'；(2) $\\{3, 7\\}, \\{1, 9\\}$; (3) 在所有 '好的集合' 中, 元素个数最少的集合是 \\{5\\}。",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_solve_291",
+      "original_id": "unit1_solve_291"
+    },
+    {
+      "id": "unit1_solve_292",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "体育老师在10米的跳台上，先向上腾空3米，接着下落15米，落在水中，此时他在水面下多少米处？如果把水面的高度记作0，那么他现在的位置记作什么？",
+      "options": [],
+      "answer": "2米；-2米",
+      "solution": "【分析】此题主要用正负数来表示具有意义相反的两种量,运用正负数意义理解向上为正，向下为负，再利用正负数加减法规则进行计算。此题主要考查正负数的意义, 正数与负数表示意义相反的两种量, 看清规定哪一个为正, 则和它意义相反的就为负。\n\n【详解】$10 + 3 - 15 = -2$ （米），记作-2米；答: 此时他在水面下 2 米处, 如果把水面的高度记作 0, 那么他现在的位置记作 - 2。",
+      "knowledge_points": [
+        "利用正负数解决实际问题",
+        "正负数的概念及辨认",
+        "正负数的意义及应用"
+      ],
+      "source": "20-21六年级下·全国·单元测试",
+      "merged_id": "unit1_solve_292",
+      "original_id": "unit1_solve_292"
+    },
+    {
+      "id": "unit1_solve_293",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "规定每分钟跳绳 90 次为标准成绩，高于标准成绩的记为正数，低于标准成绩的记为负数。某校六（1）班第一小组 8 位学生每分钟跳绳成绩分别记为 -15 次、20 次、+8 次、-6 次、-8 次、+14次、+10次和+17次。(1) 这 8 位同学中每分钟跳绳次数最少的跳了 ( ) 次,每分钟跳绳次数最多的跳了 ( ) 次, 每分钟跳得最少的比每分钟跳得最多的少 ( ) 次。  (2) 请你利用小贤的方法计算出这8位同学平均每分钟跳绳多少次。小贤：我不需要计算出这8位同学的实际成绩，就可以得到他们的平均成绩。",
+      "options": [],
+      "answer": "(1) 75; 110; 35  (2) 95 次。",
+      "solution": "【分析】（1）比较处8位同学调成成绩记录的大小值，最大的那个每分钟跳的最多。正数即用平均数加此数目；负数则用平均数减去负号后的数据。(2) 可以计算记录出来的数据, 平均值是多少, 得数为正即加平均成绩 90; 得数为负即用 90 减去负号后的数据, 据此解答即可。通常用正负数表示一组具有相反意义的量，如果规定高于标准成绩为正，那么低于标准成绩为负。平均数的公式要牢记：平均数 $=$ 总数 $\\div$ 总份数。\n\n【详解】(1) $20 > + 17 > + 14 > + 10 > + 8 > - 6 > - 8>$ -15;每分钟跳绳次数最少的是： $90 - 15 = 75$ （次）；最多的是: $90 + 20 = 110$ (次);相差: $110 - 75 = 35$ (下)\n\n(2) $(-15) + 20 + 8 + (-6) + (-8) + 14 + 10 + 17=4 0 \\\\ 4 0 \\div 8 = 5 \\\\ 9 0 + 5 = 9 5 \\text {(次)} $",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_solve_293",
+      "original_id": "unit1_solve_293"
+    },
+    {
+      "id": "unit1_solve_294",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "早上的气温是零下 $5^{\\circ} \\mathrm{C}$, 记作 -5 摄氏度, 下午的气温升高了 15 摄氏度, 应该记作?",
+      "options": [],
+      "answer": "+10 摄氏度",
+      "solution": "【分析】根据题意可知, 零下温度记作负数, 则零上温度记作正数; 下午的气温升高了 15 摄氏度, 当增加 5 摄氏度时, 温度为 0 摄氏度, 再增加 10 摄氏度时, 即为零上 10 摄氏度, 记作 +10, 据此解答即可。正负数表示一组相反意义的量, 此题中零下温度和零上温度就是一组相反意义的量, 零下温度记作负, 零上温度就记作正。\n\n【详解】答：早上的气温是零下 $5^{\\circ} \\mathrm{C}$，记作 -5 摄氏度，下午的气温升高了 15 摄氏度，变成 10 摄氏度，记作 +10 摄氏度。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_solve_294",
+      "original_id": "unit1_solve_294"
+    },
+    {
+      "id": "unit1_solve_295",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "仓库原有货物80吨，把80吨记为0吨，则后来仓库四次的进货情况是 $+6$ 吨、 $-4$ 吨、 $-12$ 吨和 $+8$ 吨。这时仓库有货物多少吨？",
+      "options": [],
+      "answer": "398吨",
+      "solution": "【分析】首先审清题意, 由 '把 80 吨记为 0 吨', 可知 +6 吨表示 $80 + 6$ 吨, -4 吨表示 $80 - 4$ 吨, ……, 因此先计算 $(+6 - 4 - 12 + 8)$ 的结果, 最后加上 $80 \\times 5$ 即可。此题解答的关键在于理解正、负数的含义，再根据题意作答。\n\n【详解】$80 \\times 5 + [(+6) + (-4) + (-12) + (+8)]$ $= 400 - 2$ $= 398$ （吨）答：这时仓库有货物 398 吨。",
+      "knowledge_points": [
+        "利用正负数解决实际问题"
+      ],
+      "source": "20-21六年级下·全国·课后作业",
+      "merged_id": "unit1_solve_295",
+      "original_id": "unit1_solve_295"
+    },
+    {
+      "id": "unit1_solve_296",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "体检中，测得 6 位同学的体重分别是 32 千克、33 千克、37 千克、36 千克、34 千克、38 千克。(1) 他们的平均体重是多少？  (2) 将平均体重记为 0 千克, 超过记为正数, 不足记为负数, 按从大到小的顺序排列。",
+      "options": [],
+      "answer": "(1) 35 千克; (2) $+3, +2, +1, -1, -2, -3$。",
+      "solution": "【分析】（1）把六个人的体重全部加起来，再除以 6 即可解答； (2) 分别计算出六个人的体重与上面计算出的平均体重的差, 多出的部分用正数表示, 少的部分用负数表示, 据此即可解答。本题考查了平均数的计算方法以及正负数的意义,比较简单。\n\n【详解】(1) $(32 + 33 + 37 + 36 + 34 + 38) \\div 6\\\\= 2 1 0 \\div 6\\\\=35 (千克)$\n\n答: 他们的平均体重是 35 千克。\n\n(2) 将平均体重记为 0 千克, 超过记为正数, 不足记为负数, 这六个人的体重可以记作:$3, - 2, + 2, + 1, - 1, + 3;$\n\n按从大到小的顺序排列为: +3, +2, +1, -1, -2, -3。",
+      "knowledge_points": [
+        "负数的认识"
+      ],
+      "source": "2014 六年级·全国·课后作业",
+      "merged_id": "unit1_solve_296",
+      "original_id": "unit1_solve_296"
+    },
+    {
+      "id": "unit1_solve_297",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "聪聪从少年宫向东行 100 米, 表示为 +100 米。(1) 他从少年宫向西行 80 米, 可以怎样表示?  (2) 聪聪现在的位置是 +36 米, 你能说出他的具体位置吗?",
+      "options": [],
+      "answer": "-80 米，他在少年宫的东边 36 米。",
+      "solution": "【分析】向东行为'正'，则向西行为'负'，直接得出结论即可。本题主要考查用正负数来表示具有意义相反的两种量。\n\n【详解】①他从少年宫向西行 80 米, 表示为 -80 米。②聪聪现在的位置是 +36 米, 说明他在少年宫的东边 36 米。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "20-21 五年级上·江苏·期中",
+      "merged_id": "unit1_solve_297",
+      "original_id": "unit1_solve_297"
+    },
+    {
+      "id": "unit1_solve_298",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "一艘潜艇在海平面以下400米处，记作-400米，一条鲨鱼在潜艇上方100米处。如果潜艇下潜100米，鲨鱼上游50米，此时潜艇和鲨鱼相距多少米？",
+      "options": [],
+      "answer": "250 米",
+      "solution": "【分析】![](images/821d7adab5418b03fb75bb3e517568939df38b4eb37d0670dc74f49a3b2ff29a.jpg)潜艇和鲨鱼最初相距 100 米, 潜艇下潜 100 米, 鲨鱼上游 50 米后的位置如图, 据此列式计算。本题考查了正负数的意义, 正负数可以表示相反意义的量。\n\n【详解】$100 + 100 + 50 = 250$ （米）答: 此时潜艇和鲨鱼相距 250 米。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20六年级下·全国·课后作业",
+      "merged_id": "unit1_solve_298",
+      "original_id": "unit1_solve_298"
+    },
+    {
+      "id": "unit1_solve_299",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "实验小学六1班为了选拔学生参加竞赛，队友希望获奖的小明进行跟踪调查，他几次测验成绩如下：95分、99分、93分、97分、89分、100分、85分、96分、99分、97分。(1) 小明这几次测验的平均分是多少?  (2) 这几次测验成绩在平均分的基础上的, 用正、负数可以怎样表示?",
+      "options": [],
+      "answer": "(1) 95 分 (2) 0 分、+4 分、-2 分、+2 分、-6 分、+5 分、-10 分、+1 分、+4 分、+2 分",
+      "solution": "【分析】（1）根据平均数 $=$ 总数 $\\div$ 份数，将所有成绩加起来，除以次数即可；(2) 以平均分为标准，高于平均分记为正，低于平均分记为负。本题考查了平均数和正负数的意义，正负数可以表示相反意义的量。\n\n【详解】(1) $(95 + 99 + 93 + 97 + 89 + 100 + 85 + 96 + 99+9 7) \\div 1 0 \\\\ = 9 5 0 \\div 1 0 \\\\ = 9 5 \\text {(分)} $\n\n答: 小明这几次测验的平均分是 95 分。\n\n(2) $99 - 95 = 4$ (分) $95 - 93 = 2$ (分)   $97 - 95 = 2$ (分)   $95 - 89 = 6$ (分)   $100 - 95 = 5$ (分) $95 - 85 = 10$ (分) $96 - 95 = 1$ (分)答: 用正、负数为0分、+4分、-2分、+2分、-6分、+5分、-10分、+1分、+4分、+2分。",
+      "knowledge_points": [
+        "平均数的意义及求法",
+        "正负数的意义及应用"
+      ],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_solve_299",
+      "original_id": "unit1_solve_299"
+    },
+    {
+      "id": "unit1_solve_300",
+      "unit": "第一单元：负数",
+      "chapter": "负数的认识",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "唱歌比赛中四位评委通过打分决定选手是否晋级，规定每位评委打10分为满分，把6分记为0分，超过的用正数表示，不足的用负数表示。如果四位评委打分的总分数达到25分选手就可以晋级。四位评委给5号选手小明的打分情况如下：一号评委 $+2$ 分，二号评委0分，三号评委-1分，四号评委 $+3$ 分。请问：小明的总分数是多少？能否晋级？",
+      "options": [],
+      "answer": "28分；能晋级",
+      "solution": "【分析】以 6 分为标准，高于 6 分记为正，低于 6 分记为负，分别求出 4 个评委打的分数，加起来，与 25 分比较。本题考查了正负数的意义和应用, 正负数可以表示相反意义的量。\n\n【详解】$6 + 2 = 8$ （分）$6 - 1 = 5$ (分)\n\n$6 + 3 = 9 \\text {(分)} \\\\ 8 + 6 + 5 + 9 \\\\ = 1 4 + 5 + 9 \\\\ = 1 9 + 8 \\\\ = 2 8 \\text {(分)} $\n\n$2 8 > 2 5$ 答: 小明的总分数是 28 分, 能晋级。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_solve_300",
+      "original_id": "unit1_solve_300"
+    },  {
+      "id": "unit1_choice_301",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "小明的期末数学成绩高于平均分 3 分记为 +3, 小亮的分数记为 -4, 说明 ( )",
+      "options": [
+        "A. 高于平均分 4 分",
+        "B. 低于平均分 4 分",
+        "C. 小明和小亮相差 4 分"
+      ],
+      "answer": "B",
+      "solution": "【分析】正负数表示一组相反意义的量，高于平均分记作正，低于平均分就记作负。本题考查正负数的意义，解答本题的关键是理解正负数表示一组相反意义的量。\n\n【详解】小明的期末数学成绩高于平均分 3 分记为 +3, 小亮的分数记为 -4 , 说明低于平均分 4 分。故答案为: B 。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "18-19 五年级上·江苏·单元测试",
+      "merged_id": "unit1_301",
+      "original_id": "unit1_choice_301"
+    },
+    {
+      "id": "unit1_choice_302",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "低于正常水位 0.34 米记为 -0.34, 高于正常水位 0.02 米记作 ( )。",
+      "options": [
+        "A. +0.02",
+        "B. -0.02",
+        "C. +0.18",
+        "D. -0.14"
+      ],
+      "answer": "A",
+      "solution": "【分析】以正常水位 0.34 米为标准，低于正常水位记为负，高于正常水位记为正，据此选择。本题考查了正负数的意义, 正负数可以表示相反意义的量。\n\n【详解】低于正常水位 0.34 米记为 -0.34，高于正常水位 0.02 米记作 +0.02。故答案为: A",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_302",
+      "original_id": "unit1_choice_302"
+    },
+    {
+      "id": "unit1_choice_303",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "数轴上 $-\\frac{1}{3}$ 在 $-\\frac{1}{9}$ 的( ) 边。",
+      "options": [
+        "A. 左",
+        "B. 右",
+        "C. 北",
+        "D. 无法确定"
+      ],
+      "answer": "A",
+      "solution": "【分析】在数轴上，越靠右数越大，越靠左数越小，比较两数大小，大数在右，小数在左，据此分析。本题考查了负数的大小比较和数轴的认识, 两负数比大小, 数值越大数越小。\n\n【详解】$-\\frac{1}{3} < -\\frac{1}{9}$ , 所以 $-\\frac{1}{3}$ 在 $-\\frac{1}{9}$ 的左边。故答案为：A",
+      "knowledge_points": [
+        "正负数在数轴上的表示",
+        "正负数的大小比较"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_303",
+      "original_id": "unit1_choice_303"
+    },
+    {
+      "id": "unit1_choice_304",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "一个温度计上的温度原来是 $6^{\\circ} \\mathrm{C}$ , 后来温度下降了 $2^{\\circ} \\mathrm{C}$ 、这时温度计上的温度是 ( )。",
+      "options": [
+        "A. $-2^{\\circ} \\mathrm{C}$",
+        "B. $8^{\\circ} \\mathrm{C}$",
+        "C. $4^{\\circ} \\mathrm{C}$",
+        "D. $6^{\\circ} \\mathrm{C}$"
+      ],
+      "answer": "C",
+      "solution": "【分析】用原来温度一下降温度 = 现在温度; 即: 由 $6^{\\circ} \\mathrm{C}$ 下降 $2^{\\circ} \\mathrm{C},$ 下降了 $(6 - 2 = 4) ^{\\circ} \\mathrm{C} $。\n\n【详解】$6 - 2 = 4$ (°C)故答案为: C",
+      "knowledge_points": [
+        "温度的应用"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_304",
+      "original_id": "unit1_choice_304"
+    },
+    {
+      "id": "unit1_choice_305",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "在 $-4, -9, -6, -0.1$ 这些数中，最大的数是（）。",
+      "options": [
+        "A. -4",
+        "B. -9",
+        "C. -6",
+        "D. -0.1"
+      ],
+      "answer": "D",
+      "solution": "【分析】根据两负数比大小，数值越大的数越小，进行选择即可。本题考查了负数的大小比较，正数 $>0>$ 负数。\n\n【详解】$-0.1 > -4 > -6 > -9$故答案为: D",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_305",
+      "original_id": "unit1_choice_305"
+    },
+    {
+      "id": "unit1_choice_306",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "数轴上, - 13 在 - 18 的 ( ) 边。",
+      "options": [
+        "A. 左",
+        "B. 右",
+        "C. 北",
+        "D. 无法确定"
+      ],
+      "answer": "B",
+      "solution": "【分析】数轴是规定了原点（0 点）、方向和单位长度的直线，在数轴上原点（0 点）的左边是负数，从原点（0 点）向左分别是 -1、-2、-3、-4、-5、-6、…右边是正数，从原点（0 点）向右分别+1、+2、+3、+4、+5、+6、…由此可见，在数轴上，-13 在 -18 的右边。数轴是规定了原点（0 点）、方向和单位长度的直线，在数轴上从左到右的方向就是数从小到大的顺序。\n\n【详解】数轴上，-13在-18的右边。故答案为: B。",
+      "knowledge_points": [
+        "正负数在数轴上的表示"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_306",
+      "original_id": "unit1_choice_306"
+    },
+    {
+      "id": "unit1_choice_307",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "两个冷库, 甲的温度是 -10 度, 乙的温度是 -12 度。（）的温度高一些。",
+      "options": [
+        "A. 甲",
+        "B. 乙",
+        "C. 一样高",
+        "D. 无法确定"
+      ],
+      "answer": "A",
+      "solution": "【分析】负数比较大小，先不考虑负号，数字部分大的数反而小。本题考查了负数的大小比较，所有的负数都比0小。\n\n【详解】$10 < 12$ , 所以 $-10 > -12$ 。故答案为: A",
+      "knowledge_points": [
+        "正负数的大小比较",
+        "温度的认识及比较"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_307",
+      "original_id": "unit1_choice_307"
+    },
+    {
+      "id": "unit1_choice_308",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "月球表面的最高温度是零上127摄氏度，记作（）；最低温度是零下183摄氏度，记作（）。①-127℃  ②+127℃  ③-183℃  ④+183℃",
+      "options": [
+        "A. ②④",
+        "B. ①④",
+        "C. ②③",
+        "D. ①③"
+      ],
+      "answer": "C",
+      "solution": "【分析】零上温度用正数表示，零下温度用负数表示，写正数时，先写正号'+'，再写出后面的数；写负数时，先写负号'-'，再写出后面的数。本题考查了正负数的写法, 正数可以省略正号不写, 负数不可以省略负号。\n\n【详解】月球表面的最高温度是零上 127 摄氏度，记作 +127°C；最低温度是零下 183 摄氏度，记作 -183°C。故答案为：C",
+      "knowledge_points": [
+        "正负数的读法和写法",
+        "温度的认识及比较"
+      ],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_308",
+      "original_id": "unit1_choice_308"
+    },
+    {
+      "id": "unit1_choice_309",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "以明明家为起点, 向东走为正，向西走为负。如果明明从家走了 $+30$ 米，又向西走了 20 米，这时明明离家的距离是（）米。",
+      "options": [
+        "A. 30",
+        "B. -30",
+        "C. 60",
+        "D. 10"
+      ],
+      "answer": "D",
+      "solution": "【分析】![](images/3501322b8f95ddeb8f83a536fcfdf4515e49dedac117b2c30ca64dce477924f5.jpg)，根据图示选择即可。本题考查了正负数的意义, 正负数表示相反意义的量。\n\n【详解】$30 - 20 = 10$ （米）故答案为: D",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_309",
+      "original_id": "unit1_choice_309"
+    },
+    {
+      "id": "unit1_choice_310",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "下面关于0的说法中，正确的是（）。",
+      "options": [
+        "A. 0 是正数",
+        "B. 0 是负数",
+        "C. 0 大于所有负数",
+        "D. 0 不是自然数"
+      ],
+      "answer": "C",
+      "solution": "【分析】比 0 大的数是正数, 比 0 小的数是负数, 据此选择。本题考查了 0 的意义, 0 是介于正数和负数中间的数, 其本身既不是正数也不是负数。\n\n【详解】A．0是正数，选项错误，0不是正数也不是负数；B. 0 是负数, 选项错误, 0 不是正数也不是负数; C. 0 大于所有负数，选项正确；D. 0 不是自然数，选项错误，0 也是自然数。故答案为: C",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_310",
+      "original_id": "unit1_choice_310"
+    },
+    {
+      "id": "unit1_choice_311",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "下列每组中的两个量不具有相反意义的是（）。",
+      "options": [
+        "A. 收入和支出",
+        "B. 浪费和节约",
+        "C. 增产和减产",
+        "D. 向北走和向东走"
+      ],
+      "answer": "D",
+      "solution": "【分析】根据相反的意义进行选择。本题考查了正负数的意义, 正负数可以表示相反意义的量。\n\n【详解】A．收入和支出，具有相反意义；B.浪费和节约，具有相反意义；C.增产和减产，具有相反意义；D.向北走和向东走，不具有相反意义。故答案为: D。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_311",
+      "original_id": "unit1_choice_311"
+    },
+    {
+      "id": "unit1_choice_312",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "小兰家本月的收入是 1200 元，支出 1250 元，本月的结余是（）。",
+      "options": [
+        "A. 50 元",
+        "B. -50 元"
+      ],
+      "answer": "B",
+      "solution": "【分析】正负数来表示具有意义相反的两种量：'收入'记为正，则'支出'就记为负；此题主要考查正负数的意义, 正数与负数表示意义相反的两种量, 看清规定哪一个为正, 则和它意义相反的就为负。\n\n【详解】小兰家本月的收入是 1200 元，支出 1250 元，1250 -1200 = 50（元），表示支出比收入多 50 元即本月的结余是 -50 元。故答案为: B",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_312",
+      "original_id": "unit1_choice_312"
+    },
+    {
+      "id": "unit1_choice_313",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "向东走 $10 \\mathrm{~m}$ 记作 + 10m, 那么向西走 $3 \\mathrm{~m}$ 记作 ( )。",
+      "options": [
+        "A. 3",
+        "B. -3",
+        "C. $3 \\mathrm{~m}$",
+        "D. -3m"
+      ],
+      "answer": "D",
+      "solution": "【分析】负数是在人为规定正方向的前提下出现的。负数常用来表示和正数意义相反的量。在选择用正数还是负数表示时，首先看是否规定了正方向。据此解答此题。此题主要考查正负数的意义，看清规定哪一个为正，则和它意义相反的就为负。\n\n【详解】如果规定向东走为正，那么向东走 $10 \\mathrm{~m}$ 记作 $+10 \\mathrm{~m}$ ，向西走就记作负数，向西走 $3 \\mathrm{~m}$ 记作 $-3 \\mathrm{~m}$ 。故答案为: D",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_313",
+      "original_id": "unit1_choice_313"
+    },
+    {
+      "id": "unit1_choice_314",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "通常，我们规定海平面的海拔高度为 $0 \\mathrm{~m}$ 。上杭一座山峰的海拔高度为 $980 \\mathrm{~m}, 980 \\mathrm{~m}$ 是指（）。",
+      "options": [
+        "A. 海平面以下的高度",
+        "B. 海平面的高度",
+        "C. 海平面以上的高度"
+      ],
+      "answer": "C",
+      "solution": "【分析】通常，我们规定海平面的海拔高度为 $0 \\mathrm{~m}$ 。高出海平面和低于海平面是两个具有相反意义的量，高出海平面的记作正数，'+'可以省略不写；低于海平面的记作负数，数字前有'-'；据此解答即可。本题是考查正、负数的意义及其作用。注意，正数前面的'+'可以省略，但负数前面的'-'不能省略。\n\n【详解】上杭一座山峰的海拔高度为 $980 \\mathrm{~m}, 980 \\mathrm{~m}$ 是指海平面以上的高度。故答案为: C",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_314",
+      "original_id": "unit1_choice_314"
+    },
+    {
+      "id": "unit1_choice_315",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "在数轴上，-6在-5的（）。",
+      "options": [
+        "A. 右边",
+        "B. 左边",
+        "C. 同一点上",
+        "D. 无法确定"
+      ],
+      "answer": "B",
+      "solution": "【分析】数轴是规定了原点（0 点）、方向和单位长度的直线，在数轴上原点（0 点）的左边是负数，从原点（0 点）向左分别是 -1、-2、-3、-4、-5、-6... 右边是正数，从原点（0 点）向右分别是 +1、+2、+3、+4、+5、+6... 由此可见，在数轴上，-6 在 -5 的左边。数轴是规定了原点（0 点）、方向和单位长度的直线，在数轴上从左到右的方向就是数从小到大的顺序。\n\n【详解】在数轴上，-6 在 -5 的左边。故答案为: B",
+      "knowledge_points": [
+        "正负数在数轴上的表示"
+      ],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_315",
+      "original_id": "unit1_choice_315"
+    },
+    {
+      "id": "unit1_choice_316",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "中等",
+      "question": "数轴上，-12 在 -18 的（）边。",
+      "options": [
+        "A. 左",
+        "B. 右",
+        "C. 无法确定"
+      ],
+      "answer": "B",
+      "solution": "【分析】数轴是规定了原点（0 点）、正方向和单位长度的直线，在数轴上原点（0 点）的左边是负数，从原点（0 点）向左分别是 -1、-2、-3、-4、-5、-6、… 右边是正数，从原点（0 点）向右分别是 +1、+2、+3、+4、+5、+6、… 由此可见，在数轴上，-12 在 -18 的右边。数轴是规定了原点（0 点）、正方向和单位长度的直线，在数轴上从左到右的方向就是数从小到大的顺序。\n\n【详解】数轴上，-12在-18的右边。故答案为: B。",
+      "knowledge_points": [
+        "正负数在数轴上的表示"
+      ],
+      "source": "18-19 五年级上·江苏·课后作业",
+      "merged_id": "unit1_316",
+      "original_id": "unit1_choice_316"
+    },
+    {
+      "id": "unit1_choice_317",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "-5、-45、7、+ 1.3、0、17、+23中正数有（）个。",
+      "options": [
+        "A. 2",
+        "B. 3",
+        "C. 4"
+      ],
+      "answer": "C",
+      "solution": "【分析】一个数大于零, 则称它是一个正数。正数有无数个, 包括正整数, 正分数。在数轴上表示正数的点都在数轴上零点的右边。0 本身不是正数。关于正负数的认识还是好理解的，带负号的是负数，正数前的正号可以省略不写。仔细观察每个数的特征即可。\n\n【详解】由正负数的意义可得：7、+1.3、17、+23都是正数，有4个。故答案为: C。",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_317",
+      "original_id": "unit1_choice_317"
+    },
+    {
+      "id": "unit1_fill_318",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "世界上最冷的地方在南极洲，绝对最低气温达零下 $94.5^{\\circ} \\mathrm{C}$ ，记作( )。世界上最热的地方在非洲撒哈拉沙漠，白天气温达零上 $55^{\\circ} \\mathrm{C}$ ，记作( )。",
+      "options": [],
+      "answer": "-94.5°C    +55°C",
+      "solution": "【分析】气温低于 $0^{\\circ} \\mathrm{C}$ , 用负数表示, 气温高于 $0^{\\circ} \\mathrm{C}$ , 用正数表示。本题考查了正负数的认识, 写正数时, 正号可以省略不写。\n\n【详解】世界上最冷的地方在南极洲，绝对最低气温达零下 $94.5^{\\circ} \\mathrm{C}$ ，记作 $-94.5^{\\circ} \\mathrm{C}$ 。世界上最热的地方在非洲撒哈拉沙漠，白天气温达零上 $55^{\\circ} \\mathrm{C}$ ，记作 $+55^{\\circ} \\mathrm{C}$ 。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_318",
+      "original_id": "unit1_fill_318"
+    },
+    {
+      "id": "unit1_fill_319",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "如果规定前进为正,那么前进 $50 \\mathrm{~m}$ 记作( ) $\\mathrm{m}$ , 后退 $20 \\mathrm{~m}$ 记作( ) $\\mathrm{m}$ 。",
+      "options": [],
+      "answer": "$+50$    $ -20$",
+      "solution": "【分析】前进记为正，那么后退记为负，据此填空。本题考查了正负数的意义, 正负数可以表示相反意义的量。\n\n【详解】如果规定前进为正，那么前进 $50 \\mathrm{~m}$ 记作 $+50 \\mathrm{~m}$ ，后退 $20 \\mathrm{~m}$ 记作 $-20 \\mathrm{~m}$ 。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_319",
+      "original_id": "unit1_fill_319"
+    },
+    {
+      "id": "unit1_fill_320",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "在有正数和负数的直线上, 所有的负数都在 0 的( )边, 所有的正数都在 0 的( )边。",
+      "options": [],
+      "answer": "左   右",
+      "solution": "【分析】在表示有正数与负数的直线上，0 的右边是正数，0 的左边是负数; 据此解答。本题考查了负数与数轴的认识, 关键是要理解在表示有正数与负数的直线上, 0 的右边是正数, 0 的左边是负数。\n\n【详解】由分析得: 在表示有正数和负数的直线上, 所有的负数都在 0 的左边, 所有的正数都在 0 的右边。",
+      "knowledge_points": [
+        "正负数在数轴上的表示"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_320",
+      "original_id": "unit1_fill_320"
+    },
+    {
+      "id": "unit1_fill_321",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "+3.6 读作( )；负七分之三写作( )。",
+      "options": [],
+      "answer": "正三点六    $-\\frac{3}{7}$",
+      "solution": "【分析】读正数时，先读'正'，后面按小数的读法继续读；写负数时，先写负号'-'，后面按分数的写法来写。本题考查了正负数的读法和写法，比0大的数是正数，比0小的数是负数。\n\n【详解】+3.6读作正三点六；负七分之三写作- $\\frac{3}{7}$ 。",
+      "knowledge_points": [
+        "正负数的读法和写法"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_321",
+      "original_id": "unit1_fill_321"
+    },
+    {
+      "id": "unit1_fill_322",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "马尔代夫是一个岛国，平均高度仅比海平面高1米，其海拔高度记作( )米；死海的海拔高度约为-392米，表示( )。",
+      "options": [],
+      "answer": "+1       死海的海拔高度比海平面低392米",
+      "solution": "【分析】通常把海平面的高度记作 0 , 高出海平面和低于海平面是两个具有相反意义的量, 高出海平面的记作 '+',低于海平面的记作 '- '，解答即可。本题是考查正、负数的意义及其作用。注意，正数前面的'+'可以省略，但负数前面的'-'不能省略。\n\n【详解】比海平面高 1 米, 其海拔高度记作 +1 米, 死海的海拔高度约为 -392 米, 表示死海的海拔高度比海平面低 392 米。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_322",
+      "original_id": "unit1_fill_322"
+    },
+    {
+      "id": "unit1_fill_323",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "一艘潜水艇所处的位置是 - 180 米，如果它先下沉 70 米，则所处的位置是( )米；然后再上升 90 米，则现在的位置是( )米。",
+      "options": [],
+      "answer": "-250    -160",
+      "solution": "【分析】首先审清题意，明确'正'和'负'所表示的意义：水平面以下记作负，水平面以上记作正，然后用潜艇的高度加上下沉的米数或上升的米数，下沉记为负数，上升记为正数，然后计算即可得解。解答此题的关键在于需要明白正数和负数的意义。\n\n【详解】$(-180) + (-70) = -250$，$-250 + 90 = -160$\n\n故本题答案为 - 250; - 160。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_323",
+      "original_id": "unit1_fill_323"
+    },
+    {
+      "id": "unit1_fill_324",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "存入银行 300 元记作( )元, 取出 500 元记作( )元。",
+      "options": [],
+      "answer": "$+300$ $-500$",
+      "solution": "【分析】在生活中，存入记为正数，取出记为负数，所以存入银行300元记作+300元，取出500元记作-500元。此题主要考查正负数的意义, 正数与负数表示意义相反的两种量, 看清规定哪一个为正, 则和它意义相反的就为负。\n\n【详解】存入银行 300 元记作 +300 元, 取出 500 元记作 -500 元。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_324",
+      "original_id": "unit1_fill_324"
+    },
+    {
+      "id": "unit1_fill_325",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "生活中的数, 比 0 大的数叫做( )数, 比 0 小的数叫做( )数。",
+      "options": [],
+      "answer": "正   负",
+      "solution": "【分析】比 0 大的数叫做正数, 比 0 小的数叫做负数。解答此题需要考生明白什么是正数，什么是负数。\n\n【详解】比 0 大的数叫做正数, 比 0 小的数叫做负数。故本题答案为: 正; 负。",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_325",
+      "original_id": "unit1_fill_325"
+    },
+    {
+      "id": "unit1_fill_326",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "$-\\frac{5}{8}$ 读作( ), +0.73 读作( )。",
+      "options": [],
+      "answer": "负八分之五   正零点七三",
+      "solution": "【分析】分数的读法是先读分母，再读'分之'，最后读分子；小数的读法是整数部分是'0'的就读作'零'，整数部分不是'0'的按照整数的读法来读，小数点读作'点'，小数部分是几就依次读出来。不管是分数还是小数，数前有'+'读作'正'，有'-'读作'负'，据此解答此题。解答此题的关键在于了解分数和小数以及正、负数的读法。属于基础题目。\n\n【详解】- $\\frac{5}{8}$ 读作：负八分之五；+0.73 读作：正零点七三。",
+      "knowledge_points": [
+        "正负数的读法和写法"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_326",
+      "original_id": "unit1_fill_326"
+    },
+    {
+      "id": "unit1_fill_327",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "比较大小。$0.5 (\\quad) - 7$  $- 3 (\\quad) - 1 0$  $- \\frac {1}{3} (\\quad) - \\frac {1}{4}$  $- \\frac {1}{3} (\\quad) 0$  $5 (\\quad) - 5$  $0 (\\quad) 0. 1$  $- 3 (\\quad) - \\frac {1}{3}$  $- \\frac {1}{2} (\\quad) 2$",
+      "options": [],
+      "answer": "$>\\quad >\\quad <\\quad <\\quad >\\quad <\\quad <\\quad <$",
+      "solution": "【分析】根据正数 $>0>$ 负数, 两负数比大小, 不管负号,数值越大的负数越小, 进行比较即可。本题考查了正负数的大小比较，在数轴上越靠右边的数越大，越靠左边的数越小。\n\n【详解】$0.5 > -7$  $-3 > -10$  $-\\frac{1}{3} < -\\frac{1}{4}$  $-\\frac{1}{3} < 0$  $5 > -5$  $0 < 0.1$  $-3 < -\\frac{1}{3}$  $-\\frac{1}{2} < 2$",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_327",
+      "original_id": "unit1_fill_327"
+    },
+    {
+      "id": "unit1_fill_328",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "二月份，妈妈在银行存入6000元，存折上应记作( )元。三月一日妈妈又取出2000元，存折上应记作( )元。",
+      "options": [],
+      "answer": "+6000 -2000",
+      "solution": "【分析】银行存入记为正，那么取出记为负，据此填空。本题考查了正负数的意义, 正负数可以表示相反意义的量。\n\n【详解】二月份，妈妈在银行存入6000元，存折上应记作 $+6000$ 元。三月一日妈妈又取出2000元，存折上应记作 $-2000$ 元。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_328",
+      "original_id": "unit1_fill_328"
+    },
+    {
+      "id": "unit1_fill_329",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "一个数从数轴上的某点出发, 先向左移动 5 个单位长度, 再向右移动两个单位长度, 这时这个点表示的数是 -1 , 那么起点表示的数是( )。",
+      "options": [],
+      "answer": "2",
+      "solution": "【分析】如图,![](images/dfe11d860a2acf8cacadf52577295433f46b27068b6131a2e0d475916246e505.jpg)，从-1向左移动两个单位长度是 -3，从 -3 向右移动 5 个单位长度是 2，据此填空。本题考查了数轴的认识, 负数在 0 的左侧, 整数在 0 的右侧, 写正数时, 正号可以省略不写。\n\n【详解】一个数从数轴上的某点出发，先向左移动 5 个单位长度，再向右移动两个单位长度，这时这个点表示的数是 -1，那么起点表示的数是 2。",
+      "knowledge_points": [
+        "正负数在数轴上的表示"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_329",
+      "original_id": "unit1_fill_329"
+    },
+    {
+      "id": "unit1_fill_330",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "在数轴上, 离原点 3 个单位长度的数分别是( )和( )。",
+      "options": [],
+      "answer": "3 -3",
+      "solution": "【分析】在数轴上，从原点往右 3 个单位长度是 3，往左 3 个单位长度是 -3。本题考查了数轴的认识, 负数在 0 的左侧, 正数在 0 的右侧。\n\n【详解】在数轴上，离原点3个单位长度的数分别是3和-3。",
+      "knowledge_points": [
+        "正负数在数轴上的表示"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_330",
+      "original_id": "unit1_fill_330"
+    },
+    {
+      "id": "unit1_fill_331",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "在-5、0、-1.5、+4、2.5中，最大的数是( )，最小的数是( )，正数和负数的分界线是( )。",
+      "options": [],
+      "answer": "+4    -5   0",
+      "solution": "【分析】根据正数 $>0>$ 负数，找到最大和最小数，其中 0 不是正数也不是负数，是正负数的分界线。本题考查了正负数的大小比较, 两负数比大小,数值越大的负数越小。\n\n【详解】在 $-5 、 0 、 -1.5 、 +4 、 2.5$ 中，最大的数是 $+4$ ,最小的数是 $-5$ , 正数和负数的分界线是 0。",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_331",
+      "original_id": "unit1_fill_331"
+    },
+    {
+      "id": "unit1_fill_332",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "+20.5 读作( ), $-\\frac{3}{4}$ 读作( )。",
+      "options": [],
+      "answer": "正二十点五     负四分之三",
+      "solution": "【分析】读正数时，先读'正'，后面按小数的读法继续读；读负数时，先读'负'，后面按分数的读法继续读。本题考查了正负数的读法，正数 $>$ 负数。\n\n【详解】+20.5 读作正二十点五，$-\\frac{3}{4}$ 读作负四分之三。",
+      "knowledge_points": [
+        "正负数的读法和写法"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_332",
+      "original_id": "unit1_fill_332"
+    },
+    {
+      "id": "unit1_fill_333",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "- $5^{\\circ} \\mathrm{C}$ 比 - $8^{\\circ} \\mathrm{C}$ 高 ( )°C, $-3^{\\circ} \\mathrm{C}$ 比 $4^{\\circ} \\mathrm{C}$ 低 ( )°C 。",
+      "options": [],
+      "answer": "3 7",
+      "solution": "【分析】-5与-8之间相差3个单位，-3与4之间相差7个单位，由此判断温度之间的差即可。负数的计算，小学生们还有些陌生，所以用'相差几个单位'来解释，更好理解。\n\n【详解】$8 - 5 = 3$ , 所以 $-5^{\\circ} \\mathrm{C}$ 比 $-8^{\\circ} \\mathrm{C}$ 高 $3^{\\circ} \\mathrm{C}$ ; $3 + 4 = 7$ , 所以 $-3^{\\circ} \\mathrm{C}$ 比 $4^{\\circ} \\mathrm{C}$ 低 $7^{\\circ} \\mathrm{C}$ 。故答案为: $3; 7$ 。",
+      "knowledge_points": [
+        "正负数的大小比较",
+        "温度的应用"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_333",
+      "original_id": "unit1_fill_333"
+    },{
+      "id": "unit1_fill_334",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "某公司5月份的营业额为200万元，6月份的营业额为180万元，比5月份减少了( )%，即比5月份负增长了( )%，可记作( )%。7月份比5月份的营业额增加了20万元，比5月份增长了( )%。8月份与5月份相比是零增长，增长率为( )%。",
+      "options": [],
+      "answer": "10  10    -10    10    0",
+      "solution": "【分析】把营业额增加看作正增长，用正数表示，则营业额下降为负增长，用负数表示，营业额零增长说明与七月份营业额相比没有变化；由此进行解答。本题考查了负数与正数的意义, 关键是要理解正负数表示具有两种相反意义的量。\n\n【详解】$(200 - 180) \\div 200 = 10\\%$，$20 \\div 200 = 10\\%$，零增长说明增长率为 $0\\%$。故答案为: $10; 10; -10; 10; 0$",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_334",
+      "original_id": "unit1_fill_334"
+    },
+    {
+      "id": "unit1_fill_335",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "某种面粉的包装袋上标有 $25 \\mathrm{~kg} \\pm 50 \\mathrm{~g}$ 的标记, 这种面粉最重是( ) , 最轻是()。",
+      "options": [],
+      "answer": "$25.05 \\mathrm{~kg} \\quad 24.95 \\mathrm{~kg}$",
+      "solution": "【分析】某种面粉的包装袋上标有 $25 \\mathrm{~kg} \\pm 50 \\mathrm{~g}$ 的标记, 所以这种面粉的标准质量是 $25 \\mathrm{~kg}$ , 最重是 $25 \\mathrm{~kg} + 50 \\mathrm{~g} = 25 \\mathrm{~kg}+ 0.05 \\mathrm{~kg} = 25.05 \\mathrm{~kg}$ , 最轻是: $25 \\mathrm{~kg} - 50 \\mathrm{~g} = 25 \\mathrm{~kg} - 0.05 \\mathrm{~kg} = 24.95 \\mathrm{~kg}$ ; 据此解答。本题考查了正负数的应用, 此题的关键是要理解当有 $a \\pm b$ 的时候, 要理解 $a$ 表示的是标准值, 最大值 $= a + b$ , 最小值 $= a - b$ 。\n\n【详解】$50 \\mathrm{~g} = 0.05 \\mathrm{~kg}$，面粉最重是: $25 \\mathrm{~kg} + 50 \\mathrm{~g} = 25 \\mathrm{~kg} + 0.05 \\mathrm{~kg} = 25.05 \\mathrm{~kg}$，最轻是: $25 \\mathrm{~kg} - 50 \\mathrm{~g} = 25 \\mathrm{~kg} - 0.05 \\mathrm{~kg} = 24.95 \\mathrm{~kg}$。故答案为: $25.05 \\mathrm{~kg}; 24.95 \\mathrm{~kg}$",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_335",
+      "original_id": "unit1_fill_335"
+    },
+    {
+      "id": "unit1_fill_336",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "某地白天最高温度是零上 $8^{\\circ} \\mathrm{C}$ , 记作 $+8^{\\circ} \\mathrm{C}$ , 夜间的最低温度比白天最高温度下降了 $10^{\\circ} \\mathrm{C}$ , 夜间最低温度记作( )°C。",
+      "options": [],
+      "answer": "-2",
+      "solution": "【分析】由题意可知: 最高温度是零上 $8^{\\circ} \\mathrm{C}$ , 最高温度与最低温度的温差是 $10^{\\circ} \\mathrm{C}, 10^{\\circ} \\mathrm{C} - 8^{\\circ} \\mathrm{C} = 2^{\\circ} \\mathrm{C}$ , 所以最低温度应该是零下 $2^{\\circ} \\mathrm{C}$ , 记作 $-2^{\\circ} \\mathrm{C}$ , 所以据此解答。本题考查了负数与正数的应用与意义, 关键是要理解正数与负数表示两个具有相反意义的量。\n\n【详解】由分析可得: $10^{\\circ} \\mathrm{C} - 8^{\\circ} \\mathrm{C} = 2^{\\circ} \\mathrm{C}$ , 所以最低温度应该是零下 $2^{\\circ} \\mathrm{C}$ , 记作 $-2^{\\circ} \\mathrm{C}$ 。故答案为: $-2^{\\circ} \\mathrm{C}$",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_336",
+      "original_id": "unit1_fill_336"
+    },
+    {
+      "id": "unit1_fill_337",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "这五个数中: 4.5, $-\\frac{3}{4}, 1, -6, -0.2$ 其中, 离 0 最近的数是( )。",
+      "options": [],
+      "answer": "-0.2",
+      "solution": "【分析】要求离 0 最近的数, 只要求出这五个数对应的点离 0 的距离大小即可; 而负数到 0 的距离为负数的数字部分;由此可得: 4.5 到 0 之间的距离为 4.5, $-\\frac{3}{4}$ 到 0 的距离为 $\\frac{3}{4}$ , 1 到 0 的距离为 1, -6 到 0 的距离为 6, -0.2 到 0 的距离为 0.2 ; 据此解答。本题考查了负数的认识, 关键是要理解负数到 0 之间的距离是负数的数字部分。\n\n【详解】由分析得：4.5 到 0 之间的距离为 4.5，$-\\frac{3}{4}$ 到 0 的距离为 $\\frac{3}{4}$ ，1 到 0 的距离为 1，-6 到 0 的距离为 6，-0.2 到 0 的距离为 0.2 而 $0 < 0.2 < \\frac{3}{4} < 1 < 4.5 < 6$ ，所以 -0.2 离 0 最近。故答案为: -0.2",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_337",
+      "original_id": "unit1_fill_337"
+    },
+    {
+      "id": "unit1_fill_338",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "一幢住宅楼共 26 层,如果把第 16 层记作 0 层, 那么第 10 层记作( )层, 那 20 层记作( )层。",
+      "options": [],
+      "answer": "-6    +4",
+      "solution": "【分析】正数和负数的意义: 正数与负数表示具有相反意义的量, 而 0 是正数与负数的分界点, 小于 0 的数是负数,大于 0 的数是正数。由题意可知: 把第 16 层记作 0 层, 所以第 10 层比 16 层小 6 所以应该记为 -6 , 20 层比 16 层大 4 所以应该记为: +4 ; 据此解答。本题考查了正负数的意义与应用, 此题的关键是要理解第 16 层记作 0 层, 而 0 是正数与负数的分界点, 所以比 16 小的层数记为负, 比 16 大的层数记为正。\n\n【详解】由分析可知: 第 16 层记作 0 层, 所以第 10 层比 16 层小 6 所以应该记为 -6, 20 层比 16 层大 4 所以应该记为:+4。故答案为：-6; +4",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_338",
+      "original_id": "unit1_fill_338"
+    },
+    {
+      "id": "unit1_fill_339",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "在 -2, 1, 8, 0, $-\\frac{1}{3}, 6, -3.5, -10\\%$ , 八个数中, 正数有( ), 负数有( )。",
+      "options": [],
+      "answer": "$1, 8, 6$ $-2, -\\frac{1}{3}, -3.5, -10\\%$",
+      "solution": "【分析】负数: 小于 0 的数叫做负数, 负数前面的符号 '-' 是负号, 所以有负号的数就是负数, 正数没有负号; 0 既不是正数, 也不是负数; 据此解答。本题考查了负数的认识, 关键是要理解负数的特点, 特别要牢记 0 既不是正数也不是负数。\n\n【详解】由分析可得：正数有：1，8，6；负数有：-2，$-\\frac{1}{3}, -3.5, -10\\%$ 。所以答案为: $1, 8, 6; -2, -\\frac{1}{3}, -3.5, -10\\%$ .",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_339",
+      "original_id": "unit1_fill_339"
+    },
+    {
+      "id": "unit1_fill_340",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "一种方便面包装袋上标着'净重 $150 \\pm 5 \\mathrm{~g}$ '，表示这种方便面的标准质量是( )g，实际每袋最少不少于( )g。",
+      "options": [],
+      "answer": "150    145",
+      "solution": "【分析】'净重 $150 \\pm 5 \\mathrm{~g}$ '表示这种方便面标准质量是 $150 \\mathrm{~g}$ ,实际每袋最多不多于 $150 + 5 = 155 \\mathrm{~g}$ , 最少不少于 $150 - 5 = 145 \\mathrm{~g}$ ; 据此解答。本题主要考查负数的意义及其应用, 理解 '净重 $150 \\pm 5 \\mathrm{~g}$ ' 是解题的关键。\n\n【详解】由分析可得：这种方便面的标准质量是(150) g, 实际每袋最少不少于 (145) g。故答案为: 150; 145",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_340",
+      "original_id": "unit1_fill_340"
+    },
+    {
+      "id": "unit1_fill_341",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "在数轴上，从0点出发，向右移动 5 个单位长度到 A 点，A 点表示的数是(); 从表示 0 的点出发向左移动 7 个单位长度到 B 点, B 点表示的数是( )。",
+      "options": [],
+      "answer": "+5   -7",
+      "solution": "【分析】在数轴上，从 0 出发向右移动几个单位长度得到的点所表示的数是几（或正几），由此找出 A 点；再从 0 点向左移动几格就是负几，据此即可找出 B 点。此题主要考查正负数的意义，明确在数轴上，0 的右面是正数，0 的左面是负数。\n\n【详解】在数轴上，从 0 点出发，向右移动 5 个单位长度A 点, A 点表示的数是 +5; 从表示 0 的点出发向左移动 7个单位长度到 B 点, B 点表示的数是 -7。",
+      "knowledge_points": [
+        "正负数在数轴上的表示"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_341",
+      "original_id": "unit1_fill_341"
+    },
+    {
+      "id": "unit1_fill_342",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "如果 -300 元表示支出 300 元, 那么 +2200 元表示( )。",
+      "options": [],
+      "answer": "收入 2200 元",
+      "solution": "【分析】如果支出记为负，那么收入记为正。本题考查了正负数的意义, 正负数可以表示相反意义的量。\n\n【详解】如果 $-300$ 元表示支出 300 元, 那么 $+2200$ 元表示收入 2200 元。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_342",
+      "original_id": "unit1_fill_342"
+    },
+    {
+      "id": "unit1_fill_343",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "负五分之三写作( ), - 25 读作( )。",
+      "options": [],
+      "answer": "$-\\frac{3}{5}$    负二十五",
+      "solution": "【分析】负数的读写按照正数的读写的方法进行读写，前面加上负号即可。此题考查负数的读写法。\n\n【详解】负五分之三写作：- $\\frac{3}{5}$ , -25 读作: 负二十五。故答案为: - $\\frac{3}{5}$ ; 负二十五",
+      "knowledge_points": [
+        "正负数的读法和写法"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_343",
+      "original_id": "unit1_fill_343"
+    },
+    {
+      "id": "unit1_fill_344",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "向东走 9 米记作 +9 米, 那么 -7 米表示( ), 0 米表示( )。",
+      "options": [],
+      "answer": "向西走7米   原地不动",
+      "solution": "【分析】向东走记为正，那么向西走记为负，据此填空。本题考查了正负数的意义, 正负数可以表示相反意义的量。\n\n【详解】向东走 9 米记作 +9 米, 那么 -7 米表示向西走 7 米, 0 米表示原地不动。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_344",
+      "original_id": "unit1_fill_344"
+    },
+    {
+      "id": "unit1_fill_345",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "在 0.5、-3、+90%、12、0，这几个数中，正数有( )；负数有( )。",
+      "options": [],
+      "answer": "$0.5$ 、 $+90\\%$ 、12    -3",
+      "solution": "【分析】比 0 小的数叫做负数, 比 0 大的数是负数。本题考查了正负数的认识, 0 不是正数也不是负数。\n\n【详解】在 $0.5 、 -3 、 +90 \\%$ 、12、0，这几个数中，正数有 $0.5 、 +90 \\%$ 、12；负数有 $-3$ 。",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_345",
+      "original_id": "unit1_fill_345"
+    },
+    {
+      "id": "unit1_fill_346",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "六一班某次数学测试平均分为 88 分, 如果比平均分高用正数表示, 比平均分低用负数表示, 那么小红考 83 分记作( )。",
+      "options": [],
+      "answer": "-5分",
+      "solution": "【分析】如果比平均分高，用成绩一平均分，结果是正数；如果比平均分低，用平均分一成绩，结果用负数表示。本题考查了正负数的应用，平均分作为标准，用正负数来表示相反的意义。\n\n【详解】$88 - 83 = 5$ （分），小红考 83 分记作 -5 分。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_346",
+      "original_id": "unit1_fill_346"
+    },
+    {
+      "id": "unit1_fill_347",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "在直线上，从表示0的点出发，向右移动3个单位长度到A点，A点表示的数是（），从表示0的点出发向左移动6个单位长度到B点，B点表示的数是（）。",
+      "options": [],
+      "answer": "$+3$   $ - 6$",
+      "solution": "【分析】根据向右移动加，向左移动减列式计算即可得解。此题主要考查了数轴的认识，解答此题的关键是要明确：从表示 0 的点出发，向右移动几个单位，就用 0 加上几；反之，就用 0 减去几即可。\n\n【详解】A点表示的数： $0 + 3 = 3$ （或 $+3$ ）B 点表示的数: $0 - 6 = -6$",
+      "knowledge_points": [
+        "正负数在数轴上的表示"
+      ],
+      "source": "14-15六年级下·全国·单元测试",
+      "merged_id": "unit1_347",
+      "original_id": "unit1_fill_347"
+    },
+    {
+      "id": "unit1_fill_348",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "某日傍晚, 黄山的气温从中午的零上 $2^{\\circ} \\mathrm{C}$ , 下降了 $7^{\\circ} \\mathrm{C}$ , 这天傍晚黄山的气温是( )。",
+      "options": [],
+      "answer": "-5°C",
+      "solution": "【分析】用下降的温度一中午温度，结果添上负号即可。本题考查了正负数的意义, 温度下降到零下用负数表示。\n\n【详解】$7 - 2 = 5$ ，这天傍晚黄山的气温是 $-5^{\\circ} \\mathrm{C}$ 。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_348",
+      "original_id": "unit1_fill_348"
+    },
+    {
+      "id": "unit1_fill_349",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "在 -16.2、23、0、-180、-10、50、1、9.07 中, ( ) 是正数; ( ) 是负数; ( ) 是整数; ( ) 既不是正数, 也不是负数。",
+      "options": [],
+      "answer": "23、50、1、9.07 -16.2、-180、-10 23、0、-180、-10、50、1 0",
+      "solution": "【分析】比 0 大的数是正数, 比 0 小的数是负数, 整数包括正整数、0 和负整数, 0 不是正数也不是负数, 据此填空。本题主要考查了正负数的认识，正数 $>0>$ 负数。\n\n【详解】在 $-16.2$ 、23、0、-180、-10、50、1、9.07中，23、50、1、9.07是正数；-16.2、-180、-10是负数；23、0、-180、-10、50、1是整数；0既不是正数，也不是负数。",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_349",
+      "original_id": "unit1_fill_349"
+    },
+    {
+      "id": "unit1_fill_350",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "在括号里填上'>''<'或'='。$0.8 (\\quad) 0. 9$   $ - 5 (\\quad) - 6$    $ 0 (\\quad) - 1$",
+      "options": [],
+      "answer": "< > >",
+      "solution": "【分析】小数比大小与整数一样，都从高位开始比；两负数比大小，数值越大的负数越小；负数都比0小。本题考查了小数及负数的大小比较，正数 $>0>$ 负数。\n\n【详解】$0.8 < 0.9$ $-5 > -6$ $0 > -1$",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_350",
+      "original_id": "unit1_fill_350"
+    },
+    {
+      "id": "unit1_fill_351",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "我国最大的湖泊是青海湖，位于青海省东部，在高出海平面3179米的山上，它的高度记作( )米。位于阿拉伯半岛的死海，水面低于海平面400米，它的深度记作( )米。",
+      "options": [],
+      "answer": "+3179   -400",
+      "solution": "【分析】通常把海平面的高度记作 0 , 高出海平面和低于海平面是两个具有相反意义的量, 高出海平面的记作 '+',低于海平面的记作 '- '，解答即可。本题是考查正、负数的意义及其作用。注意，正数前面的'+'可以省略，但负数前面的'-'不能省略。\n\n【详解】我国最大的湖泊是青海湖，位于青海省东部，在高出海平面 3179 米的山上，它的高度记作 +3179 米。位于阿拉伯半岛的死海，水面低于海平面 400 米，它的深度记作 -400 米；故答案为: +3179, -400。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_351",
+      "original_id": "unit1_fill_351"
+    },
+    {
+      "id": "unit1_fill_352",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "气温零下 $18^{\\circ} \\mathrm{C}$ 用( )表示，银行储蓄存折上显示 $-600$ ，表示( )了600元。收入 $-300$ 元的实际意义是( )。",
+      "options": [],
+      "answer": "-18°C     支出    支出300元",
+      "solution": "【分析】此题主要用正负数来表示具有意义相反的两种量:以 $0^{\\circ} \\mathrm{C}$ 为标准, $0^{\\circ} \\mathrm{C}$ 以上气温记为正, 则 $0^{\\circ} \\mathrm{C}$ 以下气温就记为负; 银行储蓄存折上 '存入' 记为正, 则 '支出' 就记为负；收入一300元的实际意义是支出300元。此题主要考查正负数的意义, 正数与负数表示意义相反的两种量, 看清规定哪一个为正, 则和它意义相反的就为负。\n\n【详解】气温零下 $18^{\\circ} \\mathrm{C}$ 用 $-18^{\\circ} \\mathrm{C}$ 表示;银行储蓄存折上显示'-600'，表示支出了600元。收入一300元的实际意义是支出300元。故答案为: $-18^{\\circ} \\mathrm{C}$ ; 支出; 支出 300 元。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_352",
+      "original_id": "unit1_fill_352"
+    },
+    {
+      "id": "unit1_fill_353",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "如果把水位升高1.2m 记作 + 1.2m, 那么水位下降 0.5m 记作( )m, 水位不升也不降记作( )m。",
+      "options": [],
+      "answer": "-0.5   0",
+      "solution": "【分析】正负数的意义: 正数与负数表示具有相反意义的量, 0 是正数与负数的分界点。则由题意可知: 水位升高1.2m 记作 + 1.2m，所以水位下降 $0.5 \\mathrm{~m}$ 记作 - 0.5m，水位不升不降记作 $0 \\mathrm{~m}$ ; 据此解答。本题考查了正负数的意义与应用, 关键是要理解正负数之间的意义, 以及 0 是正数与负数的分界点。\n\n【详解】由分析得: 水位下降 $0.5 \\mathrm{~m}$ 记作 $-0.5 \\mathrm{~m}$ , 水位不升不降记作 $0 \\mathrm{~m}$ 。故答案为: $-0.5; 0$",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_353",
+      "original_id": "unit1_fill_353"
+    },
+    {
+      "id": "unit1_fill_354",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "如果用正负数记录收支情况，爸爸今天的领到 3600 元记为( )元，充话费用去 100 元，记为( )，读作( )。",
+      "options": [],
+      "answer": "+3600    -100元   负一百元",
+      "solution": "【分析】此题主要用正负数来表示具有意义相反的两种量: 领取钱数记为正, 则花费钱数就记为负, 由此直接得出结论即可；根据负数的读法，读出即可。此题主要考查正负数的意义, 正数与负数表示意义相反的两种量, 看清规定哪一个为正, 则和它意义相反的就为负。\n\n【详解】如果用正负数记录收支情况，爸爸今天的领到3600元记为 +3600 元，充话费用去 100 元，记为 -100 元，读作：负一百元。",
+      "knowledge_points": [
+        "正负数的读法和写法",
+        "正负数的意义及应用"
+      ],
+      "source": "2020 六年级下·全国·专题练习",
+      "merged_id": "unit1_354",
+      "original_id": "unit1_fill_354"
+    },
+    {
+      "id": "unit1_fill_355",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "困难",
+      "question": "某日北京气温是 $-5^{\\circ} \\mathrm{C} \\sim 8^{\\circ} \\mathrm{C}$ ，这天的最大温差是( )°C。",
+      "options": [],
+      "answer": "13",
+      "solution": "【分析】用某日北京的最高气温减去最低气温，求出这天的最大温差是多少即可。此题主要考查了正、负数的运算，要熟练掌握运算方法。\n\n【详解】$8 - (-5) = 13 \\, (^{\\circ} \\mathrm{C})$答: 这天的最大温差是 $13^{\\circ} \\mathrm{C}$ 。答案为: 13",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "2020 六年级下·全国·专题练习",
+      "merged_id": "unit1_355",
+      "original_id": "unit1_fill_355"
+    },
+    {
+      "id": "unit1_fill_356",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "某班 5 名同学的体重分别是: 小金 $21 \\mathrm{~kg}$ , 小陆 $28 \\mathrm{~kg}$ , 小张 $25 \\mathrm{~kg}$ , 小吴 $22 \\mathrm{~kg}$ ,小沈 $24 \\mathrm{~kg}$ 。如果把他们的平均体重记为 0 , 那么这 5 名同学的体重分别记为：小金____，小陆____，小张____，小吴____，小沈____。",
+      "options": [],
+      "answer": "-3   +4    +1   -2   0",
+      "solution": "【分析】先将他们的平均体重计算出来，以平均体重为标准，超出的记为正，不足的记为负，相等的记为 0；据此解答。本题考查了负数的认识, 关键是要认真分析题意,找出题目中正数与负数的意义以及分界点。\n\n【详解】$(21 + 28 + 25 + 22 + 24) \\div 5 = 120 \\div 5 = 24 \\mathrm {(kg)}$\n\n平均体重记为 0 ;小金: $24 - 21 = 3 \\text{ (kg)}$ , 所以小金记作: -3;小陆: $28 - 24 = 4 \\text{ (kg)}$ , 所以小陆记作: +4;小张: $25 - 24 = 1(\\mathrm{~kg})$ , 所以小吴记作: +1;小吴: $24 - 22 = 2 \\text{ (kg)}$ , 所以小吴记作: -2;小沈: $24 - 24 = 0$ , 所以小吴记作: 0;故答案为: $-3; +4; +1; -2; 0$",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_356",
+      "original_id": "unit1_fill_356"
+    },
+    {
+      "id": "unit1_fill_357",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "在数轴上, 从 0 点出发,向右移动 5 个单位长度到 A 点, A 点表示的数是 5 ; 再向左移动 7 个单位长度到 B 点, B 点表示的数是_____。",
+      "options": [],
+      "answer": "-2",
+      "solution": "【分析】如图![](images/5f6d9a5226d22aea4856adee011b61b35f7bbfdf09fc18f1c4fff0c8c95c7896.jpg)从 A 点向左移动 7 个单位长度到达 -2。\n\n【详解】在数轴上，从 0 点出发，向右移动 5 个单位长度到 A 点，A 点表示的数是 5；再向左移动 7 个单位长度到 B 点，B 点表示的数是 -2。本题考查了负数的认识, 数轴上, 负数在 0 的左边。",
+      "knowledge_points": [
+        "正负数在数轴上的表示"
+      ],
+      "source": "19-20 六年级下·全国·期中",
+      "merged_id": "unit1_357",
+      "original_id": "unit1_fill_357"
+    },
+    {
+      "id": "unit1_fill_358",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "规定零摄氏度以上为正,月球表面的白天平均温度是 $126^{\\circ} \\mathrm{C}$ , 可记作 ( )°C ; 夜间平均温度是零下 $150^{\\circ} \\mathrm{C}$ , 可记作 ( )°C 。",
+      "options": [],
+      "answer": "126 - 150",
+      "solution": "【分析】以 0 摄氏度为标准, 高于 0 摄氏度记为正, 正号可以省略, 低于 0 摄氏度记为负, 据此填空。本题考查了正负数的意义，比 0 小的数是负数，0 不是正数也不是负数。\n\n【详解】规定零摄氏度以上为正，月球表面的白天平均温度是 $126^{\\circ} \\mathrm{C}$ ，可记作 $126^{\\circ} \\mathrm{C}$ ；夜间平均温度是零下 $150^{\\circ} \\mathrm{C}$ ，可记作 $-150^{\\circ} \\mathrm{C}$ 。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·期中",
+      "merged_id": "unit1_358",
+      "original_id": "unit1_fill_358"
+    },
+    {
+      "id": "unit1_fill_359",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "在直线上表示数时,负数都在 0 的( )边, 所有的正数都在 0 的( )边。",
+      "options": [],
+      "answer": "左  右",
+      "solution": "【分析】比 0 大的数记为 '+', 比 0 小的记为 '- ', 根据数轴的特征，所有的负数都在 0 的左边，所有的正数都在 0 的右边。此题考查了正、负数、0 的大小比较, 以及数轴的特征, 要熟练掌握。\n\n【详解】用直线上的点表示正数和负数的时候，正数在 0 的右边，负数在 0 的左边。",
+      "knowledge_points": [
+        "正负数在数轴上的表示"
+      ],
+      "source": "15-16 六年级下·全国·单元测试",
+      "merged_id": "unit1_359",
+      "original_id": "unit1_fill_359"
+    },
+    {
+      "id": "unit1_fill_360",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "在以向右为正的直线上，表示-2的点在表示0的点的( )侧，距离表示0的点()个单位长度。",
+      "options": [],
+      "answer": "左  两",
+      "solution": "【分析】负数在 0 的左侧，正数在 0 的右侧，-2与2距离 0 的单位长度是一样的，据此填空。本题考查了数轴的认识，在数轴上越靠右边的数越大，越靠左边的数越小。\n\n【详解】在以向右为正的直线上，表示 -2 的点在表示 0 的点的左侧，距离表示 0 的点两个单位长度。",
+      "knowledge_points": [
+        "正负数在数轴上的表示"
+      ],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_360",
+      "original_id": "unit1_fill_360"
+    },
+    {
+      "id": "unit1_fill_361",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "如果前进 $50 \\mathrm{~m}$ 记作 $+ 50 \\mathrm{~m}$ , 那么后退 $20 \\mathrm{~m}$ 记作( ) $\\mathrm{m}$ ; 如果上来 6 名乘客表示 $+ 6$ 人, 那么 -3 人表示( )。",
+      "options": [],
+      "answer": "-20   下去3名乘客",
+      "solution": "【分析】根据负数与正数表示意义相反的量，进行填空即可。本题考查了正负数的意义, 一个数前面加上负号,就变为它的相反数。\n\n【详解】如果前进 $50 \\mathrm{~m}$ 记作 $+50 \\mathrm{~m}$ , 那么后退 $20 \\mathrm{~m}$ 记作 $- 20 \\mathrm{~m}$ ; 如果上来 6 名乘客表示 +6 人, 那么 -3 人表示下去 3 名乘客。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_361",
+      "original_id": "unit1_fill_361"
+    },
+    {
+      "id": "unit1_fill_362",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "在一次食品质量检查中，如果超过标准质量 $4 \\mathrm{~g}$ ，记作 $+4 \\mathrm{~g}$ ，那么 $-3 \\mathrm{~g}$ 表示( )。",
+      "options": [],
+      "answer": "比标准质量轻 $3 \\mathrm{~g}$",
+      "solution": "【分析】根据负数与正数表示意义相反的量，进行填空即可。本题考查了正负数的意义, 一个数前面加上负号,就变为它的相反数。\n\n【详解】在一次食品质量检查中，如果超过标准质量 $4 \\mathrm{~g}$ ，记作 $+4 \\mathrm{~g}$ ，那么 $-3 \\mathrm{~g}$ 表示比标准质量轻 $3 \\mathrm{~g}$ 。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_362",
+      "original_id": "unit1_fill_362"
+    },
+    {
+      "id": "unit1_fill_363",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "将下面的数字按要求分类。1.25   $\\frac{5}{3}$   -7   +3   -2.8  -5 0  -30%   π  80。正数有( )，负数有( )。",
+      "options": [],
+      "answer": "$1.25、 \\frac{5}{3}、+3、 \\pi、 80$   ,$ -7、-2.8、 -5、 -30\\%$",
+      "solution": "【分析】负数前面都带负号 '- '，除 0 之外，所有不带负号的都是正数，据此填空。本题考查了正负数的认识, 0 既不是正数也不是负数。\n\n【详解】正数有 $1.25 、 \\frac{5}{3} 、 + 3 、 \\pi 、 80$ , 负数有 -7、-2.8、-5、-30%。",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_363",
+      "original_id": "unit1_fill_363"
+    },
+    {
+      "id": "unit1_fill_364",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "零上 $10^{\\circ} \\mathrm{C}$ 记作 + $10^{\\circ} \\mathrm{C}$ , 则零下 $5^{\\circ} \\mathrm{C}$ 记作( ), 读作( )。",
+      "options": [],
+      "answer": "-5°C    负五摄氏度",
+      "solution": "【分析】负数与正数表示意义相反的量，负数的读法，先读'负'，后面数字的读法与之前一样， ${}^{\\circ} \\mathrm{C}$ 读作摄氏度本题考查了负数的认识和读法, 一个数前面加上负号, 就变为它的相反数。\n\n【详解】零上 $10^{\\circ} \\mathrm{C}$ 记作 $+10^{\\circ} \\mathrm{C}$ , 则零下 $5^{\\circ} \\mathrm{C}$ 记作 $-5^{\\circ} \\mathrm{C}$ , 读作负五摄氏度。",
+      "knowledge_points": [
+        "正负数的读法和写法"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_364",
+      "original_id": "unit1_fill_364"
+    },
+    {
+      "id": "unit1_fill_365",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "李军向东走 50 米记作 +50 米, 那么向西走 20 米可记作( )米。",
+      "options": [],
+      "answer": "- 20",
+      "solution": "【分析】根据正负数的相反意义，向东走记为正，那么向西走记为负，据此填空。本题考查了正负数的意义, 一个数前面加上负号,表示的意义正好相反。\n\n【详解】李军向东走 50 米记作 +50 米, 那么向西走 20 米可记作 -20 米。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·山东菏泽·期中",
+      "merged_id": "unit1_365",
+      "original_id": "unit1_fill_365"
+    },
+    {
+      "id": "unit1_fill_366",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "如果上车的 10 位乘客用 +10 表示，那么 -8 表示____。",
+      "options": [],
+      "answer": "下车8位乘客",
+      "solution": "【分析】根据上车的 10 位乘客用 +10 表示，可得正数代表上车，那么负数代表下车。本题主要考查正负数表示具有相反意义的量, 关键是上车为正, 那么下车为负。\n\n【详解】上车的 10 位乘客用 +10 表示，那么 -8 表示下车。8 位乘客。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_366",
+      "original_id": "unit1_fill_366"
+    },{
+      "id": "unit1_fill_367",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "如果把小明向东走100m 记作 + 100m, 那么他又走 - 80m, 表示小明____, 这时他距离出发点有____m 远。",
+      "options": [],
+      "answer": "向西走 $80 \\mathrm{~m}$ $20$",
+      "solution": "【分析】根据正数和负数表示具有相反意义的量，向东为正，那么向西就为负。本题主要考查具有相反意义的两个量, 注意东的相反意义的量是西, -80m 代表的意思是在出发点西边 80 米。\n\n【详解】小明向东走 $100 \\mathrm{~m}$ 记作 $+100 \\mathrm{~m}$ , 那么他又走 $-80 \\mathrm{~m}$ , 表示小明向西走 $80 \\mathrm{~m}$ , 这时他距离出发点有 $20 \\mathrm{~m}$ 远。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_367",
+      "original_id": "unit1_fill_367"
+    },
+    {
+      "id": "unit1_fill_368",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "+ 8.7 读作____,  $-\\frac{3}{5}$ 读作____。",
+      "options": [],
+      "answer": "正八点七   负五分之三",
+      "solution": "【分析】正数的读法是：先读'正'，再读数；负数的读法是：先读'负'，再读数。本题考查了正负数的读法，除了读出'正'或'负'，数的读法与之前相同。\n\n【详解】+8.7读作正八点七， $-\\frac{3}{5}$ 读作负五分之三；故答案为：正八点七；负五分之三。",
+      "knowledge_points": [
+        "正负数的读法和写法"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_368",
+      "original_id": "unit1_fill_368"
+    },
+    {
+      "id": "unit1_fill_369",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "如果比标准质量多 3 千克记作 $+3 \\mathrm{~kg}$ , 那么 $-5 \\mathrm{~kg}$ 表示比标准质量_____。",
+      "options": [],
+      "answer": "少 $5 \\mathrm{kg}$",
+      "solution": "【分析】正数和负数可以表示具有相反意义的量, 正数表示多的量, 那么负数就表示少的量。本题考查了正负数的意义及应用。\n\n【详解】-5kg表示比标准质量少 $5\\mathrm{kg}$ 。故答案为: 少 $5 \\mathrm{kg}$ 。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_369",
+      "original_id": "unit1_fill_369"
+    },
+    {
+      "id": "unit1_fill_370",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "在2、4、-2、-1中，最接近0的数是____，最大的数是____。",
+      "options": [],
+      "answer": "-1 4",
+      "solution": "【分析】先计算出这些数与 0 相差越少, 那么这个数越接近 0 ; 负数比正数小。本题考查了正负数的认识和大小比较。\n\n【详解】最接近 0 的数是 -1, $-2 < -1 < 2 < 4$ , 所以最大的数是 4。故答案为: -1; 4。",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_370",
+      "original_id": "unit1_fill_370"
+    },
+    {
+      "id": "unit1_fill_371",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "在 - 5，2.9，0，125，-99， $\\frac{+2}{3}$ ，-8.3这些数中，正数有____，负数有____。",
+      "options": [],
+      "answer": "2.9、125、+ $\\frac{2}{3}$ -5、-99、-8.3",
+      "solution": "【分析】正数是大于0的数，正数前面的'+'可以省略；负数是小于 0 的数，负数前面有'-'。本题考查了正负数的辨认，负数前面一定有负号。\n\n【详解】这些数中，正数有：2.9、125、+ $\\frac{2}{3}$ ，负数有 -5、-99、-8.3。故答案为: $2.9 、 125 、 + \\frac{2}{3}; - 5 、 - 99 、 - 8.3 。$",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_371",
+      "original_id": "unit1_fill_371"
+    },
+    {
+      "id": "unit1_fill_372",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "一幢大楼有18层，地面以下有2层，如果地面以上第3层记作 $+3$ 层，那么地面以上第6层记作层，地面以下第2层记作层。",
+      "options": [],
+      "answer": "$+6$   $ - 2$",
+      "solution": "【分析】生活中，通常用正负数表示具有相反意义的两种量，如果规定地面以上为正，那么地面以下为负，据此解答。本题考查了负数意义及应用。\n\n【详解】一幢大楼有 18 层, 地面以下有 2 层, 如果地面以上第 3 层记作 +3 层, 那么地面以上第 6 层记作 +6 层, 地面以下第 2 层记作 -2 层。故答案为: $+6; -2$ 。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_372",
+      "original_id": "unit1_fill_372"
+    },
+    {
+      "id": "unit1_fill_373",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "小丽的妈妈存入银行 3000 元，记作 +3000 元，取出 2400 元，记作____元。",
+      "options": [],
+      "answer": "- 2400",
+      "solution": "【分析】生活中，通常用正负数表示具有相反意义的两种量，如果规定存入为正，那么取出为负，据此解答。本题考查了负数意义及应用。\n\n【详解】小丽的妈妈存入银行 3000 元，记作 +3000 元，取出 2400 元，记作 -2400 元。故答案为: $-2400$ 。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_373",
+      "original_id": "unit1_fill_373"
+    },
+    {
+      "id": "unit1_fill_374",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "$-\\frac{2}{5}$ 读作( )， -1.02 读作( )。",
+      "options": [],
+      "answer": "负五分之二    负一点零二",
+      "solution": "【分析】'+'读作'正'，'-'读作'负'，后面的数字按照分数或小数的读法读出来即可。此题考查负数的读法。注意正数前面的'正'可以不读，负数前面的'负'必须读出来。\n\n【详解】 $-\\frac{2}{5}$ 读作负五分之二，-1.02 读作负一点零二。",
+      "knowledge_points": [
+        "正负数的读法和写法"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_374",
+      "original_id": "unit1_fill_374"
+    },
+    {
+      "id": "unit1_fill_375",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "如果规定向东走为正，那么向东走 $5 \\mathrm{~m}$ 记作( ) $\\mathrm{m}$ ，向西走 $8 \\mathrm{~m}$ 记作( ) $\\mathrm{m}$ 。",
+      "options": [],
+      "answer": "$+5$ $- 8$",
+      "solution": "【分析】负数是在人为规定正方向的前提下出现的。负数常用来表示和正数意义相反的量。在选择用正数还是负数表示时，首先看是否规定了正方向。据此解答此题。此题主要考查正负数的意义，看清规定哪一个为正，则和它意义相反的就为负。\n\n【详解】如果规定向东走为正，那么向东走 $5 \\mathrm{~m}$ 记作 $+5 \\mathrm{~m}$ ，向西走就记作负数，向西走 $8 \\mathrm{~m}$ 记作 $-8 \\mathrm{~m}$ 。",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_375",
+      "original_id": "unit1_fill_375"
+    },
+    {
+      "id": "unit1_sort_376",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "排序题",
+      "difficulty": "中等",
+      "question": "按从小到大排列。$9.5 、 - 2. 7 、 0 、 - 4 0 、 + 1 5 、 - \\frac {7}{8}$",
+      "options": [],
+      "answer": "$-40 < -2.7 < -\\frac{7}{8} < 0 < 9.5 < +15$",
+      "solution": "【分析】根据整数 $>0>$ 负数，进行排序即可。本题考查了正负数的大小比较, 负数比较大小,不考虑负号, 数字部分大的数反而小。\n\n【详解】排序如下：$-40 < -2.7 < -\\frac{7}{8} < 0 < 9.5 < +15$",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_376",
+      "original_id": "unit1_sort_376"
+    },
+    {
+      "id": "unit1_sort_377",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "排序题",
+      "difficulty": "中等",
+      "question": "把 -2、3、- 4.5、0、21这几个数从小到大依次排列为( )<()<()><()<()。",
+      "options": [],
+      "answer": "-4.5    -2 0    3   21",
+      "solution": "【分析】根据正数 $>0>$ 负数，两负数比大小，数值越大的负数越小，进行排序即可。本题考查了正负数的大小比较，在数轴上越靠右边的数越大，越靠左边的数越小。\n\n【详解】把 $-2 、 3 、 -4.5 、 0 、 21$ 这几个数从小到大依次排列为 $-4.5 < -2 < 0 < 3 < 21$ 。",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_377",
+      "original_id": "unit1_sort_377"
+    },
+    {
+      "id": "unit1_sort_378",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "排序题",
+      "difficulty": "中等",
+      "question": "将 $+3$ 、-4、-2.5、0、+5.5按从大到小的顺序排列____。",
+      "options": [],
+      "answer": "$+5.5 > +3 > 0 > -2.5 > -4$",
+      "solution": "【分析】比较负数的大小时，负号后面的数越大，原来的数就越小；负数比0小；正数比0大。本题考查了正负数的大小比较。\n\n【详解】由分析可得，这些数按从大到小的顺序排列是 $+5.5 > +3 > 0 > -2.5 > -4$ 。",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_378",
+      "original_id": "unit1_sort_378"
+    },
+    {
+      "id": "unit1_judge_379",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "某地昨天 $3^{\\circ} \\mathrm{C}$ , 今天 $-1^{\\circ} \\mathrm{C}$ , 表示今天比昨天低 $1^{\\circ} \\mathrm{C}$ 。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】$3^{\\circ} \\mathrm{C}$ 表示零上 $3^{\\circ} \\mathrm{C}$ , 比 $0^{\\circ} \\mathrm{C}$ 高了 $3^{\\circ} \\mathrm{C}, -1^{\\circ} \\mathrm{C}$ 表示零下 $1^{\\circ} \\mathrm{C}$ , 比 0 摄氏度低了 $1^{\\circ} \\mathrm{C}$ , 两天的温差是 $3 + 1^{\\circ} \\mathrm{C}$ , 据此分析。本题考查了正负数的应用, 负数可以广泛应用于楼层、支出/收入、温度、海拔、水位、盈利、增产/减产、得分/扣分等方面。\n\n【详解】$3 + 1 = 4 (^{\\circ} \\mathrm{C})$ , 某地昨天 $3^{\\circ} \\mathrm{C}$ , 今天 $-1^{\\circ} \\mathrm{C}$ , 表示今天比昨天低 $4^{\\circ} \\mathrm{C}$ , 所以原题说法错误。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_379",
+      "original_id": "unit1_judge_379"
+    },
+    {
+      "id": "unit1_judge_380",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "$-10^{\\circ} \\mathrm{C}$ 比 $-4^{\\circ} \\mathrm{C}$ 高 $6^{\\circ} \\mathrm{C}$ 。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】负数比较大小，先不考虑负号，数字部分大的数反而小，。本题主要考查了正负数的大小比较, 正数 $>0>$ 负数。\n\n【详解】$10 - 4 = 6 (^{\\circ} \\mathrm{C}), -10^{\\circ} \\mathrm{C}$ 比 $-4^{\\circ} \\mathrm{C}$ 低 $6^{\\circ} \\mathrm{C}$ , 所以原题说法错误。",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "17-18 四年级上·辽宁·期末",
+      "merged_id": "unit1_380",
+      "original_id": "unit1_judge_380"
+    },
+    {
+      "id": "unit1_judge_381",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "$-16^{\\circ} \\mathrm{C}$ 表示温度低于 $0^{\\circ} \\mathrm{C}$ 。（）",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】根据负数的意义，比0小的数是负数，进行判断。本题考查了负数的意义，负数可以广泛应用于楼层、支出/收入、温度、海拔、水位、盈利、增产/减产、得分/扣分等方面。\n\n【详解】-16°C表示温度低于0°C，说法正确。\n\n故答案为: √",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_381",
+      "original_id": "unit1_judge_381"
+    },
+    {
+      "id": "unit1_judge_382",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "如果 $+15$ 米表示物体向北运动 15 米，那么 $-15$ 米表示物体向西运动 15 米。（）",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】如果向北运动记为正，那么向南运动记为负，据此判断。\n\n【详解】如果 $+15$ 米表示物体向北运动 15 米, 那么 $-15$ 米表示物体向南运动 15 米, 所以原题说法错误。本题考查了正负数的意义, 正负数可以表示相反意义的量。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_382",
+      "original_id": "unit1_judge_382"
+    },
+    {
+      "id": "unit1_judge_383",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "+7 读作正七, -12 读作负十二。( )",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】读正整数时，先读'正'，再按整数的读法继续读；读负数时，先读'负'，再按数的读法继续读。本题考查了正负数的读法，负数可以广泛应用于楼层、支出/收入、温度、海拔、水位、盈利、增产/减产、得分/扣分等方面。\n\n【详解】+7 读作正七，-12 读作负十二，读法正确。故答案为：√",
+      "knowledge_points": [
+        "正负数的读法和写法"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_383",
+      "original_id": "unit1_judge_383"
+    },
+    {
+      "id": "unit1_judge_384",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "在数轴上从左到右的顺序就是数从小到大的顺序。（）",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】数轴是规定了原点（0 点）、方向和单位长度的直线，在数轴上，所有负数都在原点（0 点）的左边，越往左表示的数越小，所有正数在右边，越往右表示的数越大，因此，数轴上数值从左到右的顺序就是从小到大的顺序。此题考查了数轴的认识。解答此题要明确：数轴上的点表示的数，右边的数总比左边的数大。\n\n【详解】在数轴上从左到右的顺序就是数从小到大的顺序。此说法正确。故答案为: √",
+      "knowledge_points": [
+        "正负数在数轴上的表示"
+      ],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_384",
+      "original_id": "unit1_judge_384"
+    },
+    {
+      "id": "unit1_judge_385",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "$+20^{\\circ} \\mathrm{C}$ 和 $20^{\\circ} \\mathrm{C}$ 表示的意义相同。（）",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】以 $0^{\\circ} \\mathrm{C}$ 为标准, $0^{\\circ} \\mathrm{C}$ 以上气温记为正, 则 $0^{\\circ} \\mathrm{C}$ 以下气温就记为负; 正数前面常有一个符号 '+', 通常可以省略不写。正数前面常有一个符号 '+', 通常可以省略不写。\n\n【详解】+20°C和20°C表示的意义相同都表示零上20°C。故答案为: √",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_385",
+      "original_id": "unit1_judge_385"
+    },
+    {
+      "id": "unit1_judge_386",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "苹苹的数学考试成绩是 95 分, 若记作 +3 分, 则 90 分应记作 -5 分。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】由题意知: 把 95 分记作 +3 分, 所以标准的分数是 $95 - 3 = 92$ , 比标准分数小的记为负分, 高的记为正分。 $92 - 90 = 2$ , 所以 90 分应记作 -2 分; 据此解答。本题考查了正负数的意义, 此题关键是要根据把95 分记作 +3 分, 找出标准的分数, 比标准分数小的记为负分, 高的记为正分。\n\n【详解】由分析得：标准分数为： $95 - 3 = 92$ （分） $92 - 90 = 2$ （分）所以 90 分应该记作 -2 分; 原题说法错误。故答案为: ×",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_386",
+      "original_id": "unit1_judge_386"
+    },
+    {
+      "id": "unit1_judge_387",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "因为 $5 > 2$ , 所以 -5 $> -2$ 。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】比较两个负数的大小时，数字部分大的负数反而小，据此解答。本题考查了负数比较大小的方法, 关键是要理解比较两个负数的大小时, 数字部分大的负数反而小。\n\n【详解】由分析得: 因为 $5 > 2$ , 所以 $-5 < -2$ ; 原题说法错误。故答案为: ×",
+      "knowledge_points": [
+        "正负数的大小比较"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_387",
+      "original_id": "unit1_judge_387"
+    },
+    {
+      "id": "unit1_judge_388",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "- 10 与 - 9 之间有无数个负数。（）",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】负数里也包含了负的分数、小数，所以 -10 与 -9 之间有无数个负数；据此解答。本题考查了负数的认识, 关键是要理解两个负整数之间有无数个负数。\n\n【详解】由分析得：- 10 与 -9 之间有无数个负数；所以原题说法正确。故答案为: √",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_388",
+      "original_id": "unit1_judge_388"
+    },
+    {
+      "id": "unit1_judge_389",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "中等",
+      "question": "在直线上表示正、负数时, 0 右边的数一定比 0 左边的数大。( )",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】在直线上表示正、负数时 0 右边的数是正数，左边的数是负数，正数都比负数大；据此解答。本题考查了正、负数的大小比较，关键是要理解0比负数大，正数都比负数大。\n\n【详解】在直线上表示正、负数时，0 右边的数是正数，左边的数都是负数，正数都比负数大；原题说法正确。\n\n故答案为: √",
+      "knowledge_points": [
+        "正负数在数轴上的表示",
+        "正负数的大小比较"
+      ],
+      "source": "19-20 六年级下·全国·期中",
+      "merged_id": "unit1_389",
+      "original_id": "unit1_judge_389"
+    },
+    {
+      "id": "unit1_judge_390",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "零上 $2^{\\circ} \\mathrm{C}(+ 2^{\\circ} \\mathrm{C})$ 和零下 $2^{\\circ} \\mathrm{C}(- 2^{\\circ} \\mathrm{C})$ 是两种相反意义的量。( )",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】通常用正数和负数表示具有相反意义的量。本题主要考查正负数表示具有相反意义的量, 通常增加和减少、上车和下车、零上和零下、向东和向西等,都是具有相反意义的。\n\n【详解】零上 $2^{\\circ} \\mathrm{C}$ （+2°C）和零下 $2^{\\circ} \\mathrm{C}$ （-2°C）是两种具有相反意义的量。故答案为: √",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·期中",
+      "merged_id": "unit1_390",
+      "original_id": "unit1_judge_390"
+    },
+    {
+      "id": "unit1_judge_391",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "正数与负数表示的量具有相反的意义。( )",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】正数和负数是相对的，可以说正数和负数是具有相反意义的量。本题主要考查了正负数的相对性, 也就是正数和负数是具有相反意义的, 比较简单。\n\n【详解】正数和负数可以用来表示相反意义的量，是正确的。故答案为: √",
+      "knowledge_points": [
+        "正负数的概念及辨认"
+      ],
+      "source": "19-20 六年级下·全国·期中",
+      "merged_id": "unit1_391",
+      "original_id": "unit1_judge_391"
+    },
+    {
+      "id": "unit1_solve_392",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "飞机开始在距地面 9000m的高空中飞行，之后它又进行了 3 次升降运动，分别是：第一次上升了 $30 \\mathrm{~m}$ ，第二次下降了 $50 \\mathrm{~m}$ ，第三次又上升了 -300 米时, 它在距地面多高的地方?",
+      "options": [],
+      "answer": "8680 米",
+      "solution": "【分析】在开始的距地面高度，上升就加，下降就减，据此列式解答。本题考查了正负数的意义，上升 -300 米就是下降 300 米。\n\n【详解】$9000 + 30 - 50 - 300 = 8680$ （米）\n\n答: 它在距地面 8680 米高的地方。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_392",
+      "original_id": "unit1_solve_392"
+    },
+    {
+      "id": "unit1_solve_393",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "测量5名学生的身高，测得的数据分别是：153cm, 172cm, 165cm, 168cm, 157cm。(1) 求这五名同学的平均身高。(2) 以求出的平均值为基准数 $0 \\mathrm{~cm}$ , 用正、负数表示出各次测量的数值。",
+      "options": [],
+      "answer": "(1) $163 \\mathrm{~cm}$ (2) $-10\\mathrm{cm}, +9\\mathrm{cm}, +2\\mathrm{cm}, +5\\mathrm{cm}, -6\\mathrm{cm}$",
+      "solution": "【分析】（1）根据平均数 $=$ 总数 $\\div$ 份数，列式解答即可； (2) 以平均数为标准，高于平均数记为正，低于平均数记为负。本题考查了正负数的意义和应用, 正负数可以表示相反意义的量。\n\n【详解】(1) $(153 + 172 + 165 + 168 + 157) \\div 5 = 815 \\div 5 = 163$ (厘米)\n\n答: 这五名同学的平均身高是 163 厘米。\n\n(2) $163 - 153 = 10$ （厘米）， $172 - 163 = 9$ （厘米）， $165 - 163 = 2$ （厘米）， $168 - 163 = 5$ （厘米） $163 - 157 = 6$ （厘米）\n\n答: 他们的身高以平均值为基准数 0 厘米, 分别是: $-10 \\mathrm{~cm} 、 +9 \\mathrm{~cm} 、 +2 \\mathrm{~cm} 、 +5 \\mathrm{~cm} 、 -6 \\mathrm{~cm}$ 。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_393",
+      "original_id": "unit1_solve_393"
+    },
+    {
+      "id": "unit1_solve_394",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "根据以下几个城市某天的天气预报回答问题。北京: $-6^{\\circ} \\mathrm{C} \\sim 5^{\\circ} \\mathrm{C}$ ; 上海: $13^{\\circ} \\mathrm{C} \\sim 18^{\\circ} \\mathrm{C}$ ; 天津: $-5^{\\circ} \\mathrm{C} \\sim 1^{\\circ} \\mathrm{C}$ ; 吉林: $-18^{\\circ} \\mathrm{C} \\sim 9^{\\circ} \\mathrm{C}$ ; 太原: $-9^{\\circ} \\mathrm{C} \\sim 2^{\\circ} \\mathrm{C}$ ; 石家庄: $-7^{\\circ} \\mathrm{C} \\sim 4^{\\circ} \\mathrm{C}$ 。(1) 这一天天津的最低气温是 ( )，最高气温是 ( )。(2) 最低气温最低的城市是 ( )，最高气温最高的城市是 ( )。(3) 将这几个城市的最高气温和最低气温分别按从低到高的顺序排列起来。",
+      "options": [],
+      "answer": "(1) $-5^{\\circ} \\mathrm{C}; 1^{\\circ} \\mathrm{C}$ ; (2) 吉林；上海； (3) 最低气温按从低到高: $-18^{\\circ} \\mathrm{C}< -9^{\\circ} \\mathrm{C}< -7^{\\circ} \\mathrm{C}< -6^{\\circ} \\mathrm{C}< -5^{\\circ} \\mathrm{C}< 13^{\\circ} \\mathrm{C}$ ; 最高气温按从低到高: $1^{\\circ} \\mathrm{C}< 2^{\\circ} \\mathrm{C}< 4^{\\circ} \\mathrm{C}< 5^{\\circ} \\mathrm{C}< 9^{\\circ} \\mathrm{C}< 18^{\\circ} \\mathrm{C}$ 。",
+      "solution": "【分析】根据各个城市的气温情况来判断最高、最低温度；比较气温间的大小时，可根据实际意义来比较， $-18^{\\circ} \\mathrm{C}$ 表示零下 $18^{\\circ} \\mathrm{C}, -9^{\\circ} \\mathrm{C}$ 表示零下 $9^{\\circ} \\mathrm{C}$ ，所以 $-18^{\\circ} \\mathrm{C}< -9^{\\circ} \\mathrm{C}$ ，据此判断即可。本题考查正、负数, 解答本题的关键是掌握负数的大小比较可根据实际意义进行比较。\n\n【详解】（1）天津的气温在： $-5^{\\circ} \\mathrm{C} \\sim 1^{\\circ} \\mathrm{C}$ ，所以这一天天津的最低气温是 $-5^{\\circ} \\mathrm{C}$ ，最高气温是 $1^{\\circ} \\mathrm{C}$ ；(2) 通过比较 6 个城市的最低气温, 最低的是吉林; 通过比较 6 个城市的最高气温, 最高的是上海; (3) 最低气温按从低到高: $-18^{\\circ} \\mathrm{C}< -9^{\\circ} \\mathrm{C}< -7^{\\circ} \\mathrm{C}< -6^{\\circ} \\mathrm{C}< -5^{\\circ} \\mathrm{C}< 13^{\\circ} \\mathrm{C}$ ;最高气温按从低到高: $1^{\\circ} \\mathrm{C}< 2^{\\circ} \\mathrm{C}< 4^{\\circ} \\mathrm{C}< 5^{\\circ} \\mathrm{C}< 9^{\\circ} \\mathrm{C}< 18^{\\circ} \\mathrm{C}$ 。故答案为: (1) $-5^{\\circ} \\mathrm{C}; 1^{\\circ} \\mathrm{C}$ ;(2) 吉林; 上海;\n\n(3) 最低气温按从低到高: $-18^{\\circ} \\mathrm{C}< -9^{\\circ} \\mathrm{C}< -7^{\\circ} \\mathrm{C}< -6^{\\circ} \\mathrm{C}< -5^{\\circ} \\mathrm{C}< 13^{\\circ} \\mathrm{C}$ ;\n\n最高气温按从低到高: $1^{\\circ} \\mathrm{C}<2^{\\circ} \\mathrm{C}<4^{\\circ} \\mathrm{C}<5^{\\circ} \\mathrm{C}<9^{\\circ} \\mathrm{C}<18^{\\circ} \\mathrm{C}$ 。",
+      "knowledge_points": [
+        "正负数的意义及应用",
+        "正负数的大小比较"
+      ],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_394",
+      "original_id": "unit1_solve_394"
+    },
+    {
+      "id": "unit1_solve_395",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "某种股票发行价是 13 元,第一个交易日上涨5元，记作 $+5$ 元，第二个交易日的记录为-3.5元，表示什么？经过这两个交易日，该股票每股的价格是多少元？",
+      "options": [],
+      "answer": "股票下降3.5元；14.5元",
+      "solution": "【分析】上涨记为正，那么下降记为负，用发行价 + 上涨的一下降的 = 现在股票每股价格。本题考查了正负数的应用, 正负数可以表示相反意义的量。\n\n【详解】$13 + 5 - 3.5 = 14.5$ （元）答：第二个交易日的记录为 -3.5 元，表示股票下降 3.5 元，经过这两个交易日，该股票每股的价格是 14.5 元。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·期中",
+      "merged_id": "unit1_395",
+      "original_id": "unit1_solve_395"
+    },
+    {
+      "id": "unit1_solve_396",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "(1) 一种课桌标明的高度要求是 $60 \\pm 3$ 厘米。它是什么意思?(2) 如果课桌的高度比标准高度高 $2 \\mathrm{~cm}$ 记作 $+2 \\mathrm{~cm}$ , 那么比标准高度低 $3 \\mathrm{~cm}$ 记作什么?",
+      "options": [],
+      "answer": "(1) 合格课桌的高度 $h$ 范围是 $57 \\mathrm{~cm} \\leq h \\leq 63 \\mathrm{~cm}$ ;(2) $-3 \\mathrm{~cm}$",
+      "solution": "【分析】理解'正'和'负'的相对性，明确什么是一对具有相反意义的量。在一对具有相反意义的量中，先规定其中一个为正，则另一个就用负表示。本题主要考查的是正数和负数在实际生活中的应用。\n\n【详解】（1）根据题意得： $60 + 3 = 63\\mathrm{cm},$ $60 - 3 = 57\\mathrm{cm}$ , 合格的课桌的高度 $h$ 范围是 $57\\mathrm{cm} \\leq h \\leq 63\\mathrm{cm}$ 。(2) 根据题意可得: 比标准高度高为 '+', 那么比标准高度低为 '- ' , 所以比标准高度低 $3 \\mathrm{~cm}$ 记作: -3cm。",
+      "knowledge_points": [
+        "利用正负数解决实际问题"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_396",
+      "original_id": "unit1_solve_396"
+    },
+    {
+      "id": "unit1_solve_397",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "(1) 小明向东走 $5 \\mathrm{~m}$ 记为 $+5 \\mathrm{~m}$ , 那么他走 $-2 \\mathrm{~m}$ 表示什么意思?(2) 如果小明从点 A 先走 $+12 \\mathrm{~m}$ , 再走 $-15 \\mathrm{~m}$ , 又走 $+3 \\mathrm{~m}$ , 他最后在点 A 的什么位置?",
+      "options": [],
+      "answer": "（1）向西走 $2 \\mathrm{~m}$ (2) 在点 A",
+      "solution": "【分析】用正负数可以表示具有相反意义的量, 正数表示向东走, 那么负数表示向西走。正负数是一个相对的概念, 并且表示在一个情境中成对出现的两个具有相反意义的量。\n\n【详解】（1）向东走 $5 \\mathrm{~m}$ 记作 $+5 \\mathrm{~m}$ 就是规定向东走为正,那么向西走就记作负数, 他走 $-2 \\mathrm{~m}$ 表示向西走 $2 \\mathrm{~m}$ 。(2) 如果小明从点 A 先走 +12m，再走 -15m，又走 +3m，意思是小明从点 A 向东走 12m，再向西走 15m，又向东走 3m，他最后回到点 A 的位置。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_397",
+      "original_id": "unit1_solve_397"
+    },
+    {
+      "id": "unit1_solve_398",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "在一次数学测验中，六（1）班的平均分为85分，把高于平均分的记作正数。(1) 李某得了 90 分, 应记作多少?(2) 王强被记了 -5 分, 他实际得分是多少?(3) 王兰得了 83 分, 应记作多少?(4) 李某和王强相差多少分?",
+      "options": [],
+      "answer": "(1) +5分; (2) 80分; (3) -2分; (4) 10分",
+      "solution": "【分析】规定高于平均分记为正，那么低于平均分则为负，我们运用实际的得分减去标准分，即可得到答案。解题关键是理解'正'和'负'的相对性，确定正负数是一对具有相反意义的量。\n\n【详解】(1) $90 - 85 = 5$ （分），高于平均分 5 分记作：+5 分。答: 李军得了 90 分, 应记作 +5 分。 (2) 王强被记了 -5 分表示低于平均分 5 分, $85 - 5 = 80$ (分)。 答: 王强被记了 -5 分, 他实际得分是 80 分。 (3) $85 - 83 = 2$ (分), 低于平均分则记作: $-2$ 分。 答: 王兰得了 83 分, 应记作 -2 分。 (4) $90 - 80 = 10$ (分) 答: 李军和王强相差 10 分。",
+      "knowledge_points": [
+        "利用正负数解决实际问题",
+        "正负数的意义及应用"
+      ],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_398",
+      "original_id": "unit1_solve_398"
+    },
+    {
+      "id": "unit1_solve_399",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "一条大鱼在水中所在的高度为 -50 米, 如果它再向下潜 10 米, 那么它所在的高度是多少米? 如果它从原来的位置上升 20 米, 那么它所在的位置是多少米？",
+      "options": [],
+      "answer": "-60米；-30米",
+      "solution": "【分析】本题是以海平面为标准，下潜记为负，上升记为正，-50米表示在水深50米处，现在水深+下潜距离=下潜后水深，现在水深一上升距离=上升后水深，水深添上负号就是所在位置的高度。本题考查了正负数的意义, 正负数可以表示相反意义的量。\n\n【详解】$50 + 10 = 60$ （米） $50 - 20 = 30$ (米)答: 如果它再向下潜 10 米, 那么它所在的高度是 -60 米,如果它从原来的位置上升 20 米, 那么它所在的位置是 -30 米。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20 六年级下·全国·课后作业",
+      "merged_id": "unit1_399",
+      "original_id": "unit1_solve_399"
+    },
+    {
+      "id": "unit1_solve_400",
+      "unit": "第一单元",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "中等",
+      "question": "小明家的位置记为0米，向东走为正，向西走为负，若小明从家走了 $+70$ 米，又走了 $-30$ 米，这时小明离家的距离是多少米？",
+      "options": [],
+      "answer": "40 米",
+      "solution": "【分析】向东走为正，向西走为负，小明从家走了 +70 米，是向东走了 70 米，又走了 -30 米，是又向西走了 30 米，向东走的距离一向西走的距离 = 小明离家的距离。本题考查了正负数的意义, 正负数可以表示相反意义的量。\n\n【详解】$70 - 30 = 40$ （米）答: 这时小明离家的距离是 40 米。",
+      "knowledge_points": [
+        "正负数的意义及应用"
+      ],
+      "source": "19-20六年级下·全国·课后作业",
+      "merged_id": "unit1_400",
+      "original_id": "unit1_solve_400"
+    },{
+      "id": "unit1_choice_401",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "（19-20六年级下·全国·单元测试）某日河南省四个城市的最低气温如下表。这天最冷的是（）市。",
+      "options": [
+        "A. 新乡",
+        "B. 南阳",
+        "C. 信阳",
+        "D. 郑州"
+      ],
+      "answer": "A",
+      "solution": "【解析】在负数的大小比较中, 负号后面的数越大, 这个越数小。本题考查了负数的大小比较，负数小于0，小于所有正数。【详解】因为-10<-6<-3<0, 所以最冷的是新乡。故答案为: A",
+      "knowledge_points": ["正负数的大小比较"],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_401",
+      "original_id": "unit1_choice_401"
+    },
+    {
+      "id": "unit1_choice_402",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "(19-20 六年级下·全国·单元测试) 下面直线上 A 点, 表示的数大约是 ( )。![](images/b414622d64e462712aa87b5f17c59c93f2532df6b63712c37a12e747f6e7397c.jpg)",
+      "options": [
+        "A. -2",
+        "B. -4",
+        "C. 4"
+      ],
+      "answer": "B",
+      "solution": "【解析】规定了原点、正方向和单位长度的直线叫做数轴。数轴上的数从左到右依次增大，原点左边的数都是负数，原点右边的数都是正数。考查了对数轴和负数的认识。数轴上, 正数都在 0 的右边, 负数都在 0 的左边, 数轴上每个单位长度表示1。【详解】由图示可知, 0 到 +1 是一个单位长度, 0 到 A 大约有 4 个这种单位长度, A 点在 0 的左边, 故 A 点表示的数大约是 -4。故答案为: B",
+      "knowledge_points": ["正负数在数轴上的表示"],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_402",
+      "original_id": "unit1_choice_402"
+    },
+    {
+      "id": "unit1_choice_403",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "(19-20 六年级下·山东菏泽·期末) 在直线上, 点A表示的数是 ( )。![](images/ed6c742f9f1fbd41f468578bdf3487327d864ebe30ef6ce8027ea9910e5cc41b.jpg)",
+      "options": [
+        "A. -0.1",
+        "B. 2/3",
+        "C. 4/3",
+        "D. 0.8"
+      ],
+      "answer": "C",
+      "solution": "【分析】观察数轴, 点 A 再 1 和 2 之间, 说明点 A 比 1 大,比 2 小, 据此比较选项各数, 作出选择。本题考查了数轴的认识，在数轴上，越往右数越大，越往左数越小。【详解】A．-0.1<1，不是点A表示的数；B. 2/3<1, 不是点 A 表示的数; C. 1<4/3<2, 在 1 和 2 之间, 是点 A 表示的数; D. 0.8<1, 不是点 A 表示的数。故答案为: C",
+      "knowledge_points": ["正负数在数轴上的表示"],
+      "source": "19-20 六年级下·山东菏泽·期末",
+      "merged_id": "unit1_403",
+      "original_id": "unit1_choice_403"
+    },
+    {
+      "id": "unit1_choice_404",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "(20-21 六年级下·全国·单元测试) 对于直线上点E表示的数, 下面表述中不正确的是 ( )。![](images/f6f154491a53bcaef8ab2cfbc55609c65c124f506173f26c854d3adbf68b58de.jpg)",
+      "options": [
+        "A. 一定小于 0",
+        "B. 可能是-3",
+        "C. 一定小于-2",
+        "D. 不可能大于-4"
+      ],
+      "answer": "D",
+      "solution": "【分析】从题图中可以知道, 点E在 0 的左边, 所以点E表示的数一定小于 0 , 点E与 0 的距离大于 2 与 0 的距离, 所以点E表示的数一定小于 -2 , 可能是 -3 点。E在 -4 的右边, 则点E表示的数一定大于 -4 , 所以 D 错误。对于正负数大小的比较，在数轴上比较是最直接、最容易判断的。一定要记住右边的数永远大于左边的数。【详解】由分析得：对于直线上点E表示的数, 表述中不正确的是 (不可能大于-4)。故答案为: D。",
+      "knowledge_points": ["正负数在数轴上的表示", "正负数的大小比较"],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_404",
+      "original_id": "unit1_choice_404"
+    },
+    {
+      "id": "unit1_choice_405",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "（20-21 六年级下·全国·单元测试）如图所示，与北京时间相比，东京时间早 1 小时，记为+1时；巴黎时间晚 7 小时，记为 -7 时。悉尼时间记为+2时，当北京时间是 6 月 1 日 23 时时，悉尼时间是（）。![](images/075ea6026e58e13181b50de1ff3fdf95.png)",
+      "options": [
+        "A. 6月2日1时",
+        "B. 6月1日1时",
+        "C. 6月1日21时",
+        "D. 6月2日21时"
+      ],
+      "answer": "A",
+      "solution": "【分析】悉尼时间记为+2时, 表示悉尼时间比北京时间早 2 小时, 所以用北京时间加 2 小时就是悉尼时间。当北京时间是 6 月 1 日 23 时时, 悉尼时间是 6 月 2 日 1 时。在本题中，用正负数来表示其他时区与北京的相对时间。早于北京的时间就记为正，晚于北京的时间就记为负。这种记时的方法为我们了解及计算时间提供了方便。【详解】由分析得：与北京时间相比，东京时间早1小时，记为+1时；巴黎时间晚7小时，记为-7时。悉尼时间记为+2时，当北京时间是6月1日23时时，悉尼时间是（6月2日1时）。故答案为: A。",
+      "knowledge_points": ["正负数的意义及应用"],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_405",
+      "original_id": "unit1_choice_405"
+    },
+    {
+      "id": "unit1_choice_406",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "困难",
+      "question": "（20-21 六年级下·全国·单元测试）天气预报报道某地区今天气温是0°C~4°C，明天气温是4°C~-1°C。该地区明天的最低气温和今天最低气温相比，（）。",
+      "options": [
+        "A. 上升了",
+        "B. 下降了",
+        "C. 不变",
+        "D. 无法判断"
+      ],
+      "answer": "B",
+      "solution": "【分析】以 0 为临界点, 正数都大于 0 , 负数都小于 0 , 0 不是正数也不是负数。据此比较明天和今天的最低气温即可。考查了正负数大小的比较，明确正数>0>负数是解题关键。【详解】由题意可知, 今天的最低气温是0°C, 明天的最低气温是-1°C;0°C>-1°C; 即明天的最低气温比今天降低了;故答案为: B",
+      "knowledge_points": ["正负数的大小比较"],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_406",
+      "original_id": "unit1_choice_406"
+    },
+    {
+      "id": "unit1_choice_407",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "(20-21 六年级下·全国·单元测试) 如下图, 下面选项中的数在P点左边的是 ( )。![](images/7f2a6bfbd675463b312ec2bf50643ac02c67355fe7b9cd0ff3cd7d743c39f5e6.jpg)",
+      "options": [
+        "A. -1/2",
+        "B. -9/10",
+        "C. -0.1",
+        "D. -1/3"
+      ],
+      "answer": "B",
+      "solution": "【分析】正数与负数以 0 为分界点, 正数、0 都比负数大; 负数与负数比较大小, 负号后面的数字越小, 这个负数反而越大; 反之, 负号后面的数字越大, 这个负数就越小。此题考查了正、负数大小比较的方法。要注意，在负数与负数比较大小时，负号后面的数越大这个数越小。【详解】由分析可得：-1<P<-1/2符合此范围的答案只有-9/10;故答案为: B",
+      "knowledge_points": ["正负数在数轴上的表示"],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_407",
+      "original_id": "unit1_choice_407"
+    },
+    {
+      "id": "unit1_choice_408",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "困难",
+      "question": "(20-21 六年级下·全国·单元测试) 直线上点 A 表示的数是 ( )。![](images/b9cd7a85fb541a9ab0e724eb778714368d2c18db6dff79fcc76ce2cbcfc65cfa.jpg)",
+      "options": [
+        "A. -0.5",
+        "B. -1.5",
+        "C. 0.5",
+        "D. 1.5"
+      ],
+      "answer": "C",
+      "solution": "【分析】据图可知: 每个大的单位长度表示 1 , 把 1 平均分成 2 份, 每份是 0.5 , 点 A 表示的数占 1 份, 表示 0.5 , 由此进行选择即可。本题主要考查了在数轴上表示正负数, 小数意义的应用; 明确每个单位长度表示多少, 是解答此题的关键。【详解】直线上点A表示的数是0.5；故答案为: C",
+      "knowledge_points": ["正负数在数轴上的表示", "小数的意义"],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_408",
+      "original_id": "unit1_choice_408"
+    },
+    {
+      "id": "unit1_choice_409",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "困难",
+      "question": "（20-21六年级下·全国·单元测试）王强家到学校的距离大约460米，用线段图表示（）点合适。![](images/d792a69205623419029d560f4651bc46f936d810fc0bb1bb789f576361e56659.jpg)",
+      "options": [
+        "A. A",
+        "B. B",
+        "C. C",
+        "D. D"
+      ],
+      "answer": "D",
+      "solution": "【分析】根据数轴可知, 把 600 米的线段平均分成了 6 段,每段应是 100 米, 王强家到学校的距离大约 460 米, 460 米的位置应在第四段的后面, 进而就可以在图上标注出来。此题主要考查了整数的认识, 熟悉数轴的结构是解题的关键。【详解】把 600 米的线段平均分成了 6 段, 每段应是 100 米, 460 米的位置应在第四段的后面; 故答案为: D。",
+      "knowledge_points": ["正负数在数轴上的表示"],
+      "source": "20-21六年级下·全国·单元测试",
+      "merged_id": "unit1_409",
+      "original_id": "unit1_choice_409"
+    },
+    {
+      "id": "unit1_choice_410",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "困难",
+      "question": "(20-21 四年级下·北京·期末) 观察下表分析数据, 淘气成绩应记作（）。",
+      "options": [
+        "A. -1",
+        "B. +1",
+        "C. -79",
+        "D. 79"
+      ],
+      "answer": "A",
+      "solution": "【分析】由表格可知: 以 80 为标准, 超过部分用正数表示,不足部分用负数表示。根据正负数的意义，解答此题即可，注意找到正负数之间的标准值是 80 是解题的关键。【详解】淘气的成绩是 79 低于 80 , 低于了 1 分所以淘气成绩应记作 -1。故答案为：A",
+      "knowledge_points": ["正负数的意义及应用"],
+      "source": "20-21 四年级下·北京·期末",
+      "merged_id": "unit1_410",
+      "original_id": "unit1_choice_410"
+    },
+    {
+      "id": "unit1_choice_411",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "困难",
+      "question": "（19-20 六年级下·湖南怀化·单元测试）下列图中，能正确表示 -1 的是（）。",
+      "options": [
+        "A. ![](images/162d47346772c18ec4a6f1fbb97a1da5be416cca9fa80eddaec0530b35bc61cf.jpg)",
+        "B. ![](images/0fb82d2fb97762b22bdffdd6b7bb1025ca88a0336e639912a6aa862f8c7fdc68.jpg)",
+        "C. ![](images/8ce7a2de7ae566a42ce4fbc2824d54e7815473376426b15b560e556eacbe07b2.jpg)"
+      ],
+      "answer": "C",
+      "solution": "【分析】在数轴上表示数，要包含三要素：方向、原点、单位长度，据此解答即可。本题考查数轴，解答本题的关键是掌握数轴表示数的概念。【详解】A．缺少原点，错误；B. 单位长度不等, 错误;C. 包含数轴三要素, 所以能正确表示 -1 , 正确。故答案为: C。",
+      "knowledge_points": ["正负数在数轴上的表示"],
+      "source": "19-20 六年级下·湖南怀化·单元测试",
+      "merged_id": "unit1_411",
+      "original_id": "unit1_choice_411"
+    },
+    {
+      "id": "unit1_choice_412",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "（21-22 六年级下·湖南株洲·期末）2022 年 1 月 25 日，湖南四个城市的最低气温如下表。这天气温最低的是（）。",
+      "options": [
+        "A. 长沙",
+        "B. 常德",
+        "C. 株洲",
+        "D. 岳阳"
+      ],
+      "answer": "D",
+      "solution": "【分析】根据负数比较大小的方法, 负数在数轴上原点的左边, 离原点越远则数字越小; 正数在原点右边, 正数大于负数。据此选择即可。本题考查负数的比较大小, 明确负数比较大小的方法是解题的关键。【详解】由分析可知:1>-1>-2>-3，所以气温最低是-3摄氏度，即岳阳的气温最低。故答案为: D",
+      "knowledge_points": ["正负数的大小比较"],
+      "source": "21-22 六年级下·湖南株洲·期末",
+      "merged_id": "unit1_412",
+      "original_id": "unit1_choice_412"
+    },
+    {
+      "id": "unit1_choice_413",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "（2022·甘肃金昌·小升初真题）一只野兔从兔窝出发去觅食，向东跑了 3 米（记作 +3 米）后没有发现食物，又继续向东跑了 2 米，结果仍然没有找到食物。于是就又跑了 -8 米，终于找到了食物，此时野兔的位置在兔窝（）的位置。![](images/74216827ff80e26b6bd7a070e55db39fdde66246185e44acd9d5602596c90c30.jpg)",
+      "options": [
+        "A. 向西 8 米",
+        "B. 向东 3 米",
+        "C. 向西 3 米"
+      ],
+      "answer": "C",
+      "solution": "【分析】以免窝为分界点，东和西是具有相反意义的两个量，如果向东用'+'表示，那么向西用'- '表示，野兔先向东跑 3 米记作 +3 米, 再向东跑 2 米记作 +5 米, -8 米表示野兔从 +5 米的位置向西跑了 8 米, 此时野兔在 -3 的位置, 表示兔窝向西 3 米, 据此解答。本题主要考查正负数的意义及应用, 找出相反意义的两种量是解答题目的关键。【详解】分析可知, 此时野兔的位置在兔窝向西 3 米的位置。故答案为: C",
+      "knowledge_points": ["正负数在数轴上的表示", "正负数的意义及应用"],
+      "source": "2022·甘肃金昌·小升初真题",
+      "merged_id": "unit1_413",
+      "original_id": "unit1_choice_413"
+    },
+    {
+      "id": "unit1_choice_414",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "（2021·浙江宁波·小升初真题）如图，大海中一艘潜水艇所在高度是-50m，一只海豚在潜水艇上方20m处，海豚所在的高度是（）。![](images/5f617e0fbe9958ac970b56d6466cb35ddaddafb59c07333efb23b638086c6956.jpg)",
+      "options": [
+        "A. -10m",
+        "B. -20m",
+        "C. -30m",
+        "D. -70m"
+      ],
+      "answer": "C",
+      "solution": "【分析】由题意可知, 潜水艇距离海平面有50m, 海豚在潜水艇上方20m处, 则用潜水艇到海平面的距离减去 20 即可求出海豚到海平面的距离, 且海豚在海平面的下面用负数表示即可。本题考查正负数的应用, 求出海豚到海平面的距离是解题的关键。【详解】由分析可知:50-20=30(m)所以海豚所在的高度是-30m。故答案为: C",
+      "knowledge_points": ["正负数的意义及应用"],
+      "source": "2021·浙江宁波·小升初真题",
+      "merged_id": "unit1_414",
+      "original_id": "unit1_choice_414"
+    },
+    {
+      "id": "unit1_choice_415",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "困难",
+      "question": "(20-21 六年级下·天津红桥·期末) 如下图所示 M 点所表示的数是 ( )。![](images/0e3d9f28e2ddf4b1e6f4787bdb6283c0b2f460fc943cb4381f2df27ffe5f649a.jpg)",
+      "options": [
+        "A. -1",
+        "B. -1.5",
+        "C. 1.5",
+        "D. 3/2"
+      ],
+      "answer": "B",
+      "solution": "【分析】数轴是规定了 0 点、方向和单位长度的直线，在零点左边所表示的数都是负数，在 0 点右边所表示的数都是正数，根据各点所在数轴上的位置即可确定此点所表示的数。数轴上，从左到右的顺序就是数从小到大的顺序。0是正数和负数的分界点，负数<0<正数，负号后面的数越大，这个数就越小。【详解】点M在-2和-1的中间就是-1.5。故答案为: B",
+      "knowledge_points": ["正负数在数轴上的表示"],
+      "source": "20-21 六年级下·天津红桥·期末",
+      "merged_id": "unit1_415",
+      "original_id": "unit1_choice_415"
+    },
+    {
+      "id": "unit1_choice_416",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "困难",
+      "question": "（22-23六年级下·全国·单元测试）丽丽在直线上，分别标出了几组正、负数，根据你对正、负数的理解，下面标注位置正确且合理的一组是（）。",
+      "options": [
+        "A. ![](images/73a59961bb333725c7463748076ceb5401991c5e6a0ed038c95db423d6cee315.jpg)",
+        "B. ![](images/b075c9976c15d9e19dc7ebe079c93c39e6f4eae5e196eb0bfe16eac448c6a132.jpg)",
+        "C. ![](images/9163d5f412b28fe27d7b031fa7f2c8817c66adc5ce3eadcee3197b82a0d9481d.jpg)"
+      ],
+      "answer": "C",
+      "solution": "【分析】在数轴上，0 是正数、负数的分界点，0 的右边是正数，0 的左边是负数；正数的数字前面的'+'可以省略不写，负数的数字前面的'-'不能省略。本题考查正负数的认识, 掌握正负数在数轴上的表示是解题的关键。【详解】A．数轴上，单位长度不相等，即-1到0的距离和1到0的距离应相等，标注错误； B. 数轴上, -1 应在 0 的左边, 标注错误; C. 数轴上, 单位长度一定, -2 在 0 的左边, 1 和 2 在 0 的右边，标注正确且合理。故答案为: C",
+      "knowledge_points": ["正负数在数轴上的表示"],
+      "source": "22-23六年级下·全国·单元测试",
+      "merged_id": "unit1_416",
+      "original_id": "unit1_choice_416"
+    },
+    {
+      "id": "unit1_choice_417",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "容易",
+      "question": "（19-20四年级上·辽宁·期末）如图，点A表示的数可能是（）。![](images/4ff3e02b1b0551b04482ac2363ac8163c0a32ed031dc85ec6d0c3f8931ddcc7b.jpg)",
+      "options": [
+        "A. 1/3",
+        "B. -1/3",
+        "C. -2/3"
+      ],
+      "answer": "C",
+      "solution": "【分析】原点左边的为负数，原点右边的为正数，在数轴上的数从左到右依次变大。首先根据数轴上 A 点的位置确定 A 的取值范围，这个数在 0 和 -1 之间，属于负数，再根据 A 所处的位置判断即可。本题考查的是数轴的特点, 能根据数轴的特点确定出 A 的取值范围是解答此题的关键。【详解】在数轴上，A 在 0 和 -1 之间，所以 A 是大于 -1 而小于 0 的负数，把 0 到 -1 之间平均分为 3 份，A 在靠近-1 的1/3处, 但在数轴上, -2/3<-1/3, 所以 A 点的位置为-2/3;故答案为: C。",
+      "knowledge_points": ["正负数在数轴上的表示"],
+      "source": "19-20四年级上·辽宁·期末",
+      "merged_id": "unit1_417",
+      "original_id": "unit1_choice_417"
+    },
+    {
+      "id": "unit1_choice_418",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "选择题",
+      "difficulty": "困难",
+      "question": "(22-23 六年级下·山东枣庄·期末) 下面直线上点 P 表示的数是 ( )。![](images/8257a33fc699f0f3b2d4cadc5a39cfcb54a4142ff890861a31f71d0222ad3f67.jpg)",
+      "options": [
+        "A. -2.5",
+        "B. -1.5",
+        "C. 0.5",
+        "D. 1.5"
+      ],
+      "answer": "B",
+      "solution": "【分析】在数轴上，首先确定原点 0 的位置和单位长度，且从左到右的顺序就是数从小到大的顺序，所有的负数都在 0 的左边，越往左数越小，正数都在 0 的右边，越往右数越大。点 P 表示的数所在的位置是在 -2 和 -1 之间，说明这个数大于 -2 小于 -1。据此判断即可。此题主要考查了正负数在数轴上的表示以及正负数的大小比较。【详解】A．-2.5<-2，不符合题意；B. -2<-1.5<-1, 符合题意; C. 0.5>-1, 不符合题意; D. 1.5>-1, 不符合题意;故答案为: B",
+      "knowledge_points": ["正负数在数轴上的表示", "正负数的大小比较"],
+      "source": "22-23 六年级下·山东枣庄·期末",
+      "merged_id": "unit1_418",
+      "original_id": "unit1_choice_418"
+    },
+    {
+      "id": "unit1_choice_419",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "（19-20六年级下·全国·单元测试）把下面各数分别填在相对应的圈中。-1、+0.21、-5/7、-86、6.20、-3.67、+1/10、+100.3 正数:(),负数（）；",
+      "options": [],
+      "answer": "+0.21、6.20、+1/10、+100.3 ; -1、-5/7、-86、-3.67",
+      "solution": "【分析】负数是指小于 0 的数, 负数前面带有'-'; 正数是指大于 0 的数, 正数前面的'+'可以省去。据此填空即可。本题考查了负数的认识, 注意 0 既不是正数也不是负数。【详解】由分析可得：正数有: +0.21、6.2、+1/10、+100.3;负数有：-1、-5/7、-86、-3.67。",
+      "knowledge_points": ["正负数的概念及辨认"],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_419",
+      "original_id": "unit1_fill_419"
+    },
+    {
+      "id": "unit1_choice_420",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "（19-20 六年级下·全国·单元测试）下面是某水库记录员记录的水库水位的变化情况。如果把水位上升5cm，记作+5cm，请把余下的 4 次记录表示出来。",
+      "options": [],
+      "answer": "+2cm, -3cm, -5cm, +7cm",
+      "solution": "【分析】正负数表示一组相反意义的量，水位上升和下降是一组相反意义的量。此题主要用正、负数来表示具有意义相反的两种量。基础题。【详解】由分析，填表如下：上升5cm对应+5cm；上升2cm对应+2cm；下降3cm对应-3cm；下降5cm对应-5cm；上升7cm对应+7cm",
+      "knowledge_points": ["正负数的意义及应用"],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_420",
+      "original_id": "unit1_fill_420"
+    },
+    {
+      "id": "unit1_choice_421",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "(19-20 六年级下·全国·单元测试) 写一写, 读一读。![](images/8e1e42b75cb9713e26c04bdd56fd3f15fb7bf4f3b03b21744f0bd7b58ad6f600.jpg) 黄山的最高峰莲花峰大约比海平面高1860米，海拔____米。",
+      "options": [],
+      "answer": "1860 或 +1860，读作：一千八百六十或正一千八百六十",
+      "solution": "【分析】以海平面为标准，高于海平面的高度都记作正，可以在数字前面加上'+'，也可以不加；低于海平面的高度都记作负，一定要在数字前面加上'-'。本题考查了正负数的读法、意义及应用。【详解】根据正负数的意义可知, 比海平面高 1860 米记作: 1860米或 +1860 米, 读作: 一千八百六十或正一千八百六十。",
+      "knowledge_points": ["正负数的读法和写法", "正负数的意义及应用"],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_421",
+      "original_id": "unit1_fill_421"
+    },
+    {
+      "id": "unit1_choice_422",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "(19-20 六年级下·全国·单元测试) 下表记录的是上一周小明每天做作业的时间。(1) 如果把平均每天做作业的时间作为标准（用 0 表示），超过平均时间用正数表示，不足平均时间用负数表示，请把下表填写完整。(2) 平均每天做作业的时间是____分？",
+      "options": [],
+      "answer": "20分、+5分、-5分、0分、-10分、+10分、0 ; 20",
+      "solution": "【分析】（1）把等于平均分的记作0，超出平均分的部分记作'+'，低于平均分的部分记作'-'，分别用正负数表示出每天的得分即可。(2) 把每天的得分相加后除以 5 就是平均分。本题考查正负数的实际应用，比标准数大几就用正几表示，比标准数少几就用负几表示。【详解】(1) 25-20=+5（分）；15-20=-5（分）；20-20=0(分); 10-20=-10(分); 30-20=+10(分), 填表完成。(2) (25+15+20+10+30)÷5=100÷5=20(分)",
+      "knowledge_points": ["正负数的意义及应用", "平均数的意义及求法"],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_422",
+      "original_id": "unit1_fill_422"
+    },
+    {
+      "id": "unit1_choice_423",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "(19-20 六年级下·全国·单元测试) 填上适当的数。![](images/9ee96239ce7737bf8b85da30a459f31a421c13d01f9e7df629f4e3572047ebd1.jpg)",
+      "options": [],
+      "answer": "-4;-3;3;5",
+      "solution": "【分析】规定了原点、正方向和单位长度的直线叫数轴; 原点的右边表示的是正数, 每个单位长度表示1, 2, 3……，原点的左边表示负数，每个单位长度分别是: -1, -2, -3……; 由此进行解答。本题考查了数轴的认识; 数轴是规定了原点, 单位长度和方向的一条直线。【详解】观察数轴可得：从左到右依次为-4；-3；3；5 ![](images/73e4d2ab5ce8d0cdf7df848d982c1d2a97d8885807e458547a60cc5568e43cd7.jpg) 故答案为：-4；-3；3；5",
+      "knowledge_points": ["正负数在数轴上的表示"],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_423",
+      "original_id": "unit1_fill_423"
+    },
+    {
+      "id": "unit1_choice_424",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "（19-20六年级下·全国·单元测试）实验小学501班学生的平均身高为145厘米。下表给出了该班6名同学的身高（单位：厘米）情况。试完成下表：(1) 谁最高? 谁最矮? (2) 最高与最矮的学生的身高相差多少？",
+      "options": [],
+      "answer": "（1）最高的是小山，最矮的是小亮 (2) 最高与最矮的学生身高差是 25 厘米",
+      "solution": "【分析】比平均身高 145 厘米高记为正，比平均身高 145 厘米矮记为负，则可知小斌、小丽、小樱的身高，把他们的身高进行比较则可知最高、最矮的是谁，再求出最高、最矮的学生的身高差是多少即可。本题考查正、负数的意义, 正数与负数表示相反意义的量, 看清题目是规定哪个量为正, 和它意义相反的量就记为负。【详解】（1）小斌的身高：145+2=147（厘米）小丽的身高: 145+0=145(厘米)小樱的身高: 145+3=148(厘米)165厘米>148厘米>147厘米>145厘米>144厘米>140厘米所以最高的是小山，最矮的是小亮。答: 最高的是小山, 最矮的是小亮。(2) 最高的是小山，身高是：165 厘米最矮的是小亮，身高是：140 厘米则相差：165-140=25（厘米）答: 最高与最矮的学生身高差是 25 厘米。",
+      "knowledge_points": ["正负数的意义及应用"],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_424",
+      "original_id": "unit1_fill_424"
+    },
+    {
+      "id": "unit1_choice_425",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "困难",
+      "question": "(19-20 六年级下·全国·单元测试) 请你把下面的数填入相应的圈里。36、-9、0.7、+20.4、-56、100、-13、-261、+ 4.8、109 正数:() ;负数()",
+      "options": [],
+      "answer": "36、0.7、+20.4、100、+4.8、109 ;-9、-56、-13、-261",
+      "solution": "【分析】正数都大于 0 , 正数都带有'+'号或不带任何符号; 负数都小于 0 , 负数都带有'- '号; 0 不是正数也不是负数。'+'可以省略不写；0既不是正数，也不是负数。【详解】正数: 36, 0.7, +20.4, 100, +4.8, 109; 负数: -9、-56、-13、-261",
+      "knowledge_points": ["正负数的概念及辨认"],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_425",
+      "original_id": "unit1_fill_425"
+    },
+    {
+      "id": "unit1_choice_426",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "(19-20 六年级下·全国·单元测试) 把数轴上表示的数写在括号里。![](images/7db11be2b9dba1a3828fc006f67e15cf14af4b7711572a9ad920fd00467364ce.jpg)",
+      "options": [],
+      "answer": "-4; -2; -1; 2; 4",
+      "solution": "【分析】数轴的三要素：正方向（箭头方向）、原点（0刻度）、单位长度（刻度）；正方向：一般是向上或向右；原点: 数字 0 所在的位置;单位长度: 由所表示数的大小来决定刻度之间的距离的大小。本题考查了数轴的认识, 负数在 0 的左侧, 正数在 0 的右侧。【详解】填空如下：从左到右依次为-4; -2; -1; 2; 4 ![](images/dd21190d18c932d833b98dbefcb75a7a33f8eb3902447a72571ae07f3e857129.jpg)",
+      "knowledge_points": ["正负数在数轴上的表示"],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_426",
+      "original_id": "unit1_fill_426"
+    },
+    {
+      "id": "unit1_choice_427",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "困难",
+      "question": "(19-20 六年级下·全国·单元测试) 把下面的数写在相应的圈里。(按题中数的顺序填写) 13、-2、+1、+105、-24、-9、+17、-208 正数：（）负数：（）",
+      "options": [],
+      "answer": "正数: +13、+1、+105、+17 ;负数：-2、-24、-9、-208",
+      "solution": "【分析】正数都大于 0 , 正数都带有'+'号或不带符号,负数都小于 0 , 负数都带有'- '号。本题是让学生们辨认正负数，并且提供的数目均带有正负号，这样就容易判断了。只是注意要按题中数的顺序填写。【详解】根据正负数的意义可知，正数有：+13，+1，+105，+17；负数有：-2，-24，-9，-208。故答案为：+13，+1，+105，+17；-2，-24，-9,-208。",
+      "knowledge_points": ["正负数的概念及辨认"],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_427",
+      "original_id": "unit1_fill_427"
+    },
+    {
+      "id": "unit1_choice_428",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "(19-20 六年级下·全国·单元测试) 在下面数轴上,A、B、C、D各点分别表示什么数？![](images/28310c2ec341df5adc7d706d64fcb93d09c787fbdb8125ab1ed45aec471f14c9.jpg)",
+      "options": [],
+      "answer": "-4;-1;2;3",
+      "solution": "【分析】数轴上, 0 是正数和负数的分界点, 所有的负数都在 0 的左边, 所有的正数都在 0 的右边, 数轴上从左到右的顺序就是数从小到大的顺序。数轴上，所有的负数都在 0 的左边，离零点距离越远数越小。【详解】A比-3小的数字，所以A处填-4，B处为比-2大且比 0 小的数字, C、D 处为比零大且比 4 小的数字; 所以A=-4, B=-1, C=2, D=3 ![](images/2b11efe045676aca2d878c7ca98a5e4a446eabbb7be49a7fdbe4787e77121168.jpg)",
+      "knowledge_points": ["正负数在数轴上的表示"],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_428",
+      "original_id": "unit1_fill_428"
+    },
+    {
+      "id": "unit1_choice_429",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "(19-20 六年级下·全国·单元测试) 在□里填上合适的数。![](images/2adf66e82f4237ab727f4ca62a388ed43a214237abea218f37fc91301c305a04.jpg)",
+      "options": [],
+      "answer": "2.2; 4.6; -1; -2.4 ![](images/c185b85c9c6a2fa2a62167fc310f6b2106319eb8a5f3f94a83c6303a7fd8a098.jpg)",
+      "solution": "【分析】数轴是规定了原点（(0 点）、正方向和单位长度的直线，在数轴上原点（0 点）的左边是负数，右边是正数。在这里把一个单位长度平均分成 5 份，每份表示 0.2，据此解答。数轴上越往右数越大，越往左数越小。【详解】0 右边的第一个□, 在 2 和 3 之间, 在 2 右边第 1 个小格处, 即 2.2;0 右边的第二个□, 在 4 和 5 之间, 在 4 右边第 3 个小格处,即 4.6; 0 左边的第一个口，在 0 左边第 5 个小格处，即 -1; 0 左边的第二个□, 在 -2 和 -3 之间, -2 左边第 2 个小格处, 即 -2.4。![](images/2baacf274853289c33964cd7a30fb52397c3c9e1ce8ddba39ca618ffd39c9016.jpg)",
+      "knowledge_points": ["正负数在数轴上的表示"],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_429",
+      "original_id": "unit1_fill_429"
+    },
+    {
+      "id": "unit1_choice_430",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "（19-20六年级下·全国·单元测试）下图每格表示1米，小龟刚开始的位置在0处。![](images/90ccc241f773506afdb3b39c77255cd4ac392281501df77b7e1d81a00897bb88.jpg) (1) 小龟从 0 点向东爬行 4 米记作 +4 米, 那么从 0 点向西爬行 5 米应记作_米。 (2) 如果小龟的位置是 +8 米, 说明它是向_爬行_米。 (3) 如果小龟的位置是 -6 米, 说明它是向_爬行_米。 (4) 如果小龟先向东爬行 4 米, 又向西爬行 7 米, 这时小龟的位置表示为_米。",
+      "options": [],
+      "answer": "-5 ;东;8 ;西;6;-3",
+      "solution": "【分析】正负数表示一组相反意义的量, 此题中东西方向就是一组相反意义的量, 向东记作正, 向西就记作负。本题考查正负数的意义, 解答本题的关键是掌握向东为正, 向西为负。【详解】（1）根据正负数的意义可知，小龟从 0 点向东爬行 4 米记作 +4 米，那么从 0 点向西爬行 5 米应记作：-5 米；(2) 如果小龟的位置是 +8 米, 说明它是向东爬行了 8 米; (3) 如果小龟的位置是 -6 米, 说明它是向西爬行 6 米; (4) 如果小龟先向东爬行 4 米, 又向西爬行 7 米, 7-4=3(米), 这时小龟的位置表示为: -3米。",
+      "knowledge_points": ["正负数在数轴上的表示", "正负数的意义及应用"],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_430",
+      "original_id": "unit1_fill_430"
+    },
+    {
+      "id": "unit1_choice_431",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "(19-20 六年级下·全国·单元测试) 确定兵兵的位置。(1) 如果兵兵从 0 点向东走 6 米, 记作 +6 米, 那么兵兵从 0 点向西走 5 米, 记作( )。 (2) 如果兵兵从 A 点走到 B 点, 他应向( )走( )米。 (3) 如果兵兵从 0 点出发, 先向东走 3 米, 再接着向西走 5 米, 这时兵兵距 0 点的距离是( )米, 一共走了( )米。",
+      "options": [],
+      "answer": "-5米 ;东; 9 ;2 ;8",
+      "solution": "【分析】（1）从 0 开始，向东记为正，向西记为负，据此填空；(2) A 距离 0 点 6 米, B 距离 0 点 3 米, 加起来是 A 到 B 的距离, 确定方向即可; (3) 从 0 点出发, 先向东走 3 米, 到达 B 点, 从 B 点往西数 5 米, 从第二次移动结束点确定到 0 点的距离; 两次移动距离加起来是一共走的距离。本题考查了正负数的意义, 正负数可以表示相反意义的量。【详解】（1）如果兵兵从 0 点向东走 6 米，记作 +6 米，那么兵兵从 0 点向西走 5 米，记作 -5 米。(2) 6+3=9(米), 如果兵兵从 A 点走到 B 点, 他应向东走 9 米。 (3) 如果兵兵从 0 点出发, 先向东走 3 米, 再接着向西走5米，这时兵兵距0点的距离是2米，3+5=8（米），一共走了8米。",
+      "knowledge_points": ["正负数的意义及应用"],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_431",
+      "original_id": "unit1_fill_431"
+    },
+    {
+      "id": "unit1_choice_432",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "(2014 六年级·全国·课后作业) ![](images/749274868dcf52a3c655b9b1a93f0ba8f3f23e3c0b403a2d4b7cf5b32296726c.jpg) (1) 小亮从 0 点向西行 3 米, 表示为 -3 米。那么小亮从 0 点向东行 5 米, 表示为_米. (2) 如果小亮后来的位置是 -5 米, 说明他从 0 点向 _ 行了 5 米. (3) 如果小亮要到达 +3 米的位置, 他应该从 0 点向 _行_米. (4) 如果小亮从 0 点先向西行 2 米, 又向东行 5 米, 这时小亮的位置表示为 _米, 他共行了 _米.",
+      "options": [],
+      "answer": "+5 ;西 ;东; 3 ;+3 ;7",
+      "solution": "【分析】此题主要用正负数来表示具有意义相反的两种量:向西行记为负, 则向东行就记为正, 直接得出结论即可.【详解】（1）小亮从 0 点向西行 3 米，表示为 -3 米。那么小亮从 O 点向东行 5 米，表示为 +5 米。(2) 如果小亮后来的位置是 -5 米, 说明他从 0 点向西行了 5 米. (3) 如果小亮要到达 +3 米的位置, 他应该从 0 点向东行 3 米. (4) 5-2=3（米）， 2+5=7（米），如果小亮从0点先向西行2米，又向东行5米，这时小亮的位置表示为 +3 米，他共行了7米；故答案为 +5, 西, 东, 3, +3, 7.",
+      "knowledge_points": ["正负数的意义及应用"],
+      "source": "2014 六年级·全国·课后作业",
+      "merged_id": "unit1_432",
+      "original_id": "unit1_fill_432"
+    },
+    {
+      "id": "unit1_choice_433",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "(20-21 六年级下·全国·单元测试) 下表是某超市库存大米在本周的变化情况。入库部分用正数表示，出库部分用负数表示。(1) 星期三有( )kg大米入库, 有( )kg大米出库。 (2) 星期( )没有大米入库。 (3) 若在本周星期日结束时, 该超市库存大米还有 510kg,则上周星期日结束时, 该超市库存大米有 ( )kg。",
+      "options": [],
+      "answer": "350 ;100; 四 ;300",
+      "solution": "【分析】（1）先在题中找到星期三的出、入库记录，分别是 -100kg 和 +350kg。入库部分用正数表示，出库部分用负数表示, 所以 -100kg 表示有 100kg 大米出库, +350kg 表示有 350kg大米入库。(2) 入库部分用正数表示, 出库部分用负数表示, 出、入库记录中只有星期四没有正数, 所以星期四没有大米入库。 (3) 这一个星期入库大米的总数是300+350+350+500+250+300=2050(kg), 出库大米的总数是280+300+100+400+190+320+250=1840(kg), 相当于只入库2050-1840=210(kg)大米。由题意可知本周星期日结束时，超市库存大米还有510kg，减去本周净入库的210kg，得出上周星期日结束时，该超市库存大米有510-210=300(kg)。正负数意义的实际应用，前两问较为简单，在解答第三问时，先要分别计算总入库和总出库的重量，然后依据正负数的意义求出净入库的重量，最后结合本周库存量求出上周日结束时的库存量。【详解】由分析得：(1) 星期三有 350kg 大米入库, 有 100kg 大米出库。 (2) 星期四没有大米入库。 (3) 若在本周星期日结束时, 该超市库存大米还有 510kg,则上周星期日结束时, 该超市库存大米有 300kg。",
+      "knowledge_points": ["利用正负数解决实际问题", "正负数的意义及应用"],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_433",
+      "original_id": "unit1_fill_433"
+    },{
+      "id": "unit1_choice_434",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "中等",
+      "question": "(20-21 六年级下·全国·单元测试) 看温度计填数。",
+      "options": [],
+      "answer": "-2; -15 ;+5或5",
+      "solution": "【分析】根据正负数的认识，比 0 小的数是负数，比 0 大的数是正数，进行填空，写负数时前面加负号'-'，正数前面加正号'+'，正号也可以省略不写。负数的定义：比 0 小的数叫做负数，负数与正数表示意义相反的量。【详解】第一个温度计的刻度在 0 下第 2 个刻度, 表示 -2°C;第二个温度计的刻度正好对齐 -15°C;第三个温度计的刻度正好对齐 5 , 表示 +5°C。",
+      "knowledge_points": ["温度的认识及比较", "正负数的读法和写法"],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_434",
+      "original_id": "unit1_fill_434"
+    },
+    {
+      "id": "unit1_choice_435",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "困难",
+      "question": "（20-21 六年级下·全国·单元测试）下面直线上点 A 表示的数是____，如果点 B 在 1 处，而点 C 是线段 AB 的中点，那么点 C 表示的数是____。![](images/c8cd83a863a8ed2c68a370d3670d5754d84f84764e4707569856f644714f7667.jpg)",
+      "options": [],
+      "answer": "-2 ；-0.5",
+      "solution": "【分析】观察数轴可知，每格代表 1，A 点在 0 的左边第 2 格，表示 -2，B 在 1 处，则线段 AB 之间长度是 3，点 C 是线段 AB 的中点, 3÷2=1.5, 那么点 C 表示的数是 -0.5 , 据此解答即可。解答本题的关键是要明确正负数的意义以及数轴中每格代表多少，进而找到AB的位置以及C点的位置。【详解】直线上点 A 表示的数是 -2，如果点 B 在 1 处，而点 C 是线段 AB 的中点，那么点 C 表示的数是 -0.5。",
+      "knowledge_points": ["正负数在数轴上的表示"],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_435",
+      "original_id": "unit1_fill_435"
+    },
+    {
+      "id": "unit1_choice_436",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "(20-21 六年级下·全国·单元测试) 有五袋奶糖, 第一袋 1.7kg, 第二袋 1.4kg, 第三袋 1.5kg, 第四袋 1.6kg, 第5袋 1.3kg。刚好 1.5kg 的记为 0kg，大于 1.5kg 的记为正数，小于 1.5kg 的记为负数。请把下表填完整。",
+      "options": [],
+      "answer": "+0.2kg ；- 0.1kg；0kg ；+ 0.1kg； - 0.2kg",
+      "solution": "【分析】用每袋的重量分别与 1.5kg 相减, 把高于 1.5kg 的质量记作正, 低于 1.5kg 的质量记作负, 据此答。掌握正负数的意义和表示方法是解答题目的关键。【详解】第一袋：1.7-1.5=+0.2kg；第二袋：1.4-1.5=-0.1kg；第三袋：1.5-1.5=0kg；第四袋：1.6-1.5=+0.1kg；第五袋：1.3-1.5=-0.2kg",
+      "knowledge_points": ["正负数的意义及应用"],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_436",
+      "original_id": "unit1_fill_436"
+    },
+    {
+      "id": "unit1_choice_437",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "(20-21 六年级下·全国·单元测试) 数轴上A、B、C、D、E五个点相邻之间的距离相等，A和E所表示的数如图所示，那么D所表示的数为( )，0这个数所对应的点是( )。![](images/d3a7e49380ca2986286b4a449bb863479129e6de0ca6d95f13fae8451c14abc8.jpg)",
+      "options": [],
+      "answer": "4； B",
+      "solution": "【分析】-2至6，中间相差 2+6=8 个单位长度，求出这个数轴每两个字母之间的一段表示的单位长度，确定原点0的位置，以及D表示的数即可。数轴的三要素：正方向（箭头方向）、原点（0刻度）、单位长度（刻度）。【详解】8÷4=2, 此数轴一个格表示 2, D 所表示的数为 6-2=4; 0 这个数所对应的点是 B。",
+      "knowledge_points": ["正负数在数轴上的表示"],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_437",
+      "original_id": "unit1_fill_437"
+    },
+    {
+      "id": "unit1_choice_438",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "(20-21 六年级下·全国·单元测试) 写出下面各温度计上的温度。![](images/e693728ddaa71a81cccc1209f6ae3b52.png)",
+      "options": [],
+      "answer": "-10; 0 ;15; 10",
+      "solution": "【分析】根据正负数的认识，在温度计上，0°C下边的刻度表示零下温度，用负数表示，负数前边要写负号'-'；0°C上边的温度表示零上温度，写正数时正号可以省略不写。关键是认识正负数, 比 0 大的数是正数, 比 0 小的数是负数。【详解】由分析可知：第一个温度计显示-10°C，第二个温度计显示0°C，第三个温度计显示15°C，第四个温度计显示10°C。",
+      "knowledge_points": ["正负数的意义及应用"],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_438",
+      "original_id": "unit1_fill_438"
+    },
+    {
+      "id": "unit1_choice_439",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "(20-21 六年级下·全国·单元测试) 下表记录了六 (1)班 6 名同学的体重情况, 他们的平均体重记为 0kg, 高于平均体重的记为正，低于平均体重的记为负，请根据下表中的数据把表格填写完整。",
+      "options": [],
+      "answer": "41; 38; 30 ;35; +1; -3",
+      "solution": "【分析】小红体重 33kg, 记作 -2kg, 意思是小红的体重比平均体重少 2kg, 据此可以计算出平均体重; 再将其他人的体重与平均体重对比即可, 高于平均体重记为正, 低于平均体重记为负。此题考查了正数与负数，明确比平均高记为 +，比平均低记为 - ; 弄清题意是解本题的关键。【详解】平均体重：33+2=35(kg)；小亮体重记作+6，则比平均体重高6kg，35+6=41(kg)；小明比平均体重高，36-35=1(kg)，记作+1kg；小刚体重记作+3，则比平均体重高3kg，35+3=38(kg)；小丽比平均体重低，35-32=3，记作-3kg；小晗体重记作-5，则比平均体重低5kg，35-5=30(kg)",
+      "knowledge_points": ["正负数的意义及应用"],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_439",
+      "original_id": "unit1_fill_439"
+    },
+    {
+      "id": "unit1_choice_440",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "作图题",
+      "difficulty": "困难",
+      "question": "（20-21 六年级下·全国·单元测试）在直线上表示下列各数（并标出表示各数的点）。2.5、-1/2、-5、4 ![](images/f163ef0dd16a2343a3a1b293a6e48a16b1079bd129e3a65242955248e2f6e997.jpg)",
+      "options": [],
+      "answer": "见详解",
+      "solution": "【分析】根据题图可知，每个小格代表 1，2.5 在 2 和 3 中间；1/2 在 0 的左边，即在 0 和 -1 之间；-5 在 0 往左数 5 个格的位置；4 在 0 往右数 4 个格的位置，据此解答即可。本题考查了数轴, 熟知数轴上的各点与实数是一一对应关系是解题的关键。【详解】作图如下：![](images/61f5e90afc11d622793db8b7c513f7d3557e0cc35b19ce603b31c9682e373b24.jpg)",
+      "knowledge_points": ["正负数在数轴上的表示"],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_440",
+      "original_id": "unit1_draw_440"
+    },
+    {
+      "id": "unit1_choice_441",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "(20-21 六年级下·全国·单元测试) 直线上 A 点表示的数是( )，B 点表示的数写成分数是( )，C 点表示的数写成小数是( )。![](images/198c88a74c5faf43538f8c1ec88536320fc3be569f0d4a58c2b7071bf16efc08.jpg)",
+      "options": [],
+      "answer": "-2 ;-1又1/2; 1.5",
+      "solution": "【分析】数轴是规定了原点（0 点）、方向和单位长度的一条直线。原点的左边是负数，从原点向左的每个单位长度分别是 -1、-2、-3...；右边是正数，从原点向右每个单位长度分别是 1、2、3…，把第一个单位长度平均分成 2 份，每份是 1/2。据此填表。本题是考查数轴的认识。数轴是规定了原点（0点）、方向和单位长度的一条直线。【详解】A点在0的左边第2格，表示-2；B点在-1和-2的中间，表示-1又1/2；C点在1和2的中间，表示1.5。",
+      "knowledge_points": ["正负数在数轴上的表示"],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_441",
+      "original_id": "unit1_fill_441"
+    },
+    {
+      "id": "unit1_choice_442",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "(20-21 六年级下·山东菏泽·期中) 如下图, 点 B 是 0 , C 是 0.4 , 则 A 是( )。![](images/63e6b26ed78a63460d1ee6fcb916218e7218e2d029c5e30361926338d498c5e6.jpg)",
+      "options": [],
+      "answer": "-1",
+      "solution": "【分析】数轴是规定了原点（0 点）、方向和单位长度的直线，在原点（0 点）左边的点所表示的数都是负数，右边的点表示的数都是正数；据此判断即可。此题考查的是在数轴上表示数的方法。【详解】由B、C间的长度可知，一个小格的长度是 0.4÷2=0.2，A 到 B 之间有 5 个小格，长度应是 0.2×5=1，又因 B 表示 0，A 在 B 的左边，表示的是负数，因此 A 是 -1。",
+      "knowledge_points": ["正负数在数轴上的表示"],
+      "source": "20-21 六年级下·山东菏泽·期中",
+      "merged_id": "unit1_442",
+      "original_id": "unit1_fill_442"
+    },
+    {
+      "id": "unit1_choice_443",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "（20-21 六年级下·河北秦皇岛·期末）六年一班同学进行拍球比赛，以 50 下为标准，超过的数用正数表示，不足的数用负数表示。记录如下：这组同学拍得最多的是( )，实际拍了( )下；拍得最少的是( )，实际拍了( )下，这组同学平均拍球次数( )（填'大于'或'小于'）50下。",
+      "options": [],
+      "answer": "王丽 ;58; 杨芳; 44 ;大于",
+      "solution": "【分析】由统计表知, 拍得最多的王丽, 王丽拍了: 50+8=58(下); 拍得最少的是杨芳, 杨芳拍了: 50-6=44(下);把正负数都加起来，如果得正数，则这组同学平均球次数大于50下；如果得0，则这组同学平均球次数等于50下；如果得负数，则这组同学平均拍球次数小于50下。关键是理解正负数的意义, 能够从统计表中获取知识。【详解】50+8=58(下)；50-6=44(下)；8+7+1-3-5-6=2(下)。这组同学拍得最多的是（王丽），实际拍了（58）下；拍得最少的是（杨芳），实际拍了（44）下，这组同学平均拍球次数（大于）50下。",
+      "knowledge_points": ["利用正负数解决实际问题", "正负数的意义及应用"],
+      "source": "20-21 六年级下·河北秦皇岛·期末",
+      "merged_id": "unit1_443",
+      "original_id": "unit1_fill_443"
+    },
+    {
+      "id": "unit1_choice_444",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "(19-20 六年级下·湖南怀化·单元测试) 在数轴上有三点 A、B、C 分别表示 -5, 0, 4 如下图：![](images/94c8539054300ee2d33e76e097251d7844b12a7f2d0d484f0cd00c8611bd8952.jpg) (1) 将 B 点向左移动 3 个单位长度后, B 点表示的数是( )。 (2) 将 A 点向右移动 6 个单位长度后, 再与原来 B、C 比较, 最小的点是 ( ) 最大的点是 ( )。",
+      "options": [],
+      "answer": "-3 ;B ;C",
+      "solution": "【分析】（1）B点本身代表0，向左移动3个单位长度后，左边表示负数，那么移动后的B点表示-3;(2) A 点本身代表 -5, 向右移动 6 个单位长度后, 变成了1。据此，比较 0、1 和 4 的大小，找出最小的点和最大的点即可。本题考查了正负数在数轴上的表示, 负数在 0 的左边, 正数在 0 的右边。【详解】（1）将 B 点向左移动 3 个单位长度后，B 点表示的数是 -3；(2) 将 A 点向右移动 6 个单位长度后, A 点代表了 1 。因为 4>1>0, 所以移动后的 A 点再与原来 B、C 比较, 最小的点是 B, 最大的点是 C。",
+      "knowledge_points": ["正负数在数轴上的表示"],
+      "source": "19-20 六年级下·湖南怀化·单元测试",
+      "merged_id": "unit1_444",
+      "original_id": "unit1_fill_444"
+    },
+    {
+      "id": "unit1_choice_445",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "作图题",
+      "difficulty": "困难",
+      "question": "(21-22 六年级下·全国·单元测试) 写出点A、B、C，D、E表示的数。![](images/cc41967cd59646f60193f45c114c1ce6f27e8fd8c8e723e519c71e066d86feaf.jpg)",
+      "options": [],
+      "answer": "见详解",
+      "solution": "【分析】用数轴表示数：根据数字在对应刻度上描点表示；负数在 0 的左侧，正数在 0 的右侧。本题考查了对数轴以及负数的认识, 理解在数轴上右边的数大于左边的数。【详解】A点表示-3，B点表示-1，C点表示1，D点表示3，E点表示4。![](images/7f4652e2902a8063befd485081f260ed273852e513d86acb9cb180716a922244.jpg)",
+      "knowledge_points": ["正负数在数轴上的表示"],
+      "source": "21-22 六年级下·全国·单元测试",
+      "merged_id": "unit1_445",
+      "original_id": "unit1_draw_445"
+    },
+    {
+      "id": "unit1_choice_446",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "(21-22 六年级下·全国·单元测试) 如图: 明明家在超市东面 400m 处, 芳芳家在超市西面 500m 处。![](images/bbfa71ec9933bcc9e435df8d6c8287767e6ae50e0415982079ee568bed79007a.jpg) (1) 若把芳芳家在超市的西面 500m 记作 -500m, 则明明家应该记作 ( ) m。(2) 他们俩同时从家出发相向而行, 如果明明和芳芳步行的速度都是 50 米/分, 那么( )分钟后相遇; 相遇时, 在超市的()面( )m 处, 这时的位置可以记作( )m。",
+      "options": [],
+      "answer": "+400 ;9; 西; 50; -50",
+      "solution": "【分析】（1）西面和东面是两个具有相反意义的量，如果把西面记作'-'，那么东面就记作'+' ； (2) 利用相遇问题的方法, 路程÷速度和=时间, 算出时间后, 乘芳芳的速度, 计算出芳芳走的距离, 用 500m 减去这段距离, 可求出离超市的距离, 用正负数表示出来。本题主要考查正、负数的意义及其应用，利用解决相遇问题的方法求出结果。【详解】（1）若把芳芳家在超市的西面 500m 记作 -500m, 则明明家应该记作 +400m;(2) (400+500)÷(50+50)=900÷100=9(分钟)；500-9×50=500-450=50(m)。所以相遇在超市的西面 50m 处, 这时的位置可以记作-50m。",
+      "knowledge_points": ["相遇问题", "正负数的意义及应用"],
+      "source": "21-22 六年级下·全国·单元测试",
+      "merged_id": "unit1_446",
+      "original_id": "unit1_fill_446"
+    },
+    {
+      "id": "unit1_choice_447",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "(21-22 六年级下·全国·单元测试) 下面 1 格代表10m, 英英和才才的位置都在 0 处。![](images/ac562953b03ed293c0e2e41dd4cde8fc9d0f0846991c6e08d32113413382fc65.jpg) (1) 英英从 0 处向东行 1 格表示 +10m, 那么英英从 0 处向西行 4 格表示 ( ) m。 (2) 才才从 0 处向北行 2 格表示 +20m, 那么才才从 0 处向南行 3 格表示 ( ) m。 (3) 如果才才现在的位置是 -20m, 说明他向( )行了 ( )格。 (4) 如果英英先向东行 50m, 再向西行 90m, 这时英英的位置在( )m处。",
+      "options": [],
+      "answer": "-40; -30; 南 ;2 ;-40",
+      "solution": "【分析】（1）向东和向西是两个具有相反意义的量，如果把向东规定为正，记作'+'，那么向西就记作'-'，以此作答；(2) 向北和向南是两个具有相反意义的量, 如果把向北规定为正, 记作'+', 那么向南就记作'- ' , 以此作答; (3) 现在的位置是 -20m, 说明他现在在向南的位置上走了 2 格; (4) 先向东行 50m, 记作 +50m, 再向西行 90m, 记作 -90m, 利用正负数加减法计算即可。此题首先要明确以什么为标准，规定超出标准的为正，低于标准的为负，再利用正负数加减法解答问题。【详解】（1）英英从 0 处向东行 1 格表示 +10m, 那么英英从 0 处向西行 4 格表示 -40m。(2) 才才从 0 处向北行 2 格表示 +20m，那么才才从 0 处向南行 3 格表示 -30m。 (3) 如果才才现在的位置是 -20m, 说明他向南行了 2 格。 (4) +50+(-90)=-40m, 即英英的位置。",
+      "knowledge_points": ["正负数在数轴上的表示", "正负数的意义及应用"],
+      "source": "21-22 六年级下·全国·单元测试",
+      "merged_id": "unit1_447",
+      "original_id": "unit1_fill_447"
+    },
+    {
+      "id": "unit1_choice_448",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "作图题",
+      "difficulty": "困难",
+      "question": "(21-22 六年级下·河南南阳·期中) 写出点A、B、C、D表示的数。![](images/94814b2864b9a528052fab544f5749f3239e3678a2f9c56fad59b53d158073f7.jpg)",
+      "options": [],
+      "answer": "-6; -3; -1; 4",
+      "solution": "【分析】0 是正负数的分界点, 由题干可知, 每个单位长度表示是 1 , 据此解答。此题考查的是在直线上表示数, 解答此题应注意每个单位长度表示多少。【详解】根据分析得：A点表示-6，B点表示-3，C点表示-1，D点表示4。![](images/7f1ec65a2fcb92ed20a0a4c3491df29e49d3028e77d8986559ef3aa0319345fb.jpg)",
+      "knowledge_points": ["正负数在数轴上的表示"],
+      "source": "21-22 六年级下·河南南阳·期中",
+      "merged_id": "unit1_448",
+      "original_id": "unit1_draw_448"
+    },
+    {
+      "id": "unit1_choice_449",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "(2021·浙江台州·小升初真题) 下面直线上的点分别表示 0 和 1/3, 请在括号上填入相应的分数。![](images/fa6366efb2d4a0da1c767721fbb46dd92fe7c93e93a689be8ae0123d3da7d1fd.jpg)",
+      "options": [],
+      "answer": "-2/3; 1又1/3",
+      "solution": "【分析】数轴上 0 的左边表示负数, 右边表示正数, 根据图示可知, 每小格是 1/3, 数出格数, 完成作图即可。解答本题的关键是知道每一小格代表多少。【详解】0左边第2格表示-2/3，0右边第4格表示1又1/3。如下图：![](images/28f3b09c4249c05095aafd3c2210715cd82a1f60b3a7b2a69a18b1fff213e0bd.jpg)",
+      "knowledge_points": ["认识一个整体的几分之几及应用", "正负数在数轴上的表示", "正负数的大小比较", "分数的意义"],
+      "source": "2021·浙江台州·小升初真题",
+      "merged_id": "unit1_449",
+      "original_id": "unit1_fill_449"
+    },
+    {
+      "id": "unit1_choice_450",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "（20-21 六年级下·辽宁盘锦·期末）规定海平面的海拔高度为 0m，高于海平面的为正，低于海平面的为负。珠穆朗玛峰的海拔高度为 __m，吐鲁番盆地的海拔高度为 __m。![](images/3726a7b01e869660ebdf075eee250a8683641e4f9529903d13aa56cf5a903f88.jpg)",
+      "options": [],
+      "answer": "+8848.86 ;-155",
+      "solution": "【分析】用正负数表示意义相反的两种量: 一种记作正, 则和它意义相反的就记作负。根据'高于海平面的为正, 低于海平面的为负', 直接得出结论即可。此题主要考查正负数的意义, 正数与负数表示意义相反的两种量, 看清规定哪一个为正, 则和它意义相反的就为负。【详解】根据分析可知:规定海平面的海拔高度为 0m, 高于海平面的为正, 低于海平面的为负。珠穆朗玛峰的海拔高度为: +8848.86m, 吐鲁番盆地的海拔高度为：-155m。",
+      "knowledge_points": ["正负数的意义及应用"],
+      "source": "20-21 六年级下·辽宁盘锦·期末",
+      "merged_id": "unit1_450",
+      "original_id": "unit1_fill_450"
+    },
+    {
+      "id": "unit1_choice_451",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "(22-23 六年级下·河北保定·阶段练习) 对号入座。23、460、+12、-0.57、-1/4、2/9、-1.5、0、+2.5、-2/5、+11/2、30、-0.8、+2/3 ![](images/0dd111874c92f46dbb31953e1e05257f50cfa93a483b98eb2f51e39253419722.jpg)",
+      "options": [],
+      "answer": "460、+12、2/9、 +2.5、+11/2、30、 +2/3 ;-23、-0.57、-1/4、-1.5、-2/5、-0.8。",
+      "solution": "【分析】根据正负数的意义，数的前面有'+'号或者什么符号都没有的，是正数；数的前面有'- '号的，就是负数，0既不是正数也不是负数，据此对每个数进行判断即可。【详解】由分析可得：正数有: 460, +12, 2/9, +2.5, +11/2, 30, +2/3;负数有：-23、-0.57、-1/4、-1.5、-2/5、-0.8。",
+      "knowledge_points": ["正负数的概念及辨认"],
+      "source": "22-23 六年级下·河北保定·阶段练习",
+      "merged_id": "unit1_451",
+      "original_id": "unit1_fill_451"
+    },
+    {
+      "id": "unit1_choice_452",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "困难",
+      "question": "(20-21 六年级下·全国·单元测试) 如图, 点 A 表示的数可能是 -1/3。( )![](images/c4e282313f5eb9122314f9772df883e35452ef0d67f2d8eaa2fd58e3b69517f0.jpg)",
+      "options": [
+        "A. 正确",
+        "B. 错误"
+      ],
+      "answer": "B",
+      "solution": "【分析】首先根据数轴上A点的位置确定A的取值范围，再判断即可。本题考查的是数轴的特点, 能根据数轴的特点确定出 A 的取值范围是解答此题的关键。【详解】在数轴上，A 在 0 和 -1 之间，所以 A 是大于 -1 而小于 0 的负数，把 0 到 -1 之间大约平均分为 3 份，A 在靠近 -1 处，即可能为 -2/3，所以原题表述错误。故答案为: ×。",
+      "knowledge_points": ["正负数在数轴上的表示"],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_452",
+      "original_id": "unit1_judge_452"
+    },
+    {
+      "id": "unit1_choice_453",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "计算题",
+      "difficulty": "困难",
+      "question": "(19-20 六年级下·河北·单元测试) 解不等式组：{2x+3≤3; x-(4x-1)/3<1}, 并在数轴上把解集表示出来。![](images/cfcd8268cf82b815707a6f33e278cd9224488dca1b1aff0438c65f505850aab4.jpg)",
+      "options": [],
+      "answer": "-2<x≤0 ![](images/bd6af762b2d7b8816cad3b79916b7696e523d8febbf5952531b2d0eec65d3950.jpg)",
+      "solution": "【分析】在解方程组时，先对每个不等式进行求解，然后求出公共解，并把解在数轴上表示出来即可。此题考查了方程不等式组的解法，以及在数轴上表示方程组解的方法。【详解】{2x+3≤3; x-(4x-1)/3<1}，由2x+3≤3，得 2x+3-3≤3-3，2x≤0，x≤0;......①由x-(4x-1)/3<1，得 3x-4x+1<3，即4x-3x-1+3>0，x+2>0，所以x>-2;......②由①②得:-2<x≤0。在数轴上表示为：![](images/ed1d4640654a1869559f343134016e680b77e86b29ef696e0b83e73b05a55b02.jpg)",
+      "knowledge_points": ["不等式的认识及解不等式", "正负数在数轴上的表示"],
+      "source": "19-20 六年级下·河北·单元测试",
+      "merged_id": "unit1_453",
+      "original_id": "unit1_calc_453"
+    },
+    {
+      "id": "unit1_choice_454",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "连线题",
+      "difficulty": "容易",
+      "question": "（19-20六年级下·全国·单元测试）把条件和合适的温度连起来。人的正常体温、水沸腾的温度、冰箱冷冻室的温度、水结冰时的温度 对应 -15°C、0°C、100°C、37°C",
+      "options": [],
+      "answer": "见详解 ![](images/629c65d18c1d396f63d17a3e93e7ef838faba8c10cd8582d5ad648b583a02d59.jpg)",
+      "solution": "【分析】联系生活实际连线即可。做这种题目, 要对一些常见物体的温度要有一定的估测能力, 要根据生活常识去做出选择。【详解】人的正常体温一般是37°C, 水沸腾的温度是100°C, 冰箱冷冻室的温度是-15°C, 水结冰时的温度是0°C。据此连线如下：![](images/9d6d924daecd3b096fb1e819365193d4b5c2115b185ed3ac6fd545b53cb58045.jpg)",
+      "knowledge_points": ["温度的认识及比较"],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_454",
+      "original_id": "unit1_match_454"
+    },
+    {
+      "id": "unit1_choice_455",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "作图题",
+      "difficulty": "困难",
+      "question": "（19-20六年级下·全国·单元测试）在直线上表示下面各数。-6、5、-3.5、-5.5、+2、-3/2、4、5/2、-9 ![](images/a4378235d5eeb3fc580c8e2338f651acccf688e73bbfe5f1c21a40d54b768f6b.jpg)",
+      "options": [],
+      "answer": "见详解 ![](images/c280983883a800b9f12f3ea61929f72f2a25d362e9401ae541f6102d687ff73b.jpg)",
+      "solution": "【分析】数轴上 0 的左边是负数, 0 的右边是正数, -6、5、+2、4、-9 直接表示在整数处, -3.5 在 -3 和 -4 中间, -5.5 在 -5 和 -6 的中间, -3/2 在 -1 和 -2 中间, 5/2 在 2 和 3 中间。本题主要考查数字在数轴上的表示，注意数轴从左到右，数字越来越大。【详解】![](images/6959372790b885a7e642b9af6470e05895fa34e2258c40eee1ed3b1947d4707d.jpg)",
+      "knowledge_points": ["正负数在数轴上的表示"],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_455",
+      "original_id": "unit1_draw_455"
+    },
+    {
+      "id": "unit1_choice_456",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "作图题",
+      "difficulty": "容易",
+      "question": "（19-20六年级下·全国·单元测试）在直线上表示下列各数。5 -1 -2.5 2 -4 ![](images/fa8eb6c7937c05f8b1facc8d8024afd93e90af6762fc862325efa4563b4091f0.jpg)",
+      "options": [],
+      "answer": "见详解 ![](images/662fa1c9fa9f03a07b76bc5b3adb41858a733c5a212147693f9251357a8919e4.jpg)",
+      "solution": "【分析】正数都在 0 的右边，负数都在 0 的左边，数轴上每个单位长度表示 1，由此确定各个数字在直线上的位置即可。考查了对数轴和正负数的认识，基础题。【详解】由分析可得:![](images/b36b86f05a6d391efaa10dc5f7201bab3566f337d05b1866ad6b2caee20c0c42.jpg)",
+      "knowledge_points": ["正负数在数轴上的表示"],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_456",
+      "original_id": "unit1_draw_456"
+    },
+    {
+      "id": "unit1_choice_457",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "作图题",
+      "difficulty": "容易",
+      "question": "(19-20 六年级下·全国·单元测试) 数轴上表示下面各数。4、3、-1又1/2、-2.5、+7/2、7.5 ![](images/c16ca85099e73d0b93a9305ca9203d15cbb0e9a0feff1d2d1e9922ff3e3760b8.jpg)",
+      "options": [],
+      "answer": "见详解 ![](images/50dfc16ee275a32eaf8711005ebcee34fefb29aaa7507c88e51b3ef64c897cc7.jpg)",
+      "solution": "【分析】数轴是规定了原点（0 点）、方向和单位长度的直线，在数轴上原点（0 点）的左边是负数，从原点（0 点）向左分别是 -1、-2、-3、-4、-5、-6、…右边是正数，从原点（0 点）向右分别是 +1、+2、+3、+4、+5、+6、…由此解答。【详解】-4在零的左边第4个点处；3在零的右边第3个点处；-1又1/2 在 -1 和 -2 的中间；-2.5 在 -2 和 -3 的中间；+7/2 在 3 和 4 的中间；7.5 在 7 和 8 的中间。数轴是规定了原点（0 点）、方向和单位长度的直线，在数轴上从左到右的顺序就是数从小到大的顺序。![](images/18930e68b82f7622dc75a80c4edfc3811e17bcd469fbcb03316a68b6638dec03.jpg)",
+      "knowledge_points": ["正负数在数轴上的表示"],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_457",
+      "original_id": "unit1_draw_457"
+    },
+    {
+      "id": "unit1_choice_458",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "作图题",
+      "difficulty": "容易",
+      "question": "（20-21 六年级下·全国·单元测试）在如图直线上分别表示出2又1/3和-1.5这两个数的位置。![](images/e1694a72c0213fabbc301ab0762c02fa4f749da655998d1f4fb845cd128729cb.jpg)",
+      "options": [],
+      "answer": "见详解",
+      "solution": "【分析】2又1/3在 2 和 3 之间，将 2 和 3 之间的一段平均分成 3 段，第一段处是2又1/3; -1.5 在 -1 和 -2 的中间。负数在 0 的左侧，正数在 0 的右侧。在数轴上越靠右边的数越大，越靠左边的数越小。【详解】作图如下:![](images/32744c41e90eccb8651af7271de54dcf5610f703052ffb5ffcf0f19a4d16ac50.jpg)",
+      "knowledge_points": ["正负数在数轴上的表示"],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_458",
+      "original_id": "unit1_draw_458"
+    },
+    {
+      "id": "unit1_choice_459",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "作图题",
+      "difficulty": "容易",
+      "question": "(20-21 六年级下·全国·单元测试) 在数轴上表示下列各数。![](images/89647742165586bfe7f0732f9b11a3e07b33a096ffb903db6db60d97fa2e68e0.jpg)",
+      "options": [],
+      "answer": "见详解",
+      "solution": "【分析】在数轴上，从左到右的顺序就是数从小到大的顺序；正数都在 0 的右边，负数都在 0 的左边，按照大小顺序在数轴上表示出来即可。此题考查在数轴上表示数的方法。【详解】由分析可得：![](images/ca9f69c7afbab18d9b16c0fd6e96be0a82155fd6e9a44f75299d80ae01e7d7c5.jpg)",
+      "knowledge_points": ["正负数在数轴上的表示"],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_459",
+      "original_id": "unit1_draw_459"
+    },
+    {
+      "id": "unit1_choice_460",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "作图题",
+      "difficulty": "容易",
+      "question": "(2020·山东泰安·小升初真题) 在数轴上表示下列各数。-2，0.5，5/2，3 ![](images/88cfc2b75493ce5c3736a709ebfc0cd978c5cb009bab6b0e9d6b1d0112297b02.jpg)",
+      "options": [],
+      "answer": "见详解",
+      "solution": "【分析】数轴是规定了原点（0 点）、方向和单位长度的一条直线。原点的左边是负数，从原点向左的每个单位长度分别是 -1、-2、-3...；右边是正数，从原点向右每个单位长度分别是 1、2、3...，据此解答即可。本题是考查数轴的认识。数轴是规定了原点（0点）、方向和单位长度的一条直线。【详解】![](images/4957cf066f5623ee734ecd48871e962f624f522ae5a61ce809ea960da0d51326.jpg)",
+      "knowledge_points": ["正负数在数轴上的表示"],
+      "source": "2020·山东泰安·小升初真题",
+      "merged_id": "unit1_460",
+      "original_id": "unit1_draw_460"
+    },
+    {
+      "id": "unit1_choice_461",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "作图题",
+      "difficulty": "容易",
+      "question": "(20-21 六年级下·全国·期末) 在下面直线上, 画出比-3大的数所在区域。![](images/8f2d700123f05ce17115a59ae3f7d68b7f26d7e61369dabb3b7e8f387849fa82.jpg)",
+      "options": [],
+      "answer": "见详解 ![](images/d118f2af3003d5460ba983235459391bd9ac04eb158ddbbbece2691756814a96.jpg)",
+      "solution": "【分析】根据数轴上各数的特点：右边的数总比左边的数大，据此解答。考查了数轴的认识, 解答此题要明确: 数轴上的点表示的数, 右边的数总比左边的数大。【详解】在数轴上，比 -3 大的数都在 -3 的右边。因此画图如下：![](images/e6e881a7ab5683a47b3b7eb8551fcf71597ab6fd926ecceef29d60802b64b112.jpg)",
+      "knowledge_points": ["正负数在数轴上的表示"],
+      "source": "20-21 六年级下·全国·期末",
+      "merged_id": "unit1_461",
+      "original_id": "unit1_draw_461"
+    },
+    {
+      "id": "unit1_choice_462",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "作图题",
+      "difficulty": "困难",
+      "question": "(21-22 六年级下·全国·单元测试) 在直线上表示下列各数。-5、-3.5、2、+4、-2、+1 ![](images/cca47151dcdf862d4bd22902b8decf6dd100d864c3a293c55fd23720725036d2.jpg)",
+      "options": [],
+      "answer": "见详解",
+      "solution": "【分析】用数轴表示数：根据数字在对应刻度上描点表示；负数在 0 的左侧，正数在 0 的右侧。本题考查了对数轴以及负数的认识, 理解在数轴上右边的数大于左边的数。【详解】![](images/8537ea8d535e9091290a52900431861fb2ea474bf7229f76eeeed3b5f24e82fc.jpg)",
+      "knowledge_points": ["正负数在数轴上的表示"],
+      "source": "21-22 六年级下·全国·单元测试",
+      "merged_id": "unit1_462",
+      "original_id": "unit1_draw_462"
+    },
+    {
+      "id": "unit1_choice_463",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "作图题",
+      "difficulty": "容易",
+      "question": "(21-22 六年级下·贵州黔西·期末) 在直线上表示下列各数。![](images/96f0fe6170d5ee71a9e1226924b9b3254f32b5d43337ec38e5d80ab409449935.jpg)",
+      "options": [],
+      "answer": "见详解",
+      "solution": "【分析】原点左边的为负数，原点右边的为正数，据此作图。在数轴上的数从左到右依次变大。【详解】![](images/6cd57025db1cf61aecb7bea0b61fcf21cf8d0038adbc8fa90c43a4f4c72e8abf.jpg)",
+      "knowledge_points": ["正负数在数轴上的表示"],
+      "source": "21-22 六年级下·贵州黔西·期末",
+      "merged_id": "unit1_463",
+      "original_id": "unit1_draw_463"
+    },
+    {
+      "id": "unit1_choice_464",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "（19-20 六年级下·全国·单元测试）下面的表格中列出了国外几个城市与北京的时差。（与北京时差相比，早的记为正数，晚的记为负数）(1) 北京时间上午 9:30，莫斯科是什么时间？(2) 悉尼时间晚上 10:00, 北京是什么时间?",
+      "options": [],
+      "answer": "（1）凌晨4：30；（2）晚上8：00",
+      "solution": "【分析】根据题意，以北京时间为标准，早的记为正数，晚的记为负数，列式计算即可。用北京时间 + 时差 = 所求的当地时间; 北京时间 = 当地时间 - 时差。【详解】(1) 9:30-5=4:30 答: 北京时间上午 9:30, 莫斯科是凌晨 4:30。(2) 晚上10:00=22:00；22:00-2=20:00, 即晚上 8:00 答: 悉尼时间晚上 10:00, 北京是晚上 8:00。",
+      "knowledge_points": ["正负数的意义及应用"],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_464",
+      "original_id": "unit1_solve_464"
+    },
+    {
+      "id": "unit1_choice_465",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "（19-20六年级下·全国·单元测试）下面是优优家1月份收入和支出的情况。1月3日全家旅游花费1600元；1月12日爷爷生日花费800元；1月15日父母领取工资9800元；1月18日妈妈收到稿费900元；1月29日缴纳水电费240元；1月份全家生活费2350元。(1) 用正、负数填写下表。(2) 优优家1月份实际剩余多少元？",
+      "options": [],
+      "answer": "(1) -1600; -800;+9800; +900;-240;-2350 (2) 5710元",
+      "solution": "【分析】（1）收入的钱数记作正，支出的钱数记作负，由此填表；(2) 收入与支出相互抵扣，剩余的即为所求。考查了正负数在实际生活中的应用。正负数可以表示相反意义的量；理解题意的时候正负可以理解为加减。【详解】(1)全家旅游：-1600；爷爷生日：-800;父母工资：+9800；稿费：+900；水电费：-240；生活费：-2350 (2) (-1600)+(-800)+9800+900+(-240)+(-2350)=5710(元)答: 优优家 1 月份实际剩余 5710 元。",
+      "knowledge_points": ["正负数的意义及应用"],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_465",
+      "original_id": "unit1_solve_465"
+    },
+    {
+      "id": "unit1_choice_466",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "(19-20 六年级下·全国·单元测试) 下图中每格表示10m, 棒棒刚开始的位置在 0 处。![](images/89b8b0a3a5aa4944ce4b9cfdff06da298c6b8f8d1c9d246a0231e4a0d7bed919.jpg) (1) 棒棒从 0 点向东走 40 米, 表示 +40m, 那么他从 0 点向西走 60m, 表示为 ( ) m。 (2) 如果棒棒的位置是 -20m, 说明他是向 ( ) 走了 ( ) m。 (3) 如果棒棒先向东走 30m, 又向西走 10m, 这时棒棒的位置表示为 ( ) m。",
+      "options": [],
+      "answer": "(1) -60; (2) 西; 20; (3) +20",
+      "solution": "【分析】（1）、（2）用负数可以表示具有相反意义的量，向东走用正数表示，那么向西走就用负数表示。(3) 向东走30m为+30m, 向西走10m为-10m, 30-10=20(m)即为棒棒现在的位置。正负数表示一组相反意义的量, 此题考查负数在实际生活中的应用。【详解】（1）棒棒从 0 点向东走 40 米，表示 +40m，那么他从 0 点向西走 60m，表示为 -60m；(2) 如果棒棒的位置是 -20m, 说明他是向西走了 20m; (3) 30+(-10)=20(m)",
+      "knowledge_points": ["正负数的意义及应用"],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_466",
+      "original_id": "unit1_solve_466"
+    },{
+      "id": "unit1_choice_467",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "（19-20六年级下·全国·单元测试）下表是某一天四个城市的最高气温。(1) 你会在温度计上表示出这些温度吗?![](images/8dcc01491dc440b853500ccb6b3e35e10eb9790c7f1461d4debfec0326f4d42a.jpg) (2) 把这四个城市的最高气温按从高到低的顺序排列起来。",
+      "options": [],
+      "answer": "(1) 见详解; (2) 15°C > 5°C > 0°C > -10°C",
+      "solution": "【分析】（1）以0刻度为标准，5、15在0刻度上方，-10在0刻度下方；(2) 温度是零下的, 数字越大, 温度越低, 0°C 比任何一个零下几摄氏度都高; 温度是零上的, 数字越大, 温度越高, 0°C 比任何一个零上几摄氏度都低; 据此进行比较即可。【详解】(1) 作图如下:![](images/3f00780baec19fd7e5d250d7bf2ad20b22d0833f83e4b96f6a5bad4e466b0d92.jpg)(2) 15°C > 5°C > 0°C > -10°C",
+      "knowledge_points": ["温度的认识及比较"],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_467",
+      "original_id": "unit1_solve_467"
+    },
+    {
+      "id": "unit1_choice_468",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "作图题",
+      "difficulty": "容易",
+      "question": "（19-20 六年级下·全国·单元测试）在下面的数轴上表示出各数。并在横线上按照从大到小的顺序排列。-4、0、-1 5、-3/2、1.5 ![](images/11e8acde6503653b1c497177aee06de4265899310ac0d292037dcba5cb539174.jpg)",
+      "options": [],
+      "answer": "![](images/ab413c9590df13c01ce899e959afba37c4a7b8b3d8b5a14a38cb5c34efaf88b4.jpg) 5 > 1.5 > 0 > -1 > -3/2 > -4",
+      "solution": "【分析】数轴是规定了 0 点、方向和单位长度的直线，在零点左边所表示的数都是负数，在零点右边所表示的数都是正数。数轴上，从右到左的顺序就是数从大到小的顺序。0 是正数和负数的分界点，正数 > 0 > 负数，负号后面的数越大，这个数就越小。理解正负数的意义，认识数轴就能解决此类问题。【详解】0 在 1 的左边第一个点，-1 在 0 的左边第一个点，-3/2 在 -1 和 -2 的中间，1.5 在 1 和 2 的中间。-4 在 0 的左边第四个点，5 在 0 的右边第五个点。![](images/bab7ec5ee9065dd28b7f6b1d5fb300306483fdd525ca2f2195f44029d6c595bd.jpg)观察上面的数轴，从右到左的顺序就是数从大到小的顺序。所以 5 > 1.5 > 0 > -1 > -3/2 > -4",
+      "knowledge_points": ["正负数在数轴上的表示", "正负数的大小比较"],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_468",
+      "original_id": "unit1_draw_468"
+    },
+    {
+      "id": "unit1_choice_469",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "（19-20六年级下·全国·单元测试）读文字，用正数、负数填表，并说明是亏损还是盈利，以及亏损或者盈利多少钱。佳客隆超市每个月的营业成本是12万元。今年上半年月收入分别是：1月份14万元，2月份15万元，3月份11万元，4月份10万元，5月份16万元，6月份11万元。(1) 按照例子填空。(2) 佳客隆超市今年上半年是( ) (填盈利或亏损)( )（填数字）万元。",
+      "options": [],
+      "answer": "+3; -1; -2; +4; -1 ;盈利;5",
+      "solution": "【分析】营业额比营业成本12万元高是盈利，记为正；营业额比营业成本12万元低是亏损，记为负，则可以求出每个月的计数，再根据6个月的收入来判断佳客隆超市今年上半年是亏损还是盈利即可。本题考查正、负数的意义, 正数与负数表示相反意义的量, 看清题目是规定哪个量为正, 和它意义相反的量就记为负, 本题可以直接根据盈亏计数相加减来判断盈亏。【详解】（1）盈利记为正，亏损记为负，那么佳客隆超市今年上半年的营业盈亏为：二月份: 15 - 12 = 3, 是盈利 3 万元, 记为 +3;三月份：12 - 11 = 1，是亏损1万元，记为-1；四月份：12 - 10 = 2，是亏损2万元，记为-2；五月份：16 - 12 = 4，是盈利4万元，记为 +4；六月份：12 - 11 = 1，是亏损1万元，记为-1。(2）根据营业盈亏计数可知：2 + 3 - 1 - 2 + 4 - 1 = 5 (万元)所以佳客隆超市今年上半年是盈利，盈利5万元。故答案为：(1) +3; -1; -2; +4; -1 (2) 盈利; 5。",
+      "knowledge_points": ["正负数的意义及应用"],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_469",
+      "original_id": "unit1_solve_469"
+    },
+    {
+      "id": "unit1_choice_470",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "（19-20六年级下·全国·单元测试）一辆公共汽车从起点开出后，途中经过6个停靠站，最后到达终点站。下表记录了这辆公共汽车全程载客数量的变化情况。(1) 中间第 5 站上、下车人数的情况各是多少？(2) 在中间的哪个站没有人上车？哪个站没有人下车？",
+      "options": [],
+      "answer": "（1）中间第5站上车6人，下车9人 (2) 中间第 6 站没有人上车, 中间第 3 站没有人下车",
+      "solution": "【分析】用正数表示上车的人数，用负数表示下车的人数，无上车或无下车的人数用 0 表示。本题考查正、负数的意义, 正数与负数表示相反意义的量, 看清题目是规定哪个量为正, 和它意义相反的量就记为负。【详解】（1）用正数表示上车的人数，用负数表示下车的人数，则中间第5站上车6人，下车9人。答: 中间第 5 站上车 6 人, 下车 9 人。(2) 观察表格发现中间第 3 站无人下车, 中间第 6 站无人上车。答: 中间第 6 站没有人上车, 中间第 3 站没有人下车。",
+      "knowledge_points": ["正负数的意义及应用"],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_470",
+      "original_id": "unit1_solve_470"
+    },
+    {
+      "id": "unit1_choice_471",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "（19-20六年级下·全国·单元测试）小张的银行存折上有7笔账目往来，如下表：到最后，小张存折上的金额是增加了还是减少了？增加或减少多少元？",
+      "options": [],
+      "answer": "没增加也没减少或不变",
+      "solution": "【分析】银行存折上'存入'记为正，'支出'记为负。分别计算出存入的总钱数和支出的总钱数，再作比较。即可解答本题。解答此题关键是分别计算出存入的总钱数和支出的总钱数。【详解】存入的总钱数：300 + 100 + 150 = 550 （元） 支出的总钱数：200 + 100 + 200 + 50 = 550 （元） 550 = 550 答: 小张存折上的金额既没增加也没减少。",
+      "knowledge_points": ["利用正负数解决实际问题"],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_471",
+      "original_id": "unit1_solve_471"
+    },
+    {
+      "id": "unit1_choice_472",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "（19-20六年级下·全国·单元测试）下表是某市各季度的平均气温。(1) 在温度计上标出所给的平均气温。![](images/2dc23bd0868069172c4fe8fb5b462253.png) (2) 把平均气温按从高到低排序为:（）>（）>（）>（）",
+      "options": [],
+      "answer": "（1）如图所示![](images/3442ea9e1fd90c18300360870382f5dce73f8eb0e7f1383fb0be5f187ba56f06.jpg) (2) 20; 15; -5; -10",
+      "solution": "【分析】（1）根据表格中平均气温在温度计上表示这些温度即可；(2) 正数 > 0 > 负数, 负数比较大小时, 绝对值大的负数越小, 据此判断即可。本题考查正、负数的认识和大小比较, 解答本题的关键是掌握正、负数大小比较的方法。【详解】（1）如图所示。![](images/3442ea9e1fd90c18300360870382f5dce73f8eb0e7f1383fb0be5f187ba56f06.jpg) (2) 根据正、负数的大小比较得: (20) > (15)>(-5) > (-10) 故答案为：（1）如图所示：![](images/3442ea9e1fd90c18300360870382f5dce73f8eb0e7f1383fb0be5f187ba56f06.jpg) (2) 20; 15; -5; -10 。",
+      "knowledge_points": ["温度的认识及比较", "正负数的大小比较"],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_472",
+      "original_id": "unit1_solve_472"
+    },
+    {
+      "id": "unit1_choice_473",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "（19-20六年级下·全国·单元测试）下面是林林家二月份收支情况。2月8日：妈妈领工资1600元 2月10日：交水电费180元 2月12日：林林买衣服用去60元 2月15日：爸爸领工资2200元 2月18日：去公园游玩用去50元 2月20日：妈妈买衣服用去150元 2月22日：爸爸买书报杂志用去130元 2月28日：本月伙食费合计用去820元 (1) 请你用正负数的知识填写下表。(2) 尝试计算林林家 2 月份的结余。",
+      "options": [],
+      "answer": "(1)1600,180,60,2200,50,150,130,820 (2) 2410元",
+      "solution": "【分析】（1）正负数来表示具有意义相反的两种量，'收入'记为正，则'支出'就记为负；(2) 分别计算出收入的总钱数和支出的总钱数, 再作比较即可解答本题。此题主要考查正负数的意义, 正数与负数表示意义相反的两种量, 看清规定哪一个为正则和它意义相反的就为负。【详解】(1) 填表如下：2月8日对应+1600；2月10日对应-180；2月12日对应-60；2月15日对应+2200；2月18日对应-50；2月20日对应-150；2月22日对应-130；2月28日对应-820。(2) 收入的总钱数: 1600 + 2200 = 3800 (元)支出的总钱数: 180 + 60 + 50 + 150 + 130 + 820 = 1390(元) 结余: 3800 - 1390 = 2410 (元)答：林林家2月份的结余2410元。",
+      "knowledge_points": ["利用正负数解决实际问题", "正负数的概念及辨认"],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_473",
+      "original_id": "unit1_solve_473"
+    },
+    {
+      "id": "unit1_choice_474",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "作图题",
+      "difficulty": "容易",
+      "question": "(19-20 六年级下·全国·单元测试) 在数轴上表示下列各数。-3.5、-9、3、-4、5、-5同学们做游戏，以篮球架为起点。![](images/33ad9fad207d8dde7fcd4393bd37e93a7855798d22cc3973ed4337bd3c9e4a30.jpg) (1) 将数轴上的数补充完整。(2) 文文向西 6 米记作 -6 米, 红红向 ( ) 走 4 米记作 +4 米。(3) 强强的位置是 -5，用 △ 标出他的位置。(4) 小刚先向东走 4 米又向西走 2 米, 用 ○ 标出她的最终位置。",
+      "options": [],
+      "answer": "(1) (3) (4) 作图如下: ![](images/22262f31a80f3b0d22689aff2ddd70d11d54e11b4ad7a1120f690340c38c6012.jpg) (2) 东",
+      "solution": "【分析】（1）数轴是规定了原点（0 点）、方向和单位长度的直线，在数轴上原点（0 点）的左边是负数，从原点(0 点) 向左分别是 -1、-2、-3、-4、-5、-6、…右边是正数，从原点（0 点）向右分别是 +1、+2、+3、+4、+5、+6、… 由此作图；(2) 正负数表示具有意义相反的两种量, 如果向西记作负数, 那么向东走记作正数; (3) -5 在 0 的左边第五个点处, 此处标 △ ; (4) 从数轴上可以看出小刚先向东走 4 米到 4 的位置, 又向西走 2 米, 最终在 2 的位置, 此处标 ○ 。此题考查负数的认识及在数轴上表示数, 注意:所有的正数在 0 的右边, 所有的负数在 0 的左边。【详解】（1）在数轴上原点（0 点）的左边是负数，右边是正数，-3.5 在 -3 和 -4 的中间。(2) 文文向西 6 米记作 -6 米, 那么向东走记作正数, 红红向东走 4 米记作 +4 米。(1) (3) (4) 作图如下: ![](images/88c3e0abb324603c3eeacd3ec53abc87d2caa9dd80c0d88c69329df4c7e96af3.jpg)",
+      "knowledge_points": ["正负数在数轴上的表示", "正负数的意义及应用"],
+      "source": "19-20 六年级下·全国·单元测试",
+      "merged_id": "unit1_474",
+      "original_id": "unit1_draw_474"
+    },
+    {
+      "id": "unit1_choice_475",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "（19-20六年级下·全国·单元测试）规定向右为正，在直线上表示下列各数，再比较大小。﹣6○5.5 ﹣4○﹣3 0○﹣1 ﹣4.5○0 1/2○1.5 ﹣1.5○﹣2.5",
+      "options": [],
+      "answer": "< ; < ; > ; < ; < ; >",
+      "solution": "【分析】直线的每一小段表示 1 , 由此即可在直线上表示这些数; 在直线上右边的数大于左边的数, 由此即可进行比较大小。本题考查了对数轴以及负数的认识, 关键是要理解在数轴上右边的数大于左边的数。【详解】由此可知: -6 < 5.5; -4 < -3; 0 > -1; -4.5 < 0; 1/2 <1.5; -1.5> -2.5",
+      "knowledge_points": ["正负数在数轴上的表示", "正负数的大小比较"],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_475",
+      "original_id": "unit1_judge_475"
+    },
+    {
+      "id": "unit1_choice_476",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "（19-20六年级下·全国·单元测试）下面是洋洋家三月份收支情况。3月8日：妈妈领工资3000元 3月10日：交水电费、管理费210元 3月12日：洋洋买衣服用去360元 3月15日：爸爸领工资4200元 3月18日：去公园游玩用去280元 3月20日：妈妈买衣服用去270元 3月22日：我买书报杂志用去130元 3月1~31日：本月购买食品用去1300元 (1) 请你用正负数的知识填写上表。(2) 尝试计算洋洋家三月份的结余。",
+      "options": [],
+      "answer": "(1) -210; -360; +4200; -280; -270; -130; -1300 (2) 4650元",
+      "solution": "【分析】（1）收入记为正，支出记为负，据此填空；(2) 将所有收入加起来，减去所有支出即可。本题考查了正负数的意义及应用, 正负数可以表示相反意义的量。【详解】(1) 填空如下: 3月8日对应+3000；3月10日对应-210；3月12日对应-360；3月15日对应+4200；3月18日对应-280；3月20日对应-270；3月22日对应-130；3月31日对应-1300。(2) (3000 + 4200) - (210 + 360 + 280 + 270 + 130 + 1300) = 7200 - 2550 = 4650 (元) 答：洋洋家三月份结余 4650 元。",
+      "knowledge_points": ["正负数的意义及应用", "经济问题"],
+      "source": "19-20六年级下·全国·单元测试",
+      "merged_id": "unit1_476",
+      "original_id": "unit1_solve_476"
+    },
+    {
+      "id": "unit1_choice_477",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "（17-18六年级下·全国·单元测试）下面是国外几个城市与北京的时差。当北京时间为2016年7月1日10时时，东京时间为2016年7月1日11时。(1)如果北京时间是 2016 年 9 月 1 日 12 时, 那么纽约是什么时间? (2)如果巴黎时间是 2016 年 9 月 1 日 20 时, 那么北京是什么时间?",
+      "options": [],
+      "answer": "(1) 2016年8月31日23时 (2) 2016年9月2日3时",
+      "solution": "【详解】略",
+      "knowledge_points": ["日期和时间的推算", "负数的认识"],
+      "source": "17-18六年级下·全国·单元测试",
+      "merged_id": "unit1_477",
+      "original_id": "unit1_solve_477"
+    },
+    {
+      "id": "unit1_choice_478",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "（2014 六年级·全国·课后作业）今年一月份某日几个城市的平均气温如下表：请你将这些城市的气温按从高到低的顺序排列。",
+      "options": [],
+      "answer": "12°C > 10°C > 5°C > -5°C > -18°C",
+      "solution": "【分析】根据正数大于所有的负数, 先把几个正数按照正数的大小比较方法比较, 比较负数时, 先别看 '- ' 号, 数字大的添上 '-' 号反而小, 数字小的添上 '-' 号反而大。此题考查正负数的大小比较：正数大于 0、大于所有的负数；也可以利用数轴进行正负数的大小比较，在数轴上从左到右的顺序就是数从小到大的顺序。【详解】按从高到低的顺序排列为: 12°C > 10°C > 5°C > -5°C > -18°C 。",
+      "knowledge_points": ["正负数的大小比较", "温度的认识及比较"],
+      "source": "2014 六年级·全国·课后作业",
+      "merged_id": "unit1_478",
+      "original_id": "unit1_solve_478"
+    },
+    {
+      "id": "unit1_choice_479",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "(2014 六年级·全国·课后作业) 同学们做游戏, 以篮球架为起点.文文向西走 6 米记作 -6 米, 红红向 ( ) 走 4 米记作+4米.",
+      "options": [],
+      "answer": "东",
+      "solution": "【详解】略",
+      "knowledge_points": ["数轴的认识", "负数的认识"],
+      "source": "2014 六年级·全国·课后作业",
+      "merged_id": "unit1_479",
+      "original_id": "unit1_fill_479"
+    },
+    {
+      "id": "unit1_choice_480",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "困难",
+      "question": "(20-21 六年级下·全国·单元测试) 先读、写下面各数,再在直线上表示出这些数。-4.5读作（） +2.5读作（） -3读作（）正四写作（）负二分之五写作（）正零点五写作（）",
+      "options": [],
+      "answer": "负四点五；正二点五；负三 4; -5/2; +0.5",
+      "solution": "【分析】读正、负数时，先读正、负号，再读正、负号后面的数；写正、负数时，先写正、负号，再读正、负号后面的数读数和写数的时候一定要注意数前面的符号。【详解】-4.5读作：负四点五；+2.5 读作：正二点五；-3 读作：负三;正四写作：+4;负二分之五写作：-5/2;正零点五写作：+0.5;",
+      "knowledge_points": ["正负数的读法和写法"],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_480",
+      "original_id": "unit1_fill_480"
+    },
+    {
+      "id": "unit1_choice_481",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "（20-21 六年级下·全国·单元测试）下面表格记录了某月五个城市的平均气温。(1) ( ) 的平均气温最高, ( ) 的平均气温最低。(2) 上海和北京的平均气温相差多少摄氏度? 沈阳和哈尔滨的平均气温相差多少摄氏度?",
+      "options": [],
+      "answer": "(1) 广州; 哈尔滨 (2) 14°C; 8°C",
+      "solution": "【分析】（1）正数比负数大。正数中，18 > 5，因此广州的平均气温最高；负数中，比较负数的大小可以利用直线，距离0越远的负数越小。因此-27最小，即哈尔滨的平均气温最低。(2) 5°C 表示比 0°C 高 5°C, -9°C 表示比 0°C 低 9°C, 上海和北京的平均气温相差 5 + 9 = 14 (°C)。-27°C 和 -19°C 都是 0°C 以下, 所以沈阳和哈尔滨的平均气温相差 27 - 19 = 8 (°C)。本题涉及到正负数的大小比较及简单的正负数的运算。注意计算温差时要能够灵活展开运算技能。【详解】(1) -27 < -19 < -9 < 5 < 18，因此广州的平均气温最高，哈尔滨的平均气温最低；(2) 5 + 9 = 14 (°C) 27 - 19 = 8 (°C)所以上海和北京的平均气温相差 14 摄氏度；沈阳和哈尔滨的平均气温相差 8 摄氏度。",
+      "knowledge_points": ["利用正负数解决实际问题", "正负数的大小比较"],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_481",
+      "original_id": "unit1_solve_481"
+    },
+    {
+      "id": "unit1_choice_482",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "（20-21六年级下·全国·单元测试）一只蜗牛，从10米深的井底向上爬，第一天白天向上爬4米，晚上下滑2米；第二天白天向上爬3米，晚上下滑1米；第三天白天向上爬5米，晚上下滑3米…… (1) 如果向上爬的米数记作正数, 下滑的米数记作负数,请将蜗牛每天爬行的情况用正负数在下表中表示出来。(2) 如果蜗牛每天白天向上爬 4 米, 晚上下滑 2 米, 那么这只蜗牛几天能爬出井口?",
+      "options": [],
+      "answer": "(1) +4; -2; +3; -1; +5; -3; (2) 4天",
+      "solution": "【分析】（1）向上爬记为正数，向下滑记为负数，据此填表即可；(2) 白天向上爬 4 米, 晚上下滑 2 米, 说明每天实际向上爬 4 - 2 = 2 米, 最后一天白天向上爬 4 米, 所以用井的深度减去最后一天白天向上爬的长度就是前几天一共爬的长度,这只蜗牛爬出井口需要的天数 = 前几天一共爬的长度 ÷ 2 + 1 天。本题主要考查正负数的意义，在解答第二题时要明确蜗牛每天实际向上爬的米数以及前几天一共爬的米数，一共爬的天数要加 1。【详解】(1) 如图: 第一天白天对应+4米；第一天晚上对应-2米；第二天白天对应+3米；第二天晚上对应-1米；第三天白天对应+5米；第三天晚上对应-3米。(2) 这只蜗牛每天实际向上爬的距离: 4 - 2 = 2 (米) (10 - 4) ÷ 2 + 1 = 6 ÷ 2 + 1 = 3 + 1 = 4 (天) 答: 这只蜗牛 4 天能爬出井口。",
+      "knowledge_points": ["正负数的意义及应用"],
+      "source": "20-21六年级下·全国·单元测试",
+      "merged_id": "unit1_482",
+      "original_id": "unit1_solve_482"
+    },
+    {
+      "id": "unit1_choice_483",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "困难",
+      "question": "(20-21 六年级下·全国·课后作业) 下面是明智图书馆 5 个工作日借书和还书情况的统计记录表, 借出书 1 本记作 -1 , 读者还回书 1 本记作 +1 。(1) 图书馆 10 月 5 日借出________本书，还入________本书。(2) 这 5 天中, 哪天借出书的本数最少? 借出了多少本? (3) 这 5 天中, 哪天还入书的本数最多? 还入了多少本?",
+      "options": [],
+      "answer": "(1) 45; 138 (2) 10月5日，45本 (3) 10月5日，138本",
+      "solution": "【分析】负数与正数表示意义相反的量，本题中借出用'-'表示，还入用'+'表示。(1) 观察表格中 10 月 5 日 '-' 后面的数值以及 '+' 后面的数值, 即可得出借出多少本, 还入多少本; (2) 借出用 '-' 表示, 观察表格中所有带 '-' 的数值,不看 '-', 只比较数值的大小, 找出最小的数值并找出对应的天数即可;(3) 还入用 '+' 表示, 观察表格中所有带 '+' 的数值,不看 '+', 只比较数值的大小, 找出最大的数值并找出对应的天数即可。此题考查的是正、负数的意义与应用, 正、负数大小的比较。【详解】(1) 45; 138(2) 45 < 52 < 68 < 85 < 124, 所以借出 45 本书最少, 是 10 月 5 日。答: 这 5 天中, 10 月 5 日借出书的本书最少, 借出了 45 本。(3) 43 < 81 < 96 < 108 < 138 答: 这 5 天中, 10 月 5 日还入书的本书最多, 还入了 138 本。",
+      "knowledge_points": ["正负数的意义及应用"],
+      "source": "20-21 六年级下·全国·课后作业",
+      "merged_id": "unit1_483",
+      "original_id": "unit1_solve_483"
+    },
+    {
+      "id": "unit1_choice_484",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "(20-21 六年级下·全国·单元测试) A 点表示的数是 ( ) , B 点表示的数是 ( ) 。![](images/fd8ba7d3d6e65a2cf36588effdd7de6af861d811eecea3dc1c80f25c5095634b.jpg)",
+      "options": [],
+      "answer": "5；-2",
+      "solution": "【分析】这个数轴上，一个格表示一个单位，A 在 0 右边第 5 个格子，表示 5，B 在 0 左边第 2 个格子，表示 -2。关键是掌握正负数在数轴上的表示方法, 正数在 0 的右边, 负数在 0 的左边。【详解】A点表示的数是 5, B 点表示的数是 -2。",
+      "knowledge_points": ["正负数在数轴上的表示"],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_484",
+      "original_id": "unit1_fill_484"
+    },
+    {
+      "id": "unit1_choice_485",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "（20-21六年级下·全国·单元测试）一辆公共汽车从起点站开出后，中途经过了6个停靠点，最后到达终点站。下表是售票员记录的这辆公共汽车全程载客数量的变化情况，上车人数记作正，下车人数记作负。(1) 中途的 6 个停靠点一共下车 ( ) 人。(2) 在经过第 4 站时, 售票员漏记了上车人数, 第 4 站共有 ( ) 人上车。",
+      "options": [],
+      "answer": "(1) 46; (2) 5",
+      "solution": "【分析】（1）上车人数记作正，下车人数记作负，则表格中的负数代表的是下车的人数；将数字（不带负号）相加即可；(2) 将所有上车人数相加、所有下车人数相加，作差，差额就是第 4 站上车的人数；正确理解正负数的含义是解题关键。【详解】(1) 3 + 9 + 11 + 10 + 13 = 46 （人）(2) 下车总人数: 3 + 9 + 11 + 10 + 13 + 18 = 64 (人) 上车总人数: 25 + 12 + 10 + 8 + 4 = 59 (人) 64 - 59 = 5 (人)",
+      "knowledge_points": ["正负数的意义及应用"],
+      "source": "20-21六年级下·全国·单元测试",
+      "merged_id": "unit1_485",
+      "original_id": "unit1_solve_485"
+    },
+    {
+      "id": "unit1_choice_486",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "(20-21 六年级下·全国·单元测试) 规定向东为正, 下图每格表示 10 m 。(1) 小兰开始在 0 处, 现在在 +4 处, 说明她从 0 处向 ( ) 走了 ( ) m 。",
+      "options": [],
+      "answer": "（1）东；40；",
+      "solution": "【分析】根据题意，规定向东为正，则向西为负，据此解答即可。【详解】（1）规定向东为正，所以 +4 说明向东走了4格，4 × 10 = 40 (m);(2) 40 ÷ 10 = 4, 从 0 处向西走 40 m 即向西走 4 格; (3) 从 0 处出发, 先走 -20 m, 再走 +70 m, 即先向西走 2 格再向东走 7 格, 最后在 0 处往东 5 格处。明确对于具有相反意义的量，规定其中一个量为正，则另一个为负。",
+      "knowledge_points": ["正负数的意义及应用"],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_486",
+      "original_id": "unit1_fill_486"
+    },
+    {
+      "id": "unit1_choice_487",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "判断题",
+      "difficulty": "容易",
+      "question": "（20-21 六年级下·全国·单元测试）比较大小。2又1/4 ( ) -1.5  0 ( ) -1/2  -3 ( ) 7/3",
+      "options": [],
+      "answer": "< ; > ; <",
+      "solution": "【分析】正数、0都比负数大；负数与负数比较大小，负号后面的数字越小，这个负数反而越大；反之，负号后面的数字越大，这个负数就越小。此题考查了正、负数大小比较的方法, 只要掌握方法就很好解答。但要注意, 在负数与负数比较大小时,不要认为负号后面的数越大这个数越大。【详解】由分析可得: 2又1/4 < -1.5; 0 > -1/2; -3 < 7/3",
+      "knowledge_points": ["正负数在数轴上的表示", "正负数的大小比较"],
+      "source": "20-21 六年级下·全国·单元测试",
+      "merged_id": "unit1_487",
+      "original_id": "unit1_judge_487"
+    },
+    {
+      "id": "unit1_choice_488",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "(20-21 六年级下·全国·期末)一辆公共汽车从起点站开出, 停靠站点时载客数量记录如下表。(1) 从起点站到 E 站中哪个站点没人上车? 哪个站点没人下车?(2) 公共汽车从 C 站开出时车上有多少人？(3) 从起点站到 E 站, 下车的一共有多少人?",
+      "options": [],
+      "answer": "（1）D 站点没人上车，B 站点没人下车。(2) 27人; (3) 15人",
+      "solution": "【分析】（1）根据表格我们来分析，上车为正，下车为负，没有人上车或下车，为0。(2) 问 C 站开出时车上人数, 那应该从起点到 C 站, 所有上车人数相加一所有下车人数即可。(3) 从起点到 E 站, 下车的总人数, 就应该从起点到 E 站所有下车人数相加即可。此题主要考查了从统计表中获取信息并利用它解题的能力。【详解】（1）从起点站到E站，只有D站上车人数为0，所有D站点没人上车；同理，下车B站点为0，所以B站点没人下车。(2) 15 + 10 + 3 + 5 - (2 + 4) = 33 - 6 = 27 (人) 答: 公共汽车从 C 站开出时车上有 27 人。(3) 2 + 4 + 3 + 6 = 15 （人）答: 从起点站到 E 站, 下车的一共有 15 人。",
+      "knowledge_points": ["正负数的意义及应用"],
+      "source": "20-21 六年级下·全国·期末",
+      "merged_id": "unit1_488",
+      "original_id": "unit1_solve_488"
+    },
+    {
+      "id": "unit1_choice_489",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "作图题",
+      "difficulty": "困难",
+      "question": "(19-20 六年级下·湖南怀化·单元测试) 写出A、B、C、D、E、F点表示的数，并将A点向右移动3个单位长度，F点向左移动7个单位长度，它们各表示什么数？![](images/b7ba50f023ced1fad48d2e72a62f27fa6e0f0c20a3ea5d83ab4de7aae23e41d0.jpg)",
+      "options": [],
+      "answer": "见详解",
+      "solution": "【分析】-9 和 -7 之间的是 -8，据此填出 A 表示数，同理填出 B、C、D、E、F 点表示的数。看图，A 点向右移动 3 个单位长度，变成了 -5，同理找出 F 点向左移动 7 个单位长度后表示的数。本题考查了正负数在数轴上的表示, 数轴上正数在 0 的右边, 负数在 0 的左边。【详解】![](images/1a2418a52241c96996bd95d7aa860d6173617fa95b7eaeac402b769a0087ac6d.jpg) A 点向右移动 3 个单位长度, 表示为 -5 ; F 点向左移动 7 个单位长度, 表示为 -2。",
+      "knowledge_points": ["正负数在数轴上的表示"],
+      "source": "19-20 六年级下·湖南怀化·单元测试",
+      "merged_id": "unit1_489",
+      "original_id": "unit1_draw_489"
+    },
+    {
+      "id": "unit1_choice_490",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "（19-20六年级下·湖南怀化·单元测试）开学时，幸福港小学对六（1）第一组的男生进行单杠引体向上测验，以所做的7次为标准，超过的次数用正数表示，不足的次数用负数表示，第一小组的8名男生成绩如下：1 0 -3 -2 0 0 -1 2 (1) 第一小组有多少名男生达到标准? (2) 达标学生数占全组人数的百分之几?",
+      "options": [],
+      "answer": "(1) 5 名; (2) 62.5%",
+      "solution": "【分析】（1）观察统计表，统计出其中的 0 和正数的个数,有多少个, 那么第一小组就有多少名男生达到标准;(2) 用达标人数 5 人除以第一组的总人数 8 人, 求出达标学生数占全组人数的百分之几。本题考查了正负数的应用, 如果正数表示达到标准, 那么负数表示没有达标。【详解】（1）答：第一组男生成绩达标的有：1、0、0、0、2，共5个，所以第一小组有5名男生达到标准。(2) 5 ÷ 8 = 62.5% 答：达标学生数占全组人数的 62.5%.",
+      "knowledge_points": ["求一个数是另一个数的百分之几（百分率问题）", "正负数的意义及应用"],
+      "source": "19-20六年级下·湖南怀化·单元测试",
+      "merged_id": "unit1_490",
+      "original_id": "unit1_solve_490"
+    },
+    {
+      "id": "unit1_choice_491",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "(21-22 六年级下·全国·单元测试) 如图, 以A、B、C、D、E5 位同学的平均身高为基准，用条形统计图表示其中四人的身高。![](images/42fedc87040c12271c71c559dcc78fdd98f4fa0728cc1b6f660787d644be2256.jpg) (1) 如果这 5 位同学的平均身高为 151 cm, 那么 C 同学的身高为 ( ) cm。",
+      "options": [],
+      "answer": "149",
+      "solution": "【分析】把五位同学的平均身高记作 0 ，比平均身高高的记作正数，比平均身高矮的记作负数。从图中可知，四位同学的身高比平均身高高或矮的厘米数，如果比平均身高高的厘米数大于比平均身高矮的厘米数，那么 C 同学的身高反而比平均身高矮，两者的差值就是矮的厘米数；反之亦然。C 同学的身高比平均身高矮的厘米数,用平均身高减去矮的厘米数, 就是 C 同学的身高。掌握正、负数的意义是解题的关键。【详解】比平均身高高: 3 + 4 = 7 (cm) 比平均身高矮: 2 + 3 = 5 (cm) 7 > 5 相差: 7 - 5 = 2 (cm) C 同学的身高比平均身高矮 2 cm 。如图:![](images/548722b96548f67c8524ec7fba291f4148fd2bda97b7f975335e602d23bdf8bb.jpg)(2) 151 - 2 = 149 (cm)",
+      "knowledge_points": ["正负数的意义及应用", "1格表示多个单位的单式条形统计图"],
+      "source": "21-22 六年级下·全国·单元测试",
+      "merged_id": "unit1_491",
+      "original_id": "unit1_solve_491"
+    },
+    {
+      "id": "unit1_choice_492",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "(21-22 六年级下·全国·单元测试) 8 月份某地区连降大雨, 一位水位监测员测得某大坝的水位情况如下:(1) 警戒水位是 42.00 m, 想想表中正、负数表示的意义,并完成上表。",
+      "options": [],
+      "answer": "(1) +0.35; +0.36; 0; -0.14; -0.06;",
+      "solution": "【分析】（1）首先以警戒水位为标准，规定超出标准的为正，低于标准的为负，利用正负数加减法求出结果。本题主要考查正、负数的意义及其应用，利用正负数加减法求出结果.【详解】填表如下：8月1日对应-0.20；8月2日对应+0.60；8月3日对应+0.35；8月4日对应+0.36；8月5日对应0；8月6日对应-0.14；8月7日对应-0.06。",
+      "knowledge_points": ["单式折线统计图", "正负数的意义及应用"],
+      "source": "21-22 六年级下·全国·单元测试",
+      "merged_id": "unit1_492",
+      "original_id": "unit1_fill_492"
+    },
+    {
+      "id": "unit1_choice_493",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "作图题",
+      "difficulty": "容易",
+      "question": "(21-22 六年级下·全国·单元测试) 一只小虫在一条直线上来回爬行。以出发点为 0 点, 向右爬行的长度记为正数,向左爬行的长度记为负数, 爬行的记录如下表。在直线上画出小虫的爬行过程，并把上表填写完整。（每格表示 1 cm）![](images/994e7688dfec663da9832076938d530d805f8b2418ec3f4e69415c597d41ed73.jpg)",
+      "options": [],
+      "answer": "见详解",
+      "solution": "【分析】根据正负数的意义分别计算出每次爬行后距离 0 点的距离填在表上并在数轴上表示出来即可。此题主要考查了正负数的意义，解题关键是理解'正'和'负'的相对性，明确什么是一对具有相反意义的量。在一对具有相反意义的量中，先规定其中一个为正，则另一个就用负表示。【详解】第一次爬行距离原点是 4 cm, 第二次爬行距离原点是 4+(-6)=-2 (cm), 第三次爬行距离原点是 (-2) + 5=3 (cm), 第四次爬行距离原点是 3+(-8)=-5 (cm), 第五次爬行距离原点是 (-5)+10=5 (cm), 填表如下: 第一次对应+4 cm、4 cm；第二次对应-6 cm、-2 cm；第三次对应+5 cm、3 cm；第四次对应-8 cm、-5 cm；第五次对应+10 cm、5 cm。直线上画小虫的爬行过程如下图：![](images/f80ce1e4a3e8dbc1c9b535e2bdf0151ea26eea72072c67132c643f6851bad7f4.jpg)",
+      "knowledge_points": ["正负数在数轴上的表示", "正负数的意义及应用"],
+      "source": "21-22 六年级下·全国·单元测试",
+      "merged_id": "unit1_493",
+      "original_id": "unit1_draw_493"
+    },
+    {
+      "id": "unit1_choice_494",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "(21-22 四年级上·广东揭阳·期末) 教室里原来没有人。进来1人用'+1'人表示，出去1人用'-1'表示。(1) '+30'人表示（）；'-3'人表示（）。(2) 教室里现在一共有多少人？",
+      "options": [],
+      "answer": "（1）进来30人；出去3人 (2) 50人",
+      "solution": "【分析】（1）根据题意可知，进来人数用正数表示，出去人数用负数表示。则'+30'人表示进来30人；'-3'人表示出去3人。(2) 第一次进来了 30 人, 第二次出去 12 人, 第三次出去 3 人, 第四次进来 35 人, 则现在教室里有 30 - 12 - 3 + 35 人。解决本题时先明确正负数表示的意义, 再进行解答。【详解】(1) '+30'人表示进来30人; '-3'人表示出去3人。(2) 30 - 12 - 3 + 35 = 50 (人)答: 教室里现在一共有 50 人。",
+      "knowledge_points": ["正负数的意义及应用"],
+      "source": "21-22 四年级上·广东揭阳·期末",
+      "merged_id": "unit1_494",
+      "original_id": "unit1_solve_494"
+    },
+    {
+      "id": "unit1_choice_495",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "(21-22 六年级下·全国·单元测试) 下表列出了国外几个城市与北京的时差。(带正号的数表示同一时刻比北京时间早的时数) (1) 如果现在是北京时间 11:00, 那么伦敦时间和东京时间各是多少? (2) 如果星星在北京时间 17:00 打电话给远在旧金山的姨妈, 你认为合适吗? 为什么?",
+      "options": [],
+      "answer": "(1) 3:00; 12:00 (2) 不合适; 因为北京时间 17:00 是旧金山时间的 1:00, 正处于凌晨。",
+      "solution": "【分析】（1）正数表示在北京时间向后推几个小时，即加上这个正数，负数表示在北京时间向后前推几个小时，即加上这个负数；(2) 计算出旧金山的时间，然后判断即可。此题是典型的正数与负数的实际运用问题, 应联系实际生活, 认清正数与负数所代表的实际意义。【详解】(1) 11 + (-8) = 3 ，即伦敦时间 3:00 ; 11 + (+1) = 12 ，即东京时间 12:00 (2) 17 + (-16) = 1 ，即旧金山时间 1:00 所以不合适，因为旧金山时间是凌晨 1:00，姨妈可能正在休息。",
+      "knowledge_points": ["正负数的意义及应用"],
+      "source": "21-22 六年级下·全国·单元测试",
+      "merged_id": "unit1_495",
+      "original_id": "unit1_solve_495"
+    },
+    {
+      "id": "unit1_choice_496",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "(21-22 六年级下·河北张家口·期中) 下图中, 小红家、小明家、学校、书店和超市在一条直线上, 小明家在学校东边。(单位: 千米)![](images/db5804988d0165c73b83af2718c222eb72ca5423fdfb48487875c7867809d94e.jpg) (1) 书店在学校的哪边? 距学校多少千米? (2) -3 表示什么意思?",
+      "options": [],
+      "answer": "(1) 西边; 2 千米(2) -3 表示在学校的西边，距学校3千米。",
+      "solution": "【分析】由题可知，以学校为原点，向东为正，向西为负，相邻两个单位之间的距离代表 1 千米, 据此解答。本题考查正、负数的认识以及在数轴上表示数。【详解】（1）书店在学校的西边，距学校2千米。(2) -3 表示在学校的西边，距学校3千米。",
+      "knowledge_points": ["正负数在数轴上的表示"],
+      "source": "21-22 六年级下·河北张家口·期中",
+      "merged_id": "unit1_496",
+      "original_id": "unit1_solve_496"
+    },
+    {
+      "id": "unit1_choice_497",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "作图题",
+      "difficulty": "容易",
+      "question": "(21-22 六年级下·河北保定·期末) 画出温度, 并比一比。![](images/7b3231ccb94af9132ec4df80f3ae2eba1daecbecea854ef38bc4652a610a3418.jpg)",
+      "options": [],
+      "answer": "见详解；见详解 25°C; 0°C; -10°C; -15°C",
+      "solution": "【分析】正数、负数表示两种相反意义的量；比 0°C 低的温度叫零下温度，通常在数字前面加'-'（负号）；比 0°C 高的温度叫零上温度，通常在数字前面加'+'（正号），也可以省略不写。本题考查正负数的意义及应用，掌握正、负数大小比较的方法是解题的关键。正、负数大小的比较：正数 > 0 > 负数。【详解】如图：![](images/cca881a7cc768543c1767195bfa17390d7c975657162cf302973a42e49fb5d4f.jpg)![](images/05b6a8b39d76614a0ab3ce9ebfa9fbb2afe1d3f9f2333712d25cb8bd2008a145.jpg) ![](images/1b649ce799036496018ea2dc39b13d65982c5ea5ab410df1074dced31d6b44e7.jpg) ![](images/15b708ad3830411fff946fd0fd7e966eb7f62da0a996c52dce4d56eb6ff9d4fa.jpg) 25°C > 0°C > -10°C > -15°C",
+      "knowledge_points": ["正负数的意义及应用", "正负数的大小比较"],
+      "source": "21-22 六年级下·河北保定·期末",
+      "merged_id": "unit1_497",
+      "original_id": "unit1_draw_497"
+    },
+    {
+      "id": "unit1_choice_498",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "(22-23 四年级上·辽宁·单元测试) 排一排。某市 2021 年部分日期最高气温如下表。将上面的最高气温按从低到高的顺序排列出来。( ) < ( ) < ( ) < ( ) < ( )",
+      "options": [],
+      "answer": "-13°C; -8°C; -6°C; 10°C; 25°C",
+      "solution": "【分析】用整数表示温度时, 负数 < 0 < 正数, 几个负数比较大小时,数值越大, 负数越小, 对应的温度越低, 据此判断。本题的解题关键是要理解正负数的实际含义以及正负数的大小比较。【详解】-13°C < -8°C < -6°C < 10°C < 25°C",
+      "knowledge_points": ["正负数的大小比较"],
+      "source": "22-23 四年级上·辽宁·单元测试",
+      "merged_id": "unit1_498",
+      "original_id": "unit1_solve_498"
+    },
+    {
+      "id": "unit1_choice_499",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "填空题",
+      "difficulty": "容易",
+      "question": "(21-22 四年级上·浙江丽水·期末) 看图回答问题。![](images/1039df3ba8344977fa8345cb1b6f73ed3f1323aaa2882b8c0058cddc7fd48f8a.jpg) 淘气向东走 200 米记作 +200 米, 那么他向西走 400 米记作 ( ) 米。",
+      "options": [],
+      "answer": "(1) -400",
+      "solution": "【分析】正负数是表示具有意义相反的两种量，向东记为正，则向西就记为负，向西走400米记作-400米；此题主要考查正负数的意义, 正数与负数表示意义相反的两种量, 看清规定哪一个为正, 则和它意义相反的就为负。【详解】淘气向东走 200 米记作 +200 米，那么他向西走 400 米记作 -400 米。",
+      "knowledge_points": ["东、南、西、北方向", "正负数的意义及应用", "图文问题"],
+      "source": "21-22 四年级上·浙江丽水·期末",
+      "merged_id": "unit1_499",
+      "original_id": "unit1_fill_499"
+    },
+    {
+      "id": "unit1_choice_500",
+      "unit": "第一单元：负数",
+      "chapter": "负数",
+      "question_type": "解答题",
+      "difficulty": "容易",
+      "question": "（23-24六年级下·四川绵阳·期中）小王在网店上销售文旦，计划每天销售100千克，但实际每天销售量与计划销售量相比有增减，超过计划量记为正，不足计划量记为负，如表是小王第一周文旦的销售情况：（文旦是一种水果）(1) 小王第一周销售文旦最多的一天比最少的一天多销售多少千克？(2) 小王第一周实际销售文旦的总量是多少千克？(3) 若文旦售价为 8 元/千克, 包装及快递费为 3 元/千克,则小王这一周文旦销售收入共多少元?",
+      "options": [],
+      "answer": "(1) 20 千克(2) 718 千克 (3) 3590元",
+      "solution": "【分析】（1）文旦销售最多的一天是 +13 千克，销售最少的一天是 -7 千克，因此用最多的减最少的即可；(2) 先找到文旦销售实际是超过还是不足多少千克，然后再加上计划销售的总量;(3) 先找到文旦销售后一千克的实际收入，然后再乘销售的数量即可求出销售收入。【详解】(1) 13 - (-7) = 13 + 7 = 20 （千克）故小王第一周销售文旦最多的一天比最少的一天多销售 20 千克;(2) 3 + (-5) + (-2) + 11 + (-7) + 13 + 5 + 100 × 7 = 18 + 700 = 718 (千克) 故小王第一周实际销售文旦的总量是 718 千克;(3) (8 - 3) × 718 = 5 × 718 = 3590 (元) 故小王这一周文旦销售收入共 3590 元。",
+      "knowledge_points": ["利用正负数解决实际问题", "正负数的意义及应用"],
+      "source": "23-24六年级下·四川绵阳·期中",
+      "merged_id": "unit1_500",
+      "original_id": "unit1_solve_500"
+    }
+  ]
+  
\ No newline at end of file
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@@ -0,0 +1,3390 @@
+[
+  {
+    "id": "001",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "把圆柱的侧面展开不可能得到 ( )。",
+    "options": [
+      "A. 三角形",
+      "B. 正方形",
+      "C. 平行四边形"
+    ],
+    "`answer`": "A",
+    "solution": "【分析】根据圆柱的特征可知, 圆柱的侧面是一个曲面,分析侧面用不同的方式剪开, 如沿高剪开、沿斜线剪开等,圆柱的侧面展开图可能得到一个什么样的平面图形。\n\n【详解】A．无论如何剪开，侧面展开图始终是四边形（如长方形、正方形、平行四边形），无法得到三角形，所以圆柱的侧面展开不可能得到三角形；B. 当圆柱的底面周长和高相等时, 沿高剪开, 圆柱的侧面展开图是正方形;C. 沿斜线剪开侧面时, 圆柱的侧面展开图是平行四边形。故答案为: A",
+    "knowledge_points": [
+      "圆柱的展开图"
+    ],
+    "source": "(24-25 六年级下·海南海口·单元测试)",
+    "merged_id": "001",
+    "original_id": "001"
+  },
+  {
+    "id": "002",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "圆柱的底面直径是 10厘米，如果高增加 2 厘米，侧面积增加（）平方厘米。",
+    "options": [
+      "A. 20",
+      "B. 31.4",
+      "C. 62.8",
+      "D. 314"
+    ],
+    "answer": "C",
+    "solution": "【分析】根据题意, 高增加 2 厘米圆柱的底面周长不会变,所以圆柱体增加的侧面积 = 底面周长  $\\times$  增加的高, 首先根据圆的周长公式: 圆的周长 = 圆周率  $\\times$  直径, 求出周长, 再利用公式解答即可。\n\n【详解】  $3.14 \\times 10 \\times 2= 3 1. 4 \\times 2=62.8 (平方厘米)$\n\n所以如果高增加 2 厘米，侧面积增加 62.8 平方厘米。故答案为: C",
+    "knowledge_points": [
+      "圆柱的侧面积"
+    ],
+    "source": "(2024·内蒙古包头·小升初真题)",
+    "merged_id": "002",
+    "original_id": "002"
+  },
+  {
+    "id": "003",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "下面都是圆柱形物体,求（）的表面积就是求一个底面积和侧面积之和。\n\n   ① 通风管②笔筒③厨师帽④吸管",
+    "options": [
+      "A. ①②③④",
+      "B. ①②③",
+      "C. ②③"
+    ],
+    "answer": "C",
+    "solution": "【分析】根据圆柱的特征，可知通风管的表面积相当于圆柱的侧面积；笔筒的表面积相当于圆柱的一个底面积和侧面积之和；厨师帽的表面积相当于圆柱的一个底面积和侧面积之和。面积之和；吸管的表面积相当于圆柱的侧面积；据此解答。\n\n【详解】根据分析可知, 求②笔筒、③厨师帽的表面积就是求一个底面积和侧面积之和。故答案为: C",
+    "knowledge_points": [
+      "圆柱的侧面积",
+      "圆柱的表面积"
+    ],
+    "source": "(24-25 六年级下·全国·单元测试)",
+    "merged_id": "003",
+    "original_id": "003"
+  },
+  {
+    "id": "004",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "《九章算术》是我国古代的一部数学专著，书中关于圆柱体积的计算方法，有这样一句话：'周自相乘，以高乘之，十二而一。'意思是圆柱的体积  $= \\frac{1}{12} \\times$  底面周长的平方  $\\times$  高。结合我们学过的圆柱体积的计算公式，该计算方法中圆周率的取值为（）",
+    "options": [
+      "A. 3.14",
+      "B. 3.1",
+      "C. 3"
+    ],
+    "answer": "C",
+    "solution": "【分析】将'圆柱的体积  $= \\frac{1}{12} \\times$  底面周长的平方  $\\times$  高'中的底面周长的平方写成（ $2 \\times$  圆周率  $\\times$  半径） $^{2}$ ，再转化成  $(\\frac{1}{12} \\times 4 \\times$  圆周率） $\\times$  圆周率  $\\times$  半径 $^{2} \\times$  高，根据圆柱体积  $=$  底面积  $\\times$  高  $=$  圆周率  $\\times$  半径 $^{2} \\times$  高，可知  $\\frac{1}{12} \\times 4 \\times$  圆周率  $= 1$ ，进而确定圆周率的取值。\n\n【详解】$圆柱的体积  = \\frac{1}{12} \\times  底面周长的平方 \\times 高  \\\\=  \\frac{1}{12} \\times (2 \\times  圆周率  \\times  半径）  ^{2} \\times  高 \\\\ = \\frac{1}{12} \\times 4 \\times  圆周率 ^ 2 \\times  半径^  2 \\times  高  \\\\=  (\\frac{1}{12} \\times 4 \\times  圆周率）\\times  圆周率  \\times  半径  ^{2} \\times高$\n\n因为圆柱的体积  = $ 圆周率  \\times  半径 ^2 \\times$  高,因此    $\\frac{1}{12} \\times 4 \\times  圆周率 = 1$ , 圆周率的取值为 3 。故答案为: C",
+    "knowledge_points": [
+      "圆柱的体积"
+    ],
+    "source": "(24-25 六年级下·全国·单元测试)",
+    "merged_id": "004",
+    "original_id": "004"
+  },
+  {
+    "id": "005",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "做一个圆柱形通风管要用多少铁皮就是求通风筒的 ( )。",
+    "options": [
+      "A. 容积",
+      "B. 表面积",
+      "C. 侧面积"
+    ],
+    "answer": "C",
+    "solution": "【分析】物体表面的面积之和叫做表面积，圆柱形通风管只有侧面，没有2个底面。因此，通风管的铁皮需要多少，也就是这个圆柱的侧面积。据此解答。\n\n【详解】根据分析可知：做一个圆柱形通风管要用多少铁皮就是求通风筒的侧面积。故答案为：C",
+    "knowledge_points": [
+      "圆柱的侧面积"
+    ],
+    "source": "(24-25 六年级下·海南海口·单元测试)",
+    "merged_id": "005",
+    "original_id": "005"
+  },
+  {
+    "id": "006",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "把一段重 9 千克的圆柱形钢材削成一个最大的圆锥形零件, 这个圆锥重 ( ) 千克。",
+    "options": [
+      "A. 6",
+      "B. 3",
+      "C. 4.5"
+    ],
+    "answer": "B",
+    "solution": "【分析】根据  $V_{\\text {柱}} = \\mathrm{Sh}, V_{\\text {锥}} = \\frac{1}{3} \\mathrm{Sh}$  可知, 当圆柱和圆锥等底等高时, 圆锥的体积是圆柱体积的  $\\frac{1}{3}$ ; 由此可知, 把一段重 9 千克的圆柱形钢材削成一个最大的圆锥形零件, 这个圆锥的重量是圆柱重量的  $\\frac{1}{3}$ , 根据求一个数的几分之几是多少,用圆柱的重量乘  $\\frac{1}{3}$ , 即可求出圆锥的重量。\n\n【详解】  $9 \\times \\frac{1}{3} = 3$  （千克）这个圆锥重 3 千克。故答案为: B",
+    "knowledge_points": [
+      "求一个数的几分之几的问题",
+      "圆柱与圆锥体积的关系"
+    ],
+    "source": "(24-25 六年级下·海南海口·单元测试)",
+    "merged_id": "006",
+    "original_id": "006"
+  },
+  {
+    "id": "007",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "底面直径和高相等的圆柱, 侧面沿高展开后得到的一定是一个 ( )。",
+    "options": [
+      "A. 长方形",
+      "B. 正方形",
+      "C. 平行四边形"
+    ],
+    "answer": "A",
+    "solution": "【分析】圆柱的侧面沿高展开可以是一个长方形或正方形；侧面展开是长方形时，长方形的长等于圆柱的底面周长，长方形的宽等于圆柱的高；侧面展开是正方形时，圆柱的底面周长和高相等。已知圆柱的底面直径和高相等, 根据圆的周长公式  $C = \\pi d$ , 那么  $\\pi d > d, \\pi d > h$ , 即底面周长一定大于高。据此判断选择。\n\n【详解】根据分析可得：底面直径和高相等的圆柱，侧面沿高展开后得到的一定是一个长方形。故答案为: A",
+    "knowledge_points": [
+      "长方形的概念及特点",
+      "圆柱的展开图",
+      "正方形的概念及特点",
+      "圆的周长"
+    ],
+    "source": "(23-24 六年级下·四川德阳·期末)",
+    "merged_id": "007",
+    "original_id": "007"
+  },
+  {
+    "id": "008",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "圆柱和圆锥等底等体积, 圆锥体的高是 18 厘米, 那么圆柱的高是 ( ) 厘米。",
+    "options": [
+      "A. 18",
+      "B. 12",
+      "C. 9",
+      "D. 6"
+    ],
+    "answer": "D",
+    "solution": "【分析】圆锥体积  $= \\frac{1}{3} \\times$  底面积  $\\times$  高, 圆柱体积  $=$  底面积  $\\times$  高,所以圆柱和圆锥等底等体积时, 圆锥的高是圆柱高的 3 倍,据此解答即可。\n\n【详解】根据分析可知:圆柱的高:  $18 \\div 3 = 6$  (厘米)故答案为: D",
+    "knowledge_points": [
+      "圆锥的体积（容积）",
+      "圆柱与圆锥体积的关系",
+      "圆柱的体积"
+    ],
+    "source": "(23-24 六年级下·河北保定·期末)",
+    "merged_id": "008",
+    "original_id": "008"
+  },
+  {
+    "id": "009",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "把一个底面直径为  $4 \\mathrm{~cm}$ , 高为  $5 \\mathrm{~cm}$  的圆柱, 沿底面直径切割成两半, 表面积增加了多少  $\\mathrm{cm}^{2}$ ? 正确的列式是 ( )。",
+    "options": [
+      "A.  $3.14 \\times 4 \\times 5 \\times 2$",
+      "B.  $4 \\times 5 \\times 2$",
+      "C.  $4 \\times 5$"
+    ],
+    "answer": "B",
+    "solution": "【分析】根据题意, 沿圆柱的底面直径切割成两半, 那么表面积增加了两个长等于圆柱的高, 宽等于圆柱底面直径的长方形面积, 根据长方形面积  $=$  长  $\\times$  宽, 据此解答。\n\n【详解】  $4 \\times 5 \\times 2 \\\\= 2 0 \\times 2 \\\\ = 4 0 (\\mathrm {c m} ^ {2}) $\n\n正确的列式是  $4 \\times 5 \\times 2$  。故答案为: B",
+    "knowledge_points": [
+      "长方形的面积",
+      "立体图形的切拼（圆柱）",
+      "圆柱的表面积"
+    ],
+    "source": "(23-24 六年级下·天津蓟州·期末)",
+    "merged_id": "009",
+    "original_id": "009"
+  },
+  {
+    "id": "010",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "把一个圆柱的侧面展开，不可能是（）。",
+    "options": [
+      "A．长方形",
+      "B．正方形",
+      "C．平行四边形",
+      "D．梯形"
+    ],
+    "answer": "D",
+    "solution": "【分析】圆柱的展开图可能是正方形、可能是长方形、也可能是平行四边形，根据圆柱的侧面展开图的特点，结合圆柱的侧面的几种展开图方法与展开图的特点即可进行选择。\n\n【详解】圆柱的侧面沿高展开可能是长方形或正方形，如果斜着展开是一个平行四边形，因为圆柱的上下底面是完全相同的两个圆，所以圆柱的侧面展开图不可能是梯形。故答案为：D",
+    "knowledge_points": [
+      "圆柱的展开图"
+    ],
+    "source": "(2022·河南郑州·小升初真题)",
+    "merged_id": "010",
+    "original_id": "010"
+  },
+  {
+    "id": "011",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "下列问题中, ( )需要计算物体的体积。",
+    "options": [
+      "A. 求包装一份生日礼物需要多少彩纸",
+      "B. 给大厅里的柱子刷油漆，求需要多少油漆",
+      "C. 装饰黑板的四周, 求需要多少彩带",
+      "D. 一个铁球沉入装满水的容器中, 求溢出多少水"
+    ],
+    "answer": "D",
+    "solution": "【分析】物体表面的面积之和，是表面积。围成一个封闭图形的边的长度之和，是图形的周长。物体所占空间的大小，是物体的体积。据此，再结合各个选项中的问题，选出是计算体积的即可。\n\n【详解】A．求包装一份生日礼物需要多少彩纸，需要计算礼物盒子的表面积；  B. 给大厅里的柱子刷油漆, 求需要多少油漆, 需要计算柱子的侧面积;  C. 装饰黑板的四周, 求需要多少彩带, 需要计算黑板的周长;  D. 一个铁球沉入装满水的容器中, 求溢出多少水, 需要计算铁球的体积;故答案为: D",
+    "knowledge_points": [
+      "长方体表面积的应用",
+      "周长的认识和大小比较",
+      "体积的认识",
+      "圆柱的表面积"
+    ],
+    "source": "(23-24 六年级下·北京东城·期末)",
+    "merged_id": "011",
+    "original_id": "011"
+  },
+  {
+    "id": "012",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "把一团圆柱形橡皮泥揉成与它等底面积的圆锥形，高将（）。",
+    "options": [
+      "A. 扩大到原来的 3 倍",
+      "B. 扩大到原来的 6 倍",
+      "C. 缩小到原来的  $\\frac{1}{3}$",
+      "D. 不变"
+    ],
+    "answer": "A",
+    "solution": "【分析】同一团橡皮泥揉成圆柱和圆锥，圆柱和圆锥的体积都是这团橡皮泥的体积。等体积等底面积的圆柱和圆锥，圆锥的高是圆柱高的3倍，据此分析。\n\n【详解】把一团圆柱形橡皮泥揉成与它等底面积的圆锥形,说明圆柱和圆锥等体积等底面积, 高将扩大到原来的 3 倍。故答案为: A",
+    "knowledge_points": [
+      "圆柱与圆锥体积的关系"
+    ],
+    "source": "(2021·湖北黄冈·小升初真题)",
+    "merged_id": "012",
+    "original_id": "012"
+  },
+  {
+    "id": "013",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "厨师帽的形状近似圆柱,求'做一顶帽子至少需要多少面料?'就是求( )。",
+    "options": [
+      "A. 圆柱的侧面积",
+      "B. 圆柱底面积  $x 2$",
+      "C. 圆柱侧面积 + 底面积  $\\times 1$",
+      "D. 圆柱侧面积 + 底面积  $\\times 2$"
+    ],
+    "answer": "C",
+    "solution": "【分析】据圆柱的特征可知, 圆柱的上、下底面是完全相同的两个圆, 侧面是一个曲面。根据题意可知，这个厨师帽无底，所以少一个下底面，求至少需要多少面料，就是求这个无底圆柱形厨师帽的侧面积和一个底面积的和；据此选择。\n\n【详解】根据分析可知, 求 '做一顶帽子至少需要多少面料?' 就是求圆柱的侧面积和一个底面积的和。故答案为: C",
+    "knowledge_points": [
+      "圆柱的表面积"
+    ],
+    "source": "(2023·广西柳州·小升初真题)",
+    "merged_id": "013",
+    "original_id": "013"
+  },
+  {
+    "id": "014",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一个圆柱和一个圆锥的底面直径相等，圆锥的高是圆柱的3倍，圆锥的体积是15 立方分米，圆柱的体积是多少立方分米。（）",
+    "options": [
+      "A. 5",
+      "B. 10",
+      "C. 15",
+      "D. 45"
+    ],
+    "answer": "C",
+    "solution": "【分析】一个圆柱和一个圆锥的底面直径相等, 即底面积相等, 根据圆柱的体积  $V = Sh$ , 圆锥的体积  $V = \\frac{1}{3} Sh$ ,如果圆锥的高是圆柱的 3 倍, 那么这个圆锥的体积就等于圆柱的体积,据此解答。\n\n【详解】由分析可知, 一个圆柱和一个圆锥的底面直径相等, 圆锥的高是圆柱的 3 倍, 圆锥的体积是 15 立方分米,圆柱的体积也就是15立方分米;故答案为: C",
+    "knowledge_points": [
+      "圆锥的体积（容积）",
+      "圆柱与圆锥体积的关系",
+      "圆柱的体积"
+    ],
+    "source": "(15-16 六年级·全国·阶段练习)",
+    "merged_id": "014",
+    "original_id": "014"
+  },
+  {
+    "id": "015",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "做一个圆柱形的通风管, 至少需要铁皮的面积是求圆柱 ( )。",
+    "options": [
+      "A. 侧面积十一个底面面积",
+      "B. 表面积",
+      "C. 侧面积"
+    ],
+    "answer": "C",
+    "solution": "【分析】圆柱包含 3 个面，上下两个完全相同的圆形底面和一个侧面；做通风管因为需要通风所以不需要上下两个圆形底面，所用材料需要制作的是圆柱的侧面，据此分析解答。\n\n【详解】由分析可知，做一个圆柱形的通风管，至少需要铁皮的面积是求圆柱侧面积。故答案为: C",
+    "knowledge_points": [
+      "圆柱的侧面积",
+      "圆柱的表面积"
+    ],
+    "source": "(23-24 六年级下·广东江门·期中)",
+    "merged_id": "015",
+    "original_id": "015"
+  },
+  {
+    "id": "016",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "把一个圆柱形木料削成一个最大的圆锥，削去部分的体积是原圆柱形木料体积的（）。",
+    "options": [
+      "A.  $\\frac{1}{3}$",
+      "B.  $\\frac{1}{2}$",
+      "C.  $\\frac{2}{3}$",
+      "D. 2 倍"
+    ],
+    "answer": "C",
+    "solution": "【分析】根据圆柱的体积公式  $V = S h$ , 圆锥的体积公式  $V = \\frac{1}{3} S h$  可知, 把一个圆柱形木料削成一个最大的圆锥, 那么削成的圆锥与圆柱等底等高; 把圆柱的体积看作单位 '1', 因为等底等高的圆锥的体积是圆柱体积的  $\\frac{1}{3}$ , 所以削去部分的体积相当于圆柱体积的  $(1 - \\frac{1}{3})$  。据此解答。\n\n【详解】  $1 - \\frac{1}{3} = \\frac{2}{3}$  把一个圆柱形木料削成一个最大的圆锥，削去部分的体积是原圆柱形木料体积的  $\\frac{2}{3}$ 。故答案为: C",
+    "knowledge_points": [
+      "圆柱与圆锥体积的关系"
+    ],
+    "source": "(2014·贵州贵阳·小升初真题)",
+    "merged_id": "016",
+    "original_id": "016"
+  },
+  {
+    "id": "017",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一个圆锥的体积是100 立方厘米，底面积是 50 平方厘米，它的高是（）厘米。",
+    "options": [
+      "A. 2",
+      "B.  $\\frac{2}{3}$",
+      "C. 6",
+      "D. 10"
+    ],
+    "answer": "C",
+    "solution": "【分析】圆锥的体积  $V = \\frac{1}{3} Sh$ ，则  $h = 3V \\div S$ 。\n\n【详解】  $100 \\times 3 \\div 50\\\\= 3 0 0 \\div 5 0\\\\=6 (厘米)$\n\n它的高是 6 厘米。故答案为: C",
+    "knowledge_points": [
+      "圆锥的体积（容积）"
+    ],
+    "source": "(20-21 六年级下·全国·单元测试)",
+    "merged_id": "017",
+    "original_id": "017"
+  },
+  {
+    "id": "018",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "把一支新的圆柱形铅笔削尖，笔尖（圆锥部分）的体积是削去部分的（）。",
+    "options": [
+      "A.  $\\frac{1}{3}$",
+      "B.  $\\frac{2}{3}$",
+      "C.  $\\frac{1}{2}$"
+    ],
+    "answer": "C",
+    "solution": "【分析】因为等底等高的圆锥的体积是圆柱体积的  $\\frac{1}{3}$ , 把圆柱削成最大的圆锥, 它与圆柱等底等高, 所以削去部分的体积是圆柱体积的  $(1 - \\frac{1}{3})$ , 进而求出圆锥的体积是削去部分体积的几分之几, 据此解答。\n\n【详解】  $1 - \\frac{1}{3} = \\frac{2}{3}\\\\ \\frac {1}{3} \\div \\frac {2}{3} \\\\ = \\frac {1}{3} \\times \\frac {3}{2} \\\\ = \\frac {1}{2} $\n\n笔尖（圆锥部分）的体积是削去部分的  $\\frac{1}{2}$  。故答案为: C",
+    "knowledge_points": [
+      "圆锥的体积（容积）",
+      "圆柱与圆锥体积的关系",
+      "分数与分数的除法"
+    ],
+    "source": "(2017·云南楚雄·小升初真题)",
+    "merged_id": "018",
+    "original_id": "018"
+  },
+  {
+    "id": "019",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "圆柱的底面半径和高都扩大到原来的2倍，它的体积扩大到原来的（）倍。",
+    "options": [
+      "A. 4",
+      "B. 6",
+      "C. 8"
+    ],
+    "answer": "C",
+    "solution": "【分析】圆柱的体积=底面积×高，设圆柱的底面半径为r，高为h，则扩大后的半径为2r，高为2h，分别求出变化前后的体积，即可求出体积扩大到原来的几倍，计算后再判断即可。此题主要考查圆柱的体积的计算方法。\n\n【详解】解：设圆柱的底面半径为  $r$ ，高为  $h$ ，则扩大后的半径为  $2 r$ ，高为  $2 h$  原体积： $\\pi r^2 h$  现体积： $\\pi (2 r)^{2} (2 h) = 8 \\pi r^{2} h$  体积扩大:  $8 \\pi r^{2} h \\div \\pi r^{2} h = 8$  即圆柱的体积扩大到原来的 8 倍。故答案为: C",
+    "knowledge_points": [
+      "圆柱的体积"
+    ],
+    "source": "(22-23六年级下·全国·单元测试)",
+    "merged_id": "019",
+    "original_id": "019"
+  },
+  {
+    "id": "020",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "把一个圆柱体切成相等的小扇形拼成一个近似长方体，拼成近似长方体与原来圆柱相比较（）。",
+    "options": [
+      "A. 表面积和体积都不变",
+      "B. 表面积变大, 体积不变",
+      "C. 表面积和体积都变了",
+      "D. 表面积不变, 体积变大"
+    ],
+    "answer": "B",
+    "solution": "【分析】由于把这个圆柱体切开拼成一个近似的长方体，所以形状改变，体积不变；长方体的前面和后面相当于圆柱的侧面，长方体的上下两个面相当于圆柱的两个底面，则长方体比圆柱体多了左右两个面，由此即可判断。本题主要考查立体图形的切拼, 要明确圆柱切拼成长方体的方法。\n\n【详解】由分析可知, 长方体与原来的圆柱体相比较, 体积不变, 表面积变大了。故答案为: B",
+    "knowledge_points": [
+      "立体图形的切拼",
+      "圆柱的认识及特征",
+      "长方体的认识及特征"
+    ],
+    "source": "(22-23六年级下·全国·单元测试)",
+    "merged_id": "020",
+    "original_id": "020"
+  },
+  {
+    "id": "021",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "底面周长与高都相等的长方体、正方体、圆柱体，圆锥，（）的体积最大。",
+    "options": [
+      "A. 长方体",
+      "B. 正方体",
+      "C. 圆柱体",
+      "D. 圆锥"
+    ],
+    "answer": "C",
+    "solution": "【分析】根据等底等高的圆柱体积是圆锥体积的 3 倍, 我们可知底面周长和高都相等的圆柱体体积大于圆锥的体积; 再假设圆柱、正方体、长方体的底面周长都是 12.56 厘米, 高都是 3.14 厘米, 分别依据它们的体积公式计算出各自的体积, 再进行比较, 即可解答。此题主要考查了长方体、正方体、圆柱和圆锥的体积的求法，要熟练掌握.\n\n【详解】圆柱体的底面半径为:  $12.56 \\div 3.14 \\div 2\\\\= 4 \\div 2\\\\= 2（厘米）$  \n\n圆柱的体积是：  $3.14 \\times 2^{2} \\times 3.14\\\\= 3. 1 4 \\times 4 \\times 3. 1 4 \\\\ = 1 2. 5 6 \\times 3. 1 4 \\\\ = 3 9. 4 3 8 4 \\text {(立 方 厘 米)} $\n\n正方体的棱长为:  $12.56 \\div 4 = 3.14$  (厘米)\n\n正方体的体积为:  $3.14 \\times 3.14 \\times 3.14\\\\ = 9. 8 5 9 6 \\times 3. 1 4 \\\\ = 3 0. 9 5 9 1 4 4 \\text {(立 方 厘 米)} $\n\n因为  $12.56 \\div 2 = 6.28$  所以长方体的长和宽可以是 3.15 厘米和 3.13 厘米,\n\n长方体的体积是  \n\n$3.15 \\times 3.13 \\times 3.14\\\\ = 9. 8 5 9 5 \\times 3. 1 4 \\\\ = 3 0. 9 5 8 8 3 \\text {(立 方 厘 米)} $\n\n$3 0. 9 5 8 8 3 <   3 0. 9 5 9 1 4 4 <   3 9. 4 3 8 4$ 所以圆柱体的体积最大。故答案为: C",
+    "knowledge_points": [
+      "圆锥的体积（容积）",
+      "圆柱的体积",
+      "长方体的体积",
+      "正方体的体积"
+    ],
+    "source": "(22-23六年级下·全国·单元测试)",
+    "merged_id": "021",
+    "original_id": "021"
+  },
+  {
+    "id": "022",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "把一个底面直径是 6 分米, 高是 4 分米的圆柱体的侧面沿高展开得到一个长方形, 这个长方形的长是( )分米, 宽是( )分米。",
+    "options": [],
+    "answer": "18.84 ;4",
+    "solution": "【分析】根据题意可知, 展开得到一个长方形, 长方形的长等于圆柱的底面周长, 长方形的宽等于圆柱的高, 据此根据圆的周长公式: 周长  $= \\pi \\times$  直径, 代入数据, 求出长方形的长, 据此解答。\n\n【详解】  $3.14 \\times 6 = 18.84$  （分米）把一个底面直径是 6 分米, 高是 4 分米的圆柱体的侧面沿高展开得到一个长方形，这个长方形的长是 18.84 分米，宽是 4 分米。",
+    "knowledge_points": [
+      "圆柱的展开图",
+      "圆的周长"
+    ],
+    "source": "(24-25 六年级下·海南海口·单元测试)",
+    "merged_id": "022",
+    "original_id": "022"
+  },
+  {
+    "id": "023",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个容量为 8升的圆柱形铁桶，其内底面积是4平方分米，铁桶高是( )分米。",
+    "options": [],
+    "answer": "2",
+    "solution": "【分析】先根据进率 '1 升=1 立方分米' 把 8 升换算成 8立方分米；已知圆柱形铁桶的容积和内底面积，根据圆柱的体积（容积）公式  $V = S h$  可知，圆柱的高  $h = V \\div S$  ，代入数据计算，求出铁桶的高。\n\n【详解】8升  $= 8$  立方分米 $8 \\div 4 = 2$  (分米) 铁桶高是 2 分米。",
+    "knowledge_points": [
+      "圆柱的容积",
+      "容积单位间的进率与换算（升和毫升）"
+    ],
+    "source": "(24-25 六年级下·海南海口·单元测试)",
+    "merged_id": "023",
+    "original_id": "023"
+  },
+  {
+    "id": "024",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "一个圆锥的底面积是  $6 \\mathrm{dm}^{2}$ , 高是  $3 \\mathrm{dm}$ , 它的体积是( )  $\\mathrm{dm}^{3}$  。与它等底等高的圆柱体的体积是( )dm³。",
+    "options": [],
+    "answer": "6; 18",
+    "solution": "【分析】根据圆锥的体积  $=$  底面积  $\\times$  高  $\\div 3$ , 代入数据计算即可求出圆锥的体积; 根据等底等高的圆柱的体积是圆锥体积的 3 倍, 用圆锥的体积乘 3 即可解答。\n\n【详解】  $6 \\times 3 \\div 3\\\\ = 1 8 \\div 3 \\\\ = 6 (\\mathrm {d m} ^ {3}) $\n\n$6 \\times 3 = 1 8 (\\mathrm {d m} ^ {3})$\n\n所以圆锥的体积是  $6 \\mathrm{dm}^{3}$ , 与它等底等高的圆柱体的体积是$18 \\mathrm{dm}^{3}$  。",
+    "knowledge_points": [
+      "圆锥的体积（容积）",
+      "圆柱与圆锥体积的关系"
+    ],
+    "source": "(24-25 六年级下·海南海口·单元测试)",
+    "merged_id": "024",
+    "original_id": "024"
+  },
+  {
+    "id": "025",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "一个圆锥形沙堆, 它的体积是 9 立方分米, 底面积是 5 平方分米, 它的高是( )分米。",
+    "options": [],
+    "answer": "5.4",
+    "solution": "【分析】已知圆锥形沙堆的体积和底面积, 根据圆锥的体积公式  $V = \\frac{1}{3} \\mathrm{~Sh}$  可知, 圆锥的高  $h = 3 \\mathrm{~V} \\div \\mathrm{h}$ , 代入数据计算,求出它的高。\n\n【详解】  $9 \\times 3 \\div 5\\\\= 2 7 \\div 5\\\\=5.4 (分米)$\n\n它的高是 5.4 分米。",
+    "knowledge_points": [
+      "圆锥的体积（容积）"
+    ],
+    "source": "(24-25 六年级下·海南海口·单元测试)",
+    "merged_id": "025",
+    "original_id": "025"
+  },
+  {
+    "id": "026",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "一个圆柱的底面半径是  $1 \\mathrm{dm}$ , 高  $2 \\mathrm{dm}$ , 这个圆柱的底面积是 ( )  $\\mathrm{cm}^{2}$ , 底面周长是 ( )  $\\mathrm{cm}$ , 侧面积是 ( )  $\\mathrm{cm}^{2}$ , 体积是 ( )  $\\mathrm{cm}^{3}$  。",
+    "options": [],
+    "answer": "314 ;62.8; 1256; 6280",
+    "solution": "【分析】圆柱的底面积就是半径是  $1 \\mathrm{dm}$  圆的面积，根据圆的面积公式：面积  $= \\pi \\times$  半径  $^{2}$ ，代入数据，求出圆柱的底面积；底面周长，就是半径是  $1 \\mathrm{dm}$  圆的周长，根据圆的周长公式：周长  $= \\pi \\times$  半径  $\\times 2$ ，代入数据，求出底面周长；根据圆柱的侧面积公式：侧面积  $=$  底面周长  $\\times$  高，代入数据，求出圆柱的侧面积；根据圆柱的体积公式：体积  $=$  底面积  $\\times$  高，代入数据，求出圆柱的体积，注意单位名数的统一。\n\n【详解】  $1 \\mathrm{dm} = 10 \\mathrm{~cm}; 2 \\mathrm{dm} = 20 \\mathrm{~cm}$  。  \n\n$3.14 \\times 10^{2} \\\\= 3. 1 4 \\times 1 0 0 \\\\ = 3 1 4 (\\mathrm {c m} ^ {2}) $\n\n$3.1 4 \\times 1 0 \\times 2 \\\\ = 3 1. 4 \\times 2 \\\\ = 6 2. 8 \\mathrm {(c m)} \\\\ 6 2. 8 \\times 2 0 = 1 2 5 6 (\\mathrm {c m} ^ {2}) \\\\ 3 1 4 \\times 2 0 = 6 2 8 0 (\\mathrm {c m} ^ {3}) $\n\n一个圆柱的底面半径是  $1 \\mathrm{dm}$ , 高  $2 \\mathrm{dm}$ , 这个圆柱的底面积是  $314 \\mathrm{~cm}^{2}$ , 底面周长是  $62.8 \\mathrm{~cm}$ , 侧面积是  $1256 \\mathrm{~cm}^{2}$ , 体积是  $6280 \\mathrm{~cm}^{3}$  。",
+    "knowledge_points": [
+      "圆柱的体积",
+      "圆柱的侧面积",
+      "圆的面积",
+      "圆的周长"
+    ],
+    "source": "(24-25 六年级下·海南海口·单元测试)",
+    "merged_id": "026",
+    "original_id": "026"
+  },
+  {
+    "id": "027",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "把一个圆柱体削成一个最大的圆锥体，削去的部分是6立方厘米，原来圆柱的体积是( )立方厘米，削成的圆锥的体积是( )立方厘米。",
+    "options": [],
+    "answer": "9;3",
+    "solution": "【分析】等底等高的圆锥的体积是圆柱的  $\\frac{1}{3}$ , 把圆柱的体积看作单位 '1', 削成一个最大的圆锥, 则削去部分的体积是圆柱的  $(1 - \\frac{1}{3})$ , 对应的是削去部分的体积 6 立方厘米, 求单位 '1', 用  $6 \\div (1 - \\frac{1}{3})$  解答; 求削成的圆锥的体积, 用圆柱的体积 -6 立方厘米, 即可解答。\n\n【详解】  $6 \\div (1 - \\frac{1}{3})\\\\ = 6 \\div \\frac {2}{3} \\\\ = 6 \\times \\frac {3}{2}\\\\ = 9  （立方厘米）$\n\n$9 - 6 = 3 \\text {(立 方 厘 米)}$把一个圆柱体削成一个最大的圆锥体，削去的部分是 6 立方厘米，原来圆柱的体积是 9 立方厘米，削成的圆锥的体积是3 立方厘米。",
+    "knowledge_points": [
+      "圆锥的体积（容积）",
+      "圆柱与圆锥体积的关系",
+      "圆柱的体积"
+    ],
+    "source": "(24-25六年级下·海南海口·单元测试)",
+    "merged_id": "027",
+    "original_id": "027"
+  },
+  {
+    "id": "028",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个圆柱和一个圆锥等底等高, 它们体积的和是  $28 \\mathrm{~cm}^{3}$ , 圆锥的体积是  $\\mathrm{cm}^{3}$  。",
+    "options": [],
+    "answer": "7",
+    "solution": "【分析】根据等底等高的圆柱体积是圆锥体积的 3 倍可知,把圆锥的体积看作 1 倍, 则圆柱的体积就是 3 倍, 所以 28 就是圆锥体积的  $(3 + 1)$  倍, 据此用除法求出圆锥的体积即可。\n\n【详解】  $28 \\div (3 + 1)\\\\= 2 8 \\div 4 \\\\ = 7 (\\mathrm {c m} ^ {3}) $\n\n一个圆柱和一个圆锥等底等高，它们体积的和是  $28 \\mathrm{~cm}^{3}$ ，圆锥的体积是  $7 \\mathrm{~cm}^{3}$  。",
+    "knowledge_points": [
+      "圆锥的体积（容积）",
+      "圆柱与圆锥体积的关系"
+    ],
+    "source": "(24-25 六年级下·海南海口·单元测试)",
+    "merged_id": "028",
+    "original_id": "028"
+  },
+  {
+    "id": "029",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个圆柱体的底面半径是  $2 \\mathrm{~cm}$ , 高是  $3 \\mathrm{~cm}$ , 它的侧面积是  $\\_ \\mathrm{cm}^{2}$ , 表面积是  $\\_ \\mathrm{cm}^{2}$ , 体积是  $\\_ \\mathrm{cm}^{3}$ 。",
+    "options": [],
+    "answer": "37.68; 62.8; 37.68",
+    "solution": "【分析】根据侧面积  $S = 2 \\pi r h$ ; 表面积  $S = 2 \\pi r h + 2 \\pi r^{2}$ ; 体积  $V = \\pi r^{2} h$ , 据此代入数据解答即可。\n\n【详解】  $2 \\times 3.14 \\times 2 \\times 3\\\\ = 6. 2 8 \\times 2 \\times 3 \\\\ = 1 2. 5 6 \\times 3 \\\\ = 3 7. 6 8 (\\mathrm {c m} ^ {2}) $\n\n$ 2 \\times 3. 1 4 \\times 2 \\times 3 + 2 \\times 3. 1 4 \\times 2 ^ {2} \\\\ = 6. 2 8 \\times 2 \\times 3 + 6. 2 8 \\times 4 \\\\ = 1 2. 5 6 \\times 3 + 2 5. 1 2  = 3 7. 6 8 + 2 5. 1 2 \\\\ = 6 2. 8 (\\mathrm {c m} ^ {2}) $\n\n$3.1 4 \\times 2 ^ {2} \\times 3 \\\\ = 3. 1 4 \\times 4 \\times 3 \\\\ = 1 2. 5 6 \\times 3 \\\\ = 3 7. 6 8 (\\mathrm {c m} ^ {3}) $\n\n它的侧面积是  $37.68 \\mathrm{~cm}^{2}$ , 表面积是  $62.8 \\mathrm{~cm}^{2}$ , 体积是  $37.68 \\mathrm{~cm}^{3}$  。",
+    "knowledge_points": [
+      "圆柱的体积",
+      "圆柱的侧面积",
+      "圆柱的表面积"
+    ],
+    "source": "(24-25 六年级下·海南海口·单元测试)",
+    "merged_id": "029",
+    "original_id": "029"
+  },
+  {
+    "id": "030",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "将一个体积是 24 立方分米的圆柱削成一个最大的圆锥, 削去部分的体积是( )立方分米。",
+    "options": [],
+    "answer": "16",
+    "solution": "【分析】根据题意，把一个圆柱削成一个最大的圆锥，则圆锥和圆柱等底面积等高；根据  $V_{\\text {柱}} = \\mathrm{Sh}, V_{\\text {锥}} = \\frac{1}{3} \\mathrm{Sh}$  可知，等底等高的圆锥的体积是圆柱体积的  $\\frac{1}{3}$ ，根据求一个数的几分之几是多少，用圆柱的体积乘  $\\frac{1}{3}$ ，求出这个圆锥的体积；再用圆柱的体积减去圆锥的体积，求出削去部分的体积。\n\n【详解】  $24 - 24 \\times \\frac{1}{3}\\\\= 2 4 - 8\\\\= 16$  （立方分米）\n\n削去部分的体积是 16 立方分米。",
+    "knowledge_points": [
+      "圆锥的体积（容积）",
+      "求一个数的几分之几的问题",
+      "圆柱与圆锥体积的关系",
+      "圆柱的体积"
+    ],
+    "source": "(23-24 六年级下·四川绵阳·期中)",
+    "merged_id": "030",
+    "original_id": "030"
+  },
+  {
+    "id": "031",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个圆柱的底面周长是 3 分米, 高是 3 分米, 侧面是( ), 侧面积是( )平方分米。",
+    "options": [],
+    "answer": "长方形; 9",
+    "solution": "【分析】圆柱沿高剪开，侧面是长方形，圆柱侧面积=底面周长×高，据此解答即可。\n\n【详解】侧面积： $3 \\times 3 = 9$  （平方分米）所以侧面是长方形，侧面积是9平方分米。",
+    "knowledge_points": [
+      "圆柱的展开图",
+      "圆柱的侧面积"
+    ],
+    "source": "(23-24 六年级上·河北邯郸·期中)",
+    "merged_id": "031",
+    "original_id": "031"
+  },
+  {
+    "id": "032",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个压路机的前轮是圆柱形的，轮宽 1.2 米，半径是 4 分米。前轮滚动一周，压路的面积是( )平方米。",
+    "options": [],
+    "answer": "3.0144",
+    "solution": "【分析】求压路面的面积，就是求这个压路机的前轮的侧面积，因为前轮是圆柱形，所以根据圆柱的侧面积公式：侧面积  $=$  底面周长  $\\times$  高, 代入数据, 即可解答, 注意单位名数的统一。\n\n【详解】4分米=0.4米\n\n$3.1 4 \\times 0. 4 \\times 2 \\times 1. 2\\\\= 1. 2 5 6 \\times 2 \\times 1. 2 \\\\ = 2. 5 1 2 \\times 1. 2 \\\\ = 3. 0 1 4 4 \\text {(平 方 米)} $\n\n压路的面积是 3.0144 平方米。",
+    "knowledge_points": [
+      "圆柱的侧面积"
+    ],
+    "source": "(22-23 六年级下·安徽马鞍山·期末)",
+    "merged_id": "032",
+    "original_id": "032"
+  },
+  {
+    "id": "033",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个圆柱的侧面展开图是一个正方形, 已知它的底面周长是 31.4 分米, 则这个圆柱底面直径和高的比是( )。",
+    "options": [],
+    "answer": "50:157",
+    "solution": "【分析】根据正方形的特征可知, 圆柱的底面周长等于圆柱的高, 所以根据圆周长公式:  $C = \\pi d$ , 用  $31.4 \\div 3.14$  即可求出底面直径, 进而写出圆柱底面直径和高的比, 再化简即可; 化简比根据比的基本性质作答, 即比的前项和后项同时乘或除以一个数（0 除外），比值不变。\n\n【详解】  $31.4 \\div 3.14 = 10$  （分米）\n\n$1 0: 3 1. 4 \\\\ = (1 0 \\times 5): (3 1. 4 \\times 5) \\\\ = 5 0: 1 5 7 $\n\n这个圆柱底面直径和高的比是  $50: 157$  。",
+    "knowledge_points": [
+      "圆柱的展开图",
+      "比的化简",
+      "比的意义",
+      "圆的周长"
+    ],
+    "source": "(23-24 六年级下·山东临沂·期末)",
+    "merged_id": "033",
+    "original_id": "033"
+  },{
+    "id": "034",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个等腰直角三角形的一条直角边为 $6 \\mathrm{~cm}$, 以其中一条直角边所在直线为轴旋转一周，得到的立体图形的体积是( )cm³。",
+    "options": [],
+    "answer": "226.08或$226\\frac{2}{25}$或$\\frac{5652}{25}$",
+    "solution": "【分析】根据题意, 直角三角形旋转后形成的圆锥的底面半径是 $6 \\mathrm{~cm}$, 圆锥的高是 $6 \\mathrm{~cm}$, 根据 $\\mathrm{V} = \\frac{1}{3} \\pi \\mathrm{r}^{2} \\mathrm{~h}$, 计算解答。\\n\\n【详解】 $3.14 \\times 6^{2} \\times 6 \\times \\frac{1}{3}\\\\= 3.14 \\times 36 \\times 6 \\times \\frac{1}{3}\\\\= 678.24 \\times \\frac{1}{3}\\\\= 226.08 (\\mathrm{cm}^{3})$\\n\\n故得到的立体图形的体积是 $226.08 \\mathrm{~cm}^{3}$。",
+    "knowledge_points": [
+      "圆锥的体积（容积）",
+      "圆锥的认识及特征"
+    ],
+    "source": "(23-24 六年级下·湖南娄底·期中)",
+    "merged_id": "034",
+    "original_id": "034"
+  },
+  {
+    "id": "035",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "用一张长 $4.5 \\mathrm{dm}$、宽 $2 \\mathrm{dm}$ 的长方形纸围成一个圆柱, 这个圆柱的侧面积是( ) $\\mathrm{dm}^{2}$。",
+    "options": [],
+    "answer": "9",
+    "solution": "【分析】圆柱侧面沿高展开是个长方形，长方形纸围成一个圆柱，这个圆柱的侧面积就是这张长方形纸的面积，根据长方形面积 = 长 × 宽，列式计算即可。\\n\\n【详解】 $4.5 \\times 2 = 9 (\\mathrm{dm}^{2})$ 这个圆柱的侧面积是 $9 \\mathrm{dm}^{2}$。",
+    "knowledge_points": [
+      "圆柱的侧面积"
+    ],
+    "source": "(23-24 六年级下·四川广元·期末)",
+    "merged_id": "035",
+    "original_id": "035"
+  },
+  {
+    "id": "036",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "用棱长 $6 \\mathrm{~dm}$ 的正方体木块，削成一个最大的圆锥，圆锥的底面直径是( )dm，圆锥的高是( )dm。",
+    "options": [],
+    "answer": "6;6",
+    "solution": "【分析】根据题意可知, 把这个正方体木块削成一个最大的圆锥, 也就是削成的圆锥的底面直径和高都等于正方体的棱长, 据此解答即可。\\n\\n【详解】用棱长 $6 \\mathrm{dm}$ 的正方体木块, 削成一个最大的圆锥, 圆锥的底面直径是 $6 \\mathrm{dm}$, 圆锥的高是 $6 \\mathrm{dm}$。",
+    "knowledge_points": [
+      "正方体的特征",
+      "圆锥的认识及特征",
+      "立体图形的切拼（圆锥）"
+    ],
+    "source": "(23-24 六年级下·河南郑州·期末)",
+    "merged_id": "036",
+    "original_id": "036"
+  },
+  {
+    "id": "037",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个圆柱形杯子的底面直径是 $6 \\mathrm{~cm}$, 高是 $10 \\mathrm{~cm}$, 这个杯子的容积是( ) mL。(π取3.14)",
+    "options": [],
+    "answer": "282.6",
+    "solution": "【分析】根据圆柱体积=底面积×高, 即可求出杯子的容积, $1 \\mathrm{~cm}^{3} = 1 \\mathrm{~mL}$。\\n\\n【详解】 $3.14 \\times (6 \\div 2)^{2} \\times 10\\\\= 3.14 \\times 3^{2} \\times 10\\\\= 3.14 \\times 9 \\times 10\\\\= 282.6 (\\mathrm{cm}^{3})\\\\= 282.6 (\\mathrm{mL})$\\n\\n这个杯子的容积是 $282.6 \\mathrm{~mL}$。",
+    "knowledge_points": [
+      "圆柱的容积"
+    ],
+    "source": "(23-24 六年级下·云南楚雄·期中)",
+    "merged_id": "037",
+    "original_id": "037"
+  },
+  {
+    "id": "038",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "一个圆柱, 它的一个底面直径是 $6 \\mathrm{~cm}$, 另一个底面周长是( ) $\\mathrm{cm}$。",
+    "options": [],
+    "answer": "18.84",
+    "solution": "【分析】圆柱上下两个底面是相等的两个圆，利用圆的周长公式 $C = \\pi d$, 即可计算出另一个底面周长。\\n\\n【详解】 $3.14 \\times 6 = 18.84$ (cm) 一个圆柱，它的一个底面直径是 $6 \\mathrm{~cm}$，另一个底面周长是 $18.84 \\mathrm{~cm}$。",
+    "knowledge_points": [
+      "圆柱的认识及特征",
+      "圆的周长"
+    ],
+    "source": "(23-24 六年级下·内蒙古通辽·期中)",
+    "merged_id": "038",
+    "original_id": "038"
+  },
+  {
+    "id": "039",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个圆柱和一个圆锥的体积和底面积相等，圆柱的高是 6 厘米，圆锥的高是( )厘米。",
+    "options": [],
+    "answer": "18",
+    "solution": "【分析】根据等底等高圆锥的体积是圆柱体积的 $\\frac{1}{3}$, 已知圆锥和圆柱的体积和底面积相等, 圆柱的高是 6 厘米, 那么圆锥的高是圆柱高的 3 倍, 由此解答。\\n\\n【详解】由分析可知, 圆锥的高是圆柱高的 3 倍, $6 \\times 3 = 18$ (厘米) 即圆锥的高是 18 厘米。",
+    "knowledge_points": [
+      "圆柱与圆锥体积的关系"
+    ],
+    "source": "(2014 六年级·全国·课后作业)",
+    "merged_id": "039",
+    "original_id": "039"
+  },
+  {
+    "id": "040",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "把一个圆柱形罐头盒的侧面包装纸展开，得到一个正方形，这个圆柱形罐头盒的底面半径是 5 厘米，那么它的高是( )厘米。",
+    "options": [],
+    "answer": "31.4",
+    "solution": "【分析】根据底面周长公式: $C = 2 \\pi r$, 用 $2 \\times 3.14 \\times 5$ 即可求出底面周长, 根据圆柱的特征可知, 如果侧面展开得到一个正方形, 则底面周长和高相等, 据此得出高。\\n\\n【详解】 $2 \\times 3.14 \\times 5 = 31.4$ （厘米）高是 31.4 厘米。",
+    "knowledge_points": [
+      "圆柱的展开图",
+      "圆的周长及应用",
+      "圆柱的认识及特征"
+    ],
+    "source": "(2014 六年级·全国·课后作业)",
+    "merged_id": "040",
+    "original_id": "040"
+  },
+  {
+    "id": "041",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个长方形 A4 纸, 以长边为轴旋转 $360^{\\circ}$ 后得到的图形是( )。以一个直角三角形的一条直角边为轴旋转 $360^{\\circ}$ 后得到的图形是( )。",
+    "options": [],
+    "answer": "圆柱;圆锥",
+    "solution": "【分析】根据对图形的认识可知：\\n\\n【详解】![](images/6bc25978e2b4fd619bbc3daf2b4b41662a71612fbee70822281474a0ba7ba3a6.jpg)![](images/d41dce73c3da6b27b0eaf5b92f81e02df65839504368ec0d67e947ad0db1e0ff.jpg)\\n\\n一个长方形 A4 纸, 以长边为轴旋转 $360^{\\circ}$ 后得到的图形是圆柱。以一个直角三角形的一条直角边为轴旋转 $360^{\\circ}$ 后得到的图形是圆锥。",
+    "knowledge_points": [
+      "圆柱的认识及特征",
+      "圆锥的认识及特征"
+    ],
+    "source": "(22-23 六年级下·河南洛阳·期末)",
+    "merged_id": "041",
+    "original_id": "041"
+  },
+  {
+    "id": "042",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "把一个棱长 $6 \\mathrm{~cm}$ 的正方体木块削成一个最大的圆锥体，这个圆锥体的体积是( )。",
+    "options": [],
+    "answer": "$56.52\\mathrm{~cm}^{3}$",
+    "solution": "【分析】把一个正方体木块削成一个最大的圆锥体，圆锥的底面直径和高都等于正方体棱长, 根据圆锥体积 = 底面积 × 高 ÷ 3, 列式计算即可。关键是理解正方体和圆锥之间的关系, 掌握并灵活运用圆锥体积公式。\\n\\n【详解】 $3.14 \\times (6 \\div 2)^{2} \\times 6 \\div 3\\\\= 3.14 \\times 3^{2} \\times 6 \\div 3\\\\= 3.14 \\times 9 \\times 6 \\div 3\\\\= 56.52 (\\mathrm{cm}^{3})$\\n\\n这个圆锥体的体积是 $56.52 \\mathrm{~cm}^{3}$。",
+    "knowledge_points": [
+      "圆锥的体积（容积）"
+    ],
+    "source": "(2014·全国·小升初真题)",
+    "merged_id": "042",
+    "original_id": "042"
+  },
+  {
+    "id": "043",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "等底等高的圆柱和圆锥体积的和是 48 立方分米，那么圆柱的体积( )立方分米，圆锥的体积是( )立方分米。",
+    "options": [],
+    "answer": "36;12",
+    "solution": "【分析】等底等高的圆柱的体积是圆锥的体积的 3 倍, 已知等底等高的圆柱和圆锥体积的和是 48 立方分米, 则用 $48 \\div (3 + 1)$ 即可求出圆锥的体积, 再乘 3 即可求出圆柱的体积。据此解答。此题考查了等底等高的圆柱与圆锥的体积倍数关系的灵活应用。\\n\\n【详解】 $48 \\div (3 + 1)\\\\= 48 \\div 4\\\\= 12$（立方分米）\\n\\n$12 \\times 3 = 36$ （立方分米）圆柱的体积36立方分米，圆锥的体积是12立方分米。",
+    "knowledge_points": [
+      "圆柱与圆锥体积的关系"
+    ],
+    "source": "(2014 六年级·全国·课后作业)",
+    "merged_id": "043",
+    "original_id": "043"
+  },
+  {
+    "id": "044",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "把一个长方形绕它的一条边旋转一周可以形成____，直角三角形绕它的一条直角边旋转一周可以形成____。",
+    "options": [],
+    "answer": "圆柱;圆锥",
+    "solution": "【分析】由于长方形对边平行且相等，以它的一条边为轴旋转一周它的上、下两个面是以长方形的另一条边为半径的两个完全一样的圆，与轴平行的另一条边通过旋转形成一个曲面这样就得到一个圆柱。以直角三角形的一条直角边为轴旋转一周, 它的另一条直角边绕轴旋转一周形成一个圆, 即圆锥的底面; 直角三角形的斜边经过旋转形成一个曲面, 即圆锥的侧面。掌握长方形、直角三角形通过旋转形成圆柱和圆锥的方法是解题的关键。\\n\\n【详解】把一个长方形绕它的一条边旋转一周可以形成圆柱; 直角三角形绕它的一条直角边旋转一周可以形成圆锥。\\n\\n如图: ![](images/24b22b87428d341d6aefbaa0648d7189b4d7fa999835df7470840547876dd963.jpg)![](images/1ec576060bda010cc1acaf44b5b049ca59de940b4e30145e4c68b8450a2243ef.jpg)",
+    "knowledge_points": [
+      "圆柱的认识及特征",
+      "圆锥的认识及特征"
+    ],
+    "source": "(18-19六年级下·辽宁·课后作业)",
+    "merged_id": "044",
+    "original_id": "044"
+  },
+  {
+    "id": "045",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个正方体木块的棱长是6厘米，把它削成一个最大的圆柱体，圆柱体的体积是( )立方厘米，再把这个圆柱体削成一个最大的圆锥体，圆锥体的体积约是( )立方厘米。",
+    "options": [],
+    "answer": "169.56;56.52",
+    "solution": "【分析】把正方体削成一个最大的圆柱，则这个圆柱的底面直径是6厘米，高是6厘米，根据圆柱的体积公式： $V = \\pi \\mathrm{r}^{2} \\mathrm{~h}$, 用 $3.14 \\times (6 \\div 2)^{2} \\times 6$ 即可求出圆柱的体积; 再把圆柱削成一个最大的圆锥, 则这个圆锥的底面直径是 6 厘米, 高是 6 厘米, 根据圆锥的体积公式: $\\mathrm{V} = \\frac{1}{3} \\pi \\mathrm{r}^{2} \\mathrm{~h}$, 用 $\\frac{1}{3} \\times 3.14 \\times (6 \\div 2)^{2} \\times 6$ 即可求出圆锥的体积。本题主要考查了圆柱体积公式和圆锥体积公式的应用。\\n\\n【详解】 $3.14 \\times (6 \\div 2)^{2} \\times 6\\\\= 3.14 \\times 3^{2} \\times 6 = 3.14 \\times 9 \\times 6\\\\= 169.56$ (立方厘米)\\n\\n$169.56 \\times \\frac{1}{3} = 56.52$ （立方厘米）圆柱的体积是 169.56 立方厘米, 圆锥的体积是 56.52 立方厘米。",
+    "knowledge_points": [
+      "圆锥的体积（容积）",
+      "圆柱与圆锥体积的关系",
+      "圆柱的体积"
+    ],
+    "source": "(2021·山东枣庄·小升初真题)",
+    "merged_id": "045",
+    "original_id": "045"
+  },
+  {
+    "id": "046",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个圆柱和一个圆锥等底等高，它们的体积和是 48 立方分米，圆锥的体积是( ), 圆柱的体积是( )。",
+    "options": [],
+    "answer": "12立方分米;36立方分米",
+    "solution": "【分析】等底等高的圆柱的体积是圆锥的体积的 3 倍, 已知它们的体积和是 48 立方分米, 则 $48 \\div (3 + 1)$ 即可求出圆锥的体积, 进而求出圆柱的体积; 由此即可解答。此题考查了等底等高的圆柱与圆锥的体积倍数关系的灵活应用。\\n\\n【详解】 $48 \\div (3 + 1)\\\\= 48 \\div 4\\\\= 12$（立方分米）\\n\\n$12 \\times 3 = 36$ （立方分米）圆锥的体积是 12 立方分米，圆柱的体积是 36 立方分米。",
+    "knowledge_points": [
+      "圆柱与圆锥体积的关系"
+    ],
+    "source": "(2017·河北邯郸·小升初真题)",
+    "merged_id": "046",
+    "original_id": "046"
+  },
+  {
+    "id": "047",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "圆柱有( )条高, 圆锥有( )条高。",
+    "options": [],
+    "answer": "无数;1",
+    "solution": "【分析】圆柱的两个底面之间的距离叫做高；圆柱的高有无数条。从圆锥的顶点到底面圆心的距离是圆锥的高。本题考查圆柱、圆锥的高的定义，需熟练掌握。\\n\\n【详解】如图：![](images/caf3bc3afa6315d498eb3c2e71b45e029bf572154d440a9842f21f89df8146f1.jpg)![](images/5c0b589456a3c5694a64296ae1819d664e6c9c7baf6cb39083af430f11da2dbc.jpg)\\n\\n圆柱有无数条高，圆锥有1条高。",
+    "knowledge_points": [
+      "圆柱的认识及特征",
+      "圆锥的认识及特征"
+    ],
+    "source": "(2021·云南德宏·小升初真题)",
+    "merged_id": "047",
+    "original_id": "047"
+  },
+  {
+    "id": "048",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "已知两个圆柱的高相等, 较大圆柱的半径等于较小圆柱的直径, 它们的体积共 $78.5 \\mathrm{~dm}^{3}$, 较大圆柱的体积是( ) $\\mathrm{dm}^{3}$, 较小圆柱的体积是( )dm³。",
+    "options": [],
+    "answer": "62.8;15.7",
+    "solution": "【分析】较大圆柱的半径等于较小圆柱的直径，确定两个圆柱底面半径比，圆柱体积 = $\\pi r^{2} h$，半径比的前后项平方以后的比是体积比，体积和 ÷ 总份数，求出一份数，一份数分别乘两个圆柱的对应份数，即可求出它们的体积。关键是理解比的意义，确定两个圆柱的体积比，从而求出它们的体积。\\n\\n【详解】半径比：2:1\\n\\n体积比: $2^{2}: 1^{2} = 4: 1$\\n\\n$78.5 \\div (4 + 1)\\\\= 78.5 \\div 5\\\\= 15.7 (\\mathrm{dm}^{3})$\\n\\n$15.7 \\times 4 = 62.8 (\\mathrm{dm}^{3})$ $15.7 \\times 1 = 15.7 (\\mathrm{dm}^{3})$ 较大圆柱的体积是 $62.8 \\mathrm{dm}^{3}$, 较小圆柱的体积是 $15.7 \\mathrm{dm}^{3}$。",
+    "knowledge_points": [
+      "圆柱的体积",
+      "按比分配问题"
+    ],
+    "source": "(22-23 六年级下·全国·单元测试)",
+    "merged_id": "048",
+    "original_id": "048"
+  },
+  {
+    "id": "049",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个圆柱形, 它的底面周长是 18.84 米, 这个圆柱的底面直径是( )米, 侧面展开图的长为( )米。",
+    "options": [],
+    "answer": "6;18.84",
+    "solution": "【分析】根据圆柱的底面周长是 18.84 厘米，利用圆的周长公式： $C = \\pi d$ 即可求出底面直径；展开图是一个长方形，它的长是圆柱的底面周长；据此解答。本题主要考查了圆的周长公式、圆柱的侧面展开图与圆柱的关系。\\n\\n【详解】 $18.84 \\div 3.14 = 6$ （米）这个圆柱的底面直径是 6 米; 侧面展开图的长是圆柱的底面周长，是 18.84 米。",
+    "knowledge_points": [
+      "圆柱的展开图",
+      "圆的周长及应用"
+    ],
+    "source": "(22-23 六年级下·全国·单元测试)",
+    "merged_id": "049",
+    "original_id": "049"
+  },
+  {
+    "id": "050",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "圆柱体积是圆锥体积的 3 倍。",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】根据 $V_{\\text{柱}} = \\mathrm{Sh}, V_{\\text{锥}} = \\frac{1}{3} \\mathrm{Sh}$ 可知, 当圆柱和圆锥等底等高时, 圆柱的体积是圆锥体积的 3 倍, 据此判断。\\n\\n【详解】当圆柱与圆锥等底等高时，圆柱体积是圆锥体积的3倍。若底面积或高不相等，圆柱和圆锥的体积关系无法确定，原题说法错误。故答案为: ×",
+    "knowledge_points": [
+      "圆柱与圆锥体积的关系"
+    ],
+    "source": "(2023·广东广州·小升初真题)",
+    "merged_id": "050",
+    "original_id": "050"
+  },
+  {
+    "id": "051",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "圆锥体积是圆柱体积的 $\\frac{1}{3}$。",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】只有等底等高的情况下，圆锥的体积是圆柱体积的 $\\frac{1}{3}$，据此判断即可。\\n\\n【详解】因为只有等底等高的圆锥的体积是圆柱体积的 $\\frac{1}{3}$, 所以在没有确定能否等底等高的前提条件下, 圆锥体积是圆柱体积的 $\\frac{1}{3}$, 这种说法是错误的。故答案为: ×",
+    "knowledge_points": [
+      "圆柱与圆锥体积的关系"
+    ],
+    "source": "(2015·天津静海·小升初真题)",
+    "merged_id": "051",
+    "original_id": "051"
+  },
+  {
+    "id": "052",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "一个圆柱的底面半径扩大到原来的 5 倍，高不变，这个圆柱的体积会扩大到原来的 10 倍。",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】圆柱的体积=底面积×高，圆柱的高不变，设圆柱底面半径为r，高为h，原来的体积为V，扩大后的体积为V₁，则扩大后的半径为5r，代入圆柱的体积公式，从而可以求出它的体积扩大到原来的几倍。\\n\\n【详解】设圆柱底面半径为 $r$，高为 $h$ 原来的体积: $V = \\pi r^{2} h$ 扩大后的体积:\\n\\n$\\mathrm{V}_{1} = \\pi (5 \\mathrm{r})^{2} \\mathrm{h}\\\\= \\pi \\times 5r \\times 5r \\times h\\\\= 25 \\pi r^{2} h$\\n\\n$25 \\pi r^{2} h \\div \\pi r^{2} h = 25$\\n\\n一个圆柱的底面半径扩大到原来的 5 倍, 高不变, 这个圆柱的体积会扩大到原来的 25 倍。原题干说法错误。故答案为: ×",
+    "knowledge_points": [
+      "圆柱的体积"
+    ],
+    "source": "(2022·河北石家庄·小升初真题)",
+    "merged_id": "052",
+    "original_id": "052"
+  },
+  {
+    "id": "053",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "圆柱和圆锥都有无数条高。",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】圆柱两个底面之间的距离叫做高，也就是圆柱侧面展开后得到的长方形的宽，从圆锥的顶点到底面圆心的距离是圆锥的高。据此解题。\\n\\n【详解】根据分析可知, 圆柱有无数条高, 而圆锥只有一条高。所以圆柱和圆锥都有无数条高说法错误。故答案为: ×",
+    "knowledge_points": [
+      "圆柱的认识及特征",
+      "圆锥的认识及特征"
+    ],
+    "source": "(2022·河南南阳·小升初真题)",
+    "merged_id": "053",
+    "original_id": "053"
+  },
+  {
+    "id": "054",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "圆柱的底面半径扩大到原来的 2 倍, 高缩小到原来的 $\\frac{1}{2}$, 它的体积不变。",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】圆柱的体积=底面积×高=πr²h，据此可知圆柱的底面半径扩大到原来的2倍，则底面积扩大到原来的 (2×2) 倍, 高缩小到原来的 $\\frac{1}{2}$, 则圆柱的体积先扩大到原来的 (2×2) 倍, 再缩小到 $\\frac{1}{2}$, 据此解答。\\n\\n【详解】 $2 \\times 2 \\times \\frac{1}{2}\\\\= 4 \\times \\frac{1}{2}\\\\= 2$\\n\\n圆柱的底面半径扩大到原来的 2 倍, 高缩小到原来的 $\\frac{1}{2}$, 它的体积扩大到原来的 2 倍; 原说法错误。故答案为: ×",
+    "knowledge_points": [
+      "圆柱的体积"
+    ],
+    "source": "(2022·山西太原·小升初真题)",
+    "merged_id": "054",
+    "original_id": "054"
+  },
+  {
+    "id": "055",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "圆柱体的体积与圆锥体的体积比是 $3:1$。",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】当圆柱与圆锥等底等高时，圆柱的体积是圆锥体积的 3 倍，圆锥的体积是圆柱体积的 $\\frac{1}{3}$，此时圆柱体的体积与圆锥体的体积比是 $3:1$。而题目未提及圆柱与圆锥是否等底等高，如果圆柱与圆锥不是等底等高的情况，它们的体积比就不一定是 $3:1$，所以题目的说法是错误的。\\n\\n【详解】由分析得: 圆柱体的体积与圆锥体的体积比不一定是 $3:1$。故答案为: ×",
+    "knowledge_points": [
+      "圆柱与圆锥体积的关系",
+      "比的意义"
+    ],
+    "source": "(2017·全国·小升初真题)",
+    "merged_id": "055",
+    "original_id": "055"
+  },
+  {
+    "id": "056",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "圆柱的底面半径扩大到原来的 2 倍, 高不变, 体积扩大到原来的 2 倍。",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】设圆柱的底面半径为 $r$, 高为 $h$, 扩大后圆柱底面半径为 $2 \\mathrm{r}$, 高为 $\\mathrm{h}$; 根据圆柱的体积公式: $V = \\pi r^{2} h$, 分别求出原来圆柱的体积和扩大后圆柱的体积, 再用扩大后圆柱的体积除以原来圆柱的体积, 即可解答。\\n\\n【详解】设圆柱的底面半径为 $r$, 高为 $h$, 则扩大后圆柱的底面半径为 $2r$, 高为 $h$。\\n\\n$[\\pi \\times (2r)^{2} h] \\div (\\pi r^{2} h)\\\\= [\\pi 4r^{2} h] \\div (\\pi r^{2} h)\\\\= 4$\\n\\n圆柱的底面半径扩大到原来的 2 倍, 高不变, 体积扩大到原来的 4 倍。原题干说法错误。故答案为: ×",
+    "knowledge_points": [
+      "圆柱的体积"
+    ],
+    "source": "(2024·山西太原·小升初真题)",
+    "merged_id": "056",
+    "original_id": "056"
+  },
+  {
+    "id": "057",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "高相等的圆柱和圆锥, 圆柱的体积是圆锥的 3 倍。",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】圆柱的体积=底面积×高，圆锥的体积=底面积×高× $\\frac{1}{3}$，根据圆柱和圆锥的体积公式可知：圆锥和圆柱底面积和高都相等时，圆柱的体积是圆锥体积的3倍，只有高相等时，底面积无法确定，所以也就无法确定体积的大小关系，据此解答。\\n\\n【详解】底面积和高都相等时, 圆柱的体积是圆锥体积的 3 倍。原题表述错误。故答案为:×",
+    "knowledge_points": [
+      "圆锥的体积（容积）",
+      "圆柱与圆锥体积的关系",
+      "圆柱的体积"
+    ],
+    "source": "(23-24 六年级下·四川绵阳·期末)",
+    "merged_id": "057",
+    "original_id": "057"
+  },
+  {
+    "id": "058",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "圆锥的底面半径缩小到原来的 $\\frac{1}{2}$, 高扩大到原来的 4 倍, 体积不变。",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】假设圆锥底面半径 2 厘米, 高 6 厘米, 根据圆锥体积 = 底面积 × 高 ÷ 3, 分别计算出变化前后的体积, 比较即可。\\n\\n【详解】假设圆锥底面半径2厘米，高6厘米。变化前体积：\\n\\n$3.14 \\times 2^{2} \\times 6 \\div 3\\\\= 3.14 \\times 4 \\times 6 \\div 3\\\\= 25.12$（立方厘米）\\n\\n变化后体积: $3.14 \\times (2 \\div 2)^{2} \\times (6 \\times 4) \\div 3\\\\= 3.14 \\times 1^{2} \\times 24 \\div 3\\\\= 3.14 \\times 1 \\times 24 \\div 3\\\\= 25.12$ （立方厘米）\\n\\n变化前的体积 = 变化后的体积, 体积不变, 说法正确。故答案为: √",
+    "knowledge_points": [
+      "圆锥的体积（容积）"
+    ],
+    "source": "(22-23六年级下·安徽马鞍山·期末)",
+    "merged_id": "058",
+    "original_id": "058"
+  },
+  {
+    "id": "059",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "一个圆柱的侧面沿高展开是一个正方形，则圆柱的高和底面直径相等。",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】一个圆柱的侧面沿高展开是一个长方形，长方形的长和宽分别对应圆柱的底面周长和高，如果圆柱的底面周长和高相等，则圆柱的侧面沿高展开是一个正方形，据此分析。\\n\\n【详解】一个圆柱的侧面沿高展开是一个正方形，则圆柱的高和底面周长相等，原题说法错误。故答案为: ×",
+    "knowledge_points": [
+      "圆柱的展开图"
+    ],
+    "source": "(2017·全国·小升初真题)",
+    "merged_id": "059",
+    "original_id": "059"
+  },
+  {
+    "id": "060",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "计算长方体、正方体、圆柱的体积都可以用底面积乘高。",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】长方体的体积 = 长 × 宽 × 高，其中'长 × 宽'就是长方体的底面积，所以长方体的体积 = 底面积 × 高；正方体的体积 = 棱长 × 棱长 × 棱长, 其中 '棱长 × 棱长' 就是正方体的底面积，所以正方体的体积 = 底面积 × 高；圆柱的体积 $V = \\pi r^{2} h$, 其中 '$\\pi r^{2}$' 就是圆柱的底面积, 所以圆柱的体积 = 底面积 × 高。据此判断。\\n\\n【详解】长方体的体积 = 底面积 × 高，正方体的体积=底面积×高，圆柱的体积 = 底面积 × 高，计算长方体、正方体、圆柱的体积都可以用底面积乘高。原题说法正确。故答案为: √",
+    "knowledge_points": [
+      "圆柱的体积",
+      "长方体的体积",
+      "正方体的体积"
+    ],
+    "source": "(2014·全国·小升初真题)",
+    "merged_id": "060",
+    "original_id": "060"
+  },
+  {
+    "id": "061",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "一个圆柱和圆锥的底面半径比是 $1:2$，高之比是 $2:3$，它们的体积之比是1:6。",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】由一个圆柱和圆锥的底面半径比是 $1:2$, 可以设圆柱的底面半径是 1, 圆锥的底面半径是 2; 由高之比是 $2:3$, 可以设圆柱的高是 2, 圆锥的高是 3。然后根据圆柱的体积公式 $V = \\pi r^{2} h$, 圆锥的体积公式 $V = \\frac{1}{3} \\pi r^{2} h$, 分别求出圆柱和圆锥的体积; 再根据比的意义，写出圆柱和圆锥的体积之比，并化简比。\\n\\n【详解】设圆柱的底面半径是 1, 圆锥的底面半径是 2; 圆柱的高是 2, 圆锥的高是 3。圆柱的体积: $\\pi \\times 1^{2} \\times 2 = 2\\pi$，圆锥的体积: $\\frac{1}{3} \\times \\pi \\times 2^{2} \\times 3 = 4\\pi$，$2\\pi : 4\\pi = 1:2$，圆柱与圆锥的体积之比是 $1:2$。原题说法错误。故答案为: ×",
+    "knowledge_points": [
+      "圆锥的体积（容积）",
+      "圆柱的体积",
+      "比的化简",
+      "比的意义"
+    ],
+    "source": "(22-23 六年级下·浙江绍兴·期末)",
+    "merged_id": "061",
+    "original_id": "061"
+  },
+  {
+    "id": "062",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "把一个圆柱削成最大的圆锥后体积减少了 36 立方厘米, 这个圆锥的体积是 18 立方厘米。",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】把一个圆柱削成最大的圆锥, 则原来的圆柱和削成的圆锥等底等高, 根据圆柱和圆锥的体积公式, 可知最大的圆锥是圆柱的 $\\frac{1}{3}$, 减少的体积是圆柱的 $\\frac{2}{3}$, 把原来的圆柱体积看作单位 '1', 根据分数除法的意义, 用 $36 \\div \\frac{2}{3}$ 即可求出圆柱的体积, 再减去减少的体积, 即可求出圆锥的体积。\\n\\n【详解】 $36 \\div \\frac{2}{3}\\\\= 36 \\times \\frac{3}{2}\\\\=54$ (立方厘米)\\n\\n$54 - 36 = 18$ （立方厘米）把一个圆柱削成最大的圆锥后体积减少了 36 立方厘米, 这个圆锥的体积是 18 立方厘米。原题干说法正确。故答案为: √",
+    "knowledge_points": [
+      "圆柱与圆锥体积的关系",
+      "已知一个数的几分之几是多少，求这个数"
+    ],
+    "source": "(23-24 六年级下·河北邯郸·期末)",
+    "merged_id": "062",
+    "original_id": "062"
+  },
+  {
+    "id": "063",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "圆柱的底面半径和高都扩大到原来的 2 倍, 它的体积扩大到原来的 8 倍。",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】圆柱的体积=底面积×高，设圆柱的底面半径为r，高为h，则扩大后的半径为2r，高为2h，分别求出变化前后的体积，即可求出体积扩大到原来的几倍，计算后再判断即可。\\n\\n【详解】解：设圆柱的底面半径为 $r$，高为 $h$，则扩大后的半径为 $2r$，高为 $2h$; 原体积：$\\pi \\mathrm{r}^{2} \\mathrm{~h}$，现体积：$\\pi \\times (2r)^{2} \\times (2h)\\\\= \\pi \\times 4r^{2} \\times 2h\\\\= 8\\pi r^{2} h\\\\8\\pi r^{2} h \\div \\pi r^{2} h = 8$\\n\\n圆柱的底面半径和高都扩大到原来的 2 倍, 它的体积扩大到原来的 8 倍。原题干说法正确。故答案为: √",
+    "knowledge_points": [
+      "圆柱的体积",
+      "积的变化规律（小数乘法）"
+    ],
+    "source": "(2022·湖南永州·小升初真题)",
+    "merged_id": "063",
+    "original_id": "063"
+  },
+  {
+    "id": "064",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "圆柱体、长方体、正方体的体积都可以用'底面积 × 高'来计算。",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】圆柱体、长方体、正方体的体积公式都是 $V = Sh$, 据此判断即可。\\n\\n【详解】圆柱体体积=底面积×高; 长方体体积=长×宽×高=底面积×高; 正方体体积=棱长×棱长×棱长。圆柱体、长方体、正方体的体积公式都是 $V = Sh$，都可以用底面积乘高来计算，所以原题说法正确；故答案为: √",
+    "knowledge_points": [
+      "圆柱的体积",
+      "长方体的体积",
+      "正方体的体积"
+    ],
+    "source": "(2014·全国·小升初真题)",
+    "merged_id": "064",
+    "original_id": "064"
+  },
+  {
+    "id": "065",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "侧面积相等的两个圆柱, 它们的底面周长和高一定相等。",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】根据圆柱的侧面积公式，$S = Ch$，知道圆柱的侧面积与底面周长和高有关，如果两个圆柱的侧面积相等，它们的底面周长和高不一定相等，由此做出判断。\\n\\n【详解】因为，圆柱的侧面积公式，$S = Ch$，所以，圆柱的侧面积与底面周长和高有关，如底面周长和高分别是2厘米和6厘米的圆柱的侧面积是 $2 \\times 6 = 12$ 平方厘米，底面周长与高分别是 4 厘米、3 厘米的圆柱的侧面积也是 $3 \\times 4 = 12$ 平方厘米, 由此得出，如果两个圆柱的侧面积相等，它们的底面周长不一定相等，故答案为:×",
+    "knowledge_points": [
+      "圆柱的侧面积"
+    ],
+    "source": "(22-23 六年级下·河南周口·期中)",
+    "merged_id": "065",
+    "original_id": "065"
+  },
+  {
+    "id": "066",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "长方体、正方体、圆柱体和圆锥体的体积都等于底面积乘高。",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】根据长方体体积公式：体积=底面积×高；正方体体积公式：体积=底面积×高；圆柱的体积公式：体积=底面积×高；圆锥的体积公式：体积=底面积×高× $\\frac{1}{3}$，据此解答。\\n\\n【详解】长方体、正方体、圆柱的体积都等于底面积乘高，圆锥的体积等于底面积乘高乘 $\\frac{1}{3}$。原题干说法错误。故答案为: ×",
+    "knowledge_points": [
+      "圆锥的体积（容积）",
+      "圆柱的体积",
+      "长方体的体积",
+      "正方体的体积"
+    ],
+    "source": "(2022·河北秦皇岛·小升初真题)",
+    "merged_id": "066",
+    "original_id": "066"
+  },{
+    "id": "067",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "若圆锥的体积等于圆柱体积的 $\\frac{1}{3}$, 则圆锥和圆柱一定等底等高。",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】假设圆锥的底面半径是 1 厘米，高 9 厘米；圆柱的底面半径是 3 厘米，高是 1 厘米，根据圆锥和圆柱的体积公式，分别算出圆锥的体积： $3.14 \\times 1^{2} \\times 9 \\times \\frac{1}{3} = 9.42$ （立方厘米），圆柱的体积： $3.14 \\times 3^{2} \\times 1 = 28.26$ （立方厘米）， $9.42 \\div 28.26 = \\frac{1}{3}$ ，即可得出圆锥的体积等于圆柱体积的 $\\frac{1}{3}$ ，但是圆锥和圆柱不一定等底等高，据此判断。理解并掌握等底等高圆柱与圆锥体之间的关系是解题的关键。\\n\\n【详解】假设圆锥的底面半径是 1 厘米, 高 9 厘米; 圆柱的底面半径是 3 厘米, 高是 1 厘米, 则:\\n\\n圆锥的体积: $3.14 \\times 1^{2} \\times 9 \\times \\frac{1}{3}\\\\= 3.14 \\times 9 \\times \\frac{1}{3}\\\\= 28.26 \\times \\frac{1}{3}\\\\= 9.42$ (立方厘米)\\n\\n圆柱的体积： $3.14 \\times 3^{2} \\times 1\\\\= 3.14 \\times 9\\\\= 28.26$ (立方厘米)\\n\\n$9.42 \\div 28.26 = \\frac{1}{3}$ 因此圆锥的体积等于圆柱体积的 $\\frac{1}{3}$, 则圆锥和圆柱不一定等底等高。故答案为:×",
+    "knowledge_points": [
+      "圆柱与圆锥体积的关系",
+      "圆柱的体积"
+    ],
+    "source": "(2015·广东江门·小升初真题)",
+    "merged_id": "067",
+    "original_id": "067"
+  },
+  {
+    "id": "068",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "如果圆柱体的高与底面周长相等，那么它的侧面沿高剪开再展开后是一个正方形。",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】圆柱的侧面展开图一般是长方形，长方形的长等于圆柱的底面周长，长方形的宽等于圆柱的高；特殊情况下，圆柱的侧面展开图是正方形，此时圆柱的底面周长和高相等；据此判断。本题考查圆柱侧面展开图的特点及应用, 掌握圆柱的侧面展开图与圆柱的底面周长和高之间的关系是解题的关键。\\n\\n【详解】如果圆柱体的高与底面周长相等, 那么它的侧面沿高剪开再展开后是一个正方形。原题说法正确。故答案为: √",
+    "knowledge_points": [
+      "圆柱的展开图"
+    ],
+    "source": "(2018·广西桂林·小升初真题)",
+    "merged_id": "068",
+    "original_id": "068"
+  },
+  {
+    "id": "069",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "圆锥的体积比圆柱体积小 $\\frac{2}{3}$。",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】根据圆柱的体积公式： $V = \\pi r^2 h$ ，圆锥的体积公式： $V = \\frac{1}{3}\\pi r^2 h$ ，如果圆柱和圆锥的底面积、高都相等，则圆锥的体积是圆柱的 $\\frac{1}{3}$, 则圆锥的体积比圆柱体积小 $\\frac{2}{3}$。据此解答。本题主要考查了圆柱和圆锥的体积之间的关系。\\n\\n【详解】根据分析可知, 如果圆柱和圆锥的底面积、高都相等, 则圆锥的体积比圆柱体积小 $\\frac{2}{3}$, 然而底面积和高都未知, 所以无法确定它们的体积, 无法确定它们之间的关系。原题干说法错误。故答案为:×",
+    "knowledge_points": [
+      "圆柱与圆锥体积的关系"
+    ],
+    "source": "(2015·广东江门·小升初真题)",
+    "merged_id": "069",
+    "original_id": "069"
+  },
+  {
+    "id": "070",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "判断题",
+    "difficulty": "困难",
+    "question": "圆柱的侧面展开图一定是长方形。",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】如果沿着圆柱的高展开的，圆柱的底面周长为圆柱侧面展开图的长，圆柱的高为圆柱侧面展开图的宽，如果圆柱的底面周长等于圆柱的高，那么圆柱的侧面展开图就是正方形，如果圆柱的底面周长与圆柱的高不相等，那么圆柱的侧面展开图就是长方形；如果不是沿着圆柱的高展开那么圆柱的侧面展开图可能是平行四边形或不规则图形，据此判断。此题主要考查的是圆柱的侧面展开图。\\n\\n【详解】由分析可得：圆柱的侧面展开图不一定是长方形，原题说法错误。故答案为:×",
+    "knowledge_points": [
+      "圆柱的展开图"
+    ],
+    "source": "(2016·山东菏泽·小升初真题)",
+    "merged_id": "070",
+    "original_id": "070"
+  },
+  {
+    "id": "071",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "一个圆柱体的底面半径扩大到原来的 3 倍, 高缩小到原来的 $\\frac{1}{3}$, 则体积不变。",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】根据半径扩大到原来3倍，圆的面积就扩大到原来的 9 倍, 已知把一个圆柱体底面半径扩大到原来的 3 倍, 高缩小到原来的 $\\frac{1}{3}$, 由圆柱的体积公式: $V = Sh$, 再根据积的变化规律解答。此题主要根据圆柱的体积公式和积的变化规律解决问题。\\n\\n【详解】由分析可知：一个圆柱体的底面半径扩大到原来的 3 倍, 则圆柱的底面积就扩大到原来的 9 倍, 高缩小到原来的 $\\frac{1}{3}$, 则体积就扩大到原来的 $9 \\times \\frac{1}{3} = 3$ 倍。原题干说法错误。故答案为: ×",
+    "knowledge_points": [
+      "圆柱的体积",
+      "积的变化规律（小数乘法）"
+    ],
+    "source": "(2022·甘肃武威·小升初真题)",
+    "merged_id": "071",
+    "original_id": "071"
+  },
+  {
+    "id": "072",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "把一个圆柱体削成一个最大的圆锥体, 圆锥体的体积是削去部分的一半。",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】把圆柱削成最大的圆锥，圆锥与圆柱等底等高，体积是圆柱体积的 $\\frac{1}{3}$，削去部分是圆柱体积的 $1 - \\frac{1}{3} = \\frac{2}{3}$。再用 $\\frac{1}{3} \\div \\frac{2}{3}$，求出圆锥的体积是削去部分的几分之几，再进行判断。熟练掌握圆柱与圆锥体积的关系是解题的关键。\\n\\n【详解】 $1 - \\frac{1}{3} = \\frac{2}{3}\\\\\\frac{1}{3} \\div \\frac{2}{3}\\\\= \\frac{1}{3} \\times \\frac{3}{2}\\\\= \\frac{1}{2}$\\n\\n把一个圆柱体削成一个最大的圆锥体, 圆锥体的体积是削去部分的一半。原题干说法正确。故答案为: √",
+    "knowledge_points": [
+      "圆柱与圆锥体积的关系"
+    ],
+    "source": "(21-22 六年级下·江西南昌·期中)",
+    "merged_id": "072",
+    "original_id": "072"
+  },
+  {
+    "id": "073",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "计算题",
+    "difficulty": "容易",
+    "question": "列式计算。底面半径是 $2 \\mathrm{dm}$、高是 $5 \\mathrm{dm}$、圆柱的侧面积是多少？",
+    "options": [],
+    "answer": "$62.8 \\mathrm{dm}^{2}$",
+    "solution": "【分析】根据圆柱的侧面积公式：侧面积 = 底面周长 × 高，代入数据，即可解答。\\n\\n【详解】 $3.14 \\times 2 \\times 2 \\times 5\\\\= 6.28 \\times 2 \\times 5\\\\= 12.56 \\times 5\\\\= 62.8 (\\mathrm{dm}^{2})$\\n\\n圆柱的侧面积是 $62.8 \\mathrm{dm}^{2}$。",
+    "knowledge_points": [
+      "圆柱的侧面积"
+    ],
+    "source": "(23-24六年级上·河北邯郸·期中)",
+    "merged_id": "073",
+    "original_id": "073"
+  },
+  {
+    "id": "074",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "一个圆锥的底面积是 60 平方厘米, 高 7 厘米, 这个圆锥的体积是多少立方厘米？",
+    "options": [],
+    "answer": "140 立方厘米",
+    "solution": "【分析】已知圆锥的底面积和高, 根据圆锥的体积公式 $\\mathrm{V} = \\frac{1}{3} \\mathrm{Sh}$, 代入数据计算, 求出这个圆锥的体积。\\n\\n【详解】 $\\frac{1}{3} \\times 60 \\times 7 = 140$ （立方厘米）答: 这个圆锥的体积是 140 立方厘米。",
+    "knowledge_points": [
+      "圆锥的体积（容积）"
+    ],
+    "source": "(24-25六年级下·海南海口·单元测试)",
+    "merged_id": "074",
+    "original_id": "074"
+  },
+  {
+    "id": "075",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "做一个直径 4分米, 高 8 分米的圆柱形有盖油桶至少要用铁皮多少平方分米？（铁皮厚度忽略不计）",
+    "options": [],
+    "answer": "125.6 平方分米",
+    "solution": "【分析】圆柱形有盖油桶的底面直径是 4 分米, 则半径是 $4 \\div 2 = 2$ 分米, 根据圆柱的表面积 = 侧面积 + 两个底面积, 圆柱的侧面积 = 底面周长 × 高 = 圆周率 × 直径 × 高, 底面积 = 圆周率 × 半径的平方, 据此代入相关数据解答即可。\\n\\n【详解】 $4 \\div 2 = 2$ (分米)\\n\\n$3.14 \\times 4 \\times 8 + 3.14 \\times 2^{2} \\times 2\\\\= 3.14 \\times 4 \\times 8 + 3.14 \\times 4 \\times 2\\\\= 3.14 \\times (4 \\times 8) + 3.14 \\times (4 \\times 2)\\\\= 3.14 \\times (4 \\times 8 + 4 \\times 2)\\\\= 3.14 \\times (32 + 8)\\\\= 3.14 \\times 40\\\\= 125.6$ (平方分米)\\n\\n答: 至少要用铁皮 125.6 平方分米。",
+    "knowledge_points": [
+      "圆柱的侧面积",
+      "圆柱的表面积"
+    ],
+    "source": "(24-25 六年级下·海南海口·单元测试)",
+    "merged_id": "075",
+    "original_id": "075"
+  },
+  {
+    "id": "076",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "小明家有一个长65厘米，宽25厘米的长方体鱼缸，为了装饰鱼缸，他在鱼缸内放了两块石头，其中一块为高10厘米，体积3848立方厘米的不规则石头，另一块为直径16厘米，高6厘米的圆柱形石头，现在用水管以每分钟4.5立方分米的流量向鱼缸内注水，那么至少需要多长时间能将两块石头完全淹没？（π取3）",
+    "options": [],
+    "answer": "2.5分钟",
+    "solution": "【分析】已知鱼缸内两块石头的高度分别为 10 厘米和 6 厘米, 向鱼缸内注水, 要将两块石头完全淹没, 那么水的高度是 10 厘米; 根据长方体的体积公式 $V = abh$, 求出水和两块石头的总体积; 再用总体积减去两块石头的体积，即是注水的体积；其中圆柱形石头的体积根据圆柱的体积公式 $V = \\pi r^{2} h$ 求解；最后用注水的体积除以水每分钟的流量，即可求出注水的时间。注意单位的换算：1立方分米 = 1000 立方厘米。\\n\\n【详解】将两块石头完全淹没时，水和两块石头的总体积：\\n\\n$65 \\times 25 \\times 10\\\\= 1625 \\times 10\\\\= 16250$ (立方厘米)\\n\\n圆柱形石头的体积：\\n\\n$3 \\times (16 \\div 2)^{2} \\times 6\\\\= 3 \\times 8^{2} \\times 6\\\\= 3 \\times 64 \\times 6\\\\= 1152$ (立方厘米)\\n\\n注水的体积:\\n\\n$16250 - (3848 + 1152)\\\\= 16250 - 5000\\\\= 11250$ (立方厘米)\\n\\n11250 立方厘米 = 11.25 立方分米\\n\\n注水的时间: $11.25 \\div 4.5 = 2.5$ (分钟) 答: 至少需要 2.5 分钟能将两块石头完全淹没。",
+    "knowledge_points": [
+      "圆柱的体积",
+      "长方体的体积",
+      "体积单位间的进率与换算（立方厘米、立方分米和立方米）"
+    ],
+    "source": "(24-25 六年级下·河南郑州·开学考试)",
+    "merged_id": "076",
+    "original_id": "076"
+  },
+  {
+    "id": "077",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "王老师把3个完全一样的圆柱体拼成了一个大的圆柱体。已知拼成后的圆柱体的表面积比一个小圆柱体的表面积多240平方厘米，圆柱体的底面直径是 10 厘米。拼成后的圆柱体的表面积是多少平方厘米?",
+    "options": [],
+    "answer": "517平方厘米",
+    "solution": "【分析】根据题意, 把 3 个完全一样的圆柱体拼成了一个大的圆柱体, 拼成后的圆柱体的表面积比一个小圆柱体的表面积多 240 平方厘米, 表面积多的 240 平方厘米等于原来两个小圆柱的侧面积和，据此可以求出原来每个小圆柱的侧面积，再根据圆柱的表面积 = 侧面积 + 底面积 × 2，把数据代入公式解答。\\n\\n【详解】 $240 \\div 2 \\times 3 + 3.14 \\times (10 \\div 2)^{2} \\times 2\\\\= 120 \\times 3 + 3.14 \\times 25 \\times 2\\\\= 360 + 78.5 \\times 2\\\\= 360 + 157\\\\= 517$ (平方厘米)\\n\\n答: 拼成后大圆柱的表面积是 517 平方厘米。",
+    "knowledge_points": [
+      "圆柱的侧面积",
+      "圆柱的表面积"
+    ],
+    "source": "(23-24六年级下·四川乐山·期中)",
+    "merged_id": "077",
+    "original_id": "077"
+  },
+  {
+    "id": "078",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一个圆柱形容器,从里面量，底面半径 10 厘米，高 15 厘米，容器中的水面高10 厘米。当放入一个底面半径为 5 厘米、高为 9 厘米的圆锥形铁锤，使其沉入水中时，容器中的水面会增高多少厘米?",
+    "options": [],
+    "answer": "0.75厘米",
+    "solution": "【分析】水面上升部分体积等于圆锥形铁锤的体积；根据圆锥的体积公式：体积 = 底面积 × 高 × $\\frac{1}{3}$，代入数据，求出圆锥形铁锤的体积；再根据圆柱的体积公式：体积 = 底面积 × 高；高 = 体积÷底面积，用圆锥形铁锤的体积除以圆柱形容器的底面积，求出水面上升的高度，据此解答。\\n\\n【详解】 $3.14 \\times 5^{2} \\times 9 \\times \\frac{1}{3} \\div (3.14 \\times 10^{2})\\\\= 3.14 \\times 25 \\times 9 \\times \\frac{1}{3} \\div (3.14 \\times 100)\\\\= 78.5 \\times 9 \\times \\frac{1}{3} \\div 314\\\\= 706.5 \\times \\frac{1}{3} \\div 314\\\\= 235.5 \\div 314\\\\= 0.75$ (厘米)\\n\\n答: 容器中的水面会增高 0.75 厘米。",
+    "knowledge_points": [
+      "圆锥的体积（容积）",
+      "圆柱的体积"
+    ],
+    "source": "(23-24六年级下·四川绵阳·期末)",
+    "merged_id": "078",
+    "original_id": "078"
+  },
+  {
+    "id": "079",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "有两个高度相等的容器 $A$ 和 $B$，已知 $A$ 容器半径是 6 厘米，$B$ 容器的半径是 8 厘米，现在把 $A$ 容器装满水，然后全部倒入 $B$ 容器中，测得 $B$ 容器中的水深比 $A$ 容器高的 $\\frac{3}{4}$ 低了 3 厘米。求 $A$、$B$ 两个容器的高是多少厘米？",
+    "options": [],
+    "answer": "16厘米",
+    "solution": "【分析】把容器 $A$ 的高的高度看作单位 '1', 设容器 $A$ 的高为 $x$ 厘米, 根据分数乘法的意义, 则容器 $B$ 中的水深就是 $(\\frac{3}{4}x - 3)$ 厘米, 根据等量关系: 水的体积前后没有改变, 利用圆柱的体积公式: $V = \\pi r^{2} h$, 即可列出方程解决问题。本题考查了等积变形, 关键是理解水的体积前后没有改变, 掌握相应的体积公式是解答本题的关键。\\n\\n【详解】解：设容器 $A$ 的高度为 $x$ 厘米，则容器 $B$ 中的水深就是 $(\\frac{3}{4}x - 3)$ 厘米。由题意得：\\n\\n$\\pi \\times 6^{2} \\times x = \\pi \\times 8^{2} \\times (\\frac{3}{4}x - 3)\\\\\\pi \\times 6^{2} \\times x \\div \\pi = \\pi \\times 8^{2} \\times (\\frac{3}{4}x - 3) \\div \\pi\\\\6^{2} \\times x = 8^{2} \\times (\\frac{3}{4}x - 3)\\\\36x = 64(\\frac{3}{4}x - 3)\\\\36x = 48x - 192\\\\48x - 36x = 192\\\\12x = 192\\\\x = 192 \\div 12\\\\x = 16$\\n\\n所以容器 $A$ 的高是 16 厘米。因为容器 $A、B$ 的高度相等, 所以容器 $B$ 的高度也是 16 厘米。答: A、B两个容器的高都是 16 厘米。",
+    "knowledge_points": [
+      "求一个数的几分之几的问题",
+      "圆柱的体积",
+      "列方程解含一个未知数的问题",
+      "体积的等积变形（圆柱、圆锥）"
+    ],
+    "source": "(2024 六年级·全国·竞赛)",
+    "merged_id": "079",
+    "original_id": "079"
+  },
+  {
+    "id": "080",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "有一个长方体铁块，长 8 分米，宽 4 分米，高 3 分米。把它完全铸成一个圆柱，圆柱的底面半径是 5 分米，高是多少分米？（保留一位小数）",
+    "options": [],
+    "answer": "1.2分米",
+    "solution": "【分析】铁块的体积不变，即熔铸成的圆柱的体积 = 长方体体积，要求熔铸成的圆柱体的高，先要计算出长方体的体积，运用长方体的体积 = 长 × 宽 × 高求出长方体的体积，然后根据圆柱的体积公式： $V = \\pi r^2 h$，代入数据即可解答。\\n\\n【详解】 $8 \\times 4 \\times 3\\\\= 32 \\times 3\\\\= 96$ (立方分米)\\n\\n$96 \\div (3.14 \\times 5 \\times 5)\\\\= 96 \\div 78.5\\\\\\approx 1.2$ (分米)\\n\\n答: 高是 1.2 分米。",
+    "knowledge_points": [
+      "圆柱的体积",
+      "长方体的体积",
+      "体积的等积变形（圆柱、圆锥）"
+    ],
+    "source": "(2024 六年级·全国·竞赛)",
+    "merged_id": "080",
+    "original_id": "080"
+  },
+  {
+    "id": "081",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一台压路机，前轮直径是 1 米，轮宽 1.2 米，工作时每分钟滚动 15 周。(1) 这台压路机 1 分钟前进多少米？(2) 如果每分钟前进 200 米, 一分钟前轮压过的路面是多少平方米?",
+    "options": [],
+    "answer": "(1) 47.1 米;(2) 240 平方米",
+    "solution": "【分析】（1）一台压路机，工作时每分钟滚动 15 周，则这台压路机 1 分钟前进的长度等于前轮 15 周的长度，根据圆的周长 $C = \\pi d$，求出这台压路机 1 分钟前进多少米即可。(2) 如果每分钟前进 200 米, 一分钟前轮压过的路面面积 = 前进长度 × 轮宽, 据此解答即可。\\n\\n【详解】 (1) 前进: $3.14 \\times 1 \\times 15\\\\= 3.14 \\times 15\\\\= 47.1$ (米)\\n\\n答: 这台压路机 1 分钟前进 47.1 米。\\n\\n(2) 面积: $200 \\times 1.2 = 240$ (平方米) 答: 一分钟前轮压过的路面是 240 平方米。",
+    "knowledge_points": [
+      "圆柱的侧面积",
+      "圆的周长"
+    ],
+    "source": "(20-21 六年级下·河南信阳·期末)",
+    "merged_id": "081",
+    "original_id": "081"
+  },
+  {
+    "id": "082",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一个圆锥形麦堆,量得底面周长为 12.56 米, 高 1.8 米。(1) 这个麦堆的占地面积是多少平方米? (2) 如果把这些小麦装入一个底面半径是 2 米圆柱形粮囤（从里面量），刚好装满这个粮囤。这个粮囤的高是多少米？",
+    "options": [],
+    "answer": "(1) 12.56平方米;(2) 0.6 米",
+    "solution": "【分析】（1）根据圆的周长公式： $C = 2\\pi r$，据此求出圆锥的底面半径, 再根据圆的面积公式: $S = \\pi r^{2}$, 据此进行计算即可; (2) 根据圆锥的体积公式: $V = \\frac{1}{3} \\pi r^{2} h$, 据此求出小麦的体积, 再根据圆柱的体积公式: $V = \\pi r^{2} h$, 即 $h = V \\div \\pi r^{2}$, 据此可求出这个粮囤的高。\\n\\n【详解】 (1) $12.56 \\div 3.14 \\div 2\\\\= 4 \\div 2\\\\= 2$ (米)\\n\\n$3.14 \\times 2^{2} = 12.56$ （平方米）答: 这个麦堆的占地面积是 12.56 平方米。\\n\\n(2) $\\frac{1}{3} \\times 3.14 \\times 2^{2} \\times 1.8\\\\= \\frac{1}{3} \\times 3.14 \\times 4 \\times 1.8\\\\= \\frac{1}{3} \\times 1.8 \\times 3.14 \\times 4\\\\= 0.6 \\times 3.14 \\times 4\\\\= 1.884 \\times 4\\\\= 7.536$ (立方米)\\n\\n$7.536 \\div (3.14 \\times 2^{2})\\\\= 7.536 \\div (3.14 \\times 4)\\\\= 7.536 \\div 12.56\\\\= 0.6$ (米)\\n\\n答: 这个粮囤的高是 0.6 米。",
+    "knowledge_points": [
+      "圆锥的体积（容积）",
+      "圆柱的体积",
+      "体积的等积变形（圆柱、圆锥）",
+      "圆的面积"
+    ],
+    "source": "(23-24六年级下·广东东莞·期末)",
+    "merged_id": "082",
+    "original_id": "082"
+  },
+  {
+    "id": "083",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "王大妈家有一堆晒干的圆锥形稻谷，底面半径4米，高0.9米。如果把这堆稻谷装进底面半径是2米的圆柱形粮仓，仓内稻谷高多少米？列综合算式不计算",
+    "options": [],
+    "answer": "$3.14 \\times 4^{2} \\times 0.9 \\times \\frac{1}{3} \\div (3.14 \\times 2^{2})$",
+    "solution": "【分析】先根据圆锥体积 $V = \\frac{1}{3} \\pi r^{2} h$ 求出稻谷体积, 再用稻谷体积 ÷ 圆柱形粮仓的底面积即可解答。\\n\\n【详解】 $3.14 \\times 4^{2} \\times 0.9 \\times \\frac{1}{3} \\div (3.14 \\times 2^{2})\\\\= 3.14 \\times 16 \\times 0.9 \\times \\frac{1}{3} \\div (3.14 \\times 4)\\\\= 3.14 \\times 16 \\times 0.9 \\times \\frac{1}{3} \\div 12.56\\\\= 50.24 \\times 0.9 \\times \\frac{1}{3} \\div 12.56\\\\= 45.216 \\times \\frac{1}{3} \\div 12.56\\\\= 15.072 \\div 12.56\\\\= 1.2$ (米)\\n\\n答: 仓内稻谷高 1.2 米。",
+    "knowledge_points": [
+      "圆锥的体积（容积）",
+      "圆柱的体积",
+      "体积的等积变形（圆柱、圆锥）"
+    ],
+    "source": "(23-24六年级下·河南信阳·期末)",
+    "merged_id": "083",
+    "original_id": "083"
+  },
+  {
+    "id": "084",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "计划制作 1 个没盖的圆柱形铁皮水桶，高是8分米，底面半径是2分米，制作这个水桶至少需要铁皮多少平方分米？",
+    "options": [],
+    "answer": "113.04平方分米",
+    "solution": "【分析】由于是无盖的，所以制作这个水桶需要铁皮的面积就是这个圆柱形铁皮水桶的底面积和侧面积的和；根据圆柱的表面积公式：表面积 = 底面积 + 侧面积，代入数据，即可解答。\\n\\n【详解】 $3.14 \\times 2^{2} + 3.14 \\times 2 \\times 2 \\times 8\\\\= 3.14 \\times 4 + 6.28 \\times 2 \\times 8\\\\= 12.56 + 12.56 \\times 8\\\\= 12.56 + 100.48\\\\= 113.04$ (平方分米)\\n\\n答: 制作这个水桶至少需要铁皮 113.04 平方分米。",
+    "knowledge_points": [
+      "圆柱的表面积"
+    ],
+    "source": "(23-24 六年级下·天津滨海新·期末)",
+    "merged_id": "084",
+    "original_id": "084"
+  },
+  {
+    "id": "085",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "有一个圆柱形容器，它的底面直径是 4 分米，高是 8 分米，容器里装有 $\\frac{3}{4}$ 的水，现将一个底面半径为 2 分米的圆锥放入其中（全部浸在水中），这时容器里的水位高度恰好为 8 分米，这个圆锥的高是多少分米？",
+    "options": [],
+    "answer": "6分米",
+    "solution": "【分析】把容器的高度看作单位 '1', 根据容器里装有 $\\frac{3}{4}$ 的水, 可知此时水的高度是 $(8 \\times \\frac{3}{4})$ 分米。圆锥放入其中 (全部浸在水中), 这时容器里的水位高度恰好为 8 分米, 说明容器内水面上升了 $(8 - 8 \\times \\frac{3}{4})$ 分米。由此利用圆柱的体积公式先求出容器中上升部分的水的体积, 即得出圆锥的体积, 再利用圆锥的高 = 3 × 体积÷圆锥的底面积, 即可解决问题。\\n\\n【详解】 $8 - 8 \\times \\frac{3}{4}\\\\= 8 - 6\\\\= 2$ (分米)\\n\\n$3.14 \\times (4 \\div 2)^{2} \\times 2\\\\= 3.14 \\times 4 \\times 2\\\\= 12.56 \\times 2\\\\= 25.12$ （立方分米）\\n\\n$25.12 \\times 3 \\div (3.14 \\times 2^{2})\\\\= 75.36 \\div (3.14 \\times 4)\\\\= 75.36 \\div 12.56\\\\= 6$ (分米)\\n\\n答: 这个圆锥的高是 6 分米。",
+    "knowledge_points": [
+      "圆锥的体积（容积）",
+      "圆柱的体积",
+      "不规则物体的体积算法（圆柱、圆锥）"
+    ],
+    "source": "(2024·福建莆田·小升初真题)",
+    "merged_id": "085",
+    "original_id": "085"
+  },
+  {
+    "id": "086",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一个底面直径是 12 厘米的圆柱形容器中装有一部分水，水中完全浸没了一个高是 9 厘米的圆锥形铅块。当把铅块从水中取出后, 水面下降了2厘米，这个铅块的底面积是多少平方厘米？",
+    "options": [],
+    "answer": "75.36平方厘米",
+    "solution": "【分析】水面下降的体积就是铅块的体积，用容器底面积×下降的水的高，求出下降的水的体积，即铅锥体积，再根据圆锥铅锤的底面积 = 体积 × 3 ÷ 高，列式解答即可。\\n\\n【详解】 $12 \\div 2 = 6$ （厘米）\\n\\n$3.14 \\times 6^{2} \\times 2\\\\= 3.14 \\times 36 \\times 2\\\\= 113.04 \\times 2\\\\= 226.08$ (立方厘米)\\n\\n$226.08 \\times 3 \\div 9\\\\= 678.24 \\div 9\\\\= 75.36$ (平方厘米)\\n\\n答: 这个铅块的底面积是 75.36 平方厘米。",
+    "knowledge_points": [
+      "圆锥的体积（容积）",
+      "不规则物体的体积算法(圆柱、圆锥)"
+    ],
+    "source": "(22-23六年级下·河南驻马店·期中)",
+    "merged_id": "086",
+    "original_id": "086"
+  },
+  {
+    "id": "087",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "把一个底面直径是4厘米，高是6厘米的铁制圆锥体放入盛满水的桶中，将有多少立方厘米的水溢出？",
+    "options": [],
+    "answer": "25.12立方厘米",
+    "solution": "【分析】根据题意, 溢出的水的体积等于这个圆锥体的体积, 根据圆锥的体积 $V = \\frac{1}{3} \\pi r^{2} h$, 代入数据计算即可。\\n\\n【详解】 $\\frac{1}{3} \\times 3.14 \\times (4 \\div 2)^{2} \\times 6\\\\= \\frac{1}{3} \\times 3.14 \\times 2^{2} \\times 6\\\\= \\frac{1}{3} \\times 3.14 \\times 4 \\times 6\\\\= 25.12$ (立方厘米)\\n\\n答: 将有 25.12 立方厘米的水溢出。",
+    "knowledge_points": [
+      "圆锥的体积（容积）",
+      "不规则物体的体积算法（圆柱、圆锥）"
+    ],
+    "source": "(23-24六年级下·内蒙古通辽·期中)",
+    "merged_id": "087",
+    "original_id": "087"
+  },
+  {
+    "id": "088",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一个圆锥形沙堆, 底面积是 28.26 平方米，高是 2.5 米。用这堆沙在 10 米宽的公路上铺2 厘米厚的路面，能铺多少米？",
+    "options": [],
+    "answer": "117.75 米",
+    "solution": "【分析】根据圆锥的体积公式：体积 = 底面积 × 高 × $\\frac{1}{3}$，代入数据，求出圆锥形沙堆的体积；把这堆沙铺在长方形的路面上就相当于一个长方体，只是形状改变了，但是沙的体积没有变化；根据长方体体积公式：体积 = 长 × 宽 × 高，长 = 体积 ÷ （宽 × 高），代入数据，即可解答，注意单位名数的统一。\\n\\n【详解】2 厘米 = 0.02 米\\n\\n$28.26 \\times 2.5 \\times \\frac{1}{3} \\div (10 \\times 0.02)\\\\= 70.65 \\times \\frac{1}{3} \\div 0.2\\\\= 23.55 \\div 0.2\\\\= 117.75$ (米)\\n\\n答: 能铺 117.75 米。",
+    "knowledge_points": [
+      "圆锥的体积（容积）",
+      "长方体的体积",
+      "体积的等积变形（圆柱、圆锥）"
+    ],
+    "source": "(2015·全国·小升初真题)",
+    "merged_id": "088",
+    "original_id": "088"
+  },
+  {
+    "id": "089",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一个装有水的圆柱形容器，底面直径是10厘米，高是12厘米，量得水深9.5厘米，将石头取出后，水深为7.5厘米，这块石头的积体积是多少？",
+    "options": [],
+    "answer": "157立方厘米",
+    "solution": "【分析】根据题意可知，取出石头后，下降的水的体积就是石头的体积，先求出圆柱的底面半径，然后用圆柱的底面积 × 下降的水位高度 = 下降的水的体积，也是石头的体积，据此列式解答。\\n\\n【详解】 $10 \\div 2 = 5$ （厘米）\\n\\n$3.14 \\times 5^{2} \\times (9.5 - 7.5)\\\\= 3.14 \\times 25 \\times 2\\\\= 3.14 \\times 50\\\\= 157$ (立方厘米)\\n\\n答: 这块石头的体积是 157 立方厘米。",
+    "knowledge_points": [
+      "不规则物体的体积算法（圆柱、圆锥）"
+    ],
+    "source": "(22-23 六年级下·辽宁鞍山·期中)",
+    "merged_id": "089",
+    "original_id": "089"
+  },
+  {
+    "id": "090",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一个装水的圆柱形容器的底面内直径是10厘米，一个铁块完全浸没在这个容器的水中，将铁块取出后，水面下降2厘米。这个铁块的体积是多少？",
+    "options": [],
+    "answer": "157立方厘米",
+    "solution": "【分析】只要求出下降水的体积就是这个铁块的体积，由题可知道圆柱的底面直径是 10 厘米，下降的水深是 2 厘米，运用圆柱的体积公式： $V = \\pi r^2 h$ 解答出来即可。\\n\\n【详解】 $3.14 \\times (10 \\div 2)^{2} \\times 2\\\\= 3.14 \\times 25 \\times 2\\\\= 78.5 \\times 2\\\\= 157$ (立方厘米)\\n\\n答: 这个铁块的体积是 157 立方厘米。",
+    "knowledge_points": [
+      "圆柱的体积",
+      "不规则物体的体积算法（圆柱、圆锥）"
+    ],
+    "source": "(2022·天津·小升初真题)",
+    "merged_id": "090",
+    "original_id": "090"
+  },
+  {
+    "id": "091",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一个底面半径5厘米、高18厘米的圆柱形玻璃容器里装有一部分水，水中浸没着一个高9厘米的圆锥形铅锤。把铅锤从水中取出后，水面下降了2厘米。(1) 这个圆锥形铅锤的体积是多少立方厘米？(2) 这个圆锥形铅锤的底面积约是多少平方厘米？（得数保留一位小数）",
+    "options": [],
+    "answer": "(1) 157 立方厘米;(2) 52.3 平方厘米",
+    "solution": "【分析】（1）根据下降部分水的体积等于物体的体积，下降部分水的体积 = 底面积 × 下降的高度，则用 $3.14 \\times 5^{2} \\times 2$ 即可求出圆锥形铅锤的体积；(2) 根据圆锥的体积公式: $V = \\frac{1}{3} \\mathrm{~Sh}$, 用圆锥的体积 × 3 ÷ 9 即可求出圆锥形的底面积。本题主要考查了圆柱体积公式、圆锥体积公式的灵活应用，要熟练掌握公式。\\n\\n【详解】 (1) $3.14 \\times 5^{2} \\times 2\\\\= 3.14 \\times 25 \\times 2\\\\= 157$（立方厘米）\\n\\n答: 这个圆锥形铅锤的体积是 157 立方厘米。\\n\\n(2) $157 \\times 3 \\div 9 \\approx 52.3$ （平方厘米）答: 这个圆锥形铅锤的底面积约是 52.3 平方厘米。",
+    "knowledge_points": [
+      "圆锥的体积（容积）",
+      "圆柱的体积"
+    ],
+    "source": "(22-23六年级下·河南安阳·期末)",
+    "merged_id": "091",
+    "original_id": "091"
+  },
+  {
+    "id": "092",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "一个圆柱形水池，测得底面周长是25.12米，池深3米，池上装有6个进水管，每个管每小时可以注入水6.28立方米，六管齐开，几小时可以注满水池？",
+    "options": [],
+    "answer": "4小时",
+    "solution": "【分析】根据底面周长可以求出底面半径，根据底面半径和圆柱的高可以求出水池的容积，圆柱的容积 = 底面积 × 高，用圆柱的容积除以六个水管每小时注水的体积，就可以知道需要几小时注满水池。重点考查圆柱的相关知识, 能够根据圆的周长求出半径, 会求圆柱的容积。\\n\\n【详解】半径： $25.12 \\div 3.14 \\div 2\\\\= 8 \\div 2\\\\= 4$ (米)\\n\\n$3.14 \\times 4 \\times 4 \\times 3 \\div (6.28 \\times 6)\\\\= 150.72 \\div 37.68\\\\= 4$ (小时)\\n\\n答: 4 小时可以注满水池。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆柱的容积"
+    ],
+    "source": "(22-23六年级下·贵州六盘水·期中)",
+    "merged_id": "092",
+    "original_id": "092"
+  },
+  {
+    "id": "093",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一个圆柱形的沼气池, 底面直径是 3 米, 深 2 米。在池子的四壁和下底面抹上水泥，抹水泥的面积是多少平方米？",
+    "options": [],
+    "answer": "25.905平方米",
+    "solution": "【分析】根据题意，在圆柱形沼气池的四壁和下底面抹上水泥，少上底面，那么抹水泥的面积 = 圆柱的侧面积十一个底面积，其中 $S_{\\text{侧}} = \\pi \\mathrm{dh}$， $S_{\\text{底}} = \\pi \\mathrm{r}^{2}$，代入数据计算求解。本题考查圆柱表面积公式的灵活运用，明确求抹水泥的面积是求圆柱哪些面的面积，再运用面积公式解答。\\n\\n【详解】 $3.14 \\times 3 \\times 2 + 3.14 \\times (3 \\div 2)^{2}\\\\= 3.14 \\times 6 + 3.14 \\times 2.25\\\\= 18.84 + 7.065\\\\= 25.905$ (平方米)\\n\\n答: 抹水泥的面积是 25.905 平方米。",
+    "knowledge_points": [
+      "圆柱的表面积"
+    ],
+    "source": "(18-19 六年级下·辽宁·课后作业)",
+    "merged_id": "093",
+    "original_id": "093"
+  },
+  {
+    "id": "094",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一个胶水瓶, 它的瓶身呈圆柱形（不包括瓶颈），容积为32.4立方厘米。当瓶子正放时，瓶内胶水液面高（即瓶身的高）为8厘米，瓶子倒放时，空余部分高为2厘米。请你算一算，瓶内胶水的体积是多少立方厘米？",
+    "options": [],
+    "answer": "25.92 立方厘米",
+    "solution": "【分析】由题意可知, 正放时, 空余部分的体积即为高为 2 厘米的圆柱的体积, 据此利用容积除以 $(8 + 2)$ 求出瓶子的底面积, 再利用底面积乘胶水的高度即可。解答此题的关键是理解胶水的体积包括正放时圆柱部分的体积。\\n\\n【详解】 $32.4 \\div (8 + 2) \\times 8\\\\= 32.4 \\div 10 \\times 8\\\\= 3.24 \\times 8\\\\= 25.92$ (立方厘米)\\n\\n答: 瓶内胶水的体积是 25.92 立方厘米。",
+    "knowledge_points": [
+      "不规则物体的体积算法",
+      "圆柱的容积"
+    ],
+    "source": "(2015·全国·小升初真题)",
+    "merged_id": "094",
+    "original_id": "094"
+  },
+  {
+    "id": "095",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "修建一个圆柱形的沼气池，底面直径3米，深2米。在池的四壁与下底面抹上水泥。(1) 抹水泥部分的面积是多少平方米？(2) 这个沼气池的最大容积是多少立方米?",
+    "options": [],
+    "answer": "(1) 25.905 平方米;(2) 14130 升",
+    "solution": "【分析】（1）由题意可知：抹水泥部分的面积 = 沼气池的侧面积 + 下底的面积，又因圆柱的侧面积 = 底面周长 × 高，圆锥的底面直径已知，于是可以求出其底面周长和底面积，进而可以求出抹水泥部分的面积；(2) 利用圆柱的体积公式 $v = \\pi r^{2} h$ 即可求出这个水池最多能装多少水。此题主要考查圆柱的侧面积和圆的面积及体积的计算方法。\\n\\n【详解】 (1) $3.14 \\times 3 \\times 2 + 3.14 \\times (\\frac{3}{2})^{2}\\\\= 9.42 \\times 2 + 3.14 \\times 2.25\\\\= 18.84 + 7.065\\\\= 25.905$ (平方米)\\n\\n答: 抹水泥部分的面积是 25.905 平方米。\\n\\n(2) $3.14 \\times (\\frac{3}{2})^{2} \\times 2\\\\= 3.14 \\times 2.25 \\times 2\\\\= 7.065 \\times 2\\\\= 14.13$ (立方米)\\n\\n14.13 立方米 = 14130 升\\n\\n答: 这个水池最多能装 14130 升水。",
+    "knowledge_points": [
+      "圆柱的容积",
+      "圆柱的表面积"
+    ],
+    "source": "(2022·湖北黄冈·小升初真题)",
+    "merged_id": "095",
+    "original_id": "095"
+  },
+  {
+    "id": "096",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一个水池, 有一根内直径是 20 厘米的圆柱形排水管, 水流的速度是每秒 4 米, 这根排水管 1 小时能把 450 立方米的水排完吗？",
+    "options": [],
+    "answer": "能",
+    "solution": "【分析】统一单位, 1 小时水流出的距离相等于圆柱的高, 求出底面积, 底面积 × 1 小时水流出的距离 = 流出的体积, 与 450 立方米比较即可。关键是掌握并灵活运用圆柱体积公式。\\n\\n【详解】1小时 = 3600 秒\\n\\n20 厘米 = 0.2 米\\n\\n$3.14 \\times (0.2 \\div 2)^{2} \\times (4 \\times 3600)\\\\= 3.14 \\times 0.1^{2} \\times 14400\\\\= 3.14 \\times 0.01 \\times 14400\\\\= 452.16$ (立方米)\\n\\n$452.16 > 450$ 答: 这根排水管 1 小时能把 450 立方米的水排完。",
+    "knowledge_points": [
+      "圆柱的体积"
+    ],
+    "source": "(22-23 六年级下·全国·单元测试)",
+    "merged_id": "096",
+    "original_id": "096"
+  },
+  {
+    "id": "097",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一根圆柱形木料，长8米，高减少2厘米，表面积减少18.84平方厘米，这根木料的体积是多少？",
+    "options": [],
+    "answer": "5652 立方厘米",
+    "solution": "【分析】根据题意可知, 减少的部分的面积就是高为 2 厘米圆柱的侧面积, 根据侧面积公式: 底面周长 × 高, 代入数据, 求出底面周长, 再根据圆的周长公式: 周长 = $2\\pi r$; 代入数据, 求出底面半径; 再根据圆柱的体积公式: $V = \\pi r^{2} h$, 代入数据, 即可解答。利用圆柱的体积公式解答本题, 关键明确减少的面积就是减少这个圆柱体的侧面积, 进而求出圆柱的底面半径, 解答问题。\\n\\n【详解】 $18.84 \\div 2 = 9.42$ （厘米）\\n\\n$9.42 \\div 2 \\div 3.14 = 1.5$ （厘米）\\n\\n8米=800厘米\\n\\n$3.14 \\times 1.5^{2} \\times 800\\\\= 3.14 \\times 2.25 \\times 800\\\\= 5652$ (立方厘米)\\n\\n答: 这根木料的体积是 5652 立方厘米。",
+    "knowledge_points": [
+      "圆的周长及应用",
+      "圆柱的体积",
+      "圆柱的侧面积"
+    ],
+    "source": "(22-23六年级下·全国·单元测试)",
+    "merged_id": "097",
+    "original_id": "097"
+  },
+  {
+    "id": "098",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一个圆锥体，底面周长15.7分米，高是4.5分米，体积是多少立方分米？",
+    "options": [],
+    "answer": "29.4375 立方分米",
+    "solution": "【分析】利用圆的周长公式求出底面半径, 再利用圆锥的体积公式 $V = \\frac{1}{3} \\pi r^{2} h$ 代入数据计算解答。本题考查了圆锥体积公式的应用, 关键是熟记公式。\\n\\n【详解】圆锥的底面半径: $15.7 \\div 3.14 \\div 2\\\\= 5 \\div 2\\\\= 2.5$ (分米)\\n\\n$3.14 \\times 2.5^{2} \\times 4.5 \\times \\frac{1}{3}\\\\= 3.14 \\times 6.25 \\times 4.5 \\times \\frac{1}{3}\\\\= 19.625 \\times 4.5 \\times \\frac{1}{3}\\\\= 88.3125 \\times \\frac{1}{3}\\\\= 29.4375$（立方分米）\\n\\n答: 体积是 29.4375 立方分米。",
+    "knowledge_points": [
+      "圆锥的体积（容积）"
+    ],
+    "source": "(22-23六年级下·全国·单元测试)",
+    "merged_id": "098",
+    "original_id": "098"
+  },
+  {
+    "id": "099",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一根圆柱形柱子,底面直径 4 分米, 高 6 米, 这根柱子的体积是多少立方米?",
+    "options": [],
+    "answer": "0.7536 立方米",
+    "solution": "【分析】此题就是求这个圆柱的体积，题目中已知了底面直径和高，代入公式即可求体积。此题考查了求圆柱的体积公式, 熟记公式即可解答。\\n\\n【详解】底面半径为： $4 \\div 2 = 2$ （分米）\\n\\n2分米=0.2米\\n\\n$3.14 \\times 0.2^{2} \\times 6\\\\= 3.14 \\times 0.04 \\times 6\\\\= 0.1256 \\times 6\\\\= 0.7536$ （立方米）\\n\\n答: 这根柱子的体积是 0.7536 立方米。",
+    "knowledge_points": [
+      "圆柱的体积"
+    ],
+    "source": "(22-23 六年级下·全国·单元测试)",
+    "merged_id": "099",
+    "original_id": "099"
+  },
+  {
+    "id": "100",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "制作50节长6米,底面直径 3 分米的圆柱形通风管, 至少需要多少平方平方米铁皮?",
+    "options": [],
+    "`answer`": "282.6 平方米",
+    "solution": "【分析】此题就是求 50 个底面直径为 3 分米, 长 6 米的圆柱的侧面积, 由此利用圆柱的侧面积 = 底面周长 × 高即可计算。此题考查了圆柱的侧面积公式的应用。\\n\\n【详解】3分米=0.3米\\n\\n一节圆柱形通风管需要:\\n\\n$0.3 \\times 6 \\times 3.14\\\\= 1.8 \\times 3.14\\\\= 5.652$（平方米）\\n\\n$5.652 \\times 50 = 282.6$（平方米）\\n\\n答: 至少需要 282.6 平方米铁皮。",
+    "knowledge_points": [
+      "圆柱的侧面积"
+    ],
+    "source": "(22-23六年级下·全国·单元测试)",
+    "merged_id": "100",
+    "original_id": "100"
+  },
+  {
+    "id": "101",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一个圆锥的体积是 6 立方米, 底面积是 6 平方米, 与它等体积等底面积的圆柱的高是 ( )。",
+    "options": [
+      "A. 1 米",
+      "B. 3米",
+      "C. $\\frac{1}{3}$ 米",
+      "D. $\\frac{2}{3}$ 米"
+    ],
+    "answer": "A",
+    "solution": "【分析】已知圆锥和圆柱等底面积等体积, 根据圆柱的体积 $=$ 底面积 $\\times$ 高, 用 $6 \\div 6$ 即可得知圆柱的高。此题主要考查了利用圆柱与圆锥的体积公式。\n\n【详解】$6 \\div 6 = 1$ (米)与它等体积等底面积的圆柱的高是 1 米。故答案为: A",
+    "knowledge_points": [
+      "圆柱与圆锥体积的关系",
+      "圆柱的体积"
+    ],
+    "source": "(22-23 六年级下·全国·单元测试)",
+    "merged_id": "101",
+    "original_id": "101"
+  },
+  {
+    "id": "102",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "底面直径是 $6 \\mathrm{~m}$ 的圆柱形水池, 它的占地面积是 ( ) $\\mathrm{m}^{2}$。",
+    "options": [
+      "A. 21.98",
+      "B. 28.26",
+      "C. 56.52"
+    ],
+    "answer": "B",
+    "solution": "【分析】求这个圆柱形水池的占地面积，实际上是求这个水池的底面积，先求出圆柱的底面半径，再根据圆的面积公式： $S = \\pi r^2$ ，代入数据即可求出它的占地面积。此题的解题关键是熟悉圆柱的特征，通过圆的面积公式求解。\n\n【详解】$3.14 \\times (6 \\div 2)^{2}\\\\= 3. 1 4 \\times 3 ^ {2} \\\\ = 3. 1 4 \\times 9 \\\\ = 2 8. 2 6(\\mathrm {m} ^ {2})$\n\n所以，它的占地面积是 $28.26 \\mathrm{~m}^{2}$ ;故答案为: B",
+    "knowledge_points": [
+      "圆的面积及应用",
+      "圆柱的认识及特征"
+    ],
+    "source": "(22-23 六年级下·全国·单元测试)",
+    "merged_id": "102",
+    "original_id": "102"
+  },
+  {
+    "id": "103",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "将一个正方体木块削成一个最大的圆柱, 削成的圆柱与原来正方体体积的比是 ( )。",
+    "options": [
+      "A. $2: 3$",
+      "B. $\\pi : 1$",
+      "C. $157: 200$"
+    ],
+    "answer": "C",
+    "solution": "【分析】削成体积最大的圆柱体，那么它的底面直径和高都是正方体的棱长; 设正方体的棱长为 1 , 由此求出正方体和圆柱体的体积, 据此解答。解答本题的关键是设出数据, 再根据圆柱体的底面直径和高与正方体棱长之间的关系, 求出体积即可作出判断。\n\n【详解】假设正方体的棱长是1则正方体的体积：\n\n$1 \\times 1 \\times 1 = 1\\\\1 \\div 2 = 0. 5$\n\n圆柱的体积: $3.14 \\times 0.5^{2} \\times 1\\\\= 3. 1 4 \\times 0. 2 5 \\times 1 \\\\ = 0. 7 8 5 \\times 1 \\\\ = 0. 7 8 5$\n\n$0.7 8 5: 1 \\\\ = (0. 7 8 5 \\times 1 0 0 0 \\div 5): (1 \\times 1 0 0 0 \\div 5) \\\\ = 1 5 7: 2 0 0$\n\n所以削成的圆柱与原来正方体体积的比是 $157: 200$故答案为: C",
+    "knowledge_points": [
+      "正方体的体积",
+      "比的意义",
+      "圆柱的体积"
+    ],
+    "source": "(22-23 六年级下·全国·单元测试)",
+    "merged_id": "103",
+    "original_id": "103"
+  },
+  {
+    "id": "104",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "小王从正面、左面和上面看一个几何体, 看到的图形都是正方形。这个几何体是 ( )。",
+    "options": [
+      "A. 球",
+      "B. 圆柱",
+      "C. 正方体"
+    ],
+    "answer": "C",
+    "solution": "【分析】根据球、圆柱和正方体的特征，描述出它们的三视图，从而解题。掌握球、圆柱以及正方体的特征是解题的关键。\n\n【详解】A．球从正面、左面和上面看，看到的图形都是一个圆；B. 圆柱从正面、左面看是一个长方形或正方形, 从上面看是一个圆;C. 正方体从正面、左面和上面看, 看到的图形都是正方形;所以, 小王看的几何体是正方体。故答案为: C",
+    "knowledge_points": [
+      "正方体的特征",
+      "圆柱的认识及特征",
+      "从不同位置观察单个物体"
+    ],
+    "source": "(22-23 六年级下·全国·单元测试)",
+    "merged_id": "104",
+    "original_id": "104"
+  },
+  {
+    "id": "105",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "在下面的图形中, 以任意一边为轴旋转一周, 可以得到圆柱体的是 ( )。",
+    "options": [
+      "A. 三角形",
+      "B. 长方形",
+      "C. 平行四边形"
+    ],
+    "answer": "B",
+    "solution": "【分析】圆柱是以长方形的一条边所在的直线为旋转轴,其余三边绕旋转轴旋转一周而形成的几何体，据此解答。掌握圆柱的特征是解答题目的关键。\n\n【详解】\n\n![](images/4d8c6eb3e012f2fb7d0780ed01c48b64fdbff9918ce177b8dcbb65a4023ecfe3.jpg)\n\n![](images/5f96d5018fdad3763b79b983363ea55ada19a1d1f63d0382207f3d37a530c52a.jpg)\n\n分析可知, 以长方形任意一边为轴旋转一周, 可以得到圆柱体。故答案为: B",
+    "knowledge_points": [
+      "旋转与旋转现象",
+      "圆柱的认识及特征"
+    ],
+    "source": "(22-23 六年级下·全国·单元测试)",
+    "merged_id": "105",
+    "original_id": "105"
+  },
+  {
+    "id": "106",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "压路机的前轮转动一周所压过的路面面积是指（）。",
+    "options": [
+      "A. 前轮的表面积",
+      "B. 前轮的侧面积",
+      "C. 前轮的底面积"
+    ],
+    "answer": "B",
+    "solution": "【分析】压路机的前轮是圆柱形，这个圆柱是侧躺在地面，压路机在工作时是前轮的侧面与地面接触，转动一周，所压过的面正好是圆柱的侧面，据此解答。本题考查圆柱的侧面积在实际生活中的应用。\n\n【详解】压路机的前轮转动一周所压过的路面面积是指前轮的侧面积。故答案为: B",
+    "knowledge_points": [
+      "圆柱的侧面积"
+    ],
+    "source": "(2014·全国·小升初真题)",
+    "merged_id": "106",
+    "original_id": "106"
+  },
+  {
+    "id": "107",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "把一个圆柱削成一个最大的圆锥体，削去部分的体积是圆柱体积的（）。",
+    "options": [
+      "A. 3 倍",
+      "B. $\\frac{1}{3}$",
+      "C. $\\frac{2}{3}$"
+    ],
+    "answer": "C",
+    "solution": "【分析】把一个圆柱削成一个最大的圆锥体, 圆柱和圆锥等底等高, 等底等高的圆柱和圆锥, 圆锥体积是圆柱体积的 $\\frac{1}{3}$ , 削去部分的体积是圆柱体积的 $(1 - \\frac{1}{3})$ , 据此分析。关键是理解等底等高的圆柱和圆锥体积之间的关系。\n\n【详解】$1 - \\frac{1}{3} = \\frac{2}{3}$ , 削去部分的体积是圆柱体积的 $\\frac{2}{3}$ 。故答案为: C",
+    "knowledge_points": [
+      "圆柱与圆锥体积的关系"
+    ],
+    "source": "(20-21 六年级下·湖南衡阳·期中)",
+    "merged_id": "107",
+    "original_id": "107"
+  },
+  {
+    "id": "108",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一个圆柱与一个圆锥的体积和高分别相等, 已知圆柱的底面积是 $9.42 \\mathrm{~cm}^{2}$ , 则圆锥的底面积是 ( )。",
+    "options": [
+      "A. $9.42 \\mathrm{~cm}^{2}$",
+      "B. $28.26 \\mathrm{~cm}^{2}$",
+      "C. $3.14 \\mathrm{~cm}^{2}$",
+      "D. 无法确定"
+    ],
+    "answer": "B",
+    "solution": "【分析】当圆柱和圆锥等底等高时，圆锥的体积是圆柱体积的 $\\frac{1}{3}$ ，圆柱和圆锥等体积等高时，圆锥的底面积是圆柱底面积的 3 倍，据此解答。掌握圆柱和圆锥的体积关系是解答题目的关键。\n\n【详解】$9.42 \\times 3 = 28.26 \\, (\\mathrm{cm}^{2})$所以，圆锥的底面积是 $28.26 \\mathrm{~cm}^{2}$ 。故答案为: B",
+    "knowledge_points": [
+      "圆柱与圆锥体积的关系"
+    ],
+    "source": "(20-21 六年级下·江西九江·期末)",
+    "merged_id": "108",
+    "original_id": "108"
+  },
+  {
+    "id": "109",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "冰球运动 (IceHockey)是以冰刀和冰球杆为工具在冰上进行的一种相互对抗的集体性竞技运动，冰球一般是用硬橡胶制成的圆柱体，厚2.54厘米, 直径 7.62 厘米, 重 $156 \\sim 170$ 克, 如果将 3 个冰球重叠在一起，表面积比原来减少了（）平方厘米。",
+    "options": [
+      "A. $3.14 \\times (7.62 \\div 2)^{2} \\times 2.54$",
+      "B. $3.14 \\times 7.62^{2} \\times 4$",
+      "C. $3.14 \\times (7.62 \\div 2)^{2} \\times 4$"
+    ],
+    "answer": "C",
+    "solution": "【分析】3 个冰球重叠在一起，减少的表面积是 4 个圆柱的底面积， $S = \\pi r^{2}$ ，代入数据即可。本题考查图形的组合与分割问题, 当 $n$ 个立体图形组合在一起时, 减少 $[ (n - 1) \\times 2 ]$ 个重叠面的面积。\n\n【详解】直径：7.62 厘米，半径：（7.62÷2）厘米减少的面积是 4 个底面圆形的面积，列式为：$3.1 4 \\times (7. 6 2 \\div 2) ^ {2} \\times 4$故答案为: C。",
+    "knowledge_points": [
+      "圆柱的表面积",
+      "立体图形的切拼"
+    ],
+    "source": "(2022·甘肃金昌·小升初真题)",
+    "merged_id": "109",
+    "original_id": "109"
+  },
+  {
+    "id": "110",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一个圆柱和一个圆锥的体积与底面积均相等，圆柱的高是12厘米，圆锥的高是（）厘米。",
+    "options": [
+      "A. 3",
+      "B. 6",
+      "C. 12",
+      "D. 36"
+    ],
+    "answer": "D",
+    "solution": "【分析】圆柱的体积=底面积×高，圆锥的体积= $\\frac{1}{3} \\times$ 底面积×高，那么如果一个圆柱和一个圆锥的体积与底面积均相等，则圆锥的高是圆柱高的 3 倍。据此，用圆柱的高乘 3 ，即可求出圆锥的高。本题考查了圆柱和圆锥的体积关系, 解题关键是熟记公式。\n\n【详解】$12 \\times 3 = 36$ （厘米）所以，这个圆锥的高是 36 厘米。故答案为: D",
+    "knowledge_points": [
+      "圆柱与圆锥体积的关系"
+    ],
+    "source": "(21-22 五年级下·山东泰安·单元测试)",
+    "merged_id": "110",
+    "original_id": "110"
+  },
+  {
+    "id": "111",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一个圆柱的高是 $4 \\mathrm{~cm}$ ,侧面展开是一个正方形, 那么它的侧面积是 ( ) 平方厘米。",
+    "options": [
+      "A. 16",
+      "B. $16 \\pi$",
+      "C. $64 \\pi$"
+    ],
+    "answer": "A",
+    "solution": "【分析】一个圆柱的侧面展开是一个正方形，则该圆柱的高等于圆柱的底面周长，根据圆柱的侧面积 $=$ 底面周长 $\\times$ 高，据此代入数值进行计算即可。本题考查圆柱的侧面积, 明确该圆柱的高等于圆柱的底面周长是解题的关键。\n\n【详解】$4 \\times 4 = 16$ （平方厘米）故答案为: A",
+    "knowledge_points": [
+      "圆柱的侧面积",
+      "正方形的面积"
+    ],
+    "source": "(21-22 六年级下·山西忻州·期末)",
+    "merged_id": "111",
+    "original_id": "111"
+  },
+  {
+    "id": "112",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "把一个 $60 \\mathrm{~cm}^{3}$ 的圆柱削成一个最大的圆锥, 削去部分的体积是 ( )。",
+    "options": [
+      "A. $20 \\mathrm{~cm}^{3}$",
+      "B. $40 \\mathrm{~cm}^{3}$",
+      "C. $30 \\mathrm{~cm}^{3}$",
+      "D. $10 \\mathrm{~cm}^{3}$"
+    ],
+    "answer": "B",
+    "solution": "【分析】在等底等高的情况下, 圆锥的体积是圆柱体积的 $\\frac{1}{3}$ ,把圆柱的体积看作单位 '1', 所以削去部分的体积是圆柱体积的 $(1 - \\frac{1}{3})$ , 已知圆柱的体积是 $60 \\mathrm{~cm}^{3}$ , 求一个数的几分之几是多少, 用乘法, 即可求出削去部分的体积。此题的解题关键是利用圆锥和圆柱体积之间的关系求解。\n\n【详解】$60 \\times (1 - \\frac{1}{3})\\\\= 6 0 \\times \\frac {2}{3} \\\\ = 4 0(\\mathrm {c m} ^ {3})$\n\n故答案为: B",
+    "knowledge_points": [
+      "圆柱与圆锥体积的关系"
+    ],
+    "source": "(21-22 六年级下·江西景德镇·期末)",
+    "merged_id": "112",
+    "original_id": "112"
+  },
+  {
+    "id": "113",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一个圆柱的侧面展开图是一个正方形。这个圆柱的高是底面直径的（）倍。",
+    "options": [
+      "A. $\\pi$",
+      "B. $2 \\pi$",
+      "C. $\\frac{1}{4} \\pi$",
+      "D. $\\frac{1}{2} \\pi$"
+    ],
+    "answer": "A",
+    "solution": "【分析】根据圆柱侧面展开图的特征，圆柱的侧面沿高展开是一个长方形，这个长方形的长等于圆柱的底面周长，宽等于圆柱的高，如果圆柱的侧面展开图是一个正方形，那么这个圆柱的底面周长和高相等，根据圆的周长公式： $C = \\pi d$ ，据此解答即可。此题做题的关键是要明确'圆柱的侧面展开后是长方形，长方形的长等于圆柱的底面周长，宽等于圆柱的高'，并能根据底面周长和底面直径的关系进行解答。\n\n【详解】一个圆柱的侧面展开图是一个正方形。那么这个圆柱的底面周长和高相等，由圆周率的意义， $\\frac{\\text { 圆的周长 }}{\\text { 直径 }} =$ 圆周率 $(\\pi)$ ，所以这个圆柱的高是底面直径的 $\\pi$ 倍。故答案为：A",
+    "knowledge_points": [
+      "圆柱的侧面积",
+      "圆柱的展开图"
+    ],
+    "source": "(2022·浙江金华·小升初真题)",
+    "merged_id": "113",
+    "original_id": "113"
+  },
+  {
+    "id": "114",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "用一个高9厘米的圆锥形容器盛满水，再将水倒入和它等底等高的圆柱形容器中，水的高度是（）厘米。",
+    "options": [
+      "A. 3",
+      "B. 6",
+      "C. 9",
+      "D. 27"
+    ],
+    "answer": "A",
+    "solution": "【分析】根据题意, 圆锥形容器中的水倒入等底的圆柱形容器中, 那么水的体积不变, 底面积也不变; 根据 $V_{\\text {柱}} =$ $\\mathrm{Sh}, V_{\\text {锥}} = \\frac{1}{3} \\mathrm{Sh}$ 可知, 当圆柱和圆锥等体积等底面积时, 圆柱形容器里水的高度是圆锥高度的 $\\frac{1}{3}$ 。\n\n【详解】$9 \\times \\frac{1}{3} = 3$ （厘米）故答案为: A",
+    "knowledge_points": [
+      "圆锥的体积（容积）",
+      "圆柱与圆锥体积的关系",
+      "圆柱的体积"
+    ],
+    "source": "(18-19六年级下·全国·课后作业)",
+    "merged_id": "114",
+    "original_id": "114"
+  },
+  {
+    "id": "115",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "用一个高为 15 厘米的圆锥容器盛满水，将水倒入和它等底等高的圆柱形容器里，水面的高是（）厘米。",
+    "options": [
+      "A. 15",
+      "B. 30",
+      "C. 5"
+    ],
+    "answer": "C",
+    "solution": "【分析】因为, 一个圆锥的体积是与它等底等高的圆柱体积的 $\\frac{1}{3}$ ; 所以, 一个高为 15 厘米的圆锥容器盛满水, 将水倒入和它等底等高的圆柱形容器里, 水面的高应是圆锥高的 $\\frac{1}{3}$ 。圆柱体积=底面积×高，圆锥体积=底面积×高× $\\frac{1}{3}$ 。\n\n【详解】$15 \\times \\frac{1}{3} = 5$ （厘米）故答案为: C",
+    "knowledge_points": [
+      "圆柱与圆锥体积的关系"
+    ],
+    "source": "(2018·山东临沂·小升初真题)",
+    "merged_id": "115",
+    "original_id": "115"
+  },
+  {
+    "id": "116",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一张长 $28.26 \\mathrm{~cm}$ 、宽 $15.7 \\mathrm{~cm}$ 的长方形铁皮，应该配上直径是（）的圆形铁皮就可以了做成一个容积最大的容器。",
+    "options": [
+      "A. $2.4 \\mathrm{~cm}$",
+      "B. $5 \\mathrm{~cm}$",
+      "C. $9 \\mathrm{~cm}$",
+      "D. $7 \\mathrm{~cm}$"
+    ],
+    "answer": "C",
+    "solution": "【分析】假设以 28.26 厘米为圆柱形的底面周长, 15.7 厘米为圆柱形的高，做一个圆柱形的无盖容器，求出圆柱形的体积；再假设以 15.7 厘米为圆柱形的底面周长，28.26 厘米为圆柱形的高，做一个圆柱形的无盖容器，求出圆柱形的体积，看两个体积的大小，即可得出答案。解答此题的关键是, 如何将长方形铁皮, 做成一个容积尽可能大的无盖容器, 找出它们的关系, 即可解答。\n\n【详解】$28.26 \\div 3.14 \\div 2 = 4.5$ （厘米） $3.14 \\times 4.5^{2} \\times 15.7 = 998.2845$ （立方厘米） $15.7 \\div 3.14 \\div 2 = 2.5$ （厘米） $3.14 \\times 2.5^{2} \\times 28.26 = 554.6025$ （立方厘米）因为 $998.2845 > 554.6025$所以直径是: $4.5 \\times 2 = 9$ (厘米)故答案为: C",
+    "knowledge_points": [
+      "圆柱的体积"
+    ],
+    "source": "(2018·江苏·小升初模拟)",
+    "merged_id": "116",
+    "original_id": "116"
+  },
+  {
+    "id": "117",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一个圆柱体的底面半径是3厘米，高是18.84厘米，它的侧面展开图是（）。",
+    "options": [
+      "A. 正方形",
+      "B. 长方形",
+      "C. 两个圆形和一个长方形组成"
+    ],
+    "answer": "A",
+    "solution": "【分析】计算出圆柱的底面周长再与高作比较，判断出它的侧面展开图是什么图形。圆柱的侧面展开图是一个长方形，当圆柱的底面周长与高相等，这个圆柱体的侧面展开图是一个正方形。\n\n【详解】$3.14 \\times 3 \\times 2\\\\= 9. 4 2 \\times 2\\\\=18.84 (厘米)$\n\n底面周长与高相等，所以这个圆柱体的侧面展开图是一个正方形。故答案为：A",
+    "knowledge_points": [
+      "圆柱的展开图"
+    ],
+    "source": "(2021·湖南郴州·小升初真题)",
+    "merged_id": "117",
+    "original_id": "117"
+  },
+  {
+    "id": "118",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "小亮有等底等高的圆锥和圆柱形容器各一个，他将圆柱形容器装满水后倒入圆锥形容器中，当水全部倒完后，圆锥形容器溢出 $36.2 \\mathrm{~mL}$ 水。这时圆锥形容器内还有水（） $\\mathrm{mL}$。",
+    "options": [
+      "A. 36.2",
+      "B. 54.3",
+      "C. 18.1"
+    ],
+    "answer": "C",
+    "solution": "【分析】因为圆锥的体积是与其等底等高的圆柱体体积的 $\\frac{1}{3}$ ,所以溢出的水是圆柱体体积的 $\\frac{2}{3}$ , 留在圆锥体内的水的体积是圆柱体体积的 $\\frac{1}{3}$ , 所以说, 留在圆锥体内的水的体积是溢出的水的体积的 $\\frac{1}{2}$ , 从而问题得解。本题关键是要牢记圆柱与圆锥体积公式。\n\n【详解】根据题意可得：留在圆锥体内的水的体积是溢出的水的体积的 $\\frac{1}{2}$ , $3 6. 2 \\div 2 = 1 8. 1 (\\mathrm {m L})$ 故答案为: C",
+    "knowledge_points": [
+      "圆柱与圆锥体积的关系"
+    ],
+    "source": "(2019·浙江宁波·小升初真题)",
+    "merged_id": "118",
+    "original_id": "118"
+  },
+  {
+    "id": "119",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "圆柱体的底面半径和高都扩大3倍，它的体积扩大（）倍。",
+    "options": [
+      "A. 3",
+      "B. 6",
+      "C. 9",
+      "D. 27"
+    ],
+    "answer": "D",
+    "solution": "【分析】圆柱的体积等于底面积乘高，底面半径扩大3倍，底面积扩大9倍，当底面积扩大9倍，同时高扩大3倍，体积将扩大27倍，可以举例子进行说明。本题考查的是圆柱的体积, 举例子是求解问题时常用的方法, 熟练应用公式是解决问题的前提。\n\n【详解】设原来的底面半径和高都是 1 厘米, $\\pi \\times 1 ^ {2} \\times 1 = \\pi$ 底面半径和高都扩大 3 倍后, 底面半径和高都是 3 厘米,$\\pi \\times 3 ^ {2} \\times 3 = 2 7 \\pi$ $2 7 \\pi \\div \\pi = 2 7$ 所以体积扩大 27 倍。故答案为: D",
+    "knowledge_points": [
+      "圆柱的体积"
+    ],
+    "source": "(2014·全国·小升初真题)",
+    "merged_id": "119",
+    "original_id": "119"
+  },
+  {
+    "id": "120",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "正方体、圆柱和圆锥的底面积相等，高也相等，下面说法正确的是（）。",
+    "options": [
+      "A. 圆柱的体积比正方体的体积小一些",
+      "B. 圆锥的体积是正方体体积的 $\\frac{1}{3}$",
+      "C. 圆柱的体积与圆锥的体积相等",
+      "D. 正方体的体积比圆柱的体积小一些"
+    ],
+    "answer": "B",
+    "solution": "【分析】圆柱和圆锥等底等高，那么圆柱的体积是圆锥的3倍，而正方体和圆柱等底等高，二者体积都是底面积乘高，所以体积相等。柱体的体积都可以用底面积乘高表示, 而锥体体积等于等底等高的柱体体积的 $\\frac{1}{3}$ 。\n\n【详解】$V_{\\text{正方体}} = S_{\\text{底面积}} \\times h$ $V _ {\\text {圆 柱}} = S _ {\\text {底 面 积}} \\times h \\quad V _ {\\text {圆 锥}} = \\frac {1}{3} \\times S _ {\\text {底 面 积}} \\times h$ 正方体、圆柱和圆锥等底等高，所以正方体和圆柱体积相等，都是圆锥的3倍，圆锥体积是正方体体积的 $\\frac{1}{3}$ 。故答案为: B",
+    "knowledge_points": [
+      "正方体的体积",
+      "圆锥的体积（容积）",
+      "圆柱与圆锥体积的关系",
+      "圆柱的体积"
+    ],
+    "source": "(16-17六年级下·全国·阶段练习)",
+    "merged_id": "120",
+    "original_id": "120"
+  },
+  {
+    "id": "121",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "圆柱的侧面展开图是正方形，那么圆柱的高和（）相等。",
+    "options": [
+      "A. 底面直径",
+      "B. 底面周长",
+      "C. 底面积",
+      "D. 底面半径"
+    ],
+    "answer": "B",
+    "solution": "【详解】根据圆柱特征，圆柱的侧面展开图是正方形，那么圆柱的高和底面周长相等。故答案为: B",
+    "knowledge_points": [
+      "圆柱的展开图"
+    ],
+    "source": "(19-20六年级下·北京·单元测试)",
+    "merged_id": "121",
+    "original_id": "121"
+  },
+  {
+    "id": "122",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一个圆柱的高不变，底面半径扩大到原来的2倍，它的体积就扩大到原来的（）。",
+    "options": [
+      "A. 2 倍",
+      "B. 4 倍",
+      "C. 8倍"
+    ],
+    "answer": "B",
+    "solution": "【分析】圆柱体积=底面积×高=πr²h，根据体积公式进行分析。关键是掌握圆柱体积公式，也可以用具体数据进行计算。\n\n【详解】圆柱体积 $= \\pi r^2 h\\\\ =[ \\pi (2r) ^ {2}h ] \\div (\\pi r ^ {2} h) \\\\ = [ 4 \\pi r ^ {2} h ] \\div (\\pi r ^ {2} h) \\\\ = 4$\n\n故答案为: B",
+    "knowledge_points": [
+      "圆柱的体积"
+    ],
+    "source": "(19-20六年级下·全国·期中)",
+    "merged_id": "122",
+    "original_id": "122"
+  },
+  {
+    "id": "123",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "将一个圆柱体削制成一个圆锥体，削去部分的体积是圆柱体积的（）。",
+    "options": [
+      "A. $\\frac{1}{3}$",
+      "B. $\\frac{2}{3}$",
+      "C. 2 倍",
+      "D. 不能确定"
+    ],
+    "answer": "D",
+    "solution": "【分析】题中没有说明削成的圆锥体的高和底面是什么样子的，所以削去的体积不能确定。本题考查圆柱体与圆锥体的体积关系。\n\n【详解】削成的圆锥体的形状不能确定, 因此削去的体积也不能确定, 故答案为: D。",
+    "knowledge_points": [
+      "圆柱与圆锥体积的关系"
+    ],
+    "source": "(21-22 六年级下·全国·单元测试)",
+    "merged_id": "123",
+    "original_id": "123"
+  },
+  {
+    "id": "124",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "把一根长4米的圆柱形木料锯了4次后分成若干段，表面积比原来增加了120平方分米，这根木料的横截面积是( )平方分米，体积是( )立方分米。",
+    "options": [],
+    "answer": "15; 600",
+    "solution": "【分析】圆柱形木料锯 4 次, 增加了 $4 \\times 2$ 个截面, 增加的表面积÷增加的截面个数 = 截面面积, 根据圆柱体积 = 截面面积×长, 列式计算即可。关键是掌握并灵活运用圆柱体积公式。\n\n【详解】$120 \\div (4 \\times 2)\\\\= 1 2 0 \\div 8\\\\= 15（平方分米）$\n\n4米=40分米 $15 \\times 40 = 600$ （立方分米） 这根木料的横截面积是15平方分米，体积是600立方分米。",
+    "knowledge_points": [
+      "立体图形的切拼",
+      "圆柱的体积"
+    ],
+    "source": "(22-23六年级下·全国·单元测试)",
+    "merged_id": "124",
+    "original_id": "124"
+  },
+  {
+    "id": "125",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个圆柱的侧面展开图是一个周长为 50.24 分米的正方形，这个圆柱的高是( )分米，体积是( )立方分米。",
+    "options": [],
+    "answer": "12.56 ;157.7536",
+    "solution": "【分析】圆柱的侧面展开图是正方形，说明圆柱的高 = 底面周长 = 正方形边长，根据正方形边长 = 周长 $\\div 4$ ，圆柱体积 = 底面积 $\\times$ 高，列式计算即可。关键是熟悉圆柱特征，掌握并灵活运用圆柱体积公式。\n\n【详解】$50.24 \\div 4 = 12.56$ （分米）\n\n$3.1 4 \\times (1 2. 5 6 \\div 3. 1 4 \\div 2) ^ {2} \\times 1 2. 5 6 \\\\ = 3. 1 4 \\times 2 ^ {2} \\times 1 2. 5 6 \\\\ = 3. 1 4 \\times 4 \\times 1 2. 5 6 \\\\ = 1 5 7. 7 5 3 6 \\text {(立 方 分 米)}$\n\n这个圆柱的高是 12.56 分米, 体积是 157.7536 立方分米。",
+    "knowledge_points": [
+      "圆柱的展开图",
+      "圆柱的体积"
+    ],
+    "source": "(22-23六年级下·全国·单元测试)",
+    "merged_id": "125",
+    "original_id": "125"
+  },
+  {
+    "id": "126",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个圆柱体, 侧面积是 $37.68 \\mathrm{~cm}^{2}$ , 高是 $6 \\mathrm{~cm}$ , 底面周长是( ) cm, 体积是( ) cm³。",
+    "options": [],
+    "answer": "6.28 ;18.84",
+    "solution": "【分析】根据圆柱底面周长 $=$ 侧面积 $\\div$ 高, 底面半径 $=$ 底面周长 $\\div \\pi \\div 2$ , 体积 $=$ 底面积 $\\times$ 高, 列式计算即可。关键是掌握并灵活运用圆柱体积公式，通过侧面积先求出底面周长和底面半径。\n\n【详解】$37.68 \\div 6 = 6.28$ (cm)\n\n$3.1 4 \\times (6. 2 8 \\div 3. 1 4 \\div 2) ^ {2} \\times 6 \\\\ = 3. 1 4 \\times 1 ^ {2} \\times 6 \\\\ = 3. 1 4 \\times 1 \\times 6 \\\\ = 1 8. 8 4(\\mathrm {c m} ^ {3})$\n\n底面周长是 $6.28 \\mathrm{~cm}$ , 体积是 $18.84 \\mathrm{~cm}^{3}$ 。",
+    "knowledge_points": [
+      "圆柱的侧面积",
+      "圆柱的体积"
+    ],
+    "source": "(22-23 六年级下·全国·单元测试)",
+    "merged_id": "126",
+    "original_id": "126"
+  },
+  {
+    "id": "127",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个圆锥的底面积是 3.14 平方厘米, 高是 9 厘米, 体积是( )立方厘米, 与它等底等高的圆柱的体积是( )立方厘米。",
+    "options": [],
+    "answer": "9.42; 28.26",
+    "solution": "【分析】根据圆锥体积=底面积×高÷3，圆柱体积=底面积×高，列式计算即可，等底等高的圆柱和圆锥，圆柱体积是圆锥体积的3倍。关键是掌握并灵活运用圆柱和圆锥的体积公式,理解它们体积之间的关系。\n\n【详解】$3.14 \\times 9 \\div 3 = 9.42$ （立方厘米） $3.14 \\times 9 = 28.26$ （立方厘米）一个圆锥的底面积是 3.14 平方厘米, 高是 9 厘米, 体积是9.42 立方厘米, 与它等底等高的圆柱的体积是 28.26 立方厘米。",
+    "knowledge_points": [
+      "圆锥的体积（容积）",
+      "圆柱与圆锥体积的关系",
+      "圆柱的体积"
+    ],
+    "source": "(22-23 六年级下·全国·单元测试)",
+    "merged_id": "127",
+    "original_id": "127"
+  },
+  {
+    "id": "128",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "将一张长5分米、宽 4 分米的长方形贴片卷成一个圆柱（接头不计），这个圆柱的侧面积是( )平方分米。",
+    "options": [],
+    "answer": "20",
+    "solution": "【分析】圆柱的侧面积等于长方形纸的面积，根据长方形的面积 $= \\text{长} \\times$ 宽，代入计算即可。明确长方形的面积就是圆柱的侧面积是解题关键。\n\n【详解】$5 \\times 4 = 20$ （平方分米）这个圆柱的侧面积是 20 平方分米。",
+    "knowledge_points": [
+      "长方形的面积",
+      "圆柱的侧面积"
+    ],
+    "source": "(22-23 六年级下·全国·单元测试)",
+    "merged_id": "128",
+    "original_id": "128"
+  },
+  {
+    "id": "129",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "一般情况下, 圆柱的侧面展开后是一个( )形, 圆锥侧面展开是一个( )形。",
+    "options": [],
+    "answer": "长方; 扇",
+    "solution": "【分析】根据圆柱的特征，它的上、下是完全相同的两个圆，侧面是一个曲面，侧面沿高展开是一个长方形，这个长方形的长等于圆柱的底面周长，宽等于圆柱的高；如果圆柱体的底面周长和高相等时，侧面展开是正方形，据此解答。此题主要考查圆柱与圆锥的特征和它的侧面展开图的形状。\n\n【详解】由分析可得：一般情况下，圆柱的侧面展开后是一个长方形，圆锥侧面展开是一个扇形。",
+    "knowledge_points": [
+      "圆锥的认识及特征",
+      "圆柱的展开图"
+    ],
+    "source": "(22-23 六年级下·全国·单元测试)",
+    "merged_id": "129",
+    "original_id": "129"
+  },
+  {
+    "id": "130",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个圆锥的底面积是 $9 \\mathrm{dm}^{2}$ , 高是 $6 \\mathrm{dm}$ , 它的体积是( ) $\\mathrm{dm}^{3}$ , 与它等底等高的圆柱的体积是( ) $\\mathrm{dm}^{3}$ 。",
+    "options": [],
+    "answer": "18; 54",
+    "solution": "【分析】根据 $V_{\\text {锥}} = \\frac{1}{3} \\mathrm{~Sh}$ , 求出圆锥的体积; 圆锥和圆柱等底等高时, 圆柱的体积是圆锥体积的 3 倍, 据此求出圆柱的体积。本题考查圆锥体积公式的运用, 以及等底等高的圆柱和圆锥体积之间的关系及应用。\n\n【详解】圆锥的体积：\n\n$\\frac {1}{3} \\times 9 \\times 6 \\\\ = 3 \\times 6 \\\\ = 1 8(\\mathrm {d m} ^ {3})$\n\n与它等底等高的圆柱的体积是:\n\n$1 8 \\times 3 = 5 4 (\\mathrm {d m} ^ {3})$",
+    "knowledge_points": [
+      "圆锥的体积（容积）",
+      "圆柱与圆锥体积的关系"
+    ],
+    "source": "(21-22 六年级下·黑龙江大兴安岭地·期末)",
+    "merged_id": "130",
+    "original_id": "130"
+  },
+  {
+    "id": "131",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "一个圆柱的底面积是 $10 \\mathrm{dm}^{2}$ , 它的高 $30 \\mathrm{dm}$ , 那么它的体积是( ) $\\mathrm{dm}^{3}$ 。",
+    "options": [],
+    "answer": "300",
+    "solution": "【分析】根据圆柱的体积公式: $V = S h$ , 已知圆柱的底面积是 $10 \\mathrm{~dm}^{2}$ , 它的高是 $30 \\mathrm{dm}$ , 代入到公式中, 即可求出圆柱的体积。此题的解题关键是熟练运用圆柱的体积公式求解。\n\n【详解】$10 \\times 30 = 300 \\, (\\mathrm{dm}^{3})$",
+    "knowledge_points": [
+      "圆柱的体积"
+    ],
+    "source": "(20-21 六年级下·河南洛阳·期末)",
+    "merged_id": "131",
+    "original_id": "131"
+  },
+  {
+    "id": "132",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个圆柱形水桶，从里面量，直径是 $20 \\mathrm{~cm}$ ，高是 $4 \\mathrm{dm}$ ，这个水桶的容积是( )L（考虑实际情况，得数保留整数）。",
+    "options": [],
+    "answer": "12",
+    "solution": "【分析】根据圆柱的容积公式： $V = \\pi r^2 h$ ，据此求出水桶的容积，其结果根据实际情况运用去尾法保留整数。本题考查圆柱的容积，熟记公式是解题的关键。\n\n【详解】$20 \\mathrm{~cm} = 2 \\mathrm{dm}$\n\n$3.1 4 \\times (2 \\div 2) ^ {2} \\times 4 \\\\ = 3. 1 4 \\times 1 \\times 4 \\\\ = 1 2. 5 6(\\mathrm {d m} ^ {3}) \\\\ = 1 2. 5 6 (\\mathrm {L})\\\\ \\approx 1 2 \\mathrm {(L)}$",
+    "knowledge_points": [
+      "圆柱的容积"
+    ],
+    "source": "(20-21 六年级下·广东江门·期末)",
+    "merged_id": "132",
+    "original_id": "132"
+  },
+  {
+    "id": "133",
+    "unit": "第三单元：圆柱与圆锥",
+    "chapter": "圆柱与圆锥",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "把一个棱长 $6 \\mathrm{~cm}$ 的正方体木块削成一个最大的圆柱, 这个圆柱的体积是( ) $\\mathrm{cm}^{3}$ , 表面积是( ) $\\mathrm{cm}^{2}$ 。",
+    "options": [],
+    "answer": "169.56; 169.56",
+    "solution": "【分析】由题意可知, 把一个正方体木块削成一个最大的圆柱, 则这个圆柱的底面直径和高相当于正方体的棱长,然后根据圆柱的体积公式: $V = \\pi r^{2} h$ , 圆柱的表面积公式: $S = \\pi d h + 2 \\pi r^{2}$ , 据此代入数值进行计算即可。本题考查圆柱的体积和表面积, 熟记公式是解题的关键。\n\n【详解】$3.14 \\times (6 \\div 2)^{2} \\times 6\\\\= 3. 1 4 \\times 9 \\times 6 \\\\ = 2 8. 2 6 \\times 6 \\\\ = 1 6 9. 5 6(\\mathrm {c m} ^ {3})$\n\n$3.1 4 \\times 6 \\times 6 + 3. 1 4 \\times (6 \\div 2) ^ {2} \\times 2 \\\\ = 1 1 3. 0 4 + 3. 1 4 \\times 9 \\times 2 \\\\ = 1 1 3. 0 4 + 5 6. 5 2 \\\\ = 1 6 9. 5 6(\\mathrm {c m} ^ {2})$\n\n这个圆柱的体积是 $169.56 \\mathrm{~cm}^{3}$ , 表面积是 $169.56 \\mathrm{~cm}^{2}$ 。",
+    "knowledge_points": [
+      "圆柱的表面积",
+      "圆柱的体积"
+    ],
+    "source": "(2021·浙江台州·小升初真题)",
+    "merged_id": "133",
+    "original_id": "133"
+  },  
+  
+    {
+      "id": "134",
+      "unit": "第三单元 圆柱与圆锥",
+      "question_type": "填空题",
+      "difficulty": 0.65,
+      "question": "一个正方体密封盒的棱长是6厘米，在盒内放入一个最大的圆柱，圆柱的底面积是____平方厘米；如果放入一个最大的圆锥，圆锥的体积是____立方厘米。（π取3.14）",
+      "options": [],
+      "answer": "28.26；56.52",
+      "solution": "【分析】最大圆柱的底面直径和高都等于正方体棱长6厘米，底面半径r=3厘米。先用圆的面积公式S=πr²求圆柱底面积；最大圆锥与该圆柱等底等高，高h=6厘米，用圆锥体积公式V=1/3×S×h求体积。\n【详解】圆柱的底面积：3.14×(6÷2)²=3.14×9=28.26（平方厘米）。圆锥的体积：1/3×28.26×6=28.26×2=56.52（立方厘米）。",
+      "knowledge_points": ["正方体的特征", "圆的面积及应用", "圆锥的体积（容积）"],
+      "source": "21-22 六年级下·河南南阳·期末"
+    },
+    {
+      "id": "135",
+      "unit": "第三单元 圆柱与圆锥",
+      "question_type": "填空题",
+      "difficulty": 0.65,
+      "question": "在高12厘米的圆锥形容器中装满水，再全部倒入等底的圆柱形容器中，那么水面的高是( )厘米。",
+      "options": [],
+      "answer": "4",
+      "solution": "【分析】倒入前后水的体积不变。等底圆锥体积V=1/3×S×12。倒入等底圆柱后，圆柱体积V=S×h，解得h=4。\n【详解】假设底面积为1平方厘米，则水的体积为1×12×1/3=4（立方厘米），圆柱底面积也是1平方厘米，所以水高4÷1=4（厘米）。",
+      "knowledge_points": ["圆柱的容积", "圆锥的体积（容积）", "圆柱与圆锥体积的关系"],
+      "source": "2022·河南驻马店·小升初真题"
+    },
+    {
+      "id": "136",
+      "unit": "第三单元 圆柱与圆锥",
+      "question_type": "填空题",
+      "difficulty": 0.85,
+      "question": "等底等高的圆柱体和圆锥体，它们两个体积和是64立方分米，那么圆柱体的体积是( )立方分米，圆锥体的体积是( )立方分米。",
+      "options": [],
+      "answer": "48；16",
+      "solution": "【分析】等底等高时，圆柱体积是圆锥体积的3倍。两者体积和相当于圆锥的(3+1)=4倍，先求圆锥体积，再求圆柱体积。\n【详解】圆锥体积：64÷(3+1)=64÷4=16（立方分米）。圆柱体积：16×3=48（立方分米）。",
+      "knowledge_points": ["圆锥的体积（容积）", "圆柱与圆锥体积的关系", "圆柱的体积"],
+      "source": "20-21 六年级下·黑龙江伊春·期末"
+    },
+    {
+      "id": "137",
+      "unit": "第三单元 圆柱与圆锥",
+      "question_type": "填空题",
+      "difficulty": 0.85,
+      "question": "列举两个生活中圆柱形的物体( )，它们的共同特征是( )（至少写出三条）。",
+      "options": [],
+      "answer": "见详解；见详解",
+      "solution": "【分析】生活中圆柱形物体很多，结合圆柱的特征作答。\n【详解】生活中圆柱形的物体有电线杆、铅笔（答案不唯一）；共同特征示例：1、上下两个面形状和大小完全相同，把圆柱形物体放倒可以滚动；2、有2个大小相同且相互平行的圆形底面和1个曲面侧面，侧面展开是矩形；3、可看作矩形绕一条边旋转一周形成的几何体。",
+      "knowledge_points": ["圆柱的认识及特征"],
+      "source": "21-22 六年级下·内蒙古赤峰·期末"
+    },
+    {
+      "id": "138",
+      "unit": "第三单元 圆柱与圆锥",
+      "question_type": "填空题",
+      "difficulty": 0.85,
+      "question": "一个圆柱与一个圆锥底面半径相等，高也相等，它们的体积和是80立方分米，圆锥的体积是( )立方分米，圆柱的体积是( )立方分米。",
+      "options": [],
+      "answer": "20；60",
+      "solution": "【分析】等底等高时圆柱体积是圆锥体积的3倍，总和相当于圆锥的(3+1)=4倍，先求圆锥体积，再求圆柱体积。\n【详解】圆锥体积：80÷(3+1)=80÷4=20（立方分米）；圆柱体积：20×3=60（立方分米）。",
+      "knowledge_points": ["圆柱与圆锥体积的关系"],
+      "source": "2022·安徽阜阳·小升初真题"
+    },
+    {
+      "id": "139",
+      "unit": "第三单元 圆柱与圆锥",
+      "question_type": "填空题",
+      "difficulty": 0.85,
+      "question": "一根长2dm的圆柱形木料，底面圆半径是2dm，这根木料的体积是( )dm³。",
+      "options": [],
+      "answer": "25.12",
+      "solution": "【分析】圆柱体积=底面积×高=πr²h。\n【详解】3.14×2²×2=3.14×4×2=12.56×2=25.12（dm³）。",
+      "knowledge_points": ["圆柱的体积"],
+      "source": "2022·广东梅州·小升初真题"
+    },
+    {
+      "id": "140",
+      "unit": "第三单元 圆柱与圆锥",
+      "question_type": "填空题",
+      "difficulty": 0.65,
+      "question": "把一张长8dm、宽5dm的白纸卷成一个圆柱形纸筒（接口处不计），这个纸筒的侧面积是( )dm²。",
+      "options": [],
+      "answer": "40",
+      "solution": "【分析】卷成圆柱后侧面积等于这张长方形纸的面积。\n【详解】8×5=40（dm²）。",
+      "knowledge_points": ["圆柱的侧面积"],
+      "source": "2022·甘肃陇南·小升初真题"
+    },
+    {
+      "id": "141",
+      "unit": "第三单元 圆柱与圆锥",
+      "question_type": "填空题",
+      "difficulty": 0.65,
+      "question": "将一个边长为5dm的正方形纸片卷成圆柱筒，这个圆柱的侧面积是( )dm²。",
+      "options": [],
+      "answer": "25",
+      "solution": "【分析】卷成圆柱后，底面周长=5dm，高=5dm，侧面积=周长×高。\n【详解】5×5=25（dm²）。",
+      "knowledge_points": ["圆柱的侧面积"],
+      "source": "2014 六年级·全国·课后作业"
+    },
+    {
+      "id": "142",
+      "unit": "第三单元 圆柱与圆锥",
+      "question_type": "填空题",
+      "difficulty": 0.65,
+      "question": "一个圆柱的体积是75.36立方米，与它等底等高的圆锥的体积是( )立方米。",
+      "options": [],
+      "answer": "25.12",
+      "solution": "【分析】等底等高时圆锥体积是圆柱体积的1/3。\n【详解】75.36×1/3=25.12（立方米）。",
+      "knowledge_points": ["圆锥的体积（容积）", "圆柱与圆锥体积的关系", "圆柱的体积"],
+      "source": "2014·贵州安顺·小升初真题"
+    },
+    {
+      "id": "143",
+      "unit": "第三单元 圆柱与圆锥",
+      "question_type": "填空题",
+      "difficulty": 0.65,
+      "question": "等底等高的圆柱和圆锥，它们的体积之和是36立方厘米，则圆锥的体积是( )立方厘米，圆柱的体积是( )立方厘米。",
+      "options": [],
+      "answer": "9；27",
+      "solution": "【分析】等底等高时圆柱体积是圆锥体积的3倍。设圆锥为1份，圆柱为3份，总共4份，36÷4=9为1份。\n【详解】圆锥体积：36÷(1+3)=36÷4=9（立方厘米）；圆柱体积：9×3=27（立方厘米）。",
+      "knowledge_points": ["圆锥的体积（容积）", "圆柱与圆锥体积的关系", "和差倍问题", "圆柱的体积"],
+      "source": "2015·四川成都·小升初真题"
+    },
+    {
+      "id": "144",
+      "unit": "第三单元 圆柱与圆锥",
+      "question_type": "填空题",
+      "difficulty": 0.65,
+      "question": "下图是一个等腰直角三角形，它的面积是( )cm²，以AB为轴旋转一周，形成立体图形的体积是( )cm³。![](images/412e3dab16563e992ab33b9ad66e36cce3be9f7b1a4ee46c1eb04acf08edd4e5.jpg)",
+      "options": [],
+      "answer": "4.5；28.26",
+      "solution": "【分析】等腰直角三角形两条直角边都为3厘米，面积=3×3÷2。以AB为轴旋转一周得到圆锥，AB为高3厘米，底面半径为3厘米，用圆锥体积V=1/3πr²h计算。\n【详解】面积：3×3÷2=4.5（cm²）。体积：1/3×3.14×3²×3=1/3×3.14×27=28.26（cm³）。",
+      "knowledge_points": ["三角形的面积", "圆锥的认识及特征", "圆锥的体积（容积）"],
+      "source": "2018·山东临沂·小升初真题"
+    },
+    {
+      "id": "145",
+      "unit": "第三单元 圆柱与圆锥",
+      "question_type": "填空题",
+      "difficulty": 0.65,
+      "question": "圆柱的底面半径扩大到原来的2倍，高不变，它的底面积扩大到原来的( )倍，体积扩大到原来的( )倍。",
+      "options": [],
+      "answer": "4；4",
+      "solution": "【分析】底面积S=πr²，半径扩大2倍，底面积扩大2²=4倍；体积=底面积×高，高不变，所以体积也扩大4倍。\n【详解】底面积扩大4倍，体积扩大4倍。",
+      "knowledge_points": ["圆的面积及应用", "圆柱的体积"],
+      "source": "21-22 六年级下·全国·单元测试"
+    },
+    {
+      "id": "146",
+      "unit": "第三单元 圆柱与圆锥",
+      "question_type": "填空题",
+      "difficulty": 0.65,
+      "question": "一个直角三角形的两条直角边分别长3cm和4cm，如果以4cm长的直角边所在的直线为轴旋转一周，得到一个( )，它的体积是( )cm³。",
+      "options": [],
+      "answer": "圆锥；37.68",
+      "solution": "【分析】以4cm直角边为轴旋转，得到圆锥，高h=4cm，底面半径r=3cm，体积V=1/3πr²h。\n【详解】1/3×3.14×3²×4=1/3×3.14×9×4=37.68（立方厘米）。",
+      "knowledge_points": ["圆锥的认识及特征", "圆锥的体积（容积）"],
+      "source": "2018·山东临沂·小升初真题"
+    },
+    {
+      "id": "147",
+      "unit": "第三单元 圆柱与圆锥",
+      "question_type": "填空题",
+      "difficulty": 0.65,
+      "question": "如果把一个底面直径是2分米，高是3分米的圆柱体削成一个最大的圆锥体，削去( )立方分米。如果把棱长是2分米的正方体木块削成一个最大的圆柱，这个圆柱的表面积是( )平方分米。",
+      "options": [],
+      "answer": "6.28；18.84",
+      "solution": "【分析】(1)最大圆锥与圆柱等底等高，圆锥体积是圆柱体积的1/3，削去部分占(1-1/3)=2/3；(2)正方体内最大圆柱的底面直径和高都等于棱长2dm，代入圆柱表面积公式。\n【详解】(1)削去体积=3.14×(2÷2)²×3×(1-1/3)=3.14×1×3×2/3=3.14×2=6.28（立方分米）。(2)表面积=3.14×2×2+3.14×(2÷2)²×2=12.56+6.28=18.84（平方分米）。",
+      "knowledge_points": ["圆柱的表面积", "圆柱与圆锥体积的关系", "圆柱的体积"],
+      "source": "19-20六年级下·湖南怀化·单元测试"
+    },
+    {
+      "id": "148",
+      "unit": "第三单元 圆柱与圆锥",
+      "question_type": "填空题",
+      "difficulty": 0.65,
+      "question": "把一个圆柱的侧面展开后是一个边长为62.8厘米的正方形，这个圆柱的底面半径是( )厘米。",
+      "options": [],
+      "answer": "10",
+      "solution": "【分析】侧面展开为正方形，说明底面周长C和高都等于62.8厘米。半径r=C÷π÷2。\n【详解】62.8÷3.14÷2=20÷2=10（厘米）。",
+      "knowledge_points": ["圆柱的侧面积"],
+      "source": "19-20 六年级下·湖南怀化·单元测试"
+    },
+    {
+      "id": "149",
+      "unit": "第三单元 圆柱与圆锥",
+      "question_type": "填空题",
+      "difficulty": 0.65,
+      "question": "一个圆锥的体积是1.5立方分米，与它等底等高圆柱的体积是( )立方分米。",
+      "options": [],
+      "answer": "4.5",
+      "solution": "【分析】等底等高时圆柱体积是圆锥体积的3倍。\n【详解】1.5×3=4.5（立方分米）。",
+      "knowledge_points": ["圆柱与圆锥体积的关系"],
+      "source": "19-20 六年级下·湖南怀化·单元测试"
+    },
+    {
+      "id": "150",
+      "unit": "第三单元 圆柱与圆锥",
+      "question_type": "填空题",
+      "difficulty": 0.65,
+      "question": "在一个高12厘米的圆锥体量杯内装满了水，把圆锥里的水全部倒入与它底面积相等的圆柱体量杯中，水高( )厘米。",
+      "options": [],
+      "answer": "4",
+      "solution": "【分析】水体积不变且等底，圆锥体积=1/3×S×12，倒入圆柱后S×h相等，所以h=12×1/3=4。\n【详解】12×1/3=4（厘米）。",
+      "knowledge_points": ["圆柱与圆锥体积的关系"],
+      "source": "19-20六年级下·湖南怀化·单元测试"
+    },
+    {
+      "id": "151",
+      "unit": "第三单元 圆柱与圆锥",
+      "question_type": "填空题",
+      "difficulty": 0.65,
+      "question": "把一个圆柱侧面沿高展开是一个正方形，这个圆柱的高是底面直径的( )倍。",
+      "options": [],
+      "answer": "π",
+      "solution": "【分析】侧面展开为正方形，说明底面周长C=高h。求h与直径d的比即C与d的比，C÷d=π。\n【详解】C÷d=πd÷d=π，所以圆柱的高是底面直径的π倍。",
+      "knowledge_points": ["圆柱的展开图"],
+      "source": "19-20 六年级下·湖南怀化·单元测试"
+    },
+    {
+      "id": "152",
+      "unit": "第三单元 圆柱与圆锥",
+      "question_type": "判断题",
+      "difficulty": 0.65,
+      "question": "沿圆柱的底面直径剖开将圆柱分成相等的两部分，剖面一定是长方形。( )",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】沿底面直径剖开会增加两个剖面，剖面对应底面直径与高，形状为长方形；若底面直径等于高，剖面为正方形，正方形也是长方形。\n【详解】如图![](images/224de15a165b055dbb2db6a2bf765b52b81d9962ca93b8ab24905a0193893855.jpg)，剖面为长方形（或特殊情况为正方形），所以说法正确。",
+      "knowledge_points": ["立体图形的切拼", "圆柱的认识及特征"],
+      "source": "22-23 六年级下·全国·单元测试"
+    },
+    {
+      "id": "153",
+      "unit": "第三单元 圆柱与圆锥",
+      "question_type": "判断题",
+      "difficulty": 0.94,
+      "question": "把一个圆柱削成一个圆锥，这个圆锥的体积是圆柱体积的1/3。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】只有把圆柱削成与它等底等高的圆锥时，圆锥体积才是圆柱体积的1/3；若不等底或不等高则不成立。\n【详解】削成与圆柱等底等高的圆锥才满足1/3关系，原说法不一定成立，故错误。",
+      "knowledge_points": ["圆柱与圆锥体积的关系"],
+      "source": "2017·江西吉安·小升初真题"
+    },
+    {
+      "id": "154",
+      "unit": "第三单元 圆柱与圆锥",
+      "question_type": "判断题",
+      "difficulty": 0.65,
+      "question": "圆锥的体积比与它等底等高的圆柱的体积小2倍。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】等底等高时圆锥体积=1/3圆柱体积，圆锥比圆柱少2/3，不是少2倍。\n【详解】圆锥体积比等底等高圆柱体积小2/3，原说法错误。",
+      "knowledge_points": ["圆锥的体积（容积）", "圆柱与圆锥体积的关系", "圆柱的体积"],
+      "source": "2021·河北邯郸·小升初真题"
+    },
+    {
+      "id": "155",
+      "unit": "第三单元 圆柱与圆锥",
+      "question_type": "判断题",
+      "difficulty": 0.65,
+      "question": "当圆柱、正方体、长方体的底面周长相等，高也相等时，圆柱的体积最大。( )",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】体积=底面积×高，高相等则比底面积。周长相等时圆的面积最大，其次正方形，再次长方形，所以圆柱体积最大。\n【详解】周长相等时圆面积>正方形面积>长方形面积，且高相等，所以V圆柱>V正方体>V长方体，原说法正确。",
+      "knowledge_points": ["正方体的体积、面积认识及大小的比较", "圆柱的体积", "长方体的体积"],
+      "source": "2022·山东聊城·小升初真题"
+    },
+    {
+      "id": "156",
+      "unit": "第三单元 圆柱与圆锥",
+      "question_type": "判断题",
+      "difficulty": 0.85,
+      "question": "不论沿着直角三角形的哪一条边旋转一周，都可以得到圆锥。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】直角三角形绕任意一条直角边旋转一周可得到圆锥；若绕斜边旋转则不能得到圆锥。\n【详解】绕斜边旋转得到的不是圆锥，所以原说法错误。",
+      "knowledge_points": ["旋转与旋转现象", "圆锥的认识及特征"],
+      "source": "20-21六年级下·河南信阳·期末"
+    },
+    {
+      "id": "157",
+      "unit": "第三单元 圆柱与圆锥",
+      "question_type": "判断题",
+      "difficulty": 0.65,
+      "question": "把圆柱体削成等底等高的圆锥体，则削去部分和圆锥体积的比是2:1。( )",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】等底等高时圆柱体积是圆锥体积的3倍，削去部分=圆柱-圆锥=2份，圆锥=1份，比为2:1。\n【详解】圆柱:圆锥=3:1，削去部分=3-1=2份，所以削去部分:圆锥=2:1，说法正确。",
+      "knowledge_points": ["比的意义", "圆柱与圆锥体积的关系"],
+      "source": "20-21 六年级下·河北承德·期末"
+    },
+    {
+      "id": "158",
+      "unit": "第三单元 圆柱与圆锥",
+      "question_type": "判断题",
+      "difficulty": 0.85,
+      "question": "圆柱和圆锥底面积和体积都相等，它们高的比是1:1。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】设底面积为S。圆柱体积=Sh1，圆锥体积=1/3Sh2。体积相等且底面积相等，则Sh1=1/3Sh2，得h1=1/3h2，所以高的比为1:3。\n【详解】圆柱高:圆锥高=1:3，原说法错误。",
+      "knowledge_points": ["比的意义", "圆锥的体积（容积）", "圆柱的体积"],
+      "source": "20-21 六年级下·浙江金华·期末"
+    },
+    {
+      "id": "159",
+      "unit": "第三单元 圆柱与圆锥",
+      "question_type": "判断题",
+      "difficulty": 0.85,
+      "question": "圆柱的高与底面直径相等，它的侧面展开图是一个正方形。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】圆柱侧面展开是长方形，长=底面周长，宽=高。展开为正方形需要底面周长=高，而不是直径=高。\n【详解】当高等于底面周长时展开图才是正方形，原说法错误。",
+      "knowledge_points": ["圆柱的展开图", "圆柱的认识及特征"],
+      "source": "2022·广西百色·小升初真题"
+    },
+    {
+      "id": "160",
+      "unit": "第三单元 圆柱与圆锥",
+      "question_type": "判断题",
+      "difficulty": 0.65,
+      "question": "如果圆柱和圆锥它们的体积和高分别相等，那么圆锥的底面积是圆柱底面积的3倍。( )",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】设体积为V，高为h。圆柱底面积=V/h；圆锥底面积=3V/h。两者底面积之比为3:1。\n【详解】圆锥底面积:圆柱底面积=(3V/h):(V/h)=3:1，所以圆锥底面积是圆柱底面积的3倍，说法正确。",
+      "knowledge_points": ["圆锥的体积（容积）", "圆柱的体积"],
+      "source": "2022·河南信阳·小升初真题"
+    },
+    {
+      "id": "161",
+      "unit": "第三单元 圆柱与圆锥",
+      "question_type": "判断题",
+      "difficulty": 0.65,
+      "question": "一个圆锥形沙堆的占地面积为15平方米，体积为30立方米，那么高是2米。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】圆锥体积V=底面积×高÷3。若高为2米，则体积应为15×2÷3=10立方米，不是30立方米。\n【详解】15×2÷3=10（立方米），与30不符，所以原说法错误。",
+      "knowledge_points": ["圆锥的体积（容积）"],
+      "source": "21-22 六年级下·湖南娄底·期末"
+    },
+    {
+      "id": "162",
+      "unit": "第三单元 圆柱与圆锥",
+      "question_type": "判断题",
+      "difficulty": 0.65,
+      "question": "把一个圆柱削成一个最大的圆锥，削去部分的体积是圆锥体积的2倍。( )",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】最大圆锥与圆柱等底等高，圆柱体积是圆锥体积的3倍，削去部分=3-1=2倍圆锥体积。\n【详解】圆柱:圆锥=3:1，削去部分是2份，等于圆锥体积的2倍，所以说法正确。",
+      "knowledge_points": ["圆柱与圆锥体积的关系"],
+      "source": "2018·山东·小升初真题"
+    },
+    {
+      "id": "163",
+      "unit": "第三单元 圆柱与圆锥",
+      "question_type": "判断题",
+      "difficulty": 0.85,
+      "question": "两个圆柱的底面周长相等，它们的侧面积也相等。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】圆柱侧面积=底面周长×高。底面周长相等但高不一定相等，所以侧面积不一定相等。\n【详解】高不确定，侧面积不一定相等，原说法错误。",
+      "knowledge_points": ["圆柱的侧面积"],
+      "source": "2022 六年级下·全国·专题练习"
+    },
+    {
+      "id": "164",
+      "unit": "第三单元 圆柱与圆锥",
+      "question_type": "判断题",
+      "difficulty": 0.65,
+      "question": "圆柱体的底面半径和高都扩大到原来的3倍，它的体积扩大到原来的9倍。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】圆柱体积V=πr²h。半径扩大3倍，底面积扩大9倍；高扩大3倍，体积扩大9×3=27倍。\n【详解】设原体积为πr²h，现体积为π(3r)²×3h=27πr²h，扩大27倍，不是9倍，故错误。",
+      "knowledge_points": ["圆柱的体积"],
+      "source": "2017·重庆铜梁·小升初真题"
+    },
+    {
+      "id": "165",
+      "unit": "第三单元 圆柱与圆锥",
+      "question_type": "判断题",
+      "difficulty": 0.65,
+      "question": "三个圆锥体积的和正好等于一个圆柱体的体积。( )",
+      "options": [],
+      "answer": "×",
+      "solution": "【分析】只有等底等高时圆柱体积才是圆锥体积的3倍。题干未说明等底等高，不能直接认为三个圆锥体积之和等于一个圆柱体积。\n【详解】三个圆锥体积的和与一个圆柱体的体积不一定相等，题干说法错误。",
+      "knowledge_points": ["圆柱与圆锥体积的关系"],
+      "source": "2014·全国·小升初真题"
+    },
+    {
+      "id": "166",
+      "unit": "第三单元 圆柱与圆锥",
+      "question_type": "判断题",
+      "difficulty": 0.65,
+      "question": "圆柱体的侧面展开图可能是平行四边形。( )",
+      "options": [],
+      "answer": "√",
+      "solution": "【分析】沿圆柱的高剪开侧面展开为长方形；若不是沿高剪开而是斜着剪开，侧面展开可能为平行四边形。\n【详解】如图所示：![](images/c84c8b74054d12ca030ff88bec034fa0ee4ffe80425eb957769109a4b12bed46.jpg) ![](images/a1c762ee476adc8861fe5e91810da4ad48abae5bc1afaf42d1538e34cf4f56ff.jpg) 圆柱体的侧面展开图可能是平行四边形，题干正确。",
+      "knowledge_points": ["圆柱的展开图"],
+      "source": "2019·山西·小升初真题"
+    },
+     {
+    "id": "167",
+    "unit": "第三单元 圆柱与圆锥",
+    "question_type": "判断题",
+    "difficulty": 0.85,
+    "question": "一个长方形以它的一条边为轴，旋转一周得到的旋转体是一个圆柱体。",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】长方形绕一条边旋转，形成上、下底面是两个相同圆、侧面是曲面的立体图形。\n【详解】一个长方形以一条边为轴旋转一周可得到圆柱体，故正确。",
+    "knowledge_points": ["圆柱的认识及特征"],
+    "source": "2022·海南省直辖县级单位·小升初真题"
+  },
+  {
+    "id": "168",
+    "unit": "第三单元 圆柱与圆锥",
+    "question_type": "判断题",
+    "difficulty": 0.65,
+    "question": "圆柱的底面半径扩大到原来的2倍，高也同时扩大到原来的2倍，圆柱的体积就扩大到原来的8倍。",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】底面积与半径平方成正比，半径扩大2倍，底面积扩大4倍；高扩大2倍。\n【详解】体积=底面积×高，故体积扩大4×2=8倍，正确。",
+    "knowledge_points": ["积的变化规律（小数乘法）", "圆柱的体积"],
+    "source": "18-19 六年级下·北京·单元测试"
+  },
+  {
+    "id": "169",
+    "unit": "第三单元 圆柱与圆锥",
+    "question_type": "判断题",
+    "difficulty": 0.85,
+    "question": "压路机滚筒滚动一周能压多少路面是求滚筒的表面积。",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】滚动一周压过的是一个长方形路面，对应圆柱的侧面积，不含两个底面。\n【详解】应求滚筒的侧面积，不是表面积，故错误。",
+    "knowledge_points": ["圆柱的侧面积", "圆柱的表面积"],
+    "source": "21-22 六年级下·全国·单元测试"
+  },
+  {
+    "id": "170",
+    "unit": "第三单元 圆柱与圆锥",
+    "question_type": "判断题",
+    "difficulty": 0.85,
+    "question": "一个圆柱容器的表面积一定比它的体积大。",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】表面积单位是平方单位，体积单位是立方单位，不是同类量。\n【详解】不同量不能直接比较大小，故错误。",
+    "knowledge_points": ["圆柱的表面积", "圆柱的体积"],
+    "source": "19-20 六年级下·四川·课后作业"
+  },
+  {
+    "id": "171",
+    "unit": "第三单元 圆柱与圆锥",
+    "question_type": "判断题",
+    "difficulty": 0.65,
+    "question": "圆柱的侧面展开是长方形，但不是正方形。",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】一般是长方形，但当底面周长=高时，展开图是正方形。\n【详解】“但不是正方形”不成立，故错误。",
+    "knowledge_points": ["圆柱的侧面积", "圆柱的展开图"],
+    "source": "21-22 六年级下·全国·单元测试"
+  },
+  {
+    "id": "172",
+    "unit": "第三单元 圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": 0.65,
+    "question": "一个圆柱形玻璃缸的底面半径是20厘米，缸内盛有水。将一个底面半径是10厘米、高是30厘米的圆锥形铁块完全浸没在水中（水没有溢出），玻璃缸中的水面上升多少厘米？",
+    "options": [],
+    "answer": "2.5厘米",
+    "solution": "【分析】水面上升部分体积=铁块（圆锥）体积。上升高度=圆锥体积÷玻璃缸底面积。\n【详解】h=[(1/3×3.14×10²×30)]÷(3.14×20²)=2.5（厘米）。",
+    "knowledge_points": ["圆锥的体积（容积）"],
+    "source": "18-19六年级下·浙江·单元测试"
+  },
+  {
+    "id": "173",
+    "unit": "第三单元 圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": 0.85,
+    "question": "只列式不计算。圆柱形通风管每节长2米，底面半径0.1米，做100节这样的通风管共需多少材料？",
+    "options": [],
+    "answer": "2×3.14×0.1×2×100",
+    "solution": "【分析】通风管无上下底，只求侧面积；每节侧面积=底面周长×高，再乘100。\n【详解】列式：2×3.14×0.1×2×100。",
+    "knowledge_points": ["圆柱的侧面积"],
+    "source": "21-22 六年级上·江西赣州·期末"
+  },
+  {
+    "id": "174",
+    "unit": "第三单元 圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": 0.85,
+    "question": "一个圆柱形杯子，从里面量得底面直径是6厘米，高是10厘米。把一包净含量250毫升的鲜牛奶倒入杯中，能装下吗？",
+    "options": [],
+    "answer": "能",
+    "solution": "【分析】先算杯子容积，与250毫升比较。\n【详解】V=3.14×(6÷2)²×10=282.6（立方厘米）=282.6毫升，282.6>250，能装下。",
+    "knowledge_points": ["圆柱的容积"],
+    "source": "2022·甘肃天水·小升初真题"
+  },
+  {
+    "id": "175",
+    "unit": "第三单元 圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": 0.65,
+    "question": "小乐用橡皮泥制作了一个高9厘米的圆锥体，然后沿高切成完全相同的两半。表面积比原来增加了108平方厘米。这个圆锥体用了多少体积的橡皮泥？",
+    "options": [],
+    "answer": "339.12立方厘米",
+    "solution": "【分析】增加的面积是两个切面面积，每个切面是三角形。先求底面直径，再求底面半径，最后算圆锥体积。\n【详解】切面总面积108，每个54。由三角形面积54=底×9÷2，得底=12厘米（即直径）。r=6。V=1/3×3.14×6²×9=339.12（立方厘米）。",
+    "knowledge_points": ["三角形的面积", "圆锥的体积（容积）"],
+    "source": "21-22 六年级下·河南南阳·期末"
+  },
+  {
+    "id": "176",
+    "unit": "第三单元 圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": 0.65,
+    "question": "一个圆锥形碎石堆，底面直径4米，高1.5米。用这堆碎石在8米宽的公路上铺10厘米厚的路面，能铺多少米？",
+    "options": [],
+    "answer": "7.85米",
+    "solution": "【分析】先算圆锥体积，再按长方体体积=长×宽×厚求可铺长度，注意单位统一。\n【详解】碎石体积=1/3×3.14×(4÷2)²×1.5=6.28立方米。厚10厘米=0.1米。铺设长度=6.28÷8÷0.1=7.85米。",
+    "knowledge_points": ["圆锥的体积（容积）", "体积的等积变形", "长方体的体积"],
+    "source": "2022·河南信阳·小升初真题"
+  },
+  {
+    "id": "177",
+    "unit": "第三单元 圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": 0.65,
+    "question": "王冬用橡皮泥捏成一个高1.5厘米、底面半径3厘米的圆柱。后来把剩余橡皮泥改捏成一个底面直径4厘米、高9厘米的圆锥。丢掉部分体积占原圆柱体积的百分之几？（百分号前保留一位小数）",
+    "options": [],
+    "answer": "11.1%",
+    "solution": "【分析】丢掉体积=原圆柱体积-后圆锥体积；再除以原圆柱体积。\n【详解】原圆柱V1=3.14×3²×1.5=42.39。新圆锥V2=1/3×3.14×(4÷2)²×9=37.68。丢掉=4.71。比例=4.71÷42.39≈11.1%。",
+    "knowledge_points": ["圆锥的体积（容积）", "求一个数是另一个数的百分之几（百分率问题）", "圆柱的体积"],
+    "source": "21-22 六年级下·山东济南·期末"
+  },
+  {
+    "id": "178",
+    "unit": "第三单元 圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": 0.65,
+    "question": "美术室有一块棱长2分米的正方体石膏。把这块石膏加工成一个最大的圆柱，圆柱的体积是多少立方分米？",
+    "options": [],
+    "answer": "6.28立方分米",
+    "solution": "【分析】最大圆柱的底面直径和高都等于正方体棱长2分米。\n【详解】V=3.14×(2÷2)²×2=3.14×1×2=6.28（立方分米）。",
+    "knowledge_points": ["圆柱的体积"],
+    "source": "2022·河北廊坊·小升初真题"
+  },
+  {
+    "id": "179",
+    "unit": "第三单元 圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": 0.85,
+    "question": "一堆圆锥形的沙，底面半径是2米，高是1.5米。如果每立方米沙约重1.5吨，这堆沙约重多少吨？",
+    "options": [],
+    "answer": "9.42吨",
+    "solution": "【分析】先求圆锥体积，再乘单位体积重量。\n【详解】V=1/3×3.14×2²×1.5=6.28（立方米）。重量=6.28×1.5=9.42（吨）。",
+    "knowledge_points": ["圆锥的体积（容积）"],
+    "source": "21-22六年级下·江西景德镇·期末"
+  },
+  {
+    "id": "180",
+    "unit": "第三单元 圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": 0.85,
+    "question": "还记得圆柱和圆锥的体积计算公式是如何推导出来的吗？请选择其中一个进行说明。",
+    "options": [],
+    "answer": "见详解",
+    "solution": "【分析】可用“转化”思想说明：把圆柱切拼成近似长方体。\n【详解】将圆柱切割后拼成近似长方体，长方体底面积等于圆柱底面积，高等于圆柱高，体积不变。由长方体V=底面积×高，得圆柱V=Sh。若说明圆锥，可用等底等高圆柱与圆锥体积关系推得V=1/3Sh。",
+    "knowledge_points": ["圆柱的体积"],
+    "source": "21-22六年级下·内蒙古赤峰·期末"
+  },
+  {
+    "id": "181",
+    "unit": "第三单元 圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": 0.65,
+    "question": "一个圆锥形沙堆，底面直径10米，高3米。已知每立方米沙子约重1.4吨，如果要用载重15吨的货车一次运完这堆沙子，需要多少辆这样的货车？",
+    "options": [],
+    "answer": "8辆",
+    "solution": "【分析】先求沙堆体积，再求总质量，最后除以单车载重并进一取整。\n【详解】V=1/3×3.14×(10÷2)²×3=78.5立方米。总重=78.5×1.4=109.9吨。109.9÷15≈7.33，需8辆。",
+    "knowledge_points": ["商的近似数", "圆锥的体积（容积）"],
+    "source": "21-22六年级下·湖南株洲·期末"
+  },
+  {
+    "id": "182",
+    "unit": "第三单元 圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": 0.65,
+    "question": "一个圆锥形的沙堆，占地面积是9平方米，高是1.5米。如果每立方米沙重2吨，用载重为2.5吨的车把这堆沙运走，至少要运几次？",
+    "options": [],
+    "answer": "4次",
+    "solution": "【分析】先求圆锥体积V=1/3Sh，再求总重，最后除以2.5并进一取整。\n【详解】V=1/3×9×1.5=4.5立方米。总重=4.5×2=9吨。9÷2.5=3.6，至少4次。",
+    "knowledge_points": ["商的近似数", "圆锥的体积（容积）"],
+    "source": "21-22六年级下·辽宁本溪·期末"
+  },
+  {
+    "id": "183",
+    "unit": "第三单元 圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": 0.85,
+    "question": "今年小麦大丰收，李大伯把小麦堆成一个圆锥形，小麦堆底面积是12.56平方米，高是1.5米。如果每立方米小麦质量约为700千克，这堆小麦质量约为多少千克？",
+    "options": [],
+    "answer": "4396千克",
+    "solution": "【分析】圆锥体积=1/3×底面积×高，再乘单位体积质量。\n【详解】质量=1/3×12.56×1.5×700=4396（千克）。",
+    "knowledge_points": ["圆锥的体积（容积）"],
+    "source": "2022·安徽阜阳·小升初真题"
+  },
+  {
+    "id": "184",
+    "unit": "第三单元 圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": 0.85,
+    "question": "一个圆锥形谷堆，底面直径10米，高1.5米。如果每立方米稻谷约重800千克，这堆稻谷约重多少千克？",
+    "options": [],
+    "answer": "31400千克",
+    "solution": "【分析】先算圆锥体积，再乘800千克/立方米。\n【详解】V=1/3×3.14×(10÷2)²×1.5=39.25立方米。质量=39.25×800=31400千克。",
+    "knowledge_points": ["圆锥的体积（容积）"],
+    "source": "21-22 六年级下·河南商丘·期末"
+  },
+  {
+    "id": "185",
+    "unit": "第三单元 圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": 0.85,
+    "question": "把一个底面积25平方分米、高8分米的圆柱体木料削成一个圆锥体。圆锥高是原圆柱高的3/4，底面积与原圆柱相等。削去部分体积是多少？",
+    "options": [],
+    "answer": "150立方分米",
+    "solution": "【分析】先求圆柱体积，再求圆锥高和圆锥体积，最后相减。\n【详解】圆锥高=8×3/4=6分米。圆柱体积=25×8=200。圆锥体积=1/3×25×6=50。削去部分=200-50=150（立方分米）。",
+    "knowledge_points": ["圆锥的体积（容积）", "求一个数的几分之几的问题", "圆柱的体积"],
+    "source": "2022·海南省直辖县级单位·小升初真题"
+  },
+  {
+    "id": "186",
+    "unit": "第三单元 圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": 0.85,
+    "question": "一个圆锥形钢锭，底面直径6分米，高5分米，体积多少？如果每立方分米重3千克，这个钢锭重多少千克？",
+    "options": [],
+    "answer": "47.1立方分米；141.3千克",
+    "solution": "【分析】圆锥体积V=1/3πr²h，算出体积后乘每立方分米质量。\n【详解】V=1/3×3.14×(6÷2)²×5=47.1立方分米。重量=47.1×3=141.3千克。",
+    "knowledge_points": ["圆锥的体积（容积）"],
+    "source": "2022·新疆吐鲁番·小升初真题"
+  },
+  {
+    "id": "187",
+    "unit": "第三单元 圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": 0.65,
+    "question": "把一块棱长10厘米的正方体铁块铸成一个底面直径20厘米的圆锥形铁块。这个圆锥形铁块的高约是多少？（得数保留整厘米）",
+    "options": [],
+    "answer": "10厘米",
+    "solution": "【分析】等积变形：正方体体积=圆锥体积。先求正方体体积，再用圆锥公式反求高。\n【详解】正方体体积=10×10×10=1000立方厘米。设圆锥高h：1/3×3.14×(20÷2)²×h=1000，得h≈10厘米。",
+    "knowledge_points": ["正方体的体积", "圆锥的体积（容积）", "体积的等积变形"],
+    "source": "2015·浙江·小升初真题"
+  },
+  {
+    "id": "188",
+    "unit": "第三单元 圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": 0.85,
+    "question": "小丽想准确量出圆锥的高，请你结合自己的经验说说测量方法。（可以写出来，也可以画出来）",
+    "options": [],
+    "answer": "见详解",
+    "solution": "【分析】圆锥高是顶点到底面圆心的垂直距离。\n【详解】测量方法：①把圆锥底面放平；②在顶点上方平放一块平板（与底面平行）；③用直尺竖直量平板到底面的距离，这个距离就是圆锥的高。",
+    "knowledge_points": ["长度的测量方法", "圆锥的认识及特征"],
+    "source": "21-22 六年级下·山西晋中·期末"
+  },
+  {
+    "id": "189",
+    "unit": "第三单元 圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": 0.85,
+    "question": "一个圆柱形粮囤，从里面量得底面半径2m，高3m。如果每立方米玉米重750kg，这个粮囤能装多少吨玉米？",
+    "options": [],
+    "answer": "28.26吨",
+    "solution": "【分析】先算圆柱体积，再乘单位体积重量，最后换算成吨。\n【详解】V=3.14×2²×3=37.68立方米。总重=37.68×750=28260千克=28.26吨。",
+    "knowledge_points": ["吨、千克之间的换算与比较", "圆柱的体积"],
+    "source": "21-22 六年级下·全国·期末"
+  },
+  {
+    "id": "190",
+    "unit": "第三单元 圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": 0.65,
+    "question": "一个底面半径是6cm的圆柱形玻璃器皿里装有一部分水，水中浸没着一个高9cm的圆锥形铅锤（水没溢出）。当铅锤取出后，水面下降0.5cm。这个铅锤的底面积是多少平方厘米？",
+    "options": [],
+    "answer": "18.84平方厘米",
+    "solution": "【分析】水面下降对应减少的水体积=铅锤（圆锥）体积。先用圆柱底面积×下降高度求体积，再由圆锥V=1/3Sh反求底面积S。\n【详解】V=3.14×6²×0.5=56.52立方厘米。S=V÷(1/3)÷9=56.52×3÷9=18.84平方厘米。",
+    "knowledge_points": ["不规则物体的体积算法", "圆锥的体积（容积）", "圆柱的体积"],
+    "source": "2014·全国·小升初真题"
+  },
+  {
+    "id": "191",
+    "unit": "第三单元 圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": 0.65,
+    "question": "一个圆柱和一个圆锥体积相等，高也相等，圆柱底面积是12平方厘米，圆锥的底面积是多少平方厘米？",
+    "options": [],
+    "answer": "36平方厘米",
+    "solution": "【分析】体积相等且高相等时，圆锥底面积是圆柱底面积的3倍。\n【详解】12×3=36（平方厘米）。",
+    "knowledge_points": ["圆锥的体积（容积）", "圆柱与圆锥体积的关系"],
+    "source": "21-22六年级下·全国·单元测试"
+  },
+  {
+    "id": "192",
+    "unit": "第三单元 圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": 0.65,
+    "question": "一个圆锥的体积是48cm³，高是8cm，底面积是多少cm²？",
+    "options": [],
+    "answer": "18cm²",
+    "solution": "【分析】圆锥体积公式V=1/3Sh，已知V、h，求S。\n【详解】S=48÷(1/3)÷8=144÷8=18（cm²）。",
+    "knowledge_points": ["圆锥的体积（容积）"],
+    "source": "21-22 六年级下·全国·单元测试"
+  },
+  {
+    "id": "193",
+    "unit": "第三单元 圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": 0.65,
+    "question": "一个圆柱的高是12.56分米，把它的侧面展开后是一个正方形，这个圆柱的底面半径是多少分米？",
+    "options": [],
+    "answer": "2分米",
+    "solution": "【分析】展开成正方形说明底面周长=高=12.56分米。半径r=C÷(2π)。\n【详解】r=12.56÷2÷3.14=2（分米）。",
+    "knowledge_points": ["圆柱的侧面积"],
+    "source": "21-22 六年级下·全国·单元测试"
+  },
+  {
+    "id": "194",
+    "unit": "第三单元 圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": 0.65,
+    "question": "一个圆柱削去20立方分米，正好削成一个与它等底等高的圆锥，这个圆柱的体积是多少立方分米？",
+    "options": [],
+    "answer": "30立方分米",
+    "solution": "【分析】等底等高时圆锥体积是圆柱的1/3，削去的是圆柱的2/3，已知2/3对应20。\n【详解】圆柱体积=20÷(1-1/3)=20÷(2/3)=30（立方分米）。",
+    "knowledge_points": ["圆锥的体积（容积）", "圆柱与圆锥体积的关系"],
+    "source": "21-22六年级下·全国·单元测试"
+  },
+  {
+    "id": "195",
+    "unit": "第三单元 圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": 0.65,
+    "question": "在打谷场上，有一个近似圆锥的稻谷堆，底面直径4米，高1.2米。每立方米稻谷约重735千克，这堆稻谷约有多少千克？（得数保留整千克）",
+    "options": [],
+    "answer": "3693千克",
+    "solution": "【分析】先求圆锥体积，再乘每立方米重量。\n【详解】质量=1/3×3.14×(4÷2)²×1.2×735≈3693（千克）。",
+    "knowledge_points": ["圆锥的体积（容积）"],
+    "source": "2014·全国·小升初真题"
+  },
+  {
+    "id": "196",
+    "unit": "第三单元 圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": 0.65,
+    "question": "一个圆柱的底面半径是4分米，高是10分米，它的体积是多少立方分米，侧面积是多少平方分米？",
+    "options": [],
+    "answer": "502.4立方分米；251.2平方分米",
+    "solution": "【分析】体积公式V=πr²h，侧面积公式S侧=2πrh。\n【详解】V=3.14×4²×10=502.4（立方分米）；S侧=2×3.14×4×10=251.2（平方分米）。",
+    "knowledge_points": ["圆柱的侧面积", "圆柱的体积"],
+    "source": "21-22 六年级下·全国·单元测试"
+  },
+  {
+    "id": "197",
+    "unit": "第三单元 圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": 0.65,
+    "question": "一个圆锥的底面半径是2dm，体积是12.56dm³，它的高是多少dm？",
+    "options": [],
+    "answer": "3dm",
+    "solution": "【分析】圆锥V=1/3πr²h，已知V和r，反求h。\n【详解】h=12.56÷(1/3)÷(3.14×2²)=37.68÷12.56=3（dm）。",
+    "knowledge_points": ["圆锥的体积（容积）"],
+    "source": "21-22 六年级下·全国·单元测试"
+  },
+  {
+    "id": "198",
+    "unit": "第三单元 圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": 0.4,
+    "question": "一个圆柱形储水桶，底面周长12.56分米，高3分米，盛满一桶水后倒入另一长方体水池，水池里还空着21.5%。已知长方体水池长6分米，长是宽的1又1/2倍，求水池的高是多少分米？",
+    "options": [],
+    "answer": "2分米",
+    "solution": "【分析】先由周长求圆柱底面半径，求桶内水体积；再由长和长宽关系求水池宽，进而求当前水深；已知“还空21.5%”，水占78.5%，据此求总高。\n【详解】桶半径=12.56÷3.14÷2=2分米。水体积=3.14×2²×3=37.68立方分米。水池宽=6÷1.5=4分米。当前水深=37.68÷6÷4=1.57分米。总高=1.57÷(1-21.5%)=1.57÷0.785=2分米。",
+    "knowledge_points": ["已知一个数的百分之几是多少，求这个数", "体积的等积变形", "圆柱的体积", "长方体的体积"],
+    "source": "19-20 六年级下·湖南怀化·单元测试"
+  },
+  {
+    "id": "199",
+    "unit": "第三单元 圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": 0.65,
+    "question": "一辆货车车厢是一个长方体，车厢内长4米、宽1.5米、高4米，装满一车沙后卸下，堆成一个高2米的圆锥形。圆锥底面积是多少平方米？",
+    "options": [],
+    "answer": "36平方米",
+    "solution": "【分析】沙体积不变。先求长方体体积，再由圆锥V=1/3Sh反求底面积S。\n【详解】沙体积=4×1.5×4=24立方米。S=24×3÷2=36平方米。",
+    "knowledge_points": ["圆锥的体积（容积）", "体积的等积变形", "长方体的体积"],
+    "source": "19-20 六年级下·湖南怀化·单元测试"
+  },
+  {
+    "id": "200",
+    "unit": "第三单元 圆柱与圆锥",
+    "question_type": "解答题",
+    "difficulty": 0.65,
+    "question": "一个圆锥形沙堆，底面积是18平方米，高是1.5米。如果每立方米沙重1600千克，这堆沙重多少吨？",
+    "options": [],
+    "answer": "14.4吨",
+    "solution": "【分析】先求圆锥体积，再乘单位体积重量并换算吨。\n【详解】体积=1/3×18×1.5=9立方米。质量=9×1600=14400千克=14.4吨。",
+    "knowledge_points": ["圆锥的体积（容积）"],
+    "source": "19-20 六年级下·湖南怀化·单元测试"
+  }
+    
+  
+  ]
diff --git "a/\345\205\255\345\271\264\347\272\247\351\242\230\351\233\206/\345\205\255\345\271\264\347\272\247\344\270\213/\345\205\255\345\271\264\347\272\247\344\270\213\347\254\254\344\272\214\345\215\225\345\205\203\347\231\276\345\210\206\346\225\260\346\225\260\346\215\256\351\233\206.json" "b/\345\205\255\345\271\264\347\272\247\351\242\230\351\233\206/\345\205\255\345\271\264\347\272\247\344\270\213/\345\205\255\345\271\264\347\272\247\344\270\213\347\254\254\344\272\214\345\215\225\345\205\203\347\231\276\345\210\206\346\225\260\346\225\260\346\215\256\351\233\206.json"
new file mode 100644
index 0000000000000000000000000000000000000000..307263cf185b092fefcfca15893d63267242a2fd
--- /dev/null
+++ "b/\345\205\255\345\271\264\347\272\247\351\242\230\351\233\206/\345\205\255\345\271\264\347\272\247\344\270\213/\345\205\255\345\271\264\347\272\247\344\270\213\347\254\254\344\272\214\345\215\225\345\205\203\347\231\276\345\210\206\346\225\260\346\225\260\346\215\256\351\233\206.json"
@@ -0,0 +1,9206 @@
+[
+    {
+        "id": "unit2_choice_001",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "银行存期三年的年利率为 $2.75 \\%$，小安将 20000 元存入银行，三年后，她打算把利息捐给贫困山区的孩子们，她能捐出（）元钱。",
+        "options": [
+          "A. 550",
+          "B. 20550",
+          "C. 1650",
+          "D. 21650"
+        ],
+        "answer": "C",
+        "solution": "【分析】根据利息 $=$ 本金 $\\times$ 利率 $\\times$ 存期，代入数据求出利息即可。\n\n【详解】$20000 \\times 2.75\\% \\times 3\\\\= 550 \\times 3 \\\\ = 1650元$\n\n所以她能捐出 1650 元。故答案为: C",
+        "knowledge_points": [
+          "求利息"
+        ],
+        "source": "24-25 六年级下·山东菏泽·期中",
+        "merged_id": "unit2_001",
+        "original_id": "unit2_choice_001"
+      },
+      {
+        "id": "unit2_choice_002",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "王刚想将手机里的钱全部提现。若提现超 1000 元，超出的部分需要收取 $0.1\\%$ 的手续费。王刚要付手续费 5 元，他提现了（）元。",
+        "options": [
+          "A. 1000",
+          "B. 5000",
+          "C. 6000",
+          "D. 10000"
+        ],
+        "answer": "C",
+        "solution": "【分析】已知超出部分的手续费率是 $0.1 \\%$，手续费为 5 元。因为'超出部分金额 $\\times 0.1 \\% =$ 手续费'，所以超出 1000 元部分的金额 $=$ 手续费 $\\div 0.1 \\%$。将手续费 5 元，手续费率 $0.1 \\%$ 代入，可得超出部分金额为 $5 \\div 0.1 \\% = 5 \\div 0.001 = 5000$ 元。提现总额由 1000 元（不收取手续费部分）和超出的 5000 元组成，所以提现总额就是 1000 加 5000。\n\n【详解】$5 \\div 0.1\\%\\\\= 5 \\div 0.001 \\\\ = 5000 (\\text{元})$\n\n$1000 + 5000 = 6000 \\text{(元)}$ 王刚提现了6000元。故答案为: C",
+        "knowledge_points": [
+          "已知一个数的百分之几是多少，求这个数",
+          "求税率或收入额"
+        ],
+        "source": "24-25 六年级下·福建厦门·期中",
+        "merged_id": "unit2_002",
+        "original_id": "unit2_choice_002"
+      },
+      {
+        "id": "unit2_choice_003",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "李叔叔买福利彩票中了 12 万元的奖。按规定应缴纳 $20\\%$ 的个人所得税，李叔叔实际得到（）万元的奖金。",
+        "options": [
+          "A. 2.4",
+          "B. 9.6",
+          "C. 8.6"
+        ],
+        "answer": "B",
+        "solution": "【分析】已知买福利彩票中了 12 万元的奖，按规定应缴纳 $20 \\%$ 的个人所得税，即应缴纳的个人所得税额占中奖金额的20%，把中奖金额看作单位'1'，单位'1'已知，用中奖金额乘 $20 \\%$，求出应缴纳的个人所得税，再用中奖金额减去缴纳的个人所得税，即是实际得到的奖金。\n\n【详解】$12 - 12 \\times 20\\%= 12 - 12 \\times 0.2 \\\\ = 12 - 2.4 \\\\ = 9.6 \\text{(万元)}$\n\n李叔叔实际得到9.6万元的奖金。故答案为: B",
+        "knowledge_points": [
+          "求应纳税额",
+          "求一个数的百分之几是多少"
+        ],
+        "source": "24-25 六年级下·海南海口·单元测试",
+        "merged_id": "unit2_003",
+        "original_id": "unit2_choice_003"
+      },
+      {
+        "id": "unit2_choice_004",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "成数表示（）。",
+        "options": [
+          "A. 一个数是另一个数的百分之几",
+          "B. 一个数是另一个数的几倍",
+          "C. 一个数是另一个数的十分之几"
+        ],
+        "answer": "C",
+        "solution": "【分析】成数表示一个数是另一个数的十分之几或百分之几十。据此举例解答。\n\n【详解】如：八成 $= \\frac{8}{10} = 80\\%$，七成五 $= 75\\%$。成数表示一个数是另一个数的十分之几或百分之几十。答案为：C",
+        "knowledge_points": [
+          "分数、小数、百分数与成数的互化"
+        ],
+        "source": "24-25 六年级下·海南海口·单元测试",
+        "merged_id": "unit2_004",
+        "original_id": "unit2_choice_004"
+      },
+      {
+        "id": "unit2_choice_005",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "一台电脑按九五折出售，下列说法错误的是（）。",
+        "options": [
+          "A. 比原价优惠 $5 \\%$",
+          "B. 按原价的 $95\\%$出售",
+          "C. 按原价的 $9.5\\%$ 出售"
+        ],
+        "answer": "C",
+        "solution": "【分析】几折就是现价是原价的百分之几十，据此逐项分析，进行解答。\n\n【详解】A．九五折就是现价是原价的 $95\\%$；把原价看单位'1';$1 - 95\\% = 5\\%$，比原价优惠 $5\\%$，原题干说法正确。B. 九五折就是现价是原价的 $95 \\%$，即按原价的 $95 \\%$ 出售，原题干说法正确。C. 九五折就是现价是原价的 $95 \\%$，即按原价的 $95 \\%$ 出售。按原价的 $9.5 \\%$ 出售说法错误。一台电脑按九五折出售，说法错误的是按原价的 $9.5\\%$ 出售。故答案为：C",
+        "knowledge_points": [
+          "求折扣（折扣问题）",
+          "求现价（折扣问题）"
+        ],
+        "source": "24-25 六年级下·海南海口·单元测试",
+        "merged_id": "unit2_005",
+        "original_id": "unit2_choice_005"
+      },
+      {
+        "id": "unit2_choice_006",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "李阿姨有600元国债，是2020年五年期的（五年期的年利率为 $6.34\\%$）。到期后她可以得到的利息列式为（）。",
+        "options": [
+          "A. $600 \\times 6.34 \\%$",
+          "B. $600 \\times 6.34 \\% \\times 5$",
+          "C. $600 \\times 6.34 \\% \\times 5 + 600$"
+        ],
+        "answer": "B",
+        "solution": "【分析】根据利息公式：利息 $=$ 本金 $\\times$ 利率 $\\times$ 时间，代入数据，即可求出利息，据此解答。\n\n【详解】$600 \\times 6.34\\% \\times 5\\\\= 38.04 \\times 5 \\\\ = 190.2元$\n\n李阿姨有 600 元国债，是 2020 年五年期的（五年期的年利率为 $6.34\\%$）。到期后她可以得到的利息列式为 $600 \\times 6.34\\% \\times 5$。故答案为: B",
+        "knowledge_points": [
+          "求利息"
+        ],
+        "source": "24-25 六年级下·海南海口·单元测试",
+        "merged_id": "unit2_006",
+        "original_id": "unit2_choice_006"
+      },
+      {
+        "id": "unit2_choice_007",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "折扣、成数、税率、利率都是百分数在生活中的应用。下面说法错误的是（）。",
+        "options": [
+          "A. 某商品打七五折出售，就是按原价的 $75 \\%$ 出售。",
+          "B. 农业收成，经常用'成数'表示，三成五就是 $35 \\text{‰}$。",
+          "C. 税率指应纳税额与各种收入中应纳税部分的百分比。",
+          "D. 优惠三折表示现价是原价的 $30 \\text{‰}$"
+        ],
+        "answer": "D",
+        "solution": "【分析】根据折扣、成数、税率、利率知识可知：打几折，就是按照原价的百分之几十出售；农业收成几成就是百分之几十；税率就是应纳税额与各种收入中应纳税部分的百分比，据此解答即可。\n\n【详解】A．某商品打七五折出售，就是按原价的 $75 \\%$ 出售。原题干说法正确。B. 农业收成，经常用'成数'表示，三成五就是 $35 \\text{‰}$。原题干说法正确。C. 税率指应纳税额与各种收入中应纳税部分的百分比。原题干说法正确。D. 优惠三折表示现价是原价的 $70 \\%$，原题干说法错误。折扣、成数、税率、利率都是百分数在生活中的应用。说法错误的是优惠三折表示现价是原价的 $30 \\text{%}$。故答案为: D",
+        "knowledge_points": [
+          "求应纳税额",
+          "分数、小数、百分数与成数的互化",
+          "求现价（折扣问题）"
+        ],
+        "source": "24-25 六年级下·海南海口·单元测试",
+        "merged_id": "unit2_007",
+        "original_id": "unit2_choice_007"
+      },
+      {
+        "id": "unit2_choice_008",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "爸爸有 2 万元，有两种理财方式：一种是买3年期国债，年利率是 $3.05\\%$；另一种是存银行定期3年，年利率是 $2.9\\%$。3年后，两种理财方式收益相比，（）。",
+        "options": [
+          "A. 买3年期国债收益大",
+          "B. 存银行定期 3 年收益大",
+          "C. 两种方式收益一样大",
+          "D. 无法比较"
+        ],
+        "answer": "A",
+        "solution": "【分析】根据利息 $=$ 本金 $\\times$ 利率 $\\times$ 时间，分别求出买3年期国债和存银行定期到期的利息，再进行比较，即可解答。\n\n【详解】2万 $= 20000$\n\n国债：\n\n$20000 \\times 3.05 \\% \\times 3 \\\\ = 610 \\times 3 \\\\ = 1830 (\\text{元})$\n\n存银行定期：\n\n$20000 \\times 2.9 \\% \\times 3 \\\\ = 280 \\times 3 \\\\ = 1740元$\n\n$1830 > 1740$，买 3 年期国债收益大。\n\n爸爸有 2 万元，有两种理财方式：一种是买 3 年期国债，年利率是 $3.05\\%$；另一种是存银行定期 3 年，年利率是 2.9%。3年后，两种理财方式收益相比，买3年期国债收益大。故答案为: A",
+        "knowledge_points": [
+          "求利息",
+          "选择储蓄的最佳方案"
+        ],
+        "source": "24-25 六年级下·全国·单元测试",
+        "merged_id": "unit2_008",
+        "original_id": "unit2_choice_008"
+      },
+      {
+        "id": "unit2_choice_009",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "有一块稻田，今年收稻谷 $2200 \\mathrm{~kg}$，比去年增产一成。今年比去年增产（） $\\mathrm{kg}$。",
+        "options": [
+          "A. 200",
+          "B. 2000",
+          "C. 2420"
+        ],
+        "answer": "A",
+        "solution": "【分析】已知今年收稻谷 $2200 \\mathrm{~kg}$，比去年增产一成，把去年收稻谷的质量看作单位'1'，则今年收稻谷的质量是去年的 $(1 + 10 \\%)$，单位'1'未知，用今年收稻谷的质量除以 $(1 + 10)$，求出去年收稻谷的质量;因为今年收稻谷的质量比去年增产一成，那么今年比去年增产稻谷的质量是去年的 $10\\%$，根据求一个数的百分之几是多少，用去年收稻谷的质量乘 $10\\%$，求出今年比去年增产的稻谷质量。\n\n【详解】一成 $= 10\\%$ $2200 \\div (1 + 10\\%) = 2200 \\div 1.1$ $= 2000 \\, (\\mathrm{kg})$ $2000 \\times 10\\% = 2000 \\times 0.1$ $= 200 \\, (\\mathrm{kg})$\n\n今年比去年增产 $200\\mathrm{kg}$。故答案为：A",
+        "knowledge_points": [
+          "求一个数的百分之几是多少",
+          "根据成数反求单位'1'"
+        ],
+        "source": "24-25 六年级下·全国·假期作业",
+        "merged_id": "unit2_009",
+        "original_id": "unit2_choice_009"
+      },
+      {
+        "id": "unit2_choice_010",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "李军将6000元压岁钱存入银行，定期三年，年利率是 $2.15\\%$。到期后李军应得本金和利息一共多少元？以下列式正确的是（）。",
+        "options": [
+          "A. $6000 \\times 2.15 \\% \\times 3$",
+          "B. $(6000 + 6000 \\times 2.15\\%) \\times 3$",
+          "C. $6000 + 6000 \\times 2.15\\%$",
+          "D. $6000 + 6000 \\times 2.15 \\% \\times 3$"
+        ],
+        "answer": "D",
+        "solution": "【分析】到期取款时银行多支付的钱叫利息，根据利息 $=$ 本金 $\\times$ 利率 $\\times$ 存期，用本金 + 利息即可。\n\n【详解】$6000 + 6000 \\times 2.15\\% \\times 3\\\\= 6000 + 6000 \\times 0.0215 \\times 3 \\\\ = 6000 + 387 \\\\ = 6387元$\n\n到期后李军应得本金和利息一共6387元。列式正确的是 $6000 + 6000 \\times 2.15\\% \\times 3$。故答案为: D",
+        "knowledge_points": [
+          "求利息"
+        ],
+        "source": "24-25 六年级上·北京顺义·期末",
+        "merged_id": "unit2_010",
+        "original_id": "unit2_choice_010"
+      },
+      {
+        "id": "unit2_choice_011",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "张老师买一副标价300元的乒乓球拍，下面哪种促销方法更省钱？（）",
+        "options": [
+          "A．打七折销售",
+          "B．满200元减80元",
+          "C．先打八折，在此基础上再打9折"
+        ],
+        "answer": "A",
+        "solution": "【分析】A. 打七折就是原价的 $70 \\%$，根据'原价 $\\times$ 折扣 $=$ 现价'求出现价;B. $300 > 200$，用标 300 元减去 80 元，求出现价;C. 打几折就是按原价的百分之几十销售，用原价乘 $80 \\%$,再乘 $90 \\%$ 求出现价。把三种促销方式进行比较即可解答。\n\n【详解】A. $300 \\times 70\\% = 210$ （元）B. $300 - 80 = 220$ (元)\n\nC. $300 \\times 80\\% \\times 90\\%\\\\= 340 \\times 90 \\% \\\\ = 216元$\n\n$210 < 216 < 220$所以打七折销售方法更省钱。故答案为: A",
+        "knowledge_points": [
+          "经济问题",
+          "求现价（折扣问题）"
+        ],
+        "source": "2024·四川绵阳·小升初真题",
+        "merged_id": "unit2_011",
+        "original_id": "unit2_choice_011"
+      },
+      {
+        "id": "unit2_choice_012",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "一种电视机原价1800元，现在售价是1620元，现在是打（）折出售的。",
+        "options": [
+          "A. 一",
+          "B. 九",
+          "C. 二",
+          "D. 八"
+        ],
+        "answer": "B",
+        "solution": "【分析】已知电视机 1800 元，现价是 1620 元，用现价除以原价，求出现价是原价的百分之几，再根据折扣的意义,把百分数化成折扣即可。\n\n【详解】$1620 \\div 1800 \\times 100\\%\\\\= 0.9 \\times 100 \\% \\\\ = 90 \\%$\n\n90%=九折现在是打九折出售的。故答案为: B",
+        "knowledge_points": [
+          "求折扣（折扣问题）",
+          "求一个数是另一个数的百分之几（百分率问题）"
+        ],
+        "source": "23-24 六年级上·河北邯郸·期中",
+        "merged_id": "unit2_012",
+        "original_id": "unit2_choice_012"
+      },
+      {
+        "id": "unit2_choice_013",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "某县 2022 年粮食产量为3万吨，2023年粮食产量为3.3万吨。2023年比2022年增产（）。",
+        "options": [
+          "A. 一成",
+          "B. 三成",
+          "C. 五成",
+          "D. 九成"
+        ],
+        "answer": "A",
+        "solution": "【分析】求 2023 年比 2022 年增产几成，就是求 2023 年比 2022 年增产百分之几十。求一个数比另一个数多（或少）百分之几，先求出多（或少）的具体数量，再除以单位'1'数量即可解答。据此用3.3减去3的差，再除以3即可解答。\n\n【详解】$(3.3 - 3) \\div 3\\\\= 0.3 \\div 3 \\\\ = 0.1 \\\\ = 10 \\% \\\\ = \\text{一成}$\n\n则2023年比2022年增产一成。故答案为: A",
+        "knowledge_points": [
+          "求增加或减少几成的实际问题"
+        ],
+        "source": "23-24 六年级下·北京东城·期末",
+        "merged_id": "unit2_013",
+        "original_id": "unit2_choice_013"
+      },
+      {
+        "id": "unit2_choice_014",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "甜甜妈妈花了180元买了一双打六折的童鞋，这双童鞋的原价是（）元。",
+        "options": [
+          "A. 450",
+          "B. 300",
+          "C. 288",
+          "D. 108"
+        ],
+        "answer": "B",
+        "solution": "【分析】将原价看作单位'1'，几折就是百分之几十，现价 $\\div$ 折扣 $=$ 原价，据此列式计算。\n\n【详解】$180 \\div 60\\% = 180 \\div 0.6 = 300$ （元）这双童鞋的原价是 300 元。故答案为: B",
+        "knowledge_points": [
+          "求原价（折扣问题）"
+        ],
+        "source": "23-24 六年级下·四川广元·期末",
+        "merged_id": "unit2_014",
+        "original_id": "unit2_choice_014"
+      },
+      {
+        "id": "unit2_choice_015",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "微信提现收费规则:每位注册用户享有1000元免费提现额度，超过部分收取 0.1%的手续费。杨老师是微信注册新用户，现在他需要从微信钱包中提取现金 1900 元，需要支付（）元手续费。",
+        "options": [
+          "A. 100",
+          "B. 90",
+          "C. 19",
+          "D. 0.9"
+        ],
+        "answer": "D",
+        "solution": "【分析】根据题意，杨老师是微信注册新用户，可以享有1000元免费提现额度，超过1000元的部分为（1900一1000）元，收取 $0.1\\%$ 的手续费，根据求一个数的百分之几是多少，用超过部分的金额乘 $0.1\\%$，即可求出需要支付的手续费。\n\n【详解】$(1900 - 1000) \\times 0.1\\%\\\\= 900 \\times 0.001 \\\\ = 0.9元$\n\n需要支付0.9元手续费。故答案为: D",
+        "knowledge_points": [
+          "求应纳税额",
+          "求一个数的百分之几是多少"
+        ],
+        "source": "23-24 六年级下·重庆开州·期末",
+        "merged_id": "unit2_015",
+        "original_id": "unit2_choice_015"
+      },
+      {
+        "id": "unit2_choice_016",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "聪聪的爸爸得到一笔5000元的劳务报酬，其中800元是免税的，其余部分要按 $20\\%$ 的税率缴税，这笔劳务报酬要缴税多少元？列式正确的是（）。",
+        "options": [
+          "A. $5000 \\times 20\\%$",
+          "B. $(5000 - 800) \\times 20\\%$",
+          "C. $5000 \\div 20\\%$",
+          "D. $(5000 - 800) \\div 20\\%$"
+        ],
+        "answer": "B",
+        "solution": "【分析】应缴税额 $=$ 应缴税所得额 $\\times$ 税率，应缴税所得额为 $(5000 - 800)$ 元，税率为 $20\\%$，代入数据计算即可。\n\n【详解】$(5000 - 800) \\times 20\\%\\\\= 4200 \\times 20 \\% \\\\ = 840元$\n\n这笔劳务报酬要缴税 840 元，列式正确的是（5000- 800） $\\times 20 \\text{%}$ 故答案为: B",
+        "knowledge_points": [
+          "分段计算解决纳税问题"
+        ],
+        "source": "22-23 六年级下·河南信阳·期中",
+        "merged_id": "unit2_016",
+        "original_id": "unit2_choice_016"
+      },
+      {
+        "id": "unit2_choice_017",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "下面哪一个数与三成五相等? 答案是（）。",
+        "options": [
+          "A. 3.5",
+          "B. $35\\%$",
+          "C. $0.35\\%$",
+          "D. $3.5\\%$"
+        ],
+        "answer": "B",
+        "solution": "【分析】成数表示一个数是另一个数的十分之几或百分之几十，通称'几成'。根据成数的意义可知，三成五是十分之三点五或百分之三十，据此解答。\n\n【详解】三成五 $= \\frac{3.5}{10} = \\frac{35}{100}\\\\=\\frac{35}{100} \\\\= 35\\% = 0.35$\n\n所以，$35\\%$ 或 $0.35$ 与三成五相等。故答案为: B",
+        "knowledge_points": [
+          "分数、小数、百分数与成数的互化"
+        ],
+        "source": "22-23 六年级下·广东云浮·阶段练习",
+        "merged_id": "unit2_017",
+        "original_id": "unit2_choice_017"
+      },
+      {
+        "id": "unit2_choice_018",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "某商品的标价是3000元，打八折出售后仍盈利100元，则该商品进价是（）元。",
+        "options": [
+          "A. 2100",
+          "B. 2050",
+          "C. 2300",
+          "D. 2500"
+        ],
+        "answer": "C",
+        "solution": "【分析】将标价看作单位'1'，打八折是按原价的 $80\\%$ 出售，标价 $\\times$ 折扣 $=$ 售价，售价一盈利 $=$ 进价，据此列式计算。关键是理解折扣的意义，几折就是百分之几十。\n\n【详解】$3000 \\times 80\\% - 100\\\\= 3000 \\times 0.8 - 100 \\\\ = 2400 - 100 \\\\ = 2300元$\n\n该商品进价是 2300 元。故答案为: C",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "2022·黑龙江齐齐哈尔·小升初真题",
+        "merged_id": "unit2_018",
+        "original_id": "unit2_choice_018"
+      },
+      {
+        "id": "unit2_choice_019",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "三年期国债的年利率是 $3.35\\%$，兰兰拿出9000元压岁钱购买国债，到期连本带息共（）元。",
+        "options": [
+          "A. 9301.5",
+          "B. 9100.5",
+          "C. 9904.5"
+        ],
+        "answer": "C",
+        "solution": "【分析】本金是 9000 元，年利率是 $3.35 \\%$，存期是 3 年,通过利息的计算公式: 利息 $=$ 本金 $\\times$ 利率 $\\times$ 存期，求出利息,再加上本金，即可得解。此题的解题关键是利用计算利息的公式来求解，注意连本带息指的是本金和利息。\n\n【详解】$9000 + 9000 \\times 3 \\times 3.35\\%\\\\= 9000 + 27000 \\times 0.0335 \\\\ = 9000 + 904.5\\\\ = 9904.5$ （元）\n\n即到期连本带息共 9904.5 元。故答案为: C",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "22-23 六年级下·云南楚雄·期中",
+        "merged_id": "unit2_019",
+        "original_id": "unit2_choice_019"
+      },
+      {
+        "id": "unit2_choice_020",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "爸爸有2万元钱，有两种理财方式：一种是买3年期国债，年利率为 $4\\%$；另一种是买银行1年期理财产品，年收益率是 $4\\%$，每年到期后连本带息继续购买下一年的理财产品。3年后，哪种理财方式收益更大（）。",
+        "options": [
+          "A. 国债",
+          "B. 理财产品",
+          "C. 一样大",
+          "D. 无法确定"
+        ],
+        "answer": "B",
+        "solution": "【分析】先算出爸爸2万元买3年国债到期的利息，$20000 \\times 4\\% \\times 3 = 2400$ 元，加上本金，$20000 + 2400 = 22400$ 元；爸爸买理财产品收益：第一年为： $20000 \\times 4\\% = 800$ 元，本金十利息为： $800 + 20000 = 20800$ 元；第二年为：$20800 \\times 4\\% = 832$ 元，本金 $+$ 利息为： $20800 + 832 = 21632$ 元；第三年为： $21632 \\times 4\\% = 865.28$ 元，本金 $+$ 利息为： $21632 +865.28 = 22497.28$ 元；比较 22400 和 22497.28 的大小，即可解答问题。本题考查了利息相关问题，公式：利息 $=$ 本金 $\\times$ 利率 $\\times$ 时间。\n\n【详解】三年期： $10000 \\times 4\\% \\times 3\\\\= 400 \\times 3 \\\\ = 1200 (\\text{元})$\n\n先买一年期，把本金和利息取出来合在一起，再存入一年，\n\n$10000 \\times 4 \\% \\times 1 = 400 (\\text{元}) \\\\ (10000 + 400) \\times 4 \\% \\times 1 \\\\ = 10400 \\times 4 \\% \\times 1 \\\\ = 416元$\n\n$(10000 + 400 + 416) \\times 4\\% \\times 1$ $= 10816 \\times 4\\% \\times 1$ $= 432.64$ （元）\n\n$400 + 416 + 432.64$ $= 816 + 432.64$ $= 1248.64$ （元）\n\n$1200 < 1248.64$\n\n买银行1年期理财产品收益更大。\n\n故答案为：B",
+        "knowledge_points": [
+          "选择储蓄的最佳方案"
+        ],
+        "source": "2022·重庆·小升初真题",
+        "merged_id": "unit2_020",
+        "original_id": "unit2_choice_020"
+      },
+      {
+        "id": "unit2_choice_021",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "甲、乙两家商场以同样的标价销售同样型号的数码相机。在促销活动中，甲商场先降价 $10\\%$ 后，又在降价的基础上打九折销售；乙商场打八折进行销售，则（）。",
+        "options": [
+          "A. 甲商场便宜",
+          "B. 乙商场便宜",
+          "C. 价格相同",
+          "D. 不能确定"
+        ],
+        "answer": "B",
+        "solution": "【分析】设原价是1，甲商场：先降价 $10 \\%$，那么降价后的价格就是原价的 (1- $10 \\%$)，由此用乘法求出第一次降价后的价格; 再用第一次降价后的价格乘上 $90 \\%$，就是现价;乙商场：打八折，是指现价是原价的 $80 \\%$；用原价乘上 $80 \\%$ 就是现价；比较现价，从而解决问题。找清楚两个商城的不同优惠方法，弄清楚乙商场两个单位'1'的不同是解决本题的关键。\n\n【详解】设原价是 1，则:\n\n甲商场: $1 \\times (1 - 10\\%) \\times 90 \\%\\\\= 1 \\times 90 \\% \\times 90 \\% \\\\ = 0.9 \\times 0.9 \\\\ = 0.81$\n\n乙商场:\n\n$1 \\times 80 \\% = 0.8 \\\\ 0.81 > 0.8$\n\n则两家商场调整后的价格相比，乙商场便宜。故答案为: B",
+        "knowledge_points": [
+          "经济问题",
+          "折扣问题",
+          "比一个数多/少百分之几的数是多少"
+        ],
+        "source": "2022·辽宁盘锦·小升初真题",
+        "merged_id": "unit2_021",
+        "original_id": "unit2_choice_021"
+      },
+      {
+        "id": "unit2_choice_022",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "东门中心小学今年的学生数量比去年增加二成，今年的学生数量是去年的（）。",
+        "options": [
+          "A. $80\\%$",
+          "B. $120\\%$",
+          "C. $20\\%$"
+        ],
+        "answer": "B",
+        "solution": "【分析】增加二成是指今年的学生数量比去年增加 $20 \\%$,把去年的学生数量看成单位'1'，用 $1 + 20 \\%$ 即可求出今年的学生数量是去年的百分之几。解决本题关键是理解增加二成的含义，找出单位'1'，再根据数量关系求解。\n\n【详解】$1 + 20\\% = 120\\%$ 今年的学生数量是去年的 $120 \\text{‰}$。故答案为: B",
+        "knowledge_points": [
+          "成数问题"
+        ],
+        "source": "22-23 六年级下·全国·单元测试",
+        "merged_id": "unit2_022",
+        "original_id": "unit2_choice_022"
+      },
+      {
+        "id": "unit2_choice_023",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "一种商品按原价的八折销售，然后再提价 $20\\%$，现价与原价相比（）。",
+        "options": [
+          "A. 提高 $20\\%$",
+          "B. 下降 $20\\%$",
+          "C. 下降 $4\\%$"
+        ],
+        "answer": "C",
+        "solution": "【分析】设这件商品的原价是 1，先把这件商品的原价看作单位'1'，打折后的价格是原价的 $80 \\%$；再把打折后的价格看作单位'1'，提价后的价格是打折后价格的 (1+20%)；单位'1'已知，用连乘求出现价，再与原价相比较,得出结论。本题考查百分数的应用，理解折扣的意义，区分两个单位'1'的不同，单位'1'已知，根据百分数乘法的意义列式计算。\n\n【详解】设这件商品的原价是 $1$。\n\n$1 \\times 80 \\% \\times (1 + 20 \\%) \\\\ = 1 \\times 0.8 \\times 1.2 \\\\ = 0.96$\n\n$(1 - 0.96) \\div 1 \\times 100 \\% \\\\ = 0.04 \\div 1 \\times 100 \\% \\\\ = 0.04 \\times 100 \\% \\\\ = 4 \\%$\n\n现价与原价相比下降 $4 \\%$故答案为: C",
+        "knowledge_points": [
+          "求一个数的百分之几是多少",
+          "折扣问题",
+          "比一个数多/少百分之几的数是多少"
+        ],
+        "source": "22-23 六年级下·全国·单元测试",
+        "merged_id": "unit2_023",
+        "original_id": "unit2_choice_023"
+      },
+      {
+        "id": "unit2_choice_024",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "超市卖一种轮滑鞋，售价的 $60\\%$ 是进价，售价的 $40\\%$ 是赚的钱。现在要搞促销活动，原来每双售价为150元的这种轮滑鞋，为保证一双赚的钱不少于30元，最多打（）折。",
+        "options": [
+          "A. 七",
+          "B. 七五",
+          "C. 八"
+        ],
+        "answer": "C",
+        "solution": "【分析】先把原来的售价看作单位'1'，已知售价的 $60\\%$ 是进价，用原来的售价乘 $60\\%$ 求出轮滑鞋的进价；为保证轮滑鞋赚的钱不少于 30 元，用进价加上 30 元，就是轮滑鞋的最低实际售价; 用实际售价除以原来的售价，即可得出最多打几折。本题考查折扣问题，理解进价、原价、实际售价、折扣的意义以及它们之间的关系，再根据百分数乘除法的意义进行求解。\n\n【详解】进价：\n\n$150 \\times 60 \\% \\\\ = 150 \\times 0.6 \\\\ = 90元$\n\n最低的实际售价：\n\n$90 + 30 = 120 \\text{(元)}$\n\n折扣：\n\n$120 \\div 150 \\times 100 \\% \\\\ = 0.8 \\times 100 \\% \\\\ = 80 \\%$\n\n实际售价是原售价的 $80 \\%$，也就是打八折销售。故答案为: C",
+        "knowledge_points": [
+          "求一个数是另一个数的百分之几（百分率问题）",
+          "求一个数的百分之几是多少",
+          "折扣问题"
+        ],
+        "source": "22-23 六年级下·全国·单元测试",
+        "merged_id": "unit2_024",
+        "original_id": "unit2_choice_024"
+      },
+      {
+        "id": "unit2_choice_025",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "某饮料店每杯奶茶售价30元，第二杯半价。笑笑买了两杯，她支付的钱数相当于享受了几折的优惠？（）",
+        "options": [
+          "A. 五折",
+          "B. 七折",
+          "C. 七五折"
+        ],
+        "answer": "C",
+        "solution": "【分析】已知每杯奶茶售价 30 元，买两杯，共需 $(30 \\times 2)$ 元；现在第二杯半价，买两杯，实际需 $(30 + 30 \\div 2)$ 元；用实际花费除以原价，得出百分数后再化成折扣即可。本题考查折扣问题，百分之几十几就是打几几折，掌握原价、现价、折扣之间的关系是解题的关键。\n\n【详解】原价： $30 \\times 2 = 60$ （元）\n\n现价：\n\n$30 + 30 \\div 2 \\\\ = 30 + 15 \\\\ = 45元$\n\n折扣：\n\n$45 \\div 60 \\times 100 \\% \\\\ = 0.75 \\times 100 \\% \\\\ = 75 \\%$\n\n$75\\% =$ 七五折她支付的钱数相当于享受了七五折的优惠。故答案为: C",
+        "knowledge_points": [
+          "求一个数是另一个数的百分之几（百分率问题）",
+          "折扣问题"
+        ],
+        "source": "22-23 六年级下·全国·单元测试",
+        "merged_id": "unit2_025",
+        "original_id": "unit2_choice_025"
+      },
+      {
+        "id": "unit2_choice_026",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "妈妈为青青存了4000元的教育储蓄，定期三年，年利率为 $2.85\\%$，到期时妈妈一共能取出多少钱？下面列式正确的是（）。",
+        "options": [
+          "A. $4000 \\times 3 \\times 2.85\\%$",
+          "B. $4000 + 4000 \\times 3 \\times 2.85\\%$"
+        ],
+        "answer": "B",
+        "solution": "【分析】根据'利息=本金×年利率×存期'，先求出三年后到期的利息，再加上本金，就是到期时一共能取出的钱数。正确理解利息的意义，熟练掌握利息的计算公式是解答此题的关键。\n\n【详解】$4000 + 4000 \\times 3 \\times 2.85\\%\\\\= 4000 + 12000 \\times 0.0285 \\\\ = 4000 + 342 \\\\ = 4342元$\n\n列式正确的是 $4000 + 4000 \\times 3 \\times 2.85\\%$.故答案为: B",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "22-23 六年级下·全国·单元测试",
+        "merged_id": "unit2_026",
+        "original_id": "unit2_choice_026"
+      },
+      {
+        "id": "unit2_choice_027",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "王叔叔家的果园去年共产樱桃 $3000 \\mathrm{~kg}$，今年比去年增产二成。今年产樱桃多少? 正确的列式是（）。",
+        "options": [
+          "A. $3000 \\times 20\\%$",
+          "B. $3000 \\times (1 + 2\\%)$",
+          "C. $3000 \\times (1 +20%)$"
+        ],
+        "answer": "C",
+        "solution": "【分析】把去年樱桃的产量看作单位'1'，今年比去年增产二成，即增产 $20\\%$，则今年的产量是去年的 $(1 + 20\\%)$，单位'1'已知，用去年的产量乘 $(1 + 20\\%)$，即可求出今年樱桃的产量。本题考查成数问题，几成就是百分之几十; 找出单位'1'，单位'1'已知，根据百分数乘法的意义解答。\n\n【详解】二成 $= 20\\%$ $3000 \\times (1 + 20 \\%) \\\\ = 3000 \\times 1.2 \\\\ = 3600 (\\mathrm{kg})$\n\n今年产樱桃 $3600 \\mathrm{~kg}$。故答案为: C",
+        "knowledge_points": [
+          "成数问题",
+          "比一个数多/少百分之几的数是多少"
+        ],
+        "source": "22-23 六年级下·全国·单元测试",
+        "merged_id": "unit2_027",
+        "original_id": "unit2_choice_027"
+      },
+      {
+        "id": "unit2_choice_028",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "压岁钱，年节习俗之一，其本真来由无考，长辈要给小辈压岁钱，以祝福晚辈平安度岁。压岁钱是小孩过年最盼望的礼物。压岁钱相传起源较早，妈妈建议乐乐把攒的2000元压岁钱存入银行，年利率为 $2.5\\%$，定期2年，乐乐可得到本金和利息共（）元。",
+        "options": [
+          "A. 100",
+          "B. 2100",
+          "C. 2000 和 101",
+          "D. 2101.25"
+        ],
+        "answer": "B",
+        "solution": "【分析】通过利息的计算公式：利息 $=$ 本金 $\\times$ 利率 $\\times$ 存期，求出利息，再加上本金，即可得解。此题的解题关键是利用计算利息的公式来求解，注意最后要加上本金。\n\n【详解】$2000 + 2000 \\times 2 \\times 2.5\\%\\\\= 2000 + 4000 \\times 0.025 \\\\ = 2000 + 100 \\\\ = 2100 (\\text{元})$\n\n即到期时，乐乐可得到本金和利息共2100元。故答案为: B",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "21-22 六年级上·山西阳泉·期末",
+        "merged_id": "unit2_028",
+        "original_id": "unit2_choice_028"
+      },
+      {
+        "id": "unit2_fill_029",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "润发百货 3 月份的营业额中应交税的部分是20万元，按规定缴纳 $3\\%$ 的增值税，润发百货3月份应缴纳增值税( )元。",
+        "options": [],
+        "answer": "6000",
+        "solution": "【分析】应纳税额 $=$ 应纳税部分 $\\times$ 税率，代入数据求出润发百货 3 月份应缴纳增值税多少元。\n\n【详解】20万元 $= 200000$ 元 $200000 \\times 3\\% = 6000$ （元）所以，润发百货3月份应缴纳增值税6000元。",
+        "knowledge_points": [
+          "求应纳税额"
+        ],
+        "source": "24-25 六年级下·甘肃临夏·期中",
+        "merged_id": "unit2_029",
+        "original_id": "unit2_fill_029"
+      },
+      {
+        "id": "unit2_fill_030",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "$\\frac{()}{5} = 12 \\div 15 = $ ______(填小数)=___________折 = ________成。",
+        "options": [],
+        "answer": "4;0.8;80;八;八",
+        "solution": "【分析】根据分数与除法的关系：被除数做分子，除数做分母；$12 \\div 15 = \\frac{12}{15}$；根据分数的基本性质：分数的分子分母同时乘或除以一个不为0的数，分数的大小不变；$\\frac{12}{15} = \\frac{12 \\div 3}{15 \\div 3} = \\frac{4}{5}$；再根据分数化小数的方法：用分子除以分母，得到的商就是小数；$\\frac{4}{5} = 4 \\div 5 = 0.8$；再根据小数化百分数的方法：小数点向右移动两位，加上百分号即可；$0.8 = 80\\%$；几折就是百分之几十；$80\\%$ 是八折；几成就是百分之几十；$80\\%$ 是八成，据此解答。\n\n【详解】$\\frac{4}{5} = 12 \\div 15 = 0.8 =$ 八折 $=$ 八成",
+        "knowledge_points": [
+          "求折扣（折扣问题）",
+          "百分数、小数和分数的互化",
+          "分数的基本性质",
+          "分数、小数、百分数与成数的互化"
+        ],
+        "source": "24-25 六年级下·海南海口·单元测试",
+        "merged_id": "unit2_030",
+        "original_id": "unit2_fill_030"
+      },
+      {
+        "id": "unit2_fill_031",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "$8 \\div () = 4: 5 = \\frac{16}{()} = () \\% = ()$ 成。",
+        "options": [],
+        "answer": "10;20;80;八",
+        "solution": "【分析】（1）比和除法的关系，比的前项相当于被除数，后项相当于除数，比号相当于除号，据此把 $4: 5$ 写成 $4 \\div 5$,再根据商不变的性质给被除数和除数同时乘 2 即可解答;（2）比和分数的关系，比的前项相当于分子，后项相当于分母，比号相当于分数线，据此把 $4: 5$ 写成 $\\frac{4}{5}$，再根据分数的基本性质给分子和分母同时乘 4 即可解答;（3）先用比的前项除以后项求出比值，再根据小数化百分数的方法: 把小数的小数点向右移动两位，再加上百分号即可，据此求出百分数;（4）根据百分数和成数的关系: 百分之几十就是几成求出对应的成数即可。\n\n【详解】$4:5 = 4 \\div 5 = (4 \\times 2)$ ： $(5 \\times 2) = 8:10$\n\n$4: 5 = \\frac{4}{5} = \\frac{16}{20} \\\\ 4: 5 = 4 \\div 5 = 0.8 = 80 \\% = 八成 \\\\ 8 \\div 10 = 4:5 = \\frac{16}{20} = 80 \\% = \\text{八成。}$",
+        "knowledge_points": [
+          "百分数、分数、小数和比的互化",
+          "分数、小数、百分数与成数的互化",
+          "比与分数、除法的关系"
+        ],
+        "source": "24-25 六年级下·海南海口·单元测试",
+        "merged_id": "unit2_031",
+        "original_id": "unit2_fill_031"
+      },
+      {
+        "id": "unit2_fill_032",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "前年5月1日，爸爸在银行存入20000元，存期2年，年利率是 $3.75\\%$，到期后，他从银行共取回( )元，在这次储蓄中，他得到利息是( )元。",
+        "options": [],
+        "answer": "21500;1500",
+        "solution": "【分析】利息 $=$ 本金 $\\times$ 利率 $\\times$ 存期，据此代入数据求出利息,再将利息加上本金，求出到期后，他从银行共取回多少元。\n\n【详解】$20000 \\times 3.75\\% \\times 2\\\\= 750 \\times 2 \\\\ = 1500元$\n\n$20000 + 1500 = 21500 \\text{(元)}$所以到期后，他从银行共取回 21500 元，在这次储蓄中，他得到利息是 1500 元。",
+        "knowledge_points": [
+          "求利息"
+        ],
+        "source": "24-25 六年级下·海南海口·单元测试",
+        "merged_id": "unit2_032",
+        "original_id": "unit2_fill_032"
+      },
+      {
+        "id": "unit2_fill_033",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "一件商品打八五折销售表示( )是( )的 $85\\%$。",
+        "options": [],
+        "answer": "现价;原价",
+        "solution": "【分析】打八五折就是表示现价是原价的 $85 \\%$ 销售，据此解答。\n\n【详解】根据分析可知，一件商品打八五折销售表示现价是原价的 $85\\%$。",
+        "knowledge_points": [
+          "求现价（折扣问题）"
+        ],
+        "source": "24-25 六年级下·海南海口·单元测试",
+        "merged_id": "unit2_033",
+        "original_id": "unit2_fill_033"
+      },
+      {
+        "id": "unit2_fill_034",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "某商店 3 月份的营业额是68万元。如果按营业额的 $5\\%$ 缴纳营业税，该商店3月份应缴纳营业税( )万元。",
+        "options": [],
+        "answer": "3.4",
+        "solution": "【分析】把某商店3月份的营业额看作单位'1'，按营业额的 $5 \\%$ 缴纳营业税，单位'1'已知，用营业额乘 $5 \\%$，求出应缴纳的营业税。\n\n【详解】$68 \\times 5\\%\\\\= 68 \\times 0.05 \\\\ = 3.4 \\text{(万元)}$\n\n该商店3月份应缴纳营业税3.4万元。",
+        "knowledge_points": [
+          "求应纳税额",
+          "求一个数的百分之几是多少"
+        ],
+        "source": "24-25 六年级下·海南海口·单元测试",
+        "merged_id": "unit2_034",
+        "original_id": "unit2_fill_034"
+      },
+      {
+        "id": "unit2_fill_035",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "七成 $= \\frac{()}{10} =$ （）% 九五折 $=$ （）% $90\\% =$（）成",
+        "options": [],
+        "answer": "7; 70; 95; 九",
+        "solution": "【分析】根据成数的意义，百分之几十就是几成，百分之几十十几就是几成几。根据折扣的意义，百分之几十就是几折，百分之几十十几就是几几折。\n\n【详解】七成 $= \\frac{7}{10} =$ （70）% 九五折 $= (95)\\%$ $90\\% =（九）成$",
+        "knowledge_points": [
+          "分数、小数、百分数与成数的互化"
+        ],
+        "source": "24-25 六年级下·海南海口·单元测试",
+        "merged_id": "unit2_035",
+        "original_id": "unit2_fill_035"
+      },
+      {
+        "id": "unit2_fill_036",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "林明家去年用水量是650吨，今年比去年节水二成，今年的用水量相当于去年的( )%，今年用水( )吨。",
+        "options": [],
+        "answer": "80; 520",
+        "solution": "【分析】二成 $= 20 \\%$，将去年用水量看作单位'1'，那么今年是去年的 $(1 - 20 \\%)$。将去年的用水量乘 $(1 - 20 \\%)$,即可求出今年用水多少吨。\n\n【详解】$1 - 20\\% = 80\\%\\\\650 \\times (1 - 20 \\%) \\\\ = 650 \\times 80 \\% \\\\=520 (吨)$\n\n所以，今年的用水量相当于去年的 $80\\%$，今年用水 520 吨。",
+        "knowledge_points": [
+          "分数、小数、百分数与成数的互化",
+          "比一个数多/少百分之几的数是多少"
+        ],
+        "source": "24-25 六年级下·海南海口·单元测试",
+        "merged_id": "unit2_036",
+        "original_id": "unit2_fill_036"
+      },
+      {
+        "id": "unit2_fill_037",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "一台彩电原价是3000元，打九折后现售价是原价的( )%，现销售价比原价便宜了( )%，现售价是( )元。",
+        "options": [],
+        "answer": "90 ;10; 2700",
+        "solution": "【分析】打几折就是现价是原价的百分之几十；打九折就是现价是原价的 $90 \\%$；把原价看作单位'1'，用 1 减去90%，即可求出便宜了百分之几；再用原价×90%，即可求出现价，据此解答。\n\n【详解】打九折就是现价是原价的 $90\\%$,$1 - 90 \\% = 10 \\%$ $3000 \\times 90 \\% = 2700 (\\text{元})$一台彩电原价是3000元，打九折后现售价是原价的 $90\\%$ 现销售价比原价便宜了 $10\\%$，现售价是2700元。",
+        "knowledge_points": [
+          "求现价（折扣问题）"
+        ],
+        "source": "24-25 六年级下·海南海口·单元测试",
+        "merged_id": "unit2_037",
+        "original_id": "unit2_fill_037"
+      },
+      {
+        "id": "unit2_fill_038",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "一种商品打八五折出售，也就是现价是原价的_____%。",
+        "options": [],
+        "answer": "85",
+        "solution": "【分析】一件商品打八五折出售，几折就是百分之几十，几几折就是百分之几十几，把这件商品的原价看作单位'1'，现价就是按原价的 $85\\%$ 出售；据此判断。\n\n【详解】由分析可知：一件商品打八五折出售，就是现价是原价的 $85 \\%$",
+        "knowledge_points": [
+          "求现价（折扣问题）"
+        ],
+        "source": "24-25 六年级下·海南海口·单元测试",
+        "merged_id": "unit2_038",
+        "original_id": "unit2_fill_038"
+      },
+      {
+        "id": "unit2_fill_039",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "$\\frac{3}{4} = \\frac{15}{()} =$ （）： $8=$ （）$\\% =$（）成（）。",
+        "options": [],
+        "answer": "20; 6; 75; 七; 五",
+        "solution": "【分析】根据分数的基本性质：分数的分子分母同时乘或除以一个不为 0 的数，分数的大小不变 $\\frac{3}{4} = \\frac{3 \\times 5}{4 \\times 5} = \\frac{15}{20}$；根据分数与比的关系：分子做比的前项，分母做比的后项；$\\frac{3}{4} = 3:4$；再根据比的基本性质：比的前项和后项同时乘或除以一个不为 0 的数，比值不变；$3:4 = (3 \\times 2)$ ： $(4 \\times 2) = 6:8$；再根据分数化小数的方法：用分子除以分母得到的商就是小数；$\\frac{3}{4} = 3 \\div 4 = 0.75$；再根据小数化百分数的方法：小数点向右移动两位，再加上百分号即可；$0.75 = 75\\%$；百分之几十就是几成，$75\\%$ 就是七成五，据此解答。\n\n【详解】$\\frac{3}{4} = \\frac{15}{20} = 6:8 = 75\\% =$ 七成五",
+        "knowledge_points": [
+          "百分数、分数、小数和比的互化",
+          "分数、小数、百分数与成数的互化",
+          "比与分数、除法的关系"
+        ],
+        "source": "24-25 六年级下·全国·单元测试",
+        "merged_id": "unit2_039",
+        "original_id": "unit2_fill_039"
+      },
+      {
+        "id": "unit2_fill_040",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "2019年1月1日起，我国开始实施新的个人所得税法：个人收入每月超过5000元不超过 8000 的部分，要按 $3 \\%$ 缴纳个人所得税。小强妈妈2022 年上半年每月收入 6040 元，每月应缴纳个人所得税( )元。",
+        "options": [],
+        "answer": "31.2",
+        "solution": "【分析】用小强妈妈上半年每月收入 $-5000$ 元，求出应缴纳个人所得税的钱数，再用应缴纳个人所得税的钱数 $\\times 3\\%$ 即可求出每月应缴纳个人所得税，据此解答。\n\n【详解】$(6040-5000) ×3\\%\\\\= 1040\\times 3\\% \\\\ = 31.2元$\n\n2019年1月1日起，我国开始实施新的个人所得税法：个人收入每月超过5000元不超过8000的部分，要按 $3 \\%$ 缴纳个人所得税。小强妈妈2022年上半年每月收入6040元，每月应缴纳个人所得税 31.2 元。",
+        "knowledge_points": [
+          "求应纳税额"
+        ],
+        "source": "24-25 六年级下·全国·单元测试",
+        "merged_id": "unit2_040",
+        "original_id": "unit2_fill_040"
+      },
+      {
+        "id": "unit2_fill_041",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "$( ) \\div 24 = 24 : ( ) = \\frac{()}{12} = (\\quad)\\% =七成五$。",
+        "options": [],
+        "answer": "18; 32; 9; 75",
+        "solution": "【分析】几成就是十分之几，也就是百分之几十，据此将成数化成分数和百分数。分数的分子相当于被除数、比的前项，分母相当于除数、比的后项，分数的分子和分母，同时乘或除以相同的数（0 除外），分数的大小不变，据此根据分数与除法和比的关系，继续填空。\n\n【详解】七成五 $= 75 \\% = \\frac{75}{100} = \\frac{3}{4}; 24 \\div 4 \\times 3 = 18; 24 \\div 3 \\times 4 =$ 32; $12 \\div 4 \\times 3 = 9$$18 \\div 24 = 24: 32 = \\frac{9}{12} = 75\\%$ =七成五",
+        "knowledge_points": [
+          "百分数、分数、小数和比的互化",
+          "分数的基本性质",
+          "分数、小数、百分数与成数的互化",
+          "比与分数、除法的关系"
+        ],
+        "source": "24-25 六年级上·河南商丘·期末",
+        "merged_id": "unit2_041",
+        "original_id": "unit2_fill_041"
+      },
+      {
+        "id": "unit2_fill_042",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "$\\frac{6}{()} =（ ）\\% = 3: ( ) = 七五折 = ( )$。(填小数)",
+        "options": [],
+        "answer": "8;75;4;0.75",
+        "solution": "【分析】根据折扣的意义，几几折就是百分之几十几；百分数化成小数，只要把百分号去掉，同时把小数点向左移动两位；小数化成分数，两位小数先化成分母为 100 的分数，再化简成最简分数；分数的基本性质：分数的分子和分母同时乘或除以相同的数（0 除外），分数的大小不变；分数与比的关系：分子相当于比的前项，分母相当于比的后项。后项，分数线相当于比号。\n\n【详解】七五折 $= 75\\%=75 \\% = 0.75 = \\frac{75}{100} = \\frac{3}{4}= \\frac{3 \\times 2}{4 \\times 2} = \\frac{6}{8}$ $\\frac{3}{4} = 3: 4$即 $\\frac{6}{8} = 75\\% = 3:4 =$ 七五折 $= 0.75$。",
+        "knowledge_points": [
+          "百分数、小数和分数的互化",
+          "打折的意义及应用（分数）",
+          "比与分数、除法的关系"
+        ],
+        "source": "23-24 六年级下·四川乐山·期中",
+        "merged_id": "unit2_042",
+        "original_id": "unit2_fill_042"
+      },
+      {
+        "id": "unit2_fill_043",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "红旗超市一个香粽礼盒标价 400 元，端午节超市做'满 300 元减 100 元'的优惠活动，这样一个香粽礼盒实际是打( )折出售。",
+        "options": [],
+        "answer": "七五",
+        "solution": "【分析】标价包含几个 300 元，就从标价减去几个 100 元是实际价格, $400 > 300$，因此实际钱数是 $(400 - 100)$ 元,将标价看作单位'1'，实际钱数 $\\div$ 标价 = 实际钱数是标价的百分之几，根据几折就是百分之几十，确定折扣。\n\n【详解】$(400 - 100) \\div 400\\\\= 300 \\div 400 \\\\ = 75 \\%$\n\n75%即打七五折这样一个香粽礼盒实际是打七五折出售。",
+        "knowledge_points": [
+          "求折扣（折扣问题）"
+        ],
+        "source": "2024·四川乐山·小升初真题",
+        "merged_id": "unit2_043",
+        "original_id": "unit2_fill_043"
+      },
+      {
+        "id": "unit2_fill_044",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "张叔叔为杂志社撰写了一份稿件，收入稿费3000元，其中800元是免税的，其余部分按 $20\\%$ 的税率缴税，求'这笔稿费一共要缴税多少元？'的正确列式是( )。",
+        "options": [],
+        "answer": "$(3000 - 800)\\times 20\\%$",
+        "solution": "【分析】由题意可知，这笔稿费一共要缴的税指的是从总稿费里去掉免税部分的 800 元后的钱数的 $20 \\%$，即以 3000 - 800 = 2200 元为单位'1'，根据求一个数的百分之几是多少，用乘法计算，用 $2200 \\times 20 \\%$ 即可求出缴税多少元。\n\n【详解】$(3000 - 800) \\times 20\\%\\\\= 2200 \\times 20 \\% \\\\ = 440元$\n\n这笔稿费一共要缴税440元。",
+        "knowledge_points": [
+          "求一个数的百分之几是多少",
+          "分段计算解决纳税问题"
+        ],
+        "source": "23-24 六年级下·北京朝阳·期末",
+        "merged_id": "unit2_044",
+        "original_id": "unit2_fill_044"
+      },
+      {
+        "id": "unit2_fill_045",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "爸爸花20000元买银行的3年期理财产品，年收益率为 $3.75\\%$，到期时爸爸一共可取回( )元。",
+        "options": [],
+        "answer": "22250",
+        "solution": "【分析】根据一共可取回的钱数 $=$ 本金 $+$ 本金 $\\times$ 年利率 $\\times$ 存期，据此代入数值进行计算即可。\n\n【详解】$20000 + 20000 \\times 3.75\\% \\times 3\\\\= 20000 + 2250 \\\\ = 22250元$\n\n则到期时爸爸一共可取回 22250 元。",
+        "knowledge_points": [
+          "求利息"
+        ],
+        "source": "23-24 六年级下·湖南衡阳·期中",
+        "merged_id": "unit2_045",
+        "original_id": "unit2_fill_045"
+      },
+      {
+        "id": "unit2_fill_046",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "在2024年清明节，明月山接待游客人数约为90万人次，同比增长 $20\\%$，去年清明节明月山接待游客人数约为____万人次。预计五一长假还会增长一成，按预计，五一假明月山将接待游客人数约为____万人次。",
+        "options": [],
+        "answer": "75;99",
+        "solution": "【分析】首先把去年清明节接待游客人数看作是单位'1'，今年比去年增长 $20\\%$，今年接待游客人数可以用（1+ $20\\%$）表示，要求单位'1'，用90除以（1+ $20\\%$）即可；再把今年清明节接待游客人数看作是单位'1'，五一假期再增长一成即 $10\\%$，那么五一假期接待游客人数用（1+ $10\\%$）表示，要求五一假期接待游客人数就用90乘（1+ $10\\%$）\n\n【详解】$90 \\div (1 + 20\\%)\\\\= 90 \\div 1.2\\\\= 75（万人次）$\n\n$90 \\times (1 + 10 \\%) \\\\ = 90 \\times 1.1\\\\ = 99（万人次）$\n\n去年清明节明月山接待游客人数约为 75 万人次，预计五一假明月山将接待游客人数约为 99 万人次。",
+        "knowledge_points": [
+          "求增加或减少几成的实际问题",
+          "已知比一个数多/少百分之几是多少，求这个数",
+          "比一个数多/少百分之几的数是多少"
+        ],
+        "source": "23-24 六年级下·江西宜春·期中",
+        "merged_id": "unit2_046",
+        "original_id": "unit2_fill_046"
+      },
+      {
+        "id": "unit2_fill_047",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "中百超市上个月的营业额是420万元，按营业额的 $5\\%$ 缴纳营业税，该超市上个月应缴纳营业税( )元。",
+        "options": [],
+        "answer": "210000或者21万",
+        "solution": "【分析】应纳税额 $=$ 应纳税所得额 $\\times$ 税率，题目中营业额即是应纳税所得额，求该超市上个月应缴纳营业税多少万元，就是求 420 万元的 $5 \\%$ 是多少，用乘法计算即可解答。\n\n【详解】$420 \\times 5\\% = 21$ （万元）21万元 $= 210000$ 元该超市上个月应缴纳营业税210000元。",
+        "knowledge_points": [
+          "求应纳税额"
+        ],
+        "source": "23-24 六年级下·山西忻州·期中",
+        "merged_id": "unit2_047",
+        "original_id": "unit2_fill_047"
+      },
+      {
+        "id": "unit2_fill_048",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "60 千克增加它的二成是( )千克，40 千克减少它的二成是( )千克。",
+        "options": [],
+        "answer": "72;32",
+        "solution": "【分析】几成几即百分之几十几，根据求比一个数多/少百分之几的数是多少，用乘法计算，求出未知的量。\n\n【详解】二成 $= 20\\%$\n\n$60 \\times (1 + 20 \\%)、、= 60 \\times 1.2= 72 (千克)$\n\n$40 \\times (1 - 20 \\%)\\\\= 40 \\times 0.8\\\\=32 (千克)$\n\n60千克增加它的二成是72千克，40千克减少它的二成是32千克。",
+        "knowledge_points": [
+          "求增加或减少几成的实际问题"
+        ],
+        "source": "23-24 六年级下·河南三门峡·期中",
+        "merged_id": "unit2_048",
+        "original_id": "unit2_fill_048"
+      },
+      {
+        "id": "unit2_fill_049",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "奶奶将5000元钱存入银行，定期一年，到期时支出5105元。奶奶存钱的年利率是( )。",
+        "options": [],
+        "answer": "$2.1\\%$",
+        "solution": "【分析】将到期支取的钱减去本金，求出一年的利息。年利率 $=$ 年利息 $\\div$ 本金，据此列式求出奶奶存钱的年利率。\n\n【详解】$(5105 - 5000) \\div 5000\\\\= 105 \\div 5000 \\\\ = 2.1 \\%$\n\n所以，奶奶存钱的年利率是 $2.1 \\%$",
+        "knowledge_points": [
+          "求利率或本金"
+        ],
+        "source": "23-24 六年级上·河北邯郸·期末",
+        "merged_id": "unit2_049",
+        "original_id": "unit2_fill_049"
+      },
+      {
+        "id": "unit2_fill_050",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "一件裙子打七折出售，就是按原价的( )%出售，比原价便宜了( )‰。",
+        "options": [],
+        "answer": "70;30",
+        "solution": "【分析】打七折就是按原价的 $70 \\%$ 出售，把原价看作单位'1'，现价比原价便宜 $(1 - 70\\%)$；解答即可。\n\n【详解】七折就是原价的 $70\\%$$1 - 70 \\% = 30 \\%$比原价便宜了 $30 \\%$",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "22-23 六年级下·山东菏泽·阶段练习",
+        "merged_id": "unit2_050",
+        "original_id": "unit2_fill_050"
+      },
+      {
+        "id": "unit2_fill_051",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "一件商品，打九折以后比原价少卖 15 元，这件商品现价( )元。",
+        "options": [],
+        "answer": "135",
+        "solution": "【分析】九折表示原价的 $90 \\%$，把原价看作单位'1'，现价比原价少 $(1 - 90 \\%)$，根据百分数除法的意义，用 $15 \\div (1 - 90 \\%)$ 即可求出原价，再减去 15 元，即可求出现价。\n\n【详解】九折表示原价的 $90\\%$\n\n$15 \\div (1 - 90 \\%) \\\\ = 15 \\div 10 \\% \\\\ = 150元$\n\n$150 - 15 = 135 \\text{(元)}$\n\n一件商品，打九折以后比原价少卖15元，这件商品现价135元。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "22-23 六年级上·四川达州·期中",
+        "merged_id": "unit2_051",
+        "original_id": "unit2_fill_051"
+      },
+      {
+        "id": "unit2_fill_052",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "$\\frac{7}{10} = ( ) \\% = ()填小数 = ( ) 成 = ( ) 折$。",
+        "options": [],
+        "answer": "70;0.7;七;七",
+        "solution": "【分析】直接用分子除以分母，计算出商，把分数化为小数；把小数的小数点向右移动两位，末尾再添上百分号'%'，把小数化为百分数；几折就表示十分之几，也就是百分之几十; 成数表示一个数是另一个数的十分之几，通称'几成'，据此解答。掌握小数、分数、百分数互相转化的方法以及成数和折扣的意义是解答题目的关键。\n\n【详解】$\\frac{7}{10} = 7 \\div 10 = 0.7 = 70\\% =$ 七成 $=$ 七折所以，$\\frac{7}{10} = 70\\% = 0.7 = 七成 = 七折$。",
+        "knowledge_points": [
+          "百分数、小数和分数的互化",
+          "成数问题",
+          "折扣问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit2_052",
+        "original_id": "unit2_fill_052"
+      },
+      {
+        "id": "unit2_fill_053",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "按要求改写成百分数或成数、折扣。七成( )% 六成五( )% 九五折( )% 35%( )(成数) $45\\%$ ()(折扣)",
+        "options": [],
+        "answer": "70;65;95;三成五;四五折",
+        "solution": "【分析】根据成数、折扣和百分数之间的关系：几成就是十分之几，也可以用百分数表示，几成就是百分之几十，几折就是十分之几，也就是百分之几十。打几折，即按原价的十分之几、百分之几十出售；据此解答即可。此题主要考查成数、折扣与百分数之间的互化。\n\n【详解】七成 $= 70\\%$ 六成五 $= 65\\%$ 九五折 $= 95\\%$ 35% = 三成五 (成数) 45% = 四五折 (折扣)",
+        "knowledge_points": [
+          "成数问题",
+          "折扣问题"
+        ],
+        "source": "22-23 六年级下·全国·单元测试",
+        "merged_id": "unit2_053",
+        "original_id": "unit2_fill_053"
+      },
+      {
+        "id": "unit2_fill_054",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "$30 \\div () = 0.5 = ( ) \\%=3: ( ) = \\frac{()}{10} = ( ) 成= ( ) 折$。",
+        "options": [],
+        "answer": "60;50;6;5;五;五",
+        "solution": "【分析】把 0.5 化成分数是 $\\frac{5}{10}$，根据分数与除法的关系, $\\frac{5}{10}=5÷10$；根据商不变的规律，$5 \\div 10 = 30 \\div 60$；$\\frac{5}{10}$ 化简为 $\\frac{1}{2}$，根据分数与比的关系，$\\frac{1}{2} = 1:2$，根据比的性质，1:2的前项和后项都乘3就是3:6；把0.5的小数点向右移动两位，同时添上百分号就是 $50\\%$；根据折扣的意义，$50\\%$ 就是五折；根据成数的意义，五折就是五成；据此解答。本题考查分数与除法的关系，百分数、小数、分数、比的互化，分数的基本性质，成数的意义，折扣的意义，商不变的规律。\n\n【详解】$30 \\div 60 = 0.5 = 50\\% = 3: 6 = \\frac{5}{10} =$ 五成=五折",
+        "knowledge_points": [
+          "百分数、分数、小数和比的互化",
+          "分数的基本性质及应用"
+        ],
+        "source": "22-23 六年级下·全国·单元测试",
+        "merged_id": "unit2_054",
+        "original_id": "unit2_fill_054"
+      },
+      {
+        "id": "unit2_fill_055",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "'八五折'是指现价是原价的( )，'七五折'出售，就是优惠了( )%。",
+        "options": [],
+        "answer": "$85\\%$ ;25",
+        "solution": "【分析】把原价看作单位'1'，'八五折'是指现价是原价的 $85 \\%$；'七五折'是指现价是原价的 $75 \\%$，优惠的钱数是原价的 $(1 - 75 \\%)$。本题考查折扣问题，几几折就是百分之几十几，掌握原价、现价、折扣、优惠之间的关系是解题的关键。\n\n【详解】八五折 $= 85\\%$ 七五折 $= 75\\%$ 优惠了：$1 - 75\\% = 25\\%$ '八五折'是指现价是原价的 $85 \\%$，'七五折'出售，就是优惠了 $25 \\%$。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "17-18 六年级上·全国·单元测试",
+        "merged_id": "unit2_055",
+        "original_id": "unit2_fill_055"
+      },
+      {
+        "id": "unit2_fill_056",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "微信提现手续费规定：每位用户终身享受1000元免费提现额度，超出部分按银行费率 $0.1\\%$ 收取手续费。微商李阿姨首次提现5万元，需扣除她( )元的手续费。",
+        "options": [],
+        "answer": "49",
+        "solution": "【分析】李阿姨提现 50000 元，则超出部分是 (50000-1000) 元，再用超出部分的钱乘 $0.1 \\%$，即可求出收取的手续费。弄清楚哪些钱需要收取手续费是解答本题的关键,计算要认真仔细。\n\n【详解】5万元 $= 50000$ 元\n\n$(50000 - 1000) \\times 0.1 \\% \\\\ = 49000 \\times 0.001 \\\\ = 49元$\n\n即需扣除49元手续费。",
+        "knowledge_points": [
+          "税率问题"
+        ],
+        "source": "20-21 六年级下·河南许昌·期中",
+        "merged_id": "unit2_056",
+        "original_id": "unit2_fill_056"
+      },
+      {
+        "id": "unit2_fill_057",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "$\\frac{4}{()} = \\frac{()}{15} = 0.8 = \\frac{16}{()} = ()\\% = () 折$。",
+        "options": [],
+        "answer": "5; 12; 20; 80; 八",
+        "solution": "【分析】把小数 0.8 化成分母是 10 的分数 $\\frac{8}{10}$，约分后可得 $\\frac{4}{5}$;根据分数的基本性质，把 $\\frac{4}{5}$ 的分子和分母同时乘 3，得到分母是 15 的分数 $\\frac{12}{15}$;根据分数的基本性质，把 $\\frac{4}{5}$ 的分子和分母同时乘 4，得到分子是 16 的分数 $\\frac{16}{20}$;把小数 0.8 的小数点向右移动两位添上百分号就是 $80 \\%$;根据折扣与百分数的关系，可得 $80 \\% =$ 八折。此题主要考查百分数、小数、分数之间的互化,根据分数的基本性质以及折扣的意义，求出结果。\n\n【详解】根据分析得，$\\frac{4}{5} = \\frac{12}{15} = 0.8 = \\frac{16}{20} = 80\\% =$ 八折。",
+        "knowledge_points": [
+          "百分数、小数和分数的互化",
+          "分数的基本性质及应用",
+          "折扣问题"
+        ],
+        "source": "20-21 六年级上·全国·单元测试",
+        "merged_id": "unit2_057",
+        "original_id": "unit2_fill_057"
+      },
+      {
+        "id": "unit2_judge_058",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "某商品在搞促销活动，商场按'每满100元减40元'的方式销售。因为（100一40）$\\div 100 = 60\\%$，所以这里的'每满100元减40元'实际上就是打六折销售。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】'每满100元减40元'表示，如果商品的价格低于100元，则没有这个优惠，需按原价支付；如果商品的价格正好是100元或100元的整数倍时，用原价减去40元或减去整数倍个40元，再除以原价，得出现价是原价的60%，相当于打六折；如果商品的价格大于100元但不是100元的整数倍数，每100元的部分都可以减去40元，再除以原价，得出现价是原价的百分之几，换算成折扣，不是六折。'打六折'的含义，无论商品的原价是多少，都按原价的60%进行销售。据此解答即可。\n\n【详解】如果商品的原价正好是100元，那么折扣是：\n\n$(100 - 40) \\div 100 \\times 100 \\% \\\\ = 60 \\div 100 \\times 100 \\% \\\\ = 0.6 \\times 100 \\% \\\\ = 60 \\%$\n\n60% = 六折如果商品的原价是 160 元，那么折扣是：\n\n$(160 - 40) \\div 160 \\times 100 \\% \\\\ = 120 \\div 160 \\times 100 \\% \\\\ = 0.75 \\times 100 \\% \\\\ = 75 \\%$\n\n$75\\% =$ 七五折\n\n因为不知道商品的原价，所以只能说，顾客能享受到的最大优惠相当于打六折。如果商品的原价不是 100 元或 100 元的整数倍，那么就不是打六折。原题说法错误。故答案为: ×",
+        "knowledge_points": [
+          "求折扣（折扣问题）",
+          "求一个数是另一个数的百分之几（百分率问题）"
+        ],
+        "source": "2024·四川内江·小升初真题",
+        "merged_id": "unit2_058",
+        "original_id": "unit2_judge_058"
+      },
+      {
+        "id": "unit2_judge_059",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "一种商品打六折出售，就是降低了原价的 $40\\%$ 出售。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】折扣表示现价是原价的百分之几十。六折是60%。就是按原价的 60 出售，就是降低了原价的 $1 - 60\\% =$ 40%出售。\n\n【详解】根据分析可知，一种商品打六折出售，就是降低了原价的 $40\\%$ 出售。原题干说法正确。故答案为: √",
+        "knowledge_points": [
+          "求现价（折扣问题）"
+        ],
+        "source": "24-25 六年级下·海南海口·单元测试",
+        "merged_id": "unit2_059",
+        "original_id": "unit2_judge_059"
+      },
+      {
+        "id": "unit2_judge_060",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "某品牌电视，先打'七五折'出售，后又涨价 $25\\%$，现价与原价相等。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】假设某品牌电视原价 100 元，打'七五折'出售，就是按原价的 $75 \\%$ 出售，求一个数的百分之几是多少，用乘法计算，据此求出打'七五折'是多少元；再用打'七五折'后的价钱 $\\times (1 + 25 \\%)$，即可求出现价，最后与原价比较即可。\n\n【详解】假设某品牌电视原价100元。\n\n$100 \\times 75 \\% \\times (1 + 25 \\%) \\\\ = 75 \\times 125 \\% \\\\ = 93.75元$\n\n$93.75 < 100$\n\n现价 $<$ 原价 原说法错误。故答案为: ×",
+        "knowledge_points": [
+          "求现价（折扣问题）",
+          "比一个数多/少百分之几的数是多少"
+        ],
+        "source": "23-24 六年级下·四川凉山·期末",
+        "merged_id": "unit2_060",
+        "original_id": "unit2_judge_060"
+      },
+      {
+        "id": "unit2_judge_061",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "爸爸将80000元存入银行，定期两年，年利率是 $2.25\\%$，到期后，爸爸可以取回3600元利息。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】根据'利息 $=$ 本金 $\\times$ 利率 $\\times$ 存期'，即可求出到期后可得到的利息。\n\n【详解】$80000 \\times 2.25\\% \\times 2\\\\= 80000 \\times 0.0225 \\times 2 \\\\ = 1800 \\times 2 \\\\ = 3600 (\\text{元})$\n\n到期后，爸爸可以取回 3600 元利息。原题说法正确。故答案为: √",
+        "knowledge_points": [
+          "求利息"
+        ],
+        "source": "23-24 六年级下·陕西渭南·期末",
+        "merged_id": "unit2_061",
+        "original_id": "unit2_judge_061"
+      },
+      {
+        "id": "unit2_judge_062",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "每满100元减50元就是打五折出售。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】打五折，是指现价是原价的 $50 \\%$；而'每满 100 元减 50 元'，如果商品的原价正好是 100 元的整数倍数，相当于打五折；如果商品的原价不是 100 元的整数倍时，就不是打五折了。\n\n【详解】如果商品的原价正好是100元，那么折扣是：\n\n$(100 - 50) \\div 100 \\times 100 \\% \\\\ = 50 \\div 150 \\times 100 \\% \\\\ = 0.5 \\times 100 \\% \\\\ = 50 \\%$\n\n$50 \\% = \\text{五折}$\n\n因为不知道商品的原价，所以只能说，顾客能享受到的最大优惠相当于打五折。如果顾客购买的商品不是 100 元或 100 元的整数倍，那么就不是打五折。原题说法错误。故答案为: ×",
+        "knowledge_points": [
+          "求折扣（折扣问题）"
+        ],
+        "source": "23-24 六年级下·湖北武汉·期末",
+        "merged_id": "unit2_062",
+        "original_id": "unit2_judge_062"
+      },
+      {
+        "id": "unit2_judge_063",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "一套书打九五折后，比原价便宜 6 元，这套书原价是 120 元。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】九五折也就是 $95 \\%$，我们可以设原价为 $x$ 元，利用等量关系'原价一原价 $\\times 95 \\% = 6$'列出方程，然后求解。\n\n【详解】解：设原价为 $x$ 元。\n\n$x - 95 \\%x = 6\\\\5 \\% \\mathrm{x} = 6\\\\0.05 \\mathrm{x} = 6\\\\0.05 \\mathrm{x} \\div 0.05 = 6 \\div 0.05\\\\x = 120$\n\n即这套书的原价为 120 元。故答案为: √",
+        "knowledge_points": [
+          "求原价（折扣问题）"
+        ],
+        "source": "23-24 六年级下·吉林松原·期中",
+        "merged_id": "unit2_063",
+        "original_id": "unit2_judge_063"
+      },
+      {
+        "id": "unit2_judge_064",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "一种商品，先提价25%，再打八折出售，这种商品的现价和原价相同。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】假设这种商品的价格是 1，提价 $25 \\%$，可用 $1 \\times (1 + 25 \\%)$ 算出提价后的价格。再打八折，可用提价后的价格乘 $80 \\%$，即可算出出售的价格。最后与 1 比较即可。\n\n【详解】假设商品原价为1$1 \\times (1 + 25 \\%) \\times 80 \\% \\\\ = 1 \\times 1.25 \\times 0.8 \\\\ = 1$所以，商品的现价和原价相同。故答案为: √",
+        "knowledge_points": [
+          "经济问题",
+          "含百分数的运算",
+          "求现价（折扣问题）",
+          "比一个数多/少百分之几的数是多少"
+        ],
+        "source": "23-24 六年级下·河南周口·期中",
+        "merged_id": "unit2_064",
+        "original_id": "unit2_judge_064"
+      },
+      {
+        "id": "unit2_judge_065",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "实验小学去年招生300人，今年招生240人，今年比去年少招生两成。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】成数，表示一个数是另一个数的十分之几。求今年比去年少招生几成，用去年与今年招生人数的差除以去年招生人数，结果用分母是 10 的分数表示，再转化成成数即可。\n\n【详解】$(300 - 240) \\div 300\\\\= 60 \\div 300 \\\\ = 0.2 \\\\ = \\frac{2}{10} \\\\ = \\text{二成}$\n\n故今年比去年少招生两成是正确的。故答案为: √",
+        "knowledge_points": [
+          "求增加或减少几成的实际问题"
+        ],
+        "source": "23-24 六年级下·甘肃武威·期中",
+        "merged_id": "unit2_065",
+        "original_id": "unit2_judge_065"
+      },
+      {
+        "id": "unit2_judge_066",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "一件商品先提价 $10\\%$，后来又按九折出售，现价和原价相同。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】设这件商品的原价是 1，先把这件商品的原价看作单位'1'，先提价 $10 \\%$，那么提价后的价格是原价的 $(1+$ $10 \\%)$，单位'1'已知，用乘法求出提价后的价格;再把提价后的价格看作单位'1'，又按九折出售，那么打折后的价格是提价后价格的 $90 \\%$；单位'1'已知，用乘法求出现价，再与原价相比较，得出结论。\n\n【详解】设这件商品的原价是 1。\n\n$1 \\times (1 + 10 \\%) \\times 90 \\% \\\\ = 1 \\times 1.1 \\times 0.9 \\\\ = 0.99$\n\n$0.99 < 1$\n\n现价比原价低。原题说法错误。故答案为: ×",
+        "knowledge_points": [
+          "求一个数的百分之几是多少",
+          "求现价（折扣问题）",
+          "比一个数多/少百分之几的数是多少"
+        ],
+        "source": "2019·全国·小升初真题",
+        "merged_id": "unit2_066",
+        "original_id": "unit2_judge_066"
+      },{
+        "id": "unit2_judge_067",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "一种商品先提价 $10\\%$，再打九折出售，商品价格不变。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】把商品原价看作单位'1'，设原价为100元，把原价提高 $10\\%$，则提高后的价格是原价的 $(1 + 10\\%)$，根据百分数乘法的意义，用 $100 \\times (1 + 10\\%)$ 即可求出提高后的价格，然后把提高后的价格看作单位'1'，已知提高后再打九折出售，九折表示 $90\\%$，则打折后的价格是提高后的价格的 $90\\%$，用提高后的价格 $\\times 90\\%$ 即可求出打折后的价格，然后现价与原价比较即可。本题关键是要明确每个百分率对应的单位'1'不同。\n\n【详解】假设原价是100元，\n\n九折 $= 90\\%$\n\n$100 \\times (1 + 10 \\%) \\times 90 \\% \\\\ = 100 \\times 1.1 \\times 0.9 \\\\ = 99元$\n\n$99 < 100$一种商品先提价 $10 \\%$，再打九折出售，商品价格降低。原题干说法错误。故答案为: ×",
+        "knowledge_points": [
+          "求一个数的百分之几是多少",
+          "折扣问题",
+          "比一个数多/少百分之几的数是多少"
+        ],
+        "source": "17-18 六年级上·全国·单元测试",
+        "merged_id": "unit2_067",
+        "original_id": "unit2_judge_067"
+      },
+      {
+        "id": "unit2_judge_068",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "某景点2018年春节初一到初六期间，游客约达到15万人，比去年同期大约增加了 3 万人，这样比去年同期增加了二成五。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】根据题意，用15万人减去3万人，求出去年同期游客的数量是 12 万人，把去年同期游客的数量看作单位'1'，用增加的3万人除以单位'1'的量，求出增加的百分比，再根据百分数与成数之间的关系，即可得解。此题主要考查成数问题，掌握求一个数比另一个数多百分之几的计算方法。\n\n【详解】$3 \\div (15 - 3)\\\\= 3 \\div 12 \\\\ = 0.25= 25\\% =二成五$\n\n即比去年同期增加了二成五。故答案为: √",
+        "knowledge_points": [
+          "求增加或减少几成的实际问题"
+        ],
+        "source": "21-22 六年级下·湖南湘西·期末",
+        "merged_id": "unit2_068",
+        "original_id": "unit2_judge_068"
+      },
+      {
+        "id": "unit2_judge_069",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "一件商品先打八折，再提价 $20\\%$，现价等于原价。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】设这件商品的原价是 1，先把这件商品的原价看作单位'1'，打折后的价格是原价的 $80 \\%$，根据求一个数的百分之几是多少，用乘法求出打折后的价格;再把打折后的价格看作单位'1'，提价后的价格是打折后价格的（$1 + 20\\%$）；单位'1'已知，用乘法求出现价，再与原价相比较，得出结论。本题考查百分数的实际应用，区分两个单位'1'的不同，单位'1'已知，根据百分数乘法的意义解答。\n\n【详解】设这件商品的原价是 1。\n\n$1 \\times 80 \\% \\times (1 + 20 \\%) \\\\ = 1 \\times 0.8 \\times 1.2 \\\\ = 0.96$\n\n$0.96 < 1$现价比原价低。原题说法错误。故答案为:×",
+        "knowledge_points": [
+          "求一个数的百分之几是多少",
+          "折扣问题",
+          "比一个数多/少百分之几的数是多少"
+        ],
+        "source": "18-19 六年级上·辽宁·课后作业",
+        "merged_id": "unit2_069",
+        "original_id": "unit2_judge_069"
+      },
+      {
+        "id": "unit2_judge_070",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "一种商品先升价 $20\\%$，然后又打八折销售，则现价和原价相等。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】把商品原来的价格看作单位'1'，第一次涨价后价格就是原价的（$1 + 20\\%$），用乘法可以求出第一次涨价后的价格；把第一次涨价后的价格看作单位'1'，八折相当于 $80\\%$，第二次后的价格是第一次涨价后价格的 $80\\%$，用第一次涨价后的价格乘 $80\\%$ 即可算出商品现在的售价。再比较即可得解。此题的解题关键是在理解情景问题的基础上找到'整体'也就是常说的单位'1'的量，然后分析数量关系，最后列式计算完成题目。\n\n【详解】$1 \\times (1 + 20\\%) \\times 80\\%\\\\= 1 \\times (1 + 0.2) \\times 0.8 \\\\ = 1.2 \\times 0.8 \\\\ = 0.96$\n\n$0.96 < 1$现价比原价低。故答案为: ×",
+        "knowledge_points": [
+          "折扣问题",
+          "比一个数多/少百分之几的数是多少"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit2_070",
+        "original_id": "unit2_judge_070"
+      },
+      {
+        "id": "unit2_judge_071",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "一本故事书，先涨价 $15\\%$，后来又按八五折出售，现价与原价相等。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】设这本故事书的原价是 1，先把这件商品的原价看作单位'1'，涨价后的价格是原价的 (1+15%)，单位'1'已知，用乘法计算，求出涨价后的价格;再把涨价后的价格看作单位'1'，又打八五出售，打折后的价格是涨价后价格的 $85 \\%$；单位'1'已知，用乘法求出价，再与原价相比较，得出结论。本题考查折扣问题和百分数乘法的应用，找出单位'1'，区分两个单位'1'的不同，单位'1'已知，根据百分数乘法的意义解答。\n\n【详解】设这本故事书的原价是1。\n\n$1 \\times (1 + 15 \\%) \\times 85 \\% \\\\ = 1 \\times 1.15 \\times 0.85 \\\\ = 0.9775$\n\n$0.9775 < 1$现在比原价低。原题说法错误。故答案为: ×",
+        "knowledge_points": [
+          "求一个数的百分之几是多少",
+          "折扣问题",
+          "比一个数多/少百分之几的数是多少"
+        ],
+        "source": "19-20 六年级上·全国·期中",
+        "merged_id": "unit2_071",
+        "original_id": "unit2_judge_071"
+      },
+      {
+        "id": "unit2_judge_072",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "根据国家规定写稿所得稿费超过800元的部分不足1500元按 $5\\%$ 交税，张老师写了一本1200元的稿子，需交税20元。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】由题意可知，1200 元应交税的部分为 $1200 - 800 =$ 400 元，根据应交税的部分 $\\times$ 税率 $=$ 应纳税的金额，据此判断即可。本题考查税率问题，明确应交税部分的金额是解题的关键。\n\n【详解】$(1200-800) ×5\\%\\\\= 400 \\times 5 \\% \\\\ = 20元$\n\n则需交税20元。原题干说法正确。故答案为: √",
+        "knowledge_points": [
+          "求应纳税额"
+        ],
+        "source": "20-21 六年级下·甘肃陇南·期中",
+        "merged_id": "unit2_072",
+        "original_id": "unit2_judge_072"
+      },
+      {
+        "id": "unit2_judge_073",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "某商品打'八五折'出售，就是降价 $85\\%$ 出售。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】几折就表示十分之几，也就是百分之几十，把商品原价看作单位'1'，八五折表示现价占原价的 $85\\%$，则优惠的价格占原价的 $(1 - 85\\%)$，据此解答。掌握折扣的意义，理解几折就表示现价占原价的百分之几十是解答题目的关键。\n\n【详解】八五折 $= 85\\%$ $1 - 85 \\% = 15 \\%$所以，某商品打'八五折'出售，就是降价 $15\\%$ 出售。故答案为: ×",
+        "knowledge_points": [
+          "打折的意义及应用（分数）"
+        ],
+        "source": "17-18 六年级上·全国·期末",
+        "merged_id": "unit2_073",
+        "original_id": "unit2_judge_073"
+      },
+      {
+        "id": "unit2_calc_074",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "看图列竖式计算。\n\nimages/004bb5fefe359272697a62d430eac792.png",
+        "options": [],
+        "answer": "216 个",
+        "solution": "【分析】二成就是 $20 \\%$；把苹果的个数看作单位'1'，桃子的个数是苹果的 $(1 + 20 \\%)$，求桃子的个数，用苹果的个数 $\\times$ $(1 + 20 \\%)$，即可解答。\n\n【详解】二成就是 $20 \\%$。$180 \\times (1 + 20 \\%) \\\\ = 180 \\times 120 \\% \\\\ = 216 (\\text{个})$\n\n桃子216个。",
+        "knowledge_points": [
+          "求增加或减少几成的实际问题"
+        ],
+        "source": "23-24 六年级下·云南楚雄·期中",
+        "merged_id": "unit2_074",
+        "original_id": "unit2_calc_074"
+      },
+      {
+        "id": "unit2_calc_075",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "计算题",
+        "difficulty": "容易",
+        "question": "直接写出得数。\n\n$三成 = ()\\%$ $75\\% = () 折$ $六成五 = ()\\%$ $80\\% =()成$\n\n$3.14 \\times 5 = $ () $3.14 \\times 0.1 =()$ $3.14 \\times 3^{2} =()$ $3.14 \\times 0.3 =()$",
+        "options": [],
+        "answer": "30;七五;65;八;15.7;0.314;28.26;0.942",
+        "solution": "【详解】略",
+        "knowledge_points": [
+          "成数问题",
+          "小数与整数的乘法",
+          "小数与小数的乘法",
+          "折扣问题"
+        ],
+        "source": "22-23 六年级下·山东菏泽·期中",
+        "merged_id": "unit2_075",
+        "original_id": "unit2_calc_075"
+      },
+      {
+        "id": "unit2_calc_076",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "解方程。\n\n(1) $37 - 130\\% \\mathrm{x} = 11$\n\n(2) $40\\% \\mathrm{x} + 0.5 \\mathrm{x} = 3.6$\n\n(3) $x - 75 \\% x = 0.15$\n\n(4) $1 - 30\\% \\mathrm{x} = 0.49$",
+        "options": [],
+        "answer": "(1) $x = 20$; (2) $x = 4$; (3) $x = 0.6$; (4) $x = 1.7$",
+        "solution": "【分析】（1）先化简方程，再根据等式的性质，方程两边同时加上 $1.3 \\mathrm{~x}$，再同时减去 11，最后同时除以 1.3 即可；(2) 先化简方程，再根据等式的性质，方程两边同时除以 0.9 即可;(3) 先化简方程，再根据等式的性质，方程两边同时除以 0.25 即可;(4) 先化简方程，再根据等式的性质，方程两边同时加上 $0.3 \\mathrm{~x}$，再同时减去 0.49，最后同时除以 0.3 即可;\n\n【详解】(1) $37 - 130\\% \\mathrm{x} = 11$解： $37 - 1.3\\mathrm{x} = 11$\n\n$37 - 1.3 \\mathrm{x} + 1.3 \\mathrm{x} = 11 + 1.3 \\mathrm{x} \\\\ 37 = 11 + 1.3 \\mathrm{x} \\\\ 11 + 1.3 \\mathrm{x} - 11 = 37 - 11 \\\\ 1.3 x = 26 \\\\ 1.3 x \\div 1.3 = 26 \\div 1.3$\n\n$x = 20$\n\n(2) $40\\% \\mathrm{x} + 0.5 \\mathrm{x} = 3.6$\n\n解： $0.4\\mathrm{x} + 0.5\\mathrm{x} = 3.6$\n\n$0.9 x = 3.6$\n\n$0.9 \\mathrm{x} \\div 0.9 = 3.6 \\div 0.9$\n\n$x = 4$\n\n(3) $x - 75\\% x = 0.15$\n\n解: $0.25 \\mathrm{x} = 0.15$\n\n$0.25 \\mathrm{x} \\div 0.25 = 0.15 \\div 0.25$\n\n$x = 0.6$\n\n(4) $(1 - 30\\% x) = 0.49$\n\n解: $1 - 30\\% \\mathrm{x} = 0.49$\n\n$1 - 0.3 \\mathrm{x} = 0.49$\n\n$1 - 0.3 \\mathrm{x} + 0.3 \\mathrm{x} = 0.49 + 0.3 \\mathrm{x}$\n\n$1 = 0.49 + 0.3 x$\n\n$0.49 + 0.3 x - 0.48 = 1 - 0.49$\n\n$0.3x=0.51$\n\n$0.3x÷0.3=0.51÷0.3$\n\n$x=1.7$",
+        "knowledge_points": [
+          "多位小数的不进位加法、不退位减法",
+          "除数是小数的小数除法",
+          "含百分数的运算",
+          "应用等式的性质 1 和 2 解方程"
+        ],
+        "source": "22-23 六年级下·全国·单元测试",
+        "merged_id": "unit2_076",
+        "original_id": "unit2_calc_076"
+      },
+      {
+        "id": "unit2_solve_077",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "小安的爸爸得到一笔4000元的劳务费用，其中800元是免税的，其余部分要按 $20\\%$ 的税率缴纳个人所得税，纳税后他实际获得的劳务费用是多少元？",
+        "options": [],
+        "answer": "3360元",
+        "solution": "【分析】由题意可知，应纳税部分 $= \\text{总劳务费} -$ 免税部分，税率是 $20\\%$，根据'应纳税额 $=$ 应纳税部分 $\\times$ 税率'求出小安的爸爸应该缴纳的个人所得税，实际获得的劳务费 $=$ 总劳务费一个人所得税，据此解答。\n\n【详解】$4000 - (4000 - 800) \\times 20\\%\\\\= 4000 - 3200 \\times 20 \\% \\\\ = 4000 - 640 \\\\ = 3360 (\\text{元})$\n\n答：纳税后他实际获得的劳务费用是 3360 元。",
+        "knowledge_points": [
+          "求应纳税额"
+        ],
+        "source": "24-25 六年级下·山东菏泽·期中",
+        "merged_id": "unit2_077",
+        "original_id": "unit2_solve_077"
+      },
+      {
+        "id": "unit2_solve_078",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "东方电视机厂今年电视机的产量比去年增长二成。今年生产电视机48万台，去年生产电视机多少万台？",
+        "options": [],
+        "answer": "40万台",
+        "solution": "【分析】已知今年电视机的产量比去年增长二成，把去年生产电视机的台数看作单位'1'，则今年电视机的产量是去年的 $(1 + 20 \\%)$，单位'1'未知，用今年电视机的产量除以 $(1 + 20 \\%)$，求出去年电视机的产量。\n\n【详解】二成 $= 20\\%$\n\n$48 \\div (1 + 20 \\%) \\\\ = 48 \\div (1 + 0.2) \\\\ = 48 \\div 1.2 \\\\ = 40 \\text{(万台)}$\n\n答: 去年生产电视机 40 万台。",
+        "knowledge_points": [
+          "分数、小数、百分数与成数的互化",
+          "根据成数反求单位'1'",
+          "已知比一个数多/少百分之几是多少，求这个数"
+        ],
+        "source": "24-25 六年级下·海南海口·单元测试",
+        "merged_id": "unit2_078",
+        "original_id": "unit2_solve_078"
+      },
+      {
+        "id": "unit2_solve_079",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "餐饮业需按营业额的 $5 \\%$ 缴纳营业税，一家饭店5月份缴纳了1.1万元的营业税，这家饭店5月份的营业额是多少万元？",
+        "options": [],
+        "answer": "22万元",
+        "solution": "【分析】根据题意可知，一家饭店5月份缴纳了1.1万元的营业税占营业额的 $5\\%$，把这个饭店5月份的营业额看作单位'1'，单位'1'未知，用缴纳的营业税额除以 $5\\%$，即可求出这家饭店5月份的营业额。\n\n【详解】$1.1 \\div 5\\%\\\\= 1.1 \\div 0.05 \\\\ = 22 \\text{(万元)}$\n\n答：这家饭店5月份的营业额是22万元。",
+        "knowledge_points": [
+          "已知一个数的百分之几是多少，求这个数",
+          "求税率或收入额"
+        ],
+        "source": "24-25 六年级下·海南海口·单元测试",
+        "merged_id": "unit2_079",
+        "original_id": "unit2_solve_079"
+      },
+      {
+        "id": "unit2_solve_080",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "李叔叔的公司新增了一个物联网的项目，要购买一批办公桌椅。某家具店正在打折促销办公用品，办公桌原价800元一张，现在打八折；椅子原价200元一把，现在打六折。导购员称每套办公椅（一张办公桌配一把椅子）已经让利 $30\\%$。请你帮忙分析一下，导购员的说法正确吗？为什么？",
+        "options": [],
+        "answer": "不正确，见详解",
+        "solution": "【分析】八折就是现价是原价的 $80 \\%$；六折就是现价是原价的 $60 \\%$；用办公桌的原价 $\\times 80 \\%$，求出办公桌的现价；用椅子的原价 $\\times 60 \\%$，求出椅子的现价；再用办公桌的现价十椅子的现价，求出一套桌椅的现价，再除以原来一套桌椅的原价，再乘 $100 \\%$，求出现价一套桌椅是原价一套桌椅的百分之几，再用 1 减去现价是原价的百分之几，求出每套桌椅让利百分之几，再和 $30 \\%$ 进行比较，即可解答。\n\n【详解】八折就是现价是原价的 $80 \\%$；六折就是现价是原价的 $60 \\%$。\n\n$(800 \\times 80 \\% + 200 \\times 60 \\%) \\div (800 + 200) \\times 100 \\% \\\\ = (640 + 120) \\div 1000 \\times 100 \\% \\\\ = 760 \\div 1000 \\times 100 \\% = 0.76 \\times 100 \\% \\\\ = 76 \\%$\n\n$1 - 76\\% = 24\\%$，每套桌椅让利 $24\\%$。服务员说得不正确。\n\n答: 服务员说得不正确。",
+        "knowledge_points": [
+          "求一个数是另一个数的百分之几（百分率问题）",
+          "求现价（折扣问题）"
+        ],
+        "source": "24-25 六年级下·全国·单元测试",
+        "merged_id": "unit2_080",
+        "original_id": "unit2_solve_080"
+      },
+      {
+        "id": "unit2_solve_081",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "国庆节期间，各商场开展促销活动。A商场所有商品一律七五折，B商场每满1000元减300元。陈叔叔想买一款笔记本电脑，这款电脑在两家商场的原价都是 5600 元，他去哪家商场购买更便宜一些？便宜多少钱？",
+        "options": [],
+        "answer": "B商场;100元",
+        "solution": "【分析】A商场所有商品一律七五折，七五折，即按原价的 $75 \\%$ 销售，根据求一个数的百分之几是多少，用乘法，据此求出在 A 商场购买花的钱数;B 商场每满 1000 元减 300 元，用 5600 除以 1000 求出 5600里有几个 1000，用 1000 的个数乘 300 就是在 B 商场买这款电脑便宜的钱数，用这款电脑的原价减去便宜的钱数求出现价。再把优惠后在 A 商场和 B 商场的现价进行比较，进一步求出便宜的钱数。\n\n【详解】A： $5600 \\times 75\\% = 4200$ （元）B: $5600 \\div 1000 = 5$ （个）……600（元）\n\n$5600 - 300 \\times 5 \\\\ = 5600 - 1500 \\\\ = 4100元$\n\n$4200 > 4100 \\\\ 4200 - 4100 = 100元$答: 他去 B 商场购买更便宜，便宜 100 元。",
+        "knowledge_points": [
+          "求一个数的百分之几是多少",
+          "求现价（折扣问题）"
+        ],
+        "source": "23-24 六年级下·四川内江·期末",
+        "merged_id": "unit2_081",
+        "original_id": "unit2_solve_081"
+      },
+      {
+        "id": "unit2_solve_082",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "李叔叔家买一套新房，原价110万元，实际只花了九折的钱。按规定买房要按实际房价的 $2\\%$ 缴纳契税，李叔叔家应缴纳契税多少万元？",
+        "options": [],
+        "answer": "1.98万元",
+        "solution": "【分析】已知一套新房的原价 110 万元，实际只花了九折的钱，即实际房价是原价的 $90 \\%$，根据求一个数的百分之几是多少，用原价乘 $90 \\%$ 求出实际房价;\n\n按规定买房要按实际房价的 $2 \\%$ 缴纳契税，根据百分数乘法的意义，用实际房价乘 $2 \\%$，求出应缴纳的契税。\n\n【详解】$110 \\times 90\\% \\times 2\\%\\\\= 110 \\times 0.9 \\times 0.02 \\\\ = 99 \\times 0.02 \\\\= 1.98$ （万元）\n\n答: 李叔叔家应缴纳契税 1.98 万元。",
+        "knowledge_points": [
+          "求应纳税额",
+          "求一个数的百分之几是多少",
+          "求现价（折扣问题）"
+        ],
+        "source": "23-24 六年级下·四川绵阳·期中",
+        "merged_id": "unit2_082",
+        "original_id": "unit2_solve_082"
+      },
+      {
+        "id": "unit2_solve_083",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "两年定期存款的年利率是 $2.70\\%$，妈妈存款到期后取回本金和利息共 2108 元。妈妈当时存款多少元？",
+        "options": [],
+        "answer": "2000 元",
+        "solution": "【分析】到期取款时银行多支付的钱叫利息，利息=本金 $\\times$ 利率 $\\times$ 存期，取回本金和利息总钱数 $=$ 本金 $+$ 本金 $\\times$ 利率 $\\times$ 存期 $=$ 本金 $\\times$ （1 + 利率 $\\times$ 存期），根据积 $\\div$ 因数 $=$ 另一个因数，因此本金 $=$ 取回本金和利息总钱数 $\\div$ （1 + 利率 $\\times$ 存期），据此列式解答。\n\n【详解】$2108 \\div (1 + 2.70\\% \\times 2)\\\\= 2108 \\div (1 + 0.027 \\times 2)\\\\= 2108 \\div (1 + 0.054) \\\\ = 2108 \\div 1.054 \\\\ = 2000元$\n\n答: 妈妈当时存款 2000 元。",
+        "knowledge_points": [
+          "求利率或本金"
+        ],
+        "source": "23-24 六年级上·北京昌平·期中",
+        "merged_id": "unit2_083",
+        "original_id": "unit2_solve_083"
+      },
+      {
+        "id": "unit2_solve_084",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "一本《李毓佩数学历险记》定价24元，在书店用会员卡购买可以打七折，这个价格相当于在网店购买的价格的 $80 \\%$，在网店购买这本书需要多少元？",
+        "options": [],
+        "answer": "21元",
+        "solution": "【分析】由题意可得，七折相当于 $70 \\%$，根据等量关系式:书店购买的价格 = 定价 $\\times 70 \\%$；已知一个数的百分之几是多少，求这个数，用除法，根据等量关系式: 在网店购买的价格 = 在书店购买的价格 ÷ 80%, 已知，据此解答即可。\n\n【详解】七折 $= 70\\%$ $24 \\times 70 \\% = 16.8 (\\text{元})$ $16.8 \\div 0.8 = 21 (\\text{元})$ 答：在网店购买这本书需要21元。",
+        "knowledge_points": [
+          "打折的意义及应用（分数）",
+          "求现价（折扣问题）",
+          "已知一个数的百分之几是多少，求这个数"
+        ],
+        "source": "24-25 六年级上·山西忻州·期中",
+        "merged_id": "unit2_084",
+        "original_id": "unit2_solve_084"
+      },
+      {
+        "id": "unit2_solve_085",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "六月是麦子成熟的季节，张大爷家的小麦喜迎丰收，今年他家的小麦亩产量比去年增加了一成。去年小麦的亩产量是500千克，今年小麦的亩产量是多少千克？",
+        "options": [],
+        "answer": "550千克",
+        "solution": "【分析】据题意可知，把去年小麦的亩产量看作单位'1'，几成表示的是百分之几十，比去年增加了一成，即表示比去年小麦的亩产量多 $10\\%$，根据求一个数的百分之几是多少，用乘法计算，用去年小麦的亩产量 $\\times (1 + 10 \\%)$，即可得解。\n\n【详解】一成 $= 10\\%$\n\n$500 \\times (1 + 10 \\%) \\\\ = 500 \\times 110 \\% \\\\ = 550 \\text{(千克)}$\n\n答: 今年小麦的亩产量是 550 千克。",
+        "knowledge_points": [
+          "分数、小数、百分数与成数的互化",
+          "求增加或减少几成的实际问题",
+          "比一个数多/少百分之几的数是多少"
+        ],
+        "source": "23-24 六年级下·河南安阳·期末",
+        "merged_id": "unit2_085",
+        "original_id": "unit2_solve_085"
+      },
+      {
+        "id": "unit2_solve_086",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "张阿姨到加油站给汽车加油，按原价计算需要450元，加油站推出两种优惠方式，请你帮张阿姨算一算，选择哪种方式付费更划算。方式一：在加油站APP上付费，每满100元减12元。方式二：店内支付享九五折优惠。",
+        "options": [],
+        "answer": "方式一",
+        "solution": "【分析】方式一：每满 100 元减 12 元，先用除法求出 450 元里有几个 100 元，就减去几个 12 元，即可求出方式一的现价；方式二：店内支付享九五折优惠，把原价看作单位'1'，现价是原价的 $95\\%$，单位'1'已知，用原价乘 $95\\%$，即可求出方式二的现价；最后比较两种方式的现价，找出哪种方式付费更划算。\n\n【详解】方式一：\n\n$450 \\div 100 = 4 \\text{(个)} \\dots 50 (\\text{元}) \\\\ 450 - 12 \\times 4 \\\\ = 450 - 48 \\\\ = 402元$\n\n方式二：\n\n$450 \\times 95 \\% \\\\ = 450 \\times 0.95 \\\\ = 427.5元$\n\n$402 < 427.5$\n\n答：选择方式一付费更划算。",
+        "knowledge_points": [
+          "求一个数的百分之几是多少",
+          "求现价（折扣问题）"
+        ],
+        "source": "23-24 六年级下·河南安阳·期末",
+        "merged_id": "unit2_086",
+        "original_id": "unit2_solve_086"
+      },
+      {
+        "id": "unit2_solve_087",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "'纳税光荣，人人有责'。小丽爸爸的月工资是8000元，按照个人所得税法规定：个税免征额是5000元，余额部分按7%缴纳工资薪金个人所得税。小丽爸爸每月应缴纳个人所得税多少元？",
+        "options": [],
+        "answer": "210元",
+        "solution": "【分析】从题意可知：用月工资一免征额 $=$ 余额部分（纳税部分），以余额部分为单位'1'，用余额部分 $\\times 7\\%$，即可求出个人所得税多少元。\n\n【详解】$(8000-5000) ×7\\%\\\\= 3000 \\times 7 \\% \\\\ = 210元$\n\n答: 小丽爸爸每月应缴纳个人所得税 210 元。",
+        "knowledge_points": [
+          "求应纳税额",
+          "求一个数的百分之几是多少"
+        ],
+        "source": "23-24 六年级下·湖南永州·期末",
+        "merged_id": "unit2_087",
+        "original_id": "unit2_solve_087"
+      },
+      {
+        "id": "unit2_solve_088",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "某商场售货员同时卖出两件上衣，每件都以135元售出，若按成本计算，其中一件盈利 $25\\%$，另一件亏损 $25\\%$，问这次售货员是赔了还是赚了？赚了或亏了多少元？",
+        "options": [],
+        "answer": "赔了;18元",
+        "solution": "【分析】盈利的衣服原价 $=$ 售价 $\\div$ （1+利润率），亏损的衣服原价 $=$ 售价 $\\div$ （1一利润率），求出两件衣服成本的和，与售价的和作对比即可。\n\n【详解】$135 \\div (1 + 25\\%)\\\\= 135 \\div 125 \\% \\\\ = 135 \\div 1.25 \\\\ = 108元$\n\n$135 \\div (1 - 25 \\%) \\\\ = 135 \\div 75 \\% \\\\ = 135 \\div 0.75 \\\\ = 180元$\n\n成本和: $180 + 108 = 288$ (元) 售价和: $135 + 135 = 270$ (元) $288 > 270$，成本大于售价，赔了 $288 - 270 = 18$ （元）\n\n答：这次售货员是赔了，赔了18元。",
+        "knowledge_points": [
+          "利润常见问题"
+        ],
+        "source": "2024 六年级下·全国·专题练习",
+        "merged_id": "unit2_088",
+        "original_id": "unit2_solve_088"
+      },
+      {
+        "id": "unit2_solve_089",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "某商店从江南皮革厂以每个 80 元的价格购进了 60 个皮箱，这些皮箱共卖了6300元。这个商店从这60个皮箱上共获得多少利润？利润率是多少？",
+        "options": [],
+        "answer": "1500元;$31.25\\%$",
+        "solution": "【分析】先用 $80 \\times 60 = 4800$ 元，求出进价；再根据商品利润 $=$ 商品售价一商品进价，商品利润率 $=$ 商品利润÷商品进价 $\\times 100\\%$，代入数据，即可求出利润和利润率。\n\n【详解】进价： $80 \\times 60 = 4800$ （元）利润： $6300 - 4800 = 1500$ （元）利润率：\n\n$1500 \\div 4800 \\times 100\\%\\\\= 0.3125 \\times 100 \\% \\\\ = 31.25 \\%$\n\n答：这个商店从这60个皮箱上共获得1500元的利润，利润率是 $31.25\\%$。",
+        "knowledge_points": [
+          "求一个数是另一个数的百分之几（百分率问题）",
+          "利润常见问题"
+        ],
+        "source": "2024 六年级下·全国·专题练习",
+        "merged_id": "unit2_089",
+        "original_id": "unit2_solve_089"
+      },
+      {
+        "id": "unit2_solve_090",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "资料卡：一种饮料，大瓶装每瓶1200毫升，10元一瓶；罐装每罐200毫升，2元一罐，现有两家商店出售这种饮料，并推出了不同的促销方式。甲商店：买一大瓶送一罐。乙商店：一律九折。根据资料中的信息自主选择问题并解答。(1) 乙商店一律九折，是把 ( ) 看作单位'1'的量，( )是 ( )的 ( )%。(2) 我们班共有 44 名同学，如果给每位同学配备 200 毫升饮料，购买这些饮料，去哪一家商店购买可以使所花费的钱最少? 最少花多少钱? (3) 甲商店的李老板 2023 年底将 30000 元存入银行，年利率是 $2.7 \\%$，存期三年，你能算出到期后他能得到多少利息吗?",
+        "options": [],
+        "answer": "（1）原价;现价;原价;90 (2) 甲商店; 64 元 (3) 2430 元",
+        "solution": "【分析】（1）一个整体可以用自然数 1 表示，我们通常把它叫做单位'1'。确定单位'1'，关键是看以谁为标准，谁是标准谁就是单位'1'，打折就是按照折数低价出售商品。几折就是十分之几，也就是百分之几十。(2) 每人配备饮料体积 $\\times$ 总人数 = 需要的饮料总体积，甲商店: 需要的饮料总体积 $\\div$ (大瓶容积 + 每罐容积)，求出的商是需要购买的大瓶数量，求出的余数 $\\div$ 每罐容积 = 需要购买的罐装数量，大瓶单价 $\\times$ 大瓶数量 + 罐装单价 $\\times$ 罐装数量 = 实际钱数;乙商店：尽可能的多买大瓶装，需要的饮料总体积 $\\div$ 大瓶容积，求出的商是需要购买的大瓶数量，求出的余数 $\\div$ 每罐容积 $=$ 需要购买的罐装数量，大瓶单价 $\\times$ 大瓶数量 + 罐装单价 $\\times$ 罐装数量 = 应付钱数，应付钱数 $\\times$ 折扣 = 实际钱数。据此分别计算出两家商店的实际钱数，比较即可。(3) 根据利息 $=$ 本金 $\\times$ 利率 $\\times$ 存期，列式解答即可。\n\n【详解】（1）乙商店一律九折，是把原价看作单位'1'的量，现价是原价的 $90\\%$。\n\n(2) $200 \\times 44 = 8800$ (毫升)\n\n甲商店: $8800 \\div (1200 + 200)\\\\= 8800 \\div 1400\\\\=6 (瓶) ……400 (毫升)$\n\n$400 \\div 200 = 2$ (罐)\n\n$6 \\times 10 + 2 \\times 2 \\\\ = 60 + 4 \\\\ = 64元$\n\n乙商店: $8800 \\div 1200 = 7$ (瓶) ……400 (毫升)\n\n$400 \\div 200 = 2$ （罐）\n\n$7 \\times 10 + 2 \\times 2 \\\\ = 70 + 4 \\\\ = 74元$\n\n$74 \\times 90\\% = 74 \\times 0.9 = 66.6$ （元）\n\n$64 < 66.6$\n\n答：去甲商店购买可以使所花费的钱最少，最少花64元钱。\n\n(3) $30000 \\times 2.7\\% \\times 3\\\\= 30000 \\times 0.027 \\times 3 \\\\ = 2430 (\\text{元})$\n\n答: 到期后他能得到 2430 元利息。",
+        "knowledge_points": [
+          "打折的意义及应用（分数）",
+          "单位'1'的认识与确定",
+          "求现价（折扣问题）",
+          "求利息"
+        ],
+        "source": "23-24 六年级下·河南信阳·期末",
+        "merged_id": "unit2_090",
+        "original_id": "unit2_solve_090"
+      },
+      {
+        "id": "unit2_solve_091",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "2019年5月17日，华为'备胎'芯片'海思'一夜转正，'海思'是我国半导体行业的领军者，更是我国高科技的荣耀。2019年'海思'的销售额为75亿美元，比2018年大约增长了二成五，2018年的销售额大约是多少亿美元？",
+        "options": [],
+        "answer": "60亿美元",
+        "solution": "【分析】由题意可知，2019 年'海思'的销售额比 2018 年大约增长了二成五，即 2019 年'海思'的销售额是 2018 年的 $(1 + 25 \\%),$根据已知一个数的百分之几是多少，求这个数，用除法计算，即用 75 除以 $(1 + 25 \\%)$ 即可求出 2018 年的销售额。\n\n【详解】$75 \\div (1 + 25\\%)\\\\= 75 \\div 1.25\\\\= 60$ （亿美元）\n\n答：2018 年的销售额大约是 60 亿美元。",
+        "knowledge_points": [
+          "根据成数反求单位'1'",
+          "已知比一个数多/少百分之几是多少，求这个数"
+        ],
+        "source": "23-24 六年级下·山东菏泽·期中",
+        "merged_id": "unit2_091",
+        "original_id": "unit2_solve_091"
+      },
+      {
+        "id": "unit2_solve_092",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "百货大楼搞促销活动，甲品牌鞋每满200元减100元，乙品牌鞋'折上折'，就是先打六折，在此基础上再打九五折。如两个品牌都有一双标价260元的鞋，哪个品牌的更便宜？",
+        "options": [],
+        "answer": "乙品牌",
+        "solution": "【分析】已知甲品牌鞋每满200元减100元，260里面有1个200，则用260减去100即可求出甲品牌鞋的实际价格；已知乙品牌鞋先打六折，在此基础上再打九五折，六折表示 $60\\%$，也就是原价260元的 $60\\%$，把260元看作单位'1'，根据百分数乘法的意义，用 $260 \\times 60\\%$ 即可求出打六折后的价格，再把打六折后的价格看作单位'1'，九五折示 $95\\%$，根据百分数乘法的意义，用 $260 \\times 60\\% \\times 95\\%$ 即可求出乙品牌鞋的实际价格；最后比较两个品牌的实际价格即可。本题主要考查了百分数的应用，明确折扣的含义是解答本题的关键。\n\n【详解】$260 - 100 = 160$ （元）六折 $= 60\\%$九五折 $= 95\\%$$260 \\times 60 \\% \\times 95 \\% = 148.2 (\\text{元})$\n\n$160 > 148.2$答: 乙品牌的更便宜。",
+        "knowledge_points": [
+          "求一个数的百分之几是多少",
+          "折扣问题"
+        ],
+        "source": "2022·重庆大渡口·小升初真题",
+        "merged_id": "unit2_092",
+        "original_id": "unit2_solve_092"
+      },
+      {
+        "id": "unit2_solve_093",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "书店的图书凭优惠卡可以打八折，小明用优惠卡买了一套书，节省了9.6元。小明买这套书花了多少元？",
+        "options": [],
+        "answer": "38.4元",
+        "solution": "【分析】八折表示现价是原价的 $80 \\%$，则把原价看作单位'1'，现价比原价少 $(1 - 80 \\%)$，根据百分数除法的意义,用 $9.6 \\div (1 - 80 \\%)$ 即可求出原价，进而求出现价即可。本题主要考查了百分数的应用，明确折扣的含义是解答本题的关键。\n\n【详解】$9.6 \\div (1 - 80\\%)\\\\= 9.6 \\div 20 \\% \\\\ = 48 (\\text{元})$\n\n$48 - 9.6 = 38.4 \\text{(元)}$答: 小明买这套书花了 38.4 元。",
+        "knowledge_points": [
+          "已知一个数的百分之几是多少，求这个数",
+          "折扣问题"
+        ],
+        "source": "2022·安徽蚌埠·小升初真题",
+        "merged_id": "unit2_093",
+        "original_id": "unit2_solve_093"
+      },
+      {
+        "id": "unit2_solve_094",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "我们国家规定，公民月收入在5000元以上的要缴纳个人所得税，超出1500元以内的局部纳税 $3\\%$，超出 1500 至 3000 元的局部纳税10%；小红的爸爸每月收入8000元，他每月应缴纳个人所得税多少钱？",
+        "options": [],
+        "answer": "195元",
+        "solution": "【分析】根据税额 $=$ 应纳税款 $\\times$ 税率，由题意可知：不超过1500元部分按 $3 \\%$ 的税率，超过1500元不超过3000元部分则按 $10 \\%$ 的税率。据此根据乘法的意义，用乘法分别求出1500元和不超过3000元部分的应纳税额，然后合并起来即可。这种类型属于税率问题，要理清数据关系，分步进行，做到思路清晰。\n\n【详解】$8000 - 5000 = 3000$ （元）\n\n$3000 - 1500 = 1500元\\\\ 1500 \\times 3 \\% + 1500 \\times 10 \\% \\\\ = 45 + 150 \\\\ = 195元$\n\n答: 他每月应缴纳个人所得税 195 元。",
+        "knowledge_points": [
+          "税率问题"
+        ],
+        "source": "22-23 六年级下·全国·单元测试",
+        "merged_id": "unit2_094",
+        "original_id": "unit2_solve_094"
+      },
+      {
+        "id": "unit2_solve_095",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "某超市将一种商品按增加进价的 $60 \\%$ 设为售价，然后写上'酬宾'，按售价打9折出售，结果每件商品仍获利88元，这种商品的进价是多少元？",
+        "options": [],
+        "answer": "200元",
+        "solution": "【分析】将商品的进价设为未知数 $x$ 元，将进价乘 $(1 + 60\\%)$ 再乘 $90\\%$，求出售价。根据售价减去进价等于利润,列出方程求解即可。本题考查了简易方程的应用，找出数量关系是解题的关键。\n\n【详解】解：设这种商品的进价是 $\\mathrm{x}$ 元。\n\n$(1 + 60 \\%) x \\times 90 \\% - x = 88\\\\1.44 \\mathrm{x} - \\mathrm{x} = 88 \\\\ 0.44 \\mathrm{x} = 88 \\\\ x = 88 \\div 0.44 \\\\ x = 200$\n\n答: 这种商品的进价是 200 元。",
+        "knowledge_points": [
+          "列方程解含一个未知数的问题",
+          "折扣问题"
+        ],
+        "source": "22-23 六年级下·全国·单元测试",
+        "merged_id": "unit2_095",
+        "original_id": "unit2_solve_095"
+      },
+      {
+        "id": "unit2_solve_096",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "姐姐想在网上商店买一件连衣裙，A店打六折销售，B店每满100元减40元。如果这件连衣裙的标价都是480元。(1) 在 A、B 两个网店买，各应付多少元？(2) 在哪个网店买更省钱？A、B两店的价格相差多少元？",
+        "options": [],
+        "answer": "（1）A 网店 288 元;B 网店 320 元；(2) A 网店; 32 元",
+        "solution": "【分析】（1）A店：六折 $= 60\\%$，将标价480元乘 $60\\%$ 求出在A店购买应付多少元；B店：480元里面有4个100元，可以减 $4 \\times 40 = 160$ （元），将原价480元减去160元，求出在B店购买应付多少元。(2) 根据 (1)，比较出在哪个店买更省钱，再利用减法求出价格差。本题考查了折扣问题，打几折就是按照原价的百分之几十出售。\n\n【详解】(1) A 店: $480 \\times 60\\% = 288$ （元）\n\nB店: $480 - 4 \\times 40\\\\= 480 - 160 \\\\ = 320 (\\text{元})$\n\n答: 在 A 网店买应付 288 元，在 B 网店买应付 320 元。\n\n(2) 288 元 $< 320$ 元 $320 - 288 = 32 \\text{(元)}$\n\n答: 在 A 网店买更省钱，A、B 两店的价格相差 32 元。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "22-23 六年级下·全国·单元测试",
+        "merged_id": "unit2_096",
+        "original_id": "unit2_solve_096"
+      },
+      {
+        "id": "unit2_solve_097",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "李叔叔与 2014 年 1月1日在银行存了活期储蓄1000元，如果每月的利率是 0.25%，存款3个月时，可得到利息多少元？本金和利息共多少钱？",
+        "options": [],
+        "answer": "7.5元;1007.5元",
+        "solution": "【分析】本题中，本金是 1000 元，利率是 $0.25 \\%$，时间是 3 个月，求利息，根据关系式: 利息 = 本金 $\\times$ 利率 $\\times$ 时间，求本金和利息，根据关系式: 本息 = 本金 + 利息，解决问题。此题属于利息问题，考查了关系式：利息 $=$ 本金 $\\times$ 利率 $\\times$ 时间，本息 $=$ 本金 + 利息。\n\n【详解】$1000 \\times 0.25\\% \\times 3\\\\= 2.5 \\times 3 \\\\ = 7.5元$\n\n$1000 + 7.5 = 1007.5 \\text{(元)}$\n\n答: 可得到利息 7.5 元，本金和利息一共 1007.5 元。",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "22-23 六年级下·全国·单元测试",
+        "merged_id": "unit2_097",
+        "original_id": "unit2_solve_097"
+      },
+      {
+        "id": "unit2_solve_098",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "益民五金公司去年的营业额为400万元，如果按营业额的 $3\\%$ 缴纳营业税，去年应缴营业税多少万元？",
+        "options": [],
+        "answer": "12万元",
+        "solution": "【分析】已知营业额为 400 万元，税率为 $3 \\%$，要求应缴纳营业税是多少，就是求 400 万元的 $3 \\%$ 是多少，用乘法计算。此题属于纳税问题，运用了关系式：营业额 $\\times$ 税率 $=$ 营业税。\n\n【详解】$400 \\times 3\\% = 12$ （万元）答：去年应缴营业税12万元。",
+        "knowledge_points": [
+          "税率问题"
+        ],
+        "source": "22-23 六年级下·全国·单元测试",
+        "merged_id": "unit2_098",
+        "original_id": "unit2_solve_098"
+      },
+      {
+        "id": "unit2_solve_099",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "妈妈把50000元钱存入银行，定期3年，年利率是 $3.33\\%$，到期后妈妈一共可以取出多少元钱？",
+        "options": [],
+        "answer": "54995元",
+        "solution": "【分析】根据利息 $=$ 本金 $\\times$ 利率 $\\times$ 存期，先求出到期后妈妈可以取到的利息，再根据本息和 $=$ 本金 + 利息，求出到期后妈妈一共可以取出的钱数即可。掌握利息的计算方法是解题的关键，明确到期后取出的钱是本金加上利息的总金额。\n\n【详解】$50000 + 50000 \\times 3.33\\% \\times 3 = 50000 + 4995 = 54995$ （元）答：到期后妈妈一共可以取出 54995 元钱。",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "21-22 六年级上·河南商丘·期末",
+        "merged_id": "unit2_099",
+        "original_id": "unit2_solve_099"
+      },
+      {
+        "id": "unit2_solve_100",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "同一种型号的学习机，甲商场标价4000元，现在打八折销售；乙商场标价3500元，现在降价 $10\\%$ 销售。你认为在哪个商场买划算?",
+        "options": [],
+        "answer": "在乙商场买划算",
+        "solution": "【分析】甲商场：先把原价看成单位'1'，用乘法求出它的 $80\\%$ 就是八折后的价格；乙商场：降价 $10\\%$，就是按原价的 $90\\%$ 销售，用乘法计算出售价；然后与甲商场的钱数比较即可求解。本题关键是理解打折的含义：打几折，现价就是原价的百分之几十。本题关键是理解打折的含义：打几折，现价就是原价的百分之几十。\n\n【详解】$4000 \\times 80\\% = 3200$ （元）\n\n$3500 \\times (1 - 10 \\%) \\\\ = 3500 \\times 90 \\% \\\\ = 3500 \\times 0.9 \\\\ = 3150元$\n\n$3200 > 3150$答：在乙商场买划算。",
+        "knowledge_points": [
+          "求一个数的百分之几是多少",
+          "含百分数的运算",
+          "折扣问题"
+        ],
+        "source": "21-22 六年级上·贵州毕节·期末",
+        "merged_id": "unit2_100",
+        "original_id": "unit2_solve_100"
+      },
+      {
+        "id": "unit2_choice_101",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "某面粉厂8月份销售面粉870吨，比七月份减少二成，7月份销售面粉（）吨。",
+        "options": [
+          "A. 4350",
+          "B. 1087.5",
+          "C. 725",
+          "D. 700"
+        ],
+        "answer": "B",
+        "solution": "【分析】根据'比七月份减少二成'可知：七月份销售面粉的吨数是单位'1'，求七月份销售面粉的吨数，即求单位'1'。单位'1'未知用除法解答，即已知量÷已知量所对应的分率 $=$ 单位'1'的量。8月份销售面粉的吨数所对应的分率是 $(1 - 20\\%)$，所以 $870 \\div (1 - 20\\%)$ 可求出7月份销售面粉的吨数。确定单位'1'的量是解决分数问题的关键。单位'1'已知，用乘法解答；单位'1'未知，可以列方程解答或者用除法解答。用除法解答时要注意量率对应。\n\n【详解】$870 \\div (1 - 20\\%) \\\\ = 870 \\div (1 - 0.2) \\\\ = 870 \\div 0.8 \\\\ = 1087.5 (吨)$\n\n所以7月份销售面粉1087.5吨。故答案为: B",
+        "knowledge_points": [
+          "已知比一个数多/少百分之几是多少，求这个数",
+          "根据成数反求单位'1'"
+        ],
+        "source": "21-22六年级上·河北保定·期末",
+        "merged_id": "unit2_101",
+        "original_id": "unit2_choice_101"
+      },
+      {
+        "id": "unit2_choice_102",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "王叔叔以八五折买了一辆平衡车，实际付了663元。这辆平衡车原价是（）元。",
+        "options": [
+          "A. 828",
+          "B. 780",
+          "C. 563"
+        ],
+        "answer": "B",
+        "solution": "【分析】八五折相当于 $85\\%$，根据原价 $\\times$ 折扣 $=$ 现价可得，原价 $=$ 现价 $\\div$ 折扣，即用663元除以 $85\\%$，即可求出原价。此题的解题关键是理解折扣的含义，根据原价、折扣、现价三者之间的关系，解决问题。\n\n【详解】八五折 $= 85\\%$\n\n$663 \\div 85\\% \\\\ = 663 \\div 0.85 \\\\ = 780 (元)$\n\n即这辆平衡车原价是780元。故答案为: B",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "21-22六年级上·河南洛阳·期末",
+        "merged_id": "unit2_102",
+        "original_id": "unit2_choice_102"
+      },
+      {
+        "id": "unit2_choice_103",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "某文具店开学季全场文具八折优惠。原价42元的用品，需付（）元。",
+        "options": [
+          "A. 336",
+          "B. 33.6",
+          "C. 84"
+        ],
+        "answer": "B",
+        "solution": "【分析】把文具的原价看作单位'1'，打八折优惠意思是，现价是原价的 $80\\%$，根据求一个数的百分之几是多少，用乘法计算，即可求出需付的钱数。本题考查折扣问题，掌握原价、现价、折扣之间的关系是解题的关键。\n\n【详解】$42 \\times 80\\% \\\\ = 42 \\times 0.8 \\\\ = 33.6 (元)$\n\n需付33.6元。故答案为: B",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "22-23六年级上·山西吕梁·期中",
+        "merged_id": "unit2_103",
+        "original_id": "unit2_choice_103"
+      },
+      {
+        "id": "unit2_choice_104",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "一件衣服打七折出售，现价是原价（）。",
+        "options": [
+          "A. $30\\%$",
+          "B. $40\\%$",
+          "C. $70\\%$",
+          "D. $170\\%$"
+        ],
+        "answer": "C",
+        "solution": "【分析】根据题意，把原价看作单位'1'，某衣服打'七折'出售，就是现价是原价的 $70\\%$。\n\n【详解】由分析可得：一件衣服打七折出售，意思就是现价是原价的 $70\\%$。故答案为: C",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "2022·贵州黔西·小升初真题",
+        "merged_id": "unit2_104",
+        "original_id": "unit2_choice_104"
+      },
+      {
+        "id": "unit2_choice_105",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "小莉存入银行2000元，如果每个月的利率是 $0.35\\%$，如果存4个月，可获利息（）。",
+        "options": [
+          "A. 7元",
+          "B. 2800元",
+          "C. 28元",
+          "D. 280元"
+        ],
+        "answer": "C",
+        "solution": "【分析】通过计算利息的公式：利息 $=$ 本金 $\\times$ 存期 $\\times$ 利率，本金是2000元，存期是4个月，利率是 $0.35\\%$，代入数据即可得解。此题的解题关键是掌握利息的计算方法。\n\n【详解】$2000 \\times 4 \\times 0.35\\% \\\\ = 8000 \\times 0.35\\% \\\\ = 28 (元)$\n\n故答案为: C",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "21-22六年级下·黑龙江牡丹江·期末",
+        "merged_id": "unit2_105",
+        "original_id": "unit2_choice_105"
+      },
+      {
+        "id": "unit2_choice_106",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "张叔叔每月收入10000元。在缴纳个人所得税时，扣除5000元个税免征额后，还可享受赡养老人、子女教育和房贷利息等三项专项附加扣除共3000元，剩下的工资再按 $3\\%$ 的税率缴纳个人所得税。张叔叔每月应缴纳的个人所得税是（）。",
+        "options": [
+          "A. 300元",
+          "B. 150元",
+          "C. 90元",
+          "D. 60元"
+        ],
+        "answer": "D",
+        "solution": "【分析】应纳税部分需要从每月收入中扣除（5000+3000）元，应纳税额 $=$ 应纳税部分 $\\times$ 税率，据此解答。掌握应纳税额的计算方法是解答题目的关键。\n\n【详解】$(10000 - 5000 - 3000) \\times 3\\% \\\\ = 2000 \\times 3\\% \\\\ = 60 (元)$\n\n所以，张叔叔每月应缴纳的个人所得税是60元。故答案为: D",
+        "knowledge_points": [
+          "税率问题"
+        ],
+        "source": "21-22六年级下·广东佛山·期末",
+        "merged_id": "unit2_106",
+        "original_id": "unit2_choice_106"
+      },
+      {
+        "id": "unit2_choice_107",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "妈妈买一瓶售价为200元的化妆品，其中消费税大约占售价的 $25\\%$。妈妈为此支付消费税大约多少元？列式是（）。",
+        "options": [
+          "A. $200 \\times 25\\%$",
+          "B. $200 \\div 25\\%$",
+          "C. $200 \\times (1 - 75\\%)$"
+        ],
+        "answer": "A",
+        "solution": "【分析】将化妆品售价看作单位'1'，化妆品售价×消费税税率=支付的消费税，据此分析。关键是确定单位'1'，整体数量 $\\times$ 部分对应百分率=部分数量。\n\n【详解】$200 \\times 25\\% = 50$ （元）\n\n故答案为：A",
+        "knowledge_points": [
+          "税率问题"
+        ],
+        "source": "2022·河南三门峡·小升初真题",
+        "merged_id": "unit2_107",
+        "original_id": "unit2_choice_107"
+      },
+      {
+        "id": "unit2_choice_108",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "'双十一'某网店所有商品打五折出售。李阿姨在该网店购得双肩包一个，加上邮费（邮费相当于原价的 $5\\%$）共付132元，这一双肩包的原价是（）元。",
+        "options": [
+          "A. 264",
+          "B. 240",
+          "C. 260"
+        ],
+        "answer": "B",
+        "solution": "【分析】把原价看作单位'1'，五折出售，也就是现价是原价的 $50\\%$，加上邮费（邮费相当于原价的 $5\\%$）共付132元，由此可知132元相当于原价的 $(1 - 50\\% + 5\\%)$，根据已知一个数的百分之几是多少，求这个数，用除法解答。本题首先要理解打五折的意思，打几折现价就是原价的百分之几十，邮费相当于原价的 $5\\%$；关键是求出现价加上邮费占原价的百分之几，再根据已知一个数的百分之几是多少，求这个数，用除法解答。\n\n【详解】$132 \\div (1 - 50\\% + 5\\%) \\\\ = 132 \\div 55\\% \\\\ = 132 \\div 0.55 \\\\ = 240 (元)$\n\n故答案为: B",
+        "knowledge_points": [
+          "经济问题",
+          "折扣问题"
+        ],
+        "source": "19-20六年级下·全国·期中",
+        "merged_id": "unit2_108",
+        "original_id": "unit2_choice_108"
+      },
+      {
+        "id": "unit2_choice_109",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "某超市4月份缴了10万元营业税，他们纳税的税率是 $5\\%$，该超市4月份的营业额是多少万元？列式正确的是（）。",
+        "options": [
+          "A. $10 \\times 5\\%$",
+          "B. $10 + 5\\%$",
+          "C. $10 \\div 5\\%$"
+        ],
+        "answer": "C",
+        "solution": "【分析】根据题意，营业税额是按营业额的 $5\\%$ 纳税，把营业额看作单位'1'，单位'1'未知，用营业税额除以税率，求出营业额。根据营业税额 $=$ 营业额 $\\times$ 税率，得出营业额 $=$ 营业税额 $\\div$ 税率。\n\n【详解】$10 \\div 5\\% \\\\ = 10 \\div 0.05 \\\\ = 200 (万元)$\n\n故答案为: C",
+        "knowledge_points": [
+          "求税率或收入额",
+          "已知一个数的百分之几是多少，求这个数"
+        ],
+        "source": "17-18六年级下·甘肃天水·期中",
+        "merged_id": "unit2_109",
+        "original_id": "unit2_choice_109"
+      },
+      {
+        "id": "unit2_choice_110",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "小军把800元存入银行，这里的'800元'是（）。",
+        "options": [
+          "A. 本金",
+          "B. 利息",
+          "C. 利率"
+        ],
+        "answer": "A",
+        "solution": "【分析】存入银行的钱叫做本金；取款时银行多支付的钱叫做利息；单位时间内的利息与本金的比率叫做利率；据此解答。掌握本金的意义是解答题目的关键。\n\n【详解】分析可知，小军把800元存入银行，这里的'800元'是本金。故答案为: A",
+        "knowledge_points": [
+          "求利率或本金"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit2_110",
+        "original_id": "unit2_choice_110"
+      },
+      {
+        "id": "unit2_choice_111",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "一件外套七折出售，比原来便宜了75元，这件外套原来（）元。",
+        "options": [
+          "A. 150",
+          "B. 225",
+          "C. 250"
+        ],
+        "answer": "C",
+        "solution": "【分析】按照七折出售，也就是按照原价的 $70\\%$，单位'1'是原价，比原价减少了 $30\\%$，也就是75元相当于是原价的 $30\\%$，根据公式：对应量 $\\div$ 对应百分率=单位'1'，代入数据即可求解。本题考查的是经济问题中的折扣问题，原价 $=$ 现价 $\\div$ 折扣。\n\n【详解】$1 - 70\\% = 30\\%$\n\n$75 \\div 30\\% = 250$ （元）\n\n故答案为: C",
+        "knowledge_points": [
+          "折扣问题",
+          "已知一个数的百分之几是多少，求这个数"
+        ],
+        "source": "18-19六年级上·浙江·单元测试",
+        "merged_id": "unit2_111",
+        "original_id": "unit2_choice_111"
+      },
+      {
+        "id": "unit2_choice_112",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "小王买了5000元的国家建设债券，定期3年，年利率 $5\\%$，到期时，他能取回多少钱？下面列式正确的是（）。",
+        "options": [
+          "A. $5000 \\times (1 + 5\\%) \\times 3$",
+          "B. $5000 \\times 5\\% = 5000$",
+          "C. $5000 \\times 5\\% \\times 3 + 5000$"
+        ],
+        "answer": "C",
+        "solution": "【分析】因为取回来的是本金和利息的和，利息 $=$ 本金 $\\times$ 利率 $\\times$ 时间，然后根据公式列式即可。本题主要考查百分数在利率中的应用，注意取回的钱数是本金和利息的和，有的题型还扣除利息税。\n\n【详解】利息为 $5000 \\times 5\\% \\times 3$ 元\n\n所以取回的钱数=利息+本金=$(5000 \\times 5\\% \\times 3 + 5000)$ 元。\n\n故答案为: C",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "17-18六年级下·全国·单元测试",
+        "merged_id": "unit2_112",
+        "original_id": "unit2_choice_112"
+      },
+      {
+        "id": "unit2_choice_113",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "香水的售价是240元，其中消费税为售价的 $25\\%$，妈妈为此支付消费税（）元。",
+        "options": [
+          "A. 60",
+          "B. 192",
+          "C. 48"
+        ],
+        "answer": "A",
+        "solution": "【分析】根据题意，消费税为售价的 $25\\%$，把售价看作单位'1'，用售价乘 $25\\%$，就是妈妈为此支付的消费税额。求一个数的百分之几是多少，用乘法计算。\n\n【详解】$240 \\times 25\\% = 60$ （元）\n\n故答案为: A",
+        "knowledge_points": [
+          "求一个数的百分之几是多少",
+          "税率问题"
+        ],
+        "source": "17-18六年级下·全国·期中",
+        "merged_id": "unit2_113",
+        "original_id": "unit2_choice_113"
+      },
+      {
+        "id": "unit2_choice_114",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "对于商店'买四送一'的促销方式，下面说法正确的是（）。",
+        "options": [
+          "A. 商品打八折销售",
+          "B. 商品的降价幅度比打八折要小",
+          "C. 商品的降价幅度比打八折要大",
+          "D. 买5的倍数件才打八折"
+        ],
+        "answer": "D",
+        "solution": "【分析】'买四送一'，即原来买 $(4 + 1)$ 件商品的价钱，现在只需付4件商品的价钱，用现价除以原价，即可求出折扣。掌握折扣的计算方法是解题的关键。\n\n【详解】设每件商品的价格是1。\n\n原来买5件商品需付：$5 \\times 1 = 5$\n\n现在需付：$4 \\times 1 = 4$\n\n折扣：$4 \\div 5 \\times 100\\% = 80\\%$\n\n$80\\% =$ 八折\n\n如果购买商品的件数不是5的倍数，其中是5的倍数的件数按打八折支付，多出来的件数按原价支付，那么整体商品的降价幅度要比打八折小。所以必须买5的倍数件才打八折。故答案为: D",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit2_114",
+        "original_id": "unit2_choice_114"
+      },
+      {
+        "id": "unit2_choice_115",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "高叔叔把10000元存入银行定期三年。年利率是 $2.75\\%$，到期后他将从银行得到的利息多少元？正确列式是（）。",
+        "options": [
+          "A. $10000 \\times 2.75\\% \\times 3$",
+          "B. $10000 \\times 2.75\\%$",
+          "C. $10000 - 10000 \\times 2.75\\% \\times 3$",
+          "D. $10000 + 10000 \\times 2.75\\% \\times 3$"
+        ],
+        "answer": "A",
+        "solution": "【分析】通过利息的计算公式：利息 $=$ 本金 $\\times$ 利率 $\\times$ 存期，求出利息，即可得解。此题的解题关键是利用计算利息的公式来求解。\n\n【详解】$10000 \\times 2.75\\% \\times 3 \\\\ = 275 \\times 3 \\\\ = 825 (元)$\n\n故答案为: A",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "21-22六年级下·全国·单元测试",
+        "merged_id": "unit2_115",
+        "original_id": "unit2_choice_115"
+      },
+      {
+        "id": "unit2_choice_116",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "一台平板电脑原价是2500元，现在降价500元，相当于打（）折销售。",
+        "options": [
+          "A. 九",
+          "B. 九五",
+          "C. 八",
+          "D. 五"
+        ],
+        "answer": "C",
+        "solution": "【分析】把原价看成单位'1'，用现价除以原价，求出现价是原价的百分之几，然后根据打折的含义求出是几折销售。本题关键是理解打折的含义：打几几折，现价就是原价的百分之几十几。\n\n【详解】$(2500 - 500) \\div 2500 \\\\ = 2000 \\div 2500 \\\\ = 0.8$\n\n即现价是原价的 $80\\%$，就是打八折销售。故答案为: C",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "21-22六年级下·全国·单元测试",
+        "merged_id": "unit2_116",
+        "original_id": "unit2_choice_116"
+      },
+      {
+        "id": "unit2_choice_117",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "妈妈买了6000元三年期国库券，已知三年期年利率是 $6.21\\%$，三年后妈妈可得利息多少元？正确列式为（）。",
+        "options": [
+          "A. $6000 \\times 6.21\\%$",
+          "B. $6000 + 1000 \\times 6.21\\%$",
+          "C. $6000 \\times 6.21\\% \\times 3$"
+        ],
+        "answer": "C",
+        "solution": "【分析】利用公式：利息 $=$ 本金 $\\times$ 利率 $\\times$ 存期，即可得出正确选项。此题的解题关键是掌握利息的计算方法，根据公式，列出算式，得到正确的选项。\n\n【详解】根据分析，本金为6000元，利率为 $6.21\\%$，存期为3年，代入公式：利息 $=$ 本金 $\\times$ 利率 $\\times$ 存期，可得利息为 $6000 \\times 6.21\\% \\times 3$\n\n故答案为: C",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "19-20六年级下·湖南怀化·单元测试",
+        "merged_id": "unit2_117",
+        "original_id": "unit2_choice_117"
+      },
+      {
+        "id": "unit2_choice_118",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "一种摩托车原价每台8000元，活动期间七五折优惠，活动期间摩托车需要多少元？正确列式为（）。",
+        "options": [
+          "A. $8000 \\times (1 - 75\\%)$",
+          "B. $8000 \\times 75\\%$",
+          "C. $8000 \\times 7.5\\%$"
+        ],
+        "answer": "B",
+        "solution": "【分析】根据原价 $\\times$ 折扣=现价，七五折即为 $75\\%$，然后根据求一个数的百分之几是多少，用乘法进行计算即可。本题考查求一个数的百分之几是多少，明确用乘法是解题的关键。\n\n【详解】$8000 \\times 75\\% = 6000$ （元）\n\n则活动期间摩托车需要6000元。故选: B",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "19-20六年级下·湖南怀化·单元测试",
+        "merged_id": "unit2_118",
+        "original_id": "unit2_choice_118"
+      },
+      {
+        "id": "unit2_choice_119",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "今年的产量比去年增长二成，也就是说今年的产量是去年的（）%。",
+        "options": [
+          "A. 200",
+          "B. 120",
+          "C. 20"
+        ],
+        "answer": "B",
+        "solution": "【分析】把去年的产量看作单位'1'，今年的产量比去年多 $20\\%$，今年的产量占去年的 $(1 + 20\\%)$，据此解答。分析题意找出题中的单位'1'是解答题目的关键。\n\n【详解】分析可知，二成 $= 20\\%$，$1 + 20\\% = 120\\%$，也就是说今年的产量是去年的 $120\\%$。故答案为: B",
+        "knowledge_points": [
+          "成数问题"
+        ],
+        "source": "19-20六年级下·湖南怀化·单元测试",
+        "merged_id": "unit2_119",
+        "original_id": "unit2_choice_119"
+      },
+      {
+        "id": "unit2_choice_120",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "某工厂上月纳税6万元，实际应纳税所得额为60万元，由此可知其税率为（）。",
+        "options": [
+          "A. $3\\%$",
+          "B. $5\\%$",
+          "C. $10\\%$"
+        ],
+        "answer": "C",
+        "solution": "【分析】应纳税额与各种收入中应纳税部分的比率叫做税率。此题解题的关键是理解税率的含义，通过税率的公式可求出结果。\n\n【详解】应纳税额是6万元，各种收入中应纳税部分实际就是题目中的纳税所得额60万元，税率 $= 6 \\div 60 = 0.1 = 10\\%$。故答案为: C",
+        "knowledge_points": [
+          "税率问题"
+        ],
+        "source": "18-19六年级下·广东汕头·单元测试",
+        "merged_id": "unit2_120",
+        "original_id": "unit2_choice_120"
+      },
+      {
+        "id": "unit2_choice_121",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "收入超过3500元，不超过5000元的工资，应按超出部分的 $3\\%$ 缴纳个人所得税。李阿姨的月工资是5000元，应缴纳所得税（）元。",
+        "options": [
+          "A. $5000 \\times 3\\%$",
+          "B. $3500 \\times 3\\%$",
+          "C. $(5000 - 3500) \\times 3\\%$"
+        ],
+        "answer": "C",
+        "solution": "【分析】应纳税额 $=$ 应纳税部分 $\\times$ 税率，计算出李阿姨的应纳税部分，把应纳税部分和税率代入公式即可求得应纳税额，据此解答。掌握应纳税额的计算方法是解答题目的关键。\n\n【详解】$(5000 - 3500) \\times 3\\% \\\\ = 1500 \\times 3\\% \\\\ = 45 (元)$\n\n故答案为: C",
+        "knowledge_points": [
+          "税率问题"
+        ],
+        "source": "21-22六年级下·全国·单元测试",
+        "merged_id": "unit2_121",
+        "original_id": "unit2_choice_121"
+      },
+      {
+        "id": "unit2_choice_122",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "2022年元旦，商场举办优惠大酬宾：每满400元立减100元。一台烤箱标价500元，相当于这件商品打（）折。",
+        "options": [
+          "A. 二",
+          "B. 八",
+          "C. 七五",
+          "D. 二五"
+        ],
+        "answer": "B",
+        "solution": "【分析】根据优惠活动计算出这台烤箱的现价，这件商品的折扣=这件商品的现价÷这件商品的原价，据此解答。掌握折扣的计算方法是解答题目的关键。\n\n【详解】$(500 - 100) \\div 500 \\\\ = 400 \\div 500 \\\\ = 0.8 \\\\ = 80\\%$\n\n所以，相当于这件商品打八折。故答案为: B",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "21-22六年级上·福建宁德·期末",
+        "merged_id": "unit2_122",
+        "original_id": "unit2_choice_122"
+      },
+      {
+        "id": "unit2_choice_123",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "一套沙发原价4500元，现打八折出售，如果购买这样一套沙发能省多少元（）。",
+        "options": [
+          "A. 1000",
+          "B. 900",
+          "C. 800"
+        ],
+        "answer": "B",
+        "solution": "【分析】现价 $=$ 原价 $\\times$ 折扣，据此求出现价，再用原价减去现价即可。明确现价、原价和折扣之间的关系是解答本题的关键。\n\n【详解】$4500 - 4500 \\times 80\\% \\\\ = 4500 - 3600 \\\\ = 900 (元)$\n\n故答案为: B",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "21-22六年级下·全国·假期作业",
+        "merged_id": "unit2_123",
+        "original_id": "unit2_choice_123"
+      },
+      {
+        "id": "unit2_fill_124",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "我国目前购买摩托车需要缴纳 $10\\%$ 的购置税，周先生花了5500元购买了一辆摩托车，需要缴纳( )元购置税。",
+        "options": [],
+        "answer": "550",
+        "solution": "【分析】用购买摩托车的钱数乘税率，即可求出购置税的钱数。明确购置税等于购买摩托车的钱数乘税率是解题的关键。\n\n【详解】$5500 \\times 10\\% = 550$ （元）\n\n所以，需要缴纳550元购置税。",
+        "knowledge_points": [
+          "含百分数的运算",
+          "税率问题"
+        ],
+        "source": "21-22六年级上·贵州毕节·期末",
+        "merged_id": "unit2_124",
+        "original_id": "unit2_fill_124"
+      },
+      {
+        "id": "unit2_fill_125",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "2018年11月，张阿姨把5万元存入银行，定期3年，年利率 $4.25\\%$，今年到期张阿姨可以获得本息( )元。",
+        "options": [],
+        "answer": "56375",
+        "solution": "【分析】根据利息 $=$ 本金 $\\times$ 利率 $\\times$ 时间，即可算出利息，最后求本息一共多少元，再加上原来的本金，即可解题。熟练掌握利息的求法是解题的关键。\n\n【详解】由分析可得：\n\n$50000 \\times 4.25\\% \\times 3 + 50000 \\\\ = 2125 \\times 3 + 50000 \\\\ = 6375 + 50000 \\\\ = 56375 (元)$\n\n则今年到期张阿姨可以获得本息56375元。",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "21-22六年级上·河南洛阳·期末",
+        "merged_id": "unit2_125",
+        "original_id": "unit2_fill_125"
+      },
+      {
+        "id": "unit2_fill_126",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "某电器商场举行促销活动，一台空调打八五折出售，比原价便宜了570元，这台空调原价是( )元。",
+        "options": [],
+        "answer": "3800",
+        "solution": "【分析】八五折是指现价是原价的 $85\\%$，把原价看作是单位'1'，现价就便宜了原价的 $(1 - 85\\%)$，它对应的数量是570元，由此用除法求出原价。解答本题的关键是理解打折的含义：打几折，现价就是原价的百分之几十；打几几折，现价就是原价的百分之几十。\n\n【详解】$570 \\div (1 - 85\\%) \\\\ = 570 \\div 15\\% \\\\ = 3800 (元)$\n\n所以，这台空调的原价是3800元。",
+        "knowledge_points": [
+          "折扣问题",
+          "已知一个数的百分之几是多少，求这个数"
+        ],
+        "source": "20-21六年级上·河南新乡·期末",
+        "merged_id": "unit2_126",
+        "original_id": "unit2_fill_126"
+      },
+      {
+        "id": "unit2_fill_127",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "超市里的篮球原价80元，现七五折出售，比原价便宜了( )元。",
+        "options": [],
+        "answer": "20",
+        "solution": "【分析】七五折相当于 $75\\%$，用原价 $\\times$ 折扣=现价，代入数据求出现价，再用原价减去现价，即是现价比原价便宜的钱数。此题的解题关键是理解折扣的含义，根据原价、折扣、现价三者之间的关系，解决问题。\n\n【详解】$80 - 80 \\times 75\\% \\\\ = 80 - 60 \\\\ = 20 (元)$\n\n即比原价便宜了20元。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "21-22六年级上·河南商丘·期末",
+        "merged_id": "unit2_127",
+        "original_id": "unit2_fill_127"
+      },
+      {
+        "id": "unit2_fill_128",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "商店的某种玩具打八折促销后，若要回到原价，需提价____。",
+        "options": [],
+        "answer": "$25\\%$",
+        "solution": "【分析】把商品原价看作单位'1'，现价=商品的原价×折扣，再把现价看作单位'1'，商品价格比现价提高的百分率 $=$ （原价一现价）÷现价×100%，据此解答。$80\\%$ 的单位'1'是商品的原价，商品回到原价需要提价的百分率的单位'1'是商品打完八折之后的价格，找准单位'1'是解答题目的关键。\n\n【详解】假设商品原价为1。八折 $= 80\\%$\n\n现价：$1 \\times 80\\% = 0.8$\n\n$(1 - 0.8) \\div 0.8 \\times 100\\% \\\\ = 0.2 \\div 0.8 \\times 100\\% \\\\ = 0.25 \\times 100\\% \\\\ = 25\\%$\n\n所以，需提价 $25\\%$",
+        "knowledge_points": [
+          "经济问题",
+          "求一个数比另一个数多/少百分之几",
+          "折扣问题",
+          "单位'1'的认识与确定"
+        ],
+        "source": "2022·重庆·小升初真题",
+        "merged_id": "unit2_128",
+        "original_id": "unit2_fill_128"
+      },
+      {
+        "id": "unit2_fill_129",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "李阿姨在银行存入10万元，存期为三年，年利率为 $2.75\\%$，到期后李阿姨可以从银行拿到利息( )元。",
+        "options": [],
+        "answer": "8250",
+        "solution": "【分析】根据'利息 $=$ 本金 $\\times$ 利率 $\\times$ 存期'，代入数据计算即可。本题考查利率问题，掌握利息的计算方法是解题的关键。\n\n【详解】10万元 $= 100000$ 元\n\n$100000 \\times 2.75\\% \\times 3 \\\\ = 100000 \\times 0.0275 \\times 3 \\\\ = 2750 \\times 3 \\\\ = 8250 (元)$\n\n到期后李阿姨可以从银行拿到利息8250元。",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "2021·浙江台州·小升初真题",
+        "merged_id": "unit2_129",
+        "original_id": "unit2_fill_129"
+      },
+      {
+        "id": "unit2_fill_130",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "一张飞机票的票面价是a元，打八折后是( )元，加上50元机场建设费和40元燃料附加费后一共是( )元。",
+        "options": [],
+        "answer": "$0.8a$ ; $0.8a + 90$",
+        "solution": "【分析】把飞机票的票面价看作单位'1'，实际价格占原价的 $80\\%$，实际价格 $=$ 票面价格 $\\times 80\\%$，一共的费用 $=$ 飞机票的实际价格+机场建设费+燃料附加费，据此解答。折扣表示现价占原价的十分之几，也就是百分之几十。\n\n【详解】八折 $= 80\\%$\n\n$a \\times 80\\% = 0.8a$ (元)\n\n$0.8a + 50 + 40 \\\\ = 0.8a + (50 + 40) \\\\ = 0.8a + 90$ (元)",
+        "knowledge_points": [
+          "用字母表示数",
+          "折扣问题",
+          "含有字母式子的化简与求值"
+        ],
+        "source": "21-22六年级下·浙江湖州·期末",
+        "merged_id": "unit2_130",
+        "original_id": "unit2_fill_130"
+      },
+      {
+        "id": "unit2_fill_131",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "我国《购车税法》规定：汽车购置税税率为 $10\\%$。一辆汽车裸车价a元，买这辆汽车要付( )元的购置税。张叔叔付16.5万元买了一辆这样的车，这辆车的裸车价是( )万元。",
+        "options": [],
+        "answer": "0.1a ; 15",
+        "solution": "【分析】购置税占裸车价的 $10\\%$，则汽车的购置税就是 $(10\\% \\times a)$ 元；买一辆这样的车付了16.5万元，是裸车价的 $(1 + 10\\%)$，求裸车价，用除法计算。此题主要考查了百分数的意义和用字母表示数的方法，要熟练掌握。\n\n【详解】$10\\% \\times a = 0.1a$ （元）\n\n$16.5 \\div (1 + 10\\%) \\\\ = 16.5 \\div 1.1 \\\\ = 15 (万元)$\n\n所以，我国《购车税法》规定：汽车购置税税率为 $10\\%$。一辆汽车裸车价a元，买这辆汽车要付0.1a元的购置税。张叔叔付16.5万元买了一辆这样的车，这辆车的裸车价是15万元。",
+        "knowledge_points": [
+          "用字母表示数",
+          "税率问题",
+          "含有字母式子的化简与求值"
+        ],
+        "source": "2022·重庆·小升初真题",
+        "merged_id": "unit2_131",
+        "original_id": "unit2_fill_131"
+      },
+      {
+        "id": "unit2_fill_132",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "依法纳税是每个公民应尽的义务，李叔叔得到一笔8500元的劳务费。这笔费用中800元是免税的，其余部分要按 $20\\%$ 的税率缴税。那么这笔劳务费一共要缴税( )元，李叔叔税后实际获得( )元。",
+        "options": [],
+        "answer": "1540 ; 6960",
+        "solution": "【分析】李叔叔的应纳税部分是劳务费去掉800元后的剩余部分，根据'应纳税额 $=$ 应纳税部分 $\\times$ 税率'求出李叔叔需要缴纳的税款，税后实际获得的钱数 $=$ 劳务费一应纳税额，据此解答。掌握应纳税额的计算方法是解答题目的关键。\n\n【详解】$(8500 - 800) \\times 20\\% \\\\ = 7700 \\times 20\\% \\\\ = 1540 (元)$\n\n$8500 - 1540 = 6960 (元)$",
+        "knowledge_points": [
+          "税率问题"
+        ],
+        "source": "21-22六年级下·山东济南·期末",
+        "merged_id": "unit2_132",
+        "original_id": "unit2_fill_132"
+      },
+      {
+        "id": "unit2_fill_133",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "阿姨把5000元存到银行，定期两年，如果年利率是 $2\\%$，到期可以取得本金和利息一共( )元。",
+        "options": [],
+        "answer": "5200",
+        "solution": "【分析】利息 $=$ 本金 $\\times$ 利率 $\\times$ 存期，本金 $+$ 利息即可。到期取款时银行多支付的钱叫利息。\n\n【详解】$5000 + 5000 \\times 2\\% \\times 2 \\\\ = 5000 + 200 \\\\ = 5200 (元)$",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "2022·重庆大渡口·小升初真题",
+        "merged_id": "unit2_133",
+        "original_id": "unit2_fill_133"
+      },  {
+        "id": "unit2_fill_134",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "王叔叔的果园去年产量是3.6吨，今年比去年增产二成五。今年的产量是( )吨。",
+        "options": [],
+        "answer": "4.5",
+        "solution": "【分析】根据题意，今年比去年增产二成五，把去年的产量看作单位'1'，那么今年的产量是去年的（$1 + 25\\%$），单位'1'已知，用去年的产量乘 $(1 + 25\\%)$，即可求出今年的产量。本题考查成数问题，几成几就是百分之几十几；明确求比一个数多或少百分之几的数是多少，用乘法计算。\n\n【详解】二成五 $= 25\\%$\n\n$3.6 \\times (1 + 25\\%) \\\\ = 3.6 \\times 1.25 \\\\ = 4.5 (吨)$\n\n今年的产量是4.5吨。",
+        "knowledge_points": [
+          "比一个数多/少百分之几的数是多少",
+          "成数问题"
+        ],
+        "source": "21-22六年级下·山东菏泽·期末",
+        "merged_id": "unit2_134",
+        "original_id": "unit2_fill_134"
+      },
+      {
+        "id": "unit2_fill_135",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "一件衬衫降 $10\\%$ 后，售价是135元。这件衬衫( )折出售，原来的售价是( )元。",
+        "options": [],
+        "answer": "九 ; 150",
+        "solution": "【分析】一件衣服降价 $10\\%$ 后，即是原价的 $90\\%$，相当于打九折，售价是135元，把一件衣服的原价看作单位'1'，用135除以 $(1 - 10\\%) = 90\\%$ 就是原来的售价。本题关键找准单位'1'，单位'1'不知道用除法进行解答即可。\n\n【详解】$1 - 10\\% = 90\\% =$ 九折\n\n$135 \\div (1 - 10\\%) \\\\ = 135 \\div 0.9 \\\\ = 150 (元)$",
+        "knowledge_points": [
+          "折扣问题",
+          "已知比一个数多/少百分之几是多少，求这个数"
+        ],
+        "source": "21-22六年级下·浙江温州·期末",
+        "merged_id": "unit2_135",
+        "original_id": "unit2_fill_135"
+      },
+      {
+        "id": "unit2_fill_136",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "$\\frac{40}{()}$ =80%=( )成。",
+        "options": [],
+        "answer": "50 ; 八",
+        "solution": "【分析】将 $80\\%$ 化为分数是 $\\frac{4}{5}$，根据分数的基本性质，分子和分母同时乘10，则 $\\frac{4 \\times 10}{5 \\times 10} = \\frac{40}{50}$；百分之几十表示几成，所以 $80\\%$ 表示八成。本题考查了分数和百分数的互化、分数的基本性质以及成数的意义。\n\n【详解】$\\frac{40}{50} = 80\\% =$ 八成。",
+        "knowledge_points": [
+          "百分数、小数和分数的互化",
+          "分数的基本性质及应用",
+          "成数问题"
+        ],
+        "source": "2022·山东济南·小升初真题",
+        "merged_id": "unit2_136",
+        "original_id": "unit2_fill_136"
+      },
+      {
+        "id": "unit2_fill_137",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "张叔叔买了五年期的国家建设债券20000元，年利率为 $3.81\\%$。到期后，用利息购买一辆586元的自行车后，利息还剩( )元。",
+        "options": [],
+        "answer": "3224",
+        "solution": "【分析】根据利息 $=$ 本金 $\\times$ 利率 $\\times$ 存期，据此求出利息然后再减去自行车的价钱即可解答。本题考查利率问题，明确利息 $=$ 本金 $\\times$ 利率 $\\times$ 存期是解题的关键。\n\n【详解】$20000 \\times 3.81\\% \\times 5 - 586 = 3810 - 586 = 3224$ （元）",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "21-22六年级下·山西吕梁·期末",
+        "merged_id": "unit2_137",
+        "original_id": "unit2_fill_137"
+      },
+      {
+        "id": "unit2_fill_138",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "一件衣服打八折后售价是80元，则这件衣服的原价是( )元。",
+        "options": [],
+        "answer": "100",
+        "solution": "【分析】根据题目求原价，使用公式：原价 $=$ 现价 $\\div$ 折扣，代入已知条件的数据求出结果即可。此题考查的是百分数的应用题，关键是熟记公式再计算。\n\n【详解】$80 \\div 80\\% = 100$ （元）",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "19-20五年级下·辽宁·单元测试",
+        "merged_id": "unit2_138",
+        "original_id": "unit2_fill_138"
+      },
+      {
+        "id": "unit2_fill_139",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "一种商品打'八五折'出售，也就是把这种商品优惠了____$\\%$。",
+        "options": [],
+        "answer": "15",
+        "solution": "【分析】把原价看作单位'1'，打八五折出售，也就是现价是原价的 $85\\%$，优惠部分就是现价比原价少的百分率，即优惠了 $1 - 85\\% = 15\\%$，据此解答即可。\n\n【详解】八五折 $= 85\\%$\n\n$1 - 85\\% = 15\\%$\n\n也就是这种商品优惠了 $15\\%$\n\n此题考查的目的是理解百分率与'折'数的联系及应用。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "2022·河南省直辖县级单位·小升初真题",
+        "merged_id": "unit2_139",
+        "original_id": "unit2_fill_139"
+      },
+      {
+        "id": "unit2_fill_140",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "张叔叔上个月获得一笔3000元的稿费，按照相关规定：每次收入不超过4000元的，减掉费用800元，剩余的金额需要按 $14\\%$ 的税率交纳个人所得税，张叔叔应交纳个人所得税( )元。",
+        "options": [],
+        "answer": "308",
+        "solution": "【分析】根据题意，先用收入3000元减去800元得到剩余的金额，再乘 $14\\%$ 的税率，即是应交纳的个人所得税金额。掌握'各种收入 $\\times$ 税率 $=$ 应纳税额'，关键是理解'剩余的金额需要按 $14\\%$ 的税率交纳个人所得税'中的'剩余金额'。\n\n【详解】$(3000 - 800) \\times 14\\% \\\\ = 2200 \\times 0.14 \\\\ = 308 (元)$",
+        "knowledge_points": [
+          "求一个数的百分之几是多少",
+          "税率问题"
+        ],
+        "source": "2022·广东揭阳·小升初真题",
+        "merged_id": "unit2_140",
+        "original_id": "unit2_fill_140"
+      },
+      {
+        "id": "unit2_fill_141",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "把一个棱长为4分米的正方体，削成一个最大的圆柱体，圆柱体的体积是( )。",
+        "options": [],
+        "answer": "50.24立方分米",
+        "solution": "【分析】把一个正方体削成一个最大的圆柱，那么圆柱的底面直径、高都等于正方体的棱长；根据圆柱的体积公式 $V = \\pi r^{2} h$，代入数据计算即可。掌握圆柱的体积计算公式，明确正方体削成一个最大的圆柱体，圆柱的底面直径、高与正方体棱长的关系是解题的关键。\n\n【详解】$3.14 \\times (4 \\div 2)^{2} \\times 4 \\\\ = 3.14 \\times 4 \\times 4 \\\\ = 3.14 \\times 16 \\\\ = 50.24 (立方分米)$",
+        "knowledge_points": [
+          "圆柱的体积",
+          "正方体的特征",
+          "圆柱的认识及特征"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit2_141",
+        "original_id": "unit2_fill_141"
+      },
+      {
+        "id": "unit2_fill_142",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "( )÷20= $\\frac{3}{5}$ =30÷( )=( )%=( )折。",
+        "options": [],
+        "answer": "12 ; 50 ; 60 ; 六",
+        "solution": "【分析】根据分数与除法的关系，以及它们通用的基本性质进行填空，将分数化成小数，再将小数化成百分数，根据几折就是百分之几十确定折数。分数的分子相当于被除数，分母相当于除数。\n\n【详解】$20 \\div 5 \\times 3 = 12$；$30 \\div 3 \\times 5 = 50$；$3 \\div 5 = 0.6 = 60\\% =$ 六折。",
+        "knowledge_points": [
+          "分数与除法的关系",
+          "百分数、小数和分数的互化",
+          "分数的基本性质及应用",
+          "折扣问题"
+        ],
+        "source": "18-19六年级下·吉林长春·期中",
+        "merged_id": "unit2_142",
+        "original_id": "unit2_fill_142"
+      },
+      {
+        "id": "unit2_fill_143",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "一件儿童服装原价200元，打九折后现价是( )元，现价比原价便宜( )元。",
+        "options": [],
+        "answer": "180 ; 20",
+        "solution": "【分析】一件儿童服装原价200元，打九折即按原价的 $90\\%$ 出售，根据分数乘法的意义，用原价乘现价占原价的分率，即得现价是多少，然后用原价减现价，即得比原价便宜多少钱。在商品销售中，打几折即得按原价的百分之几十出售。\n\n【详解】$200 \\times 90\\% = 180$ （元）\n\n$200 - 180 = 20$ （元）",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "2021·山东临沂·小升初真题",
+        "merged_id": "unit2_143",
+        "original_id": "unit2_fill_143"
+      },
+      {
+        "id": "unit2_fill_144",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "某电器产品，每件若以原价的九五折销售，可获利160元；若以原价的七五折销售，则亏损60元。这种电器每件的原价为( )元。",
+        "options": [],
+        "answer": "1100",
+        "solution": "【分析】假设原价为 $x$ 元，原价的九五折即 $95\\% x$ 元，减去获利的钱，等于成本价；同样，原价的七五折即 $75\\% x$ 元，加上亏损的钱，也等于成本价；可列出方程，求出结果。所以这种电器每件的原价为1100元。此题的解题关键是弄清题意，根据成本价不变，把每件的原价设为未知数 $x$，找出题中数量间的相等关系，列出包含 $x$ 的等式，解方程得到最终的结果。\n\n【详解】解：设原价为 $x$ 元，列方程：\n\n$95\\% x - 160 = 75\\% x + 60$\n\n$95\\% x - 75\\% x = 160 + 60$\n\n$0.2x = 220$\n\n$x = 1100$",
+        "knowledge_points": [
+          "经济问题",
+          "含百分数的运算",
+          "列方程解含一个未知数的问题"
+        ],
+        "source": "21-22六年级下·全国·单元测试",
+        "merged_id": "unit2_144",
+        "original_id": "unit2_fill_144"
+      },
+      {
+        "id": "unit2_fill_145",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "新世纪商场六一儿童用品八五折优惠，说明现价是原价的( )%，优惠了( )%，原价162元的书包，便宜了( )元。",
+        "options": [],
+        "answer": "85 ; 15 ; 24.3",
+        "solution": "【分析】把原价看作单位'1'，一折就是 $10\\%$，那么八五折就是现价是原价的 $85\\%$，由此可知：现价比原价少（$1 - 85\\%$）。用原价乘 $(1 - 85\\%)$，就是便宜的价钱。此题的解题关键是理解掌握'折数'与百分数之间联系及应用。\n\n【详解】八五折 $= 85\\%$\n\n$1 - 85\\% = 15\\%$\n\n$162 \\times (1 - 85\\%) \\\\ = 162 \\times 15\\% \\\\ = 24.3 (元)$",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "21-22六年级下·全国·单元测试",
+        "merged_id": "unit2_145",
+        "original_id": "unit2_fill_145"
+      },
+      {
+        "id": "unit2_fill_146",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "今年稻谷的产量是去年的 $120\\%$，今年比去年增产( )成。",
+        "options": [],
+        "answer": "二或两",
+        "solution": "【分析】把去年的产量看成单位'1'，今年的产量是去年的 $120\\%$，用 $120\\%$ 减去1，就是今年比去年增加的产量。此题的解题关键是找准本题中的单位'1'是去年的产量，直接用减法求解即可。\n\n【详解】$120\\% - 1 = 20\\%$，$20\\%$ 等于二成。",
+        "knowledge_points": [
+          "成数问题"
+        ],
+        "source": "2014六年级·全国·课后作业",
+        "merged_id": "unit2_146",
+        "original_id": "unit2_fill_146"
+      },
+      {
+        "id": "unit2_fill_147",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "亮亮把300元钱存入银行2年，按年利率 $2.15\\%$ 计算，到期时他可得到本金和利息共( )元。",
+        "options": [],
+        "answer": "312.9",
+        "solution": "【分析】通过利息的计算公式：利息 $=$ 本金 $\\times$ 利率 $\\times$ 存期，求出利息，再加上本金，即可得解。此题的解题关键是利用计算利息的公式来求解，注意本息指的是本金和利息。\n\n【详解】$300 + 300 \\times 2.15\\% \\times 2 \\\\ = 300 + 6.45 \\times 2 \\\\ = 300 + 12.9 \\\\ = 312.9 (元)$",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "21-22六年级下·全国·单元测试",
+        "merged_id": "unit2_147",
+        "original_id": "unit2_fill_147"
+      },
+      {
+        "id": "unit2_fill_148",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "$0.25 = 3:($ )=( )$\\%$ =( )折=( )。(用成数表示)",
+        "options": [],
+        "answer": "12 ; 25 ; 二五 ; 二成五",
+        "solution": "【分析】把0.25化成分数形式，然后根据小数、分数、比、百分数、折扣和成数之间的关系进行解答。本题考查小数、分数、比、百分数、折扣和成数之间的关系，明确它们的关系是解题的关键。\n\n【详解】由分析可知：$0.25 = \\frac{1}{4} = 3:12$，$0.25 = 25\\% =$ 二五折=二成五\n\n$0.25 = 3:(12) = (25)\\% = ($ 二五$)$ 折=($二成五$)。(用成数表示)",
+        "knowledge_points": [
+          "百分数、分数、小数和比的互化",
+          "百分数、小数和分数的互化",
+          "折扣问题",
+          "成数问题"
+        ],
+        "source": "19-20六年级下·湖南怀化·单元测试",
+        "merged_id": "unit2_148",
+        "original_id": "unit2_fill_148"
+      },
+      {
+        "id": "unit2_fill_149",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "一成就是( )，五成五就是( )。",
+        "options": [],
+        "answer": "$10\\%$ ; $55\\%$",
+        "solution": "【分析】成数表示一个数是另一个数的百分之几十的数，相当于百分数，可得正确结果。此题的解题关键是理解成数的概念，可得出正确的结果。\n\n【详解】根据分析，一成 $= \\frac{10}{100} = 10\\%$，不能说成 $1\\%$，五成五 $= \\frac{55}{100} = 55\\%$",
+        "knowledge_points": [
+          "成数问题"
+        ],
+        "source": "19-20六年级下·湖南怀化·单元测试",
+        "merged_id": "unit2_149",
+        "original_id": "unit2_fill_149"
+      },
+      {
+        "id": "unit2_fill_150",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "某商场的电视机先把进价提价三成，然后再打八折优惠出售，结果每台获利200元，每台电视机进价要( )元。",
+        "options": [],
+        "answer": "5000",
+        "solution": "【分析】把商场电视机的进价看作单位'1'，售价占进价的 $(1 + 30\\%) \\times 80\\%$，根据'量÷对应的百分率'求出每台电视机的进价，据此解答。计算出售价占进价的百分率是解答题目的关键。\n\n【详解】$200 \\div [(1 + 30\\%) \\times 80\\% - 1] \\\\ = 200 \\div [1.3 \\times 80\\% - 1] \\\\ = 200 \\div [1.04 - 1] \\\\ = 200 \\div 0.04 \\\\ = 5000 (元)$\n\n所以，每台电视机进价要5000元。",
+        "knowledge_points": [
+          "经济问题",
+          "折扣问题",
+          "成数问题"
+        ],
+        "source": "18-19六年级下·广东汕头·单元测试",
+        "merged_id": "unit2_150",
+        "original_id": "unit2_fill_150"
+      },
+      {
+        "id": "unit2_fill_151",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "一件商品降价出售，若打九折，仍可盈利180元，若打八折，就会亏损240元。这种商品原价( )元。",
+        "options": [],
+        "answer": "4200",
+        "solution": "【分析】设原价为 $x$ 元，若打九折，仍可盈利180元，说明原价的 $90\\%$ 减去180元等于成本，若打八折，就会亏损240元，说明原价的 $80\\%$ 加上240元等于成本，据此列出方程解答即可。本题考查折扣问题，解答本题的关键是掌握题中的等量关系以及解方程的计算方法。\n\n【详解】解：设原价为 $x$\n\n$90\\% x - 180 = 80\\% x + 240$\n\n$0.9x - 0.8x = 240 + 180$\n\n$0.1x = 420$\n\n$x = 4200$\n\n所以这种商品原价4200元。",
+        "knowledge_points": [
+          "折扣问题",
+          "列方程解含一个未知数的问题"
+        ],
+        "source": "18-19六年级下·广东汕头·单元测试",
+        "merged_id": "unit2_151",
+        "original_id": "unit2_fill_151"
+      },
+      {
+        "id": "unit2_fill_152",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "只列式不计算。李杨把1200元存入银行，存期半年，月利率 $0.15\\%$，到期后利息一共多少元( )？",
+        "options": [],
+        "answer": "$1200 \\times 0.15\\% \\times 6$",
+        "solution": "【分析】利息 $=$ 本金 $\\times$ 利率 $\\times$ 存期，据此解答即可。本题考查利率问题，解答本题的关键是掌握求利息的计算公式。\n\n【详解】由分析可得，列式如下：$1200 \\times 0.15\\% \\times 6$",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "18-19六年级下·广东汕头·单元测试",
+        "merged_id": "unit2_152",
+        "original_id": "unit2_fill_152"
+      },
+      {
+        "id": "unit2_fill_153",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "'买四送一'相当于打( )折出售。'买三送二'相当于打( )折出售。",
+        "options": [],
+        "answer": "八 ; 六",
+        "solution": "【分析】买四送一相当于买5个只需要4个的钱，买三送二相当于买5个只需要3个的钱，据此解答即可。本题考查折扣问题，解答本题的关键是掌握折扣问题的计算方法。\n\n【详解】$4 \\div (4 + 1) = 4 \\div 5 = 80\\% =$ 八折\n\n'买四送一'相当于打八折。\n\n$3 \\div (3 + 2) = 3 \\div 5 = 60\\% =$ 六折\n\n'买三送二'相当于打六折。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "18-19六年级下·广东汕头·单元测试",
+        "merged_id": "unit2_153",
+        "original_id": "unit2_fill_153"
+      },
+      {
+        "id": "unit2_fill_154",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "一件上衣280元打七五折后是( )元；一条裤子打六五折卖91元，这条裤子原来( )元。",
+        "options": [],
+        "answer": "210 ; 140",
+        "solution": "【分析】打几折表示现价是原价的百分之几，据此解答即可。本题考查折扣问题，解答本题的关键是掌握折扣问题中的数量关系。\n\n【详解】$280 \\times 75\\% = 210$ （元）\n\n$91 \\div 65\\% = 140$ （元）",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "18-19六年级下·广东汕头·单元测试",
+        "merged_id": "unit2_154",
+        "original_id": "unit2_fill_154"
+      },
+      {
+        "id": "unit2_fill_155",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "某商场二月份比一月份的营业额少60万元，一月份营业额是二月份的 $120\\%$，二月份营业额是( )万元，按营业额的 $5\\%$ 缴纳营业税，二月份应交营业税( )万元。",
+        "options": [],
+        "answer": "300 ; 15",
+        "solution": "【分析】（1）把这个商场二月份的营业额看作单位'1'，一月份营业额占二月份的 $120\\%$，根据'量÷对应的百分率'计算出二月份的营业额；(2)二月份应交营业税 $=$ 二月份的营业额 $\\times$ 税率，据此解答。根据标准量的计算方法求出二月份的营业额是解答题目的关键。\n\n【详解】(1) $60 \\div (120\\% - 1) \\\\ = 60 \\div 0.2 \\\\ = 300 (万元)$\n\n(2) $300 \\times 5\\% = 15$ （万元）",
+        "knowledge_points": [
+          "已知一个数的百分之几是多少，求这个数",
+          "税率问题"
+        ],
+        "source": "21-22六年级下·全国·单元测试",
+        "merged_id": "unit2_155",
+        "original_id": "unit2_fill_155"
+      },
+      {
+        "id": "unit2_fill_156",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "某商场9月份的营业额是40万元，如果按营业额的 $5\\%$ 缴纳营业税，这个商场9月份应纳税( )万元。",
+        "options": [],
+        "answer": "2",
+        "solution": "【分析】这个商场9月份应纳税额 $=$ 商场9月份的营业额 $\\times$ 税率，据此解答。掌握应纳税额的计算方法是解答题目的关键。\n\n【详解】$40 \\times 5\\% = 2$ （万元）\n\n所以，这个商场9月份应纳税2万元。",
+        "knowledge_points": [
+          "税率问题"
+        ],
+        "source": "21-22六年级下·全国·单元测试",
+        "merged_id": "unit2_156",
+        "original_id": "unit2_fill_156"
+      },
+      {
+        "id": "unit2_fill_157",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "超市里一种羽绒背心打八折出售，现在每件48元，这里把( )看作单位'1'，现在每件售价是( )的( )，原价是( )元。",
+        "options": [],
+        "answer": "原价 ; 原价 ; $80\\%$ ; 60",
+        "solution": "【分析】羽绒背心打八折出售，把原价看作单位'1'，羽绒背心现价是原价的 $80\\%$，现价 $=$ 原价 $\\times 80\\%$，则原价 $=$ 现价 $\\div 80\\%$，据此解答。掌握折扣的意义是解答题目的关键。\n\n【详解】原价：$48 \\div 80\\% = 60$ （元）\n\n超市里一种羽绒背心打八折出售，现在每件48元，这里把（原价）看作单位'1'，现在每件售价是（原价）的（$80\\%$），原价是（60）元。",
+        "knowledge_points": [
+          "折扣问题",
+          "已知一个数的百分之几是多少，求这个数"
+        ],
+        "source": "2022六年级上·全国·专题练习",
+        "merged_id": "unit2_157",
+        "original_id": "unit2_fill_157"
+      },
+      {
+        "id": "unit2_fill_158",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "某超市上个月的营业额是400万元，按 $5\\%$ 的税率缴纳营业税，这个超市上个月应缴纳营业税( )万元。",
+        "options": [],
+        "answer": "20",
+        "solution": "【分析】将营业额看作单位'1'，直接用营业额×税率即可。应纳税额与各种收入，比如销售额、营业额等的比率叫做税率。\n\n【详解】$400 \\times 5\\% = 20$ （万元）",
+        "knowledge_points": [
+          "税率问题"
+        ],
+        "source": "2022六年级上·全国·专题练习",
+        "merged_id": "unit2_158",
+        "original_id": "unit2_fill_158"
+      },
+      {
+        "id": "unit2_judge_159",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "一件上衣打七五折，是指这件上衣比原价便宜 $75\\%$。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】打七五折是指现价是原价的 $75\\%$，把原价看成单位'1'，现价比原价便宜的价格就是（$1 - 75\\%$），据此判断即可。理解折扣的含义，打几折是指现价是原价的百分之几十。\n\n【详解】$1 - 75\\% = 25\\%$\n\n所以，一件上衣打七五折，是指这件上衣比原价便宜 $25\\%$\n\n故答案为: ×",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "20-21六年级下·辽宁盘锦·期末",
+        "merged_id": "unit2_159",
+        "original_id": "unit2_judge_159"
+      },
+      {
+        "id": "unit2_judge_160",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "税收分为营业税和消费税两种。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】税收主要分为消费税、增值税、营业税和个人所得税等。此题考查了数学常识，要熟练掌握。\n\n【详解】由分析可得：税收主要分为消费税、增值税、营业税和个人所得税等，不止两种，故原题说法错误；\n\n故答案为：×",
+        "knowledge_points": [
+          "求应纳税额"
+        ],
+        "source": "2022·贵州黔西·小升初真题",
+        "merged_id": "unit2_160",
+        "original_id": "unit2_judge_160"
+      },
+      {
+        "id": "unit2_judge_161",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "'二成五'是十分之二点五，写成百分数是 $25\\%$（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】成数，表示一个数是另一个数的十分之几的数，也可表示一个数是另一个数的百分之几十的数，相当于百分数。例：一成就是 $10\\%$，三成五就是 $35\\%$。此题的解题关键是理解掌握成数的意义。\n\n【详解】根据分析得，二成五 $= 25\\% = 0.25 = \\frac{25}{100} = \\frac{2.5}{10}$，即'二成五'是十分之二点五，写成百分数是 $25\\%$。原说法是正确的。\n\n故答案为: √",
+        "knowledge_points": [
+          "成数问题"
+        ],
+        "source": "21-22六年级下·湖南湘潭·期末",
+        "merged_id": "unit2_161",
+        "original_id": "unit2_judge_161"
+      },
+      {
+        "id": "unit2_judge_162",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "判断题",
+        "difficulty": "困难",
+        "question": "一件衣服定价57元，售出后可获利 $50\\%$，如果按定价的七折出售，可获利1.9元。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】根据题意，定价57元售出后可获利 $50\\%$，即售价比进价高 $50\\%$，把进价看作单位'1'，售价是进价的（1+50%），单位'1'未知，用售价除以 $(1 + 50\\%)$ 求出进价；如果按定价的七折出售，即售价是定价的 $70\\%$，用定价乘70%求出售价，再与进价相减，求出获利，据此判断。掌握进价、售价、利润之间的关系，明确已知比一个数多或少百分之几是多少，求这个数，用除法计算；求一个数的百分之几是多少，用乘法计算。\n\n【详解】进价：\n\n$57 \\div (1 + 50\\%) \\\\ = 57 \\div 1.5 \\\\ = 38 (元)$\n\n七折后的售价：\n\n$57 \\times 70\\% \\\\ = 57 \\times 0.7 \\\\ = 39.9 (元)$\n\n获利：$39.9 - 38 = 1.9$ （元）\n\n故答案为: √",
+        "knowledge_points": [
+          "求一个数的百分之几是多少",
+          "折扣问题",
+          "已知比一个数多/少百分之几是多少，求这个数"
+        ],
+        "source": "21-22六年级下·河北张家口·期中",
+        "merged_id": "unit2_162",
+        "original_id": "unit2_judge_162"
+      },
+      {
+        "id": "unit2_judge_163",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "某商品先打七折再提价 $30\\%$，结果与原价相同。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】把商品的原价看作单位'1'，先打七折，即打折后的价格是原价的70%，用原价×70%；再提价30%，即现价是打折后价格的（1+30%），用打折后的价格×（1+30%）求出现价；最后把现价与原价相比较，得出结论。掌握原价、现价、折扣之间的关系是解题的关键；明确求比一个数多或少百分之几的数是多少，用乘法计算。\n\n【详解】$1 \\times 70\\% \\times (1 + 30\\%) = 0.7 \\times 1.3 = 0.91$\n\n$0.91 < 1$\n\n结果与原价不相同。\n\n故答案为: ×",
+        "knowledge_points": [
+          "折扣问题",
+          "比一个数多/少百分之几的数是多少"
+        ],
+        "source": "21-22六年级下·内蒙古通辽·期中",
+        "merged_id": "unit2_163",
+        "original_id": "unit2_judge_163"
+      },
+      {
+        "id": "unit2_judge_164",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "邮政汇5000元钱，需要交 $1\\%$ 的汇费，汇费是5元。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】根据题意，交 $1\\%$ 的汇费是指汇费占汇款钱数的 $1\\%$，用汇款的钱数乘 $1\\%$，即可求出汇费。理解汇率的含义，明确求一个数的百分之几是多少，用乘法计算。\n\n【详解】$5000 \\times 1\\% = 5000 \\times 0.01 = 50 (元)$\n\n汇费是50元。\n\n故答案为: ×",
+        "knowledge_points": [
+          "求一个数的百分之几是多少",
+          "税率问题"
+        ],
+        "source": "19-20六年级下·湖南娄底·期末",
+        "merged_id": "unit2_164",
+        "original_id": "unit2_judge_164"
+      },
+      {
+        "id": "unit2_judge_165",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "一本书打七五折出售，也就是比原价便宜 $25\\%$。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】几几折就是百分之几十几，所以七五折就是指售价是原价的 $75\\%$，把原价看作单位'1'，售价比原价便宜了（1-75%），据此解答即可。此题主要考查了'折'的意义，解答此题的关键是要明确：几几折就是百分之几十几。\n\n【详解】因为几几折就是百分之几十几，所以七五折表示售价是原价的 $75\\%$。\n\n比原价便宜了：$1 - 75\\% = 25\\%$，原说法正确。\n\n故答案为: √",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "2021·辽宁鞍山·小升初真题",
+        "merged_id": "unit2_165",
+        "original_id": "unit2_judge_165"
+      },
+      {
+        "id": "unit2_judge_166",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "在利率、本金一定的情况下，存期越长，所得的利息就越多。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】根据利息 $=$ 本金 $\\times$ 利率 $\\times$ 时间，知道在本金和利率相同的情况下，存入的时间与利息成正比，由此得出答案。解答此题的关键是根据利息的计算方法，判断利息与本金的关系。\n\n【详解】在利率、本金一定的情况下，存期越长，所得的利息就越多。\n\n故答案为: √",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "21-22六年级下·全国·单元测试",
+        "merged_id": "unit2_166",
+        "original_id": "unit2_judge_166"
+      }, {
+        "id": "unit2_judge_167",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "今年的产量比去年增长了三成，就是今年比去年增长了 $30\\%$。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】通过几成的含义来解答，几成就是百分之几十。此题的解题关键是理解成数的含义。\n\n【详解】今年的产量比去年增长了三成，就是今年比去年增长了 $30\\%$。\n\n故答案为: √",
+        "knowledge_points": [
+          "成数问题"
+        ],
+        "source": "21-22六年级下·全国·单元测试",
+        "merged_id": "unit2_167",
+        "original_id": "unit2_judge_167"
+      },
+      {
+        "id": "unit2_judge_168",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "李大爷家今年小麦的产量比去年增加两成，也就是说今年的产量是去年的 $120\\%$（）。",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】成数：工农业生产经常用'成数'表示生产增长的情况，几成就是十分之几，也可以用百分数表示，几成就是百分之几十。关键是理解成数的意义。\n\n【详解】李大爷家今年小麦的产量比去年增加两成，也就是说今年的产量是去年的 $120\\%$，说法正确。\n\n故答案为: √",
+        "knowledge_points": [
+          "成数问题"
+        ],
+        "source": "2020·江苏·小升初真题",
+        "merged_id": "unit2_168",
+        "original_id": "unit2_judge_168"
+      },
+      {
+        "id": "unit2_judge_169",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "王叔叔存在银行的钱就是本金，银行多支付的钱就是利息。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】把存入银行的钱叫本金，取款时银行多付出来的钱叫利息。\n\n【详解】把存入银行的钱叫本金，取款时银行多付出来的钱叫利息。\n\n故答案为: √",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "19-20六年级下·湖南怀化·单元测试",
+        "merged_id": "unit2_169",
+        "original_id": "unit2_judge_169"
+      },
+      {
+        "id": "unit2_judge_170",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "成数就是表示一个数是另一个数的十分之几。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】农业收成，经常用'成数'来表示。成数表示一个数是另一个数的十分之几，通称'几成'，如：'一成'就是十分之一，'二成'就是十分之二，'三成五'是十分之三点五。\n\n【详解】农业收成，经常用'成数'来表示。成数表示一个数是另一个数的十分之几，通称'几成'，如：'一成'就是十分之一，'二成'就是十分之二，'三成五'是十分之三点五。\n\n故答案为: √",
+        "knowledge_points": [
+          "成数问题"
+        ],
+        "source": "19-20六年级下·湖南怀化·单元测试",
+        "merged_id": "unit2_170",
+        "original_id": "unit2_judge_170"
+      },
+      {
+        "id": "unit2_judge_171",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "几折就是十分之几，百分之几。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】几折就是十分之几，也就是表示百分之几十。\n\n【详解】几折就是十分之几，也就是表示百分之几十。\n\n故答案为: ×",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "19-20六年级下·湖南怀化·单元测试",
+        "merged_id": "unit2_171",
+        "original_id": "unit2_judge_171"
+      },
+      {
+        "id": "unit2_judge_172",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "应纳税额与各种税收的比率叫做税率。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】应纳税额与各种收入（销售额、营业额、应纳税所得额……）的比率叫做税率。\n\n【详解】应纳税额与各种收入（销售额、营业额、应纳税所得额……）的比率叫做税率。\n\n故答案为: ×",
+        "knowledge_points": [
+          "求税率或收入额"
+        ],
+        "source": "19-20六年级下·湖南怀化·单元测试",
+        "merged_id": "unit2_172",
+        "original_id": "unit2_judge_172"
+      },
+      {
+        "id": "unit2_judge_173",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "单位时间内本金与利息的比率叫做利率。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】单位时间（如1年、1月、1日等）内的利息与本金的比率叫做利率，不是单位时间内本金与利息的比率。\n\n【详解】单位时间（如1年、1月、1日等）内的利息与本金的比率叫做利率，不是单位时间内本金与利息的比率。\n\n故答案为：×",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "18-19六年级下·广东汕头·单元测试",
+        "merged_id": "unit2_173",
+        "original_id": "unit2_judge_173"
+      },
+      {
+        "id": "unit2_judge_174",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "'买一送一'就是打五折。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】打折就是在原来售价的基础上降价销售，打五折即现价为原价的 $50\\%$，如果买的东西和送的东西价值不相同就不是打五折。掌握折扣的意义是解答题目的关键。\n\n【详解】数学来源于生活，现实生活中的'买一送一'买和送不一定是一样的东西。\n\n故答案为: ×",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "17-18六年级上·全国·单元测试",
+        "merged_id": "unit2_174",
+        "original_id": "unit2_judge_174"
+      },
+      {
+        "id": "unit2_solve_175",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "只列式，不计算。商店卖一种收音机，打七五折销售后，比原价便宜了24元，原价是多少元？",
+        "options": [],
+        "answer": "$24 \\div (1 - 75\\%)$",
+        "solution": "【分析】把原价看作单位'1'，打七五折销售就是售价是原价的 $75\\%$，那么售价比原价便宜的金额就是原价的（1- $75\\%$），已知售价比原价便宜了24元，根据已知一个数的百分之几是多少，求这个数，用除法，列式为 $24 \\div$ (1- $75\\%$）即可。本题重点考查折扣问题，明确已知一个数的百分之几是多少，求这个数，用除法是解题的关键。\n\n【详解】$24 \\div (1 - 75\\%) \\\\ = 24 \\div 25\\% \\\\ = 96 (元)$\n\n答：原价是96元。",
+        "knowledge_points": [
+          "经济问题",
+          "折扣问题",
+          "已知一个数的百分之几是多少，求这个数"
+        ],
+        "source": "20-21六年级上·湖南邵阳·期末",
+        "merged_id": "unit2_175",
+        "original_id": "unit2_solve_175"
+      },
+      {
+        "id": "unit2_solve_176",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "妈妈今年5月6日把6000元钱存入银行，存期为三年，年利率为 $4.25\\%$。到期后，妈妈能取出本金和利息一共多少元？",
+        "options": [],
+        "answer": "6765元",
+        "solution": "【分析】根据利息 $=$ 本金 $\\times$ 利率 $\\times$ 存期，求出利息，利息+本金=取出的钱，列式解答即可。到期取款时银行多支付的钱叫利息。\n\n【详解】$6000 \\times 4.25\\% \\times 3 + 6000 \\\\ = 6000 \\times 0.0425 \\times 3 + 6000 \\\\ = 765 + 6000 \\\\ = 6765 (元)$\n\n答：到期后，妈妈能取出本金和利息一共6765元。",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "21-22六年级下·四川广安·期末",
+        "merged_id": "unit2_176",
+        "original_id": "unit2_solve_176"
+      },
+      {
+        "id": "unit2_solve_177",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "王老师要为学校足球队购买80个足球，下面两个体育用品店的足球单价都是25元，但各商店的优惠方案不同。为了节省开支，王老师应到哪个店购买？请通过计算说明。甲店：满10个足球打八折优惠。乙店：购物每满200元，返现金35元。（可累积返还）",
+        "options": [],
+        "answer": "王老师应该去甲店购买。",
+        "solution": "【分析】根据题意，王老师要买80个足球，甲店满10个可以八折，八折就是按原价的 $80\\%$ 出售，计算出80个足球打八折后需要付的钱数；根据乙店的优惠，计算买80个足球，25元1个，总共需要付的钱数，再除以200，算出总钱数里有几个200元，有几个200元就可以减去几个35元，对两个店具体真实要付的钱数进行比较即可。本题考查了折扣问题，解题的关键要明确打几折就是原价的百分之几十。\n\n【详解】甲店：\n\n需要钱数：\n\n$80 \\times 25 \\times 80\\% \\\\ = 2000 \\times 80\\% \\\\ = 1600 (元)$\n\n乙店：\n\n$80 \\times 25 \\div 200 \\\\ = 2000 \\div 200 \\\\ = 10 (个)$\n\n需要钱数：\n\n$80 \\times 25 - 10 \\times 35 \\\\ = 2000 - 350 \\\\ = 1650 (元)$\n\n$1600 < 1650$\n\n所以甲店更优惠。\n\n答：王老师应该去甲店购买。",
+        "knowledge_points": [
+          "求一个数的百分之几是多少",
+          "经济问题",
+          "折扣问题"
+        ],
+        "source": "2022六年级下·全国·专题练习",
+        "merged_id": "unit2_177",
+        "original_id": "unit2_solve_177"
+      },
+      {
+        "id": "unit2_solve_178",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "商场里电器柜台举行购物节促销活动，一种品牌的电视机每台按原价的八五折出售，比原价便宜225元，这台电视机原来售价是多少元？",
+        "options": [],
+        "answer": "1500元",
+        "solution": "【分析】把电视机的原来的售价看作单位'1'，则便宜的价钱是原来售价的 $1 - 85\\%$，即225元，根据部分的量 $\\div$ 所对应的分率 $=$ 单位'1'的量，用除法解答即可。本题考查折扣问题，明确几几折就是百分之几十几是解题的关键。\n\n【详解】$225 \\div (1 - 85\\%) \\\\ = 225 \\div 15\\% \\\\ = 1500 (元)$\n\n答：这台电视机原来售价是1500元。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "20-21六年级下·河南商丘·期末",
+        "merged_id": "unit2_178",
+        "original_id": "unit2_solve_178"
+      },
+      {
+        "id": "unit2_solve_179",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "王大爷3年前将5000元存入银行，年利率 $4.2\\%$，今年正好到期，他想用取出的所有钱买一台6000元的电脑，钱够吗？",
+        "options": [],
+        "answer": "不够",
+        "solution": "【分析】要想求这笔钱能否买一台6000元钱的电脑，也就是求出本金和利息一共是多少，然后与6000元比较。此题属于利息问题，运用关系式：本息 $=$ 本金+本金 $\\times$ 利率 $\\times$ 时间，代入数据，解决问题。\n\n【详解】$5000 + 5000 \\times 3 \\times 4.2\\% \\\\ = 5000 + 630 \\\\ = 5630 (元)$\n\n$5630 < 6000$\n\n答：钱不够。",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "2022·山东日照·小升初真题",
+        "merged_id": "unit2_179",
+        "original_id": "unit2_solve_179"
+      },
+      {
+        "id": "unit2_solve_180",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "苹果大丰收，去年产量达到4.5万吨，今年比去年增产了两成，今年苹果产量是多少万吨？",
+        "options": [],
+        "answer": "5.4万吨",
+        "solution": "【分析】把去年苹果的产量看作单位'1'，那么今年的就相当于去年的（1+20%），求今年苹果的产量是多少万吨，根据百分数乘法的意义解答即可。本题考查了百分数乘法应用题，关键是确定单位'1'，解答依据是：求一个数的百分之几是多少用乘法计算。\n\n【详解】$4.5 \\times (1 + 20\\%) \\\\ = 4.5 \\times 120\\% \\\\ = 5.4 (万吨)$\n\n答：今年苹果产量是5.4万吨。",
+        "knowledge_points": [
+          "比一个数多/少百分之几的数是多少",
+          "成数问题"
+        ],
+        "source": "2022·山东日照·小升初真题",
+        "merged_id": "unit2_180",
+        "original_id": "unit2_solve_180"
+      },
+      {
+        "id": "unit2_solve_181",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "商店卖一种运动鞋，如果每双售价为150元，那么售价的 $60\\%$ 是进价。售价的 $40\\%$ 就是赚的钱。现在要搞促销活动，为保证一双鞋赚的钱不少于15元，应怎样确定折扣？",
+        "options": [],
+        "answer": "七折或者七折以上",
+        "solution": "【分析】先把原来的售价看成单位'1'，用原来的售价乘上 $60\\%$ 就是这种运动鞋的进价；为保证运动鞋赚的钱不少于15元，那么运动鞋的实际售价必须大于等于进价+15元，求出最低的实际售价，再除以原来的售价，得出实际售价是原来售价的百分之几，进而根据打折的含义求解。解答本题要注意理解进价、原价、实际售价、折扣的意义，找清楚它们的关系，再根据分数乘除法的意义进行求解。\n\n【详解】$(150 \\times 60\\% + 15) \\div 150 \\\\ = 105 \\div 150 \\\\ = 0.7 \\\\ = 70\\% \\\\ =$ 七折\n\n答：为保证一双鞋赚的钱不少于15元，应打七折或者七折以上。",
+        "knowledge_points": [
+          "折扣问题",
+          "已知一个数的百分之几是多少，求这个数"
+        ],
+        "source": "21-22六年级下·山东临沂·期末",
+        "merged_id": "unit2_181",
+        "original_id": "unit2_solve_181"
+      },
+      {
+        "id": "unit2_solve_182",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "2015年11月，刘爷爷将20000元存入银行，存期为3年，年利率为 $2.75\\%$。到期后计划将利息捐给希望工程。刘爷爷能捐多少钱？",
+        "options": [],
+        "answer": "1650元",
+        "solution": "【分析】根据利息 $=$ 本金 $\\times$ 利率 $\\times$ 存期，据此代入数值进行计算即可。本题考查利率问题，明确利息 $=$ 本金 $\\times$ 利率 $\\times$ 存期是解题的关键。\n\n【详解】$20000 \\times 2.75\\% \\times 3 \\\\ = 550 \\times 3 \\\\ = 1650 (元)$\n\n答：刘爷爷能捐1650元。",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "21-22六年级下·山东临沂·期末",
+        "merged_id": "unit2_182",
+        "original_id": "unit2_solve_182"
+      },
+      {
+        "id": "unit2_solve_183",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "小张叔叔8月份工资是5450元，另外奖金900元。工资和奖金总和扣除5000元后，所剩部分按个人所得税 $20\\%$ 交税。你能帮小张叔叔算算，他应交纳多少元税钱？",
+        "options": [],
+        "answer": "270元",
+        "solution": "【分析】先把小张叔叔8月份的工资加上奖金的金额，再减去5000元，求出所剩的部分金额，这部分的钱乘 $20\\%$，即是应缴纳的税钱。解决此题关键是先求出扣除5000元后剩的那部分钱数，也就是需要缴个人所得税的钱数，进而问题得解。\n\n【详解】$(5450 + 900 - 5000) \\times 20\\% \\\\ = 1350 \\times 20\\% \\\\ = 270 (元)$\n\n答：他应交纳270元税钱。",
+        "knowledge_points": [
+          "税率问题"
+        ],
+        "source": "21-22六年级下·湖南衡阳·期末",
+        "merged_id": "unit2_183",
+        "original_id": "unit2_solve_183"
+      },
+      {
+        "id": "unit2_solve_184",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "张阿姨想在网上商城买一部标价1500的手机，A商城'打七折'销售，B商城搞的促销活动是'每满200元减60元'。(1)在A、B两个商城买各付多少钱？哪个商城更便宜些？(2)如果要买一部2000元的手机，在哪个商城买更省钱？",
+        "options": [],
+        "answer": "（1）A：1050元；B：1080元；A商城；(2)两个商城一样省钱",
+        "solution": "【分析】（1）A商城：打七折相当于原价的 $70\\%$，用手机的标价乘折扣，求出优惠后的价格；B商城：用1500元除以200元，计算出1500元里面有几个200元，就用标价减去几个60元，即可求出优惠后的价格；比较在两个商城各付的钱的大小，即可求出在哪个商城购买更便宜。(2)A商城：打七折相当于原价的 $70\\%$，用手机的标价乘折扣，求出优惠后的价格；B商城：用2000元除以200元，计算出2000元里面有几个200元，就用标价减去几个60元，即可求出优惠后的价格；比较在两个商城各付的钱的大小，即可求出在哪个商城买更省钱。最优化问题常用比较法进行解答，分别计算出两种方案优惠后的价格，再进行比较。\n\n【详解】(1)A商城：$1500 \\times 0.7 = 1050$ (元)\n\nB商城：$1500 \\div 200 = 7$ (个)……100(元)\n\n$1500 - 7 \\times 60 = 1500 - 420 = 1080$ (元)\n\n$1050 < 1080$\n\n答：在A商城买要付1050元，在B商城买要付1080元，在A商城购买更便宜些。\n\n(2)A商城：$2000 \\times 0.7 = 1400$ (元)\n\nB商城：$2000 \\div 200 = 10$ (个)\n\n$2000 - 10 \\times 60 = 2000 - 600 = 1400$ (元)\n\n$1400 = 1400$\n\n答：两家商城优惠后的价格相等，所以在两家商城购买花的钱一样多。",
+        "knowledge_points": [
+          "折扣问题",
+          "优化问题"
+        ],
+        "source": "21-22六年级下·内蒙古赤峰·期末",
+        "merged_id": "unit2_184",
+        "original_id": "unit2_solve_184"
+      },
+      {
+        "id": "unit2_solve_185",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "为了学生的卫生安全，学校给每个学生配一个水杯，每个水杯3元。欧亚超市打九折，万客隆超市'买八送一'。学校想买180只水杯，请你当'参谋，算一算：到哪家超市购买较合算'？",
+        "options": [],
+        "answer": "到万客隆超市比较便宜",
+        "solution": "【分析】欧亚超市：打九折，是指现价是原价的 $90\\%$，先求出180只的原价，然后再乘 $90\\%$ 即可；万客隆超市：买八送一，就是买9只水杯只需付8只的钱，$180 \\div 9 = 20$，20只赠送，那么180只需付（180-20）只的钱，由此求出（180-20）只的总价就是万客隆超市应付的钱数；然后比较两个超市需要的钱数，即可求解。解决本题关键是分清楚两个超市不同的优惠方法，找出计算现价的方法，从而得解。\n\n【详解】$3 \\times 180 \\times 90\\% \\\\ = 540 \\times 90\\% \\\\ = 486 (元)$\n\n$180 \\div 9 = 20$\n\n$(180 - 20) \\times 3 \\\\ = 160 \\times 3 \\\\ = 480 (元)$\n\n$480 < 486$\n\n答：到万客隆超市比较便宜。",
+        "knowledge_points": [
+          "经济问题",
+          "折扣问题"
+        ],
+        "source": "2015·广东·小升初真题",
+        "merged_id": "unit2_185",
+        "original_id": "unit2_solve_185"
+      },
+      {
+        "id": "unit2_solve_186",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "妈妈帮小智把4000元压岁钱存入银行，存期一年，年利率为 $2.15\\%$。到期后，小智留了利息的 $20\\%$ 买书本，其余的利息买了水果去看望爷爷奶奶。小智买水果用了多少钱？",
+        "options": [],
+        "answer": "68.8元",
+        "solution": "【分析】根据'利息 $=$ 本金 $\\times$ 利率 $\\times$ 存期'求出存款到期后得到的利息，把存款利息看作单位'1'，买水果的钱数占存款利息的（$1 - 20\\%$），最后用乘法求出买水果的钱数，据此解答。掌握利息的计算方法是解答题目的关键。\n\n【详解】$4000 \\times 2.15\\% \\times 1 \\times (1 - 20\\%) \\\\ = 4000 \\times 2.15\\% \\times 1 \\times 0.8 \\\\ = 86 \\times 0.8 \\\\ = 68.8 (元)$\n\n答：小智买水果用了68.8元。",
+        "knowledge_points": [
+          "求一个数的百分之几是多少",
+          "利率问题"
+        ],
+        "source": "21-22六年级下·四川南充·期末",
+        "merged_id": "unit2_186",
+        "original_id": "unit2_solve_186"
+      },
+      {
+        "id": "unit2_solve_187",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "外国语学校要买60个足球，现有万达、国贸两个商场可以选择，两个商场足球的单价都是25元，但每个商店的优惠方法不同。万达店：一律打八折销售。国贸店：购物每满300元，返还现金50元。为了节省费用，应到哪个商场购买？",
+        "options": [],
+        "answer": "万达店",
+        "solution": "【分析】根据单价 $\\times$ 数量=总价，求出60个足球的价钱，根据原价 $\\times$ 折扣=现价，求出万达店的现价；再求出60个足球的总价里面有几个300就减去几个50即可求出国贸店的现价，最后对比即可。本题考查折扣问题，明确几折即百分之几十是解题的关键。\n\n【详解】$25 \\times 60 \\times 80\\% \\\\ = 1500 \\times 80\\% \\\\ = 1200 (元)$\n\n$25 \\times 60 \\div 300 \\\\ = 1500 \\div 300 \\\\ = 5 (个)$\n\n$25 \\times 60 - 5 \\times 50 \\\\ = 1500 - 250 \\\\ = 1250 (元)$\n\n答：为了节省费用，应到万达店购买。",
+        "knowledge_points": [
+          "经济问题",
+          "折扣问题"
+        ],
+        "source": "21-22六年级下·广东东莞·期末",
+        "merged_id": "unit2_187",
+        "original_id": "unit2_solve_187"
+      },
+      {
+        "id": "unit2_solve_188",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "江淮村有一个种粮大户，去年收稻谷 $26000 \\mathrm{~kg}$，今年比去年增产了一成五。这个种粮大户今年比去年多收多少稻谷？",
+        "options": [],
+        "answer": "3900千克",
+        "solution": "【分析】增产一成五，也就是指今年产量比去年多 $15\\%$，把去年的产量看成单位'1'，用去年的产量乘 $15\\%$ 就是今年比去年多收多少稻谷。本题关键是理解几成几的含义，几成几就是百分之几十几。\n\n【详解】$26000 \\times 15\\% = 3900$ （千克）\n\n答：今年比去年多收3900千克稻谷。",
+        "knowledge_points": [
+          "成数问题"
+        ],
+        "source": "18-19六年级下·安徽·期中",
+        "merged_id": "unit2_188",
+        "original_id": "unit2_solve_188"
+      },
+      {
+        "id": "unit2_solve_189",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "王老师存了10万元3年期的教育储蓄，年利息是 $5.22\\%$。到期后，可以从银行得到本金和利息一共多少元？",
+        "options": [],
+        "answer": "115660元",
+        "solution": "【分析】根据'利息 $=$ 本金 $\\times$ 利率 $\\times$ 存期'表示出到期可以得到的利息，最后加上本金即可。掌握利息的计算方法是解答题目的关键。\n\n【详解】10万元 $= 100000$ 元\n\n$100000 \\times 3 \\times 5.22\\% + 100000 \\\\ = 300000 \\times 5.22\\% + 100000 \\\\ = 15660 + 100000 \\\\ = 115660 (元)$\n\n答：可以从银行得到本金和利息一共115660元。",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "18-19六年级上·浙江·课后作业",
+        "merged_id": "unit2_189",
+        "original_id": "unit2_solve_189"
+      },
+      {
+        "id": "unit2_solve_190",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "小红妈妈的服装店实行薄利多销的原则，一般在进价的基础上提高二成后作为销售价。照这样计算，一件进价为320元的衣服应标价多少元？",
+        "options": [],
+        "answer": "384元",
+        "solution": "【分析】二成是 $20\\%$，所以用进价320元乘 $(1 + 20\\%)$ 可以求出这件衣服的标价。本题考查了成数问题，几成就是百分之几十，所以二成是 $20\\%$。\n\n【详解】$320 \\times (1 + 20\\%) \\\\ = 320 \\times 120\\% \\\\ = 384 (元)$\n\n答：一件进价为320元的衣服应标价384元。",
+        "knowledge_points": [
+          "成数问题"
+        ],
+        "source": "21-22六年级下·全国·单元测试",
+        "merged_id": "unit2_190",
+        "original_id": "unit2_solve_190"
+      },
+      {
+        "id": "unit2_solve_191",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "李叔叔买福利彩票，中奖80000元，根据税法规定，他应按照 $20\\%$ 的税率缴纳个人所得税，李叔叔实得奖金多少万元？",
+        "options": [],
+        "answer": "6.4万元",
+        "solution": "【分析】把所中奖金80000元看作单位'1'，则实得奖金就占单位'1'的 $1 - 20\\% = 80\\%$，要计算实得奖金可列式：$80000 \\times (1 - 20\\%)$。解题时需要充分理解'税率'的含义：税率是 $20\\%$，是指有 $20\\%$ 的奖金需要上缴给国家，所以在计算时要减去相应的百分率。\n\n【详解】根据税率的意义可得：\n\n$80000 \\times (1 - 20\\%) \\\\ = 80000 \\times 0.8 \\\\ = 64000 (元) \\\\ = 6.4 (万元)$\n\n答：李叔叔实得奖金6.4万元。",
+        "knowledge_points": [
+          "税率问题"
+        ],
+        "source": "21-22六年级下·全国·单元测试",
+        "merged_id": "unit2_191",
+        "original_id": "unit2_solve_191"
+      },
+      {
+        "id": "unit2_solve_192",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "赵刚在中国人民银行存入两万元，存期5年，年利率 $3.25\\%$，到期后赵刚应取回多少钱？",
+        "options": [],
+        "answer": "23250元",
+        "solution": "【分析】通过利息的计算公式：利息 $=$ 本金 $\\times$ 利率 $\\times$ 存期，求出利息，再加上本金，即可得解。此题的解题关键是利用计算利息的公式来求解，注意本息指的是本金和利息。\n\n【详解】$20000 + 20000 \\times 3.25\\% \\times 5 \\\\ = 20000 + 650 \\times 5 \\\\ = 20000 + 3250 \\\\ = 23250 (元)$\n\n答：到期后赵刚应取回23250元。",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "21-22六年级下·全国·单元测试",
+        "merged_id": "unit2_192",
+        "original_id": "unit2_solve_192"
+      },
+      {
+        "id": "unit2_solve_193",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "张老师买了一套新房，原价68万元，实际只花了九五折的钱。按规定，买房要按实际房价的 $2\\%$ 缴纳契税，应缴纳契税多少元？",
+        "options": [],
+        "answer": "12920元",
+        "solution": "【分析】九五折是指现价是原价的 $95\\%$，把原价看成单位'1'，先用乘法求出现价；再把现价看成单位'1'，运用乘法求出现价的 $2\\%$，即为买房缴纳的契税。解答此题的关键是分清两个单位'1'的区别，再根据已知单位'1'的量，求它的百分之几用乘法计算。\n\n【详解】$68 \\times 95\\% \\times 2\\% \\\\ = 64.6 \\times 2\\% \\\\ = 1.292 (万元)$\n\n1.292万元=12920元\n\n答：应缴纳契税12920元。",
+        "knowledge_points": [
+          "折扣问题",
+          "税率问题"
+        ],
+        "source": "19-20六年级下·湖南怀化·单元测试",
+        "merged_id": "unit2_193",
+        "original_id": "unit2_solve_193"
+      },
+      {
+        "id": "unit2_solve_194",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "王叔叔月工资7200元，扣除5000元个税免征额后的部分需要按 $3\\%$ 的税率缴纳个人所得税，他缴个人所得税后应得多少元？",
+        "options": [],
+        "answer": "7134元",
+        "solution": "【分析】应纳税额与各种收入（销售额、营业额、应纳税所得额……）的比率叫做税率，根据税率的概念可知，5000元是免征个税的，所以王叔叔的工资里只有2200元才需要缴纳个人所得税，乘税率就是他应该缴纳的个人所得税。再用月工资减去个人所得税即得结果。此题的解题关键是理解税率的概念，掌握应纳税所得额、应纳税额、税率三者之间的数量关系，才能得出正确的结果。\n\n【详解】根据分析可知，应纳税所得额 $= 7200 - 5000 = 2200$ 元，应纳税额 $=$ 应纳税所得额 $\\times$ 税率 $= 2200 \\times 3\\% = 66$ 元。\n\n减去个人所得税后 $= 7200 - 66 = 7134$ 元。\n\n答：他缴个人所得税后应得7134元。",
+        "knowledge_points": [
+          "税率问题"
+        ],
+        "source": "19-20六年级下·湖南怀化·单元测试",
+        "merged_id": "unit2_194",
+        "original_id": "unit2_solve_194"
+      },
+      {
+        "id": "unit2_solve_195",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "小敏存了一笔存款，定期三年，年利率是 $5.00\\%$，三年后得到本金和利息一共11500元。小敏存入多少元？（方程解）",
+        "options": [],
+        "answer": "10000元",
+        "solution": "【分析】设小敏存入 $x$ 元，根据利息 $=$ 本金 $\\times$ 利率 $\\times$ 存期，表示出利息，再根据本金和利息一共11500元，列出方程解答即可。本题考查列方程解决问题，解答本题的关键是掌握利息的计算公式。\n\n【详解】解：设小敏存入 $x$ 元。\n\n$x + x \\times 5.00\\% \\times 3 = 11500$\n\n$1.15x = 11500$\n\n$x = 10000$\n\n答：小敏存入10000元。",
+        "knowledge_points": [
+          "列方程解含一个未知数的问题",
+          "利率问题"
+        ],
+        "source": "18-19六年级下·广东汕头·单元测试",
+        "merged_id": "unit2_195",
+        "original_id": "unit2_solve_195"
+      },
+      {
+        "id": "unit2_solve_196",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "李叔叔领上个月工资，按规定超过3500的部分按 $5\\%$ 缴纳个人所得税，李叔叔纳税75元，实领工资多少元？",
+        "options": [],
+        "answer": "4925元",
+        "solution": "【分析】缴纳的个人所得税 $\\div$ 税率=超过3500元的部分，超过3500元的部分+3500元是应得工资，应得工资一个人所得税=实领工资，据此分析。应纳税额与各种收入的比率叫税率。\n\n【详解】$75 \\div 5\\% + 3500 \\\\ = 1500 + 3500 \\\\ = 5000 (元)$\n\n$5000 - 75 = 4925$ （元）\n\n答：实领工资4925元。",
+        "knowledge_points": [
+          "税率问题"
+        ],
+        "source": "18-19六年级下·广东汕头·单元测试",
+        "merged_id": "unit2_196",
+        "original_id": "unit2_solve_196"
+      },
+      {
+        "id": "unit2_solve_197",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "妈妈把3万元存入银行，定期三年，年利率为 $3.96\\%$。到期后她可以取出多少钱？",
+        "options": [],
+        "answer": "33564元",
+        "solution": "【分析】到期后妈妈可以取出的钱数为本金和利息的和，利用'利息=本金 $\\times$ 利率 $\\times$ 存期'求出妈妈到期后得到的利息，最后求出本息即可。掌握利息的计算方法是解答题目的关键。\n\n【详解】3万元 $= 30000$ 元\n\n$30000 + 30000 \\times 3.96\\% \\times 3 \\\\ = 30000 + 1188 \\times 3 \\\\ = 30000 + 3564 \\\\ = 33564 (元)$\n\n答：到期后她可以取出33564元。",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "21-22六年级下·全国·单元测试",
+        "merged_id": "unit2_197",
+        "original_id": "unit2_solve_197"
+      },
+      {
+        "id": "unit2_solve_198",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "希望小学有学生2000人，有三成五的同学没有参加意外保险，参加意外保险的人数是多少人？",
+        "options": [],
+        "answer": "1300人",
+        "solution": "【分析】把希望小学的学生总数看作单位'1'，参加意外保险的人数=希望小学总人数 $\\times$ （1-35%），据此解答。计算出参加意外保险人数占总人数的百分率是解答题目的关键。\n\n【详解】三成五 $= 35\\%$\n\n$2000 \\times (1 - 35\\%) \\\\ = 2000 \\times 0.65 \\\\ = 1300 (人)$\n\n答：参加意外保险的人数是1300人。",
+        "knowledge_points": [
+          "求一个数的百分之几是多少",
+          "成数问题"
+        ],
+        "source": "21-22六年级下·全国·单元测试",
+        "merged_id": "unit2_198",
+        "original_id": "unit2_solve_198"
+      },
+      {
+        "id": "unit2_solve_199",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "叔叔购得三年期国库券4000元，年利率 $4.18\\%$，到期后可得利息多少元？",
+        "options": [],
+        "answer": "501.6元",
+        "solution": "【分析】用本金4000元乘年利率 $4.18\\%$，先求出一年的利息，再将其乘3，求出三年到期后的利息。本题考查了百分数的应用，年利息 $=$ 本金 $\\times$ 年利率。\n\n【详解】$4000 \\times 4.18\\% \\times 3 = 501.6$ （元）\n\n答：到期后可获得利息501.6元。",
+        "knowledge_points": [
+          "含百分数的运算",
+          "利率问题"
+        ],
+        "source": "22-23六年级上·湖南长沙·期末",
+        "merged_id": "unit2_199",
+        "original_id": "unit2_solve_199"
+      },
+      {
+        "id": "unit2_solve_200",
+        "unit": "第二单元：百分数",
+        "chapter": "百分数（2）",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "某林业局前年春季收山菜产量为2.8万吨，去年比前年增产四成。去年山菜产量是多少万吨？",
+        "options": [],
+        "answer": "3.92万吨",
+        "solution": "【分析】根据题意可知，'前年的产量 $\\times (1 + 40\\%) =$ 去年的产量'，据此解答即可。本题考查了成数的知识点，明确去年和前年产量的关系是解答本题的关键。\n\n【详解】四成 $= 40\\%$\n\n$2.8 \\times (1 + 40\\%) \\\\ = 2.8 \\times 1.4 \\\\ = 3.92 (万吨)$\n\n答：去年山菜产量是3.92万吨。",
+        "knowledge_points": [
+          "成数问题"
+        ],
+        "source": "16-17六年级上·吉林延边·期末",
+        "merged_id": "unit2_200",
+        "original_id": "unit2_solve_200"
+      }, {
+        "id": "unit2_choice_201",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "(21-22 六年级下·全国·假期作业) 1997 年末, 某银行储蓄存款金额达 115 亿元, 1998 年末增加到 147 亿元, 计算增长了百分之几的正确列式是 ( )。",
+        "options": [
+          "A. $(147 - 115) \\div 115$",
+          "B. (147-115) $\\div 147$",
+          "C. $147 \\div (147 - 115)$"
+        ],
+        "answer": "A",
+        "solution": "【分析】要计算增长了百分之几，确定基数为 1997 年末存款额 115 亿元，则增长率 $=$ （1998 年末存款额 -1997 年末存款额） $\\div$ 1997 年末存款额 $\\times 100\\%$ ，据此可得出答案。本题主要考查的是增长率问题, 解题的关键是找出增长基数, 运用增长率解答, 进而得出答案。\n\n【详解】由题意可得，计算增长率式子为：（147- 115）÷115。因此，本题答案为：A。",
+        "knowledge_points": [
+          "求一个数比另一个数多/少百分之几"
+        ],
+        "source": "21-22六年级下·全国·假期作业",
+        "merged_id": "unit2_201",
+        "original_id": "unit2_choice_201"
+      },
+      {
+        "id": "unit2_choice_202",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "（20-21六年级下·湖北随州·期末）一种练习本的单价是0.8元，李老师要买100个这样的练习本，选择（）种购买方式比较合算。",
+        "options": [
+          "A. 一律九折",
+          "B. 买5赠1",
+          "C. 满 50 元打八折"
+        ],
+        "answer": "C",
+        "solution": "【分析】由分析可知, 根据单价 $\\times$ 数量 = 总价, 八折即 $80 \\%$ , 九折即 $90 \\%$ , 买 5 赠 1 即 0.8 元可以买 6 本, 按照各选项的买法分别求出价钱, 然后比较即可。本题考查折扣问题, 明确几折即为百分之几十是解题的关键。\n\n【详解】A. $0.8 \\times 100 \\times 90 \\%\\\\= 80 \\times 90 \\%\\\\= 72$ (元)\n\nB. $0.8 \\times 80 + 0.8 \\times 4 \\\\= 64 + 3.2\\\\ = 67.2$ (元)\n\nC. $0.8 \\times 100 \\times 80 \\%$ $= 80 \\times 0.8$ $= 64$ (元)\n\n$72 > 67.2 > 64$ ，所以选择 C 种购买方式比较合算。故选：C",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "20-21六年级下·湖北随州·期末",
+        "merged_id": "unit2_202",
+        "original_id": "unit2_choice_202"
+      },
+      {
+        "id": "unit2_choice_203",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "(18-19 六年级上·江苏·单元测试) 一件衣服原价 150 元,现价 120 元, 这件衣服是 ( ) 折出售。",
+        "options": [
+          "A. 六",
+          "B. 二",
+          "C. 八",
+          "D. 四"
+        ],
+        "answer": "C",
+        "solution": "【分析】 '折扣 $=$ 现价 $\\div$ 原价' 据此解答即可。明确'折扣、现价和原价'之间的关系是解答本题的关键。\n\n【详解】 $120 \\div 150 = 0.8 =$ 八折;故答案为: C。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "18-19六年级上·江苏·单元测试",
+        "merged_id": "unit2_203",
+        "original_id": "unit2_choice_203"
+      },
+      {
+        "id": "unit2_choice_204",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "（18-19六年级下·吉林长春·期中）王老师把4000元存入银行，定期3年，年利率是 $4.50\\%$ ，到期一共可以取回（）元。",
+        "options": [
+          "A. 540",
+          "B. 4540",
+          "C. 4432"
+        ],
+        "answer": "B",
+        "solution": "【分析】利息 = 本金 $\\times$ 利率 $\\times$ 存期，根据公式计算出到期取回的利息，再加上本金就是一共可以取回的钱数。掌握公式：利息 $=$ 本金 $\\times$ 利率 $\\times$ 存期，这是解决此题的关键。\n\n【详解】 $4000 + 4000 \\times 4.50\\% \\times 3$ $= 4000 + 540$ $= 4540$ （元） 故答案为：B",
+        "knowledge_points": [
+          "求一个数的百分之几是多少",
+          "利率问题"
+        ],
+        "source": "18-19六年级下·吉林长春·期中",
+        "merged_id": "unit2_204",
+        "original_id": "unit2_choice_204"
+      },
+      {
+        "id": "unit2_choice_205",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "（17-18 六年级下·全国·单元测试）六（2）班有四成的学生是女生，那么男生占全班人数的（）。",
+        "options": [
+          "A. $\\frac{2}{3}$",
+          "B. $40\\%$",
+          "C. $\\frac{3}{5}$",
+          "D. 五成"
+        ],
+        "answer": "C",
+        "solution": "【分析】此题主要考查了成数的认识，四成 $= 40 \\%$ , 据此化成分数, 然后把全班人数看作单位 '1', 用单位 '1'一女生占全班人数的分率 = 男生占全班人数的分率, 据此列式解答。主要考查了成数与百分数、分数的相互转化，几成相当于十分之几，也就是百分之几十。\n\n【详解】四成 $= 40\\% = \\frac{2}{5}$ ,$1 - \\frac {2}{5} = \\frac {3}{5} 。$ 故答案为: C。",
+        "knowledge_points": [
+          "成数问题"
+        ],
+        "source": "17-18六年级下·全国·单元测试",
+        "merged_id": "unit2_205",
+        "original_id": "unit2_choice_205"
+      },
+      {
+        "id": "unit2_choice_206",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "（18-19六年级下·湖南永州·期中）一件商品原价100元，现在便宜25元，相当于打（）出售。",
+        "options": [
+          "A. 二五折",
+          "B. 七五折",
+          "C. 六折",
+          "D. 八折"
+        ],
+        "answer": "B",
+        "solution": "【分析】用原价减去便宜的钱数求出现价，用现价除以原价求出现价是原价的百分之几，根据百分数确定折扣即可。本题的关键是求出现价，然后用公式现价 $\\div$ 原价，算出折扣数。\n\n【详解】 $(100 - 25) \\div 100 \\times 100\\%\\\\= 75 \\div 100 \\times 100 \\% \\\\ = 75 \\% $\n\n75%就是七五折。故答案为: B",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "18-19六年级下·湖南永州·期中",
+        "merged_id": "unit2_206",
+        "original_id": "unit2_choice_206"
+      },
+      {
+        "id": "unit2_choice_207",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "（19-20 五年级下·广东河源·期末）一双运动鞋七折出售，现价是 140 元，原价是多少元？列式为（）",
+        "options": [
+          "A. $140 \\times \\frac{7}{10}$",
+          "B. $140 \\div \\frac{7}{10}$",
+          "C. $140 \\times (1 - \\frac{7}{10})$"
+        ],
+        "answer": "B",
+        "solution": "【分析】根据'七折销售，'知道现价是原价的 $70 \\%$ , 由原价 $\\times 70 \\% =$ 现价, 可得现价 $\\div 70 \\% =$ 原价, 据此解答。关键是理解'打折'的意义，几折就是原价的百分之几十。\n\n【详解】列式为: $140 \\div 70\\% = 140 \\div \\frac{7}{10}$ ,故选: B。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "19-20五年级下·广东河源·期末",
+        "merged_id": "unit2_207",
+        "original_id": "unit2_choice_207"
+      },
+      {
+        "id": "unit2_choice_208",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "(20-21 六年级下·广东广州·期中) 一件商品原价为 a 元,第一次打八折, 第二次涨价 $20 \\%$ , 现价为 b 元。那 a, b 的大小关系是 ( )。",
+        "options": [
+          "A. a $>\\mathrm{~b}$",
+          "B. $a = b$",
+          "C. $a < b$",
+          "D. 无法比较"
+        ],
+        "answer": "A",
+        "solution": "【分析】现价 $= \\text{原价} \\times 80 \\% \\times (1 + 20 \\%)$ ，据此求出现价，再与原价比较即可。明确打折和提价的单位 '1'不一致是解答本题的关键。\n\n【详解】 $a \\times 80\\% \\times (1 + 20\\%)\\\\= 0. 8 \\mathrm {a} \\times 1. 2 \\\\ = 0. 9 6 \\mathrm {a}; $\n\n$\\mathrm {b} = 0. 9 6 \\mathrm {a} < \\mathrm {a};$故答案为: A。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "20-21六年级下·广东广州·期中",
+        "merged_id": "unit2_208",
+        "original_id": "unit2_choice_208"
+      },
+      {
+        "id": "unit2_choice_209",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "（20-21 六年级下·全国·单元测试）李叔叔买了一个台灯，原价 280 元，现在只花了 9 折的钱，比原价便宜了（）元。",
+        "options": [
+          "A. 252",
+          "B. 250",
+          "C. 28",
+          "D. 9"
+        ],
+        "answer": "C",
+        "solution": "【分析】首先弄清折数与百分数的关系，一折是 $10 \\%$ , 根据现在只花了九折的钱, 确定把原价看作单位 '1', 要求比原价便宜了多少元, 需先求出比原价便宜了 (1 - 90%) ,再根据一个数乘百分数的意义解答。解答此题的关键是找单位'1'，进一步发现比单位'1'多或少百分之几，由此解决问题。\n\n【详解】 $280 \\times (1 - 90\\%)\\\\= 280\\times 10\\% \\\\ = 2 8 0 \\times 0. 1 \\\\ = 2 8 \\text {(元)} ; $\n\n故答案为: C。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_209",
+        "original_id": "unit2_choice_209"
+      },
+      {
+        "id": "unit2_choice_210",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "(2022·河北邢台·小升初真题) 一年前王老师把 3000元钱存入了银行，定期2年。年利息按 $2.25\\%$ 计算，到期可得本金和税后利息（税率为 $20\\%$ ）一共（）元。",
+        "options": [
+          "A. 3000",
+          "B. 3108",
+          "C. 108",
+          "D. 3135"
+        ],
+        "answer": "B",
+        "solution": "【分析】税后利息 $=$ 本金 $\\times$ 利率 $\\times$ 存款年限 $\\times$ (1 - 20%)，本息 $=$ 本金 + 利息，据此解答即可。熟练掌握求利息的公式是解答本题的关键。\n\n【详解】 $3000 + 3000 \\times 2.25\\% \\times 2 \\times (1 - 20\\%)\\\\= 3000 + 135 \\times 80 \\% \\\\ = 3 0 0 0 + 1 0 8 \\\\ = 3 1 0 8 \\text {(元)} ; $\n\n故答案为: B。",
+        "knowledge_points": [
+          "税率问题",
+          "利率问题"
+        ],
+        "source": "2022·河北邢台·小升初真题",
+        "merged_id": "unit2_210",
+        "original_id": "unit2_choice_210"
+      },
+      {
+        "id": "unit2_choice_211",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "（20-21六年级下·全国·单元测试）'个人所得税起征点调至3500元，一级（1500元以内）税率降至 $3\\%$ 。'这是人代会二次审议的草案结果。细心的小王马上计算出自己要缴纳的税收为36.9元，请问现在小王每月的收入为（）元。",
+        "options": [
+          "A. 4730",
+          "B. 4720",
+          "C. 4820",
+          "D. 4830"
+        ],
+        "answer": "A",
+        "solution": "【分析】根据'一级（1500元以内）税率降至 $3\\%$ ，缴纳的税收为36.9元'，先求出交税的部分，然后加上3500元即可。此题解答的关键在于根据税率和税款求出交税的部分。\n\n【详解】 $36.9 \\div 3\\% + 3500$ $= 1230 + 3500$ $= 4730$ (元)\n\n故答案为: A。",
+        "knowledge_points": [
+          "税率问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_211",
+        "original_id": "unit2_choice_211"
+      },
+      {
+        "id": "unit2_choice_212",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "（18-19六年级下·云南曲靖·期末）王阿姨在银行存入1000元，作为一年期定期储蓄，到期后王阿姨获得利息33元，则一年期的年利率是（）。",
+        "options": [
+          "A. $3.00\\%$",
+          "B. $3.30\\%$",
+          "C. $3.25\\%$",
+          "D. $3.50\\%$"
+        ],
+        "answer": "B",
+        "solution": "【分析】根据利息的计算公式：利息 $=$ 本金 $\\times$ 利率 $\\times$ 时间，可得：利息 $\\div$ 时间 $\\div$ 本金 $=$ 利率；据此列式计算。掌握利息的计算公式是解题的关键。\n\n【详解】 $33 \\div 1 \\div 1000 = 3.30\\%$ 故选：B。",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "18-19六年级下·云南曲靖·期末",
+        "merged_id": "unit2_212",
+        "original_id": "unit2_choice_212"
+      },
+      {
+        "id": "unit2_choice_213",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "(20-21 六年级下·全国·单元测试) 下列说法正确的是 ( )。",
+        "options": [
+          "A. 一件商品打六折出售，就是指这件商品的现价比原价便宜 $60\\%$",
+          "B. 商品开展 '买四送一' 活动, 作为顾客能享受到的最大优惠相当于打八折",
+          "C. 一件商品先提价 $10 \\%$ , 再打九折, 价格不变",
+          "D. '打八折'与'每满100元减20元'的含义相同"
+        ],
+        "answer": "B",
+        "solution": "【分析】折扣：用于商品，现价是原价的是十分之几或百分之几，叫做折扣，通称'打折'，据此分析解答。考查了打折的灵活应用，学生应掌握。\n\n【详解】A．一件商品打六折出售，就是指这件商品的现价是原价的 $60 \\%$ , 现价比原价便宜 $1 - 60 \\% = 40 \\%$ , 原说法错误; B. 商品开展 '买四送一' 活动, 即花费买 4 件物品的钱数,能买到 $(4 + 1)$ 件物品, 即现价是原价: $4 \\div (4 + 1) =$ $80 \\%$ , 即打八折, 作为顾客能享受到的最大优惠相当于打八折, 原说法正确; C. 一件商品先提价 $10 \\%$ , 把这件商品的价格看作单位 '1', 提价后是这件商品的 $1 + 10 \\% = 1.1$ , 再打九折, 现价为这件商品的 $1.1 \\times 90 \\% = 0.99$ , 价格降低了, 原说法错误; D. 例如原价 120 元的商品 '打八折' 后价格为: $120 \\times 80 =$ 96（元），'每满100元减20元'商品价格为： $120 - 20 =$ 100（元），两者不相等，所以'打八折'与'每满100元减20元'的含义不相同，原说法错误。故答案为: B",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_213",
+        "original_id": "unit2_choice_213"
+      },
+      {
+        "id": "unit2_choice_214",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "（20-21 六年级下·全国·单元测试）每年的 4 月 23 日是世界读书日，这一天三味书屋的图书全部打九五折出售。小贤买了一套《平凡的世界》，比原来少花了 4.5 元，这套书的原价是多少元？下面列式不正确的是（）。",
+        "options": [
+          "A. $4.5 \\div (1 - 95\\%)$",
+          "B. 设这套书的原价是 $x$ 元, $x - 95 \\% x = 4.5$",
+          "C. 设这套书的原价是 $x$ 元, $(1 - 95 \\%) x = 4.5$",
+          "D. 设这套书的原价是 $x$ 元, $95 \\% x = 4.5$"
+        ],
+        "answer": "D",
+        "solution": "【分析】打九五折出售，就是现价是原价的 $95 \\%$ , 据此分析解答。此题考查了折扣问题, 找准数量关系, 认真解答即可。\n\n【详解】A．把原价看作单位'1'，则原价的 $(1 - 95\\%)$ 是4.5元，根据除法的意义，即可知原价为 $4.5 \\div (1 - 95\\%)$ ； B. 如果设这套书的原价是 $x$ 元, 则现价是 $95 \\% \\mathrm{x}$ , 原价一现价 $= 4.5$ 元, 据此列方程为 $x - 95 \\% x = 4.5$ ; C. 如果设这套书的原价是 $x$ 元, 则现价比原价少 $(1-95 \\%) ,$ 乘原价就是比原价少的钱数, 即 $(1 - 95 \\%) x = 4.5 ;$ D. 如果设这套书的原价是 $x$ 元, $95 \\% x$ 表示的是现价, 方程不成立。故选择：D",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_214",
+        "original_id": "unit2_choice_214"
+      },
+      {
+        "id": "unit2_choice_215",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "（20-21六年级下·全国·单元测试）2020年上半年，为防控新冠肺炎疫情，人们响应政府号召，尽量避免外出，网上购物数量增多。某县城4月份的快递数量比去年同期增长了2万件，达到了4.5万件。用成数表示，该县城2020年4月份的快递数量比去年同期增长了（）。",
+        "options": [
+          "A. 约三成一",
+          "B. 约七成",
+          "C. 八成",
+          "D. 约四成四"
+        ],
+        "answer": "C",
+        "solution": "【分析】根据题意可知, 增长的百分率 $=$ 增长的件数 $\\div$ 去年同期件数, 据此解答。此题考查了成数问题，明确百分之几十就是几成。注意单位'1'是去年同期快递数量。\n\n【详解】 $2 \\div (4.5 - 2)\\\\= 2 \\div 2. 5 \\\\ = 80 \\% =八成$\n\n故选择：C",
+        "knowledge_points": [
+          "成数问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_215",
+        "original_id": "unit2_choice_215"
+      },
+      {
+        "id": "unit2_choice_216",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "（20-21 六年级下·全国·单元测试）某种商品因换季准备打折出售。按定价打七五折出售将赔本 25 元；按定价打九折出售，将盈利 35 元。这种商品的定价是（）元。",
+        "options": [
+          "A. 300",
+          "B. 400",
+          "C. 600",
+          "D. 900"
+        ],
+        "answer": "B",
+        "solution": "【分析】设这件商品定价为 $x$ 元，打七五折赔钱，赔本 25 元，就是比进价少了 25 元， $x \\times 75\\% + 25$ 元，才是进价；要是打九折，即 $x \\times 90\\%$ ，还盈利 35 元，比进价多了 35 元，进价就是 $x \\times 90\\% - 35$ 元，进价不变，根据进价不变，可列方程: $90\\% \\mathrm{x} - 35 = 75\\% \\mathrm{x} + 25$ , 解方程即可。本题考查折扣问题，利用进价不变的原则，解决定价的问题。\n\n【详解】解：设商品定价为 $x$ 元\n\n$90 \\% \\mathrm {x} - 35 = 75 \\% \\mathrm {x} + 25$\n\n$90 \\% \\mathrm {x} - 75 \\% \\mathrm {x} = 25 + 35$\n\n$15 \\% \\mathrm {x} = 60$\n\n$x = 6 0 \\div 1 5$\n\n$x = 4 0 0$\n\n这种商品定价为 400 元。故答案选：B",
+        "knowledge_points": [
+          "利润与折扣的综合问题",
+          "列方程解含一个未知数的问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_216",
+        "original_id": "unit2_choice_216"
+      },
+      {
+        "id": "unit2_choice_217",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "（20-21六年级下·全国·单元测试）张阿姨买了一套总价150万元的住房，按规定要缴纳 $1.5\\%$ 的房屋契税，张阿姨应缴纳房屋契税（）元。",
+        "options": [
+          "A. 2.25",
+          "B. 225",
+          "C. 2250",
+          "D. 22500"
+        ],
+        "answer": "D",
+        "solution": "【分析】由题意知：用总价150万元乘契税率 $1.5\\%$ 可得要缴纳的房契税，据此解答。理解房屋契税等于房子总价乘契税率是解答本题的关键。\n\n【详解】 $1500000 \\times 1.5\\% = 22500$ （元）故答案为: D",
+        "knowledge_points": [
+          "税率问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_217",
+        "original_id": "unit2_choice_217"
+      },
+      {
+        "id": "unit2_choice_218",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "（20-21 六年级下·全国·单元测试）一件衣服原价 280 元，现在打七折出售现在买可以节省（）元。",
+        "options": [
+          "A. 30",
+          "B. 70",
+          "C. 84",
+          "D. 196"
+        ],
+        "answer": "C",
+        "solution": "【分析】打七折出售则现价是原价的 $70 \\%$ , 也就是比原价便宜 $30 \\%$ ; 据此解答。本题主要考查折扣问题, 解题时注意所求是节省多少元。\n\n【详解】 $280 \\times (1 - 70\\%)\\\\= 2 8 0 \\times 0. 3 \\\\ = 8 4 \\text {(元)} $\n\n故答案为: C",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_218",
+        "original_id": "unit2_choice_218"
+      },
+      {
+        "id": "unit2_choice_219",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "（20-21 六年级下·全国·单元测试）下面不相等的一组是（）。",
+        "options": [
+          "A. 七成五与 $75\\%$",
+          "B. 三折与 0.3",
+          "C. 二成与 $\\frac{1}{2}$",
+          "D. 八五折与 $85\\%$"
+        ],
+        "answer": "C",
+        "solution": "【分析】几成几就是百分之几十几，几成就是百分之几十；几折就是百分之几十，几几折就是百分之几十几，据此选择。此题考查了成数、折扣与百分数的互化，属于基础类题目。\n\n【详解】A. 七成五 $= 75\\%$ B. 三折 $= 30\\% = 0.3$ C. 二成 $= 20\\% = \\frac{1}{5} \\neq \\frac{1}{2}$ D. 八五折 $= 85\\%$ 故选择：C",
+        "knowledge_points": [
+          "成数问题",
+          "折扣问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_219",
+        "original_id": "unit2_choice_219"
+      },
+      {
+        "id": "unit2_choice_220",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "（20-21六年级下·全国·单元测试）李叔叔购买了五年期的国家建设债券20000元，年利率是 $3.81\\%$ ，到期时李叔叔应得到利息（）元。",
+        "options": [
+          "A. 3810",
+          "B. 762",
+          "C. 1524",
+          "D. 1955"
+        ],
+        "answer": "A",
+        "solution": "【分析】利息 = 本金 $\\times$ 利率 $\\times$ 存期，把题中数据代入公式计算即可。掌握利息的计算方法是解答题目的关键。\n\n【详解】 $20000 \\times 3.81\\% \\times 5\\\\= 7 6 2 \\times 5 \\\\ = 3 8 1 0 \\text {(元)} $\n\n故答案为: A",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_220",
+        "original_id": "unit2_choice_220"
+      },
+      {
+        "id": "unit2_choice_221",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "（2020·湖北孝感·小升初真题）妈妈把5万元钱存入银行，年利率为 $2.1\\%$ ，2年后，妈妈能取回多少万元？下面计算方法正确的是（）。",
+        "options": [
+          "A. $5 \\times 2.1\\% \\times 2$",
+          "B. $5 \\times (1 + 2.1\\% \\times 2)$",
+          "C. $5 \\times 2.1\\% + 5$"
+        ],
+        "answer": "B",
+        "solution": "【分析】取回的钱包括本金和利息，利息 $=$ 本金 $\\times$ 利率 $\\times$ 存期，据此分析。关键是掌握利息的公式，B 选项根据乘法分配律进行了简算。\n\n【详解】 $5 + 5 \\times 2.1\\% \\times 2\\\\= 5 \\times (1 + 2.1 \\% \\times 2)$\n\n故答案为: B",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "2020·湖北孝感·小升初真题",
+        "merged_id": "unit2_221",
+        "original_id": "unit2_choice_221"
+      },
+      {
+        "id": "unit2_choice_222",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "(20-21 六年级下·全国·单元测试) 李强承包一块地,前年收获粮食 5.6 吨, 去年比前年增产三成, 求去年收获粮食多少吨? 正确的算式是 ( )。",
+        "options": [
+          "A. $5.6 \\times (1 + 30\\%)$",
+          "B. $5.6 \\times (1 + 3\\%)$",
+          "C. $5.6 \\div (1 + 30\\%)$"
+        ],
+        "answer": "A",
+        "solution": "【分析】把前年粮食产量看作单位 '1', 去年粮食产量是前年的 $(1 + 30 \\%) ,$ 去年粮食的产量 $=$ 前年粮食的产量 $\\times$ $(1 + 30 \\%) ,$ 据此解答。找准题目中的单位 '1', 求比一个数多百分之几的数是多少的计算方法: 这个数 $\\times$ (1 + 百分率)。\n\n【详解】三成 $= 30\\%$\n\n$5.6 \\times (1 + 30 \\%) \\\\ = 5. 6 \\times 1. 3 \\\\ = 7. 2 8 (\\text {吨}) $\n\n故答案为：A",
+        "knowledge_points": [
+          "成数问题",
+          "单位 '1'的认识与确定",
+          "比一个数多/少百分之几的数是多少"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_222",
+        "original_id": "unit2_choice_222"
+      },
+      {
+        "id": "unit2_choice_223",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "（2020·山东泰安·小升初真题）商店以80元一件的价格购进一批衬衫，并以 $25\\%$ 的利润率出售，过了一段时间发现还剩下150件，于是打九折出售，又过了一段时间发现一共卖掉了总量的 $90\\%$ ，于是将最后几件按进货价出售，最后商店共获利 2300 元, 则商店一共进了多少件衬衫? ( )",
+        "options": [
+          "A. 180 件",
+          "B. 200件",
+          "C. 240 件",
+          "D. 300件"
+        ],
+        "answer": "B",
+        "solution": "【分析】以 $25\\%$ 的利润率出售部分的售价 $+9$ 折出售部分的售价 $+$ 进价出售部分的售价一总进价 $= 2300$ ，据此列出方程解答即可。用方程解决问题的关键是找到等量关系, 几折就是百分之几十, 求一个数的百分之几是多少用乘法。\n\n【详解】解：设商店一共进了 $\\mathbf{X}$ 件衬衫。\n\n$(x - 150) \\times 80 \\times (1 + 25 \\%) + (150 - 0.1 x) \\times 80 \\times (1 +25 \\%) \\times 90 \\% + 0.1 \\mathrm {x} \\times 80 - 80 \\mathrm {x} = 2300$\n\n$(x - 1 5 0) \\times 1 0 0 + (1 5 0 - 0. 1 x) \\times 9 0 + 0. 1 x \\times 8 0 - 8 0 x = 2 3 0 0$\n\n$1 0 0 x - 1 5 0 0 0 + 1 3 5 0 0 - 9 x + 8 x - 8 0 x = 2 3 0 0$\n\n$1 9 x = 3 8 0 0 \\\\ x = 2 0 0 $\n\n所以商店一共进了 200 件衬衫。故答案为: B。",
+        "knowledge_points": [
+          "列方程解含一个未知数的问题",
+          "利润常见问题"
+        ],
+        "source": "2020·山东泰安·小升初真题",
+        "merged_id": "unit2_223",
+        "original_id": "unit2_choice_223"
+      },
+      {
+        "id": "unit2_choice_224",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "（20-21六年级下·全国·单元测试）某种商品的进价为800元，出售时标价1200元，后来由于该商品积压，商店准备打折出售，但是需要保持利润率不低于 $5\\%$ ，则最多可以打（）折。",
+        "options": [
+          "A. 五",
+          "B. 六",
+          "C. 七",
+          "D. 八"
+        ],
+        "answer": "C",
+        "solution": "【分析】当利润率为 $5 \\%$ 时，商品的利润 $=$ 商品的进价 $\\times$ 利润率，折扣 $=$ 现价 $\\div$ 原价 $\\times 100\\%$ ，百分之几十也就是几折，据此解答。解答本题的关键是要明确就是求折扣，用现价 $\\div$ 标价 $=$ 折扣，现价即为进价加上利润。\n\n【详解】利润： $800 \\times 5\\% = 40$ （元）\n\n折扣： $(800 + 40) \\div 1200 \\times 100\\%\\\\= 840 \\div 1200 \\times 100 \\% \\\\ = 0.7 \\times 100 \\% \\\\ = 70 \\% $\n\n70%=七折 故答案为: C",
+        "knowledge_points": [
+          "经济问题",
+          "利润问题",
+          "折扣问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_224",
+        "original_id": "unit2_choice_224"
+      },
+      {
+        "id": "unit2_choice_225",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "(20-21 六年级下·全国·单元测试) 某村去年生产油菜籽 120 吨，比前年增产一成五，前年生产油菜籽多少吨？正确的算式是（）。",
+        "options": [
+          "A. $120 \\times 15\\%$",
+          "B. $120 \\times (1 +15\\%)$",
+          "C. $120 \\div (1 + 15 \\%)$"
+        ],
+        "answer": "C",
+        "solution": "【分析】一成五 $= 15 \\%$ , 即去年比前年增产 $15 \\%$ , 根据 '前年产量 $\\times (1 + 15 \\%) =$ 去年产量' 解答即可。解答本题的关键是要明确成数的含义，理解去年比前年增产 $15 \\%$ .\n\n【详解】 $120 \\div (1 + 15\\%)$ ;故答案为: C。",
+        "knowledge_points": [
+          "成数问题",
+          "已知比一个数多/少百分之几是多少，求这个数。"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_225",
+        "original_id": "unit2_choice_225"
+      },
+      {
+        "id": "unit2_fill_226",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（20-21六年级上·广东佛山·期末）李强将80000元钱存入银行，定期五年，年利率是 $4.25\\%$ ，到期后可得利息( )元，一共可取出( )元。",
+        "options": [],
+        "answer": "17000；97000",
+        "solution": "【分析】根据利息的计算公式：利息 $=$ 本金 $\\times$ 利率 $\\times$ 存期，计算出到期后得到的利息；取出的钱数 $=$ 本金 $+$ 利息，据此解答。掌握利息的计算方法是解答题目的关键。\n\n【详解】 (1) $80000 \\times 4.25\\% \\times 5\\\\= 3 4 0 0 \\times 5 \\\\ = 1 7 0 0 0 \\text {(元)} $\n\n(2) $80000 + 17000 = 97000$ (元)",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "20-21六年级上·广东佛山·期末",
+        "merged_id": "unit2_226",
+        "original_id": "unit2_fill_226"
+      },
+      {
+        "id": "unit2_fill_227",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "（20-21六年级上·广东佛山·期中）海洋馆门票原价120元，现在七五折优惠，现在去海洋馆能节省( )元。",
+        "options": [],
+        "answer": "30",
+        "solution": "【分析】根据题意可知，现价是原价的 $75 \\%$ , 则现价比原价便宜 $1 - 75 \\% = 25 \\%$ , 再根据百分数乘法的意义解答即可。明确折扣的含义是解答本题的关键。\n\n【详解】 $120 \\times (1 - 75\\%)\\\\= 120\\times 25\\% \\\\ = 3 0 \\text {(元)} $",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "20-21六年级上·广东佛山·期中",
+        "merged_id": "unit2_227",
+        "original_id": "unit2_fill_227"
+      },
+      {
+        "id": "unit2_fill_228",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（20-21 六年级上·广东韶关·期末）小明在商店购买一双打 8.5 折的运动鞋，付款 170 元，这双鞋原价( )。",
+        "options": [],
+        "answer": "200元",
+        "solution": "【分析】打 8.5 折即现价是原价的 $85 \\%$ , 把原价看作单位 '1', 即现价占原价的分率为 $85 \\%$ , 对应付款 170 元, 运用除法即可求出这双鞋原价。解答本题关键是理解打折的含义，打几折现价就是原价的百分之几十。\n\n【详解】 $170 \\div 85\\% = 200$ （元）则这双鞋原价是200元。",
+        "knowledge_points": [
+          "已知一个数的百分之几是多少，求这个数",
+          "折扣问题"
+        ],
+        "source": "20-21六年级上·广东韶关·期末",
+        "merged_id": "unit2_228",
+        "original_id": "unit2_fill_228"
+      },
+      {
+        "id": "unit2_fill_229",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "（19-20六年级下·山东济南·期末）某景区今年五一的游客数比去年减少了二成五，这个成数改写成百分数是( )%。",
+        "options": [],
+        "answer": "25",
+        "solution": "【分析】根据几成就是百分之几十，写成百分数即可。关键是理解成数的意义，工农业生产经常用'成数'表示生产增长的情况。\n\n【详解】某景区今年五一的游客数比去年减少了二成五，这个成数改写成百分数是 $25\\%$",
+        "knowledge_points": [
+          "分数、小数、百分数与成数的互化"
+        ],
+        "source": "19-20六年级下·山东济南·期末",
+        "merged_id": "unit2_229",
+        "original_id": "unit2_fill_229"
+      },
+      {
+        "id": "unit2_fill_230",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "（20-21六年级下·全国·期末）李大爷去年种的水稻产量是1000千克，今年改种了'杂交水稻之父'袁隆平院士研制的第二代超级杂交水稻后，产量比去年增加二成。今年的产量是( )千克。",
+        "options": [],
+        "answer": "1200",
+        "solution": "【分析】根据题意可知，'去年产量 $\\times (1 + 20\\%) =$ 今年的产量'，据此解答即可。熟练掌握百分数乘法的意义是解答本题的关键。\n\n【详解】 二成 $= 20\\%$\n\n$1000 \\times (1 + 20 \\%) \\\\ = 1 0 0 0 \\times 1. 2 \\\\ = 1 2 0 0 \\text {(千 克)} $",
+        "knowledge_points": [
+          "成数问题"
+        ],
+        "source": "20-21六年级下·全国·期末",
+        "merged_id": "unit2_230",
+        "original_id": "unit2_fill_230"
+      },
+      {
+        "id": "unit2_fill_231",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "(20-21 六年级下·云南曲靖·期末) 某酒店 6 月份的营业额是80万元，如果按营业额的 $3\\%$ 缴纳营业税，应缴纳营业税( )万元。",
+        "options": [],
+        "answer": "2.4",
+        "solution": "【分析】用营业额乘营业税税率即可求出需要缴纳的税额,据此解答即可。明确 '营业额 $\\times$ 税率 $=$ 缴纳的税额' 是解答本题的关键。\n\n【详解】 $80 \\times 3\\% = 2.4$ （万元）",
+        "knowledge_points": [
+          "税率问题"
+        ],
+        "source": "20-21六年级下·云南曲靖·期末",
+        "merged_id": "unit2_231",
+        "original_id": "unit2_fill_231"
+      },
+      {
+        "id": "unit2_fill_232",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "(20-21 六年级下·云南玉溪·期末) 新的个人所得税起征点为 5000 元。杨叔叔本月工资是 8000 元，若超过起征点部分按照 $3\\%$ 的得税率缴税，他本月应缴纳个人所得税()元。",
+        "options": [],
+        "answer": "90",
+        "solution": "【分析】根据题意可知，杨叔叔工资超出5000元的部分需要纳税，即3000元需要纳税，用3000元乘缴税税率即可。明确需要纳税的金额是解答本题的关键。\n\n【详解】 $(8000 - 5000) \\times 3\\%\\\\= 3000 \\times 3 \\% \\\\ = 9 0 \\text {(元)} $",
+        "knowledge_points": [
+          "税率问题"
+        ],
+        "source": "20-21六年级下·云南玉溪·期末",
+        "merged_id": "unit2_232",
+        "original_id": "unit2_fill_232"
+      },
+      {
+        "id": "unit2_fill_233",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（20-21 六年级下·河北石家庄·期末）一种商品打八五折销售，'八五折'表示原价的( )%；如果这种商品的原价是 500 元，那么付款时只需付( )元。",
+        "options": [],
+        "answer": "85；425",
+        "solution": "【分析】几折就是百分之几十；原价×折扣=现价。打折就是按照折数低价出售商品。同种商品，折数越低，价格越低。\n\n【详解】 $500 \\times 85\\% = 425$ （元）一种商品打八五折销售，'八五折'表示原价的 $85 \\%$ ；如果这种商品的原价是 500 元，那么付款时只需付 425 元。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "20-21六年级下·河北石家庄·期末",
+        "merged_id": "unit2_233",
+        "original_id": "unit2_fill_233"
+      },{
+        "id": "unit2_fill_234",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "（20-21六年级下·广东梅州·期末）李明把1000元钱存入银行，定期两年，年利率是 $2.79\\%$ 。如果不缴纳利息税，两年后连本加利息应得钱数是( )元。",
+        "options": [],
+        "answer": "1055.8",
+        "solution": "【分析】'利息 $=$ 本金 $\\times$ 利率 $\\times$ 存期'据此求出利息，再加上本金即可。熟练掌握求利息的公式是解答本题的关键。\n\n【详解】 $1000 \\times 2.79\\% \\times 2 + 1000\\\\= 5 5. 8 + 1 0 0 0 \\\\ = 1 0 5 5. 8 \\text {(元)} $",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "20-21六年级下·广东梅州·期末",
+        "merged_id": "unit2_234",
+        "original_id": "unit2_fill_234"
+      },
+      {
+        "id": "unit2_fill_235",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（2021·河南信阳·小升初真题）小丽把 1000 元压岁钱存入银行，存期两年，年利率 $3.25\\%$ ，到期能得到利息( )元。",
+        "options": [],
+        "answer": "65",
+        "solution": "【分析】由题意可知, 要求利息, 根据利息 $=$ 本金 $\\times$ 利率 $\\times$ 存期, 据此可解答。本题考查利率问题, 明确利息的算法是解题的关键。\n\n【详解】 $1000 \\times 3.25\\% \\times 2\\\\= 3 2. 5 \\times 2 \\\\ = 6 5 \\text {(元)} $",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "2021·河南信阳·小升初真题",
+        "merged_id": "unit2_235",
+        "original_id": "unit2_fill_235"
+      },
+      {
+        "id": "unit2_fill_236",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(2022·黑龙江鸡西·小升初真题) 李叔叔把 5000 元钱存入银行，定期2年，如果年利率是 $3.75\\%$ ，到期后，李叔叔一共可以取回( )元。",
+        "options": [],
+        "answer": "5375",
+        "solution": "【分析】根据'利息 $=$ 本金 $\\times$ 利率 $\\times$ 时间'，求出利息，再加上本金，就是李叔叔一共可以取回的钱。熟练掌握求利息的方式是解答本题的关键。\n\n【详解】 $5000 \\times 3.75\\% \\times 2 + 5000\\\\= 1 8 7. 5 \\times 2 + 5 0 0 0 \\\\ = 3 7 5 + 5 0 0 0 \\\\ = 5 3 7 5 (\\text {元}) $",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "2022·黑龙江鸡西·小升初真题",
+        "merged_id": "unit2_236",
+        "original_id": "unit2_fill_236"
+      },
+      {
+        "id": "unit2_fill_237",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（19-20六年级下·云南曲靖·期末）明明买一件玩具，商家做活动打八折出售，便宜了60元，这个玩具原价是()元。",
+        "options": [],
+        "answer": "300",
+        "solution": "【分析】将原价看作单位'1'，打八折出售，便宜了 $1 - 80\\%$ ，便宜了60元，用便宜的钱数 $\\div$ 对应百分率 $=$ 原价。关键是理解折扣的意义，几折就是百分之几十。\n\n【详解】 $60 \\div (1 - 80\\%)\\\\= 6 0 \\div 0. 2 \\\\ = 3 0 0 \\text {(元)} $",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "19-20六年级下·云南曲靖·期末",
+        "merged_id": "unit2_237",
+        "original_id": "unit2_fill_237"
+      },
+      {
+        "id": "unit2_fill_238",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（19-20六年级下·河南洛阳·期末）李叔叔把20000元存入银行，定期两年，年利率 $2.25\\%$ ，到期后可取回本金和利息共( )元。",
+        "options": [],
+        "answer": "20900",
+        "solution": "【分析】审题需要加强注意，问：取回本金和利息共多少元？首先利用利息公式：本金×年利率×存期，进行代入，算出利息之后，再加上本金即可。本题考查利息公式，利息 $=$ 本金 $\\times$ 年利率 $\\times$ 存期，同时要注意问题问什么，如果问'本息和'、'本金和利息共多少'，一定要记住最后加本金。\n\n【详解】 $20000 \\times 2.25\\% \\times 2\\\\= 4 5 0 \\times 2 \\\\ = 9 0 0 (\\text {元}) $\n\n$900 + 20000 = 20900$（元）答: 到期后可取回本金和利息共 20900 元。",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "19-20六年级下·河南洛阳·期末",
+        "merged_id": "unit2_238",
+        "original_id": "unit2_fill_238"
+      },
+      {
+        "id": "unit2_fill_239",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "（2024·河北沧州·小升初真题）把8000元存入银行2年，年利率是 $2.1\\%$ ，到期后可得利息____元。",
+        "options": [],
+        "answer": "336",
+        "solution": "【分析】到期后可得的利息 = 存入银行的钱数 $\\times$ 年份数 $\\times$ 年利率，据此代入数据作答即可。掌握利息计算公式是解答本题的关键。\n\n【详解】 $8000 \\times 2 \\times 2.1\\%\\\\= 16000 \\times 2.1 \\% \\\\ = 3 3 6 \\text {(元)} $",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "2024·河北沧州·小升初真题",
+        "merged_id": "unit2_239",
+        "original_id": "unit2_fill_239"
+      },
+      {
+        "id": "unit2_fill_240",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "（20-21六年级下·山东临沂·期中）某商品打七折销售，就表示现价是原价的( )%，现价比原价降低了（ ）%。",
+        "options": [],
+        "answer": "70；30",
+        "solution": "【分析】打几折表示现价是原价的百分之几十，据此解答即可。本题考查折扣问题, 解答本题的关键是掌握打折的含义。\n\n【详解】某商品打七折销售，就表示现价是原价的 $70 \\%$ , 现价比原价降低了 $30 \\%$ 。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "20-21六年级下·山东临沂·期中",
+        "merged_id": "unit2_240",
+        "original_id": "unit2_fill_240"
+      },
+      {
+        "id": "unit2_fill_241",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（20-21六年级下·全国·单元测试）一家饭店2月份的营业额中应纳税的部分是50万元。如果按应纳税部分的 $3\\%$ 纳增值税，这家饭店2月份应缴纳增值税( )万元。",
+        "options": [],
+        "answer": "1.5",
+        "solution": "【分析】应缴纳增值税 $=$ 应纳税部分营业额 $\\times$ 税率, 代入数据计算即可。税率是表示税额和各种收入的比值。正确掌握 '税率'的概念和各部分之间的关系并灵活运用是解答本题的关键。\n\n【详解】 $50 \\times 3\\% = 1.5$ （万元）",
+        "knowledge_points": [
+          "税率问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_241",
+        "original_id": "unit2_fill_241"
+      },
+      {
+        "id": "unit2_fill_242",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(20-21 六年级下·全国·单元测试) 妈妈买了一部新手机，原价4000元，因店庆优惠只花了九折的钱，妈妈因此省了（ ）元。",
+        "options": [],
+        "answer": "400",
+        "solution": "【分析】首先弄清折数与百分数的关系，一折 $= 10\\%$ ，根据现在只花了九折的钱，确定把原价看作单位'1'，要求比原价便宜了多少元，需先求出比原价便宜了百分之几（1-90%），再根据一个数乘百分数的意义解答。解答此题的关键是找单位'1'，进一步发现比单位'1'多或少百分之几，由此解决问题。\n\n【详解】 $4000 \\times (1 - 90\\%)\\\\= 4000 \\times 10 \\% \\\\ = 4 0 0 0 \\times 0. 1 \\\\ = 4 0 0 \\text {(元)} $",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_242",
+        "original_id": "unit2_fill_242"
+      },
+      {
+        "id": "unit2_fill_243",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "（20-21六年级下·全国·单元测试）一件400元的羽绒服，打折后卖240元，现价是原价的( )%，比原价便宜了( )%。",
+        "options": [],
+        "answer": "60；40",
+        "solution": "【分析】用现价除以原价即可求出折扣；把原价看作单位'1'，用 1 - 现价所占的百分率即可求出比原价便宜百分之几。这道题考查了百分数的应用题, 熟练掌握折扣的含义是解答本题的关键。\n\n【详解】 $240 \\div 400 = 60\\%$ ; $1 - 60 \\% = 40 \\%$",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_243",
+        "original_id": "unit2_fill_243"
+      },
+      {
+        "id": "unit2_fill_244",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "（20-21六年级下·全国·单元测试）2016年3月，王爷爷把5000元钱存入银行，存期为5年，年利率为 $4.75\\%$ 。到期时王爷爷一共能取回( )元钱。",
+        "options": [],
+        "answer": "6187.5",
+        "solution": "【分析】本题中，本金是5000元，利率是 $4.75 \\%$ , 存期是 5 年, 要求到期后王爷爷共能取回多少元, 求的是本金和利息的和, 根据关系式: 本息 $=$ 本金 + 本金 $\\times$ 利率 $\\times$ 存期, 解决问题。此题属于利息问题，运用关系式：本息 $=$ 本金 + 本金 $\\times$ 利率 $\\times$ 存期解答即可。\n\n【详解】 $5000 + 5000 \\times 4.75\\% \\times 5\\\\= 5 0 0 0 + 1 1 8 7. 5 \\\\ = 6 1 8 7. 5 \\text {(元)} $",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_244",
+        "original_id": "unit2_fill_244"
+      },
+      {
+        "id": "unit2_fill_245",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(20-21 六年级下·全国·单元测试) 比 50 千克多二成是( )千克; 120 米比( )米少 $40\\%$",
+        "options": [],
+        "answer": "60；200",
+        "solution": "【分析】首先理解'成'的概念，'二成'就是 $20\\%$ ，把50千克看作单位'1'，增加50千克的 $20\\%$ ，增加后的数量是50千克的 $(1 + 20\\%)$ ，所以列式为： $50 \\times (1 + 20\\%)$ ；把要求的长度看成单位'1'，它的 $(1 - 40\\%)$ 就是120米，由此用除法求出这个长度。解答此类问题, 首先找准单位 '1', 进一步理清解答思路, 列式的顺序, 从而较好的解答问题。\n\n【详解】 $50 \\times (1 + 20\\%)\\\\= 50 \\times 120 \\%\\\\= 60（千克）$\n\n$120 \\div (1 - 40 \\%) \\\\ = 120 \\div 60 \\% \\\\ = 2 0 0 \\text {(米)} $",
+        "knowledge_points": [
+          "成数问题",
+          "已知比一个数多/少百分之几是多少，求这个数"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_245",
+        "original_id": "unit2_fill_245"
+      },
+      {
+        "id": "unit2_fill_246",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "(20-21 六年级下·全国·单元测试) 按规律填数。100%, 0.9, $\\frac{4}{5}$ , ( ) (成数), ( ) (百分数), ( )(小数), ( ) (分数)。",
+        "options": [],
+        "answer": "七成；$60\\%$；0.5；$\\frac{2}{5}$",
+        "solution": "【分析】根据给出的数列，发现依次减少 0.1 ，再根据题目中的要求化成需要的数即可。关键是根据给出的数列，找出数与数之间变化的规律，再由规律解决问题。\n\n【详解】 $\\frac{4}{5} - 0.1 = 0.7 =$ 七成;\n\n$0.7 - 0. 1 = 0. 6 = 60 \\%; \\\\ 0. 6 - 0. 1 = 0. 5 \\\\ 0. 5 - 0. 1 = 0. 4 = \\frac {2}{5}; $",
+        "knowledge_points": [
+          "分数、小数、百分数与成数的互化"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_246",
+        "original_id": "unit2_fill_246"
+      },
+      {
+        "id": "unit2_fill_247",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(2019·全国·小升初模拟) $\\frac{3}{5} = ( ) :20 = ( ) \\div 15 = ( ) \\% = ( )折$。",
+        "options": [],
+        "answer": "12；9；60；六",
+        "solution": "【分析】从已知的 $\\frac{3}{5}$ 入手, 根据分数与除法和比的关系, 以及它们通用的基本性质, 进行填空, 用分子 $\\div$ 分母, 化成小数, 再将小数化成百分数, 写出折数即可。分数的分子相当于被除数，分母相当于除数。\n\n【详解】 $20 \\div 5 \\times 3 = 12; 15 \\div 5 \\times 3 = 9; 3 \\div 5 = 0.6 = 60\\% =$ 六折 $\\frac{3}{5} = \\frac{12}{12}: 20 = \\frac{19}{15} = \\frac{60\\%}{60\\%} = \\frac{1}{6}$ 六折。",
+        "knowledge_points": [
+          "百分数、分数、小数和比的互化"
+        ],
+        "source": "2019·全国·小升初模拟",
+        "merged_id": "unit2_247",
+        "original_id": "unit2_fill_247"
+      },
+      {
+        "id": "unit2_fill_248",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(19-20 六年级下·陕西宝鸡·期末) 李大爷在 3 年前把5000元存入银行，定期3年，年利率是 $5.2\\%$ 。今年我国西南几省发生特大旱灾，他准备在这笔存款到期时将获得的利息( )元全部捐给灾区（利息税率为 $5\\%$ ）。",
+        "options": [],
+        "answer": "741",
+        "solution": "【分析】根据定期时间和年利率，先计算出 5000 元 3 年的利息总额，再乘上 $95\\%$ 算出李大爷 3 年可获得的实际利息。本题考查了百分数的应用, 利息 = 本金 $\\times$ 利率 $\\times$ 时间 $\\times$ (1 - 利息税)。\n\n【详解】 $5000 \\times 3 \\times 5.2\\% \\times (1 - 5\\%)\\\\= 15000 \\times 5.2 \\% \\times 95 \\% \\\\ = 7 4 1 \\text {(元)} $\n\n所以，李大爷有 741 元利息可捐给灾区。",
+        "knowledge_points": [
+          "税率问题",
+          "利率问题"
+        ],
+        "source": "19-20六年级下·陕西宝鸡·期末",
+        "merged_id": "unit2_248",
+        "original_id": "unit2_fill_248"
+      },
+      {
+        "id": "unit2_fill_249",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（20-21 六年级下·全国·单元测试）李经理的月工资是 6500 元，扣除 5000 元个税免征额后的部分需要按 $3\\%$ 的税率缴纳个人所得税，不考虑其他因素，李经理每月实际得到( )元工资。",
+        "options": [],
+        "answer": "6455",
+        "solution": "【分析】加纳的个人所得税 $=$ 需缴纳个人所得税部分的工资 $\\times$ 税率, 实际所得工资 = 月工资一个人所得税即可。此题考查了税率问题，明确交税部分的工资金额是解题关键。\n\n【详解】 $(6500-5000) ×3\\%= 1500 \\times 3 \\% \\\\ = 4 5 \\text {(元)} $\n\n$6500 - 45 = 6455（元）$李经理每月实际得到6455元工资。",
+        "knowledge_points": [
+          "税率问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_249",
+        "original_id": "unit2_fill_249"
+      },
+      {
+        "id": "unit2_fill_250",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "（20-21 六年级下·全国·单元测试）一块稻田去年产水稻 3.4 吨，今年增产一成，这块稻田今年水稻( )吨。",
+        "options": [],
+        "answer": "3.74",
+        "solution": "【分析】根据学过的成数，把去年稻田的产量看作单位'1'今年增产一成，增加了 $10\\%$ 的产量，，今年的产量是 $1 + 10\\%$ ，去年的产水稻是3.4吨，这块稻田今年水稻的吨数： $3.4 \\times (1 + 10\\%)$ ，即可算出。本题考查百分数成数的问题，注意几成就是百分之几十。\n\n【详解】 $3.4 \\times (1 + 10\\%)\\\\= 3. 4 \\times 1. 1 \\\\ = 3. 7 4 (\\text {吨}) $",
+        "knowledge_points": [
+          "成数问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_250",
+        "original_id": "unit2_fill_250"
+      },
+      {
+        "id": "unit2_fill_251",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "（20-21六年级下·全国·单元测试）学校开展'节水进校园'活动，十月份的用水量比九月份节约二成，也就是说十月份的用水量比九月份节约( )%，十月份的用水量是九月份的( )%。",
+        "options": [],
+        "answer": "20；80",
+        "solution": "【分析】把九月份的用水量看作单位'1'，那么十月份的用水量比九月份节约二成，也就是节约了 $20\\%$ ，十月份用水量是： $1 - 20\\%$ ，十月份的用水量是九月份的（ $1 - 20\\%$ ）÷1，即可算出。本题考查利用成数解答问题, 关键是几成就是百分之几十。\n\n【详解】十月份用水量比九月份节约 $20\\%$\n\n$(1 - 20 \\%) \\div 1 \\\\ = 80 \\% \\div 1 \\\\ = 80 \\% $",
+        "knowledge_points": [
+          "成数问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_251",
+        "original_id": "unit2_fill_251"
+      },
+      {
+        "id": "unit2_fill_252",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "（20-21六年级下·全国·单元测试）王奶奶把20000元存入银行，定期三年到期时她一共从银行取出22250元，存款的年利率是( )%。",
+        "options": [],
+        "answer": "3.75",
+        "solution": "【分析】由题意知: 先求出利息 $22250 - 20000 = 2250$(元), 再根据利息公式的推导公式: 利率 $=$ 利息 $\\div$ (本金 $\\times$ 期限) $\\times 100 \\%$ , 据此解答。求出利息后，利用利息公式的推导公式'利率 $=$ 利息 $\\div$ （本金 $\\times$ 期限） $\\times 100\\%$ '，是解答本题的关键。\n\n【详解】利息为： $22250 - 20000 = 2250$ （元）\n\n利率是： $2250 \\div (20000 \\times 3) \\times 100\\%\\\\= 2250 \\div 60000 \\times 100 \\% \\\\ = 3.75 \\% $",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_252",
+        "original_id": "unit2_fill_252"
+      },
+      {
+        "id": "unit2_fill_253",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（20-21六年级下·全国·单元测试）佳友超市2020年10月份的营业额中的应纳税部分是450万元，如果按 $6\\%$ 的税率缴纳增值税，那么佳友超市应缴纳增值税( )万元。",
+        "options": [],
+        "answer": "27",
+        "solution": "【分析】根据应缴税款 $=$ 应纳税部分 $\\times$ 税率，据此解答。掌握应缴税款的计算方法即税款 = 应纳税部分 $\\times$ 税率, 是解答此题的关键。\n\n【详解】 $450 \\times 6\\% = 27$ （万）",
+        "knowledge_points": [
+          "税率问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_253",
+        "original_id": "unit2_fill_253"
+      },
+      {
+        "id": "unit2_fill_254",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(20-21 六年级下·全国·单元测试) 小贤的爸爸把 3000元存入银行，定期2年，年利率是 $2.10\\%$ ，到期时可获得利息( )元，一共能取回( )元。",
+        "options": [],
+        "answer": "126；3126",
+        "solution": "【分析】根据利息 $=$ 本金 $\\times$ 时间 $\\times$ 利率，代入数据计算即可；一共可以取回的钱数 = 本金 + 利息，据此解答。此题考查了利率问题, 掌握利息计算公式是解题关键。\n\n【详解】 $3000 \\times 2 \\times 2.10\\%\\\\= 6 0 0 0 \\times 0. 0 2 1\\\\= 1 2 6 (\\text {元})$\n\n$1 2 6 + 3 0 0 0 = 3 1 2 6 \\text {(元)}$到期时可获得利息 126 元，一共能取回 3126 元。",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_254",
+        "original_id": "unit2_fill_254"
+      },
+      {
+        "id": "unit2_fill_255",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "(18-19 六年级下·全国·期中)六一儿童节当天, 某游乐场门票打八五折, 表示现价是原价的( )%; 某儿童文具店所有学习用品一律打九折, 相当于降价( )%。",
+        "options": [],
+        "answer": "85；10",
+        "solution": "【分析】由题意知：某游乐场门票打八五折，是以原价为单位 '1', 现价相当于原价的 $85 \\%$ ; 学习用品一律打九折,是以原价为单位 '1', 现价相当于原价的 $90 \\%$ , 也就是降价了 $10 \\%$ 。据此解答。找准单位 '1', 理解打几折就表示现价是原价的百分之多少是解答的关键。\n\n【详解】某游乐场门票打八五折，表示现价是原价的（85）%；某儿童文具店所有学习用品一律打九折，相当于降价（90）%。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "18-19六年级下·全国·期中",
+        "merged_id": "unit2_255",
+        "original_id": "unit2_fill_255"
+      },
+      {
+        "id": "unit2_fill_256",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(20-21 六年级下·全国·单元测试) ( ) : $():15 = 4 \\div ( ) =0.8 = \\frac{()}{20} = ( ) 折 = ( ) 成$",
+        "options": [],
+        "answer": "12；5；16；80；八；八",
+        "solution": "【分析】等式中的结果都等于0.8，根据等于0.8结算其它的未知数。利用比例的基本性质的，比号就是除号，一个数比 15 就是一个数除以 15，等于 0.8，用 $0.8 \\times 15$ ，算出这个数；4 除以一个数等于 0.8，就用 4 除以 0.8，算出。把 0.8 化成分母是 20 的分数，求出分子即可；0.8 换成百分数即可；百分之几十就是打几折；百分之几十就是几成，即可解答。本题考查比例、除法、小数、分数、百分数、折扣、成数之间的关系，关键是比的基本性质，商不变的性质，分数和除法的转换的熟练运用。\n\n【详解】 $12: 15 = 4 \\div 5 = 0.8 = \\frac{16}{20} = 80\\% =$ 八折 $=$ 八成",
+        "knowledge_points": [
+          "百分数、分数、小数和比的互化",
+          "成数问题",
+          "比的基本性质",
+          "折扣问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_256",
+        "original_id": "unit2_fill_256"
+      },
+      {
+        "id": "unit2_fill_257",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（20-21六年级下·全国·单元测试）服装店老板买进500双袜子，每双进价3元，原定零售价是4元。因为太贵，没人买，老板决定按零售价八折出售，卖了 $60\\%$ ，剩下的又按原零售价的七折售完。请你算一下，卖完着500双袜子时（）。",
+        "options": [],
+        "answer": "盈利20元",
+        "solution": "【分析】计划零售价 4 元, 则打八折后的价格是 $4 \\times 80\\%$ 元,所以卖了 300 双, 可卖 $4 \\times 80\\% \\times 300$ 元, 同理可知, 剩下的 $500 - 200$ 双可卖 $4 \\times 70\\% \\times (500 - 300)$ 元, 将所卖钱数相加后与成本价比较即得全部卖出后是盈利还是亏本。售价 $=$ 标价 $\\times$ 折扣, 打几折相当于乘上百分之几十,分别求出售价以及所对应的数量是解题的关键。\n\n【详解】成本 $= 500 \\times 3 = 1500$ 元；八折售价 $= 4 \\times 80\\% = 3.2$ 元；七折售价 $= 4 \\times 70\\% = 2.8$ 元卖 $60\\% = 500 \\times 0.6 \\times 3.2 = 960$ 元；卖剩下的 $= 500 \\times 0.4 \\times 2.8 = 560$ 元一共盈利 $= 960 + 560 - 1500 = 20$ 元 500 双袜子卖完盈利 20 元。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_257",
+        "original_id": "unit2_fill_257"
+      },
+      {
+        "id": "unit2_fill_258",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "(20-21 六年级下·全国·单元测试) 某商场在 '六一'期间益智类玩具打'六六折'促销，也就是把这类商品优惠了 ___%。",
+        "options": [],
+        "answer": "34",
+        "solution": "【分析】打'六六折'意思是现价是原价的 $66 \\%$ , 便宜了原价的 $34 \\%$ , 据此解答即可。明确六六折的含义是解答本题的关键。\n\n【详解】 六六折 $= 66\\%$ 1- 66% = 34%, 商品优惠了 34%。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_258",
+        "original_id": "unit2_fill_258"
+      },
+      {
+        "id": "unit2_fill_259",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（20-21 六年级下·全国·单元测试）原价是 1200 元的商品，打九折出售，售价是______ 元，比原价便宜___元。",
+        "options": [],
+        "answer": "1080；120",
+        "solution": "【分析】现价 $=$ 原价 $\\times$ 折扣, 现价比原价便宜的钱数 $=$ 原价一现价，代入数值计算即可。几折就是百分之几十，求一个数的百分之几是多少用乘法计算。\n\n【详解】 现价 $= 1200 \\times 90\\%\\\\= 1 2 0 0 \\times 0. 9 \\\\ = 1 0 8 0 (\\text {元}) $\n\n$1200 - 1080 = 120$ （元）所以现在售价是 1080 元，比原价便宜 120 元。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_259",
+        "original_id": "unit2_fill_259"
+      },
+      {
+        "id": "unit2_fill_260",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(20-21 六年级下·全国·单元测试) 元旦节, 如果游乐园门票八折优惠，门票现价是原价的 _%, 文具店的学习用品按九折出售，原来买50元的学习用品，可以节省_元。",
+        "options": [],
+        "answer": "80；5",
+        "solution": "【分析】折扣，把一个商品打折出售，几折就是百分之几十, 即现价是原价的百分之几十; 节省的钱数 = 原价 $\\times$ (1 - 折扣数), 据此解答即可。解答本题的关键是要明确折扣的含义，理解折扣、现价和原价之间的关系。\n\n【详解】门票现价是原价的 $80\\%$\n\n$50 \\times (1 - 90 \\%) \\\\ = 50 \\times 10 \\% \\\\ = 5 (\\text {元}) $",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_260",
+        "original_id": "unit2_fill_260"
+      },
+      {
+        "id": "unit2_fill_261",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(20-21 六年级下·全国·单元测试) _: 16 = $\\frac{3}{4}$ = 27 \\div _= _\\% = _折。",
+        "options": [],
+        "answer": "12；36；75；七五",
+        "solution": "【分析】根据分数的基本性质：分数的分子和分母同时乘或除以同一个不为 0 的数, 分数的大小不变; 分数与除法的关系: 被除数 $\\div$ 除数 $= \\frac{\\text{被除数}}{\\text{除数}}$ 。此题考查的是分数与除法、比和除法之间的关系以及小数和百分数的互化方法。\n\n【详解】 $\\frac{3}{4} = \\frac{3 \\times 4}{4 \\times 4} = \\frac{12}{16} = 12:16$ ; $\\frac {3}{4} = \\frac {4 \\times 9}{3 \\times 9} = \\frac {2 7}{3 6} = 2 7 \\div 3 6;$ $\\frac{3}{4} = 3\\div 4 = 0.75 = 75\\%;$\n\n$0.75=$ 七五折 12:16= $\\frac{3}{4} = 27 \\div 36 = 75\\% =$ 七五折",
+        "knowledge_points": [
+          "百分数、小数和分数的互化",
+          "分数的基本性质及应用",
+          "分数与除法的关系"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_261",
+        "original_id": "unit2_fill_261"
+      },
+      {
+        "id": "unit2_judge_262",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "（2021·山东临沂·小升初真题）今年某品牌汽车的销售量比去年增加两成，今年汽车的销售量是去年的 $120 \\%$ 。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】去年销售量是单位'1'，今年销售量占去年的 $1 + 20\\%$ ，据此分析。关键是理解成数的意义，成数就是百分之几十。\n\n【详解】 $1 + 20\\% = 120\\%$ ，所以原题说法正确。",
+        "knowledge_points": [
+          "求一个数是另一个数的百分之几（百分率问题）",
+          "成数问题。"
+        ],
+        "source": "2021·山东临沂·小升初真题",
+        "merged_id": "unit2_262",
+        "original_id": "unit2_judge_262"
+      },
+      {
+        "id": "unit2_judge_263",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "（2024·河北沧州·小升初真题）商店按 $5\\%$ 的税率交营业税 20 元，则营业额是 2 万元。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】根据营业税 $=$ 营业额 $\\times$ 税率，可以求得营业税，然后再和 20 元比较即可。本题考查了税率有关的知识，本题根据 $5 \\%$ 的税率和营业税 20 元，还可以先求出营业额再和 2 万元比较。\n\n【详解】 $20000 \\times 5\\% = 1000$ （元） $\\neq 20$ 元，原题说法错误；故答案为：×。",
+        "knowledge_points": [
+          "税率问题"
+        ],
+        "source": "2024·河北沧州·小升初真题",
+        "merged_id": "unit2_263",
+        "original_id": "unit2_judge_263"
+      },
+      {
+        "id": "unit2_judge_264",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "（20-21六年级下·全国·单元测试）东方饭店2月份的营业额是90万元，按规定应缴纳 $5\\%$ 的营业税，这家酒店缴纳营业税后的收入是4.5万元。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】由题意知，营业额为90万元，税率是 $5 \\%$ , 要求除去营业税后的收入是多少钱, 根据 '税后的收入 = 营业额 $\\times$ (1 - 税率)', 列式解答。此题解答的关键是根据'税后的收入 $=$ 营业额 $\\times$ (1 - 税率）'，这一关系式列式。\n\n【详解】 $90 \\times (1 - 5\\%)\\\\= 90 \\times 95 \\% \\\\ = 8 5. 5 (\\text {万 元}); $\n\n这家酒店缴纳营业税后的收入是85.5万元。故判断错误。",
+        "knowledge_points": [
+          "求税率或收入额"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_264",
+        "original_id": "unit2_judge_264"
+      },
+      {
+        "id": "unit2_judge_265",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "（19-20六年级下·河南信阳·期中）一件小家电原价200元，现价170元，是按八五折出售的。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】题目给出了商品的原价和现价，用现价除以原价，得到相应的折扣。本题考查的是百分数应用题中的折扣问题，在折扣问题中，折扣 $=$ 现价 $\\div$ 原价。\n\n【详解】 $170 \\div 200 \\times 100\\% = 85\\%$ 该商品是按八五折出售的，题干阐述正确，答案为：√。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "19-20六年级下·河南信阳·期中",
+        "merged_id": "unit2_265",
+        "original_id": "unit2_judge_265"
+      },
+      {
+        "id": "unit2_solve_266",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "(19-20六年级下·山东济南·期末) 商店卖一种玩具,如果每个售价为200元，那么售价的 $70\\%$ 是进价，售价的30%就是赚的钱。现在要搞促销活动，为保证一个玩具赚的钱不少于30元，应该怎样确定折扣？",
+        "options": [],
+        "answer": "八五折",
+        "solution": "【分析】先用售价 200 元乘 $70 \\%$ , 求出进价。再将进价加上 30 元, 求出为保证一个玩具赚的钱不少于 30 元的最低定价,最后利用最低定价除以原售价, 求出折扣。本题考查了折扣问题, 能正确理解题意并列式是解题的关键。\n\n【详解】 $200 \\times 70\\% = 140$ （元）\n\n$(1 4 0 + 3 0) \\div 2 0 0 \\\\ = 1 7 0 \\div 2 0 0 \\\\ = 85 \\% $\n\n$85\\% =$ 八五折\n\n答: 为保证一个玩具赚的钱不少于 30 元, 折扣应为八五折。",
+        "knowledge_points": [
+          "求一个数是另一个数的百分之几（百分率问题）",
+          "折扣问题"
+        ],
+        "source": "19-20六年级下·山东济南·期末",
+        "merged_id": "unit2_266",
+        "original_id": "unit2_solve_266"
+      },{
+        "id": "unit2_solve_267",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "(19-20 六年级下·河南洛阳·期末) 李叔叔得到了 5000元劳务费用。其中1000元是免税的，其余部分按 $20\\%$ 的税率缴税。这笔劳务费用一共要缴税多少元？",
+        "options": [],
+        "answer": "800元",
+        "solution": "【分析】用劳务费减去免税金额，求出缴税部分，再乘税率即可，据此列式解答。求一个数的百分之几是多少用乘法, 纳税是每个公民应尽的义务。\n\n【详解】 $(5000 - 1000) \\times 20\\%\\\\=64000\\times 20\\%\\\\=8 0 0 \\text {(元)}$\n\n答: 这笔劳务费用一共要缴税 800 元。",
+        "knowledge_points": [
+          "税率问题"
+        ],
+        "source": "19-20六年级下·河南洛阳·期末",
+        "merged_id": "unit2_267",
+        "original_id": "unit2_solve_267"
+      },
+      {
+        "id": "unit2_solve_268",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（2022·吉林四平·小升初真题）小丽家买了一套售价为32万元的普通商品房。他们选择一次付清房款，可以按九六折优惠价付款。\n\n(1) 打折后房子的总价是多少元？\n(2) 买这套房子还要按照实际购房价的 $1.5\\%$ 缴纳契税,契税是多少元?",
+        "options": [],
+        "answer": "(1) 307200 元; (2) 4608 元",
+        "solution": "【分析】（1）由题意可知，一次付清房款，按原价 $96 \\%$ 付款，根据'现价 $=$ 原价 $\\times$ 折扣'进行解答即可；\n\n(2) 用实际房价乘 $1.5\\%$ 即可求出应缴纳的契税。解答本题的关键是要明确'现价、原价和折扣'之间的关系，求出实际的房价；再根据百分数乘法的意义，求出应缴纳的契税。\n\n【详解】九六折 $= 96\\%$ (1) 32万元 $= 320000$ 元 $320000 \\times 96\\% = 307200$ （元）答: 打折后房子的总价是 307200 元。(2) $307200 \\times 1.5\\% = 4608$ （元）答: 契税是 4608 元。",
+        "knowledge_points": [
+          "求一个数的百分之几是多少",
+          "折扣问题"
+        ],
+        "source": "2022·吉林四平·小升初真题",
+        "merged_id": "unit2_268",
+        "original_id": "unit2_solve_268"
+      },
+      {
+        "id": "unit2_solve_269",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "(20-21 六年级下·广东汕头·期末) 书店有一套百科全书原价240元，现按六折出售，买一套可以便宜多少元？",
+        "options": [],
+        "answer": "96元",
+        "solution": "【分析】根据'现价 $=$ 原价 $\\times$ 折扣'求出现价，再用原价减去现价即可。熟练掌握'现价、原价、折扣'之间的关系是解答本题的关键。\n\n【详解】六折 $= 60\\%$ $240 - 240 \\times 60 \\% \\\\ = 2 4 0 - 1 4 4 \\\\ = 9 6 \\text {(元)} ; $\n\n答: 买一套可以便宜 96 元。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "20-21六年级下·广东汕头·期末",
+        "merged_id": "unit2_269",
+        "original_id": "unit2_solve_269"
+      },
+      {
+        "id": "unit2_solve_270",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "（20-21六年级下·云南玉溪·期末）妈妈花360元买了一台打八折出售的微波炉，这台微波炉原价是多少元？妈妈比平时购买省了多少元？",
+        "options": [],
+        "answer": "450元；90元",
+        "solution": "【分析】将原价看作单位'1'，现价÷折扣 $=$ 原价，原价-现价 $=$ 省的钱数。关键是理解折扣的意义，几折就是百分之几十。\n\n【详解】 $360 \\div 80\\% = 450$ （元）$450 - 360 = 90$ (元) 答: 这台微波炉原价是 450 元, 妈妈比平时购买省了 90 元。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "20-21六年级下·云南玉溪·期末",
+        "merged_id": "unit2_270",
+        "original_id": "unit2_solve_270"
+      },
+      {
+        "id": "unit2_solve_271",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "(20-21 六年级下·新疆省直辖县级单位·期末) 李明用2000元买了2年的国债券，年利率为 $4.5\\%$ 。到期后他把利息捐给了'希望工程'。李明给'希望工程'捐了多少元？",
+        "options": [],
+        "answer": "180元",
+        "solution": "【分析】'利息 $=$ 本金 $\\times$ 存期 $\\times$ 利率' 据此解答即可。熟练掌握求利息的公式是解答本题的关键。\n\n【详解】 $2000 \\times 2 \\times 4.5\\%\\\\= 4000 \\times 4.5 \\% \\\\ = 1 8 0 \\text {(元)}; $\n\n答: 李明给 '希望工程' 捐了 180 元。",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "20-21六年级下·新疆省直辖县级单位·期末",
+        "merged_id": "unit2_271",
+        "original_id": "unit2_solve_271"
+      },
+      {
+        "id": "unit2_solve_272",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "（19-20六年级下·河南南阳·期末）商店卖一种自行车，如果每辆售价2000元，那么售价的 $60\\%$ 是进价，现在要搞促销活动，为保证每辆自行车赚的钱不少于300元，最低应打几折？",
+        "options": [],
+        "answer": "七五折",
+        "solution": "【分析】用 $2000 \\times 60\\%$ 先求出自行车的进价，搞促销活动每辆自行车赚的钱不少于 300 元，则用进价加上 300 元即为搞促销期间的最低价格，再根据 '现价÷原价 = 折扣' 解答即可。解答本题的关键是求先出自行车的进价，进而求出活动期间的现价，再根据现价、折扣、原价之间的关系解答。\n\n【详解】 $(2000 \\times 60\\% + 300) \\div 2000\\\\= 1 5 0 0 \\div 2 0 0 0\\\\= 0. 7 5\\\\0.75 =七五折$;\n\n答: 最低应打七五折。",
+        "knowledge_points": [
+          "求一个数的百分之几是多少",
+          "折扣问题"
+        ],
+        "source": "19-20六年级下·河南南阳·期末",
+        "merged_id": "unit2_272",
+        "original_id": "unit2_solve_272"
+      },
+      {
+        "id": "unit2_solve_273",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "（2021·江西九江·小升初真题）六年级61名学生去游乐园玩，每张门票30元，暑假期间有优惠促销，请你参考一下，哪种购票方式最划算？\n\n(1) 30人以上可购团体票，每张按九折出售。\n(2) 买 9 张送 1 张, 不满 9 张不赠送。\n(3) 每满 500 元返还 50 元。",
+        "options": [],
+        "answer": "方案一购票方式最划算",
+        "solution": "【分析】先按每一种方案分别算出总价，再进行比较即可知道哪种购票方式最划算。方案一每张按九折出售，即单价按 $90\\%$ 计算；方案二是可赠送部分的票价加上不可赠送部分的票价就是购票总价；方案三是先算出不返还时的总价，再减去返还的价钱就是购票的总价。理解好题意，掌握折扣知识，会计算赠送及返还后的购票总价，这是解决此题的关键。\n\n【详解】 (1) $61 \\times 30 \\times 90\\%\\\\= 1830\\times 90\\% \\\\ = 1 6 4 7 \\text {(元)} $\n\n(2) $(61 - 6) \\times 30\\\\=55\\times30\\\\=1650$\n\n(3) $61 \\times 30 - 50 \\times 3\\\\ = 1 8 3 0 - 1 5 0 \\\\ = 1 6 8 0 \\text {(元)} $\n\n$1 6 4 7 < 1 6 5 0 < 1 6 8 0$答：方案一购票方式最划算。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "2021·江西九江·小升初真题",
+        "merged_id": "unit2_273",
+        "original_id": "unit2_solve_273"
+      },
+      {
+        "id": "unit2_solve_274",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（20-21六年级下·湖北黄冈·期中）某商品的进价是800元，标价是1100元。商店要求以获利不低于 $10\\%$ 的售价打折出售，则最低可以打几折出售此商品？",
+        "options": [],
+        "answer": "八折",
+        "solution": "【分析】先求出不低于 $10\\%$ 的售价，用进价的 $10\\%$ 即是最低的利润。然后用进价加上利润即是最低要卖的价格，用最低的价格除以标价乘 $100\\%$ 即为所求。本题考查折扣问题, 明确折扣和百分数的关系是解题的关键。\n\n【详解】 $800 + 800 \\times 10\\%\\\\= 8 0 0 + 8 0 \\\\ = 8 8 0 \\text {(元)} $\n\n$880 \\div 1100 \\times 100 \\% \\\\ = 0.8 \\times 100 \\% \\\\ = 80 \\% $\n\n$80 \\% = \\text{八折}$ 答: 最低可以打八折出售此商品。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "20-21六年级下·湖北黄冈·期中",
+        "merged_id": "unit2_274",
+        "original_id": "unit2_solve_274"
+      },
+      {
+        "id": "unit2_solve_275",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "(18-19六年级下·江苏南京·期末) 张叔叔在商场买了一双标价400元的运动鞋，购买时商场正好搞活动，该运动鞋打八折并且购物满300元返还现金40元。张叔叔买的这双鞋实际打了几折？",
+        "options": [],
+        "answer": "七折",
+        "solution": "【分析】打八折的意思是降到原价的 $80\\%$ ，我们先求出张叔叔实际花了多少钱，再求实际花的钱数是原价的百分之几，进而求出打了几折。本题关键是对打折的理解，打折是将商品的原有价格进行百分比的换算后卖出，打几折就是原价的百分之几十。\n\n【详解】 $400 \\times 80\\% = 320$ （元） $3 2 0 - 4 0 = 2 8 0 \\text {(元)}$\n\n $280 \\div 400 \\times 100 \\% \\\\ = 0.7 \\times 100 \\% \\\\ = 70 \\% $\n\n70%就是打七折答: 张叔叔买的这双鞋实际打了七折。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "18-19六年级下·江苏南京·期末",
+        "merged_id": "unit2_275",
+        "original_id": "unit2_solve_275"
+      },
+      {
+        "id": "unit2_solve_276",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（19-20六年级下·浙江·期中）丽丽妈妈的服装店实行薄利多销的原则，一般在进价的基础上提高二成后作为销售价。照这样计算，一件进价为220元的衣服应标价多少元？",
+        "options": [],
+        "answer": "264元",
+        "solution": "【分析】丽丽妈妈的服装店实行薄利多销的原则，一般在进价的基础上提高二成后作为销售价。照这样计算，求一件进价为220元的衣服应标价多少元，是把进价看做单位'1'，售价比进价多 $20\\%$ ，已知进价，求售价，也就是已知单位'1'，求部分量，用乘法计算。列式为 $220 \\times (1 + 20\\%)$ 。解答此题的关键是理解成数，一成是 $10\\%$ ，二成就是 $20\\%$ 。\n\n【详解】 $220 \\times (1 + 20\\%)\\\\= 2 2 0 \\times 1. 2 \\\\ = 2 6 4 \\text {(元)} $\n\n答: 一件进价为 220 元的衣服应标价 264 元。",
+        "knowledge_points": [
+          "成数问题"
+        ],
+        "source": "19-20六年级下·浙江·期中",
+        "merged_id": "unit2_276",
+        "original_id": "unit2_solve_276"
+      },
+      {
+        "id": "unit2_solve_277",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（2020·湖北孝感·小升初真题）新华书店的图书凭会员卡可以打八折，小玲用会员卡买了一本书，省了2.4元。这本书原价多少元？",
+        "options": [],
+        "answer": "12元",
+        "solution": "【分析】根据题意, 把原价看作单位 '1', 则 2.4 元占原价的 $(1 - 80 \\%$ , 求单位 '1', 用除法计算。本题主要考查百分数的应用，关键找到单位'1'，利用数量关系做题。\n\n【详解】八折 $= 80\\%$ $2.4 \\div (1 - 80\\%) \\\\ = 2. 4 \\div 0. 2 \\\\ = 1 2 \\text {(元)} $\n\n答: 这本书原价 12 元。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "2020·湖北孝感·小升初真题",
+        "merged_id": "unit2_277",
+        "original_id": "unit2_solve_277"
+      },
+      {
+        "id": "unit2_solve_278",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（2020·北京东城·小升初真题）小新准备在网上书店买一套精装版《中国儿童百科全书》，原价300元。网上书店搞促销活动，打八折销售，现在买这套图书应付多少钱？",
+        "options": [],
+        "answer": "240元",
+        "solution": "【分析】打八折是指现价是原价的 $80 \\%$ , 把原价看成单位 '1', 用原价乘 $80 \\%$ 即可求出现价。本题考查打折问题, 关键是理解打折的含义: 打几折, 现价就是原价的百分之几十。\n\n【详解】 $300 \\times 80\\% = 240$ （元） 答: 现在买这套图书应付 240 元钱。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "2020·北京东城·小升初真题",
+        "merged_id": "unit2_278",
+        "original_id": "unit2_solve_278"
+      },
+      {
+        "id": "unit2_solve_279",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（20-21六年级下·广东肇庆·期中）爸爸将10万元存入银行，年利率是 $2.75\\%$ ，到期得到利息1.375万元，爸爸存了几年？",
+        "options": [],
+        "answer": "5年",
+        "solution": "【分析】由'利息 $=$ 本金 $\\times$ 利率 $\\times$ 时间'可知：时间 $=$ 利息÷本金÷利率，代入数据计算即可。本题主要考查利率问题，明确：利息、本金、利率、时间之间的关系是解题的关键。\n\n【详解】 $1.375 \\div 10 \\div 2.75\\%\\\\= 0. 1 3 7 5 \\div 0. 0 2 7 5 \\\\ = 5 (\\text {年}) $\n\n答: 爸爸存了 5 年。",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "20-21六年级下·广东肇庆·期中",
+        "merged_id": "unit2_279",
+        "original_id": "unit2_solve_279"
+      },
+      {
+        "id": "unit2_solve_280",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "（20-21 六年级下·全国·单元测试）某地去年早稻产量是 500 吨，晚稻产量 600 吨，晚稻产量比早稻增产几成？",
+        "options": [],
+        "answer": "二成",
+        "solution": "【分析】此题应把早稻的产量看作单位 '1', 然后根据求一个数比另一个数多百分之几, 用 ' (大数一小数) ÷ 单位 '1'的量解答即可。解答此题的关键是先判断单位 '1', 然后根据求一个数比另一个数多百分之几, 用 ' (大数一小数) ÷ 单位 '1'的量解答即可。\n\n【详解】 $(600-500) ÷ 500\\\\= 1 0 0 \\div 5 0 0 \\\\ = 20 \\%; $\n\n$20\\% =$ 二成;答: 晚稻产量比早稻增产二成。",
+        "knowledge_points": [
+          "成数问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_280",
+        "original_id": "unit2_solve_280"
+      },
+      {
+        "id": "unit2_solve_281",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "（20-21六年级下·全国·单元测试）妈妈存入银行10000元，定期两年，年利率是 $2.43\\%$ ，到期后，妈妈一共能取回多少元？",
+        "options": [],
+        "answer": "10486元",
+        "solution": "【分析】在此题中，本金是 10000 元，时间是 2 年，利率是 $2.43 \\%$ , 求到期后一共可以取回多少钱, 取回的应是本金和利息, 根据关系式: 本息 = 本金 + 本金 $\\times$ 利率 $\\times$ 时间, 解决问题即可。此题属于存款利息问题，根据关系式'本息=本金+本金×利率×时间'，即可解决问题。\n\n【详解】 $10000 + 10000 \\times 2.43\\% \\times 2\\\\= 1 0 0 0 0 + 4 8 6 \\\\ = 1 0 4 8 6 \\text {(元)}; $\n\n答: 妈妈一共能取回 10486 元钱。",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_281",
+        "original_id": "unit2_solve_281"
+      },
+      {
+        "id": "unit2_solve_282",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "（20-21 六年级下·全国·单元测试）利用收集到的存款利率算一算：甲用 2000 元先存一年定期，到期后连本带息再存一年定期；乙用 2000 元直接存了二年定期，哪种存款方式到期后获得的利息多？（银行的利率分别为：定期一年 $3.25\\%$ ，定期两年 $3.75\\%$ ）",
+        "options": [],
+        "answer": "直接存两年定期",
+        "solution": "【分析】根据本金及存期分别按两种储蓄办法计算分析即能确定哪种办法得到的利息多一些:甲的方法：先存入一年期的，年利率是 $3.25\\%$ ，等一年到期时再把本金和利息取出来合在一起，再存入银行一年。则先存一年后可得利息 $2000 \\times 3.25\\% \\times 1 = 65$ 元，第二年本金和利息共有 $2000 + 65$ 元，加在一起再存一年可得利息（2000 +65） $\\times 3.25\\% \\times 1 \\approx 67.11$ 元，两年共得利息 $65 + 67.11 = 132.11$ 元。乙的方法：年利率是 $3.75 \\%$ , 则存两年后可得利息: $2000 \\times 3.75 \\% \\times 2 = 150$ 元;150 元 $> 132.11$ 元, 所以直接存两年定期的方式到期后获得的利息多。根据利息 $=$ 本金 $\\times$ 年利率 $\\times$ 时间, 按两种方法分别进行分析计算得出结果是完成本题的关键。\n\n【详解】甲的方法可得利息：$2000 \\times 3.25 \\% \\times 1 = 65 (\\text{元}) $ $ (2000 + 65) \\times 3.25 \\% \\times 1 \\approx 67.11 (\\text{元}) $两年共得利息: $65 + 67.11 = 132.11$ (元)乙的方法可得利息：$2000 \\times 3.75 \\% \\times 2 = 150 (\\text{元}) $ $1 5 0 元 > 1 3 2. 1 1 元 $所以乙的存款方式得到的利息多一些。答：直接存两年定期的方式到期后获得的利息多。",
+        "knowledge_points": [
+          "选择储蓄的最佳方案"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_282",
+        "original_id": "unit2_solve_282"
+      },
+      {
+        "id": "unit2_solve_283",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（20-21六年级下·全国·单元测试）为庆祝'国际六一儿童节'，某体育用品店开展优惠大酬宾活动：凡进店购买体育用品一律六一折，张敏购买一个原价是220元的篮球和一个原价是280元的足球要多少元？",
+        "options": [],
+        "answer": "305元",
+        "solution": "【分析】六一折就是原价的 $61 \\%$ , 分别求出打折后篮球和足球的价格, 再相加解答即可。求一个数的百分之几是多少用乘法计算, 注意打几折就是百分之几十。\n\n【详解】 $220 \\times 61\\% + 280 \\times 61\\%\\\\= 1 3 4. 2 + 1 7 0. 8 \\\\ = 3 0 5 \\text {(元)} $\n\n答: 需要 305 元。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_283",
+        "original_id": "unit2_solve_283"
+      },
+      {
+        "id": "unit2_solve_284",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "（20-21六年级下·全国·单元测试）爸爸买了一个智能手机，原价1300元，现在只花了九折的钱，比原价便宜了多少钱？",
+        "options": [],
+        "answer": "130元",
+        "solution": "【分析】九折即现价是原价的 $90 \\%$ , 把原价看作单位'1'，则便宜的分率为（ $1 - 90\\%$ ），已知原价为1300元，运用乘法即可求出便宜了多少钱。解答本题的关键是找准单位 '1', 求出便宜的分率,再根据求一个数的几分之几是多少用乘法计算即可。\n\n【详解】 $1300 \\times (1 - 90\\%)\\\\= 1300 \\times 10 \\% \\\\ = 1 3 0 (\\text {元}); $\n\n答: 比原价便宜了 130 元。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_284",
+        "original_id": "unit2_solve_284"
+      },
+      {
+        "id": "unit2_solve_285",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "(20-21 六年级下·全国·单元测试) 一件商品按照 $30\\%$的利润出售，后来又打八折，最后的利润是520元，那么这件商品的成本价是多少元？",
+        "options": [],
+        "answer": "13000元",
+        "solution": "【分析】根据题干，设这件商品的成本是 $x$ 元，把成本价看做单位'1'，定价是 $(1 + 30\\%)$ x 元，八折是指现价是定价的 $80\\%$ ，根据：售价一成本 $=$ 利润，列出方程即可解答问题。解答此题关键是设出这件商品的成本价，从而得出定价、售价以及与利润之间的等量关系，列出方程即可解答问题。\n\n【详解】解：设这件商品的成本是 $x$ 元。\n\n$(1 + 30 \\%) x \\times 0.8 - x = 520$\n\n$1.0 4 \\mathrm {x} - \\mathrm {x} = 5 2 0$\n\n$0.0 4 \\mathrm {x} = 5 2 0$\n\n$x = 13000$\n\n答: 这件商品的成本价是 13000 元。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_285",
+        "original_id": "unit2_solve_285"
+      },
+      {
+        "id": "unit2_solve_286",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（19-20六年级·广东佛山·期中）李老师获得一笔劳务费，按规定：超出800元的部分要按 $20\\%$ 的税率缴纳个人所得税。李老师缴税120元，他实际收入是多少元？",
+        "options": [],
+        "answer": "1280元",
+        "solution": "【分析】根据题意可知: '超出 800 元的钱数 $\\times 20 \\% = 120$ ' 据此求出超出 800 元的钱数, 再加上 800 元即可求出总收入,最后减去缴纳的税款即可求出实际收入。解答本题的关键是理解'实际收入'，求出收入的总钱数后，还需要减去缴纳的税款。\n\n【详解】 $120 \\div 20\\% + 800 - 120\\\\= 6 0 0 + 8 0 0 - 1 2 0 \\\\ = 1 2 8 0 (\\text {元}) $\n\n答: 他实际收入是 1280 元。",
+        "knowledge_points": [
+          "税率问题"
+        ],
+        "source": "19-20六年级·广东佛山·期中",
+        "merged_id": "unit2_286",
+        "original_id": "unit2_solve_286"
+      },
+      {
+        "id": "unit2_solve_287",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（20-21六年级下·全国·单元测试）幸福村今年苹果大丰收，一共收苹果3.25万吨。今年苹果产量比去年增加了二成五。\n\n(1) 幸福村去年收苹果多少万吨？\n(2）王叔叔家也获得了丰收，为了方便运输他想买一辆3.3万元的农用三轮车。按规定，购买家用运输车需要缴纳 $10\\%$ 的车辆购置税，同时获得政府惠农补贴2000元。王叔叔买这辆车实际要花多少元？",
+        "options": [],
+        "answer": "(1) 2.6万吨 (2) 34300元",
+        "solution": "【分析】（1）把去年产量看作单位'1'，则今年产量是去年的 $(1 + 20\\%)$ , 已知今年的产量是 3.25 万吨, 根据分数除法的意义, 解答即可。(2) 王叔叔实际花的钱数 = 三轮车的价格 $\\times$ (1 + 10%) - 补贴的钱数, 据此解答。此题考查了有关成数和税率的综合应用，找准数量关系认真解答即可。\n\n【详解】 (1) $3.25 \\div (1 + 25\\%)\\\\= 3. 2 5 \\div 1. 2 5\\\\= 2.6（万吨）$\n\n答: 幸福村去年收苹果 2.6 万吨。\n\n(2) $33000 \\times (1 + 10\\%) - 2000\\\\= 3 3 0 0 0 \\times 1. 1 - 2 0 0 0 \\\\ = 3 6 3 0 0 - 2 0 0 0 \\\\ = 3 4 3 0 0 \\text {(元)} $\n\n答: 王叔叔买这辆车实际要花 34300 元。",
+        "knowledge_points": [
+          "成数问题",
+          "税率问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_287",
+        "original_id": "unit2_solve_287"
+      },
+      {
+        "id": "unit2_solve_288",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（20-21六年级下·全国·单元测试）店购进80个书包，每个成本为60元，商店按成本价提高 $40\\%$ 作为售价。当售出一半后，剩下的打八折全部售完。该商店共获利多少元？",
+        "options": [],
+        "answer": "1248元",
+        "solution": "【分析】售出一般，也就是40个书包，其单价为成本价×(1+40%)，另一半的单价为 $84 \\times 80\\%$ ，分别用单价 $\\times$ 数量，求出两部分卖出的总价，再减去成本价即可。此题考查了折扣问题，分别求出售出的两部分单价是解题关键。\n\n【详解】 $60 \\times (1 + 40\\%) \\times 40\\\\= 8 4 \\times 4 0 \\\\ = 3 3 6 0 (\\text {元}) \\\\ 84 \\times 80 \\% \\times 40 \\\\ = 6 7. 2 \\times 4 0 \\\\ = 2 6 8 8 \\text {(元)} \\\\ (3 3 6 0 + 2 6 8 8) - 6 0 \\times 8 0 \\\\ = 6 0 4 8 - 4 8 0 0 \\\\= 1248$ （元）\n\n答：该商店共获利1248元。",
+        "knowledge_points": [
+          "折扣问题",
+          "比一个数多/少百分之几的数是多少"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_288",
+        "original_id": "unit2_solve_288"
+      },
+      {
+        "id": "unit2_solve_289",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "(20-21 六年级下·全国·单元测试) 母亲节当天, 爱潮服装店的衣服全部打八折出售，持有贵宾卡还可以在打折的基础上再享受 $5 \\%$ 的优惠，乐乐的妈妈持有贵宾卡，买了一件原价580元的连衣裙，她实际付款多少元？",
+        "options": [],
+        "answer": "440.8元",
+        "solution": "【分析】把八折即按原价的 $80 \\%$ 出售，用原价 $\\times 80 \\%$ 求出折后价，再优惠 $5 \\%$ ，则实际付款钱数是折后价的（1- $5 \\%$ ），根据分数乘法的意义，用乘法解答即可。此题考查了折扣问题，明确打几折就是按原价的百分之几十出售，求一个数的百分之几是多少用乘法。\n\n【详解】 $580 \\times 80\\% \\times (1 - 5\\%)\\\\= 4 6 4 \\times 0. 9 5 \\\\ = 4 4 0. 8 \\text {(元)} $\n\n答: 她实际付款 440.8 元。",
+        "knowledge_points": [
+          "求一个数的百分之几是多少",
+          "求现价（折扣问题）"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_289",
+        "original_id": "unit2_solve_289"
+      },
+      {
+        "id": "unit2_solve_290",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（2021六年级下·全国·专题练习）小刚的爸爸参加一项研究活动，得到劳务费1500元。按照国家规定，个人劳务费收入1000元及以内的，交纳 $3\\%$ 个人所得税，1000元以上的部分，交纳 $20\\%$ 的个人所得税，小刚的爸爸缴纳个人所得税以后，实际得到多少钱？",
+        "options": [],
+        "answer": "1370元",
+        "solution": "【分析】依据个人所得税缴纳标准，先计算出小刚爸爸应缴纳的税，再利用劳务费减去税得到他实际得到的钱。本题考查了百分数的应用, 灵活运用 '应纳税部分 $\\times$ 税率 $=$ 应纳税额' 是解题的关键。\n\n【详解】 $1500 - 1000 \\times 3\\% - (1500 - 1000) \\times 20\\%\\\\= 1 5 0 0 - 3 0 - 1 0 0 \\\\ = 1 3 7 0 (\\text {元}) $\n\n答: 小刚的爸爸实际得到 1370 元钱。",
+        "knowledge_points": [
+          "税率问题"
+        ],
+        "source": "2021六年级下·全国·专题练习",
+        "merged_id": "unit2_290",
+        "original_id": "unit2_solve_290"
+      },
+      {
+        "id": "unit2_solve_291",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（20-21六年级下·全国·单元测试）东方商场搞促销，服饰类每满300元减120元，时代广场服饰类打五六折。\n\n(1) 如果妈妈想买一件相同的标价 400 元的外衣, 到哪个商场买更省钱？\n(2) 如果爸爸想买一件相同的标价 600 元的外衣, 到哪个商场买更省钱?",
+        "options": [],
+        "answer": "（1）时代广场；（2）东方商场",
+        "solution": "【分析】分别算出两个商场需要的实际价格，比较即可，东方广场：先算出衣服标价包含几个 300 元，就用标价减几个 120 元；时代广场：直接用标价乘折扣即可。打折就是按照折数低价出售商品。几折就是十分之几，也就是百分之几十。同种商品，折数越低，价格越低。\n\n【详解】（1）东方商场： $400 \\div 300 = 1$ ……100（元）$400 - 120 = 280$ （元） 时代广场： $400 \\times 65\\% = 260$ （元）260元<280元 答：到时代广场买更省钱。\n\n(2) 东方商场: $600 \\div 300 = 2$\n\n$6 0 0 - 1 2 0 \\times 2 \\\\ = 6 0 0 - 2 4 0 \\\\ = 3 6 0 (\\text {元}) $\n\n时代广场： $600 \\times 65\\% = 390$ （元） 360元<390元 答：到东方商场买更省钱。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_291",
+        "original_id": "unit2_solve_291"
+      },
+      {
+        "id": "unit2_solve_292",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（20-21六年级下·全国·单元测试）某品牌的电视机搞促销活动，在A商场按'每满500元减100元'的方式销售，在B商场打八折销售。李老师准备购买一台标价4800元的彩电。\n\n(1) 在 A、B 两个商场买, 各应付多少元？\n(2) 在哪个商场买更省钱? A、B 两个商场的价格相差多少钱?",
+        "options": [],
+        "answer": "(1) A 商场 3900 元, B 商场 3840 元 (2) B商场; 60元",
+        "solution": "【分析】（1）A商场，先算出彩电标价里有几个500元，就用标价减去几个 100 元; B 商场, 直接用标价 $\\times$ 折扣即可。(2) 比较两个商场的实际钱数，求差即可。关键是理解折扣的含义，几折就是百分之几十。\n\n【详解】 (1) A: $4800 \\div 500 = 9$ （个）……300（元）\n\n$4 8 0 0 - 1 0 0 \\times 9 \\\\ = 4 8 0 0 - 9 0 0 \\\\ = 3 9 0 0 (\\text {元}) $\n\nB: $4800 \\times 80\\% = 3840$ （元）答: 在 A 商场买应付 3900 元, 在 B 商场买应付 3840 元。\n\n(2) $3900 > 3840$ $3 9 0 0 - 3 8 4 0 = 6 0 \\text {(元)}$ 答: 在 B 商场买更省钱, A、B 两个商场的价格相差 60 元。",
+        "knowledge_points": [
+          "求一个数的百分之几是多少",
+          "折扣问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_292",
+        "original_id": "unit2_solve_292"
+      },
+      {
+        "id": "unit2_solve_293",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（20-21六年级下·全国·单元测试）张老师把5000元钱存入银行，定期三个月，年利率是 $3.33\\%$ ，他到期时从银行里拿到本金和利息共多少元钱？（国家规定，个人在银行存款所得的利息要按 $5\\%$ 的税率缴纳利息税",
+        "options": [],
+        "answer": "5039.54元",
+        "solution": "【分析】根据利息 $=$ 本金 $\\times$ 利率 $\\times$ 存期, 先求出存期, 再求出利息, 按 $5 \\%$ 缴纳利息税, 实际得到利息的 $1 - 5 \\%$ ，用利息 $\\times$ 交税后对应百分率, 求出实得利息, 再加上本金即可。到期取款时银行多支付的钱叫利息，纳税是每个公民应尽的义务。\n\n【详解】 $5000 \\times 3.33\\% \\times (3 \\div 12)\\\\= 1 6 6. 5 \\times 0. 2 5 \\\\ = 4 1. 6 2 5 \\text {(元)} $\n\n$41.625 \\times (1 - 5 \\%) \\approx 39.54 (\\text{元}) \\\\ 3 9. 5 4 + 5 0 0 0 = 5 0 3 9. 5 4 \\text {(元)} $\n\n答: 他到期时从银行里拿到本金和利息共 5039.54 元。",
+        "knowledge_points": [
+          "税率问题",
+          "利率问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_293",
+        "original_id": "unit2_solve_293"
+      },
+      {
+        "id": "unit2_solve_294",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "(20-21 六年级下·全国·单元测试) 李阿姨把 5000 元存入银行，存款方式为活期，年利率是 $0.35\\%$ 。存入4个月，把钱全部取出。李阿姨一共取出本金和利息多少钱？",
+        "options": [],
+        "answer": "5005.83元",
+        "solution": "【分析】利息 = 本金 $\\times$ 利率 $\\times$ 存期，计算利息时，存款的利率是年利率，计算时所乘时间应是年；存款的利率是月利率，计算时所乘时间单位应是月。李阿姨的 5000 元存的是活期, 年利率是 $0.35 \\%$ ,本题容易将存期 4 个月当作 4 年计算。\n\n【详解】 $5000 + 5000 \\times 0.35\\% \\times \\frac{4}{12}\\\\=5 0 0 0 + 5. 8 3 \\\\ = 5 0 0 5. 8 3 \\text {(元)} $\n\n答: 李阿姨一共取出本金和利息 5005.83 元。",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_294",
+        "original_id": "unit2_solve_294"
+      },
+      {
+        "id": "unit2_solve_295",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "(20-21 六年级下·全国·单元测试) 2016 年 3 月, 王爷爷把5000元钱存入银行，存期为5年，年利率为 $4.75\\%$ 。到期时王爷爷一共能取回多少元钱？",
+        "options": [],
+        "answer": "6187.5元",
+        "solution": "【分析】可按利息公式：利息 $=$ 本金 $\\times$ 利率 $\\times$ 时间，先计算出利息，再加上本金，就是到期时一共能取回的钱数。首先要对本金、利息和利率这些名词有所了解，其次在应用公式时，注意题目里的数字与公式中量的对应。\n\n【详解】 $5000 + 5000 \\times 4.75\\% \\times 5\\\\= 5 0 0 0 + 5 0 0 0 \\times 0. 2 3 7 5 \\\\ = 5 0 0 0 + 1 1 8 7. 5 \\\\ = 6 1 8 7. 5 \\text {(元)} $\n\n答: 到期时王爷爷一共能取回 6187.5 元钱。",
+        "knowledge_points": [
+          "求一个数的百分之几是多少",
+          "利率问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_295",
+        "original_id": "unit2_solve_295"
+      },
+      {
+        "id": "unit2_solve_296",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "(20-21 六年级下·全国·单元测试) 2015 年 8 月王奶奶将2000元钱存入银行，存期为5年，年利率为 $4.75\\%$ 。到期取回时，王奶奶可得到利息多少元？",
+        "options": [],
+        "answer": "475元",
+        "solution": "【分析】根据利息 $=$ 本金 $\\times$ 年利率 $\\times$ 存期，列式解答即可。关键是掌握利息公式，取款时银行多支付的钱叫利息。\n\n【详解】 $2000 \\times 4.75 \\% \\times 5\\\\= 9 5 \\times 5 \\\\ = 4 7 5 \\text {(元)} $\n\n答: 王奶奶可得到利息 475 元。",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_296",
+        "original_id": "unit2_solve_296"
+      },
+      {
+        "id": "unit2_solve_297",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（20-21六年级下·全国·单元测试）妈妈把8000元存入银行，定期2年，年利率是 $2.10\\%$ 。到期后应取多少钱？",
+        "options": [],
+        "answer": "8336元",
+        "solution": "【分析】分析题意，明确到期后取回的总钱数 = 本金 + 利息，利息 = 本金 × 利率 × 存期。本题容易误把到期后应取多少钱，理解成取得多少利息，忘记加本金。\n\n【详解】 $8000 \\times 2 \\times 2.10\\% + 8000\\\\= 3 3 6 + 8 0 0 0 \\\\ = 8 3 3 6 (\\text {元}) $\n\n答: 到期后应取 8336 元。",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_297",
+        "original_id": "unit2_solve_297"
+      },
+      {
+        "id": "unit2_solve_298",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（20-21六年级下·全国·单元测试）有一块小麦地，今年收小麦 $2200\\mathrm{kg}$ ，比去年增产一成，今年比去年增产多少千克？",
+        "options": [],
+        "answer": "$200 \\mathrm{~kg}$",
+        "solution": "【分析】2200千克比去年增产1成，那么2200千克就是去年的（ $1 + 10\\%$ ），用 $2200 \\div (1 + 10\\%)$ 即可求出去年的产量，再乘上 $10\\%$ 即可求出增产的千克数。分析题意，在题干中找到出现单位'1'的句子，确定单位'1'，'比去年增产一成'，是把去年收成看作单位'1'。解决此题要先求出去年收小麦的质量，然后求今年比去年增产的质量。\n\n【详解】 $2200 \\div (1 + 10\\%) \\times 10\\%\\\\= 2200 \\div 110 \\% \\times 10 \\% \\\\ = 2000 \\times 10 \\% \\\\ = 2 0 0 (\\mathrm {k g}) $\n\n答: 今年比去年增产 $200 \\mathrm{~kg}$ 。",
+        "knowledge_points": [
+          "成数问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_298",
+        "original_id": "unit2_solve_298"
+      },
+      {
+        "id": "unit2_solve_299",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "（20-21 六年级下·全国·单元测试）苏州人民商场为了促销，推出两种促销方式：方式①：所有商品打7折销售；方式②：一次购物每满200元送50元现金。杨老师要购买标价为630元和780元的商品各一件，现有四种购买方案：方案一：630元和780元的商品均按促销方式①购买；方案二：630元的商品按促销方式①购买，780元的商品按促销方式②购买;方案三：630元的商品按促销方式②购买，780元的商品按促销方式①购买；方案四：630元和780元的商品均按促销方式②购买。请你通过计算给杨老师提出最合理的购买方案。",
+        "options": [],
+        "answer": "方案一最合理",
+        "solution": "【分析】方案一：用总价直接乘 $70\\%$ 求出售价；方案二：780元满3个200元，共送3个50元，630元按照7折销售，分别计算出售价再相加；方案三：630元满3个200元，共送3个50元，780元按照7折销售，分别计算出售价再相加；方案四：把两件商品原价相加，然后计算出共有几个200元，确定送了几个 50 元，用原价减去送的钱数即可求出售价。分别计算后选择购买方案即可。打折就是按照折数低价出售商品。几折就是十分之几，也就是百分之几十。同种商品，折数越低，价格越低。\n\n【详解】方案一： $(630 + 780) \\times 0.7\\\\= 1 4 1 0 \\times 0. 7 \\\\ = 9 8 7 \\text {(元)} $\n\n方案二： $780 \\div 200 = 3 \\dots 180$ 元， $3 \\times 50 = 150$ 元\n\n$7 8 0 - 1 5 0 + 6 3 0 \\times 0. 7 \\\\ = 6 3 0 + 4 4 1 \\\\ = 1 0 7 1 \\text {(元)} $\n\n方案三： $630 \\div 200 = 3 \\dots 30$ 元， $3 \\times 50 = 150$ 元\n\n$6 3 0 - 1 5 0 + 7 8 0 \\times 0. 7 \\\\ = 4 8 0 + 5 4 6 \\\\ = 1 0 2 6 \\text {(元)} $\n\n方案四： $(630 + 780) \\div 200 = 7 \\dots 10$ 元， $7 \\times 50 = 350$ 元\n\n$630 + 780 - 350 = 1060 \\text{ (元)}$ $987 < 1026 < 1060 < 1071$ 答：选择方案一最合理。",
+        "knowledge_points": [
+          "优化问题",
+          "折扣问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_299",
+        "original_id": "unit2_solve_299"
+      },
+      {
+        "id": "unit2_solve_300",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "（20-21六年级下·全国·单元测试）一部手机的进价是2500元，加价两成售出，则售价是多少元？十一活动促销九折优惠，优惠后便宜多少元？",
+        "options": [],
+        "answer": "3000元；300元",
+        "solution": "【分析】几成即十分之几、百分之几十，即将两成化为百分数是 $20\\%$ ，售价 $=$ 原价 $\\times (1 + 20\\%)$ ，代入数值计算即可；打几折，即按原价的十分之几、百分之几十出售，即将九折化为百分数是 $90 \\%$ , 优惠后的便宜的钱数 = 售价 $\\times$ (1- $90 \\%$ ), 计算即可。明确折扣、成数的含义是解答本题的关键。\n\n【详解】 $2500 \\times (1 + 20\\%)\\\\= 2 5 0 0 \\times 1. 2 \\\\ = 3 0 0 0 (\\text {元}) \\\\ 3000 \\times (1 - 90 \\%) \\\\ = 3 0 0 0 \\times 0. 1 \\\\ = 3 0 0 \\text {(元)} $\n\n答: 售价是 3000 元, 优惠后便宜 300 元。",
+        "knowledge_points": [
+          "成数问题",
+          "折扣问题",
+          "比一个数多/少百分之几的数是多少"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_300",
+        "original_id": "unit2_solve_300"
+      },{
+        "id": "unit2_choice_301",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "(20-21 六年级下·全国·单元测试)一台电扇去年售价 180 元, 今年夏天售价 216 元, 今年比去年增加 ( )。",
+        "options": [
+          "A. 二成",
+          "B. 八成三",
+          "C. 一成"
+        ],
+        "answer": "A",
+        "solution": "【分析】A比B多百分之几的计算方法：（A一B)  $\\div \\mathrm{B} \\times 100 \\%$ , 几成就是十分之几; 也就是百分之几十, 据此解答。今年比去年增加百分之几，把去年风扇的价格看作单位'1'。\n\n【详解】  $(216 - 180) \\div 180 \\times 100\\%\\\\ = 36 \\div 180 \\times 100 \\% \\\\ = 0.2 \\times 100 \\% \\\\ = 20 \\% $\n\n20%就是二成，今年比去年增加二成。故答案为: A",
+        "knowledge_points": [
+          "成数问题",
+          "求一个数比另一个数多/少百分之几"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_301",
+        "original_id": "unit2_choice_301"
+      },
+      {
+        "id": "unit2_choice_302",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "(20-21 六年级下·全国·单元测试) 2015 年 11 月, 李爷爷把5000元钱存入银行，存期为3年，年利率为  $2.75\\%$ 。到期支取时，李爷爷可以取回（）元。",
+        "options": [
+          "A. 412.5",
+          "B. 5000",
+          "C. 5412.5"
+        ],
+        "answer": "C",
+        "solution": "【分析】到期支取时，李爷爷可以取回的钱数 = 存入银行的钱数 + 存入银行的钱数 × 年利率 × 存期，据此代入数据作答即可。\n\n【详解】  $5000 + 5000 \\times 2.75\\% \\times 3\\\\ = 5 0 0 0 + 4 1 2. 5 \\\\ = 5 4 1 2. 5 \\text {(元)} $\n\n故答案为: C",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_302",
+        "original_id": "unit2_choice_302"
+      },
+      {
+        "id": "unit2_choice_303",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "(20-21 六年级下·全国·课后作业)一台电冰箱的原价是 2400 元, 现在按八折出售, 求现价多少元? 列式是 ( )。",
+        "options": [
+          "A.  $2400 \\div 80\\%$",
+          "B.  $2400 \\times 80\\%$",
+          "C.  $2400 \\times (1 - 80\\%)$"
+        ],
+        "answer": "B",
+        "solution": "【分析】八折出售的意思就是现价是原价的  $80 \\%$ , 以原价为单位 '1', 用原价乘  $80 \\%$  即可求出现价。在日常生活中，折扣是较为常见的现象，要准确理解其含义，运用百分数运算求得结果。\n\n【详解】根据数量关系列式为： $2400 \\times 80\\%$ 。故答案为: B",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "20-21六年级下·全国·课后作业",
+        "merged_id": "unit2_303",
+        "original_id": "unit2_choice_303"
+      },
+      {
+        "id": "unit2_choice_304",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "（20-21六年级下·全国·单元测试）王老师要为学校买40个小足球，甲、乙、丙三个商店的小足球的单价都是25元，但优惠的方式不同。甲商店买10送2，不足10个不送；乙商店全场八折；丙商店每满100元返回现金25元。王老师选择（）商店最合算。",
+        "options": [
+          "A. 甲",
+          "B. 乙",
+          "C. 丙",
+          "D. 无法确定"
+        ],
+        "answer": "C",
+        "solution": "【分析】根据各商店的优惠规则，分别求出在每家商店购买 40 个小足球所需的费用，再进行比较。到甲商店买小足球的费用:  $(30 + 10 - 3 \\times 2) \\times 25 = 850$  （元）; 到乙商店买小足球的费用:  $40 \\times 25 \\times 80\\% = 800$  （元）;到丙商店买小足球的费用：$4 0 \\times 2 5 - 4 0 \\times 2 5 \\div 1 0 0 \\times 2 5 = 7 5 0 \\quad (\\text {元}) 。$$8 5 0 > 8 0 0 > 7 5 0, \\text {选 择 丙 商 店 最 合 算 。}$本题涉及到折扣、满返等优惠方式, 故要先弄明白这些方式的具体含义, 再着手解答。注意丙商店的实际花费就是买小足球总价减去返回的现金。\n\n【详解】由分析得:甲商店:\n\n$ (3 0 + 1 0 - 3 \\times 2) \\times 2 5 \\\\ = (4 0 - 6) \\times 2 5 \\\\ = 3 4 \\times 2 5 \\\\ = 8 5 0 \\quad (\\text {元}) $\n\n乙商店:\n\n$ 40 \\times 25 \\times 80 \\% \\\\ = 1 0 0 0 \\times 0. 8 \\\\= 8 0 0 \\quad (\\text {元})$\n\n丙商店：\n\n$ 4 0 \\times 2 5 - 4 0 \\times 2 5 \\div 1 0 0 \\times 2 5 \\\\ = 1 0 0 0 - 2 5 0 \\\\ = 7 5 0 \\quad (\\text {元}) $\n\n$8 5 0 > 8 0 0 > 7 5 0$ 故答案为: C。",
+        "knowledge_points": [
+          "折扣问题",
+          "经济问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_304",
+        "original_id": "unit2_choice_304"
+      },
+      {
+        "id": "unit2_choice_305",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "（20-21六年级下·全国·单元测试）用某移动支付平台提现时，平台将收取相当于提现金额的  $0.1\\%$  的手续费。小贤的哥哥参加工作以后，每个月都会用该移动支付平台给妈妈转账 2000 元。一年后，妈妈将这笔钱全部提取出来，该移动支付平台共收取手续费（）元。",
+        "options": [
+          "A. 2",
+          "B. 24",
+          "C. 1976",
+          "D. 2024"
+        ],
+        "answer": "B",
+        "solution": "【分析】每个月转账2000元，一年总共转账$2 0 0 0 \\times 1 2 = 2 4 0 0 0$（元），提现金额就是24000元，平台将收取相当于提现金额的  $0.1\\%$  的手续费，根据百分数乘法计算即可。此题主要考查了百分数乘法的计算问题。\n\n【详解】  $12 \\times 2000 \\times 0.1\\% = 24$  （元）故答案为: B",
+        "knowledge_points": [
+          "求一个数的百分之几是多少",
+          "税率问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_305",
+        "original_id": "unit2_choice_305"
+      },
+      {
+        "id": "unit2_choice_306",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "（2014·四川绵阳·小升初真题）商品A的单价打九折后和商品B的单价相等。下面说法不正确的是（）。",
+        "options": [
+          "A. B 的单价是 A 的  $90\\%$",
+          "B. A 的单价比 B 多  $10\\%$",
+          "C. B 的单价比 A 少  $10\\%$",
+          "D. A 的单价是 B 的  $\\frac{10}{9}$"
+        ],
+        "answer": "B",
+        "solution": "【分析】根据题意可知, 商品 A 的单价的  $90 \\% =$  商品 B 的单价, 即商品 B 的单价是商品 A 的  $90 \\%$ , 选项 A 说法正确; 把商品 B 的单价看作单位 '1', 则商品 A 的单价是商品 B的  $1 \\div 90 \\% = \\frac{10}{9}$ , 即商品 A 比商品 B 多  $\\frac{10}{9} - 1 = \\frac{1}{9}$ ,所以选项 D 说法正确, 选项 B 说法错误; 把商品 A 的单价看作单位 '1', 商品 B 的单价比商品 A 的单价少1 - 90% = 10%, 选项 C 说法正确。由已知商品 A 的单价打九折后和商品 B 的单价相等, 能够推理出单价 A 与单价 B 之间的多种数量关系, 在求一个单价比另一个单价多或少百分之几时, 可应用 (大一小)  $\\div$  单位 '1' 来解答。\n\n【详解】由分析得：A. 商品 A 单价  $\\times 90 \\% =$  商品 B 单价, 所以 B 的单价是 A 的  $90 \\%$ , A 正确;  B. 把 B 的单价看作单位 '1', 则 A 的单价为  $1 \\div 90 \\% = \\frac{10}{9}$ ,\n\n$ (\\frac {1 0}{9} - 1) \\div 1 \\\\ = \\frac {1}{9} \\div 1 \\\\ \\approx 11.11\\%,$\n\nA 的单价比 B 多  $11.11 \\%$ , B 错误;  C. 把商品 A 的单价看作单位 '1',$1 - 90\\% = 10\\%$B 的单价比 A 少  $10 \\%$ , C 正确;  D. 因为商品 A 单价  $\\times 90 \\% =$  商品 B 单价, 把商品 B 的单价看作单位 '1', 则商品 A 的单价是商品 B 的\n\n1÷90% =  $\\frac{10}{9}$ , 所以 A 的单价是 B 的  $\\frac{10}{9}$ , D 正确。故答案为: B",
+        "knowledge_points": [
+          "折扣问题",
+          "经济问题",
+          "求一个数比另一个数多/少百分之几"
+        ],
+        "source": "2014·四川绵阳·小升初真题",
+        "merged_id": "unit2_306",
+        "original_id": "unit2_choice_306"
+      },
+      {
+        "id": "unit2_choice_307",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "（20-21六年级下·全国·单元测试）平安路小学去年组织学生体检，检查出近视的学生共145人。今年组织学生体检，检查出近视的学生人数比去年少二成。今年检查出近视的学生有（）人。",
+        "options": [
+          "A. 29",
+          "B. 116",
+          "C. 182",
+          "D. 725"
+        ],
+        "answer": "B",
+        "solution": "【分析】知道学校去年的近视人数是 145 , 以及今年与去年相比近视人数少二成, 我们可以得出今年近视的人数是去年的  $(1 - 20\\%)$ , 由  $145 \\times (1 - 20\\%)$  即可得出今年近视的人数。'成数'的实际应用，几成就是十分之几，也是百分之几十，明白'成数'的意义，是解答本题的关键。\n\n【详解】由分析得：\n\n$145 \\times (1 - 20 \\%) \\\\ = 1 4 5 \\times 0. 8 \\\\ = 1 1 6 （人) $\n\n故答案为: B。",
+        "knowledge_points": [
+          "成数问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_307",
+        "original_id": "unit2_choice_307"
+      },
+      {
+        "id": "unit2_choice_308",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "（13-14五年级下·全国·课后作业）王老师把3000元存入银行，定期2年，年利率按  $2.25\\%$  计算，到期可得本息共（）元。",
+        "options": [
+          "A. 3135",
+          "B. 108",
+          "C. 3108"
+        ],
+        "answer": "A",
+        "solution": "【分析】根据本息  $=$  本金 + 本金  $\\times$  年利率  $\\times$  时间, 本金 3000 元, 时间两年, 年利率  $2.25 \\%$  。代入解答即可。此题重点考查学生对'本息=本金+本金×利率×时间'这一关系式的掌握与运用情况。\n\n【详解】  $3000 + 3000 \\times 2.25\\% \\times 2\\\\ = 3 0 0 0 + 1 3 5 \\\\ = 3 1 3 5 \\text {(元)} $\n\n故答案为: A",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "13-14五年级下·全国·课后作业",
+        "merged_id": "unit2_308",
+        "original_id": "unit2_choice_308"
+      },
+      {
+        "id": "unit2_choice_309",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "（19-20六年级下·河北唐山·期末）李叔叔一月份的工资收入是7800元，其中3500元是免税的，其余部分要按3% 的税率缴纳个人所得税，计算李叔叔一月份应缴纳个人所得税的正确列式是（）。",
+        "options": [
+          "A.  $7800 \\times 3\\%$",
+          "B.  $3500 \\times 3\\%$",
+          "C.  $(7800 - 3500) \\times 3\\%$",
+          "D. $(7800 - 3500) \\times (1 - 3 \\%)$"
+        ],
+        "answer": "C",
+        "solution": "【分析】李叔叔一月份应缴纳个人所得税  $= ($  月份的工资收入  $-3500)$  ×税率，据此列式。个人所得税  $=$  应纳税款  $\\times$  税率, 解题的关键是分析出应纳税款为  $(7800 - 3500)$  元。\n\n【详解】根据分析可知，李叔叔一月份应缴纳个人所得税的正确列式是  $(7800 - 3500) \\times 3\\%$  。故答案为: C",
+        "knowledge_points": [
+          "税率问题"
+        ],
+        "source": "19-20六年级下·河北唐山·期末",
+        "merged_id": "unit2_309",
+        "original_id": "unit2_choice_309"
+      },
+      {
+        "id": "unit2_choice_310",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "（19-20六年级下·河南郑州·期末）李伟将压岁钱2000元存入银行，存期三年，年利率是  $2.75\\%$  。到期后，银行支付的利息是（）元。",
+        "options": [
+          "A. 55",
+          "B. 165",
+          "C. 2165"
+        ],
+        "answer": "B",
+        "solution": "【分析】根据利息  $=$  本金  $\\times$  利率  $\\times$  存期，列式计算即可。本题考查了利率问题, 取款时银行多支付的钱叫利息。\n\n【详解】  $2000 \\times 2.75 \\% \\times 3 = 165$  （元）故答案为: B",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "19-20六年级下·河南郑州·期末",
+        "merged_id": "unit2_310",
+        "original_id": "unit2_choice_310"
+      },
+      {
+        "id": "unit2_choice_311",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "(19-20 六年级下·内蒙古呼伦贝尔·单元测试) 五成改写成百分数是（）。",
+        "options": [
+          "A.  $50\\%$",
+          "B.  $15\\%$",
+          "C.  $5\\%$",
+          "D.  $5.5\\%$"
+        ],
+        "answer": "A",
+        "solution": "【分析】根据几成就是十分之几，也就是百分之几十，据此把成数改写成百分数即可此题考查成数与百分数的互化方法, 基础知识,掌握即可。\n\n【详解】五成即十分之五，百分之五十，改写成百分数是  $50\\%$ ;故答案为：A",
+        "knowledge_points": [
+          "成数问题"
+        ],
+        "source": "19-20六年级下·内蒙古呼伦贝尔·单元测试",
+        "merged_id": "unit2_311",
+        "original_id": "unit2_choice_311"
+      },
+      {
+        "id": "unit2_choice_312",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "（19-20 六年级下·内蒙古呼伦贝尔·单元测试）某品牌的旅游鞋搞促销活动，在 A 商场按'每满 100 元减 40 元'的方式销售，在 B 商场打六折销售。妈妈准备给小丽买一双标价120元的这种品牌的旅游鞋。在A、B商场各买应付（）元。",
+        "options": [
+          "A. 75; 90",
+          "B. 78; 80",
+          "C. 80; 72",
+          "D. 90; 100"
+        ],
+        "answer": "C",
+        "solution": "【分析】A 商场按'每满 100 元减 40 元'的方式销售，120 元里含 1 个 100，所以可以减 40 元；B 商场打六折销售，六折即现价是原价的  $60\\%$ ；据此分别计算即可。理解折扣的含义是解答本题的关键。几折就是百分之几十；注意 A 商场 '每满 100 元减 40 元'，不满 100的部分不减。\n\n【详解】120 元里含 1 个 100，所以可以减 40 元，在 A 商场需要： $120 - 40 = 80$ （元）B 商场需要:  $120 \\times 60\\% = 72$  (元)故答案为: C",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "19-20六年级下·内蒙古呼伦贝尔·单元测试",
+        "merged_id": "unit2_312",
+        "original_id": "unit2_choice_312"
+      },
+      {
+        "id": "unit2_choice_313",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "（19-20六年级下·内蒙古呼伦贝尔·单元测试）某品牌的裙子搞促销活动，在A商场打五折销售，在B商场按'每满 100 元减 50 元'的方式销售。妈妈要买一条标价230 元的这种品牌的裙子。在 A、B 商场各买应付（）元。",
+        "options": [
+          "A. 115; 120",
+          "B. 120; 130",
+          "C. 115; 130",
+          "D. 130; 125"
+        ],
+        "answer": "C",
+        "solution": "【分析】A 商场：裙子标价  $\\times$  折扣 = 应付钱数; B 商场：看230 元里面包含几个 100 元，就减去几个 50 元。本题考查了折扣问题，几折就是百分之几十。\n\n【详解】A商场： $230 \\times 50\\% = 115$  （元）B 商场:  $230 - 50 \\times 2= 2 3 0 - 1 0 0= 130（元）$故答案为: C",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "19-20六年级下·内蒙古呼伦贝尔·单元测试",
+        "merged_id": "unit2_313",
+        "original_id": "unit2_choice_313"
+      },
+      {
+        "id": "unit2_choice_314",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "（2019·福建厦门·小升初真题）某村前年生产粮食500吨，去年粮食丰收，生产粮食600吨，去年粮食增产（）。",
+        "options": [
+          "A. 一成",
+          "B. 四成",
+          "C. 二成",
+          "D. 十成"
+        ],
+        "answer": "C",
+        "solution": "【分析】几成就是十分之几、百分之几十，把前年粮食生产总量看作单位'1'，求出去年比前年粮食增产百分之几十，然后把百分数化为成数即可。本题重点要理解成数的意义及成数与百分数之间的互化。\n\n【详解】  $(600 - 500) \\div 500\\\\= 1 0 0 \\div 5 0 0 \\\\ = 0. 2 \\\\ = 20 \\% $\n\n20%即二成故答案为: C",
+        "knowledge_points": [
+          "成数问题"
+        ],
+        "source": "2019·福建厦门·小升初真题",
+        "merged_id": "unit2_314",
+        "original_id": "unit2_choice_314"
+      },
+      {
+        "id": "unit2_choice_315",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "（19-20六年级下·全国·单元测试）公司今年的利润比去年减少三成五，意思是（）。",
+        "options": [
+          "A. 今年的利润是去年的  $35 \\%$",
+          "B. 今年的利润比去年多  $35 \\%$",
+          "C. 今年的利润比去年少  $35 \\%$",
+          "D. 去年的利润比今年多  $35 \\%$"
+        ],
+        "answer": "C",
+        "solution": "【分析】成数转化为百分数时，几成几就是百分之几十几，所以三成五 = 35%；由题意'公司今年的利润比去年减少三成五'，也就是说今年的利润比去年少  $35\\%$ ；据此解答。本题考查了百分数的应用, 关键是要掌握成数转化为百分数的方法。\n\n【详解】三成五  $= 35 \\%$ , 所以今年的利润比去年少  $35 \\%$  。故选: C",
+        "knowledge_points": [
+          "成数问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit2_315",
+        "original_id": "unit2_choice_315"
+      },
+      {
+        "id": "unit2_fill_316",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(20-21 六年级下·全国·单元测试) 王叔叔买了一台手提电脑, 现打八五折后 6800 元, 这台电脑原价( )元。",
+        "options": [],
+        "answer": "8000",
+        "solution": "【分析】打几折就是按照原价的百分之几十进行出售，本题中原价  $\\times 85 \\% =$  现价，即可得出原价  $=$  现价  $\\div 85\\%$  ，代入数值计算即可。本题的关键是掌握折扣的计算方法，打几折就是按照原价的百分之几十。\n\n【详解】  $6800 \\div 85\\%\\\\= 6 8 0 0 \\div 0. 8 5 \\\\ = 8 0 0 0 \\text {(元)} $\n\n所以这台电脑原价 8000 元。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_316",
+        "original_id": "unit2_fill_316"
+      },{
+        "id": "unit2_fill_317",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（20-21 六年级下·全国·单元测试）某种商品的原价是 800 元，打八折销售，现在买一件这样的商品可以便宜( )元。如果改为'每满 400 元减 200 元'，这件商品相当于打( )折。",
+        "options": [],
+        "answer": "160；五",
+        "solution": "【分析】几折就是百分之几十，表示现价是原价的百分之几十；便宜的钱数 = 原价 × (1 - 折扣数)；折扣 = 现价 ÷ 原价，现价 = 原价 - 原价里面有几个 400 × 200，据此解答即可。解答本题的关键是要明确折扣的含义，理解折扣、现价和原价之间的关系。\n\n【详解】  $800 \\times (1 - 80\\%)\\\\ = 800 \\times 20 \\% \\\\ = 1 6 0 \\text {(元)} \\\\ 8 0 0 - (8 0 0 \\div 4 0 0) \\times 2 0 0 \\\\ = 8 0 0 - 4 0 0 \\\\ = 4 0 0 \\text {(元)} \\\\ 400 \\div 800 = 0.5 = 50 \\% \\\\ 50 \\% = \\text{五折} $",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_317",
+        "original_id": "unit2_fill_317"
+      },
+      {
+        "id": "unit2_fill_318",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（20-21六年级下·全国·单元测试）王叔叔将10000元钱存入银行，定期三年，年利率为  $3.75\\%$  ，到期后一共可取回( )元。",
+        "options": [],
+        "answer": "11125",
+        "solution": "【分析】利息 = 本金  $\\times$  利率  $\\times$  时间, 到期后一共可取回的钱数 = 本金 + 利息。本题考查百分数在利率实际问题中的计算, 注意到期取出的是本金和利息的和。\n\n【详解】  $10000 \\times 3 \\times 3.75\\%\\\\= 30000 \\times 3.75 \\%\\\\=1125(元)$\n\n$= 1125$  （元）\n\n$1 1 2 5 + 1 0 0 0 0 = 1 1 1 2 5 \\text {(元)}$",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_318",
+        "original_id": "unit2_fill_318"
+      },
+      {
+        "id": "unit2_fill_319",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "(20-21 六年级下·全国·单元测试) 一件衣服打完五折后售价是 135 元, 这件衣服原价( )元。",
+        "options": [],
+        "answer": "270",
+        "solution": "【分析】五折的意思就是售价是原价的  $50 \\%$ , 根据分数除法的意义, 用售价除以  $50 \\%$  即可求出原价。明确折扣的含义是解答本题的关键。\n\n【详解】  $135 \\div 50\\% = 270$  （元）",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_319",
+        "original_id": "unit2_fill_319"
+      },
+      {
+        "id": "unit2_fill_320",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(20-21 六年级下·全国·单元测试)  ( ): 200 =( )÷14 = ( ) \\% = 0.35 = $\\frac {()}{()}$ =( )（用成数表示）。",
+        "options": [],
+        "answer": "70；40；35；$\\frac{7}{20}$；三成五",
+        "solution": "【分析】比的前项  $=$  比的后项  $\\times$  比值；除数  $=$  被除数÷商；小数化成百分数，只需将 0.35 的小数点向后移动两位，在末尾加上百分号即可；小数化成分数，原来有几位小数，就在 1 的后面写几个零作分母, 把原来的小数去掉小数点作分子, 能约分的要约分。成数表示一个数是另一个数的十分之几, 俗称 '几成'; '几成' 就是十分之几, 也就是百分之几十。理解成数的意义，掌握比中和除法中各部分之间的关系。\n\n【详解】  $0.35 \\times 200 = 70; 14 \\div 0.35 = 40; 0.35 = 35\\%$ ;  $0.35 =\\frac {35}{100} = \\frac {35 \\div 5}{100 \\div 5} = \\frac {7}{20}; 35 \\% = \\text{三成五。}$所以:  $70:200 = 14 \\div 40 = 35\\% = 0.35 = \\frac{7}{20} =$  三成五（用成数表示）。",
+        "knowledge_points": [
+          "成数问题",
+          "求比值",
+          "百分数、分数、小数和比的互化"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_320",
+        "original_id": "unit2_fill_320"
+      },
+      {
+        "id": "unit2_fill_321",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "（20-21六年级下·全国·单元测试）商店进了一种新款自行车，如果每辆售价300元，那么售价的  $60\\%$  是成本，售价的  $40\\%$  是利润。现在要搞促销活动，为保证每辆自行车的利润不少于 60 元, 至多打( )折。",
+        "options": [],
+        "answer": "八",
+        "solution": "【分析】由题意我们可以很容易求出每辆自行车的成本是[ 300 \\times 60\\% = 180 ] （元）。根据成本和最低利润60元，可以知道在利润最低时的售价为  $60 + 180 = 240$  （元）。而保证最低利润的售价与原售价的比值即为商店对这款自行车至多打的折扣。关键是在充分理解题意的基础上利用百分数的意义求出现在的售价，再结合折扣的意义，确定打几折。\n\n【详解】由分析得:$ (300 \\times 60 \\% + 60) \\div 300 \\\\ = 2 4 0 \\div 3 0 0 \\\\ = 0. 8 \\\\ = 80 \\% =八折$",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_321",
+        "original_id": "unit2_fill_321"
+      },
+      {
+        "id": "unit2_fill_322",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（20-21六年级下·全国·单元测试）甲、乙两家文具店以同样的定价出售一种作业本，甲店打七五折出售，乙店买四送一。王新要买10本这样的作业本，到( )店买更省钱。",
+        "options": [],
+        "answer": "甲",
+        "solution": "【分析】'乙店买四送一'，这样买 10 本只需要付 8 本的钱，相当于打八折。定价相同时，七五折  $<$  八折，所以到甲店买更省钱。对于甲店打七五折容易理解，而乙店优惠方式是买四送一，这意味着买2份4本，即花8本的钱，可得到10本这样的练习本，据此得到甲店更省钱。\n\n【详解】由分析得：假设每本作业本 a 元，\n\n甲文具店:\n\n$10 \\times a \\times 75 \\%\\\\= 1 0 a \\times 0. 7 5 \\\\ = 7. 5 \\mathrm {a} (\\text {元}) $\n\n乙文具店:\n\n$1 0 \\div (4 + 1) \\times 4 \\times a \\\\ = 2 \\times 4 \\times a \\\\ = 8 \\mathrm {a} (\\text {元}) $\n\n$7.5 \\mathrm {a} <   8 \\mathrm {a}$所以到甲文具店买更省钱。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_322",
+        "original_id": "unit2_fill_322"
+      },
+      {
+        "id": "unit2_fill_323",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（20-21六年级下·全国·单元测试）小丽将2000元存入银行，定期三个月，年利率是  $1.35\\%$  ，到期时可以取回( )元。",
+        "options": [],
+        "answer": "2006.75",
+        "solution": "【分析】三个月即一年的  $\\frac{1}{4}$ , 先求出一年的利息, 再乘  $\\frac{1}{4}$  即可得到三个月的利息, 列式计算为[ 2000 \\times 1.35\\% \\times \\frac{1}{4} = 6.75 ] （元），最后加上本金就是到期时可以取回的钱数，即\n\n$6.75 + 2000 = 2006.75$  （元）。注意审题，这里存钱的期限是三个月，而不是一年。所以先计算三个月能够得到的利息，再加上本金，就是到期时可以取回的钱。\n\n【详解】由分析得:\n\n$ 2000 \\times 1.35 \\% \\times \\frac{1}{4} + 2000 \\\\ = 2 7 \\times 0. 2 5 + 2 0 0 0 \\\\ = 6. 7 5 + 2 0 0 0 \\\\ = 2 0 0 6. 7 5 \\text {(元)} $",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_323",
+        "original_id": "unit2_fill_323"
+      },
+      {
+        "id": "unit2_fill_324",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（20-21六年级下·全国·单元测试）一种商品先以盈利二成来定价，再打九折出售仍能盈利( )%。",
+        "options": [],
+        "answer": "8",
+        "solution": "【分析】分析题意，假设进价为1，根据'以盈利二成来定价'，可得定价为  $1 \\times (1 + 20\\%)$ ；由于'打九折出售'，可得此时售价为  $1 \\times (1 + 20\\%) \\times 90\\% = 1.08$ ；求盈利百分之几，即（售价一进价）÷进价×100%，据此列式计算即可。此题主要考查了折扣问题，关键在于找准单位'1'，根据分数除法的意义，进而解决问题。\n\n【详解】  $1 \\times (1 + 20\\%) \\times 90\\% = 1.08$   $(1.08 - 1) \\div 1 = 8 \\%$",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_324",
+        "original_id": "unit2_fill_324"
+      },
+      {
+        "id": "unit2_fill_325",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（18-19六年级上·上海·期末）某商品季节性打折，按照原价打七折，现在的售价是420元，则原价是( )元。",
+        "options": [],
+        "answer": "600",
+        "solution": "【分析】七折是指现价是原价的  $70 \\%$ , 把原价看成单位 '1', 根据分数除法的意义, 用现价 420 元除以  $70 \\%$  即可求出原价。本题关键是理解打折的含义：打几折，现价就是原价的百分之几十；打几几折，现价就是原价的百分之几十。\n\n【详解】  $420 \\div 70\\% = 600$  （元）",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "18-19六年级上·上海·期末",
+        "merged_id": "unit2_325",
+        "original_id": "unit2_fill_325"
+      },
+      {
+        "id": "unit2_fill_326",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（18-19六年级上·上海·期末）某商店10月的销售额为20万元，如果11月销售额的增长率为  $10\\%$  ，那么11月的销售额为( )万元。",
+        "options": [],
+        "answer": "22",
+        "solution": "【分析】由题意可知，11月份销售额比10月份多了  $10\\%$  ，11月份销售额就是10月份的  $(1 + 10\\%)$  ，求11月份的销售额就用  $20 \\times (1 + 10\\%)$  。求已知比较量比单位 '1'的量多百分之几或少百分之几，先转化为比较量是单位 '1'的量的百分之几，再求比较量，根据乘法进行计算。\n\n【详解】  $20 \\times (1 + 10\\%)\\\\ = 20 \\times 110 \\% \\\\ = 2 2 \\text {(万 元)} $",
+        "knowledge_points": [
+          "比一个数多/少百分之几的数是多少"
+        ],
+        "source": "18-19六年级上·上海·期末",
+        "merged_id": "unit2_326",
+        "original_id": "unit2_fill_326"
+      },
+      {
+        "id": "unit2_fill_327",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "(18-19 六年级上·上海·期末) 一件商品打七五折后售价为 45 元, 那么这件商品的原价为( )元。",
+        "options": [],
+        "answer": "60",
+        "solution": "【分析】七五折是指现价是原价的  $75 \\%$ , 即 '原价  $\\times 75 \\% =$  现价', 由此解答即可。明确折扣的含义是解答本题的关键。\n\n【详解】 七五折  $= 75\\%$ $45 \\div 75 \\% = 60$  （元）",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "18-19六年级上·上海·期末",
+        "merged_id": "unit2_327",
+        "original_id": "unit2_fill_327"
+      },
+      {
+        "id": "unit2_fill_328",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（20-21六年级上·全国·期末）某购物中心按  $5\\%$  的税率交营业税，上个月上交营业税是84万元，购物中心上个月营业额是( )万元。",
+        "options": [],
+        "answer": "1680",
+        "solution": "【分析】根据：营业税  $=$  营业额  $\\times$  税率，可得：营业额  $=$  营业税÷税率，代入数据计算即可。税额问题有固定的计算方法，营业税  $=$  营业额  $\\times$  税率，找清数据与问题，代入公式计算即可。\n\n【详解】  $84 \\div 5\\% = 1680$  （万元）",
+        "knowledge_points": [
+          "税率问题"
+        ],
+        "source": "20-21六年级上·全国·期末",
+        "merged_id": "unit2_328",
+        "original_id": "unit2_fill_328"
+      },
+      {
+        "id": "unit2_fill_329",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（20-21六年级上·江苏·课后作业）一个小家电原价200元，打八折出售一段时间后，又涨价  $20\\%$  ，现在的售价是( )元。",
+        "options": [],
+        "answer": "192",
+        "solution": "【分析】将原价看成单位'1'，打八折出售就是按原价的  $80\\%$  出售，则出售价是  $200 \\times 80\\%$  元；再将出售价(200×80%) 元看成单位 '1', 涨价 20%, 相当于是出售价的  $(1 + 20\\%)$ , 求现在的售价用乘法。本题主要考查百分数应用题, 解题时注意单位 '1' 的变化。\n\n【详解】  $200 \\times 80\\% \\times (1 + 20\\%)\\\\= 1 6 0 \\times 1. 2 \\\\ = 1 9 2 \\text {(元)} $",
+        "knowledge_points": [
+          "折扣问题",
+          "比一个数多/少百分之几的数是多少",
+          "求一个数的百分之几是多少"
+        ],
+        "source": "20-21六年级上·江苏·课后作业",
+        "merged_id": "unit2_329",
+        "original_id": "unit2_fill_329"
+      },
+      {
+        "id": "unit2_fill_330",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(20-21 六年级上·四川成都·期中) 光明饭店今年一月份的营业额是40万元，按规定要缴纳  $5\\%$  的营业税，还要按营业税的  $7\\%$  缴纳城市维护建设税，那么，这个饭店一月份需缴纳营业税( )元和城市维护建设税( )元。",
+        "options": [],
+        "answer": "20000；1400",
+        "solution": "【分析】营业税  $=$  营业额  $\\times 5\\%$ ；城市维护建设税  $=$  营业税  $\\times 7\\%$ ，据此解答。明确税款  $=$  收入额  $\\times$  税率, 注意两次单位 '1'是不同的。\n\n【详解】  $40 \\times 5\\% = 2$  （万元），2万元  $= 20000$  元这个饭店一月份需缴纳营业税20000元。$20000 \\times 7\\% = 1400$  （元），城市维护建设税1400元。",
+        "knowledge_points": [
+          "税率问题"
+        ],
+        "source": "20-21六年级上·四川成都·期中",
+        "merged_id": "unit2_330",
+        "original_id": "unit2_fill_330"
+      },
+      {
+        "id": "unit2_fill_331",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "(20-21 六年级上·全国·单元测试) 百分之零点一二写作( )，二五折改写成百分数是( )，它含有( )个  $1 \\%$",
+        "options": [],
+        "answer": "$0.12\\%$；$25\\%$；25",
+        "solution": "【分析】百分数常常不写成分母是 100 的分数形式, 而是在原来的分子后面添加上 '百分号%'来表示。写百分数时,先写分子, 再写百分号。几折就是百分之几十, 百分数的计数单位是  $1 \\%$ , 分子是几就有几个计数单位。表示一个数是另一个数的百分之几的数，叫百分数，也叫百分比、百分率。\n\n【详解】百分之零点一二写作  $0.12\\%$  ，二五折改写成百分数是  $25 \\%$ , 它含有 25 个  $1\\%$ 。",
+        "knowledge_points": [
+          "折扣问题",
+          "百分数的读法和写法"
+        ],
+        "source": "20-21六年级上·全国·单元测试",
+        "merged_id": "unit2_331",
+        "original_id": "unit2_fill_331"
+      },
+      {
+        "id": "unit2_fill_332",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "(2018·广西桂林·小升初真题) 一个圆锥的体积是 24 立方厘米, 底面积是 8 平方厘米, 它的高是( )。",
+        "options": [],
+        "answer": "9",
+        "solution": "【详解】略",
+        "knowledge_points": [
+          "体积公式",
+          "圆锥的认识及特征"
+        ],
+        "source": "2018·广西桂林·小升初真题",
+        "merged_id": "unit2_332",
+        "original_id": "unit2_fill_332"
+      },
+      {
+        "id": "unit2_fill_333",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "(17-18 六年级下·全国·单元测试) 八五折 =( )%   $65\\%$ =( )折   二成五 = ( )%  40% = ( )成",
+        "options": [],
+        "answer": "85；六五；25；四",
+        "solution": "【分析】根据折扣、成数的概念直接写出答案即可.\n\n【详解】 八五折  $= 85 \\%$ ,  $65 \\%$  = 六五折, 二成五 =25%, 40% = 四成.故答案为 85 ; 六五; 25 ; 四.",
+        "knowledge_points": [
+          "百分数简单应用题"
+        ],
+        "source": "17-18六年级下·全国·单元测试",
+        "merged_id": "unit2_333",
+        "original_id": "unit2_fill_333"
+      }, {
+        "id": "unit2_fill_334",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(18-19 六年级上·浙江·课后作业) 某市场二月份比一月份的营业额少60万元，一月份营业额是二月份营业额的150%，按营业额的3%缴纳营业税，一月份应缴纳营业税( )万元.",
+        "options": [],
+        "answer": "5.4",
+        "solution": "【详解】设二月份营业额是  $x$  万元, 则一月份营业额为  $1.5 x$  万元.1.5x-x=60解得x=120万元．所以一月份营业额为180万元.一月份应缴纳营业税  $180 \\times 3 \\% = 5.4$  （万元）.",
+        "knowledge_points": [
+          "百分数的四则运算"
+        ],
+        "source": "18-19六年级上·浙江·课后作业",
+        "merged_id": "unit2_334",
+        "original_id": "unit2_fill_334"
+      },
+      {
+        "id": "unit2_fill_335",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(20-21 六年级上·全国·单元测试)  ( ) $\\div$ ( ) = 0.6 = ( ):  15 =  ( )折。",
+        "options": [],
+        "answer": "3；5；9；六",
+        "solution": "【分析】根据题意可知，先把已知的小数化成分数，然后再根据分数、比、除法的关系：比的前项相当于除法中的被除数，分数中的分子；比的后项相当于除法中的除数，分数中的分母；比号相当于除法中的除号，分数中的分数线；比值相当于除法中的商，分数的分数值，据此解答，几折就是百分之几十，据此解答。考查了除法、小数、比及百分数的应用这几者间的互化，它们既有联系又有区别，可灵活应用它们之间的联系来解题。\n\n【详解】  $3 \\div 5 = 0.6 = 9:15 =$  六折。",
+        "knowledge_points": [
+          "百分数、分数、小数和比的互化"
+        ],
+        "source": "20-21六年级上·全国·单元测试",
+        "merged_id": "unit2_335",
+        "original_id": "unit2_fill_335"
+      },
+      {
+        "id": "unit2_fill_336",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(2022·吉林·小升初真题) 一种商品打八折出售, '八折'表示原价的( )%，如果这种商品原价100元，现在便宜了( )元.",
+        "options": [],
+        "answer": "80；20",
+        "solution": "【详解】略",
+        "knowledge_points": [
+          "百分数的四则运算"
+        ],
+        "source": "2022·吉林·小升初真题",
+        "merged_id": "unit2_336",
+        "original_id": "unit2_fill_336"
+      },
+      {
+        "id": "unit2_fill_337",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(18-19 六年级下·全国·课后作业) 取款时银行除还给本金外, 另外付给的钱叫( )。",
+        "options": [],
+        "answer": "利息",
+        "solution": "【详解】人们将钱存入银行，到期取款时除了可以取出本金外，还可以得到利息。",
+        "knowledge_points": [
+          "百分数的实际应用题"
+        ],
+        "source": "18-19六年级下·全国·课后作业",
+        "merged_id": "unit2_337",
+        "original_id": "unit2_fill_337"
+      },
+      {
+        "id": "unit2_fill_338",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（19-20六年级下·河南郑州·期末）近几年我市快递业务量逐年递增，预计今年将同比增长近两成，'两成'改写成百分数是( )%。周叔叔去快递公司应聘，该公司每日基本工资80元，另外每送一件快递再加0.5元。如果周叔叔每天送n件快递，一天可以拿到工资( )元。（1天工资=基本工资+送快递另加的费用）",
+        "options": [],
+        "answer": "20；$80 + 0.5 \\mathrm{n}$",
+        "solution": "【分析】根据几成就是百分之几十，将'两成'进行改写；1天工资=基本工资+送快递另加的费用，送快递另加的费用=送的件数×每件价格，用字母表示出一天工资，化简即可。本题考查了成数和字母表示数，字母可以表示任意的数, 也可以表示特定含义的公式, 用字母将数量关系表示出来。\n\n【详解】两成  $= 20\\%$  $8 0 + 0. 5 \\times n = 8 0 + 0. 5 n \\text {(元)}$故答案为:  $20; 80 + 0.5 \\mathrm{n}$",
+        "knowledge_points": [
+          "用字母表示数",
+          "成数问题"
+        ],
+        "source": "19-20六年级下·河南郑州·期末",
+        "merged_id": "unit2_338",
+        "original_id": "unit2_fill_338"
+      },
+      {
+        "id": "unit2_fill_339",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(19-20 六年级下·内蒙古呼伦贝尔·单元测试) 某县前年秋粮产量为 2.8 万吨, 去年比前年增产三成。去年秋粮产量是( )万吨。",
+        "options": [],
+        "answer": "3.64",
+        "solution": "【分析】增产三成就是增加原来产量的  $30 \\%$ , 即去年产量是前年的  $1 + 30 \\% = 130 \\%$ , 把前年产量看作单位 '1', 运用分数乘法意义即可解答。此题考查了成数的含义：几成就是百分之几十。\n\n【详解】  $2.8 \\times (1 + 30\\%)\\\\= 2.8 \\times 130 \\% \\\\ = 3. 6 4 (\\text {万 吨}) $\n\n答: 去年秋粮产量为 3.64 万吨。",
+        "knowledge_points": [
+          "成数问题"
+        ],
+        "source": "19-20六年级下·内蒙古呼伦贝尔·单元测试",
+        "merged_id": "unit2_339",
+        "original_id": "unit2_fill_339"
+      },
+      {
+        "id": "unit2_fill_340",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "(19-20 六年级下·内蒙古呼伦贝尔·单元测试) 在银行存款的方式有多种，如活期、整存整取、零存整取等。存入银行的钱叫做( )；取款时银行多支付的钱叫做( )；单位时间（如1年、1月、1日等）内的利息与本金的比率叫做( )。利息的计算公式是：利息  $=$  ( )。",
+        "options": [],
+        "answer": "本金；利息；利率；本金×利率×时间",
+        "solution": "【分析】存入银行的钱叫本金，取款时银行多支付的钱叫做利息，利息与本金的比率叫做利率。利息=本金×利率×时间。此题考查学生对利息、利率、本金的理解。牢记利息 = 本金  $\\times$  利率  $\\times$  时间。\n\n【详解】在银行存款的方式有多种，如活期、整存整取、零存整取等。存入银行的钱叫做本金；取款时银行多支付的钱叫做利息；单位时间（如1年、1月、1日等）内的利息与本金的比率叫做利率。利息的计算公式是：利息  $=$ 本金×利率×时间。",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "19-20六年级下·内蒙古呼伦贝尔·单元测试",
+        "merged_id": "unit2_340",
+        "original_id": "unit2_fill_340"
+      },
+      {
+        "id": "unit2_fill_341",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "（19-20六年级下·内蒙古呼伦贝尔·单元测试）妈妈在邮局给奶奶汇2000元钱，需要交  $1\\%$  的汇费。汇费是（）元。",
+        "options": [],
+        "answer": "20",
+        "solution": "【分析】汇费  $=$  汇款总额  $\\times$  汇率。代入数据计算即可。考查对汇费的理解和含百分数的计算，基础题。\n\n【详解】  $2000 \\times 1\\% = 20$  （元）故答案为: 20",
+        "knowledge_points": [
+          "百分数的其他问题"
+        ],
+        "source": "19-20六年级下·内蒙古呼伦贝尔·单元测试",
+        "merged_id": "unit2_341",
+        "original_id": "unit2_fill_341"
+      },
+      {
+        "id": "unit2_fill_342",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（19-20六年级下·内蒙古呼伦贝尔·单元测试）2015年11月，张爷爷把儿子寄来的8000元钱存入银行，存期为3年，年利率为  $2.75\\%$  。到期支取时，张爷爷可得到利息( )元。到期时张爷爷一共能取回( )元。",
+        "options": [],
+        "answer": "660；8660",
+        "solution": "【分析】根据利息  $=$  本金  $\\times$  利率  $\\times$  时间, 本息  $=$  本金 + 利息,代入数据解答即可。此题考查利息问题, 熟练掌握利息的计算公式是解题关键。\n\n【详解】  $8000 \\times 2.75\\% \\times 3 = 660$  （元） $8 0 0 0 + 6 6 0 = 8 6 6 0 \\text {(元)}$  故答案为: 660; 8660。",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "19-20六年级下·内蒙古呼伦贝尔·单元测试",
+        "merged_id": "unit2_342",
+        "original_id": "unit2_fill_342"
+      },
+      {
+        "id": "unit2_fill_343",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(19-20 六年级下·内蒙古呼伦贝尔·单元测试) 2015 年11月，王奶奶把5000元钱存入银行两年。年利率为2.10%。到期时可以取回( )元。",
+        "options": [],
+        "answer": "5210",
+        "solution": "【分析】根据利息  $=$  本金  $\\times$  利率  $\\times$  时间, 据此求出利息, 再加上本金即可。此题考查的目的是理解利息的意义，掌握利息的计算方法，明确：本息  $=$  本金 + 利息。\n\n【详解】  $5000 \\times 2.10\\% \\times 2 + 5000\\\\= 2 1 0 + 5 0 0 0 \\\\ = 5 2 1 0 (\\text {元}) $\n\n故答案为: 5210",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "19-20六年级下·内蒙古呼伦贝尔·单元测试",
+        "merged_id": "unit2_343",
+        "original_id": "unit2_fill_343"
+      },
+      {
+        "id": "unit2_fill_344",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "（19-20六年级下·内蒙古呼伦贝尔·单元测试）一家饭店10月份的营业额中应纳税的部分是30万元。如果按应纳税部分的  $3\\%$  缴纳增值税，这家饭店10月份应缴纳增值税( )万元。",
+        "options": [],
+        "answer": "0.9",
+        "solution": "【分析】应缴纳增值税  $=$  应纳税部分营业额  $\\times$  税率, 代入数据计算即可。税率是表示税额和各种收入的比值。正确掌握'税率'的概念和各部分之间的关系并灵活运用是解答本题的关键。\n\n【详解】  $30 \\times 3\\% = 0.9$  （万元）故答案为: 0.9",
+        "knowledge_points": [
+          "税率问题"
+        ],
+        "source": "19-20六年级下·内蒙古呼伦贝尔·单元测试",
+        "merged_id": "unit2_344",
+        "original_id": "unit2_fill_344"
+      },
+      {
+        "id": "unit2_fill_345",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "（19-20六年级下·内蒙古呼伦贝尔·单元测试）农业收成，经常用'成数'来表示。例如，报纸上写道：'今年我省油菜籽比去年增产二成'。成数表示一个数是另一个数的十分之几，通称'几成'。例如，'一成'就是十分之一，改写成百分数是( )；'二成'就是十分之二，改写成百分数是( )；'三成五'是十分之三点五，改写成百分数就是( )。",
+        "options": [],
+        "answer": "$10\\%$；$20\\%$；$35\\%$",
+        "solution": "【分析】根据成数与百分数的关系，填空即可。基础题, 掌握成数的含义即可。几成就是百分之几十, 几成几就是百分之几十几。\n\n【详解】'一成'就是十分之一，改写成百分数是10%；'二成'就是十分之二，改写成百分数是  $20\\%$ ；'三成五'是十分之三点五，改写成百分数就是  $35\\%$ 。故答案为： $10\\% ; 20\\% ; 35\\%$",
+        "knowledge_points": [
+          "分数、小数、百分数与成数的互化"
+        ],
+        "source": "19-20六年级下·内蒙古呼伦贝尔·单元测试",
+        "merged_id": "unit2_345",
+        "original_id": "unit2_fill_345"
+      },
+      {
+        "id": "unit2_fill_346",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(19-20六年级下·内蒙古呼伦贝尔·单元测试) 某市2012年出境旅游人数为15000人次，比上一年增长两成。该市2011年出境旅游人数为( )人次。",
+        "options": [],
+        "answer": "12500",
+        "solution": "【分析】两成就是  $20\\%$  ，把2011年出境旅游人数看成单位'1'，2012年出境旅游人数是2011年出境旅游人数的（ $1 + 20\\%$ ），要求2011年出境旅游人数，用除法解答。解决本题关键是找准单位 '1', 找到 15000 对应的分率, 然后根据已知一个数的百分之几是多少, 求这个数,用除法求解。\n\n【详解】  $15000 \\div (1 + 20\\%)\\\\= 1 5 0 0 0 \\div 1. 2 \\\\ = 1 2 5 0 0 \\text {(人 次)} $\n\n答：2011年出境旅游人数为12500人次。",
+        "knowledge_points": [
+          "成数问题",
+          "已知比一个数多/少百分之几是多少，求这个数"
+        ],
+        "source": "19-20六年级下·内蒙古呼伦贝尔·单元测试",
+        "merged_id": "unit2_346",
+        "original_id": "unit2_fill_346"
+      },
+      {
+        "id": "unit2_fill_347",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(19-20 六年级下·内蒙古呼伦贝尔·单元测试) 爸爸给小雨买了一辆自行车，原价180元，现在商店打八五折出售。买这辆车用了多少钱？$1 8 0 \\times ($ )$ = ($ )$（\\text {元})$  买这辆车少花了（ ）元。",
+        "options": [],
+        "answer": "$85\\%$；153；27",
+        "solution": "【分析】八五折就是原价的  $85 \\%$ , 现价  $=$  原价  $\\times 85 \\%$ , 少花的钱  $=$  原价一现价。据此列式解答。现价  $=$  原价  $\\times$  折扣。\n\n【详解】  $180 \\times 85\\% = 153$  （元）  $1 8 0 - 1 5 3 = 2 7 \\text {(元)}$  买这辆车少花了27元。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "19-20六年级下·内蒙古呼伦贝尔·单元测试",
+        "merged_id": "unit2_347",
+        "original_id": "unit2_fill_347"
+      },
+      {
+        "id": "unit2_fill_348",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "（19-20六年级下·内蒙古呼伦贝尔·单元测试）商店有时降价出售商品，叫做打折扣销售，俗称'打折'。几折就表示十分之几，也就是百分之几十。例如，打九折出售，就是按原价的( )出售。",
+        "options": [],
+        "answer": "$90\\%$",
+        "solution": "【分析】根据折扣的含义填空即可。几折就表示十分之几，也就是百分之几十；注意打九折是现价是原价的  $90 \\%$ , 比原价便宜了  $10 \\%$  。\n\n【详解】商店有时降价出售商品，叫做打折扣销售，俗称'打折'。几折就表示十分之几，也就是百分之几十。例如，打九折出售，就是按原价的  $90\\%$  出售。\n\n故答案为： $90\\%$",
+        "knowledge_points": [
+          "求现价 (折扣问题)"
+        ],
+        "source": "19-20六年级下·内蒙古呼伦贝尔·单元测试",
+        "merged_id": "unit2_348",
+        "original_id": "unit2_fill_348"
+      },
+      {
+        "id": "unit2_fill_349",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（19-20六年级下·内蒙古呼伦贝尔·单元测试）李阿姨的月工资是7500元，扣除5000元个税免征额后的部分需要按  $3\\%$  的税率缴纳个人所得税。她应缴个人所得税( )元。",
+        "options": [],
+        "answer": "75",
+        "solution": "【分析】（工资一免征额）  $\\times$  税率  $=$  应缴纳的个人所得税。本题考查了税率，纳税是每个公民应尽的义务。\n\n【详解】$ (7500-5000) \\times 3\\%\\\\= 2 5 0 0 \\times 0. 0 3 \\\\ = 7 5 \\text {(元)} $",
+        "knowledge_points": [
+          "税率问题"
+        ],
+        "source": "19-20六年级下·内蒙古呼伦贝尔·单元测试",
+        "merged_id": "unit2_349",
+        "original_id": "unit2_fill_349"
+      },
+      {
+        "id": "unit2_fill_350",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（19-20六年级下·全国·单元测试）王师傅本月的收入是5400元，超出5000元部分要按  $3\\%$  的税率，应缴纳个人所得税( )元。",
+        "options": [],
+        "answer": "12",
+        "solution": "【分析】扣除 5000 元个税免征额后的部分是  $5400 - 5000 =$  400 元, 也就是说应缴纳税额部分应是 400 元, 然后代入关系式：应缴纳税额部分  $\\times$  税率  $=$  个人所得税，计算即可。此题解答的关键是掌握关系式：应缴纳税额部分  $\\times$  税率  $=$  个人所得税。\n\n【详解】 $(5400—5000) \\times 3\\% \\\\= 400 \\times 3 \\% \\\\ = 1 2 \\text {(元)} $\n\n答: 他应缴个人所得税 12 元。",
+        "knowledge_points": [
+          "税率问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit2_350",
+        "original_id": "unit2_fill_350"
+      },
+      {
+        "id": "unit2_fill_351",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(19-20 六年级下·全国·单元测试) ( )÷( )=0.5=( )%=( )成=( )折。",
+        "options": [],
+        "answer": "1；2；50；五；五",
+        "solution": "【分析】小数转化为百分数时, 小数点的位置向右移动两位, 位数不够的用 0 补齐,  $0.5 = 50 \\%$ ; 百分数转化为成数和折数时, 百分之几十就是几成或几折; 据此解答。本题考查了小数转化为百分数以及百分数的应用,关键是要掌握小数转化为百分数的方法。\n\n【详解】由分析得： $1 \\div 2 = 0.5 = 50\\% =$  五成=五折",
+        "knowledge_points": [
+          "百分数、小数和分数的互化"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit2_351",
+        "original_id": "unit2_fill_351"
+      },
+      {
+        "id": "unit2_fill_352",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（19-20六年级下·全国·单元测试）李宁登山包的原价是90元，现在打七折，现价是( )元。",
+        "options": [],
+        "answer": "63",
+        "solution": "【分析】七折出售的意思就是售价是原价的  $70 \\%$ , 根据分数乘法的意义用原价乘  $70 \\%$  即可求出现价。本题考查了折扣问题，几折就是百分之几十。\n\n【详解】  $90 \\times 70\\% = 63$  （元）",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit2_352",
+        "original_id": "unit2_fill_352"
+      },
+      {
+        "id": "unit2_fill_353",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "（19-20六年级下·全国·单元测试）打'七折'就是按原价的( )%出售。",
+        "options": [],
+        "answer": "70",
+        "solution": "【分析】打七折，是指现价是原价的  $70\\%$  ，由此求解。本题考查了折扣问题，几折就是百分之几十。\n\n【详解】按原价打七折，就是按原价的  $70\\%$  出售。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit2_353",
+        "original_id": "unit2_fill_353"
+      },
+      {
+        "id": "unit2_fill_354",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（19-20六年级下·全国·单元测试）张阿姨上个月领到的工资是6800元，超过5000元的部分要缴纳  $3 \\%$  的个人所得税，张阿姨上个月要缴纳( )元个人所得税。",
+        "options": [],
+        "answer": "54",
+        "solution": "【分析】个人所得税  $= (工资 - 5000) \\times 3\\%$ , 据此求解。此题主要考查税率问题, 本题的关键是求出需要缴纳税率的部分。\n\n【详解】 $(6800-5000) \\times 3\\%\\\\ = 1800 \\times 3 \\% \\\\ = 5 4 \\text {(元)} $\n\n所以张阿姨上个月要缴纳54元个人所得税。",
+        "knowledge_points": [
+          "税率问题",
+          "含百分数的运算"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit2_354",
+        "original_id": "unit2_fill_354"
+      },
+      {
+        "id": "unit2_fill_355",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(19-20 六年级下·全国·单元测试) 一件衣服现在售价 180 元, 比原价便宜了 20 元, 这件衣服是打了( )折。",
+        "options": [],
+        "answer": "九",
+        "solution": "【分析】几折就是百分之几十，打几折就是现价是原价的百分之几十。掌握折扣的求法是解决本题的关键。\n\n【详解】  $180 \\div (180 + 20)\\\\= 1 8 0 \\div 2 0 0 \\\\ = 90 \\% \\\\ = \\mathrm {九 折} $\n\n所以这件衣服是打了九折。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit2_355",
+        "original_id": "unit2_fill_355"
+      },
+      {
+        "id": "unit2_fill_356",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(17-18 六年级上·全国·单元测试) 一件商品标价 500元，参加端午'满300元减100元'优惠促销活动，这件商品实际上是打( )折出售。",
+        "options": [],
+        "answer": "八",
+        "solution": "【分析】'满 300 元减 100 元'，即活动期间购买这件商品需要花： $500 - 100 = 400$ （元），用现价除以原价，即可求出这件商品实际是在打几折出售。解答本题的关键是理解打折的含义，打几折，现价就是原价的十分之几或百分之几十。\n\n【详解】  $(500 - 100) \\div 500\\\\ = 4 0 0 \\div 5 0 0 \\\\ = 80 \\% $\n\n80%也就是八折，所以这件商品实际打八折出售。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "17-18六年级上·全国·单元测试",
+        "merged_id": "unit2_356",
+        "original_id": "unit2_fill_356"
+      },
+      {
+        "id": "unit2_fill_357",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "(19-20 六年级下·全国·单元测试) 天天副食商店六月份营业额为5000元，按营业额的  $3\\%$  缴纳营业税，天天副食商店这个月要缴纳营业税( )元。",
+        "options": [],
+        "answer": "150",
+        "solution": "【分析】用营业额  $\\times$  税率  $=$  缴纳的营业税。本题考查了税率问题, 应纳税额与各种收入的比率叫税率。\n\n【详解】  $5000 \\times 3\\% = 150$  （元）",
+        "knowledge_points": [
+          "税率问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit2_357",
+        "original_id": "unit2_fill_357"
+      },
+      {
+        "id": "unit2_fill_358",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（19-20六年级下·全国·单元测试）刘强把1000元零花钱存入银行，存期一年，按年利率  $1.75\\%$  计算，到期后能从银行取出( )元。",
+        "options": [],
+        "answer": "1017.5",
+        "solution": "【分析】根据利息  $=$  本金  $\\times$  利率  $\\times$  存期, 求出利息, 再加上本金, 就是到期后刘强取出的钱。本题考查利率问题, 解答本题的关键是熟练掌握利息的计算公式。\n\n【详解】  $1000 \\times 1.75\\% \\times 1 + 1000\\\\ = 1 7. 5 + 1 0 0 0 \\\\ = 1 0 1 7. 5 \\text {(元)} $",
+        "knowledge_points": [
+          "求一个数的百分之几是多少",
+          "利率问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit2_358",
+        "original_id": "unit2_fill_358"
+      },
+      {
+        "id": "unit2_fill_359",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(19-20 六年级下·全国·单元测试)  $2 \\div 5 = \\frac{6}{($ )}$ =8: ($ )$ = ($ )$ %= ($ )$ 成。",
+        "options": [],
+        "answer": "15；20；40；四",
+        "solution": "【分析】从已知的  $2 \\div 5$  入手，求出商，用小数表示，小数点向右移动两位添上百分号就是百分数，根据几成就是百分之几十填上成数，根据分数与除法和比的关系及分数的基本性质继续填空。本题考查了百分数、分数、比的相互转化，分数的分子相当于比的前项、被除数，分数的分母相当于比的后项、除数。\n\n【详解】  $2 \\div 5 = 0.4 = 40\\% =$  四成，  $2 \\div 5 = \\frac{2}{5}$  ，  $6 \\div 2 \\times 5 =15,  8 \\div 2 \\times 5 = 20$",
+        "knowledge_points": [
+          "百分数、小数和分数的互化",
+          "百分数、分数、小数和比的互化"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit2_359",
+        "original_id": "unit2_fill_359"
+      },
+      {
+        "id": "unit2_fill_360",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "（19-20六年级下·全国·单元测试）某商场一件衣服原价360元，现在商家打八折出售，现价是原价的( )%，现价( )元。",
+        "options": [],
+        "answer": "80；288",
+        "solution": "【分析】打八折出售，就是按原价的  $80\\%$  进行销售；原价  $\\times$  折扣 = 现价。本题考查了折扣问题, 打几折就是原价的百分之几十。\n\n【详解】  $360 \\times 0.8 = 288$ （元）某商场一件衣服原价360元，现在商家打八折出售，现价是原价的  $80\\%$  ，现价288元。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit2_360",
+        "original_id": "unit2_fill_360"
+      },
+      {
+        "id": "unit2_fill_361",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "(19-20 六年级下·全国·单元测试) 现价  $= ($ )$\\times($ )",
+        "options": [],
+        "answer": "原价；打折数",
+        "solution": "【分析】打几折表示现价是原价的百分之几十，求一个数的百分之几用乘法，据此解答即可。本题考查折扣问题, 解答本题的关键是掌握打几折表示现价是原价的百分之几十。\n\n【详解】现价  $=$  原价  $\\times$  打折数。故答案为：原价；打折数。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit2_361",
+        "original_id": "unit2_fill_361"
+      },
+      {
+        "id": "unit2_judge_362",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "（20-21 六年级下·全国·单元测试）华东农场的稻谷产量是华西农场的  $120 \\%$ , 华东农场比华西农场多二成。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】将华西农场的稻谷产量看作'1'即  $100 \\%$ , 则华东农场的稻谷产量为  $120 \\%$ , 华东农场的稻谷产量一华西农场的稻谷产量表示华东农场比华西农场多的产量。百分数化成成数, 去掉百分号, 再除以 10。此题考查的是成数，几成就是百分之几十。\n\n【详解】  $120\\% - 100\\% = 20\\%$  ， $20\\%$  表示两成。即华东农场的稻谷产量是华西农场的  $120 \\%$ , 华东农场比华西农场多二成说法正确。故答案为:  √",
+        "knowledge_points": [
+          "分数、小数、百分数与成数的互化",
+          "求增加或减少几成的实际问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_362",
+        "original_id": "unit2_judge_362"
+      },
+      {
+        "id": "unit2_judge_363",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "(20-21 六年级下·全国·单元测试) 六成五就是  $6.5\\%$  。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】 几成就是百分之几十，六成五就是  $65\\%$  。掌握成数的意义是关键。\n\n【详解】六成五就是  $65\\%$  ，原题说法错误。故答案为: ×",
+        "knowledge_points": [
+          "成数问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_363",
+        "original_id": "unit2_judge_363"
+      },
+      {
+        "id": "unit2_judge_364",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "（2021·广东阳江·小升初真题）一种商品打八折出售就是比原价降低了  $20\\%$ 。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】打八折销售是指现价是原价的  $80 \\%$ , 把原价看作单位 '1', 现价比原价便宜了  $1 - 80 \\%$ , 由此求解。本题关键是：理解打折的含义：打几几折，现价就是原价的百分之几十几。\n\n【详解】八折出售的商品就是现价是原价的  $80\\%$ ;比原价降低： $1 - 80\\% = 20\\%$ ；现价比原价便宜了  $20\\%$ 。故答案为:  √",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "2021·广东阳江·小升初真题",
+        "merged_id": "unit2_364",
+        "original_id": "unit2_judge_364"
+      },
+      {
+        "id": "unit2_judge_365",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "（2020五年级上·全国·专题练习）一本书原价a元，现在九折出售，现价比原价便宜了0.1a元。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】九折是指现价是原价的  $90\\%$  ，把原价看成单位'1'，现价就比原价便宜了（ $1 - 90\\%$ ），用原价乘上这个分率就是便宜的钱数。做这类用字母表示数的题目时，解题关键是根据已知条件，把未知的数用字母正确的表示出来，然后根据题意列式计算即可得解。\n\n【详解】  $\\mathrm{a} \\times (1 - 90\\%)\\\\= a \\times 0. 1 \\\\ = 0. 1 \\mathrm {a} (\\text {元}) $\n\n所以现价比原价便宜了  $0.1 \\mathrm{a}$  元, 原题说法正确。故答案为:  √ 。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "2020五年级上·全国·专题练习",
+        "merged_id": "unit2_365",
+        "original_id": "unit2_judge_365"
+      },
+      {
+        "id": "unit2_judge_366",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "（20-21六年级上·北京·单元测试）某店营业额是40万元，按营业额的  $5\\%$  缴纳营业税，应缴2万元。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】此题主要考查了纳税的知识，营业额  $\\times$  营业税的税率  $=$  营业税，据此列式解答。税收取之于民，用之于民，理解税率的意义并灵活应用营业税的公式，是解题关键。\n\n【详解】  $40 \\times 5 \\% = 2$  （万元），原题说法正确。故答案为：√。",
+        "knowledge_points": [
+          "税率问题"
+        ],
+        "source": "20-21六年级上·北京·单元测试",
+        "merged_id": "unit2_366",
+        "original_id": "unit2_judge_366"
+      },
+      {
+        "id": "unit2_judge_367",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "(2018·福建厦门·小升初模拟) 存期和本金不变, 如果利率上调, 那么所得利息将增加。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】根据利息的计算公式是：利息  $=$  本金  $\\times$  利率  $\\times$  时间，即利息的多少取决于本金、利率和时间这三个量的变化，当本金不变，存期不变时，利率上调，所得利息将增加；利率下调，所得利息将减少。据此解答。此题主题考查是利息的计算公式：利息  $=$  本金  $\\times$  利率  $\\times$  时间。\n\n【详解】利息  $=$  本金  $\\times$  利率  $\\times$  时间, 由此可知存期和本金不变, 如果利率上调, 那么所得利息将增加。说法是正确的。故答案为:√ 。",
+        "knowledge_points": [
+          "求利息"
+        ],
+        "source": "2018·福建厦门·小升初模拟",
+        "merged_id": "unit2_367",
+        "original_id": "unit2_judge_367"
+      },
+      {
+        "id": "unit2_judge_368",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "（19-20六年级下·内蒙古呼伦贝尔·单元测试）利率是由国家制定的，永远不变。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】利率是随着经济状况而随时做出更改的。利率是会变的。\n\n【详解】利率是由国家制定的，永远不变。此说法错误。故答案：×。",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "19-20六年级下·内蒙古呼伦贝尔·单元测试",
+        "merged_id": "unit2_368",
+        "original_id": "unit2_judge_368"
+      },
+      {
+        "id": "unit2_judge_369",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "(19-20 六年级下·内蒙古呼伦贝尔·单元测试) 一种商品打'五五折'出售，也就是把这种商品优惠了  $5.5\\%$  。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】一种商品打'五五折'出售，即按原价的  $55 \\%$  出售，将原价当作单位'1'，根据分数减法的意义，打折后比原价优惠了  $1 - 55 \\%$ , 据此判断即可。理解和掌握折扣的含义是解题关键。在商品销售中，几折即是按原价的百分之几十出售；几几折即是按原价的百分之几十几出售。\n\n【详解】由分析可知：一种商品打'五五折'出售，也就是把这种商品优惠了  $1 - 55\\% = 45\\%$  ，题干说法错误。故答案为: ×",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "19-20六年级下·内蒙古呼伦贝尔·单元测试",
+        "merged_id": "unit2_369",
+        "original_id": "unit2_judge_369"
+      },
+      {
+        "id": "unit2_judge_370",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "(19-20 六年级下·内蒙古呼伦贝尔·单元测试) 应纳税额与本金的比率叫利率。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】根据税率、利率的定义进行判断，应纳税额与各种收入的比率叫做税率；利率又称利息率，表示一定时期内利息与本金的比率；据此判断即可。正确掌握 '税率' '利率'的概念, 是解答本题的关键。\n\n【详解】题干混淆了税率和利率，一定时期内利息与本金的比率叫利率；题干说法错误。故答案为:×",
+        "knowledge_points": [
+          "利率问题",
+          "税率问题"
+        ],
+        "source": "19-20六年级下·内蒙古呼伦贝尔·单元测试",
+        "merged_id": "unit2_370",
+        "original_id": "unit2_judge_370"
+      },
+      {
+        "id": "unit2_judge_371",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "（19-20六年级下·全国·单元测试）爸爸把5000元钱存入银行，存期三年，年利率是  $2.25\\%$  ，到期后从银行取回 $5000 \\times 2.25 \\% \\times 3$  元。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】利息  $=$  本金  $\\times$  利率  $\\times$  存期，到期后取得的钱是本金  $+$  利息，据此判断即可。本题考查利率问题，解答本题的关键是掌握到期后取得的钱是本金 + 利息。\n\n【详解】利息： $5000 \\times 2.25\\% \\times 3$  到期后取得： $5000 \\times 2.25\\% \\times 3 + 5000$   故答案为: ×",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit2_371",
+        "original_id": "unit2_judge_371"
+      },
+      {
+        "id": "unit2_judge_372",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "（19-20六年级下·全国·单元测试）今年稻谷增产三成，就是增产  $30\\%$ 。（ ）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】成数代表一个数是另一个数的十分之几，改写成百分数就是百分之几十，据此解答即可。本题考查成数, 解答本题的关键是熟练掌握成数的意义。\n\n【详解】三成  $= 30 \\%$ , 所以今年稻谷增产三成, 就是增产  $30 \\%$ , 本题说法正确。故答案为: √",
+        "knowledge_points": [
+          "成数问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit2_372",
+        "original_id": "unit2_judge_372"
+      },
+      {
+        "id": "unit2_solve_373",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "（20-21六年级下·全国·单元测试）某小学要买60个篮球，现有A、B、C三个商店可以选择，这三个商店篮球的单价都是25元，但各个商店的优惠办法不同，通过计算，说明在哪个商店购买比较合算。A店：打八折销售。B店：每买10个赠送2个，不足10个不赠送。C店：购物每满300元，返还现金50元。",
+        "options": [],
+        "answer": "A店购买比较合算",
+        "solution": "【分析】此题主要考查了最佳方案的设计，A店，原价×折扣=现价，据此求出买60个篮球的现价；B店，根据条件'每买10个赠送2个，不足10个不赠送'可知，买60个篮球，只需要付50个篮球的价钱，另外10个是送的，据此列式解答；C店，单价×数量=总价，先求出一共要付的钱数，然后依据条件'购物每满300元，返还现金50元'可知，要付的钱数中有几个300，就返几个50元，最后用付出的钱数一返还的现金=实际支付的钱数，最后对比，哪种方案最省钱，就选哪种。解答本题的关键是在 B 店和 C 店上，B 店一定要明确 '每买 10 个赠送 2 个, 不足 10 个不赠送'的含义, 也是买 12 个只需要花 10 元的钱; C 店 '购物每满 300 元, 返还现金50元'是每满300元，就返回50元。\n\n【详解】A店： $25 \\times 60 \\times 80\\%\\\\= 1500\\times 80\\% \\\\ = 1 2 0 0 (\\text {元}); $\n\nB店:  $60 \\div (10 + 2)\\\\= 6 0 \\div 1 2 \\\\= 5 （组）$；\n\n$5 \\times (1 0 \\times 2 5) \\\\ = 5 \\times 2 5 0 \\\\ = 1 2 5 0 \\text {(元)} ; $\n\nC店： $25 \\times 60 = 1500$  （元）；\n\n$1 5 0 0 \\div 3 0 0 \\times 5 0 \\\\ = 5 \\times 5 0 \\\\ = 2 5 0 \\text {(元)}  $\n\n$1500 - 250 = 1250（元)$；\n\n$1 2 0 0 <   1 2 5 0$\n\n答: 在 A 店购买比较合算。",
+        "knowledge_points": [
+          "折扣问题",
+          "优化问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_373",
+        "original_id": "unit2_solve_373"
+      },
+      {
+        "id": "unit2_solve_374",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（20-21六年级下·全国·单元测试）某商店按  $15\\%$  的利润定价，然后又按定价打九折出售，结果每件还赚70元，这一商品的成本价是多少元？",
+        "options": [],
+        "answer": "2000元",
+        "solution": "【分析】这一商品的成本价 = 每件赚的钱数 ÷ 赚的钱数是成本价的几分之几，其中赚的钱数是成本价的几分之几 = (1 + 原来的利润率) × 打的折扣数 - 1。解答本题的关键是明确成本价、定价、现价、折扣和利润的含义；本题也可以列方程解答，将成本价看成单位'1'，设为  $x$  元，定价是成本价的  $(1 + 15\\%)$ ，即  $(1 + 15\\%) x$ ，又打九折，则现价为  $(1 + 15\\%) x \\times 90\\%$ ，现在的售价减去成本价就是70元，由此列方程解答即可。\n\n【详解】  $70 \\div [ (1 + 15\\%) \\times 90\\% - 1]\\\\= 7 0 \\div 0. 0 3 5 \\\\ = 2 0 0 0 \\text {(元)} $\n\n答: 这一商品的成本价是 2000 元。",
+        "knowledge_points": [
+          "折扣问题",
+          "利润问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_374",
+        "original_id": "unit2_solve_374"
+      },
+      {
+        "id": "unit2_solve_375",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（20-21六年级下·全国·单元测试）小明的爸爸12月的工资收入为7800元，按照个人所得税的政策，扣除5000元后，剩下的部分按照  $3\\%$  纳税。小明爸爸的税后实际收入是多少元？",
+        "options": [],
+        "answer": "7716元",
+        "solution": "【分析】小明的爸爸工资收入  $-5000$  元  $=$  需要纳税的工资;需要纳税的工资  $\\times (1 - 3\\%) =$  纳税部分的税后工资; 5000元 + 纳税部分的税后工资 = 小明爸爸的税后实际收入。此题根据应纳税额  $=$  收入  $\\times$  税率来解答, 注意超过 5000 元的部分需要缴税。\n\n【详解】  $5000 + (7800 - 5000) \\times (1 - 3\\%)\\\\ = 5000 + 2800 \\times 97 \\% \\\\ = 5 0 0 0 + 2 7 1 6 \\\\ = 7 7 1 6 \\text {(元)} $\n\n答: 小明爸爸的税后实际收入是 7716 元。",
+        "knowledge_points": [
+          "求一个数的百分之几是多少",
+          "税率问题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_375",
+        "original_id": "unit2_solve_375"
+      },
+      {
+        "id": "unit2_solve_376",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（2022·甘肃天水·小升初真题）张村有一块试验田种水稻，今年换种新品种后，产量是 2600 千克，今年产量比去年增产三成，去年的产量是多少千克？",
+        "options": [],
+        "answer": "2000千克",
+        "solution": "【分析】根据条件'今年换种新品种后，产量是2600千克，今年产量比去年增产三成'可知，去年的产量  $\\times (1 + 30\\%) =$  今年的产量，已知今年的产量，要求去年的产量，用除法计算，据此列式解答。此题主要考查了成数的应用，明确去年产量和今年产量的关系是解答本题的关键。\n\n【详解】 三成  $= 30\\%$ ;\n\n$ 2600 \\div (1 + 30 \\%) \\\\ = 2 6 0 0 \\div 1. 3 \\\\ = 2 0 0 0 \\text {(千 克)} $\n\n答: 去年的产量是 2000 千克。",
+        "knowledge_points": [
+          "成数问题"
+        ],
+        "source": "2022·甘肃天水·小升初真题",
+        "merged_id": "unit2_376",
+        "original_id": "unit2_solve_376"
+      },
+      {
+        "id": "unit2_solve_377",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（20-21六年级下·全国·单元测试）某中介公司为顾客买卖房屋，会按房屋售价的  $2\\%$  收取中介费。李叔叔通过该中介公司购买了一套房屋，中介公司收取中介费16000元。按规定买房还需要按房屋售价的  $1.5\\%$  缴纳契税。李叔叔购买这套房屋最终共花费多少钱？",
+        "options": [],
+        "answer": "828000元",
+        "solution": "【分析】由题意可知，李叔叔购买房屋最终共花费的钱数 = 房屋售价 + 中介费 + 契税，根据分数的除法的意义，即可计算出房屋售价，再根据百分数乘法计算出契税。此题主要考查了分数除法的意义、百分数乘法、房屋契税的计算。\n\n【详解】  $16000 \\div 2\\% = 800000$  （元）  $800000 \\times 1.5 \\% = 12000\\quad (\\text{元})$   $8 0 0 0 0 0 + 1 6 0 0 0 + 1 2 0 0 0 = 8 2 8 0 0 0 \\quad (\\text {元})$答：李叔叔购买这套房屋最终共花费828000元。",
+        "knowledge_points": [
+          "百分数的实际应用题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_377",
+        "original_id": "unit2_solve_377"
+      },
+      {
+        "id": "unit2_solve_378",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（20-21六年级下·全国·单元测试）某地区去年二手房的价格比前年的价格上涨了一成，今年的价格比去年又上涨了一成。这两年以来共涨价百分之几？",
+        "options": [],
+        "answer": "$21\\%$",
+        "solution": "【分析】假设前年二手房的价格为 1 , 先根据条件 '去年二 手房的价格比前年的价格上涨了一成', 可得去年二手房 的价格为  $1 \\times (1 + 10 \\%) = 1.1$ , 再根据条件 '今年的价 格比去年又涨了一成', 可得今年二手房的价格为$1.1 \\times (1 + 10\\%) = 1.21$ ，所以这两年以来共涨价(1.21 - 1)  $\\div 1 = 21 \\%$  ，注意本题中两次涨价的单位'1'是不同的。此题主要考查了百分数的应用，关键是找准'单位 1', 利用数量关系进行计算。\n\n【详解】假设前年二手房的价格为1，去年二手房的价格：  $1 \\times (1 + 10\\%) = 1.1$今年二手房的价格：  $1.1 \\times (1 + 10\\%) = 1.21$ $(1.21 - 1) \\div 1 = 21 \\%$答：这两年以来共涨价  $21\\%$ 。",
+        "knowledge_points": [
+          "成数问题",
+          "百分数的实际应用题"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_378",
+        "original_id": "unit2_solve_378"
+      },
+      {
+        "id": "unit2_solve_379",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "（20-21六年级下·全国·单元测试）张爷爷于2017年3月在银行存了10000元钱，定期三年，年利率是2.75%。2019年3月，张爷爷家装修，急需这10000元钱作为装修的费用。银行规定：定期存款提前取出，利息按活期存款的年利率  $0.35\\%$  计算。若不提取这笔存款，则张爷爷也可以向银行贷款10000元钱，等存款到期取出后一次性还清贷款，贷款年利率为  $4.35\\%$  。从经济角度考虑，请你替张爷爷算一算提前取出存款和贷款哪种方法更合算。",
+        "options": [],
+        "answer": "贷款更合算",
+        "solution": "【分析】根据题意可以知道，要想求出哪种方法更合算，必须先分别计算出提前取出的利息、到期取出的利息和贷款的利息。已知存款 10000 元钱，三年期的年利率是2.75%, 若存 2 年就取出, 需按活期存款的年利率  $0.35 \\%$  计算, 依据 '本金  $\\times$  利率  $\\times$  存期  $=$  利息', 可得两者的利息;而贷款 10000 元钱的年利率是  $4.35 \\%$ , 依据上面的公式还可求出贷款应该付给银行的利息。通过比较提前取出的利息、到期取出的利息与贷款利息差值的大小关系, 即可解答题目。题意较为复杂，尤其是向银行贷款这一方式，要计算出三年到期后能取出的利息与贷款要付给银行的利息的差值。通过本题能锻炼学生的处理实际问题的能力，及增加对'利率'的认识。\n\n【详解】  $10000 \\times 2 \\times 0.35\\%\\\\ = 2 0 0 0 0 \\times 0. 0 0 3 5 \\\\ = 7 0 (元) $\n\n$10000 \\times 2.75 \\% \\times 3 \\\\ = 3 0 0 0 0 \\times 0. 0 2 7 5 \\\\ = 8 2 5 (元) $\n\n$ 10000 \\times 4.35 \\% = 435\\quad (\\text{元}) \\\\ 8 2 5 - 4 3 5 = 3 9 0 (元) $\n\n$70 <   390$ 答：贷款更合算。",
+        "knowledge_points": [
+          "选择储蓄的最佳方案"
+        ],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit2_379",
+        "original_id": "unit2_solve_379"
+      },
+      {
+        "id": "unit2_solve_380",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（18-19六年级上·上海·期末）某机械厂今年第二季度的工业总产值是2400万元，比第一季度增长了  $20\\%$  ，预计第三季度的增长率在第二季度的基础上将提高4个百分点。(1) 问第一季度的工业总产值是多少万元？(2) 问第三季度的工业总产值是多少万元？",
+        "options": [],
+        "answer": "(1) 2000万元；(2) 2496 万元",
+        "solution": "【分析】（1）由题意可知，把第一季度工业总产值看作单位'1'，第二季度的工业总产值是2400万元，比第一季度增长了  $20\\%$  ，第二季度是第一季度的  $(1 + 20\\%)$  ，求第一季度工业总产值是多少用  $2400 \\div (1 + 20\\%)$  ；(2) 第三季度的增长率在第二季度的基础上将提高 4 个百分点, 第三季度就是第二季度的  $1 + 4\\% = 104 \\%$ , 用第二季度的工业总产值  $\\times (1 + 4 \\%)$  解答即可。此题解答的关键是找准单位 '1'的量，求单位 '1' 的量用比较量除以对应的百分数；求比较量用单位 '1' 的量乘比较量对应的分率。\n\n【详解】 (1)  $2400 \\div (1 + 20\\%)\\\\ = 2 4 0 0 \\div 1. 2 \\\\ = 2 0 0 0 \\text {(万 元)} $\n\n答：第一季度的工业总产值是 2000 万元。\n\n(2)  $2400 \\times (1 + 4\\%)\\\\ = 2 4 0 0 \\times 1. 0 4 \\\\ = 2 4 9 6 \\text {(万 元)} $\n\n答：第三季度的工业总产值是2496万元。",
+        "knowledge_points": [
+          "比一个数多/少百分之几的数是多少",
+          "已知比一个数多/少百分之几是多少，求这个数"
+        ],
+        "source": "18-19六年级上·上海·期末",
+        "merged_id": "unit2_380",
+        "original_id": "unit2_solve_380"
+      },
+      {
+        "id": "unit2_solve_381",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "（2014 六年级·全国·课后作业）小明带 24 个自制的纪念品去伦敦奥运会卖。早上每个纪念品卖 7 英镑，卖出的纪念品不到总数的一半。下午他对每个纪念品的价格进行打折，折后的价格仍是一个整数。下午他卖完了剩下的纪念品，全天共收入 120 英镑。那么早上他卖了多少个纪念品？",
+        "options": [],
+        "answer": "8个",
+        "solution": "【分析】假设上午把全部都卖完，那么共可得收入  $24 \\times 7 = 168$  英镑，但是实际只有 120 英镑，所以多了 168 - 120 = 48 英镑；多出来的钱是因为把下午便宜卖的纪念品也当成了 7 英镑来卖的，所以 48 英镑等于每一个多卖的钱乘下午卖的个数; 将 48 分解, 即  $48 = 2 \\times 2 \\times 2 \\times 2 \\times 3$ , 又因为下午卖的个数超过 12 个, 所以只能搭配成  $48 = 16 \\times 3$ , 即下午卖了 16 个, 每个价钱是  $7 - 3 = 4$  英镑, 所以上午卖了 24 - 16 = 8 个; 据此解答.\n\n【详解】  $24 \\times 7 - 120 = 48$  （英镑），$48 = 2 \\times 2 \\times 2 \\times 2 \\times 3$ , 又因为下午卖的个数超过 12 个,所以只能搭配成  $48 = 16 \\times 3$ , 即下午卖了 16 个, 每个价钱是 7 - 3 = 4 英镑,所以上午卖了  $24 - 16 = 8$  个;答: 早上他卖了 8 个纪念品.",
+        "knowledge_points": [
+          "折扣问题",
+          "百分数的实际应用题"
+        ],
+        "source": "2014六年级·全国·课后作业",
+        "merged_id": "unit2_381",
+        "original_id": "unit2_solve_381"
+      },
+      {
+        "id": "unit2_solve_382",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（18-19六年级上·浙江·课后作业）王芳的妈妈3年前买的金融债券，年利率是  $2.89\\%$  ，到期时共得本金和利息54335元．3年前，王芳妈妈买金融债券多少元？",
+        "options": [],
+        "answer": "50000元",
+        "solution": "【详解】设：3年前，王芳妈妈买金融债券  $x$  元.\n\n$54335 - x = x \\times 2.89 \\% \\times 3$   解得  $x = 50000$  答: 3 年前, 王芳妈妈买金融债券 50000 元.",
+        "knowledge_points": [
+          "百分数复合应用题"
+        ],
+        "source": "18-19六年级上·浙江·课后作业",
+        "merged_id": "unit2_382",
+        "original_id": "unit2_solve_382"
+      },
+      {
+        "id": "unit2_solve_383",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（2014·全国·小升初真题）土豆每千克售价 2.4 元，一菜农为了让市民多买土豆，按原价打了折扣。已知买 25 千克土豆就少花 6 元。这个菜农按原价的百分之几出售？",
+        "options": [],
+        "answer": "$90\\%$",
+        "solution": "【详解】略",
+        "knowledge_points": [
+          "百分数的实际应用题"
+        ],
+        "source": "2014·全国·小升初真题",
+        "merged_id": "unit2_383",
+        "original_id": "unit2_solve_383"
+      },
+      {
+        "id": "unit2_solve_384",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "(2020·河北保定·小升初真题)甲、乙两种商品成本共2200 元，甲商品按  $20\\%$  的利润定价，乙商品按  $15\\%$  的利润定价。后来因市场需求，两种商品按定价打九折出售，结果可获利131元，问甲、乙两种商品成本各多少元？",
+        "options": [],
+        "answer": "甲商品成品是1200元；乙商品成品是1000元",
+        "solution": "【分析】设甲成本为  $x$  元, 则乙为  $2200 - x$  元, 分别把甲、乙商品定价后的价钱求出, 然后根据一个数乘分数的意义,求出后来都按定价的  $90\\%$  打折出售的总价钱, 继而根据 '按定价的  $90\\%$  打折出售的总价钱一成本价 = 获利钱数'列出方程, 解答即可。解答此题的关键是先设出要求的量，进而判断出单位'1'，根据题意，找出数量间的相等关系式，然后根据关系式，进行解答即可。\n\n【详解】  $(1 + 20\\%) x\\times 90\\% + (1 + 15\\%) = (2200 -x) \\times 90 \\% - 2200 = 131$\n\n解:  $1.08 \\mathrm{x} + 1.035 \\times 2200 - 1.035 \\mathrm{x} - 2200 = 131$\n\n$0.0 4 5 x = 1 3 1 + 2 2 0 0 - 2 2 7 7$\n\n$x = 5 4 \\div 0. 0 4 5$\n\n$x = 1 2 0 0 \\\\ 2 2 0 0 - 1 2 0 0 = 1 0 0 0 (元) $\n\n答: 甲商品成本是 1200 元, 乙商品成本是 1000 元。",
+        "knowledge_points": [
+          "折扣问题",
+          "列方程解含两个未知数的问题",
+          "百分数的意义"
+        ],
+        "source": "2020·河北保定·小升初真题",
+        "merged_id": "unit2_384",
+        "original_id": "unit2_solve_384"
+      },
+      {
+        "id": "unit2_solve_385",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "(2022·河北唐山·小升初真题) 某工厂去年用电 350 万千瓦时, 今年比去年节电二成五, 今年用电多少万千瓦时?",
+        "options": [],
+        "answer": "262.5 万千瓦时",
+        "solution": "【分析】由题意可知, 把去年用电看做单位 '1', 二成五即  $25 \\%$ , 今年用电是去年用电总量的  $1 - 25 \\%$ , 据此列式解答即可。找准单位 '1', 理解成数与百分数的关系是解题关键, 求一个数的百分之几是多少用乘法计算即可。\n\n【详解】  $350 \\times (1 - 25\\%)\\\\= 350 \\times 75 \\% \\\\ = 2 6 2. 5 \\text {(万 千 瓦 时)} $\n\n答：今年用电262.5万千瓦时。",
+        "knowledge_points": [
+          "成数问题"
+        ],
+        "source": "2022·河北唐山·小升初真题",
+        "merged_id": "unit2_385",
+        "original_id": "unit2_solve_385"
+      },
+      {
+        "id": "unit2_solve_386",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "(19-20 六年级下·内蒙古呼伦贝尔·单元测试) 爸爸买了一个随身听, 原价 160 元, 现在只花了九折的钱, 比原价便宜了多少钱?",
+        "options": [],
+        "answer": "16元",
+        "solution": "【分析】九折即现价是原价的  $90 \\%$  。便宜的钱数是原价的  $(1 - 90 \\%$ ; 求一个数的百分之几十多少, 用乘法计算即可。理解折扣的含义是解题关键。几折就是现价是原价的百分之几十。\n\n【详解】  $160 \\times (1 - 90\\%)\\\\= 160 \\times 10 \\% \\\\ = 1 6 \\text {(元)} $\n\n答: 比原价便宜了 16 元。",
+        "knowledge_points": [
+          "折扣问题",
+          "求一个数的百分之几是多少",
+          "含百分数的运算"
+        ],
+        "source": "19-20六年级下·内蒙古呼伦贝尔·单元测试",
+        "merged_id": "unit2_386",
+        "original_id": "unit2_solve_386"
+      },
+      {
+        "id": "unit2_solve_387",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "(19-20 六年级下·全国·单元测试) 奶奶在 2012 年 2 月 5 日把 8000 元钱存入银行, 定期 3 年, 如果年利率是4.25%，到期时，林奶奶可取回多少钱？",
+        "options": [],
+        "answer": "9020元",
+        "solution": "【分析】取回的钱包括本金和利息，根据利息  $=$  本金  $\\times$  利率  $\\times$  存期，求出利息，再加上本金即可。本题考查了利率问题，到期取款时银行多支付的钱叫利息。\n\n【详解】  $8000 + 8000 \\times 4.25\\% \\times 3\\\\ = 8 0 0 0 + 1 0 2 0 \\\\ = 9 0 2 0 (\\text {元}) $\n\n答: 林奶奶可取回 9020 元钱。",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit2_387",
+        "original_id": "unit2_solve_387"
+      },
+      {
+        "id": "unit2_solve_388",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（19-20六年级下·全国·单元测试）苏宁家电商场周年店庆搞促销销售豆浆机，采用'折上折'方式销售，即先打七折，在此基础上再打九五折。国美商场购物每满150元减65元现金。如果两家豆浆机标价都是380元，在苏宁家电和国美商场各应付多少钱？在哪家商场购买更省钱？",
+        "options": [],
+        "answer": "252.7元，250元；国美",
+        "solution": "【分析】分别求出两家商场购买豆浆机的实际费用，比较即可。本题主要考查了折扣问题，几折就是百分之几十。\n\n【详解】苏宁： $380 \\times 70\\% \\times 95\\% = 252.7$  （元）国美:  $380 \\div 150 = 2$  （份）………80（元）\n\n$ 3 8 0 - 6 5 \\times 2 \\\\ = 3 8 0 - 1 3 0 \\\\ = 2 5 0 \\text {(元)} $\n\n$2 5 0 <   252. 7$\n\n答：苏宁家电应付252.7元，国美商场应付250元，在国美商场购买更省钱。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit2_388",
+        "original_id": "unit2_solve_388"
+      },
+      {
+        "id": "unit2_solve_389",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "(19-20 六年级下·全国·单元测试) 一条裤子 200 元,打折后便宜20元，该上衣打多少折出售？",
+        "options": [],
+        "answer": "九折",
+        "solution": "【分析】一条裤子 200 元，打折后便宜 20 元，即打折后价格是（200- 20）元，求出打折后的价格是原价的百分之几十，换成打折数即可。本题考查折扣问题, 解答本题的关键是掌握打几折表示现价是原价的百分之几十。\n\n【详解】  $(200 - 20) \\div 200\\\\ = 1 8 0 \\div 2 0 0 \\\\ = 90 \\% $\n\n$90\\% =$  九折答: 该上衣打九折出售。",
+        "knowledge_points": [
+          "折扣问题",
+          "求一个数是另一个数的百分之几(百分率问题)"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit2_389",
+        "original_id": "unit2_solve_389"
+      },
+      {
+        "id": "unit2_solve_390",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "(19-20 六年级下·全国·单元测试) 某商场今年一月份的营业额是850万，二月份的营业额比一月份多  $20\\%$  ，如果按营业额的  $5\\%$  缴纳营业税，这个商场二月份要缴纳营业税多少万元？",
+        "options": [],
+        "answer": "51万元",
+        "solution": "【分析】二月份的营业额比一月份多  $20 \\%$ , 是以一月份营业额为单位 '1', 二月份营业额占一月份的  $(1 + 20 \\%)$ ,一月份营业额  $\\times$  二月份对应百分率 = 二月份营业额, 再乘税率即可。本题考查了税率问题, 应纳税额与各种收入的比率叫税率。\n\n【详解】  $850 \\times (1 + 20\\%) \\times 5\\%\\\\ = 8 5 0 \\times 1. 2 \\times 0. 0 5 \\\\ = 5 1 \\text {(万 元)} $\n\n答：这个商场二月份要缴纳营业税51万元。",
+        "knowledge_points": [
+          "税率问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit2_390",
+        "original_id": "unit2_solve_390"
+      },
+      {
+        "id": "unit2_solve_391",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（19-20六年级下·全国·单元测试）元旦假期，甲、乙、丙三个旅行社推出'家庭组团'优惠活动，三家旅行社原价都是每人260元的近郊游，优惠方法如下：甲旅行社：成人全价，儿童六折，乙旅行社：成人、儿童一律九折，丙旅行社：满700元减80元，欣宇和爸爸、妈妈一家三口去近郊游，选择哪家旅行社最便宜？",
+        "options": [],
+        "answer": "甲旅行社",
+        "solution": "【分析】分别求出三家旅行社的实际费用，比较即可。本题考查了折扣问题，几折就是百分之几十。\n\n【详解】甲旅行社：\n\n$ 260 \\times 2 + 260 \\times 60 \\% \\\\ = 5 2 0 + 1 5 6 \\\\ = 6 7 6 \\text {(元)} $\n\n乙旅行社： $3 \\times 260 \\times 90\\% = 702$  （元）丙旅行社:  $260 \\times 3 = 780$  (元)  $780 > 700$\n\n$ 7 8 0 - 8 0 = 7 0 0 \\text {(元)} \\\\ 6 7 6 <   7 0 0 <   7 0 2 $\n\n答：选择甲旅行社最便宜。",
+        "knowledge_points": [
+          "万以内数的大小比较",
+          "折扣问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit2_391",
+        "original_id": "unit2_solve_391"
+      },
+      {
+        "id": "unit2_solve_392",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（19-20六年级下·全国·单元测试）'全国旅游队服务管理系统'数据显示，2019年10月1日至7日，出境跟团游客总数量约为139.9万人，比上一年增长一成二。2018年同期出境游客人数约为多少万人次？（得数保留整数）",
+        "options": [],
+        "answer": "125万人",
+        "solution": "【分析】上一年，即2018年出境游客人数为单位'1'，2019年出境游客人数占2018年的  $1 + 12\\%$  ，用2019年出境游客人数÷对应百分率 = 2018 年出境游客人数，据此列式解答。本题考查了成数问题，几成就是百分之几十。\n\n【详解】  $139.9 \\div (1 + 12\\%)\\\\= 1 3 9. 9 \\div 1. 1 2 \\\\ \\approx 125$  （万人）\n\n答：2018年同期出境游客人数约为125万人次。",
+        "knowledge_points": [
+          "成数问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit2_392",
+        "original_id": "unit2_solve_392"
+      },
+      {
+        "id": "unit2_solve_393",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "(19-20六年级下·全国·单元测试) 刘勇家要买一套总价80万元的住房，如果一次性交清房款买房可以享受九五折优惠，买这套房子还要按照实际房价的  $1.5\\%$  缴纳契税。如果刘勇家一次性交清房款，要缴纳契税多少元？",
+        "options": [],
+        "answer": "11400元",
+        "solution": "【分析】用总价×折扣×契税即可。本题考查了折扣和税率问题，几折就是百分之几十。\n\n【详解】  $80 \\times 95\\% \\times 1.5\\% = 1.14$  （万元） 1.14 万元  $= 11400$  元答: 如果刘勇家一次性交清房款, 要缴纳契税 11400 元。",
+        "knowledge_points": [
+          "折扣问题",
+          "税率问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit2_393",
+        "original_id": "unit2_solve_393"
+      },
+      {
+        "id": "unit2_solve_394",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（19-20六年级下·全国·单元测试）杨伯伯购买了2000元国家建设债券，定期5年，如果年利率为  $3\\%$  ，到期时他可以获得本金和利息多少元？",
+        "options": [],
+        "answer": "2300元",
+        "solution": "【分析】先根据利息  $=$  本金  $\\times$  利率  $\\times$  时间, 计算出利息, 再加上本金。本题考查本息和的计算，先计算出利息，再加上本金即可。\n\n【详解】  $2000 \\times 3\\% \\times 5 + 2000\\\\= 3 0 0 + 2 0 0 0 \\\\ = 2 3 0 0 (\\text {元}) $\n\n答: 到期时他可以获得本金和利息 2300 元。",
+        "knowledge_points": [
+          "利率问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit2_394",
+        "original_id": "unit2_solve_394"
+      },
+      {
+        "id": "unit2_solve_395",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "(19-20 六年级下·全国·单元测试) 小明家经营一家水果店，上月营业额40000元，如果按营业额的  $5\\%$  缴纳营业税，还剩营业额多少元？",
+        "options": [],
+        "answer": "38000元",
+        "solution": "【分析】先求出营业税，再求出剩下的营业额即可。本题考查税率问题，解答本题的关键是掌握营业税  $=$  营业额  $\\times$  税率。\n\n【详解】营业税： $40000 \\times 5\\% = 2000$  （元）剩余营业额:  $40000 - 2000 = 38000$  (元)答：还剩营业额 38000 元。",
+        "knowledge_points": [
+          "税率问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit2_395",
+        "original_id": "unit2_solve_395"
+      },
+      {
+        "id": "unit2_solve_396",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "（19-20六年级下·全国·单元测试）商店对某种商品作调价，按原价（标价）的8折出售，此时商品的利润率是 $10\\%$  ，此商品的进价为1600元，商品的原价是多少？",
+        "options": [],
+        "answer": "2200元",
+        "solution": "【分析】按原价（标价）的8折出售，表示现价是原价的  $80\\%$ ；商品的利润率是  $10\\%$ ，表示利润是进价的  $10\\%$ ，求出利润，利润加上进价就是现价，再根据现价求出原价即可。本题考查折扣问题，解答本题关键是掌握利润率  $=$  利润  $\\div$  进价  $\\times 100\\%$  ，商品的利润率是  $10\\%$  ，表示利润是商品进价的  $10\\%$  。\n\n【详解】现价：\n\n$1600 \\times 10 \\% + 1600 \\\\ = 1 6 0 + 1 6 0 0 \\\\ = 1 7 6 0 \\text {(元)} $\n\n原价： $1760 \\div 80\\% = 2200$  （元）答: 商品的原价是 2200 元。",
+        "knowledge_points": [
+          "折扣问题",
+          "利润问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit2_396",
+        "original_id": "unit2_solve_396"
+      },
+      {
+        "id": "unit2_solve_397",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "(19-20 六年级下·全国·单元测试) 某商品的进价是2000元，标价为3000元，商店要求以利润率不低于  $5\\%$  的售价打折出售，售货员最低可以打几折出售此商品？",
+        "options": [],
+        "answer": "打七折",
+        "solution": "【分析】把商品原价看作单位'1'，以利润率不低于  $5 \\%$  的售价出售，就是以不低于进价的  $1 + 5 \\% = 105 \\%$  的单价出售，求出最低的出售单价，再用最低单价除以标价即可解答。本题考查折扣问题，解答本题的关键是理解以利润率不低于  $5 \\%$  的售价出售，就是以不低于进价的  $1+5 \\%=105\\%$  的单价出售。\n\n【详解】最低售价：\n\n$2000 \\times (1 + 5 \\%) \\\\ = 2000 \\times 105 \\% \\\\ = 2 1 0 0 \\text {(元)} $\n\n$2100 \\div 3000 = 70 \\%$以标价的  $70 \\%$  出售就是打七折。答: 售货员最低可以打七折出售此商品。",
+        "knowledge_points": [
+          "折扣问题",
+          "利润问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit2_397",
+        "original_id": "unit2_solve_397"
+      },
+      {
+        "id": "unit2_solve_398",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（19-20六年级下·全国·单元测试）某种商品因换季准备打折出售，如果按定价的七五折出售，将赔25元，而按定价的九折出售，将赚20元，这种商品的定价为多少元？",
+        "options": [],
+        "answer": "300元",
+        "solution": "【分析】本题中唯一不变的是商品的成本，可根据成本相等列方程求解。设这种商品的定价为  $x$  元，根据'成本 = 售价 - 利润'可用  $x$  表示出商品的成本，即可列出方程求解。本题主要考查折扣问题，理解成本不变，灵活应用'成本 = 售价 - 利润'这个关系式是解决此题的关键。\n\n【详解】解：设这种商品的定价为  $x$  元。\n\n$0.7 5 x + 2 5 = 0. 9 x - 2 0 \\\\ 0. 1 5 \\mathrm {x} = 4 5 \\\\ x = 3 0 0 $\n\n答：设这种商品的定价为 300 元。",
+        "knowledge_points": [
+          "折扣问题",
+          "利润问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit2_398",
+        "original_id": "unit2_solve_398"
+      },
+      {
+        "id": "unit2_solve_399",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "（19-20六年级下·全国·单元测试）一种矿泉水，零售每瓶卖2元，生产厂家为感谢广大顾客对产品的厚爱，特开展'买四赠一'大酬宾活动，生产厂家的做法优惠了百分之几？",
+        "options": [],
+        "answer": "$20\\%$",
+        "solution": "【分析】要求生产厂家的做法优惠了百分之几，根据买四赠一可知，花 4 瓶的钱可以买到 5 瓶，可以先求现在一瓶的单价，再看比原来每瓶降低了多少元，最后用降低的价钱除以原来每瓶的单价即可。本题考查了百分数的实际应用，关键是要理解'买四赠一'的含义。\n\n【详解】现在每瓶的单价： $2 \\times 4 \\div 5 = 1.6$  （元）优惠的百分比： $(2 - 1.6) \\div 2 = 0.2 = 20\\%$答: 生产厂家的做法优惠了  $20 \\%$ .",
+        "knowledge_points": [
+          "百分数的其他问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit2_399",
+        "original_id": "unit2_solve_399"
+      },
+      {
+        "id": "unit2_solve_400",
+        "unit": "第二单元：百分数(二)",
+        "chapter": "百分数综合应用",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "(19-20 六年级下·全国·单元测试) '国庆'商场促销,一套服装打八五折出售是 1020 元, 这套西服原价多少元?",
+        "options": [],
+        "answer": "1200元",
+        "solution": "【分析】根据折扣的含义，打'八五折'出售表示现价是原价的  $85\\%$  ，根据'现价  $=$  原价  $\\times$  折扣'这个关系式可知，求原价用除法，代入数据即可解答。理解折扣的含义，灵活应用'现价  $=$  原价  $\\times$  折扣'这个关系式是解决本题的关键。\n\n【详解】 八五折  $= 85\\%$  $1020 \\div 85 \\% = 1200 (\\text{元})$ 答: 这套西服原价是 1200 元。",
+        "knowledge_points": [
+          "折扣问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit2_400",
+        "original_id": "unit2_solve_400"
+      },
+     {
+    "id": "unit2_choice_401",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "王爷爷将2万元人民币以整存整取的方式在银行存六个月，根据现行的存款利率（见下表），到期后应得利息（）元。\n\n整存整取\n\n| 存期 | 三个月 | 六个月 | 一年 | …… |\n| ----------- | ------ | ------ | ---- | ---- |\n| 年利率（%） | 1.35 | 1.55 | 1.75 | …… |",
+    "options": [
+      "A. 155",
+      "B. 175",
+      "C. 310",
+      "D. 350"
+    ],
+    "answer": "A",
+    "solution": "【分析】计算公式：到期利息收入 = 本金 × 年利率 ÷ 2 ，结合公式计算即可。注意本题中，整存整取的期限为六个月，即一年的一半，故计算时要除以 2 。本题不是十分易于理解，看上去年利率与存起六个月有些冲突。实际上只要记住银行的利率一般是按年利率来计算的，而至于存期，则取决于各人不同的存款方式。\n\n【详解】$20000 \\times 1.55\\% \\div 2\\\\= 3 1 0 \\div 2 \\\\ = 1 5 5 (\\text {元})$\n\n故答案为: A。",
+    "knowledge_points": [
+      "利率问题"
+    ],
+    "source": "20-21六年级上·浙江·期末",
+    "merged_id": "unit2_401",
+    "original_id": "unit2_choice_401"
+  },
+  {
+    "id": "unit2_choice_402",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "明明买 10 罐促销饮料, 相当于该商品打了 ( ) 折。\n\n可乐：'五一'促销买四赠一\n\n明天同学们来我家，买10罐这种饮料就够了。",
+    "options": [
+      "A. 七五",
+      "B. 八",
+      "C. 八五"
+    ],
+    "answer": "B",
+    "solution": "【分析】'买四赠一'是指用买4件商品的钱可以得到5件商品，用 4 除以 5 求出实际的价格是原价的百分之几，再根据打折的含义求解。本题关键是理解打折的含义，打几折现价就是原价的百分之几十。\n\n【详解】$4 \\div (4 + 1)\\\\= 4 \\div 5 \\\\ = 0. 8 \\\\ = 80 \\%$ 80%相当于打八折。故答案为: B",
+    "knowledge_points": [
+      "折扣问题"
+    ],
+    "source": "21-22 六年级下·河北保定·期末",
+    "merged_id": "unit2_402",
+    "original_id": "unit2_choice_402"
+  },
+  {
+    "id": "unit2_choice_403",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "下表是某影城的宣传信息。笑笑一家三口去看电影，票价共计144元。笑笑一家看的是哪一场次的电影？（）\n\n| 定价：60 元/场 | 上午场 | 下午场 | 夜场 |\n| -------------- | ------ | ------ | ---- |\n| 折扣 | 四折 | 六折 | 八折 |",
+    "options": [
+      "A. 上午场",
+      "B. 下午场",
+      "C. 夜场"
+    ],
+    "answer": "C",
+    "solution": "【分析】根据单价 × 数量 = 总价，据此先求出笑笑一家三口需要花费的总钱数，然后根据原价 × 折扣 = 现价，分别求出上午场、下午场和夜场的现价，最后与 144 元进行对比即可。本题考查折扣问题, 明确几折即百分之几十是解题的关键。\n\n【详解】上午场： $60 \\times 3 \\times 40\\%\\\\= 180 \\times 40 \\% \\\\ = 7 2 \\text {(元)}$\n\n下午场： $60 \\times 3 \\times 60\\%\\\\= 180 \\times 60 \\%\\\\= 108（元）$\n\n夜场： $60 \\times 3 \\times 80\\%\\\\= 180 \\times 80 \\% \\\\ = 1 4 4 (\\text {元})$\n\n故答案为: C",
+    "knowledge_points": [
+      "经济问题",
+      "折扣问题"
+    ],
+    "source": "2022·河南焦作·小升初真题",
+    "merged_id": "unit2_403",
+    "original_id": "unit2_choice_403"
+  },
+  {
+    "id": "unit2_choice_404",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一件衬衣打6折，现价比原价降低（）。",
+    "options": [
+      "A. 6元",
+      "B. $60\\%$",
+      "C. $40\\%$",
+      "D. $12.5\\%$"
+    ],
+    "answer": "C",
+    "solution": "【分析】打6折表示现价是原价的 $60\\%$ ，据此解答即可。本题考查折扣问题, 解答本题的关键是掌握打几折表示现价是原价的百分之几十。\n\n【详解】一件商品打六折，就是按原价的 $60\\%$ 出售，现价比原价降低 $40\\%$ 。故答案为: C。",
+    "knowledge_points": [
+      "求一个数比另一个数多/少百分之几",
+      "折扣问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_404",
+    "original_id": "unit2_choice_404"
+  },
+  {
+    "id": "unit2_choice_405",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "杭州出游人数比去年增加三成，就是说杭州出游人数（）。",
+    "options": [
+      "A. 是去年的 $30 \\%$",
+      "B. 是去年的 $70 \\%$",
+      "C. 是去年的 $130 \\%$"
+    ],
+    "answer": "C",
+    "solution": "【分析】三成即 $30 \\%$ , 杭州出游人数比去年增加三成, 是把去年的出游人数看做单位 '1', 今年是去年的 (1 + $30 \\%$ ), 据此解答即可。理解成数的含义是解题关键。一般地，是、比、占等字样后面的量是单位'1'。\n\n【详解】由分析可得，杭州出游人数比去年增加三成，就是说杭州出游人数是去年的 $1 + 30\\% = 130\\%$ ;故答案为: C",
+    "knowledge_points": [
+      "成数问题"
+    ],
+    "source": "19-20 六年级下·全国·单元测试",
+    "merged_id": "unit2_405",
+    "original_id": "unit2_choice_405"
+  },
+  {
+    "id": "unit2_choice_406",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "八折就是 ( )。",
+    "options": [
+      "A. $8\\%$",
+      "B. $80\\%$",
+      "C. $0.8\\%$"
+    ],
+    "answer": "B",
+    "solution": "【分析】八折是把原价看成单位'1'，那么现价是原价的 $80\\%$ ，由此求解。理解折扣的含义是解答本题的关键。\n\n【详解】由分析可得，八折就是 $80\\%$ 。故选: B。",
+    "knowledge_points": [
+      "折扣问题"
+    ],
+    "source": "19-20 六年级下·全国·单元测试",
+    "merged_id": "unit2_406",
+    "original_id": "unit2_choice_406"
+  },
+  {
+    "id": "unit2_choice_407",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "某服装店实行薄利多销的原则，一般在进价的基础上提高两成后作为销售价，照这样计算，一件进价640元的衣服应标价多少元？正确列式是（）。",
+    "options": [
+      "A. $640 \\times 20\\%$",
+      "B. $640 \\div (1 + 20\\%)$",
+      "C. $640 \\times (1 +20\\%)$"
+    ],
+    "answer": "C",
+    "solution": "【分析】在进价的基础上提高二成后作为销售价，是把进价看成单位'1'，标价比进价多 $20\\%$ ，标价是进价的 $(1 + 20\\%)$ ，用进价乘上这个百分数就是标价。理解成数的含义，找出单位'1'，再根据分数乘法的意义进行求解。\n\n【详解】由分析可得，标价应该是： $640 \\times (1 + 20\\%)\\\\= 640 \\times 120 \\% \\\\ = 7 6 8 \\text {(元)}$\n\n故答案为: C",
+    "knowledge_points": [
+      "成数问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_407",
+    "original_id": "unit2_choice_407"
+  },
+  {
+    "id": "unit2_choice_408",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一种吹风机原来要 180 元，现在打七五折出售，现在买可以节约（）元。",
+    "options": [
+      "A. 30",
+      "B. 45",
+      "C. 60"
+    ],
+    "answer": "B",
+    "solution": "【分析】用原价减去实际售价，就是现在买这种吹风机少花的钱数，即节约的钱数。明确七五折就是实际售价是原价的 $75 \\%$ 是解题关键。\n\n【详解】$180 - 180 \\times 75\\%\\\\= 1 8 0 - 1 3 5 \\\\ = 4 5 \\text {(元)}$\n\n故答案为: B",
+    "knowledge_points": [
+      "折扣问题"
+    ],
+    "source": "19-20 六年级下·全国·单元测试",
+    "merged_id": "unit2_408",
+    "original_id": "unit2_choice_408"
+  },
+  {
+    "id": "unit2_choice_409",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "王老师将 3000 元钱存入银行，定期三年，年利率是 $5.00\\%$ 。存款到期时他应拿到的利息的算式是（）。",
+    "options": [
+      "A. $3000 \\times 5.00 \\% \\times 3 \\times 20 \\%$",
+      "B. $3000 \\times 5.00 \\% \\times 3 \\times (1 -20\\%)$",
+      "C. $3000 \\times 5.00 \\% \\times 3$",
+      "D. $3000 \\times 5.00$"
+    ],
+    "answer": "C",
+    "solution": "【分析】根据关系式: 利息 = 本金 × 利率 × 时间, 代入数据即可求解。考查对利息公式的理解和掌握：利息 = 本金 × 利率 × 时间，基础题。\n\n【详解】由分析可得，存款到期时王老师应拿到的利息为： $3000 \\times 5.00\\% \\times 3$故答案为: C",
+    "knowledge_points": [
+      "利率问题"
+    ],
+    "source": "19-20 六年级下·全国·单元测试",
+    "merged_id": "unit2_409",
+    "original_id": "unit2_choice_409"
+  },
+  {
+    "id": "unit2_choice_410",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "李老师每月工资5800元，如果按国家'超出5000元的部分应按 $3\\%$ 的税率缴纳个人所得税'的规定，李老师每月应缴纳的个人所得税为（）元。",
+    "options": [
+      "A. $5000 \\times 3\\%$",
+      "B. $5800 \\times 3\\%$",
+      "C. (5800-5000) ×3%"
+    ],
+    "answer": "C",
+    "solution": "【分析】由题意可得，超出5000元的部分应按 $3 \\%$ 的税率缴纳个人所得税，则需要纳税的部分是（5800－5000）元，求它的 $3 \\%$ 用乘法计算，据此解答即可。找准需要纳税的钱数是解题关键。求一个数的百分之几是多少，用乘法计算。\n\n【详解】由分析可得，李老师每月应缴纳的个人所得税为：(5800-5000) ×3%;故答案为: C",
+    "knowledge_points": [
+      "税率问题"
+    ],
+    "source": "19-20 六年级下·全国·单元测试",
+    "merged_id": "unit2_410",
+    "original_id": "unit2_choice_410"
+  },
+  {
+    "id": "unit2_choice_411",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "一件衣服原价150元，现价105元，这件衣服按（）折销售。",
+    "options": [
+      "A. 7",
+      "B. 6",
+      "C. 8"
+    ],
+    "answer": "A",
+    "solution": "【分析】用现价除以原价，求出现价是原价的百分之几十，再由打折的含义求解。本题关键是理解打折的含义，打几折现价就是原价的百分之几十。\n\n【详解】$105 \\div 150 = 0.7 = 70\\% =$ 七折;故答案为: A",
+    "knowledge_points": [
+      "折扣问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_411",
+    "original_id": "unit2_choice_411"
+  },
+  {
+    "id": "unit2_choice_412",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一款商品8月上市销售，10月降价 $20\\%$ ，在10月价格的基础上，12月又降价50%甩卖。这款商品 12 月的价格与 8 月的价格比较，相当于（）。",
+    "options": [
+      "A. 打三折",
+      "B. 打四折",
+      "C. 打五折",
+      "D. 打七折"
+    ],
+    "answer": "B",
+    "solution": "【分析】原价为单位'1'，降价 $20 \\%$ 后的价格是原价的（1 $-20 \\%$ , 又降价 $50 \\%$ 后的价格是第一次降价后价格的（1 $-50 \\%$ , 根据分数乘法的意义求出现价是原价的百分之几,然后确定折扣即可。本题考查了百分数的应用, 关键是要找出题目中的单位 '1' 以及存在的数量关系, 还要理解百分之几十就是几折。\n\n【详解】$1 \\times (1 - 20\\%) \\times (1 - 50\\%)\\\\= 0. 8 \\times 0. 5 \\\\ = 40 \\%$\n\n相当于打四折。故答案为: B。",
+    "knowledge_points": [
+      "折扣问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_412",
+    "original_id": "unit2_choice_412"
+  },
+  {
+    "id": "unit2_choice_413",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "把 $38\\%$ 改写成成数，正确的是（）。",
+    "options": [
+      "A. 三成八",
+      "B. 十二成",
+      "C. 九成九",
+      "D. 二成五"
+    ],
+    "answer": "A",
+    "solution": "【分析】工农业生产经常用'成数'表示生产增长的情况, 几成就是十分之几, 也可以用百分数表示, 几成就是百分之几十; 几成几就是百分之几十几。此题是对成数的基础知识的考查，明确几成几就是百分之几十几。\n\n【详解】$38 \\%$ 就是三成八;故选: A。",
+    "knowledge_points": [
+      "成数问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_413",
+    "original_id": "unit2_choice_413"
+  },
+  {
+    "id": "unit2_choice_414",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "算一算, 填一填。\n\n![](images/a06bec041803e1d492dd238df9f91be88d3790d46ded1078f75fbb3c41f615f6.jpg)原价：560元 折扣：八五折 现价：（ ）元\n\n![](images/6250db36db3e7bdc805e6ac3d0fdb9834e0919fa595347380c034427ea325360.jpg)原价：（ ）元 折扣：八折 现价：320元\n\n![](images/4c674256211115ebefd6b6017571b73007904ae61cdb790fe7f0b8fd7ec136d6.jpg)去年： $3500 \\mathrm{~kg}$今年：增产两成是( )kg\n\n![](images/073ebf979b90923d011d685c1e4cba73b0625c887195bbaf92d14f10b60c0c9e.jpg)去年：（)kg今年：减产二成四是 $6080 \\mathrm{~kg}$",
+    "options": [],
+    "answer": "476 ;400 ;4200 ;8000",
+    "solution": "【分析】打几折表示现价是原价的百分几十；成数表示一个数是另一个数的十分之几，改成百分数就是百分几十，据此解答即可。本题考查折扣和成数问题, 解答本题的关键是掌握单位 '1'的量 = 部分量 ÷ 对应分率这个数量关系式。\n\n【详解】（1）现价： $560\\times 85\\% = 476$ （元）；\n\n(2) 原价: $320 \\div 80\\% = 400$ （元）；\n\n(3) 今年: $3500 \\times 20\\% + 3500 = 7 0 0 + 3 5 0 0 = 4 2 0 0$ (千克)\n\n(4) 去年: $6080 \\div (1 - 24\\%)= 6080 \\div 76 \\% = 6 0 8 0 \\div 0. 7 6=8000$ (千克)",
+    "knowledge_points": [
+      "成数问题",
+      "已知一个数的百分之几是多少，求这个数",
+      "折扣问题"
+    ],
+    "source": "19-20 六年级下·全国·单元测试",
+    "merged_id": "unit2_414",
+    "original_id": "unit2_choice_414"
+  },
+  {
+    "id": "unit2_choice_415",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "算出下面各物品打折后出售的价钱。(单位: 元)\n\n![](images/e838eab0f1d308854add0769fe959f7ef6ac1c7a2cdd18744ac24a98321f8263.jpg)六五折 原价：80.00 现价：（）\n\n![](images/cff3880f7e17dc818fd56be39fd85ef7761189d95dd7a7fa7ee2ef77b1a9af08.jpg)七折原价：105.00 现价：（）\n\n![](images/39aec8fa2edc1ace4bec2e8b20115dbf5baa26e1f818aee45b952d83d36e991a.jpg)八八折原价：35.00 现价：（）",
+    "options": [],
+    "answer": "52 73.5 30.8",
+    "solution": "【分析】几折就表示原价的十分之几，也就是百分之几十。几几折就是原价的百分之几十几；据此计算填空即可。求一个数的百分之几是多少，用乘法计算。\n\n【详解】六五折 $= 65\\%$ ， $80 \\times 65\\% = 52$ （元）七折 $= 70\\%$ ， $105 \\times 70\\% = 73.5$ （元）八八折 $= 88\\%$ ， $35 \\times 88\\% = 30.8$ （元）故答案为：52；73.5；30.8",
+    "knowledge_points": [
+      "折扣问题"
+    ],
+    "source": "19-20 六年级下·内蒙古呼伦贝尔·单元测试",
+    "merged_id": "unit2_415",
+    "original_id": "unit2_choice_415"
+  },
+  {
+    "id": "unit2_choice_416",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "按要求把下面的表格填完整。\n\n| 行业 | 娱乐业 | 服务业 | 交通运输业 |\n| ----------------------- | ------ | ------ | ---------- |\n| 营业额中应纳税的部分/元 | 8000 | | 3000 |\n| 税率 | 20% | 6% | |\n| 应纳税额/元 | | 360 | 150 |",
+    "options": [],
+    "answer": "见详解",
+    "solution": "【分析】应纳税额 = 应纳税部分 × 税率, 应纳税部分 = 应纳税额 ÷ 税率, 税率 = 应纳税额 ÷ 应纳税部分。据此, ——列式求出各空, 再填空即可。\n\n【详解】$8000 \\times 20\\% = 1600$ （元）$360 \\div 6\\% = 6000$ （元） $150 \\div 3000 = 5\\%$如表:\n\n| 行业 | 娱乐业 | 服务业 | 交通运输业 |\n| ----------------------- | ------ | ------ | ---------- |\n| 营业额中应纳税的部分/元 | 8000 | 6000 | 3000 |\n| 税率 | 20% | 6% | 5% |\n| 应纳税额/元 | 1600 | 360 | 150 |",
+    "knowledge_points": [
+      "求税率或收入额"
+    ],
+    "source": "22-23六年级下·辽宁鞍山·阶段练习",
+    "merged_id": "unit2_416",
+    "original_id": "unit2_choice_416"
+  },
+  {
+    "id": "unit2_choice_417",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "国庆期间，聪聪一家三口去电影院观看《万里归途》，他们上午去比晚上去一共节省了27元，每张电影票的原价是( )元。\n\n| 优惠信息 | 上午场 | 下午场 | 夜场 |\n| -------- | ------ | ------ | ---- |\n| 折扣 | 七折 | 八折 | 九折 |",
+    "options": [],
+    "answer": "45",
+    "solution": "【分析】几折就表示十分之几，也就是百分之几十，七折就是 $70 \\%$ , 八折就是 $80 \\%$ , 九折就是 $90 \\%$ , 先用一共节省的 27 元除以人数 3 人, 求出每人节约的钱数, 再求出上午与晚上打折的差是 (90 - 70%) , 再根据已知一个数的百分之几是多少, 求这个数用除法, 据此列式为 $27 \\div 3 \\div (90 \\% - 70 \\%)$ 。\n\n【详解】$27 \\div 3 \\div (90\\% - 70\\%)\\\\= 9 \\div 0. 2 \\\\ = 4 5 \\text {(元)}$\n\n所以每张电影票的原价是 45 元。",
+    "knowledge_points": [
+      "求原价（折扣问题）",
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "23-24六年级下·四川遂宁·期末",
+    "merged_id": "unit2_417",
+    "original_id": "unit2_choice_417"
+  },
+  {
+    "id": "unit2_choice_418",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "填写下面的表格。\n\n| 征税企业 | 税率 | 营业额中应纳税部分 | 应纳税额 |\n| -------- | ---- | ------------------ | -------- |\n| 酒店 | 5% | 7.8万元 | 0.39万元 |\n| 客运公司 | 3% | | 4.2万元 |\n| 音乐茶吧 | | 12万元 | 1.2万元 |",
+    "options": [],
+    "answer": "见详解",
+    "solution": "【分析】根据应纳税额 = 营业额中应纳税部分 × 税率, 代入数据计算即可。\n\n【详解】酒店： $7.8 \\times 5\\% = 0.39$ （万元）客运公司：已知的是应纳税额和税率，则营业额中应纳税部分 = 应纳税额 ÷ 税率$4.2 \\div 3\\% = 140$ （万元）音乐茶吧：税率 = 应纳税额 ÷ 营业额中应纳税部分$1.2 \\div 12 \\times 100\\% = 10\\%$\n\n| 征税企业 | 税率 | 营业额中应纳税部分 | 应纳税额 |\n| -------- | ---- | ------------------ | -------- |\n| 酒店 | 5% | 7.8万元 | 0.39万元 |\n| 客运公司 | 3% | 140万元 | 4.2万元 |\n| 音乐茶吧 | 10% | 12万元 | 1.2万元 |",
+    "knowledge_points": [
+      "求应纳税额",
+      "求税率或收入额"
+    ],
+    "source": "24-25 六年级下·全国·单元测试",
+    "merged_id": "unit2_418",
+    "original_id": "unit2_choice_418"
+  },
+  {
+    "id": "unit2_choice_419",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "某电影大世界的影片告示如下表所示，张老师一家三口去看了某一场次的电影《长津湖》，总票价节省了40.5元，那么张老师一家看的是( )场次的电影，优惠后的总票价是( )元。\n\n| 项目 | 内容 |\n| -------- | ------------ |\n| 片名 | 《长津湖》 |\n| 票价 | 45 元 |\n| 优惠方式 | 上午场：六折 |\n| | 下午场：七折 |\n| | 晚场：不优惠 |",
+    "options": [],
+    "answer": "下午 ;94.5",
+    "solution": "【分析】张老师家一共3个人，每张票价是45元，则总价是 135 元。总票价节省了 40.5 元，即现价就是 94.5 元，再根据优惠的百分数 = 现价 ÷ 原价 × 100%，再将百分数后转化为几折，最后对比表格即可。\n\n【详解】$45 \\times 3 = 135$ （元） $135 - 40.5 = 94.5$ （元） $94.5 \\div 135 \\times 100\\% = 70\\%$ 70%=七折 则张老师一家看的是下午场次的电影，优惠后的总票价是94.5元。",
+    "knowledge_points": [
+      "求折扣（折扣问题）"
+    ],
+    "source": "24-25六年级下·全国·单元测试",
+    "merged_id": "unit2_419",
+    "original_id": "unit2_choice_419"
+  },
+  {
+    "id": "unit2_choice_420",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "请你把它们的原价写出来。\n\n![](images/1f495879e545d96c65c84ece9c0aa495d3e42514b4e1c82f66d284ee5bc1d792.jpg)八折 原价：（ ）现价：14.4元\n\n![](images/0a49bbf4e5bccd6da688a06d38b1e037184ae3698318f5718cc2cc02a0dca814.jpg)九折 原价：（ ） 现价：7.2元\n\n![](images/f36bd768f92d2218de720aa691e5dd6e74694180776f1445563a8d874133034c.jpg)七折 原价：（ ）现价：8.4元",
+    "options": [],
+    "answer": "18; 8; 12",
+    "solution": "【分析】打八折，表示现价是原价的 $80 \\%$ , 已知现价是 14.4 元, 根据 '已知一个数的百分之几是多少, 求这个数,用除法计算', 用 14.4 除以 $80 \\%$ 即可求出汉堡的原价;打九折, 表示现价是原价的 $90 \\%$ , 已知现价是 7.2 元, 用 7.2 除以 $90 \\%$ 可以求出薯条的原价;打七折, 表示现价是原价的 $70 \\%$ , 已知现价是 8.4 元, 用 8.4 除以 $70 \\%$ 可以求出鸡腿的原价。\n\n【详解】$14.4 \\div 80\\%\\\\= 1 4. 4 \\div 0. 8 \\\\ = 1 8 \\text {(元)}\\\\$\n\n$7.2 \\div 90 \\% \\\\ = 7. 2 \\div 0. 9\\\\= 8 \\text {(元)} $\n\n$\\\\ 8.4 \\div 70 \\% \\\\ = 8. 4 \\div 0. 7 \\\\ = 1 2 \\text {(元)}$\n\n它们的原价依次是 18 元、8 元、12 元。",
+    "knowledge_points": [
+      "求原价（折扣问题）",
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "24-25六年级下·全国·单元测试",
+    "merged_id": "unit2_420",
+    "original_id": "unit2_choice_420"
+  },
+  {
+    "id": "unit2_choice_421",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "把表格填写完整。\n\n| 小数 | 分数 | 百分数 | 成数 | 折扣 |\n| ---- | ---- | ------ | ------ | ------ |\n| 0.8 | 4/5 | | | |\n| | | 66% | | |\n| | 9/20 | | | |\n| | | | 三成五 | |\n| | | | | 八五折 |",
+    "options": [],
+    "answer": "见详解",
+    "solution": "【分析】分数化成小数，用分子除以分母即可；小数化成百分数，小数点向右移动两位，同时在数的后面添上百分号；小数化成分数, 原来有几位小数, 就在 1 的后面写几个零作分母, 把原来的小数去掉小数点作分子, 再化简成最简分数;百分数化成小数，只要把百分号去掉，同时把分子的小数点向左移动两位；根据成数的意义，百分之几十就是几成，百分之几十十几就是几成几。根据折扣的意义，百分之几十就是几折，百分之几十几就是几几折。\n\n【详解】$0.8 = \\frac{8}{10} = \\frac{4}{5}$ , $0.8 = 80\\%$ , $80\\% =$ 八成, $80\\% =$ 八折 $66\\% = 0.66, 66\\% = \\frac{66}{100} = \\frac{33}{50}, 66\\% =$ 六成六, $66\\% =$ 六六折 $\\frac{9}{20} = 9 \\div 20 = 0.45, 0.45 = 45\\%$ , $45\\% =$ 四成五, $45\\% =$ 四五折三成五 $= \\frac{3.5}{10} = \\frac{35}{100} = 35\\%$ ， $35\\% =$ 三五折， $35\\% = 0.35$ ， $\\frac{35}{100} =$ $\\frac{7}{20}$八五折 $= 85 \\%$ , $85 \\% =$ 八成五, $85 \\% = 0.85, 0.85 = \\frac{85}{100} = \\frac{17}{20}$ 填表如下:\n\n| 小数 | 分数 | 百分数 | 成数 | 折扣 |\n| ---- | ----- | ------ | ------ | ------ |\n| 0.8 | 4/5 | 80% | 八成 | 八折 |\n| 0.66 | 33/50 | 66% | 六成六 | 六六折 |\n| 0.45 | 9/20 | 45% | 四成五 | 四五折 |\n| 0.35 | 7/20 | 35% | 三成五 | 三五折 |\n| 0.85 | 17/20 | 85% | 八成五 | 八五折 |",
+    "knowledge_points": [
+      "打折的意义及应用（分数）",
+      "分数、小数、百分数与成数的互化"
+    ],
+    "source": "24-25 六年级下·海南海口·单元测试",
+    "merged_id": "unit2_421",
+    "original_id": "unit2_choice_421"
+  },
+  {
+    "id": "unit2_choice_422",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "算出下面各物品打折后的价格。\n\n(1)![](images/92c7b8e62792ab5d8b58ecf88ed16007266ecce6f827696e933cde908fdb2751.jpg)原价：90.00，现价：（ ）。\n\n(2)![](images/90b24b4397e1a2e6574589a07930cae49f93221d02bdccb437ac1cf05030c163.jpg)七折原价：38.00，现价：（ ）。\n\n(3)![](images/6b9f0dd154be95649fe67522633a60ad9d7e99d76e3c9e3b5ee3f8b1005e4411.jpg)八折 原价：65.00，现价：（ ）。\n\n(4)![](images/d03eea002e4657fe79450afd4bdcf1248b01d464d7d7cee6c07860f775b361e8.jpg)八八折 原价：40.00，现价：（ ）。",
+    "options": [],
+    "answer": "76.5；26.6；52；35.2",
+    "solution": "【分析】用原价 × 折扣 = 现价，直接列式计算即可。本题考查了折扣问题，几折就是百分之几十。\n\n【详解】$90 \\times 85\\% = 76.5$ （元）$38 \\times 70 \\% = 26.6 (\\text{元})$$65 \\times 80\\% = 52$ （元） $40 \\times 88\\% = 35.2$ （元）",
+    "knowledge_points": [
+      "折扣问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_422",
+    "original_id": "unit2_choice_422"
+  },
+  {
+    "id": "unit2_choice_423",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "一本词典打七折后卖 35 元。这本词典原价是( )元。",
+    "options": [],
+    "answer": "50",
+    "solution": "【分析】词典原价是单位'1'，现价是原价的 $70 \\%$ , 用现价 ÷ 对应百分率 = 原价。本题考查了折扣问题，几折就是百分之几十。\n\n【详解】$35 \\div 70\\% = 50$ （元）",
+    "knowledge_points": [
+      "折扣问题"
+    ],
+    "source": "19-20 六年级下·全国·单元测试",
+    "merged_id": "unit2_423",
+    "original_id": "unit2_choice_423"
+  },
+  {
+    "id": "unit2_choice_424",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "2019年5月11日，爸爸把5000元钱存入银行。定期一年。年利率是 $1.75\\%$ 。到期后，爸爸可得到利息( )元，到期时爸爸一共能取回( )元钱。",
+    "options": [],
+    "answer": "87.5 ;5087.5",
+    "solution": "【分析】根据本金 × 利率 × 存期 = 利息, 用利息 + 本金 = 一共取回的钱。本题考查了利率问题，取款时银行多支付的钱叫利息。\n\n【详解】$5000 \\times 1.75\\% \\times 1 = 87.5$ （元） $5000 + 87.5 = 5087.5$ （元）",
+    "knowledge_points": [
+      "利率问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_424",
+    "original_id": "unit2_choice_424"
+  },
+  {
+    "id": "unit2_choice_425",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "某学校去年用水500吨，今年比去年节水一成五，今年用水( )吨。",
+    "options": [],
+    "answer": "425",
+    "solution": "【分析】去年用水量是单位'1'，今年用水量占去年的 $1 - 15\\%$ ，用去年用水量 × 今年用水量占去年的百分比即可。本题考查了成数问题，几成就是百分之几十。\n\n【详解】$500 \\times (1 - 15\\%)\\\\= 5 0 0 \\times 0. 8 5 \\\\ = 4 2 5 \\text {(吨)}$",
+    "knowledge_points": [
+      "成数问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_425",
+    "original_id": "unit2_choice_425"
+  },
+  {
+    "id": "unit2_choice_426",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "成数表示一个数是另一个数的( )，通称'几成'。例如'四成五'是( )，改写成百分数就是( )。",
+    "options": [],
+    "answer": "十分之几或百分之几十; $\\frac{4.5}{10}$ ;$45\\%$",
+    "solution": "【分析】几成就是十分之几，也可以用百分数表示，几成就是百分之几十。本题考查了成数问题，工农业生产经常用'成数'表示生产增长的情况。\n\n【详解】成数表示一个数是另一个数十分之几或百分之几十，通称'几成'。例如'四成五'是 $\\frac{4.5}{10}$ ，改写成百分数就是 $45\\%$ 。",
+    "knowledge_points": [
+      "成数问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_426",
+    "original_id": "unit2_choice_426"
+  },
+  {
+    "id": "unit2_choice_427",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "一件商品打八折出售，八折 $= (\\quad) = (\\quad)\\%$ 。",
+    "options": [],
+    "answer": "$\\frac{8}{10}$ 80",
+    "solution": "【分析】根据几折就是十分之几，也就是百分之几十，进行填空即可。本题考查了折扣问题，同种商品，折数越低，价格越低。\n\n【详解】一件商品打八折出售，八折 $= \\frac{8}{10} = 80\\%$ 。",
+    "knowledge_points": [
+      "折扣问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_427",
+    "original_id": "unit2_choice_427"
+  },
+  {
+    "id": "unit2_choice_428",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "打几折就是( )是( )的( )。",
+    "options": [],
+    "answer": "现价 原价 百分之几十",
+    "solution": "【分析】按照折扣的概念解答即可。本题考查折扣问题, 解答本题的关键是掌握打几折表示现价是原价的百分之几十。\n\n【详解】打几折表示现价是原价的百分之几十。故答案为：现价；原价；百分之几十。",
+    "knowledge_points": [
+      "折扣问题"
+    ],
+    "source": "19-20 六年级下·全国·单元测试",
+    "merged_id": "unit2_428",
+    "original_id": "unit2_choice_428"
+  },
+  {
+    "id": "unit2_choice_429",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "打五折就是( )占( )的 $50 \\%$",
+    "options": [],
+    "answer": "现价 ;原价",
+    "solution": "【分析】打几折表示现价是原价的百分之几十，据此解答即可。本题考查折扣问题, 解答本题关键是掌握打几折表示现价是原价的百分之几十。\n\n【详解】打五折表示现价是原价的 $50 \\%$ 。故答案为：现价；原价。",
+    "knowledge_points": [
+      "折扣问题"
+    ],
+    "source": "19-20 六年级下·全国·单元测试",
+    "merged_id": "unit2_429",
+    "original_id": "unit2_choice_429"
+  },
+  {
+    "id": "unit2_choice_430",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "商场店庆搞活动促销活动，酸奶'买四送一'，这实际是打( )折销售；某品牌饼干'买一送一'，这实际是打( )折销售。",
+    "options": [],
+    "answer": "八 五",
+    "solution": "【分析】'买四送一'即买5件只需要付4件的钱；设每件商品的价格是1，求出原来5件的价格，再求出4件的价格；然后用4件的价格除以5件的价格，求出现在用的钱数是原来的百分之几十，再根据打折的含义求解即可；同理，'买一送一'即买2件只需要付1件的钱，折扣同上即可求解。理解买四送一即买 5 件只需要付 4 件的钱; 买一送一即买 2 件只需要付 1 件的钱是解题关键。\n\n【详解】设每件商品的价格是 1 , 原来 5 件需要 5 , 4 件需要 4, 折扣为: $4 \\div 5 = 0.8 = 80\\% =$ 八折;'买一送一'：折扣为： $1 \\div 2 = 0.5 = 50\\%$ = 五折",
+    "knowledge_points": [
+      "折扣问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_430",
+    "original_id": "unit2_choice_430"
+  },
+  {
+    "id": "unit2_choice_431",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "李叔叔准备买一部手机，原价4500元，手机专卖店打九折优惠，在此基础上再享受 $5 \\%$ 的优惠，他买这部手机实际要付( )元。",
+    "options": [],
+    "answer": "3847.5",
+    "solution": "【分析】九折就是现价是原价的 $90 \\%$ , 把原价看做单位 '1', 现价是 $4500 \\times 90 \\%$ , 在此基础上在享受 $5 \\%$ 的优惠,这个 $5 \\%$ 对应的单位 '1'是 $4500 \\times 90 \\%$ , 据此计算即可。理解折扣的含义是解题关键。本题还要注意前后两个单位'1'不同。\n\n【详解】$4500 \\times 90\\% \\times (1 - 5\\%)\\\\= 4500 \\times 90 \\% \\times 95 \\% \\\\ = 3 8 4 7. 5 \\text {(元)}$\n\n故答案为: 3847.5",
+    "knowledge_points": [
+      "折扣问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_431",
+    "original_id": "unit2_choice_431"
+  },
+  {
+    "id": "unit2_choice_432",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "$4 \\div 5 = \\frac{()}{5} = \\frac{8}{()} =$ （） $\\% =$ （）折 $=$ （）成。",
+    "options": [],
+    "answer": "4；10；80；八；八",
+    "solution": "【分析】根据分数与除法的关系、分数的基本性质、分数小数百分数的互相转化即折扣、成数的含义，直接填空即可。考查了小数、分数、百分数的互相转化，明确几折就是百分之几十，几成也是百分之几十。\n\n【详解】根据分数与除法的关系， $4 \\div 5 = \\frac{4}{5}$ ；根据分数的基本性质， $\\frac{4}{5} = \\frac{4 \\times 2}{5 \\times 2} = \\frac{8}{10}$ ；4÷5=0.8=80%, 80%=八折=八成。故答案为: 4; 10; 80; 八; 八",
+    "knowledge_points": [
+      "折扣问题",
+      "成数问题"
+    ],
+    "source": "19-20 六年级下·全国·单元测试",
+    "merged_id": "unit2_432",
+    "original_id": "unit2_choice_432"
+  },
+  {
+    "id": "unit2_choice_433",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "聪聪和明明带了150元钱去看电影，买完票后还剩54元。根据表中的价格信息，他们看的是( )场的电影。票价: 60元 上午场: 六折 ; 中午场: 五折 ;下午场: 八折 ; 晚场: 不打折;",
+    "options": [],
+    "answer": "下午场",
+    "solution": "【分析】用（带的钱一剩的钱）÷2=每张票实际价格，再用每张票实际价格÷票价60元即可。本题考查了折扣问题，几折就是百分之几十。\n\n【详解】$(150 - 54) \\div 2\\\\= 9 6 \\div 2 \\\\ = 4 8 \\text {(元)}$\n\n$48 \\div 60 = 0.8 = 80 \\% = 八折$",
+    "knowledge_points": [
+      "折扣问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_433",
+    "original_id": "unit2_choice_433"
+  }, {
+    "id": "unit2_choice_434",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "一台取暖器的原价是280 元，现在的售价是 252 元，这台取暖器是打( )折出售的。",
+    "options": [],
+    "answer": "九",
+    "solution": "【分析】用现在售价÷原价即可。本题考查了折扣问题，几折就是百分之几十。\n\n【详解】$252 \\div 280 = 0.9 = 90\\% =$ 九折",
+    "knowledge_points": [
+      "折扣问题"
+    ],
+    "source": "2018·山东临沂·小升初真题",
+    "merged_id": "unit2_434",
+    "original_id": "unit2_choice_434"
+  },
+  {
+    "id": "unit2_choice_435",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一折 $= (\\quad)\\%$ 七三折 $= (\\quad)\\%$ 四成五 $= (\\quad)\\%$",
+    "options": [],
+    "answer": "10 73 45",
+    "solution": "【分析】几折表示十分之几, 就是百分之几十; 几成表示百分之几十, 据此解答即可。本题考查折扣和成数，解答本题的关键是掌握几折表示十分之几，就是百分之几十；几成表示百分之几十。\n\n【详解】一折是 $10\\%$ ；七三折是 $73\\%$ ；四成五是 $45\\%$ 。故答案为：10；73；45。",
+    "knowledge_points": [
+      "成数问题",
+      "折扣问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_435",
+    "original_id": "unit2_choice_435"
+  },
+  {
+    "id": "unit2_choice_436",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "填空题",
+    "difficulty": "困难",
+    "question": "某牛奶原定价为5元/瓶，甲、乙、丙、丁四个商店以不同的销售方式促销。甲：打九折出售 乙：买四送一 丙：满80元减10元,丁：买够百元打七五折,如果买 15 瓶，去( )商场最省钱。如果买 16 瓶，去( )商场最省钱。如果买20瓶，去( )商场最省钱。",
+    "options": [],
+    "answer": "乙; 乙 ;丙",
+    "solution": "【分析】甲：打九折出售，即用单价乘买的瓶数再乘 0.9 就得到了花的钱数。乙：买四送一，5 瓶为一组，可以先算出有多少组，即送几瓶，然后就可以算出多少瓶需要花钱，总价也就可以算出。丙：满 80 元减 10 元，算出一共花的钱数，满 80 元少付 10 元即可。丁：买够百元打七五折，一共花的钱数够百元，乘 $75\\%$ 即可，如果不够百元，就没有优惠。据此分析各种情况即可。本题考查对买卖活动中打折等各种促销方式的理解，算出每种促销方式下的最终售价是解决本题的关键。\n\n【详解】(1) 甲店:买 15 瓶: $5 \\times 15 \\times 0.9 = 67.5$ (元)买 16 瓶: $5 \\times 16 \\times 0.9 = 72$ (元)买20瓶： $5 \\times 20 \\times 0.9 = 90$ （元）(2) 乙店:买 15 瓶: $15 \\div (4 + 1) = 3$ （组）; $(15 - 3) \\times 5 = 60$(元)\n\n买 16 瓶: $16 \\div (4 + 1) = 3$ （组）……1（瓶）；$(16-3) \\times5 = 65$ (元)买 20 瓶: $20 \\div (4 + 1) = 4$ （组）; $(20 - 4) \\times 5 = 80$(元)\n\n(3) 丙店:买15瓶： $5\\times 15 = 75$ （元）买 16 瓶: $5 \\times 16 = 80$ (元); $80 - 10 = 70$ (元)买 20 瓶: $5 \\times 20 = 100$ (元); $100 - 10 = 90$ (元)\n\n(4) 丁店:买 15 瓶: $5 \\times 15 = 75$ (元)买 16 瓶: $5 \\times 16 = 80$ (元)买20瓶： $5\\times 20 = 100$ （元）； $100\\times 75\\% = 75$ （元）综上：买 15 瓶去乙店；买 16 瓶去乙店；买 20 瓶去丙店最省钱。",
+    "knowledge_points": [
+      "折扣问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_436",
+    "original_id": "unit2_choice_436"
+  },
+  {
+    "id": "unit2_choice_437",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "王叔叔买了1000元的国家债券，定期3年，如果年利率是 $2.89\\%$ ，到期时的利息是( )元，本金和利息共( )元。",
+    "options": [],
+    "answer": "86.7 ;1086.7",
+    "solution": "【分析】利息 = 本金 × 利率 × 存款时间，由此代入数据即可求得到期利息，本金 + 利息代入数据即可得解。此题是考查了公式：利息 = 本金 × 利率 × 存款时间的实际应用。\n\n【详解】$1000 \\times 2.89\\% \\times 3 = 86.7$ （元）1000＋86.7＝1086.7（元)；故答案为: 86.7; 1086.7\n\n$1000 + 86.7 = 1086.7$ （元）；",
+    "knowledge_points": [
+      "利率问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_437",
+    "original_id": "unit2_choice_437"
+  },
+  {
+    "id": "unit2_choice_438",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "某店2月份营业收入是4600元，若按 $18\\%$ 的税率缴税，税后收入是( )元。",
+    "options": [],
+    "answer": "3772",
+    "solution": "【分析】按营业额的 $18 \\%$ 缴税，就是从营业额中扣除 $18 \\%$ , 把营业额看作单位 '1', 扣除 $18 \\%$ , 那么还剩 1-18% = 82%, 即求 4600 元的 83% 是多少, 用乘法计算。考查税率问题，实际上就是'已知一个数，求它的百分之几是多少'，用乘法计算即可。\n\n【详解】$4600 \\times (1 - 18\\%)\\\\= 4600 \\times 82 \\% \\\\ = 3 7 7 2 \\text {(元)}$\n\n故答案为: 3772",
+    "knowledge_points": [
+      "税率问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_438",
+    "original_id": "unit2_choice_438"
+  },
+  {
+    "id": "unit2_choice_439",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一件商品原价500元，现打七五折销售，便宜了( )元。",
+    "options": [],
+    "answer": "125",
+    "solution": "【分析】根据折扣的意义知，打几折就是原价的十分之几或百分之几十，把原价看作是单位'1'，便宜了原价的（1 -75%），单位'1'已知是500元，用乘法计算即可。本题主要考查了学生对折扣知识的掌握，重点是求出现在便宜的价格是原价的百分之几，再根据求一个数的百分之几是多少，用乘法计算。\n\n【详解】$500 \\times (1 - 75\\%)\\\\= 500\\times 25\\% \\\\ = 1 2 5 \\text {(元)}$\n\n故答案为: 125。",
+    "knowledge_points": [
+      "折扣问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_439",
+    "original_id": "unit2_choice_439"
+  },
+  {
+    "id": "unit2_choice_440",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "三折 $= (\\quad)\\%$ ，五五折 $= (\\quad)\\%$ ，四成五 $= (\\quad)\\%$ ，十成 $= (\\quad)$ 。",
+    "options": [],
+    "answer": "30； 55 ；45； 100%",
+    "solution": "【分析】根据折扣和成数的意义知，几折就是十分之几、百分之几十，几几折就是百分之几十几；几成就是十分之几、百分之几十，据此填空即可。此题主要是考查百分数、折扣、成数之间的关系及转化，属于基础知识。利用它们之间的关系即可进行转化。\n\n【详解】三折 $= 30 \\%$ , 五五折 $= 55 \\%$ , 四成五 $= 45 \\%$ , 十成 $= 100\\%$ ;故答案为: $30; 55; 45; 100\\%$",
+    "knowledge_points": [
+      "成数问题",
+      "折扣问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_440",
+    "original_id": "unit2_choice_440"
+  },
+  {
+    "id": "unit2_choice_441",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一种商品，标价400元，商场开展优惠活动'满300元减100'元，这件商品实际打( )折出售。",
+    "options": [],
+    "answer": "七五",
+    "solution": "【分析】根据优惠政策，计算出商品的实际售价。用实际售价除以商品原价就是折扣数。理解'满300元减100'，就满300元才减，不满部分不减是解题关键。\n\n【详解】400里面包含一个300，所以商品实际售价是：400-100=300（元）折扣为: $300 \\div 400 = 0.75 = 75\\%$ = 七五折",
+    "knowledge_points": [
+      "求折扣（折扣问题）"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_441",
+    "original_id": "unit2_choice_441"
+  },
+  {
+    "id": "unit2_choice_442",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一件商品的成本200元，按成本提高 $20\\%$ 标价，然后打九折出售，这件商品的售价是( )元。",
+    "options": [],
+    "answer": "216",
+    "solution": "【分析】九折就是 $90 \\%$ , 根据标价 = 进价 × (1 + 提高率)计算出标价, 由标价 × 90 \\% 就是售价。本题考查了销售问题在实际生活中的运用，标价 = 进价（1 + 提高率），售价 = 标价 × 折扣率，解答时弄清销售问题的数量关系是关键。\n\n【详解】商品的标价为：200（1+20%）=240元； 商品的售价为： $240 \\times 90\\% = 216$ 元； 故答案为：216",
+    "knowledge_points": [
+      "折扣问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_442",
+    "original_id": "unit2_choice_442"
+  },
+  {
+    "id": "unit2_choice_443",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个书包原价是240元，现在打八五折销售，现价比原价少( )%，现价比原价便宜了( )元。",
+    "options": [],
+    "answer": "15 36",
+    "solution": "【分析】打八五折销售，就是把原价看做单位'1'，现价是原价的 $85\\%$ ，比原价少 $1 - 85\\%$ ，求便宜了多少元，就是求原价的 $1 - 85\\%$ 是多少，用乘法计算即可。理解折扣的含义是解答本题的关键。\n\n【详解】$1 - 85\\% = 15\\%$ \n\n$240 \\times (1 - 85 \\%) \\\\ = 240 \\times 15 \\% \\\\ = 3 6 \\text {(元)}$\n\n故答案为: 15; 36",
+    "knowledge_points": [
+      "折扣问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_443",
+    "original_id": "unit2_choice_443"
+  },
+  {
+    "id": "unit2_choice_444",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个帽子打八折优惠了6元，这个帽子的原价是( )元。",
+    "options": [],
+    "answer": "30",
+    "solution": "【分析】八折即现价是原价的 $80 \\%$ , 那么便宜了 $1 - 80 \\% =$ $20 \\%$ , 便宜的 6 元对应的分率是 $20 \\%$ , 据此可解。理解折扣的含义是解题关键。折扣是现在是原价的百分之多少，原价是单位'1'。\n\n【详解】$6 \\div (1 - 80\\%)\\\\= 6 \\div 20 \\% \\\\ = 3 0 \\text {(元)}$\n\n故答案为: 30",
+    "knowledge_points": [
+      "折扣问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_444",
+    "original_id": "unit2_choice_444"
+  },
+  {
+    "id": "unit2_choice_445",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "李阿姨准备买一部手机，原价2500元，现在专卖店打八折优惠，在此基础上再享受 $5 \\%$ 的优惠，她买这部手机实际要付( )元。",
+    "options": [],
+    "answer": "1900",
+    "solution": "【分析】手机原价是 2500 元，打完八折后，再享受 $5 \\%$ 的优惠，这个 $5 \\%$ 是以打完八折后的价格为单位 '1'，所以实际支付的钱是打完八折后价格的 $1 - 5 \\% = 95 \\%$ 本题考查打折问题，解答本题的关键是理解 $5 \\%$ 是以打完八折后的价格为单位'1'，实际支付的钱是打完八折后价格的 $1 - 5\\% = 95\\%$ 。\n\n【详解】打八折： $2500 \\times 80\\% = 2000$ （元）实际支付：\n\n$2000 \\times (1 - 5 \\%) \\\\ = 2000 \\times 95 \\% \\\\ = 1 9 0 0 (\\text {元})$",
+    "knowledge_points": [
+      "求一个数的百分之几是多少",
+      "折扣问题"
+    ],
+    "source": "19-20 六年级下·全国·单元测试",
+    "merged_id": "unit2_445",
+    "original_id": "unit2_choice_445"
+  },
+  {
+    "id": "unit2_choice_446",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "张老师出版一本书获得稿费6800元，按规定，超出800元的部分应缴纳 $14\\%$ 的个人所得税，张老师实际得到( )元。",
+    "options": [],
+    "answer": "5960",
+    "solution": "【分析】6800元超出800元的部分是 $6800 - 800 = 6000$ 元，对6000元进行缴纳 $14\\%$ 的个人所得税，实际上是求6000元的 $14\\%$ 是多少，用稿费减去缴纳的个人所得税就是张老师实际上获得的钱。本题考查税率问题，解答本题的关键是求出需缴纳个人所得税部分。\n\n【详解】个人所得税：\n\n$(6800 - 800) \\times 14 \\% \\\\ = 6000 \\times 14 \\% \\\\ = 8 4 0 \\text {(元)}$\n\n实际得到钱： $6800 - 840 = 5960$ （元）",
+    "knowledge_points": [
+      "税率问题",
+      "含百分数的运算"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_446",
+    "original_id": "unit2_choice_446"
+  },
+  {
+    "id": "unit2_choice_447",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "一种篮球原价180元,现在按原价的七五折出售。这种篮球现价每只( )元，每只便宜了( )元。",
+    "options": [],
+    "answer": "135;45",
+    "solution": "【分析】按原价的七五折出售就是将原价看作单位'1'，现价是原价的 $75 \\%$ .打几几折的意思就是现价是原价的百分之几十几。\n\n【详解】现价： $180 \\times 75\\% = 135$ （元）每只便宜了: $180 - 135 = 45$ (元)答：现价每只135元，每只便宜了45元。",
+    "knowledge_points": [
+      "折扣问题"
+    ],
+    "source": "19-20六年级下·全国·期中",
+    "merged_id": "unit2_447",
+    "original_id": "unit2_choice_447"
+  },
+  {
+    "id": "unit2_choice_448",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "某市下半年汽车销售量比上半年增加一成，下半年的汽车销售量是上半年汽车销售量 $110\\%$ 。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】上边年销售量是单位'1'，下半年销售量是上半年的 $1 + 10\\%$ ，据此分析。本题考查了成数问题，几成就是百分之几十。\n\n【详解】$1 + 10\\% = 110\\%$ 故答案为: √",
+    "knowledge_points": [
+      "成数问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_448",
+    "original_id": "unit2_choice_448"
+  },
+  {
+    "id": "unit2_choice_449",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "银行存款在2008年10月前要缴纳利息税，所以往往会使本金减少。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】到期取款时包括本金和利息，缴纳利息税减少的是利息。本题考查了利率问题，取款时银行多支付的钱是利息。\n\n【详解】银行存款在 2008 年 10 月前要缴纳利息税, 本金不变, 所以原题说法错误。",
+    "knowledge_points": [
+      "利率问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_449",
+    "original_id": "unit2_choice_449"
+  },
+  {
+    "id": "unit2_choice_450",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "某件商品打六五折出售，就是比原价降低 $65\\%$ 出售。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】打折就是按照折数低价出售商品。本题考查了折扣问题，几折就是十分之几，也就是百分之几十，同种商品，折数越低，价格越低。\n\n【详解】某件商品打五六折出售，就是按原价的 $65\\%$ 出售，所以原题说法错误。",
+    "knowledge_points": [
+      "折扣问题"
+    ],
+    "source": "2022·河北唐山·小升初真题",
+    "merged_id": "unit2_450",
+    "original_id": "unit2_choice_450"
+  },
+  {
+    "id": "unit2_choice_451",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "税收是国家收入的主要来源之一，因此，每个公民都有依法纳税的义务。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】国家建设离不开税收，纳税是每个公民应尽的义务。纳税是根据税法的有关规定，按照一定的比率把集体或个人收入的一部分缴纳给国家。\n\n【详解】税收是国家收入的主要来源之一，因此，每个公民都有依法纳税的义务，说法正确。故答案为: √",
+    "knowledge_points": [
+      "税率问题"
+    ],
+    "source": "2022·宁夏石嘴山·小升初真题",
+    "merged_id": "unit2_451",
+    "original_id": "unit2_choice_451"
+  },
+  {
+    "id": "unit2_choice_452",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "王老师到商场买足球, 一个足球打八折后是 72 元, 原价是 80 元。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】八折即现价是原价的 $80 \\%$ , 一个足球打八折后是 72 元, 把足球原价看做单位 '1', 原价的 $80 \\%$ 是 72 元, 用 72 除以对应分率 $80 \\%$ 即可求得原价。明确八折的含义是解题关键。已知部分量, 求单位 '1'用除法计算。\n\n【详解】$72 \\div 80\\% = 90$ （元）足球原价是 90 元，原题计算错误。故答案为:×",
+    "knowledge_points": [
+      "已知一个数的百分之几是多少，求这个数",
+      "折扣问题"
+    ],
+    "source": "19-20 六年级下·全国·单元测试",
+    "merged_id": "unit2_452",
+    "original_id": "unit2_choice_452"
+  },
+  {
+    "id": "unit2_choice_453",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "税率是指应纳税额与各种收入（销售额、营业额……）中应纳税部分的比率。",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】根据税率的概念判断即可。本题考查税率，解答本题关键是掌握税率的概念。\n\n【详解】应纳税额与各种收入（销售额、营业额……）中应纳税部分的比率叫做税率，所以本题说法正确。故答案为: √",
+    "knowledge_points": [
+      "税率问题"
+    ],
+    "source": "19-20 六年级下·全国·单元测试",
+    "merged_id": "unit2_453",
+    "original_id": "unit2_choice_453"
+  },
+  {
+    "id": "unit2_choice_454",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "本金越多, 利息也越多。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】根据关系式：利息 = 本金 × 利率 × 时间，同等条件下，即在利率、时间都相同的条件下，本金越多，所得的利息也越多，据此判断。考查对利息公式的理解和掌握：利息 = 本金 × 利率 × 时间，利息的多少由本金、利率和时间三方面共同决定。\n\n【详解】因为题干没有给出利率和时间，所以本金越多利息也越多说法错误。故答案为: ×",
+    "knowledge_points": [
+      "利率问题"
+    ],
+    "source": "13-14 五年级下·全国·课后作业",
+    "merged_id": "unit2_454",
+    "original_id": "unit2_choice_454"
+  },
+  {
+    "id": "unit2_choice_455",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "打七折的意思就是按原价的 $30\\%$ 出售，也就是降价 $70\\%$ 出售。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】根据折扣的含义，折扣是现价是原价的百分之几，据此判断即可。考查对折扣含义的理解，基础题，较简单。\n\n【详解】打七折的意思就是按原价的 $70\\%$ 出售，也就是降价 $30 \\%$ 出售。题干说法错误。故答案为: ×",
+    "knowledge_points": [
+      "折扣问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_455",
+    "original_id": "unit2_choice_455"
+  },
+  {
+    "id": "unit2_choice_456",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "按照规定, 买房子要缴纳 $1.5\\%$ 的契税，笑笑家新买了一套30万元的住房，她家要缴纳 4500 元的契税。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】买房子要缴纳 $1.5 \\%$ 的契税, 即契税占住房总价的 $1.5 \\%$ , 用乘法计算出来即可判断。求一个数的百分之几是多少，用乘法计算。\n\n【详解】$30 \\times 1.5\\% = 0.45$ （万元） $= 4500$ 元 故答案为: √",
+    "knowledge_points": [
+      "税率问题"
+    ],
+    "source": "19-20 六年级下·全国·单元测试",
+    "merged_id": "unit2_456",
+    "original_id": "unit2_choice_456"
+  },
+  {
+    "id": "unit2_choice_457",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "看图列式计算。求原价。\n\n![](images/ef87f63322faf0725e7f7eb3593016bf7b4cf98af8183f90642d37976f0305cd.jpg)打九折 比原价便宜2元",
+    "options": [],
+    "answer": "20元",
+    "solution": "【分析】已知一个数的百分之几是多少求这个数用除法计算，打九折相当于现价是原价的 $90 \\%$ , 那么便宜的部分就是原价的 $1 - 90 \\%$ , 用 2 除以 $1 - 90 \\%$ 即可求出原价。\n\n【详解】$2 \\div (1 - 90\\%)\\\\= 2 \\div 0. 1 \\\\ = 2 0 \\text {(元)}$\n\n原价是 20 元。",
+    "knowledge_points": [
+      "求原价（折扣问题）"
+    ],
+    "source": "22-23六年级下·甘肃庆阳·阶段练习",
+    "merged_id": "unit2_457",
+    "original_id": "unit2_choice_457"
+  },
+  {
+    "id": "unit2_choice_458",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "看图列式计算。求原价。\n\n![](images/9d40a9a8d5ecc56f3b4cdd0893ca602fa738b3d0a40659456f672ad51091a588.jpg)打七五折比原价便宜60元",
+    "options": [],
+    "answer": "$60 \\div (1 - 0.75)$ ; 240 元",
+    "solution": "【分析】根据图文，结合现价 = 原价 × 折扣这一公式可知，把原价看作单位 '1', 七五折等于 0.75 , 则用 60 除以 (1- 0.75) 即可求出答案。\n\n【详解】$60 \\div (1 - 0.75)\\\\= 6 0 \\div 0. 2 5 \\\\ = 2 4 0 \\text {(元)}$\n\n这条裙子的原价是 240 元。",
+    "knowledge_points": [
+      "求原价 (折扣问题)"
+    ],
+    "source": "22-23 六年级下·广东广州·期中",
+    "merged_id": "unit2_458",
+    "original_id": "unit2_choice_458"
+  },
+  {
+    "id": "unit2_choice_459",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "看图列式计算。\n\n![](images/bde0908542208ac33618727e47b2a3a37662493528e8286b173502c3b0656b85.jpg)",
+    "options": [],
+    "answer": "$240 \\times (1 + 30\\%)$ ; 312 棵",
+    "solution": "【分析】三成就是 $30 \\%$ ；把杨树的棵数看作单位 '1'，柳树的棵数是杨树的 $(1 + 30 \\%$ , 用杨树的棵数 × $(1+$ $30 \\%$ ) ，即可求出柳树的棵数，据此解答。\n\n【详解】三成就是 $30 \\%$ 。\n\n$240 \\times (1 + 30 \\%) \\\\ = 2 4 0 \\times 1. 3 \\\\ = 3 1 2 \\text {(棵)}$\n\n柳树有312棵。",
+    "knowledge_points": [
+      "求增加或减少几成的实际问题"
+    ],
+    "source": "24-25 六年级下·全国·单元测试",
+    "merged_id": "unit2_459",
+    "original_id": "unit2_choice_459"
+  },
+  {
+    "id": "unit2_choice_460",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "困难",
+    "question": "国家公民的工资薪金必须依法缴纳个人所得税。2011年9月1日实施的个税法：工资薪金不超过3500元免征个税，超额部分征收按7级超额累进税率计算，如下表：\n\n| 全月应纳税所得额 | 税率 | 速算扣除数 |\n| -------------------- | ---- | ---------- |\n| 不超过1500元 | 3% | 0 |\n| 超过1500元至4500元 | 10% | 105 |\n| 超过4500元至9000元 | 20% | 555 |\n| 超过9000元至35000元 | 25% | x |\n| 超过55000元至80000元 | 35% | 5505 |\n| 超过80000元 | 45% | 13505 |\n\n(1) 王老师 2016 年 1 月的工资是 8500 元, 他应缴多少元的个人所得税?\n(2) 李老师 2016 年 1 月缴税 1870 元, 他这个月的工资是多少元?",
+    "options": [],
+    "answer": "（1）445元；（2）15000元",
+    "solution": "【分析】（1）工资一免税部分 = 纳税部分， $8500 - 1500 = 3500 = 5000$ （元），王老师需要交三档的税，超出 4500 的部分 × 20% + 超出 1500 元不超过 4500 元的部分 × 10% + 1500 元 × 3% = 应缴个人所得税；(2) 用缴纳的税款一第一档满额税款一第二档满额税款一第三档满额税款 = 第四档缴纳的税款, 李老师工资超过 9000 元, 用第四档缴纳的税款 ÷ 税率 = 第四档缴纳的税款,用免税部分 +9000 元 + 第四档缴纳的税款 = 本月工资。本题考查了税率问题, 纳税是每个公民应尽的义务。\n\n【详解】(1) $8500 - 3500 = 5000$ （元）\n\n$(5000 - 4500)\\times 20\\% +(4500 - 1500)\\times 10\\% +1500\\times 3\\% \\\\ = 500\\times 0.2 + 3000\\times 0.1 + 45\\\\ = 100 + 300 + 45\\\\ = 445 (\\text{元}) $\n\n答: 他应缴 445 元的个人所得税。\n\n(2) $1870 - 1500 \\times 3\\%\\\\= 1 8 7 0 - 4 5 \\\\ = 1 8 2 5 (\\text {元})$\n\n$1825 - (4500 - 1500) \\times 10 \\% \\\\ = 1 8 2 5 - 3 0 0 0 \\times 0. 1 \\\\ = 1 8 2 5 - 3 0 0 \\\\ = 1 5 2 5 \\text {(元)}$\n\n$1 5 2 5 - (9 0 0 0 - 4 5 0 0) \\times 2 0 \\\\ = 1 5 2 5 - 4 5 0 0 \\times 0. 2 \\\\ = 1 5 2 5 - 9 0 0 \\\\ = 6 2 5 \\text {(元)}$\n\n$625 \\div 25\\% = 2500 \\text{ (元)}$ $3500 + 9000 + 2500 = 15000$ （元）答：他这个月的工资是15000元。",
+    "knowledge_points": [
+      "税率问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_460",
+    "original_id": "unit2_choice_460"
+  },
+  {
+    "id": "unit2_choice_461",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一台液晶电视机打九折销售，售价为7650元，比原来便宜多少钱？\n\n![](images/851121e1756b3363731e93e2016daae30ee64517b8939ca76b72d6e5093d880f.jpg)",
+    "options": [],
+    "answer": "850元",
+    "solution": "【分析】用售价 ÷ 折扣 = 原价, 原价一售价 = 便宜的钱, 据此列式解答。本题考查了折扣问题，几折就是百分之几十。\n\n【详解】$7650 \\div 90\\% - 7650\\\\= 8 5 0 0 - 7 6 5 0 \\\\ = 8 5 0 (\\text {元})$\n\n答: 比原来便宜 800 元钱。",
+    "knowledge_points": [
+      "折扣问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_461",
+    "original_id": "unit2_choice_461"
+  },
+  {
+    "id": "unit2_choice_462",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "学校要买15个足球和20个排球，一共应付多少钱？\n\n![](images/7d0f818e2f9d423a03cf010a3ce770495c2b5c9162016ec726122f13b341f9ed.jpg)65元\n\n![](images/7ba1eb465c270499e460fc1b31dd04e8f183298c3d4efaafadbda732eeb5b90e.jpg)40元\n\n每种球买10个以上可以打8折.",
+    "options": [],
+    "answer": "1420元",
+    "solution": "【分析】个数够打折标准，用足球个数 × 单价 + 排球个数 × 单价，求出总价钱，用总价钱 × 折扣即可。本题考查了折扣问题，几折就是百分之几十。\n\n【详解】$15 \\times 65 + 20 \\times 40\\\\= 9 7 5 + 8 0 0\\\\=1775(元)$\n\n$= 1775$ （元）\n\n$1775 \\times 80\\% = 1420$ （元）答：一共应付1420元钱。",
+    "knowledge_points": [
+      "含百分数的运算",
+      "折扣问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_462",
+    "original_id": "unit2_choice_462"
+  },
+  {
+    "id": "unit2_choice_463",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "困难",
+    "question": "小明一家三口准备去吃烤鱼，有下面两种不同的消费方式。\n\n| 消费方式一 | 消费方式二 |\n| ------------------------------------------------------------ | ------------------------------------------- |\n| 团购代金券：每张59元，可抵100元消费；每桌限用2张，不足部分现金补齐。 | 全场八折；持会员卡者在八折基础上再打8.5折。 |\n\n小明一家持有该店会员卡。如果他们一家三口吃烤鱼人均消费预计80元，那么上述哪种消费方式更优惠？请通过计算来说明。",
+    "options": [],
+    "answer": "消费方式一更优惠",
+    "solution": "【分析】本题是和折扣相关的应用题，解答此题先要求出全家预计消费总额，再结合不同的消费方式求出两种消费方式的具体花费。根据59元1张的代金券可抵100元消费，每桌限用2张，先计算消费200元实际付款多少元，然后再加上一家人消费超过200元的部分，得出消费方式一需要多少钱；八折优惠，持有会员卡的顾客在八折的基础上再打八五折，就是说消费总额 × 80 \\% × 85 \\% = 一家人按消费方式二的应付款。通过比较，选择更优惠的消费方式即可。在本题中，两种消费方式的规则稍显复杂，因此，一定要把题目读懂读透，再着手解答。对于消费方式一，消费金额分为两部分，一是团购两张代金券的花费，二是不足部分补足的现金。\n\n【详解】全家预计消费： $80 \\times 3 = 240$ （元）\n\n消费方式一：\n\n$2 \\times 5 9 + (2 4 0 - 1 0 0 \\times 2)\\\\=11 8 + 4 0\\\\= 1 5 8 \\quad (\\text {元})$\n\n消费方式二：\n\n$240\\times 80\\% \\times 85\\% \\\\ =1 9 2 \\times 0. 8 5 \\\\ =163.2 (元)\\\\1 6 3. 2 > 1 5 8$\n\n答：消费方式一更优惠。",
+    "knowledge_points": [
+      "经济问题",
+      "折扣问题"
+    ],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit2_463",
+    "original_id": "unit2_choice_463"
+  },
+  {
+    "id": "unit2_choice_464",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "王大伯要买一台4500元的冰箱，甲、乙、丙三家店的促销活动如下，同一冰箱原价相同，你建议王大伯去哪家店买冰箱更划算？甲店：每满 1000元减100元。乙店：超出4000元的部分打五折。丙店：一律九五折，且折后满4000元再减200元。",
+    "options": [],
+    "answer": "丙",
+    "solution": "【分析】甲店：先看原价里有几个 1000 元，就从总价里减去几个 100 元；乙店：可先用原价减去 4000 元，再用超出的部分乘 $50\\%$ ，最后加上原价 4000 元；丙店：先用原价乘 $95\\%$ ，再看折后是否满 4000 元，如果满的话，就再减去 200 元；最后再比较大小即可。本题综合考查了有关满减、折扣等百分数的实际应用，所以一定要读懂题意，并在此基础上进行运算及比较。\n\n【详解】甲：4500元里面有4个1000元\n\n$4 5 0 0 - 4 \\times 1 0 0 \\\\ = 4 5 0 0 - 4 0 0 \\\\ = 4 1 0 0 \\text {(元)}$\n\n乙: $4000 + (4500 - 4000) \\times 50\\%\\\\= 4000 + 500 \\times 50 \\% \\\\ = 4 0 0 0 + 2 5 0\\\\= 4250 (元)$\n\n丙： $4500 \\times 95\\% = 4275$ （元） $4275 - 200 = 4075$ （元） 因为 4075 元 $< 4100$ 元 $< 4250$ 元, 所以丙店价格 $<$ 甲店价格 $<$ 乙店价格。答：王大伯去丙店买冰箱更划算。",
+    "knowledge_points": [
+      "经济问题",
+      "折扣问题",
+      "优化问题"
+    ],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit2_464",
+    "original_id": "unit2_choice_464"
+  },
+  {
+    "id": "unit2_choice_465",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "李老师要买 30 本《没头脑和不高兴》。三家书店的标价都是每本10元，李老师到哪家书店购买最划算？最少需要多少元？\n\n| 甲书店 | 乙书店 | 丙书店 |\n| ---------- | --------------------------- | --------------- |\n| 全部打九折 | 每买5本送1本（不满5本不送） | 每满100元减12元 |",
+    "options": [],
+    "answer": "乙书店；250元",
+    "solution": "【分析】根据题意，分别求出三家书店需要的钱数，找出用钱最少的一家。此题考查了折扣问题，明确打几折就是按原价的百分之几十出售，明确三家书店的优惠规则是解题关键。\n\n【详解】$30 \\times 10 = 300$ （元）甲: $300 \\times 90\\% = 270$ （元）\n\n乙: $30 \\div (5 + 1) \\times 5 \\times 10\\\\= 5 \\times 5 \\times 1 0 \\\\ = 2 5 0 \\text {(元)}$\n\n丙: $300 \\div 100 = 3$\n\n$3 0 0 - 3 \\times 1 2 \\\\ = 3 0 0 - 3 6 \\\\ = 2 6 4 \\text {(元)}$\n\n$2 5 0 < 2 6 4 < 2 7 0$\n\n答: 李老师到乙书店购买最划算。最少需要 250 元。",
+    "knowledge_points": [
+      "折扣问题",
+      "优化问题"
+    ],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit2_465",
+    "original_id": "unit2_choice_465"
+  },
+  {
+    "id": "unit2_choice_466",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "下面是中国交通银行2020年10月的存款利率表。张老师将4万元存入交通银行，存期为三年。到期时取出的利息够买一台价值3250元的空调吗？\n\n| 存款方式 | 活期 | 整存整取 | 整存整取 | 整存整取 | 整存整取 | 整存整取 |\n| --------- | ---- | -------- | -------- | -------- | -------- | -------- |\n| 存期 | —— | 三个月 | 六个月 | 一年 | 二年 | 三年 |\n| 年利率(%) | 0.30 | 1.35 | 1.55 | 1.75 | 2.25 | 2.75 |",
+    "options": [],
+    "answer": "够",
+    "solution": "【分析】由统计表可知：存期三年的年利率为 $2.75 \\%$ , 根据利息 = 本金 × 利率 × 时间, 代入数据求出利息, 再与空调价格比较即可。本题主要考查利率问题，确定利率是解题的关键。\n\n【详解】$40000 \\times 2.75\\% \\times 3\\\\= 1 1 0 0 \\times 3\\\\= 3 3 0 0 (\\text {元})$\n\n$3 3 0 0 > 3 2 5 0$\n\n答: 到期时取出的利息够买一台价值 3250 元的空调。",
+    "knowledge_points": [
+      "利率问题"
+    ],
+    "source": "20-21六年级下·全国·单元测试",
+    "merged_id": "unit2_466",
+    "original_id": "unit2_choice_466"
+  },{
+    "id": "unit2_choice_467",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "下面是 A、B 两个医疗器械公司同种防护服的促销方式。原价均为480元/套。\n\n| A公司 | B公司 |\n| -------- | ----------------- |\n| 八折促销 | 每满200元优惠40元 |\n\n如果要买这样的一套防护服，在A、B两个公司买，各应付多少钱？选择哪个公司更省钱？",
+    "options": [],
+    "answer": "A公司应付384元；B公司应付400元；选择选择 A公司更省钱。",
+    "solution": "【分析】A 公司打八折代表示现价是原价的百分之八十；B公式每满 200 元优惠 40 元，原价 480 元里有 2 个 200 元，所以共优惠 80 元，据此解答即可。本题考查折扣问题, 解答本题的关键是理解打几折表示现价是原价的百分之几十。\n\n【详解】A： $480\\times 80\\% = 384$ （元）B: $480 - 40 - 40 = 400$ (元) $384 < 400$ , 选择 A 公司更省钱。答：A公司应付384元；B公司应付400元；选择选择A公司更省钱。",
+    "knowledge_points": [
+      "折扣问题"
+    ],
+    "source": "19-20 六年级下·山东济南·期末",
+    "merged_id": "unit2_467",
+    "original_id": "unit2_choice_467"
+  },
+  {
+    "id": "unit2_choice_468",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "下图是一件商品价格标签的一部分，这件商品的原价是多少？\n\n![](images/343608aa70af36d81d4e38ece425da52a9ba10100481f4e8b08bf4fa338cec2e.jpg)8折后现价200元",
+    "options": [],
+    "answer": "250元",
+    "solution": "【分析】8折即 $80 \\%$ , 根据原价 × 折扣 = 现价, 则现价 ÷ 折扣 = 原价, 代入数值即可。本题考查折扣问题，明确8折即 $80\\%$ 是解题的关键。\n\n【详解】$200 \\div 80\\% = 250$ （元）答: 这件商品的原价是 250 元。",
+    "knowledge_points": [
+      "折扣问题"
+    ],
+    "source": "21-22六年级上·江西赣州·期末",
+    "merged_id": "unit2_468",
+    "original_id": "unit2_choice_468"
+  },
+  {
+    "id": "unit2_choice_469",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "两个书店的《中国故事》每本标价都是 10 元,请你算一算：王老师到哪家书店购买比较便宜？最少要花多少钱？\n\n王老师:我们班级要买24本《中国故事》 大众书店 每本按原价 按9折出售。 求知书店每买5本赠送1本、不满5本不赠送。",
+    "options": [],
+    "answer": "求知书店，200元。",
+    "solution": "【分析】方案一：计算出原来买 24 本书的总价，乘折扣就是大众书店买书的价钱；方案二：把 24 本买 5 送 1 看成 4 组，就是一共买了多少本，再乘每本的单价，即是求知书店买书的价钱。【详解】方案一： $24 \\times 10 \\times 90\\% \\div 216$ 元；方案二: $24 \\div (5 + 1) = 4$ 组, $4 \\times 5 = 20$ 本, $20 \\times 10 = 200$ 元;$216 > 200$ , 求知书店比较便宜。答: 王老师到求知书店购买比较便宜, 最少要花 200 元。此题的解题关键是掌握折扣和原价之间的关系,理解买几送几问题中实际购买的数量, 才能解决问题。",
+    "knowledge_points": [
+      "折扣问题"
+    ],
+    "source": "19-20 六年级下·湖南怀化·单元测试",
+    "merged_id": "unit2_469",
+    "original_id": "unit2_choice_469"
+  },
+  {
+    "id": "unit2_choice_470",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "学校篮球队要买50个篮球，张老师去了甲、乙、丙三家专卖店，单价都是75元。但促销方式各不相同。\n\n| 甲专卖店 | 乙专卖店 | 丙专卖店 |\n| -------- | -------- | ----------------- |\n| 买十送二 | 打八折 | 每满100返现金20元 |\n\n请你帮张老师算一算，去哪家专卖店买最合算？（请写出计算过程）",
+    "options": [],
+    "answer": "去乙专卖店最划算",
+    "solution": "【分析】甲专卖店买十送二，50个篮球里有多少个十二，就意味着有多少个二要送，计算出实际购买的数量，再乘单价，计算出甲专卖店优惠后的价格；乙专卖店打八折，就是原价的 $80\\%$ ，根据一个数乘百分数的意义，用乘法计算出乙专卖店优惠后的价格；丙专卖店每满 100 元返现金 20 元，总价 3750 元里有几个 100，就能减几个 20 元，据此计算出丙专卖店优惠后的价格；再进行比较即可得出更省钱的店。本题先理解各专卖店优惠的办法，再根据这些办法求出在各专卖店实际花的钱数，进而比较得解。\n\n【详解】甲专卖店： $50\\div (10 + 2) = 50\\div 12 = 4$ （组）…2（个）\n\n$(5 0 - 4 \\times 2) \\times 7 5\\\\=(5 0 - 8) \\times 7 5\\\\= 4 2 \\times 7 5\\\\=3150 (元)$\n\n乙专卖店: $50 \\times 75 \\times 80\\% = 3750 \\times 0.8 = 3000$ (元)\n\n丙专卖店： $50 \\times 75 \\div 100 = 3750 \\div 100 = 37$ （个）……50（元）\n\n$3 7 5 0 - 3 7 \\times 2 0\\\\= 3 7 5 0 - 7 4 0\\\\=3010 (元)$\n\n$3000 < 3010 < 3150$答：经比较，去乙专卖店购买最划算。",
+    "knowledge_points": [
+      "含百分数的运算",
+      "折扣问题"
+    ],
+    "source": "21-22六年级下·全国·单元测试",
+    "merged_id": "unit2_470",
+    "original_id": "unit2_choice_470"
+  },
+  {
+    "id": "unit2_choice_471",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "某市商业银行存款利率如下表：\n\n| | 活期 | 整存整取 | 整存整取 | 整存整取 | 整存整取 | 整存整取 | 整存整取 |\n| --------- | ---- | -------- | -------- | -------- | -------- | -------- | -------- |\n| 存期 | —— | 三个月 | 半年 | 一年 | 二年 | 三年 | 五年 |\n| 年利率(%) | 0.46 | 2.73 | 2.99 | 3.25 | 4.03 | 4.88 | 5.23 |\n\n李阿姨今年3月把20000元钱存入该银行，存期2年，到期时可以取回多少钱？",
+    "options": [],
+    "answer": "21612元",
+    "solution": "【分析】由题意可知，存2年的年利率是 $4.03 \\%$ , 根据利息 = 本金 × 利率 × 存期, 据此求出利息, 然后再加上本金即可。本题考查利率问题，明确利息 = 本金 × 利率 × 存期是解题的关键。\n\n【详解】$20000 \\times 4.03\\% \\times 2 + 20000\\\\= 1 6 1 2 + 2 0 0 0 0 \\\\ = 2 1 6 1 2 \\text {(元)}$\n\n答: 到期时可以取回 21612 元。",
+    "knowledge_points": [
+      "利率问题"
+    ],
+    "source": "2022·河南安阳·小升初真题",
+    "merged_id": "unit2_471",
+    "original_id": "unit2_choice_471"
+  },
+  {
+    "id": "unit2_choice_472",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "兰兰将350元存入银行，存期两年，年利率 $2.10\\%$ ，两年后，她取出的钱可以买哪种品牌的学习机呢？\n\n![](images/7861e49e8eeac1a6b8e8d168452063cd095d38317cdf458dda2cf0951cb3348b.jpg)",
+    "options": [],
+    "answer": "乙品牌",
+    "solution": "【分析】要求两年后，她可以买哪种品牌的学习机，应先求出两年后的本息共多少元。本题中，本金是350元，利率是 $2.10\\%$ ，时间是2年，根据关系式：本息 = 本金 + 本金 × 利率 × 时间，求出本息，然后看看可以买哪种品牌的学习机，解决问题。此题属于利息问题，此类利息问题一般都有固定公式，完成此类问题根据公式代入数据计算即可。\n\n【详解】$350 + 350 \\times 2.10\\% \\times 2\\\\= 3 5 0 + 7. 3 5 \\times 2 \\\\ = 3 5 0 + 1 4. 7 \\\\ = 3 6 4. 7 \\text {(元)}$\n\n360 元 $< 364.7$ 元 $< 370$ 元\n\n答: 她可以买乙种品牌的学习机。",
+    "knowledge_points": [
+      "经济问题",
+      "利率问题"
+    ],
+    "source": "21-22六年级下·广东云浮·期末",
+    "merged_id": "unit2_472",
+    "original_id": "unit2_choice_472"
+  },
+  {
+    "id": "unit2_choice_473",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "根据国家税务法规定，个人所得税征收标准为：个人月收入5000元以下的不征税，超过5000元的，超过部分按如表的标准征收个人所得税。李叔叔6月份工资是9500元，需要交税多少元？\n\n| 每月应纳税所得额 | 税率/% |\n| -------------------------------- | ------ |\n| 不超过3000元的部分 | 3 |\n| 超过3000元，不超过12000元的部分 | 10 |\n| 超过12000元，不超过25000元的部分 | 20 |",
+    "options": [],
+    "answer": "240元",
+    "solution": "【分析】根据个人所得税缴纳办法，李叔叔的收入超过 5000 元的部分为 $9500 - 5000 = 4500$ 元，因为 $4500 = 3000 + 1500$ ，所以分两段缴税即可。本题主要考查个人所得税纳税问题，关键分清总收入和超过个人所得税收税起点的关系。\n\n【详解】$9500 - 5000 = 4500$ （元）\n\n$4 5 0 0 = 3 0 0 0 + 1 5 0 0 \\\\ 3000 \\times 3 \\% + 1500 \\times 10 \\% \\\\ = 9 0 + 1 5 0 \\\\ = 2 4 0 \\text {(元)}$\n\n答：需要交税240元。",
+    "knowledge_points": [
+      "税率问题"
+    ],
+    "source": "21-22六年级下·辽宁盘锦·期末",
+    "merged_id": "unit2_473",
+    "original_id": "unit2_choice_473"
+  },
+  {
+    "id": "unit2_choice_474",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "王老师要给学校买7个大花盆。请帮王老师选择一家更优惠的店铺购买，并算一算需要多少钱？\n\n![](images/0336f48db5cac22875e5dd07e14d4febfe4d5e18af99dccfbca61302197c0c57.jpg) 200元/个\n\n甲店：打九折\n\n乙店：买四送一",
+    "options": [],
+    "answer": "1200元",
+    "solution": "【分析】甲店：用花盆的单价乘花盆的数量，求出买7个大花盆的原价；九折 $= 90\\%$ ，利用原价 × 折扣 = 现价，把数据代入求甲店优惠后的价格。乙店：买四送一，要先计算7个里面有多少组（4+1），就送多少组花盆，用要买的花盆数减去送的花盆数，再乘花盆的单价，即可求出乙店优惠后的价格。再比较两家商店优惠后的价格，选择一家更实惠的商店购买即可。最优化问题常用比较法进行解答，分别计算出两种方案优惠后的价格，再进行比较。\n\n【详解】甲店： $200 \\times 7 \\times 90\\%\\\\= 1400 \\times 90 \\% \\\\ = 1 2 6 0 \\text {(元)}$\n\n即在甲店购买需要 1260 元。\n\n乙店: $7 \\div (4 + 1)\\\\= 7 \\div 5 \\\\\\approx 1$ (组)\n\n$2 0 0 \\times (7 - 1) \\\\ = 2 0 0 \\times 6 \\\\ = 1 2 0 0 (\\text {元})$\n\n即在乙店购买需要 1200 元。$1 2 0 0 < 1 2 6 0$答：选择在乙店购买，需要 1200 元。",
+    "knowledge_points": [
+      "折扣问题",
+      "优化问题"
+    ],
+    "source": "21-22六年级下·云南玉溪·期末",
+    "merged_id": "unit2_474",
+    "original_id": "unit2_choice_474"
+  },
+  {
+    "id": "unit2_choice_475",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "妈妈到超市买了下面的物品各一件。这些物品都按八折出售，共要付多少钱？\n\n![](images/50c1d290a392cdf50784241e1ab1e2a76e57afc8b29b3c7480a0117ac7806afe.jpg) 34元\n\n![](images/f59f8d1664b0c670c1a74dfcfe8379d6220f7fc0d334e8ea557573e82826c651.jpg)26元",
+    "options": [],
+    "answer": "48元",
+    "solution": "【分析】'按八折出售'是指现在的商品价格是原价的 $80 \\%$ , 即现价 = 原价 × 80 \\% 。先求台灯打折后的价格, 即 $34 \\times 80 \\%$ ; 再求水壶打折后的价格, 即 $26 \\times 80 \\%$ ; 最后求两件物品的现价和。打几折就是按原价的百分之几十出售。在解决问题时，直接把折扣写成百分数列式。\n\n【详解】$34 \\times 80\\% + 26 \\times 80\\%\\\\= 3 4 \\times 0. 8 + 2 6 \\times 0. 8 \\\\ = (3 4 + 2 6) \\times 0. 8 \\\\ = 6 0 \\times 0. 8 \\\\ = 4 8 \\text {(元)}$\n\n答: 共要付 48 元。",
+    "knowledge_points": [
+      "折扣问题"
+    ],
+    "source": "22-23六年级上·甘肃张掖·期中",
+    "merged_id": "unit2_475",
+    "original_id": "unit2_choice_475"
+  },
+  {
+    "id": "unit2_choice_476",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "李阿姨到银行存款 50000 元, 定期 3 年, 到期后她一共可以取回多少钱?\n2017年中国建行城乡居民存款挂牌利率表\n\n| 存期 | 年利率（%） |\n| ---- | ----------- |\n| 一年 | 1.75 |\n| 二年 | 2.25 |\n| 三年 | 2.75 |\n| 五年 | 2.75 |",
+    "options": [],
+    "answer": "54125元",
+    "solution": "【分析】已知存款本金为 50000 元，定期 3 年，从利率表中可知，定期三年的年利率是 $2.75\\%$ ；先根据'利息 = 本金 × 利率 × 存期'，求出到期后可得到的利息，再加上本金，即是到期后一共可以取回的钱数。\n\n【详解】$50000 + 50000 \\times 2.75\\% \\times 3\\\\= 5 0 0 0 0 + 5 0 0 0 0 \\times 0. 0 2 7 5 \\times 3 \\\\ = 5 0 0 0 0 + 4 1 2 5 \\\\ = 5 4 1 2 5 \\text {(元)}$\n\n答: 到期后她一共可以取回 54125 元。",
+    "knowledge_points": [
+      "求利息"
+    ],
+    "source": "23-24 六年级下·河北邯郸·期末",
+    "merged_id": "unit2_476",
+    "original_id": "unit2_choice_476"
+  },
+  {
+    "id": "unit2_choice_477",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "学校奖励优秀少先队员，需要12元一支的钢笔12支，到哪家购买合算些？\n\n| 兴华商店 | 睿智商店 | 利民商店 |\n| ---------- | ---------------- | ------------- |\n| 买5支送1支 | 所有商品一律九折 | 满100元减20元 |",
+    "options": [],
+    "answer": "：去兴华商店买合算。",
+    "solution": "【分析】兴华商店买 5 支送 1 支, 即将 $5 + 1 = 6$ 支看作一组, 12 里有几个 6 就送几支, 用单价 × 实际购买数量, 即可求出兴华商店优惠后的价格。睿智商店一律九折，就是原价的 $90 \\%$ , 用单价 × 数量 × 90 \\% , 即可计算出去睿智商店购买的优惠后的价格。利民商店满 100 元减 20 元，先用单价 × 数量求出总价，再看总价里有几个 100 元，就能减去几个 20 元，据此计算出利民商店优惠后的价格。三家比较，即可得出最省钱最合算的店。\n\n【详解】兴华: $12 - 12 \\div (5 + 1)\\\\= 1 2 - 1 2 \\div 6 \\\\ = 1 2 - 2 \\\\ = 1 0 (\\text {支})$\n\n$1 2 \\times 1 0 = 1 2 0 (\\text {元})$\n\n睿智： $12 \\times 12 \\times 90\\%\\\\= 144 \\times 90 \\% \\\\ = 1 2 9. 6 \\text {(元)}$\n\n利民: $12 \\times 12 = 144$ (元) $144 \\div 100 = 1$ （个）……44元 $144 - 20 = 124$ (元) $120 < 124 < 129.6$ 答：所以去兴华商店买合算。",
+    "knowledge_points": [
+      "经济问题",
+      "求现价（折扣问题）"
+    ],
+    "source": "23-24 六年级上·河北邯郸·期中",
+    "merged_id": "unit2_477",
+    "original_id": "unit2_choice_477"
+  },
+  {
+    "id": "unit2_choice_478",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "今年母亲节那天，小乐一家来到'小妹湘菜馆'就餐（消费情况见菜单），小乐还将这次消费的金额绘制成了扇形统计图。![](images/3371f0e26acaadde184b4f6728658a89084a24cef9fb15c4800846a92d6674ab.jpg)![](images/251cce411b3b0b7827ae45bf3084627b68fde380e9a1b1b93fb4d37e2e811e89.jpg)\n\n(1) 点心的金额处不小心被弄破了一个洞, 请你算出这份点心的价格。\n(2) 母亲节这天 '小妹湘菜馆' 八八折优惠, 小乐妈妈带了一张 30 元的消费券, 可抵现金 30 元。结账时先抵券再打折, 小乐家应支付多少元?",
+    "options": [],
+    "answer": "(1) 48元 (2) 325.6元",
+    "solution": "【分析】（1）以总金额为单位'1'，用米饭的金额除以米饭金额占总金额的分率即可求出就餐花的总金额，用求得的总金额乘点心占总金额的分率即可求出点心的价格；(2) 用就餐花的总金额减去 30 元，再乘 $88\\%$ 即可求出小乐家应支付的金额。\n\n【详解】(1) $16 \\div 4\\% = 400$ （元） $400 \\times 12 \\% = 48 (\\text{元})$ 答: 这份点心的价格是 48 元。(2) 八八折 $= 88\\%$\n\n$(16 \\div 4 \\% -30) \\times 88 \\% \\\\ = (400 - 30) \\times 88 \\% \\\\ = 370 \\times 88 \\% \\\\ = 3 2 5. 6 \\text {(元)}$\n\n答: 小乐家应支付 325.6 元。",
+    "knowledge_points": [
+      "统计图表的综合应用",
+      "求一个数的百分之几是多少",
+      "已知一个数的百分之几是多少，求这个数",
+      "求现价（折扣问题）"
+    ],
+    "source": "23-24六年级下·湖南湘潭·期末",
+    "merged_id": "unit2_478",
+    "original_id": "unit2_choice_478"
+  },
+  {
+    "id": "unit2_choice_479",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "妈妈有10万元人民币，要按照定期存入银行，想存两年。银行存款利率表2024年12月\n\n| 存款期限（整存整取） | 年利率 |\n| -------------------- | ------ |\n| 1年定期 | 1.1% |\n| 2年定期 | 1.2% |\n\n请你设计出2种存款方案，并计算出每种方案所得利息各是多少。",
+    "options": [],
+    "answer": "方案一：可以先存一年，到期后再连本带息再存一年。方案二：存两年定期。方案一所得利息是 2212.1 元，方案二所得利息是 2400 元。（方案不唯一）",
+    "solution": "【分析】想存两年，方案一：可以先存一年，到期后再连本带息再存一年。方案二：存两年定期。根据利息 = 本金 × 利率 × 时间，代入数据计算即可。单位转化为元再计算。\n\n【详解】方案一：可以先存一年，到期后再连本带息再存一年。\n\n10万元 $= 100000$ 元\n\n$100000 \\times 1.1\\% \\times 1 = 1100$ （元）\n\n$(1100 + 100000) \\times 1.1\\% \\times 1\\\\=101100×1.1\\%×1\\\\=1112.1 (元)$\n\n$1100 + 1112.1 = 2212.1$ （元）\n\n方案二：存两年定期。\n\n$100000 \\times 1.2\\% \\times 2\\\\=1200×2\\\\=2400 (元) (方案不唯一)$\n\n答: 方案一所得利息是 2212.1 元, 方案二所得利息是 2400 元。",
+    "knowledge_points": [
+      "求利息",
+      "含百分数的运算"
+    ],
+    "source": "24-25六年级上·北京通州·期末",
+    "merged_id": "unit2_479",
+    "original_id": "unit2_choice_479"
+  },
+  {
+    "id": "unit2_choice_480",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "水果批发市场某种水果的标价是20元/袋，为了促销，又提出了如下优惠条件。\n\n| 数量/袋 | 10~30 | 31~50 | 51及以上 |\n| ------- | ------ | ----- | ------------------------------------------------------------ |\n| 折扣 | 九五折 | 九折 | 八五折甲、乙两人都要购买这种水果，甲有600元，乙有1000元，他们两人最多分别可以购买这种水果多少袋？ |",
+    "options": [],
+    "answer": "甲33袋;乙58袋",
+    "solution": "【分析】如果不优惠，根据总价÷单价 = 数量，可得甲最少可以买 $600 \\div 20 = 30$ （袋），乙最少可以买 $1000 \\div 20 = 50(袋)$, 优惠后买的袋数会更多, 则甲购买时享受九折优惠, 乙享受八五折优惠。九折优惠, 表示现价是原价的90%, 根据 '求一个数的百分之几是多少, 用乘法计算',用 20 乘 $90 \\%$ 可以求出甲购买每袋水果的现价, 再用 600 除以每袋水果的现价, 即可求出甲最多可以买多少袋; 打八五折，表示现价是原价的 $85 \\%$ , 用 20 乘 $85 \\%$ 乙购买每袋水果的现价, 再用 1000 除以每袋水果的现价, 即可求出乙最多可以买多少袋。结果要用 '去尾法' 取整数值。\n\n【详解】$600 \\div 20 = 30$ （袋）$1000 \\div 20 = 50$ （袋）九折 $= 90\\%$ 八五折=85%\n\n甲: $600 \\div (20 \\times 90\\%)\\\\= 6 0 0 \\div 1 8 \\\\ \\approx 33 (袋)$\n\n乙: $1000 \\div (20 \\times 85\\%)\\\\= 1 0 0 0 \\div 1 7 \\\\\\approx 58$ (袋)\n\n答: 甲最多可以购买这种水果 33 袋, 乙最多可以购买这种水果 58 袋。",
+    "knowledge_points": [
+      "求一个数的百分之几是多少",
+      "经济问题",
+      "含百分数的运算",
+      "求现价（折扣问题）"
+    ],
+    "source": "24-25 六年级下·全国·单元测试",
+    "merged_id": "unit2_480",
+    "original_id": "unit2_choice_480"
+  },
+  {
+    "id": "unit2_choice_481",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "实验小学举行书法比赛，准备给获得优秀奖的学生奖励一个水杯，每个水杯15元，三家商场有不同的促销活动。\n\n| 甲商场 | 乙商场 | 丙商场 |\n| ---------------- | -------- | --------------- |\n| 商品一律九折优惠 | 买八送一 | 每满100元减15元 |\n\n学校想买 180 个水杯，请你算一算，到哪家商场购买最合算？",
+    "options": [],
+    "answer": "丙商场",
+    "solution": "【分析】甲商场：九折优惠，表示现价是原价的 $90 \\%$ , 用原价乘 $90 \\%$ 可以求出一个水杯的现价, 再根据单价 × 数量 = 总价, 用一个水杯的现价乘 180 , 即可求出一共要花多少钱。乙商场: 买八送一, 即花 8 个水杯的价钱可以得到 9 个水杯。把这 9 个水杯看作一组, 用 180 除以 9 可以求出组数。 $15 \\times 8=120（元）$，每组水杯120元，用120乘组数，即可求出一共要花多少钱。丙商场：根据单价 × 数量，用 15 乘 180 求出水杯的总价，再除以 100 表示总价里面有几个 100，里面有几个 100，就从总价中减去几个 15，即可求出一共需要多少元。最后把三个商场需要的钱数比较即可解答。\n\n【详解】甲商场： $15 \\times 90\\% \\times 180\\\\= 1 3. 5 \\times 1 8 0 \\\\ = 2 4 3 0 \\text {(元)}$\n\n乙商场: $15 \\times 8 = 120$ (元)\n\n$1 8 0 \\div (8 + 1) \\times 1 2 0 \\\\ = 1 8 0 \\div 9 \\times 1 2 0 \\\\ = 2 0 \\times 1 2 0 \\\\ = 2 4 0 0 \\text {(元)}$\n\n丙商场: $15 \\times 180 = 2700$ (元)\n\n$2 7 0 0 \\div 1 0 0 \\times 1 5\\\\= 2 7 \\times 1 5 \\\\ = 4 0 5 \\text {(元)}$\n\n$2 7 0 0 - 4 0 5 = 2 2 9 5 \\text {(元)} \\\\ 2 2 9 5 < 2 4 0 0 < 2 4 3 0$\n\n答：到丙商场购买最合算。",
+    "knowledge_points": [
+      "求一个数的百分之几是多少",
+      "经济问题",
+      "带有小括号的混合运算",
+      "求现价（折扣问题）"
+    ],
+    "source": "24-25六年级下·全国·单元测试",
+    "merged_id": "unit2_481",
+    "original_id": "unit2_choice_481"
+  },
+  {
+    "id": "unit2_choice_482",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "我们国家坚持 '房子是用来住的，不是用来炒的'定位，因地制宜，精准施策，把'稳房价'落到实处。下图是根据某网站统计的镇海区2021年1-5月份新房平均价格统计表。\n\n| 月份 | 1月 | 2月 | 3月 | 4月 | 5月 |\n| --------------------- | ----- | ----- | ----- | ----- | ----- |\n| 新房均价（元/平方米） | 14750 | 14650 | 18624 | 18624 | 19625 |\n\n(1) 如果想清楚地看出房价的变化趋势，选用________统计图比较合适。\n(2) 小宁家以 5 月份新房的平均价格首次购买了一套面积是100平方米的住房。按照国家规定，需要缴纳一定的契税(如下表)。那么小宁家需要缴纳契税多少元?\n\n| 套数 | 面积 | 税率 | 税额 |\n| ---------- | --------------- | ---- | ----------- |\n| 首套 | 90 m²（含）以下 | 1% | 总房价×1% |\n| 首套 | 90 m²以上 | 1.5% | 总房价×1.5% |\n| 二套 | 90 m²（含）以下 | 1% | 总房价×1% |\n| 二套 | 90 m²以上 | 2% | 总房价×2% |\n| 三套及以上 | 不区分面积 | 3% | 总房价×3% |",
+    "options": [],
+    "answer": "(1) 折线;(2) 29437.5元",
+    "solution": "【分析】（1）条形统计图能清楚地看出数量的多少；折线统计图能清楚地看出数量的增减变化情况；扇形统计图能反应各个部分占总体的百分之几；根据不同统计图的特点进行选择。(2) 将房屋面积乘单价，先求出房屋总价。再根据应缴税部分 × 税率 = 应交税款，代入数据即可得到结果。\n\n【详解】（1）如果想清楚地看出房价的变化趋势，选用折线统计图比较合适。\n\n(2) $19625 \\times 100 \\times 1.5 \\%\\\\= 1962500 \\times 1.5 \\%\\\\= 2 9 4 3 7. 5 \\text {(元)}$\n\n答: 小宁家需要缴纳契税 29437.5 元。",
+    "knowledge_points": [
+      "单式折线统计图",
+      "求应纳税额"
+    ],
+    "source": "25-26 六年级上·全国·单元测试",
+    "merged_id": "unit2_482",
+    "original_id": "unit2_choice_482"
+  },
+  {
+    "id": "unit2_choice_483",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "李老师得到一笔6000元的稿费，其中1200元是免税的，其余部分要按 $14\\%$ 的税率缴税。这笔稿费一共要缴税多少元？",
+    "options": [],
+    "answer": "672元",
+    "solution": "【分析】用稿费一免税部分 = 纳税部分, 再纳税部分 × 税率 = 缴纳的税款。本题考查了税率问题, 依法纳税是每个公民应尽的义务。\n\n【详解】$(6000 - 1200) \\times 14\\%\\\\= 4 8 0 0 \\times 0. 1 4 \\\\ = 6 7 2 \\text {(元)}$\n\n答: 这笔稿费一共要缴税 672 元。",
+    "knowledge_points": [
+      "税率问题"
+    ],
+    "source": "19-20 六年级下·全国·单元测试",
+    "merged_id": "unit2_483",
+    "original_id": "unit2_choice_483"
+  },
+  {
+    "id": "unit2_choice_484",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "妈妈带依依到文具店买了下面的学习用品各一件。这些学习用品都按九折出售，共要付多少钱？\n\n![](images/e6ab005bf2a1dc4b955ca8b414dd4775ae99cc7c25bc4b502c12c365d7deff55.jpg)42元\n\n![](images/63ee5a940ba29405f8338f32c4eaad8140650155c44e8b7474a8349e6095dae3.jpg)28元\n\n![](images/85258f2b4534d4ec899ed7ad43925733d3daa4aa2fc2e92fbec7f56a69c92e80.jpg)14元\n\n![](images/a0478b58d0383d4667928f56a824cf1ed0d35b6eb660847fd88223d5e729a43d.jpg)10元",
+    "options": [],
+    "answer": "84.6元",
+    "solution": "【分析】先将四种学习用品的价钱加起来，用总价钱 × 折扣 = 应付价钱。本题考查了折扣问题，打折就是按照折数低价出售商品，几折就是百分之几十。\n\n【详解】$(42 + 28 + 14 + 10) \\times 90\\%\\\\= 9 4 \\times 0. 9 \\\\ = 8 4. 6 \\text {(元)}$\n\n答：共要付 84.6 元钱。",
+    "knowledge_points": [
+      "折扣问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_484",
+    "original_id": "unit2_choice_484"
+  },
+  {
+    "id": "unit2_choice_485",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "某品牌电脑十一期间开展促销活动，甲商场按'每1000元减100元'的方式促销；乙商场打九折销售。爸爸准备买一台标价5200元的该品牌电脑：(1) 在甲乙两个商场买各应付多少元？(2) 去哪个商场买更省钱？",
+    "options": [],
+    "answer": "（1）甲商场：4700元，乙商场：4680元；(2)乙商场",
+    "solution": "【分析】'每 1000 元减 100 元'即满 1000 元减 100 元，不满部分不减；九折即按原价的 $90\\%$ ，据此根据电脑标价 5200 元，分别计算出在甲乙两个商场购买分别需要付的钱数；对比两个钱数，钱数越少则在对应那个商场购买更省钱。理解'每1000元减100元'和九折的含义是解题关键，计算过程要认真仔细。\n\n【详解】(1) 甲商场: $5200 \\div 1000 = 5 \\dots 200$ 根据优惠政策 '每 1000 元减 100 元', 可减 5 个 100 即 500 元;$5200 - 500 = 4700$ （元）乙商场： $5200 \\times 90\\% = 4680$ （元）答: 在甲商场需要付 4700 元, 在乙商场需要付 4680 元。(2) $4680 < 4700$所以在乙商场更省钱。答: 去乙商场购买更省钱。",
+    "knowledge_points": [
+      "折扣问题",
+      "优化问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_485",
+    "original_id": "unit2_choice_485"
+  },
+  {
+    "id": "unit2_choice_486",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "水果店运进400千克苹果，按每千克8元出售。卖 $80\\%$ 掉后，剩下的打八折全部售出，这些苹果共卖多少元？",
+    "options": [],
+    "answer": "3072元",
+    "solution": "【分析】水果店运进 400 千克苹果, 其中 $80 \\%$ 按每千克 8 元售出, 根据分数乘法的意义, 按 8 元出售的有 $400 \\times 80 \\%$ 千克, 可卖 $400 \\times 80 \\% \\times 8$ 元; 根据分数减法的意义, 打八折出售的占全部的 $1 - 80 \\%$ , 即为 $400 \\times (1 - 80 \\%)$ 千克, 打八折即按原价的 $80 \\%$ 出售, 可卖 $400 \\times (1 - 80 \\%) \\times 8 \\times 80 \\%$ 元,然后相加即得共卖多少元。完成本题要认真分析条件中所给数量之间的关系,然后列出正确算式。\n\n【详解】$400 \\times 80\\% \\times 8 + 400 \\times (1 - 80\\%) \\times 8 \\times 80\\%\\\\= 400 \\times 80 \\% \\times 8 + 400 \\times 20 \\% \\times 8 \\times 80 \\% \\\\ = 2 5 6 0 + 5 1 2 \\\\ = 3 0 7 2 \\text {(元)}$\n\n答: 这些苹果共卖 3072 元。",
+    "knowledge_points": [
+      "折扣问题",
+      "含百分数的运算"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_486",
+    "original_id": "unit2_choice_486"
+  },
+  {
+    "id": "unit2_choice_487",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "某品牌的童装搞促销活动，在A商场按'每满100元减30元'的方式销售，在B商场打七折销售。妈妈要给小小买一件标价240元的这种品牌的童装。(1) 在 A、B 两个商场买, 各应付多少钱? (2) 在哪个商场买更省钱？A、B两商场的价格相差多少钱？",
+    "options": [],
+    "answer": "(1) 180 元; 168 元 (2) B 商场; 12 元",
+    "solution": "【分析】（1）A商场：'每满 100 元减 30 元'，240 元可以减去 2 个 30 元，用 240 元减去 $30 \\times 2$ 元就是 A 商场应付的钱数；B 商场：打七折，是指现价是原价的 $70 \\%$ , 把原价看成单位 '1', 用原价乘 $70 \\%$ 就是现价。(2) 最后比较即可求出哪个商场更省钱，用减法求出A、B 两商场的价格相差多少钱。本题关键是理解打折以及'每满 100 减 30 元'的含义，分别求出现价，从而得解。\n\n【详解】(1) A 商场: $240 - 30 \\times 2\\\\= 2 4 0 - 6 0 \\\\ = 1 8 0 \\text {(元)}$\n\nB 商场: $240 \\times 70\\% = 168$ （元）答: A 商场买付 180 元, B 商场买付 168 元。(2) $168 < 180$ $180 - 168 = 12$ （元） 答: 在 B 商场买更省钱, A、B 两商场的价格相差 12 元。",
+    "knowledge_points": [
+      "折扣问题",
+      "优化问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_487",
+    "original_id": "unit2_choice_487"
+  },
+  {
+    "id": "unit2_choice_488",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "某羽绒服店搞换季清仓活动，甲品牌羽绒服满800元减300元，乙品牌羽绒服'折上折'，就是先打八折，在此基础上再打八五折。如果两个品牌都有一件标价950元的羽绒服，哪个品牌的更便宜？",
+    "options": [],
+    "answer": "乙品牌",
+    "solution": "【分析】甲品牌，满 800 元减 300 元，那么原价 950 元的羽绒服，只需要 950-300 元；乙品牌，'折上折'，先打八折，在此基础上再打八五折，先把原价看成单位'1'，用原价乘 $80\\%$ ，就是八折后的价格，再把八折后的价格看成单位'1'，再乘 $85\\%$ ，就是现价;比较两种品牌的现价即可求解。本题主要考查折扣的计算, 注意打几折, 就是标价乘百分之几十。\n\n【详解】甲品牌: $950 > 800$ , 所以减 300 元, $950 - 300 =$ 650 (元)。\n\n乙品牌： $950 \\times 80\\% \\times 85\\%\\\\= 760 \\times 85 \\% \\\\ = 6 4 6 \\text {(元)}$\n\n$6 4 6 < 6 5 0$答：乙品牌的更便宜。",
+    "knowledge_points": [
+      "折扣问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_488",
+    "original_id": "unit2_choice_488"
+  },
+  {
+    "id": "unit2_choice_489",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "![](images/caa987d1aec49aaf809623b4fe27517a89f3fbfb801ccd83a2a9a5f8f1ca487a.jpg)\n\n(1) 一件西服原价 850 元，现在九折出售，比原价便宜了多少钱？\n(2) 张阿姨买了一件羊毛衫，省了35元。这件羊毛衫原价多少钱？",
+    "options": [],
+    "answer": "(1) 85元；(2) 350元",
+    "solution": "【分析】（1）原价是单位'1'，现价比原价便宜了 $1 - 90\\%$ ，用原价 × 便宜的百分率 = 便宜的价钱；(2) 原价是单位 '1', 现价比原价省了 $1 - 90 \\%$ , 用省了的钱 ÷ 省了的百分率 = 原价。本题考查了折扣问题，几折就是百分之几十。\n\n【详解】(1) $850 \\times (1 - 90\\%)\\\\= 8 5 0 \\times 0. 1 \\\\ = 8 5 \\text {(元)}$\n\n答: 比原价便宜了 85 元钱。\n\n(2) $35 \\div (1 - 90\\%)\\\\= 3 5 \\div 0. 1 \\\\ = 3 5 0 \\text {(元)}$\n\n答: 这件羊毛衫原价 350 元钱。",
+    "knowledge_points": [
+      "折扣问题",
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_489",
+    "original_id": "unit2_choice_489"
+  },
+  {
+    "id": "unit2_choice_490",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "某航空公司规定:乘坐飞机的每位旅客可免费携带20千克行李，超过20千克的部分，每千克按机票票价的 $1.5\\%$ 购买行李票。李俊从某地飞往北京，票价1500元，他带了26千克行李，应付行李票款多少元？",
+    "options": [],
+    "answer": "135元",
+    "solution": "【分析】已知超过20千克的部分每千克要按飞机票原价的1.5%购买行李票，那么1千克应付行李费 $1500 \\times 1.5\\%$ 元；行李超过 20 千克的部分是 $26 - 20 = 6$ （千克），因此，应付行李费 $1500 \\times 1.5\\% \\times (26 - 20)$ ，解决问题。此题解答的关键是求出行李超过20千克的部分，然后按'每千克要按飞机票原价的 $1.5\\%$ 购买行李票'来解决问题。\n\n【详解】$1500 \\times 1.5\\% \\times (26 - 20)\\\\= 1500 \\times 1.5 \\% \\times 6 \\\\ = 1 3 5 (\\text {元})$\n\n答: 应付行李票款 135 元。",
+    "knowledge_points": [
+      "含百分数的运算"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_490",
+    "original_id": "unit2_choice_490"
+  },
+  {
+    "id": "unit2_choice_491",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "如今网购团购已经走进我们的生活，小明的爸爸、妈妈准备星期天带他去吃火锅。妈妈说，她在网上发现团购代金券，59元一张，可抵100元消费，每桌限用两张，不足部分用现金补齐。爸爸打电话订座位时，服务员告诉他不用代金券的话可以享受七折优惠。聪明的你替小明算一算，若一家三口吃火锅人均消费预计为 80 元，上述哪种消费方式更优惠？（通过计算说明）",
+    "options": [],
+    "answer": "用团购代金券更优惠",
+    "solution": "【分析】使用团购代金券时，59元可抵100元，相当于每100元减41元，总消费里面有几个100元，就减几个41元（最多减2个），即可算出抵扣后的费用。然后再计算出享受七折优惠后的费用，两者进行比较。本题关键是理解折扣的意义及代金券的使用方法,分别计算出两种方案优惠后的费用进行比较即可。\n\n【详解】使用团购代金券优惠后的费用：$8 0 \\times 3 - (1 0 0 - 5 9) \\times 2 \\\\ = 2 4 0 - 8 2 \\\\ = 1 5 8 \\text {(元)}$\n\n享受七折优惠后的费用:七折 $= 70\\%$ $80 \\times 3 \\times 70\\% = 168$ （元） $158 < 168$ 答：使用团购代金券更优惠。",
+    "knowledge_points": [
+      "经济问题",
+      "优化问题"
+    ],
+    "source": "2022·河北邯郸·小升初真题",
+    "merged_id": "unit2_491",
+    "original_id": "unit2_choice_491"
+  },
+  {
+    "id": "unit2_choice_492",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "某楼盘原来准备以每平方米9000元的平均价格对外销售，现在为了加快资金周转。房地产开发商对价格进行两次下调后，决定以每平方米8000元的平均价格开盘销售。(1) 王伯伯准备以开盘平均价格购买一套 $100 \\mathrm{~m}^{2}$ 的房子,开发商还给予下列两种优惠方案以供选择。方案一：打九八折。方案二：不打折。送三年物业管理费。（物业管理费是每平方米每月1.5元）请你帮王伯伯选择一种方案，并说明理由。(2）优惠后王伯伯买这套房还要按照实际房价的 $1.5\\%$ 缴纳契税，契税是多少元？",
+    "options": [],
+    "answer": "（1）王伯伯选择方案一，通过计算可知方案一比方案二更优惠。(2) 11760元",
+    "solution": "【分析】（1）方案一：打九八折就是原价的 $98 \\%$ , 用原价乘 $98 \\%$ 即可求出现价。方案二：物业管理费是每平方米每月 1.5 元，先求出 $100 \\mathrm{~m}^{2}$ 的房子三年物业管理费；再从买房的总价减去三年物业管理费。即为实际付的钱数。最后比较两种方案哪个更优惠。(2) 把实际房价看作单位 '1', 单位 '1'的 $1.5\\%$ 用乘法。此题考查百分数的实际应用。要理解折扣的意义；掌握求单位'1'的百分之几是多少用乘法解答。\n\n【详解】（1）方案一： $8000 \\times 98\\% \\times 100 = 784000$ （元）\n\n方案二: $8000 \\times 100 - 1.5 \\times 100 \\times 12 \\times 3\\\\= 8 0 0 0 0 0 - 5 4 0 0 \\\\ = 7 9 4 6 0 0 \\text {(元)}$\n\n因为： $784000 < 794600$ ，方案一比方案二更优惠，所以王伯伯选择方案一。答: 王伯伯选择方案一。(2) $784000 \\times 1.5\\% = 11760$ （元）答: 契税是 11760 元。",
+    "knowledge_points": [
+      "经济问题",
+      "优化问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_492",
+    "original_id": "unit2_choice_492"
+  },
+  {
+    "id": "unit2_choice_493",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "困难",
+    "question": "下表是工商银行2018年存款利率表：\n\n| 存款时间 | 利率 % |\n| -------- | ------ |\n| 三个月 | 1.35 |\n| 半年 | 1.55 |\n| 一年 | 1.75 |\n| 二年 | 2.25 |\n| 三年 | 2.75 |\n| 五年 | 2.75 |\n\n小明妈妈将 20000 元存入银行，存期三年，小华妈妈同样将 20000 元存入银行，存期 1 年，到期后连本带利自动转存，请你帮忙计算一下，三年后，他们谁得到的利息多？",
+    "options": [],
+    "answer": "小明妈妈得到的利息多",
+    "solution": "【分析】两人的本金都是 20000 元，小明妈妈存的时间是 3 年，对应的利率是 $2.75 \\%$ ; 小华妈妈每次存一年, 利率是 $1.75 \\%$ , 存 3 次, 每一次的本金是上一年到期后的本息和。根据利息 = 本金 × 利率 × 时间, 分别计算出两人所得的利息进行比较即可。本题考查两种不同的存钱方式所获得的利息的计算方法，对于到期后自动转存这种方式，利息要分开计算后再求和，明确每一次的本金就是上一年的本息和是解答此题的关键。\n\n【详解】小明妈妈得到的利息： $20000 \\times 2.75\\% \\times 3 = 1650$ （元）小华妈妈得到的利息分3次计算（取整数）：\n\n第一年的利息： $20000 \\times 1.75\\% \\times 1 = 350$ （元）本金和利息的和: $20000 + 350 = 20350$ (元)第二年的利息： $20350 \\times 1.75\\% \\times 1 \\approx 356$ （元）本金和利息的和: $20350 + 356 = 20706$ (元)第三年的利息： $20706 \\times 1.75 \\% \\times 1 \\approx 362$ （元）三年的利息和: $350 + 356 + 362 = 1068$ (元)$1068 < 1650$答: 三年后小明妈妈得到的利息多。",
+    "knowledge_points": [
+      "选择储蓄的最佳方案"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_493",
+    "original_id": "unit2_choice_493"
+  },
+  {
+    "id": "unit2_choice_494",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "小明把2000元存入银行，整存整取一年，年利率为 $4\\%$ ，到期时小明可得利息多少元？",
+    "options": [],
+    "answer": "80元",
+    "solution": "【分析】利息 = 本金 × 利率 × 时间，据此解答。本题考查利率问题，熟记利息计算公式即可。\n\n【详解】$2000 \\times 4\\% \\times 1 = 80$ （元）答：到期时小明可得利息 80 元。",
+    "knowledge_points": [
+      "利率问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_494",
+    "original_id": "unit2_choice_494"
+  },
+  {
+    "id": "unit2_choice_495",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "体育彩票的兑奖规定：奖金超过一万元的部分要缴纳 $20\\%$ 的个人所得税。某人幸运地中了一个500万元的巨奖，他要缴纳的税款是多少万元？",
+    "options": [],
+    "answer": "99.8万元",
+    "solution": "【分析】超过一万元的奖金要按获奖额的 $20 \\%$ 缴纳个人所得税，先用 500 万减去 1 万，求出需要缴税部分的奖金，再乘上税率 $20 \\%$ 即可求出他应缴纳的税款是多少万元。此题解答的关键是求出超过 1 万元的部分, 然后运用关系式: 税款 = 总收入 × 税率, 解决问题。\n\n【详解】$(500 - 1) \\times 20\\%\\\\= 499 \\times 20 \\% \\\\ = 9 9. 8 \\text {(万 元)}$\n\n答: 他应缴纳的税款是 99.8 万元。",
+    "knowledge_points": [
+      "税率问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_495",
+    "original_id": "unit2_choice_495"
+  },
+  {
+    "id": "unit2_choice_496",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "某工厂十月份的销销售额是1400万元，如果按销售额的 $8\\%$ 缴纳营业税，十月份应缴纳营业税多少万元？",
+    "options": [],
+    "answer": "112 万元",
+    "solution": "【分析】$8 \\%$ 的单位'1'是销售额，即1400万元，根据分数乘法的意义，即可求出十月份应缴纳的营业税。此题属于简单的百分数乘法应用题，利用应缴纳营业税 = 销售额 × 税率解决问题。\n\n【详解】$1400 \\times 8\\% = 112$ （万元）答：十月份应缴纳营业税112万元。",
+    "knowledge_points": [
+      "税率问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_496",
+    "original_id": "unit2_choice_496"
+  },
+  {
+    "id": "unit2_choice_497",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "体育用品上的排球打九折销售，军军买了一个排球便宜了25元，这个排球原价多少元？",
+    "options": [],
+    "answer": "250元",
+    "solution": "【分析】九折就是现价是原价的 $90 \\%$ , 原价是单位 '1',那么便宜了 $1 - 90 \\%$ , 即 25 元对应的分率, 求单位 '1'用除法，据此计算即可。明确折扣的含义，找准25元对应的分率是解题关键。\n\n【详解】$25 \\div (1 - 90\\%)= 25 \\div 10 \\% = 2 5 0 \\text {(元)}$答: 这个排球原价 250 元。",
+    "knowledge_points": [
+      "折扣问题",
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_497",
+    "original_id": "unit2_choice_497"
+  },
+  {
+    "id": "unit2_choice_498",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "春节期间，有三个家庭计划去青岛旅游。福美旅行社的收费标准是：如果买4张全票，则其余人按半价优惠；平安旅行社的收费标准是：家庭旅游算团体票，按原价的七五折优惠。已知这两家旅行社的全票价格均为 1000 元。请你为以下三个家庭选择较为实惠的旅行社，并计算每个家庭应付的钱数。\n\n| 家庭\\人数 | 大人 | 孩子 | 合计 |\n| ---------- | ---- | ---- | ---- |\n| 张家 | 4 | 3 | 7 |\n| 李家 | 6 | 4 | 10 |\n| 王家 | 3 | 1 | 4 |",
+    "options": [],
+    "answer": "张家选择平安旅行比较实惠，需 5250 元；李家选择福美旅行社比较实惠，需 7000 元；王家选择平安旅行比较实惠，需 3000 元。",
+    "solution": "【分析】算出张家、李家、王家在福美旅行社、平安旅行社各需支付的钱，进行比较选哪家旅行社较实惠即可。本题考查优化和打折问题, 解答本题的关键是算出张家、李家、王家在福美旅行社、平安旅行社各需支付的钱, 进行比较, 选择更实惠的方案即可。\n\n【详解】福美旅行社：\n\n(1) 张家：$4 \\times 1 0 0 0 + 3 \\times 1 0 0 0 \\times 0. 5 \\\\ = 4 0 0 0 + 1 5 0 0 \\\\ = 5 5 0 0 \\text {(元)}$\n\n(2）李家：$4 \\times 1 0 0 0 + 6 \\times 1 0 0 0 \\times 0. 5 \\\\ = 4 0 0 0 + 3 0 0 0 \\\\ = 7 0 0 0 (\\text {元})$\n\n(3) 王家: $4 \\times 1000 = 4000$ (元)平安旅行社:\n\n(1) 张家:\n\n$7 \\times 1 0 0 0 \\times 0. 7 5 \\\\ = 7 0 0 0 \\times 0. 7 5 \\\\ = 5 2 5 0 (\\text {元})$\n\n(2) 李家:\n\n$1 0 \\times 1 0 0 0 \\times 0. 7 5 \\\\ = 1 0 0 0 0 \\times 0. 7 5 \\\\ = 7 5 0 0 \\text {(元)}$\n\n(3) 王家:\n\n$4 \\times 1 0 0 0 \\times 0. 7 5\\\\= 4 0 0 0 \\times 0. 7 5 \\\\ = 3 0 0 0 (\\text {元})$\n\n张家： $5500 > 5250$ ，所以张家选择平安旅行比较实惠，需 5250 元。李家： $7000 < 7500$ ，所以李家选择福美旅行社比较实惠，需7000元。王家： $4000 > 3000$ ，所以王家选择平安旅行比较实惠，需 3000 元。答：张家选择平安旅行比较实惠，需5250元；李家选择精美旅行社比较实惠，需7000元；王家选择平安旅行比较实惠，需3000元。",
+    "knowledge_points": [
+      "折扣问题",
+      "优化问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_498",
+    "original_id": "unit2_choice_498"
+  },
+  {
+    "id": "unit2_choice_499",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "商场新进一批服装，比进价提高 $20\\%$ 后销售，一件也没卖出去，后来打八折销售，全部售完，结果亏损 200 元。这批服装的进价是多少元？",
+    "options": [],
+    "answer": "5000元",
+    "solution": "【分析】20%是以进价为单位'1'，则售价是 $1 + 20\\%$ ，打八折后销售是 $(1 + 20\\%) \\times 80\\% = 96\\%$ ，则亏损的钱是进价的 $1 - 96\\% = 4\\%$ ，根据单位'1'的量=部分量÷对应分率，求出进价多少元即可。本题考查打折问题，解决本题的关键是找出亏损的200元占进价的百分之几。\n\n【详解】售价： $1 + 20\\%$打折后售价： $(1 + 20\\%) \\times 80\\% = 96\\%$亏损的钱是进价的： $1 - 96\\% = 4\\%$进价: $200 \\div 4\\% = 5000$ (元)答: 这批服装的进价是 5000 元。",
+    "knowledge_points": [
+      "已知一个数的百分之几是多少，求这个数",
+      "折扣问题"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_499",
+    "original_id": "unit2_choice_499"
+  },
+  {
+    "id": "unit2_choice_500",
+    "unit": "第二单元：百分数（二）",
+    "chapter": "百分数（二）",
+    "question_type": "解答题",
+    "difficulty": "困难",
+    "question": "六年级二班有 $70\\%$的同学参加了英语小组，有 $70\\%$ 的同学参加了数学小组，两个小组都参加的占 $60\\%$ ，另外有8人这两个小组都没参加。六年级二班有学生多少人？",
+    "options": [],
+    "answer": "40人",
+    "solution": "【分析】有 $70 \\%$ 的同学参加了英语小组, 有 $70 \\%$ 的同学参加了数学小组，两个小组都参加的占全班人数的 $60\\%$ ，可得参加了学习小组的人数是全班人数的 $70\\% + 70\\% - 60\\% =80%$, 则没参加的人数是全班的 $1 - 80\\% = 20\\%$ , 再根据单位 '1'的量 = 部分量 ÷ 对应分率，求出六年级二班的学生人数即可。本题考查百分数的实际应用，解答本题的关键是求出参加了一个或两个小组的人数是全班人数的 $70\\% + 70\\% - 60\\% = 80\\%$ 。\n\n【详解】参加了学习小组的人数是全班人数的： $70\\% + 70\\% - 60\\% = 80\\%$ 没参加的人数是全班的： $1 - 80\\% = 20\\%$全班人数: $8 \\div 20\\% = 40$ （人）答: 六年级二班有学生 40 人。",
+    "knowledge_points": [
+      "已知一个数的百分之几是多少，求这个数"
+    ],
+    "source": "19-20六年级下·全国·单元测试",
+    "merged_id": "unit2_500",
+    "original_id": "unit2_choice_500"
+  }
+]
\ No newline at end of file
diff --git "a/\345\205\255\345\271\264\347\272\247\351\242\230\351\233\206/\345\205\255\345\271\264\347\272\247\344\270\213/\345\205\255\345\271\264\347\272\247\344\270\213\347\254\254\344\272\224\345\215\225\345\205\203\351\270\275\345\267\242\346\225\260\346\215\256\351\233\206.json" "b/\345\205\255\345\271\264\347\272\247\351\242\230\351\233\206/\345\205\255\345\271\264\347\272\247\344\270\213/\345\205\255\345\271\264\347\272\247\344\270\213\347\254\254\344\272\224\345\215\225\345\205\203\351\270\275\345\267\242\346\225\260\346\215\256\351\233\206.json"
new file mode 100644
index 0000000000000000000000000000000000000000..308436d387fa715665ea5181fc353be9896ee36d
--- /dev/null
+++ "b/\345\205\255\345\271\264\347\272\247\351\242\230\351\233\206/\345\205\255\345\271\264\347\272\247\344\270\213/\345\205\255\345\271\264\347\272\247\344\270\213\347\254\254\344\272\224\345\215\225\345\205\203\351\270\275\345\267\242\346\225\260\346\215\256\351\233\206.json"
@@ -0,0 +1,8542 @@
+
+[
+    {
+        "id": "unit5_choice_001",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "（2024·广东东莞·小升初真题）有红、白、蓝三种颜色的卡片各5张，至少随意抽取（）张卡片，才能保证取到两张不同颜色的卡片。",
+        "options": [
+          "A. 3",
+          "B. 4",
+          "C. 5",
+          "D. 6"
+        ],
+        "answer": "D",
+        "solution": "【分析】运用最不利原则，先考虑最糟糕的抽取情况，就是先把一种颜色的卡片全部抽完，因为每种颜色卡片有 5 张，所以先把一种颜色（比如红色）的 5 张卡片全部抽出来，此时再抽一张，无论抽到白色还是蓝色卡片，都能保证取到两张不同颜色的卡片，所以至少要抽取  $(5 + 1)$  张卡片，据此解答。\n\n【详解】根据分析：$5 + 1 = 6$  (张) 即至少取 6 张卡片, 才能保证取到两张不同颜色的卡片。故答案为: D",
+        "knowledge_points": ["最不利原则"],
+        "source": "2024·广东东莞·小升初真题",
+        "merged_id": "unit5_001",
+        "original_id": "unit5_choice_001"
+      },
+      {
+        "id": "unit5_choice_002",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "(24-25 六年级下·全国·单元测试) 小华的 3 次数学考试成绩总分为 290 分（成绩都为整数），3 次成绩中至少有一次成绩不低于（）分。",
+        "options": [
+          "A. 99",
+          "B. 98",
+          "C. 97",
+          "D. 96"
+        ],
+        "answer": "C",
+        "solution": "【分析】把 3 次数学成绩看作 3 个抽屉, 290 分看做 290 个元素, 利用抽屉原来最差情况: 要使每次成绩最小, 只要使每个抽屉的元素尽量平均, 据此解答。\n\n【详解】  $290 \\div 3 = 96$  （分）…2（分） $96 + 1 = 97$  (分) 小华的 3 次数学考试成绩总分为 290 分 (成绩都为整数),3次成绩中至少有一次成绩不低于97分。故答案为: C",
+        "knowledge_points": ["最不利原则"],
+        "source": "24-25 六年级下·全国·单元测试",
+        "merged_id": "unit5_002",
+        "original_id": "unit5_choice_002"
+      },
+      {
+        "id": "unit5_choice_003",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "(24-25 六年级下·全国·单元测试) 某幼儿园大班共有 30名小朋友，老师最少要准备（）件玩具，才能保证有一个小朋友手中至少有 3 件玩具。",
+        "options": [
+          "A. 60",
+          "B. 61",
+          "C. 91",
+          "D. 101"
+        ],
+        "answer": "B",
+        "solution": "【分析】根据鸽巢原理考虑最不利情况：假设每个小朋友都刚好有 2 件玩具, 一共有  $30 \\times 2 = 60$  件玩具, 若此时再增加 1 件玩具, 无论怎样分配, 一定有一个小朋友手中有 3 件玩具，据此解答。\n\n【详解】  $30 \\times 2 + 1\\\\= 6 0 + 1 \\\\ = 6 1 \\text {(件)} $\n\n所以，老师最少要准备 61 件玩具，才能保证有一个小朋友手中至少有 3 件玩具。故答案为: B",
+        "knowledge_points": ["最不利原则"],
+        "source": "24-25 六年级下·全国·单元测试",
+        "merged_id": "unit5_003",
+        "original_id": "unit5_choice_003"
+      },
+      {
+        "id": "unit5_choice_004",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "（24-25 六年级下·全国·单元测试）一副扑克牌有 54 张，去掉大小王后，最少要抽取（）张牌，才能保证其中至少有 2 张牌的点数相同。",
+        "options": [
+          "A. 9",
+          "B. 13",
+          "C. 14",
+          "D. 27"
+        ],
+        "answer": "C",
+        "solution": "【分析】本题考查了抽屉原理，一副扑克牌有 54 张，去掉大小王后，还剩下  $54 - 2 = 52$  （张）扑克牌；扑克牌有红桃、黑桃、方块、梅花四种花色, 每种花色有  $52 \\div 4 = 13$  (张),相当于有 13 个抽屉, 要保证其中至少有 2 张牌的点数相同,则每个抽屉放入一张扑克牌, 最少再放入一张扑克牌即可。\n\n【详解】  $(54 - 2) \\div 4\\\\= 5 2 \\div 4 \\\\ = 1 3 (\\text {张}) $\n\n$13 + 1 = 14$  (张)",
+        "knowledge_points": ["鸽巢问题初步"],
+        "source": "24-25 六年级下·全国·单元测试",
+        "merged_id": "unit5_004",
+        "original_id": "unit5_choice_004"
+      },
+      {
+        "id": "unit5_choice_005",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "（24-25 六年级下·全国·单元测试）李明和好朋友玩'剪刀、石头、布'的游戏，他至少要出（）次，才能保证有 4 次的手势是相同的。",
+        "options": [
+          "A. 5",
+          "B. 7",
+          "C. 10",
+          "D. 13"
+        ],
+        "answer": "C",
+        "solution": "【分析】根据题意，李明和好朋友玩'剪刀、石头、布'的游戏，有3种不同的手势。运气最差的情况，李明每种手势各出了3次，即出了  $3 \\times 3 = 9$  次，此时再出任意一种手势，即第10次都会出现有4次手势是相同。\n\n【详解】  $3 \\times 3 + 1\\\\= 9 + 1 \\\\ = 1 0 \\text {(次)}$\n\n他至少要出 10 次, 才能保证有 4 次的手势是相同的。故答案为: C",
+        "knowledge_points": ["最不利原则"],
+        "source": "24-25 六年级下·全国·单元测试",
+        "merged_id": "unit5_005",
+        "original_id": "unit5_choice_005"
+      },
+      {
+        "id": "unit5_choice_006",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "(24-25 六年级下·重庆九龙坡·开学考试) 在一个不透明的纸箱里有除颜色不同, 其他全部相同的小球 15 个, 其中蓝球 4 个, 红球 5 个, 白球 6 个, 要想确保摸出 2 个同色的小球, 至少要摸 ( )。",
+        "options": [
+          "A. 2 次",
+          "B. 3 次",
+          "C. 4 次",
+          "D. 6 次"
+        ],
+        "answer": "C",
+        "solution": "【分析】纸箱里有同样大小的篮球 4 个, 红球 5 个, 白球 6 个, 最坏的情况是, 红球、篮球、白球各摸出一个, 此时只要再任意摸出一个球, 摸出的球一定有 2 个同色的, 即至少要摸出  $3 + 1 = 4$  个。此题考查了抽屉原理在实际问题中的灵活应用。\n\n【详解】  $3 + 1 = 4$  （个）要想确保摸出 2 个同色的小球, 至少要摸 4 次。故答案为: C",
+        "knowledge_points": ["最不利原则"],
+        "source": "24-25 六年级下·重庆九龙坡·开学考试",
+        "merged_id": "unit5_006",
+        "original_id": "unit5_choice_006"
+      },
+      {
+        "id": "unit5_choice_007",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "(24-25 六年级上·吉林长春·期末) 一个袋子里有红、白、蓝三种颜色的球各 5 个, 至少摸出 ( ) 个球, 可以保证有两个球颜色相同。",
+        "options": [
+          "A. 4",
+          "B. 5",
+          "C. 6",
+          "D. 10"
+        ],
+        "answer": "A",
+        "solution": "【分析】由题意可知, 有红、白、蓝三种颜色的球, 要保证至少有 2 个颜色相同, 最坏的情况是每种颜色各摸出 1 ,即摸出 3 个, 此时只要再任摸一个, 即摸出  $3 + 1 = 4$  个就能保证至少有 2 个球颜色相同, 据此解答。\n\n【详解】  $3 + 1 = 4$  （个）一个袋子里有红、白、蓝三种颜色的球各 5 个, 至少摸出 4 个球, 可以保证有两个球颜色相同。故答案为: A",
+        "knowledge_points": ["最不利原则"],
+        "source": "24-25 六年级上·吉林长春·期末",
+        "merged_id": "unit5_007",
+        "original_id": "unit5_choice_007"
+      },
+      {
+        "id": "unit5_choice_008",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "(23-24 六年级下·四川成都·期末) 一个盒子里有 5 个红球, 3 个白球和 4 个蓝球, 至少需要摸 ( ) 个球才能保证有 2 个不同颜色的球。",
+        "options": [
+          "A. 4",
+          "B. 5",
+          "C. 6",
+          "D. 8"
+        ],
+        "answer": "C",
+        "solution": "【分析】抽屉原理的题目，利用最不利原则，最倒霉情况是一种颜色球都拿完，即 5 个红球都拿完，即至少需要摸 5 +1 = 6（个）球才能保证有 2 个不同颜色的球。据此解答。\n\n【详解】  $5 + 1 = 6$  （个）所以至少需要摸 6 个球才能保证有 2 个不同颜色的球。故答案为: C",
+        "knowledge_points": ["最不利原则"],
+        "source": "23-24 六年级下·四川成都·期末",
+        "merged_id": "unit5_008",
+        "original_id": "unit5_choice_008"
+      },
+      {
+        "id": "unit5_choice_009",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "(23-24 六年级下·四川绵阳·期末) 一个盒子里有黄、白乒乓球各 5 个, 要想使取出的乒乓球保证至少有 2 个白色的,则至少取出的个数是 ( )。",
+        "options": [
+          "A. 3",
+          "B. 6",
+          "C. 7",
+          "D. 10"
+        ],
+        "answer": "C",
+        "solution": "【分析】考虑最倒霉的情况，取出的前 5 个都是黄乒乓球，再取两个，一定是 2 个白乒乓球，据此解答。\n\n【详解】  $5 + 2 = 7$  （个）一个盒子里有黄、白乒乓球各 5 个, 要想使取出的乒乓球保证至少有 2 个白色的, 则至少取出的个数是 7 。故答案为: C",
+        "knowledge_points": ["最不利原则"],
+        "source": "23-24 六年级下·四川绵阳·期末",
+        "merged_id": "unit5_009",
+        "original_id": "unit5_choice_009"
+      },
+      {
+        "id": "unit5_choice_010",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "（2024·浙江湖州·小升初真题）盒子里装有 1 个红球，3 个黄球和 4 个白球（这些球除颜色外完全相同）。至少摸出（）个球，才能保证取到两个颜色相同的球。",
+        "options": [
+          "A. 3",
+          "B. 4",
+          "C. 5",
+          "D. 6"
+        ],
+        "answer": "B",
+        "solution": "【分析】把 3 种不同颜色看作 3 个抽屉, 把不同颜色的球看作元素, 从最不利情况考虑, 先从每个抽屉摸出 1 个球, 一共是 3 个球, 再摸出 1 个球, 不论是什么颜色, 总有一个抽屉里的球和它同色, 据此解答。\n\n【详解】  $3 + 1 = 4$  （个）因此至少摸出4个球，才能保证取到两个颜色相同的球。故答案为: B",
+        "knowledge_points": ["最不利原则"],
+        "source": "2024·浙江湖州·小升初真题",
+        "merged_id": "unit5_010",
+        "original_id": "unit5_choice_010"
+      },
+      {
+        "id": "unit5_choice_011",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "（23-24六年级下·湖北鄂州·期末）希望小学书法兴趣小组的30名同学中，年龄最大的12岁，最小的6岁，最少从中挑选（）名学生，就一定能找到两个年龄相同的学生。",
+        "options": [
+          "A. 7",
+          "B. 8",
+          "C. 12",
+          "D. 13"
+        ],
+        "answer": "B",
+        "solution": "【分析】根据最不利原理，先挑选出7名同学，他们的年龄分别是 6、7、8、9、10、11、12 岁, 则再挑出一名同学一定能找到两个年龄相同的学生。\n\n【详解】  $7 + 1 = 8$  （名）则最少从中挑选 8 名学生, 就一定能找到两个年龄相同的学生。故答案为: B",
+        "knowledge_points": ["最不利原则"],
+        "source": "23-24六年级下·湖北鄂州·期末",
+        "merged_id": "unit5_011",
+        "original_id": "unit5_choice_011"
+      },
+      {
+        "id": "unit5_choice_012",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "(23-24 六年级下·湖南湘西·期末) 某班 52 名同学按学号依次轮流当值日生班长, 本学期共 22 周, 每人至少当 ( ) 次。",
+        "options": [
+          "A. 4",
+          "B. 3",
+          "C. 2",
+          "D. 1"
+        ],
+        "answer": "C",
+        "solution": "【分析】在本题中, 一周有 5 天上学, 因此本学期的总天数是  $(22 \\times 5)$ , 学生数是 52, 用除法计算并对商和余数进行分析即可得解。\n\n【详解】  $22 \\times 5 \\div 52\\\\= 1 1 0 \\div 5 2\\\\=2（次）……6（天）$\n\n每人当 2 次, 还有 6 天, 即前 6 个学号的学生每人当 3 次,其余同学每人当 2 次, 最少当 2 次。故答案为: C",
+        "knowledge_points": ["最不利原则"],
+        "source": "23-24 六年级下·湖南湘西·期末",
+        "merged_id": "unit5_012",
+        "original_id": "unit5_choice_012"
+      },
+      {
+        "id": "unit5_choice_013",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "（23-24六年级下·湖北武汉·期末）六（1）班有36名同学，按学号依次轮流当值日班长，这学期有22周，每人至少轮到（）次。",
+        "options": [
+          "A. 2",
+          "B. 3",
+          "C. 4",
+          "D. 5"
+        ],
+        "answer": "B",
+        "solution": "【分析】在本题中，一周有 5 天上学，因此本学期的总天数是  $(22 \\times 5)$ , 学生数是 36, 用除法计算并对商和余数进行分析即可得解。\n\n【详解】  $22 \\times 5 \\div 36\\\\= 1 1 0 \\div 3 6\\\\=3（次）……2（天）$\n\n每人轮 3 次, 还余 2 天。则前两号同学轮 4 次, 后面 34 号同学轮 3 次。因此，每人至少轮到3次。故答案为: B",
+        "knowledge_points": ["鸽巢问题初步"],
+        "source": "23-24六年级下·湖北武汉·期末",
+        "merged_id": "unit5_013",
+        "original_id": "unit5_choice_013"
+      },
+      {
+        "id": "unit5_choice_014",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "(23-24 六年级下·河南许昌·期中) 25 个鸡蛋最多放进( ) 个碗中才能保证有一个碗中至少放进 7 个鸡蛋。",
+        "options": [
+          "A. 7",
+          "B. 6",
+          "C. 5",
+          "D. 4"
+        ],
+        "answer": "D",
+        "solution": "【分析】要使碗的数量最多，就要使每个碗里的鸡蛋的个数最少。数最少, 可以使其中一个碗放 7 个鸡蛋, 剩下的每个都放 7  $-1 = 6$  个鸡蛋, 据此用除法解法。\n\n【详解】  $(25 - 1) \\div (7 - 1)\\\\= 2 4 \\div 6 \\\\ = 4 (\\text {个}) $\n\n25 个鸡蛋最多放进 4 个碗中才能保证有一个碗中至少放进 7 个鸡蛋。故答案为: D",
+        "knowledge_points": ["最不利原则"],
+        "source": "23-24 六年级下·河南许昌·期中",
+        "merged_id": "unit5_014",
+        "original_id": "unit5_choice_014"
+      },
+      {
+        "id": "unit5_choice_015",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "（2024六年级下·四川·专题练习）六（1）班有37名同学，至少有（）人出生的月份相同。",
+        "options": [
+          "A. 2",
+          "B. 4",
+          "C. 6",
+          "D. 8"
+        ],
+        "answer": "B",
+        "solution": "【分析】把一年12个月看作12个抽屉，把37人看作37个元素, 那么每个抽屉需要放  $37 \\div 12 = 3$  (个) 元素, 还剩余 1 个, 因此, 至少有 4 名同学同一个月出生, 据此解答。\n\n【详解】  $37 \\div 12 = 3$  （个）……1（个）$3 + 1 = 4$  （个）至少有4人出生的月份相同。故答案为: B",
+        "knowledge_points": ["鸽巢问题初步"],
+        "source": "2024六年级下·四川·专题练习",
+        "merged_id": "unit5_015",
+        "original_id": "unit5_choice_015"
+      },
+      {
+        "id": "unit5_choice_016",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "(22-23 六年级下·云南昆明·期末) 把红、橙、黄、绿、青、蓝、紫七种颜色的珠子（珠子的大小、形状完全相同）各 10 颗放到一个袋子里。至少取出几颗才能保证取到两颗颜色相同的珠子？（）",
+        "options": [
+          "A. 7 颗",
+          "B. 8 颗",
+          "C. 10颗",
+          "D. 11 颗"
+        ],
+        "answer": "B",
+        "solution": "【分析】把红、橙、黄、绿、青、蓝、紫七种颜色看做 7 个抽屉, 利用抽屉原理, 考虑最差情况, 摸出 7 个球, 分别是红、橙、黄、绿、青、蓝、紫不同的颜色, 再任意摸出 1 个球即可。\n\n【详解】  $7 + 1 = 8$所以，至少取出 8 颗才能保证取到两颗颜色相同的珠子。故答案为：B",
+        "knowledge_points": ["最不利原则"],
+        "source": "22-23 六年级下·云南昆明·期末",
+        "merged_id": "unit5_016",
+        "original_id": "unit5_choice_016"
+      },
+      {
+        "id": "unit5_choice_017",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "(21-22 六年级上·湖南郴州·期末) 六 (一) 班有 50 人,在一次数学测试中, 全班同学都及格了 (60 分及格, 100 分满分, 都是整数分), 至少一定有 ( ) 个人的分数是相同的。",
+        "options": [
+          "A. 9",
+          "B. 10",
+          "C. 2"
+        ],
+        "answer": "C",
+        "solution": "【分析】抽屉原理（鸽巢原理）：把  $\\mathrm{m}$  个物体放进  $\\mathrm{n}$  个抽屉里  $(\\mathrm{m} > \\mathrm{n} > 1)$ ,  $\\mathrm{m} \\div \\mathrm{n} = \\mathrm{a} \\dots \\dots \\mathrm{b}$ , 不管怎么放总有一个抽屉至少放进  $(\\mathrm{a} + 1)$  个物体。由题意可知, 一共有  $100 - 60 + 1 = 41$  (个) 分数, 即抽屉数是 41 个; 六 (一) 班有 50 人, 即物体数是 50 人; 用  $50 \\div 41$  求出商几余几, 再用商数 +1 求出至少数。解决抽屉原理问题, 要分清 '要放的物体数和抽屉数'。\n\n【详解】  $100 - 60 + 1\\\\= 4 0 + 1 \\\\ = 4 1 \\text {(个)} $\n\n$5 0 \\div 4 1 = 1 \\text {(人)} \\dots \\dots 9 \\text {(人)} \\\\ 1 + 1 = 2 \\text {(人)} $\n\n所以至少一定有 2 个人的分数是相同的。故答案为: C",
+        "knowledge_points": ["鸽巢问题进阶"],
+        "source": "21-22 六年级上·湖南郴州·期末",
+        "merged_id": "unit5_017",
+        "original_id": "unit5_choice_017"
+      },
+      {
+        "id": "unit5_choice_018",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "（21-22六年级下·河北秦皇岛·期末）把红、黄、绿三种颜色的鞋带各一双混在一起，如果闭上眼睛拿，最少拿出几根才能保证一定有一双同色的鞋带？（）",
+        "options": [
+          "A. 2 根",
+          "B. 3 根",
+          "C. 4 根",
+          "D. 5 根"
+        ],
+        "answer": "C",
+        "solution": "【分析】根据抽屉原理，如果取出的头 3 根分别是 3 种不同的颜色，那么第 4 根取出后，能得到一双同色的鞋带。据此解题。本题考查了抽屉原理, 关键是要从最差情况去考虑。\n\n【详解】  $3 + 1 = 4$  （根）所以，如果闭上眼睛拿，最少拿出 4 根才能保证一定有一双同色的鞋带。故答案为: C",
+        "knowledge_points": ["鸽巢问题进阶"],
+        "source": "21-22六年级下·河北秦皇岛·期末",
+        "merged_id": "unit5_018",
+        "original_id": "unit5_choice_018"
+      },
+      {
+        "id": "unit5_choice_019",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "（22-23六年级下·黑龙江哈尔滨·期末）教室内有30名学生，至少有（）名学生是同一个月出生的。",
+        "options": [
+          "A. 2",
+          "B. 3",
+          "C. 4"
+        ],
+        "answer": "B",
+        "solution": "【分析】一年有 12 个月, 那么可以看作是 12 个抽屉, 30 名学生看做 30 个元素, 把 30 名学生平均分配在 12 个抽屉中:  $30 \\div 12 = 2$  (名) ……6 (名), 那么每个抽屉都有 2 名学生, 那么剩下的 6 名, 无论放到哪个抽屉都会出现 3 名学生在同一个抽屉里。此题属于典型的抽屉原理习题, 解答此类题的关键是找出把谁看作 '抽屉个数', 把谁看作 '物体个数',然后根据抽屉原理解答即可。\n\n【详解】  $30 \\div 12 = 2$  （名）……6（名）  $2 + 1 = 3 \\text {(名)}$  即至少有 3 名学生是同一个月出生的。故答案为: B",
+        "knowledge_points": ["鸽巢问题初步"],
+        "source": "22-23六年级下·黑龙江哈尔滨·期末",
+        "merged_id": "unit5_019",
+        "original_id": "unit5_choice_019"
+      },
+      {
+        "id": "unit5_choice_020",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "(22-23 六年级下·河南信阳·期末) 20 本书放在 6 层书架上, 总有一层至少放 ( ) 本书。",
+        "options": [
+          "A. 4",
+          "B. 3",
+          "C. 5",
+          "D. 2"
+        ],
+        "answer": "A",
+        "solution": "【分析】把 20 本书看作被分放物体, 6 层书架看作 6 个抽屉, 被分放物体的数量  $\\div$  抽屉的数量 = 平均每个抽屉分放物体的数量……剩下物体的数量, 一个抽屉里至少分放物体的数量 = 平均每个抽屉分放物体的数量 +1 , 据此解答。本题主要考查抽屉原理的应用, 明确被分放物体的数量和抽屉的数量是解答题目的关键。\n\n【详解】  $20 \\div 6 = 3$  （本）……2（本） $3 + 1 = 4$  （本）所以，总有一层至少放 4 本书。故答案为：A",
+        "knowledge_points": ["鸽巢问题初步"],
+        "source": "22-23 六年级下·河南信阳·期末",
+        "merged_id": "unit5_020",
+        "original_id": "unit5_choice_020"
+      },
+      {
+        "id": "unit5_choice_021",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "(20-21 六年级下·天津南开·期末) 某班至少有学生 ( ) 人, 总有不少于 4 人出生在同一个月。",
+        "options": [
+          "A. 49",
+          "B. 37",
+          "C. 48",
+          "D. 无法确定"
+        ],
+        "answer": "B",
+        "solution": "【分析】一年中共有 12 个月, 将这 12 个月当作 12 个抽屉,根据抽屉原理可知, 每个抽屉里放 3 个元素, 共需要  $3 \\times 12 = 36$  个元素, 加上 1 个元素即可。本题考查了抽屉原理的灵活利用, 要从平均分的情况入手解答。\n\n【详解】  $3 \\times 12 + 1 = 36 + 1=37 (人)$则某班至少有学生 37 人, 总有不少于 4 人出生在同一个月。故答案为: B",
+        "knowledge_points": ["鸽巢问题进阶"],
+        "source": "20-21 六年级下·天津南开·期末",
+        "merged_id": "unit5_021",
+        "original_id": "unit5_choice_021"
+      },
+      {
+        "id": "unit5_choice_022",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "（14-15六年级下·河南郑州·阶段练习）一个盒子里装有白乒乓球5个和黄乒乓球若干个，则至少应取出（）个。",
+        "options": [
+          "A. 5",
+          "B. 6",
+          "C. 7"
+        ],
+        "answer": "C",
+        "solution": "【分析】考虑最倒霉的情况，取出的前 5 个都是白乒乓球, 再取两个, 一定是 2 个黄乒乓球, 据此分析。关键是构造物体和抽屉, 也就是找到代表物体和抽屉的量, 然后依据抽屉原则进行分析。\n\n【详解】  $5 + 2 = 7$  （个）至少应取出 7 个。故答案为: C",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "14-15六年级下·河南郑州·阶段练习",
+        "merged_id": "unit5_022",
+        "original_id": "unit5_choice_022"
+      },
+      {
+        "id": "unit5_choice_023",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "(22-23 六年级下·全国·单元测试) 13 个苹果放在四个篮子里, 总有一个篮子里至少有 ( ) 个苹果。",
+        "options": [
+          "A. 1",
+          "B. 2",
+          "C. 4"
+        ],
+        "answer": "C",
+        "solution": "【分析】根据抽屉原理, 把 4 个篮子看作 4 个抽屉, 要使每个篮子里的苹果尽量少, 要尽量平均分, 即  $13 \\div 4 = 3$  …1余下的一个苹果需要放在随机的一个篮子中, 所以总有一个篮子里至少有 4 个苹果, 由此即可解决问题。此题主要考查抽屉原理及其应用。\n\n【详解】  $13 \\div 4 = 3$  （个）……1（个）$3 + 1 = 4$  （个）所以至少有 4 个苹果放进一个篮子里。故答案为: C",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "22-23 六年级下·全国·单元测试",
+        "merged_id": "unit5_023",
+        "original_id": "unit5_choice_023"
+      },
+      {
+        "id": "unit5_choice_024",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "(22-23 六年级下·全国·单元测试) 书架分上、中、下三层, 婷婷把新买的 10 本书放入书架, 放书最多的一层至少要放入 ( ) 本书。",
+        "options": [
+          "A. 2",
+          "B. 3",
+          "C. 4"
+        ],
+        "answer": "C",
+        "solution": "【分析】把上、中、下三层看作3个抽屉，把新买的10本书看作10个元素，那么每个抽屉需要放  $10 \\div 3 = 3$  （本）……1（本），所以每个抽屉需要放3本，剩下的1本不论怎么放，总有一个抽屉里至少有：  $3 + 1 = 4$  （本），所以，放书最多的一层至少要放入 4 本书; 据此解答。此题属于典型的抽屉原理习题, 根据抽屉原理进行解答即可。\n\n【详解】  $10 \\div 3 = 3$  （本）…1（本） $3 + 1 = 4$  (本)故答案为: C",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "22-23 六年级下·全国·单元测试",
+        "merged_id": "unit5_024",
+        "original_id": "unit5_choice_024"
+      },
+      {
+        "id": "unit5_choice_025",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "（22-23六年级下·全国·单元测试）20个孩子参加6个兴趣小组，至少有一个兴趣小组的人不少于（）人。",
+        "options": [
+          "A. 4",
+          "B. 3",
+          "C. 5"
+        ],
+        "answer": "A",
+        "solution": "【分析】20 个学生参加 6 个兴趣小组, 平均每个小组先放 3 人, 还剩下 2 人, 这 2 人无论放在哪个小组, 至少有一个兴趣小组的人不少于  $(3 + 1)$  人。本题考查鸽巢问题（抽屉问题），至少数=物体数除以抽屉数的商+1（有余数的情况下）。\n\n【详解】  $20 \\div 6 = 3$  （人）……2（人） $3 + 1 = 4$  （人）至少有一个兴趣小组的人不少于4人。故答案为：A",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "22-23六年级下·全国·单元测试",
+        "merged_id": "unit5_025",
+        "original_id": "unit5_choice_025"
+      },
+      {
+        "id": "unit5_choice_026",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "(22-23 六年级下·全国·单元测试) 把 25 枝月季花插到 4 个花瓶中, 总有一个花瓶至少插 ( ) 枝月季花。",
+        "options": [
+          "A. 8",
+          "B. 7",
+          "C. 6"
+        ],
+        "answer": "B",
+        "solution": "【分析】把 4 个花瓶看作 4 个抽屉, 25 枝月季花看作 25 个元素, 把 25 枝花插到 4 个花瓶中, 利用抽屉原理最差情况: 要使花瓶里花的朵数最少, 只要使每个抽屉的元素数尽量平均，即每个花瓶中插 6 枝还剩 1 枝，所以总有一个花瓶插  $6 + 1 = 7$  （枝）。在此类抽屉问题中, 至少数 = 物体数除以抽屉数的商 +1。\n\n【详解】  $25 \\div 4 = 6$  （枝）……1（枝） $6 + 1 = 7$  (枝)故答案为: B。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "22-23 六年级下·全国·单元测试",
+        "merged_id": "unit5_026",
+        "original_id": "unit5_choice_026"
+      },
+      {
+        "id": "unit5_choice_027",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "（22-23六年级下·全国·单元测试）把红、黄、蓝三种颜色的小球各12个放到一个盒子里，要保证一次摸到两个同色的小球，一次至少要摸出（）个小球。",
+        "options": [
+          "A. 13",
+          "B. 4",
+          "C. 5"
+        ],
+        "answer": "B",
+        "solution": "【分析】由于袋子里共有红、黄、蓝三种颜色的球各 12 个,如果一次取 3 个, 最差情况为红、黄、蓝三种颜色各一个,所以只要再多取一个球, 就能保证取到两个颜色相同的球。据此解答。解决抽屉原理问题的关键是根据最差原理对问题进行分析。\n\n【详解】  $3 + 1 = 4$  （个）故答案为: B",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "22-23六年级下·全国·单元测试",
+        "merged_id": "unit5_027",
+        "original_id": "unit5_choice_027"
+      },
+      {
+        "id": "unit5_choice_028",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "（22-23六年级下·全国·单元测试）会议室里坐着1至6年级的班干部各5人，一位不熟悉这些学生的老师要想喊出的人一定有2名同年级的学生，最少要喊出（）人。",
+        "options": [
+          "A. 5",
+          "B. 6",
+          "C. 7"
+        ],
+        "answer": "C",
+        "solution": "【分析】由于会议室里共有 1 至 6 年级，共六个年级，如果一次喊 6 人, 运气最差的情况为 1 至 6 年级各出来一个人,所以只要再多喊一个人, 就能保证喊出的人一定有 2 名同年级的学生, 据此解答。本题考查鸽巢问题（抽屉问题），采用最不利原则（运气最差原则）来解题。\n\n【详解】  $6 + 1 = 7$  （人）即最少要喊出 7 人。故答案为: C",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "22-23六年级下·全国·单元测试",
+        "merged_id": "unit5_028",
+        "original_id": "unit5_choice_028"
+      },
+      {
+        "id": "unit5_choice_029",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "(22-23 六年级下·全国·单元测试) 红、黄、蓝三种糖果各 10 个混合装在袋子里, 一次至少拿 ( ) 个, 才能保证一定有 2 个是同颜色的糖果。",
+        "options": [
+          "A. 2",
+          "B. 3",
+          "C. 4"
+        ],
+        "answer": "C",
+        "solution": "【分析】把三种颜色看作3个抽屉，把三种糖果各10个看作元素，从最不利情况考虑，每个抽屉先放1个，共需要3个，再取出1个不论是什么颜色，总有一个抽屉里的糖果和它同色，据此解答即可。此题考查了利用抽屉原理解决实际问题的灵活应用，关键是从最差情况考虑。\n\n【详解】  $3 + 1 = 4$  （个）故答案为: C",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "22-23 六年级下·全国·单元测试",
+        "merged_id": "unit5_029",
+        "original_id": "unit5_choice_029"
+      },
+      {
+        "id": "unit5_choice_030",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "（22-23六年级下·全国·单元测试）箱子中有质地、型号完全相同的红、黄、白三种颜色的袜子各8只。至少拿出（）只，可以保证凑成两双颜色不相同的袜子。",
+        "options": [
+          "A. 5",
+          "B. 8",
+          "C. 11"
+        ],
+        "answer": "C",
+        "solution": "【分析】从最不利的情况考虑，如果取出的头 8 只袜子是同一种颜色，再取 2 只是剩下的两种颜色的各一只，然后再取 1 只，可以保证凑成两双颜色不相同的袜子，据此解答即可。此题考查了利用抽屉原理解决实际问题的灵活应用，关键是从最差情况考虑。\n\n【详解】  $8 + 2 + 1 = 11$  （只）至少拿出 11 只, 可以保证凑成两双颜色不相同的袜子。故答案为: C",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "22-23六年级下·全国·单元测试",
+        "merged_id": "unit5_030",
+        "original_id": "unit5_choice_030"
+      },
+      {
+        "id": "unit5_choice_031",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "(22-23六年级下·全国·单元测试) 把 7 支铅笔放进三个笔盒里, 总有一个笔盒至少放进 ( ) 支笔。",
+        "options": [
+          "A. 2",
+          "B. 3",
+          "C. 4"
+        ],
+        "answer": "B",
+        "solution": "【分析】把7枝铅笔放进3个笔盒中， $7 \\div 3 = 2$ （支）……1（支），即平均每个笔盒放2支，还余1支，根据抽屉原理可知，总有一个笔盒里至少放  $2 + 1 = 3$  支。在此类抽屉问题中，至少数 = 物体数除以抽屉数的商 +1（有余数的情况下）。\n\n【详解】  $7 \\div 3 = 2$  （支）…1（支）$2 + 1 = 3$  (支) 所以总有一个笔盒至少放进 3 支笔。故答案为: B",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "22-23六年级下·全国·单元测试",
+        "merged_id": "unit5_031",
+        "original_id": "unit5_choice_031"
+      },
+      {
+        "id": "unit5_choice_032",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "（22-23六年级下·全国·单元测试）六年级甲班59名同学中至少有（）名同学是同一个月份出生的。",
+        "options": [
+          "A. 4",
+          "B. 5",
+          "C. 6"
+        ],
+        "answer": "B",
+        "solution": "【分析】把59名同学看作被分放物体，一年中的12个月份看作抽屉数，被分放物体的数量÷抽屉的数量 = 平均每个抽屉分放物体的数量……剩下物体的数量，一个抽屉里至少分放物体的数量 = 平均每个抽屉分放物体的数量 + 1，据此解答。本题主要考查利用抽屉原理解决实际问题, 找出被分放物体数和抽屉数是解答题目的关键。\n\n【详解】一年一共有 12 个月。$5 9 \\div 1 2 = 4 \\dots 1 1$ $4 + 1 = 5$  (名)所以，至少有 5 名同学是同一个月份出生的。故答案为: B",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "22-23六年级下·全国·单元测试",
+        "merged_id": "unit5_032",
+        "original_id": "unit5_choice_032"
+      },
+      {
+        "id": "unit5_choice_033",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "(22-23 六年级下·全国·单元测试) 盒子里有 2 个黑球, 3 个黄球, 5 个绿球, 任意拿出 6 个, 一定有一个 ( )。",
+        "options": [
+          "A. 黑球",
+          "B. 黄球",
+          "C. 绿球"
+        ],
+        "answer": "C",
+        "solution": "【分析】根据抽屉原理进行分析, 考虑最倒霉的情况, 拿出的前 5 个球是 2 个黑球和 3 个黄球, 再拿一个, 一定是绿球, 据此分析。关键是构造物体和抽屉, 也就是找到代表物体和抽屉的量, 然后依据抽屉原则进行计算。\n\n【详解】  $2 + 3 + 1 = 6$  （个）至少拿出 6 个球, 可以保证拿出 1 个绿球, 反过来, 任意拿出 6 个, 一定有一个绿球。故答案为: C。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "22-23 六年级下·全国·单元测试",
+        "merged_id": "unit5_033",
+        "original_id": "unit5_choice_033"
+      }, {
+        "id": "unit5_choice_067",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "(20-21 六年级下·浙江台州·期末) 实验小学六 (1) 班共有 45 名同学, 他们中至少有( )名同学的生日在同一月。",
+        "options": [],
+        "answer": "4",
+        "solution": "【分析】把 45 名同学看作被分放物体, 12 个月看作 12 个抽屉, 被分放物体的数量  $\\div$  抽屉的数量 = 平均每个抽屉分放物体的数量……剩下物体的数量, 一个抽屉里至少分放物体的数量 = 平均每个抽屉分放物体的数量 + 1 , 据此解答。本题主要考查抽屉原理的应用, 准确找出被分放物体数和抽屉数是解答题目的关键。\n\n【详解】  $45 \\div 12 = 3 \\dots 9$   $3 + 1 = 4$  (名)   所以，他们中至少有 4 名同学的生日在同一月。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21 六年级下·浙江台州·期末",
+        "merged_id": "unit5_067",
+        "original_id": "unit5_choice_067"
+      }, {
+        "id": "unit5_choice_034",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "(22-23 六年级下·全国·单元测试) 运动会上, 在 5 分钟投篮比赛中, 六年 (1) 班的 10 名同学共投中了 82 个,总有一名队员至少投中 ( ) 个球。",
+        "options": [
+          "A. 7",
+          "B. 8",
+          "C. 9"
+        ],
+        "answer": "C",
+        "solution": "【分析】将 10 名同学看作 10 个抽屉, 用 82 个球除以 10 ,求出商和余数, 将商加上 1, 即可求出总有一名队员至少投中几个球。本题考查了抽屉原理, 能根据题意正确列式是解题关键。\n\n【详解】  $82 \\div 10 = 8$  （个）……2（个） $8 + 1 = 9 \\text {(个)}$  所以，总有一名队员至少投中 9 个球。 故答案为: C",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "22-23 六年级下·全国·单元测试",
+        "merged_id": "unit5_034",
+        "original_id": "unit5_choice_034"
+      },
+      {
+        "id": "unit5_choice_035",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "（22-23六年级下·全国·单元测试）一个口袋里装有红、黄、蓝3种不同颜色的小球各10个，要摸出的球一定有2个同色的，最少要摸（）个。",
+        "options": [
+          "A. 10",
+          "B. 11",
+          "C. 4",
+          "D. 以上都不对"
+        ],
+        "answer": "C",
+        "solution": "【分析】因总共有红、黄、蓝三种颜色，如果运气最糟糕，就是摸出的 3 个是不同颜色的，这时，只要再摸出一个，不论是什么颜色的，就一定有两个球是同色的。据此解答。根据抽屉原理中的最差情况进行分析是完成本题的关键。\n\n【详解】  $3 + 1 = 4$  （个）即要摸出的球一定有 2 个同色的, 最少要摸 4 个。故答案为: C",
+        "knowledge_points": ["鸽巢问题进阶"],
+        "source": "22-23六年级下·全国·单元测试",
+        "merged_id": "unit5_035",
+        "original_id": "unit5_choice_035"
+      },
+      {
+        "id": "unit5_choice_036",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "(22-23六年级下·全国·单元测试) 会议室里坐着 1 至 6 年级的班干部各 5 人, 一位不熟悉这些学生的老师要想喊出的人一定有 2 名同年级的学生, 最少要喊出 ( ) 人。",
+        "options": [
+          "A. 5",
+          "B. 6",
+          "C. 7",
+          "D. 以上都不对"
+        ],
+        "answer": "C",
+        "solution": "【分析】由于会议室里共有 1 至 6 年级共六个年级的人数,如果一次喊 6 人, 最差情况为 1 至 6 年级各一个人, 所以只要再多喊一个人, 就能保证喊出的人一定有 2 名同年级的学生。据此解答。解决抽屉原理问题的关键是根据最差原理对问题进行分析。\n\n【详解】  $6 + 1 = 7$  （人）即最少要喊出 7 人。故答案为: C",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "22-23六年级下·全国·单元测试",
+        "merged_id": "unit5_036",
+        "original_id": "unit5_choice_036"
+      },
+      {
+        "id": "unit5_choice_037",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "(22-23 六年级下·全国·单元测试) 把 10 本书放进 3 个抽屉, 不管怎么放, 总有一个抽屉至少放进 ( ) 本书。",
+        "options": [
+          "A. 4",
+          "B. 3",
+          "C. 5",
+          "D. 以上都不对"
+        ],
+        "answer": "A",
+        "solution": "【分析】把 10 本书放进 3 个抽屉, 平均每个抽屉先放 3 本,还剩下 1 本, 不管怎么放, 总有一个抽屉至少放进 4 本书。本题考查鸽巢问题，采用最不利原则（运气最差原则）来解题。\n\n【详解】  $10 \\div 3 = 3$  （本）……1（本） $3 + 1 = 4$  (本) 把 10 本书放进 3 个抽屉, 不管怎么放, 总有一个抽屉至少放进 4 本书。故答案为: A",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "22-23 六年级下·全国·单元测试",
+        "merged_id": "unit5_037",
+        "original_id": "unit5_choice_037"
+      },
+      {
+        "id": "unit5_choice_038",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "(22-23 六年级下·全国·单元测试) 有白色手套和黑色手套各5只（不分左右手），如果蒙上眼睛，至少拿出（）只，才能使拿出的手套中一定有一双是同色的。",
+        "options": [
+          "A. 4",
+          "B. 5",
+          "C. 3",
+          "D. 以上都不对"
+        ],
+        "answer": "C",
+        "solution": "【分析】因只有两种颜色，所以考虑到最差情况，就是拿出的 2 只是不同颜色的，这时，只要再拿出一只，不论是什么颜色的，就一定有一双是同色的。据此解答。根据抽屉原理中的最差情况进行分析是完成本题的关键。\n\n【详解】  $2 + 1 = 3$  （只）即至少拿出 3 只, 才能使拿出的手套中一定有一双是同色的。故答案为: C",
+        "knowledge_points": ["最不利原则"],
+        "source": "22-23 六年级下·全国·单元测试",
+        "merged_id": "unit5_038",
+        "original_id": "unit5_choice_038"
+      },
+      {
+        "id": "unit5_choice_039",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "(21-22 六年级上·江西赣州·期末) 将 ( ) 本书放入 3 个抽屉, 放得最多的抽屉至少放 5 本。",
+        "options": [
+          "A. $13 \\sim 15$",
+          "B. $12 \\sim 16$",
+          "C. $12 \\sim 15$"
+        ],
+        "answer": "A",
+        "solution": "【分析】将书本放得最多的抽屉至少是 5 本书, 则另外两个抽屉中此时应放 4 本书, 此时是书本最少的情况; 书本最多的情况: 每个抽屉都是 5 本, 据此可得出答案。本题主要考查的是抽屉原理, 解题的关键是熟练掌握抽屉原理并合理运用, 进而得出答案。\n\n【详解】书本最少的情况：\n\n$4 \\times 2 + 5\\\\= 8 + 5\\\\=13 (本)$\n\n书本最多的情况是:  $5 \\times 3 = 15$  (本)。故书本数应为  $13 \\sim 15$  本。故答案为: A",
+        "knowledge_points": ["鸽巢问题进阶"],
+        "source": "21-22 六年级上·江西赣州·期末",
+        "merged_id": "unit5_039",
+        "original_id": "unit5_choice_039"
+      },
+      {
+        "id": "unit5_choice_040",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "（21-22六年级下·内蒙古呼伦贝尔·期末）袋子里有红、黄、黑、白四种颜色的珠子各15颗，闭着眼睛从袋子里摸珠子，要想摸出颜色相同的5颗珠子，至少摸出（）颗才能保证达到目的。",
+        "options": [
+          "A. 15",
+          "B. 16",
+          "C. 17"
+        ],
+        "answer": "C",
+        "solution": "【分析】考虑最差情况，前面 16 次摸出四种颜色的珠子各 4 颗，那么下一次再摸出 1 颗珠子，就能保证此时摸出了 5 颗颜色相同的珠子。本题考查了抽屉原理, 能熟练考虑最不利情况是解题的关键。\n\n【详解】  $4 \\times 4 + 1\\\\= 1 6 + 1\\\\= 17 (颗)$\n\n所以，至少摸出 17 颗才能保证达到目的。故答案为: C",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "21-22六年级下·内蒙古呼伦贝尔·期末",
+        "merged_id": "unit5_040",
+        "original_id": "unit5_choice_040"
+      },
+      {
+        "id": "unit5_choice_041",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "(24-25 六年级下·全国·单元测试) 从六年级学生中任意挑选 13 名学生, 那么在这 13 名学生中至少有( )名是同月出生的。",
+        "options": [],
+        "answer": "2",
+        "solution": "【分析】先建立抽屉，因为一年有 12 个月，所以相当于有12 个抽屉，先取出 12 名学生的出生月份，最不利的情况是这 12 名学生的出生月份都不同即每个抽屉里放一名学生，然后还剩 1 名学生，无论放在哪个抽屉里，都可以保证有两名学生；所以至少有 2 名学生是同月出生的。\n\n【详解】  $13 \\div 12 = 1$  （名）……1（名）$1 + 1 = 2$  （名） 从六年级学生中任意挑选 13 名学生, 那么在这 13 名学生中至少有 2 名是同月出生的。",
+        "knowledge_points": ["鸽巢问题初步"],
+        "source": "24-25 六年级下·全国·单元测试",
+        "merged_id": "unit5_041",
+        "original_id": "unit5_choice_041"
+      },
+      {
+        "id": "unit5_choice_042",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "（2024·陕西商洛·小升初真题）杭州亚运会成为亚运史上规模最大、项目最多、覆盖面最广的一届运动会。中国一共派出了886名运动员参赛，这些运动员中至少有____人是同一个月生日。",
+        "options": [],
+        "answer": "74",
+        "solution": "【分析】一年有 12 个月, 把 12 个月看作 12 个 '抽屉', 886 名运动员相当于要放进这 12 个 '抽屉' 里的 '物品', 先用运动员总数 886 除以月数 12 , 即  $886 \\div 12 = 73$  (人) ……10 (人), 这里 73 表示如果平均分配, 每个月分配到 73 个人, 余数是 10 表示分完后还剩下 10 个人。剩下的这 10 人, 无论放到 12 个月中的哪一个月, 都会使得那个月的人数至少  $(73 + 1)$  人, 即这些运动员中至少有  $(73 + 1)$  人是同一个月生日, 据此解答。\n\n【详解】一年有 12 个月。\n\n$8 8 6 \\div 1 2 = 7 3 (\\text {人}) \\dots 1 0 (\\text {人})$\n\n$7 3 + 1 = 7 4 \\text {(人)}$\n\n即这些运动员中至少有 74 人是同一个月生日。",
+        "knowledge_points": ["鸽巢问题进阶"],
+        "source": "2024·陕西商洛·小升初真题",
+        "merged_id": "unit5_042",
+        "original_id": "unit5_choice_042"
+      },
+      {
+        "id": "unit5_choice_043",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "(2024·江西吉安·小升初真题)有红、黄、蓝、紫 4 种颜色的帽子各 5 顶放入一个盒子里。要保证取出的帽子有 2 种颜色, 至少应取出( )顶帽子; 要保证取出的帽子中至少有2顶是同色的，至少应取出( )顶帽子。",
+        "options": [],
+        "answer": "6 ;5",
+        "solution": "【分析】已知有红、黄、蓝、紫 4 种颜色的帽子各 5 顶放入一个盒子里，考虑最不利原则，把一种颜色的帽子 5 顶全部取完，再任意取一顶，一定有 2 种颜色的帽子；考虑最不利原则，把 4 种颜色的帽子各取 1 顶，再任意取 1 顶，则至少有 2 顶帽子是同色的。\n\n【详解】  $5 + 1 = 6$  （顶）$4 + 1 = 5$  (顶)要保证取出的帽子有 2 种颜色, 至少应取出 (6) 顶帽子;要保证取出的帽子中至少有 2 顶是同色的，至少应取出(5) 顶帽子。",
+        "knowledge_points": ["最不利原则"],
+        "source": "2024·江西吉安·小升初真题",
+        "merged_id": "unit5_043",
+        "original_id": "unit5_choice_043"
+      },
+      {
+        "id": "unit5_choice_044",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "(2024·山东菏泽·小升初真题) 口袋里有 6 个红球和 4个黄球，它们的大小和形状都相同，现从中任意摸出一个球，则摸出红球的可能性是( )，要保证摸出2个红球，至少一次要摸出( )个球。",
+        "options": [],
+        "answer": "$\\frac{3}{5};6$",
+        "solution": "【分析】红球有 6 个, 合计有  $(6 + 4)$  个球, 求摸出红球的可能性, 用红球的个数除以口袋里面球的个数即可;要保证摸出 2 个红球, 考虑最不利原则, 把 4 个黄球全部摸出后, 再任意摸 2 个, 必定能摸出 2 个红球, 即至少一次性摸出  $(4 + 2)$  个。\n\n【详解】  $$6 \\div (6 + 4)\\\\= 6 \\div 1 0 \\\\ = \\frac {3}{5} $$\n\n$4 + 2 = 6$  （个）\n\n口袋里有 6 个红球和 4 个黄球, 它们的大小和形状都相同,现从中任意摸出一个球, 则摸出红球的可能性是  $\\frac{3}{5}$ , 要保证摸出 2 个红球, 至少一次要摸出 6 个球。",
+        "knowledge_points": ["约分的认识及应用", "最不利原则", "求一个数占另一个数几分之几", "简单事件发生的可能性求解"],
+        "source": "2024·山东菏泽·小升初真题",
+        "merged_id": "unit5_044",
+        "original_id": "unit5_choice_044"
+      },
+      {
+        "id": "unit5_choice_045",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "(24-25 六年级下·海南海口·单元测试) 一个袋子里有红、蓝、绿球各 6 个, 要保证取出的两个同颜色的球, 至少要取____个球: 要保证取出两个不同颜色的球, 至少要取____个球。",
+        "options": [],
+        "answer": "4;7",
+        "solution": "【分析】根据鸽巢原理，要保证取出的两个球是同颜色的，用颜色数加 1 即可解答；要保证取出的两个球是不同颜色的，需要考虑最坏的情况。最坏的情况是取出的球都是同一种颜色的。袋子里每种颜色有 6 个球, 因此用  $6 + 1$  列式解答, 根据鸽巢原理, 则至少有两个球是不同颜色的。\n\n【详解】  $3 + 1 = 4$  （个）$6 + 1 = 7$  (个) 所以要保证取出的两个同颜色的球，至少要取 4 个球，要保证取出两个不同颜色的球，至少要取 7 个球。",
+        "knowledge_points": ["最不利原则"],
+        "source": "24-25 六年级下·海南海口·单元测试",
+        "merged_id": "unit5_045",
+        "original_id": "unit5_choice_045"
+      },
+      {
+        "id": "unit5_choice_046",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "(24-25 六年级上·四川成都·期末) 把红、黄、蓝、白四种颜色的小球各10个放到一个袋子里，至少取( )个球，可以保证取到两个相同的颜色。",
+        "options": [],
+        "answer": "5",
+        "solution": "【分析】由题意可知, 袋子中的球共有 4 种颜色, 要保证取到两个颜色相同的球, 最差的情况是, 取了 4 个球, 每种颜色各一个, 此时只要再任取一个球, 就能保证取到两个颜色相同的球, 即  $4 + 1 = 5$  个。\n\n【详解】  $4 + 1 = 5$  （个）所以至少取 5 个球, 可以保证取到两个相同的颜色。",
+        "knowledge_points": ["最不利原则"],
+        "source": "24-25 六年级上·四川成都·期末",
+        "merged_id": "unit5_046",
+        "original_id": "unit5_choice_046"
+      },
+      {
+        "id": "unit5_choice_047",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "(24-25 六年级上·四川资阳·期末) 在一个不透明的袋子里有 8 颗蓝珠子, 7 颗红珠子, 6 颗绿珠子, 至少取出总颗数的( )，才能保证有 3 颗颜色相同的珠子。",
+        "options": [],
+        "answer": "$\\frac{1}{3}$",
+        "solution": "【分析】要想保证有 3 颗颜色相同的珠子考虑最不利的情况,每种颜色的珠子都取了 2 颗, 此时再任意取一颗珠子一定有3颗颜色相同的珠子，那么至少要取出  $(2 \\times 3 + 1)$  颗珠子，最后求出至少取出珠子的数量占珠子总数量的分率，据此解答。\n\n【详解】 $(2 \\times 3 + 1) \\div (8 + 7 + 6)\\\\= (6 + 1) \\div 2 1 \\\\ = 7 \\div 2 1 \\\\ = \\frac {1}{3}$\n\n所以，至少取出总颗数的  $\\frac{1}{3}$ ，才能保证有 3 颗颜色相同的珠子。",
+        "knowledge_points": ["最不利原则", "求一个数占另一个数几分之几"],
+        "source": "24-25 六年级上·四川资阳·期末",
+        "merged_id": "unit5_047",
+        "original_id": "unit5_choice_047"
+      },
+      {
+        "id": "unit5_choice_048",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "(2023·陕西西安·小升初真题) 有红、黄、白三种颜色的小球各 10 个, 混合放在一个布袋中, 一次至少摸出( )个, 才能保证有 6 个小球是同色的。",
+        "options": [],
+        "answer": "16",
+        "solution": "【分析】根据题意, 袋子里有红、黄、白三种颜色的小球各 10 个, 运气最差的情况为每种颜色的小球各摸出 5 个, 再摸出一个任意颜色的球就能保证有 6 个小球是同色的。\n\n【详解】  $5 \\times 3 + 1\\\\= 1 5 + 1 \\\\ = 1 6 \\text {(个)} $\n\n填空如下：一次至少摸出（16）个，才能保证有 6 个小球是同色的。",
+        "knowledge_points": ["最不利原则"],
+        "source": "2023·陕西西安·小升初真题",
+        "merged_id": "unit5_048",
+        "original_id": "unit5_choice_048"
+      },
+      {
+        "id": "unit5_choice_049",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "(2009 四年级·全国·竞赛) 一只黑色口袋中装有形状、大小、材料相同的红球、黄球、白球各 3 个, 至少要摸出____个球，才能保证三种颜色的球都能拿到。",
+        "options": [],
+        "answer": "7",
+        "solution": "【分析】通过利用抽屉原理中的最不利原则。最不利的情况是：先将所有的红色有球摸完，需要摸出 3 个球；后来将所有黄色的球全部摸出，摸出 3 个球，最后只剩下 1 种白色的球, 只需要摸出 1 个就能保证三种颜色的球都能拿到。\n\n【详解】就是将任意两种颜色的球全部摸出，第三种颜色的球摸出一种即可。3＋3＋1＝7（个）至少要摸出 7 个球，才能保证三种颜色的球都能拿到。$3 + 3 + 1 = 7$  （个）",
+        "knowledge_points": ["最不利原则"],
+        "source": "2009 四年级·全国·竞赛",
+        "merged_id": "unit5_049",
+        "original_id": "unit5_choice_049"
+      },
+      {
+        "id": "unit5_choice_050",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "（23-24六年级下·浙江杭州·期末）有红、黄、蓝三种颜色的筷子（这些筷子除了颜色不同外，其他都相同）各3根混在一起。塘塘闭上眼睛，从中至少取出( )根才能保证一定有2根同色的筷子；至少取出( )根才能保证有2双不同色的筷子（指一双筷子为其中一种颜色，另一双筷子为另一种颜色）。",
+        "options": [],
+        "answer": "4;6",
+        "solution": "【分析】考虑最不利的情况，红、黄、蓝各拿一根，再拿一根，无论什么颜色，都可保证一定有 2 根同色的筷子；根据前面的分析，拿 4 根能保证一定有 2 根同色的筷子，假设前 4 根是 2 根红, 1 根黄, 1 根蓝; 再拿 2 根, 无论是红黄、红蓝、蓝蓝、蓝黄, 还是黄黄, 都可再组成一双同色筷子,据此解答。\n\n【详解】  $3 + 1 = 4$  (根)$4 + 2 = 6$  (根)有红、黄、蓝三种颜色的筷子（这些筷子除了颜色不同外,其他都相同）各3根混在一起。塘塘闭上眼睛，从中至少取出 4 根才能保证一定有 2 根同色的筷子; 至少取出 6 根才能保证有 2 双不同色的筷子 (指一双筷子为其中一种颜色, 另一双筷子为另一种颜色）。",
+        "knowledge_points": ["最不利原则"],
+        "source": "23-24六年级下·浙江杭州·期末",
+        "merged_id": "unit5_050",
+        "original_id": "unit5_choice_050"
+      },
+      {
+        "id": "unit5_choice_051",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "（23-24六年级下·山东济宁·期末）六（一）班有41名同学，至少有( )人是同一个月出生的。",
+        "options": [],
+        "answer": "4",
+        "solution": "【分析】一年一共有 12 个月, 用 41 名同学除以 12 可得出有余数的除数, 得到的商是几就有几个人同一月出生, 若有余数则再加 1 , 据此可得出答案。\n\n【详解】  $41 \\div 12 = 3$  （人）……5（人） $3 + 1 = 4$  （人）则至少有 4 人是同一个月出生的。",
+        "knowledge_points": ["鸽巢问题初步"],
+        "source": "23-24六年级下·山东济宁·期末",
+        "merged_id": "unit5_051",
+        "original_id": "unit5_choice_051"
+      },
+      {
+        "id": "unit5_choice_052",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "（23-24六年级下·贵州黔西·期末）望谟县位于黔西南州东部，因布依方言'王母'谐音而得名。2024年贵州望谟'三月三'第二届乡村山地马拉松设置四个项目，共计1350人，总有一个项目至少有( )人参跑。",
+        "options": [],
+        "answer": "338",
+        "solution": "【分析】根据抽屉原理，把四个项目看成四个抽屉，将总人数平均分成4份后，剩余的人数也会选择项目，那么肯定会有一个项目中的人数会至少增加一人。据此解答。\n\n【详解】  $1350 \\div 4 = 337$  （人）……2（人）$337 + 1 = 338$  （人）所以，总有一个项目至少有 338 人。",
+        "knowledge_points": ["鸽巢问题初步"],
+        "source": "23-24六年级下·贵州黔西·期末",
+        "merged_id": "unit5_052",
+        "original_id": "unit5_choice_052"
+      },
+      {
+        "id": "unit5_choice_053",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "(23-24 六年级下·湖北十堰·期末) 在 1、2、3、…、20中至少要取出( )个不同的数, 才能保证其中一定有一个数是合数。",
+        "options": [],
+        "answer": "10",
+        "solution": "【分析】只有 1 和它本身两个因数的数叫作质数。除了 1 和它本身外还有别的因数的数叫作合数。最不利原则是指考虑所有可能情况中，最不利于某件事情发生的情况。在这道题里，最不利的情况就是先把不是合数的数都取出来，然后再多取一个就能保证有合数。在 1 到 20 中，质数有 2、3、5、7、11、13、17、19 共 8 个, 1 既不是质数也不是合数, 所以先取出这 9 个数, 再取 1 个数就一定是合数。\n\n【详解】把 1、2、3、5、7、11、13、17、19 全部取出，即 9 个；$9 + 1 = 10$  （个）即在  $1 、 2 、 3 、 \\cdots 、 20$  中至少要取出 10 个不同的数, 才能保证其中一定有一个数是合数。",
+        "knowledge_points": ["质数与合数的认识", "最不利原则"],
+        "source": "23-24 六年级下·湖北十堰·期末",
+        "merged_id": "unit5_053",
+        "original_id": "unit5_choice_053"
+      },
+      {
+        "id": "unit5_choice_054",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "(23-24六年级下·广东东莞·期末）学校航模小组有32人，航模小组至少有( )人的生日是同一个月；把蓝、绿、红、黄4种颜色的球各6个放进1个袋子里，至少取( )个球可以保证取到两个颜色相同的球。",
+        "options": [],
+        "answer": "3 ;5",
+        "solution": "【分析】由题意可知, 一年有 12 个月, 则  $32 \\div 12 = 2$  (人)……8 (人), 所以至少有  $2 + 1 = 3$  人的生日是同一个月; 先取 4 个球, 分别是蓝、绿、红、黄球各一个, 再取一个球无论是什么颜色就可以保证取到两个颜色相同的球。\n\n【详解】  $32 \\div 12 = 2$  （人）……8（人）$2 + 1 = 3 \\text {(人)}$ $4 + 1 = 5 \\text {(个)}$则学校航模小组有 32 人, 航模小组至少有 3 人的生日是同一个月; 把蓝、绿、红、黄 4 种颜色的球各 6 个放进 1 个袋子里, 至少取 5 个球可以保证取到两个颜色相同的球。",
+        "knowledge_points": ["鸽巢问题初步"],
+        "source": "23-24六年级下·广东东莞·期末",
+        "merged_id": "unit5_054",
+        "original_id": "unit5_choice_054"
+      },
+      {
+        "id": "unit5_choice_055",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "（23-24六年级下·湖南岳阳·期末）五一劳动节，有51名老人在广场上载歌载舞，欢度节日，那么至少有____名老人生日在同一个月。",
+        "options": [],
+        "answer": "5",
+        "solution": "【分析】一年有 12 个月, 把这 12 个月看作 12 个抽屉, 把 51 名老人看作 51 个整体,  $51 \\div 12 = 4 \\dots 1$ , 由此利用抽屉原理可知, 每个抽屉有 4 名, 还余下 1 名, 不管放哪个抽屉里, 一定至少有  $4 + 1 = 5$  名老人相同生日。\n\n【详解】  $51 \\div 12 = 4$  （名）……1名）$4 + 1 = 5$  (名) 根据分析可得，至少有 5 名老人生日在同一个月。",
+        "knowledge_points": ["鸽巢问题初步"],
+        "source": "23-24六年级下·湖南岳阳·期末",
+        "merged_id": "unit5_055",
+        "original_id": "unit5_choice_055"
+      },
+      {
+        "id": "unit5_choice_056",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "（2024六年级下·全国·专题练习）今年是2023年，至少要有( )人，才能保证其中至少有2个人在同一天过生日。",
+        "options": [],
+        "answer": "366",
+        "solution": "【分析】2023年是平年，全年一共365天，把所有人看作物品, 365 天看作抽屉, 要保证至少有 2 个人在同一个抽屉,那么可以每个抽屉先放一个人, 再在某一个抽屉中多放一个人。\n\n【详解】  $(2 - 1) \\times 365 + 1\\\\= 1 \\times 3 6 5 + 1 \\\\ = 3 6 5 + 1 \\\\ = 3 6 6 (\\text {人}) $\n\n至少要有 366 人。",
+        "knowledge_points": ["鸽巢问题初步"],
+        "source": "2024六年级下·全国·专题练习",
+        "merged_id": "unit5_056",
+        "original_id": "unit5_choice_056"
+      },
+      {
+        "id": "unit5_choice_057",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "（22-23五年级下·全国·单元测试）桌上放有同样的3支铅笔和30块橡皮。来了一群学生，每人从这60个文具中拿一个或两个，至少有5人拿到的文具完全相同，这群学生至少有( )人。",
+        "options": [],
+        "answer": "21",
+        "solution": "【分析】每人拿走 1 个或者 2 个, 则只有 1 支铅笔, 1 块橡皮, 2 支铅笔, 2 块橡皮, 1 支铅笔和 1 块橡皮 5 种不同的情况; 根据鸽巢原理, 假设每种情况都有 4 个人, 只要再多 1 个人则保证至少有 5 人拿到的文具完全相同。此题考查了利用抽屉原理解决实际问题的灵活应用，关键是确定有几种情况。\n\n【详解】  $5 \\times 4 + 1\\\\= 2 0 + 1 \\\\ = 2 1 \\text {(人)} \\\\$\n\n这群学生至少有 21 人。",
+        "knowledge_points": ["鸽巢问题进阶"],
+        "source": "22-23五年级下·全国·单元测试",
+        "merged_id": "unit5_057",
+        "original_id": "unit5_choice_057"
+      },
+      {
+        "id": "unit5_choice_058",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "（20-21六年级下·天津南开·期末）在某班学生中，有 10 人都订阅了《小朋友》《少年报》《儿童时代》三种报刊中的一种或者几种。那么，这 10 个人中至少有( )个人所定的报刊种类完全相同。",
+        "options": [],
+        "answer": "2",
+        "solution": "【分析】先求出每人订阅一种、两种、三种报刊一共有几种订阅方法, 把学生的总人数看作被分放物体的数量, 订阅方法看作抽屉的数量, 被分放物体的数量  $\\div$  抽屉的数量 = 平均每个抽屉分放物体的数量……剩下物体的数量, 一个抽屉里至少分放物体的数量 = 平均每个抽屉分放物体的数量 +1 , 据此解答。本题主要考查抽屉问题, 准确求出抽屉数是解答题目的关键。\n\n【详解】每人订阅一种：《小朋友》或《少年报》或《儿童时代》；每人订阅两种：《小朋友》和《少年报》、《小朋友》和《儿童时代》、《少年报》和《儿童时代》；每人订阅三种：《小朋友》、《少年报》和《儿童时代》。  $3 + 3 + 1 = 7$  （种）$1 0 \\div 7 = 1 \\dots 3$   $1 + 1 = 2$  （人）所以，这 10 个人中至少有 2 个人所定的报刊种类完全相同。",
+        "knowledge_points": ["鸽巢问题进阶"],
+        "source": "20-21六年级下·天津南开·期末",
+        "merged_id": "unit5_058",
+        "original_id": "unit5_choice_058"
+      },
+      {
+        "id": "unit5_choice_059",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "(19-20 六年级下·全国·单元测试) 给正方体的六个面涂上不同的 4 种颜色, 不论怎样涂, 至少有( )个面的颜色相同。",
+        "options": [],
+        "answer": "2",
+        "solution": "【分析】根据题意，给正方体的六个面涂上不同的 4 种颜色，先把这 4 种颜色涂在正方体的 4 个面上，还剩下 2 个面，这 2 个面无论涂哪种颜色，至少有 2 个面的颜色相同。本题考查鸽巢问题（抽屉问题），根据'至少数=物体数÷抽屉的个数+1（有余数的情况下）'解答。\n\n【详解】  $6 \\div 4 = 1$  （个）……2（个） $1 + 1 = 2$  （个）不论怎样涂，至少有 2 个面的颜色相同。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_059",
+        "original_id": "unit5_choice_059"
+      },
+      {
+        "id": "unit5_choice_060",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "（22-23六年级下·全国·单元测试）把10支铅笔放入4个文具盒中，总有一个文具盒中至少放入了( )支铅笔。如果把这些铅笔放进3个文具盒中，总有一个文具盒中至少放入了( )支铅笔。",
+        "options": [],
+        "answer": "3 ;4",
+        "solution": "【分析】把 10 支铅笔放进 4 个文具盒中,  $10 \\div 4 = 2$  (支) ……2 (支), 即平均每个文具盒放 2 支, 还余 2 支, 根据抽屉原理可知, 总有一个文具盒里至少放  $2 + 1 = 3$  (支)。把 10 支铅笔放进 3 个文具盒中,  $10 \\div 3 = 3$  (支) …… 1(支), 即平均每个文具盒放 3 支, 还余 1 支, 根据抽屉原理可知，总有一个文具盒里至少放  $3 + 1 = 4$  （支）。在此类抽屉问题中，至少数 = 物体数除以抽屉数的商 +1（有余数的情况下），没有余数的情况下，至少数 = 平均数。\n\n【详解】  $10 \\div 4 = 2$  （支）…2（支）  $2 + 1 = 3$  (支) 总有一个文具盒中至少放入了 3 支铅笔。$10 \\div 3 = 3$  (支) ……1 (支) $3 + 1 = 4$  (支)总有一个文具盒中至少放入了 4 支铅笔。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "22-23六年级下·全国·单元测试",
+        "merged_id": "unit5_060",
+        "original_id": "unit5_choice_060"
+      },
+      {
+        "id": "unit5_choice_061",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "(22-23 六年级下·全国·单元测试) 红、白、黄、黑四种颜色的玻璃球各 6 个放到一个袋里。闭着眼睛从中取球,至少取( )个球, 可以保证取到两个颜色相同的球。",
+        "options": [],
+        "answer": "5",
+        "solution": "【分析】由于袋子里共有红、白、黄、黑四种颜色的球各 6 个, 如果一次取 4 个, 最差情况为红、白、黄、黑四种颜色各一个, 所以只要再多取一个球, 就能保证取到两个颜色相同的球。据此解答。解决抽屉原理问题的关键是根据最差原理对问题进行分析。\n\n【详解】  $4 + 1 = 5$  （个）即至少取 5 个球, 可以保证取到两个颜色相同的球。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "22-23 六年级下·全国·单元测试",
+        "merged_id": "unit5_061",
+        "original_id": "unit5_choice_061"
+      },
+      {
+        "id": "unit5_choice_062",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "（22-23五年级上·浙江温州·期中）一个不透明的袋子里装有6颗白珠子，3颗红珠子，2颗蓝珠子，1颗黑珠子，珠子颜色不同、形状大小相同，一次摸出( )颗珠子才能保证至少有两颗白珠子。",
+        "options": [],
+        "answer": "8",
+        "solution": "【分析】考虑最差情况，把红珠子、蓝珠子和黑珠子都摸完，再加上两个白珠子，那么就可以保证至少有两颗白珠子。此题考查了抽屉原理, 能熟练考虑最不利情况是解答的关键。\n\n【详解】  $3 + 2 + 1 + 2\\\\= 5 + 1 + 2\\\\= 8  (颗)$\n\n先把所有不符合情况全部摸出，再加上需要达到目标的数量，所以一次摸出8颗珠子才能保证至少有两颗白珠子。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "22-23五年级上·浙江温州·期中",
+        "merged_id": "unit5_062",
+        "original_id": "unit5_choice_062"
+      },
+      {
+        "id": "unit5_choice_063",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "（20-21 六年级下·山东济宁·期末）把 11 把拖把发给 5 个小组，总有一个小组至少分( )把拖把。",
+        "options": [],
+        "answer": "3",
+        "solution": "【分析】把 5 个小组可以看作是 5 个抽屉, 11 把拖把看作 11 个元素, 考虑最差情况: 把 11 个元素平均分配在 5 个抽屉中:  $11 \\div 5 = 2$  (把) ……1 (把), 那么每个抽屉都有 2 把,那么剩下的 1 把, 无论放到哪个抽屉都会出现 3 把在同一个抽屉里。此题属于典型的抽屉原理习题, 解答此类题的关键是找出把谁看作 '抽屉个数', 把谁看作 '物体个数',然后根据抽屉原理解答即可。\n\n【详解】  $11 \\div 5 = 2$  （把）……1（把） $2 + 1 = 3$  (把)",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21 六年级下·山东济宁·期末",
+        "merged_id": "unit5_063",
+        "original_id": "unit5_choice_063"
+      },
+      {
+        "id": "unit5_choice_064",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "(20-21 六年级下·河南洛阳·期末) 某数学兴趣小组有 13 名学生, 他们中至少有( )个人是同一月出生的。",
+        "options": [],
+        "answer": "2",
+        "solution": "【分析】抽屉原则一：如果把  $(n + 1)$  个我要放在  $n$  个抽屉里，那么必有一个抽屉中至少放有 2 个物体。关键是构造物体和抽屉，也就是找到代表物体和抽屉的量，然后依据抽屉原则进行计算。\n\n【详解】  $13 \\div 12 = 1$  ……1 $1 + 1 = 2$  （人） 某数学兴趣小组有 13 名学生, 他们中至少有 2 个人是同一月出生的。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21 六年级下·河南洛阳·期末",
+        "merged_id": "unit5_064",
+        "original_id": "unit5_choice_064"
+      },
+      {
+        "id": "unit5_choice_065",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "（2022·云南文山·小升初真题）把32个鸡蛋放进6个盒子里，总有一个盒子里至少放进( )个鸡蛋。",
+        "options": [],
+        "answer": "6",
+        "solution": "【分析】把32个鸡蛋看作被分放物体，6个盒子看作6个抽屉, 被分放物体的数量  $\\div$  抽屉的数量 = 平均每个抽屉分放物体的数量……剩下物体的数量, 一个抽屉里至少分放物体的数量 = 平均每个抽屉分放物体的数量 +1 , 据此解答。掌握抽屉原理的解题方法是解答题目的关键。\n\n【详解】  $32 \\div 6 = 5 \\dots 2$ $5 + 1 = 6$  (个)所以，总有一个盒子里至少放进 6 个鸡蛋。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "2022·云南文山·小升初真题",
+        "merged_id": "unit5_065",
+        "original_id": "unit5_choice_065"
+      },
+      {
+        "id": "unit5_choice_066",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "（2021·安徽铜陵·小升初真题）六年级有 100 名同学订阅 A、B、C 三种杂志。如果他们都只订阅了其中一种，至少有（）名同学订阅的杂志种类相同；如果他们订阅了其中的一种或两种杂志，至少有（）名同学订阅的杂志种类相同。",
+        "options": [],
+        "answer": "34;17",
+        "solution": "【分析】（1）如果他们都只订阅了其中一种，则有A、B、C三种订阅方式；用除法求出100里有多少个3，商是33，还余1名同学，那么这1名同学无论订阅哪种杂志，都会出现有一种杂志至少有（ $33 + 1$ ）名同学订阅；(2) 如果他们订阅了其中的一种或两种杂志, 则会出现A、B、C、AB、AC、BC，一共6种不同的订阅方式；用除法求出100里有多少个6，商是16，还余4名同学，那么这4名同学无论选取哪种订阅方式，都会出现有一种杂志种类至少有（ $16 + 1$ ）名同学订阅。本题考查鸽巢问题，采用最不利原则来解题。\n\n【详解】 (1)  $100 \\div 3 = 33$  （名）……1（名）$3 3 + 1 = 3 4 \\text {(名)}$如果他们都只订阅了其中一种，至少有 34 名同学订阅的杂志种类相同；(2) 如果他们订阅了其中的一种或两种杂志, 共有 6 种不同的订阅方式; $100\\div6=16(名)\\dots4 （名）$$16 + 1 = 17$  (名)如果他们订阅了其中的一种或两种杂志，至少有 17 名同学订阅的杂志种类相同。$100 \\div 6 = 16$  （名）……4（名）",
+        "knowledge_points": ["鸽巢问题进阶"],
+        "source": "2021·安徽铜陵·小升初真题",
+        "merged_id": "unit5_066",
+        "original_id": "unit5_choice_066"
+      },
+      {
+        "id": "unit5_choice_068",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "(20-21 六年级下·广东潮州·期末) 把 5 个椰子放进两个筐里, 总有一个筐里至少放进( )个椰子。",
+        "options": [],
+        "answer": "3",
+        "solution": "【分析】利用抽屉原理最差情况，要使筐里的数量尽量少，要尽量平均分，把5个椰子放进两个筐里， $5 \\div 2 = 2$  个……1个，即平均每个筐里放入2个后，还有1个没有放入，即至少有一个筐要放入  $2 + 1 = 3$  个椰子，据此解答。此题考查了抽屉原理解决问题的灵活运用, 关键是从最差情况考虑。\n\n【详解】  $5 \\div 2 = 2$  （个）……1（个） $2 + 1 = 3$  (个)",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21 六年级下·广东潮州·期末",
+        "merged_id": "unit5_068",
+        "original_id": "unit5_choice_068"
+      },
+      {
+        "id": "unit5_choice_069",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "(21-22 六年级下·河北衡水·期末) 袋子里有红、黄、蓝球各 4 个, 至少随意拿出( )个, 才能保证有两个颜色相同的球。",
+        "options": [],
+        "answer": "4",
+        "solution": "【分析】考虑最倒霉的情况，拿出的前 3 个球都是不同颜色的球，再拿一个，无论是什么颜色，都可保证有两个颜色相同的球，据此分析。关键是构造物体和抽屉, 也就是找到代表物体和抽屉的量, 然后依据抽屉原则进行计算。\n\n【详解】  $3 + 1 = 4$  （个）",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "21-22 六年级下·河北衡水·期末",
+        "merged_id": "unit5_069",
+        "original_id": "unit5_choice_069"
+      },
+      {
+        "id": "unit5_choice_070",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "(2024·河北张家口·小升初真题) 18 只鸽子飞回 5 个鸽舍, 至少有 4 只要飞进同一个鸽舍。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】根据题意，先将 18 只鸽子平均放进 5 个鸽舍里，\n\n每个鸽舍平均放 3 只, 还剩下 3 只, 这 3 只鸽子, 无论飞进哪个鸽舍里, 总有一个鸽舍至少有 4 只鸽子。\n\n【详解】  $18 \\div 5 = 3$  （只）……3（只） $3 + 1 = 4 \\text {(只)}$所以至少有 4 只鸽子要飞进同一个鸽舍。\n\n原题说法正确。故答案为:  √",
+        "knowledge_points": ["鸽巢问题初步"],
+        "source": "2024·河北张家口·小升初真题",
+        "merged_id": "unit5_070",
+        "original_id": "unit5_choice_070"
+      },
+      {
+        "id": "unit5_choice_071",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "(22-23 六年级下·陕西商洛·期末) 盒子里有大小相同的红球、白球、蓝球各 4 个, 要想摸出的球一定有 2 个是同色的, 至少要摸出 4 个球。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】假设运气最差的情况，先摸出的 3 个球的颜色都不一样，此时再任意摸出 1 个，就有 2 个同色的球，所以至少要摸出  $(3 + 1)$  个球。\n\n【详解】  $3 + 1 = 4$  （个）所以要想摸出的球一定有 2 个是同色的，至少要摸出 4 个球。原题说法正确。故答案为: √",
+        "knowledge_points": ["最不利原则"],
+        "source": "22-23 六年级下·陕西商洛·期末",
+        "merged_id": "unit5_071",
+        "original_id": "unit5_choice_071"
+      },
+      {
+        "id": "unit5_choice_072",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "（2023·陕西西安·小升初真题）六年级的 34 位同学中，至少有 3 位同学在同一个月过生日。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】一年有 12 个月, 把 12 个月看作 12 个抽屉, 34 位同学看作物体个数, 根据抽屉原理得:  $34 \\div 12 = 2 \\dots 10$ ;则至少有:  $2 + 1 = 3$  (位) 在同一个月过生日。\n\n【详解】建立抽屉：一年有 12 个月分别看做 12 个抽屉  $34 \\div 12 = 2$  （位）………10（位）$2 + 1 = 3$  （位）至少有 3 位同学在同一个月过生日，原题说法正确。故答案为:  √",
+        "knowledge_points": ["最不利原则"],
+        "source": "2023·陕西西安·小升初真题",
+        "merged_id": "unit5_072",
+        "original_id": "unit5_choice_072"
+      },
+      {
+        "id": "unit5_choice_073",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "（2024·四川绵阳·小升初真题）把红、黄、蓝、绿四种颜色的袜子各 10 只放到一个袋子里，至少取出 11 只袜子，可以保证取出一双颜色相同的袜子。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】根据题意, 有四种颜色的袜子, 要保证取到两只颜色相同的袜子, 考虑最坏的情况, 红、黄、蓝、绿都各取 1 只, 那么这时, 只要再取出任意一只颜色的袜子, 就可以保证取到一双颜色相同的袜子。\n\n【详解】通过分析可得：$4 + 1 = 5$  （只），至少取出 5 只袜子，可以保证取出一双颜色相同的袜子。原题说法错误。故答案为:  ×",
+        "knowledge_points": ["最不利原则"],
+        "source": "2024·四川绵阳·小升初真题",
+        "merged_id": "unit5_073",
+        "original_id": "unit5_choice_073"
+      },
+      {
+        "id": "unit5_choice_074",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "（23-24六年级下·山东临沂·期末）袋子中有大小相同的白色、黄色和红色乒乓球各4个，一次至少摸出4个才能保证其中有两个同色的。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】根据最不利原则考虑, 是白色、黄色和红色乒乓球各摸出一个, 此时再摸出 1 个, 无论是什么颜色, 一定有两个球同色, 所以一次至少需要摸出 4 个球才能保证其中有两个同色的。\n\n【详解】袋子中有大小相同的白色、黄色和红色乒乓球各 4 个, 一次至少摸出 4 个才能保证其中有两个同色的, 说法正确。故答案为:  √",
+        "knowledge_points": ["最不利原则"],
+        "source": "23-24 六年级下·山东临沂·期末",
+        "merged_id": "unit5_074",
+        "original_id": "unit5_choice_074"
+      },
+      {
+        "id": "unit5_choice_075",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "(23-24 六年级下·广东云浮·期末) 用三种颜色给正方体的 6 个面涂色（每个面只涂一种颜色），至少有两个面涂色相同。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】根据抽屉原则可知，把正方形的6个面看作6个抽屉。 三种颜色看作三个球。用  $6 \\div 3$ , 商表示每个抽屉至少放几个球, 也就是至少有几个面涂色相同。\n\n【详解】  $6 \\div 3 = 2$  （个）用三种颜色给正方体的 6 个面涂色（每个面只涂一种颜色），至少有两个面涂色相同，说法正确。故答案为: √",
+        "knowledge_points": ["鸽巢问题初步"],
+        "source": "23-24 六年级下·广东云浮·期末",
+        "merged_id": "unit5_075",
+        "original_id": "unit5_choice_075"
+      },
+      {
+        "id": "unit5_choice_076",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "(23-24 六年级下·河北邢台·期末) 袋子里有黑、白、红 3 种颜色的球各 4 个, 闭眼摸出 4 个颜色相同的球, 1 次一定可以摸出来。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】袋子里有黑、白、红 3 种颜色的球各 4 个, 闭眼摸出 4 个球, 考虑最差的情况, 1 次至少可以摸出 2 个颜色相同的球, 但是不一定可以 1 次摸出 4 个颜色相同的球, 据此判断即可。本题考查鸽巢问题, 解答本题的关键是掌握根据最不利原则解决问题的方法。\n\n【详解】根据分析可得，闭眼摸出 4 个颜色相同的球，1 次一定可以摸出来，说法错误。故答案为：×",
+        "knowledge_points": ["最不利原则"],
+        "source": "23-24 六年级下·河北邢台·期末",
+        "merged_id": "unit5_076",
+        "original_id": "unit5_choice_076"
+      },
+      {
+        "id": "unit5_choice_077",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "(23-24 六年级下·河南新乡·期末) 15 个人里至少有 3 个人是同一个属相。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】一共有 12 个不同的属相, 尽可能地让每个人的属相不同, 将 15 人除以 12 , 求出商和余数。将商加上 1 , 求出至少有几个人是同一属相。\n\n【详解】  $15 \\div 12 = 1 \\dots 3$   $1 + 1 = 2$  （人）所以，15 个人里至少有 2 个人是同一个属相，原题说法错误。故答案为: ×",
+        "knowledge_points": ["鸽巢问题初步"],
+        "source": "23-24 六年级下·河南新乡·期末",
+        "merged_id": "unit5_077",
+        "original_id": "unit5_choice_077"
+      },
+      {
+        "id": "unit5_choice_078",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "（23-24六年级下·广东江门·期中）把5只兔子放进2个笼子里，无论怎样放，总有一个笼子里面至少要放进3只兔子。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】本题考查了利用鸽巢原理解决实际问题的灵活应用，解答此题的关键在于从最差情况去考虑。5只兔子放进2个笼子，平均每个笼子放2只，还剩1只，剩下的一只无论放在哪个笼子里，总有1个笼子里至少要放进3只兔子。\n\n【详解】  $5 \\div 2 = 2$  （只）……1（只）$2 + 1 = 3$  （只）把 5 只兔子放进 2 个笼子里, 无论怎样放, 总有一个笼子里面至少要放进 3 只兔子。故答案为:  √",
+        "knowledge_points": ["鸽巢问题初步"],
+        "source": "23-24六年级下·广东江门·期中",
+        "merged_id": "unit5_078",
+        "original_id": "unit5_choice_078"
+      },
+      {
+        "id": "unit5_choice_079",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "(23-24 六年级下·湖南怀化·期中) 6 只小鸡装入 5 个笼子, 至少有一个笼子放小鸡 3 只。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】根据题意，先将 6 只小鸡平均放入 5 个笼子里，每个笼子里放入 1 只，还剩下 1 只，这 1 只小鸡无论放入哪个笼子里，总有一个笼子里至少有 2 只小鸡，据此判断。\n\n【详解】  $6 \\div 5 = 1$  （只）……1（只） $1 + 1 = 2 \\quad (\\text {只})$至少有一个笼子放小鸡 2 只。原题说法错误。故答案为: ×",
+        "knowledge_points": ["鸽巢问题初步"],
+        "source": "23-24 六年级下·湖南怀化·期中",
+        "merged_id": "unit5_079",
+        "original_id": "unit5_choice_079"
+      },
+      {
+        "id": "unit5_choice_080",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "(23-24六年级下·河南信阳·期中) 六 (1) 班数学兴趣小组 15 名同学, 至少有 2 人的出生月份相同。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】一年有 12 个月, 那么可以看作是 12 个抽屉, 15 名同学看做 15 个元素, 考虑最差情况: 把 15 名同学平均分配在 12 个抽屉中:  $15 \\div 12 = 1$  (个) ……3 (个), 那么每个抽屉都有 1 人, 那么剩下的 3 人, 无论放到哪个抽屉都会出现 2 个人在同一个抽屉里。\n\n【详解】  $15 \\div 12 = 1$  （名）……3（名）$1 + 1 = 2$  (名) 至少有 2 人的出生月份相同，原题说法正确。故答案为:  √",
+        "knowledge_points": ["鸽巢问题初步"],
+        "source": "23-24六年级下·河南信阳·期中",
+        "merged_id": "unit5_080",
+        "original_id": "unit5_choice_080"
+      },
+      {
+        "id": "unit5_choice_081",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "困难",
+        "question": "(21-22 六年级下·辽宁抚顺·期中) 任意 25 名小学生中,至少有 5 人所在年级是相同的。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】把 6 个年级看作是 6 个抽屉, 25 名小学生看做 25 个元素, 根据抽屉原理: 把 25 名小学生平均分配在 6 个抽屉中:  $25 \\div 6 = 4$  (人) ……1 (人), 那么每个抽屉都有 4 人,那么剩下的 1 人, 无论放到哪个抽屉都会出现 5 人在同一个抽屉里。此题属于典型的抽屉原理习题, 解答此类题的关键是找出把谁看作 '抽屉个数', 把谁看作 '物体个数',然后根据抽屉原理解答即可。\n\n【详解】  $25 \\div 6 = 4$  （人）……1（人）$4 + 1 = 5 \\text {(人)}$即至少有 5 人所在年级是相同的，所以原题说法正确。故答案为:  √",
+        "knowledge_points": ["鸽巢问题初步"],
+        "source": "21-22 六年级下·辽宁抚顺·期中",
+        "merged_id": "unit5_081",
+        "original_id": "unit5_choice_081"
+      },
+      {
+        "id": "unit5_choice_082",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "（20-21六年级下·河南信阳·期末）学校把转入的18名新生分到3个年级6个班里，总有一个班至少分到3名同学。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】把 6 个班看作 6 个抽屉, 把 18 名新生看作物体的个数, 根据抽屉原理进行解答即可。此题属于典型的抽屉原理习题, 解答此类题的关键是找出把谁看作抽屉个数，把谁看作物体个数，然后根据抽屉原理解答即可。\n\n【详解】  $18 \\div 6 = 3$  （个）即总有一个班至少分到 3 名同学。故答案为: √",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·河南信阳·期末",
+        "merged_id": "unit5_082",
+        "original_id": "unit5_choice_082"
+      },{
+        "id": "unit5_choice_083",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "（20-21六年级下·陕西商洛·期末）盒子里有同样大小的红球、黑球和白球各10个，要保证摸出的球一定有2个同色的，至少要摸出11个球。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】把这三种颜色看作三个抽屉，考虑最差情况：摸出3个球，每种颜色的球摸出1个，则再任意摸出一个，即可得出至少有一个抽屉出现两个球颜色相同。本题考查抽屉原理在实际问题中的灵活应用。\n\n【详解】根据分析可得：  $3 + 1 = 4$  （个）盒子里有同样大小的红球、黑球和白球各 10 个, 要保证摸出的球一定有 2 个同色的, 至少要摸出 4 个球。原题干说法错误。故答案为:×",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·陕西商洛·期末",
+        "merged_id": "unit5_083",
+        "original_id": "unit5_choice_083"
+      },
+      {
+        "id": "unit5_choice_084",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "(21-22 六年级下·陕西西安·期末) 把 20 个苹果放进 3 个果篮, 总有一个果篮中至少要放进 8 个苹果。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】从最坏的情况分析, 3 个果篮目前尽可能的平均放,即  $20 \\div 3 = 6$  （个）……2（个），即每个果篮放 6 个苹果，还剩下 2 个苹果，这两个苹果任意放 2 个果盘里，即总有一个果盘至少放  $6 + 1 = 7$  （个），据此判断。此题考查的是抽屉原理, 一定要从从最不利情况考虑。\n\n【详解】由分析可知：$2 0 \\div 3 = 6 (\\text {个}) \\dots \\dots 2 (\\text {个})$  $6 + 1 = 7 \\text {(个)}$总有一个果篮中至少要放进 7 个苹果。故答案为:  ×",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "21-22 六年级下·陕西西安·期末",
+        "merged_id": "unit5_084",
+        "original_id": "unit5_choice_084"
+      },
+      {
+        "id": "unit5_choice_085",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "（24-25六年级下·全国·单元测试）为了发展和培养同学们的能力，学校开设了航模、科技、漫画三个社团，规定每个学生最多可以参加其中的两个社团（也可不参加）。那么，至少有多少名学生，才能保证有不少于30名学生参加社团的情况完全相同？",
+        "options": [],
+        "answer": "204名",
+        "solution": "【分析】根据题意，学生参加社团的情况有：不参加社团的；只参加其中的一个社团的，有航模、科技、漫画3种；参加其中的两个社团的，有航模和科技、航模和漫画、科技和漫画3种。一共有  $1 + 3 + 3 = 7$  种情况。把这7种情况看作7个抽屉，从最不利情况考虑，每个抽屉需要放  $30 - 1 = 29$  （名）学生，共需要  $29 \\times 7 = 203$  （名），再增加1个学生不论参加什么社团，总有一个抽屉的学生数量是  $29 + 1 = 30$  （名），所以至少有  $203 + 1 = 204$  （名）学生，才能保证有不少于30名学生参加社团的情况完全相同。\n\n【详解】通过分析可得：\n\n$1 + 3 + 3 = 7 \\\\ (3 0 - 1) \\times 7 + 1 \\\\ = 2 9 \\times 7 + 1 \\\\ = 2 0 3 + 1 \\\\ = 2 0 4 \\text {(名)} $\n\n答: 至少有 204 名学生, 才能保证有不少于 30 名学生参加社团的情况完全相同。",
+        "knowledge_points": ["鸽巢问题初步"],
+        "source": "24-25六年级下·全国·单元测试",
+        "merged_id": "unit5_085",
+        "original_id": "unit5_choice_085"
+      },
+      {
+        "id": "unit5_choice_086",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "(2024 六年级下·全国·专题练习) 有 5050 张数字卡片,其中一张上写着 1, 2 张上写着 2, 3 张上写着 3, …, 100 张上写着 100。现在要从中抽取若干张，为了确保抽出的卡片中至少有 10 张以上的数字完全相同，至少要抽取多少张卡片？",
+        "options": [],
+        "answer": "865 张",
+        "solution": "【分析】从最不利的情况考虑，先把数量不足 10 张的  $1 - 9$  全部取完，再把剩下的数字都分别取了 9 张，最后再取 1 张就能确保抽出的卡片中至少有 10 张以上的数字完全相同。\n\n【详解】  $(1 + 2 + 3 + 4 + \\dots + 9) + (110 - 10 + 1) \\times 9 + 1\\\\= (1 + 9) \\times 9 \\div 2 + (1 1 0 - 1 0 + 1) \\times 9 + 1 \\\\ = 1 0 \\times 9 \\div 2 + 9 1 \\times 9 + 1 \\\\ = 4 5 + 8 1 9 + 1 \\\\ = 8 6 5 \\text {(张)} $\n\n答: 至少要抽取 865 张卡片。",
+        "knowledge_points": ["最不利原则"],
+        "source": "2024 六年级下·全国·专题练习",
+        "merged_id": "unit5_086",
+        "original_id": "unit5_choice_086"
+      },
+      {
+        "id": "unit5_choice_087",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "（2024 六年级下·全国·专题练习）一把钥匙只能开一把锁，现有 8 把钥匙和 8 把相配的锁，最多要试验几次能保证全部的钥匙和锁相匹配？",
+        "options": [],
+        "answer": "28次",
+        "solution": "【分析】从最不利的情况考虑，用 8 把钥匙去试第一把锁，最不利的情况是实验了 7 次，前 6 次都没有打开，第 7 次无论打开与否，都能确定这把锁匹配的钥匙；以此类推，第二把锁最多实验 6 次，第三把锁最多实验 5 次，……最后一把锁最多实验 1 次，据此用加法求出总次数。\n\n【详解】  $7 + 6 + 5 + 4 + 3 + 2 + 1 = 28$  （次）答: 最多要试验 28 次能保证全部的钥匙和锁匹配。",
+        "knowledge_points": ["最不利原则"],
+        "source": "2024 六年级下·全国·专题练习",
+        "merged_id": "unit5_087",
+        "original_id": "unit5_choice_087"
+      },
+      {
+        "id": "unit5_choice_088",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "(2024 六年级下·全国·专题练习) 按照星座学说, 根据出生时间不同，有十二个不同星座，请问至少找多少个同学，才能保证有四个人是同一个星座？",
+        "options": [],
+        "answer": "37个",
+        "solution": "【分析】把同学看作物品，星座看作抽屉，要保证至少有 4 个人在同一个抽屉，那么可以每个抽屉先放 3 个人，再在某一个抽屉中多放一个人。\n\n【详解】  $(4 - 1) \\times 12 + 1\\\\= 3 \\times 1 2 + 1 \\\\ = 3 6 + 1 \\\\ = 3 7 \\text {(个)}$\n\n答: 至少找 37 个同学, 才能保证有四个人是同一个星座。",
+        "knowledge_points": ["鸽巢问题初步"],
+        "source": "2024 六年级下·全国·专题练习",
+        "merged_id": "unit5_088",
+        "original_id": "unit5_choice_088"
+      },
+      {
+        "id": "unit5_choice_089",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "(23-24六年级下·山东菏泽·期中) 希望小学六年级准备开展'中华好诗词'活动，六（1）班有45名学生，男、女\n\n女生的人数比是  $3: 2$ , 从中随机选取, 至少选出多少人才能保证选出的学生中男、女生都有?",
+        "options": [],
+        "answer": "28人",
+        "solution": "【分析】根据题意可知，男、女生的人数比是  $3: 2$  ，由此可知，男生人数大于女生人数；男、女生的人数比是  $3: 2$  ，即男生和女生人数分成了  $3 + 2 = 5$  份，用六（1）班人数÷总份数，求出 1 份是多少，进而求出男生人数，如果必须保证选中的人有男有女，那么要作最坏的打算，即全是男生，把男生全部选完了，再选一定是女生，所以用男生人数 +1,即可解答。\n\n【详解】男、女生的人数比是  $3: 2$ , 男生人数  $>$  女生人数。  $3 + 2 = 5$  (份)\n\n男生： $45 \\div 5 \\times 3\\\\= 9 \\times 3 \\\\ = 2 7 \\text {(人)} $\n\n$2 7 + 1 = 2 8 \\text {(人)}$答: 至少选出 28 人才能保证选出的学生中男、女生都有。",
+        "knowledge_points": ["比的应用", "最不利原则"],
+        "source": "23-24六年级下·山东菏泽·期中",
+        "merged_id": "unit5_089",
+        "original_id": "unit5_choice_089"
+      },
+      {
+        "id": "unit5_choice_090",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "(23-24 六年级下·全国·课后作业) 7 名学生去图书馆借书, 图书馆有 A、B、C 三类图书, 每名学生最多可以借两本不同类的书，最少可以借一本，那么至少有几名学生所借书的种类完全相同？",
+        "options": [],
+        "answer": "2名",
+        "solution": "【分析】根据题意可知, 有 6 种不同的借书方式, 用 7 除以 6 可知商为 1 , 余数也为 1 , 用  $1 + 1$  即可知道至少有 2 名学生所借书的种类完全相同。\n\n【详解】  $7 \\div 6 = 1$  (组) ……1 (名)  $1 + 1 = 2$  （名） 答: 至少有 2 名学生所借书的种类完全相同。",
+        "knowledge_points": ["鸽巢问题初步"],
+        "source": "23-24 六年级下·全国·课后作业",
+        "merged_id": "unit5_090",
+        "original_id": "unit5_choice_090"
+      },
+      {
+        "id": "unit5_choice_091",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "（2023 六年级下·福建莆田·竞赛）请你证明：任意 5 个自然数中，必可找出 3 个数，使这三个数的和是 3 的倍数。",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】根据 3 的倍数的特征, 如果这三个数被 3 除, 余数的和是 3 的倍数, 那这三个数的和就是 3 的倍数。自然数被 3 除所得的余数有 0、1、2 三种情况, 据此把全体自然数分成 3 类, 即构成 3 个抽屉, 然后根据抽屉原理分情况讨论。结合3的倍数特征和抽屉原理进行分析，在有些抽屉问题中，'抽屉'和'物品'不是很明显，需要精心构造'抽屉'和'物品'。\n\n【详解】证明：自然数被 3 除所得的余数有 0、1、2 三种情况，据此把全体自然数分成 3 类，即构成 3 个抽屉。如果任选的 5 个自然数中：(1) 如果 3 个抽屉都有数, 根据抽屉原理,  $5 \\div 3 = 1 \\dots 2$ , 那要么 5 个数在 3 个抽屉中的分配方案必为 2 个, 2 个, 1 个; 每个抽屉中各取 1 个数, 余数的和为  $0 + 1 + 2 = 3$ , 因此, 它们的和一定是 3 的倍数。(2) 如果至少有一个抽屉没有数, 即将 5 个数放进 2 个抽屉或者 1 个抽屉, 根据抽屉原理,  $5 \\div 2 = 2 \\dots 1$ , 那么至少必有 3 个数在同一个抽屉, 在同一个抽屉里的三个数的和一定是 3 的倍数。所以，任意 5 个自然数中，必可找出 3 个数，使这三个数的和是 3 的倍数。",
+        "knowledge_points": ["3的倍数特征", "鸽巢问题进阶"],
+        "source": "2023 六年级下·福建莆田·竞赛",
+        "merged_id": "unit5_091",
+        "original_id": "unit5_choice_091"
+      },
+      {
+        "id": "unit5_choice_092",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "（2024·河南郑州·小升初真题）任意给出 3 个不同的自然数，其中一定有 2 个数的和是偶数，请说明理由。",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】把任意给出 3 个不同的自然数中, 是由偶数还是奇数组成的情况罗列出来, 再根据: 偶数十偶数 = 偶数, 偶数十奇数 = 奇数, 奇数十奇数 = 偶数, 解答即可。\n\n【详解】任意给出3个不同的自然数可能是：(1) 由 3 个奇数组成, 任意取 2 个数都是奇数, 奇数十奇数 = 偶数;  (2) 由 2 个奇数和一个偶数组成, 其中 2 个奇数的和: 奇数 + 奇数 = 偶数;  (3) 由 2 个偶数和一个奇数组成, 其中 2 个偶数的和: 偶数 + 偶数 = 偶数;  (4) 由 3 个偶数组成, 任意取 2 个数都是偶数, 偶数十偶数 = 偶数;所以任意给出 3 个不同的自然数, 其中一定有 2 个数的和是偶数。",
+        "knowledge_points": ["鸽巢问题进阶", "奇数与偶数的认识", "运算性质（奇数和偶数）"],
+        "source": "2024·河南郑州·小升初真题",
+        "merged_id": "unit5_092",
+        "original_id": "unit5_choice_092"
+      },
+      {
+        "id": "unit5_choice_093",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "(22-23 六年级下·全国·课前预习) 38 名学生进行答题游戏，每人答2道题，规定答对一题得2分，不答不得分，答错扣1分，则至少有几名学生的成绩相同？",
+        "options": [],
+        "answer": "7名",
+        "solution": "【分析】抽屉原理:  $m \\div n = a \\dots b (m > n > 1)$ , 把  $m$  个物体放进  $\\mathrm{n}$  个抽屉里, 不管怎么放总有一个抽屉至少放进 (a)(1) 个物体。2 道题全答对可得  $2 \\times 2 = 4$  (分); 1 道题答对, 另 1 道题不答, 可得  $2 \\times 1 = 2$  (分); 1 道题答对, 另 1道题答错, 可得  $2 \\times 1 - 1 \\times 1 = 1$  (分); 2 道题全不答可得 0分; 1 道题不答, 另 1 道题答错可得 -1 分; 2 道题全答错可得 -2 分。即物体数是 38, 抽屉数为 6。解决抽屉原理问题，要分清'要放的物体数和抽屉数'。\n\n【详解】  $38 \\div 6 = 6$  （名）…2（名）$6 + 1 = 7$  (名)答: 至少有 7 名学生的成绩相同。",
+        "knowledge_points": ["鸽巢问题进阶"],
+        "source": "22-23 六年级下·全国·课前预习",
+        "merged_id": "unit5_093",
+        "original_id": "unit5_choice_093"
+      },
+      {
+        "id": "unit5_choice_094",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "（22-23六年级下·全国·单元测试）学校开设了书法、舞蹈、棋类、乐器四个课外学习班，每个学生最多可以参加两个（可以不参加）学习班。某班有52名同学，至少有几名同学参加课外学习班的情况完全相同？",
+        "options": [],
+        "answer": "5人",
+        "solution": "【分析】本题同学参加情况共 11 种, 不参加、书法、舞蹈、 棋类、乐器、书法和舞蹈、书法和棋类、书法和乐器、舞 蹈和棋类、舞蹈和乐器、棋类和乐器; 这里可以把这 11 个情况看做 11 个抽屉, 考虑最差情况, 每个抽屉的人数尽量平均,  $52 \\div 11 = 4$  (人) ……8 (人), 每个抽屉都有 4 人,还剩下 8 人, 由此即可利用抽屉原理解决问题。此题考查了抽屉原理在实际问题中的灵活应用;根据题干, 找出学生参加学习班的所有可能情况, 是解决本题的关键。\n\n【详解】  $52 \\div 11 = 4$  （人）……8（人）$4 + 1 = 5$  （人）答：至少有5名同学参加课外学习班的情况完全相同。",
+        "knowledge_points": ["鸽巢问题进阶"],
+        "source": "22-23六年级下·全国·单元测试",
+        "merged_id": "unit5_094",
+        "original_id": "unit5_choice_094"
+      },
+      {
+        "id": "unit5_choice_095",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "（22-23六年级下·全国·单元测试）把104粒花生分给15只小猴，每只小猴都要分到花生，那么至少有两只小猴分得的花生一样多，为什么？",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】考虑最不利原则，假设前 13 只小猴分得的花生各不相同，从 1 一直加到 13 为 91 粒，还剩下 2 只小猴子分 13 粒花生，不管怎么分，至少有 2 只小猴分得的花生一样多。本题考查鸽巢问题, 采用最不利原则进行分析是解题的关键。\n\n【详解】假设前13只小猴分得的花生各不相同，共有：\n\n$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 1 0 + 1 1 + 1 2 + 1 3 \\\\ = (1 + 1 3) \\times 1 3 \\div 2 \\\\ = 1 4 \\times 1 3 \\div 2 \\\\ = 9 1 \\text {(粒)} $\n\n还剩下花生:  $104 - 91 = 13$  (粒) 还有小猴:  $15 - 13 = 2$  (只) 不管怎么分, 至少有 2 只小猴分得的花生一样多。\n\n答: 至少有 2 只小猴分得的花生一样多, 因为前 13 只小猴分得的花生各不相同后, 剩下的 2 只小猴不管怎么分剩下的 13 粒花生, 分得的花生粒数都只能是  $1 \\sim 12$  粒, 这样至少有 2 只小猴分得的花生一样多。",
+        "knowledge_points": ["最不利原则"],
+        "source": "22-23六年级下·全国·单元测试",
+        "merged_id": "unit5_095",
+        "original_id": "unit5_choice_095"
+      },
+      {
+        "id": "unit5_choice_096",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "（22-23六年级下·全国·单元测试）小悦，冬冬和阿奇到费叔叔家玩，费叔叔拿出许多巧克力来招待他们，他们一数，共有19块巧克力，如果把这些巧克力分给他们三人试说明：一定有人至少拿到7块巧克力，但不一定有人拿到8块。",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】把 3 人看作是 3 个抽屉, 19 块巧克力看做 19 个元素, 考虑最差情况: 把 19 块巧克力平均分配在 3 个抽屉中:  $19 \\div 3 = 6​$  (块) ……1 (块), 那么每个抽屉都有 6 块, 那么剩下的 1 块, 无论放到哪个抽屉都会出现 7 块在同一个抽屉里。此题属于典型的抽屉原理习题, 解答此类题的关键是找出把谁看作 '抽屉个数', 把谁看作 '物体个数',然后根据抽屉原理解答即可。\n\n【详解】  $19 \\div 3 = 6$  （块）……1（块）$6 + 1 = 7$  (块) 答: 所以一定有人至少拿到 7 块巧克力, 那么此时其他两个人分得 6 块, 所以不能保证一定有人拿到 8 块。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "22-23六年级下·全国·单元测试",
+        "merged_id": "unit5_096",
+        "original_id": "unit5_choice_096"
+      },
+      {
+        "id": "unit5_choice_097",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "(22-23 六年级下·全国·单元测试) 某单位购进 92 箱桔子, 每箱至少 110 个, 至多 138 个, 现将桔子数相同的作为一组, 箱子数最多的一组至少有几箱?",
+        "options": [],
+        "answer": "4箱",
+        "solution": "【分析】每箱装的个数在  $110 \\sim 138$  个, 从最不利的情况考虑, 最多有  $138 - 110 + 1 = 29$  种装箱情况, 把 29 种装箱情况看作 29 个抽屉, 把 92 箱看作 92 个元素, 那么每个抽屉需要放  $92 \\div 29 = 3$  (箱) ……5 (箱), 所以每个抽屉放剩下的 5 箱, 再不论怎么放, 总有一个抽屉里至少有:  $3 + 1 = 4$  箱, 所以, 现将桔子数相同的作为一组, 箱子数最多的一组至少有 4 箱, 据此解答。此题属于典型的抽屉原理习题, 解答此类题的关键是找出把谁看作 '抽屉个数', 把谁看作 '物体个数',然后根据抽屉原理解答即可。\n\n【详解】根据分析可得，  $138 - 110 + 1 = 29$  （种）$92 \\div 29 = 3$  （箱）……5（箱）$3 + 1 = 4$  (箱)答: 箱子数最多的一组至少有 4 箱。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "22-23 六年级下·全国·单元测试",
+        "merged_id": "unit5_097",
+        "original_id": "unit5_choice_097"
+      },
+      {
+        "id": "unit5_choice_098",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "(22-23 六年级下·全国·单元测试) 生活实践题。(1) 上学期有 18 名留守儿童插班进入实验小学就读, 将 18 名留守儿童编入 5 个班, 总有一个班至少要编入 4 名。为什么？(2) 18 名留守儿童来自全国的 4 个省份, 至少有 5 名来自同一个省份。为什么?  (3) 把 50 本图书分给 18 名留守儿童, 总有一名至少分到 3 本图书。为什么?",
+        "options": [],
+        "answer": "(1) (2) (3)见详解;",
+        "solution": "【分析】（1）5个班可以看作是5个抽屉，18名留守儿童看作18个元素，考虑最差情况：把18名留守儿童平均分配在5个抽屉中： $18 \\div 5 = 3$ （名）……3（名），那么每个抽屉都有3名，那么剩下的3名，无论放到哪个抽屉都会出现至少4名留守儿童在同一个抽屉里。(2) 4 个省份可以看作是 4 个抽屉, 18 名留守儿童看作 18 个元素, 考虑最差情况: 把 18 名留守儿童平均分配在 4 个抽屉中： $18 \\div 4 = 4$ （名）……2（名），那么每个抽屉都有4名，那么剩下的2名，无论放到哪个抽屉都会出现至少5名留守儿童在同一个抽屉里。(3) 18 名留守儿童可以看作是 18 个抽屉, 50 本图书看做 50 个元素, 考虑最差情况: 把 50 本图书平均分配在 18 个抽屉中:  $50 \\div 18 = 2$  (本) ……14 (本), 那么每个抽屉都有 2 本, 那么剩下的 14 本, 无论放到哪个抽屉都会出现至少 3 本图书在同一个抽屉里。此题属于典型的抽屉原理习题, 解答此类题的关键是找出把谁看作 '抽屉个数', 把谁看作 '物体个数',然后根据抽屉原理解答即可。\n\n【详解】 (1)  $18 \\div 5 = 3$  （名）……3（名）$3 + 1 = 4$  (名)答: 所以将 18 名留守儿童编入 5 个班, 总有一个班至少要编入 4 名。(2)  $18 \\div 4 = 4$  （名）……2（名）$4 + 1 = 5$  (名)答: 18 名留守儿童来自全国的 4 个省份, 所以至少有 5 名来自同一个省份。(3)  $50 \\div 18 = 2$  (本) ……14 (本) $2 + 1 = 3$  (本)答: 所以把 50 本图书分给 18 名留守儿童, 总有一名至少分到 3 本图书。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "22-23 六年级下·全国·单元测试",
+        "merged_id": "unit5_098",
+        "original_id": "unit5_choice_098"
+      },
+      {
+        "id": "unit5_choice_099",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "(2021·陕西渭南·小升初真题) 把 20 个西瓜放进 9 个筐里, 无论怎么放, 总有一个筐里至少放了 3 个西瓜, 为什么?",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】把 20 个西瓜看作被分放物体, 9 个筐看作抽屉,\n\n被分放物体的数量÷抽屉的数量 = 平均每个抽屉分放物体的数量……剩下物体的数量，一个抽屉里至少分放物体的数量 = 平均每个抽屉分放物体的数量 + 1，据此解答。本题主要考查应用抽屉原理解决实际问题, 准确找出抽屉数和被分放物体数是解答题目的关键。\n\n【详解】  $20 \\div 9 = 2$  （个）……2（个）  $2 + 1 = 3$  （个）答: 把 20 个西瓜放进 9 个筐里, 无论怎么放, 总有一个筐里至少放了 3 个西瓜。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "2021·陕西渭南·小升初真题",
+        "merged_id": "unit5_099",
+        "original_id": "unit5_choice_099"
+      },
+      {
+        "id": "unit5_choice_100",
+        "unit": "第五单元:鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "（21-22六年级下·山东济南·期末）'六一'儿童节，李老师拿133个小礼物发给班里的所有学生，如果至少有一名学生拿到了4个小礼物，那么，李老师班里最多有多少名学生？",
+        "options": [],
+        "answer": "44名",
+        "solution": "【分析】从最不利的情况考虑：只有一名学生拿到了 4 个小礼物，其他学生每人拿到了 3 个小礼物，那么小礼物的总个数减 1 刚好是 3 的倍数，此时学生的总人数 = （礼物总个数 -1）÷3，据此解答。本题主要考查鸽巢原理的应用, 从最不利情况考虑问题是解答题目的关键。\n\n【详解】  $(133 - 1) \\div 3\\\\= 1 3 2 \\div 3 \\\\ = 4 4 \\text {(名)} $\n\n答: 李老师班里最多有 44 名学生。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "21-22六年级下·山东济南·期末",
+        "merged_id": "unit5_100",
+        "original_id": "unit5_choice_100"
+      },
+      {
+        "id": "unit5_choice_101",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "把13本书放进4个抽屉里,不管怎么放,总有一个抽屉里至少放进( )本书。",
+        "options": [
+          "A. 3",
+          "B. 4",
+          "C. 5"
+        ],
+        "answer": "B",
+        "solution": "【分析】考虑最差情况:13本数平均分配给4个抽屉:$13 \\div 4 = 3 \\dots 1$,那么每个抽屉都有3本书,剩下的1本无论放到哪个抽屉,都会出现1个抽屉里面有4本书,据此解答。此题考查了抽屉原理的灵活应用,根据抽屉原理解答出正确结果,即可判断。\n【详解】$13 \\div 4 = 3$(本)……1(本) $3 + 1 = 4 \\text{(本)}$所以总有一个抽屉里至少放进4本书。故答案为:B",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "2022·天津·小升初真题",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_101",
+        "original_id": "unit5_choice_101"
+      },
+      {
+        "id": "unit5_choice_102",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "把$m+1$个物体放进$m$个抽屉里,总有一个抽屉里至少放进( )个物体($m$为大于0的自然数)。",
+        "options": [
+          "A. m",
+          "B. $m+1$",
+          "C. 2",
+          "D. 1"
+        ],
+        "answer": "C",
+        "solution": "【分析】根据鸽巢原理:如果把$(n+1)$个物体放在$n$个抽屉里,那么必有一个抽屉里至少放有两个物体;据此解答即可。本题主要考查了鸽巢原理的应用,关键是要理解巢原理。\n【详解】由分析可得:把$(\\mathrm{m}+1)$个物体放进$\\mathrm{m}$个抽屉里,总有一个抽屉放进2个物体。故答案为:C",
+        "knowledge_points": ["用字母表示数", "鸽巢问题"],
+        "source": "19-20六年级下·四川乐山·期末",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_102",
+        "original_id": "unit5_choice_102"
+      },
+      {
+        "id": "unit5_choice_103",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "13人中,( )有2个人在同一个月出生。",
+        "options": [
+          "A. 一定",
+          "B. 可能",
+          "C. 不可能"
+        ],
+        "answer": "A",
+        "solution": "【分析】一年只有12个月,13个人中哪怕平均分配一个人一个月过生日,也会多出来一个人,需要分配到任意一个月过生日,这就满足了一定出现的条件,当然,极端条件下也会出现十三个人都在一个月过生日,但这同样满足了其中有两个人在同一个月过生日的条件。据此解答。熟悉鸽巢问题的原理,能够把题目里的条件与鸽巢原理的元素相对应,是解题关键。\n【详解】根据分析得:$13 \\div 12 = 1 \\dots \\dots 1(\\text{人})$ $1+1=2$(人)13人中,一定有2个人在同一个月出生。故答案为:A",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21五年级上·广东广州·期中",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_103",
+        "original_id": "unit5_choice_103"
+      },
+      {
+        "id": "unit5_choice_104",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "幼儿园老师给10个孩子分香蕉,无论怎么分总有一个孩子至少得到2根香蕉,老师至少拿来( )根香蕉。",
+        "options": [
+          "A. 20",
+          "B. 21",
+          "C. 11"
+        ],
+        "answer": "C",
+        "solution": "【分析】根据抽屉原理,把10个孩子看作10个抽屉,要使每个孩子手里的香蕉尽量少,要尽量平均分,假设每个孩子手里都有1根香蕉,其中至少有1个孩子得到了2根香蕉,则$10+1=11$(根),由此即可解决问题。此题主要考查抽屉原理及其应用。注意逆向思考。\n【详解】假设每个孩子手里都有1根香蕉,其中至少有1个孩子得到了2根香蕉,则:$10 \\times 1 + 1 \\\\ = 10 + 1 \\\\ = 11 \\text{(根)}$故答案为:C",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "21-22六年级下·河北张家口·期中",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_104",
+        "original_id": "unit5_choice_104"
+      },
+      {
+        "id": "unit5_choice_105",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "有9本书分别放入4个抽屉,至少有一个抽屉放( )本。",
+        "options": [
+          "A. 2",
+          "B. 3",
+          "C. 4"
+        ],
+        "answer": "B",
+        "solution": "【分析】被分放物体的总个数$\\div$抽屉个数=平均每个抽屉放物体的个数……剩下物体的个数,每个抽屉里至少放物体的个数=平均每个抽屉放物体的个数+1,据此解答。找出题中的被分放物体的数量和抽屉数量是解答题目的关键。\n【详解】$9 \\div 4 = 2 \\dots \\dots 1$ $2+1=3$(本)所以,至少有一个抽屉放3本书。故答案为:B",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20六年级下·湖南怀化·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_105",
+        "original_id": "unit5_choice_105"
+      },
+      {
+        "id": "unit5_choice_106",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "向东小学六(2)班有学生50人,六(2)班至少有( )人是在同一个月出生的。",
+        "options": [
+          "A. 4",
+          "B. 5",
+          "C. 6"
+        ],
+        "answer": "B",
+        "solution": "【分析】将12个月看成12个'抽屉',将50名学生看成50个元素,根据抽屉问题解答即可。本题主要考查抽屉问题的简单应用。\n【详解】$50 \\div 12 = 4$(人)……2(人) $4+1=5 \\text{(人)}$所以六(2)班至少有5人是在同一个月出生的。故答案为:B",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·假期作业",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_106",
+        "original_id": "unit5_choice_106"
+      },
+      {
+        "id": "unit5_choice_107",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "从一个口袋中摸球,如果每次摸4个,总有2个颜色相同,那么口袋中球的颜色最多有( )。",
+        "options": [
+          "A. 2种",
+          "B. 3种",
+          "C. 4种"
+        ],
+        "answer": "B",
+        "solution": "【分析】从最坏的情况考虑,又摸出1个才有2个颜色相同,说明前面摸出的3个球颜色都不相同,也就是最多有3种颜色。本题考查了抽屉原理问题之一,它的解答思路是:要从最不利情况考虑,准确地建立抽屉和确定元素的总个数,然后根据'至少数=抽屉的个数+1'解答。\n【详解】$4-1=3$(种)故答案为:B。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_107",
+        "original_id": "unit5_choice_107"
+      },
+      {
+        "id": "unit5_choice_108",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "有6种颜色的小球,从中至少取出( )个才能保证有5个球颜色相同。",
+        "options": [
+          "A. 6",
+          "B. 7",
+          "C. 11",
+          "D. 25"
+        ],
+        "answer": "D",
+        "solution": "【分析】此题主要考查了抽屉原理的应用,考虑最差情况:摸出$4 \\times 6 = 24$个球,分别是6种不同的颜色的球各4个,那么再任意摸出1个球,一定可以保证有5个球颜色相同,据此列式解答。从最坏的情况考虑是解决此题的关键。\n【详解】$6 \\times 4 + 1 \\\\ = 24 + 1 \\\\ = 25 \\text{(个)}$有6种颜色的小球,从中至少取出25个才能保证有5个球颜色相同。故答案为:D",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_108",
+        "original_id": "unit5_choice_108"
+      },
+      {
+        "id": "unit5_choice_109",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "任意15个中国人,至少有( )个人的属相一样。",
+        "options": [
+          "A. 2",
+          "B. 3",
+          "C. 4",
+          "D. 5"
+        ],
+        "answer": "A",
+        "solution": "【分析】一共有12个属相,从不利的情况考虑,如果15个人中有12个人分别是这12个属相,那么剩下的人无论是哪个属相都能保证至少有2个人的属相一样。本题考查鸽巢问题,解答本题的关键是掌握利用最不利原则。\n【详解】$15 \\div 12 = 1$……3 $1+1=2$(人)故答案为:A。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_109",
+        "original_id": "unit5_choice_109"
+      },
+      {
+        "id": "unit5_choice_110",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "把红、黄、蓝、白、绿五种颜色的球各10个放到一个袋子里,要保证取到两个颜色相同的球,至少要取出几个球?( )",
+        "options": [
+          "A. 6",
+          "B. 5",
+          "C. 4",
+          "D. 3"
+        ],
+        "answer": "A",
+        "solution": "【分析】要想保证2个球颜色相同,考虑最不利的情况,把每种颜色的球都取一遍,那么再取一个就能保证2个球颜色相同。根据最坏原理进行分析是完成本题的关键。\n【详解】$5+1=6$(个)故答案为:A。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_110",
+        "original_id": "unit5_choice_110"
+      },
+      {
+        "id": "unit5_choice_111",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "六(1)班有50名同学,至少( )人的生日在同一个月。",
+        "options": [
+          "A. 4",
+          "B. 5",
+          "C. 6",
+          "D. 12"
+        ],
+        "answer": "B",
+        "solution": "【分析】从最坏的情况考虑,如果每个月都有4个人过生日,那么剩下的2人无论在哪个月生日,都至少有5个人的生日在同一个月。此题考查简单的抽屉问题,至少数$=$商+1(有余数的情况下)。\n【详解】一年有12个月。$50 \\div 12 = 4(\\text{个}) \\dots \\dots 2(\\text{个})$ $4+1=5 \\text{(个)}$所以至少5个人的生日在同一个月。故答案为:B。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_111",
+        "original_id": "unit5_choice_111"
+      },
+      {
+        "id": "unit5_choice_112",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "袋子中有红、黄、蓝球各4个,至少任意拿出( )个球,才能保证某种颜色的球有2个。",
+        "options": [
+          "A. 3",
+          "B. 4",
+          "C. 5",
+          "D. 7"
+        ],
+        "answer": "B",
+        "solution": "【分析】把3种不同颜色看作3个抽屉,从最不利情况考虑,每个抽屉先放1个球一共需要3个球,再取出1个不论是什么颜色,总有一个抽屉里的球和它同色,所以至少要取出:$3+1=4$(个),据此解答。把多于$n+1$个的物体放进$n$个抽屉里,至少有一个抽屉里放了2个物体。\n【详解】$3+1=4$(个)故答案为:B。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·课后作业",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_112",
+        "original_id": "unit5_choice_112"
+      },
+      {
+        "id": "unit5_choice_113",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "一个口袋里装有红、黄、蓝3种不同颜色的小球各10各,要摸出的球一定有2个同色的,最少要摸( )个。",
+        "options": [
+          "A. 10",
+          "B. 11",
+          "C. 4"
+        ],
+        "answer": "C",
+        "solution": "【分析】把3种不同颜色看作3个抽屉,把3种不同颜色的球看作元素,从最不利情况考虑,每个抽屉先放1个球,共需要3个,再取出1个不论是什么颜色,总有一个抽屉里的球和它同色,所以至少要取出:$3+1=4$(个),据此解答。本题考查鸽巢问题,解答本题的关键是掌握抽屉原理解决问题的方法。\n【详解】根据分析可得,$3+1=4 \\text{(个)}$所以要摸出的球一定有2个同色的,最少要摸4个。故答案为:C。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·课后作业",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_113",
+        "original_id": "unit5_choice_113"
+      },
+      {
+        "id": "unit5_choice_114",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "把7本书放进3个抽屉里,不管怎么放,总有一个抽屉里至少放进( )本书。",
+        "options": [
+          "A. 1",
+          "B. 2",
+          "C. 3",
+          "D. 4"
+        ],
+        "answer": "C",
+        "solution": "【分析】把7本书放进3个抽屉里,$7 \\div 3 = 2$(本)……1(本),平均每个抽屉放入2本后还余一本书没有放入,至少有一个抽屉里要放进$2+1=3$本书。此题主要考查利用抽屉原理解决实际问题,至少数=物体数除以抽屉数的商+1(有余数的情况下)。\n【详解】$7 \\div 3 = 2$(本)……1(本) $2+1=3$(本)所以,总有一个抽屉至少会放进3本书。故答案为:C",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_114",
+        "original_id": "unit5_choice_114"
+      },
+      {
+        "id": "unit5_choice_115",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "把红、黄、蓝、绿四种同样大小的小球各5个放在同一箱子里,一次至少要摸出( )个球才能保证摸出2个红球。",
+        "options": [
+          "A. 5",
+          "B. 20",
+          "C. 17"
+        ],
+        "answer": "C",
+        "solution": "【分析】有红、黄、蓝、绿四种同样大小的小球各5个,最坏的情况是先摸出的全是黄、蓝、绿球$5 \\times 3 = 15$个,则剩下的全是红球,因此只要再多摸出2个,就一定有2个红球,所以一次至少要摸出$15+2=17$个球,才能保证有2个红球。此题考查了利用抽屉原理解决实际问题的灵活应用,关键是从最差情况考虑。\n【详解】$5 \\times 3 + 2 \\\\ = 15 + 2 \\\\ = 17(个)$;故答案为:C。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_115",
+        "original_id": "unit5_choice_115"
+      },
+      {
+        "id": "unit5_choice_116",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "一个盒子里装有同样大小的红球、黄球、白球各3个。至少取出( )个球,才能保证取到两个颜色相同的球。",
+        "options": [
+          "A. 3",
+          "B. 4",
+          "C. 5"
+        ],
+        "answer": "B",
+        "solution": "【分析】把3种不同颜色看作3个抽屉,把不同颜色的球看作元素,从最不利情况考虑,每个抽屉需要先放1个球,共需要3个,再取出1个不论是什么颜色,总有一个抽屉里的球和它同色,所以至少要取出:$3+1=4$(个),据此解答。本题考查了抽屉原理问题之一,它的解答思路是:要从最不利情况考虑,准确地建立抽屉和确定元素的总个数,然后根据'至少数=抽屉的个数+1'解答。\n【详解】$3+1=4$(个)至少取出4个球,才能保证取到两个颜色相同的球。故选:B。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_116",
+        "original_id": "unit5_choice_116"
+      },
+      {
+        "id": "unit5_choice_117",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "某小学有61名学生在4月份出生,至少有( )名学生在同一天过生日。",
+        "options": [
+          "A. 2",
+          "B. 3",
+          "C. 4",
+          "D. 5"
+        ],
+        "answer": "B",
+        "solution": "【分析】4月份有30天,$61 \\div 30 = 2$(名)……1(名),最坏的情况是,每天都有2名学生过生日的话,还余1名,余下的1名无论是哪天过生日,这样一天都至少有$2+1=3$名在同一天过生日,据此解答即可。在此类抽屉问题中,至少数=物体数除以抽屉数所得的商+1(有余数的情况下)。\n【详解】$61 \\div 30 = 2$(名)……1(名);$2+1=3$(名);故答案为:B。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_117",
+        "original_id": "unit5_choice_117"
+      },
+      {
+        "id": "unit5_choice_118",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "袋子里有红、黄、蓝、绿四种颜色的球各5个,至少要摸( )个球才能保证摸出的球中有两个颜色相同。",
+        "options": [
+          "A. 4",
+          "B. 5",
+          "C. 8",
+          "D. 10"
+        ],
+        "answer": "B",
+        "solution": "【分析】由题意可知,有红、黄、蓝、绿四种颜色的球,要保证至少有2个颜色相同,最坏的情况是每种颜色各摸出1个,即摸出4个,此时只要再任摸出一个,即摸出$4+1=$5个就能保证至少有2个球颜色相同。此题考查了利用抽屉原理解决实际问题的灵活应用,关键是从最差情况考虑。\n【详解】$4+1=5$(个)至少要摸5个球才能保证摸出的球中有两个颜色相同。故选:B。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_118",
+        "original_id": "unit5_choice_118"
+      },
+      {
+        "id": "unit5_choice_119",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "某班39名同学,其中至少有( )名同学出生日期的月份相同。",
+        "options": [
+          "A. 3",
+          "B. 4",
+          "C. 5",
+          "D. 6"
+        ],
+        "answer": "B",
+        "solution": "【分析】一年有12个月,那么把这12个月看作12个抽屉,要求至少有多少名同学在同一个月过生日,可以考虑最差情况:39名尽量平均分配在12个抽屉中,利用抽屉原理即可解答。解答此类题的关键是:找出把谁看作'抽屉个数',把谁看作'物体个数'。\n【详解】$39 \\div 12 = 3$(名)……3(名);$3+1=4$(名);故答案为:B。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_119",
+        "original_id": "unit5_choice_119"
+      },
+      {
+        "id": "unit5_choice_120",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "把红、黄、蓝、白、黑五种颜色的球各8个放到一个袋子里,至少取( )个球,就能保证取到两个颜色相同的球。",
+        "options": [
+          "A. 2",
+          "B. 6",
+          "C. 9"
+        ],
+        "answer": "B",
+        "solution": "【分析】考虑最差情况,取出5个球分别为红、黄、蓝、白、黑五种颜色,再取一个就能保证取到两个颜色相同的球,据此解答即可。本题考查了抽屉原理,要从最差的情况入手考虑。\n【详解】$5+1=6$(个);故答案为:$\\mathrm{B}$。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20六年级下·陕西宝鸡·期末",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_120",
+        "original_id": "unit5_choice_120"
+      },
+      {
+        "id": "unit5_choice_121",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "某小学共有368名学生,该小学里至少有( )名学生在同一天过生日。",
+        "options": [],
+        "answer": "2",
+        "solution": "【分析】一年最多有366天,$368 \\div 366 = 1$(名)……2(名),最坏的情况是,每天都有1名学生过生日的话,还余2名学生,根据抽屉原理,总有至少$1+1=2$名学生在同一天过生日。在此类抽屉问题中,至少数=物体数除以抽屉数所得的商+1(有余数的情况下)。\n【详解】$368 \\div 366 = 1$(名)……2(名) $1+1=2$(名)即该小学里至少有2名学生在同一天过生日。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "2022·河北廊坊·小升初真题",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_121",
+        "original_id": "unit5_choice_121"
+      },
+      {
+        "id": "unit5_choice_122",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "王老师给全家人买衣服,有红、黄、蓝三种颜色,但结果总是至少有两人的颜色一样,她家里至少有( )口人。",
+        "options": [],
+        "answer": "4/四",
+        "solution": "【分析】把颜色的种类看作'抽屉',把人数看作物体的个数,根据抽屉原理得出:人数至少比颜色的种类多1时,才能至保证少有两个人的颜色一样;据此解答。本题考查鸽巢原理,解答此类题的关键是找出把谁看作抽屉个数,把谁看作物体个数。\n【详解】根据分析:$3+1=4 \\text{(口)}$所以她家里至少有4口人。",
+        "knowledge_points": ["鸽巢问题初步"],
+        "source": "2022·山东菏泽·小升初真题",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_122",
+        "original_id": "unit5_choice_122"
+      },
+      {
+        "id": "unit5_choice_123",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "新乡市6月份的天气有晴天、阴天、小雨、多云四种情况,新乡市6月份有( )天是同一种天气。",
+        "options": [],
+        "answer": "8",
+        "solution": "【分析】四种天气就是四个抽屉,6月有30天,从最坏的情况考虑,假如四种天气各有7天,则剩下的2天无论是什么天气,都至少有8天是同一种天气。此题的解题关键是灵活运用抽屉原理解决实际的问题。\n【详解】$30 \\div 4 = 7$(天)……2(天) $7+1=8$(天)",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "21-22六年级下·河南新乡·期末",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_123",
+        "original_id": "unit5_choice_123"
+      },
+      {
+        "id": "unit5_choice_124",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "抽屉里面放了3双颜色不同的袜子,在不看颜色的情况下,至少取出( )只袜子,才能保证一定取出1双颜色相同的袜子。",
+        "options": [],
+        "answer": "4或四",
+        "solution": "【分析】考虑最不利情况:取出颜色不同的袜子各1只,需要先取出3只袜子,此时再任意取出1只袜子,一定有1双颜色相同的袜子,据此解答。本题主要考查利用抽屉原理解决实际问题,考虑最不利情况是解答题目的关键。\n【详解】分析可知,在不看颜色的情况下,至少取出4只袜子,才能保证一定取出1双颜色相同的袜子。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "21-22六年级下·湖北武汉·期末",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_124",
+        "original_id": "unit5_choice_124"
+      },
+      {
+        "id": "unit5_choice_125",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "小明把红、黄、蓝、白四种颜色的球各8个放到一个袋子里。他至少要取( )个球,才可以保证取到两个颜色相同的球。",
+        "options": [],
+        "answer": "5",
+        "solution": "【分析】题目中已知鸽巢数量(4种颜色即4个鸽巢)和分的结果(保证一个鸽巢里至少有2个同色的),求要分放物体的数量,用鸽巢数加1来计算。已知鸽巢数量和分的结果,求要分放物体的数量,可以用'鸽巢数$+1=$分放物体的数量'来计算。解答本题要注意,各种颜色小球的数量并不参与运算。\n【详解】4种颜色即4个鸽巢,保证一个鸽巢里至少有2个同色的,至少要取的球的个数是:$4+1=5$(个)。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "2020·湖北黄冈·小升初真题",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_125",
+        "original_id": "unit5_choice_125"
+      },
+      {
+        "id": "unit5_choice_126",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "把13本书放进4个抽屉,总有一个抽屉里至少有( )本书。",
+        "options": [],
+        "answer": "4",
+        "solution": "【分析】把13本书放进4个抽屉,$13 \\div 4 = 3$(本)……1(本),即平均每个抽屉放入3本后,还余一本书没有放入,即至少有一个抽屉里要放进$3+1=4$本书。在此类抽屉问题中,至少数=物体数除以抽屉数的商+1(有余数的情况下)。\n【详解】$13 \\div 4 = 3$(本)……1(本) $3+1=4 \\text{(本)}$",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "2022·广西百色·小升初真题",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_126",
+        "original_id": "unit5_choice_126"
+      },
+      {
+        "id": "unit5_choice_127",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "一副扑克牌有四种花色(大小王除外)、每种花色有13张、从中任意抽牌、最少抽( )张牌,才能保证5张牌是同一花色。",
+        "options": [],
+        "answer": "17",
+        "solution": "【分析】根据最不利情况考虑,每种花色都抽了4张,共16张,再抽1张,才能保证5张牌是同一花色。本题考查鸽巢问题,解答本题的关键是掌握鸽巢问题中的数量关系。\n【详解】$4 \\times 4 + 1 \\\\ = 16 + 1 \\\\ = 17(张)$",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "2022·四川凉山·小升初真题",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_127",
+        "original_id": "unit5_choice_127"
+      },
+      {
+        "id": "unit5_choice_128",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "有11只鸽子飞进4个鸽舍,至少有( )只鸽子要飞进同一个鸽舍。",
+        "options": [],
+        "answer": "3",
+        "solution": "【分析】把4个鸽舍看作4个抽屉,把11只鸽子看作11个元素,那么每个抽屉需要放$11 \\div 4 = 2$(只)……3(只),所以每个抽屉有2只,剩下的3只鸽子不论怎么放,总有一个抽屉里至少有:$2+1=3$(只),所以,至少有3只鸽子要飞进同一个鸽舍;据此解答。此题属于典型的抽屉原理习题,根据抽屉原理进行解答即可。\n【详解】$11 \\div 4 = 2$(只)……3(只) $2+1=3(\\text{只})$所以至少有3只鸽子要飞进同一个鸽舍里。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "21-22六年级下·全国·假期作业",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_128",
+        "original_id": "unit5_choice_128"
+      },
+      {
+        "id": "unit5_choice_129",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "生日在3月份的任意32名同学中,至少有( )人的生日是同一天。",
+        "options": [],
+        "answer": "2",
+        "solution": "【分析】3月份有31天,把这31天看作31个抽屉,把32名同学看作32个元素,利用抽屉原理,至少有2人的生日是同一天。根据抽屉原理,至少在同一抽屉里相同物体的个数=物体总个数÷抽屉的个数+1。\n【详解】$32 \\div 31 = 1$(人)……1(人) $1+1=2 \\text{(人)}$",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "21-22六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_129",
+        "original_id": "unit5_choice_129"
+      },
+      {
+        "id": "unit5_choice_130",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "六(1)班有40名学生,年龄最大的13岁年龄最小的12岁,那么其中必有( )名同学是同年同月出生的。",
+        "options": [],
+        "answer": "2",
+        "solution": "【分析】12岁、13岁共2个年龄段,每个年龄段12个月,因此两个年龄段共24个月。这40个学生分别在这24个月出生,先平均每个月放1名学生,那么还余下16名学生,无论放在哪一个月,都会有2名同学是同年同月出生的。本题考查鸽巢问题,采用最不利原则解题。\n【详解】两个年龄段共有月份:$12 \\times 2 = 24$(个) $40 \\div 24 = 1$(名)……16(名) $1+1=2$(名)所以其中必有2名同学是同年同月出生的。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "2022六年级下·全国·专题练习",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_130",
+        "original_id": "unit5_choice_130"
+      },
+      {
+        "id": "unit5_choice_131",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "袋子中有红、黄、蓝三色球各15个,从中依次取出球,如果保证取到两种颜色的球,至少需要取( )个。",
+        "options": [],
+        "answer": "16",
+        "solution": "【分析】利用抽屉原理,考虑最差情况:取出15个球,都是同一种颜色的球,此时再任意取出1个球,一定是另一种颜色的球,此时即可保证取到两种颜色的球。此题考查了利用抽屉原理解决实际问题的灵活应用,要注意考虑最差情况。\n【详解】$15+1=16$(个)",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "2021·云南曲靖·小升初真题",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_131",
+        "original_id": "unit5_choice_131"
+      },
+      {
+        "id": "unit5_choice_132",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "把15个学生分到6个组,总有一个组至少有( )人。",
+        "options": [],
+        "answer": "3",
+        "solution": "【分析】考虑最差情况,$15 \\div 6 = 2$(个)$\\dots 3$(个),说明每个组平均分2个学生,还余下3个,$2+1=3$,总有一个组至少有3人,据此解答即可。本题考查了抽屉原理的灵活利用,解答本题时要从最差情况入手考虑。\n【详解】$15 \\div 6 = 2$(个)……3(个);$2+1=3$(个)。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·湖南娄底·期末",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_132",
+        "original_id": "unit5_choice_132"
+      },
+      {
+        "id": "unit5_choice_133",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "把10本书放进9个抽屉里,不管怎么放,总有一个抽屉里至少放进( )本书。",
+        "options": [],
+        "answer": "2",
+        "solution": "【分析】抽屉原则一:如果把$(n+1)$个物体放在$n$个抽屉里,那么必有一个抽屉中至少放有2个物体。关键是构造物体和抽屉,也就是找到代表物体和抽屉的量,然后依据抽屉原则进行分析。\n【详解】把10本书放进9个抽屉里,不管怎么放,总有一个抽屉里至少放进2本书。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·新疆·期末",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_133",
+        "original_id": "unit5_choice_133"
+      }, {
+        "id": "unit5_choice_134",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "把7本书放进3个抽屉里，不管怎么放，总有一个抽屉里至少放进( )本书。",
+        "options": [],
+        "answer": "3",
+        "solution": "【分析】如果把 n 个物体放在 m 个抽屉里, 其中 n > m, 那么必有一个抽屉至少有:(1) 当 n 不能被 m 整除时, k = [n/m] + 1 个物体。 (2) 当 n 能被 m 整除时, k = n/m 个物体。此题考查鸽巢原理应用。\n【详解】把 7 本书放进 3 个抽屉里，平均每个抽屉放 2 本还剩下一本书，不管放在哪个抽屉都会使那个抽屉里面书的本书比 2 本多一本，所以总有一个抽屉至少放进 3 本书。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "2015·全国·小升初真题",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_134",
+        "original_id": "unit5_choice_134"
+      },
+      {
+        "id": "unit5_choice_135",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "妈妈准备了7只信封，在每只信封里都放了钱共100元，要求每一只信封里都放整元数，而且都不相同，那么钱放得最多的一只信封里至少放____元。",
+        "options": [],
+        "answer": "18",
+        "solution": "【分析】这题有多种方法, 只要每一袋的数不同就可以了,但题中要求'最多的一袋至少放多少', 那么必须是这 7 袋的数是非常接近的, 把 100 分成接近的数, 所以每袋是十几元, 根据个位数的和是 30 元, 结果是: 11 + 12 + 13 + 14+ 15 + 17 + 18 = 100 (元), 最多的一袋至少是 18 元。抽屉原则二：如果把 n 个物体放在 m 个抽屉里，其中 n > m, 那么必有一个抽屉至少有: ① k = [n ÷ m] + 1 个物体: 当 n 不能被 m 整除时; ② k = n ÷ m 个物体: 当 n 能被 m 整除时。\n【详解】11 + 12 + 13 + 14 + 15 + 17 + 18 = 100 （元）妈妈准备了 7 只信封，在每只信封里都放了钱，共 100 元，要求每一只信封里都放整元数，而且都不相同，那么钱放得最多的一只信封里至少放 18 元。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_135",
+        "original_id": "unit5_choice_135"
+      },
+      {
+        "id": "unit5_choice_136",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "从一副扑克牌（54 张）中抽出____张来，才能保证一定有一张是黑桃。",
+        "options": [],
+        "answer": "42",
+        "solution": "【分析】一副扑克牌 4 种花色加两个王，抽出红桃，方块，梅花各 13 张，再加上 2 张大小王后，只剩下黑桃了，最后再抽一张黑桃，就能保证一定有一张是黑桃。典型的'鸽巢问题'，解题的关键是要从最糟糕的情况去考虑，也可称为'最不利原则'，从而保证某种情况出现。\n【详解】13 × 3 + 1 + 2\\= 3 9 + 1 + 2 \\ = 4 0 + 2 \\ = 4 2 (张)。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21 六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_136",
+        "original_id": "unit5_choice_136"
+      },
+      {
+        "id": "unit5_choice_137",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "把同样大小的红、黑、白三种颜色的球各 9 个放在同一个盒子里, 要想摸出的球一定有 2 个同色的, 至少要摸出_个球。",
+        "options": [],
+        "answer": "4",
+        "solution": "【分析】考虑最不利原理，要想摸出的球一定有 2 个同色的，最不利的情况是每种颜色都摸一个出来，然后再任意摸一个出来，那么必定有 2 个球是同色的。根据最坏原理进行分析是完成本题的关键。\n【详解】3 + 1 = 4 （个）",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21 六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_137",
+        "original_id": "unit5_choice_137"
+      },
+      {
+        "id": "unit5_choice_138",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "4 支铅笔放入 3 个笔筒中，不管怎么放，总有一个笔筒里至少要放____支铅笔。",
+        "options": [],
+        "answer": "2",
+        "solution": "【分析】从最不利的情况考虑，如果每个笔筒中各放 1 支铅笔，那么剩下的 1 支无论放进哪个笔筒里都总有一个笔筒里至少放 2 支铅笔。此题考查简单的抽屉问题，在此类题目中，至少数 = 商十余数。\n【详解】4 ÷ 3 = 1 (支) ……1 (支), 1 + 1 = 2 (支) 所以: 总有一个笔筒里至少要放 2 支铅笔。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21 六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_138",
+        "original_id": "unit5_choice_138"
+      },
+      {
+        "id": "unit5_choice_139",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "有红色和蓝色的水彩笔各 5 支，如果闭着眼睛，至少拿出____支，才能保证拿出的水彩笔中有 2 支是同色的。",
+        "options": [],
+        "answer": "3",
+        "solution": "【分析】考虑最坏的情况，即第一次拿红色，第2次拿蓝色，第三次不管是拿红色还是蓝色，都有两支是同色的。此题考查简单的抽屉问题, 根据最坏原理进行分析是完成本题的关键。\n【详解】2 + 1 = 3 （支）所以至少拿出 3 支, 才能保证拿出的水彩笔有 2 支是同色的。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21 六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_139",
+        "original_id": "unit5_choice_139"
+      },
+      {
+        "id": "unit5_choice_140",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "希望小学共有368名学生，其中六年级有48名。希望小学至少有( )名学生的生日是同一天，六年级中至少有( )名学生是同一个月出生的。",
+        "options": [],
+        "answer": "2;4",
+        "solution": "【分析】（1）平年有 365 天，闰年有 366 天，即使是闰年，将 366 天当作抽屉，368 ÷ 366 = 1（名）……2（名），即平均每天有一名学生过生日的话，还余 2 名学生，根据抽屉原理可知，至少有 1 + 1 = 2 名学生的生日是同一天； (2) 一年有 12 个月, 那么把这 12 个月看作 12 个抽屉, 要求至少有多少名同学在同一个月过生日, 可以考虑最差情况: 48 名尽量平均分配在 12 个抽屉中, 利用抽屉原理即可解答。此题考查了利用抽屉原理解决实际问题的灵活应用, 关键是从最差情况考虑, 至少数 = 物体数除以抽屉数的商 +1 (有余数的情况下)。\n【详解】(1) 368 ÷ 366 = 1 （名）……2（名）1 + 1 = 2 (名)希望小学至少有 2 名学生的生日是同一天。(2) 建立抽屉: 一年有 12 个月分别看作 12 个抽屉, 48 ÷ 12 = 4 (名)六年级中至少有 4 名学生是同一个月出生的。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_140",
+        "original_id": "unit5_choice_140"
+      },
+      {
+        "id": "unit5_choice_141",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "某小学六一班有50名学生，这个班至少有( )名学生是在同一个月出生的；若转走2名学生，这时，班上至少有( )名学生是在同一个月出生。",
+        "options": [],
+        "answer": "5; 4",
+        "solution": "【分析】一年一共有 12 个月, 将 50 名学生先尽量平均分到 12 个月中, 50 ÷ 12 = 4 … 2, 每个月 4 个人还多两人, 所以至少有 4 + 1 = 5 名学生在同一个月出生; 转走 2 名, 剩下 50 - 2 = 48 (名), 48 ÷ 12 = 4, 刚好每个月 4 个人, 所以至少 4 名学生在同一个月出生。本题主要考查了抽屉原理，需要学生充分理解抽屉原理的计算方法。\n【详解】50 ÷ 12 = 4 （名）……2（名）4 + 1 = 5 (名) 所以，至少有 5 名学生同一个月出生；50 - 2 = 48 (名)48 ÷ 12 = 4 (名)所以，转走 2 名后，至少有 4 人在同一个月出生。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_141",
+        "original_id": "unit5_choice_141"
+      },
+      {
+        "id": "unit5_choice_142",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "奋发小学六（1）班有 55 个同学参加智力游戏，若任意分成四组，则必然有一组的女生多于 2 人，又知参与者中任意 10 人中必有男生，则参与者中女生的人数是____人。",
+        "options": [],
+        "answer": "9",
+        "solution": "【分析】若任意分成四组，则必然有一组的女生多于2人，则说明女生至少是 2 × 4 + 1 = 9 人, 又因为参与者中任意 10 人中必有男生, 所以女生最多是 9 人, 由此可以确定女生人数是 9 人, 据此即可解答问题。此题考查了抽屉原理在实际问题中的灵活应用。\n【详解】根据题干分析可得，若任意分成四组，则必然有一组的女生多于2人，则说明女生至少是 2 × 4 + 1 = 9 （人），又因为参与者中任何10人必有男生，所以女生最多是9人；",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21 六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_142",
+        "original_id": "unit5_choice_142"
+      },
+      {
+        "id": "unit5_choice_143",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "一个盒子里放着 3 个红球、4 个蓝球、5 个黄球（这些球除颜色不同外，其余完全相同，如果从盒子中摸出的球保证含有三种颜色，那么至少要摸出_个球。",
+        "options": [],
+        "answer": "10",
+        "solution": "【分析】从最极端情况考虑：假设把4个蓝球和5个黄球都摸出, 这时只剩下 3 个红球, 再摸一个球, 一定是红球, 就能确保含有三种颜色; 由此解答即可。此题属于抽屉原理，应从最极端情况分析，进而得出结论。\n【详解】4 + 5 + 1 = 10 （个）",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21 六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_143",
+        "original_id": "unit5_choice_143"
+      },
+      {
+        "id": "unit5_choice_144",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "10 只鸽子飞回 3 个鸽舍, 至少有( )只鸽子要飞进同一个鸽舍里。",
+        "options": [
+          "A. 3",
+          "B. 4",
+          "C. 5"
+        ],
+        "answer": "B",
+        "solution": "【分析】将 3 个鸽舍看出 3 个抽屉, 10 只鸽子为 10 个元素, 10 ÷ 3 = 3 （个）……1（个），3 + 1 =4（个），据此解答即可。本题考查了抽屉原理, 至少数 = 元素总个数 ÷ 抽屉的个数 +1 (有余数的情况下)。\n【详解】10 只鸽子飞回 3 个鸽舍, 至少有 4 只鸽子要飞进同一个鸽舍里。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "2022·贵州黔东南·小升初真题",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_144",
+        "original_id": "unit5_choice_144"
+      },
+      {
+        "id": "unit5_choice_145",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "从 1 ~ 8 这 8 个自然数中, 至少要取出( )个不同的数, 才能保证其中必定有两个数成倍数关系。",
+        "options": [
+          "A. 3",
+          "B. 4",
+          "C. 5"
+        ],
+        "answer": "C",
+        "solution": "【分析】取出的数不含倍数关系的有(2,3,5,7)、(3,4,5,7)、(3,5,7,8)等最多四个数，而当第5个数出现时，立刻就有两个数成倍数关系。本题是抽屉原理试题, 设置好抽屉数, 用最不利原则分析问题是解答本题的关键。\n【详解】由分析知:4 + 1 = 5 (个)",
+        "knowledge_points": ["鸽巢问题", "因数和倍数的认识"],
+        "source": "20-21 六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_145",
+        "original_id": "unit5_choice_145"
+      },
+      {
+        "id": "unit5_choice_146",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "在 13 名 2020 年出生的儿童中，至少有( )人是同一个月出生的。",
+        "options": [],
+        "answer": "2",
+        "solution": "【分析】一年有 12 个月, 共有 13 名儿童, 若每个月都有一名儿童出生, 则还剩一名儿童, 剩余的这名儿童无论是几月份出生, 总有一个月份, 至少有 2 名儿童出生。抽屉原理问题的重点是建立抽屉，关键是在考虑最差情况的基础上得出均份数（商）；然后根据：至少数 = 商 + 1 （在有余数的情况下）。\n【详解】13 ÷ 12 = 1 （人）……1（人）1 + 1 = 2  (人)",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21 六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_146",
+        "original_id": "unit5_choice_146"
+      },
+      {
+        "id": "unit5_choice_147",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "把 7 只鸽子放进 3 个鸽笼，总有一个鸽笼至少放进( )只鸽子。",
+        "options": [],
+        "answer": "3",
+        "solution": "【分析】7 ÷ 3 = 2 （只）……1（只），把7只鸽子平均放进3个鸽笼里，每个鸽笼里放2只，还剩1只。剩余的1只无论如何放，总有一个鸽笼至少有 2 + 1 = 3 （只）鸽子。抽屉原理问题的重点是建立抽屉，关键是在考虑最差情况的基础上得出均份数（商）；然后根据：至少数 = 商 + 1 （在有余数的情况下）。\n【详解】7 ÷ 3 = 2 （只）……1（只）2 + 1 = 3 (只)",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21 六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_147",
+        "original_id": "unit5_choice_147"
+      },
+      {
+        "id": "unit5_choice_148",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "小红参加象棋比赛, 胜一盘得 3 分, 平一盘得 1 分, 负一盘不得分, 小红已得了 7 分, 她至少下了( )盘。",
+        "options": [
+          "A. 2",
+          "B. 3",
+          "C. 4"
+        ],
+        "answer": "B",
+        "solution": "【分析】小红已得了 7 分, 要使得小红下的盘数尽可能少,那么获胜的盘数就尽可能多, 胜 2 盘是 6 分, 还差 1 分, 平局得 1 分, 刚好是 7 分。本题考查的是数的分拆, 要把 7 拆成尽可能少的数,那么每个数就尽可能大。\n【详解】胜2盘，平1盘，2 × 3 + 1 = 7 （分）所以小红至少下了 3 盘。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21 六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_148",
+        "original_id": "unit5_choice_148"
+      },
+      {
+        "id": "unit5_choice_149",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "盒子里有同样大小的红球和黄球各5个，想摸出的球一定有2个同色的，最少要摸出（）个；一个袋子里装了同样大小的红、黑、白玻璃球各2个，要保证取出的玻璃球三种颜色都有，至少要取出（）个，要保证取出的玻璃球中至少有两种颜色，至少应取出（）个。",
+        "options": [],
+        "answer": "3 5 3",
+        "solution": "【分析】想摸出的球一定有 2 个同色的, 可以是两个红球,也可以是两个黄球, 先取出一个红球, 一个黄球, 然后再任取一个, 一定符合要求; 要保证取出的玻璃球三种颜色都有, 可以先把两种颜色的球取完, 完后再取一个球, 一定符合要求; 要保证取出的玻璃球中至少有两种颜色, 可以先把一种颜色的球取完, 再取一个, 一定符合要求。本题考查的是最不利原则，先找出不符合要求的最大数量，然后再加上 1，得出符合要求的最低要求。\n【详解】盒子里有同样大小的红球和黄球各 5 个, 想摸出的球一定有 2 个同色的, 最少要摸出 3 个球; 袋子里装了同样大小的红、黑、白玻璃球各 2 个, 要保证取出的玻璃球三种颜色都有, 至少要取出 5 个球; 袋子里装了同样大小的红、黑、白玻璃球各 2 个, 要保证取出的玻璃球中至少有两种颜色, 至少应取出 3 个球。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_149",
+        "original_id": "unit5_choice_149"
+      },
+      {
+        "id": "unit5_choice_150",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "13只鸽子飞进4个鸽笼，总有一个鸽笼至少飞进4只鸽子。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】13只鸽子飞进4个鸽笼，13÷ 4 = 3(只)…1(只), 即平均每个鸽笼飞入 3 只鸽子后, 还有 1 只鸽子没有飞入, 因此总有一个鸽笼至少飞进 3 + 1 = 4 只, 据此解答。此题的是典型的抽屉问题，至少数=物体数除以抽屉数的商 +1 （有余数的情况下）。\n【详解】13 ÷ 4 = 3 （只）……1（只）3 + 1 = 4 (只) 故答案为: √",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "2022·云南昭通·小升初真题",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_150",
+        "original_id": "unit5_choice_150"
+      },
+      {
+        "id": "unit5_choice_151",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "11 只鸽子飞进 4 个鸽笼, 总有一个鸽笼里至少飞进 5 只鸽子。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】在此类抽屉问题中，至少数 = 被分配的物体数 ÷ 抽屉数的商 +1 （有余数的情况下）。在本题中，被分配的物体数是 11，抽屉数是 4，据此计算即可。抽屉原理问题的解答思路是：要从最不利情况考虑，准确地建立抽屉和确定元素的总个数，然后根据'至少数 = 元素的总个数 ÷ 抽屉的个数 +1 （有余数的情况下）'解答。\n【详解】11 ÷ 4 = 2 （只）……3（只）2 + 1 = 3 (只) 11 只鸽子飞进 4 个鸽笼, 总有一个鸽笼里至少飞进 3 只鸽子。原题说法错误。故答案为: ×",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "2022·河北秦皇岛·小升初真题",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_151",
+        "original_id": "unit5_choice_151"
+      },
+      {
+        "id": "unit5_choice_152",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "把10名学生分到4个小组，至少有2人要分进同一个小组。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】要求至少数，用物体数除以抽屉数，求出商，用商 +1 就是至少数，据此解答即可。本题考查鸽巢问题，解答本题的关键是掌握解决鸽巢问题的方法。\n【详解】10 ÷ 4 = 2 （名）…2（名）2 + 1 = 3  (名)即至少有 3 名要分进同一个小组; 所以原题说法错误。故答案为: ×。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "21-22六年级下·全国·假期作业",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_152",
+        "original_id": "unit5_choice_152"
+      },
+      {
+        "id": "unit5_choice_153",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "把10个苹果分给7个小朋友，其中有一个小朋友至少会分到3个。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】把 7 个小朋友看作 7 个抽屉, 把 10 个苹果看作 10 个元素, 从最不利情况考虑, 每个抽屉先放 1 个, 共需要 7 个, 余下的这 3 个苹果无论放在哪个抽屉里, 总有一个抽屉里的有 1 + 1 = 2 (个), 据此解答。此题的解题关键是掌握鸽巢问题的解题原理, 构造物体和抽屉。\n【详解】10 ÷ 7 = 1 （个）……3（个）1 + 1 = 2 (个)故答案为: ×",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_153",
+        "original_id": "unit5_choice_153"
+      },
+      {
+        "id": "unit5_choice_154",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "5人进入4间客房，至少有2人住进同一间客房。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】被分放物体的数量÷抽屉的数量 = 平均每个抽屉分放物体的数量……剩下物体的数量，一个抽屉里至少分放物体的数量 = 平均每个抽屉分放物体的数量 + 1，据此解答。掌握鸽巢原理问题的解题方法是解答题目的关键。\n【详解】5 ÷ 4 = 1 … 1   1 + 1 = 2 (人)  所以，5人进入4间客房，至少有2人住进同一间客房。故答案为: √",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20六年级下·湖南怀化·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_154",
+        "original_id": "unit5_choice_154"
+      },
+      {
+        "id": "unit5_choice_155",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "因为 21 ÷ 6 = 3 ……3，所以21个梨放进6个盘子里，总有1个盘子至少放进6个梨。（ ）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】每个盘子里放梨的个数 = 每个盘子里平均放的个数 + 1, 据此解答。本题主要考查'鸽巢问题', 又称'抽屉原理',用梨的总个数除以盘子数量先求出商, 结果有余数时, 至少数 = 商 +1; 结果没有余数时: 至少数 = 商。\n【详解】21 ÷ 6 = 3 （个）……3（个）3 + 1 = 4 (个) 所以，总有 1 个盘子至少放进 4 个梨。故答案为: ×",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21 六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_155",
+        "original_id": "unit5_choice_155"
+      },
+      {
+        "id": "unit5_choice_156",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "有 7 本书放入 2 个抽屉, 有一个抽屉至少放 4 本书。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】假设每个抽屉都放进 3 本书, 那么余下的一本放进任意一个抽屉, 至少有一个抽屉里有 4 本书, 由此判断即可。此题考查简单的抽屉问题，在此类题目中，至少数 = 商十余数。\n【详解】7 ÷ 2 = 3 … 1 , 余下的一本无论放进哪个抽屉里,都有一个抽屉至少放 4 本书, 原题说法正确。故正确答案为: √。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21 六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_156",
+        "original_id": "unit5_choice_156"
+      },
+      {
+        "id": "unit5_choice_157",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "某地五月份天气有晴、阴、小雨三种天气, 至少有 11 天是同一种天气。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】因为每年的五月份都有 31 天, 假设其中有 10 天晴天、10 天阴天, 有 10 天小雨, 另外一天必和其中的一种天气一样。所以, 至少有 11 天天气一样。此题考查简单的抽屉问题, 分清 31 天看作物体总个数, 三种天气情况看作 3 个抽屉, 解答方法为: 至少数 = 商 +1 (有余数的情况下)。\n【详解】31 ÷ 3 = 10 （天）……1（天）；10 + 1 = 11 （天）。故答案为: √ 。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21 六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_157",
+        "original_id": "unit5_choice_157"
+      },
+      {
+        "id": "unit5_choice_158",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "任意13人中，至少有2人是在同一个月出生的。（ ）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】一年有 12 个月, 把这 12 个月看做 12 个抽屉, 把13 个人看做 13 个元素, 由此利用抽屉原理即可解答。此题考查了利用抽屉原理解决实际问题的灵活应用，这里要注意考虑最差情况。\n【详解】13 ÷ 12 = 1 （人）……1（人）；1 + 1 = 2 （人），至少有 2 人是同一个月出生的。故答案为: √",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_158",
+        "original_id": "unit5_choice_158"
+      },
+      {
+        "id": "unit5_choice_159",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "困难",
+        "question": "要保证从一副完整的扑克牌（54 张）中，抽到一张黑桃至少要抽取 42 张。()",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】一副扑克牌有 54 张, 每种花色都有 13 张牌, 把这四种花色看作四个抽屉, 考虑最差情况: 红桃、方块、梅花、大小王都全部抽出, 则再任意抽出一张, 必定是黑桃,据此即可解答问题。此题考查了利用抽屉原理解决实际问题的灵活应用，关键是从最差情况考虑。\n【详解】13 × 3 + 2 + 1\\= 3 9 + 2 + 1 \\ = 4 2 (张)即要抽出 42 张来, 才能保证一定有一张黑桃; 所以原题说法正确。故答案为: √",
+        "knowledge_points": ["最不利原则"],
+        "source": "20-21 六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_159",
+        "original_id": "unit5_choice_159"
+      },
+      {
+        "id": "unit5_choice_160",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "一个袋子中装有红、黄、白三种颜色的球各 8 个，至少要摸出 8 个球才能保证摸出的球中至少有 4 个球的颜色相同。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】由题意可知, 盒子里装有红、黄、白三种颜色的球, 要保证至少有四个小球的颜色相同, 最坏的情况是每种颜色各取出 3 个, 即取出 9 个中, 3 个红色, 3 个黄色的, 3 个白球, 此时只要再任取一个, 即取出 3 × 3 + 1 = 10 个就能保证至少有四个小球的颜色相同。此题考查了利用抽屉原理解决实际问题的灵活应用，关键是从最差情况考虑。\n【详解】3 × 3 + 1\\= 9 + 1 \\ = 1 0 (个);至少摸出 10 个才能保证有 4 个小球的颜色相同, 所以原题说法错误。故答案为:×。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21 六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_160",
+        "original_id": "unit5_choice_160"
+      },
+      {
+        "id": "unit5_choice_161",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "老师把 36 副羽毛球拍分给 5 个班，至少有 7 副羽毛球拍分给同一个班。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】把 5 个班看作 5 个抽屉, 把 36 副羽毛球拍看作 36 个元素, 从最不利情况考虑, 36 ÷ 5 = 7 (副) …… 1 (副), 每个抽屉先放 7 副, 共需要 35 副, 余这 1 副无论放在那个抽屉里, 总有一个抽屉里的有 7 + 1 = 8 (副), 据此解答。抽屉原理问题的思路是：根据'至少数 = 元素的总个数 ÷ 抽屉的个数 +1 （有余数的情况下）'解答。\n【详解】36 ÷ 5 = 7 （副）…1（副）7 + 1 = 8 (副)即至少有 8 副羽毛球拍分给同一个班, 所以原题说法错误。故答案为: × 。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21 六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_161",
+        "original_id": "unit5_choice_161"
+      },
+      {
+        "id": "unit5_choice_162",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "把 7 支钢笔放进 2 个笔盒中，总有一个笔盒至少要放进 4 支钢笔。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】把 7 支钢笔放进 2 个文具盒中, 7 ÷ 2 = 3 (支) …… 1 (支), 即平均每个文具盒里放 3 支, 还余 1 支, 根据抽屉原理可知, 总有一个文具盒里至少放 3 + 1 = 4 支。在此类抽屉问题中，至少数 = 物体数除以抽屉数的商 +1（有余数的情况下）。\n【详解】7 ÷ 2 = 3 （支）……1（支）3 + 1 = 4 (支)即总有一个文具盒至少放进 4 枝钢笔; 所以原题说法正确。故答案为:√ 。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21 六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_162",
+        "original_id": "unit5_choice_162"
+      },
+      {
+        "id": "unit5_choice_163",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "某校六年级共有 368 名同学, 至少有两个人的生日是同一天。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】一年最多有 366 天, 将这 366 天当作 366 个抽屉,由于 368 ÷ 366 = 1 (人) ……2 (人), 即平均每天有 1 人过生日, 还余 2 人, 根据抽屉原理可知, 至少有 1 + 1 = 2 人是同一天过生日。在此类抽屉问题中，至少数 = 被分配的物体数除以抽屉数的商 +1（有余的情况下）。\n【详解】368 ÷ 366 = 1 （人）……2（人）；1 + 1 = 2 (人) ;即至少 2 人同一天过生日，所以原题说法正确。故答案为: √ 。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21 六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_163",
+        "original_id": "unit5_choice_163"
+      },
+      {
+        "id": "unit5_choice_164",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "六（1）班有学生52人，全班至少有5人在同一个月过生日。这种说法对吗？为什么？",
+        "options": [],
+        "answer": "对；原因见详解",
+        "solution": "【分析】一年有 12 个月, 把月份看作抽屉数, 把学生人数看作被分放物体数, 被分放物体的数量 ÷ 抽屉的数量 = 平均每个抽屉分放物体的数量……剩下物体的数量, 一个抽屉里至少分放物体的数量 = 平均每个抽屉分放物体的数量 + 1 , 据此解答。找准抽屉的数量和被分放物体的数量是解答此类问题的关键。\n【详解】52 ÷ 12 = 4 （人）……4（人）4 + 1 = 5 (人)答: 全班至少有 5 人在同一个月过生日, 所以这种说法对。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "2022·山东菏泽·小升初真题",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_164",
+        "original_id": "unit5_choice_164"
+      },
+      {
+        "id": "unit5_choice_165",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "有外形相同的红、黄、绿三色球各10个。混合放入同一布袋中。一次至少摸几个球，才能保证有两种颜的同色球各一对？",
+        "options": [],
+        "answer": "13 个",
+        "solution": "【分析】由题意可知, 袋中有红、黄、绿 3 种颜色的球, 要保证有两个球是同色球, 最差情况是一次摸出的 3 个球中, 红、黄、绿 3 种颜色各一个, 此时只要再任意摸出一个即摸出 4 个球, 就能保证有两个球是同色球。最坏的打算是摸出 10 个, 都是同一种颜色的, 那再摸 2 个,又是 2 种颜色, 那再摸一个, 就能保证有两种颜色的同色球各一对, 进而计算得出结论。根据抽屉原理中的最差情况进行分析是完成本题的关键。\n【详解】10 + 2 + 1 = 13 （个）答: 一次至少摸 13 个球, 才能保证有两种颜色的球各一对。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "21-22六年级下·全国·假期作业",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_165",
+        "original_id": "unit5_choice_165"
+      },
+      {
+        "id": "unit5_choice_166",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "某校六年级有 320 人,这些同学中，至少有多少名同学在同一月过生日？为什么？",
+        "options": [],
+        "answer": "至少有 27 名同学在同一月过生日, 因为无论怎么样剩余的同学都会在 12 个月其中一个月里生日。",
+        "solution": "【分析】因一年有 12 个月, 320 ÷ 12 = 26 （名）……8（名），最差情况是 26 名在一个月过生日，还余 8 名，根据抽屉原理，至少 26 + 1 = 27 人在同一个月过生日。在此抽屉问题中，至少数=物体数除以抽屉数的商 +1 (有余的情况下)。\n【详解】320 ÷ 12 = 26 （名）……8（名）剩下的 8 名同学, 无论怎么样都会在 12 个月其中一个月里生日2 6 + 1 = 2 7 (名)答: 至少有 27 名同学在同一月过生日。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "21-22 六年级下·全国·期末",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_166",
+        "original_id": "unit5_choice_166"
+      },
+      {
+        "id": "unit5_choice_167",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "小东家有三种花纹不同的筷子,小东吃早饭时要去拿一双花纹一样的筷子。假如他闭上眼睛,至少要拿几根筷子,才能保证拿到一双花纹相同的筷子？",
+        "options": [],
+        "answer": "4根",
+        "solution": "【分析】从最不利的情况考虑,如果前3次刚好拿出三种花纹的筷子各1根,那么再拿出1根无论是什么花纹,都能保证拿到一双花纹相同的筷子。根据抽屉原理中的'最不利原则'进行分析是完成本题的关键。\n【详解】3+1=4(根)答:至少要拿4根筷子,才能保证拿到一双花纹相同的筷子。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "2022六年级下·全国·专题练习",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_167",
+        "original_id": "unit5_choice_167"
+      },
+      {
+        "id": "unit5_choice_168",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "六年一班有55个学生,每个学生参加篮球、足球、排球中的两项活动,那么至少多少人参加的活动项目相同？",
+        "options": [],
+        "answer": "19人",
+        "solution": "【分析】由题意可知,每个学生可以选择参加篮球和足球,篮球和排球,足球和排球,一共3种不同的选择方案,把55个学生看作被分放物体数,3种不同的选择方案看作抽屉数,被分放物体的数量÷抽屉的数量=平均每个抽屉分放物体的数量……剩下物体的数量,一个抽屉里至少分放物体的数量=平均每个抽屉分放物体的数量+1,据此解答。准确找出被分放物体数量和抽屉数量是解答题目的关键。\n【详解】分析可知,被分放物体的数量为55,抽屉的数量为3。55÷3=18(人)……1(人) 18+1=19(人)答:至少19人参加的活动项目相同。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "2022六年级下·全国·专题练习",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_168",
+        "original_id": "unit5_choice_168"
+      },
+      {
+        "id": "unit5_choice_169",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "7个小朋友相约去看电影,共有《哈利·波特》、《驯龙高手》、《功夫熊猫》三部电影可选择,每个小朋友可选一个电影组合(不重复的两部电影)观看,至少有几个小朋友选的电影组合相同？",
+        "options": [],
+        "answer": "3个",
+        "solution": "【分析】先列出所有可能的两组电影组合,再用抽屉原理将7个小朋友分配。本题考查抽屉原理。\n【详解】每个小朋友的观影方式有3种:《哈利·波特》和《驯龙高手》、《哈利·波特》和《功夫熊猫》、《驯龙高手》和《功夫熊猫》,相当于3个抽屉。将7个小朋友看成苹果,根据平均分配的思想:7÷3=2(个)……1(个),根据抽屉原理:2+1=3(个)。答:至少有3个小朋友选的电影组合相同。",
+        "knowledge_points": ["搭配问题", "鸽巢问题"],
+        "source": "2022六年级下·全国·专题练习",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_169",
+        "original_id": "unit5_choice_169"
+      },
+      {
+        "id": "unit5_choice_170",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "11个苹果放进3个抽屉,苹果最多的一个抽屉里至少有几个苹果？",
+        "options": [],
+        "answer": "4个",
+        "solution": "【分析】根据抽屉原理,要使每个抽屉里的苹果尽量少,要尽量平均分,即11÷3=3(个)……2(个),由此即可解决问题。在此类抽屉问题中,至少数=物体数除以抽屉数的商+1(有余数的情况下)。\n【详解】11÷3=3(个)……2(个) 3+1=4(个)答:苹果最多的一个抽屉里至少有4个苹果。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "2022六年级下·全国·专题练习",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_170",
+        "original_id": "unit5_choice_170"
+      },
+      {
+        "id": "unit5_choice_171",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "刘渊参加飞镖比赛,投了7镇,成绩是57环,刘渊至少有一镖不低于9环,对吗?为什么?",
+        "options": [],
+        "answer": "对,理由见详解",
+        "solution": "【分析】不低于就是大于等于,因为57÷7=8…1,就是说至少有一镖大于等于9环。如果都小于九环,成绩就会小于等于56环,据此即可解答。此题也可用用假设法:若7镇都低于9环,最多环数是7×8=56(环),所以至少一镖要大于等于9。\n【详解】57÷7=8…1  8+1=9(环) 7×8=56(环)答:所以至少有一镖大于等于9环。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "21-22六年级下·内蒙古通辽·期中",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_171",
+        "original_id": "unit5_choice_171"
+      },
+      {
+        "id": "unit5_choice_172",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "把红、黄、蓝、黑、白五种颜色的筷子各9根放在一个盒子里。至少取多少根才能保证一定有2根颜色相同的筷子？",
+        "options": [],
+        "answer": "6根",
+        "solution": "【分析】把5种不同颜色看作5个抽屉,把不同颜色的筷子看作元素,从最不利情况考虑,每个抽屉需要先放1根筷子,共需要5根,再取出1根不论是什么颜色,总有一个抽屉里的筷子和它同色,所以至少要取出:5+1=6(根),据此解答。\n【详解】5+1=6(根)答:至少取6根才能保证一定有2根颜色相同的筷子。\n【点睛】本题考查了抽屉原理问题之一,它的解答思路是:要从最不利情况考虑,准确地建立抽屉和确定元素的总个数,然后根据'至少数=抽屉的个数+1'解答。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "2021·云南德宏·小升初真题",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_172",
+        "original_id": "unit5_choice_172"
+      },
+      {
+        "id": "unit5_choice_173",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "在100张卡片上不重复地编写上1~100,请问至少要随意抽出几张卡片才能保证所抽出卡片上的数相乘后之乘积可被4整除？",
+        "options": [],
+        "answer": "52张",
+        "solution": "【分析】若想确保若干个数的乘积能被4整除,就要先抽出这些数中所有的奇数,再抽2张偶数卡片即可,1~100中所有的奇数有50个,若一开始就抽中的50张奇数卡片,则还需要抽出2张偶数卡片,它们之积才能被4整除。本题考查的是最不利原则,解题的关键是需要找出能被4整除的数的特征,从1~100中的数抽取,即可解答。\n【详解】50+2=52(个)先取出1~100中所有的奇数,一共50个;至少还需要取出两个偶数,共52个数,这52个数的乘积一定可以被4整除。答:至少要随意抽出52张卡片。",
+        "knowledge_points": ["最不利原则"],
+        "source": "20-21六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_173",
+        "original_id": "unit5_choice_173"
+      },
+      {
+        "id": "unit5_choice_174",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "五年级有47名学生参加一次数学竞赛,成绩都是整数,满分是100分,已知3名学生的成绩在60分以下,其余学生的成绩均在75~95分之间,问:至少有几名学生的成绩相同？",
+        "options": [],
+        "answer": "3名",
+        "solution": "【分析】此题主要考查了抽屉原理的应用,解题的关键是弄清抽屉数量,根据条件'成绩都是整数,已知3名学生的成绩在60分以下,其余学生的成绩均在75~95分之间',可以计算出75~95之间的整数有几个,也就是有几个抽屉,然后用总人数-3=剩下的学生总数,将剩下的学生总数放入抽屉中,根据抽屉原理的解题方法:a个物体放入n个抽屉,如果a÷n=b…c,那么有一个抽屉至少放(b+1)个物体,据此解答。关键是构造物体和抽屉,也就是找到代表物体和抽屉的量,然后依据抽屉原则进行计算。\n【详解】75~95之间的整数有95-75+1=21(个) 47-3=44(名) 44÷21=2…2 2+1=3(名)答:至少有3名学生的成绩相同。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_174",
+        "original_id": "unit5_choice_174"
+      },
+      {
+        "id": "unit5_choice_175",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "夏令营有500个学生参加,请问在这些学生中,至少有多少人在同一天过生日？至少有多少人在同一个月过生日？",
+        "options": [],
+        "answer": "至少2人同一天过生日;至少42人同一月过生日。",
+        "solution": "【分析】(1)一年多有366天,假如每天都有1人过生日,那么余下的人数无论在哪一天过生日都能保证至少有2人在同一天过生日;(2)一年有12个月,假如每个月都有41人过生日,那么余下的人数无论在哪一月过生日,都能保证至少有42人同一个月过生日。本题考查鸽巢问题,解答本题的关键是掌握抽屉原理解决问题。\n【详解】500÷366=1…134 1+1=2(人) 500÷12=41…8 41+1=42(人)答:至少2人同一天过生日;至少42人同一月过生日。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·课后作业",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_175",
+        "original_id": "unit5_choice_175"
+      },
+      {
+        "id": "unit5_choice_176",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "叔叔参加飞镖比赛,投了5镖,成绩是41环。张叔叔至少有一镖不低于9环。为什么？",
+        "options": [],
+        "answer": "因为叔叔投了5镖,成绩是41环,从最不利情况考虑,叔叔前4镖都投8环,第5镖至少要投9环才能保证环数是41环,即张叔叔至少有一镖不低于9环。",
+        "solution": "【分析】不低于就是大于等于,因为41÷5=8…1,就是说至少有一镖大于等于9环。如果都小于九环,成绩就会小于等于40环,据此即可解答。本题考查鸽巢问题,解答本题的关键是掌握抽屉原理。\n【详解】41÷5=8…1 8+1=9(环)答:因为叔叔投了5镖,成绩是41环,从最不利情况考虑,叔叔前4镖都投8环,第5镖至少要投9环才能保证环数是41环,即张叔叔至少有一镖不低于9环。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·课后作业",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_176",
+        "original_id": "unit5_choice_176"
+      },
+      {
+        "id": "unit5_choice_177",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "纸箱里杂乱地放着黑、白、红、绿、黄五种颜色的袜子各50只,规格都相同。在黑暗中至少要取出多少只袜子,才能保证有15双颜色相同的袜子？",
+        "options": [],
+        "answer": "146只",
+        "solution": "【分析】15双就是30只,考虑最不利原则,五种颜色,每种都摸到29只,怎么办呢,那就随便再摸一只,因为不管摸到什么色,都可以跟前面的29相加,到30了,这样就能保证有15双颜色相同的袜子。本题考查鸽巢问题,解答本题的关键是掌握抽屉原理解决问题。\n【详解】5×29+1=145+1=146(只)答:在黑暗中至少要取出146只袜子,才能保证有15双颜色相同的袜子。",
+        "knowledge_points": ["最不利原则"],
+        "source": "20-21六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_177",
+        "original_id": "unit5_choice_177"
+      },
+      {
+        "id": "unit5_choice_178",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "7个小朋友乘6只小船游玩,至少要有多少个小朋友坐在同一只小船里,为什么？",
+        "options": [],
+        "answer": "2个",
+        "solution": "【分析】把6只船看做6个抽屉,考虑最差情况:7个小朋友,最差情况是:每只船上分的人相等,7÷6=1(个)……1(个);那剩下1人,随便分给哪一只船,都会使得一只船分得1+1=2人,据此解答。抽屉原理问题的重点是建立抽屉,关键是在考虑最差情况的基础上得出均份数(商);然后根据:至少数=商+1(在有余数的情况下)求解。\n【详解】7÷6=1(个)……1(个);1+1=2(个);答:至少要2个小朋友坐在同一只小船里。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_178",
+        "original_id": "unit5_choice_178"
+      },
+      {
+        "id": "unit5_choice_179",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "一副扑克牌有54张,最少要抽几张牌,方能保证其中至少有3张牌有相同的点数？",
+        "options": [],
+        "answer": "29张",
+        "solution": "【分析】建立抽屉:一副扑克牌有54张,大小王不相同,那么(54-2)÷4=13,所以一共有13+2=15个抽屉;分别是:1、2、3、…K、小王、大王,由此利用抽屉原理考虑最差情况,即可进行解答。此题考查了利用抽屉原理解决实际问题的灵活应用,关键是从最差情况考虑。\n【详解】建立抽屉:54张牌,根据点数特点可以分别看作15个抽屉,考虑最差情况:小王、大王先抽取,剩下的每个抽屉都抽取了2张牌,共抽出13×2=26张牌,此时再任意抽取1张,就有3张牌点数相同,所以最少要抽取:2+13×2+1=2+26+1=29(张)答:最少要抽29张牌,方能保证其中至少有3张牌有相同的点数。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_179",
+        "original_id": "unit5_choice_179"
+      },
+      {
+        "id": "unit5_choice_180",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "在一个直径为2m的圆形花坛周围放上7盆花,那么至少有2盆花之间的距离不超过1米,为什么?(提示:可以通过计算后画图说明)",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】根据题干分析可得,7盆花一共有7-1=6个间隔,根据抽屉原理,从最差情况考虑:使每个间隔的长度尽量的平均,则每个间隔的长度最少是6.28÷6=1.04米,由此即可解答。此题原型属于抽屉原理,关键是根据7盆花求出间隔数是6,即得出6个抽屉,再利用抽屉原理即可解答。\n【详解】2×3.14=6.28(米);7-1=6(个);每个间隔平均是6.28÷6=1.04(米);把这6个间隔看作6个抽屉,把7盆花放在6个抽屉里,总能保证至少有一个抽屉里有两盆花,即至少有2盆花的距离不超过1米。![](images/49b1f30d3e7fb475b527289519a2d1026994a1db900ab7ee6c2ac674afb5f85d.jpg)",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_180",
+        "original_id": "unit5_choice_180"
+      },
+      {
+        "id": "unit5_choice_181",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "数学竞赛,填空题8道,答对1道,得4分,未答对。得0分;问答题6道。答对1道。得7分,未答对,得0分。参赛人数400人。至少有多少人的总分相同?",
+        "options": [],
+        "answer": "8人",
+        "solution": "【分析】假设填空题对的是x道,问答题对的是y道,总分应为4x+7y,0≤x≤8,0≤y≤6,且x,y为整数,y=0,1,2,3,总分分别有9种不同情况,y=4,5,6,总分有7种情况(要与之前不同,即x≠0,1),即共有4×9+3×7=57种情况,所以一共有57种分值,即57个抽屉,据此解答即可。此题考查了抽屉原理的基本解决方法,关键是找到抽屉的数量。\n【详解】400÷57=7(人)……1(人) 7+1=8(人)答:至少有8人的总分相同。",
+        "knowledge_points": ["鸽巢问题初步"],
+        "source": "20-21六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_181",
+        "original_id": "unit5_choice_181"
+      },
+      {
+        "id": "unit5_choice_182",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "将红、绿、黄三种颜色的筷子各5根混放在一起,如果闭上眼睛,最少拿多少根筷子就一定能保证拿出的筷子里至少有两根是同色的？请说明你的理由。",
+        "options": [],
+        "answer": "4根;见详解",
+        "solution": "【分析】从最不利的情况考虑,如果取出的头3根分别是3种颜色中的各1根,那么第4根肯定能与头3根中的一根配成颜色相同的一双,据此解答即可。此题考查了利用抽屉原理解决实际问题的灵活应用,关键是从最差情况考虑。\n【详解】3+1=4(根);答:最少拿4根筷子就一定能保证拿出的筷子里至少有两根是同色的。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_182",
+        "original_id": "unit5_choice_182"
+      },
+      {
+        "id": "unit5_choice_183",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "袋子里有4只红手套,2只黑手套,2只紫手套。一次摸出几只手套才能保证至少有一只红手套？",
+        "options": [],
+        "answer": "5只",
+        "solution": "【分析】根据题干,最坏的情况是取出4只手套:2只黑手套,2只紫手套,此时剩下的全是红色手套,再任意取出1只,就能保证至少有一只红手套。\n【详解】2+2+1=5(只);答:一次摸出5只手套,才能保证至少有一只红手套。\n【点睛】此题主要考查了抽屉原理的灵活应用,要注意考虑最不利情况。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_183",
+        "original_id": "unit5_choice_183"
+      },
+      {
+        "id": "unit5_choice_184",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "(1、8)、(2、7)、(3、6)、(4、5)这四组数,至少取出几个数,才能保证其中有两个数的和是9?",
+        "options": [],
+        "answer": "5个",
+        "solution": "【分析】根据题干,考虑最差情况:每组数据中都取了1个数,即此时取出了4个数,那么再任意取出1个数字,即可得出两个数的和是9,据此即可解答问题。此题考查了利用抽屉原理解决实际问题的灵活应用,关键是从最差情况考虑。\n【详解】4+1=5(个);答:至少取出5个数字,才能保证其中有两个数的和是9。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_184",
+        "original_id": "unit5_choice_184"
+      },
+      {
+        "id": "unit5_choice_185",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "一个口袋里分别有4个红球,7个黄球,8个黑球,为保证取出的球中有6个球颜色相同,则至少要取多少个小球？",
+        "options": [],
+        "answer": "15个",
+        "solution": "【分析】考虑最'坏'的情况,先取出4个红球,5个黄球,5个黑球,这样再取一个,不论取出的是黄球还是黑球,将有6个球颜色相同。本题考查的是最不利原则,不符合要求的最大数量加上去,得到符合要求的最小数量。\n【详解】4+5+5+1=15(个)答:至少要取15个小球。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "2021六年级·全国·竞赛",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_185",
+        "original_id": "unit5_choice_185"
+      },
+      {
+        "id": "unit5_choice_186",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "一个玻璃瓶里一共装有44个弹珠,其中:白色的2个,红色的3个,绿色的4个,蓝色的5个,黄色的6个,棕色的7个,黑色的8个,紫色的9个,如果要求每次从中取出1个弹珠,从而得到2个相同颜色的弹珠,请问最少需要取几次？",
+        "options": [],
+        "answer": "9次",
+        "solution": "【分析】总共有8种颜色的弹珠,要取出2个相同颜色的弹珠,最倒霉的情况就是前面8次取出的弹珠颜色都不一样,每种颜色各一个,这样第9次,不论取什么,一定可以保证有2个相同颜色的弹珠。本题考查的是抽屉问题,求解此类问题,就要按照最不利于事件发生的情况考虑问题。\n【详解】8+1=9(次)答:最少需要取9次。",
+        "knowledge_points": ["最不利原则"],
+        "source": "2021六年级·全国·竞赛",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_186",
+        "original_id": "unit5_choice_186"
+      },
+      {
+        "id": "unit5_choice_187",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "在1米长的直尺上任意点五个点,请你说明这五个点中至少有两个点的距离不大于25厘米。",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】5个点最多把1米长的直尺分成4段,要想使每一段都尽量长,应采取平均分的办法,把1米长的直尺平均划分成四段,每一段25厘米,把这四段看成四个抽屉。本题考查的是抽屉原理,抽屉原理注意应用的是最不利原则和平均原则。\n【详解】当把五个点随意放入四个抽屉时,根据抽屉原理,一定有一个抽屉里面有两个或两个以上的点,落在同一段上的这两点间的距离一定不大于25厘米,所以结论成立。",
+        "knowledge_points": ["抽屉原理"],
+        "source": "2021六年级·全国·竞赛",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_187",
+        "original_id": "unit5_choice_187"
+      },
+      {
+        "id": "unit5_choice_188",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "有49个小孩,每人胸前有一个号码,号码从1到49各不相同。现在请你挑选若干个小孩,排成一个圆圈,使任何相邻两个小孩的号码数的乘积小于100,那么你最多能挑选出多少个孩子？",
+        "options": [],
+        "answer": "18个",
+        "solution": "【分析】找出1~49中,乘积小于100的两个数的组合,根据任何相邻两个小孩的号码数的乘积小于100这一要求,选择出合适的数进行构造。本题考查的是抽屉原理的问题,抽屉原理是最不利原则和平均原则的体现。\n【详解】将1至49中相乘小于100的两个数,按被乘数分成9组,如下:(1×2)、(1×3)、(1×4)、…、(1×49)(2×3)、(2×4)、(2×5)、…、(2×49)……(8×9)、(8×10)、(8×11)、(8×12)(9×10)、(9×11)因为每个数只能与左右两个数相乘,也就是每个数作为被乘数或乘数最多两次,所以每一组中最多会有两对数出现在圆圈中,最多可以取出18个数对,共18×2=36次,但是每个数都出现两次,故出现了18个数。例如:(10×9)、(9×11)、(1×8)、(8×12)、(12×7)、(7×13)、(13×6)、(6×14)、(14×5)、(5×15)、(15×4)、(4×16)、(16×3)、(3×17)、(17×2)、(2×18)、(18×1)、(1×10),共出现1~18号,共18个孩子。若随意选取出19个孩子,那么共有19个号码,由于每个号码数要与旁边两数分别相乘,则会形成19个相乘的数对。那么在9组中取出19个数时,有19=9×2+1,由抽屉原则知,必有三个数对落入同一组中,这样某个数字会在数对中出现三次(或三次以上),由分析知,这是不允许的。故最多挑出18个孩子。答:最多能挑选出18个孩子。",
+        "knowledge_points": ["最不利原则"],
+        "source": "2021六年级·全国·竞赛",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_188",
+        "original_id": "unit5_choice_188"
+      },
+      {
+        "id": "unit5_choice_189",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "从1,2,3,…,99,100这100个数中任意选出51个数。证明:(1)在这51个数中,一定有两个数互质;(2)在这51个数中,一定有两个数的差等于50;(3)在这51个数中,一定存在9个数,它们的最大公约数是大于1。",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】第(1)问,任意相邻的两个非零自然数互质,可以把1~100这100个数每两个相邻的数分成1组,总共50组;第(2)问,把差是50的两个数分成1组,同样可以分成50组;第(3)问,最大公因数大于1,可能是2或3,那么按照2的倍数、3的倍数、既不是2又不是3的倍数分成3组。本题考查的是抽屉原理,求解问题的关键是准确构造出抽屉,问题不同,构造抽屉的形式也不相同。\n【详解】(1)证明:我们将1~100分成(1,2),(3,4),(5,6),(7,8),…,(99,100)这50组,每组内的数相邻,而相邻的两个自然数互质;将这50组数作为50个抽屉,同一个抽屉内的两个数互质,而现在51个数,放进50个抽屉,则必定有两个数在同一抽屉,于是这两个数互质,问题得证。(2)证明:我们将1~100分成(1,51),(2,52),(3,53),…,(50,100)这50组,每组内的数相差50;将这50组数视为抽屉,则现在有51个数放进50个抽屉内,则必定有2个数在同一抽屉,那么这两个数的差为50,问题得证。(3)证明:我们将1~100按2的倍数、3的奇数倍、既不是2又不是3的倍数的情况分组,有(2,4,6,8,…,98,100),(3,9,15,21,27,…,93,99),(5,7,11,13,17,19,23,…,95,97)这三组;第一、二、三组分别有50、17、33个元素,最不利的情况是下,51个数中有33个元素在第三组,那么剩下的18个数分到第一、二两组内,那么至少有9个数在同一组;所以这9个数的最大公因数为2或3或它们的倍数,显然大于1,问题得证。",
+        "knowledge_points": ["抽屉原理"],
+        "source": "2021六年级·全国·竞赛",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_189",
+        "original_id": "unit5_choice_189"
+      },
+      {
+        "id": "unit5_choice_190",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "从1、2、3、4、…、19、20这20个自然数中,至少任选几个数,就可以保证其中一定包括两个数,它们的差是12。",
+        "options": [],
+        "answer": "13个",
+        "solution": "【分析】把相差是12的两个数分成1组,构造抽屉,然后利用最不利原则进行求解。本题考查的是抽屉原理,解题的关键是如何准确地构造出抽屉。\n【详解】在这20个自然数中,差是12的有以下8对:{20,8},{19,7},{18,6},{17,5};{16,4},{15,3},{14,2},{13,1};另外还有4个不能配对的数{9},{10},{11};{12},共制成12个抽屉(每个括号看成一个抽屉);从12个抽屉中各选一个数,不能满足两个数的差是12,但只要再取1个数,一定可以找到两个数的差是12;答:至少任选13个数。",
+        "knowledge_points": ["抽屉原理"],
+        "source": "2021六年级·全国·竞赛",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_190",
+        "original_id": "unit5_choice_190"
+      },
+      {
+        "id": "unit5_choice_191",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "证明:在从1开始的前10个奇数中任取6个,一定有2个数的和是20。",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】从1开始的前10个奇数是1、3、5、7、9、11、13、15、17、19,把和是20的两个数分成1组,总共可以分成5组,那么抽屉数是5。本题考查的是抽屉问题,没有抽屉的情况下,就要根据题目的要求,合理地构造出抽屉,然后求解问题。\n【详解】证明:构造抽屉:(1,19),(3,17),(5,15),(7,13),(9,11);任取6个数,一定有两个数处在同一个抽屉中,其和是20;所以在从1开始的前10个奇数中任取6个,一定有2个数的和是20。",
+        "knowledge_points": ["抽屉原理"],
+        "source": "2021六年级·全国·竞赛",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_191",
+        "original_id": "unit5_choice_191"
+      },
+      {
+        "id": "unit5_choice_192",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "幼儿园买来许多牛、马、羊、狗塑料玩具,每个小朋友任意选择两件,但不能是同样的,问:至少有多少个小朋友去拿,才能保证有两人所拿玩具相同？",
+        "options": [],
+        "answer": "7个",
+        "solution": "【分析】从四种玩具中挑选不同的两件,所有的搭配有以下6组:牛、马;牛、羊;牛、狗;马、羊;马、狗;羊、狗,把每一组搭配看作一个'抽屉',共6个抽屉,每种拿玩具的方式先安排一人,然后再多一个人,一定能保证有两人所拿玩具相同。本题考查的是抽屉原理,首先要枚举出所有拿玩具的方法,确定抽屉数。\n【详解】有6种不同的拿玩具的方式;考虑最不利原则,前6个人的方式各不相同,那么第7个人的方式一定与前面的一个人相同;答:至少有7个小朋友去拿,才能保证有两人所拿玩具相同。",
+        "knowledge_points": ["抽屉原理"],
+        "source": "2021六年级·全国·竞赛",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_192",
+        "original_id": "unit5_choice_192"
+      },
+      {
+        "id": "unit5_choice_193",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "一副扑克牌有54张,最少要抽取几张牌,方能使其中至少有2张牌有相同的点数？",
+        "options": [],
+        "answer": "16张",
+        "solution": "【分析】如果不算大、小王,每个点数的牌各取1张,可以取出13张牌,再取1张,便一定有两张相同点数的牌,14张加上大、小王,则需要16张牌。【点睛】根据最不利原则,先求出不符合要求的最大数量,加上1,即为符合要求的最低数量。\n【详解】13+1+2=14+2=16(张)答:最少要抽取16张牌。",
+        "knowledge_points": ["最不利原则"],
+        "source": "2021六年级·全国·竞赛",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_193",
+        "original_id": "unit5_choice_193"
+      },
+      {
+        "id": "unit5_choice_194",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "班上有28名小朋友,老师至少拿几本书,随意分给小朋友,才能保证至少有一个小朋友能得到不少于两本书？",
+        "options": [],
+        "answer": "29本",
+        "solution": "【分析】要保证至少有一个小朋友能得到不少于两本书,可以给每个小朋友都先分1本书,现在是不符合要求的,但只要再拿一本书分给任意一个小朋友,就可以保证至少有一个小朋友能得到不少于两本书。本题考查的是最不利原则,可以先找出不符合要求的最大数量,加上1即为符合要求的最小数量。\n【详解】28×1=28(本) 28+1=29(本)答:老师至少拿29本书。",
+        "knowledge_points": ["抽屉原理"],
+        "source": "2021六年级·全国·竞赛",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_194",
+        "original_id": "unit5_choice_194"
+      },
+      {
+        "id": "unit5_choice_195",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "6只鸽子要飞进5个笼子,每个笼子里都必须有1只,一定有一个笼子里有2只鸽子。对吗？",
+        "options": [],
+        "answer": "对",
+        "solution": "【分析】6只鸽子要飞进5个笼子,可以先让每个笼子飞进1只,这样每个笼子各有1只,第6只鸽子不论飞进哪一个笼子中,一定可以保证有一个笼子里有2只鸽子。本题考查的是抽屉原理的问题,题目明确给出了抽屉数和苹果数,直接求解即可。\n【详解】6只鸽子相当于是苹果,5个笼子相当于是抽屉;6÷5=1(只)…1(只) 1+1=2(只)答:一定有一个笼子里有2只鸽子是对的。",
+        "knowledge_points": ["抽屉原理"],
+        "source": "2021六年级·全国·竞赛",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_195",
+        "original_id": "unit5_choice_195"
+      },
+      {
+        "id": "unit5_choice_196",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "一次数学考试,六(2)班最高分是98分,最低分是76分,每人的得分都是整数,并且班上一定至少有3名学生得分相同,六(2)班至少有多少名学生？",
+        "options": [],
+        "answer": "47名",
+        "solution": "【分析】从76到98,一共有98-76+1=23(个)整数,要使人数最少且一定至少有3名学生得分相同,考虑最不利的情况,每个分数都有2名学生,则再多1名学生,必定保证有3名学生的得分相同,所以这个班至少有23×2+1=47(名)学生。此题考查了利用抽屉原理解决实际问题的灵活应用,关键是从最差情况考虑。\n【详解】98-76+1=22+1=23(个) 23×(3-1)+1=23×2+1=46+1=47(名)答:六(2)班至少有47名学生。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_196",
+        "original_id": "unit5_choice_196"
+      },
+      {
+        "id": "unit5_choice_197",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "袋子里有红色、白色、蓝色手套各5只。(不分左右手,一双手套为一种颜色)(1)至少要拿出多少只,才能使拿出的手套中一定有两双是同颜色的?(2)至少要拿出多少只,才能使拿出的手套中一定有两双是不同颜色的?",
+        "options": [],
+        "answer": "(1)10只;(2)8只",
+        "solution": "【分析】(1)'一定有两双是同颜色的',也就是说有4只手套是同颜色的。把红色、白色、蓝色看作3个抽屉,根据最不利原则考虑,每个抽屉都放3只同颜色的手套,如果再放一只,无论放到哪个抽屉里,都能够保证有4只,即一定有两双是同颜色的。(2)根据最不利原则考虑,如果先拿出5只相同颜色的手套,再拿出两只不同颜色的手套,那么只要再拿出一只,无论是什么颜色,都能保证有两双不同颜色的手套。根据抽屉原则,要准确建立三个抽屉的,求出最差取法的总只数是解答的关键。同时注意'两双',不是两只,否则整个题就会全错。\n【详解】3×3+1=9+1=10(只)答:至少要拿出10只,才能使拿出的手套中一定有两双是同颜色的。(2)5+2+1=7+1=8(只)答:至少要拿出8只,才能使拿出的手套中一定有两双是不同颜色的。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_197",
+        "original_id": "unit5_choice_197"
+      },
+      {
+        "id": "unit5_choice_198",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "一个袋子里装有白帽子和黄帽子各7顶,闭着眼睛摸。(1)从袋中至少摸出几顶帽子,才能保证有2顶同色的帽子?(2)从袋中至少摸出几顶帽子,才能保证有2种颜色的帽子?",
+        "options": [],
+        "answer": "(1)3顶;(2)8顶",
+        "solution": "【分析】(1)根据最不利原则考虑,先摸出2顶不同颜色的帽子,此时只要再任意摸出一顶,就能保证有2顶是同色的,即至少要摸出2+1=3(顶)。(2)根据最不利原则考虑,先摸出7顶同种颜色的帽子,此时只要再任意摸出一顶,就能保证摸出2种颜色的帽子,即至少要摸7+1=8(顶)。本题考查了抽屉原理,抽屉原理的解答思路,从最不利情况考虑,准确地建立抽屉和确定元素的总个数。\n【详解】(1)2+1=3(顶)答:从袋中至少摸出3顶帽子,才能保证有2顶同色的帽子。(2)7+1=8(顶)答:从袋中至少摸出8顶帽子,才能保证有2种颜色的帽子。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_198",
+        "original_id": "unit5_choice_198"
+      },
+      {
+        "id": "unit5_choice_199",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "任意给出4个不同的自然数,其中必有两个数的差是3的倍数。为什么？",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】自然数除以3,其结果只有三种情况:整除、余数为1或余数为2。任意给出4个不同的自然数,最不理想情况下,其中3个数除以3的结果都不相同,则第4个数出现时,就会有两个数除以3的余数相同,或是都能被3整除,用这两个数的差除以3可以整除,即必有两个数的差是3的倍数。解答本题的关键是明确任意自然数除以3的余数中有三种情况即余数为0、1或2,且余数相同的两个不同自然数的差必定是3的倍数,如7和4,除以3的余数都是1,它们的差是7-4=3,是3的倍数。\n【详解】根据题意:自然数除以3,其余数只有三种情况:0、1、或2;而4个非零自然除以3,其中就会有两个数除以3的余数相同(即同是0,1或2),用这两个数的差除以3的余数就是0,所以任意给出4个不同的自然数,其中必有两个数的差是3的倍数。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_199",
+        "original_id": "unit5_choice_199"
+      },
+      {
+        "id": "unit5_choice_200",
+        "unit": "第五单元:鸽巢",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "把5双手套(手套分左、右手)放进暗箱里,要保证取出的手套中至少有1双,至少要取出几只手套？",
+        "options": [],
+        "answer": "6只",
+        "solution": "【分析】考虑最不利的情况,取出的5只手套要么都是左手,要么都是右手,此时是不符合要求的最大数量,但只要再取1个,一定是满足要求的。本题考查的是抽屉原理,要求符合要求的最小数量,可以先求出不符合要求的最大数量。\n【详解】先取出5个左手的手套或5个右手的手套,此时不符合要求;5+1=6(只)答:至少要取出6只手套。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·单元测试",
+        "create_time": "2024-11-03 00:00:00",
+        "merged_id": "unit5_200",
+        "original_id": "unit5_choice_200"
+      },
+      {
+        "id": "unit5_choice_201",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "一个盒子里装有红、黄、蓝、绿四种颜色的球各 100 个，从中至少取（）个球才能保证有 2 个球颜色相同。",
+        "options": [
+          "A. 4",
+          "B. 5",
+          "C. 6",
+          "D. 101"
+        ],
+        "answer": "B",
+        "solution": "【分析】将红、黄、蓝、绿 4 种颜色视为 4 个抽屉, 要保证有一个抽屉至少有 2 个球, 分的物体个数要比抽屉数多 1 ,所以至少取 (4 + 1) 个球, 才能保证有 2 个同色球。本题考查了抽屉原理的运用。设置好抽屉数，要从最不利的情况思考。\n【详解】由分析得：4 + 1 = 5 (个)故答案为: B",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21 六年级下·全国·单元测试",
+        "merged_id": "unit5_201",
+        "original_id": "unit5_choice_201"
+      },
+      {
+        "id": "unit5_choice_202",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "某班有50名学生，他们都参加了课外兴趣小组。活动内容有美术、声乐、书法，每名学生只参加1个兴趣小组，班级中至少有（）名学生参加的兴趣小组相同。",
+        "options": [
+          "A. 16",
+          "B. 17",
+          "C. 18",
+          "D. 19"
+        ],
+        "answer": "B",
+        "solution": "【分析】把 50 名学生看作 50 个苹果, 美术、声乐、书法三个兴趣小组看作 3 个抽屉, 将 50 个苹果平均分成 3 份,每个抽屉放 16 个, 还剩 2 个, 则至少有 (16 + 1) 名学生参加的兴趣小组相同。此题属于典型的抽屉问题，确定好抽屉个数以及放入的总个数是解题关键。\n【详解】50 ÷ 3 = 16 …… 2, 16 + 1 = 17 （名）班级中至少有 17 名学生参加的兴趣小组相同。故选择：B",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit5_202",
+        "original_id": "unit5_choice_202"
+      },
+      {
+        "id": "unit5_choice_203",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "下列问题可以运用 '抽屉原理' 解决的是 ( )。",
+        "options": [
+          "A. 在一条线段的两个端点之间再点上 3 个点, 以每两点为端点的线段共有多少条",
+          "B. A 地到 B 地有 2 条路, B 地到 C 地有 3 条路, A 地到 C 地有多少种不同的走法",
+          "C. 把 4 名男生分到 3 个小组做游戏, 至少有几名男生要分到同一个小组",
+          "D. 抽屉里有 3 个红球、4 个蓝球, 任意摸一个球, 摸到什么球的可能性大"
+        ],
+        "answer": "C",
+        "solution": "【分析】A. 可按握手问题进行解答;B. 属于乘法原理问题; C. 属于抽屉原理; D. 属于可能性问题。本题主要考查了学生根据抽屉原理解答问题的能力。\n【详解】由题意可知, 线段上有 5 个点, 每个点与其他的 5 - 1 = 4 (个) 点可组成线段, 因为每两个点之间要重复一次,所以线段共有 5 × (5 - 1) ÷ 2 = 10 (条); 可按握手问题的方法进行解答。由题意可知, A 地到 C 地的走法有 2 × 3 = 6 (种), 属于搭配问题。每个小组分 1 名男生，剩下的 1 名男生不论分到哪一组，至少有 2 名男生要分到一个小组。属于抽屉原理。蓝球数量多于红球，所以摸到蓝球的可能性大，属于可能性问题。故答案为: C",
+        "knowledge_points": ["搭配问题", "鸽巢问题"],
+        "source": "20-21 六年级下·全国·单元测试",
+        "merged_id": "unit5_203",
+        "original_id": "unit5_choice_203"
+      },
+      {
+        "id": "unit5_choice_204",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "给正方体的6个面涂上3种颜色（每个面涂1种颜色），不论怎么涂，至少有（）个面的颜色相同。",
+        "options": [
+          "A. 2",
+          "B. 3",
+          "C. 4",
+          "D. 5"
+        ],
+        "answer": "A",
+        "solution": "【分析】给正方体 6 个面分别涂上不同的 3 种颜色, 将 3 种颜色看成抽屉, 将 6 个面当元素, 因为 6 > 3, 根据抽屉原理可知, 不管怎么涂至少有两个面涂的颜色相同。把多于 n 个的物体放到 n 个抽屉里, 则至少有一个抽屉里的东西不少于两件。\n【详解】由分析可知：不论怎么涂，至少有 6 ÷ 3 = 2 个面的颜色相同。故答案为：A。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "2021·广东汕尾·小升初真题",
+        "merged_id": "unit5_204",
+        "original_id": "unit5_choice_204"
+      },
+      {
+        "id": "unit5_choice_205",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "跳绳比赛分为男生组和女生组，六（1）班派出 3 名同学参加跳绳比赛。下列说法正确的是（）。",
+        "options": [
+          "A. 每个组至少有 1 名六 (1) 班的同学",
+          "B. 男生组一定有 2 名六 (1) 班的同学",
+          "C. 总有一个组至少有 2 名六 (1) 班的同学",
+          "D. 以上说法都不正确"
+        ],
+        "answer": "C",
+        "solution": "【分析】本题是道抽屉原理问题，按照方法逐个分析即可解答问题。此题考查了利用抽屉原理解决实际问题的灵活应用。\n【详解】A.有可能3名同学都是男生或者都是女生，也就是说女生组或者男生组可能1名六（1）班同学也没有；B.同理。C.把男生组和女生组看作2个抽屉, 3 ÷ 2 = 1 (名) ……1 (名), 把3名同学看作3个苹果, 将3个苹果平均放进2个抽屉, 每个抽屉里放1个, 还剩1个。剩余的1个无论怎么放, 总有一个抽屉里至少有2个苹果, 即总有一个组至少有2名六 (1) 班的同学。故答案为: C",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21 六年级下·全国·单元测试",
+        "merged_id": "unit5_205",
+        "original_id": "unit5_choice_205"
+      },
+      {
+        "id": "unit5_choice_206",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "有 10 张数字卡片分别写着数字 1 ~ 10, 至少要抽出 ( ) 张才能保证既有奇数又有偶数。",
+        "options": [
+          "A. 3",
+          "B. 4",
+          "C. 5",
+          "D. 6"
+        ],
+        "answer": "D",
+        "solution": "【分析】1 ~ 10 这 10 个数, 有 5 个奇数, 5 个偶数, 可以先取出 5 个奇数或 5 个偶数, 此时是不符合要求的最大数量,只要再任取 1 个, 一定能保证既有奇数又有偶数。本题考查的抽屉原理, 抽屉原理主要是应用平均原则来求解问题。\n【详解】先取出 5 个奇数或 5 个偶数, 此时不符合要求; 5 + 1 = 6 (张)至少要抽出6张才能保证既有奇数又有偶数。",
+        "knowledge_points": ["奇数与偶数的认识", "鸽巢问题初步"],
+        "source": "20-21 六年级下·全国·单元测试",
+        "merged_id": "unit5_206",
+        "original_id": "unit5_choice_206"
+      },
+      {
+        "id": "unit5_choice_207",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "希望小学的学生中年龄最小的是6岁，最大的是13岁，任选（）位同学就能保证其中一定有两位同学的年龄相同。",
+        "options": [
+          "A. 6",
+          "B. 7",
+          "C. 8",
+          "D. 9"
+        ],
+        "answer": "D",
+        "solution": "【分析】最小的是 6 岁, 最大的是 13 岁, 年龄可能是 6 岁、 7 岁、 8 岁、 9 岁、 10 岁、 11 岁、 12 岁、 13 岁, 共 8 种可能, 即抽屉数是 8。本题考查的是抽屉原理，这里的抽屉数是学生的年龄情况，总共8种可能，所以抽屉数是8。\n【详解】年龄可能是 6 岁、7 岁、8 岁、9 岁、10 岁、11 岁、12 岁、13 岁, 共 8 种可能;每个年龄先取 1 名学生, 取出 8 名学生, 此时没有两位同学的年龄相同;再取 1 名学生, 一定有两位同学的年龄相同;8 + 1 = 9 （名）故答案选：D。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit5_207",
+        "original_id": "unit5_choice_207"
+      },
+      {
+        "id": "unit5_choice_208",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "王老师在福利幼儿园工作，儿童节她给孩子们买衣服，有红、黄、白三种颜色，结果总是至少有 2 个孩子的衣服颜色一样，她至少给（）个孩子买了衣服。",
+        "options": [
+          "A. 2",
+          "B. 3",
+          "C. 4",
+          "D. 5"
+        ],
+        "answer": "C",
+        "solution": "【分析】有红、黄、白三种颜色, 那么抽屉数是 3 , 总是至少有 2 个孩子的衣服颜色一样, 那么可以先让每种颜色的衣服各买一件, 总共 3 件, 此时有 3 个孩子, 在此基础上, 只有再买一件, 一定有 2 个孩子的衣服颜色一样。本题考查的是抽屉原理, 已知抽屉数求总数, 可以利用最不利原则, 求出不符合要求的最大数量, 再加上 1 即可。\n【详解】2 - 1 = 1  1 × 3 = 3 (个) 3 + 1 = 4 (个)所以王老师至少给 4 个孩子买了衣服;故答案选：C。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21 六年级下·全国·单元测试",
+        "merged_id": "unit5_208",
+        "original_id": "unit5_choice_208"
+      },
+      {
+        "id": "unit5_choice_209",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "下列说法正确的是 ( )。",
+        "options": [
+          "A. 任意 3 个不同的自然数, 其中一定有两个数的和是偶数",
+          "B. 六 (2) 班有 42 名学生, 至少有 4 人的生日在 9 月份",
+          "C. 盒子里有同样大小的红、黄、蓝三种颜色的球各 5 个,要想摸出的球一定有 2 个是同色的, 至少要摸出 6 个球",
+          "D. 以上说法均不正确"
+        ],
+        "answer": "A",
+        "solution": "【分析】对于抽屉数已知的情况，直接用总数除以抽屉数，如果抽屉数位置，需要先找出抽屉数是多少，然后再进行求解。求解抽屉原理的问题, 关键是求出抽屉数和苹果数, 然后用苹果数除以抽屉数, 根据是否有余数, 进行判断。\n【详解】A．任意3个不同的自然数，可能是3个奇数、3个偶数、1奇2偶、1偶2奇这四种情况，这4种情况都可以保证有两个数的和是偶数，正确； B. 一年有 12 个月, 42 ÷ 12 = 3 … 6, 3 + 1 = 4 至少有 4 人的生日在同一月, 但不能保证是 9 月份, 错误; C. 红、黄、蓝三种颜色的球各 5 个, 可以先各取 1 个, 然后再取 1 个, 那么一定有 2 个是同色的, 至少取出 4 个就可以了, 错误; D. A 选项正确, D 选项错误;故答案选：A。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21 六年级下·全国·单元测试",
+        "merged_id": "unit5_209",
+        "original_id": "unit5_choice_209"
+      },
+      {
+        "id": "unit5_choice_210",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "某地 1 月份的天气有晴、阴、多云、雨、雪五种情况, 总有一种天气至少有 ( ) 天。",
+        "options": [
+          "A. 5",
+          "B. 6",
+          "C. 7",
+          "D. 8"
+        ],
+        "answer": "C",
+        "solution": "【分析】1月份有31天，在这里是总数，晴、阴、多云、雨、雪五种天气情况，是抽屉数，31除以5，根据是否有余数进行判断。求解抽屉原理的问题, 关键是找出总数和抽屉数,然后用总数除以抽屉数, 根据是否有余数进行判断。\n【详解】31 ÷ 5 = 6 … 1  6 + 1 = 7 (天) 总有一种天气至少有 7 天;故答案选 C。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21 六年级下·全国·单元测试",
+        "merged_id": "unit5_210",
+        "original_id": "unit5_choice_210"
+      },
+      {
+        "id": "unit5_choice_211",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "有红色、白色的筷子各 10 根, 至少拿出 ( ) 根筷子, 才能保证有一双是同色的。",
+        "options": [
+          "A. 3",
+          "B. 4",
+          "C. 11",
+          "D. 12"
+        ],
+        "answer": "A",
+        "solution": "【分析】要保证有一双是同色的，可以红色、白色先各取一支，此时只要再取 1 支，一定有 2 支是同色的。本题考查的是最不利原则, 首先要找出不符合要求的最大数量, 在此基础上加上 1 即可。\n【详解】红色、白色先各取一支，取出2支，此时不符合要求；2 + 1 = 3 (支) 至少拿出 3 根筷子, 才能保证有一双是同色的;故答案选：A。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21 六年级下·全国·单元测试",
+        "merged_id": "unit5_211",
+        "original_id": "unit5_choice_211"
+      },
+      {
+        "id": "unit5_choice_212",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "某小学有 6 个年级，每个年级有 8 个班。一天放学，8 位小朋友一起走出校门。下列说法中正确的是（）。",
+        "options": [
+          "A. 他们中至少有 2 人的出生月份相同",
+          "B. 他们中至少有2人是同一年级的",
+          "C. 他们中至少有 2 人的属相相同",
+          "D. 他们中至少有 2 人是同一班级的"
+        ],
+        "answer": "B",
+        "solution": "【分析】这个小学总共有 48 个班级，A 选项，总数是 8 个学生，抽屉数是 12 个月份；B 选项，总数是 8 个学生，抽屉数是 6 个年级；C 选项，总数是 8 个学生，抽屉数是 12 个属相；D 选项，总数是 8 个学生，抽屉数是48 个班级。本题考查的是抽屉原理, 求解问题的关键是确定抽屉数是多少。\n【详解】A．8位小朋友的出生月份可以互不相同，不能保证至少有2人的出生月份相同，错误； B. 8 ÷ 6 = 1 … 2, 1 + 1 = 2, 至少有 2 人是同一年级的, 正确; C. 8位小朋友的属相可以互不相同, 不能保证至少有2人的属相相同, 错误; D. 8位小朋友的班级可以互不相同, 不能保证至少有2人是同一班级的，错误; 故答案选：B",
+        "knowledge_points": ["鸽巢问题初步"],
+        "source": "2022·河南三门峡·小升初真题",
+        "merged_id": "unit5_212",
+        "original_id": "unit5_choice_212"
+      },
+      {
+        "id": "unit5_choice_213",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "把21块糖分给6个人，总有一个人至少分得（）块。",
+        "options": [
+          "A. 3",
+          "B. 4",
+          "C. 5",
+          "D. 6"
+        ],
+        "answer": "B",
+        "solution": "【分析】21 块糖是总数, 6 个人是抽屉数, 21 除以 6 , 商 3 , 余数是 3 , 只要有余数, 不论余数是多少, 商加上 1 即为所求。本题考查的是抽屉原理, 抽屉原理其实是平均原则和最不利原则的体现。\n【详解】21 ÷ 6 = 3 … 3 3 + 1 = 4 (块) 故答案选: B。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit5_213",
+        "original_id": "unit5_choice_213"
+      },
+      {
+        "id": "unit5_choice_214",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "学校将新购置的 40 张桌子分给 6 个班，总有一个班至少分得（）张桌子。",
+        "options": [
+          "A. 5",
+          "B. 6",
+          "C. 7",
+          "D. 8"
+        ],
+        "answer": "C",
+        "solution": "【分析】40 张桌子分给 6 个班, 那么 6 个班是抽屉数, 总数除以抽屉数, 根据是否有余数做出选择。本题考查的是抽屉原理，对于此类问题，首先要确定抽屉数和整数分别是多少，然后用总数除以抽屉数，根据是否有余数进行判断。\n【详解】40 ÷ 6 = 6 … 4 6 + 1 = 7 (张) 所以总有一个班至少分得 7 张桌子; 故答案选：C。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21 六年级下·全国·单元测试",
+        "merged_id": "unit5_214",
+        "original_id": "unit5_choice_214"
+      },
+      {
+        "id": "unit5_choice_215",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "袋中有 60 粒大小相同的弹珠, 每 15 粒是同一种颜色, 为保证取出的弹珠中一定有 2 粒是同色的, 至少要取出粒数是 ( )。",
+        "options": [
+          "A. 4",
+          "B. 5",
+          "C. 6",
+          "D. 7"
+        ],
+        "answer": "B",
+        "solution": "【分析】60粒弹珠，只有四个颜色，如果取出的颜色都不一样最多可以取四个，再多取一个一定会和已经取出来的弹珠颜色重复，依此得出答案。抽屉问题, 四个颜色为四个抽屉, 至少有 2 个颜色一样的, 只需要 4 + 1 个弹珠即可。\n【详解】弹珠共 4 个颜色;如果取出的弹珠颜色都不重复，最多可以取出 4 粒；所以一定有 2 个颜色重复的弹珠数是: 4 + 1 = 5 (个)故答案为: B。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "18-19 三年级上·上海·课后作业",
+        "merged_id": "unit5_215",
+        "original_id": "unit5_choice_215"
+      },
+      {
+        "id": "unit5_choice_216",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "从 49 名学生中选一名班长，小红、小明和小华为候选人。统计37票后的结果是：小红15票，小明10票，小华12票，小红至少再得（）张票才能保证票数最多当选为班长。",
+        "options": [
+          "A. 7",
+          "B. 5",
+          "C. 6",
+          "D. 4"
+        ],
+        "answer": "B",
+        "solution": "【分析】由题可知 49 名学生, 有 37 张选票, 还可以有 49 - 37 = 12 张选票。又知小红 15 票, 小明 10 票, 小华 12 票, 小红比小华多 3 张, 如果三人再各得 4 张选票, 小红当选; 若小红得 4 张, 小明不得, 小华得 8 张, 小华选票 > 小红选票, 小红不当选; 若小红得 5 张, 小明不得, 小华得 7 张, 小红选票 > 小华选票, 小红必当选; 据此解答。解答本题需抓住小红得票最多这一点进行分析。\n【详解】由题可知还剩选票: 49 - 37 = 12 (张),如果把这 12 张平均得, 每人得: 12 ÷ 3 = 4 (张),小红 15 + 4 = 19 （张），小明 10 + 4 = 14 （张），小华 12 + 4 = 16 （张），小红当选。如小红得 4 张, 小明不得, 小华得 8 张, 15 + 4 < 12 + 8, 小华当选, 小红不当选。如小红得 5 张, 小明不得, 小华得 7 张, 15 + 5 > 12 + 7, 小红当选。所以小红至少得 5 张才能保证得票最多当选为班长。故答案为: B",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "2020·江苏盐城·小升初真题",
+        "merged_id": "unit5_216",
+        "original_id": "unit5_choice_216"
+      },
+      {
+        "id": "unit5_choice_217",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "一群小学生中最多有6人同一年级，那么这群小学生最多有（）人。",
+        "options": [
+          "A. 6",
+          "B. 36",
+          "C. 31",
+          "D. 11"
+        ],
+        "answer": "B",
+        "solution": "【分析】小学有 6 个年级, 看作 6 个抽屉。因为一群小学生中最多有 6 人同一年级, 则每个抽屉里最多放 6 人, 据此解答即可。本题考查抽屉问题, 解答本题的关键是理解小学生中最多有 6 人同一年级, 则每个抽屉里最多放 6 人, 再求这群小学生最多有几人即可。\n【详解】每个抽屉里放最多放6人，则：6 × 6 = 36 (人)这群小学生最多有36人故答案为: B",
+        "knowledge_points": ["鸽巢问题进阶"],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_217",
+        "original_id": "unit5_choice_217"
+      },
+      {
+        "id": "unit5_choice_218",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "把一些书放进5个抽屉里，总有一个抽屉至少放2本书。这些书可能有（）本。",
+        "options": [
+          "A. 5",
+          "B. 2",
+          "C. 8",
+          "D. 4"
+        ],
+        "answer": "C",
+        "solution": "【分析】抽屉原则一：如果把 (n + 1) 个物品放在 n 个抽屉里，那么必有一个抽屉中至少放有 2 个物体。本题考查了抽屉问题, 一个抽屉先放一本数, 只要总数多与 5 本, 无论剩下的怎么放, 都有一个抽屉至少放 2 本。\n【详解】A． 5 = 5 ，选项不对；B. 2 < 5 ，选项不对；C. 8 > 5 ，选项正确；D. 4 < 5 ，选项不对；故答案为: C。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_218",
+        "original_id": "unit5_choice_218"
+      },
+      {
+        "id": "unit5_choice_219",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "把黑、白围棋子各 10 粒混在一起，要保证摸出的棋子中有 3 粒同色，至少要摸出（）粒。",
+        "options": [
+          "A. 3",
+          "B. 6",
+          "C. 5",
+          "D. 20"
+        ],
+        "answer": "C",
+        "solution": "【分析】黑、白 2 种颜色看作是 2 个抽屉, 利用抽屉原理来解答即可。本题考查抽屉问题, 解答本题关键在于理解考虑最差的情况每种颜色棋子各摸 2 粒, 共 4 粒, 再多摸 1 粒,无论是什么颜色, 都能保证摸出的棋子中有 3 粒同色。\n【详解】黑、白 2 种颜色看作是 2 个抽屉, 考虑最差的情况,每个抽屉摸出 2 粒棋子, 2 × 2 = 4 粒, 则再摸 1 粒, 无论从哪个抽屉里摸出, 都会出现摸出的棋子中有 3 粒同色:2 × 2 = 4 (粒)4 + 1 = 5 (粒)所以至少要摸出 5 粒, 才能保证摸出的棋子中有 3 粒同色。故正确答案为: C。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_219",
+        "original_id": "unit5_choice_219"
+      },
+      {
+        "id": "unit5_choice_220",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "张老师给15个小朋友分糖果，其中至少有一个小朋友分到3颗。这批糖果至少有（）颗。",
+        "options": [
+          "A. 15",
+          "B. 3",
+          "C. 31",
+          "D. 18"
+        ],
+        "answer": "C",
+        "solution": "【分析】15 个小朋友看作是 15 个抽屉，利用抽屉原理来解答即可。本题考查抽屉问题，解答本题关键在于理解考虑最差的情况 15 个小朋友每个小朋友分得 2 颗糖果，共 30 颗糖果，再多分 1 颗，无论分给哪个小朋友，都能保证至少有一个小朋友分到 3 颗。\n【详解】15 个小朋友看作是 15 个抽屉，考虑最差的情况，每个抽屉分得 2 颗糖果，共 15 × 2 = 30 颗，则再分 1 颗，无论分到哪个抽屉里，都会出现至少有一个小朋友分到 3 颗： 15 × 2 = 30 （颗）30 + 1 = 31 (颗)所以这批糖果至少有 31 颗, 才能保证至少有一个小朋友分到 3 颗。故正确答案为: C。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_220",
+        "original_id": "unit5_choice_220"
+      },
+      {
+        "id": "unit5_choice_221",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "李奶奶对一群小学生说：'你们当中至少有两人同一个月生日。'这群小学生至少有（）人。",
+        "options": [
+          "A. 2",
+          "B. 3",
+          "C. 20",
+          "D. 13"
+        ],
+        "answer": "D",
+        "solution": "【分析】一年有 12 个月, 看作 12 个抽屉, 每个抽屉里有 1 人, 共 12 人, 则再增加 1 人, 无论在哪个月出生, 都保证至少有两人同一个月生日。本题考查抽屉问题，解答本题的关键是理解一年有 12 个月，考虑最差的情况，这群小学生中 12 人的生日分布在 12 个月，只要再加上 1 人，无论在哪个月出生，都保证至少有两人同一个月生日。\n【详解】12 + 1 = 13 （人）所以这群学生至少有 13 人, 才能保证至少有两人同一个月生日。故答案为: D",
+        "knowledge_points": ["鸽巢问题进阶"],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_221",
+        "original_id": "unit5_choice_221"
+      },
+      {
+        "id": "unit5_choice_222",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "将一些梨放入3个水果盘里，每个盘子里放的个数都不相同，放得最多的盘子放5个，这些梨一共有（）个。",
+        "options": [
+          "A. 8 - 12",
+          "B. 12 - 16",
+          "C. 12 - 15",
+          "D. 10 - 14"
+        ],
+        "answer": "A",
+        "solution": "【分析】每个盘子里放的个数都不相同，放得最多的盘子放 5 个，所以其余两个盘子里最多放 4 个和 3 个、最少放 1 个和 2 个，据此解答。解答本题的关键是理解其余两个盘子里最多放 4 个和 3 个、最少放 1 个和 2 个。\n【详解】最多： 5 + 4 + 3 = 12 （个）最少: 1 + 2 + 5 = 8 (个)故答案为: A",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_222",
+        "original_id": "unit5_choice_222"
+      },
+      {
+        "id": "unit5_choice_223",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "一个盒子里红、黄、蓝、白、绿色球若干，至少摸出（）个球，才能保证有5个同色的球。",
+        "options": [
+          "A. 6",
+          "B. 20",
+          "C. 21",
+          "D. 25"
+        ],
+        "answer": "C",
+        "solution": "【分析】考虑最坏情况：假设前 20 个都摸出红、黄、蓝、白、绿各 4 个，再摸一个只能是 5 种颜色中的 1 个，据此可以推出：至少取 (5 × 4 + 1) 个，才能保证有 5 个同色的球，据此解答。本题主要考查抽屉问题的实际应用，解题的关键是从极端情况入手。\n【详解】5 × 4 + 1 = 20 + 1 = 21 (个)\n\n故答案为: C",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_223",
+        "original_id": "unit5_choice_223"
+      },
+      {
+        "id": "unit5_choice_224",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "把31个桃子最多放进（）个盘子里，才能保证有一个盘子里至少放进6个桃子。",
+        "options": [
+          "A. 3",
+          "B. 4",
+          "C. 5",
+          "D. 6"
+        ],
+        "answer": "D",
+        "solution": "【分析】可以利用公式：（分的物品总数 - 1）÷（其中一个抽屉里至少有的物品个数 - 1）=a……b（a、b均为自然数，且 b<a），则 a 就是所求的抽屉数。本题主要考查抽屉原理的逆应用，理解抽屉原理是解题的关键。\n【详解】(31 - 1) ÷ (6 - 1) = 30 ÷ 5 = 6 (个)\n\n把 31 个桃子最多放进 6 个盘子里, 才能保证有一个盘子里至少放进 6 个桃子。故答案为: D",
+        "knowledge_points": ["鸽巢问题进阶"],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_224",
+        "original_id": "unit5_choice_224"
+      },
+      {
+        "id": "unit5_choice_225",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "抽屉里有 8 个红球,5个黄球，至少一次摸出（）个一定会摸到黄球。",
+        "options": [
+          "A. 5",
+          "B. 6",
+          "C. 8",
+          "D. 9"
+        ],
+        "answer": "D",
+        "solution": "【分析】考虑最坏情况：一次摸出 8 个球，摸出的都是红球,则再摸出一个一定是红球, 据此即可解答。此考查抽屉原理，要注意考虑最差情况。\n【详解】8 + 1 = 9 （个）故答案为: D",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_225",
+        "original_id": "unit5_choice_225"
+      },
+      {
+        "id": "unit5_choice_226",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "任意取 ( ) 个不同的自然数, 才能保证至少有两个数的差为 9 的倍数。",
+        "options": [
+          "A. 9",
+          "B. 11",
+          "C. 10",
+          "D. 13"
+        ],
+        "answer": "C",
+        "solution": "【分析】根据余数相同的两数之差一定能被除数整除，也就是两个数除以 9 的余数相同, 它们的差一定是 9 的倍数,可找出除数是 9 的余数的所有情况:0、1、2、3、4、5、6、7、8，共9种，这样可以把它们看成9个抽屉，利用抽屉原理来解决。本题考查抽屉问题, 具体是把多于 m 个物品放到 m 个抽屉里, 至少有一个抽屉里的物品不止一个, 分的物品个数至少要比抽屉数多 1 , 就能满足至少有一个抽屉里的物品不止一个。解题关键在于理解余数相同的两数之差一定能被除数整除, 再把题目转化成抽屉问题, 根据题目信息判断有几个抽屉, 然后根据分的物体个数比抽屉数多 1 解答即可。\n【详解】一个自然数除以9，余数可能出现的情况为：0、1、2、3、4、5、6、7、8，这样自然数就被分成9类，把它们看成9个抽屉，我们考虑最不利原则取了这9类数，没有两个数的余数相同，当再取第10个数时，必定与之前取的9个数中某一数除以9余数相同，就满足了至少有两个数的差为9的倍数，因此至少要取10个数才能保证必然有两个数在同一抽屉里（也就是有两个数除以9余数相同），也就是它们的差是9的倍数。故答案为: C",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_226",
+        "original_id": "unit5_choice_226"
+      },
+      {
+        "id": "unit5_choice_227",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "小东玩掷骰子游戏，要保证掷出的骰子点数一定有两次是相同的小东至少应掷( )次。",
+        "options": [],
+        "answer": "7",
+        "solution": "【分析】骰子有 1、2、3、4、5、6 这 6 种不同的点数，小东掷骰子时，可以假设前面 6 次，分别掷出1、2、3、4、5、6点各一次，那么第7次结束后，一定有两次是相同点数。本题考查的是最不利原则，所谓最不利原则，通俗讲就是最倒霉的情况。\n【详解】6 + 1 = 7 （次）",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21 六年级下·全国·单元测试",
+        "merged_id": "unit5_227",
+        "original_id": "unit5_choice_227"
+      },
+      {
+        "id": "unit5_choice_228",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "把若干本书放进2个抽屉中，不管怎么放，总有一个抽屉至少放进 5 本书，这些书至少有( )本。",
+        "options": [],
+        "answer": "9",
+        "solution": "【分析】把若干本书放进 2 个抽屉中, 不管怎么放, 总有一个抽屉至少放进 5 本书, 那么不符合要求的最多的数量是每个抽屉里都放 4 本书, 这样再增加一本书, 不论在哪一个抽屉里, 一定是符合要求的。本题考查的是抽屉原理中的最不利原则，先找出不符合要求的最大数量，再加 1 就是符合要求的最小数量。\n【详解】5 - 1 = 4 (本) 4 × 2 = 8 (本)  8 + 1 = 9 (本)",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit5_228",
+        "original_id": "unit5_choice_228"
+      },
+      {
+        "id": "unit5_choice_229",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "在任意的 37 人中,至少有( )人是同一属相。",
+        "options": [],
+        "answer": "4",
+        "solution": "【分析】总共有 12 个属相, 那么抽屉数是 12, 总数是 37, 37 除以 12, 根据是否有余数进行判断。本题考查的是抽屉原理, 这里的抽屉数隐含在题目中，类似的例子还有'若干个人中，至少有多少人出生在同一月'的等。\n【详解】37 ÷ 12 = 3 … 1 3 + 1 = 4 （人） 至少有 4 人是同一属相。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21 六年级下·全国·单元测试",
+        "merged_id": "unit5_229",
+        "original_id": "unit5_choice_229"
+      },
+      {
+        "id": "unit5_choice_230",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "81 只鸽子飞回 20 个鸽房, 总有一个鸽房至少飞进( )只鸽子。",
+        "options": [],
+        "answer": "5",
+        "solution": "【分析】本题中抽屉数是 20, 总数是 81, 直接用 81 除以 20, 根据是否有余数进行判断。求解抽屉原理问题时，不论余数是多少，只有有余数，就是商加 1。\n【详解】81 ÷ 20 = 4 … 1 4 + 1 = 5 所以总有一个鸽房至少飞进 5 只鸽子。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21 六年级下·全国·单元测试",
+        "merged_id": "unit5_230",
+        "original_id": "unit5_choice_230"
+      },
+      {
+        "id": "unit5_choice_231",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "将 15 名学生分到 6个班级，总有一个班级至少分到( )名学生。",
+        "options": [],
+        "answer": "3",
+        "solution": "【分析】15名学生是总数，6个班级是抽屉数，15除以6，上2余3，只要有余数，不论余数是多少，商加上1即为所求。本题考查的是抽屉原理, 给出了抽屉数和整数,利用公式直接计算即可。\n【详解】15 ÷ 6 = 2 … 3 2 + 1 = 3 （名）",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21 六年级下·全国·单元测试",
+        "merged_id": "unit5_231",
+        "original_id": "unit5_choice_231"
+      },
+      {
+        "id": "unit5_choice_232",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "给一个长方体木块的6个面分别涂上红、黄两种颜色中的一种，则无论如何涂至少都有（ )个面的颜色相同。",
+        "options": [],
+        "answer": "3",
+        "solution": "【分析】长方体木块有 6 个面, 红、黄两种颜色进行染色,抽屉数是 2 , 总数是 6 , 6 除以 2 得到 3 , 没有余数, 那么至少都有 3 个面的颜色相同。本题中明确给出了总数和抽屉数，根据公式直接计算即可，在没有余数的情况下，商就是所求的结果。\n【详解】6 ÷ 2 = 3 （个）至少都有 3 个面的颜色相同。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit5_232",
+        "original_id": "unit5_choice_232"
+      },
+      {
+        "id": "unit5_choice_233",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "15 个学生要分到 6 个班, 至少( )个人要分进同一个班。",
+        "options": [],
+        "answer": "3",
+        "solution": "【分析】把 6 个班看作 6 个 '抽屉', 把 15 个人看作 '物体的个数', 根据抽屉原理进行解答即可。此题属于典型的抽屉原理习题, 解答此类题的关键是找出把谁看作 '抽屉个数', 把谁看作 '物体个数',然后根据抽屉原理解答即可。\n【详解】15 ÷ 6 = 2 … 3 （人） 2 + 1 = 3 （人）故答案为 3 。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "2014 六年级·全国·课后作业",
+        "merged_id": "unit5_233",
+        "original_id": "unit5_choice_233"
+      },{
+        "id": "unit5_choice_234",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "梁老师在给班上同学们分组，若想要一定有两个同学的生日在同一个月份，则这组至少有( )名同学。",
+        "options": [],
+        "answer": "13",
+        "solution": "【分析】一年有 12 个月，可以把这 12 个月看做 12 个抽屉，由此即可利用抽屉原理解答。\n【详解】建立抽屉：一年有 12 个月分别看做 12 个抽屉，把学生看做元素,考虑最差情况：每个抽屉都有 1 个元素，即一共有 12 名同学，此时再多一个同学无论放到哪个抽屉都会出现一个抽屉有 2 个元素,12 + 1 = 13 （名），答：这组至少有 13 个同学.此题考查了抽屉原理解决实际问题的灵活应用, 正确建立抽屉是解决本题的关键.",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "2014 六年级·全国·课后作业",
+        "merged_id": "unit5_234",
+        "original_id": "unit5_choice_234"
+      },
+      {
+        "id": "unit5_choice_235",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "从 1, 2, 3…, 12 这十二个数字中，任意取出 7 个数，其中两个数之差是 6 的至少有( )对。",
+        "options": [],
+        "answer": "1",
+        "solution": "【分析】将 1 ~ 12 这十二个数，组成构造 6 个抽屉：(1, 7), (2, 8), (3, 9), (4, 10),(5, 11), (6, 12); 任取 7 个数, 必定有两个数差在同一组中, 这一对数的差为 6。此题属于典型的抽屉原理习题, 解答此类题的关键是找出把谁看作 '抽屉个数', 把谁看作 '物体个数',然后根据抽屉原理解答即可。\n【详解】将 1 ~ 12 这十二个数，组成构造6个抽屉：(1, 7), (2, 8), (3, 9), (4, 10),(5, 11), (6, 12); 根据抽屉原理可得: 任意取出 7 个数, 其中两个数之差是 6 的至少有 1 对;故答案为: 1。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "2014 六年级·全国·课后作业",
+        "merged_id": "unit5_235",
+        "original_id": "unit5_choice_235"
+      },
+      {
+        "id": "unit5_choice_236",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "有 12 张扑克牌打乱后反扣在桌面上，其中有 5 张红桃，7 张黑桃，至少要摸出( )张扑克牌，才能保证一定能摸到红桃。",
+        "options": [],
+        "answer": "8",
+        "solution": "【分析】根据题干，从最不利情况分析：假设摸出 7 张全部是黑桃，此时再摸出 1 张，必定是红桃，据此即可解答问题。此题做题的关键是从最极端情况进行分析，进而通过分析得出问题答案。\n【详解】7 + 1 = 8 (张)至少要摸出 8 张扑克牌, 才能保证一定能摸到红桃。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20- 21 四年级上·江苏·单元测试",
+        "merged_id": "unit5_236",
+        "original_id": "unit5_choice_236"
+      },
+      {
+        "id": "unit5_choice_237",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "袋中有 7 个黄球, 5 个白球, 至少摸出( )个球, 才能保证其中一定有一个是黄球。",
+        "options": [],
+        "answer": "6",
+        "solution": "【分析】要保证其中一定有一个是黄球，如果很不巧，摸了 5 次全部都是白球，那么第 6 次摸出的球就一定是黄球。解决本题的关键是要看清题干信息中的'一定'。\n【详解】根据分析：至少摸出6个球，才能保证其中一定有一个是黄球。",
+        "knowledge_points": ["鸽巢问题", "事件的确定性与不确定性"],
+        "source": "20-21 四年级上·江苏·单元测试",
+        "merged_id": "unit5_237",
+        "original_id": "unit5_choice_237"
+      },
+      {
+        "id": "unit5_choice_238",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "袋中有 4 个红球, 5 个黄球, 6 个黑球。那么, 任意摸出 1 个球, 摸到黑球的可能性是( )；至少摸出( )个球, 才能保证有一个是红球。",
+        "options": [],
+        "answer": "2/5; 12",
+        "solution": "【分析】简单事件发生的可能性的计算步骤：1.列出所有可能发生的结果，并判定每个结果发生的可能性都相等2、确定所有可能发生的结果个数 n 和其中出现所求事件的结果个数 m; 3、用所求事件出现的可能结果个数 ÷ 所有可能发生的结果个数 = m/n 。考虑最差的情况，将所有的黄球和黑球都摸出来，再摸一个，一定是红球。本题考查了简单的可能性求解和抽屉问题, 求可能结果的个数均等比例分配, 而且只有在每个结果发生的可能性都相等的条件下才能进行均等比例分配。\n【详解】6 ÷ (4 + 5 + 6) = 6 ÷ 15 = 2/5\n\n5 + 6 + 1 = 12 (个)任意摸出 1 个球, 摸到黑球的可能性是 2/5; 至少摸出 12 个球,才能保证有一个是红球。",
+        "knowledge_points": ["简单事件发生的可能性求解", "鸽巢问题"],
+        "source": "2019·浙江杭州·小升初真题",
+        "merged_id": "unit5_238",
+        "original_id": "unit5_choice_238"
+      },
+      {
+        "id": "unit5_choice_239",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "六 (3) 班有一些同学今年都是 11 岁, 若要这些同学中有同月出生的, 这些同学至少有( )人。",
+        "options": [],
+        "answer": "13",
+        "solution": "【分析】1年有12个月，将12个月看成12个抽屉，1个抽屉1人，再来1人进入任意1个抽屉都有两名同学同月生日，据此解答。本题主要考查抽屉问题的应用，解题的关键是找到代表物体和抽屉的量，然后依据抽屉原则进行计算。\n【详解】12 + 1 = 13 （人）答: 若要这些同学中有同月出生的, 这些同学至少有 13 人。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_239",
+        "original_id": "unit5_choice_239"
+      },
+      {
+        "id": "unit5_choice_240",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "把 21 只小白兔放进5 个笼子里, 至少有( )只小白兔要放进同一个笼子里。",
+        "options": [],
+        "answer": "5",
+        "solution": "【分析】物体个数÷鸽巢个数=商……余数，至少个数=商+1。鸽巢原理是一个重要而又基本的组合原理，在解决数学问题时有非常重要的作用。\n【详解】21 ÷ 5 = 4 （只）……1（只）",
+        "knowledge_points": ["抽屉原理"],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_240",
+        "original_id": "unit5_choice_240"
+      },
+      {
+        "id": "unit5_choice_241",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "把 9 盒水彩笔分发给 5 个同学, 总有一个同学至少分到( )盒水彩笔。",
+        "options": [],
+        "answer": "2",
+        "solution": "【分析】因为是至少得到几盒水彩笔，所以考虑最差的情况，先5个人平均分1盒，那么还剩4盒需要分配，分给4个小朋友，所以总有一个同学最少要有 2 盒水彩笔。本题的关键是根据抽屉原理，在考虑最差情况的基础上得出平均数，然后根据至少数 = 平均数 +1（在有余数的情况下）。\n【详解】9 ÷ 5 = 1 （盒）……4（盒）1 + 1 = 2 (盒)所以总有一个同学至少分到 2 盒水彩笔。",
+        "knowledge_points": ["抽屉原理"],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_241",
+        "original_id": "unit5_choice_241"
+      },
+      {
+        "id": "unit5_choice_242",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "把 5 块糖果分给 4 个小朋友，总有一个小朋友至少分得( )块糖果。",
+        "options": [],
+        "answer": "2",
+        "solution": "【分析】因为是至少得到几颗糖，所以考虑最差的情况，先4个人平均分1块，那么还剩1块需要分配，分给1个小朋友，所以总有一个小朋友最少要有2块糖果。本题的关键是根据抽屉原理，在考虑最差情况的基础上得出平均数，然后根据至少数 = 平均数 +1（在有余数的情况下）。\n【详解】5 ÷ 4 = 1 （块）……1（块） 1 + 1 = 2 (块) 所以总有一个小朋友至少分到 2 块糖果。",
+        "knowledge_points": ["抽屉原理"],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_242",
+        "original_id": "unit5_choice_242"
+      },
+      {
+        "id": "unit5_choice_243",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "学校成立了书法、音乐、跳绳三个兴趣小组，每人至少参加一个小组，也可以参加两个或三个小组。六（3）班有47人，至少有（ ）人参加的兴趣小组相同。",
+        "options": [],
+        "answer": "7",
+        "solution": "【分析】同学们参加的兴趣小组有：（书法）、（音乐）、（跳绳）、（书法、音乐）、（书法、跳绳）、（音乐、跳绳）、（书法、音乐、跳绳），共7种可能，根据抽屉原理分析即可。本题考查了抽屉问题，抽屉原则二：如果把n个物体放在m个抽屉里，其中n＞m，那么必有一个抽屉至少有：当n不能被m整除时,k＝[n/m]+1个物体\n【详解】47 ÷ 7 = 6 （人）……5（人） 6 + 1 = 7 (人)",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_243",
+        "original_id": "unit5_choice_243"
+      },
+      {
+        "id": "unit5_choice_244",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "有黑色、白色、红色手套各4只，至少要拿出( )只（拿的时候不许看）才能使拿出的手套中一定有两只是同色的；至少要拿出( )只（拿的时候不许看）才能使拿出的手套能配成两双不同颜色的手套。",
+        "options": [],
+        "answer": "4 7",
+        "solution": "【分析】假如拿出的前 3 只手套, 黑色、白色、红色各 1 只,第 4 只无论拿出的是什么颜色, 都有一只与其颜色相同; 将一种颜色的手套全部拿出，另两种颜色一种拿一只，再拿一只无论什么颜色都能组成一双同色手套。本题考查了抽屉问题，要考虑极端的情况。\n【详解】3 + 1 = 4 （只）\n\n4 + 2 + 1 = 7 (只)",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_244",
+        "original_id": "unit5_choice_244"
+      },
+      {
+        "id": "unit5_choice_245",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "李叔叔参加射击比赛, 5 枪打出 46 环, 他至少有一枪不低于( )环。",
+        "options": [],
+        "answer": "10",
+        "solution": "【分析】抽屉原则二：如果把 n 个物体放在 m 个抽屉里，其中 n > m，那么必有一个抽屉至少有：当 n 不能被 m 整除时，k = [n/m] + 1 个物体。本题考查了抽屉问题, 将 5 枪看成 5 个抽屉, 46 环看成 46 个物体, 46 个物体放入 5 个抽屉, 至少有一个抽屉不低于 10 个。\n【详解】46 ÷ 5 = 9 （环）……1（环） 9 + 1 = 10 (环)",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_245",
+        "original_id": "unit5_choice_245"
+      },
+      {
+        "id": "unit5_choice_246",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "有黄、白两种颜色的乒乓球各8个，至少取出（）个就可以保证取出的球中有4个颜色相同。",
+        "options": [],
+        "answer": "7",
+        "solution": "【分析】最倒霉的情况, 取出的前 6 个乒乓球, 3 个黄色, 3 个白色, 第 7 个无论什么颜色都可以组成 4 个颜色相同的球。本题考查了抽屉问题，这种问题要考虑全面。\n【详解】3 + 3 + 1 = 7 （个）至少取出7个就可以保证取出的球中有4个颜色相同。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_246",
+        "original_id": "unit5_choice_246"
+      },
+      {
+        "id": "unit5_choice_247",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "抽屉里放着同样长短的红、黄铅笔各 4 支, 至少要摸出( )支铅笔, 才能保证至少有 1 支红铅笔。",
+        "options": [],
+        "answer": "5",
+        "solution": "【分析】考虑最倒霉的情况, 前 4 支摸出的都是黄铅笔, 第五支一定是红铅笔。本题考查了抽屉问题, 运气好摸一次即可, 要想保证有 1 支红笔得考虑最倒霉的情况。\n【详解】4 + 1 = 5 (支)至少要摸出 5 支铅笔, 才能保证至少有 1 支红铅笔。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_247",
+        "original_id": "unit5_choice_247"
+      },
+      {
+        "id": "unit5_choice_248",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "实验小学有 3200 人,六年级有 560 人, 该小学中至少有( )人的生日在同一天,六年级中至少有( )人的生日在同一个星期。",
+        "options": [],
+        "answer": "9 11",
+        "solution": "【分析】抽屉原则二：如果把 n 个物体放在 m 个抽屉里，其中 n > m，那么必有一个抽屉至少有：当 n 不能被 m 整除时，k = [n/m] + 1 个物体。本题考查了鸽巢问题, 当 n 能被 m 整除时, k = n/m 个物体。\n【详解】3200 ÷ 365 = 8 （人）……280（人）8 + 1 = 9 (人) 560 ÷ 52 = 10 (人) … 40 (人) 10 + 1 = 11 (人)\n\n实验小学有 3200 人, 六年级有 560 人, 该小学中至少有 9 人的生日在同一天, 六年级中至少有 11 人的生日在同一个星期。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_248",
+        "original_id": "unit5_choice_248"
+      },
+      {
+        "id": "unit5_choice_249",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "三 (2) 班有 49 名学生, 至少有( )名小朋友的生日在同一个月。",
+        "options": [],
+        "answer": "5",
+        "solution": "【分析】抽屉原则二：如果把 n 个物体放在 m 个抽屉里，其中 n > m ，那么必有一个抽屉至少有：(1) 当 n 不能被 m 整除时, k = [n/m] + 1 个物体。 (2) 当 n 能被 m 整除时, k = n/m 个物体。本题考查了鸽巢问题，也叫抽屉问题。\n【详解】49 ÷ 12 = 4 （名）……1（名）4 + 1 = 5三 (2) 班有 49 名学生, 至少有 5 名小朋友的生日在同一个月。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_249",
+        "original_id": "unit5_choice_249"
+      },
+      {
+        "id": "unit5_choice_250",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "5 只小鸡装入 4 个笼子, 至少有一个笼子放小鸡 2 只。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】5 只小鸡装入 4 个笼子, 总数是 5 只, 抽屉数是 4 个, 5 除以 4 , 商 1 余 1 , 有余数, 所以至少有一个笼子放小鸡 2 只。本题考查的是抽屉原理，用总数除以抽屉数，只要有余数，不论余数是多少，商加上1就是结果。\n【详解】5 ÷ 4 = 1 … 1 1 + 1 = 2 (只)5 只小鸡装入 4 个笼子, 至少有一个笼子放小鸡 2 只; 题干阐述正确, 答案为: √。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_250",
+        "original_id": "unit5_choice_250"
+      },
+      {
+        "id": "unit5_choice_251",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "因为 8 ÷ 3 = 2 … 2 所以8个球放进3个口袋中，其中至少有一个口袋放进了4个球。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】因为 8 ÷ 3 = 2 … 2, 此时每个口袋里放了 2 个球,余下的 2 个球无论是放进一个口袋, 还是分开放进两个口袋,都满足至少有一个口袋放进 3 个球。本题考查抽屉问题, 解决本题的关键是理解 '平均分'的思路, 利用公式 a ÷ n = b … c, 总有一个抽屉至少可以放 (b + 1) 个物体 (a 是物体个数, n 是抽屉个数) 来解决。\n【详解】8 ÷ 3 = 2 … 2 2 + 1 = 3 (个) 所以 8 个球放进 3 个口袋中, 其中至少有一个口袋放进了 3 个球, 因此题目说法错误。故答案为: ×",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_251",
+        "original_id": "unit5_choice_251"
+      },
+      {
+        "id": "unit5_choice_252",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "小华说: '我们组一共有 14 个人, 一年有 12 个月, 所以一定有人和我在同一个月过生日。'( )",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】假设把 14 个人平均分配给 12 个月, 每个月份有一个人过生日, 剩下的 2 人不管放哪里, 总有一个月份里至少有 2 个人过生日。在抽屉原理中，总有一个抽屉至少放了若干个苹果，但这个抽屉可不是固定的，而是任意的。\n【详解】把 14 个人看作 14 个苹果, 12 个月份看作 12 个抽屉, 因为 14 ÷ 12 = 1 （人）……2（人）1 + 1 = 2 （人），故一定有两个人在同一个月份过生日，只是这个月份不是指定的，故不一定有人和小华在同一个月过生日。故答案为×。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_252",
+        "original_id": "unit5_choice_252"
+      },
+      {
+        "id": "unit5_choice_253",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "小明说：'我们学校有700名学生，至少有3人同一天过生日。'（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】一年按 365 天算, 把 365 天看作 365 个抽屉, 利用抽屉原理解答即可。本题考查抽屉问题, 解答本题的关键是掌握一年按 365 天算, 把 365 天看作 365 个抽屉, 700 名学生放进 365 个抽屉，至少有 1 人同一天过生日。\n【详解】把 700 名学生放进 365 个抽屉, 则:700 ÷ 365 = 1 … 335此时 365 个抽屉里各有 1 人, 余下的 335 人, 按最差的情况考虑, 平均分到 335 个抽屉里, 则至少有 30 个抽屉里放 1 人, 所以至少有 1 人同一天过生日, 因此原题说法错误。故答案为: ×。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_253",
+        "original_id": "unit5_choice_253"
+      },
+      {
+        "id": "unit5_choice_254",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "4 只鸽子飞进A、B、C三个鸽舍，其中B鸽舍中至少飞进两只鸽子。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】抽屉原理：将 n + 1 个苹果放入 n 个抽屉中，则必有一个抽屉中至少有 2 个苹果。假设把 4 只鸽子平均分，每个鸽舍放 1 只，剩下的 1 只不管放哪里，总有一个鸽舍至少放了 2 只鸽子。本题属于抽屉原理, 也叫 '鸽巢问题'。想要求得任意一个抽屉至少有几个苹果, 前提是要把苹果平均分配。\n【详解】把 A、B、C 三个鸽舍看作三个抽屉, 4 只鸽子看作 4 个苹果, 因为 4 ÷ 3 = 1 (只) ……1 (只), 1 + 1 = 2(只), 故任意一个鸽舍中至少飞进两只鸽子。而不一定是 B 鸽舍至少飞进两只鸽子。故答案为 ×。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_254",
+        "original_id": "unit5_choice_254"
+      },
+      {
+        "id": "unit5_choice_255",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "投掷一枚硬币 3 次,至少有两次出现同一面。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】假设把投掷硬币的次数——3次平均分配给硬币的正反两面，每个面出现1次，剩下的1次不管哪个面朝上，总有一面至少出现了2次。投掷一枚硬币 3 次, 至少有两次出现同一面。这正符合把 n + 1 个苹果放到 n 个抽屉中, 总有一个抽屉至少有 2 个苹果的原理。\n【详解】把 3 次看作苹果数, 硬币的两面看作 2 个抽屉, 因为 3 ÷ 2 = 1 (次) ……1 (次), 1 + 1 = 2 (次), 所以投掷 3 次硬币时, 至少有两次出现同一面。故答案安为 √。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_255",
+        "original_id": "unit5_choice_255"
+      },
+      {
+        "id": "unit5_choice_256",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "有一样大小的白、黄、绿、粉色球各 5 个, 放在一个盒子里, 至少一次拿到 5 个, 才能保证拿到 2 个颜色不同的球。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】最坏打算: 5 个全是同一种颜色, 则再拿出 1 个即可保证拿到 2 个颜色不同的球; 据此解答。本题主要考查抽屉原理的应用, 解答时采用最坏打算。\n【详解】由分析可得：至少拿 5 + 1 = 6 个，才能保证拿到2个颜色不同的球。故答案为: ×",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_256",
+        "original_id": "unit5_choice_256"
+      },
+      {
+        "id": "unit5_choice_257",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "从一副没有大、小王的扑克牌中, 任意抽出五张, 一定有花色相同的。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】一副没有大、小王的扑克牌中有四种花色，将其视为 4 个抽屉，1 个抽屉 1 张花色，则再抽取 1 张则一定存在与所抽取花色相同的抽屉；据此解答。本题主要考查抽屉原理的应用，解题的关键是找到代表物体和抽屉的量，然后依据抽屉原则进行计算。\n【详解】5 ÷ 4 = 1 （张）……1（张）即最少有 1 + 1 = 2 张花色相同的。故答案为: √",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_257",
+        "original_id": "unit5_choice_257"
+      },
+      {
+        "id": "unit5_choice_258",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "在从1开始的连续19 个奇数中任取 6 个, 一定有两个数的和是 20。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】即在1、3、5、7、9、11、13、15、17、19、21、23、25、27、29、31、33、35、37这19个数中选6个数可以将这6个数分成（1，19），(3，17)，(5，15)，(7，13)，(9, 11) 这 5 组每组的 2 个数和为 20 , 将其视为 5 个抽屉。任选 6 个数由抽屉原理可知一定存在 6 个数中任意两个数的和不是 20 ; 据此解答。本题也可通过举例法进行验证，比如选27、29、31、33、35、37这六个数。\n【详解】由分析可得：在从 1 开始的连续 19 个奇数中任取 6 个，不一定有两个数的和是 20。故答案为: ×",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_258",
+        "original_id": "unit5_choice_258"
+      },
+      {
+        "id": "unit5_choice_259",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "把 10 个红球、 9 个黄球、 8 个绿球、 3 个蓝球混合后放到一个布袋里, 一次至少摸出多少个球才能保证有 2 个红球?",
+        "options": [],
+        "answer": "22 个",
+        "solution": "【分析】考虑最不利原则，可以先把除红球外的 9 个黄球、8 个绿球、3 个蓝球先全部取出来，取出来了 20 个球，里面一个红球有没有，再取出 2 个球即可保证有 2 个红球。本题考查的是最不利原则，在求解问题的时候，先要找出不符合要求的最大数量。\n【详解】9 + 8 + 3 + 2 = 22 （个）答: 至少摸出 22 个球才能保证有 2 个红球。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21 六年级下·全国·单元测试",
+        "merged_id": "unit5_259",
+        "original_id": "unit5_choice_259"
+      },
+      {
+        "id": "unit5_choice_260",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "六（1）班有 43 名同学订报纸，每人至少订一种报纸最多可订三种报纸。已知报纸有 A 、 B 、 C 三种。至少有几人订的报纸完全相同？",
+        "options": [],
+        "answer": "7人",
+        "solution": "【分析】每人至少订一种报纸最多可订三种报纸，可以先列举出订报纸的方式，方式的数量即为抽屉数，然后用 43 除以抽屉数，根据是否有余数，进行判断。本题考查的是抽屉原理, 也就是鸽巢问题, 用苹果数除以抽屉数, 如果没有余数, 结果就是商, 如果有余数, 商加 1 是结果。\n【详解】订报纸的方式:只订A，只订B，只订C，订A和B，订A和C，订B和C，订A、B和C，共7种订报纸的方式；\n\n43 ÷ 7 = 6 … 1 6 + 1 = 7 (个)答: 至少有 7 人订的报纸完全相同。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21 六年级下·全国·单元测试",
+        "merged_id": "unit5_260",
+        "original_id": "unit5_choice_260"
+      },
+      {
+        "id": "unit5_choice_261",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "一副除去大小王的扑克牌，共有4种花色，每种花色有13张。要保证抽出的牌中，4种花色的牌都有，至少要抽出多少张牌？",
+        "options": [],
+        "answer": "40 张",
+        "solution": "【分析】4种花色的牌都有，可以先把其它3种花色的牌全部取完，这是不符合要求的最大数量，再加上1，即可求出符合要求的最低数量。本题考查的是最不利原则，不符合要求的最大数量加 1 就是符合要求的最小数量。\n【详解】13 × 3 = 39 （张） 39 + 1 = 40 （张） 答: 至少要抽出 40 张牌。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·单元测试",
+        "merged_id": "unit5_261",
+        "original_id": "unit5_choice_261"
+      },
+      {
+        "id": "unit5_choice_262",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "把 165 本书分给六(3) 班的学生, 如果总有人至少分到 5 本书, 那么六 (3) 班最多有多少人？",
+        "options": [],
+        "answer": "41人",
+        "solution": "【分析】本题相当于是求抽屉数，如果没有余数，那么 5 就是商，用被除数 165 除以 5，得到的就是抽屉数；如果有余数，那么余数至少是 1，而此时商是 4，164 除以 4 得到抽屉数，然后进行比较。抽屉原理其实就是平均原则和最不利原则的体现,要使得数量最多的一个不那么突出, 那么就尽可能平均分。\n【详解】165 ÷ 5 = 33 （人）\n\n(165 - 1) ÷ (5 - 1) = 164 ÷ 4 = 41 (人)\n\n显然最多有 41 人;答: 六 (3) 班最多有 41 人。",
+        "knowledge_points": ["鸽巢问题进阶"],
+        "source": "20-21 六年级下·全国·单元测试",
+        "merged_id": "unit5_262",
+        "original_id": "unit5_choice_262"
+      },
+      {
+        "id": "unit5_choice_263",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "植树节, 育才小学有41名老师和381名学生参加义务植树活动。参加植树的老师至少有 4 人是同一个月出生的。参加植树的学生至少有2人的生日是同一天。他们说得对吗？",
+        "options": [],
+        "answer": "对",
+        "solution": "【分析】每年有 12 个月是固定的，每年 365 天或 366 天，用 41 除以 12，用 381 除以 365 或 366，根据是否有余数进行判断。本题考查的是抽屉原理，解决此类问题，首先要找出抽屉数和总数分别是多少。\n【详解】41 ÷ 12 = 3 … 5 3 + 1 = 4 （人）所以参加植树的老师至少有 4 人是同一个月出生的;\n\n381 ÷ 365 = 1 … 16 381 ÷ 366 = 1 … 15 1 + 1 = 2 （人） 不论这一年是多少天，参加植树的学生至少有 2 人的生日是同一天； 答: 他们说得对。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21 六年级下·全国·单元测试",
+        "merged_id": "unit5_263",
+        "original_id": "unit5_choice_263"
+      },
+      {
+        "id": "unit5_choice_264",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "幼儿园某班有32名小朋友，现有各种玩具108个，把这些玩具全部分给这32名小朋友，总有一名小朋友至少得到多少个玩具？",
+        "options": [],
+        "answer": "4 个",
+        "solution": "【分析】108 ÷ 32 = 3 （个）……12（个），将108个玩具平均分给32名小朋友，每名小朋友分到3个，还剩12个。把剩余的12个继续分给32名小朋友中的某几名，总有一名小朋友至少得到 (3 + 1) 个玩具。此题考查了利用抽屉原理解决实际问题的灵活应用，关键是从最差情况考虑。\n【详解】108 ÷ 32 = 3 （个）……12（个） 3 + 1 = 4 (个) 答：总有一名小朋友至少得到 4 个玩具。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "20-21六年级下·全国·课后作业",
+        "merged_id": "unit5_264",
+        "original_id": "unit5_choice_264"
+      },
+      {
+        "id": "unit5_choice_265",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "希望小学有 367 人，请问有没有两个学生的生日是同一天？为什么？",
+        "options": [],
+        "answer": "有;见详解。",
+        "solution": "【分析】如果这一年为闰年，即有366天，367 ÷ 366 = 1 …… 1 1 + 1 = 2 (人)(2)如果这一年为闰年, 即有 365 天, 367 ÷ 365 = 1 … 2, 1 + 1 = 2 (人)所以不管是闰年还是平年，都至少有两个学生的生日是同一天的。\n【详解】如果这一年为闰年，即有366天，367 ÷ 366 = 1 …… 1 1 + 1 = 2 (人)(2)如果这一年为闰年, 即有 365 天, 367 ÷ 365 = 1 … 2, 1 + 1 = 2 (人)所以不管是闰年还是平年，都至少有两个学生的生日是同一天的。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "2014 六年级·全国·课后作业",
+        "merged_id": "unit5_265",
+        "original_id": "unit5_choice_265"
+      },
+      {
+        "id": "unit5_choice_266",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "把 9 本书放进 2 个抽屉里，总有一个抽屉至少放进 5 本书，为什么？",
+        "options": [],
+        "answer": "平均每个抽屉放 4 本后，还余 1 本，这 1 本无论放到哪个抽屉，那个抽屉都会有5本。",
+        "solution": "【分析】把 9 本书放进 2 个抽屉, 9 ÷ 2 = 4 (本) ……1 (本), 即平均每个抽屉放 4 本后, 还余 1 本, 所以至少有一个抽屉至少要放: 4 + 1 = 5 本; 据此即可解答。在此类抽屉问题中，至少数 = 物体数除以抽屉数的商 +1（有余数的情况下）。\n【详解】9 ÷ 2 = 4 （本）……1（本） 4 + 1 = 5 （本）所以把 9 本书放进 2 个抽屉里, 总有一个抽屉至少要放 5 本。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "2014 六年级·全国·课后作业",
+        "merged_id": "unit5_266",
+        "original_id": "unit5_choice_266"
+      }, {
+        "id": "unit5_choice_267",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "一副扑克牌(大、小王除外)，有四种花色，每种花色有 13 张，从中任意抽牌，至少要抽几张，才能保证有四张牌是同一花色的？",
+        "options": [],
+        "answer": "13 张",
+        "solution": "【分析】3 × 4 + 1 = 13 (张)\n【详解】3 × 4 + 1 = 13 (张)",
+        "knowledge_points": ["抽屉原理"],
+        "source": "2014 六年级·全国·课后作业",
+        "merged_id": "unit5_267",
+        "original_id": "unit5_choice_267"
+      },
+      {
+        "id": "unit5_choice_268",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "红、黄、蓝三种颜色的球各 6 个，混合后放在一个布袋里，一次至少摸出几只，才能保证有两只是同色的？",
+        "options": [],
+        "answer": "一次至少摸出 4 个",
+        "solution": "【分析】把 3 种不同颜色看作 3 个抽屉, 把不同颜色的球看作元素, 从最不利情况考虑, 每个抽屉先放 1 个球, 共需要 3 个, 再取出 1 个不论是什么颜色, 总有一个抽屉里的球和它同色, 所以至少要取出: 3 + 1 = 4 (个), 据此解答。\n【详解】3 + 1 = 4 （个）；答: 一次至少摸出 4 个, 才能保证有两个是同色的。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "2014 六年级·全国·课后作业",
+        "merged_id": "unit5_268",
+        "original_id": "unit5_choice_268"
+      },
+      {
+        "id": "unit5_choice_269",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "从 1、2、3、…、1998、1999这些自然数中，最多可以取多少个数，才能使其中每两个数的差不等于4？",
+        "options": [],
+        "answer": "1000 个",
+        "solution": "【分析】把这组数据先划分成四组公差为 4 的等差数列, 则差是 4 的数都在同一个数列之中, 由此即可进行推理解答。本题难度较大, 关键是掌握满足条件的数的特征,然后有的放矢的进行解答。注意不要漏解。\n【详解】把 1, 2, 3...1998, 1999 这 1999 个数分成四组公差是 4 的等差的数列,1, 5, 9, 13...1993, 1997 - - - - 共 500 个数; 2, 6, 10, 14...1994, 1998 - - - - 共500个数; 3, 7, 11, 15...1995, 1999 - - - - 共 500 个数; 4, 8, 12, 16...1992, 1996 - - - - 共 499 个数;我们发现：1. 四行中每一行中任意相邻两数相差为 4，不相邻两数相差不可能是 4；而分属不同两行的任意两个数相差不可能为 4 , 因为如果相差为 4 的话, 两数将被归为一行, 这显然与事实矛盾;故我们用这样的方法来选符合规定的数：前三行每隔一个数选一个，每行最多可选250个数；第四行先选4，再隔一个数字选一个，可选出250个，最终得到 250 × 4 = 1000 个数。答：最多可以取1000个数，才能使其中每两个数的差不等于4。",
+        "knowledge_points": ["抽屉原理", "等差数列"],
+        "source": "2014 六年级·全国·课后作业",
+        "merged_id": "unit5_269",
+        "original_id": "unit5_choice_269"
+      },
+      {
+        "id": "unit5_choice_270",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "学校图书馆有科普读物、故事书、连环画三种图书。每个学生从中借阅两本。那么至少要多少个学生借阅才能保证其中一定有2个人所借阅的图书属于同一种类。",
+        "options": [],
+        "answer": "4个",
+        "solution": "【分析】首先把科普读物、故事书、连环画三种图书任意两本排列，一共有（科普读物，故事书）, (科普读物,连环画), (故事书, 连环画) 三种情况, 看做三个抽屉,只要学生数比抽屉多 1 就可以达到要求。此题属于典型的抽屉原理习题, 解答此类题的关键是找出把谁看作 '抽屉个数', 把谁看作 '物体个数',然后根据抽屉原理解答即可。\n【详解】按（科普读物，故事书），（科普读物，连环画），（故事书，连环画）三种情况，构造三个抽屉， 3 + 1 = 4 （个），答: 至少要 4 个学生借阅才能保证其中一定有 2 个人所借阅的图书属于同一种类。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "2014六年级·全国·课后作业",
+        "merged_id": "unit5_270",
+        "original_id": "unit5_choice_270"
+      },
+      {
+        "id": "unit5_choice_271",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "王老师借来了历史、文艺和科普三种书若干本。每个同学从中任意借一本或两本，那么至少要几个同学借阅才能保证一定有两人借的图书一样？",
+        "options": [],
+        "answer": "10 个",
+        "solution": "【分析】每个学生从中任意借1本，有3种借法，借两本，那么一共有6种借法：历史两本、文艺两本、科普两本、历史和文艺各一本、历史和科普各一本、文艺和科普各一本；所以一共有 3 + 6 = 9 种不同的借法，把9种借法看作9个抽屉，把学生数看作元素，从最不利情况考虑，每个抽屉先放1个元素，共需要9个，再取出1个不论是哪一种借法，总有一个抽屉里和他相同，所以至少要有： 9 + 1 = 10 （个），据此解答。\n【详解】每个学生从中任意借两本，那么一共有 9 种借法：只借 1 本，有三种情况；借两本：历史两本、文艺两本、科普两本、历史和文艺各一本、历史和科普各一本、文艺和科普各一本， 9 + 1 = 10 （个）；答：那么至少要 10 个学生才能保证一定有两人接到的图书是一样的。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "2014 六年级·全国·课后作业",
+        "merged_id": "unit5_271",
+        "original_id": "unit5_choice_271"
+      },
+      {
+        "id": "unit5_choice_272",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "给一个七边形的 7 条边分别涂上红、黑两种颜色，不论怎么涂，至少有 4 条边涂的颜色相同。为什么？",
+        "options": [],
+        "answer": "7 ÷ 2 = 3 （种） … 1 （条） 3 + 1 = 4 （种）",
+        "solution": "【分析】7 ÷ 2 = 3 （种） … 1 （条） 3 + 1 = 4 （种）\n【详解】略",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "2014 六年级·全国·课后作业",
+        "merged_id": "unit5_272",
+        "original_id": "unit5_choice_272"
+      },
+      {
+        "id": "unit5_choice_273",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "体育用品仓库里有许多足球、排球和篮球，某班 50 名同学来仓库拿球，规定每个人至少拿 1 个球，至多拿 2 个球，问至少有几名同学所拿的球种类是一致的？",
+        "options": [],
+        "answer": "至少有 6 名同学所拿的球种类是一致的",
+        "solution": "【分析】根据抽屉问题的解答步骤：可先找抽屉。由题意可知，拿一个球有 3 种可能，拿两个一样球的有 3 种可知，同时拿两个不同球的也有 3 种可能，即具体为拿球的配组方式有以下 9 种：（足），（排），（篮）；（足，足），（排，排），（篮，篮）；（足，排），（足，篮），（排，篮）。把这 9 种配组方式看作 9 个抽屉。因为 50 ÷ 9 = 5 … 5 ，所以至少有 5 + 1 = 6 （名）同学所拿的球的种类是完全一样的。先根据排列组合的有关知识求出抽屉的个数是完成本题的关键。\n【详解】由题意可知，拿球的配组方式有：\n\n3 + 3 + 3 = 9 (种) , 50 ÷ 9 = 5 (名) … 5 个 5 + 1 = 6 (名)答: 至少有 6 名同学所拿的球种类是一致的。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "2014 六年级·全国·课后作业",
+        "merged_id": "unit5_273",
+        "original_id": "unit5_choice_273"
+      },
+      {
+        "id": "unit5_choice_274",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "一次考试有 10 道题,每道题的评分标准是：回答完全正确得5分，回答不完全正确得3分；回答错误或不回答得0分。至少有多少人参加考试，才能保证至少有3人得分相同？试说明原因。",
+        "options": [],
+        "answer": "至少有 91 人参加考试, 才能保证至少有 3 人得分相同",
+        "solution": "【分析】最低得分为 0 分, 最高得分为 50 分, 分数在 0 ~ 50 分之间, 由于 1 分, 2 分, 4 分, 7 分, 47 分, 49 分都不可能出现, 所以共有 45 种得分情况, 求至少有多少人参加考试, 才能保证至少有 3 人得分相同, 最坏的打算是每种得分情况都有 2 人, 那么再有 1 个, 才能保证至少有 3 人得分相同, 从而得出问题答案.\n【详解】最低得分为 0 分, 最高得分为 50 分, 分数在 0 ~ 50 分之间, 由于 1 分, 2 分, 4 分, 7 分, 47 分, 49 分都不可能出现, 所以共有 45 种得分情况,至少: 45 × 2 + 1 = 91 （人）;答: 至少有 91 人参加考试, 才能保证至少有 3 人得分相同.",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "2014 六年级·全国·课后作业",
+        "merged_id": "unit5_274",
+        "original_id": "unit5_choice_274"
+      },
+      {
+        "id": "unit5_choice_275",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "口袋里装有42个红球、15个黄球、20个绿球、14个白球和9个黑球。至少要摸出多少个球，才能保证其中有15个球的颜色是相同的？",
+        "options": [],
+        "answer": "66个",
+        "solution": "【分析】红球、黄球、绿球数目都超过了 15 个, 白球和黑球数目没超过15个，考虑最不利的情况，先把9个黑球、14 个白球全部摸出，再把红球、黄球、绿球各摸出 14 个,红球、黄球、绿球还有剩余，只要在它们中再摸出 1 个，就能保证其中有 15 个球的颜色是相同的。本题考查抽屉问题，解答本题的关键是理解考虑最不利的情况，当把白球、黑球全部摸出后，再把红球、黄球、绿球各摸出14个，此时只要在剩余的红球、黄球、绿球中任意摸出1个，就能保证其中有15个球的颜色是相同的。\n【详解】考虑最不利的情况，先把9个黑球全部摸出，14个白球全部摸出，此时仍然没有 15 个球的颜色是相同的,继续摸出 14 个黄球, 14 个绿球, 14 个红球, 此时已摸出65 个球, 只要在剩余的红球、黄球、绿球中任意摸出 1 个,就能保证其中有 15 个球的颜色是相同的, 则至少摸出: 9 + 14 + 14 + 14 + 14 + 1 = 66 (个)答: 至少要摸出 66 个球, 才能保证其中有 15 个球的颜色是相同的。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_275",
+        "original_id": "unit5_choice_275"
+      },
+      {
+        "id": "unit5_choice_276",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "参加数学竞赛的210名学生中，能否保证有18名或18名以上的学生出生的月份相同？为什么？",
+        "options": [],
+        "answer": "能。因为 210 ÷ 12 = 17 … 6, 余下的 6 人一定会和前面其他人的出生月份相同, 所以至少有 18 名学生出生的月份相同。",
+        "solution": "【分析】一年有 12 个月, 把 12 个月看作 12 个抽屉, 根据抽屉原理来解答即可。本题考查抽屉问题, 解答本题的关键是理解一年有 12 个月, 考虑最差的情况, 这 210 名学生出生月份分到 12 个月, 每个月都有 17 人, 余下 6 人, 无论在哪个月出生,都保证至少有 18 人在同一个月出生。\n【详解】一年有 12 个月, 看作 12 个抽屉, 把 210 名学生分到 12 个抽屉, 每个抽屉里有 17 人, 余下 6 人无论在哪个月出生, 都保证至少有 18 人出生的月份相同, 用算式表示为: 210 ÷ 12 = 17 … 6 17 + 1 = 18 （人）答: 能, 因为 210 ÷ 12 = 17 … 6, 余下的 6 人一定会和前面其他人的出生月份相同, 所以至少有 18 名学生出生的月份相同。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_276",
+        "original_id": "unit5_choice_276"
+      },
+      {
+        "id": "unit5_choice_277",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "一个布袋中有60块大小、形状相同的木块，编上号码1、2、3、4的各有15块。一次至少要摸出多少块木块，才能保证其中至少有3块的号码相同？",
+        "options": [],
+        "answer": "9 块",
+        "solution": "【分析】把号码 1、2、3、4 的木块看作 4 个抽屉，利用抽屉原理解答即可。本题考查抽屉问题, 解答本题关键在于理解考虑最差的情况下, 每个抽屉摸出 2 块木块, 共有 8 块, 再多摸出 1 块, 无论在哪个抽屉里摸出, 都能保证至少有 3 块的号码相同。\n【详解】把号码 1、2、3、4 的木块看作 4 个抽屉，考虑最差的情况，每个抽屉摸出 2 块木块，则共有 2 × 4 = 8 块，只要再摸出 1 块，无论在哪个抽屉里摸出，都满足至少有一个抽屉摸出 3 块，也就是保证至少有 3 块的号码相同，用算式表示为：2 × 4 = 8 (块)8 + 1 = 9 (块)答: 一次至少要摸出 9 块木块, 才能保证其中至少有 3 块的号码相同。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_277",
+        "original_id": "unit5_choice_277"
+      },
+      {
+        "id": "unit5_choice_278",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "学校田径运动会，六年级男生共有26名学生报名参加 50m 、 100m 和 200m 这三项中的一项、两项或三项。这26名学生中参加项目完全相同的至少有几人？",
+        "options": [],
+        "answer": "4人",
+        "solution": "【分析】报名参加 50 m 、 100 m 和 200 m 这三项中的一项、两项或三项, 共有 7 种情况, 把这 7 种情况看作 7 个抽屉,利用抽屉原理解答即可。本题考查抽屉原理, 解答本题的关键是找到抽屉的数量, 也就是报名参加 50 m 、 100 m 和 200 m 这三项中的一项、两项或三项, 共有 7 种情况, 把这 7 种情况看作 7 个抽屉。\n【详解】每名学生报名参加 50 m 、 100 m 和 200 m 这三项中的一项、两项或三项, 共有 7 种情况, 把这 7 种情况看作 7 个抽屉, 26 名学生分到 7 个抽屉, 每个抽屉分到 3 人, 余下的 5 人无论到哪个抽屉, 都满足至少有 3 + 1 = 4 人在同一个抽屉, 也就是至少有 4 人报名参加项目完全相同, 用算式表达是:26 ÷ 7 = 3 … 5 3 + 1 = 4 (人)答: 这 26 名学生中参加项目完全相同的至少有 4 人。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_278",
+        "original_id": "unit5_choice_278"
+      },
+      {
+        "id": "unit5_choice_279",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "从自然数 10—20 里,至少取多少个数，才能保证其中一定有一个数是 6 的倍数？",
+        "options": [],
+        "answer": "10 个",
+        "solution": "【分析】自然数 10—20 里, 6 的倍数有 2 个, 不是 6 的倍数. 有: 11 - 2 = 9 个, 根据最不利原则考虑即可。本题考查抽屉问题, 解答本题的关键是理解自然数 10—20 里共有 11 个数, 其中 6 的倍数有 2 个, 不是 6 的倍数有 9 个。\n【详解】自然数 10—20 里, 6 的倍数有 2 个, 不是 6 的倍数. 有: 11 - 2 = 9 (个)考虑最差的情况，取了 9 个数都不是 6 的倍数，再多取 1 个数，就满足至少有一个数是 6 的倍数，则至少取： 9 + 1 = 10 （个）。答: 至少取 10 个数, 才能保证其中一定有一个数是 6 的倍数。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_279",
+        "original_id": "unit5_choice_279"
+      },
+      {
+        "id": "unit5_choice_280",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "把一些桃子放进了3个盘子里，总有一个盘子里至少有3个桃子，这些桃子一定是7个。这句话对吗？为什么？",
+        "options": [],
+        "answer": "不对。因为 3 个盘子, 总有一个盘子里至少有 3 个桃子, 桃子数是不定的, 最少是 7 个而非一定是 7 个。",
+        "solution": "【分析】利用'鸽巢原理'进行分析即可。3个盘子即是3个抽屉，考虑最差的情况，每个抽屉放2个桃，共 3 × 2 = 6 个，则再分1个，无论分到哪个抽屉里，都会出现至少有一个盘子里有3个桃，此时桃子数最少为： 3 × 2 + 1 = 7 （个）。本题主要考查对'鸽巢原理'的理解和应用。\n【详解】不对。因为 3 个盘子, 总有一个盘子里至少有 3 个桃子, 桃子数是不定的, 最少是 7 个, 不是一定是 7 个。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_280",
+        "original_id": "unit5_choice_280"
+      },
+      {
+        "id": "unit5_choice_281",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "任意连续的6个非0自然数中，一定有一个数是6的倍数，为什么？",
+        "options": [],
+        "answer": "因为自然数除以 6 所得的余数只有 6 种情况: 正好整除, 余数是 1, 余数是 2, 余数是 3, 余数是 4, 余数是 5, 把 6 种余数情况看作 6 个抽屉, 考虑最坏的情况 6 个连续的自然数被平均分到这 6 个抽屉, 则每个抽屉里都有一个数, 所以至少有一个数能被 6 整除, 因此连续的 6 个非 0 自然数中一定有一个数能被 6 整除, 这个数就是 6 的倍数。",
+        "solution": "【分析】因为任意一个非 0 自然数除以 6 会出现 6 种情况:正好整除, 余数是 1 , 余数是 2 , 余数是 3 , 余数是 4 , 余数是 5 , 把 6 种余数情况看作 6 个抽屉, 根据抽屉原理解答即可。本题考查抽屉问题，解答本题关键是掌握任意一个非0自然数除以6会出现6种情况：正好整除，余数是1，余数是2，余数是3，余数是4，余数是5。\n【详解】因为自然数除以 6 所得的余数只有 6 种情况: 正好整除, 余数是 1, 余数是 2, 余数是 3, 余数是 4, 余数是 5, 把 6 种余数情况看作 6 个抽屉, 考虑最坏的情况 6 个连续的自然数被平均分到这 6 个抽屉, 则每个抽屉里都有一个数, 所以至少有一个数能被 6 整除, 因此连续的 6 个非 0 自然数中一定有一个数能被 6 整除, 这个数就是 6 的倍数。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_281",
+        "original_id": "unit5_choice_281"
+      },
+      {
+        "id": "unit5_choice_282",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "10个人分组打扫卫生，要保证其中一个组至少有4人，最多可以分成几组？",
+        "options": [],
+        "answer": "3组",
+        "solution": "【分析】抽屉原则二：如果把 n 个物体放在 m 个抽屉里，其中 n > m ，那么必有一个抽屉至少有：(1) 当 n 不能被 m 整除时, k = [n/m] + 1 个物体。 (2) 当 n 能被 m 整除时, k = n/m 个物体。本题考查了抽屉问题, 构造物体和抽屉, 也就是找到代表物体和抽屉的量, 然后依据抽屉原则进行计算。\n【详解】10 ÷ 4 = 2 (组) ……2 (人) 2 + 1 = 3 (组) 答: 最多可以分成 3 组。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_282",
+        "original_id": "unit5_choice_282"
+      },
+      {
+        "id": "unit5_choice_283",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "一场篮球比赛，A队得 62 分，A 队上过场的队员有 7 人。一定有一名队员得分不低于多少分？",
+        "options": [],
+        "answer": "9 分",
+        "solution": "【分析】利用'鸽巢原理'进行分析即可。本题主要考查对'鸽巢原理'的理解和应用。\n【详解】62 ÷ 7 = 8 （分）……6（分）如果每名队员得了 8 分, 那么就有 6 分没有分配, 所以至少有一名队员得分不低于 9 分。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_283",
+        "original_id": "unit5_choice_283"
+      },
+      {
+        "id": "unit5_choice_284",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "六（1）班45名同学分成6个组玩'老鹰捉小鸡'游戏，总有一个组至少有多少人？",
+        "options": [],
+        "answer": "8人",
+        "solution": "【分析】因为是每组至少有几人，所以考虑最差的情况，把45个人平均分6组，那么还剩3人需要分配，分给3个组，所以总有一个组最少要有8人。本题的关键是根据抽屉原理，在考虑最差情况的基础上得出平均数，然后根据至少数 = 平均数 +1（在有余数的情况下）。\n【详解】45 ÷ 6 = 7 （人）……3（人） 7 + 1 = 8 （人）答: 总有一个组至少有 8 人。",
+        "knowledge_points": ["抽屉原理"],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_284",
+        "original_id": "unit5_choice_284"
+      },
+      {
+        "id": "unit5_choice_285",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "六 (3) 班同学分成5 个组进行跳绳比赛，不管怎么分，总有一个组至少有 10 人。六（3）班至少有学生多少人？",
+        "options": [],
+        "answer": "46人",
+        "solution": "【分析】根据抽屉原理，至少数 = 平均数 +1（在有余数的情况下）。因为每组至少有 10 人， 10 - 1 = 9 （人），总人数分成了 5 组，每组 9 人，总人数就是 9 × 5 + 1 = 46 （人）。此题考查的是对抽屉原理的问题的理解。\n【详解】(10 - 1) × 5 + 1 = 9 × 5 + 1 = 45 + 1 = 46 (人)\n\n答: 六 (3) 班至少有学生 46 人。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_285",
+        "original_id": "unit5_choice_285"
+      },
+      {
+        "id": "unit5_choice_286",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "有红、黄、蓝、白 4 色的小球各 10 个, 混合放在一个布袋里。一次摸出小球 9个，其中至少有几个小球的颜色是相同的？",
+        "options": [],
+        "answer": "3 个",
+        "solution": "【分析】红、黄、蓝、白 4 种颜色看作是 4 个抽屉, 利用抽屉原理来解答即可。本题考查抽屉问题, 关键在于理解 '平均分' 的思路, 总结规律: 至少数 = 被分物体个数 ÷ 抽屉个数 +1。\n【详解】把红、黄、蓝、白 4 种颜色看作是 4 个抽屉, 9 个球往抽屉里面放, 考虑最差的情况, 每个抽屉摸出 2 个球, 2 × 4 = 8 个, 则余下 1 个球, 无论从哪个抽屉里摸出, 都会出现至少有 3 个小球的颜色相同: 9 ÷ 4 = 2 …… 1 2 + 1 = 3 (个)故答案为: 3 个。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_286",
+        "original_id": "unit5_choice_286"
+      },
+      {
+        "id": "unit5_choice_287",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "池塘里有 6 只青蛙跳到4片荷叶上，总有一片荷叶上至少有2只青蛙。为什么？",
+        "options": [],
+        "answer": "如果每片荷叶上跳上 1 只青蛙, 那么余下是 2 只无论跳到哪片荷叶上总有一片荷叶上至少有 2 只青蛙。",
+        "solution": "【分析】这是一道比较简单的抽屉问题，把这道题转化成抽屉问题解答即可。本题考查抽屉问题, 具体是把多于 k n (k 是正整数) 个物体任意分放进 n 个空抽屉, 那么一定有一个抽屉中放进了至少 (k + 1) 个物体。解决本题的关键是理解'平均分'的思路，利用公式 a ÷ n = b 。这个数是一个抽屉，总有一个抽屉至少可以放（ b + 1 ）个物体（ a 是物体个数， n 是抽屉个数）来解决。\n【详解】6 只青蛙跳到 4 片荷叶上, 按平均分的方法, 4 片荷叶各有一只青蛙共 4 只, 还余下 2 只青蛙要跳到荷叶上,无论这余下的 2 只青蛙, 是同时跳到一片荷叶上, 还是分开跳到两片荷叶上, 总会满足总有一片荷叶上至少有 2 只青蛙,用算式表达就是: 6 ÷ 4 = 1 …… 2 。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_287",
+        "original_id": "unit5_choice_287"
+      },
+      {
+        "id": "unit5_choice_288",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "实验小学合唱队有60人，年龄最大是12岁，年龄最小是6岁，他们当中至少有几人的年龄相同？",
+        "options": [],
+        "answer": "9人",
+        "solution": "【分析】6 到 12 岁有 6 岁、7 岁、8 岁、9 岁、10 岁、11 岁、12 岁, 共 7 种不同的年龄, 每种年龄对应 8 人, 余 4 人, 根据抽屉原理, 至少有 8 + 1 人年龄相同。抽屉原理问题的解答思路是：要从最不利情况考虑，准确地建立抽屉和确定元素的总个数，然后根据'至少数 = 元素的总个数 ÷ 抽屉的个数 +1（有余数的情况下）'解答。\n【详解】60 ÷ 7 = 8 （人）……4（人） 8 + 1 = 9 (人) 答: 他们当中至少有 9 人的年龄相同。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_288",
+        "original_id": "unit5_choice_288"
+      },
+      {
+        "id": "unit5_choice_289",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "某次数学竞赛, 六(1) 班有 4 名同学参加, 总分为 365 分, 则一定至少有一名学生的得分不低于 90 分。为什么?",
+        "options": [],
+        "answer": "平均每人得 91 分还余 1 分, 余下的 1 分无论分给哪一名学生, 都会出现 92 分。",
+        "solution": "【分析】考虑最差情况: 4 名同学的得分尽量的平均, 则每人得分是: 365 ÷ 4 = 91 (分) … 1 (分), 余下的 1 分无论分给哪一名学生, 都会出现 92 分, 据此即可解答.抽屉原理问题的解答思路是：要从最不利情况考虑，准确地建立抽屉和确定元素的总个数，然后根据'至少数 = 元素的总个数 ÷ 抽屉的个数 +1 （有余数的情况下）'解答。\n【详解】365 ÷ 4 = 91 （分）……1（分） 90 + 1 = 91 （分）答: 平均每人得 91 分还余 1 分, 余下的 1 分无论分给哪一名学生, 都会出现 92 分。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_289",
+        "original_id": "unit5_choice_289"
+      },
+      {
+        "id": "unit5_choice_290",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "从 1 ~ 20 这20个自然数中，至少要取出几个不同的数，才能保证其中一定有一个数是2的倍数？",
+        "options": [],
+        "answer": "11个",
+        "solution": "【分析】1 ~ 20 中2的倍数有：2、4、6、8、10、12、14、16、18、20，共10个2的倍数，不是2的倍数有：1、3、5、7、9、11、13、15、17、19，共10个不是2的倍数，最不利的情况是取出的前10个都不是 2 的倍数, 再取一个一定是 2 的倍数。本题考查了抽屉原理, 抽屉原理的解答思路, 从最不利情况考虑, 准确地建立抽屉和确定元素的总个数。\n【详解】10 + 1 = 11 （个）答: 至少要取出 11 个不同的数。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_290",
+        "original_id": "unit5_choice_290"
+      },
+      {
+        "id": "unit5_choice_291",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "书架上有6本故事书, 8本文艺书, 10本连环画。(1) 从书架上取书, 要想取出的书一定有 4 本是同一种类的, 至少要取出多少本? (2) 要想取出的书一定有 3 个种类, 至少要取出多少本?",
+        "options": [],
+        "answer": "(1) 10本 (2) 19本",
+        "solution": "【分析】（1）从最不利情况考虑，每种都取出 3 本，再取一本，无论是哪种书，一定有 4 本是同一种类的，据此分析；(2) 从最不利情况考虑, 连续取出 10 本连环画, 又连续取出 8 本文艺书, 再取一本一定是故事书, 一定有 3 个种类,据此分析。本题考查了抽屉原理, 抽屉原理的解答思路, 从最不利情况考虑, 准确地建立抽屉和确定元素的总个数。\n【详解】(1) 3 × 3 + 1 = 9 + 1 = 10 (本)\n\n答: 至少要取出 10 本。\n\n(2) 10 + 8 + 1 = 19 (本)答: 至少要取出 19 本。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_291",
+        "original_id": "unit5_choice_291"
+      },
+      {
+        "id": "unit5_choice_292",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "把60本图书分给25名留守儿童，总有一名留守儿童至少分到几本？为什么？",
+        "options": [],
+        "answer": "3 本；抽屉原则二：如果把 n 个物体放在 m 个抽屉里，其中 n > m ，那么必有一个抽屉至少有：当 n 不能被 m整除时, k = [n/m] + 1 个物体。",
+        "solution": "【分析】根据抽屉原理进行分析, 抽屉原理的解答思路,从最不利情况考虑, 准确地建立抽屉和确定元素的总个数。本题考查了抽屉原理， [n/m] 表示 n ÷ m 的整数商。\n【详解】60 ÷ 25 = 2 （本）……10（本） 2 + 1 = 3 (本)\n\n答: 总有一名留守儿童至少分到 3 本, 抽屉原则二: 如果把 n 个物体放在 m 个抽屉里, 其中 n > m, 那么必有一个抽屉至少有: 当 n 不能被 m 整除时, k = [n/m] + 1 个物体。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_292",
+        "original_id": "unit5_choice_292"
+      },
+      {
+        "id": "unit5_choice_293",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "六（2）班有48人，每人至少订一份刊物，现有甲、乙、丙三种刊物，每人有几种选择方式？这个班订相同刊物的至少有多少人？",
+        "options": [],
+        "answer": "7种选择方式；7人",
+        "solution": "【分析】现有甲、乙、丙三种刊物, 每人至少订一份刊物,则有甲、乙、丙、甲乙、甲丙、乙丙、甲乙丙 7 种选择方式。7 种选择方式看作 7 个 '抽屉', 48 看作 '物体个数', 根据抽屉原理 48 ÷ 7 = 6 人……6 人, 这个班订相同刊物的至少有 6 + 1 = 7 人。此题要理清什么是'抽屉'，什么是'物品'，解题的关键是制造'抽屉'，确定假设的'物品'，根据'抽屉少，物品多'转化为抽屉原理来解。\n【详解】有甲、乙、丙、甲乙、甲丙、乙丙、甲乙丙 7 种选择方式。\n\n48 ÷ 7 = 6 (人) … 6 (人)  6 + 1 = 7 (人) 答: 有 7 种选择方式。这个班订相同刊物的至少有 7 人。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_293",
+        "original_id": "unit5_choice_293"
+      },
+      {
+        "id": "unit5_choice_294",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "某次数学竞赛有9名学生参加，总分是825分，则至少有一名学生不低于多少分？",
+        "options": [],
+        "answer": "92分",
+        "solution": "【分析】此题属于简单的抽屉问题，把9名学生看作9个'抽屉'，825看作'物体个数'，825÷9=91分……6分，则至少有一名学生不低于 91 + 1 = 92 分。此题考查简单的抽屉问题，解答方法为：至少数 = 商 + 1 （有余数的情况下）。\n【详解】825 ÷ 9 = 91 （分）……6（分） 91 + 1 = 92 (分) 答: 则至少有一名学生不低于 92 分。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_294",
+        "original_id": "unit5_choice_294"
+      },
+      {
+        "id": "unit5_choice_295",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "某地元月份的天气有晴、阴、多云、雨、雪这五种情况，至少有多少天是同一种天气？",
+        "options": [],
+        "answer": "7天",
+        "solution": "【分析】元月份有 31 天，晴、阴、多云、雨、雪这五种情况看作 5 个抽屉， 31 ÷ 5 = 6 （天）……1（天），即平均每种天气情况有 6 天，还余 1 天，所以至少有 6 + 1 = 7 （天）是同一种天气。此题考查简单的抽屉问题, 分清 31 天看作物体总个数, 五种天气情况看作 5 个抽屉, 解答方法为: 至少数 = 商 +1 (有余数的情况下)。\n【详解】31 ÷ 5 = 6 （天）……1（天） 6 + 1 = 7 (天) 答: 至少有 7 天是同一种天气。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_295",
+        "original_id": "unit5_choice_295"
+      },
+      {
+        "id": "unit5_choice_296",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "1998 年 3 月份出生的任意32名同学中，至少有2人是同一天出生的，为什么？",
+        "options": [],
+        "answer": "因为 3 月份有 31 天, 32 ÷ 31 = 1 … 1, 所以一定至少有 2 人是同一天出生的。",
+        "solution": "【分析】3月份有31天，把31天看作31个抽屉，32名同学看作物体总个数； 32 ÷ 31 = 1 （人）……1（人），即平均每天出生1人的话，还余1人，所以至少有 1 + 1 = 2 （人）是同一天出生的。此题考查简单的抽屉问题, 如果把 (n + 1) 个物体放在 n 个抽屉里, 那么必有一个抽屉中至少放有 2 个物体。\n【详解】因为 3 月份有 31 天, 32 ÷ 31 = 1 (人) …… 1 (人),即平均每天出生 1 人的话, 还余 1 人, 所以至少有 1 + 1 = 2 (人) 是同一天出生的。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_296",
+        "original_id": "unit5_choice_296"
+      },
+      {
+        "id": "unit5_choice_297",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "有4个运动员练习投篮，一共投进了35个球，一定有1个运动员至少投进几个球？",
+        "options": [],
+        "answer": "9 个",
+        "solution": "【分析】此题考查简单的抽屉问题，4个运动员看作4个抽屉，一共投进35个球看作物体总个数； 35 ÷ 4 = 8 （个）……3（个），即平均每个运动员进8个球的话，还余3个球，所以一定有一个运动员至少投进 8 + 1 = 9 个球。此题考查简单的抽屉问题，解答方法为：至少数 = 商 + 1 （有余数的情况下）。\n【详解】35 ÷ 4 = 8 （个）……3（个） 8 + 1 = 9 (个) 答: 一定有 1 个运动员至少投进 9 个球。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_297",
+        "original_id": "unit5_choice_297"
+      },
+      {
+        "id": "unit5_choice_298",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "六（2）班有45人，男生、女生的人数比是3:2，随机选取，至少选多少人才能保证选出的人中男生和女生都有？",
+        "options": [],
+        "answer": "28人",
+        "solution": "【分析】根据比例的知识可知男生女生各有 27 人、18 人,所以如果必须保证选中的人有男有女, 那么要作最坏的打算即全是男生, 把 27 位男生都选完了, 再选一定是女生,所以至少选 (27+1) 人即 28 人。此类题此题主要考查了鸽巢原理的运用, 要从最坏的情况考虑。\n【详解】男生人数: 45 × 3/(3 + 2) = 27 (人)女生人数: 45 × 2/(3 + 2) = 18 (人) 27 + 1 = 28 (人) 答: 至少选 28 人才能保证选出的人中男生和女生都有。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_298",
+        "original_id": "unit5_choice_298"
+      },
+      {
+        "id": "unit5_choice_299",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "几个要好的朋友去A、B、C三个景点游玩，每人只游览其中两个景点，不管他们怎样安排游览方案，都至少有4个人游览的景点完全相同。请问至少有几人去游玩？",
+        "options": [],
+        "answer": "10人",
+        "solution": "【分析】我们可以根据鸽巢原理公式倒着推, 即如果把 n 个物体放在 m 个鸽巢里, 其中 n > m, 那么必有一个鸽巢至少有: k = (n ÷ m) + 1 个物体 (当 n 不能被 m 整除时)。此题把游玩的总人数看成分放的物体总数 n 。游览方案有以下 3 种: AB、AC、BC, 把 3 种游览方案看成 3 个鸽巢数 m。至少有 4 个人游览景点相同, 就是要使其中一个鸽巢里至少有 4 人, 则游玩的总人数至少要比鸽巢数的 (4-1) 倍多 1 个。运用逆推法解决鸽巢问题。\n【详解】游览方案有以下3种：AB、AC、BC。\n\n(4 - 1) × 3 + 1 = 3 × 3 + 1 = 9 + 1 = 10 (人) 。\n\n答: 至少有 10 人去游玩。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_299",
+        "original_id": "unit5_choice_299"
+      },
+      {
+        "id": "unit5_choice_300",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "容易",
+        "question": "袋子里有同样大小的红、白、黄、蓝颜色的球各5个，至少取出多少个球，可以保证取到两个颜色相同的球？",
+        "options": [],
+        "answer": "5个",
+        "solution": "【分析】从最坏的情况考虑，假如前四次摸到四种颜色，那么再摸一次无论是什么颜色都能保证有两个颜色相同的球。此类题此题主要考查了鸽巢原理的运用, 要从最坏的情况考虑。\n【详解】袋子里有 4 种颜色的球, 只要摸出的球比它们的颜色种数多 1 , 就能保证有两个球同色。 4 + 1 = 5 (个)答: 至少取出 5 个球, 可以保证取到两个颜色相同的球。",
+        "knowledge_points": ["鸽巢问题"],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_300",
+        "original_id": "unit5_choice_300"
+      },{
+        "id": "unit5_choice_301",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "5个自然数的和是偶数，那么这5个自然数中至少有()个自然数是偶数。",
+        "options": [
+          "A. 1",
+          "B. 2",
+          "C. 3",
+          "D. 4"
+        ],
+        "answer": "A",
+        "solution": "【分析】根据抽屉原理和奇偶性，进行选择即可。奇数+奇数=偶数，偶数+偶数=偶数。\n【详解】5个自然数的和是偶数，那么这5个自然数中至少有1个自然数是偶数。故答案为:A",
+        "knowledge_points": [
+          "鸽巢问题",
+          "运算性质（奇数和偶数）"
+        ],
+        "source": "18-19六年级下·全国·课后作业",
+        "merged_id": "unit5_301",
+        "original_id": "unit5_choice_301"
+      },
+      {
+        "id": "unit5_choice_302",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "10只鸽子飞进4个鸽笼中，则至少有一个鸽笼中的鸽子不少于()只。",
+        "options": [
+          "A. 1",
+          "B. 2",
+          "C. 3",
+          "D. 4"
+        ],
+        "answer": "C",
+        "solution": "【分析】抽屉原则二：如果把n个物体放在m个抽屉里，其中n>m，那么必有一个抽屉至少有：当n不能被m整除时，k=[n/m]+1个物体。本题考查了抽屉问题，[n/m]表示n÷m的整数商。\n【详解】10÷4=2……2，2+1=3(只)故答案为:C",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_302",
+        "original_id": "unit5_choice_302"
+      },
+      {
+        "id": "unit5_choice_303",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "饲养员给10只猴子分苹果，其中至少要有一只猴子得到7个苹果，饲养员至少要拿来()个苹果。",
+        "options": [
+          "A. 70",
+          "B. 59",
+          "C. 61",
+          "D. 11"
+        ],
+        "answer": "C",
+        "solution": "【分析】假如10只猴子每只都得到6个苹果，再拿1个苹果即可保证至少有一只猴子得到7个苹果。本题考查了抽屉问题，考虑这类问题要全面。\n【详解】(7-1)×10+1=6×10+1=60+1=61(个)\n故答案为:C。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_303",
+        "original_id": "unit5_choice_303"
+      },
+      {
+        "id": "unit5_choice_304",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "李老师给学生买衣服，有红、黄、白3种颜色，结果总是至少有3个学生的衣服颜色一样，她至少给()个学生买衣服。",
+        "options": [
+          "A. 7",
+          "B. 8",
+          "C. 9"
+        ],
+        "answer": "A",
+        "solution": "【分析】把颜色的种类看作'抽屉'，把学生的数量看作物体的个数，根据抽屉原理得出：学生的个数至少比颜色的种类的(3—1)倍多1时，才能保证至少有3个学生的颜色一样；据此解答。运用逆推法解决抽屉问题。\n【详解】(3—1)×3+1=6+1=7(个)\n答:老师至少给7个学生买衣服。故选:A。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_304",
+        "original_id": "unit5_choice_304"
+      },
+      {
+        "id": "unit5_choice_305",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "某幼儿园有160名小朋友，至少有()名小朋友同一个月出生。",
+        "options": [
+          "A. 13",
+          "B. 14",
+          "C. 15"
+        ],
+        "answer": "B",
+        "solution": "【分析】把12个月看作12个抽屉，把160名小朋友看作物体总个数。根据抽屉原理，如果把n个物体放在m个抽屉里，其中n>m，那么必有一个抽屉至少有k=(n÷m)+1个物体(当n不能被m整除时)。据此列式计算即可。此题把160人看作物体总个数，12个月看作12个抽屉，解答方法为:至少数=商+1(有余数的情况下)。\n【详解】160÷12=13(人)……4(人)，13+1=14(人)答:至少有14人的属相相同。故选B。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_305",
+        "original_id": "unit5_choice_305"
+      },
+      {
+        "id": "unit5_choice_306",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "李阿姨给孩子买衣服，有红、黄、绿三种颜色，但结果总是至少有两个孩子的颜色一样，李阿姨至少有()个孩子。",
+        "options": [
+          "A. 2",
+          "B. 3",
+          "C. 4"
+        ],
+        "answer": "C",
+        "solution": "【分析】根据鸽巢原理(一)：如果把(n+1)个物体放在n个抽屉里，那么必有一个抽屉里至少放有两个物体；建立正确的抽屉，进行解答即可。本题主要考查了鸽巢原理的应用，关键是要认真分析题意，熟练运用鸽巢原理建立正确的抽屉。\n【详解】将红、黄、蓝三种颜色的衣服看作三个抽屉，因为李阿姨家的孩子总会有两个孩子的衣服的颜色一样，那说明至少有一个抽屉里至少有两件衣服，也就是说李阿姨家至少有3+1=4(个)孩子。故答案为:C。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "2014六年级·全国·课后作业",
+        "merged_id": "unit5_306",
+        "original_id": "unit5_choice_306"
+      },
+      {
+        "id": "unit5_choice_307",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "李叔叔给正方体的六个面涂上不同的颜色，结果至少有两个面的颜色是一致的，颜料的颜色至少有()种。",
+        "options": [
+          "A. 3",
+          "B. 4",
+          "C. 5"
+        ],
+        "answer": "A",
+        "solution": "【分析】正方体有6个面，总是至少有两个面的颜色是一致的，根据鸽巢原理考虑最差的情况：6个面中每两个面的颜色都相同，这时颜料的颜色是最少的，所以至少有6÷2=3(种)；据此解答即可。本题主要考查了鸽巢原理的应用，关键是要仔细分析题意，熟练运用鸽巢原理考虑最差的情况进行解答。\n【详解】由分析可得：考虑最差的情况：6个面中每两个面的颜色都相同，此时所需的颜料颜色至少为6÷2=3(种)；故选：A。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_307",
+        "original_id": "unit5_choice_307"
+      },
+      {
+        "id": "unit5_choice_308",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "聪聪玩掷骰子游戏，要保证掷出来的点子有两次相同，他至少要掷()次。",
+        "options": [
+          "A. 6",
+          "B. 7",
+          "C. 8"
+        ],
+        "answer": "B",
+        "solution": "【分析】一个骰子的点数有1,2,3,4,5,6，共6种情况，考虑最差的情况：前面每一个点都掷出来一次，这时只要再掷出一次，不管是哪个点，都能与前面的点数相同，从而能够保证掷出来的点子有两次相同。本题主要考查了鸽巢原理的应用，关键是要认真分析题意，根据鸽巢原理考虑最差的情况时再多进行一次即可解答。\n【详解】考虑最差的情况:6个点数都被掷出来一次，再掷出来一次不管是哪个点都能与前面的6个点中的某一点相同，所以要掷的次数为:6+1=7(次)故选:B",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "2014六年级·全国·课后作业",
+        "merged_id": "unit5_308",
+        "original_id": "unit5_choice_308"
+      },
+      {
+        "id": "unit5_choice_309",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "把12颗荔枝放入5个果盘里，总有一个果盘里至少放入()颗。",
+        "options": [
+          "A. 2",
+          "B. 3",
+          "C. 4"
+        ],
+        "answer": "B",
+        "solution": "【分析】将5个果盘看作5个抽屉，用荔枝总数除以5得出商，根据鸽巢原理(二):不能整除时至少数=商+1，能整除时至少数=商；据此解答。本题主要考查了鸽巢原理的应用，关键是要认真分析题意，选取合适的鸽巢原理进行解答。\n【详解】12÷5=2(颗)……2(颗)，2+1=3(颗)答:总有一个果盘里至少放入3颗。故选:B",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_309",
+        "original_id": "unit5_choice_309"
+      },
+      {
+        "id": "unit5_choice_310",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "把9只兔子放进4个笼子，总有一个笼子至少要放入()只兔子。",
+        "options": [
+          "A. 2",
+          "B. 3",
+          "C. 4"
+        ],
+        "answer": "B",
+        "solution": "【分析】将4个笼子看作四个抽屉，用兔子总数除以4得出商，根据鸽巢原理(二):不能整除时至少数=商+1，能整除时至少数=商；据此解答即可。本题主要考查了鸽巢原理的应用，关键是要理解鸽巢原理(二)中不能整除时至少数=商+1，能整除时至少数=商。\n【详解】9÷4=2(只)……1(只)，2+1=3(只)答:总有一个笼子至少要放入3只兔子。故选：B",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_310",
+        "original_id": "unit5_choice_310"
+      },
+      {
+        "id": "unit5_choice_311",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "给一个正方体木块的6个面分别涂上蓝黄两种颜色，不论怎么涂至少有()个面图的颜色相同。",
+        "options": [
+          "A. 2",
+          "B. 3",
+          "C. 4"
+        ],
+        "answer": "B",
+        "solution": "【分析】此题根据抽屉原理，把两种颜色看作两个抽屉，把6个面看作6个苹果，那么不管怎么涂至少有三个面的颜色相同。本题也可以分类讨论：如果1个面涂黄色，则5面涂绿色；如果2个面涂黄色，则4个面涂绿色；如果3个面涂黄色，则3个面涂绿色；如果4个面涂黄色，则2个面涂绿色；如果5个面涂黄色，则1个面涂绿色；所以不论怎么涂至少有3个面涂的颜色相同。\n【详解】6÷2=3(个)所以无论怎么涂至少有3个面图的颜色相同。故答案为:B",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_311",
+        "original_id": "unit5_choice_311"
+      },
+      {
+        "id": "unit5_choice_312",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "一个布袋里有黑、白、灰三种颜色的袜子各10只，最少要摸()只才能保证其中至少有2只颜色不相同的袜子。",
+        "options": [
+          "A. 13",
+          "B. 14",
+          "C. 11"
+        ],
+        "answer": "C",
+        "solution": "【分析】极端思想：用最不利的摸法先摸出同种颜色的袜子，再摸出1只颜色的袜子，都能保证一定有2只袜子是不同色的。考查了学生分析问题的能力，利用极端思想是解决此题的好方法。\n【详解】考虑最差情况：摸出10只袜子，都是同一种颜色，那么再任意摸出1只袜子，一定可以保证有2只袜子的颜色不相同，即：10+1=11(只)答:最少要摸11只才能保证其中至少有2只颜色不相同的袜子。故答案为:C",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_312",
+        "original_id": "unit5_choice_312"
+      },
+      {
+        "id": "unit5_choice_313",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "阳光幼儿园有157名小朋友，至少有()名小朋友同一个月出生。",
+        "options": [
+          "A. 12",
+          "B. 13",
+          "C. 14",
+          "D. 15"
+        ],
+        "answer": "C",
+        "solution": "【分析】每年有12个月，用157除以12，假如每个月都有13人出生，那么余下的人数无论在哪个月出生，都至少有14人是同一个月出生的。物体个数÷鸽巢个数=商……余数；至少个数=商+1，注意关键是找到鸽巢个数:12。\n【详解】157÷12=13(名)……1(名)，13+1=14(名)答:至少有14名小朋友同一个月出生。故答案为:C",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_313",
+        "original_id": "unit5_choice_313"
+      },
+      {
+        "id": "unit5_choice_314",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "某班有男生25人，女生18人，下面说法正确的是()。",
+        "options": [
+          "A. 至少有2名男生是在同一个月出生的",
+          "B. 至少有2名女生是在同一个月出生的",
+          "C. 全班至少有5个人是在同一个月出生的",
+          "D. 以上选项都有误"
+        ],
+        "answer": "B",
+        "solution": "【分析】一年有12个月，把这12个月看做12个抽屉，把男女生的人数看做元素，由此利用抽屉原理逐项即可解答。此题属于典型的抽屉原理习题，解答此类题的关键是找出把谁看作'抽屉个数'，把谁看作'物体个数'，然后根据抽屉原理解答即可。\n【详解】A.25÷12=2(人)……1(人)，2+1=3(人)至少有3名男生是在同一个月出生的;B.18÷12=1(人)……6(人)，1+1=2(人)即，至少有2名女生的生日在同一个月；C.(25+18)÷12=3(人)……7(人)，3+1=4(人)全班至少有4个人是在同一个月出生的;故答案为:B",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_314",
+        "original_id": "unit5_choice_314"
+      },
+      {
+        "id": "unit5_choice_315",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "箱子中有5个红球，4个白球，至少要取出()个才能保证两种颜色的球都有，至少要取()个才能保证有2个白球。应选()。",
+        "options": [
+          "A. 6、7",
+          "B. 7、6",
+          "C. 5、4",
+          "D. 1、1"
+        ],
+        "answer": "A",
+        "solution": "【分析】把两种颜色分别看做2个抽屉，利用抽屉原理，考虑最差情况，即可解答问题。此题考查了抽屉原理在实际问题中的灵活应用。\n【详解】建立抽屉：把两种颜色分别看做2个抽屉：(1)根据抽屉原理:考虑最差情况，5个红球全部取出来，那么再任意取出1个都是白球，5+1=6(个)所以至少取出6个球才能保证两种颜色的球都有;(2)根据抽屉原理:考虑最差情况:取出5个红球和1个白球，那么再任意取出1个球，就会出现2个白球，5+1+1=7(个)所以至少取出7个球才能保证有2个白球;箱子中有5个红球，4个白球，至少要取出6个才能保证两种颜色的球都有，至少要取7个才能保证有2个白球。故答案为：A。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_315",
+        "original_id": "unit5_choice_315"
+      },
+      {
+        "id": "unit5_choice_316",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "在学校科技比赛中，有31名同学报名参加了航模、海模和创意制作三个项目的比赛，总有一个项目至少有()名同学参加。",
+        "options": [
+          "A. 9",
+          "B. 10",
+          "C. 11",
+          "D. 12"
+        ],
+        "answer": "C",
+        "solution": "【分析】本着尽量平均分配的原则是本题的关键思路。此类'至少'题型只要进行除法计算，再将商加上1就可以得到结果。\n【详解】将此问题看作鸽巢问题。31名同学相当于31只鸽子，3个比赛项目相当于3个鸽巢，将31名同学平均分配到3个比赛项目中，每项比赛10名，还剩1名，剩余的1名无论分到哪个项目中，总会有一个项目至少有11名同学。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_316",
+        "original_id": "unit5_choice_316"
+      },
+      {
+        "id": "unit5_choice_317",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "学校篮球队4名队员练习投篮，共投进50个球。总有一名队员至少投进()个球。",
+        "options": [
+          "A. 10",
+          "B. 11",
+          "C. 12",
+          "D. 13"
+        ],
+        "answer": "D",
+        "solution": "【分析】本着尽量平均分配的原则是本题的关键思路。此类'至少'题型只要进行除法计算，再将商加上1就可以得到结果。\n【详解】将此问题看作鸽巢问题。4名队员相当于4个鸽巢，50个进球相当于50只鸽子，将50个进球平均分配给4名队员，每名队员进12个球，还剩2个进球，剩余的2个进球无论分给哪名队员，总会有一名队员至少进13个球。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_317",
+        "original_id": "unit5_choice_317"
+      },
+      {
+        "id": "unit5_choice_318",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "学校彩旗队有红、黄、蓝、绿四种颜色的旗子各12面，至少取()面彩旗，可以保证取到2面颜色相同的旗子。",
+        "options": [
+          "A. 4",
+          "B. 5",
+          "C. 10",
+          "D. 11"
+        ],
+        "answer": "B",
+        "solution": "【分析】可以用假设的方法，逐一排除即可。取比种类数多1就可保证有2面颜色相同的旗子。\n【详解】旗子共有4种颜色，最不理想的情况下，取4面旗子，颜色各不相同，再取1面无论是什么颜色，都能保证取到2面颜色相同的旗子。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_318",
+        "original_id": "unit5_choice_318"
+      },
+      {
+        "id": "unit5_fill_319",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "把17支铅笔放进4个文具盒里，至少有一个文具盒放___支。",
+        "options": [],
+        "answer": "5",
+        "solution": "【分析】把17支笔放进4个文具盒中，17÷4=4(支)……1支，即平均每个文具盒放4支，还余1支，根据鸽巢原理可知，总有一个文具盒里至少放4+1=5支。在此类鸽巢问题中，铅笔数相当于鸽子数，文具盒相当于鸽巢数，至少数=鸽子数除以鸽巢数的商+1(有余数的情况下)。\n【详解】17÷4=4(支)……1支，4+1=5(支)答:至少有一个文具盒不少于5支。故答案为:5。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_319",
+        "original_id": "unit5_fill_319"
+      },
+      {
+        "id": "unit5_fill_320",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "数学兴趣小组有28人，至少有()人的属相相同。",
+        "options": [],
+        "answer": "3",
+        "solution": "【分析】把12种属相看作12个抽屉，根据抽屉原理，如果把n个物体放在m个抽屉里，其中n>m，那么必有一个抽屉至少有k=(n÷m)+1个物体(当n不能被m整除时)。据此列式计算即可。此题把28人看作物体总个数，12种属相看作12个抽屉。\n【详解】28÷12=2(人)……4(人)，2+1=3(人)答:至少有3人的属相相同。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_320",
+        "original_id": "unit5_fill_320"
+      },
+      {
+        "id": "unit5_fill_321",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "9个人住进5个房间，至少有()个人要住进同一个房间。",
+        "options": [],
+        "answer": "2",
+        "solution": "【分析】此题属于典型的抽屉原理的习题，应明确房间数即'抽屉'；人数即'物体个数'；把多于n个的物体放到n个抽屉里，则至少有一个抽屉里有2个或2个以上的物体。解答此类题的关键是找出把谁看作'抽屉个数'，把谁看作'物体个数'。\n【详解】9÷5=1(人)……4(人)，1+1=2(人)答:至少有2个人要住进同一个房间。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_321",
+        "original_id": "unit5_fill_321"
+      },
+      {
+        "id": "unit5_fill_322",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "有20张扑克牌(不同花色的5,6,7,8,9各4张)，洗一下牌反扣在桌子上，至少摸出()张才能保证有两张牌的颜色(红或黑)是相同的，至少摸出()张才能保证四种花色的牌都有，至少摸出()张才能保证有三张同一花色的。",
+        "options": [],
+        "answer": "3；16；11",
+        "solution": "【分析】从最坏的情况考虑，假如前2次摸到2种颜色1红和1黑，那么再摸一次无论是什么颜色都能保证有两张牌的颜色(红或黑)是相同的。从最坏的情况考虑，假如3种花色的牌5，6，7，8，9各3张都摸完了，就剩同一种的花色，那么再摸一次能保证四种花色的牌都有。从最坏的情况考虑，假如2种花色的牌5，6，7，8，9各2张都摸完了，那么再摸一次能保证能保证有三张同一花色的。此类题此题主要考查了鸽巢原理的运用，要从最坏的情况考虑。\n【详解】有2种颜色的扑克牌，只要摸出的扑克牌数比它们的颜色种数多1，就能保证有有两张牌的颜色(红或黑)是相同的。2+1=3(张)答:至少摸出3张才能保证有两张牌的颜色(红或黑)是相同的。有四种花色的牌，每种花色的牌各有5张，3种花色的牌有15张，只要摸出的扑克牌数比15多1，就能保证四种花色的牌都有。5×3+1=15+1=16(张)答:至少摸出16张才能保证四种花色的牌都有。有四种花色的牌，每种花色的牌各有5张，2种花色的牌有10张，只要摸出的扑克牌数比10多1，就能保证有三张同一花色的。5×2+1=10+1=11(张)答:至少摸出11张才能保证有三张同一花色的。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_322",
+        "original_id": "unit5_fill_322"
+      },
+      {
+        "id": "unit5_fill_323",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "小王训练射击，共射6发，成绩是55环，小王至少有()发不低于10环。",
+        "options": [],
+        "answer": "1",
+        "solution": "【分析】此题属于简单的抽屉原理的习题，6发看作6个抽屉，55看作物体总个数；55÷6=9(环)……1(环)，即平均每发9环，还余1环，所以必须至少有1个十环，其他5个9环，6发子弹才可以55环。简单抽屉问题解答方法为:至少数=商+1(有余数的情况下)。\n【详解】55÷6=9(环)……1(环)，9+1=10(环)答:小王至少有1发不低于10环。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_323",
+        "original_id": "unit5_fill_323"
+      },
+      {
+        "id": "unit5_fill_324",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "一副扑克牌(去掉大、小王)共52张，至少摸出()张牌，就能保证有两张牌的花色相同；至少摸出()张牌，才能保证至少有两种花色；至少摸出()张牌，才能保证有四种花色的牌都有。",
+        "options": [],
+        "answer": "5；14；40",
+        "solution": "【分析】(1)52张扑克牌，共有4种花色，要保证有两张牌的花色相同；考虑最差的情况：前面摸出的花色各不一样，即红桃，方块，黑桃，梅花各一张，此时再任意摸出一张牌不管是哪种花色都能和之前的四种花色中的任一种相同;(2)扑克牌每一种花色的牌有13张，考虑最差的情况:前面摸出的13张牌都是同一种花色，此时再任意摸出一张牌。无论是哪种花色，这时只需再摸出一张牌，就能保证至少有两种花色;(3)考虑最差的情况:3种花色的牌都被摸出，此时再任意摸出一张牌。无论是哪种花色，都能保证四种花色的都有；据此解答即可。本题主要考查了鸽巢原理的应用，关键是要认真分析题意，熟练运用鸽巢原理。\n【详解】由分析可得：(1)4+1=5(张)答:至少摸出5张牌，就能保证有两张牌的花色相同。(2)13+1=14(张)答:至少摸出14张牌，才能保证至少有两种花色。(3)13×3+1=40(次)答:至少摸出40张牌，才能保证有四种花色的牌都有。故答案为:5,14,40。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_324",
+        "original_id": "unit5_fill_324"
+      },
+      {
+        "id": "unit5_fill_325",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "盒子里有同样大小的白球和黄球各5个，要想从盒子里摸出的球一定有2个是同色的，至少要摸出()个球。",
+        "options": [],
+        "answer": "3",
+        "solution": "【分析】由题意可知，盒子里有同样大小的白球和黄球各5个，要想从盒子里摸出的球一定有2个是同色的；由于只有两种颜色，考虑最差的情况:前面两次取的球的颜色都不一样即一白一黄，这时根据鸽巢原理，再摸出一个球不管这个球的颜色是白还是黄，都和之前的颜色重复；由此即可进行解答。本题主要考查了鸽巢原理的应用，关键是要理解鸽巢原理(一):如果把(n+1)个物体放在n个抽屉里，那么必有一个抽屉里至少放有两个物体。\n【详解】考虑最差的情况：前面两次取的球的颜色都不一样即一白一黄，这时再摸出一个球都能和之前的颜色重复，所以至少要摸出2+1=3(个)球。故答案为:3",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "2021·甘肃陇南·小升初真题",
+        "merged_id": "unit5_325",
+        "original_id": "unit5_fill_325"
+      },
+      {
+        "id": "unit5_fill_326",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "三个小朋友做游戏，至少有()个小朋友的性别相同。",
+        "options": [],
+        "answer": "2",
+        "solution": "【分析】鸽巢原理(一)：如果把(n+1)个物体放在n个抽屉里，那么必有一个抽屉里至少放有两个物体。由于性别只有男与女两种情况，把这两种情况看作两个抽屉，根据鸽巢原理，把三个小朋友的性别放在这两个抽屉里，有一个抽屉里至少存在两种性别；据此解答即可。本题主要考查了鸽巢原理的应用，关键是要认真分析题意，建立正确的抽屉，再运用合适的鸽巢原理进行解答。\n【详解】由分析可得：三个小朋友做游戏，至少有2个小朋友的性别相同。故答案为:2。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_326",
+        "original_id": "unit5_fill_326"
+      },
+      {
+        "id": "unit5_fill_327",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "6只子飞进5个鸽舍，总有一个鸽舍里飞进()只鸽子。",
+        "options": [],
+        "answer": "2",
+        "solution": "【分析】根据鸽巢原理(一)：如果把(n+1)个物体放在n个抽屉里，那么必有一个抽屉里至少放有两个物体；进行解答即可。本题主要考查鸽巢原理的简单应用，关键是要认真分析题意，理解鸽巢原理并灵活应用。\n【详解】由分析可得:6只子飞进5个鸽舍，总有一个鸽舍里飞进2只鸽子。故答案为:2",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_327",
+        "original_id": "unit5_fill_327"
+      },
+      {
+        "id": "unit5_fill_328",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "把(m+1)个物体放进m个抽屉里，总有一个抽屉放进()个物体。",
+        "options": [],
+        "answer": "2",
+        "solution": "【分析】根据鸽巢原理：如果把(n+1)个物体放在n个抽屉里，那么必有一个抽屉里至少放有两个物体；据此解答即可。本题主要考查了鸽巢原理的应用，关键是要理解巢原理(一):如果把(n+1)个物体放在n个抽屉里，那么必有一个抽屉里至少放有两个物体。\n【详解】由分析可得:把(m+1)个物体放进m个抽屉里，总有一个抽里放进2个物体。故答案为:2",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_328",
+        "original_id": "unit5_fill_328"
+      },
+      {
+        "id": "unit5_fill_329",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "把4个玻璃球放入下面的三个三角形，总有()个三角形里放入2个玻璃球。",
+        "options": [],
+        "answer": "1",
+        "solution": "【分析】将三个三角形看成是三个抽屉，根据抽屉原理可知将4个玻璃球放入三个抽屉中，总有一个抽屉里有两个玻璃球；据此解答即可。本题主要考查了鸽巢问题的简单应用，关键是要理解鸽巢原理(一):如果把(n+1)个物体放在n个抽屉里，那么必有一个抽屉里至少放有两个物体。\n【详解】由分析可得：把4个玻璃球放入下面的三个三角形，总有1个三角形里放入2个玻璃球;故答案为:1。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_329",
+        "original_id": "unit5_fill_329"
+      },
+      {
+        "id": "unit5_fill_330",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "学校图书馆里有A、B、C、D四类书，规定每个同学最多可以借1本书，在借书的5名同学中，可以保证至少()个人所借书的类型是一样的。",
+        "options": [],
+        "answer": "2",
+        "solution": "【分析】根据鸽巢原理(一)：如果把(n+1)个物体放在n个抽屉里，那么必有一个抽屉里至少放有两个物体。我们可将四类书看成是4个抽屉，所以把在借书的5名同学中所借的书，放在4个抽屉中，必将会有一个抽屉至少有两本书，也就是说至少有两个人所借的书的类型是一样的；据此解答。本题主要考查了鸽巢原理的应用，关键是要认真分析题意，建立正确的抽屉，熟练运用鸽巢原理。\n【详解】由分析可得:将四类书看成是4个抽屉，所以把在借书的5名同学中所借的书，放在4个抽屉中，必将会有一个抽屉至少有两本书，也就是说至少有两个人所借的书的类型是一样的。故答案为:2。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_330",
+        "original_id": "unit5_fill_330"
+      },
+      {
+        "id": "unit5_fill_331",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "把5个苹果放进4个抽屉里，总有一个抽屉里至少有()个苹果。",
+        "options": [],
+        "answer": "2",
+        "solution": "【分析】根据鸽巢原理(一)：如果把(n+1)个物体放在n个抽屉里，那么必有一个抽屉里至少放有两个物体；进行解答即可。本题主要考查了鸽巢问题的简单应用，关键是要理解并熟练运用鸽巢原理。\n【详解】由分析可得：把5个苹果放进4个抽屉里，总有一个抽屉里至少有2个苹果。故答案为:2",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_331",
+        "original_id": "unit5_fill_331"
+      },
+      {
+        "id": "unit5_fill_332",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "我用10分钟做完了11道数学题，那么在某一分钟内我至少做了()道题。",
+        "options": [],
+        "answer": "2",
+        "solution": "【分析】根据鸽巢原理(一)：把每一分钟看作是一个抽屉，则一共有10个抽屉，把11道数学题放入其中，则至少有一个抽屉要放两道题；据此解答即可。本题主要考查了鸽巢问题的应用，关键是要理解鸽巢原理(一):如果把(n+1)个物体放在n个抽屉里，那么必有一个抽屉里至少放有两个物体；并据此建立正确的抽屉。\n【详解】10分钟可看作是10个抽屉，把11道数学题放入其中，则至少有一个抽屉要放两道题；也就是说在某一分钟内我至少做了2道题目。故答案为:2",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_332",
+        "original_id": "unit5_fill_332"
+      },
+      {
+        "id": "unit5_fill_333",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "任意10个正整数，每一个除以9，其中必有()个余数相同。",
+        "options": [],
+        "answer": "2",
+        "solution": "【分析】鸽巢原理(一)：如果把(n+1)个物体放在n个抽屉里，那么必有一个抽屉里至少放有两个物体。一个正整数，除以9的余数有0,1,2,3,…,8；一共有9种情况，把这9种情况看作9个抽屉，任取10个正整数，放入这9个抽屉中，则必然有一个抽屉要放两个；据此解答。本题主要考查了鸽巢问题的应用，关键是要认真分析题意，建立正确的抽屉进行解答。\n【详解】除以9的余数有0,1,2,3,…,8；一共有9种情况，把这9种情况看作9个抽屉，任取10个正整数，放入这9个抽屉中，则至少有一个抽屉里至少有两个数，即其中必有两个正整数除以9的余数相同。故答案为:2",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_333",
+        "original_id": "unit5_fill_333"
+      }, {
+        "id": "unit5_fill_334",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "一个口袋里装着5个红球和5个蓝球，这10个球大小形状完全一样，至少要从口袋里取出()个球，才能保证其中有两个球颜色相同；至少要从口袋里取出()个球，才能保证其中有两个球颜色不同。",
+        "options": [],
+        "answer": "3；6",
+        "solution": "【分析】鸽巢原理(一)：如果把(n+1)个物体放在n个抽屉里，那么必有一个抽屉里至少放有两个物体。\n由题意可知：(1)要使取出的两个球颜色完全相同，考虑最差的情况，前面两次取出来的球一红一蓝，根据鸽巢原理则只需再抽取一次不管是红还是蓝都能保证取出的球有两个球颜色完全相同；(2)要使取出的球其中有两个球颜色不同，考虑最差的情况，前5次取出的球都是同一种颜色(都是红或都是蓝)，这时根据鸽巢原理，只要再取一次就能保证取出的球其中有两个球颜色不同；据此解答即可。本题主要考查了鸽巢问题的应用，关键是要认真分析题意，熟练掌握鸽巢原理并灵活运用。\n【详解】由分析可知：(1)考虑最差的情况：前面两次一红一蓝，再取一次不管是红还是蓝都有两个球颜色相同，所以要取的次数为:2+1=3(次)(2)考虑最差的情况，前5次取出的球都是同一种颜色(都是红或都是蓝)，则只要再取一次就能保证取出的球其中有两个球颜色不同，所以要取的次数为:5+1=6(次)故答案为:3;6",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_334",
+        "original_id": "unit5_fill_334"
+      },
+      {
+        "id": "unit5_fill_335",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "幸福小学学生年龄最小的6岁，最大的13岁，从学校里任选()位同学才能保证其中有两位同学的年龄相同。",
+        "options": [],
+        "answer": "9",
+        "solution": "【分析】由题意可知：学生的年龄有6岁，7岁，……13岁，一共有8个年龄，把每个年龄看成是一个抽屉，将选取的同学们的年龄放入抽屉中，考虑最差的情况：前面8个同学的年龄都不相同，这时根据鸽巢原理，则只需再多选一位同学就能保证至少有一个抽屉里的年龄是有两个，也就是说有两个年龄相同的学生；据此解答。本题主要考查了鸽巢原理的应用，关键是要理解鸽巢原理(一):如果把(n+1)个物体放在n个抽屉里，那么必有一个抽屉里至少放有两个物体。\n【详解】分析可知：(13-6)+1=8，所以有8个抽屉。考虑最差的情况：前面8个同学的年龄都不相同，则需再选一位同学就能保证有两个年龄相同的学生，所以所选的同学为：8+1=9(位)故答案为:9",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "2014六年级·全国·课后作业",
+        "merged_id": "unit5_335",
+        "original_id": "unit5_fill_335"
+      },
+      {
+        "id": "unit5_fill_336",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "任意13个人中，必然有()人是在同一个月出生的。",
+        "options": [],
+        "answer": "2",
+        "solution": "【分析】一年有12个月，根据鸽巢原理(一):如果把(n+1)个物体放在n的抽屉里，那么必有一个抽屉里至少放有两个物体，可得:在任意13人中必有2人是在同一个月出生的。本题主要考查了鸽巢问题的简单应用，关键是要理解鸽巢原理(一):如果把(n+1)个物体放在n个抽屉里，那么必有一个抽屉里至少放有两个物体。\n【详解】由分析得:任意13个人中，必然有2人是在同一个月出生的。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "2014六年级·全国·课后作业",
+        "merged_id": "unit5_336",
+        "original_id": "unit5_fill_336"
+      },
+      {
+        "id": "unit5_fill_337",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "已知a/6=b/8=c/10=d/12=2，a+b+c+d=()。",
+        "options": [],
+        "answer": "72",
+        "solution": "【分析】根据题意可得4个分数的值都等于2，然后分别计算出a、b、c和d的值。本题关键是根据分数的性质求出4个分子，最后求出4个分子的和。\n【详解】因为a/6=b/8=c/10=d/12=2，a=6×2=12;b=8×2=16;c=10×2=20;d=12×2=24，所以a+b+c+d=12+16+20+24=28+20+24=48+24=72。",
+        "knowledge_points": [],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_337",
+        "original_id": "unit5_fill_337"
+      },
+      {
+        "id": "unit5_fill_338",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "图形按一定的比放大或缩小，图形的()不变，对应边的长度发生变化，图形的大小也随之发生变化。",
+        "options": [],
+        "answer": "形状",
+        "solution": "【分析】根据图形的放大和缩小的特点回答即可。本题考查图形的放大和缩小，注意按比例放大缩小的是图形的边长。\n【详解】图形按一定的比放大或缩小，图形的形状不变，图形的大小也随之发生变化。",
+        "knowledge_points": [],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_338",
+        "original_id": "unit5_fill_338"
+      },
+      {
+        "id": "unit5_fill_339",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "在1/500的设计图上，量得正方形花圃的边长是4cm，这个花圃的实际面积是()。",
+        "options": [],
+        "answer": "400m²",
+        "solution": "【分析】比例尺是图上距离与实际距离的比值，已知正方形边长的图上距离是4cm，图上距离除以比例尺得到实际距离，再根据正方形的面积=边长×边长，求出花圃的实际面积。本题考查比例尺的应用，本题注意要先求出花圃边长的实际距离后，最后求出花圃的实际面积。\n【详解】4÷1/500=4×500=2000cm=20(m)，20×20=400(m²)",
+        "knowledge_points": [],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_339",
+        "original_id": "unit5_fill_339"
+      },
+      {
+        "id": "unit5_fill_340",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "圆锥的底面积、高和体积这三个量，当底面积一定时，体积和高成()比例；当体积一定时，底面积和高成()比例。",
+        "options": [],
+        "answer": "正；反",
+        "solution": "【分析】相关的两个量，它们的积一定，那么它们成反比例；相关的两个量，它们的比值一定，那么它们成正比例。本题的关键判定相关的两个量积和商的定值，最后根据正反比例的定义，分辨正反比例即可。\n【详解】(1)因为圆锥体积/高=1/3×底面积(一定)，所以当底面积一定时，体积和高成正比例。(2)因为底面积×高=圆锥体积×3(一定)，所以体积一定时，底面积和高成反比例。",
+        "knowledge_points": [
+          "正比例和反比例"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_340",
+        "original_id": "unit5_fill_340"
+      },
+      {
+        "id": "unit5_fill_341",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "在比例3:8=15:40中，如果外项40增加16，内项15应增加()。",
+        "options": [],
+        "answer": "6",
+        "solution": "【分析】根据比例的基本性质两内项之积等于两外项之积。外项40增加16即56，所以外项之积3×56=168，所以内项之积也是168，一个内项是8，所以另个内项是21，所以内项15应增加6。本题考查比例的基本性质的应用，解题关键是求出变化后的两外项之积即内项之积，已知一个内项8，用除法求出另一个内项，最后求出内项增加的数值。\n【详解】3×(40+16)÷8-15=3×56÷8-15=168÷8-15=21-15=6",
+        "knowledge_points": [],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_341",
+        "original_id": "unit5_fill_341"
+      },
+      {
+        "id": "unit5_fill_342",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "若两个外项分别是3和16，且比值为1/2，这个比例是()。",
+        "options": [],
+        "answer": "3:6=8:16",
+        "solution": "【分析】因为两个外项分别是3和16，比的前项是3，比值是1/2，那么后项就是6。比的后项是16，比值是1/2，那么前项是8。本题解题关键是前后比的比值是1/2，从而求出比例的两个内项。注意最后可用比例的基本性质检验对错。\n【详解】3÷1/2=6;16×1/2=8，两个外项分别是3和16，且比值为1/2，这个比例是3:6=8:16。",
+        "knowledge_points": [],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_342",
+        "original_id": "unit5_fill_342"
+      },
+      {
+        "id": "unit5_fill_343",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "红旗小学六(5)班有15人，至少有___人是同一个月出生的。",
+        "options": [],
+        "answer": "2",
+        "solution": "【分析】每年有12个月，用15除以12，假如每个月都有1人出生，那么余下的人数无论在哪个月出生，都至少有2人是同一个月出生的。物体个数÷鸽巢个数=商……余数；至少个数=商+1，注意关键是找到鸽巢个数:12。\n【详解】15÷12=1(人)……3(人)，1+1=2(人)答:至少有2人是同一个月出生的。故答案为:2",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_343",
+        "original_id": "unit5_fill_343"
+      },
+      {
+        "id": "unit5_fill_344",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "小东玩掷骰子游戏，要保证掷出的骰子点数至少有两次是相同的，小东至少应掷___次。",
+        "options": [],
+        "answer": "7",
+        "solution": "【分析】骰子能掷出的结果只有6种，掷7次的话必有2次相同；即把骰子的出现的六种情况看作'抽屉'，把掷出的次数看作'物体的个数'，要保证至少有两次相同，那么物体个数应比抽屉数至少多1；进行解答即可。此题属于典型的抽屉原理习题，解答此类题的关键是找出把谁看作'抽屉个数'，把谁看作'物体个数'，然后根据抽屉原理解答即可。\n【详解】6+1=7(次)要保证掷出的骰子点数至少有两次是相同的，小东至少应掷7次。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_344",
+        "original_id": "unit5_fill_344"
+      },
+      {
+        "id": "unit5_fill_345",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "一副扑克牌(包括大、小王)。最少拿出___张。才能保证四种花色全都有。",
+        "options": [],
+        "answer": "42",
+        "solution": "【分析】此题应从最极端情况分析，因为每一种花色的扑克牌有13张，假设前13×3=39次都摸出前三种花色的扑克牌(即把前三种花色的牌取完)，又摸2次是2张花牌，再摸1张只能是第四种花色，进而得出结论。此题做题的关键是从最极端情况进行分析，进而通过分析得出问题答案。\n【详解】13×3+2+1=39+2+1=42(张)\n答:最少要拿42张，才能保证四种花色都有。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_345",
+        "original_id": "unit5_fill_345"
+      },
+      {
+        "id": "unit5_fill_346",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "一个布袋里有红色、黄色、蓝色袜子各10只，最少要拿出()只才能保证至少有2双颜色不相同的袜子。",
+        "options": [],
+        "answer": "13",
+        "solution": "【分析】把三种颜色看做三个抽屉，考虑最差情况：其中一个抽屉10只全部摸出，则此时再任意摸出2只，最差情况是剩下的两个抽屉各摸出一只，则再次摸出1只，无论放到哪个抽屉，都能保证其中至少有2双颜色不相同的袜子。此题考查了抽屉原理在实际问题中的灵活应用，要注意考虑最差情况。\n【详解】10+2+1=13(只)故答案为:13。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_346",
+        "original_id": "unit5_fill_346"
+      },
+      {
+        "id": "unit5_fill_347",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "从1~50中至少取出___个不同的数，才能保证其中一定有2的倍数。",
+        "options": [],
+        "answer": "26",
+        "solution": "【分析】首先找到1,2……50中为2的倍数的数字，从而得到不为2的倍数的数字的个数，加上1即可得到答案。\n【详解】从1~50中2的倍数有25个，取出25个不能保证有一个是2的倍数，应取出:25+1=26(个)从1~50中至少取出26个不同的数，才能保证其中一定有2的倍数。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_347",
+        "original_id": "unit5_fill_347"
+      },
+      {
+        "id": "unit5_fill_348",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "从8个盘子里拿出17个苹果，无论怎么拿，我们一定能找到拿出苹果最多的那个盘子，从它里面至少拿出()个苹果。",
+        "options": [],
+        "answer": "3",
+        "solution": "【分析】把8个盘子看作8个抽屉，把17个苹果看作17个元素，那么每个抽屉需要拿17÷8=2(个)……1(个)，所以每个抽屉需要拿2个，剩下的1个再不论怎么拿，总有一个抽屉里至少有:2+1=3(个)，所以，最多的抽屉有3个元素，据此解答。本题考查了抽屉原理即把m个元素任意放入n(n≤m)个集合，则一定有一个集合至少要有k个元素，其中k=m÷n(当n能整除m时)或k=m÷n+1(当n不能整除m时)。\n【详解】17÷8=2(个)……1(个)，2+1=3(个)所以最多的抽屉里面有3个苹果；答:拿出苹果最多的抽屉，从它里面至少拿出3个苹果。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_348",
+        "original_id": "unit5_fill_348"
+      },
+      {
+        "id": "unit5_fill_349",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "1001只鸽子飞进50个巢，无论怎么飞，我们一定能找到一个飞进鸽子最多的巢，它里面至少有___只鸽子。",
+        "options": [],
+        "answer": "21",
+        "solution": "【分析】因为50只鸽笼看作50个抽屉，1001鸽子看作1001个元素，依据抽屉原理，将1001个元素放入到50个抽屉里，由此解答即可。此题考查了抽屉原理在实际问题中的灵活应用。\n【详解】1001÷50=20(只)……1(只)，20+1=21(只)。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_349",
+        "original_id": "unit5_fill_349"
+      },
+      {
+        "id": "unit5_fill_350",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "将2016颗黑子，201颗白子排成一条直线，至少会有___颗黑子连在一起。",
+        "options": [],
+        "answer": "10",
+        "solution": "【分析】2016颗黑子，被201颗白子分成了202个线段，每个线段平均有黑子2016÷202≈9.98(颗)，由于黑子必须是整数，因此最少会出现10颗黑子连在一起。考查了学生分析问题的能力，排成直线，201颗白子把这条直线分成了202段。\n【详解】2016÷202≈9.98(颗)，9+1=10(颗)",
+        "knowledge_points": [
+          "鸽巢问题",
+          "植树问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_350",
+        "original_id": "unit5_fill_350"
+      },
+      {
+        "id": "unit5_fill_351",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "在2个盒子里放入11块橡皮，总有一个盒子里至少放进___块橡皮。",
+        "options": [],
+        "answer": "6",
+        "solution": "【分析】物体个数是11，鸽巢个数是2，根据物体个数÷鸽巢个数=商……余数；至少个数=商+1解答。鸽巢原理是一个重要而又基本的组合原理，在解决数学问题时有非常重要的作用。\n【详解】11÷2=5(块)……1(块)，5+1=6(块)所以总有一个盒子里至少放进6块橡皮。故答案为:6。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_351",
+        "original_id": "unit5_fill_351"
+      },
+      {
+        "id": "unit5_fill_352",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "47名同学参加考试，成绩都是整数，满分100分。有3名同学的成绩在60分以下，其余同学的成绩都在75~95分之间，至少有___名同学的分数相同。",
+        "options": [],
+        "answer": "3",
+        "solution": "【分析】既然是问'至少有几名学生的成绩相同'，说明应以成绩为抽屉，学生为物品，除3名成绩在60分以下的学生外，其余成绩均在75~95分之间，75~95共有21个不同分数，将这21个分数作为21个抽屉，把47-3=44(个)学生作为物品，44÷21=2……2，根据抽屉原理2，至少有1个抽屉至少有3件物品，即这47名学生中至少有3名学生的成绩是相同的。此题考查了利用抽屉原理解决实际问题的灵活应用，关键是构造合适的抽屉。\n【详解】75~95分的有：47-3=44(个)，44÷21=2(人)……2(人)，2+1=3(人)答:至少有3名学生的成绩相同。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_352",
+        "original_id": "unit5_fill_352"
+      },
+      {
+        "id": "unit5_fill_353",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "在鸽巢问题中，'鸽巢'至少放入物体数的求法是用物体数除以()数，当除得的商没有余数时，至少放入的物体数就等于()；当除得的商有余数时，至少放入的物体数就等于()。",
+        "options": [],
+        "answer": "鸽巢；商；商+1",
+        "solution": "【分析】原理1:把多于n个的物体放到n个抽屉里，则至少有一个抽屉里的东西不少于两件；原理2:把多于m(m乘n)个的物体放到n个抽屉里，则至少有一个抽屉里有不少于m+1的物体。此题考查了利用抽屉原理解决实际问题的灵活应用。\n【详解】在鸽巢问题中，'鸽巢'至少放入物体数的求法是用物体数除以鸽巢数，当除得的商没有余数时，至少放入的物体数就等于商；当除得的商有余数时，至少放入的物体数就等于商+1。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_353",
+        "original_id": "unit5_fill_353"
+      },
+      {
+        "id": "unit5_fill_354",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "一个袋子里装有同样的白帽子和黄帽子各7顶，闭着眼睛，从袋中至少摸出()顶帽子，可以保证有2顶是同色的；至少摸出()顶帽子，才能保证摸出2种颜色的帽子。",
+        "options": [],
+        "answer": "3；8",
+        "solution": "【分析】帽子有两种颜色，最不理想的情况下，摸出2顶帽子有2种颜色，再摸出1顶，无论是什么颜色都能保证有2顶是同色的。两种颜色的帽子各有7顶，最不理想的情况下，摸出7顶都是同种颜色的，再摸出1顶，一定与前7顶颜色不同，保证有2种颜色的帽子。此类数字较小的题目可以采用假设的方法来解决问题。\n【详解】1+2=3(顶)答:至少摸出3顶帽子，可以保证有2顶是同色的。7+1=8(顶)答:至少摸出8顶帽子，才能保证摸出2种颜色的帽子。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_354",
+        "original_id": "unit5_fill_354"
+      },
+      {
+        "id": "unit5_fill_355",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "一个盒子里装有黄、白乒乓球各5个，要想使取出的乒乓球中至少有2个黄乒乓球，至少应取出乒乓球()个。",
+        "options": [],
+        "answer": "7",
+        "solution": "【分析】可以用假设的方法解决。要想至少抽到几个同色球，则要在另一种球都可能抽到的情况下加上此种颜色至少抽到的数量。\n【详解】假如取5个球有可能都是同色的白球，要想保证至少有两个黄色乒乓球，则要在白色抽取完的基础上再加上2个黄色乒乓球的数量。故答案为：7。",
+        "knowledge_points": [],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_355",
+        "original_id": "unit5_fill_355"
+      },
+      {
+        "id": "unit5_judge_356",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "容易",
+        "question": "某校有367名2012年出生的小朋友，他们至少有两位小朋友的生日相同。()",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】首先判断2012年是平年还是闰年，再根据这一年的天数和小朋友的总人数进行分析求解。总结：把m个物体任意分放进n个鸽巢中(m和n是非0自然数，且m>n)，那么一定有一个鸽巢中至少放进了2个物体。\n【详解】2012÷4=503，所以2012年是闰年，这年有366天；367÷366=1(人)……1(人)，1+1=2(人)所以他们至少有两位小朋友的生日相同。原题说法正确。故答案为:√。",
+        "knowledge_points": [
+          "平年、闰年的认识及判定方法",
+          "鸽巢问题",
+          "年、月、日的认识及换算"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_356",
+        "original_id": "unit5_judge_356"
+      },
+      {
+        "id": "unit5_judge_357",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "11只鸽子飞进了4个鸽笼，至少有一个鸽笼飞进了3只鸽子。",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】把4个鸽笼看作4个抽屉，把11只鸽子看作11个元素，那么每个抽屉需要放11÷4=2(个)……3(个)，所以每个抽屉需要放2个，剩下的3个不论怎么放，总有一个抽屉里至少有:2+1=3(个)，所以，至少有一个鸽笼要飞进3只鸽子，据此解答。抽屉原理问题的解答思路是：准确地建立抽屉和确定元素的总个数，然后根据'至少数=元素的总个数÷抽屉的个数+1(有余数的情况下)'解答。\n【详解】11÷4=2(个)……3(只)，2+1=3(只)答:至少有一个鸽笼要飞进3只鸽子。故答案为:√。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "17-18六年级下·河北石家庄·期末",
+        "merged_id": "unit5_357",
+        "original_id": "unit5_judge_357"
+      },
+      {
+        "id": "unit5_judge_358",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "367人中必有2人的生日相同。()",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】平年有365天，闰年有366天。用367除以365或366，假如每个月都有1人出生，那么余下的人数无论在哪个月出生，都至少有2人是同一天出生的。考查了鸽巢问题，注意根据平年、闰年分情况讨论。\n【详解】367÷365=1(人)……2(人)，1+1=2(人);367÷366=1(人)……1(人)，1+1=2(人)即至少有2个人的生日是同一天，所以原题说法正确。故答案为:√",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_358",
+        "original_id": "unit5_judge_358"
+      },
+      {
+        "id": "unit5_judge_359",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "在366人当中，一定有2人是同一天出生的。()",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】平年有365天，闰年有366天。在闰年中，如果根据极端思想考虑，每天都有一个人出生。极端思想就是从最差的情况考虑，注意：年份可能是平年或闰年。\n【详解】根据分析，在闰年中，每天都有一个人出生，所以原题说法错误。故答案为:×",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_359",
+        "original_id": "unit5_judge_359"
+      },
+      {
+        "id": "unit5_calc_360",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "下面各题，怎样简便就怎样算。\n5/8×2/3+3/8÷3/2\n(9.3×5/6-7.3)÷9/4\n24/5-(2.8-5/7)+2/7\n13/5×8×5/13×12.5%\n8/9×[5/8-(3/4-1/6)]\n(2/13+4/7)×7×13",
+        "options": [],
+        "answer": "2/3；0.2；3；1；1/27；66",
+        "solution": "【分析】5/8×2/3+3/8÷3/2，除法改乘法，用乘法分配律进行简算；(9.3×5/6-7.3)÷9/4，先算乘法，再算减法，最后算除法；24/5-(2.8-5/7)+2/7，先去括号，用结合律进行简算；13/5×8×5/13×12.5%，用乘法交换结合律进行简算；8/9×[5/8-(3/4-1/6)]，先算小括号里的减法，再算中括号里的减法，最后算乘法；(2/13+4/7)×7×13，用乘法分配律进行简算。本题考查了分数的简便计算及小数和分数、分数和百分数的简便计算，整数的运算定律同样适用。\n【详解】5/8×2/3+3/8÷3/2=(5/8+3/8)×2/3=1×2/3=2/3\n(9.3×5/6-7.3)÷9/4=(7.75-7.3)×4/9=0.45×4/9=0.2\n24/5-(2.8-5/7)+2/7=(4.8-2.8)+(5/7+2/7)=2+1=3\n13/5×8×5/13×12.5%=(13/5×5/13)×(8×0.125)=1×1=1\n8/9×[5/8-(3/4-1/6)]=8/9×(5/8-7/12)=8/9×1/24=1/27\n(2/13+4/7)×7×13=2/13×7×13+4/7×7×13=14+52=66",
+        "knowledge_points": [
+          "分数的四则混合运算",
+          "整数、小数、分数、百分数的简便运算",
+          "分数乘法运算律"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_360",
+        "original_id": "unit5_calc_360"
+      },
+      {
+        "id": "unit5_calc_361",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "计算题",
+        "difficulty": "中等",
+        "question": "解比例。\n12:x=4/7:2.8\n5%:1/4=x:0.25\n21/x=35/8",
+        "options": [],
+        "answer": "x=58.8；x=0.05；x=4.8",
+        "solution": "【分析】根据比例的基本性质，写成两数相乘的形式，再根据等式的性质计算即可。本题考查了解比例，比例的两内项积=两外项积。\n【详解】12:x=4/7:2.8\n解:4/7x=12×2.8\n4/7x×7/4=12×2.8×7/4\nx=58.8\n\n5%:1/4=x:0.25\n解:1/4x=0.05×0.25\n1/4x×4=0.05×0.25×4\nx=0.05\n\n21/x=35/8\n解:35x=168\n35x÷35=168÷35\nx=4.8",
+        "knowledge_points": [],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_361",
+        "original_id": "unit5_calc_361"
+      },
+      {
+        "id": "unit5_solve_362",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "一个口袋里装有红球、白球、黄球各5个，这15个球除颜色不同外形状都一样，至少要从口袋里摸出几个球才能保证其中有两个球的颜色相同？至少要从口袋里摸出几个球才能保证其中有两个球的颜色不相同？",
+        "options": [],
+        "answer": "4个；6个",
+        "solution": "【分析】(1)由题意可知，一共有三种颜色的球，考虑最差的情况：三种颜色的球都被摸出来了个，这时根据鸽巢原理再摸出一个球，不管这个球的颜色是什么都能和之前三种颜色中的某一种相同，也就能保证其中有两个球的颜色相同；(2)由题意可知，三种颜色的球各有5个，考虑最差的情况，前面5次摸出来的5个球都是同一种颜色，这时根据鸽巢原理再摸出一个球，而这个球是不可能与之前球的颜色相同，也就是说这样能够保证有两个球的颜色不相同；据此解答。本题主要考查了鸽巢原理的应用，关键是要认真分析题意，熟练掌握鸽巢原理并灵活运用。\n【详解】由分析可得：(1)3+1=4(个)答:至少要从口袋里摸出4个球才能保证其中有两个球的颜色相同。(2)5+1=6(个)答:至少要从口袋里摸出6个球才能保证其中有两个球的颜色不相同",
+        "knowledge_points": [],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_362",
+        "original_id": "unit5_solve_362"
+      },
+      {
+        "id": "unit5_solve_363",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "任意给出5个不同的自然数，其中至少有两个数的差是4的倍数。为什么？",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】一个数除以4的余数有0,1,2,3，这4种情况，根据同余原理：两数被某数除得相同余数，则这两个数的差一定能被这个数整除；再根据鸽巢原理5个不同的自然数除以4所得余数，必有两个是相同的；据此解答即可。本题主要考查鸽巢原理的应用，关键是要认真分析题意，熟练掌握鸽巢原理并灵活运用。\n【详解】由分析可得：一个数除以4的余数有0,1,2,3这4种情况，5个不同的自然数除以4所得的余数，必有两个是相同的；而这两个余数之差一定能被4整除；也就是说任意给出5个不同的自然数，其中至少有两个数的差是4的倍数。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_363",
+        "original_id": "unit5_solve_363"
+      },
+      {
+        "id": "unit5_solve_364",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "从1~10这10个数中，任意选6个数，其中一定有两个数的和是11，你能说说其中的道理吗？",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】由题意可知，从1~10这10个数中，任意选6个数，根据鸽巢原理将1~10分成5组(1,10),(2,9),(3,8),(4,7),(5,6)，将这5组数看成是5个抽屉，从每个抽屉中抽出一个数，得到的这5个数；它们中的任意两个数之和不等于11，而第6个数必定是这5个抽屉中另一个数，它能和其在同一个抽屉里的数之和等于11；据此解答。本题主要考查了鸽巢原理的应用，关键是认真分析题意，建立正确的抽屉，运用鸽巢原理进行解答。\n【详解】将1~10分成5组(1，10),(2，9),(3,8),(4,7),(5,6)，因为从这5组数中每一组任选一个数得到5个数，这5个数它们中的两个数之和不等于11，而第6个数必定是这5组数中的一组的另一个数，能够和它同一组的数的和等于11，所以1~10这10个数中任选6个数，其中一定有两个数的和是11。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_364",
+        "original_id": "unit5_solve_364"
+      },
+      {
+        "id": "unit5_solve_365",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "六(2)班有46名同学，其中至少有多少名同学在同一个月过生日？为什么？",
+        "options": [],
+        "answer": "4名；因为46÷12=3……10，一年12个月，平均最少3个同学在同一个月，剩下的不管在哪个月，都最少有4个，所以有4名。",
+        "solution": "【分析】将12个月份看成是12个抽屉，根据鸽巢原理(二):不能整除时至少数=商+1，能整除时至少数=商，进行解答即可。本题主要考查了鸽巢原理的应用，关键是要认真分析题意，建立正确的抽屉，熟练地运用适当的鸽巢原理进行解答。\n【详解】由分析可得：46÷12=3……10，3+1=4所以至少有4名同学在同一个月过生日。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_365",
+        "original_id": "unit5_solve_365"
+      },
+      {
+        "id": "unit5_solve_366",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "学校把11名新转入的学生分进2个班，总有一个班至少有几名学生？为什么？",
+        "options": [],
+        "answer": "6名；因为11÷2=5……1，余的1个不管去哪个班，哪个班就是6人，所以总有有一个班有6名。",
+        "solution": "【分析】将两个班看成是两个抽屉，根据鸽巢原理(二):不能整除时至少数=商+1，能整除时至少数=商，进行解答即可；进行解答即可。本题主要考查了鸽巢原理的应用，关键是要认真分析题意，建立正确的抽屉，运用抽屉原理进行解答。\n【详解】由分析可得:11÷2=5……1，所以余的1个不管去哪个班，哪个班就是6人，所以总有一个班至少有6名学生。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_366",
+        "original_id": "unit5_solve_366"
+      },{
+        "id": "unit5_solve_367",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "新兴镇上设置了3个信箱，现在有16封信要发出去，不管这些信怎样投，必有一个信箱里至少要投进6封信。你知道为什么吗？",
+        "options": [],
+        "answer": "16÷3=5……1，也就是说将16个信封平均装进3个信箱，还差一封没有装，所以必然有一个信箱要装6封。",
+        "solution": "【分析】将3个信箱看成是3个抽屉，再根据鸽巢原理中的抽屉原则二:不能整除时至少数=商+1，能整除时至少数=商，进行解答即可。本题主要考查了鸽巢原理的应用，关键是要理解鸽巢原理中的抽屉原则二:如果将n个物体放进m个抽屉(其中n>m)，那么必有一个抽屉至少有:①k=n/m+1个物体(n不能被m整除是);②k=n/m个物体(n能被m整除时)。\n【详解】16÷3=5(封)……1(封)，5+1=6(封)也就是说，将16个信封平均装进3个信箱，还差一封没有装，所以必然有一个信箱要装6封。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_367",
+        "original_id": "unit5_solve_367"
+      },
+      {
+        "id": "unit5_solve_368",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "将14个气球挂在教室的4面墙上，至少有一面墙上要挂4个气球，为什么？",
+        "options": [],
+        "answer": "14÷4=3……2，多的2个气球，可以挂在两面墙上，所以至少有一面墙有4个气球。",
+        "solution": "【分析】将教室的四个面看作是四个抽屉，根据鸽巢原理(二):不能整除时至少数=商+1，能整除时至少数=商，进行解答即可。本题主要考查了鸽巢原理的应用，关键是要认真分析题意，建立正确的抽屉，选取适当的鸽巢原理进行解答。\n【详解】由分析可得:14÷4=3……2，多的2个气球，可以挂在两面墙上，所以至少有一个面有4个气球。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_368",
+        "original_id": "unit5_solve_368"
+      },
+      {
+        "id": "unit5_solve_369",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "把一块长与宽的比为5:3的长方形土地，用1/500的比例尺画在图纸上，得到长方形的周长是32cm，这块长方形土地的实际面积是多少平方米？",
+        "options": [],
+        "answer": "1500平方米",
+        "solution": "【分析】因为长方形的周长是32cm，所以长和宽的和是16cm，再按照长与宽的比为5:3，计算出图纸上的长和宽，再把长和宽按照1/500的比例尺，计算出实际的长和宽，最后按照长方形=长×宽，计算出实际面积。本题的易错点是要先根据长方形的周长求出长和宽的和后，再按比分配，计算出长和宽。\n【详解】(32÷2)×5/8=16×5/8=10(厘米)\n(32÷2)×3/8=16×3/8=6(厘米)\n10÷1/500=10×500=5000(cm)=50(米)\n6÷1/500=6×500=3000(cm)=30(米)\n50×30=1500(平方米)\n答：长方形土地的实际面积是1500平方米。",
+        "knowledge_points": [],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_369",
+        "original_id": "unit5_solve_369"
+      },
+      {
+        "id": "unit5_solve_370",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "分别写着3、5、8的数字卡片各12张。如果从中任选两张组成一个两位数，至少组合成几次一定会出现两个相同的两位数？",
+        "options": [],
+        "answer": "10次",
+        "solution": "【分析】3、5、8进行组合会有9组不同的两位数，假设前9次出现的都是不同的两位数，那么第10次组成的数字一定和前9次的数字中的一个相等。解决这类有多种可能的题目，需要先根据题意把所有可能按顺序列出，再解答问题。\n【详解】因为可以组成的数字为：35、38、53、58、85、83、33、55、88共9组两位数，假设前9次抽到的数都不相同，那么至少组合10次一定会出现两个相同的两位数。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_370",
+        "original_id": "unit5_solve_370"
+      },
+      {
+        "id": "unit5_solve_371",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "甲、乙两地相距360km，一辆汽车从甲地开往乙地，4小时行了160km。照这样计算，到达乙地还需要几小时？(用比例解)",
+        "options": [],
+        "answer": "5小时",
+        "solution": "【分析】因为汽车前后的速度不变，路程/时间=速度(一定)，所以路程和时间成正比例。4小时行了160km，剩下200km。设到达乙地还需要x小时，在按照正比例关系列方程。本题考查比例的应用，关键是汽车前后的速度不变，所以前160km的速度等于后200km的速度。\n【详解】解：设到达乙地还需要x小时。\n160:4=(360-160):x\n160x=4×(360-160)\n160x=4×200\n160x=800\nx=800÷160\nx=5\n答:到达乙地还需要5小时。",
+        "knowledge_points": [],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_371",
+        "original_id": "unit5_solve_371"
+      },
+      {
+        "id": "unit5_solve_372",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "一些孩子在海洋球里玩耍，他们把海洋球分成许多堆。其中有一个孩子发现，从海洋球堆中任意选出六堆，其中至少有两堆海洋球数之差是5的倍数。你说他的结论对吗？为什么？",
+        "options": [],
+        "answer": "答：原题说法正确。我们把6堆海洋球数看作任意6个自然数，它们被5除，其余数不外乎是0、1、2、3、4五种可能，如果把每一种余数看作一个抽屉，那么余数相同的两数就在同一抽屉里，根据'抽屉原理'，6个自然数被5除后，必有两个余数相同，显然两数之差是5的倍数。",
+        "solution": "【分析】此题主要利用'抽屉原理'解决简单的实际问题，任何一个正整数除以5所得的余数只有5种情况：余0(整数)、余1、余2、余3、余4。所以对于任意的六个正整数A、B、C、D、E、F除以5最多可以有5个不同的余数。此题是主要考查利用'抽屉原理'解决简单的实际问题，属于比较困难的题目，应该适当增加些此类题目的训练，提供自身运算速度和运算正确率。\n【详解】答：原题说法正确。我们把6堆海洋球数看作任意6个自然数，它们被5除，其余数不外乎是0、1、2、3、4五种可能，如果把每一种余数看作一个抽屉，那么余数相同的两数就在同一抽屉里，根据'抽屉原理'，6个自然数被5除后，必有两个余数相同，显然两数之差是5的倍数。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_372",
+        "original_id": "unit5_solve_372"
+      },
+      {
+        "id": "unit5_solve_373",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "某班有49个学生，最大的12岁，最小的9岁，是否一定有两个学生，他们是同年同月出生的？",
+        "options": [],
+        "answer": "一定有两个学生是同年同月出生的。",
+        "solution": "【分析】首先确定全班学生的年龄状况有4种情况，即9岁、10岁、11岁、12岁，而每年又有12个月，所以:就有12×4=48(种)情况，看作48个抽屉，根据'抽屉原理'解答即可。本题主要考查了学生对用'抽屉原理'解决简单的实际问题这个知识点的掌握情况，解答此题的关键是：巧用年龄段构造抽屉，进一步利用'抽屉原理'解答问题即可。\n【详解】12×4=48(种)，49÷48=1(人)……1(人)，1+1=2(人)即一定在同一个抽屉里有两个人，故必有2人是同年同月出生的；答:一定有两个学生他们是同年同月出生的。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_373",
+        "original_id": "unit5_solve_373"
+      },
+      {
+        "id": "unit5_solve_374",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "某班有44名学生，他们都订阅了甲、乙、丙三种报刊中的若干种(每名学生订了其中的一种、两种或三种)。至少有几名学生订阅的报刊是完全相同的？",
+        "options": [],
+        "answer": "7名",
+        "solution": "【分析】先求出订阅报刊的情况，再根据'抽屉原理'解答：'至少数=元素的总个数÷抽屉的个数+1(有余数的情况下)'进行解答即可。此题关键运用了'抽屉原理'的解题思路：要从最不利的情况考虑，准确建立抽屉和确定元素的总个数进行解答。\n【详解】学生订阅报刊的情况共有7种。44÷7=6(名)……2(名)，6+1=7(名)",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_374",
+        "original_id": "unit5_solve_374"
+      },
+      {
+        "id": "unit5_solve_375",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "食堂有5种不同的菜和3种不同的主食，每人只能买一种菜和一种主食，在16名同学中，一定至少有两名同学所买的菜和主食都是相同的。为什么？",
+        "options": [],
+        "answer": "原题说法正确。因为菜和主食有5×3=15(种)搭配方式；6÷15=1(名)……1(名)，1+1=2(名)所以至少有两名同学所买的菜和主食都是相同的。",
+        "solution": "【分析】首先根据乘法原理，可得菜和主食一共有5×3=15(种)不同的搭配方式，而学生的人数是16名，学生的人数比菜和主食的搭配方式的种数多1，所以根据'抽屉原理'，可得一定至少有两名同学所买的菜和主食都是相同的，据此判断即可。此题主要考查了'抽屉原理'的应用，解答此题的关键是判断出一共有多少种不同的搭配方式。\n【详解】原题说法正确。因为菜和主食有5×3=15(种)搭配方式；6÷15=1(名)……1(名)，1+1=2(名)所以至少有两名同学所买的菜和主食都是相同的。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_375",
+        "original_id": "unit5_solve_375"
+      },
+      {
+        "id": "unit5_solve_376",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "国王让阿凡提在8×8的国际象棋棋盘的每个格子里放米粒。结果每个格子里至少放一粒米，无论怎么放都至少有3个格子里的米粒一样多，那么至多有多少个米粒？",
+        "options": [],
+        "answer": "1055个",
+        "solution": "【分析】如果不满足条件，最多只有两个格子中的米粒数一样多，则64个格子里至少有1+1+2+2+3+3+...+32+32=1056个米粒。如果少于1056个米粒，那必然有三个格子里的米粒数一样多，因此至多有1055个米粒。此题考查了学生分析、解决问题的能力，注意计算要细心。\n【详解】8×8=64(个)\n64÷2=32(个)\n1+1+2+2+3+3+...+32+32=(1+32)×32÷2×2=33×32÷2×2=33×32=1056(个)\n1056-1=1055(个)\n答:至多有1055个米粒。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_376",
+        "original_id": "unit5_solve_376"
+      },
+      {
+        "id": "unit5_solve_377",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "希望小学有36人乘车外出春游，最多乘几辆车才能保证至少有一辆车上的人数不少于8？",
+        "options": [],
+        "answer": "5辆",
+        "solution": "【分析】保证至少有一辆车上的人数不少于8，那么其余车上最少有7人，把总人数减去1人，保证每辆车7人，这样加上1人就能保证至少有一辆车上的人数不少于8。本题考查了鸽巢问题和分配问题，分析出其余车辆最少的人数是解题的关键。\n【详解】根据分析可得：\n(36-1)÷(8-1)=35÷7=5(辆)\n答:最多乘5辆车才能保证至少有一辆车上的人数不少于8。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_377",
+        "original_id": "unit5_solve_377"
+      },
+      {
+        "id": "unit5_solve_378",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "三年级二班有43名同学，班上的'图书角'至少要准备多少本课外书，才能保证有的同学可以同时借两本书？",
+        "options": [],
+        "answer": "44本",
+        "solution": "【分析】43名同学看做43个抽屉，要保证至少有1个学生拿到2本课外书，则课外书的数量应该比学生数多1，即43+1=44本，据此即可解答。此题考查了抽屉原理，关键是找抽屉(43名同学)。\n【详解】43+1=44(本)答:至少要准备44本课外书，才能保证有的同学可以同时借两本书。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_378",
+        "original_id": "unit5_solve_378"
+      },
+      {
+        "id": "unit5_solve_379",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "有5050张数字卡片，其中1张上面写着数字'1'，2张上面写着数字'2'，3张上面写着数字'3'，……，99张上面写着数字'99'，100张上面写着数字'100'。现在要从中任意取出若干张，为了确保抽出的卡片中至少有10张完全相同的数字，至少要抽出多少张卡片？",
+        "options": [],
+        "answer": "865张",
+        "solution": "【分析】根据极端思想从最不利的情况解答此题。解决此类问题利用极端思想是一种有效的方法。计算时，要认真，不要出错。\n【详解】最不利情形是写着1到9的全抽了，写着10到100的各抽了9张，则只要再任抽一张，就能保证抽出的卡片至少有10张的数字完全相同，至少要抽:\n(1+2+...+9)+(100-10+1)×9+1=45+819+1=45+820=865(张)\n答:至少要从中抽出865张，才能确保在抽出的卡片中至少10张卡片上的数字完全相同。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_379",
+        "original_id": "unit5_solve_379"
+      },
+      {
+        "id": "unit5_solve_380",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "一组中的8名同学一共投进63个球，一定有一名同学至少投进几个球？",
+        "options": [],
+        "answer": "8个",
+        "solution": "【分析】本着尽量平均分配的原则是本题的关键思路。此类'至少'题型只要进行除法计算，再将商加上1就可以得到结果。\n【详解】63÷8=7(个)……7(个)，7+1=8(个)答:一定有一名同学至少投进8个球。",
+        "knowledge_points": [],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_380",
+        "original_id": "unit5_solve_380"
+      },
+      {
+        "id": "unit5_solve_381",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "六(1)班的55人进行投篮比赛，如果分成6个组，总有一个组至少有多少人？",
+        "options": [],
+        "answer": "10人",
+        "solution": "【分析】把55人平均分成6个组，每组有9人，还剩1人，剩余的1人，无论分到那个组，总有1组至少是(9+1)人，据此解答。此类'至少'题型只要进行除法计算，再将商加上1就可以得到结果。\n【详解】55÷6=9(人)……1(人)，9+1=10(人)答:总有一个组至少有10人。",
+        "knowledge_points": [],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_381",
+        "original_id": "unit5_solve_381"
+      },
+      {
+        "id": "unit5_solve_382",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "体育老师把35个毽子分给4个班，总有一个班至少分到多少个毽子？",
+        "options": [],
+        "answer": "9个",
+        "solution": "【分析】本着尽量平均分配的原则是本题的关键思路。此类'至少'题型只要进行除法计算，再将商加上1就可以得到结果。\n【详解】35÷4=8(个)……3(个)，8+1=9(个)答:总有一个班至少分到9个毽子。",
+        "knowledge_points": [],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_382",
+        "original_id": "unit5_solve_382"
+      },
+      {
+        "id": "unit5_solve_383",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "在一次世界极限运动会中，意大利、法国、美国、加拿大分别有7名运动员参赛。(1)至少几人报名参加滑板街道赛，可以保证有两人来自同一个国家？(2)至少有几人参加极限单车比赛，可以保证有来自两个国家的运动员？",
+        "options": [],
+        "answer": "(1)至少5人。(2)至少有8人。",
+        "solution": "【分析】可采用假设的方法解决。比种类数多1即可保证有两者属于同一类别，比单类人数多1即可保证有分属不同类别的对象。\n【详解】(1)一共有4个不同国家，按照最不利原则，先报名的4位运动员分别来自4个不同的国家，这时再有1位运动员报名，无论来自哪个国家，这个项目都会有2名运动员来自同一个国家。答:至少5人报名参加滑板街道赛，可以保证有两人来自同一个国家。(2)每个国家有7名运动员参赛，按照最不利原则，先报名的7位运动员都来自同一个国家，当再有1位运动员报名时，无论来自其他三国中的哪个国家，这个项目都会有2个不同国家的运动员参赛。答:至少有8人参加极限单车比赛，可以保证有来自两个国家的运动员。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_383",
+        "original_id": "unit5_solve_383"
+      },
+      {
+        "id": "unit5_solve_384",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "有5名同学参加科技比赛，团体总分为426分，则总有一名同学的得分不低于多少分？(得分为整数)",
+        "options": [],
+        "answer": "86",
+        "solution": "【分析】本着尽量平均分配的原则是本题的关键思路。此类'至少'题型只要进行除法计算，再将商加上1就可以得到结果。\n【详解】426÷5=85(分)……1(分)，85+1=86(分)答:总有一名同学的得分不低于86分。",
+        "knowledge_points": [],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_384",
+        "original_id": "unit5_solve_384"
+      },
+      {
+        "id": "unit5_solve_385",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "四个连续的自然数分别被3除后，必有两个余数相同，请说明理由。",
+        "options": [],
+        "answer": "在有余数的除法中余数要比除数小，一个数除以3，无论商是几，余数只有三种情况，余0(正好除尽)、余1、余2，将此问题看作鸽巢问题。4个连续的自然数相当于4只鸽子，余数的3种情况相当于3个鸽巢，将4个连续的自然数平均分配到3种余数情况中，分别余0、余1、余2后，还剩1个自然数，剩余的自然数除以3无论余几，总会与之前出现相同的余数。",
+        "solution": "【分析】除以3的两个数若相差的是3的倍数则余数相同。根据除法算式的关系分析即可。\n【详解】有四个连续的自然数，则有两个数相差3，他们除以3会有相同的余数。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_385",
+        "original_id": "unit5_solve_385"
+      },
+      {
+        "id": "unit5_solve_386",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "任意给出3个不同的自然数，其中一定有2个数的差是偶数，请说明理由。",
+        "options": [],
+        "answer": "把自然数分为奇数和偶数，从奇数和偶数中抽出3个数，可以看作鸽巢问题。奇数和偶数看作2个鸽巢。抽出的3个数看作3只鸽子，3只鸽子放进2个鸽巢里，无论怎么放，都有一个鸽巢里有2只鸽子。也就是说3个数中一定有两个数是偶数或者是奇数。而两个奇数或两个偶数的差一定是偶数。",
+        "solution": "【分析】同为奇数或同为偶数的两个数的差是偶数。此题转化为鸽巢问题，把奇数和偶数分别看作2个鸽巢，3个不同的自然数看作3只鸽子。\n【详解】3个不同的自然数其中必有两个数同为奇数或者两个同为偶数，这样的两个数的差即是偶数。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_386",
+        "original_id": "unit5_solve_386"
+      },
+      {
+        "id": "unit5_solve_387",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "桂苑学校六年级每位同学都订了《数学小灵通》《小学生作文》《英语天地》《科学画报》四种书刊中的两种，他们当中至少有34人订阅的书刊种类相同。你知道六年级至少有多少人吗？",
+        "options": [],
+        "answer": "199人",
+        "solution": "【分析】每位同学都订阅了《数学小灵通》《小学生作文》《英语天地》《科学画报》四种报刊中的两种，由此可得一共有6种不同的订阅方法，这6种不同的订阅方法看做6个抽屉，根据抽屉原理，考虑最差情况：每个抽屉都有33个同学，则一共有6×33=198个同学，如果再有1个同学，无论他采用哪种方法订阅，都会出现一个抽屉里的34为同学出现，据此解答。解决此题关键是先求出两种报刊共有几种不同的订法，进而根据抽屉问题的解答方法解答。\n【详解】每个同学都订阅四种报刊中的两种，共有的方法有:\n3+2+1=6(种)\n6×(34-1)+1=6×33+1=198+1=199(人)\n答:六年级同学至少有199人。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_387",
+        "original_id": "unit5_solve_387"
+      },
+      {
+        "id": "unit5_solve_388",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "30个标有号码的小球，其中号码是1、2、3的各有10个。至少取出多少个，才能保证有两个号码相同的小球？至少取出多少个，才能保证有3个不同号码的小球？",
+        "options": [],
+        "answer": "4个；21个",
+        "solution": "【分析】(1)从最坏情况考虑，假如前面取的3个小球分别是1、2、3号球各一个，然后再取任意一个球，都能和前面相对应一个球号码相同的小球，所以至少要取(3+1=4)个小球;(2)从最坏情况考虑，假如前面取的10个1号球，10个2号球，然后再取任意一个球，就取到3个不同号码的小球，至少要取(10×2+1=21)个小球，才能保证有3个不同号码的小球。抽屉原理问题的解答思路是：要从最不利情况考虑，准确地建立抽屉和确定元素的总个数，然后根据'至少数=元素的总个数÷抽屉的个数+1(有余数的情况下)'解答。\n【详解】3+1=4(个)答:至少取出4个，才能保证有两个号码相同的小球。10+10+1=21(个)答:至少取出21个，才能保证有3个不同号码的小球。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_388",
+        "original_id": "unit5_solve_388"
+      },
+      {
+        "id": "unit5_solve_389",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "饲养员给10只猴子分苹果，其中至少要有一只猴子分到7个苹果，饲养员至少要拿来少个苹果？",
+        "options": [],
+        "answer": "61个",
+        "solution": "【分析】把10只猴子看做10个抽屉，苹果的个数看做元素，利用抽屉原理最差情况:每个抽屉里先放6个共需要6×10=60个，再任意放一个，就能保证至少要有一只猴子分到7个苹果。此题考查了利用抽屉原理解决实际问题的灵活应用，关键是从最差情况考虑。\n【详解】1+6×10=1+60=61(个)\n答:饲养员至少要拿来61个苹果。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_389",
+        "original_id": "unit5_solve_389"
+      },
+      {
+        "id": "unit5_solve_390",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "新学期开始了，六(3)班班主任把全班学生分成6个小组，每个小组有7人，那么每个组中至少有多少名学生的性别相同？",
+        "options": [],
+        "answer": "4名",
+        "solution": "【分析】因为每个小组有7人，性别只有两种，把两种性别看作两个抽屉，把7个人看作是7个元素，利用抽屉原理最差情况:要使性别相同的人数最少，只要使每个抽屉里的元素数尽量平均，即可解答。此题考查了利用抽屉原理解决实际问题的灵活应用，关键是从最差情况考虑。\n【详解】7÷2=3(名)……1(名)，3+1=4(名)答:每个小组中至少有4名学生的性别相同。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_390",
+        "original_id": "unit5_solve_390"
+      },
+      {
+        "id": "unit5_solve_391",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "42只鸽子飞进5个笼子里，可以保证总有一个笼子中至少有几只鸽子？",
+        "options": [],
+        "answer": "9只",
+        "solution": "【分析】此题考虑最差情况:每个笼子飞回的鸽子尽量平均，由此利用抽屉原理即可解答。抽屉原理问题的解答思路是：要从最不利情况考虑，准确地建立抽屉和确定元素的总个数，然后根据'至少数=元素的总个数÷抽屉的个数(商)+1(有余数的情况下)'解答。\n【详解】42÷5=8(只)……2(只)，8+1=9(只)答:可以保证总有一个笼子中至少有9只鸽子。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_391",
+        "original_id": "unit5_solve_391"
+      },
+      {
+        "id": "unit5_solve_392",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "把13个练习本全部发给5名同学，那么总有一名同学至少发到几个练习本？",
+        "options": [],
+        "answer": "3个",
+        "solution": "【分析】把5名同学看作5个抽屉，把13本练习本看作13个元素，从最不利情况考虑，因为13÷5=2……3，每个抽屉需要放2个元素，再取出1本，总有一个抽屉里有:2+1=3(个)，据此解答。抽屉原理问题的解答思路是：要从最不利情况考虑，准确地建立抽屉和确定元素的总个数，然后根据'至少数=元素的总个数÷抽屉的个数(商)+1(有余数的情况下)'解答。\n【详解】13÷5=2(个)……3(个)，2+1=3(个)答:总有一名同学至少发到3个练习本。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_392",
+        "original_id": "unit5_solve_392"
+      },
+      {
+        "id": "unit5_solve_393",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "某校五年级学生共有380人，年龄最大的与年龄最小的相差不到1岁，不用去查看学生的出生日期，这380名学生中至少有几名学生是同年同月同日出生的？",
+        "options": [],
+        "answer": "2名",
+        "solution": "【分析】平年有365天，闰年有366天，由于求至少有多少同年同月同日生，可按天数多的闰年计算，把366天看作'抽屉'，把380人看作'物体个数'，380÷366=1(名)……14(名)，即平均每天有一个学生出生的话，还余1名学生，根据抽屉原理可知，至少有1+1=2名学生的生日是同一天。此题考查了利用抽屉原理解决实际问题的灵活应用。\n【详解】380÷366=1(名)……14(名)，1+1=2(名)答:这380名学生中至少有2名学生是同年同月同日出生的。",
+        "knowledge_points": [
+          "鸽巢问题初步"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_393",
+        "original_id": "unit5_solve_393"
+      },
+      {
+        "id": "unit5_solve_394",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "有规格相同的6种颜色的袜子各20只，混装在箱内，从箱内至少取出多少只袜子才能保证有3双袜子？",
+        "options": [],
+        "answer": "11只",
+        "solution": "【分析】预想运气最差情况，假想先取6只颜色都不一样，再取1只肯定能配1双；再取1只跟刚刚那个颜色一样，等于配齐6种颜色，跟着再取1只又配了1双；继续取1只跟刚刚配好那双颜色一样，又配齐了6种颜色，还得再取1只，就3双了;所以从箱内至少取出11只袜子才能保证有3双袜子。抽屉原理问题的解答思路是：要从最不利情况考虑，准确地建立抽屉和确定元素的总个数。\n【详解】6+1+1+1+2=11(只)答:从箱内至少要取出11只袜子才能保证有3双袜子。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_394",
+        "original_id": "unit5_solve_394"
+      },
+      {
+        "id": "unit5_solve_395",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "一个班有40名学生，现在有课外书125本。把这些书分给这个班的学生，是否一定有人会得到4本或4本以上的课外书？",
+        "options": [],
+        "answer": "是，一定有人会得到。",
+        "solution": "【分析】把40名学生看做40个抽屉，125本看做125个元素，利用抽屉原理最差情况:要使每个抽屉的数量最少，只要使每个抽屉的元素数尽量平均，即可解答。此题考查了利用抽屉原理解决实际问题的灵活应用，关键是从最差情况考虑。\n【详解】125÷40=3(本)……5(本)，3+1=4(本)答:把这些书分给这个班的学生，一定有人会得到4本或4本以上的课外书。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_395",
+        "original_id": "unit5_solve_395"
+      },
+      {
+        "id": "unit5_solve_396",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "盒子里有同样大小的红、黄、蓝、白球各12个。要想摸出的球中一定有4个同色的，至少要摸出几个球？",
+        "options": [],
+        "answer": "13个",
+        "solution": "【分析】每个球都摸到，且数量均匀则同色的数量就少，若摸出12个有可能每种颜色都有3个，再摸一个就必然有一个颜色有4个。做此类题目要具备一定的假设和列举能力，要想4个同色，则要考虑所有颜色都有3个的前提下再摸一下就能保住有4个同色。\n【详解】3×4=12(个)，12+1=13(个)答:至少要摸出13个球。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_396",
+        "original_id": "unit5_solve_396"
+      },
+      {
+        "id": "unit5_solve_397",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "红星小学六(1)班有学生50人，他们中至少有几名学生的属相是一样的？",
+        "options": [],
+        "answer": "5名",
+        "solution": "【分析】12年为一周期共有12个属相。此类'至少'题目都是先用除法计算再用加法计算，直接用商加上1即可得到最多的一组的数量。\n【详解】50÷12=4(名)……2(名)，4+1=5(名)答:他们中至少有5名学生的属相是一样的。",
+        "knowledge_points": [],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_397",
+        "original_id": "unit5_solve_397"
+      },
+      {
+        "id": "unit5_solve_398",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "把165本日记本分给六(1)班学生，如果其中至少有1人分到5本日记本，那么这个班最多有多少人？",
+        "options": [],
+        "answer": "41人",
+        "solution": "【分析】要使得人数最多，则只需计算只有1人分到5本日记本。1人分到5本日记本，其余人分到4本即可得出结果。\n【详解】5-1=4(本)，165-1=164(本)，164÷4=41(人)答:这个班最多有41人。",
+        "knowledge_points": [],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_398",
+        "original_id": "unit5_solve_398"
+      },
+      {
+        "id": "unit5_solve_399",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "六(1)班有41名学生，他们做了210只千纸鹤，要把这些千纸鹤分给全班的学生，会不会有人得到6只或6只以上的千纸鹤？",
+        "options": [],
+        "answer": "会",
+        "solution": "【分析】尽可能平均分配就可以得到'至少'最多的一组数量。此类题目都是先用除法计算再用加法计算，直接用商加上1即可得到最多的一组的数量。\n【详解】210÷41=5(只)……5(只)，5+1=6(只)答:会有人得到6只或6只以上的千纸鹤。",
+        "knowledge_points": [],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_399",
+        "original_id": "unit5_solve_399"
+      },
+      {
+        "id": "unit5_solve_400",
+        "unit": "第五单元：鸽巢问题",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "6只猴子在5棵树上玩耍，至少有2只猴子在同一棵树上玩耍。为什么？",
+        "options": [],
+        "answer": "如果每1棵树上有一只猴子，还剩1只，剩下的1只在任何一棵树上玩耍，都至少有2只猴子在同一棵树上玩耍。",
+        "solution": "【分析】若1只猴子在一棵树上玩耍则需要6棵树。此类题目数量较小，用假设的方法便可轻松解决。\n【详解】6÷5=1(只)……1(只)，1+1=2(只)故至少有2只猴子在同一棵树上玩耍。",
+        "knowledge_points": [],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_400",
+        "original_id": "unit5_solve_400"
+      },
+      {
+        "id": "unit5_choice_401",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "将 22 枚棋子放到下图中, 则至少有一个小正方形中有 ( ) 枚棋子。\n\n![](images/f1e0ca963dce2ca722075a6d84fb2a2a645bcf54649c6da99aa1711c8e2abb20.jpg)",
+        "options": [
+          "A. 5",
+          "B. 6",
+          "C. 7"
+        ],
+        "answer": "B",
+        "solution": "【分析】把 4 个小正方形看作 4 个抽屉, 22 枚棋子看作 22 个物体, 利用抽屉原理: 把多于 (mn + 1) 个物体放到 n 个抽屉里 (其中 m 和 n 是正整数), 则至少有一个抽屉里有不少于 (m + 1) 个物体。据此解答。\n\n【详解】$22 \\div 4 = 5$ （枚）…2（枚） $5 + 1 = 6$ （枚）至少有一个小正方形中有 6 枚棋子。故答案为: B",
+        "knowledge_points": [
+          "最不利原则"
+        ],
+        "source": "23-24 六年级下·河南焦作·期末",
+        "merged_id": "unit5_401",
+        "original_id": "unit5_choice_401"
+      },
+      {
+        "id": "unit5_choice_402",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "九龙坡区今年五月份的天气有 4 种情况 (如图), 五月份至少有 ( ) 天是同一种天气。\n\n![](images\\ceb0bd67c0b7576af0a2c65b15d0ee70.png)",
+        "options": [
+          "A. 6",
+          "B. 7",
+          "C. 8",
+          "D. 9"
+        ],
+        "answer": "C",
+        "solution": "【分析】五月份共有 31 天, 一共有 4 种天气情况, 考虑最不利的情况, 列式: $31 \\div 4$, 将 31 种情况平均分成 4 份, 商就是每种天气的天数, 如果有余数, 那么某一种天气会至少增加一天。\n\n【详解】$31 \\div 4 = 7 \\dots 3$ $7 + 1 = 8$ （天） 五月份至少有8天是同一种天气。故答案为: C",
+        "knowledge_points": [
+          "最不利原则"
+        ],
+        "source": "23-24六年级下·重庆九龙坡·期末",
+        "merged_id": "unit5_402",
+        "original_id": "unit5_choice_402"
+      },
+      {
+        "id": "unit5_choice_403",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "把 18 枚棋子放入如图所示的三角形内, 一定有一个小三角形中至少放入 ( )枚棋子。\n\n![](images/2d7dd9aada09ab400217f54707ce3723f7f90b86a486840c7e978e2d673d9f49.jpg)",
+        "options": [
+          "A. 3",
+          "B. 4",
+          "C. 5",
+          "D. 6"
+        ],
+        "answer": "C",
+        "solution": "【分析】抽屉原则二：如果把 n 个物体放在 m 个抽屉里，其中 n > m，那么必有一个抽屉至少有：（1）当 n 不能被 m 整除时，$k = [\\frac{n}{m}] + 1$ 个物体；（2）当 n 能被 m 整除时，$k =\\frac{n}{m}$ 个物体。\n\n【详解】$18 \\div 4 = 4$ （枚） $\\dots 2$ （枚）$4 + 1 = 5$ (枚)一定有一个小三角形中至少放入 5 枚棋子。故答案为: C",
+        "knowledge_points": [
+          "鸽巢问题初步"
+        ],
+        "source": "23-24 六年级下·四川南充·期末",
+        "merged_id": "unit5_403",
+        "original_id": "unit5_choice_403"
+      },
+      {
+        "id": "unit5_choice_404",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "如图, 把 40 个★图案放在小方格内，不管怎么放，总有一个小方格至少要放入（）个★。\n\n![](images/92e9d44d9a10694ae602a78a5b4668401b018dbd66edb9985790e91da5de7839.jpg)",
+        "options": [
+          "A. 1",
+          "B. 2",
+          "C. 3",
+          "D. 4"
+        ],
+        "answer": "C",
+        "solution": "【分析】从图中可知, 共有 $4 \\times 4 = 16$ 个小方格; 将 40 个 $\\star$ 图案平均放到 16 个小方格内, 每个小方格里放 2 个, 还剩下 8 个, 这 8 个 $\\star$ 图案, 无论放在哪个小方格里, 总有一个小方格至少有 3 个 $\\star$ 图案。\n\n【详解】$4 \\times 4 = 16$ （个）$40 \\div 16 = 2$ (个)……8（个）$ $2 + 1 = 3$ (个)总有一个小方格至少要放入 3 个★。故答案为: C",
+        "knowledge_points": [
+          "鸽巢问题初步"
+        ],
+        "source": "23-24 六年级下·重庆开州·期末",
+        "merged_id": "unit5_404",
+        "original_id": "unit5_choice_404"
+      },
+      {
+        "id": "unit5_choice_405",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "4 个同学练习投篮,一共投进 30 个球，投进球数最多的同学至少投进（）个球。",
+        "options": [
+          "A. 8",
+          "B. 9",
+          "C. 10"
+        ],
+        "answer": "A",
+        "solution": "【分析】4个同学共投进30个球，$30\\div 4 = 7$ （个）…2(个), 即平均每个同学进 7 个球的话, 还余 2 个球, 所以投进最多的同学至少投进 $7 + 1 = 8$ (个) 球。此题关键运用了'抽屉原理'的解题思路：要从最不利的情况考虑，准确建立抽屉和确定元素的总个数，然后根据'至少数 = 元素的总个数 ÷ 抽屉的个数 +1 （有余数的情况下）'解答。\n\n【详解】$30 \\div 4 = 7$ （个）……2（个）$7 + 1 = 8$ （个）故答案为：A。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_405",
+        "original_id": "unit5_choice_405"
+      },
+      {
+        "id": "unit5_choice_406",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "要在 $20 \\mathrm{~m}$ 长的阳台上放 11 盆花，不管怎么放，（）花之间的距离不超过 $2 \\mathrm{~m}$。",
+        "options": [
+          "A. 刚好 2 盆",
+          "B. 至少 2 盆",
+          "C. 至少 3 盆"
+        ],
+        "answer": "B",
+        "solution": "【分析】通过观察题干可知，依据抽屉原题和不封闭路线的植树问题知识即可解答。此题问题原型属于 '抽屉原理', 关键是理解 '抽屉原理' 的意义: 将多于 n 件的物品任意放到 n 个抽屉中, 那么至少有一个抽屉中的物品不少于 2 件。\n\n【详解】11盆花之间有 $11 - 1 = 10$ 个间隔，每个间隔平均是 $20 \\div 10 = 2$ （米），把这10个间隔看作10个抽屉，把11盆花放在10个抽屉里，总能保证至少有一个抽屉里有2盆花，即至少有2盆花的距离不超过2米。故答案为: B。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_406",
+        "original_id": "unit5_choice_406"
+      },
+      {
+        "id": "unit5_choice_407",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "妈妈用13个纽扣钉了3件上衣，其中有1件上衣至少钉了（）个纽扣。",
+        "options": [
+          "A. 4",
+          "B. 5",
+          "C. 6"
+        ],
+        "answer": "B",
+        "solution": "【分析】根据'抽屉原理'，从最坏的情况考虑，即可解答。此题考查了'抽屉原理'解决实际问题的灵活应用，关键在于考虑最坏的情况，然后得出相应的答案。\n\n【详解】$13 \\div 3 = 4$ （个）……1（个）$4 + 1 = 5 \\text {(个)}$故答案为: B",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_407",
+        "original_id": "unit5_choice_407"
+      },
+      {
+        "id": "unit5_choice_408",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "如果从放有 10 个球的瓶中任意取出 3 个球一定有红球, 则原来瓶中红球至少有 ( )。",
+        "options": [
+          "A. 8",
+          "B. 9",
+          "C. 10"
+        ],
+        "answer": "A",
+        "solution": "【分析】任意取出 3 个, 要想保证取出的球中有红球, 则球的总个数与红球的差必须小于 3 , 也就是红球的个数加 3 要比球的总个数多, 即红球的个数加 3 大于 10 , 所以红球最少要有 8 个。正确理解可能性是解决此题的关键, 因为必须有红球, 所以球的总个数与红球的差必须小于任意取出的 3 个球, 可以转化成 $10 - ( ) < 3$ 的形式解决。\n\n【详解】球的总个数与红球的差必须小于 3 , 因为共 10 个球, 所以红球最少要有 8 个。故答案为: A",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_408",
+        "original_id": "unit5_choice_408"
+      },
+      {
+        "id": "unit5_choice_409",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "李老师参加射击比赛，射了3场，成绩是25环，他至少有一场的成绩不低于（）环。",
+        "options": [
+          "A. 5",
+          "B. 6",
+          "C. 7",
+          "D. 8"
+        ],
+        "answer": "A",
+        "solution": "【分析】要求最低成绩，则其他场次的环数要尽可能多。已知总数求可能的最低数, 则可假设其他场次都是满分便可得到结果。\n\n【详解】若有两场 10 环，则另一场最少是 5 环。故答案为: A。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_409",
+        "original_id": "unit5_choice_409"
+      },
+      {
+        "id": "unit5_choice_410",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "王老师把 36 根跳绳分给 5 个班, 有一个班至少要分到 ( ) 根。",
+        "options": [
+          "A. 6",
+          "B. 7",
+          "C. 8",
+          "D. 9"
+        ],
+        "answer": "C",
+        "solution": "【分析】36 根跳绳若平均分给 5 个班则每班 7 根还剩余 1 根。此类题目都是先用除法计算再用加法计算，若除的得商为 1，直接用商加上 1 即可得到最多的一组的数量。\n\n【详解】$36 \\div 5 = 7$ （根）……1（根）$7 + 1 = 8$ (根)故答案为: C",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_410",
+        "original_id": "unit5_choice_410"
+      },
+      {
+        "id": "unit5_choice_411",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "把9个苹果放在2个盘子里，总有一个盘子里至少要放（）个。",
+        "options": [
+          "A. 2",
+          "B. 3",
+          "C. 4",
+          "D. 5"
+        ],
+        "answer": "D",
+        "solution": "【分析】9 个苹果放在 2 个盘子里，要想多的那盘尽可能少只能一盘 4 个一盘 5 个。此类题目都是先用除法计算再用加法计算，若除的得商为 1 , 直接用商加上 1 即可得到最多的一组的数量。\n\n【详解】$9 \\div 2 = 4$ （个）……1（个） $4 + 1 = 5 \\text {(个)}$ 故答案为: D。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_411",
+        "original_id": "unit5_choice_411"
+      },
+      {
+        "id": "unit5_choice_412",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "容易",
+        "question": "10个人坐9把椅子，不管怎么坐，总有一把椅子上至少坐（）人。",
+        "options": [
+          "A. 3",
+          "B. 2",
+          "C. 1",
+          "D. 4"
+        ],
+        "answer": "B",
+        "solution": "【分析】考虑最差情况, $10 \\div 9 = 1$ (人) $\\dots \\dots 1$ (人), 表示平均每把椅子上坐 1 人, 还余下 1 人, $1 + 1 = 2$ (人), 所以总有一把椅子上至少坐 2 人, 据此解答即可。本题考查了抽屉原理的灵活利用，解答本题时要考虑最差情况。\n\n【详解】$10 \\div 9 = 1$ （人）……1（人）$1 + 1 = 2 \\text {(人)}$总有一把椅子上至少坐 2 人。故答案为: B",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_412",
+        "original_id": "unit5_choice_412"
+      },
+      {
+        "id": "unit5_choice_413",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "金都小区有35位大妈跳广场舞，她们来自不同的8幢楼，至少有（）位大妈来自同一幢楼。",
+        "options": [
+          "A. 3",
+          "B. 4",
+          "C. 5",
+          "D. 6"
+        ],
+        "answer": "C",
+        "solution": "【详解】略",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_413",
+        "original_id": "unit5_choice_413"
+      },
+      {
+        "id": "unit5_choice_414",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "困难",
+        "question": "从 $1 、 2 、 3 、 4$ 、…、12 这 12 个自然数中至少选 ( ) 个, 就可以保证其中一定包括两个数, 它们的差是 7 。",
+        "options": [
+          "A. 7",
+          "B. 10",
+          "C. 9",
+          "D. 8"
+        ],
+        "answer": "D",
+        "solution": "【分析】略",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_414",
+        "original_id": "unit5_choice_414"
+      },
+      {
+        "id": "unit5_choice_415",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "一副扑克牌（不包括两张王牌）有4种花色，每种花色各有13张。现从中任意抽（）张牌，才能保证有4张牌是同一种花色。",
+        "options": [
+          "A. 12",
+          "B. 13",
+          "C. 15",
+          "D. 16"
+        ],
+        "answer": "B",
+        "solution": "【分析】把 4 种花色看作 4 个抽屉，考虑最差情况：抽 12 张扑克牌，每个抽屉都有 3 张，那么再任意抽出 1 张无论放到哪个抽屉，都会出现一个抽屉里面有 4 张牌是同花色的；据此解答。此题考查了鸽巢问题, 关键能够结合抽屉原理去解答。\n\n【详解】根据分析，\n\n$3 \\times 4 + 1\\\\ = 1 2 + 1 \\\\ = 1 3 \\text {(张)} $\n\n所以，现从中任意抽 13 张牌, 才能保证有 4 张牌是同一种花色;故答案为: B",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_415",
+        "original_id": "unit5_choice_415"
+      },
+      {
+        "id": "unit5_choice_416",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "2009 年某地一年新生婴儿有 368 人, 他们中同一天出生的有 ( ) 人。",
+        "options": [
+          "A. 2",
+          "B. 3",
+          "C. 4",
+          "D. 10人以上"
+        ],
+        "answer": "A",
+        "solution": "【分析】略",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_416",
+        "original_id": "unit5_choice_416"
+      },
+      {
+        "id": "unit5_choice_417",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "选择题",
+        "difficulty": "中等",
+        "question": "一个袋子里有红、白、蓝三种颜色的球各 10 个, 至少拿出 ( ) 个才能保证有 3 个球的颜色相同。",
+        "options": [
+          "A. 3",
+          "B. 6",
+          "C. 21",
+          "D. 7"
+        ],
+        "answer": "D",
+        "solution": "【分析】最坏的打算是每种球都摸出 2 个, 那么摸了 6 个,那再摸一个, 就能得到 3 个颜色相同, 进而计算得出结论。\n\n【详解】$2 \\times 3 + 1\\\\= 6 + 1 \\\\ = 7 (\\text {个}) $\n\n答: 至少拿出 7 球才能保证有 3 个颜色的球是同色;故选：D",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_417",
+        "original_id": "unit5_choice_417"
+      },
+      {
+        "id": "unit5_choice_418",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如图, 要想摸出的笔一定是2种颜色，最少要摸出( )支。\n\n![](images/62334aa95bb0f974e919d2d72484eacc92060cc146bed1279b2481add288d9e8.jpg)",
+        "options": [],
+        "answer": "6",
+        "solution": "【分析】根据题意, 盒子里有红、白两种颜色的笔各 5 支,运气最差的情况为先摸出的 5 支都是同一种颜色的笔, 再从盒子中任意摸出一支, 一定是另一种颜色中的笔, 此时会出现两支不同颜色的笔。\n\n【详解】$5 + 1 = 6$ (次)要想摸出的笔一定是 2 种颜色, 最少要摸出 6 次。",
+        "knowledge_points": [
+          "最不利原则"
+        ],
+        "source": "24-25 六年级下·全国·单元测试",
+        "merged_id": "unit5_418",
+        "original_id": "unit5_choice_418"
+      },
+      {
+        "id": "unit5_choice_419",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "给一个正方体木块的 6 个面分别涂上红、黄、蓝三种颜色, 不论如何涂至少有 ( )个面的颜色相同。",
+        "options": [],
+        "answer": "2/两",
+        "solution": "【分析】把正方体木块的 6 个面看作被分放物体, 红、黄、蓝三种颜色看作 3 个抽屉, 被分放物体的数量: 抽屉的数量 = 平均每个抽屉分放物体的数量……剩下物体的数量, 若能整除, 一个抽屉里至少分放物体的数量 = 平均每个抽屉分放物体的数量, 若不能整除, 一个抽屉里至少分放物体的数量 = 平均每个抽屉分放物体的数量 +1 , 据此解答。\n\n【详解】$6 \\div 3 = 2$ （个）所以，给一个正方体木块的 6 个面分别涂上红、黄、蓝三种颜色，不论如何涂至少有 2 个面的颜色相同。",
+        "knowledge_points": [
+          "鸽巢问题初步"
+        ],
+        "source": "24-25 六年级下·全国·单元测试",
+        "merged_id": "unit5_419",
+        "original_id": "unit5_choice_419"
+      },
+      {
+        "id": "unit5_choice_420",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "实验小学学生的属相有如图几种。那么至多选出( )名学生, 就一定能找到属相相同的两名学生。\n\n![](images/73b814573feefb548b16122d0fbd2f0627781475bdde2da727662305fa6c60ff.jpg)",
+        "options": [],
+        "answer": "11",
+        "solution": "【分析】所有学生的属相一共有 10 种, 把这 10 种属相看作 10 个抽屉。从最不利情况考虑, 每个抽屉需要放 1 名不同属相的学生, 共需要选出 $1 \\times 10 = 10$ (名) 学生, 再增加 1 名不论什么属相, 总有一个抽屉的学生和他属相相同, 即一定能找到属相相同的两名学生, 所以至多要选出 $10 + 1 = 11$ (名)。\n\n【详解】通过分析可得：$1 0 + 1 = 1 1 \\text {(名)}$则至多选出 11 名学生, 就一定能找到属相相同的两名学生。",
+        "knowledge_points": [
+          "最不利原则"
+        ],
+        "source": "24-25 六年级下·全国·单元测试",
+        "merged_id": "unit5_420",
+        "original_id": "unit5_choice_420"
+      },
+      {
+        "id": "unit5_choice_421",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "端午节是中国首个入选世界非遗的节日，各地都有包粽子的习俗。小明、倩倩两家制作了三种口味粽子的数量如表。从小明家的粽子里任意拿一个吃，吃到( )口味的可能性最大。从倩倩家粽子里至少从中拿出( )个才能保证有 2 个粽子的口味是相同的。\n\n| 种类 | 小明  | 倩倩  |\n| ---- | ----- | ----- |\n| 豆沙 | 15 个 | 15 个 |\n| 红枣 | 10 个 | 15 个 |\n| 花生 | 5 个  | 15 个 |",
+        "options": [],
+        "answer": "豆沙; 4",
+        "solution": "【分析】根据小明家里粽子的种类和数量进行判断, 数量多, 吃到的可能性就越大; 债倩家粽子的种类数量相同,所以每一种都拿出一次, 再拿一次就会出现口味相同的,据此解答。\n\n【详解】从小明家的粽子里任意拿一个吃，吃到豆沙口味的可能性最大。$3 + 1 = 4$ （个），因此从倩倩家粽子里至少从中拿出 4 个才能保证有 2 个粽子的口味是相同的。",
+        "knowledge_points": [
+          "最不利原则",
+          "判断事件发生的可能性的大小"
+        ],
+        "source": "2024·山西晋中·小升初真题",
+        "merged_id": "unit5_421",
+        "original_id": "unit5_choice_421"
+      },
+      {
+        "id": "unit5_choice_422",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "四（1）班的小朋友举行投掷飞镖比赛，靶心如图。投中中心圆得5分，投中其余部分得2分，脱靶（飞镖不在靶心上）不得分。每人投掷3次，那么班上至少有( )个小朋友参加比赛，才能保证至少有5人得分相同。\n\n![](images/d1547af53cde17cc67bb384f9db8efcd554ee742284f52fe79f06cc5fe1dc61a.jpg)",
+        "options": [],
+        "answer": "41",
+        "solution": "【分析】先将得分不同的所有情况列举出来，然后考虑极端情况，每种情况都有 4 个人得分相同，此时再加 1 人，即可得出保证至少有 5 人得分相同的人数。本题考查了鸽巢问题, 可以先列举出所有可能的情况, 再从极端的情况进行考虑。\n\n【详解】得分不同的所有情况：$0 + 0 + 0$ 、$0 + 0 + 2$ 、$0 + 0 + 5$ 、$0 + 2 + 2$ 、$0 + 2 + 5$ 、$0 + 5 + 5$ 、$2 + 2 + 2$ 、$2 + 2 + 5$ 、$2 + 5 + 5$ 、$5 + 5 + 5$ ；一共有10种情况；\n\n$1 0 \\times 4 + 1 \\\\ = 4 0 + 1 \\\\ = 4 1 \\text {(种)} $\n\n班上至少有 41 个小朋友参加比赛, 才能保证至少有 5 人得分相同",
+        "knowledge_points": [
+          "最不利原则"
+        ],
+        "source": "2010四年级·全国·竞赛",
+        "merged_id": "unit5_422",
+        "original_id": "unit5_choice_422"
+      },
+      {
+        "id": "unit5_choice_423",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "西湖的游船承载着游人的欢声笑语，穿越历史的长河，划过美丽的湖面，将古今的美好记忆串联在一起。如表是贝贝和妈妈乘坐游船时行驶的路程与时间的情况记录。\n\n| 时间/分 | 1    | 3    | 10   | 15   | ...  |\n| ------- | ---- | ---- | ---- | ---- | ---- |\n| 路程/m  | 135  | 405  | 1350 | 2025 | ...  |\n\n(1)这艘游船行驶的路程和行驶的时间成( )比例关系; 游船 45 分钟可以行驶( )km。\n(2)贝贝和妈妈所乘的这艘游船上一共有 8 人，他们参加采摘莲蓬活动，要求在规定时间内采摘够 33 个莲蓬，不管怎样分配，总有 1 个人至少要采摘()个莲蓬。",
+        "options": [],
+        "answer": "(1) 正 ;6.075(2)5",
+        "solution": "【分析】（1）速度 = 路程 ÷ 时间，速度一定，也即路程与时间的比值一定，则这艘游船行驶的路程和行驶的时间成正比例关系，先求出行驶的速度，再求出行驶路程；(2) 根据最不利情况考虑, 用莲蓬总数除以人数, 求出每个人平均摘莲蓬数量, 再加上余数, 就是 1 个人至少要摘的数量。\n\n【详解】(1) 速度: $135 \\div 1 = 135$ (米/分)$135 \\times 45 = 6075$ (米) 6075 米 = 6.075 千米这艘游船行驶的路程和行驶的时间成正比例关系；游船 45 分钟可以行驶 6.075 千米。(2) $33 \\div 8 = 4$ （个）……1（个）$4 + 1 = 5$ （个）贝贝和妈妈所乘的这艘游船上一共有 8 人, 他们参加采摘莲蓬活动, 要求在规定时间内采摘够 33 个莲蓬, 不管怎样分配, 总有 1 个人至少要采摘 5 个莲蓬。",
+        "knowledge_points": [
+          "正比例的意义及辨识",
+          "最不利原则"
+        ],
+        "source": "23-24 六年级下·湖南衡阳·期末",
+        "merged_id": "unit5_423",
+        "original_id": "unit5_choice_423"
+      },
+      {
+        "id": "unit5_choice_424",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "礼堂里有 253 人开会，这 253 人中至少有( )人的属相相同。温馨小提示：属相一共十二种，也就是12个抽屉",
+        "options": [],
+        "answer": "22",
+        "solution": "【分析】属相一共有 12 种, 根据抽屉原理, 把 12 种属相看作 12 个抽屉, 把 253 人看作 253 个元素, 那么每个抽屉需要放 $253 \\div 12 = 21$ (个) 元素, 还剩余 1 个, 余下的 1 个无论怎么放, 总有一个抽屉至少放 22 个元素, 因此至少有 22人属相相同。据此解答。\n\n【详解】$253 \\div 12 = 21$ （人）……1（人）$2 1 + 1 = 2 2 \\text {(人)}$礼堂里有 253 人开会, 这 253 人中至少有 22 人属相同。",
+        "knowledge_points": [
+          "鸽巢问题初步"
+        ],
+        "source": "2024 六年级下·全国·专题练习",
+        "merged_id": "unit5_424",
+        "original_id": "unit5_choice_424"
+      },
+      {
+        "id": "unit5_choice_425",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "将 15 枚棋子放入下图的小三角形中, 总有一个小三角形中至少放( )枚棋子。\n\n![](images/bf57381c47311c53eab6c1572a1d787f343af7d5db7d82e6efb4811e3f495280.jpg)",
+        "options": [],
+        "answer": "2",
+        "solution": "【分析】本着尽量平均分配的原则是本题的关键思路。此类'至少'题型只要进行除法计算，再将商加上1就可以得到结果。\n\n【详解】题图中小三角形有9个，$15 \\div 9 = 1$ （枚）……6(枚), 将 15 枚棋子平均放进 9 个小三角形中, 每个小三角形中放 1 枚, 剩余 6 枚。剩余的 6 枚继续放进小三角形中,无论怎么放，都能保证总有一个小三角形中至少放 2 枚棋子。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_425",
+        "original_id": "unit5_choice_425"
+      },
+      {
+        "id": "unit5_choice_426",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "把7个盒子装进2个抽屉中，不管怎么放，总有一个抽屉里至少装进( )个盒子。",
+        "options": [],
+        "answer": "4",
+        "solution": "【分析】把 7 个盒子装进 2 个抽屉中, 按照平均分配的原则,那么多出来的总要放进一个抽屉。要想每一个抽屉尽可能的少, 则要采取平均分配的原则, 多出来的余数同样也要尽可能分给不同的对象。\n\n【详解】$7 \\div 2 = 3$ （个）……1（个）$3 + 1 = 4$ （个）故答案为: 4。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_426",
+        "original_id": "unit5_choice_426"
+      },
+      {
+        "id": "unit5_choice_427",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "在 366 个 1999 年出生的儿童中，至少有( )个人是同一天出生的。",
+        "options": [],
+        "answer": "2",
+        "solution": "【分析】要求至少有几个人是同一天出生的，先要判断出 1999 年是平年，全年有 365 天；$366 \\div 365 = 1 \\dots 1$，剩下的这个同学无论生日在哪天，都至少有 $1 + 1 = 2$ 个同学生日在同一天。本题是简单的抽屉原理的应用: 要把 a 个物体放进 n 个抽屉里, 如果 $a \\div n = b \\dots c$, (c≠0), 那么有 1 个抽屉至少可以放 $b + 1$ 个物体。\n\n【详解】$366 \\div 365 = 1$ （个）……1（个）$1 + 1 = 2 \\quad (\\text {个})$故答案为: 2。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_427",
+        "original_id": "unit5_choice_427"
+      },
+      {
+        "id": "unit5_choice_428",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "抽屉里放着红、黄、绿三种颜色的球各3个，一次至少摸出( )个球，才能保证每种颜色的球至少有一个。",
+        "options": [],
+        "answer": "7",
+        "solution": "【分析】根据'抽屉原理'，从最坏的情况考虑，即可解答。此题考查了'抽屉原理'解决实际问题的灵活应用，关键在于考虑最坏的情况，然后得出相应的答案。\n\n【详解】根据题意, 红黄绿的球各有 3 个, 一共有 9 个球,要想保证每种颜色的球至少有一个, 那么要确定拿出 7 个球。故答案为: 7。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_428",
+        "original_id": "unit5_choice_428"
+      },
+      {
+        "id": "unit5_choice_429",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "图书馆里放有六年级数学课本上、下册各5本，至少摸出( )本，才能保证一定有一本下册书；至少摸出( )本，才能保证有两本同样的书。",
+        "options": [],
+        "answer": "6 ;3",
+        "solution": "【分析】上、下册书各 5 本, 假如前 5 次均为上册书, 那么就得至少摸出 6 本就能保证一定有下册书, 要有两本同样的书, 至少要摸出 2 本。本题考查的内容是概率的基本性质的认识, 是一类常见的基础性题型, 在本题中, 要注意 '一定' 和 '可能' 这样的词汇, 考生在平时的练习中, 多注意总结。\n\n【详解】图书馆里放有六年级数学课本上、下册各 5 本, 至少摸出 6 本, 才能保证一定有一本下册书; 至少摸出 3 本,才能保证有两本同样的书。故答案为: 6 ; 3。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_429",
+        "original_id": "unit5_choice_429"
+      },
+      {
+        "id": "unit5_choice_430",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "六年级一班有 25 个同学是同一年出生的，那么至少有( )个同学是同一个月出生的。",
+        "options": [],
+        "answer": "3",
+        "solution": "【分析】把一年 12 个月看作 12 个抽屉, 把 25 人看作 25 个元素, 根据 '抽屉原理' 可知, 那么每个抽屉需要放 $25 \\div 12 = 2$ (个) ……1 (人), 余下的 1 人无论是哪个月出生的, 这个月都至少有 $2 + 1 = 1$ (个) 同学出生, 据此解答。本题主要考查了是 '抽屉原理' : 把 $m$ 个元素任意放入 $n (n \\leq m)$ 个集合, 则一定有一个集合至少要有 $k$ 个元素, 其中 $k = m \\div n$ (当 $n$ 能整除 $m$ 时) 或 $k = m \\div n + 1$ (当n 不能整除 m 时)。\n\n【详解】$25 \\div 12 = 2$ （个）……1（人）$2 + 1 = 3 \\text {(个)}$ 答: 至少有 4 个同学同月出生的。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_430",
+        "original_id": "unit5_choice_430"
+      },
+      {
+        "id": "unit5_choice_431",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "张老师参加飞镖比赛, 投了 5 镖, 成绩是 42 环, 张老师至少有一镖不低于( )环。",
+        "options": [],
+        "answer": "9",
+        "solution": "【分析】根据题意知飞镖上的环数是 10 环, 所以投了 5 镖就是 50 环, 而投了 5 镖, 成绩是 42 环, 所以 $42 \\div 5 = 8 \\dots$ 2 , 所以至少有一镖不低于 9 环。本题主要考查了学生对'简单除法运算中有余数的认识以及在实际生活中的应用'知识点的基本掌握情况。\n\n【详解】$42 \\div 5 = 8 \\dots 2$, 余数是 2 , 所以 $8 + 1 = 9$, 即张叔叔至少有一镖不低于 9 环, 要是低于 9 环, 就得不到 42 环的成绩。故答案为: 9。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_431",
+        "original_id": "unit5_choice_431"
+      },
+      {
+        "id": "unit5_choice_432",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "有红、黄、蓝三种颜色的弹珠各10颗混在一起（大小、形状均相同），蒙上眼睛，至少摸出( )颗才能保证一定有 3 颗颜色相同的弹珠。",
+        "options": [],
+        "answer": "7",
+        "solution": "【分析】由题意可知, 盒子里装有红、黄、蓝三种颜色的球，要保证至少有 3 个小球的颜色相同，最坏的情况是每种颜色各取出 2 个, 即取出 7 个中, 2 个红色, 2 个黄色, 2 个蓝色，此时只要再任取一个，即取出 $2 \\times 3 + 1 = 7$ 个就能保证至少有 3 个小球的颜色相同。此题属于抽屉问题，'抽屉原理'问题的解答思路是：要从最不利情况考虑，准确地建立抽屉和确定元素的总个数，关键是找出'最坏情况'，然后进行分析，继而解答得出结论。\n\n【详解】$2 \\times 3 + 1 = 7$ （个）故答案为: 7。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_432",
+        "original_id": "unit5_choice_432"
+      },
+      {
+        "id": "unit5_choice_433",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "把7个盒子装进3个抽屉中，不管怎么装，总有一个抽屉里至少有（）个盒子。",
+        "options": [],
+        "answer": "3",
+        "solution": "【分析】把 7 个盒子装进 3 个抽屉中, $7 \\div 3 = 2$ (个) ……1 (个), 即平均每个抽屉放入 2 个后, 还余一个盒子没有放入, 即至少有一个抽屉里要放进 $2 + 1 = 3$ 个盒子。本题考查了运用 '抽屉原理' 解决简单的实际问题。把 m 个元素任意放入 n (n≤m) 个集合, 则一定有一个集合至少要有 k 个元素, 其中 $k = m \\div n$ (当 n 能整除 m 时)或 $k = m \\div n + 1$ (当 n 不能整除 m 时)。\n\n【详解】$7 \\div 3 = 2$ （个）……1（个）$2 + 1 = 3$ （个）。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_433",
+        "original_id": "unit5_choice_433"
+      }, {
+        "id": "unit5_choice_434",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "将红、黄、蓝三种颜色的帽子各3顶放入一个箱子里，要保证取出的帽子至少有两种颜色，至少应取出( )顶帽子；要保证三种颜色都有，则至少应取出( )顶；要保证取出的帽子中至少有两个是同色的，则至少应取出( )顶。",
+        "options": [],
+        "answer": "4 7 4",
+        "solution": "【详解】略",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_434",
+        "original_id": "unit5_choice_434"
+      },
+      {
+        "id": "unit5_choice_435",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "一个不透明的盒子里装了同样大小的红、黑、白、黄玻璃球各4个，要保证取出的玻璃球有两个相同颜色的，至少应取出( )个；要使取出的玻璃球中至少有两种颜色，至少应取出( )个。",
+        "options": [],
+        "answer": "5 5",
+        "solution": "【详解】略",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_435",
+        "original_id": "unit5_choice_435"
+      },
+      {
+        "id": "unit5_choice_436",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "如果实验小学六(2) 班有 38 名学生, 那么在六 (2) 班中至少有( )个人出生在同一月。",
+        "options": [],
+        "answer": "4",
+        "solution": "【详解】略",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_436",
+        "original_id": "unit5_choice_436"
+      },
+      {
+        "id": "unit5_choice_437",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "晚饭后, 26 名阿姨在院子里跳健美操, 她们中至少有( )个人的属相相同。",
+        "options": [],
+        "answer": "3",
+        "solution": "【详解】略",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_437",
+        "original_id": "unit5_choice_437"
+      },
+      {
+        "id": "unit5_choice_438",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "容易",
+        "question": "开学时，六年级转来了11名同学，把他们分到5个班里，总有一个班至少分进( )名同学。",
+        "options": [],
+        "answer": "3",
+        "solution": "【分析】略",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_438",
+        "original_id": "unit5_choice_438"
+      },
+      {
+        "id": "unit5_choice_439",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "盒子里有同样大小的红球和黄球各 8 个, 要想摸出的球一定有两种颜色的, 最少要摸出( )个球。",
+        "options": [],
+        "answer": "9",
+        "solution": "【分析】略",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_439",
+        "original_id": "unit5_choice_439"
+      },
+      {
+        "id": "unit5_choice_440",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "若 $x = \\frac{8}{y}$, 则 $x$ 与 $y$ 成 ( )比例关系, 若 $x = 8y$, 则 $x$ 与 $y$ 成 ( )比例关系。",
+        "options": [],
+        "answer": "反正",
+        "solution": "【分析】略",
+        "knowledge_points": [
+          "辨识正比例的量与反比例的量"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_440",
+        "original_id": "unit5_choice_440"
+      },
+      {
+        "id": "unit5_choice_441",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "$( ) \\div 5 = \\frac{10}{()}= 0. 4 = (): 1 5 。$",
+        "options": [],
+        "answer": "2；25；6",
+        "solution": "【分析】略",
+        "knowledge_points": [
+          "小数、分数、百分数、比的互相转换"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_441",
+        "original_id": "unit5_choice_441"
+      },
+      {
+        "id": "unit5_choice_442",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "六 (1) 班有 39 人,总有一个月中至少有( )人过生日。",
+        "options": [],
+        "answer": "4",
+        "solution": "【分析】略",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_442",
+        "original_id": "unit5_choice_442"
+      },
+      {
+        "id": "unit5_choice_443",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "把 9 只小兔装进 4个笼子里，总有一个笼子里有( )只小兔。",
+        "options": [],
+        "answer": "3",
+        "solution": "【分析】略",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_443",
+        "original_id": "unit5_choice_443"
+      },
+      {
+        "id": "unit5_choice_444",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "把线段比例尺![](images/f022638f30f43f3445bf1aa2f45d34799989e2ce5224f4e7cb0c981b5f7f8d87.jpg)改写成数字比例尺是( )。",
+        "options": [],
+        "answer": "1: 400000",
+        "solution": "【分析】略",
+        "knowledge_points": [
+          "比例尺的意义"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_444",
+        "original_id": "unit5_choice_444"
+      },
+      {
+        "id": "unit5_choice_445",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "六年级共有370名学生，一定有两人的生日是同一天。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【详解】略",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_445",
+        "original_id": "unit5_choice_445"
+      },
+      {
+        "id": "unit5_choice_446",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "某小学有1千多名学生，从学生中最少选取( )人，才能使得这些人中有两人属相相同。",
+        "options": [],
+        "answer": "13",
+        "solution": "【详解】略",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_446",
+        "original_id": "unit5_choice_446"
+      },
+      {
+        "id": "unit5_choice_447",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "抽屉理有 4 支红铅笔和3支蓝铅笔，如果闭着眼睛摸，一次必须拿( )支，才能保证至少有1支蓝铅笔。",
+        "options": [],
+        "answer": "5",
+        "solution": "【详解】略",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_447",
+        "original_id": "unit5_choice_447"
+      },
+      {
+        "id": "unit5_choice_448",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "把 5 支圆珠笔放进 4 个文具盒中，不管怎么放，总有一个文具盒里至少放进( )支圆株笔。",
+        "options": [],
+        "answer": "2",
+        "solution": "【详解】略",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_448",
+        "original_id": "unit5_choice_448"
+      },
+      {
+        "id": "unit5_choice_449",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "一副扑克牌共 54 张,其中 $1 \\sim 13$ 点各有 4 种, 还有两张皇牌, 至少要取出( )张才能保证其中必有 4 张牌的点数相同。",
+        "options": [],
+        "answer": "42",
+        "solution": "【分析】由于 $3 \\times 13 + 2 + 1 = 42$, 取出 42 张牌, 其中必有 4 张点数相同。如果只取 41 张, 那么其中可能有 3 张 A, 3 张 2, 3 张 3, …, 3 张 K 及两张王牌, 没有 4 张一样的点数相同。所以, 至少要取 42 张, 才能保证其中必有 4 张牌的点数相同。\n\n【详解】抽屉原理",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_449",
+        "original_id": "unit5_choice_449"
+      },
+      {
+        "id": "unit5_choice_450",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "把 $m$ 个物体任意分放进 $n$ 个抽屉里（$m > n, n$ 是非 0 的自然数），那么一定有一个抽屉中至少放进了( )个物体。",
+        "options": [],
+        "answer": "2",
+        "solution": "【分析】略",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_450",
+        "original_id": "unit5_choice_450"
+      },
+      {
+        "id": "unit5_choice_451",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "有黑色、白色、蓝色手套各5只（不分左右手），至少要拿出( )只（拿的时候不许看颜色），才能使拿出的手套中一定有两双是同颜色的。",
+        "options": [],
+        "answer": "10",
+        "solution": "【分析】略",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "18-19六年级下·全国·单元测试",
+        "merged_id": "unit5_451",
+        "original_id": "unit5_choice_451"
+      },
+      {
+        "id": "unit5_choice_452",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "某个年级有202人参加考试，满分为100分，且得分都为整数，总得分为10101 分, 则至少有( )人得分相同。",
+        "options": [],
+        "answer": "3",
+        "solution": "【详解】略",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_452",
+        "original_id": "unit5_choice_452"
+      },
+      {
+        "id": "unit5_choice_453",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "制作123456789这样 10 张卡片，想一想，至少要抽出_张卡片才能保证摸出的球至少有3种不同的颜色？试一试。",
+        "options": [],
+        "answer": "6",
+        "solution": "【分析】把奇偶两种数看做 2 个抽屉, 10 张卡片看做 10 个元素, 奇数和偶数各有 5 张, 利用抽屉原理最差情况: 把其中一种数取出, 再任取一张就能保证既有偶数又有奇数,即可解答。\n\n【详解】$5 + 1 = 6$ (张)答: 至少要抽出 6 张卡片才能保证既有偶数又有奇数。故答案为: $6_\\circ$",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_453",
+        "original_id": "unit5_choice_453"
+      },
+      {
+        "id": "unit5_choice_454",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "6 个苹果放进 5 个盘子中，总有一个盘子至少放_个苹果。",
+        "options": [],
+        "answer": "2",
+        "solution": "【分析】把 5 个盘子看作 5 个抽屉, 把 6 个苹果看作 6 个元素，从最不利情况考虑，每个抽屉先放 1 个，共需要 5 个，余这一个苹果无论放在那个抽屉里，总有一个抽屉里的有$1 + 1 = 2$ （个），据此解答。\n\n【详解】$6 \\div 5 = 1$ （个）…1（个）$1 + 1 = 2 \\quad (\\text {个})$即总有一个盘子放 2 个苹果。答: 总有 1 个盘子至少要放 2 个苹果。故答案为: $20\\mathrm{~g}$",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_454",
+        "original_id": "unit5_choice_454"
+      },
+      {
+        "id": "unit5_choice_455",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "困难",
+        "question": "在每个格子中任意画上符号 '☆' 和 '△', 则至少有_列的符号是完全一样的。\n\n![](images/0e89fd65074d57176aa6594b373be7284e2bb11860b01da35e74421490109583.jpg)",
+        "options": [],
+        "answer": "3",
+        "solution": "【分析】因为每列的填写的只能是下列 4 种之一: ☆△、△☆、△△、☆☆。一共有 9 列, 考虑最差的情况, $9 \\div 4 = 2 \\dots1$先把 4 种不同的方法填写 2 遍, 最后还剩下 1 列, 这一列无论是哪种方法, 都会使得有 3 列的符号是完全一样的,据此即可解答问题。\n\n【详解】每列的填写方法一共有下列 4 种情况:01、10、11、00。考虑最差的情况, $9 \\div 4 = 2$ (列) $\\cdots 1$ (列)$2 + 1 = 3$ (列)答: 至少有 3 列的符号是完全一样的。故答案为: $3_\\circ$",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_455",
+        "original_id": "unit5_choice_455"
+      },
+      {
+        "id": "unit5_choice_456",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "在 3 个篮子里装 7个苹果，总有一个篮子至少要装入( )个苹果。",
+        "options": [],
+        "answer": "3",
+        "solution": "【分析】根据抽屉原理, 把 3 个篮子看作 3 个抽屉, 把 7 个苹果看作 7 个元素, 要使每个篮子里的苹果尽量少, 要尽量平均分, 即 $7 \\div 3 = 2 \\dots 1$, 由此即可解决问题。\n\n【详解】$7 \\div 3 = 2$ （个）…1（个），$2 + 1 = 3$ （个），答: 总有一个篮子里至少放进 3 个苹果。故答案为: $3_\\circ$",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_456",
+        "original_id": "unit5_choice_456"
+      },
+      {
+        "id": "unit5_choice_457",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "填空题",
+        "difficulty": "中等",
+        "question": "把 5 份快餐分给 4 个工人, 有一个人至少得到_____份。",
+        "options": [],
+        "answer": "2",
+        "solution": "【分析】把 4 个工人看作 4 个抽屉, 把 5 份快餐看作 5 个元素, 从最不利情况考虑, 每个抽屉先放 1 个, 共需要 4 个,余下一份快餐无论放在那个抽屉里, 总有一个抽屉里的有 $1 + 1 = 2$ (个), 据此解答。\n\n【详解】$5 \\div 4 = 1$ （份）……1（份）$1 + 1 = 2$ （份）",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20 四年级下·北京·单元测试",
+        "merged_id": "unit5_457",
+        "original_id": "unit5_choice_457"
+      },
+      {
+        "id": "unit5_choice_458",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "小明玩掷骰子游戏，要保证掷出的骰子的点数至少有两次相同，他最少应掷6次。（ ）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】根据'抽屉原理'：'至少数=抽屉的个数+1'进行解答即可。此题关键在于理解'抽屉原理'并熟练运用。\n\n【详解】把骰子的 6 个点数分别看作 6 个抽屉, 按最不利原则, 6 次掷出分别为 6 个不同点数, 然后再掷出 1 次, 就会和前 6 次当中的任意一个点数相同。$6 + 1 = 7$ (次)所以原题说法错误。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_458",
+        "original_id": "unit5_choice_458"
+      },
+      {
+        "id": "unit5_choice_459",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "口袋里有红、黄、白、黑四种颜色的手套（不分左右）各10只，至少从口袋里取出4只才能保证配成一双。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】根据'抽屉原理'：'至少数=抽屉的个数+1'进行解答即可。此题关键在于理解'抽屉原理'并熟练运用。\n\n【详解】口袋里的 4 种颜色的手套, 按最坏的情况, 先摸出的是 4 只四种颜色各一只, 再摸出一只都可以使得这种颜色的手套有 2 只, 配成一双。$4 + 1 = 5$ (只)所以原题说法错误。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_459",
+        "original_id": "unit5_choice_459"
+      },
+      {
+        "id": "unit5_choice_460",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "把21支铅笔分给9个小朋友，如果每人至少分得2支铅笔，那么最多的可以分到5支铅笔。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】把 21 支铅笔平均分给 9 个小朋友, 每人得 2 支,还剩余 3 支, 把剩余的全部给其中一个小朋友即可解答。此题关键在于理解'抽屉原理'并熟练运用。\n\n【详解】$21 \\div 9 = 2$ （支）……3（支）$2 + 3 = 5$ (支) 所以原题说法正确。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_460",
+        "original_id": "unit5_choice_460"
+      },
+      {
+        "id": "unit5_choice_461",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "把10封信投入3个信箱里，至少有4封信被投入同一个信箱里。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】根据'抽屉原理'：'至少数=元素的总个数÷抽屉的个数+1（有余数的情况下）'进行解答即可。此题关键运用了'抽屉原理'的解题思路：要从最不利的情况考虑，准确建立抽屉和确定元素的总个数进行解答。\n\n【详解】$10 \\div 3 = 3 \\dots 1$ $3 + 1 = 4$ 故至少有 4 封信投入同一个信箱。所以原题说法正确。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_461",
+        "original_id": "unit5_choice_461"
+      },
+      {
+        "id": "unit5_choice_462",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "把9块水果糖分给4名同学，其中有一名同学最少能分到3块。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】先平均分配, 多出的再必须给到某位同学得到的就是得到最多的同学最少能分得的数量。在平均分配的原则上再将余数也尽可能公平分配是此类题的分配原则。\n\n【详解】$9 \\div 4 = 2$ （块）……1（块）$2 + 1 = 3$ (块)故答案为: √",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_462",
+        "original_id": "unit5_choice_462"
+      },
+      {
+        "id": "unit5_choice_463",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "6 只小鸟飞回 5 个鸟巢, 至少有 2 只小鸟要飞进同一个鸟巢里。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】只有 5 个鸟巢, 其中 5 只鸟飞往不同的鸟巢, 则剩余的那只鸟只能飞往已经有一只鸟的鸟巢。本题数量较小，用假设的方法即可解决。\n\n【详解】$6 \\div 5 = 1$ （只）……1（只）$1 + 1 = 2 \\quad (\\text {只})$故答案为: √",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_463",
+        "original_id": "unit5_choice_463"
+      },
+      {
+        "id": "unit5_choice_464",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "学校举行数学智力竞赛, 31 名同学分成 6 组, 其中有一组至少有 5 名同学参加。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】每一组都是 5 名同学也只有 30 名同学。此类'至少'题目都是先用除法计算再用加法计算，若除的得商为1，直接用商加上1即可得到最多的一组的数量。\n\n【详解】$31 \\div 6 = 5$ （名）……1（名）$5 + 1 = 6 \\text {(名)}$故答案为: ×",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_464",
+        "original_id": "unit5_choice_464"
+      },
+      {
+        "id": "unit5_choice_465",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "把 8 本书分别放进 3 个抽屉里, 至少有一个抽屉放 4 本书。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】要具有推理能力。\n\n【详解】比如一个抽屉放了 6 本书, 另两个抽屉各放了一本书。故答案为:× ",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_465",
+        "original_id": "unit5_choice_465"
+      },
+      {
+        "id": "unit5_choice_466",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "盒子里有 8 个黄球, 5 个红球, 不放回摸至少摸 9 次一定会摸到红球。( )",
+        "options": [],
+        "answer": "√",
+        "solution": "【分析】略",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_466",
+        "original_id": "unit5_choice_466"
+      },  {
+        "id": "unit5_choice_467",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "用两个骰子玩游戏，至少掷9次，才能保证掷骰子的点数和与前面出现过的点数和相等。（）",
+        "options": [],
+        "answer": "×",
+        "solution": "【分析】略",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_467",
+        "original_id": "unit5_choice_467"
+      },
+      {
+        "id": "unit5_choice_468",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "某班有49名学生，班级中至少有5人是同一个月出生。（）",
+        "options": [],
+        "answer": "√",
+        "solution": "【详解】略故答案为: √",
+        "knowledge_points": [
+          "鸽巢问题初步"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_468",
+        "original_id": "unit5_choice_468"
+      },
+      {
+        "id": "unit5_choice_469",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "判断题",
+        "difficulty": "中等",
+        "question": "把 5 块糖分给 3 个小朋友, 有两种分法。( )",
+        "options": [],
+        "answer": "×",
+        "solution": "【详解】略故答案为: ×",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_469",
+        "original_id": "unit5_choice_469"
+      },
+      {
+        "id": "unit5_choice_470",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "连线题",
+        "difficulty": "中等",
+        "question": "如图盒子里有同样大小的球，要想摸出的球一定有 2 个相同的号码，至少要摸出几个球？\n\n![](images\\d649f02ea899ddd713bd6195ac37ed2a.png)",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】第一个盒子有 3 种不同的号码, 根据最不利原则,最不利的情况是摸3个球号码都不相同，那么再摸出1个不论什么颜色，总有一个和前面摸出的同色，所以至少要摸出 $3 + 1 = 4$ （个）球；同理，第二个盒子有5种不同的号码，最不利的情况是摸5个球号码都不相同，那么再摸出1个不论什么颜色，总有一个和前面摸出的同色，所以至少要摸出 $5 + 1 = 6$ （个）球；第三个盒子有4种不同的号码，最不利的情况是摸4个球号码都不相同，那么再摸出1个不论什么颜色，总有一个和前面摸出的同色，所以至少要摸出 $4 + 1 = 5$ （个）球。\n\n【详解】通过分析可得：![](images/df81975e9d3aafece8995ff545eb1b166eda5e6f45c95c1086e25b690cad0ea7.jpg)",
+        "knowledge_points": [
+          "最不利原则"
+        ],
+        "source": "24-25 六年级下·全国·单元测试",
+        "merged_id": "unit5_470",
+        "original_id": "unit5_choice_470"
+      },
+      {
+        "id": "unit5_choice_471",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "先按 3: 1 把下面的图形放大, 再把放大后的图形按 1: 2 缩小。\n\n![](images/0ae71340c628c2bbd36c153649bc4922f67e56fb9975e506b8f5960a80d3898b.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】![](images/df53184edd4e1f85d1af31cc44ed196ee866dbb8a0b9279c1a49cfed101a29a2.jpg)",
+        "knowledge_points": [
+          "图形的放大与缩小"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_471",
+        "original_id": "unit5_choice_471"
+      },
+      {
+        "id": "unit5_choice_472",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "作图题",
+        "difficulty": "中等",
+        "question": "如图，在小方格里最多放入一个'☆'，要想使得同一行、同一列或对角线上的三个小方格都不同时出现三个'☆'，那么在这九个小方格里最多能放入多少个'☆'，应该怎么画呢？\n\n![](images/71669d6e5e2e0906c338fc77f55e9b987b861079529ff6fce561ec23d4e7f6f6.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】因为每行每列不能出现 3 个 $\\star$, 且使九小方格里的 $\\star$ 最多, 所以每行每列都有 2 个 $\\star$, 只要保证正方形的对角线上的 $\\star$ 不出现 3 个即可; 据此解答。\n\n【详解】根据分析如图：\n\n| ☆    | ☆    |      |\n| ---- | ---- | ---- |\n| ☆    |      | ☆    |\n|      | ☆    | ☆    |",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_472",
+        "original_id": "unit5_choice_472"
+      },
+      {
+        "id": "unit5_choice_473",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "6 个人进行射击训练,共射中 121 环, 必定有 1 个人至少射中 21 环, 为什么?\n\n![](images/94d22d654f8e68b209f57d4b2a388bbf58b1404c28c231a515f64557745b3ffe.jpg)",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】把 6 个人看作 6 个抽屉, 把 121 环看作 121 个元素,从最不利情况考虑, 每人射击 20 环, 共射击 $20 \\times 6 = 120$ (环), 剩下 1 环无论放在哪个抽屉, 总有一个抽屉是 20 $+1 = 21$ (环), 据此解答。\n\n【详解】$121 \\div 6 = 20$ （环）……1（环）$2 0 + 1 = 2 1 \\text {(环)}$ 答：因为如果每个人都射中不超过 20 环，那么 6 个人最多只能射中 120 环，但题目给出的总环数是 121 环，超过了 120 环，所以必定有 1 人至少射中 21 环。",
+        "knowledge_points": [
+          "最不利原则"
+        ],
+        "source": "24-25 六年级下·全国·单元测试",
+        "merged_id": "unit5_473",
+        "original_id": "unit5_choice_473"
+      },
+      {
+        "id": "unit5_choice_474",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "王老师准备给班上的35名同学发糖，要保证有人能拿到6颗糖，王老师至少要准备多少颗糖？温馨小提示：可以先给每个人分5颗糖",
+        "options": [],
+        "answer": "176颗",
+        "solution": "【分析】把同学人数看作抽屉，要保证至少有 6 颗糖在同一个抽屉，那么可以每个抽屉先放 5 颗糖，再在某一个抽屉中多放一颗糖即可。\n\n【详解】$(6 - 1) \\times 35 + 1\\\\ = 5 \\times 3 5 + 1 \\\\ = 1 7 5 + 1 \\\\ = 1 7 6 \\text {(颗)} $\n\n答: 王老师至少要准备 176 颗糖。",
+        "knowledge_points": [
+          "鸽巢问题初步"
+        ],
+        "source": "2024六年级下·全国·专题练习",
+        "merged_id": "unit5_474",
+        "original_id": "unit5_choice_474"
+      },
+      {
+        "id": "unit5_choice_475",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "某班有48位同学参加跳绳比赛，在规定的时间内，最多的同学跳了175次，最少的同学跳了160次，那么在该班中至少要挑出多少位同学，从中必能选出3位在规定的时间内跳绳次数相同的同学？\n\n![](images/770a729ff46b608039995539bca4c2eb614d9dbfc5d18fb15cc57f2f907964e8.jpg)",
+        "options": [],
+        "answer": "33位",
+        "solution": "【分析】在 160 次到 175 次之间共有 16 种不同的跳绳次数,把每个跳绳次数看作 1 个抽屉, 共有 16 个抽屉。最坏的情况是每个抽屉里放 2 个相同的跳绳次数, 就必须选出 $16 \\times 2 = 32$ (位) 同学。如果再选一位同学, 不管他跳其中哪种次数, 放入相应的抽屉中, 这个抽屉中便有 3 个相同的跳绳次数，所以至少要挑出 33 位同学，才能保证从中必能选出 3 位在规定的时间内跳绳次数相同的同学。\n\n【详解】$(175 - 160 + 1) \\times 2 + 1\\\\= (1 5 + 1) \\times 2 + 1\\\\=1 6 \\times 2 + 1\\\\= 3 2 + 1\\\\=33 (位)$\n\n答: 在该班中至少要挑出 33 位同学, 从中必能选出 3 位在规定的时间内跳绳次数相同的同学。",
+        "knowledge_points": [
+          "最不利原则"
+        ],
+        "source": "23-24六年级下·全国·课后作业",
+        "merged_id": "unit5_475",
+        "original_id": "unit5_choice_475"
+      },
+      {
+        "id": "unit5_choice_476",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "给下面每个格子涂上红色或蓝色, 至少有两列的涂色相同。为什么?\n\n![](images/d2934e857ac2a1449d809a50ab97e8a2e0adb38da70dc0a3c850747dfb68e34a.jpg)\n\n如果只涂两行的话，结论有什么变化呢？",
+        "options": [],
+        "answer": "见详解",
+        "solution": "【分析】格子有 9 列 3 行, 每个格子涂上红色或蓝色, 每列的涂法共有 8 种情况: 红红红、红红蓝、红蓝红、红蓝蓝、蓝蓝蓝、蓝蓝红、蓝红蓝、蓝红红。把这 8 种情况看成 8 个鸽巢, 9 列格子看成 9 个物体, 根据抽屉原则一: 如果把 $(n + 1)$ 个物体放在 $n$ 个抽屉里, 那么必有一个抽屉中至少放有 2 个物体。如果只涂两行，每列的涂法共有 4 种情况：红红、红蓝、蓝蓝、蓝红。把这 4 种情况看成 4 个鸽巢，9 列格子看成 9 个物体，根据抽屉原则二：如果把 n 个物体放在 m 个抽屉里，其中 $n > m$，那么必有一个抽屉至少有：（1）当 n 不能被 m 整除时，$k = [\\frac{n}{m}] + 1$ 个物体。（2）当 n 能被 m 整除时，$k = \\frac{n}{m}$ 个物体。\n\n【详解】$9 \\div 8 = 1 \\dots 1$ $1 + 1 = 2 \\quad (\\text {列})$ 每个格子涂上红色或蓝色，至少有两列的涂色相同。$9 \\div 4 = 2 \\dots \\dots 1$ $2 + 1 = 3 \\quad (\\text {列})$ 如果只涂两行，至少有三列的涂色相同。",
+        "knowledge_points": [
+          "鸽巢问题进阶"
+        ],
+        "source": "23-24 六年级下·全国·课后作业",
+        "merged_id": "unit5_476",
+        "original_id": "unit5_choice_476"
+      },
+      {
+        "id": "unit5_choice_477",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "军训中小海练习打靶, 中靶 31 发, 他至少有几发子弹打中相同的环数?\n\n![](images/54c2c6ad8a2616e6934a139d7812fe486d3a8477cd5ac31202ff69203068fe6d.jpg)",
+        "options": [],
+        "answer": "8发",
+        "solution": "【分析】总共只有 4 种环数结果, 尽可能均匀分配便可得到至少有几发子弹打中相同的环数。此类'至少'题目都是先用除法计算再用加法计算，直接用商加上1即可得到最多的一组的数量。\n\n【详解】$31 \\div 4 = 7$ （发）……3（发）$7 + 1 = 8 \\quad (\\text {发})$ 答: 至少有 8 发子弹打中相同的环数。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_477",
+        "original_id": "unit5_choice_477"
+      },
+      {
+        "id": "unit5_choice_478",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "文具盒里有同样规格的8支红色笔、5支蓝色笔、15支黑色笔。闭上眼睛从文具盒里至少取出14支笔，才能保证取出的笔中有黑色笔，为什么？",
+        "options": [],
+        "answer": "原题说法正确。按最不利原则，假设先取出的 $8 + 5 = 13$ （支）笔分别是 8 支红色笔和 5 支蓝色笔，然后再取一支，一定是黑笔。\n\n$8 + 5 + 1 = 1 4 \\text {(支)}$ 所以至少取出 14 支笔, 才能保证取出的笔中有黑色笔。",
+        "solution": "【分析】此题根据'抽屉原理'最不利原则即可解答。此题主要考查'抽屉原理'解决简单的实际问题能力，按最不利原则考虑即可。\n\n【详解】答：原题说法正确。按最不利原则，假设先取出的 $8 + 5 = 13$ (支) 笔分别是 8 支红色笔和 5 支蓝色笔, 然后再取一支, 一定是黑笔。$8 + 5 + 1 = 14$ (支) 所以至少取出 14 支笔, 才能保证取出的笔中有黑色笔。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_478",
+        "original_id": "unit5_choice_478"
+      },
+      {
+        "id": "unit5_choice_479",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "小红家的杏熟了,她把一些杏分成了几堆，准备送给她的邻居尝尝。她发现任意选出 5 堆, 其中至少有两堆杏的个数差是 4 的倍数。她的结论正确吗？为什么？",
+        "options": [],
+        "answer": "正确；理由：把 5 堆杏的个数看成任意的 5 个自然数，它们被 4 除，所得余数有 0、1、2、3 四种可能。所以5 个自然数被 4 除后至少有两个余数是相同的, 所以这两个数之差是 4 的倍数。",
+        "solution": "【详解】略",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_479",
+        "original_id": "unit5_choice_479"
+      },
+      {
+        "id": "unit5_choice_480",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "某班有23名同学订报纸，至少订一种报纸，最多可订三种报纸。已知报纸有A、B、C三种。\n\n![](images/d637219379752c2f9ae1ed1f7a3ba1f8878221f20b0a394f55540b8754169bc5.jpg)",
+        "options": [],
+        "answer": "4人",
+        "solution": "【详解】$23 \\div 7 = 3$ （人）……2（人）$3 + 1 = 4 \\text {(人)}$ 答: 至少有 4 人订的报纸是完全相同的。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_480",
+        "original_id": "unit5_choice_480"
+      },
+      {
+        "id": "unit5_choice_481",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "在一个纸箱里装有大小、质地都相同的红、白、蓝、黑四种颜色的袜子各3只,闭着眼睛从中随意拿。(1) 至少拿出多少只才能确保有一双同色的袜子? (2) 至少拿出多少只才能确保有两双同一种颜色的袜子?(同色的 2 只算一双)",
+        "options": [],
+        "answer": "(1) 5只 (2) 7只",
+        "solution": "【详解】略",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_481",
+        "original_id": "unit5_choice_481"
+      },
+      {
+        "id": "unit5_choice_482",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "前进小学六年级有320人，男生和女生人数的比正好是 $1: 1$，至少随机选出多少人，才能保证选取的学生中既有男生又有女生？",
+        "options": [],
+        "answer": "$320 \\div (1 + 1) + 1 = 161$ （人）",
+        "solution": "【详解】略",
+        "knowledge_points": [
+          "鸽巢问题",
+          "比的意义"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_482",
+        "original_id": "unit5_choice_482"
+      },
+      {
+        "id": "unit5_choice_483",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "一个盒子里放了质地、形状、大小都相同的红、黄、绿三种颜色的粉笔各8支，当你蒙上眼睛去盒子中取粉笔时，为了确保自己取出的粉笔中至少有5支颜色相同，应至少取出多少支粉笔？",
+        "options": [],
+        "answer": "13支",
+        "solution": "【详解】$(5 - 1) \\times 3 + 1\\\\ = 1 2 + 1 \\\\ = 1 3 (\\text {支}) $\n\n答: 应至少取出 13 支粉笔。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_483",
+        "original_id": "unit5_choice_483"
+      },
+      {
+        "id": "unit5_choice_484",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "一个盒子里有大小、质地都相同的4个红球，5个白球，要想摸出的球一定有2个颜色不同，至少要摸出几个球？",
+        "options": [],
+        "answer": "6 个",
+        "solution": "【详解】略",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_484",
+        "original_id": "unit5_choice_484"
+      },
+      {
+        "id": "unit5_choice_485",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "根据图中信息，用比例知识解决问题。\n\n![](images/9fbaf829cbb57b9adea00149a9d348898e327468aa8e3cfdeb1e97ef67b50424.jpg)\n\n(1) 这辆汽车行驶 $480 \\mathrm{~km}$, 需要多长时间?\n(2) 9 小时可以行驶多少千米?",
+        "options": [],
+        "answer": "(1) 6小时; (2) 720千米",
+        "solution": "【分析】略",
+        "knowledge_points": [
+          "正比例、反比例的应用"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_485",
+        "original_id": "unit5_choice_485"
+      },
+      {
+        "id": "unit5_choice_486",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "一辆汽车从东城开往西城，每小时行驶 $68 \\mathrm{~km}, 5$ 小时到达，返回时因是上坡路，每小时比原来慢 $\\frac{2}{17}$，返回时用了多少小时？",
+        "options": [],
+        "answer": "$\\frac{17}{3}$ 小时",
+        "solution": "【分析】略",
+        "knowledge_points": [
+          "分数四则复合应用题",
+          "行程问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_486",
+        "original_id": "unit5_choice_486"
+      },
+      {
+        "id": "unit5_choice_487",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "某市南北长 $60 \\mathrm{~km}$,在比例尺是 1:250000 得地图上，长度约是多少厘米？如果在这幅地图上量得东西长是18厘米，那么东西的实际距离是多少千米？",
+        "options": [],
+        "answer": "24 厘米；45 千米",
+        "solution": "【分析】略",
+        "knowledge_points": [
+          "图上距离与实际距离的换算"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_487",
+        "original_id": "unit5_choice_487"
+      },
+      {
+        "id": "unit5_choice_488",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "上午 9 时, 测得光明塔在阳光下的影长是9.6米，同时测得一栋高7米得楼房影长为2.4米，求光明塔得实际高度是多少米？（用比例解）",
+        "options": [],
+        "answer": "28米",
+        "solution": "【分析】略",
+        "knowledge_points": [
+          "正比例、反比例的应用"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_488",
+        "original_id": "unit5_choice_488"
+      },
+      {
+        "id": "unit5_choice_489",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "幼儿园某班有32名小朋友，现有各种玩具108个，把这些玩具全部分给小朋友，是否总会有一名小朋友至少得到4个玩具？",
+        "options": [],
+        "answer": "总会有一名小朋友至少得到 4 个玩具",
+        "solution": "【分析】把 32 名小朋友看做 32 个抽屉, 108 个玩具看做108 个元素, 利用抽屉原理最差情况: 要使得到玩具数最少,只要使每个抽屉的元素数尽量平均，即可解答。\n\n【详解】$108 \\div 32 = 3$ （个）…12（个）$3 + 1 = 4$ （个）即总会有一名小朋友至少得到 4 个玩具。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_489",
+        "original_id": "unit5_choice_489"
+      },
+      {
+        "id": "unit5_choice_490",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "口袋里有大小、质地、形状完全相同的红球、黄球、黑球、粉球和白球各 15个，想要不放回地摸出 6 个同色的球，最多需要摸出多少个球？",
+        "options": [],
+        "answer": "26 个",
+        "solution": "【分析】球的颜色一共有 5 种, 最坏的结果, 要摸出(5+1) 个才有同一种颜色的 2 个; 再摸 5 个必同前面 2 个颜色相同的；再摸 5 个，必有同前面 3 个颜色相同的；再摸5 个, 必须同前面 4 个颜色相同的; 再摸 5 个, 必有与前面5 个颜色相同的; 最后再摸 1 个, 必须同前面 5 个颜色相同的。\n\n【详解】$5 \\times 5 + 1\\\\ = 2 5 + 1 \\\\ = 2 6 (\\text {个}) $\n\n答: 最多需要摸出 26 个球。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_490",
+        "original_id": "unit5_choice_490"
+      },
+      {
+        "id": "unit5_choice_491",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "黑色、白色、黄色的筷子各有 8 根，混杂放在一个盒子里，想从这些筷子中取出颜色不同的两双筷子，至少要取多少根才能保证达到要求？",
+        "options": [],
+        "answer": "11 根",
+        "solution": "【详解】10 根筷子, 可能是 8 根黑色, 1 根白色, 1 根黄色,其中没有颜色不同的两双筷子。如果取 11 根, 那么由于 11 >3, 其中必有两根同色组成一双, 不妨设这一双是黑色的,去掉这两根, 余下 9 根, 其中黑色的至多有 6 根, 因而白、黄两色的筷子至少有 3 根, 3 根中必有 2 根同色组成一双。这样就得到颜色不同的两双筷子。所以至少要取11根。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "2014 六年级·全国·课后作业",
+        "merged_id": "unit5_491",
+        "original_id": "unit5_choice_491"
+      },
+      {
+        "id": "unit5_choice_492",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "扑克牌里学数学：一副扑克牌（取出两张王牌）。(1) 在剩下的 52 张牌中任意抽出 9 张, 至少有多少张是同花色的? (2) 扑克牌一共有 4 种花色, 每种花色都有 13 张牌, 问至少要抽出几张牌才能保证有一张是红桃? (3) 至少要抽出多少张才能保证有 5 张牌是同一花色的?",
+        "options": [],
+        "answer": "(1) 3 张 (2) 40 张 (3) 17 张",
+        "solution": "【详解】(1) $9 \\div 4 = 2$ （张）……1（张） $2 + 1 = 3 \\text {(张)}$ 答: 至少有 3 张是同花色的。\n\n(2) $13 \\times 3 + 1\\\\= 3 9 + 1 \\\\ = 4 0 \\text {(张)} $\n\n答: 至少要抽出 40 张牌才能保证有一张是红桃。\n\n(3) $4 \\times 4 + 1\\\\ = 1 6 + 1 \\\\ = 1 7 \\text {(张)} $\n\n答: 至少要抽出 17 张才能保证有 5 张牌是同一花色的。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_492",
+        "original_id": "unit5_choice_492"
+      },
+      {
+        "id": "unit5_choice_493",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "幼儿园小朋友分苹果、梨、橘子这三种水果。如果每个小朋友任意拿两个水果，那么至少几个小朋友拿过后才一定能出现两人拿的水果是相同的？",
+        "options": [],
+        "answer": "7 个",
+        "solution": "【详解】拿二个水果情况有如下 6 种情况 (作为 6 个抽屉):(苹果、苹果), (苹果, 梨), (苹果, 橘子), (梨, 梨), (梨, 橘子), (橘子, 橘子)。假设有 6 人, 分别拿了其中的一种, 那么再多 1 人, 拿的只能是这 6 种中一种情况, 即: $6 + 1 = 7$ (个)答: 至少 7 个小朋友拿过后才一定能出现两人拿的水果是相同的。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_493",
+        "original_id": "unit5_choice_493"
+      },
+      {
+        "id": "unit5_choice_494",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "在如下图的盒子中，小华蒙着眼睛往外摸球，至少要摸出多少个，才能保证摸出的球至少有3种不同的颜色？\n\n![](images/21d967665c89992b5f95310029f2bb858067b742675f127616be6cbfb52fcb1b.jpg)",
+        "options": [],
+        "answer": "10 个",
+        "solution": "【分析】盒子中有：黄色球 4 个，蓝色球 4 个，绿色球 5 个，红色球 3 个，共 16 个球；要保证摸出的球至少有 3 种不同的颜色，根据抽屉原理最不利原则，要把 2 种颜色数量多的球摸出后，再多摸出 1 个即可；据此解答。此题考查了抽屉原理的应用，关键能够理解最不利原则。\n\n【详解】$5 + 4 + 1\\\\ = 9 + 1 \\\\ = 1 0 (\\text {个}) $\n\n答: 至少要摸出 10 个, 才能保证摸出的球至少有 3 种不同的颜色。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_494",
+        "original_id": "unit5_choice_494"
+      },
+      {
+        "id": "unit5_choice_495",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "某班同学为地震灾区小朋友捐献图书，所捐图书共分为故事书、科技树和教辅资料书三类，捐书的情况是：有捐一本的，有捐两本的，还有捐三本的。问至少要有几位同学来捐书才能保证一定有两位同学所捐书的类型相同？（每种类型的书最多捐一本）",
+        "options": [],
+        "answer": "8 位",
+        "solution": "【详解】$7 + 1 = 8$ （位）答: 至少要有 8 位同学来捐书才能保证一定有两位同学所捐书的类型相同。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_495",
+        "original_id": "unit5_choice_495"
+      },
+      {
+        "id": "unit5_choice_496",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "笔盒里有4支圆珠笔和 3 支钢笔（一样粗细），如果闭上眼睛拿笔，一次至少拿几支笔才能保证有 1 支是钢笔?",
+        "options": [],
+        "answer": "$4 + 1 = 5$ (支)",
+        "solution": "【详解】略",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_496",
+        "original_id": "unit5_choice_496"
+      },
+      {
+        "id": "unit5_choice_497",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "小李家到学校有几条路可以走，那一条最近？\n\n![](images/b8d7b742688122e4f2d88d5fe5801919a9948f480114089a12f80729824e74e2.jpg)",
+        "options": [],
+        "answer": "4条；小李家--小东家--学校",
+        "solution": "【详解】小李家到学校有4条路可以走：(1) 小李家--小东家--学校, 长度是: $300 + 850 = 1150​$ (米)。 (2) 小李家--小明家--学校, 长度是: $1150 + 260 = 1410​$ (米)。 (3) 小李家--小明家--小东家--学校, $1150 + 980 + 850=2130+850=2980(米)​$。(4) 小李家--小东家--小明家--学校, $300 + 980 + 260 =​$ $1280 + 260 = 1540​$ (米)。 $1150 < 1410 < 1540 < 2980​$答: 小李家到学校有 4 条路可以走, 第一条最近。",
+        "knowledge_points": [
+          "两点间线段最短与两点间的距离"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_497",
+        "original_id": "unit5_choice_497"
+      },
+      {
+        "id": "unit5_choice_498",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "新兵训练时，战士小王10枪命中了81环，那么小王至少有一枪是打中了9环，你认为说得对吗？为什么？\n\n![](images/735c8aa1953322ee5405972e47a4a346aa49ad78cf70652d342d8ab0c1723a3d.jpg)",
+        "options": [],
+        "answer": "说得对；见详解",
+        "solution": "【分析】抽屉原理问题的解答思路是：要从最不利情况考虑，准确地建立抽屉和确定元素的总个数，然后根据'至少数 = 元素的总个数 ÷ 抽屉的个数（商） +1 （有余数的情况下）'解答。\n\n【详解】$81 \\div 10 = 8$ （环）…1（环）$8 + 1 = 9$ (环)所以，他至少有一枪打中了 9 环，所以说得对。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20六年级下·全国·单元测试",
+        "merged_id": "unit5_498",
+        "original_id": "unit5_choice_498"
+      },
+      {
+        "id": "unit5_choice_499",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "困难",
+        "question": "一次数学竞赛出了10 道选择题, 评分标准为: 基础分 10 分, 每道题答对得 3分, 答错扣 1 分, 不答不得分, 问: 要保证至少有 4 人得分相同，至少需要多少人参加竞赛？",
+        "options": [],
+        "answer": "115人",
+        "solution": "【详解】因为最高可得 $4 \\times 10 = 40$ (分)最低是倒扣：$10 - 1 \\times 10 = 0$ （分）共有 $40 + 1 = 41$ （种）不同分数。答对与答错之间的分数差是 $3 + 1 = 4$ 分；答对一题和空一题之间差 3 分, 所以最高分 40 分, 对 9 题的情况下, 最高分40-3=37分，最低分40-3-1=36分，中间的38分和39分不会出现；依次列举可得 35 分也不会出现。$41-3=38 \\quad38 \\times 3 + 1 = 115$ （人）为了保证至少有 4 人得分相同, 那么参加考试的学生至少有115人。答: 参加考试的学生至少有 115 人。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_499",
+        "original_id": "unit5_choice_499"
+      },
+      {
+        "id": "unit5_choice_500",
+        "unit": "第五单元：鸽巢原理",
+        "chapter": "鸽巢问题",
+        "question_type": "解答题",
+        "difficulty": "中等",
+        "question": "从 13 个连续的自然数中，一定可以找到两个数，它们的差是12的倍数。任意取多少个连续的自然数，才能保证至少有两个自然数的差是7的倍数？",
+        "options": [],
+        "answer": "8 个",
+        "solution": "【分析】因为余数相同的两数之差一定能被除数整除，此题可以先找出除以 7 的余数的所有情况分别为：0、1、2、3、4、5、6，这样就可以把它们看做7个抽屉，利用抽屉原理即可解决问题。\n\n【详解】自然数除以 7 的余数为: $0 、 1 、 2 、 3 、 4 、 5 、 6$,因此 7 就把自然数分成了 7 类,即：除以7余0、1、2、3、4、5、6，因此，可以把它看成是7个抽屉，至少要有 8 个数, 才能必然有一个抽屉里有两个数, 而这两个数除以 7 的余数相同, 也就是差是 7 的倍数,答: 根据上述分析, 至少任意取 8 个连续的自然数, 就能保证其中必有两个数, 它们的差是 7 的倍数。",
+        "knowledge_points": [
+          "鸽巢问题"
+        ],
+        "source": "19-20 六年级下·全国·单元测试",
+        "merged_id": "unit5_500",
+        "original_id": "unit5_choice_500"
+      }
+]
\ No newline at end of file
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new file mode 100644
index 0000000000000000000000000000000000000000..90d995cde3fba9312bc5787451baa08852bc64d1
--- /dev/null
+++ "b/\345\205\255\345\271\264\347\272\247\351\242\230\351\233\206/\345\205\255\345\271\264\347\272\247\344\270\213/\345\205\255\345\271\264\347\272\247\344\270\213\347\254\254\345\233\233\345\215\225\345\205\203\346\257\224\344\276\213.json"
@@ -0,0 +1,6740 @@
+[
+  {
+    "id": "unit4_choice_001",
+    "unit": "第四单元：比例",
+    "chapter": "正比例的意义及辨识",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "圆的面积与 ( ) 成正比例关系。",
+    "options": [
+      "A. 半径",
+      "B. 圆周率",
+      "C. 半径的平方"
+    ],
+    "answer": "C",
+    "solution": "根据成正比例的两个量的比值一定，结合圆的面积公式：S = πr² 计算即可。本题主要考查成正比例的量的判断, 关键是理解正比例的意义。圆的面积 ÷ 半径的平方 = π (一定), 圆的面积与半径的平方成正比例关系。",
+    "knowledge_points": [
+      "正比例的意义及辨识"
+    ],
+    "source": "18-19六年级下·全国·期中"
+  },
+  {
+    "id": "unit4_choice_002",
+    "unit": "第四单元：比例",
+    "chapter": "图上距离与实际距离的换算",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "在比例尺是1:6000000 的地图上，量得南京到北京的图上距离是15cm，南京到北京的实际距离大约是（）。",
+    "options": [
+      "A. 800千米",
+      "B. 90千米",
+      "C. 900千米",
+      "D. 80千米"
+    ],
+    "answer": "C",
+    "solution": "根据实际距离 = 图上距离 ÷ 比例尺, 据此进行计算即可。本题考查比例尺，明确图上距离、实际距离和比例尺之间的关系是解题的关键。15 ÷ 1/6000000 = 90000000(cm) = 900(km)则南京到北京的实际距离大约是 900km。",
+    "knowledge_points": [
+      "图上距离与实际距离的换算"
+    ],
+    "source": "21-22六年级下·浙江·单元测试"
+  },
+  {
+    "id": "unit4_choice_003",
+    "unit": "第四单元：比例",
+    "chapter": "反比例的意义及辨识",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "平行四边形的面积一定,它的底和高（）。",
+    "options": [
+      "A. 成正比例",
+      "B. 成反比例",
+      "C. 不成比例",
+      "D. 不确定"
+    ],
+    "answer": "B",
+    "solution": "平行四边形面积=底×高，乘积一定的两个量成反比例关系，据此解题。本题考查了反比例和平行四边形的面积, 掌握反比例的意义以及平行四边形的面积公式是解题的关键。底 × 高 = 平行四边形面积（一定）所以，平行四边形的面积一定，它的底和高成反比例。",
+    "knowledge_points": [
+      "反比例的意义及辨识",
+      "平行四边形的面积"
+    ],
+    "source": "14-15六年级·全国·课后作业"
+  },
+  {
+    "id": "unit4_choice_004",
+    "unit": "第四单元：比例",
+    "chapter": "正比例的意义及辨识",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "下面四个句子中，描述正确的有（）个。\n(1) 含有未知数的等式叫方程。\n(2) 单价一定，总价与数量成正比例。\n(3) 黄金比的比值约等于 0.618。\n(4) 得数是 1 的两个数, 互为倒数。",
+    "options": [
+      "A. 1",
+      "B. 2",
+      "C. 3",
+      "D. 4"
+    ],
+    "answer": "C",
+    "solution": "根据方程、正比例、黄金比、倒数的定义，解答此题即可。熟练掌握方程、正比例、黄金比、倒数的定义,是解答此题的关键。含有未知数的等式叫方程，正确；总价÷数量 = 单价（一定），单价一定，总价与数量成正比例，正确；黄金比的比值约等于 0.618，正确；乘积等于 1 的两个数, 互为倒数, 题干说法错误。",
+    "knowledge_points": [
+      "正比例的意义及辨识",
+      "倒数的认识及应用",
+      "方程的认识",
+      "求比值"
+    ],
+    "source": "2023六年级下·江苏连云港·专题练习"
+  },
+  {
+    "id": "unit4_choice_005",
+    "unit": "第四单元：比例",
+    "chapter": "比例尺应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "在比例尺是 1:8 的图纸上，甲、乙两个圆的直径比是 2:3 ，那么甲、乙两个圆实际的直径比是（）。",
+    "options": [
+      "A. 1:8",
+      "B. 2:3",
+      "C. 4:9",
+      "D. 9:4"
+    ],
+    "answer": "B",
+    "solution": "根据比例尺的意义，令甲乙两圆的图上直径为 2d，3d，根据比例尺可得实际圆的直径分别是16d，24d，由此利用比例尺进行计算，即可选择正确答案。令甲乙两圆的图上直径为 2d, 3d, 根据比例尺可得实际甲乙两圆的直径分别是 2d × 8 = 16d, 3d × 8 = 24d,此题考查了利用比例尺解决实际问题的方法, 关键是根据比例尺可得实际圆的直径分别是 16d, 24d。16d:24d = (16d ÷ 8d): (24d ÷ 8d) = 2: 3。甲、乙两个圆实际的直径比是 2: 3。",
+    "knowledge_points": [
+      "比例尺应用"
+    ],
+    "source": "2014·全国·小升初真题"
+  },
+  {
+    "id": "unit4_choice_006",
+    "unit": "第四单元：比例",
+    "chapter": "正比例的意义及辨识",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "下面每组的两个量中,成正比例的量是 ( )。",
+    "options": [
+      "A. 书的总页数一定, 未读的页数和已读的页数",
+      "B. 正方体的表面积与它的棱长",
+      "C. 一个人的身高和体重",
+      "D. 速度一定, 路程和时间"
+    ],
+    "answer": "D",
+    "solution": "两个相关联的量，若它们的乘积一定，则它们成反比例；若它们的比值一定，则它们成正比例，据此判断即可。本题考查正反比例的判定，明确正反比例的定义是解题的关键。A．未读页数+已读页数=书的总页数（一定），是和一定，所以书的总页数一定，未读页数与已读页数不成比例；B. 因为正方体的表面积: S = 6a², 所以 S ÷ a² = 6 (一定),即正方体的表面积和棱长的平方成正比例, 所以正方体的表面积与棱长不成正比例; C. 人的身高和体重虽是两种相关联的量, 但是它们的乘积或比值都不一定, 所以不成比例; D. 因为路程÷时间 = 速度 (一定), 是比值一定, 符合正比例的意义, 所以速度一定, 路程和时间成正比例;",
+    "knowledge_points": [
+      "正比例的意义及辨识"
+    ],
+    "source": "22-23六年级下·广东东莞·期中"
+  },
+  {
+    "id": "unit4_choice_007",
+    "unit": "第四单元：比例",
+    "chapter": "反比例的应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "修一条路, 若每天修 180m, 则 40 天可以完成; 若每天修 200m, 则 x 天可以完成。正确的比例是 ( )。",
+    "options": [
+      "A. 180x = 200 × 40",
+      "B. 200x = 180 × 40",
+      "C. 40/x = 180/200"
+    ],
+    "answer": "B",
+    "solution": "根据工作总量 = 工作效率 × 工作时间; 由于这条路的长度不变, 工作效率和工作时间成反比例, 设 x 天可以完成, x 天修的长度等于 40 天修的长度; 列方程: 200x = 180 × 40, 据此解答。解答本题的关键是求出工作效率和工作时间成什么比例，进而进行解答。根据分析可知, 修一条路, 若每天修 180m, 则 40 天可以完成; 若每天修 200m, 则 x 天可以完成。正确的比例是 200x = 180 × 40。",
+    "knowledge_points": [
+      "反比例的应用"
+    ],
+    "source": "21-22六年级下·江苏·单元测试"
+  },
+  {
+    "id": "unit4_choice_008",
+    "unit": "第四单元：比例",
+    "chapter": "比例的意义",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "下列数中, ( ) 不能与 1/2、1/3、1/6 组成比例。",
+    "options": [
+      "A. 1/4",
+      "B. 1",
+      "C. 12",
+      "D. 1/9"
+    ],
+    "answer": "C",
+    "solution": "表示两个比相等的式子叫做比例, 据此把选项中的数与 1/2、1/3、1/6 组成两个比, 比值相等即可组成比例。根据比例的意义即可解答。把选项中的数分别与三个数其中的一个组成比，剩下的两个数也组成比，再判断比值是否相等。A. 1/4: 1/2 = 1/2, 1/6: 1/3 = 1/2, 两个比比值相等, 即 1/4 能与 1/2, 1/3, 1/6 组成比例;B. 1:1/2 = 2, 1/3:1/6 = 2, 两个比比值相等, 即 1 能与 1/2、1/3、1/6组成比例;C. 12 和 1/2、1/3、1/6 组成的两个比, 比值不相等, 则 12 不能与 1/2、1/3、1/6 组成比例; D. 1/3: 1/9 = 3, 1/2: 1/6 = 3, 两个比比值相等, 即 1/9 能与 1/2、1/3、1/6 组成比例。",
+    "knowledge_points": [
+      "比例的意义"
+    ],
+    "source": "21-22六年级下·辽宁·单元测试"
+  },
+  {
+    "id": "unit4_choice_009",
+    "unit": "第四单元：比例",
+    "chapter": "比例的基本性质",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "对于两个比'3.6:1.8和 0.5:0.25',因为 1.8 × 0.5 = 0.9, 3.6 × 0.25 = 0.9，所以这两个比可以组成比例。这是根据（）来判断的。",
+    "options": [
+      "A. 比的意义",
+      "B. 比例的意义",
+      "C. 比例的基本性质"
+    ],
+    "answer": "C",
+    "solution": "比例的基本性质：两内项之积等于两外项积。因为 1.8 × 0.5 = 0.9, 3.6 × 0.25 = 0.9，符合比例的基本性质，据此判断。利用比例的基本性质, 是判断两个比能否成比例的有效方法。因此掌握比例的基本性质是解答本题的关键。因为 1.8 × 0.5 = 0.9, 3.6 × 0.25 = 0.9根据比例的基本性质，可以判定 3.6: 1.8 和 0.5: 0.25 成比例。",
+    "knowledge_points": [
+      "比例的基本性质"
+    ],
+    "source": "22-23六年级下·江苏·单元测试"
+  },
+  {
+    "id": "unit4_choice_010",
+    "unit": "第四单元：比例",
+    "chapter": "比例的基本性质",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "如果 2a = 5b（a、b都是非0自然数），用2、5、a、b组成正确的比例式是（）。",
+    "options": [
+      "A. 2: a = 5: b",
+      "B. 2: b = 5: a",
+      "C. 2: 5 = a: b"
+    ],
+    "answer": "B",
+    "solution": "比例的基本性质：在比例里，两个外项的积等于两个内项的积。根据比例的基本性质把各选项的比例式改写成两数相乘的形式，与原式 2a = 5b 相比较，算式一致的，就是正确的比例式。掌握比例的基本性质是解题的关键。A. 2: a = 5: b, 那么 2b = 5a, 不符合题意;B. 2: b = 5: a, 那么 2a = 5b, 符合题意; C. 2: 5 = a: b, 那么 2b = 5a, 不符合题意;",
+    "knowledge_points": [
+      "比例的基本性质"
+    ],
+    "source": "22-23六年级下·江苏·单元测试"
+  },
+  {
+    "id": "unit4_choice_011",
+    "unit": "第四单元：比例",
+    "chapter": "正比例的意义及辨识",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "货车每次拉煤3吨，煤的总量和拉煤的次数（）。",
+    "options": [
+      "A. 成正比例",
+      "B. 成反比例",
+      "C. 不成比例"
+    ],
+    "answer": "A",
+    "solution": "判断两种相关联的量成不成比例，成什么比例，就看这两种量是对应的比值一定，还是对应的乘积一定，如果是比值一定，就成正比例，如果是乘积一定，就成反比例，如果是其它的量一定或乘积、比值不一定，就不成比例。此题属于辨识正反比例关系，就看这两个量是对应的比值一定，还是乘积一定，再做判断。煤的总量 ÷ 货车拉煤的次数 = 每次拉煤的吨数（一定），则货车每次拉煤 3 吨，煤的总量和拉煤的次数成正比例。",
+    "knowledge_points": [
+      "反比例的意义及辨识",
+      "正比例的意义及辨识"
+    ],
+    "source": "22-23六年级下·全国·单元测试"
+  },
+  {
+    "id": "unit4_choice_012",
+    "unit": "第四单元：比例",
+    "chapter": "正比例的意义及辨识",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "用四根木条制作一个平行四边形框架，双手将它的两个对角慢慢向两边拉动，这个变化过程中平行四边形的面积和高（）。",
+    "options": [
+      "A. 成正比例关系",
+      "B. 成反比例关系",
+      "C. 不成比例"
+    ],
+    "answer": "A",
+    "solution": "判断两个相关联的量之间成什么比例，就看这两种量是对应的比值一定，还是对应的乘积一定，如果是比值一定，就成正比例，如果是乘积一定，就成反比例，如果是其它的量一定或乘积、比值不一定，就不成比例。本题考查正反比例的判定，明确正反比例的定义是解题的关键。因为在这个变化过程中平行四边形的底不变，根据平行四边形的面积÷高=底（一定），它们的比值不变，所以平行四边形的面积和高成正比例。",
+    "knowledge_points": [
+      "反比例的意义及辨识",
+      "平行四边形的面积",
+      "正比例的意义及辨识"
+    ],
+    "source": "22-23六年级下·全国·单元测试"
+  },
+  {
+    "id": "unit4_choice_013",
+    "unit": "第四单元：比例",
+    "chapter": "比例尺的意义",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "一个零件长 0.6cm,画在一幅图上长 18cm。这幅图的比例尺是 ( )。",
+    "options": [
+      "A. 30:1",
+      "B. 1:30",
+      "C. 1:3"
+    ],
+    "answer": "A",
+    "solution": "比例尺 = 图上距离 : 实际距离, 把题中数据代入公式计算, 据此解答。掌握比例尺的意义是解答题目的关键。18cm: 0.6cm = 180: 6 = (180 ÷ 6): (6 ÷ 6) = 30: 1",
+    "knowledge_points": [
+      "比例尺的意义"
+    ],
+    "source": "22-23六年级下·全国·单元测试"
+  },
+  {
+    "id": "unit4_choice_014",
+    "unit": "第四单元：比例",
+    "chapter": "图上距离与实际距离的换算",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "在比例尺是1:200000 的地图上, 量得小华家到外婆家的距离是 5cm, 小华 8:30 从家出发, 10:30 走到外婆家, 小华平均每时行 ( )。",
+    "options": [
+      "A. 20km",
+      "B. 15km",
+      "C. 5km"
+    ],
+    "answer": "C",
+    "solution": "图上距离 ÷ 比例尺 = 实际距离, 据此代入数据求出小华家到外婆家的实际距离, 再根据路程 ÷ 时间 = 速度, 求出小华平均每时行的路程。明确图上距离、实际距离、比例尺三者间的关系是解题的关键。5 ÷ 1/200000 = 1000000 (cm), 1000000cm = 10km, 10时30分-8时30分=2时, 10 ÷ 2 = 5 (km)。小华平均每时行 5km。",
+    "knowledge_points": [
+      "图上距离与实际距离的换算",
+      "基础行程问题"
+    ],
+    "source": "22-23六年级下·全国·单元测试"
+  },
+  {
+    "id": "unit4_choice_015",
+    "unit": "第四单元：比例",
+    "chapter": "图形的放大与缩小",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "将一个平面图形按 4:1 放大, 变为原来图形 4 倍的是 ( )。",
+    "options": [
+      "A. 图形角的度数",
+      "B. 图形的面积",
+      "C. 图形的周长"
+    ],
+    "answer": "C",
+    "solution": "一个图形按 4:1 放大后, 就是把这个图形的各边长放大 4 倍, 也就是各边乘 4 , 周长扩大 4 倍; 所得到的新图形的各边都是原图形的 4 倍, 它的面积将是原图形的 4² 倍,即 16 倍; 图形的形状不变, 即角的度数不变。据此解答即可。此题是考查图形放大与缩小的意义, 如果把一个图形按一定的比放大, 图形周长扩大的倍数与图形边长放大的倍数相同; 图形面积扩大的倍数等于图形边长放大倍数的平方; 对应角大小不变。把一个图形按 4:1 放大，则图形角的度数不变，周长扩大 4 倍，面积扩大了 4 × 4 = 16 倍。",
+    "knowledge_points": [
+      "图形的放大与缩小"
+    ],
+    "source": "22-23六年级下·全国·单元测试"
+  },
+  {
+    "id": "unit4_choice_016",
+    "unit": "第四单元：比例",
+    "chapter": "比例的基本性质",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "一个比例的两个内项互为倒数，其中一个外项是最小的合数，另一个外项是（）。",
+    "options": [
+      "A. 2",
+      "B. 1/2",
+      "C. 1/4"
+    ],
+    "answer": "C",
+    "solution": "互为倒数的两个数的乘积是 1, 最小的合数是 4 ,据此解答即可。本题考查倒数的意义, 明确互为倒数的两个数的乘积是 1 是解题的关键。1 ÷ 4 = 1/4，所以另一外项是 1/4。",
+    "knowledge_points": [
+      "倒数的认识及应用",
+      "比例的基本性质"
+    ],
+    "source": "2024·新疆吐鲁番·小升初真题"
+  },
+  {
+    "id": "unit4_choice_017",
+    "unit": "第四单元：比例",
+    "chapter": "正比例的意义及辨识",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "下列各题中的两个量一定'成正比例关系'的是（）。",
+    "options": [
+      "A. 时间一定, 路程和速度",
+      "B. 在一幅地图上的图上距离和比例尺",
+      "C. 小明的身高和体重"
+    ],
+    "answer": "A",
+    "solution": "判断两个相关联的量之间成什么比例，就看这两个量是对应的比值一定，还是对应的乘积一定；如果是比值一定，就成正比例；如果是乘积一定，则成反比例。此题属于辨识成正、反比例的量，就看这两个量是对应的比值一定，还是对应的乘积一定，再做判断。A．根据路程÷速度 = 时间，时间一定，即路程和速度的比值一定，符合正比例的意义，所以路程和速度成正比例；B. 根据实际距离 = 图上距离 ÷ 比例尺, 题干中并没有说实际距离一定, 即图上距离与比例尺之间的比值也不确定,所以图上距离和比例尺不成比例; C. 人的身高和体重虽是两种相关联的量, 但是它们的乘积或比值都不一定, 所以不成比例;",
+    "knowledge_points": [
+      "反比例的意义及辨识",
+      "正比例的意义及辨识"
+    ],
+    "source": "22-23六年级下·全国·单元测试"
+  },
+  {
+    "id": "unit4_choice_018",
+    "unit": "第四单元：比例",
+    "chapter": "图上距离与实际距离的换算",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "在比例尺为1:500000 的地图上，量得一正方形的实验基地边长是 1.2 厘米，实际上这个基地的周长是（）千米。",
+    "options": [
+      "A. 24",
+      "B. 240000",
+      "C. 2400"
+    ],
+    "answer": "A",
+    "solution": "根据实际距离 = 图上距离 ÷ 比例尺, 代入图上距离和比例尺的数据, 求出这个正方形的实际边长, 再根据正方形的周长公式, 即可求出这个基地的周长。此题的解题关键是掌握图上距离和实际距离之间的换算以及正方形的周长公式。1.2 ÷ 1/500000 = 600000（厘米）= 6（千米），6 × 4 = 24（千米）",
+    "knowledge_points": [
+      "图上距离与实际距离的换算",
+      "正方形的周长"
+    ],
+    "source": "22-23六年级下·全国·单元测试"
+  },
+  {
+    "id": "unit4_choice_019",
+    "unit": "第四单元：比例",
+    "chapter": "反比例的意义及辨识",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "下面说法正确的是 ( )。",
+    "options": [
+      "A. 一个人的年龄和身高成正比例",
+      "B. 路程一定，速度和时间成正比例",
+      "C. 总价一定, 单价和数量成反比例"
+    ],
+    "answer": "C",
+    "solution": "判断两个相关联的量之间成什么比例，就看这两个量是对应的比值一定，还是对应的乘积一定；如果是比值一定，就成正比例；如果是乘积一定，则成反比例。据此解答。此题属于辨识成正、反比例的量，就看这两个量是对应的比值一定，还是对应的乘积一定，再做判断。A．通常在生长期，人的身高是随着年龄的增长而增长，但是生长期过了后，骨膜会闭合，停止长高；即人的身高与年龄的比值是不一定的，所以一个人的年龄和身高不成比例；B. 根据 '路程 = 速度 × 时间', 路程一定, 速度和时间的乘积一定，符合反比例的意义，所以速度和时间成反比例；C. 根据 '总价 = 单价 × 数量', 总价一定, 单价和数量的乘积一定，符合反比例的意义，所以单价和数量成反比例；",
+    "knowledge_points": [
+      "反比例的意义及辨识",
+      "正比例的意义及辨识"
+    ],
+    "source": "22-23六年级下·全国·单元测试"
+  },
+  {
+    "id": "unit4_choice_020",
+    "unit": "第四单元：比例",
+    "chapter": "图形的放大与缩小",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一个长方形, 按 3: 1 放大后, 面积扩大到原来的 ( ) 倍。",
+    "options": [
+      "A. 3",
+      "B. 1",
+      "C. 9"
+    ],
+    "answer": "C",
+    "solution": "我们可以假设原长方形的长为 2，宽为 1，根据图形放大与缩小的意义，按 3:1 放大后的长方形的长为 6，长为 3，分别求出原长方形、放大后的长方形的面积，进而求出面积扩大到原来了多少倍。本题考查了图形的放大, 图形的放大或缩小是指对应边放大或缩小。设原长方形的长为 2 , 宽为 1 ,其面积是: 2 × 1 = 2，按 3: 1 放大后长方形的长为 6,长为 3,其面积是: 6 × 3 = 18，18 ÷ 2 = 9，即面积扩大到原来的 9 倍。",
+    "knowledge_points": [
+      "图形的放大与缩小",
+      "长方形的面积"
+    ],
+    "source": "22-23六年级下·全国·单元测试"
+  },
+  {
+    "id": "unit4_choice_021",
+    "unit": "第四单元：比例",
+    "chapter": "正比例的应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "龙龙制作了一个摩天轮模型，模型高度与摩天轮的实际高度之比是 3: 400 ，现测得模型的高度为 27cm, 则摩天轮的实际高度是 ( ) 米。",
+    "options": [
+      "A. 27",
+      "B. 33",
+      "C. 36",
+      "D. 42"
+    ],
+    "answer": "C",
+    "solution": "设摩天轮的实际高度是 x 米, 根据摩天轮的实际高度与模型高度的比值是一定, 即两种量成正比例, 由此设出未知数, 列比例解答问题。本题考查了正反比例应用题, 关键是得出摩天轮实际高度与模型高度的比值是一定的。解：设摩天轮的实际高度是 x 米，由题意得：27 厘米 = 0.27 米，0.27: x = 3: 400，3x = 0.27 × 400，3x ÷ 3 = 0.27 × 400 ÷ 3，x = 108 ÷ 3，x = 36。摩天轮的实际高度是36米。",
+    "knowledge_points": [
+      "正比例的应用"
+    ],
+    "source": "20-21六年级下·陕西榆林·期中"
+  },
+  {
+    "id": "unit4_choice_022",
+    "unit": "第四单元：比例",
+    "chapter": "比例的意义",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "与62:3能组成比例的比是 ( )。",
+    "options": [
+      "A. 2: 5",
+      "B. 12: 0.2",
+      "C. 310: 15"
+    ],
+    "answer": "C",
+    "solution": "表示两个比相等的式子叫比例, 据此分别求出题干和各选项中比的比值, 找到与题干比值相等的即可。关键是理解比例的意义，找到比值相等的两个比。62: 3 = 62/3，A. 2: 5 = 2/5; B. 12: 0.2 = 120/2 = 60; C. 310: 15= 310/15 = 62/3，能组成比例的比是310：15，62: 3 = 310: 15, 与 62: 3 能组成比例的比是 310: 15。",
+    "knowledge_points": [
+      "比例的意义"
+    ],
+    "source": "20-21六年级下·湖北十堰·期中"
+  },
+  {
+    "id": "unit4_choice_023",
+    "unit": "第四单元：比例",
+    "chapter": "反比例的意义及辨识",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "下列说法错误的是 ( )。",
+    "options": [
+      "A. 故事书的单价一定，买故事书的本数与总钱数成正比例",
+      "B. 用方砖铺教室地面（面积一定），每块方砖的面积与所用方砖的块数成反比例",
+      "C. 六 (2) 班总人数一定, 男生和女生的人数成反比例"
+    ],
+    "answer": "C",
+    "solution": "判断两个相关联的量之间成什么比例，就看这两个量是对应的比值一定，还是对应的乘积一定；如果是比值一定，就成正比例；如果是乘积一定，则成反比例。据此解答。此题属于辨识成正、反比例的量, 就看这两个量是对应的比值一定, 还是对应的乘积一定, 再做判断。A．总钱数÷本数=单价（一定），符合正比例的意义，买故事书的本数与总钱数成正比例，选项说法正确；B. 每块方砖面积 × 块数 = 教室面积（一定），符合反比例的意义, 每块方砖的面积与所用方砖的块数成反比例, 选项说法正确;C. 男生人数 + 女生人数 = 总人数（一定），是和一定，男生和女生的人数不成比例关系，选项说法错误；此题属于辨识成正、反比例的量, 就看这两个量是对应的比值一定, 还是对应的乘积一定, 再做判断。",
+    "knowledge_points": [
+      "反比例的意义及辨识",
+      "正比例的意义及辨识"
+    ],
+    "source": "2022·河南漯河·小升初真题"
+  },
+  {
+    "id": "unit4_choice_024",
+    "unit": "第四单元：比例",
+    "chapter": "比例的基本性质",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "下面各组四个数不可以组成比例的是 ( )。",
+    "options": [
+      "A. 4、5、12和15",
+      "B. 1.6、6.4、2和5",
+      "C. 1/2、1/3、1/6 和 1/4"
+    ],
+    "answer": "B",
+    "solution": "判断四个数能不能组成比例，可根据比例的意义看这四个数组成的两个比的比值是否相等，也可以利用比例的基本性质，看这四个数中是否存在两个数的积等于另两个数的积，用每组数据的最大数 × 最小数，与中间两个数的乘积进行比较，相等的可以组成比例。此题的解题关键是灵活运用比例的基本性质求解。A. 4 × 15 = 60, 5 × 12 = 60, 60 = 60，能组成比例；B. 1.6 × 2 = 3.2, 6.4 × 5 = 32, 3.2 ≠ 32, 不能组成比例; C. 1/2 × 1/6 = 1/12, 1/3 × 1/4 = 1/12, 1/12 = 1/12, 能组成比例;",
+    "knowledge_points": [
+      "比例的基本性质"
+    ],
+    "source": "21-22六年级下·湖北十堰·期中"
+  },
+  {
+    "id": "unit4_choice_025",
+    "unit": "第四单元：比例",
+    "chapter": "反比例的意义及辨识",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "式子 (m - 10)/x = y, 且 x 和 y 都不为 0, 当 m 一定时, x 和 y ( )。",
+    "options": [
+      "A. 成反比例关系",
+      "B. 成正比例关系",
+      "C. 不成比例关系",
+      "D. 以上都不对"
+    ],
+    "answer": "A",
+    "solution": "两个相关联的量，若它们的乘积一定，则它们成反比例；若它们的比值一定，则它们成正比例。本题考查正反比例的判定，明确正反比例的定义是解题的关键。因为 (m - 10)/x = y, 所以 xy = m - 10, 当 m 一定时, m - 10 是一个定值。所以 x 和 y 的乘积一定, 故 x 和 y 成反比例。",
+    "knowledge_points": [
+      "反比例的意义及辨识",
+      "正比例的意义及辨识"
+    ],
+    "source": "21-22六年级下·浙江湖州·期末"
+  },
+  {
+    "id": "unit4_fill_026",
+    "unit": "第四单元：比例",
+    "chapter": "比例的意义",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "12 的因数有( )，选出 4 个组成一个比例是( )。",
+    "options": [],
+    "answer": "1、2、3、4、6、12;1: 2 = 3: 6",
+    "solution": "根据求一个数的因数的方法，求出 12 的因数，再结合比例的意义，若两组比的比值相等，则这两组比可以组成比例。据此解答即可。本题考查求一个数的因数, 结合比例的意义是解题的关键。12的因数有：1、2、3、4、6、12因为 1: 2 = 1 ÷ 2 = 1/2, 3: 6 = 3 ÷ 6 = 1/2，则可以组成的比例是 1:2 = 3:6。",
+    "knowledge_points": [
+      "比例的意义",
+      "因数和倍数的求法"
+    ],
+    "source": "2014五年级·全国·课后作业"
+  },
+  {
+    "id": "unit4_fill_027",
+    "unit": "第四单元：比例",
+    "chapter": "正比例的意义及辨识",
+    "question_type": "填空题",
+    "difficulty": "困难",
+    "question": "判断下面的两个量成不成比例, 如果成分别成什么比例, 写在括号里。打疫苗时，每小时打疫苗的人数一定，打疫苗的总人数与所用时间( )。",
+    "options": [],
+    "answer": "成正比例",
+    "solution": "两种相关联的量，如果它们的比值或商一定，则这两种量成正比例关系。据此解答。两种相关联的量，若其比值一定，两种量成正比例关系。打疫苗的总人数 ÷ 所用的时间 = 每小时打疫苗的人数（一定），因此打疫苗的总人数与所用时间成正比例。",
+    "knowledge_points": [
+      "正比例的意义及辨识"
+    ],
+    "source": "22-23六年级下·山西吕梁·期中"
+  },
+  {
+    "id": "unit4_fill_028",
+    "unit": "第四单元：比例",
+    "chapter": "正比例的意义及辨识",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "汽车车轮的直径一定，汽车行驶的路程与车轮转动的周数成( )比例。",
+    "options": [],
+    "answer": "正",
+    "solution": "判断两个相关联的量之间成什么比例，就看这两个量是对应的比值一定，还是对应的乘积一定；如果是比值一定，就成正比例；如果是乘积一定，则成反比例。据此解答。此题属于辨识成正、反比例的量，就看这两个量是对应的比值一定，还是对应的乘积一定，再做判断。根据 π × 车轮的直径 = 车轮的周长, 车轮周长 × 车轮转动的周数 = 行驶的路程, 可得: 行驶的路程/车轮转动的周数 = π × 车轮的直径, 汽车车轮的直径一定时, 那么汽车行驶的路程与车轮转动的周数的比值也是一定的, 所以汽车行驶的路程与车轮转动的周数成正比例。",
+    "knowledge_points": [
+      "反比例的意义及辨识",
+      "正比例的意义及辨识"
+    ],
+    "source": "2016·全国·小升初真题"
+  },
+  {
+    "id": "unit4_fill_029",
+    "unit": "第四单元：比例",
+    "chapter": "比例的基本性质",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "在比例 35: 10 =21:6 中，如果将第一个比的后项增加 30，第二个比的后项应该加上( )才能使比例成立。",
+    "options": [],
+    "answer": "18",
+    "solution": "先算出第一个比的后项增加 30 后两内项的积, 即 (10 + 30) × 21 = 840; 根据比例的基本性质可知, 两外项的积也是 840, 用 840 除以 35 求出第二个比的后项是 24 ;最后用 24 减去 6 求出第二个比的后项应该加几。明确比例的基本性质是解决此题的关键。(10 + 30) × 21 ÷ 35 - 6 = 40 × 21 ÷ 35 - 6 = 840 ÷ 35 - 6 = 24 - 6 = 18。所以第二个比的后项应该加上 18 才能使比例成立。",
+    "knowledge_points": [
+      "比例的基本性质"
+    ],
+    "source": "16-17六年级下·全国·单元测试"
+  },
+  {
+    "id": "unit4_fill_030",
+    "unit": "第四单元：比例",
+    "chapter": "比例的基本性质",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "在一个比例中，两个外项互为倒数，其中一个内项是2.5，另一个内项是()。",
+    "options": [],
+    "answer": "2/5",
+    "solution": "互为倒数的两个数的乘积为 1 ; 再根据比例的基本性质，内项积等于外项积，据此计算即可。本题考查比例的基本性质, 明确内项积等于外项积是解题的关键。因为两个外项互为倒数，所以两个内项也互为倒数，1 ÷ 2.5 = 2/5，则另一个内项是 2/5。",
+    "knowledge_points": [
+      "倒数的认识及应用",
+      "比例的基本性质"
+    ],
+    "source": "19-20六年级上·河北·单元测试"
+  },
+  {
+    "id": "unit4_fill_031",
+    "unit": "第四单元：比例",
+    "chapter": "反比例的意义及辨识",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "从兰州到北京, 火车所行的时间与速度成( )比例。",
+    "options": [],
+    "answer": "反",
+    "solution": "判断两个相关的量之间成什么比例，就看这两个量是对应的比值一定，还是乘积一定，如果比值一定，成正比例，如果乘积一定，成反比例，据此解答。熟练掌握正比例意义及辨识和反比例意义及辨识是解答本题的关键。因为兰州到北京的距离一定，速度 × 火车所行的时间 = 兰州到北京的路程（一定），所以火车所行的时间与速度成反比例。从兰州到北京，火车所行的时间与速度成反比例。",
+    "knowledge_points": [
+      "反比例的意义及辨识",
+      "正比例的意义及辨识"
+    ],
+    "source": "22-23六年级下·辽宁·单元测试"
+  },
+  {
+    "id": "unit4_fill_032",
+    "unit": "第四单元：比例",
+    "chapter": "比的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "黄老师买了 6kg 荔枝和 4kg 樱桃, 买两种水果所花的钱数同样多, 荔枝和樱桃的单价之比是( ), 如果荔枝的单价是 24 元, 那么樱桃的单价是( )元。",
+    "options": [],
+    "answer": "2:3;36",
+    "solution": "分别把买两种水果的钱看成单位'1'，再用解比例的方法求解即可；可以设樱桃的单价为 x，已知买两种水果所花的钱数同样多，根据'单价 × 数量 = 总价'这一公式列比例求解即可。本题考查了比例的基本性质和比例的应用。（1）分别把买两种水果的钱看成单位'1'，即荔枝和樱桃的单价之比为：1/6: 1/4 = 2:3。(2) 解: 设樱桃的单价是 x 元。2/3 = 24/x，2x = 3 × 24，2x ÷ 2 = 3 × 24 ÷ 2，x = 72 ÷ 2，x = 36。樱桃的单价是 36 元。",
+    "knowledge_points": [
+      "比的应用"
+    ],
+    "source": "22-23六年级下·全国·课前预习"
+  },
+  {
+    "id": "unit4_fill_033",
+    "unit": "第四单元：比例",
+    "chapter": "反比例的意义及辨识",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个三角形的面积是 10cm², 它的底和高成( )比例。",
+    "options": [],
+    "answer": "反",
+    "solution": "判断两个相关联的量之间成什么比例，就看这两个量是对应的比值一定，还是对应的乘积一定，如果是比值一定，就成正比例，如果是乘积一定，则成反比例。据此解答。掌握正反比例的意义是解答此题的关键。三角形的底 × 高 = 面积 × 2，因为三角形的面积为 10cm²，所以三角形的底 × 高 = 20cm²（一定），三角形的底和高乘积一定，所以底和高成反比例。",
+    "knowledge_points": [
+      "反比例的意义及辨识",
+      "正比例的意义及辨识"
+    ],
+    "source": "20-21六年级下·吉林长春·期中"
+  },{
+    "id": "unit4_fill_034",
+    "unit": "第四单元：比例",
+    "chapter": "比例的基本性质",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "在一个比例中，两个内项互为倒数，一个外项是最小的合数，另一个外项是( )。",
+    "options": [],
+    "answer": "1/4",
+    "solution": "两个内项互为倒数，即两个内项的乘积是 1。根据比例的基本性质可知：在比例里，两个外项的积等于两个内项的积。据此可知：两个外项的乘积是 1。最小的合数是 4，所以另一个外项是 1 ÷ 4。明确比例的基本性质是解决此题的关键。1 ÷ 4 = 1/4，所以在一个比例中，两个内项互为倒数，一个外项是最小的合数，另一个外项是 1/4。",
+    "knowledge_points": [
+      "倒数的认识及应用",
+      "比例的基本性质",
+      "质数与合数"
+    ],
+    "source": "20-21六年级下·湖北十堰·期中"
+  },
+  {
+    "id": "unit4_fill_035",
+    "unit": "第四单元：比例",
+    "chapter": "正比例的应用",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "成语'立竿见影'用数学的眼光来看，这是应用了比例中同时同地竿高和影长成正比例关系。身高 1.2m 的小芮在阳光照射下的影子长 2.1m，同时同地量得妈妈的影子长 2.8m，妈妈的身高是（)m。",
+    "options": [],
+    "answer": "1.6",
+    "solution": "同一时刻，不同物体的实际高度和它的影长的比值是一定的，即物体的实际高度和它的影长成正比例。设妈妈的身高是 xm，根据题意，小芮的高度：小芮的影长 = 妈妈的身高：妈妈的影长，据此列出比例并解答。本题考查正比例的应用。明确'同一时刻，物体的实际高度和它的影长成正比例'是解题的关键。解：设这棵树高 x 米，x: 2.8 = 1.2: 2.1，2.1x = 1.2 × 2.8，2.1x = 3.36，x = 3.36 ÷ 2.1，x = 1.6。即妈妈的身高是 1.6m。",
+    "knowledge_points": [
+      "正比例的意义及辨识",
+      "正比例的应用"
+    ],
+    "source": "21-22六年级下·湖北十堰·期中"
+  },
+  {
+    "id": "unit4_fill_036",
+    "unit": "第四单元：比例",
+    "chapter": "比例的基本性质",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "在一个比例中, 两个外项互为倒数, 一个内项是 10 , 另一个内项是( )。",
+    "options": [],
+    "answer": "0.1",
+    "solution": "比例的基本性质：在比例里，两个外项的积等于两个内项的积。两个外项互为倒数，它们的乘积是 1，则两个内项的乘积也是 1，用 1 除以 10 即可求出另一个内项。本题考查了倒数的意义和比例的基本性质。根据倒数的意义，明确'两个外项等于两个内项的积'是解题的关键。1 ÷ 10 = 0.1，即另一个内项是 0.1。",
+    "knowledge_points": [
+      "倒数的认识及应用",
+      "比例的基本性质"
+    ],
+    "source": "21-22六年级下·湖北十堰·期中"
+  },
+  {
+    "id": "unit4_fill_037",
+    "unit": "第四单元：比例",
+    "chapter": "图形的放大与缩小",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个长方形长10.5cm，宽8.5cm，按1:5缩小后的长方形的面积是( )cm²。",
+    "options": [],
+    "answer": "3.57",
+    "solution": "按照 1:5 缩小后，长和宽都变成原来的 1/5，先求出缩小后的长和宽，再计算面积，长方形的面积 = 长 × 宽。此题考查图形的放大与缩小的应用, 掌握长方形的面积公式也是解题的关键。10.5 × 1/5 = 2.1 (cm)，8.5 × 1/5 = 1.7 (cm)，2.1 × 1.7 = 3.57 (cm²)",
+    "knowledge_points": [
+      "图形的放大与缩小",
+      "长方形的面积",
+      "分数乘小数",
+      "小数与小数的乘法"
+    ],
+    "source": "20-21六年级下·湖北鄂州·期中"
+  },
+  {
+    "id": "unit4_fill_038",
+    "unit": "第四单元：比例",
+    "chapter": "比例的基本性质",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "如果 6x = 8y, 那么 x : y = ( ) : ( )。",
+    "options": [],
+    "answer": "4;3",
+    "solution": "利用比例的基本性质，在比例里，两个外项的积等于两个内项的积。把 6 和 x 看作比例的两个外项，把 8 和 y 看作比例的两个内项，据此即可解答。此题的解题关键是灵活运用比例的基本性质求解。根据分析得, 如果 6x = 8y, 那么 x: y = 8: 6, 化简后可得 x: y = 4: 3。",
+    "knowledge_points": [
+      "比例的基本性质"
+    ],
+    "source": "22-23六年级上·河北保定·期末"
+  },
+  {
+    "id": "unit4_fill_039",
+    "unit": "第四单元：比例",
+    "chapter": "图上距离与实际距离的换算",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "两地之间的铁路长 260km, 画在比例尺是 1:2000000 的地图上, 应画()cm。",
+    "options": [],
+    "answer": "13",
+    "solution": "比例尺和实际距离已知，依据'图上距离 = 实际距离 × 比例尺'即可求出在图上应画的长度。此题主要考查图上距离、实际距离和比例尺的关系, 解答此题关键是掌握图上距离 ÷ 实际距离 = 比例尺。260km = 26000000cm，26000000 × 1/2000000 = 13 (cm)",
+    "knowledge_points": [
+      "图上距离与实际距离的换算"
+    ],
+    "source": "20-21六年级上·重庆合川·期末"
+  },
+  {
+    "id": "unit4_fill_040",
+    "unit": "第四单元：比例",
+    "chapter": "比例的基本性质",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "在比例里两个外项都是 3 的倍数，和为 15，两个内项的积是( )。",
+    "options": [],
+    "answer": "36或54",
+    "solution": "比例的两内项积 = 两外项积, 将两外项和拆成两个3的倍数相加的形式，再求出两个加数的积即可。关键是掌握并灵活运用比例的基本性质。15 = 3 + 12 = 6 + 9，3 × 12 = 36，6 × 9 = 54。在比例里两个外项都是 3 的倍数, 和为 15 , 两个内项的积是 36 或 54。",
+    "knowledge_points": [
+      "比例的基本性质"
+    ],
+    "source": "20-21六年级下·贵州贵阳·期末"
+  },
+  {
+    "id": "unit4_fill_041",
+    "unit": "第四单元：比例",
+    "chapter": "图上距离与实际距离的换算",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "在 1:30000000 的地图上, 量得 A、B 两地的图上距离是 5.3 厘米, AB 两地的实际距离是( )千米。",
+    "options": [],
+    "answer": "1590",
+    "solution": "根据图上距离：实际距离 = 比例尺的变式，实际距离 = 图上距离 ÷ 比例尺，据此解答即可。本题考查比例尺，明确图上距离：实际距离 = 比例尺是解题的关键。5.3 ÷ 1/30000000 = 159000000（厘米）= 1590（千米）则AB两地的实际距离是1590千米。",
+    "knowledge_points": [
+      "图上距离与实际距离的换算"
+    ],
+    "source": "2021·湖南郴州·小升初真题"
+  },
+  {
+    "id": "unit4_fill_042",
+    "unit": "第四单元：比例",
+    "chapter": "比例的基本性质",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "如果 3a = 5b (a · b ≠ 0), 那么 a:b =",
+    "options": [],
+    "answer": "5:3",
+    "solution": "比例的两个内项之积等于两个外项之积，这叫比例的基本性质。根据比例的基本性质，a 在外项，将 3 放到外项；b 在内项，将 5 放到内项即可。本题关键是熟记比例的基本性质, 且能够掌握比的意义。根据分析可知,如果 3a = 5b (a · b ≠ 0), 那么 a:b = 5:3。",
+    "knowledge_points": [
+      "比例的基本性质",
+      "比的意义"
+    ],
+    "source": "17-18六年级下·全国·期末"
+  },
+  {
+    "id": "unit4_fill_043",
+    "unit": "第四单元：比例",
+    "chapter": "比例尺的意义",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "图上 30 厘米的距离表示实际距离 60 千米, 这幅地图的数值比例尺是( )。",
+    "options": [],
+    "answer": "1:200000或1/200000",
+    "solution": "根据比例尺 = 图上距离：实际距离解答即可。本题考查了比例尺的意义, 注意求比例尺时先统一单位。60千米 = 6000000 厘米，30厘米：6000000厘米 = (30 ÷ 30): (6000000 ÷ 30) = 1: 200000。所以这幅地图的数值比例尺是 1: 200000。",
+    "knowledge_points": [
+      "比例尺的意义"
+    ],
+    "source": "2022·浙江金华·小升初真题"
+  },
+  {
+    "id": "unit4_fill_044",
+    "unit": "第四单元：比例",
+    "chapter": "正比例的意义及辨识",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "若 A 的 2/5 等于 B 的 35% (A、B 均不为 0), 则 A 与 B 的最简单的整数比为( ), A 与 B 成( )比例。",
+    "options": [],
+    "answer": "7:8;正",
+    "solution": "由题意可知, A × 2/5 = B × 35%, 然后根据比例的基本性质, 内项积等于外项积, 求出 A 与 B 的比, 再进行化简即可; 两个相关联的量, 若它们的比值一定, 则它们成正比例; 若它们的乘积一定, 则它们成反比例。本题考查正反比例的判断, 明确正反比例的定义是解题的关键。因为 A × 2/5 = B × 35%, 所以 A: B = 35%: 2/5 = 7/20: 2/5 = 7: 8；因为 A: B = 7: 8 = 7/8, 它们的比值一定, 所以 A 与 B 成正比例。",
+    "knowledge_points": [
+      "反比例的意义及辨识",
+      "正比例的意义及辨识",
+      "比的化简",
+      "比例的基本性质"
+    ],
+    "source": "2022·河南周口·小升初真题"
+  },
+  {
+    "id": "unit4_fill_045",
+    "unit": "第四单元：比例",
+    "chapter": "反比例的意义及辨识",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "如果 x/5 = 8/y, 则 x 与 y 成( )比例, xy 的值是( )。",
+    "options": [],
+    "answer": "反;40",
+    "solution": "判断两个相关联的量之间成什么比例，就看这两个量是对应的比值一定，还是对应的乘积一定；如果是比值一定，就成正比例；如果是乘积一定，则成反比例。根据比例的基本性质，两个外项的乘积等于两个内项的乘积，由此可把比例 x/5 = 8/y 转换成方程，再判断 x 与 y 成什么比例及求出 xy 的值。此题的解题关键是熟练运用比例的基本性质，再去辨识成正、反比例的量，就看这两个量是对应的比值一定，还是对应的乘积一定，再做判断。由 x/5 = 8/y 可得 x × y = 8 × 5，所以 xy = 40。x 和 y 的乘积一定, 符合反比例的意义, 所以 x 和 y 成反比例。",
+    "knowledge_points": [
+      "反比例的意义及辨识",
+      "正比例的意义及辨识",
+      "比例的基本性质"
+    ],
+    "source": "21-22六年级下·湖南衡阳·期末"
+  },
+  {
+    "id": "unit4_fill_046",
+    "unit": "第四单元：比例",
+    "chapter": "正比例的意义及辨识",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "如果 y = 8x, 那么 x 和 y 成( )比例, 原因是( )。",
+    "options": [],
+    "answer": "正;比值一定",
+    "solution": "两个相关联的量，若它们的比值一定，则它们成正比例；若它们的乘积一定，则它们成反比例。本题考查正反比例的判定，明确正反比例的定义是解题的关键。因为 y = 8x, 所以 y ÷ x = 8, 它们的比值一定, 所以它们成正比例。",
+    "knowledge_points": [
+      "反比例的意义及辨识",
+      "正比例的意义及辨识"
+    ],
+    "source": "2022·浙江金华·小升初真题"
+  },
+  {
+    "id": "unit4_fill_047",
+    "unit": "第四单元：比例",
+    "chapter": "图上距离与实际距离的换算",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "在比例尺是 1:3000000的地图上，量得 A、B 两地之间的距离是 5 厘米，A、B 两地实际相距( )千米。一辆轿车和一辆客车同时从 A 地出发开往B地，轿车每小时行驶60千米，客车每小时行驶45千米，当轿车到达B地时，客车距离B地还有( )千米。",
+    "options": [],
+    "answer": "150; 37.5",
+    "solution": "先求两地的实际距离是多少千米，根据'实际距离 = 图上距离：比例尺'代入数值求出实际距离，然后根据时间 = 路程 ÷ 速度，求出轿车和客车行驶的时间，进而求出轿车行驶的路程，再求客车距离 B 地还有多少千米。此题主要考查比例尺、图上距离、实际距离三者的关系式: 比例尺 = 图上距离 : 实际距离, 灵活变形列式解决问题。5 ÷ 1/3000000 = 15000000（厘米），15000000 厘米 = 150 千米，150 - 150 ÷ 60 × 45 = 150 - 2.5 × 45 = 150 - 112.5 = 37.5 (千米)",
+    "knowledge_points": [
+      "图上距离与实际距离的换算",
+      "基础行程问题"
+    ],
+    "source": "2021·江苏苏州·小升初真题"
+  },
+  {
+    "id": "unit4_fill_048",
+    "unit": "第四单元：比例",
+    "chapter": "比例尺的意义",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "一幅图的( )和( )的比, 叫做这幅图的比例尺。",
+    "options": [],
+    "answer": "图上距离;实际距离",
+    "solution": "根据比例尺的定义可知，一幅图的图上距离和实际距离的比，叫做这幅图的比例尺。",
+    "knowledge_points": [
+      "比例尺的意义"
+    ],
+    "source": "18-19六年级上·四川·课后作业"
+  },
+  {
+    "id": "unit4_fill_049",
+    "unit": "第四单元：比例",
+    "chapter": "反比例的意义及辨识",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "AB 两城间的铁路长170千米，在一幅比例尺是 1:5000000 的地图上，这条铁路的图上距离是( )厘米。一列动车沿此铁路从 A 城开往 B 城，所用的时间与行驶的平均速度成( )比例关系。",
+    "options": [],
+    "answer": "3.4 ;反",
+    "solution": "根据图上距离 = 实际距离 × 比例尺, 据此求出这条铁路的图上距离; 两个相关联的量, 若它们的乘积一定,则它们成反比例; 若它们的比例一定, 则它们成正比例。本题考查比例尺, 明确图上距离 = 实际距离 × 比例尺是解题的关键。170千米 = 17000000 厘米，17000000 × 1/5000000 = 3.4（厘米）因为行驶的平均速度 × 所用的时间 = 铁路的长度（一定），它们的乘积一定，所以所用的时间与行驶的平均速度成反比例关系。",
+    "knowledge_points": [
+      "反比例的意义及辨识",
+      "正比例的意义及辨识",
+      "图上距离与实际距离的换算"
+    ],
+    "source": "2022·浙江杭州·小升初真题"
+  },
+  {
+    "id": "unit4_fill_050",
+    "unit": "第四单元：比例",
+    "chapter": "比例的意义",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "24 的因数有( )，从中选出 4 个组成一个比例( )。",
+    "options": [],
+    "answer": "1、2、3、4、6、8、12、24 ; 1: 2 = 4: 8（答案不唯一)",
+    "solution": "列乘法算式找因数，按照从小到大的顺序，一组一组地写出所有积是这个数的乘法算式，乘法算式中的两个乘数就是这个数的因数。表示两个比相等的式子，叫做比例。关键是会找一个数的因数，理解比例的意义。24 的因数有：1、2、3、4、6、8、12、24，从中选出 4 个组成一个比例 1: 2 = 4: 8。",
+    "knowledge_points": [
+      "比例的意义",
+      "因数和倍数的求法"
+    ],
+    "source": "2022·新疆克拉玛依·小升初真题"
+  },
+  {
+    "id": "unit4_fill_051",
+    "unit": "第四单元：比例",
+    "chapter": "比例尺的意义",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一种微型零件长 3 毫米,画在图纸上的长是 6 厘米, 这幅图纸的比例尺是( )。",
+    "options": [],
+    "answer": "20:1",
+    "solution": "根据比例尺 = 图上距离 : 实际距离, 代入数据计算即可, 注意单位的换算: 1 厘米 = 10 毫米。掌握比例尺的意义是解题的关键, 解答时注意单位的换算。6厘米：3毫米 = (6 × 10) = 60毫米：3毫米 = 60:3 = (60 ÷ 3) : (3 ÷ 3) = 20 : 1",
+    "knowledge_points": [
+      "比例尺的意义"
+    ],
+    "source": "2022·广东韶关·小升初真题"
+  },
+  {
+    "id": "unit4_fill_052",
+    "unit": "第四单元：比例",
+    "chapter": "反比例的意义及辨识",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "路程一定，车轮周长和转数成( )比例。",
+    "options": [],
+    "answer": "反",
+    "solution": "判断两种相关联的量成不成比例，成什么比例，就看这两种量是对应的比值一定，还是对应的乘积一定，如果是比值一定，就成正比例，如果是乘积一定，就成反比例，如果是其它的量一定或乘积、比值不一定，就不成比例。此题属于辨识正反比例关系, 就看这两个量是对应的比值一定, 还是乘积一定, 再做判断。车轮周长 × 转数 = 路程（一定），则路程一定时，车轮周长和转数成反比例。",
+    "knowledge_points": [
+      "反比例的意义及辨识",
+      "正比例的意义及辨识"
+    ],
+    "source": "2022·广西百色·小升初真题"
+  },
+  {
+    "id": "unit4_judge_053",
+    "unit": "第四单元：比例",
+    "chapter": "反比例的意义及辨识",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "圆锥的体积一定, 它的底面积和高成反比例。( )",
+    "options": [],
+    "answer": "√",
+    "solution": "判断两种相关联的量之间成什么比例，就看这两个量是对应的比值（商）一定，还是对应的乘积一定；如果是比值（商）一定，这两种相关联的量成正比例；如果是乘积一定，这两种相关联的量成反比例。掌握正、反比例的意义及辨识方法是解题的关键。根据圆锥的体积公式 V = 1/3Sh 可知, 圆锥的体积一定, 即乘积一定, 那么它的底面积和高成反比例。原题说法正确。",
+    "knowledge_points": [
+      "反比例的意义及辨识",
+      "正比例的意义及辨识",
+      "圆锥的体积（容积）"
+    ],
+    "source": "2018·山东临沂·小升初真题"
+  },
+  {
+    "id": "unit4_judge_054",
+    "unit": "第四单元：比例",
+    "chapter": "反比例的意义及辨识",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "汽车的速度和时间成反比例关系。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "两种相关联的量，一种量变化，另一种量也随着变化，如果这两种量中相对应的两个数的乘积一定，这两种量就叫作成反比例的量，它们的关系叫作反比例关系，如果用字母 x 和 y 表示两种相关联的量, 用 k 表示它们的乘积（一定），反比例关系可以用 xy = k 表示，据此解答。本题主要考查反比例关系的辨识，掌握反比例关系的意义是解答题目的关键。路程不一定时，汽车的速度和时间不成比例关系；路程一定时，速度 × 时间 = 路程（一定），此时汽车的速度和时间成反比例关系。",
+    "knowledge_points": [
+      "反比例的意义及辨识"
+    ],
+    "source": "17-18六年级下·全国·单元测试"
+  },
+  {
+    "id": "unit4_judge_055",
+    "unit": "第四单元：比例",
+    "chapter": "比例的基本性质",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "在比例里, 两个外项的积除以两个内项的积，商是 1。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "比例的基本性质：两外项之积等于两内项之积；两个相同的数（0 除外）相除，商是 1。据此解答。本题主要考查了比例的性质, 熟练掌握比例的基本性质是解题的关键。根据分析可知，两个外项的积除以两个内项的积，商是 1。例如：1:2 = 2:4，(1 × 4) ÷ (2 × 2) = 4 ÷ 4 = 1。所以原题干说法正确。",
+    "knowledge_points": [
+      "比例的基本性质"
+    ],
+    "source": "2021·河北保定·小升初真题"
+  },
+  {
+    "id": "unit4_judge_056",
+    "unit": "第四单元：比例",
+    "chapter": "比例的基本性质",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "在比例里，两个内项互为倒数，那么两个外项一定互为倒数。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "比例的基本性质：两外项之积等于两内项之积，倒数的定义：乘积是 1 的两个数互为倒数，例如：如果 a、b 不为 0，a × b = 1，则 a 是 b 的倒数，b 是 a 的倒数。据此解答。本题考查了比例的基本性质和倒数的认识。根据分析可知，在比例里，两个内项互为倒数，那么两个外项一定互为倒数。例如: 4: 5 = 0.2: 0.25。原题干说法正确。",
+    "knowledge_points": [
+      "倒数的认识及应用",
+      "比例的基本性质"
+    ],
+    "source": "2015·天津静海·小升初真题"
+  },
+  {
+    "id": "unit4_judge_057",
+    "unit": "第四单元：比例",
+    "chapter": "比例的基本性质",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "如果把比例写成分数的形式, 等号两端的分子和分母分别交叉相乘, 它们的积相等。( )",
+    "options": [],
+    "answer": "√",
+    "solution": "根据比例基本性质的内容直接解答。此题考查学生对比例基本性质的内容理解和记忆。在比例里，两个外项的积等于两个内项的积，这叫做比例的基本性质；如果把比例写成分数形式，等号两端的分子和分母分别交叉相乘，它们的积相等。",
+    "knowledge_points": [
+      "比例的基本性质"
+    ],
+    "source": "21-22六年级下·江苏·单元测试"
+  },
+  {
+    "id": "unit4_judge_058",
+    "unit": "第四单元：比例",
+    "chapter": "反比例的应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "总工作量一定, 已经完成的工作量和没有完成的工作量成反比例。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "此题中的三个量关系式为：已经完成的工作量 + 剩下的工作量 = 一项工程的总量，但是已经完成的工作量与剩下的工作量的比值和乘积都不一定，所以它们不成比例，由此即可判断。此题主要利用正比例与反比例的意义解决问题,关键看两种量的关系式, 是否符合 y/x = k (一定) 成正比例,符合 xy = k (一定) 成反比例。因为已经完成的工作量与剩下的工作量的比值和乘积都不一定，所以已经完成的工作量与剩下的工作量不成比例。",
+    "knowledge_points": [
+      "反比例的应用"
+    ],
+    "source": "21-22六年级下·浙江·单元测试"
+  },
+  {
+    "id": "unit4_judge_059",
+    "unit": "第四单元：比例",
+    "chapter": "比例的意义",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "15:5 和 1.2:0.4 能组成比例。( )",
+    "options": [],
+    "answer": "√",
+    "solution": "分别算出 15:5 和 1.2:0.4 的比值，进行比较判断。本题考查成比例的意义, 可以通过计算内项积与外项积进行比较, 也可以分别算出比值进行比较判断。15:5 = 15 ÷ 5 = 3，1.2: 0.4 = 1.2 ÷ 0.4 = 3。15:5 和 1.2: 0.4 能组成比例, 原题说法正确。",
+    "knowledge_points": [
+      "比例的意义",
+      "比与比例的区别与联系",
+      "求比值"
+    ],
+    "source": "22-23六年级下·全国·单元测试"
+  },
+  {
+    "id": "unit4_judge_060",
+    "unit": "第四单元：比例",
+    "chapter": "比例的基本性质",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "在一个比例里, 两个内项的积除以两个外项的积, 商是 1。( )",
+    "options": [],
+    "answer": "√",
+    "solution": "比例的基本性质：在比例里，两个外项的积等于两个内项的积；据此解答。本题考查比例的基本性质的应用。如：6:4 = 3:2两个内项的积是 4 × 3 = 12, 两个外项的积是 6 × 2 = 12; 12 ÷ 12 = 1。所以，在一个比例里，两个内项的积除以两个外项的积，商是 1。原题说法正确。",
+    "knowledge_points": [
+      "比例的基本性质"
+    ],
+    "source": "2018·山东临沂·小升初真题"
+  },
+  {
+    "id": "unit4_judge_061",
+    "unit": "第四单元：比例",
+    "chapter": "反比例的意义及辨识",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "李老师的钱数一定,购买《快乐成长》的本数和单价成反比例。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "两种相关联的量，一种量变化，另一种量也随着变化，如果这两种量相对应的两个数的比值（也就是商）一定，这两种量就叫做成正比例的量，它们的关系叫做正比例关系；两种相关联的量，一种量变化，另一种量也随着变化，如果这两种量相对应的两个数的积一定，这两种量就叫做成反比例的量，它们的关系叫做反比例关系。此题考查了正、反比例的判断, 关键能够理解概念。根据：总价 = 单价 × 数量，已知李老师的钱数一定，也就是购买《快乐成长》的本数和单价乘积一定，所以本数和单价成反比例，原题说法正确；",
+    "knowledge_points": [
+      "反比例的意义及辨识",
+      "正比例的意义及辨识"
+    ],
+    "source": "20-21六年级下·湖北十堰·期中"
+  },
+  {
+    "id": "unit4_judge_062",
+    "unit": "第四单元：比例",
+    "chapter": "图上距离与实际距离的换算",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "在比例尺是1:1000000 的地图上，图上距离 2cm，表示实际距离20km。()",
+    "options": [],
+    "answer": "√",
+    "solution": "比例尺表示图上距离与实际距离的比, 比例尺 = 图上距离: 实际距离, 则实际距离 = 图上距离 ÷ 比例尺, 把题中数据代入公式计算, 据此解答。掌握图上距离和实际距离换算的方法是解答题目的关键。2 ÷ 1/1000000 = 2 × 1000000 = 2000000 (cm)，2000000cm = 20km。所以，在比例尺是 1: 1000000 的地图上，图上距离 2cm，表示实际距离 20km。",
+    "knowledge_points": [
+      "图上距离与实际距离的换算"
+    ],
+    "source": "20-21六年级下·湖北鄂州·期中"
+  },
+  {
+    "id": "unit4_judge_063",
+    "unit": "第四单元：比例",
+    "chapter": "比例尺的意义",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "图上距离一定比相对应的实际距离短。（ ）",
+    "options": [],
+    "answer": "×",
+    "solution": "地图相当大，要画在纸上就要将其缩小，此时，图上距离一定比相对应的实际距离要短；有的零件比较小，画在纸上时要将其适当放大，此时，图上距离一定比相对应的实际距离要长。比例尺分为缩小比例尺和放大比例尺，在看地图时，应用的是缩小比例尺；在研究机器较小的零件时，应用的是放大比例尺。根据分析得, 图上距离有时比相对应的实际距离要长, 有时比相对应的实际距离要短。原题说法错误。",
+    "knowledge_points": [
+      "比例尺的意义"
+    ],
+    "source": "2014·四川成都·小升初真题"
+  },
+  {
+    "id": "unit4_judge_064",
+    "unit": "第四单元：比例",
+    "chapter": "图形的放大与缩小",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "图形的放大与缩小就是把原图形各边加上或减去一个相同的数。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "把图形按照 n:1 放大, 就是将图形的每一条边放大到原来的 n 倍, 放大后图形与原图形对应边长的比是n:1; 把图形按照 1:n 缩小, 就是将图形的每一条边缩小到原来的 1/n, 缩小后图形与原图形对应边长的比是 1:n。掌握图形的放大与缩小的概念是解题的关键。图形放大或缩小的倍数是指对应边放大或缩小的倍数。",
+    "knowledge_points": [
+      "图形的放大与缩小"
+    ],
+    "source": "21-22六年级上·河北保定·期末"
+  },
+  {
+    "id": "unit4_judge_065",
+    "unit": "第四单元：比例",
+    "chapter": "比例尺的意义",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "一幅图的比例尺是 1: 2000000, 也就是图上 1cm 表示实际距离 20km。( )",
+    "options": [],
+    "answer": "√",
+    "solution": "根据比例尺的意义可知，比例尺是 1: 2000000 表示图上 1cm 相当于实际 2000000cm，根据进率 ' 1km = 1000000cm' 换算单位即可。掌握比例尺的意义以及长度单位的换算是解题的关键。2000000cm = 20km，一幅图的比例尺是 1: 2000000，也就是图上 1cm 表示实际距离 20km",
+    "knowledge_points": [
+      "比例尺的意义"
+    ],
+    "source": "21-22六年级下·黑龙江大兴安岭地·期末"
+  },
+  {
+    "id": "unit4_judge_066",
+    "unit": "第四单元：比例",
+    "chapter": "正比例的意义及辨识",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "六 (3) 班有 45 人,其中 13 人喜欢围棋, 喜欢围棋的人数和不喜欢围棋的人数成比例。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "判断两种相关联的量之间成什么比例，就看这两个量是对应的比值（商）一定，还是对应的乘积一定；如果是比值（商）一定，这两种相关联的量成正比例；如果是乘积一定，这两种相关联的量成反比例；如果既不是比值一定，也不是乘积一定，则这两种相关联的量不成比例。掌握正、反比例的意义及辨识方法是解题的关键。喜欢围棋的人数 + 不喜欢围棋的人数 = 45 人（一定），和一定，所以喜欢围棋的人数和不喜欢围棋的人数不成比例。原题说法错误。",
+    "knowledge_points": [
+      "反比例的意义及辨识",
+      "正比例的意义及辨识"
+    ],
+    "source": "21-22六年级下·四川广安·期末"
+  },
+  {
+    "id": "unit4_judge_067",
+    "unit": "第四单元：比例",
+    "chapter": "反比例的应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "圆柱的体积一定, 圆柱的高与底面积成正比例关系。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "判断两种相关联的量之间成什么比例，就看这两个量是对应的比值（商）一定，还是对应的乘积一定；如果是比值（商）一定，这两种相关联的量成正比例；如果是乘积一定，这两种相关联的量成反比例。掌握正、反比例的意义及辨识方法是解题的关键。底面积 × 高 = 圆柱的体积（一定）乘积一定，那么圆柱的高与底面积成反比例关系。原题说法错误。",
+    "knowledge_points": [
+      "反比例的应用",
+      "正比例的应用"
+    ],
+    "source": "2021·四川绵阳·小升初真题"
+  },
+  {
+    "id": "unit4_judge_068",
+    "unit": "第四单元：比例",
+    "chapter": "比例的基本性质",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "如果甲数的 1/3 等于乙数的 5/8, 那么甲数与乙数的比为 8: 15。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "根据条件'甲数的 1/3 等于乙数的 5/8'可得：甲数 × 1/3 = 乙数 × 5/8。依据比例的基本性质，在比例里，两外项之积等于两内项之积，如果甲是一个外项，则 1/3 是另一个外项，如果乙是一个内项，则 5/8 是另一个内项，据此写出比，然后化简比即可。考查了比例的基本性质两外项之积等于两内项之积，同时考查了比的化简。如果甲数 × 1/3 = 乙数 × 5/8;则甲数：乙数 = 5/8: 1/3 = 5/8: 1/3 = (5/8 × 24): (1/3 × 24) = 15: 8。所以甲数与乙数的比为 15: 8, 原题说法错误;",
+    "knowledge_points": [
+      "比的化简",
+      "比例的基本性质",
+      "比的意义"
+    ],
+    "source": "21-22六年级下·山西忻州·期末"
+  },
+  {
+    "id": "unit4_judge_069",
+    "unit": "第四单元：比例",
+    "chapter": "正比例的意义及辨识",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "'一只青蛙一张嘴，两只眼睛四条腿；两只青蛙两张嘴，四只眼睛八条腿……'儿歌中青蛙的只数和对应的腿数成正比例。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "根据 x ÷ y = k（一定），x 和 y 成正比例关系，进行分析。关键是理解正比例的意义，商一定是正比例关系。对应腿数÷青蛙只数 = 4（一定），儿歌中青蛙的只数和对应的腿数成正比例，说法正确。",
+    "knowledge_points": [
+      "正比例的意义及辨识"
+    ],
+    "source": "2014·全国·小升初真题"
+  },
+  {
+    "id": "unit4_judge_070",
+    "unit": "第四单元：比例",
+    "chapter": "反比例的意义及辨识",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "小明看一本书, 看过的页数与剩下的页数成反比例。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "判断两种相关联的量是否成反比例，就看这两种量是否是对应的乘积一定，如果是乘积一定，就成反比例，如果不是乘积一定或乘积不一定，就不成反比例。据此进行判断。此题属于根据正、反比例的意义，辨识两种相关联的量是否成反比例，就看这两种量是否是对应的乘积一定，再做出判断。看过的页数 + 剩下的页数 = 这本书的总页数（一定），是对应的'和'一定，不是'乘积'一定，所以看过的页数与剩下的页数不成比例，原题说法错误；",
+    "knowledge_points": [
+      "反比例的意义及辨识"
+    ],
+    "source": "2014·全国·小升初真题"
+  },
+  {
+    "id": "unit4_judge_071",
+    "unit": "第四单元：比例",
+    "chapter": "正比例的意义及辨识",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "若圆柱的底面积一定，则圆柱的体积和高成正比例。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "两个相关联的量，若它们的比值一定，则它们成正比例；若它们的乘积一定，则它们成反比例。解答此题的主要依据是：若两个量的商一定，则这两个量成正比例。因为圆柱的体积=底面积×高, 则 圆柱的体积/高 = 底面积（一定），它们的比值一定。所以说圆柱的体积和其高成正比例关系。",
+    "knowledge_points": [
+      "反比例的意义及辨识",
+      "正比例的意义及辨识"
+    ],
+    "source": "2022·河南鹤壁·小升初真题"
+  },
+  {
+    "id": "unit4_judge_072",
+    "unit": "第四单元：比例",
+    "chapter": "反比例的意义及辨识",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "圆的周长一定，直径和圆周率成反比例。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "判断两种相关联的量之间成什么比例，就看这两个量是对应的比值（商）一定，还是对应的乘积一定；如果是比值（商）一定，这两种相关联的量成正比例；如果是乘积一定，这两种相关联的量成反比例；如果既不是比值一定，也不是乘积一定，则这两种相关联的量不成比例。掌握正、反比例的意义及辨识方法是解题的关键。C = πd，因为圆周率是一个定值，不会发生变化，所以直径和圆周率不成比例。",
+    "knowledge_points": [
+      "反比例的意义及辨识",
+      "正比例的意义及辨识"
+    ],
+    "source": "2014·全国·小升初真题"
+  },
+  {
+    "id": "unit4_calc_073",
+    "unit": "第四单元：比例",
+    "chapter": "解比例",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "解比例。3/4: 2/5 = x: 4/15；0.75: x = 25/8；x/2 = 2/5: 7/8",
+    "options": [],
+    "answer": "x = 1/2; x = 6/25; x = 32/35",
+    "solution": "（1）先把比例方程改写成 2/5x = 3/4 × 4/15，然后方程两边同时除以 2/5，求出方程的解；(2) 先把比例方程改写成 25x = 0.75 × 8，然后方程两边同时除以 25，求出方程的解；(3) 先把比例方程改写成 7/8x = 2 × 2/5, 然后方程两边同时除以 7/8, 求出方程的解。(1) 3/4: 2/5 = x: 4/15，解: 2/5x = 3/4 × 4/15，2/5x = 1/5，2/5x ÷ 2/5 = 1/5 ÷ 2/5，x = 1/5 × 5/2，x = 1/2。(2) 0.75: x = 25: 8，解: 25x = 0.75 × 8，25x = 6，25x ÷ 25 = 6 ÷ 25，x = 6/25。(3) x/2 = 2/5: 7/8，解: x: 2 = 2/5: 7/8，7/8x = 2 × 2/5，7/8x = 4/5，7/8x ÷ 7/8 = 4/5 ÷ 7/8，x = 4/5 × 8/7，x = 32/35",
+    "knowledge_points": [
+      "解比例"
+    ],
+    "source": "22-23六年级下·全国·单元测试"
+  },
+  {
+    "id": "unit4_calc_074",
+    "unit": "第四单元：比例",
+    "chapter": "解比例",
+    "question_type": "计算题",
+    "difficulty": "容易",
+    "question": "解比例。x/20 = 0.4/16；2/5: 3/10 = x: 5/6",
+    "options": [],
+    "answer": "x = 0.5; x = 10/9",
+    "solution": "（1）根据比例的基本性质，两内项之积等于两外项之积，把比例转化成方程后，再根据等式的性质2，方程左右两边同时除以16，解出方程。(2) 根据比例的基本性质, 两内项之积等于两外项之积,把比例转化成方程后, 再根据等式的性质 2, 方程左右两边同时除以 3/10, 解出方程。x/20 = 0.4/16，解: 16x = 20 × 0.4，16x = 8，x = 8 ÷ 16，x = 0.5。2/5: 3/10 = x: 5/6，解: 3/10x = 2/5 × 5/6，3/10x = 1/3，x = 1/3 ÷ 3/10，x = 1/3 × 10/3，x = 10/9",
+    "knowledge_points": [
+      "比例的基本性质",
+      "解比例",
+      "应用等式的性质2解方程"
+    ],
+    "source": "22-23六年级下·山东·单元测试"
+  },
+  {
+    "id": "unit4_calc_075",
+    "unit": "第四单元：比例",
+    "chapter": "解比例",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "求 x 的得数。7.5/12 = 25/x；5/8: 3/7 = x: 1/14；1.5: 2.5 = 12: x",
+    "options": [],
+    "answer": "x = 40; x = 5/48; x = 20",
+    "solution": "7.5/12 = 25/x, 根据比例的基本性质, 先写成 7.5x = 12 × 25 的形式, 两边同时 ÷7.5 即可;5/8: 3/7 = x: 1/14, 根据比例的基本性质, 先写成 3/7x = 5/8 × 1/14 的形式,两边同时 ×7/3 即可;1.5: 2.5 = 12: x, 根据比例的基本性质, 先写成 1.5x = 2.5 × 12 的形式, 两边同时 ÷1.5 即可。7.5/12 = 25/x，解: 7.5x = 12 × 25，7.5x ÷ 7.5 = 300 ÷ 7.5，x = 40。5/8: 3/7 = x: 1/14，解: 3/7x = 5/8 × 1/14，3/7x × 7/3 = 5/112 × 7/3，x = 5/48。1.5: 2.5 = 12: x，解: 1.5x = 2.5 × 12，1.5x ÷ 1.5 = 30 ÷ 1.5，x = 20",
+    "knowledge_points": [
+      "解比例"
+    ],
+    "source": "20-21六年级下·湖北十堰·期中"
+  },
+  {
+    "id": "unit4_calc_076",
+    "unit": "第四单元：比例",
+    "chapter": "解比例",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "解方程。(1) 96 - x = 64；(2) 1/5 = 5x/8；(3) 20%x + 1/2x = 280；(4) 1.6x: 6/13 = 26/9: 0.5",
+    "options": [],
+    "answer": "(1) x = 32;(2) x = 0.32;(3) x = 400; (4) x = 5/3",
+    "solution": "（1）根据等式的性质，方程两边同时加上 x，再同时减去 64 即可；(2) 根据比例的基本性质, 把方程转化为 5x × 5 = 8, 再化简方程, 最后根据等式的性质, 方程两边同时除以 25 即可;(3) 先化简方程, 再根据等式的性质, 方程两边同时除以 0.7 即可;(4) 根据比例的基本性质, 把方程转化为 0.5 × 1.6x = 6/13 × 26/9, 再化简方程, 最后根据等式的性质, 方程两边同时除以 0.8 即可。(1) 96 - x = 64，解：96 - x + x = 64 + x，64 + x = 96，64 + x - 64 = 96 - 64，x = 32。(2) 1/5 = 5x/8，解: 5x × 5 = 8，25x = 8，25x ÷ 25 = 8 ÷ 25，x = 0.32。(3) 20%x + 1/2x = 280，解: 0.7x = 280，0.7x ÷ 0.7 = 280 ÷ 0.7，x = 400。(4) 1.6x: 6/13 = 26/9: 0.5，解: 0.5 × 1.6x = 6/13 × 26/9，0.8x = 4/3，0.8x ÷ 0.8 = 4/3 ÷ 0.8，x = 5/3",
+    "knowledge_points": [
+      "含百分数的运算",
+      "分数与整数的除法",
+      "解比例",
+      "除数是小数的小数除法"
+    ],
+    "source": "21-22六年级下·广西桂林·期末"
+  },
+  {
+    "id": "unit4_solve_077",
+    "unit": "第四单元：比例",
+    "chapter": "应用比例尺画图",
+    "question_type": "作图题",
+    "difficulty": "中等",
+    "question": "春光小学的操场是一个长方形，长 120 米，宽 80 米，用 1:2000 的比例尺画出春光小学操场的平面图。",
+    "options": [],
+    "answer": "见详解",
+    "solution": "要求图上的长和宽各是多少，根据公式'图上距离 = 实际距离 × 比例尺'，代入数据，进行解答，即可得出结论。考查了应用比例尺画图, 能够根据比例尺正确进行计算, 注意单位的统一。120米=12000厘米，80米=8000厘米，12000 × 1/2000 = 6（厘米），8000 × 1/2000 = 4（厘米）",
+    "knowledge_points": [
+      "应用比例尺画图"
+    ],
+    "source": "20-21六年级下·福建福州·期中"
+  },
+  {
+    "id": "unit4_solve_078",
+    "unit": "第四单元：比例",
+    "chapter": "正比例的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一本课外书有 275 页, 亮亮前 3 天看了 75 页, 照这样的速度, 他看完这本课外书需要多少天? (用比例解答)",
+    "options": [],
+    "answer": "11天",
+    "solution": "根据'照这样的速度'可知，看的页数：看的天数 = 每天看的页数（一定），比值一定，则看的页数和看的天数成正比例关系，据此列出正比例方程，并求解。关键是判断题目中的两种相关联的量成什么比例关系，据此列出相应的比例方程。解：设他看完这本课外书需要 x 天。275: x = 75: 3，75x = 275 × 3，75x = 825，x = 825 ÷ 75，x = 11。答: 他看完这本课外书需要 11 天。",
+    "knowledge_points": [
+      "正比例的意义及辨识",
+      "解比例",
+      "正比例的应用"
+    ],
+    "source": "22-23六年级下·四川宜宾·期中"
+  },
+  {
+    "id": "unit4_solve_079",
+    "unit": "第四单元：比例",
+    "chapter": "反比例的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "给一间房间铺地砖,如果用边长 7 分米的方砖, 需要 100 块。如果改用面积是 25 平方分米的方砖来铺, 需要多少块? (用比例解)",
+    "options": [],
+    "answer": "196块",
+    "solution": "根据题意可知, 一块方砖的面积 × 块数 = 房间的面积 (一定), 积一定, 则一块方砖的面积与块数成反比例关系, 据此列出反比例方程, 并求解。关键是判断题目中的两种相关联的量成什么比例关系，据此列出相应的比例方程。解：设需要 x 块。25x = 7 × 7 × 100，25x = 4900，x = 4900 ÷ 25，x = 196。答：需要 196 块。",
+    "knowledge_points": [
+      "反比例的意义及辨识",
+      "反比例的应用"
+    ],
+    "source": "22-23六年级下·湖南衡阳·期中"
+  },
+  {
+    "id": "unit4_solve_080",
+    "unit": "第四单元：比例",
+    "chapter": "比例的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "王阿姨和李阿姨制作一批仿真花，王阿姨做一束用5分钟，李阿姨做一束用9分钟。如果两人同时开始做，李阿姨制作110束仿真花时，王阿姨制作了多少束？（用比例知识解答）",
+    "options": [],
+    "answer": "198束",
+    "solution": "由题意可知, 设王阿姨制作了 x 束, 根据二人用的时间相等列比例解答。本题考查了利用反比例知识解决问题, 需准确理解题意。解：王阿姨制作了 x 束。x × 5 = 110 × 9，5x = 990，5x ÷ 5 = 990 ÷ 5，x = 198。答: 王阿姨制作了 198 束。",
+    "knowledge_points": [
+      "比例的应用"
+    ],
+    "source": "22-23六年级下·河北保定·期中"
+  },
+  {
+    "id": "unit4_solve_081",
+    "unit": "第四单元：比例",
+    "chapter": "正比例的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一辆小汽车 3 时行驶了 195km, 照这样的速度行驶, 这辆车从 A 地开往 B 地共行驶了9时，两地相距多少千米？（用比例解）",
+    "options": [],
+    "answer": "585千米",
+    "solution": "速度一定，这辆汽车行驶的路程和所用的时间成正比例，据此列比例解答。本题解题的关键是能够判断出题中的两种相关联的量成正比例。解：设两地相距 x 千米。195: 3 = x: 9，3x = 1755，x = 585。答: 两地相距 585 千米。",
+    "knowledge_points": [
+      "解比例",
+      "正比例的应用"
+    ],
+    "source": "22-23六年级下·河南周口·期中"
+  },
+  {
+    "id": "unit4_solve_082",
+    "unit": "第四单元：比例",
+    "chapter": "按比分配问题",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "甲、乙两队人数的比是 5: 4, 在一次调整中, 如果从甲队调 22 人到乙队, 则乙、甲两队人数的比是3:1，甲、乙两个原有多少人？",
+    "options": [],
+    "answer": "甲、乙两队原来人数分别是40人，32人",
+    "solution": "由题意可知, 把甲、乙两队的总人数看作单位 '1', 则 22 人所对应的分率为 (3/(3 + 1) - 4/(5 + 4)), 根据除法的意义, 用除法即可求出甲、乙两队的总人数, 再根据按比例分配问题分别求出、乙两个原有多少人。本题考查按比分配问题，求出甲、乙两队的总人数是解题的关键。22 ÷ (3/(3 + 1) - 4/(5 + 4)) × 5/(5 + 4) = 22 ÷ 11/36 × 5/9 = 72 × 5/9 = 40 (人)，40 × 4/5 = 32 (人)。答: 甲、乙两队原来人数分别是 40 人, 32 人。",
+    "knowledge_points": [
+      "比的应用",
+      "按比分配问题"
+    ],
+    "source": "22-23六年级上·四川·单元测试"
+  },
+  {
+    "id": "unit4_solve_083",
+    "unit": "第四单元：比例",
+    "chapter": "反比例的应用",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "世界上最粗的树是 '百骑大栗树'，它生长在地中海西西里岛的埃特纳火山的山坡上。据悉，它的树干大约需要40个身高 1.35m 的小学生伸开双臂才能围住，换成身高 1.8m 的成年人，大约需要多少个成年人伸开双臂才能围住？（人双臂展开的长度约等于人的身高）（用比例知识解答）",
+    "options": [],
+    "answer": "30个",
+    "solution": "根据'百骑大栗树'的树干总米数是不变的，可得伸开双臂人的身高与人的个数成反比例，即人双臂展开的长度乘人数的乘积是相等的, 设大约需要 x 个成年人伸开双臂才能围住，列式即可解答。本题主要考查了反比例应用题, 关键是得出伸开双臂人的身高与人的个数成反比例。解：设大约需要 x 个成年人伸开双臂才能围住。1.8x = 40 × 6.35，1.8x = 54，x = 30。答: 大约需要 30 个成年人伸开双臂才能围住。",
+    "knowledge_points": [
+      "反比例的应用"
+    ],
+    "source": "2022·湖北十堰·小升初真题"
+  },
+  {
+    "id": "unit4_solve_084",
+    "unit": "第四单元：比例",
+    "chapter": "正比例的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "100 千克黄豆可以榨豆油 13 千克, 照这样计算, 要榨豆油 6.5 吨, 需黄豆多少吨? (用比例解决问题)",
+    "options": [],
+    "answer": "50吨",
+    "solution": "根据题意, 知道每榨 1 千克的油所需的黄豆一定,即黄豆的重量和油的重量成正比例, 由此列式解答即可。解答此题的关键是，根据题意，正确判断出两种相关联的量成什么比例，找出对应量，注意单位统一，列式解答即可。解：设需黄豆 x 吨。100/13 = x/6.5，13x = 100 × 6.5，13x = 650，x = 650 ÷ 13，x = 50。答: 需黄豆 50 吨。",
+    "knowledge_points": [
+      "正比例的应用"
+    ],
+    "source": "21-22六年级下·江苏·单元测试"
+  },
+  {
+    "id": "unit4_solve_085",
+    "unit": "第四单元：比例",
+    "chapter": "图上距离与实际距离的换算",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "在比例尺是1:500000 的地图上，量得甲、乙两城之间的距离是 3.2 厘米。一辆汽车以 80 千米/时的速度在上午 9 时从甲城开出，到达乙城的时间是几时？",
+    "options": [],
+    "answer": "9时12分",
+    "solution": "根据实际距离 = 图上距离 ÷ 比例尺, 求得甲、乙两城间的实际距离, 根据时间 = 路程 ÷ 速度, 求出汽车从甲城到乙城的时间, 然后用起始时间加上经过时间即可得到达乙城的时间是几时, 据此解答。本题主要考查图上距离和实际距离的换算以及行程问题。3.2 ÷ 1/500000 = 3.2 × 500000 = 1600000 (厘米)，1600000 厘米 = 16 千米，16 ÷ 80 = 0.2 (时)，0.2 × 60 = 12 (分)，9时+12分=9时12分。答: 到达乙城的时间是 9 时 12 分。",
+    "knowledge_points": [
+      "图上距离与实际距离的换算",
+      "基础行程问题",
+      "时、分、秒时间的推算"
+    ],
+    "source": "22-23六年级下·全国·单元测试"
+  },
+  {
+    "id": "unit4_solve_086",
+    "unit": "第四单元：比例",
+    "chapter": "图上距离与实际距离的换算",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "朝阳小学的操场是一个长方形，长120米，宽75米，用1:3000的比例尺画成平面图，长和宽各是多少厘米？",
+    "options": [],
+    "answer": "长：4厘米；宽：2.5厘米",
+    "solution": "要求图上的长和宽各是多少，根据公式'图上距离 = 实际距离 × 比例尺'，代入数据，进行解答，即可得出结论。此题做题的关键是：根据图上距离、实际距离和比例尺三者的关系进行解答。120米=12000厘米，75米=7500厘米，12000 × 1/3000 = 4（厘米），7500 × 1/3000 = 2.5（厘米）答：长和宽分别是4厘米，2.5厘米。",
+    "knowledge_points": [
+      "图上距离与实际距离的换算"
+    ],
+    "source": "22-23六年级下·江苏·单元测试"
+  },
+  {
+    "id": "unit4_solve_087",
+    "unit": "第四单元：比例",
+    "chapter": "正比例的应用",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "一种汽车采用了节油技术，2个月节省汽油46千克，照这样计算，一年能节省汽油多少千克？（用比例解）",
+    "options": [],
+    "answer": "276千克",
+    "solution": "由题意可知：每个月节省汽油的质量是一定的，即汽油的质量与时间的比值是一定的，则汽油的质量与时间成正比例，据此即可列比例求解。解答此题的关键是弄清楚哪两种量成何比例, 进而列比例求解。解：设一年能节省汽油 x 千克，46: 2 = x: 12，2x = 46 × 12，2x = 552，x = 552 ÷ 2，x = 276。答: 一年能节省汽油 276 千克。",
+    "knowledge_points": [
+      "正比例的应用"
+    ],
+    "source": "20-21六年级下·广东阳江·期中"
+  },
+  {
+    "id": "unit4_solve_088",
+    "unit": "第四单元：比例",
+    "chapter": "图上距离与实际距离的换算",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "在一幅 1:20000000的地图上量得甲乙两地机场的距离是8厘米，一架民航客机以每小时640千米的速度从甲地机场飞往乙地机场，需要飞行多少小时？",
+    "options": [],
+    "answer": "2.5小时",
+    "solution": "先根据'图上距离÷比例尺=实际距离'求出甲乙两地的路程；再根据'路程÷速度=时间'求出从甲地飞往乙地需要飞行的时间。明确比例尺的意义是解决此题的关键。8 ÷ 1/2000000 = 8 × 2000000 = 16000000（厘米），160000000 厘米 = 1600 千米，1600 ÷ 640 = 2.5（小时）。答：需要飞行2.5小时。",
+    "knowledge_points": [
+      "图上距离与实际距离的换算",
+      "基础行程问题"
+    ],
+    "source": "20-21六年级下·湖北鄂州·期中"
+  },
+  {
+    "id": "unit4_solve_089",
+    "unit": "第四单元：比例",
+    "chapter": "反比例的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一间长 4.8 米, 宽3.6 米的房间，用边长 0.15 米的正方形砖铺地面、需要 768块，在长6米、宽4.8米的房间里，如果用同样的砖来铺，要几块？（用比例解）",
+    "options": [],
+    "answer": "1280块",
+    "solution": "房间的面积是一定的，每块砖的面积和块数成反比例，由此列式解答即可。解答此题的关键是弄清题意，先判断哪两种相关联的量成何比例，再找准对应量。解：设用同样的砖来铺，要 x 块。(0.15 × 0.15) × x = 6 × 4.8，0.0225x = 28.8，0.0225 × 0.0225 = 28.8 ÷ 0.0225，x = 1280。答: 要 1280 块。",
+    "knowledge_points": [
+      "反比例的应用"
+    ],
+    "source": "21-22五年级下·浙江温州·期中"
+  },
+  {
+    "id": "unit4_solve_090",
+    "unit": "第四单元：比例",
+    "chapter": "图上距离与实际距离的换算",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "在比例尺为1:5000000 的地图上，量得甲乙两地的距离是 8 厘米，一列客车以每小时 80 千米的速度从甲地到乙地，需要多少小时？",
+    "options": [],
+    "answer": "5小时",
+    "solution": "根据实际距离 = 图上距离 ÷ 比例尺, 代入数据求出从甲地到乙地的距离, 换算成千米后, 再利用路程 ÷ 速度 =时间，代入数据即可求出这列客车从甲地到乙地需要多少小时。此题的解题关键是掌握图上距离和实际距离之间的换算以及路程、速度、时间三者之间的关系，从而解决问题。8 ÷ 1/5000000 =8 × 5000000= 40000000（厘米）= 400  (千米)，400 ÷ 80 = 5 (小时)答: 需要 5 小时。",
+    "knowledge_points": [
+      "比例尺应用",
+      "图上距离与实际距离的换算",
+      "基础行程问题"
+    ],
+    "source": "22-23六年级上·河南周口·期末"
+  },
+  {
+    "id": "unit4_solve_091",
+    "unit": "第四单元：比例",
+    "chapter": "比例的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "小玲打印一份文件，如果按114个/分钟的打字速度，需要35分钟完成，如果要求必须30分钟完成，每分钟应该打多少个字？（用比例知识解决）",
+    "options": [],
+    "answer": "133个",
+    "solution": "字数一定，根据工作量 = 工作时间 × 工作效率列反比例式解答。列比例式解决问题的关键是分析出题目中的等量关系。解：设每分钟应该打 x 个字。30x = 114 × 35，30x ÷ 30 = 3990 ÷ 30，x = 133。答: 每分钟应该打 133 个字。",
+    "knowledge_points": [
+      "比例的应用"
+    ],
+    "source": "2022·云南文山·小升初真题"
+  },
+  {
+    "id": "unit4_solve_092",
+    "unit": "第四单元：比例",
+    "chapter": "正比例的应用",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "要测量一棵树的高度，量得树的影长是 10.2 米, 同时有一根长 4.8 米的标杆直立在地面上，量得影长是1.6米，这棵树高多少米？（用比例解决)",
+    "options": [],
+    "answer": "30.6 米",
+    "solution": "同一时刻，不同物体的实际高度和它的影长的比值是一定的，即物体的实际高度和它的影长成正比例。设这棵树高 x 米，根据题意，树的高度：树的影长 = 标杆的高度：标杆的影长，据此列出比例并解答。本题考查正比例的应用。明确'同一时刻，物体的实际高度和它的影长成正比例'是解题的关键。解：设这棵树高 x 米，x: 10.2 = 4.8: 1.6，1.6x = 10.2 × 4.8，1.6x = 48.96，x = 48.96 ÷ 1.6，x = 30.6。答: 这棵树有 30.6 米高。",
+    "knowledge_points": [
+      "正比例的意义及辨识",
+      "正比例的应用"
+    ],
+    "source": "2021·河南驻马店·小升初真题"
+  },
+  {
+    "id": "unit4_solve_093",
+    "unit": "第四单元：比例",
+    "chapter": "反比例的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "刘叔叔从 A 地出发,每分钟走50米，33分钟能走到B地，然后沿原路返回，每分钟走 75 米。刘叔叔从 A 地到 B 地再返回的平均速度是每分钟多少米？",
+    "options": [],
+    "answer": "每分钟60米",
+    "solution": "A、B两地之间的总路程不变，则速度与时间成反比例关系，把刘叔叔从B地到A地需要的时间设为未知数，根据'路程 = 速度 × 时间'列方程求出返回A地需要的时间，刘叔叔往返一次的平均速度 = 往返的总路程÷（去时需要的时间 + 返回时需要的时间），据此解答。本题主要考查应用比例知识解决实际问题, 往返的总路程是去时路程的 2 倍。解：设刘叔叔从 B 地到 A 地需要 x 分钟。75x = 50 × 33，75x = 1650，x = 1650 ÷ 75，x = 22，(50 × 33 × 2) ÷ (33 + 22) = 3300 ÷ 55 = 60 (米/分钟)。答: 刘叔叔从 A 地到 B 地再返回的平均速度是每分钟 60 米。",
+    "knowledge_points": [
+      "反比例的应用",
+      "基础行程问题",
+      "带有小括号的混合运算"
+    ],
+    "source": "21-22五年级下·广东揭阳·期末"
+  },
+  {
+    "id": "unit4_solve_094",
+    "unit": "第四单元：比例",
+    "chapter": "正比例的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "去年我市全年节水8340万立方米，相当于减少碳排放8.34吨。如果节水9200万立方米，相当于减少碳排放多少吨？",
+    "options": [],
+    "answer": "9.2吨",
+    "solution": "设如果节水 9200 万立方米，相当于减少碳排放 x 吨，根据减少的碳排放吨数：节水体积 = 每立方米相当于减少的碳排放量，列出正比例算式解答即可。关键是确定比例关系，用比例解决问题只要等号两边的比统一即可。解：设如果节水 9200 万立方米，相当于减少碳排放 x 吨。x: 9200 = 8.34: 8340，8340x = 9200 × 8.34，8340x ÷ 8340 = 76728 ÷ 8340，x = 9.2。答: 如果节水 9200 万立方米, 相当于减少碳排放 9.2 吨。",
+    "knowledge_points": [
+      "比例的应用",
+      "正比例的应用"
+    ],
+    "source": "2022·浙江温州·小升初真题"
+  },
+  {
+    "id": "unit4_solve_095",
+    "unit": "第四单元：比例",
+    "chapter": "正比例的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "印刷厂装订车间原计划装订 5300 本书，开始 7 天装订了 2100 本，余下的书每天装订400本，照这样计算，完成任务共用了多少天？（用比例解)",
+    "options": [],
+    "answer": "15天",
+    "solution": "可以设还需要 x 天完成, 根据余下的书每天装订 400 本, 知道工作效率一定, 工作量和工作时间成正比例关系, 由此列式解答即可。解答此题的关键是根据题意判断出哪两种相关联的量成何比例，由此即可解答。解：可以设还需要 x 天完成，(5300 - 2100): x = 400: 1，3200: x = 400: 1，400x = 3200 × 1，400x ÷ 400 = 3200 ÷ 400，x= 8，8 + 7 = 15 (天)。答: 完成任务共用了 15 天。",
+    "knowledge_points": [
+      "列方程解含一个未知数的问题",
+      "工程问题",
+      "解比例",
+      "正比例的应用"
+    ],
+    "source": "21-22六年级下·湖北黄冈·期末"
+  },
+  {
+    "id": "unit4_solve_096",
+    "unit": "第四单元：比例",
+    "chapter": "相遇问题",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "快车从甲站开往乙站需要8小时，慢车从乙站开往甲站需要10小时。两车同时从两站分别开出。相向而行，相遇时，慢车行了240千米，甲乙两站之间的距离是多少千米？（用比例解）",
+    "options": [],
+    "answer": "540千米",
+    "solution": "设甲乙两站之间的距离为 x 千米, 根据行驶时间相同 (一定), 路程与速度成正比例; 快车行完全程用了 8 小时, 速度为 1/8; 慢车行完全程用了 10 小时, 速度为 1/10; 列出比例式, 再解答即可。此题应先判断行驶的路程与速度成什么比例, 再列式解答。解：甲乙两站之间的距离是 x 千米。240: (x - 240) = 1/10: 1/8，1/10(x - 240) = 240 × 1/8，1/10x - 24 = 30，1/10x = 30 + 24，1/10x = 54，x = 54 ÷ 1/10，x = 540。答: 甲乙两站之间的距离是 540 千米。",
+    "knowledge_points": [
+      "比例的应用",
+      "相遇问题"
+    ],
+    "source": "21-22六年级下·陕西商洛·期末"
+  },
+  {
+    "id": "unit4_solve_097",
+    "unit": "第四单元：比例",
+    "chapter": "正比例的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "某地区出产的花生出量非常高, 100 千克花生可以榨油 20 千克。照这样计算, 榨 10 吨花生油, 要用花生多少吨? (用比例解)",
+    "options": [],
+    "answer": "50吨",
+    "solution": "由题意可得, 某地区出产的花生的出油率是一定的, 则榨的花生油的重量与花生的重量成正比例关系, 据此即可列比例求解。本题考查了用比例解决问题, 注意找出两种相关联的量是成正比例还是反比例。解：设榨 10 吨花生油，要用花生 x 吨。20: 100 = 10: x，20x = 100 × 10，20x = 1000，20x ÷ 20 = 1000 ÷ 20，x = 50。答: 榨 10 吨花生油, 要用花生 50 吨。",
+    "knowledge_points": [
+      "列方程解含一个未知数的问题",
+      "比例的应用",
+      "解比例",
+      "正比例的应用"
+    ],
+    "source": "2022·山东济南·小升初真题"
+  },
+  {
+    "id": "unit4_solve_098",
+    "unit": "第四单元：比例",
+    "chapter": "比的应用",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "学校图书馆里的科技书与故事书的比是 3: 5, 科技书有 600 本, 故事书有多少本?",
+    "options": [],
+    "answer": "1000本",
+    "solution": "根据'科技书：故事书 = 3:5'，设故事书有 x 本，列出比例，解出比例中的未知项即可。本题考查应用比例的知识解决问题。根据比例的基本性质, 把含有未知项的比例转化成 '内项之积 = 外项之积'的形式, 解比例问题就变成了一般的解方程问题,体现了转化思想。解：设故事书有 x 本，由题意得，600: x = 3: 5，3x = 600 × 5，x = 600 × 5 ÷ 3，x = 1000。答: 故事书有 1000 本。",
+    "knowledge_points": [
+      "比的应用",
+      "解比例"
+    ],
+    "source": "21-22六年级下·辽宁鞍山·期末"
+  },
+  {
+    "id": "unit4_solve_099",
+    "unit": "第四单元：比例",
+    "chapter": "图上距离与实际距离的换算",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "在一张比例尺是1:5000000 的地图上，量得 AB 两地的距离为 4.5 厘米，则AB两地的实际距离是多少千米？",
+    "options": [],
+    "answer": "225千米",
+    "solution": "根据比例尺的意义可知，比例尺 = 图上距离：实际距离，所以实际距离 = 图上距离 ÷ 比例尺，代入数据，即可求出 AB 两地的实际距离。此题的解题关键是掌握图上距离和实际距离之间的换算，解答时要注意单位换算。4.5 ÷ 1/500000 = 2250000（厘米），22500000 厘米 = 225 千米答: AB 两地的实际距离是 225 千米。",
+    "knowledge_points": [
+      "图上距离与实际距离的换算"
+    ],
+    "source": "21-22六年级下·吉林通化·期末"
+  },
+  {
+    "id": "unit4_solve_100",
+    "unit": "第四单元：比例",
+    "chapter": "正比例的应用",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一根5米长的竹竿，某一时刻测得它在阳光下的影子长度为3.5米，同一时刻测得旁边的旗杆影子长为10.5米，这根旗杆高多少米？",
+    "options": [],
+    "answer": "15米",
+    "solution": "同一时刻阳光照射角度相同，因此竹竿高度与竹竿影长的比，和旗杆高度与旗杆影长的比相等，即高度与影长成正比例关系，即竹竿高度∶竹竿影长＝旗杆高度∶旗杆影长。设旗杆高x米，已知竹竿高5米、影长3.5米，旗杆影长10.5米，代入比例关系可得：5∶3.5＝x∶10.5，然后根据比例的基本性质解比例即可。解：设这根旗杆高x米。5∶3.5＝x∶10.5，3.5x＝10.5×5，3.5x＝52.5，x＝52.5÷3.5，x＝15。答：这根旗杆高15米。",
+    "knowledge_points": [
+      "解比例",
+      "比例的应用",
+      "正比例的意义及辨识"
+    ],
+    "source": "24-25六年级上·广西贵港·期中"
+  }, {
+    "id": "unit4_choice_101",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "下列表述中，成反比例关系的是（）。",
+    "options": [
+      "A. 三角形的高不变，它的底和面积",
+      "B. 平行四边形的面积一定，它的底和高",
+      "C. 圆的面积是 $24 \\mathrm{~m}^{2}$，它的半径与圆周率",
+      "D. 你的年龄一定，你的身高与体重"
+    ],
+    "answer": "B",
+    "solution": "判断两个相关联的量之间成什么比例，就看这两个量是对应的比值一定，还是对应的乘积一定，如果是比值一定，就成正比例，如果是乘积一定，则成反比例。据此解答。此题属于辨识成正、反比例的量，就看这两个量是对应的比值一定，还是对应的乘积一定，再作判断。A. 三角形的面积 $\\times 2 \\div$ 底 = 高（一定），商一定，所以三角形的底和面积成正比例关系，所以 A 不符合；B. 平行四边形的底 × 高 = 平行四边形的面积（一定），乘积一定，所以它的底和高成反比例，所以 B 符合；C. 圆周率是一个定值，不是变量，所以圆的半径与圆周率不成比例，所以 C 不符合；D. 一个人的身高和年龄不成比例，通常在生长期，人的身高是随着年龄的增长而增长，但是生长期过了后，骨膜会闭合，停止长高，所以 D 不符合。故答案为: B",
+    "knowledge_points": [
+      "反比例的意义及辨识",
+      "正比例的意义及辨识"
+    ],
+    "source": "2022·安徽黄山·小升初真题"
+  },
+  {
+    "id": "unit4_choice_102",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "在比例尺是 $1: 4000000$ 的地图上，量得 A、B 两港距离为 9 厘米，一艘货轮于上午 6 时以每小时 24 千米的速度从 A 开向 B 港，到达 B 港的时间是（）。",
+    "options": [
+      "A. 15点",
+      "B. 17 点",
+      "C. 21 点"
+    ],
+    "answer": "C",
+    "solution": "先依据'实际距离=图上距离÷比例尺'求出两地的实际距离，再根据'路程÷速度=时间'求出货轮从A地到B地需要的时间，进而可以求出到达B地的时刻。此题主要考查图上距离、实际距离和比例尺的关系以及基本的数量关系'路程÷速度 = 时间'。$9 \\div \\frac{1}{4000000} = 36000000$ （厘米）36000000 厘米 = 360 千米 $360 \\div 24 = 15$ (小时)6时+15时=21时所以，一艘货轮于上午 6 时以每小时 24 千米的速度从 A 开向 B 港，到达 B 港的时间是 21 时。故答案为: C",
+    "knowledge_points": [
+      "时、分、秒时间的推算",
+      "基础行程问题",
+      "图上距离与实际距离的换算"
+    ],
+    "source": "2022·河南漯河·小升初真题"
+  },
+  {
+    "id": "unit4_choice_103",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "在一幅地图上，量得A、B两城市的距离是7厘米，而A、B两城市之间的实际距离是350千米，这幅地图的比例尺是（）。",
+    "options": [
+      "A. $1: 50$",
+      "B. $1: 5000$",
+      "C. $1: 50000$",
+      "D. $1: 5000000$"
+    ],
+    "answer": "D",
+    "solution": "根据比例尺的意义，图上距离：实际距离 = 比例尺，换算单位后，代入即可求出这幅地图的比例尺。此题的解题关键是理解掌握比例尺的意义。$7 厘米：350 千米\\\\ = 7 \\text {厘 米}: 3 5 0 0 0 0 0 \\text {厘 米} \\\\ = 7: 3 5 0 0 0 0 0 0 \\\\ = (7 \\div 7): (3 5 0 0 0 0 0 \\div 7) \\\\ = 1: 5 0 0 0 0 0 0 $ 故答案为: D",
+    "knowledge_points": [
+      "比例尺的意义"
+    ],
+    "source": "18-19 六年级下·江苏·期中"
+  },
+  {
+    "id": "unit4_choice_104",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "把边长3厘米的正方形，按2∶1的比放大，放大后的面积是（    ）平方厘米。",
+    "options": [
+      "A. 6",
+      "B. 12",
+      "C. 36"
+    ],
+    "answer": "C",
+    "solution": "根据题意，把边长 3 厘米的正方形，按 2:1 的比放大，即正方形的边长放大到原来的 2 倍，再根据正方形的面积 $= \\text{边长} \\times \\text{边长}$，求出放大后的面积。明确图形的放大或缩小是指围成图形的每条线段按比例放大或缩小，以及掌握正方形的面积公式是解题的关键。$3 \\times 2 = 6$ （厘米） $6 \\times 6 = 3 6 (\\text {平 方 厘 米})$ 故答案为: C.",
+    "knowledge_points": [
+      "图形的放大与缩小",
+      "正方形的面积"
+    ],
+    "source": "2014三年级·全国·课后作业"
+  },
+  {
+    "id": "unit4_choice_105",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "将一个底是 $6 cm$，高是 $8 cm$ 的三角形，按 1:2 的比缩小后，再按 5:1 的比放大后，得到的三角形面积是原三角形（）倍。",
+    "options": [
+      "A. $\\frac{5}{2}$",
+      "B. $\\frac{2}{5}$",
+      "C. $\\frac{5}{4}$",
+      "D. $\\frac{25}{4}$"
+    ],
+    "answer": "D",
+    "solution": "用原来三角形的底除以 2，再乘 5 求出得到的三角形的底，用同样的方法求出得到的三角形的高。然后用得到三角形的面积除以原来三角形的面积即可。此题考查图形放大与缩小的方法的灵活应用。$6 \\div 2 \\times 5\\\\= 3 \\times 5\\\\ = 1 5 (cm) \\\\ 8 \\div 2 \\times 5 \\\\ = 4 \\times 5 \\\\ = 2 0 (cm)$ $ (1 5 \\times 2 0 \\div 2) \\div (6 \\times 8 \\div 2) \\\\ = 1 5 0 \\div 2 4 \\\\ = \\frac {2 5}{4} $ 故答案为: D",
+    "knowledge_points": [
+      "图形的放大与缩小",
+      "比的应用",
+      "三角形的面积"
+    ],
+    "source": "21-22 六年级下·全国·单元测试"
+  },
+  {
+    "id": "unit4_choice_106",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "下面所给例子中（）不成比例。",
+    "options": [
+      "A. 速度一定时，路程和所用时间",
+      "B. 单价一定时，数量和总价",
+      "C. 长方形的面积和宽",
+      "D. 圆锥底面积一定时，体积和高"
+    ],
+    "answer": "C",
+    "solution": "判断两个相关联的量之间成什么比例，就看这两个量是对应的比值一定，还是对应的乘积一定；如果是比值一定，就成正比例；如果是乘积一定，则成反比例；据此依次分析即可。此题属于辨识成正、反比例的量，就看这两个量是对应的比值一定，还是对应的乘积一定，再做判断。A. 因为路程÷时间 = 速度（一定），是比值一定，所以速度一定，路程与时间成正比例；B. 总价 $\\div$ 数量 = 单价 (一定)，数量和总价成正比例；C. 因为长方形的面积 $\\div$ 宽 = 长，题中没有说明长是否一定，所以长方形的面积和宽不成正比例；D. 圆锥的体积 $\\div$ 高 $=$ 底面积 $\\div 3$ (一定)，体积和高成正比例。故答案为: C",
+    "knowledge_points": [
+      "反比例的意义及辨识",
+      "正比例的意义及辨识"
+    ],
+    "source": "21-22 六年级下·全国·单元测试"
+  },
+  {
+    "id": "unit4_choice_107",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "要建长 $40 \\mathrm{~m}$、宽 $20 \\mathrm{~m}$ 的厂房，在比例尺是 $1: 500$ 的图纸上，长要画（）。",
+    "options": [
+      "A. $5 cm$",
+      "B. $8 cm$",
+      "C. $7 cm$",
+      "D. $6 cm$"
+    ],
+    "answer": "B",
+    "solution": "本题已告诉实际距离和比例尺，根据图上距离 $=$ 实际距离 × 比例尺，即可求出长和宽的图上距离。本题是考查根据实际距离和比例尺求图上距离。$40 \\mathrm{~m} = 4000 cm$     $4 0 0 0 \\times \\frac {1}{5 0 0} = 8 (cm)$   故答案为: B",
+    "knowledge_points": [
+      "图上距离与实际距离的换算"
+    ],
+    "source": "15-16 六年级下·全国·课后作业"
+  },
+  {
+    "id": "unit4_choice_108",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "正方形的周长和边长（）。",
+    "options": [
+      "A. 不成比例",
+      "B. 成正比例",
+      "C. 成反比例"
+    ],
+    "answer": "B",
+    "solution": "判断两种相关联的量成不成比例，成什么比例，就看这两种量是对应的比值一定，还是对应的乘积一定，如果是比值一定，就成正比例，如果是乘积一定，就成反比例，如果是其它的量一定或乘积、比值不一定，就不成比例。此题属于辨识正反比例关系，就看这两个量是对应的比值一定，还是乘积一定，再做判断。正方形的周长 $= \\text{边长} \\times 4$，则正方形的周长 $\\div$ 边长 $= 4$ (一定)，所以正方形的周长和边长成正比例，故答案为: B",
+    "knowledge_points": [
+      "正比例的意义及辨识"
+    ],
+    "source": "2014·全国·小升初真题"
+  },
+  {
+    "id": "unit4_choice_109",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "阳光小区的草坪长120米，宽80米，把它的平面图画在作业本上，选用比例尺（）比较合适。",
+    "options": [
+      "A. $\\frac{1}{200}$",
+      "B. $\\frac{1}{2000}$",
+      "C. $\\frac{1}{20000}$"
+    ],
+    "answer": "B",
+    "solution": "草坪长 120 米，宽 80 米，可以按照不同的比例尺分别求出图上距离是多少，然后根据作业本的尺寸判断是否合理。比例尺指的是图上距离与实际距离之比，计算的时候注意单位换算。A. $120 \\div 200 = 0.6$ （米） $= 60$ （厘米），不合适；B. $120 \\div 2000 = 0.06$ （米） $= 6$ （厘米），合适；C. $120 \\div 20000 = 0.006$ （米） $= 0.6$ （厘米），不合适；故答案为：B",
+    "knowledge_points": [
+      "图上距离与实际距离的换算"
+    ],
+    "source": "19-20六年级下·全国·单元测试"
+  },
+  {
+    "id": "unit4_choice_110",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一种糖水，糖和糖水的质量比是 $1: 10$，现在有糖 20 克，要配制这种糖水，需加水（）克。",
+    "options": [
+      "A. 180",
+      "B. 200",
+      "C. 100"
+    ],
+    "answer": "A",
+    "solution": "把需要加水的质量设为未知数，根据糖和糖水的质量比写出比例，再解比例求出未知数，据此解答。本题主要考查用比例解决问题，根据糖和糖水的质量比写出比例是解答题目的关键。解：设需加水 $x$ 克。$2 0: (2 0 + x) = 1: 1 0\\\\2 0 + x = 2 0 \\times 1 0\\\\2 0 + x = 2 0 0\\\\x = 2 0 0 - 2 0\\\\x = 1 8 0$ 所以，需加水 180 克。故答案为: A",
+    "knowledge_points": [
+      "比例的应用"
+    ],
+    "source": "21-22 六年级下·河南驻马店·期中"
+  },
+  {
+    "id": "unit4_choice_111",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "甲数的 $\\frac{4}{5}$ 等于乙数的 $\\frac{2}{3}$（甲数、乙数不为 0），那么甲数与乙数的比是（）。",
+    "options": [
+      "A. $\\frac{2}{3}: \\frac{4}{5}$",
+      "B. 6:5",
+      "C. 5:6",
+      "D. $\\frac{4}{5}: \\frac{2}{3}$"
+    ],
+    "answer": "C",
+    "solution": "由题意可知：甲数 $\\times \\frac{4}{5} = i$ 乙数 $\\times \\frac{2}{3}$，再逆运用比例的基本性质，即两内项之积等于两外项之积，即可求出二者的比。此题主要依据比例的基本性质解决问题。甲数 $\\times \\frac{4}{5} = 6$ 乙数 $\\times \\frac{2}{3}$，甲数：乙数 $\\frac{2}{3}: \\frac{4}{5} = 5: 6$。故答案为: C",
+    "knowledge_points": [
+      "比例的基本性质"
+    ],
+    "source": "2022·广东惠州·小升初真题"
+  },
+  {
+    "id": "unit4_choice_112",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一个正方形的边长是 10 厘米，如果按照 2:1 放大，放大后的图形的面积是（）平方厘米。",
+    "options": [
+      "A. 200",
+      "B. 400",
+      "C. 1000"
+    ],
+    "answer": "B",
+    "solution": "按照 2:1 放大，就是把正方形的边长扩大 2 倍，再根据正方形的面积公式解答即可。本题考查图形的放大与缩小，解答本题的关键是掌握放大比例尺的概念。$10 \\times 2 = 20$ （厘米） $20 \\times 20 = 400$ （平方厘米）故答案为: B",
+    "knowledge_points": [
+      "图形的放大与缩小",
+      "正方形的面积"
+    ],
+    "source": "19-20 六年级下·浙江·课后作业"
+  },
+  {
+    "id": "unit4_choice_113",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "把一个长 4 厘米，宽 2 厘米的长方形按 4:1 放大，所得到的图形的面积是（）平方厘米。",
+    "options": [
+      "A. 32",
+      "B. 72",
+      "C. 128"
+    ],
+    "answer": "C",
+    "solution": "长方形按 4:1 放大，则这个长方形的长和宽都放大 4 倍，据此先求出放大后的长方形的长和宽，再利用长方形的面积公式求出它的面积即可。本题考查了图形的放大，图形按一定的比例放大，图形的每条边都放大相应的倍数。$4 \\times 4 \\times (2 \\times 4)\\\\= 1 6 \\times 8\\\\= 128$ （平方厘米）故答案为: C",
+    "knowledge_points": [
+      "图形的放大与缩小"
+    ],
+    "source": "2021·吉林四平·小升初真题"
+  },
+  {
+    "id": "unit4_choice_114",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "在下面关联的量中，不成反比例的量是（）。",
+    "options": [
+      "A. 路程一定，速度和时间。",
+      "B. 减数一定，被减数和差。",
+      "C. 总价一定，单价和数量。"
+    ],
+    "answer": "B",
+    "solution": "两种相关联的量，一种量变化，另一种量也随着变化，如果这两种量中相对应的两个数的乘积一定，这两种量就叫做成反比例的量，它们的关系叫做反比例关系，据此解答。成反比例关系的两种量的乘积一定。A. 速度 × 时间 = 路程（一定），路程一定，速度和时间成反比例关系；B. 被减数一差 $=$ 减数 (一定)，被减数和差不成反比例关系；C. 单价 × 数量 = 总价 (一定)，总价一定，单价和数量成反比例关系。故答案为: B",
+    "knowledge_points": [
+      "反比例的意义及辨识"
+    ],
+    "source": "19-20 六年级下·湖南怀化·单元测试"
+  },
+  {
+    "id": "unit4_choice_115",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "在一幅地图上，实际距离是图上距离的 12000 倍，则比例尺是（）。",
+    "options": [
+      "A. $1: 12$",
+      "B. $1: 12000$",
+      "C. $12000: 1$"
+    ],
+    "answer": "B",
+    "solution": "比例尺 = 图上距离：实际距离，实际距离 = 图上距离 $\\times 12000$，据此解答。掌握比例尺的意义是解答题目的关键。分析可知，图上距离：$实际距离=图上距离：（图上距离 $\\times 12000$ ）= 1: 12000$所以比例尺是 $1: 12000$故答案为: B",
+    "knowledge_points": [
+      "比例尺的意义"
+    ],
+    "source": "19-20 六年级下·湖南怀化·单元测试"
+  },
+  {
+    "id": "unit4_choice_116",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "六（1）班人数的 $\\frac{2}{3}$ 等于六（2）班的 $\\frac{3}{4}$，则六（1）班与六（2）班人数的最简整数比是（）。",
+    "options": [
+      "A. 9:8",
+      "B. $8: 9$",
+      "C. $\\frac{2}{3}: \\frac{3}{4}$",
+      "D. $\\frac{3}{4}: \\frac{2}{3}$"
+    ],
+    "answer": "A",
+    "solution": "根据六（1）班人数的 $\\frac{2}{3}$ 等于六（2）班的 $\\frac{3}{4}$，得六(1) 班人数 $\\times \\frac{2}{3} =$ 六 (2) 班 $\\times \\frac{3}{4}$，再逆用比例的基本性质(在比例里，两个内项的积等于两个外项的积) 解决问题。此题考查的是比例的应用，解决此题关键是根据题意写出数量关系等式，再灵活利用比例的基本性质解决问题。因为六（1）班人数 $\\times \\frac{2}{3} =$ 六（2）班 $\\times \\frac{3}{4}$所以六（1）班与六（2）班人数的比是：$\\frac {3}{4}: \\frac {2}{3} = 9: 8$故选：A。",
+    "knowledge_points": [
+      "比例的基本性质",
+      "比的化简"
+    ],
+    "source": "20-21六年级上·江西赣州·单元测试"
+  },
+  {
+    "id": "unit4_choice_117",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一个长 4 厘米、宽 3 厘米的长方形按 3:1 放大，得到的图形的面积是（）平方厘米。",
+    "options": [
+      "A. 12",
+      "B. 21",
+      "C. 108",
+      "D. 36"
+    ],
+    "answer": "C",
+    "solution": "长方形按 3:1 放大，则这个长方形的长和宽都放大 3 倍，据此先求出放大后的长方形的长和宽，再利用长方形的面积公式求出它的面积即可。本题考查了图形的放大，图形按一定的比例放大，图形的每条边都放大相应的倍数。长: $4 \\times 3 = 12$ (厘米) 宽: $3 \\times 3 = 9$ (厘米)面积: $12 \\times 9 = 108$ (平方厘米) 故答案为: C",
+    "knowledge_points": [
+      "图形的放大与缩小",
+      "长方形的面积"
+    ],
+    "source": "20-21 六年级下·江苏泰州·期中"
+  },
+  {
+    "id": "unit4_choice_118",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "下列能与 $\\frac{1}{2}: \\frac{1}{3}$ 组成比例的是（）。",
+    "options": [
+      "A. $2: 3$",
+      "B. $\\frac{1}{3}: \\frac{1}{4}$",
+      "C. $3 : 2$",
+      "D. $\\frac{3}{4}: \\frac{1}{3}$"
+    ],
+    "answer": "C",
+    "solution": "比例：表示两个比相等的式子叫做比例。可分别化简每个选项中的比，以及题干中的比，相等的就可以组成比例。比是比例的一部分，本题涉及到比的化简，就要应用比的性质；最后再结合比例的意义进行选择。$\\frac{1}{2}: \\frac{1}{3} = (6 \\times \\frac{1}{2}) : (6 \\times \\frac{1}{3}) = 3: 2$   B. $\\frac{1}{3}: \\frac{1}{4} = (12 \\times \\frac{1}{3}) : (12 \\times \\frac{1}{4}) = 4: 3$  D. $\\frac{3}{4}: \\frac{1}{3} = (12 \\times \\frac{3}{4}) : (12 \\times \\frac{1}{3}) = 9: 4$  由此可见，能与 $\\frac{1}{2}: \\frac{1}{3}$ 组成比例的是选项 C。故答案为: C。",
+    "knowledge_points": [
+      "比例的意义"
+    ],
+    "source": "20-21 六年级下·全国·单元测试"
+  },
+  {
+    "id": "unit4_choice_119",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "下列 X 和 Y 成反比例关系的是（）。",
+    "options": [
+      "A. $Y= 3 + X$",
+      "B. $X + Y = \\frac{5}{6}$",
+      "C. $X = \\frac{5}{6} Y$",
+      "D. $Y = \\frac{6}{X}$"
+    ],
+    "answer": "D",
+    "solution": "两个相关联的量，当比值一定时成正比例关系；当乘积一定时成反比例关系，据此解答即可。明确正、反比例的意义是解答本题的关键。A．Y-X=3，差一定，不成反比例关系；B．X+Y= $\\frac{5}{6}$，和一定，不成反比例关系；C． $\\frac{X}{Y}=\\frac{5}{6}$，比值一定，成正比例关系；D. XY=6（一定），乘积一定，成反比例关系；故答案为: D。",
+    "knowledge_points": [
+      "反比例的意义及辨识",
+      "正比例的意义及辨识"
+    ],
+    "source": "2016·全国·小升初真题"
+  },
+  {
+    "id": "unit4_choice_120",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一幅地图上，用 20 厘米表示 380 千米，则该地图的比例尺为（）。",
+    "options": [
+      "A. $1: 19000000$",
+      "B. $1: 1900000 $",
+      "C. $1: 190000$",
+      "D. $1: 19000$"
+    ],
+    "answer": "B",
+    "solution": "比例尺 = 图上距离：实际距离，据此解答。此题考查了比例尺的意义，换算单位时注意数清 0 的个数。380千米 $= 38000000$ 厘米则该地图的比例尺为: $20: 38000000 = 1: 1900000$。故选择：B",
+    "knowledge_points": [
+      "比例尺的意义"
+    ],
+    "source": "20-21 六年级下·辽宁大连·期末"
+  },
+  {
+    "id": "unit4_choice_121",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "困难",
+    "question": "一幅图的比例尺是（）。",
+    "options": [
+      "A. 图上距离：实际距离",
+      "B. 实际距离：图上距离",
+      "C. 是一把尺子"
+    ],
+    "answer": "A",
+    "solution": "根据比例尺的意义：图上距离和实际距离的比，叫做这幅图的比例尺，即图上距离：实际距离。故答案为: A",
+    "knowledge_points": [
+      "比例尺的意义"
+    ],
+    "source": "20-21 六年级下·广东深圳·期末"
+  },
+  {
+    "id": "unit4_choice_122",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "下列各数中，能与 $\\frac{1}{2}$，6，$\\frac{1}{3}$ 组成比例的是（）。",
+    "options": [
+      "A. 2",
+      "B. 3",
+      "C. 4"
+    ],
+    "answer": "C",
+    "solution": "根据比例的基本性质：在比例里，两内项之积等于两外项之积；将四个数中间的最小数与最大数同时作外项或内项，将最小数与最大数相乘，剩下的两个数相乘，如果积相等，就能组成比例，据此解答。此题考查了比例的认识及组成比例的判断。A. $\\frac{1}{2} \\times 2 = 1, \\frac{1}{3} \\times 6 = 2, 1 \\neq 2, 2$ 不能与 $\\frac{1}{2}, 6, \\frac{1}{3}$ 组成比例；B. $\\frac{1}{2} \\times 3 = \\frac{3}{2}, \\frac{1}{3} \\times 6 = 2, \\frac{3}{2} \\neq 2, 3$ 不能与 $\\frac{1}{2}, 6, \\frac{1}{3}$ 组成比例；C. $\\frac{1}{2} \\times 4 = 2, \\frac{1}{3} \\times 6 = 2, 2 = 2, 4$ 能与 $\\frac{1}{2}, 6, \\frac{1}{3}$ 组成比例；故答案为: C。",
+    "knowledge_points": [
+      "比例的基本性质"
+    ],
+    "source": "19-20 六年级下·贵州铜仁·期末"
+  },
+  {
+    "id": "unit4_choice_123",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "下列各式（a、b 均不为 0），a 和 b 成反比例的是（）。",
+    "options": [
+      "A. $a \\times 8 = \\frac{b}{5}$",
+      "B. $a \\times \\frac{1}{3} = \\frac{1}{b}$",
+      "C. $2a - 5 = b$"
+    ],
+    "answer": "B",
+    "solution": "如果 a 和 b 成反比例，则 a 和 b 的积一定。据此把各选项的式子进行转化。本题考查反比例的辨认，把原式转化为 a 和 b 相乘或相除的形式是解题的关键。A. $a \\times 8 = \\frac{b}{5}$，则 $a \\times 8 = b \\times \\frac{1}{5}, a \\div b = \\frac{1}{5} \\div 8 = \\frac{1}{40}$ (一定)，a 和 b 的商一定，则 a 和 b 成正比例；B. $a \\times \\frac{1}{3} = \\frac{1}{b}$，则 $\\frac{a}{3} = \\frac{1}{b}, a b = 3$ (一定)，a 和 b 的积一定，则 a 和 b 成反比例；C. $2a - 5 = b$，则 $2a - b = 5, a$ 和 $b$ 的商和积都不一定，则 $a$ 和 $b$ 不成比例。故答案为: B",
+    "knowledge_points": [
+      "反比例的意义及辨识",
+      "正比例的意义及辨识"
+    ],
+    "source": "20-21 六年级下·山西忻州·期末"
+  },
+  {
+    "id": "unit4_choice_124",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "下列等式中，$x$ 与 $y(x 、 y$ 均不为 0）成反比例关系的是（）。",
+    "options": [
+      "A. $x + y = 30$",
+      "B. $xy = 30$",
+      "C. $30 = \\frac{x}{y}$"
+    ],
+    "answer": "B",
+    "solution": "根据 $x y = k$ (一定)，$x$ 和 $y$ 成反比例关系，进行选择。关键是理解反比例的意义，积一定是反比例关系。A．和的关系，不成比例关系；B. 乘积一定，成反比例关系；C. 比值一定，成正比例关系。故答案为: B",
+    "knowledge_points": [
+      "反比例的意义及辨识"
+    ],
+    "source": "20-21 六年级下·湖北随州·期末"
+  },
+  {
+    "id": "unit4_choice_125",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "能与 $\\frac{1}{3}: \\frac{1}{2}$ 组成比例的是（）。",
+    "options": [
+      "A. $\\frac{1}{2}: \\frac{1}{3}$",
+      "B. 2:3",
+      "C. 3:2",
+      "D. $\\frac{1}{3}: \\frac{1}{4}$"
+    ],
+    "answer": "B",
+    "solution": "比例是表示两个比相等的式子，求出各比的比值即可解答。掌握比例的意义是关键。$\\frac{1}{3}: \\frac{1}{2} = 2: 3 = \\frac{2}{3}$  A. $\\frac{1}{2}: \\frac{1}{3} = 3: 2 = \\frac{3}{2}$；    B. $2: 3 = \\frac{2}{3}$；    C. $3: 2 = \\frac{3}{2}$；    D. $\\frac{1}{3}: \\frac{1}{4} = \\frac{4}{3}$；$\\frac {1}{3}: \\frac {1}{2} = 2: 3 = \\frac {2}{3};$ 故答案为: B。",
+    "knowledge_points": [
+      "比例的意义"
+    ],
+    "source": "19-20 六年级下·四川·课后作业"
+  },
+  {
+    "id": "unit4_choice_126",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "比例尺一定，图上距离与实际距离成（）。",
+    "options": [
+      "A. 正比例",
+      "B. 反比例",
+      "C. 不成比例"
+    ],
+    "answer": "A",
+    "solution": "如果用字母 x 和 y 表示两种相关联的量，用 k 表示它们的比值，正比例关系可以用以下关系式表示：y: x = k（一定）；如果用字母 x 和 y 表示两种相关联的量，用 k 表示它们的积，反比例关系可以用下面关系式表示：xy = k（一定），据此判断。本题主要考查的是比例尺、正比例和反比例，比例尺 = 图上距离 ÷ 实际距离，解题的关键是判断是否符合正比例的定义，进一步求解。因为图上距离 $\\div$ 实际距离 $=$ 比例尺，所以当比例尺一定，图上距离与实际距离成正比例。故答案选择 A。",
+    "knowledge_points": [
+      "反比例的意义及辨识",
+      "正比例的意义及辨识",
+      "比例尺的意义"
+    ],
+    "source": "2014 六年级·全国·课后作业"
+  },
+  {
+    "id": "unit4_fill_127",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "一个比例里，两个内项互为倒数，其中一个外项是8，另一个外项是（）。",
+    "options": [],
+    "answer": "$\\frac{1}{8}$",
+    "solution": "根据比例的基本性质，两内项之积等于两外项之积，已知两个内项互为倒数，互为倒数的两个数的乘积为1，所以两个外项的乘积也等于1，用1除以其中一个外项，即可得解。此题的解题关键是灵活运用比例的基本性质以及倒数的定义来求解。根据分析得，另一个外项是：$1 \\div 8 = \\frac{1}{8}$。",
+    "knowledge_points": [
+      "比例的基本性质",
+      "倒数的认识及应用"
+    ],
+    "source": "2018·河北秦皇岛·小升初真题"
+  },
+  {
+    "id": "unit4_fill_128",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "圆的面积和圆的直径（）（成正比例、不成比例）。",
+    "options": [],
+    "answer": "不成比例",
+    "solution": "判断两个相关联的量之间成什么比例，就看这两个量是对应的比值一定，还是对应的乘积一定；如果是比值一定，就成正比例；如果是乘积一定，则成反比例。解答此题的关键是，根据题意，判定所求的两种相关联的量是比值一定，还是乘积一定。圆的面积 $S = (\\frac{d}{2})^{2}$，可推出 $4 S = \\pi d^{2}$；已知 $\\pi$ 一定，可 $\\pi d$ 不一定，根据正、反比例的意义可得圆的直径和圆的面积不成比例。",
+    "knowledge_points": [
+      "反比例的意义及辨识",
+      "正比例的意义及辨识"
+    ],
+    "source": "21-22 六年级下·全国·单元测试"
+  },
+  {
+    "id": "unit4_fill_129",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "如果 $y = 15 x$，x 和 y 成_____比例；如果 $y = \\frac{15}{x}$，x 和 y 成_____比例。",
+    "options": [],
+    "answer": "正；反",
+    "solution": "如果 $x$ 和 $y$ 的比值一定，它们成正比例关系；如果x 和 y 的乘积一定，它们成反比例关系。判断两个相关联的量成怎样的比例关系，就看它们是比值（商）一定，还是乘积一定。如果 $y = 15 x$，那么 $y: x = 15$ (一定)，它们成正比例关系；如果 $y = \\frac{15}{x}$，那么 $x y = 15$ (一定)，它们成反比例关系。",
+    "knowledge_points": [
+      "反比例的意义及辨识",
+      "正比例的意义及辨识"
+    ],
+    "source": "2022·内蒙古呼伦贝尔·小升初真题"
+  },
+  {
+    "id": "unit4_fill_130",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "24 的因数有____，选择其中四个组成比例是_____。",
+    "options": [],
+    "answer": "1、2、3、4、6、8、12、24；$1:2 = 3:6$",
+    "solution": "一个数的因数的个数是有限的，最小的是 1，最大的是它本身，比例的意义是：表示两个比相等的式子叫做比例。由此解答。本题考查因数、比例的意义，解答本题的关键是掌握求一个数的因数的方法和比例的意义。24的因数有：1、2、3、4、6、8、12、24；1:2=3:6 或 2:4=6:12 等；(答案不唯一)。",
+    "knowledge_points": [
+      "比例的意义",
+      "因数和倍数的求法"
+    ],
+    "source": "2014 五年级·全国·课后作业"
+  },
+  {
+    "id": "unit4_fill_131",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "边长是 2 厘米的正方形按 3:1 的比放大后，得到的图形与放大前的图形的面积比（）。",
+    "options": [],
+    "answer": "9:1",
+    "solution": "把边长 2 厘米的正方形按 3:1 放大后，得到的新正方形的边长是 $3 \\times 2 = 6$ 厘米，再利用正方形的面积公式求出放大前后的两个正方形的面积，然后再求面积比即可解答。本题主要是考查图形的放大与缩小。本题也可这样理解：一个正方形的边长扩大或缩小 $n$ 倍，它的面积将会扩大或缩小 $n^{2}$ 倍。$3 \\times 2 = 6$ （厘米）$ (6 \\times 6): (2 \\times 2) \\\\ = 3 6: 4 \\\\= 9:1$",
+    "knowledge_points": [
+      "图形的放大与缩小",
+      "正方形的面积",
+      "比的化简"
+    ],
+    "source": "2022·山西晋中·小升初模拟"
+  },
+  {
+    "id": "unit4_fill_132",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "18 的因数有____，选出其中四个数组成一个比例是_____。",
+    "options": [],
+    "answer": "1、2、3、6、9、18；$1: 2 = 3: 6$",
+    "solution": "一个数的因数的个数是有限的，最小的因数是 1，最大的因数是它本身；比例的意义是：表示两个比相等的式子叫做比例；由此解答。掌握求一个数的因数的方法和比例的意义是解题的关键。18 的因数有：1，2，3，6，9，18；组成一个比例是：$1:2 = 3:6$（答案不唯一）。",
+    "knowledge_points": [
+      "比例的意义",
+      "因数和倍数的求法"
+    ],
+    "source": "2019·河南焦作·小升初真题"
+  },
+  {
+    "id": "unit4_fill_133",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "如果 $y = 6: x$，那么 $x$ 与 $y$ 成（）比例。",
+    "options": [],
+    "answer": "反",
+    "solution": "如果两种量中相对应的两个数的乘积一定，这两种量就叫做成反比例的量。本题考查反比例的意义。y=6:x，所以xy=6（乘积一定），所以x与y成反比例。",
+    "knowledge_points": [
+      "反比例的意义及辨识"
+    ],
+    "source": "21-22 六年级下·全国·单元测试"
+  },{
+    "id": "unit4_fill_134",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "困难",
+    "question": "在一个比例中，两个外项的积是 12，其中一个内项是 15，另一个内项是（）。",
+    "options": [],
+    "answer": "0.8",
+    "solution": "比例的基本性质：两个外项的积等于两个内项的积。本题考查比例的基本性质的应用。$12 \\div 15 = 0.8$。",
+    "knowledge_points": [
+      "比例的基本性质"
+    ],
+    "source": "21-22 六年级下·全国·单元测试"
+  },
+  {
+    "id": "unit4_fill_135",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "在一个比例中，两个内项正好互为倒数，已知一个外项是 $\\frac{7}{8}$，则另一个外项是（）。",
+    "options": [],
+    "answer": "$\\frac{8}{7}$",
+    "solution": "比例的基本性质：两个外项的积等于两个内项的积；互为倒数的两个数乘积为 $1$。本题考查倒数的意义和比例的基本性质的应用。$1 \\div \\frac{7}{8} = \\frac{8}{7}$",
+    "knowledge_points": [
+      "比例的基本性质",
+      "倒数的认识及应用"
+    ],
+    "source": "2022·河南许昌·小升初真题"
+  },
+  {
+    "id": "unit4_fill_136",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "在 $\\mathrm{A} \\times \\mathrm{B} = \\mathrm{C}$ 中，当 B 一定时，A 和 C 成（）比例，当 C 一定时，A 和 B 成（）比例。",
+    "options": [],
+    "answer": "正；反",
+    "solution": "（1）两种相关联的量，一种量变化，另一种量也随着变化，如果这两种量中相对应的两个数的比值一定，这两种量就叫做成正比例的量，它们的关系叫做正比例关系。如果用字母 $y$ 和 $x$ 表示两种相关联的量，用 $k$ 表示它们的比值（一定），正比例关系可以用 $\\frac{x}{y} = k$ 表示；(2) 两种相关联的量，一种量变化，另一种量也随着变化，如果这两种量中相对应的两个数的乘积一定，这两种量就叫做成反比例的量，它们的关系叫做反比例关系，如果用字母 $x$ 和 $y$ 表示两种相关联的量，用 $k$ 表示它们的乘积（一定），反比例关系可以用 $x y = k$ 表示，据此解答。掌握正反比例关系的判断方法是解答题目的关键。（1）由 $A \\times B = C$ 可知，$B = C \\div A$，则 $C \\div A = B$（一定），此时 A 和 C 成正比例；(2) $\\mathrm{A} \\times \\mathrm{B} = \\mathrm{C}$（一定），A 和 B 成反比例。",
+    "knowledge_points": [
+      "反比例的意义及辨识",
+      "正比例的意义及辨识"
+    ],
+    "source": "20-21 六年级上·吉林长春·期末"
+  },
+  {
+    "id": "unit4_fill_137",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "有 16 名学生要分到 6 个班，至少有一个班分进（）名学生。",
+    "options": [],
+    "answer": "3",
+    "solution": "被分放物体的数量÷抽屉的数量 = 平均每个抽屉分放物体的数量……剩下物体的数量，一个抽屉里至少分放物体的数量 = 平均每个抽屉分放物体的数量 + 1，据此解答。找准被分放物体的数量和抽屉的数量是解答鸽巢问题的关键。$16 \\div 6 = 2$（名）……4（名） $2 + 1 = 3$ (名) 所以，至少有一个班分进 3 名学生。",
+    "knowledge_points": [
+      "鸽巢问题"
+    ],
+    "source": "19-20 六年级下·湖南怀化·单元测试"
+  },
+  {
+    "id": "unit4_fill_138",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "有 4 只鸽子，要飞进3个鸽巢里，至少有（）只鸽子飞进同一个鸽巢里；如果有9只鸽子飞进4个鸽巢，至少有（）只鸽子飞进同一个鸽巢里。",
+    "options": [],
+    "answer": "2；3",
+    "solution": "被分放物体的数量÷抽屉的数量 = 平均每个抽屉分放物体的数量……剩下物体的数量，每份抽屉至少分放物体的数量 = 平均每个抽屉分放物体的数量 + 1，据此解答。准确找出被分放物体的数量和抽屉的数量是解答题目的关键。(1) $4 \\div 3 = 1 \\dots 1$   $1 + 1 = 2$（只）  所以，至少有 2 只鸽子飞进同一个鸽巢里。(2) $9 \\div 4 = 2 \\dots 1$   $2 + 1 = 3$ (只)  所以，至少有 3 只鸽子飞进同一个鸽巢里。",
+    "knowledge_points": [
+      "鸽巢问题"
+    ],
+    "source": "19-20 六年级下·湖南怀化·单元测试"
+  },
+  {
+    "id": "unit4_fill_139",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "根据 $7 x =0.9y(y\\neq 0)$，写出 $\\mathrm{x:y} = (\\quad):(\\quad)$。",
+    "options": [],
+    "answer": "0.9；7",
+    "solution": "根据比例的基本性质，x 在外项，将 7 放到外项，y 在内项，将 0.9 放到内项即可。熟练掌握比例的基本性质是解题的关键。根据 $7 × = 0.9 \\mathrm{y} (y \\neq 0)$，写出 $×: \\mathrm{y} = 0.9: 7$。",
+    "knowledge_points": [
+      "比例的基本性质"
+    ],
+    "source": "19-20 六年级下·湖南怀化·单元测试"
+  },
+  {
+    "id": "unit4_fill_140",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "速度一定，路程和时间成（）比例；购买课本的总价和数量成（）比例；一批纸总页数一定，装订练习本本数和每本练习本的页数成（）比例。",
+    "options": [],
+    "answer": "正；正；反",
+    "solution": "根据 $x \\div y = k$（一定），$x$ 和 $y$ 成正比例关系；$xy = k$（一定），$x$ 和 $y$ 成反比例关系，进行分析。关键是理解正比例和反比例的意义，商一定是正比例关系，积一定是反比例关系。路程÷时间 = 速度，速度一定，路程和时间成正比例；总价÷数量 = 单价，购买课本的总价和数量成正比例；每本练习本页数 × 本数 $=$ 总页数，一批纸总页数一定，装订练习本本数和每本练习本的页数成反比例。",
+    "knowledge_points": [
+      "反比例的意义及辨识",
+      "正比例的意义及辨识"
+    ],
+    "source": "19-20六年级下·湖南怀化·单元测试"
+  },
+  {
+    "id": "unit4_fill_141",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "在比例尺是 $1: 2000000$ 的地图上，测得两地的距离是 38 厘米，这两地的实际距离是（）千米。",
+    "options": [],
+    "answer": "760",
+    "solution": "根据实际距离 $=$ 图上距离 $\\div$ 比例尺，列式计算即可。关键是掌握图上距离与实际距离的换算方法。$38 \\times 2000000 = 76000000$（厘米）$= 760$（千米）",
+    "knowledge_points": [
+      "图上距离与实际距离的换算"
+    ],
+    "source": "2014 六年级·全国·课后作业"
+  },
+  {
+    "id": "unit4_fill_142",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "在 $x: \\frac{1}{2} = \\frac{1}{3}: 2$ 中，$x = (\\quad)$。",
+    "options": [],
+    "answer": "$\\frac{1}{12}$",
+    "solution": "在比例中，两个外项的乘积等于两个内项的乘积，据此解比例求出未知数的值。掌握根据比例的基本性质解比例的方法是解答题目的关键。$x:\\frac{1}{2} = \\frac{1}{3}:2$ 解: $2 × = \\frac{1}{2} \\times \\frac{1}{3}\\\\ 2 x = \\frac {1}{6} \\\\ x = \\frac {1}{6} \\div 2 \\\\ x = \\frac {1}{1 2}$",
+    "knowledge_points": [
+      "解比例"
+    ],
+    "source": "19-20 六年级下·湖南怀化·单元测试"
+  },
+  {
+    "id": "unit4_fill_143",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "把 $4 \\times 8 = 2 \\times 16$ 可以改写成四个不同比例，请写出任意两个不同的比例（），（）。",
+    "options": [],
+    "answer": "$2: 4 = 8: 16$；$2: 8 = 4: 16$",
+    "solution": "根据比例的基本性质，比例的两内项积 $=$ 两外项积，将 4 和 8 都放到内项，2 和 16 都放到外项即可。关键是掌握比例的基本性质，表示两个比相等的式子叫比例。写的比例如下（答案不唯一）：$2:4 = 8:16$、$2:8 = 4:16$",
+    "knowledge_points": [
+      "比例的基本性质",
+      "比例的意义"
+    ],
+    "source": "19-20 六年级下·湖南怀化·单元测试"
+  },
+  {
+    "id": "unit4_fill_144",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "在 $4: 9 = 36: 81$ 中，4 和 81 是比例的（），9 和 36 是比例的（）。",
+    "options": [],
+    "answer": "外项；内项",
+    "solution": "组成比例的四个数，叫做比例的项，两端的两项叫做比例的外项，中间的两项叫做比例的内项；据此解答。掌握比例的各部分名称是解答题目的关键。由分析得：在 $4: 9 = 36: 81$ 中，4 和 81 是比例的 (外项)，9 和 36 是比例的 (内项)。",
+    "knowledge_points": [
+      "比例的意义"
+    ],
+    "source": "19-20 六年级下·湖南怀化·单元测试"
+  },
+  {
+    "id": "unit4_fill_145",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "在一个比例中，两个外项的积除以两个内项的积，商是（）。",
+    "options": [],
+    "answer": "1",
+    "solution": "由比例的基本性质可知，在比例中，两个内项的乘积与两个外项的乘积相等，据此解答。掌握比例的基本性质是解答题目的关键。在一个比例中，两个外项的积除以两个内项的积，商是 $1$。",
+    "knowledge_points": [
+      "比例的基本性质"
+    ],
+    "source": "19-20 六年级下·湖南怀化·单元测试"
+  },
+  {
+    "id": "unit4_fill_146",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "在 $4: 6 = 20: 30$ 中，那么 $6 \\times 20 = (\\quad) \\times (\\quad)$。",
+    "options": [],
+    "answer": "4；30",
+    "solution": "根据比例的两内项积 $=$ 两外项积，进行填空。关键是掌握比例的基本性质。在 $4: 6 = 20: 30$ 中，那么 $6 \\times 20 = 4 \\times 30$。",
+    "knowledge_points": [
+      "比例的基本性质"
+    ],
+    "source": "19-20 六年级下·湖南怀化·单元测试"
+  },
+  {
+    "id": "unit4_fill_147",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "在一个比例中，两个外项互为倒数，其中一个内项是最小的质数，另一个内项是__。",
+    "options": [],
+    "answer": "$\\frac{1}{2}$",
+    "solution": "比例的性质是指在比例里，两内项的积等于两外项的积；根据两个外项互为倒数，可知两个内项也互为倒数，又因为一个内项是最小的质数2，所以另一个内项是2的倒数 $\\frac{1}{2}$。此题考查比例性质的运用，也考查了倒数的意义及运用。最小的质数是 2；在一个比例中，两个外项互为倒数，其中一个内项是最小的质数，另一个内项是 $\\frac{1}{2}$。",
+    "knowledge_points": [
+      "质数与合数",
+      "比例的基本性质",
+      "倒数的认识及应用"
+    ],
+    "source": "2014·全国·小升初真题"
+  },
+  {
+    "id": "unit4_fill_148",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "比的前项、后项同时乘或除以相同的数（0 除外），比值不变。这叫做比的（）。两个比的内项的积（）两个外项的积。",
+    "options": [],
+    "answer": "基本性质；等于",
+    "solution": "比的前项和后项同时乘或除以相同的数（0 除外），比值不变，这叫做比的基本性质，如：$2:3 =$ (2×2)：(3×2) =4:6，则 $2: 3 = 4: 6, 2 \\times 6 = 3 \\times 4 = 12$，在比例中，两个内项的乘积等于两个外项的乘积。",
+    "knowledge_points": [
+      "比的基本性质",
+      "比例的基本性质"
+    ],
+    "source": "20-21 六年级上·湖南怀化·单元测试"
+  },
+  {
+    "id": "unit4_fill_149",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "晚上散步，人远离路灯，影子越（），国庆节期间，小明在北京天安门旅游，上午11点的影子比早上8点的影子（）。（填'长'或'短'）",
+    "options": [],
+    "answer": "长；短",
+    "solution": "物体距光源越远，影子越长，反之影子就越短；太阳光线与地面越斜，地上影子越长，与地面垂直时，影子最短，所以早晨到中午，物体的影子越来越短，从中午到傍晚物体的影子越来越长。只要平时多留心观察，不难发现：物体距光源越远，影子越长，反之影子就越短；从早上到中午，地面上物体的影子越来越短，从中午到太阳落山前，物体影子越来越长。物体距光源越远，影子越长，反之影子就越短；太阳光线与地面越斜，地上影子越长，与地面垂直时，影子最短；如图（粗线段为物体的影子）![](images\\68b48499a23df3514cbf219752e8a56d.png)所以晚上散步，人远离路灯，影子越长，国庆节期间，小明在北京天安门旅游，上午11点的影子比早上8点的影子短；",
+    "knowledge_points": [
+      "比例尺应用"
+    ],
+    "source": "20-21六年级上·广东韶关·期末"
+  },
+  {
+    "id": "unit4_fill_150",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "困难",
+    "question": "吴媛和施燕从学校同时出发到图书馆去，当吴媛走了一半时，施燕离图书馆还有786米，速度不变，吴媛到图书馆时，施燕还有全程的 $\\frac{1}{5}$ 没走，学校到图书馆有（）米。",
+    "options": [],
+    "answer": "1310",
+    "solution": "由题意可知：设学校到图书馆的距离是 $x$ 米，当吴媛走了一半时，施燕离图书馆还有 786 米，即吴媛走了 $\\frac{1}{2} x$ 米时，施燕走了 $x - 786$ 米，速度不变，吴媛到图书馆时，施燕还有全程的 $\\frac{1}{5}$ 没走，即当吴媛走了 $x$ 米时，施燕走了 $\\frac{4}{5} x$ 米，利用比例的意义进行解答即可。本题考查用比例解决问题，明确两次走的路程是解题的关键。解：设学校到图书馆的距离为 $x$ 米。$\\frac {1}{2} x: (x - 7 8 6) = x: (1 - \\frac {1}{5}) x \\\\ \\frac {1}{2} x: (x - 7 8 6) = x: \\frac {4}{5} x \\\\  \\frac {1}{2} x: (x - 7 8 6) = 5: 4 \\\\ 2 x = 5 x - 3 9 3 0 \\\\ 3 x = 3 9 3 0 \\\\x = 1 3 1 0$ 则学校到图书馆有 1310 米。",
+    "knowledge_points": [
+      "正比例的应用"
+    ],
+    "source": "2025·重庆江北·小升初真题"
+  },
+  {
+    "id": "unit4_fill_151",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "站得越高，看的范围越（）；林林在马路边离路灯越近，他的影子就越（）。",
+    "options": [],
+    "answer": "大；短",
+    "solution": "站得高，看得远'，所以一般来说，站得越高，看到的范围就越大，由此即可解答影子在与光的来源相反的方向，人与灯的水平之间的夹角越大，影子越短。据此解答。解答此类问题要结合生活经验，观察物体时，站得越高，看的越远，所以看的范围就越大。根据一般生活经验可得：站得越高，看到的范围就越大，晚上在路灯下散步，走向路灯时，影子在人与灯的相反方向，离路灯越近影子越短。",
+    "knowledge_points": [
+      "观察的范围（视野与盲区）"
+    ],
+    "source": "20-21 六年级上·广东深圳·期中"
+  },
+  {
+    "id": "unit4_fill_152",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个电子零件实际长度是 $2 mm$，画在图纸上的长度是 $4 cm$。这张图纸的比例尺是（）。",
+    "options": [],
+    "answer": "20:1",
+    "solution": "根据图上距离：实际距离 $=$ 比例尺，进行分析。比例尺没有单位名称。为了方便，通常把比例尺的前项化作 1（图上距离大于实际距离的，常把后项化为 1）。4 厘米：2 毫米 = 40 毫米：2 毫米 = 20：1",
+    "knowledge_points": [
+      "比例尺的意义"
+    ],
+    "source": "19-20 六年级下·江苏·单元测试"
+  },
+  {
+    "id": "unit4_fill_153",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "在一幅地图上，用 $4 cm$ 表示实际 $140 km$，这幅地图的比例尺是（）。",
+    "options": [],
+    "answer": "1:3500000",
+    "solution": "根据图上距离：实际距离 $=$ 比例尺，进行分析。比例尺没有单位名称。为了方便，通常把比例尺的前项化作 1（图上距离大于实际距离的，常把后项化为 1）。4厘米：140千米 $= 4$ 厘米：14000000厘米 =1:3500000",
+    "knowledge_points": [
+      "比例尺的意义"
+    ],
+    "source": "20-21 六年级下·天津·期末"
+  },
+  {
+    "id": "unit4_fill_154",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "如果 $a 、 b$ 都是非 0 的自然数，且 $a:b = 2:1$，那么 $a$ 和 $b$ 的最大公因数是（）。",
+    "options": [],
+    "answer": "b",
+    "solution": "根据比例尺的基本性质：内项积 $=$ 外项积，即 $a =2b$，由于 a 和 b 成倍数关系，两个数成倍数关系，则最大公因数是较小的数，由此即可判断。本题主要考查比例的基本性质以及最大公因数的求法，熟练掌握比例的基本性质并灵活运用。由分析可知 $a:b = 2:1$，则 $a = 2b$  a 和 b 的最大公因数是: b",
+    "knowledge_points": [
+      "公因数与最大公因数",
+      "比例的基本性质"
+    ],
+    "source": "2021·辽宁沈阳·小升初真题"
+  },
+  {
+    "id": "unit4_fill_155",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个比例里，两个外项的积是 1，其中一个内项是 $\\frac{3}{7}$，另一个内项是（）。",
+    "options": [],
+    "answer": "$\\frac{7}{3}$",
+    "solution": "用外项积 $\\div$ 其中一个内项 $=$ 另一个内项，据此列式计算即可。本题考查了比例的基本性质，比例的两外项积 = 1两内项积。$1 \\div \\frac{3}{7} = \\frac{7}{3}$",
+    "knowledge_points": [
+      "比例的基本性质"
+    ],
+    "source": "18-19 六年级下·山西太原·期末"
+  },
+  {
+    "id": "unit4_fill_156",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "在一个比例中，两个内项互为倒数，一个外项是最小的质数，另一个外项是（）。",
+    "options": [],
+    "answer": "$\\frac{1}{2}$",
+    "solution": "根据比例的基本性质，两个内项的积等于两个外项的积解答即可。熟练掌握比例的基本性质并能灵活利用是解答本题的关键。最小的质数是 2；$1 \\div 2 = \\frac {1}{2}$",
+    "knowledge_points": [
+      "比例的基本性质",
+      "倒数的认识及应用"
+    ],
+    "source": "2014·全国·小升初真题"
+  },
+  {
+    "id": "unit4_fill_157",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "在比例里，两个外项的积是 4，一个内项是 $\\frac{2}{3}$，另一个内项是_____。",
+    "options": [],
+    "answer": "6",
+    "solution": "在比例里，两个外项的积等于两个内项的积。据此用外项的积除以一个内项即可求出另一个内项。根据比例的基本性质即可解答。$4 \\div \\frac{2}{3} = 6$",
+    "knowledge_points": [
+      "比例的基本性质"
+    ],
+    "source": "20-21 六年级下·全国·假期作业"
+  },
+  {
+    "id": "unit4_fill_158",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "比例尺一定时，图上距离和实际距离成____关系。",
+    "options": [],
+    "answer": "正比例",
+    "solution": "两种相关联的量，一种量变化，另一种量也随着变化，如果这两种量相对应的两个数的比值（也就是商）一定，这两种量就叫做成正比例的量，它们的关系叫做成正比例关系，由此判定即可。本题考查重点：两种量成正比例与成反比例的辨识。图上距离：实际距离 = 比例尺，图上距离和实际距离是两种相关联的量，图上距离随实际距离的变化而变化，比例尺一定，也就是图上距离和实际距离的比值一定，所以图上距离和实际距离成正比例关系。",
+    "knowledge_points": [
+      "反比例的意义及辨识",
+      "正比例的意义及辨识"
+    ],
+    "source": "2021·河北沧州·小升初真题"
+  },
+  {
+    "id": "unit4_fill_159",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "张强的爸爸想带家人自驾游，在一幅比例尺为 $1: 5000000$ 的地图上，张强量得要去的地方距出发地为 $4.8$ 厘米，这段路程的实际距离是（）千米，计划三小时到达目的地，小车的平均速度是（）千米/时。",
+    "options": [],
+    "answer": "240；80",
+    "solution": "由题意可知，已知比例尺和图上距离，根据实际距离 = 图上距离 $\\div$ 比例尺即可求出；已知路程和时间，求速度。根据速度 = 路程 $\\div$ 时间即可。本题考查比例尺的应用，掌握比例尺 = 图上距离 $\\div$ 实际距离是解题的关键。$4.8 \\times 5000000 = 24000000$（厘米）$= 240$（千米） $240 \\div 3 = 80$（千米/时）",
+    "knowledge_points": [
+      "图上距离与实际距离的换算"
+    ],
+    "source": "20-21六年级下·湖北黄冈·期中"
+  },
+  {
+    "id": "unit4_fill_160",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "在一张图纸上，图上距离 $12 cm$ 表示实际距离 $4 mm$，这张图纸的比例尺是（）。",
+    "options": [],
+    "answer": "30:1",
+    "solution": "根据图上距离：实际距离 $=$ 比例尺，进行分析。比例尺没有单位名称。为了方便，通常把比例尺的前项化作 1（图上距离大于实际距离的，常把后项化为 1）。12 厘米：4 毫米 = 120 毫米：4 毫米 = 30:1",
+    "knowledge_points": [
+      "比例尺的意义"
+    ],
+    "source": "19-20六年级下·浙江金华·期末"
+  },
+  {
+    "id": "unit4_fill_161",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "在一幅比例尺是1:3000000 的地图上，量得甲、乙两地间的公路长 $4.8 cm$，一辆汽车每小时行 40 千米，从甲地到乙地需要（）小时。",
+    "options": [],
+    "answer": "3.6",
+    "solution": "根据图上距离÷比例尺 = 实际距离，求出甲乙两地实际距离，用实际距离÷速度 = 时间。关键是理解比例尺的含义，理解速度、时间、路程之间的关系。$4.8 \\times 3000000 = 14400000$（厘米）$= 144$（千米） $144 \\div 40 = 3.6$（小时）",
+    "knowledge_points": [
+      "基础行程问题",
+      "图上距离与实际距离的换算"
+    ],
+    "source": "2017 六年级下·天津河北·学业考试"
+  },
+  {
+    "id": "unit4_fill_162",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个零件的高是 $5 mm$，在图纸上的高是 $2 cm$，那么这幅图纸的比例尺（）。",
+    "options": [],
+    "answer": "4:1",
+    "solution": "比例尺 = 图上距离：实际距离，已知图上距离和实际距离，即可求出这幅图纸的比例尺。此题主要考查比例尺的计算方法，解答时要注意单位的换算。$5 mm = 0.5 cm$  $2 cm: 0.5 cm = 4: 1$",
+    "knowledge_points": [
+      "比例尺的意义"
+    ],
+    "source": "2018·山东临沂·小升初真题"
+  },
+  {
+    "id": "unit4_judge_163",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "长方形的宽一定，它的面积和长成正比例。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "根据 $x \\div y = k$（一定），$x$、$y$、$k \\neq 0$，$x$ 和 $y$ 成正比例关系，进行分析。关键是理解正比例的意义，商一定是正比例关系。长方形的面积÷长 = 宽（一定），长方形的宽一定，它的面积和长成正比例，说法正确。故答案为: √",
+    "knowledge_points": [
+      "正比例的意义及辨识"
+    ],
+    "source": "2022·安徽黄山·小升初真题"
+  },
+  {
+    "id": "unit4_judge_164",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "小麦每公顷产量一定，小麦的总产量与公顷数成正比例关系。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "判断两种相关联的量之间成什么比例，就看这两个量是对应的比值（商）一定，还是对应的乘积一定；如果是比值（商）一定，这两种相关联的量成正比例；如果是乘积一定，这两种相关联的量成反比例。掌握正、反比例的意义及辨识方法是解题的关键。小麦的总产量 $\\div$ 公顷数 = 小麦每公顷产量（一定），商一定，小麦的总产量与公顷数成正比例关系，原题说法正确。故答案为: √",
+    "knowledge_points": [
+      "反比例的意义及辨识",
+      "正比例的意义及辨识"
+    ],
+    "source": "21-22 六年级下·河南洛阳·期末"
+  },
+  {
+    "id": "unit4_judge_165",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "能与 $\\frac{1}{4} : \\frac{2}{3}$ 组成比例的比有无数个。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "根据比例的意义，可知所有比值与 $\\frac{1}{4}: \\frac{2}{3}$ 相等的比都能与它组成比例。据此解题。本题考查了比例的意义，两个或两个以上比值相等的比，可以组成比例。$\\frac{1}{4}:\\frac{2}{3} = \\frac{1}{4}\\div \\frac{2}{3} = \\frac{3}{8}$   所以，所有比值是 $\\frac{3}{8}$ 的比都能与 $\\frac{1}{4}: \\frac{2}{3}$ 组成比例，那么这样的比有无数个。故答案为: √",
+    "knowledge_points": [
+      "比例的意义"
+    ],
+    "source": "21-22 六年级下·全国·假期作业"
+  },
+  {
+    "id": "unit4_judge_166",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "如果 $7A =3B$（A、B都不等于0），那么 $\\mathrm{A:B} = 3:7$。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "根据比例的性质，把所给的等式 $7 \\mathrm{~A} = 3 \\mathrm{~B}$（A 和 B 都不等于 0），改写成一个外项是 A，一个内项是 B 的比例，则和 A 相乘的数 7 就作为比例的另一个外项，和 B 相乘的数 3 就作为比例的另一个内项，据此写出比例。根据比例的基本性质：在比例里，两外项之积等于两内项之积，据此把相乘的两个数同时作外项或内项，据此解答。如果 $7 \\mathrm{~A} = 3 \\mathrm{~B}$ (A、B都不等于0)，那么 $\\mathrm{A}: \\mathrm{B} =$ 3:7，原题说法正确。故答案为: √",
+    "knowledge_points": [
+      "比例的基本性质"
+    ],
+    "source": "21-22 六年级下·全国·单元测试"
+  }, {
+    "id": "unit4_judge_167",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "在比例尺中，图上距离总是小于实际距离。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "根据比例尺实际需要，可分放大和缩小两种比例尺，放大比例尺，图上距离要比实际距离大，据此就可作答。此题主要考查对比例尺的认识，掌握比例尺可分扩大和缩小两种比例尺以及它们的意义是解题关键。因为放大型的比例尺，图上距离要比实际距离大，所以'在比例尺中，图上距离总是小于实际距离'的说法是错误的。故答案为:×",
+    "knowledge_points": [
+      "比例尺的意义"
+    ],
+    "source": "2017·贵州铜仁·小升初真题"
+  },
+  {
+    "id": "unit4_judge_168",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "在比例中，两个内项互为倒数，那么两个外项的积是 $1$。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "乘积是 1 的两个数互为倒数；在比例中，两个内项互为倒数，那么它们的乘积是 1；根据比例的基本性质可知，两个外项的积等于两个内项的积，所以两个外项的积也是 1。掌握比例的基本性质以及倒数的定义是解题的关键。在比例中，两个内项互为倒数，那么两个外项的积是 1；原题说法正确。故答案为: √",
+    "knowledge_points": [
+      "比例的基本性质",
+      "倒数的认识及应用"
+    ],
+    "source": "17-18 六年级下·江苏盐城·期中"
+  },
+  {
+    "id": "unit4_judge_169",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "如果 $7 \\mathrm{a} = 8 \\mathrm{~b}$，那么 $\\mathrm{a}: \\mathrm{b} = 7: 8$。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "由比例的基本性质可知，两个内项的乘积等于两个外项的乘积，则当 a 为比例的外项时，7 也为比例的外项，当 b 为比例的内项时，8 也为比例的内项，据此解答。掌握比例的基本性质是解答题目的关键。分析可知，$7a = 8b$，则 $a:b = 8:7$。故答案为：×",
+    "knowledge_points": [
+      "比例的基本性质"
+    ],
+    "source": "21-22 六年级下·广西防城港·期中"
+  },
+  {
+    "id": "unit4_judge_170",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "圆柱的体积一定，它的底面积与高成反比例关系。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "判断两种相关联的量之间成什么比例，就看这两个量是对应的比值（商）一定，还是对应的乘积一定；如果是比值（商）一定，这两种相关联的量成正比例；如果是乘积一定，这两种相关联的量成反比例。掌握正、反比例的意义及辨识方法是解题的关键。底面积 $x$ 高 $=$ 圆柱的体积（一定）乘积一定，则底面积与高成反比例关系。原题说法正确。故答案为: √。",
+    "knowledge_points": [
+      "反比例的意义及辨识"
+    ],
+    "source": "2022·陕西西安·小升初真题"
+  },
+  {
+    "id": "unit4_judge_171",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "一个正方形按 3:1 放大后，周长和面积都扩大到原来的 3 倍。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "图形放大或缩小后，对应边长的比相等，周长的比也相等，但面积比不相等，原图形按 $3: 1$ 放大后，面积扩大到原来的 $3^{2}$ 倍，据此解答。原图形按 n:1 放大后，周长扩大到原来的 n 倍，面积扩大到原来的 $n^{2}$ 倍。一个正方形按 3:1 放大后，现在周长：原来的周长 $= 3:1 = 3$，现在的面积：原来的面积 $= 3^{2}:1^{2} = 9:1 = 9$，所以周长扩大到原来的 3 倍，面积扩大到原来的 9 倍。故答案为：×",
+    "knowledge_points": [
+      "正方形的周长",
+      "图形的放大与缩小",
+      "正方形的面积"
+    ],
+    "source": "2018·山东临沂·小升初真题"
+  },
+  {
+    "id": "unit4_judge_172",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "路程一定，小汽车行驶的速度与行驶的时间成反比例关系。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "判定两种相关联的量是否成正、反比例，要看这两种量是对应的比值一定，还是对应的乘积一定，如果是比值一定就成正比例；如果是乘积一定就成反比例。此题属于根据正、反比例的意义，判断两种相关联的量是成正比例还是成反比例，就看两种量是对应的比值一定，还是对应的乘积一定，再做出解答。由分析可知:若路程一定，则$速度 \\times时间 = 路程$ (一定)，乘积一定，则速度和时间成反比例。所以原题干是说法正确。故答案为: √",
+    "knowledge_points": [
+      "反比例的意义及辨识",
+      "正比例的意义及辨识"
+    ],
+    "source": "17-18六年级下·四川南充·期末"
+  },
+  {
+    "id": "unit4_judge_173",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "如果 $3^{a} = 8^{b}$，那么 $a: b = 8: 3$。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "由比例的基本性质可知，在比例中，两个内项的乘积等于两个外项的乘积，据此解答。掌握比例的基本性质是解答题目的关键。$a:b = 8:3$ 解: $3 \\times a = 8 \\times b\\\\3 a = 8 b$ 所以，如果 $3a = 8b$，那么 $a:b = 8:3$。故答案为: √",
+    "knowledge_points": [
+      "比例的基本性质",
+      "解比例"
+    ],
+    "source": "19-20 六年级下·湖南怀化·单元测试"
+  },
+  {
+    "id": "unit4_judge_174",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "三角形的高不变，它的面积和底成正比例。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "两种相关联的量，一种量变化，另一种量也随着变化，如果这两种量中相对应的两个数的比值一定，这两种量就叫做成正比例的量，它们的关系叫做正比例关系。如果用字母 $y$ 和 $x$ 表示两种相关联的量，用 $k$ 表示它们的比值 (一定)，正比例关系可以用 $\\frac{x}{y} = k$ 表示；据此解答。掌握两种相关联的量成正比例关系的判断方法是解答题目的关键。由三角形的面积计算公式可知，三角形的高 = 三角形的面积 $\\times 2 \\div$ 三角形的底，三角形的面积 $\\div$ 三角形的底 $=$ $\\frac{\\text { 三角形的高 }}{2}$ (一定)，则三角形的高不变，它的面积和底成正比例。故答案为: √。",
+    "knowledge_points": [
+      "正比例的意义及辨识"
+    ],
+    "source": "2014 六年级·全国·课后作业"
+  },
+  {
+    "id": "unit4_judge_175",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "$1.2: 0.75$ 和 $8: 5$ 不能组成比例。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "比例的基本性质：两内项之积等于两外项之积；由题意可得：$1.2 \\times 5 = 0.75 \\times 8$，所以 $1.2:0.75$ 和 $8:5$ 可以组成比例；据此解答。本题关键是要掌握比例的基本性质并灵活运用。$1.2 \\times 5 = 0.75 \\times 8$，所以 $1.2: 0.75$ 和 $8: 5$ 可以组成比例，原题说法错误。故答案为: ×",
+    "knowledge_points": [
+      "比例的基本性质"
+    ],
+    "source": "20-21 六年级下·河北秦皇岛·期末"
+  },
+  {
+    "id": "unit4_judge_176",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "1 只排球重量相当于 90 只乒乓球重量，3 只排球重量相当于 270 只乒乓球重量。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "根据比例的意义即可解决问题。比例的意义为本题考查重点。因为 $1: 90 = 3: 270$ 所以原题干说法正确。故答案为: √",
+    "knowledge_points": [
+      "比例的意义"
+    ],
+    "source": "20-21 六年级下·全国·假期作业"
+  },
+  {
+    "id": "unit4_judge_177",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "在 $a b - 3 = 15$ 中，因为有减法，所以 a 与 b 不成比例。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "相关联的两个量，如果它们的乘积一定，则这两个量成反比例关系。本题考查反比例，解答本题的关键是掌握反比例的含义。ab-3=15，所以有：ab=18，因此a与b成反比例，因此题干表述错误。故答案为: ×",
+    "knowledge_points": [
+      "反比例的意义及辨识"
+    ],
+    "source": "20-21 六年级下·湖北随州·期末"
+  },
+  {
+    "id": "unit4_judge_178",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "一个正方形按 4:1 放大，它的周长和面积都是原来的 4 倍。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "根据正方形的边长比 $= \\text{周长比}$，边长平方以后的比 $=$ 面积比，进行分析。本题考查了图形的放大与缩小，图形放大或缩小后，对应边长的比相等，周长的比相等，但面积的比不相等。$4^{2}: 1^{2} = 16: 1$，一个正方形按 $4: 1$ 放大后，周长扩大到原来的 4 倍，面积扩大到原来的 16 倍，所以原题说法错误。故答案为:×",
+    "knowledge_points": [
+      "图形的放大与缩小"
+    ],
+    "source": "2022·河北唐山·小升初真题"
+  },
+  {
+    "id": "unit4_judge_179",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "圆的直径一定，圆的周长和圆周率成正比例。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "两种相关联的量，一种量变化，另一种量也随着变化，且它们的商一定，则这两种量成正比例，据此解答。本题考查正比例的辨认，根据正比例的意义即可解答，要注意圆周率是固定不变的。圆的周长 $\\div$ 圆周率 $=$ 直径，但是圆周率是一个固定的数，所以圆的直径一定，圆的周长和圆周率不成正比例。故答案为: ×",
+    "knowledge_points": [
+      "正比例的意义及辨识"
+    ],
+    "source": "2014·全国·小升初真题"
+  },
+  {
+    "id": "unit4_calc_180",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "求未知数 $x$。$1 0: (x + 3) = \\frac {2}{3}$；$\\frac {3}{4} x - \\frac {2}{5} x = \\frac {2 1}{1 0}$",
+    "options": [],
+    "answer": "$x = 12; x = 6$",
+    "solution": "根据比例的基本性质把解比例转化为解方程。根据等式的性质解方程。等式的性质 1：等式两边同时加上或减去同一个数，等式仍然成立；等式的性质 2：等式两边同时乘或除以同一个不是零的数，等式仍然成立；注意写'解'字和'='要对齐。$10:(x + 3) = \\frac{2}{3}$ 解: $(x + 3) \\times 2 = 10 \\times 3$ $2 x + 6 = 3 0\\\\2 x = 3 0 - 6\\\\2 x = 2 4\\\\x = 1 2$ $\\frac {3}{4} x - \\frac {2}{5} x = \\frac {2 1}{1 0}$ 解: $(\\frac{3}{4} - \\frac{2}{5}) x = \\frac{21}{10}$ $ \\frac {7}{2 0} x = \\frac {2 1}{1 0} \\\\ x = \\frac {2 1}{1 0} \\div \\frac {7}{2 0} \\\\ x = \\frac {2 1}{1 0} \\times \\frac {2 0}{7} \\\\ x = 6$",
+    "knowledge_points": [
+      "应用等式的性质1和2解方程",
+      "解比例"
+    ],
+    "source": "18-19 六年级下·广东汕头·单元测试"
+  },
+  {
+    "id": "unit4_calc_181",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "解比例。$3 5: x = 7: 2$；$1. 2 5: 0. 2 5 = x: 1. 6$；$\\frac {3}{5} = \\frac {9}{x}$；$\\frac {3}{4}: 3 = x: 1 2$",
+    "options": [],
+    "answer": "$x = 10; x = 8; x = 1 5; x = 3$",
+    "solution": "在比例里，两个外项的乘积等于两个内项的乘积，据此计算。(1) $35: x = 7: 2$ 解: $7 × = 35 \\times 2$ $7 x = 7 0\\\\x = 7 0 \\div 7\\\\x = 1 0$ (2) $1.25: 0.25 = x: 1.6$ 解：$0.25× = 1.25 \\times 1.6$ $0.2 5 x = 2\\\\x = 2 \\div 0. 2 5\\\\x = 8$ (3) $\\frac{3}{5} = \\frac{9}{x}$ 解: $3 x = 5 \\times 9\\\\3 x = 4 5\\\\x = 4 5 \\div 3\\\\x = 1 5$ (4) $\\frac{3}{4}: 3 = x: 12$ 解: $3 × = 12 \\times \\frac{3}{4}\\\\3 x = 9\\\\ x = 9 \\div 3 \\\\ x = 3$",
+    "knowledge_points": [
+      "解比例"
+    ],
+    "source": "19-20 六年级下·湖南怀化·单元测试"
+  },
+  {
+    "id": "unit4_calc_182",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "解方程或比例。$\\frac {x}{6} = \\frac {2 . 5}{3}$；$\\frac {1}{4}: x = 5: 0. 4$；$2. 5 x - 1. 9 \\times 3 = 6. 2$",
+    "options": [],
+    "answer": "$x = 5; x = 0.02; x = 4.76$",
+    "solution": "$\\frac{x}{6} = \\frac{2.5}{3}$，解比例，原式化为: $3x = 2.5 \\times 6$，再用 $2.5 \\times 6$ 的积除以 3，即可解答。$\\frac{1}{4}: x = 5: 0.4$，解比例，原式化为：$5 x = 0.4 \\times \\frac{1}{4}$，再用 $0.4 \\times \\frac{1}{4}$ 的积除以 5，即可解答；2.5x-1.9×3=6.2，先算出1.9×3的和，再用6.2-1.9×3的差除以2.5，即可解答。$\\frac{x}{6} = \\frac{2.5}{3}$ 解: $3 x = 2.5 \\times 6\\\\3 x = 1 5\\\\ x = 1 5 \\div 3 \\\\ x = 5$ $ \\frac {1}{4}: x = 5: 0. 4$ 解: $5 \\mathrm{~x} = 0.4 \\times \\frac{1}{4}5 x = 0. 1 \\\\ x = 0. 1 \\div 5 \\\\ x = 0. 0 2 \\\\ 2. 5 x - 1. 9 \\times 3 = 6. 2$ 解：$2.5× - 5.7 = 6.2$ $2.5 x = 6. 2 + 5. 7 \\\\ 2. 5 x = 1 1. 9 \\\\ x = 1 1. 9 \\div 2. 5 \\\\ x = 4. 7 6$",
+    "knowledge_points": [
+      "应用等式的性质 1 和 2 解方程",
+      "解比例"
+    ],
+    "source": "20-21 六年级下·山西忻州·期末"
+  },
+  {
+    "id": "unit4_calc_183",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "解方程。$2 x + 3 \\times 0. 9 = 2 4. 7$；$\\frac {2}{3}: \\frac {5}{6} = \\frac {1 . 2}{x}$；$\\frac {2}{3} x \\div 0. 2 5 = 1 2$",
+    "options": [],
+    "answer": "$x = 11; x = \\frac{3}{2}; x = \\frac{9}{2}$",
+    "solution": "第一题先计算 $3 \\times 0.9$，将其转化为 $2 × + 2.7 = 24.7$，再左右两边同时减去 2.7，将其转化为 $2 × = 22$，再左右两边同时除以 2 即可；第二题根据比例的基本性质可知 $\\frac{2}{3} x = \\frac{5}{6} \\times 1.2$，再左右两边同时除以 $\\frac{2}{3}$ 即可；第三题左右两边同时乘 0.25，将其转化为 $\\frac{2}{3} x = 3$，再左右两边同时除以 $\\frac{2}{3}$ 即可。$2 × + 3 \\times 0.9 = 24.7$ 解：$2× + 2.7 = 24.7\\\\2 x + 2. 7 - 2. 7 = 2 4. 7 - 2. 7\\\\2 x = 2 2\\\\2 x \\div 2 = 2 2 \\div 2\\\\x = 1 1$ 解: $\\frac{2}{3} x = \\frac{5}{6} \\times 1.2\\\\ \\frac{2}{3} × \\div 0.25 \\times 0.25 = 12 \\times 0.25\\\\ \\frac {2}{3}: \\frac {5}{6} = \\frac {1 . 2}{x} \\\\ \\frac {2}{3} x \\div \\frac {2}{3} = \\frac {5}{6} \\times 1. 2 \\div \\frac {2}{3} \\\\ x = \\frac {3}{2}$ $\\frac {2}{3} x \\div 0. 2 5 = 1 2$ 解: $\\frac {2}{3} x = 3 \\\\ \\frac {2}{3} x \\div \\frac {2}{3} = 3 \\div \\frac {2}{3} \\\\ x = \\frac {9}{2}$",
+    "knowledge_points": [
+      "应用等式的性质1和2解方程",
+      "解比例"
+    ],
+    "source": "2021·湖北十堰·小升初真题"
+  },
+  {
+    "id": "unit4_solve_184",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "甲、乙两列火车同时从 A、B 两地相对开出，相遇时，甲、乙两车所行的路程比为 $4:5$。已知甲车每小时行 80 千米，乙车行完全程要 7 小时。求 A、B 两地相距多少千米?",
+    "options": [],
+    "answer": "700千米",
+    "solution": "时间一定时，路程和速度成正比，即速度比和时间比相等，据此可知甲、乙两车的速度比也为 $4: 5$。用$80 \\div 4 \\times 5$ 即可求出乙车的速度，再乘乙车行完全程需要的时间即可解答。明确时间一定时，速度比和时间比相等是解答本题的关键。时间一定时，路程和速度成正比例，则甲、乙两车的速度之比也为 $4: 5$，$ 8 0 \\div 4 \\times 5 \\times 7 \\\\ = 2 0 \\times 5 \\times 7 \\\\ = 1 0 0 \\times 7 \\\\ = 7 0 0 (\\text {千 米})$ 答：A、B 两地相距 700 千米。",
+    "knowledge_points": [
+      "正比例的意义及辨识",
+      "比的应用",
+      "相遇问题"
+    ],
+    "source": "21-22 六年级下·江西吉安·期末"
+  },
+  {
+    "id": "unit4_solve_185",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "学校把制作爱心贺卡的任务按5:4分配给六年级和五年级。六年级实际制作了108张贺卡，超过原分配任务的 $20\\%$，原计划五年级制作多少张爱心贺卡？",
+    "options": [],
+    "answer": "72 张",
+    "solution": "根据题意，六年级实际制作了108张贺卡，超出原分配任务的 $20\\%$，求出六年级原计划制作的爱心贺卡，把六年级原计划制作的爱心贺卡总数看作单位'1'，超出 $20\\%$，实际制作了 $1 + 20\\%$，用 $108 \\div (1 + 20\\%)$，求出六年级原计划制作的爱心贺卡的数量。设五年级原计划制作 $x$ 张爱心贺卡，六年级制作爱心卡片：五年级制作爱心卡片 $= 5:4$，列方程：$[108 \\div (1 + 20\\%)]：x = 5:4$，根据比例的基本性质，两内项之积等于两外项之积，解比例，即可解答。本题考查方程的实际应用，根据比例的基本性质，列方程，解比例。解：设五年级原计划制作 $x$ 张爱心贺卡。$ [ 108 \\div (1 + 20 \\%) ]: x = 5: 4 \\\\ [ 1 0 8 \\div 1. 2 ]: x = 5: 4 \\\\ 9 0: x = 5: 4 \\\\ 5 x = 9 0 \\times 4 \\\\ 5 x = 3 6 0 \\\\ x = 3 6 0 \\div 5 \\\\ \\mathrm {x = 7 2}$ 答：原计划五年级制作 72 张爱心贺卡。",
+    "knowledge_points": [
+      "解比例",
+      "已知比一个数多/少百分之几是多少，求这个数",
+      "列方程解含一个未知数的问题"
+    ],
+    "source": "21-22六年级下·河南洛阳·期末"
+  },
+  {
+    "id": "unit4_solve_186",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一列动车从 A 城开往 B 城前 3 小时行了 540 千米，照这样的速度，动车还要行驶 4 小时才能到达 B 城，A 城和 B 城相距多远？（用比例的方法解答）",
+    "options": [],
+    "answer": "1260千米",
+    "solution": "由题意可知，该动车的速度不变，则路程和时间成正比例关系，据此列比例即可。本题考查用比例解决实际问题，明确速度不变，则路程和时间成正比例关系是解题的关键。解：设A城到B城相距 $x$ 千米。$x: (4 + 3) = 5 4 0: 3\\\\3 x = 5 4 0 \\times (4 + 3)\\\\3 x = 3 7 8 0\\\\x = 1 2 6 0$ 答：A 城和 B 城相距 1260 千米。",
+    "knowledge_points": [
+      "比例的应用",
+      "基础行程问题"
+    ],
+    "source": "21-22 六年级下·贵州遵义·期末"
+  },
+  {
+    "id": "unit4_solve_187",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一张照片长 4 厘米，宽 3 厘米，如果按 4:1 的比把这张照片放大，放大后照片的长、宽分别是多少厘米？如果要使放大后照片的宽是9厘米，那么放大后照片的长应是多少厘米？",
+    "options": [],
+    "answer": "16 厘米；12 厘米",
+    "solution": "根据图形放大与缩小的方法可知：按 4:1 放大后，长是 $4 \\times 4 = 16$ (厘米)，宽是 $3 \\times 4 = 12$ (厘米)；由题意可知：放大前后的长和宽的比值是一定的，即放大前后的长和宽成正比例，由此列出比例解决问题。此题考查图形放大与缩小的方法的灵活应用，解答时要注意实际的长与原来的长数的对应。$4 \\times 4 = 16$（厘米）$3 \\times 4 = 12$ (厘米) 解：设放大后照片的长是 $x$ 厘米。$4: 3 = x: 9\\\\3 x = 4 \\times 9\\\\3 x = 3 6\\\\ 3 x \\div 3 = 3 6 \\div 3 \\\\ \\mathrm {x = 1 2}$ 答：放大后照片的长是 16 厘米，宽是 12 厘米。如果要使放大后照片的宽是 9 厘米，那么放大后照片的长应是 12 厘米。",
+    "knowledge_points": [
+      "图形的放大与缩小",
+      "解比例",
+      "比的应用"
+    ],
+    "source": "21-22 六年级下·全国·单元测试"
+  },
+  {
+    "id": "unit4_solve_188",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "兰州到乌鲁木齐的铁路线大约长 1900 千米，如果画在 $1:40000000$ 的地图上，应画多少厘米?",
+    "options": [],
+    "answer": "4.75 厘米",
+    "solution": "由比例尺的意义可知，比例尺 $=$ 图上距离 $\\div$ 实际距离，则图上距离 $=$ 实际距离 × 比例尺，最后把单位转化为厘米，据此解答。掌握图上距离和实际距离换算的方法是解答题目的关键。$1900 \\times \\frac{1}{40000000} = 0.0000475$（千米）0.0000475 千米 = 4.75 厘米  答：应画 4.75 厘米。",
+    "knowledge_points": [
+      "图上距离与实际距离的换算"
+    ],
+    "source": "21-22 六年级下·宁夏吴忠·期中"
+  },
+  {
+    "id": "unit4_solve_189",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "贵阳花果园小区1号楼的实际高度为 $75 \\mathrm{~m}$，小区的销售处有这些楼的模型，其中 1 号楼的高度与模型高度的比是 $250: 1$。这个模型高度是多少cm？（用比例解）",
+    "options": [],
+    "answer": "$30 cm$",
+    "solution": "先把 $75 \\mathrm{~m}$ 转化为 $7500 cm$，把这个模型的高度设为未知数，再根据'实际高度：模型高度 $= 250: 1$'列出比例式，并解比例求出未知数的值，据此解答。本题主要考查用比例解决问题，计算过程中注意单位的换算。$75 \\mathrm{~m} = 7500 cm$  解：设这个模型高度是 $x cm$。$7 5 0 0: x = 2 5 0: 1\\\\2 5 0 x = 7 5 0 0 \\times 1\\\\2 5 0 x = 7 5 0 0\\\\x = 7 5 0 0 \\div 2 5 0\\\\x = 3 0$ 答：这个模型高度是 $30 cm$。",
+    "knowledge_points": [
+      "比例的应用"
+    ],
+    "source": "21-22六年级下·贵州遵义·期中"
+  },
+  {
+    "id": "unit4_solve_190",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "李师傅加工一批零件，计划每分钟做8个，因任务紧迫，实际每分钟做10个，结果比计划少用45分钟，这批零件一共多少个？",
+    "options": [],
+    "answer": "1800个",
+    "solution": "因为每分钟加工零件的个数 × 加工的时间 = 零件的总数（一定），所以，工作效率和工作时间成反比例。本题考查用比例解决问题，明确零件的总数不变，是解题的关键。解：设计划用 $x$ 分钟，则实际用 $(x - 45)$ 分钟。$8 x = 1 0 \\times (x - 4 5)\\\\8 x = 1 0 x - 4 5 0\\\\2 x = 4 5 0\\\\x = 2 2 5$ $8 \\times 2 2 5 = 1 8 0 0 (\\text {个})$ 答：这批零件一共 1800 个。",
+    "knowledge_points": [
+      "比例的应用",
+      "工程问题"
+    ],
+    "source": "2022六年级下·全国·专题练习"
+  },
+  {
+    "id": "unit4_solve_191",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "今年三毛和二毛的年龄比是 7:5，五年后，三毛与二毛的年龄比是 13:10，问两人今年各几岁？",
+    "options": [],
+    "answer": "21岁；15岁",
+    "solution": "设今年三毛 $7 \\mathrm{~x}$ 岁，二毛 $5 \\mathrm{~x}$ 岁，根据（三毛今年年龄 +5）：（二毛今年年龄 +5）=13:10，列出比例求出 $×$ 的值，再根据 $×$ 的值求出 $7 \\mathrm{~x}$ 和 $5 \\mathrm{~x}$ 的值，就是三毛和二毛年龄。关键是找到比例关系，用比例解决问题只要等号两边的比统一即可。解：设今年三毛 $7 \\mathrm{~x}$ 岁，二毛 $5 \\mathrm{~x}$ 岁。$(7 x + 5): (5 x + 5) = 1 3: 1 0\\\\1 0 (7 x + 5) = 1 3 (5 x + 5)\\\\7 0 x + 5 0 = 6 5 x + 6 5\\\\5 x \\div 5 = 1 5 \\div 5\\\\x = 3$ $7 \\times 3 = 2 1 (\\text {岁})$   $5 \\times 3 = 1 5 (\\text {岁})$ 答：今年三毛 21 岁，二毛 15 岁。",
+    "knowledge_points": [
+      "比例的应用"
+    ],
+    "source": "2022 六年级下·全国·专题练习"
+  },
+  {
+    "id": "unit4_solve_192",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "甲乙两个仓库，堆放物品的质量比是3:7，甲仓库运进6吨，乙仓库运出4吨后，甲乙仓库堆放的物品的质量比是3:5，求甲乙仓库原来各堆放多少吨物品？",
+    "options": [],
+    "answer": "21吨；49吨",
+    "solution": "设原来甲仓库有物品 $3 \\mathrm{~x}$ 吨，乙仓库就有 $7 \\mathrm{~x}$ 吨，根据（甲仓库原有物品质量 +6）：（乙仓库原有物品质量 -4）=3:5，列出比例求出 $×$ 的值，再分别求出 $3 \\mathrm{~x}$ 和 $7 \\mathrm{~x}$ 的值，就是甲乙仓库原来各堆放物品的质量。关键是找到比例关系，用比例解决问题只要等号两边的比统一即可。解：设原来甲仓库有物品 $3 \\mathrm{~x}$ 吨，乙仓库就有 $7 \\mathrm{~x}$ 吨。$(3 x + 6): (7 x - 4) = 3: 5\\\\3 (7 x - 4) = 5 (3 x + 6)\\\\ 2 1 x - 1 2 = 1 5 x + 3 0 \\\\ 6 x \\div 6 = 4 2 \\div 6 \\\\ x = 7$ 甲: $3 \\times 7 = 21$ (吨)；乙: $7 \\times 7 = 49$ (吨) 答：甲乙仓库原来堆放21吨、乙仓库原来堆放49吨物品。",
+    "knowledge_points": [
+      "比例的应用"
+    ],
+    "source": "2022六年级下·全国·专题练习"
+  },
+  {
+    "id": "unit4_solve_193",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "学校有一块长120米，宽90米的长方形草地，请你选择合适的比例尺在下面的空白处画出草地的平面图。① $1: 30000$ ②  $1: 4000$ m ③ $1: 600$ (1) 选择第_种比例尺，改写成线段比例尺是。(2) 绘制草地平面图。",
+    "options": [],
+    "answer": "(1) ②；40米 (2) 见详解",
+    "solution": "（1）先把操场实际的长和宽的单位换算成厘米作单位的数，再利用实际距离乘比例尺看看图上距离是多少，图上距离与纸张大小合适就可以选择，过小，过大都不行；(2) 根据 (1) 题的图上距离结果画长方形即可。本题考查了比例尺在生活中的应用。(1) 120 米 $\\dot{1} 2000$ 厘米   90米 $\\angle 9000$ 厘米  $12000 \\times \\frac{1}{30000} = 0.4$（厘米）$9000 \\times \\frac{1}{30000} = 0.3$（厘米）图上距离太小，画图太小，因此①的比例尺不合适；$12000 \\times \\frac{1}{4000} = 3$（厘米）$9000 \\times \\frac{1}{4000} = 2.25$（厘米）这个数值适合画图，这个比例尺合适；$12000 \\times \\frac{1}{600} = 20$（厘米）$9000 \\times \\frac{1}{600} = 15$（厘米）这个比例尺太大，不适合画图。因此选择②的比例尺合适，改为线段比例尺是 $\\frac{40 \\mathrm{~米}}{2}$；(2) 作图如下:![](images\\9ce22afa8407a0c772ab7eb00d9192e9.png)",
+    "knowledge_points": [
+      "应用比例尺画图",
+      "比例尺的意义",
+      "比例尺应用",
+      "图上距离与实际距离的换算"
+    ],
+    "source": "20-21六年级下·江西萍乡·期中"
+  },
+  {
+    "id": "unit4_solve_194",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一辆货车前往武汉灾区运送救灾物资，3小时行驶了45千米。从出发地到灾区 150 千米，按照这样的速度，全程需要多少小时？（列比例解答）",
+    "options": [],
+    "answer": "10小时",
+    "solution": "用路程÷时间 = 速度这个数量关系，再结合速度不变这个条件，可以列出比例方程。设时间为未知数 $x$。本题考查用比例方程解决实际问题，找准数量关系就能列出方程。解：设全程需要 $x$ 小时。$4 5: 3 = 1 5 0: x\\\\4 5 x = 1 5 0 \\times 3\\\\4 5 x = 4 5 0\\\\4 5 x \\div 4 5 = 4 5 0 \\div 4 5\\\\x = 1 0$ 答：全程需要 10 小时。",
+    "knowledge_points": [
+      "比例的应用"
+    ],
+    "source": "21-22 六年级下·全国·单元测试"
+  },
+  {
+    "id": "unit4_solve_195",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "路路在一幅比例尺是 1:15000000 的地图上量得重庆到贵阳的铁路长约 3.1厘米，重庆到贵阳的铁路实际长度约为多少千米？",
+    "options": [],
+    "answer": "465千米",
+    "solution": "根据实际距离 $=$ 图上距离 $\\div$ 比例尺，列式解答即可。关键是掌握图上距离与实际距离的换算方法。$3.1 \\times 15000000 = 46500000$（厘米）46500000 厘米 = 465 千米答：重庆到贵阳的铁路实际长度约为 465 千米。",
+    "knowledge_points": [
+      "图上距离与实际距离的换算"
+    ],
+    "source": "19-20 六年级下·湖南怀化·单元测试"
+  },
+  {
+    "id": "unit4_solve_196",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "师傅 8 分钟加工30 个零件，徒弟每分钟加工 3 个零件，师徒二人合作完成一批零件，两人完成加工任务后共得工钱3600元。按照加工零件的数量分工钱，师徒两人各得工钱多少元？",
+    "options": [],
+    "answer": "师傅2000元；徒弟1600元",
+    "solution": "先求出师傅每分钟加工的零件数: $30 \\div 8 = \\frac{15}{4}$(个)，再求出师傅和徒弟的工效之比，$\\frac{15}{4}: 3 = 5: 4$，再把两人完成加工任务后共得工钱 3600 元按 5:4 进行分配即可。此题主要考查工作时间、工作效率、工作总量三者之间的数量关系，解答此题的关键是理解把所得的工钱按工作效率的比进行分配。$30 \\div 8 = \\frac{15}{4}$（个），$ \\frac {1 5}{4}: 3 = 5: 4, \\\\ 3 6 0 0 \\times \\frac {5}{5 + 4} = 2 0 0 0 (\\text {个}) \\\\ 3 6 0 0 \\times \\frac {4}{5 + 4} = 1 6 0 0 (\\text {个})$ 答：师傅得工钱 2000 元，徒弟得工钱 1600 元。",
+    "knowledge_points": [
+      "比例的应用"
+    ],
+    "source": "19-20 六年级上·河南驻马店·期中"
+  },
+  {
+    "id": "unit4_solve_197",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "谢老师买回两筐鸡蛋，甲筐有 100 个，乙筐有 60 个，从甲筐取出多少个鸡蛋放入乙筐后，甲乙两筐鸡蛋个数的比是 $7: 9$？",
+    "options": [],
+    "answer": "30 个",
+    "solution": "根据题意，设从甲筐取出 $×$ 个鸡蛋放入乙筐后，甲乙两筐鸡蛋个数为 $(100 - x): (60 + x) = 7: 9$，进而解比例解决问题。灵活利用比例的数量关系列比例式并解比例是解决本题的关键。解：设从甲筐取出 $x$ 个鸡蛋放入乙筐后，甲乙两筐鸡蛋个数的比是 $7: 9$ $ (1 0 0 - x) : (6 0 + x) = 7: 9 \\\\ 6 0 \\times 7 + 7 x = 1 0 0 \\times 9 - 9 x \\\\ 1 6 x = 9 0 0 - 4 2 0 \\\\ 1 6 x = 4 8 0 \\\\ x = 3 0$ 答：从甲筐取出 30 个鸡蛋放入乙筐后，甲乙两筐鸡蛋个数的比是 $7: 9$。",
+    "knowledge_points": [
+      "比例的应用"
+    ],
+    "source": "21-22 六年级上·全国·单元测试"
+  },
+  {
+    "id": "unit4_solve_198",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一个精密零件的长是3毫米，画在图纸上的长是9厘米，这幅图的比例尺是多少？",
+    "options": [],
+    "answer": "30:1",
+    "solution": "根据图上距离：实际距离 $=$ 比例尺，进行分析。关键是理解比例尺的意义，比例尺没有单位名称，为了方便，通常把比例尺的前项化作 1（图上距离大于实际距离的，常把后项化为 1）。9 厘米：3 毫米 = 90 毫米：3 毫米 = 30:1答：这幅图的比例尺是 $30: 1$。",
+    "knowledge_points": [
+      "比例尺的意义"
+    ],
+    "source": "20-21六年级下·山西忻州·期末"
+  },
+  {
+    "id": "unit4_solve_199",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一列火车为灾区运送救灾物资，1.2小时行驶了 $108\\mathrm{km}$，按照这样的速度，2.5小时能行驶多少千米？（用比例知识解答）",
+    "options": [],
+    "answer": "225千米",
+    "solution": "由题意可知这列火车的速度不变，根据路程 $=$ 速度 × 时间，可知速度一定，路程随时间增加而增加，是正比例关系，据此可解出本题答案。本题主要考查的用比例解决实际问题，解题的关键是找出正比例关系，速度一定，路程和时间成正比例关系。设 2.5 小时火车能行驶 $x$ 千米，则可列比例：$ x: 2. 5 = 1 0 8: 1. 2 \\\\ 1. 2 x = 2. 5 \\times 1 0 8 \\\\ x = 2 2 5$ 答：2.5 小时火车能行驶 225 千米。",
+    "knowledge_points": [
+      "正比例的应用",
+      "列方程解含一个未知数的问题"
+    ],
+    "source": "20-21 六年级下·湖北随州·期末"
+  },
+  {
+    "id": "unit4_solve_200",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "小明想在暑假期间练习钢笔字，他计划4天写8页。照这样的速度，暑假60天，他一共能写多少页？（用比例解）",
+    "options": [],
+    "answer": "120页",
+    "solution": "暑假每一天的练字速度是一致的，据此列出比例，再解比例即可。本题考查了比例的应用，能根据题意列比例解比例是解题的关键。解：设他一共能写 $x$ 页。$ 8: 4 = x: 6 0 \\\\ 4 x = 8 \\times 6 0 \\\\ x = 4 8 0 \\div 4 \\\\ x = 1 2 0$ 答：他一共能写 120 页。",
+    "knowledge_points": [
+      "比例的应用",
+      "解比例"
+    ],
+    "source": "20-21六年级下·河南濮阳·期末"
+  },{
+    "id": "unit4_choice_201",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "3 毫米精密零件画在图纸上是 30 厘米, 图纸比例尺是 ( )。",
+    "options": [
+      "A. $1: 10$",
+      "B. $1: 100$",
+      "C. $10: 1$",
+      "D. $100: 1$"
+    ],
+    "answer": "D",
+    "solution": "【分析】图上距离与实际距离的比叫做比例尺, 把图上距离换算成毫米, 然后写出图上距离与实际距离的比并化成后项是 1 的比即可。本题主要考查的比例尺的概念及计算, 解题的关键是将图上距离和实际距离化为相同单位, 再得出比例尺。\n【详解】30 厘米 : 3 毫米 = 300 毫米 : 3 毫米 = 100 : 1。故答案为: D。",
+    "knowledge_points": ["比例尺的意义"],
+    "source": "18-19 六年级下·全国·期末",
+    "merged_id": "unit4_201",
+    "original_id": "unit4_choice_201"
+  },
+  {
+    "id": "unit4_choice_202",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "已知 $mn = c, \\frac{c}{b} =$ a (a, b, c, m, n都是大于0的自然数), 那么下面比例中正确的是 ( )。",
+    "options": [
+      "A. $\\frac{m}{n} = \\frac{b}{a}$",
+      "B. $\\frac{n}{m} = \\frac{a}{b}$",
+      "C. $\\frac{n}{a} = \\frac{m}{b}$",
+      "D. $\\frac{m}{a} = \\frac{b}{n}$"
+    ],
+    "answer": "D",
+    "solution": "【分析】根据已知条件 $mn = c, \\frac{c}{b} = a$ , 可得出 $mn = ab$ , 再根据比例的基本性质 (在比例里, 两个外项的积等于两个内项的积。) 对四个选项进行分析, 即可得出答案。本题考查比例的基本性质, 根据比例的基本性质进行解答。\n【详解】mn=c， $\\frac{c}{b} = a$ $\\mathrm {m n} = \\mathrm {c}, \\mathrm {a b} = \\mathrm {c}, \\text {即} \\mathrm {m n} = \\mathrm {a b},$$\\frac {m}{a} = \\frac {b}{n} \\text {或} \\frac {n}{a} = \\frac {b}{m} 。$故答案选：D",
+    "knowledge_points": ["比例的基本性质"],
+    "source": "2022·四川绵阳·小升初真题",
+    "merged_id": "unit4_202",
+    "original_id": "unit4_choice_202"
+  },
+  {
+    "id": "unit4_choice_203",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "能与 $\\frac{1}{3}: \\frac{2}{9}$ 组成比例的比是 ( )。",
+    "options": [
+      "A. $\\frac{2}{3}: \\frac{1}{9}$",
+      "B. $18: 27$",
+      "C. $3: 2$"
+    ],
+    "answer": "C",
+    "solution": "【分析】去选择谁能与 $\\frac{1}{3}: \\frac{2}{9}$ 组成比例, 可以通过比例的基本性质: 外项积等于内项积, ——来判定即可。本题也可以通过比例的定义来判定，比例的定义：表示两个比相等的式子。把比进行化简，最终比相等的就是正确选项。\n【详解】假设： $\\frac{1}{3}: \\frac{2}{9} = \\frac{2}{3}: \\frac{1}{9}$ ，但是 $\\frac{1}{3} \\times \\frac{1}{9} \\neq \\frac{2}{9} \\times \\frac{2}{3}$ ，所以 A 不是。假设： $\\frac{1}{3}: \\frac{2}{9} = 18:27$ ，但是 $\\frac{1}{3} \\times 27 \\neq \\frac{2}{9} \\times 18$ ，所以 B 不是。假设: $\\frac{1}{3}: \\frac{2}{9} = 3: 2, \\frac{1}{3} \\times 2 = \\frac{2}{9} \\times 3 = \\frac{2}{3}$ , 所以 C 满足。故答案为: C",
+    "knowledge_points": ["比例的基本性质", "比例的意义"],
+    "source": "20-21 六年级下·全国·期末",
+    "merged_id": "unit4_203",
+    "original_id": "unit4_choice_203"
+  },
+  {
+    "id": "unit4_choice_204",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "要建一个长30米，宽20米的长方形广场，在比例尺是 $1: 500$ 的图纸上，宽要画（）。",
+    "options": [
+      "A. $4 cm$",
+      "B. $6 cm$",
+      "C. $12 cm$"
+    ],
+    "answer": "A",
+    "solution": "【分析】据公式：图上距离：实际距离 = 比例尺，那么图上距离 = 实际距离 $\\times$ 比例尺，据此计算即可。知道通过比例尺推算出求图上距离是解决此题的关键。\n【详解】 $20 \\times \\frac{1}{500} = 0.04$ （米）0.04 米=4 厘米故答案选：A",
+    "knowledge_points": ["图上距离与实际距离的换算"],
+    "source": "2021·云南玉溪·小升初模拟",
+    "merged_id": "unit4_204",
+    "original_id": "unit4_choice_204"
+  },
+  {
+    "id": "unit4_choice_205",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "在一幅比例尺是 1:200000 的地图上，量得甲、乙两地之间的距离是14.4cm，如果将甲、乙两地之间的距离画在另一幅比例尺是 $1: 500000$ 的地图上, 应画 ( ) $cm$ 。",
+    "options": [
+      "A. 5.76",
+      "B. 7.2",
+      "C. 11.52"
+    ],
+    "answer": "A",
+    "solution": "【分析】根据图上距离：比例尺 = 实际距离，由第一幅图可求出实际距离，再用实际距离 $\\times$ 另一幅图的比例尺即可。熟练掌握实际距离和图上距离的求法是解答此题的关键。\n【详解】 $14.4 \\div \\frac{1}{200000} \\times \\frac{1}{500000}\\\\= 2 8 8 0 0 0 0 \\times \\frac {1}{5 0 0 0 0 0} \\\\ = 5. 7 6 \\text {(厘 米)} $\n故答案为: A",
+    "knowledge_points": ["图上距离与实际距离的换算"],
+    "source": "20-21六年级下·黑龙江大兴安岭地·期中",
+    "merged_id": "unit4_205",
+    "original_id": "unit4_choice_205"
+  },
+  {
+    "id": "unit4_choice_206",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "下列各题中两种量成反比例关系的是 ( )。",
+    "options": [
+      "A. 购买面值 1.5 元的邮票, 邮票枚数与总价",
+      "B. 三角形面积一定, 底和高",
+      "C. 车轮直径一定, 车轮行驶的路程和转数",
+      "D. 如果 $x = 3y$ , x 和 y"
+    ],
+    "answer": "B",
+    "solution": "【分析】判断两个相关的量之间成什么比例，就看着两个量之间是对应的比值一定，还是对应的乘积一定，如果比值一定，就成正比例，如果乘积一定就成反比例，据此分析即可。本题考查反比例的意义, 根据反比例的意义解答问题。\n【详解】A．总价 $=$ 邮票的面值 $\\times$ 邮票的枚数，邮票的面值 $=$ 总价：邮票数量（一定），邮票枚数与总价成正比例； B. 三角形的底 $\\times$ 高 = 面积 $\\times 2$ (一定), 是乘积一定, 三角形面积一定, 底和高成反比例; C. 车轮行驶的路程: 转数 = 车轮的周长 (一定), 周长一定则直径一定, 是比值一定, 车轮行驶的路程和转数成正比例; D. 如果 $x = 3 y, x: y = 3, x$ 和 $y$ 比值一定, $x$ 和 $y$ 成正比例关系。故答案选：B",
+    "knowledge_points": ["正比例的意义及辨识", "反比例的意义及辨识"],
+    "source": "20-21 六年级下·全国·课后作业",
+    "merged_id": "unit4_206",
+    "original_id": "unit4_choice_206"
+  },
+  {
+    "id": "unit4_choice_207",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "已知 $5x = 7 y, x$ 与 $y $ () 。",
+    "options": [
+      "A. 不成比例",
+      "B. 成正比例",
+      "C. 成反比例"
+    ],
+    "answer": "B",
+    "solution": "【分析】根据题意可知两种相关联的量，一种量变化，另一种量也随着变化，如果这两种量中相对应的两个数的比值一定，这两种量就叫做成正比例的量。此题考查正反比例的辨别, 要学活灵活运用比例的基本性质。\n【详解】 $5 x = 7 y$ $\\frac{x}{y} = \\frac{7}{5}$ （比值一定），成正比例。故答案为: B。",
+    "knowledge_points": ["正比例的意义及辨识", "比例的基本性质", "反比例的意义及辨识"],
+    "source": "20-21 六年级下·全国·课后作业",
+    "merged_id": "unit4_207",
+    "original_id": "unit4_choice_207"
+  },
+  {
+    "id": "unit4_choice_208",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "如果 $A \\times 2 = B \\div 3$ ，那么 $A: B = ()$ 。",
+    "options": [
+      "A. $2: 3$",
+      "B. $3: 2$",
+      "C. $1: 6$"
+    ],
+    "answer": "C",
+    "solution": "【分析】把除法转化成乘法后，根据比例的基本性质写出比例，然后把后面的比化成最简整数比即可做出选择。灵活运用比例的基本性质。\n【详解】 $A \\times 2 = B \\div 3$ , 则 $A \\times 2 = B \\times \\frac{1}{3}$ , 那么 $A: B = \\frac{1}{3}: 2 = 1: 6$ 。故答案为: C",
+    "knowledge_points": ["比例的基本性质"],
+    "source": "2015·广东江门·小升初真题",
+    "merged_id": "unit4_208",
+    "original_id": "unit4_choice_208"
+  },
+  {
+    "id": "unit4_choice_209",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "圆柱的体积一定, 它的高和 ( ) 成反比例。",
+    "options": [
+      "A. 底面半径",
+      "B. 底面积",
+      "C. 底面周长"
+    ],
+    "answer": "B",
+    "solution": "【分析】根据 $x y = k$ （一定）， $x$ 和 $y$ 成反比例关系，进行选择。关键是理解反比例的意义，掌握圆柱体积公式。\n【详解】圆柱底面积 $x$ 高 $=$ 体积（一定），则圆柱的高和底面积成反比例。故答案为: B",
+    "knowledge_points": ["反比例的意义及辨识"],
+    "source": "19-20 六年级下·四川·课后作业",
+    "merged_id": "unit4_209",
+    "original_id": "unit4_choice_209"
+  },
+  {
+    "id": "unit4_choice_210",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "困难",
+    "question": "把一个直径 5 毫米的手表零件，画在图纸上直径是 10 厘米，这幅图纸的比例尺是（）。",
+    "options": [
+      "A. 20:1",
+      "B. $1: 20$",
+      "C. $2: 1$",
+      "D. $1 : 2$"
+    ],
+    "answer": "A",
+    "solution": "【分析】根据比例尺的意义解答, 即比例尺是图上距离与实际距离的比。本题主要考查了比例尺的意义, 注意图上距离与实际距离的单位要统一。\n【详解】10 厘米 $= 100$ 毫米 $100: 5 = 20: 1$ ;故答案为: A。",
+    "knowledge_points": ["比例尺的意义"],
+    "source": "20-21 六年级下·江苏·单元测试",
+    "merged_id": "unit4_210",
+    "original_id": "unit4_choice_210"
+  },
+  {
+    "id": "unit4_choice_211",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "困难",
+    "question": "如果 $2 \\mathrm{x} = 3 \\mathrm{y}$ , 那么下面选项中正确的是 ( )。",
+    "options": [
+      "A. $2: x = 3: y$",
+      "B. $x:y = 2:3$",
+      "C. $\\frac{x}{y} = \\frac{2}{3}$",
+      "D. $\\frac{x}{y}= \\frac {3}{2}$"
+    ],
+    "answer": "D",
+    "solution": "【分析】逆用比例的基本性质作答, 即在比例里, 两个外项的积等于两个内项的积。本题主要是灵活利用比例的基本性质解决问题。\n【详解】因为 $x 、 y$ 都不为零, 且 $2 x = 3 y$ ; 所以 $x: y = 3: 2$ ;即 $\\frac{x}{y} = \\frac{3}{2}$ 或 $\\frac{x}{3} = \\frac{y}{2}$ ; 所以选项中正确的是: $\\frac{x}{y} = \\frac{3}{2}$ 。故答案为: D。",
+    "knowledge_points": ["比例的基本性质"],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_211",
+    "original_id": "unit4_choice_211"
+  },
+  {
+    "id": "unit4_choice_212",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "下面各组中的两个比不能组成比例的是（）。",
+    "options": [
+      "A. 8:7 和 16:14",
+      "B. 0.6:0.2 和 3:1",
+      "C. 19:110 和 10:9"
+    ],
+    "answer": "C",
+    "solution": "【分析】分别求出每个选项中两个比的比值，比值相等的话，就可以组成比例，否则就不能组成比例。表示两个比相等的式子叫做比例，也可以假设可以组成比例，然后根据比例的基本性质进行验证。\n【详解】A. $8: 7 = \\frac{8}{7}, 16: 14 = \\frac{8}{7}$ , 比值相等, 可以组成比例; B. $0.6: 0.2 = 3, 3: 1 = 3$ , 比值相等, 可以组成比例;C. $19: 110 = \\frac{19}{110}, 10: 9 = \\frac{10}{9}$ , 比值不相等, 不能组成比例; 故答案选: C。",
+    "knowledge_points": ["比例的意义"],
+    "source": "13-14六年级下·全国·课后作业",
+    "merged_id": "unit4_212",
+    "original_id": "unit4_choice_212"
+  },
+  {
+    "id": "unit4_choice_213",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "关于正比例的判断，有以下四种说法，不正确的是（）。",
+    "options": [
+      "A. 订同一份杂志的钱数和份数成正比例。",
+      "B. 正方形的面积和它的边长成正比例。",
+      "C. 八小时内, 工人做零件的个数和做每个零件所用时间成正比例。",
+      "D. 平行四边形的面积与和它等底等高的三角形的面积成正比例。"
+    ],
+    "answer": "C",
+    "solution": "【分析】根据正比例的定义，一一判定各个选项中两个数量间的关系，再选出正确选项即可。本题考查了正比例, 明确正比例的定义是解题的关键。\n【详解】A．钱数等于份数乘单价，同一份杂志的单价一定，份数增加时钱数也增加。所以，订同一份杂志的钱数和份数成正比例；B. 正方形的面积等于边长乘边长, 边长增加时面积也增加,所以, 正方形的面积和它的边长成正比例; C. 工作时间一定时, 做每个零件用时越短, 能做的零件个数越多, 所以, 八小时内, 工人做零件的个数和做每个零件所用时间成反比例; D. 平行四边形的面积等于等底等高三角形面积的 2 倍, 所以, 平行四边形的面积与和它等底等高的三角形的面积成正比例。故答案为: C",
+    "knowledge_points": ["正比例的意义及辨识"],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_213",
+    "original_id": "unit4_choice_213"
+  },
+  {
+    "id": "unit4_choice_214",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "把 $8 \\times 2.5 = 0.4 \\times 50$ 改写成比例是 ( )。",
+    "options": [
+      "A. $8: 2.5 = 0.4: 50$",
+      "B. $8: 0.4 = 2.5: 50 $",
+      "C. $ 8: 0.4 = 50: 2.5$",
+      "D. $8: 50 = 2.5: 0.4$"
+    ],
+    "answer": "C",
+    "solution": "【分析】根据比例的基本性质，只要 8 和 2.5 同时在内项或外项，0.4 和 50 同时在内项或外项即可。关键是掌握比例的基本性质，比例的两内项积 =两外项积。\n【详解】A． $8:2.5 = 0.4:50$ ，8和2.5没有同时在内项或外项，错误；B. $8: 0.4 = 2.5: 50, 8$ 和 $2.5$ 没有同时在内项或外项，错误; C. $8: 0.4 = 50: 2.5, 8$ 和 2.5 同时在外项, 0.4 和 50 同时在内项,正确; D. $8: 50 = 2.5: 0.4$ , 8 和 2.5 没有同时在内项或外项, 错误。故答案为: C。",
+    "knowledge_points": ["比例的基本性质"],
+    "source": "19-20 六年级下·河南信阳·期末",
+    "merged_id": "unit4_214",
+    "original_id": "unit4_choice_214"
+  },
+  {
+    "id": "unit4_choice_215",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "困难",
+    "question": "下面比中, 能与 $\\frac{1}{2}: \\frac{1}{3}$ 组成比例的是 ( )。",
+    "options": [
+      "A. $2: 3$",
+      "B. $3: 2$",
+      "C. $1: 6$"
+    ],
+    "answer": "B",
+    "solution": "【分析】可先将 $\\frac{1}{2}: \\frac{1}{3}$ 进行化简, 得出最简比, 再与各个选项相比较, 可得出结论。表示两个比相等的式子叫比例, 所以要选择与题目里所给的比相等的式子, 化简的方法不唯一, 也可按分数计算来化简。\n【详解】 $\\frac{1}{2}: \\frac{1}{3} = (\\frac{1}{2} \\times 6): (\\frac{1}{3} \\times 6) = 3: 2$故答案为: B。",
+    "knowledge_points": ["比例的意义"],
+    "source": "19-20 六年级下·河南南阳·期末",
+    "merged_id": "unit4_215",
+    "original_id": "unit4_choice_215"
+  },
+  {
+    "id": "unit4_choice_216",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "两个圆柱形容器内原来的水面高度都是 $8 cm$ , 它们的底面直径都是 $10 cm$ 。①号容器内放一个小球后, 水面高度为 $10 cm$ 。②号容器内放入一个小球和一个大球后, 水面高度为 $15 cm$ 。两个容器内的小球完全相同, 小球的体积与大球的体积的比是 ( )。",
+    "options": [
+      "A. 2:5",
+      "B. 2:7",
+      "C. $5: 7$",
+      "D. 5:12"
+    ],
+    "answer": "A",
+    "solution": "【分析】①号容器内放一个小球后，水面高度为 $10 cm$ ， $10 - 8 = 2$ （厘米），则小球的体积等于高为 2 厘米的圆柱的体积；②号容器内放入一个小球和一个大球后, 水面高度为 (厘米), 则大球的体积等于高为 5 厘米的圆柱的体积。两个圆柱形容器的底面直径都是 , 则底面积相等。圆柱的体积 高 = 底面积, 底面积一定, 圆柱的体积和高成正比例关系, 则两个圆柱的体积比等于它们高的比，据此解答。本题考查正比例的应用, 理解 '底面积一定, 两个圆柱的体积比等于高的比' 是解题的关键。\n【详解】 $10 - 8 = 2$ （厘米） $15 - 10 = 5$ (厘米) 小球的体积与大球的体积的比是 $2: 5$ 。 故答案为: A。",
+    "knowledge_points": ["正比例的应用", "圆柱的体积"],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_216",
+    "original_id": "unit4_choice_216"
+  },
+  {
+    "id": "unit4_choice_217",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "把一个图形先按 $2: 1$ 放大，再把放大后的图形按 $1: 3$ 缩小，最后得到的图形的面积是原图形的（）。",
+    "options": [
+      "A. $\\frac{9}{4}$",
+      "B. $\\frac{4}{9}$",
+      "C. $\\frac{1}{4}$",
+      "D. $\\frac{2}{3}$"
+    ],
+    "answer": "B",
+    "solution": "【分析】设原图形的面积是 1, 按 $2: 1$ 放大后就是 $1 \\times 2 \\times 2 =4,$ 再把图形按 $1:3$ 缩小后就是 $4 \\times \\frac{1}{3} \\times \\frac{1}{3} = \\frac{4}{9}$ , 得到的图形面积是原图形的: $\\frac{4}{9} \\div 1$ , 解答即可。此题主要考查学生对图形放大与缩小的理解与应用。\n【详解】假设原图形的面积是 1。按 $2: 1$ 放大后的图形面积: $1 \\times 2 \\times 2 = 4$ ;按 $1: 3$ 缩小后的图形面积: $4 \\times \\frac{1}{3} \\times \\frac{1}{3} = \\frac{4}{9}$ ;得到的图形面积是原图形: $\\frac{4}{9} \\div 1 = \\frac{4}{9}$故答案为: B",
+    "knowledge_points": ["图形的放大与缩小"],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_217",
+    "original_id": "unit4_choice_217"
+  },
+  {
+    "id": "unit4_choice_218",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "一辆汽车 3 小时行驶 $126 \\mathrm{~km}$ , 照这样的速度行驶 168 千米, 需要多少小时? 设需要 $x$ 小时, 下列方程正确的是 ( )。",
+    "options": [
+      "A. $\\frac{126}{3} = \\frac{x}{168}$",
+      "B. $126 \\times 3 = 168 x$",
+      "C. $3 x = 168 \\times 126$",
+      "D. $\\frac{126}{3} = \\frac{168}{x}$"
+    ],
+    "answer": "D",
+    "solution": "【分析】因为路程÷时间 = 速度（一定），所以路程和时间成正比例，设需要 x 小时，据此列比例解答。此题考查的目的是理解掌握路程、速度、时间三者之间的关系及应用，以及正比例的意义及应用。\n【详解】设：需要 $x$ 小时；$\\frac {1 2 6}{3} = \\frac {1 6 8}{x}$ 故答案为: D",
+    "knowledge_points": ["比例的应用"],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_218",
+    "original_id": "unit4_choice_218"
+  },
+  {
+    "id": "unit4_choice_219",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "下列描述中的两种相关联的量成正比例关系的有（）个，成反比例关系的有（）个。①圆柱的体积一定, 它的底面积和高。②一盒饼干, 吃掉的个数和剩下的个数。③树苗的成活率一定, 成活的树苗和树苗的总棵数。④《小学生数学报》订阅的总价钱和订阅的数量。",
+    "options": [
+      "A. 1; 2",
+      "B. 2; 1",
+      "C. 3; 2",
+      "D. 4; 1"
+    ],
+    "answer": "B",
+    "solution": "【分析】判断两个相关联的量之间成什么比例，就看这两个量是对应的比值一定，还是对应的乘积一定；如果是比值一定，就成正比例；如果是乘积一定，则成反比例。此题属于辨识成正、反比例的量，就看这两个量是对应的比值一定，还是对应的乘积一定，再做判断。\n【详解】①圆柱的底面积 $x$ 高 $=$ 体积（一定），是乘积一定，所以它的底面积和高成反比例。② 一盒饼干, 吃掉的个数 + 剩下的个数 = 饼干的总个数 (一定), 是和一定, 所以吃掉的个数和剩下的个数不成比例。③ 成活的树苗÷树苗的总棵树 = 成活率（一定），比值一定，成正比例关系。(4)《小学生数学报》订阅的总价钱 $\\div$ 订阅的数量 $=$ 单价（一定），比值一定，成正比例关系。成正比例关系的有③④，2个；成反比例关系的有①，1个。故选择：B",
+    "knowledge_points": ["正比例的意义及辨识", "反比例的意义及辨识"],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_219",
+    "original_id": "unit4_choice_219"
+  },
+  {
+    "id": "unit4_choice_220",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "困难",
+    "question": "下面的数中, 不能与 2、3、4 组成比例的是 ( )。",
+    "options": [
+      "A. $\\frac{3}{2}$",
+      "B. $\\frac{8}{3}$",
+      "C. 3",
+      "D. 6"
+    ],
+    "answer": "C",
+    "solution": "【分析】表示两个比相等的式子叫作比例，据此选择。此题考查了比例的意义，属于基础类题目。\n【详解】A. $4: 2 = 3:\\frac{3}{2}$ , 可以组成比例。B. $4: 3 = \\frac{8}{3}: 2$ , 可以组成比例。C. 不能与 2、3、4 组成比例。D. $4: 2 = 6: 3$ , 可以组成比例。故选择：C",
+    "knowledge_points": ["比例的意义"],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_220",
+    "original_id": "unit4_choice_220"
+  },
+  {
+    "id": "unit4_choice_221",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "下面的比中，能与 $\\frac{1}{4}: \\frac{1}{6}$ 组成比例的是（）。",
+    "options": [
+      "A. $\\frac{1}{6}: \\frac{1}{4}$",
+      "B. $\\frac{1}{3}: \\frac{1}{2}$",
+      "C. $\\frac{1}{2}: \\frac{1}{3}$",
+      "D. 2:3"
+    ],
+    "answer": "C",
+    "solution": "【分析】比例是指表示两个比相等的式子，因此可以用求比值的方法，先求出 $\\frac{1}{4}: \\frac{1}{6}$ 的比值，进而求出每一个选项中比的比值，再根据比值相等，就能组成比例，比值不相等，就不能组成比例。解决此题也可以根据比例的性质'两外项的积等于两内项的积'，等于能组成比例，不等于就不能组成比例。\n【详解】 $\\frac{1}{4}: \\frac{1}{6} = 3: 2$ A. $\\frac{1}{6}: \\frac{1}{4} = 2: 3$ , 所以 A 不能; B. $\\frac{1}{3}: \\frac{1}{2} = 2: 3$ , 所以 B 不能;C. $\\frac{1}{2}: \\frac{1}{3} = 3: 2$ , 所以 C 能;D. $2: 3$ , 不能。故答案为: C",
+    "knowledge_points": ["比例的意义"],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_221",
+    "original_id": "unit4_choice_221"
+  },
+  {
+    "id": "unit4_choice_222",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "小云看一本故事书, 他每天看的页数跟所用的天数 ( )。",
+    "options": [
+      "A. 成正比例",
+      "B. 成反比例",
+      "C. 不成比例",
+      "D. 以上说法都不对"
+    ],
+    "answer": "B",
+    "solution": "【分析】根据 $x \\div y = k$ （一定）， $x$ 和 $y$ 成正比例关系； $xy = k$ （一定）， $x$ 和 $y$ 成反比例关系，进行辨识。关键是理解正比例和反比例的意义，商一定是正比例关系，积一定是反比例关系。\n【详解】每天看的页数 $\\times$ 天数 $=$ 故事书总页数（一定），他每天看的页数跟所用的天数成反比例关系。故答案为: B",
+    "knowledge_points": ["反比例的意义及辨识"],
+    "source": "2014 六年级·全国·课后作业",
+    "merged_id": "unit4_222",
+    "original_id": "unit4_choice_222"
+  },
+  {
+    "id": "unit4_choice_223",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "困难",
+    "question": "下列两种量成正比例的是（）。",
+    "options": [
+      "A. 铺地总面积一定，每块砖的面积和砖的块数",
+      "B. 全班人数一定，每组人数和组数。",
+      "C. 正方形边长和周长",
+      "D. 圆的半径和面积"
+    ],
+    "answer": "C",
+    "solution": "【分析】判定两种相关联的量是否成正、反比例，要看这两种量是对应的比值一定，还是对应的乘积一定，如果是比值一定就成正比例；如果是乘积一定就成反比例；如果不是比值或乘积一定，就不成比例。\n【详解】A.因为：每块砖的面积 $\\times$ 铺地砖块数 $=$ 铺地面积（一定），所以每块砖的面积与铺地砖块数成反比例;B.每组人数 $\\times$ 组数 = 全班人数 (一定)是乘积一定, 所以全班人数一定, 每组人数和组数成反比例;C.正方形的周长：边长 $= 4$ （一定），所以正方形边长和周长成正比例;D. 圆面积 $\\div$ 圆的半径 $= \\pi \\times$ 圆的半径, 圆的半径是变量, 所以 $(\\pi \\times$ 圆的半径) 就不一定, 是乘积不一定, 所以圆的半径与圆面积不成比例关系。故答案为: C",
+    "knowledge_points": ["正比例的意义及辨识", "反比例的意义及辨识"],
+    "source": "20-21 六年级下·江苏·阶段练习",
+    "merged_id": "unit4_223",
+    "original_id": "unit4_choice_223"
+  },
+  {
+    "id": "unit4_fill_224",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "a 与 b 的比是 3:4, b是c的 $\\frac{2}{5}$ , 则 $a:c = 6$ ( ) , a比c少 $\\frac{()}{()}$ 。",
+    "options": [],
+    "answer": "$3: 10; \\frac{7}{10}$",
+    "solution": "【分析】b是c的 $\\frac{2}{5}$ , 即 $b = \\frac{2}{5} c$ , 又a与b的比是3:4, 将b用 $\\frac{2}{5} c$ 进行等量代换, 所以 $a:\\frac{2}{5} c = 3:4$ , 化简即可得 $a:c$ 的值;求a比c少几分之几, 先求出 (c-a), 再除以c即可。本题属于求多个数的连比和一个数比另一个数少几分之几的问题，要掌握等量代换的方法。\n【详解】因为 $b = \\frac{2}{5} c, a:b = 3:4$ , 所以\n$\\mathrm {a}: \\frac {2}{5} \\mathrm {c} = 3: 4 \\\\ \\frac {6}{5} \\mathrm {c} = 4 \\mathrm {a} \\\\ \\mathrm {a : c = 3 : 1 0} $\na 为 3 份, c 为 10 份则 a 比 c 少几分之几列式为:\n$(1 0 - 3) \\div 1 0 \\\\ = 7 \\div 1 0 \\\\ = \\frac {7}{1 0} $",
+    "knowledge_points": ["比例的基本性质", "等量代换", "求比值"],
+    "source": "19-20 六年级下·陕西西安·期末",
+    "merged_id": "unit4_224",
+    "original_id": "unit4_fill_224"
+  },
+  {
+    "id": "unit4_fill_225",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一幅地图的比例尺是 1:400000, 把它改成线段比例尺是( ), 已知 AB 两地的实际距离是 24 千米, 在这幅地图上应画( )厘米。",
+    "options": [],
+    "answer": "见详解；6",
+    "solution": "【分析】将数值比例尺的后项看成厘米为单位的数，换算成千米为单位的数，用1厘米的线段表示这个千米数即可；根据实际距离÷线段比例尺一厘米表示的千米数即可。比例尺没有单位名称。为了方便，通常把比例尺的前项化作 1（图上距离大于实际距离的，常把后项化为 1）。\n【详解】400000 厘米 = 4 千米![](images\\057f9aea2bf91c5bddf4d588193b2514.png)$24 \\div 4 = 6$ (厘米)",
+    "knowledge_points": ["图上距离与实际距离的换算", "比例尺的意义"],
+    "source": "19-20 六年级下·江西赣州·期末",
+    "merged_id": "unit4_225",
+    "original_id": "unit4_fill_225"
+  },
+  {
+    "id": "unit4_fill_226",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "如果 $4 \\mathrm{a} = 5 \\mathrm{~b}$ , 那么 $\\mathrm{a}: \\mathrm{b} = (\\quad): (\\quad)$ , $\\mathrm{a}$ 相当于 $\\mathrm{b}$ 的 $(\\quad) \\%$ 。",
+    "options": [],
+    "answer": "5 4 125",
+    "solution": "【分析】在比例里，两个外项的积等于两个内项的积，所以 $a:b = 5:4$ ； $a$ 相当于 $b$ 的百分之几，就是把 $b$ 看作单位'1'， $5 \\div 4 \\times 100\\% = 125\\%$ .掌握比例的基本性质以及理清把'相当于'后面的数看作单位'1'是解题的关键。\n【详解】 $5 \\div 4 \\times 100\\% = 125\\%$a:b=5:4, a 相当于 b 的 $125 \\%$",
+    "knowledge_points": ["比例的基本性质", "求一个数是另一个数的百分之几 (百分率问题)"],
+    "source": "19-20 六年级下·江西赣州·期末",
+    "merged_id": "unit4_226",
+    "original_id": "unit4_fill_226"
+  },
+  {
+    "id": "unit4_fill_227",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "甲乙两地相距50千米，在一幅地图上量得图上距离是2.5厘米，这幅图的比例尺是( )。",
+    "options": [],
+    "answer": "1:2000000",
+    "solution": "【分析】图上距离：实际距离 = 比例尺，据此解答。要注意先统一单位，再写出比例尺。本题考查求比例尺，根据比例尺的意义即可解答。\n【详解】50千米 $= 5000000$ 厘米 2.5:5000000 = 1:2000000",
+    "knowledge_points": ["比例尺的意义"],
+    "source": "19-20六年级下·江西赣州·期末",
+    "merged_id": "unit4_227",
+    "original_id": "unit4_fill_227"
+  },
+  {
+    "id": "unit4_fill_228",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个精密零件，在比例尺是 20:1 的图纸上，量得它的长度是 5 厘米，这个精密零件的实际长度是( )毫米。",
+    "options": [],
+    "answer": "2.5",
+    "solution": "【分析】根据图上距离 $\\div$ 比例尺 = 实际距离，进行换算即可。关键是理解比例尺的意义，掌握图上距离与实际距离的换算方法。\n【详解】5 厘米 $= 50$ 毫米 $50 \\div 20 = 2.5$ (毫米)",
+    "knowledge_points": ["图上距离与实际距离的换算"],
+    "source": "20-21 六年级下·山东德州·期中",
+    "merged_id": "unit4_228",
+    "original_id": "unit4_fill_228"
+  },
+  {
+    "id": "unit4_fill_229",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "$2: 3 = 4: 6$ , 如果第一个比的后项加 5 , 那么第二个比的后项应该加_____。",
+    "options": [],
+    "answer": "10",
+    "solution": "【分析】先把第一个比的后项加上 5 , 然后计算出两个内项的积, 那么两个外项的积也是这个数, 用两个外项的积除以第一个比的前项即可求出第二个比的后项, 用这个后项减去原来的后项即可求出第二个比的后项应该加的数。熟练运用比例的基本性质是关键。\n【详解】 $3 + 5 = 8\\\\8 \\times 4 = 3 2 \\\\ 3 2 \\div 2 = 1 6 \\\\ 1 6 - 6 = 1 0 $",
+    "knowledge_points": ["比例的基本性质"],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_229",
+    "original_id": "unit4_fill_229"
+  },
+  {
+    "id": "unit4_fill_230",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个长方形操场，长 110 米，宽 90 米。把它画在比例尺是 $\\frac{1}{1000}$ 的图纸上，长应画() 厘米，宽应画() 厘米。",
+    "options": [],
+    "answer": "11; 9",
+    "solution": "【分析】由比例尺 $=$ 图上距离 $\\div$ 实际距离可知, 图上距离 $=$ 实际距离 $\\times$ 比例尺, 据此解答。明确实际距离、图上距离和比例尺之间的关系是解答本题的关键。\n【详解】110米 $= 11000$ 厘米，90米 $= 9000$ 厘米 长: $11000 \\times \\frac{1}{1000} = 11$ (厘米) 宽: $9000 \\times \\frac{1}{1000} = 9$ (厘米)",
+    "knowledge_points": ["图上距离与实际距离的换算"],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_230",
+    "original_id": "unit4_fill_230"
+  },
+  {
+    "id": "unit4_fill_231",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "在速度、路程和时间这三个量中，如果()一定，()和()成正比例；如果()一定，()和()成反比例。",
+    "options": [],
+    "answer": "速度 ;路程; 时间 ;路程;速度 ;时间",
+    "solution": "【分析】根据速度、路程、时间之间的关系，确定相关联的两个量的商一定还是积一定，如果商一定就成正比例，如果积一定就成反比例，否则不成比例。知道正比例和反比例的定义以及知道速度、路程、时间之间的关系是解决此题的关键。\n【详解】路程÷时间 = 速度, 如果速度一定, 路程和时间成正比例;速度 $\\times$ 时间 $=$ 路程，路程一定，速度和时间成反比例。",
+    "knowledge_points": ["正比例的意义及辨识", "反比例的意义及辨识"],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_231",
+    "original_id": "unit4_fill_231"
+  },
+  {
+    "id": "unit4_fill_232",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "困难",
+    "question": "若两个比的比值$\\frac{a}{b}$和$\\frac{c}{d}$互为倒数，则a、b、c、d这四个数组成的比例是(      )。",
+    "options": [],
+    "answer": "a:b=d:c;a:d=b:c;c:b=d:a;c:d=b:a(答案写一个即可)",
+    "solution": "【分析】 $\\frac{a}{b}$ 和 $\\frac{c}{d}$ 互为倒数，即 $\\frac{a}{b} \\times \\frac{c}{d} = 1$ ，则 $\\frac{ac}{bd} = 1$ ，即 $ac = bd$ ，根据比例的基本性质改写成比例即可。此题考查了倒数和比例基本性质的灵活运用。\n【详解】因为 $\\frac{a}{b}$ 和 $\\frac{c}{d}$ 互为倒数，即 $\\frac{a}{b} \\times \\frac{c}{d} = 1$ ，则 $\\frac{ac}{bd} = 1$ ，即 $ac = bd$ ; 所以 $a:b = d:c, a:d = b:c, c:b = d:a, c:d = b:a$ 。",
+    "knowledge_points": ["比例的基本性质", "倒数的认识及应用"],
+    "source": "20-21六年级下·全国·单元测试",
+    "merged_id": "unit4_232",
+    "original_id": "unit4_fill_232"
+  },
+  {
+    "id": "unit4_fill_233",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "困难",
+    "question": "如果 $4 \\mathrm{x} = 5 \\mathrm{y}$ , 那么$\\mathrm {x}: \\mathrm {y} = (\\quad): (\\quad), \\mathrm {x}: 5 = (\\quad): (\\quad) _ {\\circ}$",
+    "options": [],
+    "answer": "5 ;4 ;y ;4",
+    "solution": "【分析】逆用比例的基本性质, 把 $4 \\mathrm{x} = 5 \\mathrm{y}$ 改写成比例的形式, 使相乘的两个数 $\\mathrm{x}$ 和 4 做比例的外项, 则相乘的另两个数 $\\mathrm{y}$ 和 5 就做比例的内项, 由此求出 $\\mathrm{x}$ 与 $\\mathrm{y}$ 的比; 再把相乘的两个数 $\\mathrm{x}$ 和 4 做比例的外项, 则相乘的另两个数 5 和 $\\mathrm{y}$ 就做比例的内项, 由此求出 $\\mathrm{x}$ 与 5 的比。解答此题的关键是比例基本性质的逆运用, 要注意: 相乘的两个数要做外项就都做外项, 要做内项就都做内项。\n【详解】x:y=5:4;x:5=y:4",
+    "knowledge_points": ["比例的基本性质"],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_233",
+    "original_id": "unit4_fill_233"
+  },{
+    "id": "unit4_fill_234",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "困难",
+    "question": "用 4、15、5 和 12 组成的比例可能是( )。",
+    "options": [],
+    "answer": "$5: 4 = 15: 12$",
+    "solution": "【分析】根据比例的基本性质：两内项之积等于两外项之积，进行解答即可。熟练掌握比例的基本性质是解答本题的关键。\n【详解】因为 $5 \\times 12 = 4 \\times 15$ ;所以 $5: 4 = 15: 12$ ;(答案不唯一)。",
+    "knowledge_points": ["比例的基本性质"],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_234",
+    "original_id": "unit4_fill_234"
+  },
+  {
+    "id": "unit4_fill_235",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "困难",
+    "question": "若 $5 \\mathrm{~A} = 6 \\mathrm{~B}$ (A, B均不为 0), 则 $\\mathrm{A}: \\mathrm{B} = (\\quad): (\\quad)$ 。",
+    "options": [],
+    "answer": "6 ;5",
+    "solution": "【分析】根据比例的性质，把所给的等式 $5 \\mathrm{~A} = 6 \\mathrm{~B}$ （A，B均不为0），改写成一个外项是A，一个内项是B的比例，则和A相乘的数5就作为比例的另一个外项，和B相乘的数6就作为比例的另一个内项，据此写出比例。本题主要是灵活利用比例的基本性质解决问题。\n【详解】 $5 \\mathrm{~A} = 6 \\mathrm{~B}$ ;则 $\\mathrm{A}: \\mathrm{B} = 6: 5$ 。",
+    "knowledge_points": ["比例的基本性质"],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_235",
+    "original_id": "unit4_fill_235"
+  },
+  {
+    "id": "unit4_fill_236",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一个长 $5 cm$ 、宽 $3 cm$ 的长方形按 3:1 放大, 得到的图形的面积是( ) $cm^{2}$ 。",
+    "options": [],
+    "answer": "135",
+    "solution": "【分析】把长 $5 cm$ 、宽 $3 cm$ 的长方形按照 $3: 1$ 放大, 那么长和宽分别是 15 厘米和 9 厘米, 然后计算放大后的面积。用比例尺对图形进行放大时, 边长放大 $n$ 倍, 面积放大 $n^{2}$ 倍。\n【详解】 $5 \\times 3 = 15$ (cm)\n$3 \\times 3 = 9 \\quad (\\mathrm {c m}) \\\\ 1 5 \\times 9 = 1 3 5 (\\mathrm {c m} ^ {2}) $\n得到的图形的面积是 $135 cm^{2}$ 。",
+    "knowledge_points": ["图形的放大与缩小"],
+    "source": "2017·重庆·小升初真题",
+    "merged_id": "unit4_236",
+    "original_id": "unit4_fill_236"
+  },
+  {
+    "id": "unit4_fill_237",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "如果 $y = 15 x$ 。那么 x 和 y 成()_比例关系; 如果 $y = \\frac{15}{x}$ (x, y 均不为 0), 那么 x 和 y 成()_比例关系。",
+    "options": [],
+    "answer": "正 ;反",
+    "solution": "【分析】根据 $x \\div y = k$ （一定）， $x$ 和 $y$ 成正比例关系； $xy = k$ （一定）， $x$ 和 $y$ 成反比例关系，将 $y = 15x$ 和 $y = \\frac{15}{x}$ 转化后，进行辨识即可。关键是理解正比例和反比例的意义，商一定是正比例关系，积一定是反比例关系。\n【详解】根据 $y = 15 x$ 可得 $y \\div x = 15$ （一定），所以 $x$ 和 $y$ 成正比例关系；根据 $y = \\frac{15}{x}$ 可得 $x y = 15$ （一定），所以 $x$ 和 $y$ 成反比例关系。",
+    "knowledge_points": ["正比例的意义及辨识", "反比例的意义及辨识"],
+    "source": "19-20 六年级·广东佛山·期中",
+    "merged_id": "unit4_237",
+    "original_id": "unit4_fill_237"
+  },
+  {
+    "id": "unit4_fill_238",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "分子一定, 分母和分数值成( )比例; 如果 $x = 7 y$ , $(x, y$ 都不为 0 $)$ 则 $x$ 和 $y$ 成( )比例。",
+    "options": [],
+    "answer": "反;正",
+    "solution": "【分析】判断两种相关联的量成什么比例，就看这两种量是对应的比值一定，还是对应的乘积一定，如果是比值一定，就成正比例，如果是乘积一定，就成反比例。此题属于辨识成正、反比例的量，就看这两种量是对应的比值一定，还是对应的乘积一定，再做出判断。\n【详解】因为：分数值 $\\times$ 分母 $=$ 分子（一定），所以分母和分数值成反比例；如果 $x = 7 y$ ，那么 $\\frac{x}{y} = 7$ ，则 $x$ 和 $y$ 成正比例。",
+    "knowledge_points": ["正比例的意义及辨识", "反比例的意义及辨识"],
+    "source": "19-20 六年级下·四川绵阳·期末",
+    "merged_id": "unit4_238",
+    "original_id": "unit4_fill_238"
+  },
+  {
+    "id": "unit4_fill_239",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "a 与 b 成正比例关系,如果 $a = 4$ , 则 $b = 60$ ; 如果 $a = (\\quad)$ , 则 $b = 150$ 。",
+    "options": [],
+    "answer": "10",
+    "solution": "【分析】因为 a 与 b 成正比例关系，所以 a 与 b 的比值不变，据此列出正比例算式，求出 $a$ 的值即可。x:y=k（一定），x和y成正比例关系。\n【详解】 $a: 150 = 4: 60$\n解: $60 \\mathrm{a} \\div 60 = 600 \\div 60\\\\a = 1 0$",
+    "knowledge_points": ["正比例的意义及辨识", "解比例"],
+    "source": "19-20 六年级下·陕西宝鸡·期末",
+    "merged_id": "unit4_239",
+    "original_id": "unit4_fill_239"
+  },
+  {
+    "id": "unit4_fill_240",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "困难",
+    "question": "已知 $8 \\mathrm{a} = 9 \\mathrm{~b}$ (a, b均不为 0), 则 $\\mathrm{a}: \\mathrm{b} = (\\quad): (\\quad)$ 。",
+    "options": [],
+    "answer": "9 8",
+    "solution": "【分析】逆用比例的基本性质, 把 $8 \\mathrm{a} = 9 \\mathrm{~b}$ 改写成比例的形式, 使相乘的两个数 9 和 b 做比例的内项, 则相乘的另两个数 8 和 a 就做比例的外项。本题考查比例基本性质的逆运用，要注意：相乘的两个数要做外项就做外项，要做内项就都做内项。本题考查比例基本性质的逆运用，要注意：相乘的两个数要做外项就做外项，要做内项就都做内项。\n【详解】如果 $8a = 9b$ ，那么 $a:b = 9:8$ 。",
+    "knowledge_points": ["比例的基本性质"],
+    "source": "18-19 六年级下·云南曲靖·期末",
+    "merged_id": "unit4_240",
+    "original_id": "unit4_fill_240"
+  },
+  {
+    "id": "unit4_fill_241",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "已知 $15 \\times 3 = 5 \\times 9$ ，根据比例的基本性质写出两个比例：（ ）、（）。",
+    "options": [],
+    "answer": "$15: 5 = 9: 3 (答案不唯一)$;$15: 9 = 5: 3 (答案不唯一)$",
+    "solution": "【分析】根据比例的基本性质, 只要 15 和 3 或 5 和 9 同时在比例的内项或外项即可。关键是掌握比例的基本性质，比例的两内项积 $=$ 两外项积。\n【详解】已知 $15 \\times 3 = 5 \\times 9$ , 根据比例的基本性质写出两个比例: $15: 5 = 9: 3; 15: 9 = 5: 3.$",
+    "knowledge_points": ["比例的基本性质"],
+    "source": "19-20六年级下·贵州六盘水·期末",
+    "merged_id": "unit4_241",
+    "original_id": "unit4_fill_241"
+  },
+  {
+    "id": "unit4_fill_242",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "在 $3: 2, \\frac{1}{2}: \\frac{2}{3}$ , $0.6: 0.4$ 中选两个比组成比例是____。",
+    "options": [],
+    "answer": "$3: 2 = 0.6: 0.4$",
+    "solution": "【分析】比值相等的两个比可以组成比例, 可以分别求出这三个比的比值, 找出比值相等的两个比即可。本题选出来组成比例的两个比是唯一的，但是组 成的比例并不唯一。\n【详解】 $3: 2 = 3 \\div 2 = 1.5$ ; $06: 0. 4 = 0. 6 \\div 0. 4 = 1. 5;$可以组成比例:$3: 2 = 0. 6: 0. 4 $。",
+    "knowledge_points": ["求比值", "比例的意义"],
+    "source": "20-21 六年级下·江苏·单元测试",
+    "merged_id": "unit4_242",
+    "original_id": "unit4_fill_242"
+  },
+  {
+    "id": "unit4_fill_243",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "困难",
+    "question": "如果 $5 \\mathrm{~x} = 9 \\mathrm{y}$ , 那么\n$\\mathrm {x}: \\mathrm {y} = (\\quad): (\\quad) _ {\\circ}$",
+    "options": [],
+    "answer": "9 ；5",
+    "solution": "【分析】比例的基本性质：在比例中，两个外项的积等于两个内项的积。例如 $1.5: 3 = 2: 4$ , 可得: $1.5 \\times 4 = 3 \\times 2$ ; 应用比例的基本性质, 能将乘积式转化为比例式。应用比例的基本性质, 可使比例式与乘积式相互转化。转化时, 注意找准内项、外项。\n【详解】因为: $5 \\mathrm{x} = 9 \\mathrm{y}$ , 可把数字 5 和 $\\mathrm{x}$ 看作外项, 数字 9 和 $\\mathrm{y}$ 看作内项, 则可得比例式: $\\mathrm{x}: \\mathrm{y} = 9: 5$ 。",
+    "knowledge_points": ["比例的基本性质"],
+    "source": "19-20 六年级下·云南玉溪·期中",
+    "merged_id": "unit4_243",
+    "original_id": "unit4_fill_243"
+  },
+  {
+    "id": "unit4_fill_244",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "在一幅比例尺是1:500000的地图上量得 $A 、 B$ 两地的距离是 $3.2 c m$ , 在另一幅地图上, 量得这两地的距离是 $2 c m$ , 另一幅地图的比例尺是( )。",
+    "options": [],
+    "answer": "1:800000",
+    "solution": "【分析】根据实际距离 $=$ 图上距离 $\\div$ 比例尺, 求出两地的实际距离, 再根据比例尺 = 图上距离 : 实际距离, 解答即可。此题考查了比例尺的应用，明确图上距离、实际距离和比例尺之间的关系是解题关键。另外注意比例尺的前项一般都是 1。\n【详解】 $3.2 \\div \\frac{1}{500000} = 1600000$ （厘米）另一幅地图的比例尺是 $2: 1600000$ ，化简得 $1: 800000$ 。",
+    "knowledge_points": ["图上距离与实际距离的换算", "比例尺的意义"],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_244",
+    "original_id": "unit4_fill_244"
+  },
+  {
+    "id": "unit4_fill_245",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "甲数的 $\\frac{2}{3}$ 等于乙数的 $\\frac{2}{9}$ , 甲: 乙 $6 (\\quad): (\\quad)$ 。若甲数是 30 , 则乙数是()。",
+    "options": [],
+    "answer": "13 ；90",
+    "solution": "【分析】甲数 $\\times \\frac{2}{3} =$ 乙数 $\\times \\frac{2}{9}$ , 根据比例的基本性质, 两内项积等于两外项积, 所以可以写成比例甲: 乙 $= \\frac{2}{9}: \\frac{2}{3}$ , 化简得甲: 乙 $= 1: 3$ , 已知甲数是 30 , 根据比例的基本性质, 即可求出乙数。此题主要考查了比例的基本性质, 要学会灵活运用。\n【详解】由分析可知, 甲: $乙 \\div 1: 3$ , 甲数是 30 , 则$乙数 =3 \\times 3 0 \\div 1 = 9 0$",
+    "knowledge_points": ["比例的基本性质"],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_245",
+    "original_id": "unit4_fill_245"
+  },
+  {
+    "id": "unit4_fill_246",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "困难",
+    "question": "若 $8a \\div 6 = b$ , 则 $a$ 与 $b$ 成 ( )比例关系。若 $x: 4 = 5: y$ , 则 $x$ 与 $y$ 成 ( )比例关系。",
+    "options": [],
+    "answer": "正;反",
+    "solution": "【分析】 $8 \\mathrm{a} \\div 6 = \\mathrm{b}$ 可化为 $8 \\mathrm{a} = 6 \\mathrm{~b}$ , 在变成 $\\frac{a}{b} = \\frac{6}{8}$ , 即 $\\mathrm{a}$ 和 $\\mathrm{b}$ 对应的比值一定, 符合正比例的意义, $\\mathrm{a}$ 与 $\\mathrm{b}$ 成正比例关系; $\\mathrm{x}: 4 = 5: \\mathrm{y}$ , 可化为 $\\mathrm{xy} = 4 \\times 5$ , 即 $\\mathrm{xy} = 20$ , $\\mathrm{x}$ 和 $\\mathrm{y}$ 对应的乘积一定, 符合反比例的意义, $\\mathrm{x}$ 与 $\\mathrm{y}$ 成反比例, 即可解答。本题考查正比例和反比例, 根据正比例意义和反比例的意义进行判断。\n【详解】 $8a \\div 6 = b$ $8 a = 6 b$ $\\frac{a}{b} = \\frac{6}{8}$ , 即 a 与 b 对应的比值一定, 根据正比例的意义, a 与b 成正比例关系; $x: 4 = 5: y$ $x y = 4 \\times 5$ $\\mathrm{xy} = 20$ ，即 $\\mathrm{x}$ 与 $\\mathrm{y}$ 对应的乘积一定，根据反比例的意义， $\\mathrm{x}$ 与 $\\mathrm{y}$ 成反比例关系。",
+    "knowledge_points": ["正比例的意义及辨识", "反比例的意义及辨识"],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_246",
+    "original_id": "unit4_fill_246"
+  },
+  {
+    "id": "unit4_fill_247",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "写出比值是 1.5 的两个比: ( )和( )。组成一个比例是( )。",
+    "options": [],
+    "answer": "$3: 2; 6: 4 ;3: 2 = 6: 4$",
+    "solution": "【分析】依据比例的意义，即表示两个比相等的式子，叫做比例，比值已知，从而可以写出符合要求的比，进而组成比例。本题考查了比的意义, 解答此题的关键是比例的意义, 要注意答案的不唯一性。\n【详解】如果比值为 1.5，可以写出最简单的一个比 $3:2$ ，然后利用比的性质，前项和后项同时扩大 2 倍得到 $6:4$ ，所以组成一个比例是 $3:2 = 6:4$",
+    "knowledge_points": ["比例的基本性质", "比例的意义"],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_247",
+    "original_id": "unit4_fill_247"
+  },
+  {
+    "id": "unit4_fill_248",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "在一个比例式里，两个内项的积是最小的合数，一个外项是 0.4，另一个外项是（ ）；已知其中的一个内项是 0.5，这个比例是（ ）。",
+    "options": [],
+    "answer": "10 ; $0.4:0.5 = 8:10$",
+    "solution": "【分析】首先根据最小的合数是 4 , 可得两个内项的积是 4 ; 再根据比例的基本性质, 可得两个外项的积也是 4 , 用 4 除以 0.4 , 求出另一个外项是多少; 然后用两个内项的积除以其中的一个内项, 求出另一个内项是多少;最后根据比例的基本性质，如果把0.4看作比的一个外项，0.5 看作比的一个内项，那么比的另一个外项是 10，比的另一个内项是 8 , 构造出比例即可。此题主要考查了比例的意义和基本性质的应用，要熟练掌握, 解答此题的关键是要明确, 在比例中, 两个外项的积等于两个内项的积。\n【详解】因为两个内项的积是最小的合数，最小的合数是 4,所以两个内项的积是 4 ,所以两个外项的积也是 4 ,另一个外项是：$4 \\div 0. 4 = 1 0$另一个内项是：$4 \\div 0. 5 = 8$这个比例是：$0.4: 0.5 = 8: 10$ （构造的比例不唯一）。",
+    "knowledge_points": ["比例的基本性质"],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_248",
+    "original_id": "unit4_fill_248"
+  },
+  {
+    "id": "unit4_fill_249",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "如果 $\\frac{M}{8} = N$ , 那么 M 和 N 成( )比例, 如果 $\\frac{8}{M} = N$ , 那么 M 和 N 成( )比例。",
+    "options": [],
+    "answer": "正;反",
+    "solution": "【分析】根据 $x \\div y = k$ （一定）， $x$ 和 $y$ 成正比例关系； $xy = k$ （一定）， $x$ 和 $y$ 成反比例关系，根据运算技巧将 $M$ 和 $N$ 转化到等式的一侧，再看 $M$ 和 $N$ 的关系即可。关键是熟悉正比例和反比例的特点，商一定是正比例，积一定是反比例。\n【详解】 $\\frac{M}{8} = N$\n解: $\\frac{M}{8} \\div N \\times 8 = N \\div N \\times 8$\n$M \\div N = 8, M$ 和 $N$ 成正比例。\n$\\frac {8}{M} = N$\n解: $\\frac{8}{M} \\times M = N \\times M$\n$N \\times M = 8, \\mathrm{M}$ 和 $\\mathrm{N}$ 成反比例。",
+    "knowledge_points": ["正比例的意义及辨识", "反比例的意义及辨识"],
+    "source": "2021 六年级下·全国·专题练习",
+    "merged_id": "unit4_249",
+    "original_id": "unit4_fill_249"
+  },
+  {
+    "id": "unit4_judge_250",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "判断题",
+    "difficulty": "困难",
+    "question": "一个圆锥形的谷堆，它的底面半径是 $8 \\mathrm{~m}$ ，把它画在图上量得半径长为 $3.2 cm$ ，这幅图的比例尺是 $1: 250$ 。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】根据比例尺 = 图上距离 : 实际距离, 一个圆锥形的谷堆, 它的底面半径是 $8 \\mathrm{~m}$ , 把它画在图上量得半径长为 $3.2 cm$ , 即图上距离为 $3.2 cm$ , 实际距离 $8 \\mathrm{~m} = 800 cm$ , 代入公式求得比例尺。本题借助圆锥的半径，实际上考查比例尺的相关知识，关键是牢记比例尺的计算公式。\n【详解】 $8 \\mathrm{~m} = 800 cm$ 这幅图的比例尺是： 3.2:800=1:250故答案为: √",
+    "knowledge_points": ["比例尺的意义"],
+    "source": "2021·陕西渭南·小升初真题",
+    "merged_id": "unit4_250",
+    "original_id": "unit4_judge_250"
+  },
+  {
+    "id": "unit4_judge_251",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "一个机器零件长4毫米，画在图纸上长8厘米，这幅图的比例尺是 $2:1$ 。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】根据比例尺 = 图上距离 : 实际距离, 把数代入并化简即可 (要注意先统一单位)。本题主要考查比例尺的意义，熟练掌握比例尺的意义并灵活运用。\n【详解】8 厘米 $= 80$ 毫米比例尺 $= 80$ 毫米：4毫米 $= 20:1$故答案为: ×",
+    "knowledge_points": ["比例尺的意义"],
+    "source": "19-20六年级下·河南洛阳·期末",
+    "merged_id": "unit4_251",
+    "original_id": "unit4_judge_251"
+  },
+  {
+    "id": "unit4_judge_252",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "判断题",
+    "difficulty": "困难",
+    "question": "由 $2.5 \\times 4=5 \\times 2$ 可以写出比例 $2.5: 2 = 4: 5$ 。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】根据比例的性质，两内项积等于两外项积，转化成比例即可。此题考查了比例的基本性质，要学会灵活运用。\n【详解】因为 $2.5 \\times 4 = 5 \\times 2$ , 如果 2.5 是外项, 则 4 也是外项, 5 和 2 是内项, 则组成的比例是 $2.5:5 = 2:4$ 或 $2.5:2 = 5:4$ , 原题说法错误。故答案为: ×",
+    "knowledge_points": ["比例的基本性质"],
+    "source": "20-21 六年级下·黑龙江大兴安岭地·期中",
+    "merged_id": "unit4_252",
+    "original_id": "unit4_judge_252"
+  },
+  {
+    "id": "unit4_judge_253",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "判断题",
+    "difficulty": "困难",
+    "question": "两个量成正比例,那么一个量变大，另一个量会变小。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】正比例的比值一定，由于两个量的比值一定，那么只要一个量变大，另一个量也会变大，据此解答即可。熟记并理解正比例的意义是解答本题的关键。\n【详解】两个量成正比例，那么一个量变大，另一个量也会变大，原题说法错误；故答案为：×",
+    "knowledge_points": ["正比例的意义及辨识"],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_253",
+    "original_id": "unit4_judge_253"
+  },
+  {
+    "id": "unit4_judge_254",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "判断题",
+    "difficulty": "困难",
+    "question": "在比例尺是 $\\frac{1}{16000000}$ 的地图上，是用图上距离1厘米表示实际距离160千米。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】根据'实际距离 $=$ 图上距离 $\\div$ 比例尺'，代入数值计算即可。理清图上距离、比例尺和实际距离三者的关系以及单位的统一是关键。\n【详解】 $1 \\div \\frac{1}{16000000} = 16000000$ （厘米） $16000000$ 厘米 $= 160$ 千米图上距离1厘米表示实际距离160千米。故答案:√ 。",
+    "knowledge_points": ["图上距离与实际距离的换算"],
+    "source": "14-15六年级下·宁夏石嘴山·期中",
+    "merged_id": "unit4_254",
+    "original_id": "unit4_judge_254"
+  },
+  {
+    "id": "unit4_judge_255",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "在比例里, 两个外项的积等于两个内项的积。( )",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】根据比例的基本性质进行判断。\n【详解】在比例里，两个外项的积等于两个内项的积，这叫比例的基本性质，所以原题说法正确。关键是掌握比例的基本性质。",
+    "knowledge_points": ["比例的基本性质"],
+    "source": "16-17 六年级下·甘肃·课后作业",
+    "merged_id": "unit4_255",
+    "original_id": "unit4_judge_255"
+  },
+  {
+    "id": "unit4_judge_256",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "判断题",
+    "difficulty": "困难",
+    "question": "把一个圆按 5:1 放大，就是把这个圆的面积扩大到原来的 5 倍。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】因为把一个圆按 5:1 放大, 所以一个圆周长扩大 5 倍, 半径也扩大 5 倍, 而根据圆的面积 $S = \\pi r^{2}$ 可得, 圆的面积与半径的平方成正比, 所以当圆的半径扩大 5 倍, 面积扩大 $5^{2}$ 倍, 由此得出答案。此题主要是利用圆的周长与半径、面积的关系及积的变化规律解决问题。\n【详解】把一个圆按 5:1 放大, 圆的半径扩大到原来的 5 倍,面积扩大到原来的 $5^{2} = 25$ 倍, 原题说法错误。故答案为: × 。",
+    "knowledge_points": ["图形的放大与缩小"],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_256",
+    "original_id": "unit4_judge_256"
+  },
+  {
+    "id": "unit4_judge_257",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "判断题",
+    "difficulty": "困难",
+    "question": "在比例尺为1:4000000 的地图上量得南京到扬州的距离约为 3.6 厘米,南京到扬州的实际距离大约是 144 千米。( )",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】要求南京到扬州的实际距离是多少千米，根据'图上距离÷比例尺 = 实际距离'，代入数值计算即可。熟记图上距离、比例尺和实际距离三者的关系是解答本题的关键。\n【详解】 $3.6 \\div \\frac{1}{4000000} = 14400000$ （厘米）；14400000 厘米 = 144 千米，原题说法正确。故答案为: 。√",
+    "knowledge_points": ["比例尺应用", "图上距离与实际距离的换算"],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_257",
+    "original_id": "unit4_judge_257"
+  },
+  {
+    "id": "unit4_judge_258",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "判断题",
+    "difficulty": "困难",
+    "question": "一个圆形, 按 $3: 1$放大后，周长扩大到原来的3倍，面积扩大到原来的9倍。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】因为圆的周长和半径成正比, 所以一个圆周长扩大 3 倍, 半径也扩大 3 倍, 而根据圆的面积 $S = \\pi r^{2}$ 可得, 圆的面积与半径的平方成正比, 所以当圆的半径扩大 3 倍, 面积扩大 $3^{2}$ 倍, 由此得出答案。此题主要是利用圆的周长与半径、面积的关系及积的变化规律解决问题。\n【详解】当一个圆的周长扩大到原来的 3 倍, 圆的半径扩大 3 倍, 面积扩大 $3^{2} = 9$ 倍, 原题说法正确; 故答案为: √",
+    "knowledge_points": ["图形的放大与缩小", "圆的面积及应用", "圆的周长及应用"],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_258",
+    "original_id": "unit4_judge_258"
+  },
+  {
+    "id": "unit4_judge_259",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "判断题",
+    "difficulty": "困难",
+    "question": "正比例与反比例的图象都是一条直线。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】根据正、反比例的意义，正比例是相对应的两个数的比值一定，反比例是相对应的两个数的乘积一定，解答即可。此题考查的目的是理解掌握正、反比例的意义及应用。\n【详解】正比例的图象的一条直线，而反比例的图象是曲线。原题说法是错误的。故答案为: ×",
+    "knowledge_points": ["正比例的意义及辨识", "反比例的意义及辨识"],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_259",
+    "original_id": "unit4_judge_259"
+  },
+  {
+    "id": "unit4_judge_260",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "判断题",
+    "difficulty": "困难",
+    "question": "一幅图的比例尺是 $1:5000 \\mathrm{~m}$ 。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】比例尺是一个比，是没有单位的。据此判断即可。本题考查了比例尺, 明确比例尺的概念和特点是解题的关键。\n【详解】比例尺是一个比，表示图上距离与实际距离的倍比关系，不能加单位名称。所以判断错误。",
+    "knowledge_points": ["比例尺的意义"],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_260",
+    "original_id": "unit4_judge_260"
+  },
+  {
+    "id": "unit4_judge_261",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "判断题",
+    "difficulty": "困难",
+    "question": "把一条 $20 \\mathrm{~m}$ 长的绳子, 剪去一部分, 剪去的和剩下的成反比例关系。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】剪去的和剩下的绳子，两者是和一定，积不一定。据此判断即可。牢记并理解反比例关系的判断方法：两种相关联的量，一种量变化，另一种量也随着变化，这两种量中对应的两个数的积一定，两种量是成反比例的量，它们成反比例关系。\n【详解】只有两个数的积一定时，才可能存在反比例关系。剪去的和剩下的绳子只有和的关系，所以二者不存在反比例关系。所以判断错误。",
+    "knowledge_points": ["反比例的意义及辨识"],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_261",
+    "original_id": "unit4_judge_261"
+  },
+  {
+    "id": "unit4_judge_262",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "若 $5 \\mathrm{x} = 6 \\mathrm{y}$ , 则 $\\mathrm{x}: \\mathrm{y} = 5: 6$ 。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】根据比例的基本性质，在改写比例时， $\\mathrm{x}$ 作外项，和 $\\mathrm{x}$ 相乘的 5 一定也作外项。据此判断即可。把等式 $a x = b y$ (a, b, x, y 均不为 0) 改写成比例时, 相乘的 2 个字母必须同时作比例的外项或内项。\n【详解】若 $5 \\mathrm{x} = 6 \\mathrm{y}$ , 则 $\\mathrm{x}: \\mathrm{y} = 6: 5$ 故答案为: ×",
+    "knowledge_points": ["比例的基本性质", "比例的意义"],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_262",
+    "original_id": "unit4_judge_262"
+  },
+  {
+    "id": "unit4_judge_263",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "判断题",
+    "difficulty": "困难",
+    "question": "$10: 2 = 5$ 是比例。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】根据比例的定义，直接判断题干的正误即可。本题考查了比例的定义, 属于概念性题目, 明确比例的定义和特点是解题的关键。\n【详解】比例等号的左右两边都必须是一个比, 所以 $10: 2 = 5$ 不是比例。所以判断错误。",
+    "knowledge_points": ["比例的意义"],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_263",
+    "original_id": "unit4_judge_263"
+  },
+  {
+    "id": "unit4_judge_264",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "判断题",
+    "difficulty": "困难",
+    "question": "三角形的面积一定，它的底与高成正比例。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】三角形的底 $\\times$ 高 $=$ 面积 $\\times 2$ (一定), 乘积一定, 说明底与高成反比例关系, 据此解答即可。明确正反比例的意义和区别是解答本题的关键，两个相关联的量，乘积一定时，成反比例关系；比值一定时，成正比例关系。\n【详解】三角形的底 $\\times$ 高 $=$ 面积 $\\times 2$ （一定），所以三角形的底与高成反比例关系，原题说法错误；故答案为：×。",
+    "knowledge_points": ["正比例的意义及辨识", "反比例的意义及辨识"],
+    "source": "2014 六年级·全国·课后作业",
+    "merged_id": "unit4_264",
+    "original_id": "unit4_judge_264"
+  },
+  {
+    "id": "unit4_judge_265",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "车轮周长一定, 车轮行驶的路程和转数成正比例关系。( )",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】根据 $x \\div y = k$ （一定）， $x$ 和 $y$ 成正比例关系，进行辨识即可。关键是理解正比例的意义，商一定是正比例关系。\n【详解】车轮行驶的路程 $\\div$ 转数 = 车轮周长（一定），车轮行驶的路程和转数成正比例关系，所以原题说法正确。",
+    "knowledge_points": ["正比例的意义及辨识"],
+    "source": "19-20 六年级下·河南南阳·期末",
+    "merged_id": "unit4_265",
+    "original_id": "unit4_judge_265"
+  },
+  {
+    "id": "unit4_judge_266",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "如果长方形的面积一定，则长方形的长和宽成反比例。（）",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】根据 $x y = k$ (一定), $x$ 和 $y$ 成反比例关系, 进行分析。关键是理解反比例的意义，积一定是反比例关系。\n【详解】长方形的长 $\\times$ 宽 = 面积（一定），长和宽成反比例，所以原题说法正确。",
+    "knowledge_points": ["反比例的意义及辨识"],
+    "source": "2016·全国·小升初真题",
+    "merged_id": "unit4_266",
+    "original_id": "unit4_judge_266"
+  },{
+    "id": "unit4_judge_267",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "判断题",
+    "difficulty": "困难",
+    "question": "看一本红楼梦, 平均每天读的页数和读的天数成反比例。( )",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】根据题意可知，红楼梦的总页数是固定不变的，'平均每天读的页数 $\\times$ 读的天数 = 总页数（一定）' 据此解答即可。明确成反比例关系的两个量乘积一定是解答本题的关键。\n【详解】'平均每天读的页数 $\\times$ 读的天数 = 总页数（一定）'两个相关联的量乘积一定，所以成反比例关系，原题说法正确；故答案为: √。",
+    "knowledge_points": ["反比例的意义及辨识"],
+    "source": "2021 六年级下·全国·专题练习",
+    "merged_id": "unit4_267",
+    "original_id": "unit4_judge_267"
+  },
+  {
+    "id": "unit4_calculate_268",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "解比例。\n$\\frac {6 . 5}{x} = \\frac {3 . 2 5}{8} \\quad \\frac {2}{3}: \\frac {1}{4} = \\frac {1}{5}: x \\\\ 7 \\frac {1}{2}: x = 3: 4 \\quad 0.375: x = \\frac {1}{20}: 60 \\% $",
+    "options": [],
+    "answer": "$x = 16; x = \\frac{3}{40}; x = 10; x = 4.5$",
+    "solution": "【分析】（1）根据比例的基本性质，原式化成 $3.25 \\mathrm{x} = 6.5 \\times 8$ ，再根据等式的性质，方程两边同时除以 3.25 求解；(2) 根据比例的基本性质, 原式化成 $\\frac{2}{3} x = \\frac{1}{4} \\times \\frac{1}{5}$ , 再根据等式的性质, 方程两边同时除以 $\\frac{2}{3}$ 求解; (3) 根据比例的基本性质, 原式化成 $3 \\mathrm{~x} = 7 \\frac{1}{2} \\times 4$ , 再根据等式的性质, 方程两边同时除以 3 求解;(4) 根据比例的基本性质, 原式化成 $\\frac{1}{20} \\mathrm{x} = 0.375 \\times 60 \\%$ , 再根据等式的性质, 方程两边同时除以 $\\frac{1}{20}$ 求解。\n【详解】 $\\frac{6.5}{x} = \\frac{3.25}{8}$\n解: $3.25 \\mathrm{x} = 6.5 \\times 8$\n$3.25x \\div 3.25 = 52 \\div 3.25$\n$x = 16;$\n$\\frac {2}{3}: \\frac {1}{4} = \\frac {1}{5}: x$\n解: $\\frac{2}{3} x = \\frac{1}{4} \\times \\frac{1}{5}\\\\ \\frac {2}{3} x \\div \\frac {2}{3} = \\frac {1}{20} \\div \\frac {2}{3}\\\\x = \\frac {3}{40};$\n$7 \\frac {1}{2}: x = 3: 4$\n解: $3 x = 7 \\frac{1}{2} \\times 4\\\\3 x \\div 3 = 30 \\div 3\\\\x = 10$\n$0.375: x = \\frac{1}{20}:60\\%$\n解: $\\frac{1}{20} \\mathrm{x} = 0.375 \\times 60 \\%\\\\\\frac {1}{20} x \\div \\frac {1}{20} = 0.225 \\div \\frac {1}{20} \\\\ x = 4.5$",
+    "knowledge_points": ["解比例"],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_268",
+    "original_id": "unit4_calculate_268"
+  },
+  {
+    "id": "unit4_calculate_269",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "求未知数 $x$ 。\n$\\mathrm {x} + 20 \\% \\mathrm {x} = 0.45 \\quad 72 - 4 \\mathrm {x} = 60 \\quad \\mathrm {x}: 2.5 = 4: \\frac {5}{7}$",
+    "options": [],
+    "answer": "$x = 0.375; x = 3; x = 14$",
+    "solution": "【分析】（1）先化简方程的左边，变成 $1.2 \\mathrm{x} = 0.45$ ，再把方程两边同时除以 1.2 即可；(2) 先把方程的两边同时加 $4 \\mathrm{~x}$ , 再同时减去 60 , 最后同时除以 4 即可; (3) 先根据比例的性质把比例方程转化成简易方程, 再把方程的两边同时除以 $\\frac{5}{7}$ 即可。\n【详解】 (1) $x + 20\\% x = 0.45$\n解: $1.2 \\mathrm{~x} = 0.45\\\\ 1.2 \\mathrm {x} \\div 1.2 = 0.45 \\div 1.2\\\\ x = 0.375$\n(2) $72 - 4x = 60$\n解： $72 - 4\\mathrm{x} + 4\\mathrm{x} = 60 + 4\\mathrm{x}\\\\4 x + 60 - 60 = 72 - 60\\\\4 x = 12\\\\4 x \\div 4 = 12 \\div 4\\\\x = 3$\n(3) $x: 2.5 = 4: \\frac{5}{7}$\n解: $\\frac{5}{7} x = 2.5 \\times 4\\\\\\frac {5}{7} x = 10\\\\\\frac {5}{7} x \\div \\frac {5}{7} = 10 \\div \\frac {5}{7}\\\\x = 14$",
+    "knowledge_points": ["解比例", "应用等式的性质 1 和 2 解方程"],
+    "source": "20-21 六年级下·辽宁·单元测试",
+    "merged_id": "unit4_269",
+    "original_id": "unit4_calculate_269"
+  },
+  {
+    "id": "unit4_calculate_270",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "解方程。\n$\\frac {0.5}{3} = \\frac {9}{x}$",
+    "options": [],
+    "answer": "$x = 54$",
+    "solution": "【分析】根据比例的基本性质，两外项之积等于两内项之积，据此先将分数方程写成一般方程，再解方程即可。\n【详解】 $\\frac{0.5}{3} = \\frac{9}{x}$\n解: $0.5 \\mathrm{x} = 3 \\times 9\\\\x = \\frac {3 \\times 9}{0.5} \\\\ x = 54$",
+    "knowledge_points": ["解比例", "应用等式的性质 2 解方程"],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_270",
+    "original_id": "unit4_calculate_270"
+  },
+  {
+    "id": "unit4_calculate_271",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "计算题",
+    "difficulty": "困难",
+    "question": "求未知数 $x$ 。\n$2 x - 1.2 = 4.8\\\\x: 1.2 = 3: 4\\\\\\frac {3}{4} x - 0.25x = \\frac {5}{8}$",
+    "options": [],
+    "answer": "$x = 3; x = 0.9; x = \\frac{5}{4}$",
+    "solution": "【分析】第一题方程左右两边同时加上 1.2，将其转化为 $2 \\mathrm{~x} = 6$ ，再左右两边同时除以 2 即可；第二题根据比例的基本性质可知 $4 \\mathrm{~x} = 1.2 \\times 3$ , 再左右两边同时除以4即可;第三题先化简方程为 $\\frac{1}{2} x = \\frac{5}{8}$ , 再左右两边同时除以 $\\frac{1}{2}$ 即可。\n【详解】 $2 \\mathrm{x} - 1.2 = 4.8$\n解: $2 \\mathrm{x} - 1.2 + 1.2 = 4.8 + 1.2\\\\2 x = 6\\\\2 x \\div 2 = 6 \\div 2\\\\x = 3;$\n$x: 1.2 = 3: 4$\n解: $4 \\mathrm{x} = 1.2 \\times 3\\\\4 x \\div 4 = 3.6 \\div 4\\\\x = 0.9;$\n$\\frac {3}{4} x - 0.25x = \\frac {5}{8}$\n解: $\\frac{1}{2} x = \\frac{5}{8}\\\\\\frac {1}{2} x \\div \\frac {1}{2} = \\frac {5}{8} \\div \\frac {1}{2}\\\\x = \\frac {5}{4}$",
+    "knowledge_points": ["解比例", "应用等式的性质 1 和 2 解方程"],
+    "source": "19-20 六年级下·陕西宝鸡·期末",
+    "merged_id": "unit4_271",
+    "original_id": "unit4_calculate_271"
+  },
+  {
+    "id": "unit4_calculate_272",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "求未知数 $x$ 。\n$\\frac {4}{7} \\mathrm {x} = 0.4 \\mathrm {x} + 12 \\quad \\frac {1}{2} \\mathrm {x} + \\frac {2}{3} \\mathrm {x} = \\frac {7}{10} \\quad 0.8: \\mathrm {x} = \\frac {4}{3}: 0.2$",
+    "options": [],
+    "answer": "$x = 70; x = \\frac{3}{5}; x = 0.12$",
+    "solution": "【分析】根据等式的性质以及解比例的方法，直接计算即可。\n【详解】 $\\frac{4}{7} x = 0.4 x + 12$\n解: $\\frac{4}{7} x - \\frac{2}{5} x = 12$\n$\\frac {6}{35} x = 12\\\\x = 12 \\div \\frac {6}{35}\\\\x = 70;$\n$\\frac {1}{2} x + \\frac {2}{3} x = \\frac {7}{10}$\n解: $\\frac{7}{6} x = \\frac{7}{10}\\\\x = \\frac {7}{10} \\div \\frac {7}{6}\\\\x = \\frac {3}{5};$\n$0.8: x = \\frac {4}{3}: 0.2$\n解: $\\frac{4}{3} \\mathrm{x} = 0.8 \\times 0.2\\\\\\frac {4}{3} x = 0.16 \\\\ x = 0.16 \\div \\frac {4}{3} \\\\ x = 0.12$",
+    "knowledge_points": ["解比例", "应用等式的性质 1 和 2 解方程"],
+    "source": "18-19 六年级下·陕西商洛·期末",
+    "merged_id": "unit4_272",
+    "original_id": "unit4_calculate_272"
+  },
+  {
+    "id": "unit4_calculate_273",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "解方程。\n$x + \\frac {1}{5} x = 186$\n$5 x - \\frac {5}{6} = \\frac {5}{12}$\n$12: 16 = \\frac {1}{4}: x$",
+    "options": [],
+    "answer": "$x = 155; x = \\frac{1}{4}; x = \\frac{1}{3}$",
+    "solution": "【分析】根据比例的基本性质把解比例转化为解方程; 根据等式的性质解方程。等式的性质 1: 等式两边同时加上或减去同一个数, 等式仍然成立;等式的性质 2: 等式两边同时乘或除以同一个不是零的数,等式仍然成立;注意写'解'字和'='要对齐。\n【详解】 $x + \\frac{1}{5} x = 186$\n解: $\\frac{6}{5} x = 186\\\\x = 186 \\div \\frac {6}{5}\\\\x = 186 \\times \\frac {5}{6}\\\\x = 155$\n$5 x - \\frac {5}{6} = \\frac {5}{12}$\n解: $5 x=\\frac{5}{12} + \\frac{5}{6}\\\\5 x= \\frac {5}{4}\\\\x =\\frac {1}{4}$\n$12: 16 = \\frac {1}{4}: x$\n解: $12 \\mathrm{~x} = 16 \\times \\frac{1}{4}\\\\12 x = 4\\\\x = \\frac {1}{3}$",
+    "knowledge_points": ["解比例", "应用等式的性质 1 和 2 解方程"],
+    "source": "19-20 六年级下·陕西宝鸡·期末",
+    "merged_id": "unit4_273",
+    "original_id": "unit4_calculate_273"
+  },
+  {
+    "id": "unit4_calculate_274",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "已知 $a:b = 3:4$ ,$b:c=4:2\\frac{2}{3}$ , 求 $a:b:c$ 。",
+    "options": [],
+    "answer": "$9: 12: 8$",
+    "solution": "【分析】因为 $a:b = 3:4, b:c = 4:2 \\frac{2}{3}$ , 两比例式中 $b$ 的值相同, 所以直接比即可。\n【详解】a:b=3:4，b:c=4:$2\\frac{2}{3}$\n$\\mathrm {a : b : c} \\\\ = 3: 4: 2\\frac {2}{3} \\\\ = 3: 4: \\frac {8}{3} \\\\ = 9: 12: 8$",
+    "knowledge_points": ["比例的基本性质"],
+    "source": "19-20 六年级上·上海闵行·期末",
+    "merged_id": "unit4_274",
+    "original_id": "unit4_calculate_274"
+  },
+  {
+    "id": "unit4_calculate_275",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "计算题",
+    "difficulty": "中等",
+    "question": "求未知数 $x$ 。\n$4 + 0.7 x = 102 \\quad 3 x - \\frac {3}{2} x = 10 \\quad x: \\frac {5}{14} = 21: \\frac {5}{8}$",
+    "options": [],
+    "answer": "$x = 140; x = \\frac{20}{3}; x = 12$",
+    "solution": "【分析】 $4 + 0.7 \\mathrm{x} = 102$ 根据等式的性质 1 , 两边同时减去 4 之后得的结果再根据等式的性质 2 两边同时除以 0.7 即可算出 $x$的值；3x - $\\frac{3}{2} x = 10$ 利用乘法分配律把 $3 x - \\frac{3}{2} x$ 化简下, 化简后是 $\\frac{3}{2} x = 10$ , 之后再两边同时除以 $\\frac{3}{2}$ 即可;$\\mathrm{x}: \\frac{5}{14} = 21: \\frac{5}{8}$ 利用比例的基本性质, 外项积 $=$ 内项积, 即 $\\frac{5}{8} \\mathrm{x} = 21 \\times \\frac{5}{14}$ , 算出 $21 \\times \\frac{5}{14}$ 的结果, 两边再同时除以 $\\frac{5}{8}$ 即可。本题主要考查解方程的计算，熟练掌握等式的性质 1 和等式的性质 2 解方程，同时关于比例的基本性质，内项积 = 外项积。\n【详解】 $4 + 0.7 \\mathrm{x} = 102$\n解: $0.7 \\mathrm{x} = 102 - 4\\\\0.7 x = 98\\\\x = 98 \\div 0.7\\\\x = 140$\n$3 x - \\frac {3}{2} x = 10$\n解: $\\frac{3}{2} x = 10\\\\x = 10 \\div \\frac {3}{2}\\\\x = \\frac {20}{3}$\n$x: \\frac {5}{14} = 21: \\frac {5}{8}$\n解: $\\frac{5}{8} x = 21 \\times \\frac{5}{14}\\\\\\frac {5}{8} x = \\frac {15}{2} \\\\ x = \\frac {15}{2} \\div \\frac {5}{8} \\\\ x = 12$",
+    "knowledge_points": ["解比例", "应用等式的性质1和2解方程"],
+    "source": "20-21 六年级下·浙江嘉兴·期末",
+    "merged_id": "unit4_275",
+    "original_id": "unit4_calculate_275"
+  },
+  {
+    "id": "unit4_solve_276",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "小兰的身高是 $1.5 \\mathrm{~m}$ , 她的影子长是 $2.4 \\mathrm{~m}$ 。如果同一时间, 同一地点测得一棵树的影子长 $4 \\mathrm{~m}$ , 这棵树有多高?",
+    "options": [],
+    "answer": "2.5 米",
+    "solution": "【分析】已知小兰的身高是 $1.5 \\mathrm{~m}$ , 影子长为 $2.4 \\mathrm{~m}$ ; 且同一时间、同一地点测得一棵树的影子长 $4 \\mathrm{~m}$ , 则要求这棵树有多高, 可假设这棵树 $x \\mathrm{~m}$ 高, 列方程为: $x: 4 = 1.5: 2.4$ 。解答本题的依据是：同一时间、同一地点，物体的身高和影长成正比例，故可按正比例关系列方程。\n【详解】解：设这棵树高 $\\mathrm{xm}$ ，由题意得，\n$x: 4 = 1.5: 2.4 \\\\ 2.4 \\mathrm {x} = 1.5 \\times 4 \\\\ 2.4 x = 6 \\\\ x = 2.5$\n答: 这棵树有 2.5 米高。",
+    "knowledge_points": ["比例的应用"],
+    "source": "2015·天津静海·小升初真题",
+    "merged_id": "unit4_276",
+    "original_id": "unit4_solve_276"
+  },
+  {
+    "id": "unit4_solve_277",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "网通公司为光明小区安装电话，如果每天安装25部，18天可以装完。如果想提前3天完成，平均每天要多装多少部？",
+    "options": [],
+    "answer": "5 部",
+    "solution": "【分析】根据题意可知，这些电话机的总部数一定，也就是每天安装的部数与所用的天数的积一定，因此每天安装的部数与所用天数成反比例，可以设平均每天安装 $x$ 部可以提前 3 天完成任务，即 $(18 - 3) x = 25 \\times 18$ ，由此解方程即可，求出 $x$ 再减去原来每天安装的部数即可知道平均每天要多装多少部。本题主要考查比例应用题, 解题的关键是判断题目中相关联的两个量成什么比例, 乘积一定是反比例, 比值一定是正比例, 由此解答。\n【详解】解：设平均每天安装 $x$ 部可以提前 3 天完成任务。\n$(18 - 3) \\mathrm {x} = 25 \\times 18\\\\15 x = 450\\\\x = 450 \\div 15\\\\x = 30$\n$30 - 25 = 5 \\text {(部)}$答: 提前 3 天完成任务, 平均每天要多装 5 部。",
+    "knowledge_points": ["列方程解含一个未知数的问题", "反比例的应用"],
+    "source": "2022·甘肃陇南·小升初真题",
+    "merged_id": "unit4_277",
+    "original_id": "unit4_solve_277"
+  },
+  {
+    "id": "unit4_solve_278",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "在比例尺是1:6000000的地图上，量得两地相距6厘米，一列客车和一列慢车同时从两地出发相向而行，2小时相遇。已知客车与货车的速度比为5:4，货车每小时行多少千米？",
+    "options": [],
+    "answer": "80千米",
+    "solution": "【分析】两地的实际距离 $=$ 两地的图上距离 $\\div$ 比例尺, 然后进行单位换算, 即 1 千米 $= 100000$ 厘米, 那么货车每小时行的距离 $=$ 两地的实际距离 $\\div$ 相遇经过的时间 $\\times$货车速度所占的份数 $\\frac{1}{\\text {辆 车 速 度 占 的 份 数 和 }}$ ，据此代入数据作答即可。根据实际距离、图上距离和比例尺之间的关系求出实际距离是解答本题的关键，进而求出两车的速度和，再根据按比例分配的知识点解答即可。\n【详解】 $6 \\div \\frac{1}{6000000} = 36000000$ （厘米） $= 360$ 千米\n$360 \\div 2 \\times \\frac {4}{4 + 5} \\\\ = 180 \\times \\frac {4}{9} \\\\ = 80 \\text {(千 米)}$\n答: 货车每小时行 80 千米。",
+    "knowledge_points": ["图上距离与实际距离的换算", "按比分配问题"],
+    "source": "2024·河北张家口·小升初真题",
+    "merged_id": "unit4_278",
+    "original_id": "unit4_solve_278"
+  },
+  {
+    "id": "unit4_solve_279",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一间教室，用面积是 $0.16 \\mathrm{~m}^{2}$ 的方砖铺地，需要 275 块，如果改用边长是 $0.5 \\mathrm{~m}$ 的方砖铺地，需要方砖多少块？（用比例解答）",
+    "options": [],
+    "answer": "176 块",
+    "solution": "【分析】方砖的面积 $= \\text{边长} \\times$ 边长, 等量关系式: 第二种方砖的面积 $\\times$ 需要方砖的块数 = 第一种方砖的面积 $\\times$ 需要方砖的块数, 据此列方程解答。分析题意找出等量关系式是解答题目的关键。\n【详解】解：设需要方砖 $x$ 块。\n$(0.5 \\times 0.5) x = 0.16 \\times 275\\\\0.25 \\mathrm{x} = 44\\\\ \\mathrm{x} = 44 \\div 0.25\\\\ \\mathrm{x} = 176$\n答: 需要方砖 176 块。",
+    "knowledge_points": ["反比例的应用"],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_279",
+    "original_id": "unit4_solve_279"
+  },
+  {
+    "id": "unit4_solve_280",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "在一幅比例尺是1:2000 的平面图上，量得一块长方形地的长是 6 厘米，宽是 4 厘米。这块长方形地的实际面积是多少平方米？",
+    "options": [],
+    "answer": "9600 平方米",
+    "solution": "【分析】已知比例尺和图上距离，要求实际距离，依据图上距离 $\\div$ 比例尺 $=$ 实际距离，最后用实际的长 $\\times$ 宽= 实际的长方形面积，据此列式解答。此题主要考查了比例尺的应用，牢记图上距离、实际距离和比例尺之间的关系是解答的关键。\n【详解】长: $6 \\div \\frac{1}{2000} = 12000$ (厘米) 12000 厘米=120 米 宽: $4 \\div \\frac{1}{2000} = 8000$ (厘米) 8000 厘米 = 80 米 $120 \\times 80 = 9600$ （平方米）答: 这块长方形地的实际面积是 9600 平方米。",
+    "knowledge_points": ["长方形的面积", "图上距离与实际距离的换算", "厘米和米之间的进率与换算"],
+    "source": "20-21六年级下·全国·单元测试",
+    "merged_id": "unit4_280",
+    "original_id": "unit4_solve_280"
+  },
+  {
+    "id": "unit4_solve_281",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一棵树高 12 米, 它的影长是 15 米，如果同一时间地点测得小明的身高是 1.6米，它的影子长多少米？（用比例解答）",
+    "options": [],
+    "answer": "2 米",
+    "solution": "【分析】树高：它的影长 $=$ 小明的身高：它的影子长，根据比例的基本性质解比例。熟悉比例的定义与比例的基本性质是解决此题的关键。\n【详解】 $12: 15 = 1.6:x\\\\12 x = 15 \\times 1.6 \\\\ 12 x = 24 \\\\ x = 24 \\div 12 \\\\ x = 2$\n答: 它的影子长 2 米。",
+    "knowledge_points": ["比例的应用"],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_281",
+    "original_id": "unit4_solve_281"
+  },
+  {
+    "id": "unit4_solve_282",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "在一个比例尺为1:1000000 的地图上，量得甲乙两地的距离是 $20 cm$ ，一辆汽车以每小时80千米的速度从甲地开往乙地需要多长时间？",
+    "options": [],
+    "answer": "2.5小时",
+    "solution": "【分析】根据图上距离÷比例尺 = 实际距离，求出实际距离，再根据路程 $\\div$ 速度 $=$ 时间，列式解答。此题主要考查比例尺的意义及已知比例尺和图上距离求实际距离，解题时注意单位的换算。\n【详解】 $20 \\div \\frac{1}{1000000}\\\\=20000000 (厘米)$\n20000000 厘米 = 200 千米 $200 \\div 80 = 2.5$ （小时） 答: 从甲地开往乙地需要 2.5 小时。",
+    "knowledge_points": ["比例尺应用"],
+    "source": "20-21 六年级下·广东河源·期中",
+    "merged_id": "unit4_282",
+    "original_id": "unit4_solve_282"
+  },
+  {
+    "id": "unit4_solve_283",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一辆汽车上午4小时行驶 320 千米，下午 3 小时行驶 240 千米。(1) 上午行驶的路程和时间的比是几比几? 下午呢? 这两个比能组成比例吗？为什么？(2) 上下午行驶路程的比和上下午行驶时间的比也能组成比例吗？",
+    "options": [],
+    "answer": "（1）80:1；80:1；这两个比能组成比例，320:4=240:3，因为它们之比都是 $80: 1$ 。(2）能",
+    "solution": "【分析】（1）先分别表示出上午、下午行驶的路程和时间的比是几比几，再判断这两个比能不能组成比例； \n(2) 先分别表示出上下午行驶路程的比和上下午行驶时间的比, 再判断这两个比能不能组成比例。此题考查了根据比例意义判断两个比能不能组成比例。\n【详解】（1）上午行驶的路程和时间的比是 $320:4 = 80:1$ ;下午行驶的路程和时间的比是 $240: 3 = 80: 1$ ;这两个比能组成比例, $320: 4 = 240: 3$ , 因为它们之比都是 $80: 1$ ;(2) 路程比是 $320: 240 = 4: 3$ ;时间比是 $4: 3$ ;即也能组成比例;",
+    "knowledge_points": ["比例的意义"],
+    "source": "20-21六年级下·全国·单元测试",
+    "merged_id": "unit4_283",
+    "original_id": "unit4_solve_283"
+  },
+  {
+    "id": "unit4_solve_284",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "在一幅比例尺是1:2000000的地图上，量得甲、乙两个城市之间高速公路的距离是 $2.5 cm$ ，在另一幅比例尺是 $1:5000000$ 的地图上，这条公路的图上距离是多少？",
+    "options": [],
+    "answer": "1 厘米",
+    "solution": "【分析】根据第一幅图的比例尺和甲乙间高速公路的图上距离，先计算出甲乙间高速公路的实际距离，再根据第二幅图的比例尺，计算出这条公路的图上距离。本题考查了比例尺, 比例尺等于图上距离比实际距离。\n【详解】实际距离： $2.5 \\div \\frac{1}{2000000} = 5000000$ （厘米）图上距离： $5000000 \\times \\frac{1}{5000000} = 1$ （厘米）答：这条公路的图上距离是 1 厘米。",
+    "knowledge_points": ["图上距离与实际距离的换算"],
+    "source": "19-20 六年级下·河北唐山·期中",
+    "merged_id": "unit4_284",
+    "original_id": "unit4_solve_284"
+  },
+  {
+    "id": "unit4_solve_285",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "某工程队修一条公路，6 天铺了 $228 \\mathrm{~m}$ 。照这样计算，还要用 12 天完成全部的工程，这条公路一共长多少米？",
+    "options": [],
+    "answer": "$684 \\mathrm{~m}$",
+    "solution": "【分析】先计算出修完这条公路一共需要的时间，再根据题意找出修路时间和修路长的比例关系，据此列比例解比例即可。本题考查了比例的应用, 能找出修路时间和修路长的比例关系是解题的关键。\n【详解】解：设这条公路一共长 $x$ 米。\n$\\frac {228}{6} = \\frac {x}{12 + 6}\\\\x = \\frac {228 \\times 18}{6}\\\\x = 684$\n答: 这条公路一共长 $684 \\mathrm{~m}$ 。",
+    "knowledge_points": ["解比例", "比例的应用"],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_285",
+    "original_id": "unit4_solve_285"
+  },
+  {
+    "id": "unit4_solve_286",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "在比例尺为 $1: 1000$ 的图纸上，正方体的棱长是 $5 cm$ 。这个正方体的体积是多少立方米？",
+    "options": [],
+    "answer": "$125000 \\mathrm{~m}^{3}$",
+    "solution": "【分析】根据比例尺先将这个正方体的实际棱长计算出来,再根据正方体的体积公式求出它的体积即可。本题考查了比例尺和正方体的体积, 比例尺等于图上距离比实际距离, 正方体的体积等于棱长乘棱长乘棱长。\n【详解】 $5 \\div \\frac{1}{1000} = 5000 \\text{(cm)}$$5000 \\times 5000 \\times 5000 = 125000000000 (\\mathrm {c m} ^ {3}) = 125000 (\\mathrm {m} ^ {3})$答: 这个正方体实际的体积是 $125000 \\mathrm{~m}^{3}$ 。",
+    "knowledge_points": ["正方体的体积", "比例尺应用"],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_286",
+    "original_id": "unit4_solve_286"
+  },
+  {
+    "id": "unit4_solve_287",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "甲、乙两地的实际距离是 $7500 \\mathrm{~m}$ , 在比例尺为 $1: 30000$ 的地图上, 两地的图上距离是多少厘米?",
+    "options": [],
+    "answer": "$25 cm$",
+    "solution": "【分析】先统一图上和实际的距离单位，实际距离应先转化为以厘米为单位的数，再代入公式求图上距离。在用比例尺解决实际问题时, 一定要将图上距离和实际距离的单位统一。\n【详解】 $7500 \\mathrm{~m} = 750000 cm$ $750000 \\times \\frac{1}{30000} = 25 \\text{(cm)}$答: 两地的图上距离为 $25 cm$ 。",
+    "knowledge_points": ["比例尺应用", "图上距离与实际距离的换算", "分数与整数的乘法"],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_287",
+    "original_id": "unit4_solve_287"
+  },
+  {
+    "id": "unit4_solve_288",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "在比例尺是1:5000000的地图上，量得 A、B 两地的距离是 16 厘米，甲、乙两列火车同时从 A、B 两地同时出发，相向而行。5 小时后相遇，已知甲列火车每小时行 90 千米，乙列火车每小时行驶多少千米？",
+    "options": [],
+    "answer": "70千米",
+    "solution": "【分析】先根据比例尺, 求出两地之间的实际距离, 再根据路程 $\\div$ 时间 = 速度和, 用两地距离除以相遇时间求出甲、乙两车速度和, 进而求出乙火车的速度。关键是理解速度、时间、路程之间的关系，实际距离 = 图上距离 ÷ 比例尺。\n【详解】 $16 \\times 5000000 = 80000000$ （厘米） $80000000$ 厘米 $= 800$ 千米\n$800 \\div 5 - 90$ \n$= 160 - 90$ \n=70 (千米)\n答: 乙列火车每小时行驶 70 千米。",
+    "knowledge_points": ["图上距离与实际距离的换算", "相遇问题"],
+    "source": "18-19 六年级下·云南曲靖·期末",
+    "merged_id": "unit4_288",
+    "original_id": "unit4_solve_288"
+  },
+  {
+    "id": "unit4_solve_289",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "淘气和笑笑收集的邮票张数的比是 $5: 3$ ，淘气收集了 35 张邮票，笑笑收集了多少张？（用比例方法解答）",
+    "options": [],
+    "answer": "21 张",
+    "solution": "【分析】设笑笑收集了 $x$ 张, 根据淘气收集的张数: 笑笑收集的张数 $= 5: 3$ , 列出比例解答即可。用比例解决问题时，左右两边的比只要统一即可。\n【详解】解：设笑笑收集了 $x$ 张。\n$35:x=5:3\\\\5 \\mathrm{x} = 35 \\times 3\\\\5 \\mathrm{x} \\div 5 = 105 \\div 5\\\\x = 21$\n答: 笑笑收集了 21 张。",
+    "knowledge_points": ["比例的应用"],
+    "source": "2023·广东广州·小升初真题",
+    "merged_id": "unit4_289",
+    "original_id": "unit4_solve_289"
+  },
+  {
+    "id": "unit4_solve_290",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "在一幅比例尺是1:5000000 的地图上，量得甲、乙两个城市之间高速公路的距离是 $6 cm$ 。一辆汽车以 50 千米/时速度从乙城开往甲城,需多长时间到达？",
+    "options": [],
+    "answer": "6小时",
+    "solution": "【分析】依据图上距离÷比例尺 = 实际距离，求出路程，再根据路程÷速度 = 时间解答。本题的关键是根据实际距离 $=$ 图上距离 $\\div$ 比例尺,先求出实际距离。\n【详解】 $6 \\div \\frac{1}{5000000}= 6 \\times 5000000 \\\\ = 30000000 (\\mathrm {c m}) \\\\ 30000000 \\mathrm {c m} = 300 \\mathrm {k m} \\\\ 300 \\div 50 = 6 (\\text {小 时}) $\n答: 需 6 小时到达。",
+    "knowledge_points": ["图上距离与实际距离的换算", "基础行程问题"],
+    "source": "19-20六年级下·河南南阳·期末",
+    "merged_id": "unit4_290",
+    "original_id": "unit4_solve_290"
+  },
+  {
+    "id": "unit4_solve_291",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "只列式不计算。400g 大豆榨油 48g，要榨出 $2 \\mathrm{~kg}$ 豆油需要多少千克大豆(用比例的方法解决问题)",
+    "options": [],
+    "answer": "解：设需要 $x$ 千克大豆。\n$400: 48 = x: 2$",
+    "solution": "【分析】设需要 $\\mathrm{x}$ 千克大豆，根据大豆质量：榨出油的质量=要求的大豆质量：榨出油的质量，列出比例即可。用比例解决问题只要左右两边的比统一即可。\n【详解】解：设需要 $\\mathbf{X}$ 千克大豆。根据分析, 比例式为: $400: 48 = x: 2$",
+    "knowledge_points": ["比例的应用"],
+    "source": "17-18 六年级下·陕西宝鸡·期末",
+    "merged_id": "unit4_291",
+    "original_id": "unit4_solve_291"
+  },
+  {
+    "id": "unit4_solve_292",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "一辆汽车从甲地开往乙地，180千米用了3小时，照这样的速度，乙地到丙地长 540 千米, 需要几小时才能到达? (用比例方法解)",
+    "options": [],
+    "answer": "9小时",
+    "solution": "【分析】设需要 $\\mathrm{x}$ 小时才能到达，根据路程÷时间 $=$ 速度(一定), 列出正比例算式, 解答即可。关键是确定比例关系，商一定是正比例关系。\n【详解】解：设需要 $x$ 小时才能到达。\n$540: x = 180: 3\\\\180 x \\div 180 = 540 \\times 3 \\div 180\\\\x = 9$\n答: 需要 9 小时才能到达。",
+    "knowledge_points": ["正比例的应用"],
+    "source": "19-20六年级下·福建厦门·期末",
+    "merged_id": "unit4_292",
+    "original_id": "unit4_solve_292"
+  },
+  {
+    "id": "unit4_solve_293",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "小红读一本故事书,如果每天读30页，8天可以读完，小红想6天读完，那么平均每天要多读多少页？（用比例知识解答）",
+    "options": [],
+    "answer": "40页",
+    "solution": "【分析】设平均每天要多读 $x$ 页, 根据每天读的页数 $\\times$ 天数 = 总页数 (一定), 列出反比例算式解答即可。关键是确定比例关系，积一定是反比例关系。\n【详解】解：设平均每天要多读 $x$ 页。\n$6 x = 30 \\times 8\\\\6 x \\div 6 = 240 \\div 6\\\\x = 40$\n答: 平均每天要多读 40 页。",
+    "knowledge_points": ["反比例的应用"],
+    "source": "19-20六年级下·河南南阳·期末",
+    "merged_id": "unit4_293",
+    "original_id": "unit4_solve_293"
+  },
+  {
+    "id": "unit4_solve_294",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "用方砖给一间教室铺地。如果用边长 $40 cm$ 的方砖铺, 需要 500 块; 如果改用边长为 $8 \\mathrm{~dm}$ 的方砖来铺, 需要多少块? (用比例解)",
+    "options": [],
+    "answer": "125 块",
+    "solution": "【分析】教室的面积是一定的，方砖面积 $\\times$ 方砖个数 = 教室面积，据此列比例解答。此题考查了用比例解决实际问题，先确定两个量之间是比值一定还是乘积一定再解答。\n【详解】解：设需要 $x$ 块。\n40 厘米 = 4 分米\n$4 \\times 4 \\times 500 = 8 \\times 8 x\\\\64 x = 8000\\\\x = 125$\n答: 需要 125 块。",
+    "knowledge_points": ["反比例的应用"],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_294",
+    "original_id": "unit4_solve_294"
+  },
+  {
+    "id": "unit4_solve_295",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "打字员打一篇文稿， $\\frac{2}{3}$ 小时完成了这篇文稿的 $\\frac{5}{6}$ 。照这样计算，还要多少小时才能打完这篇文稿？",
+    "options": [],
+    "answer": "$\\frac{2}{15}$ 小时",
+    "solution": "【分析】 $\\frac{2}{3}$ 小时完成了这篇文稿的 $\\frac{5}{6}$ , 可知完成这篇文稿一共需要 $\\frac{2}{3} \\div \\frac{5}{6}$ 小时, 再减去已经用的时间即可。此题考查了分数除法的应用，明确已知一个数的几分之几是多少，求这个数用除法，先求出一共需要的时间是解题关键。\n【详解】 $\\frac{2}{3} \\div \\frac{5}{6} - \\frac{2}{3}\\\\= \\frac {4}{5} - \\frac {2}{3} \\\\ = \\frac {2}{15} \\quad (\\text {小 时}) $\n答: 还要 $\\frac{2}{15}$ 小时才能打完这篇文稿。",
+    "knowledge_points": ["分数除法的应用", "分数的四则混合运算"],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_295",
+    "original_id": "unit4_solve_295"
+  },
+  {
+    "id": "unit4_solve_296",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "在一幅比例尺是1:500000的地图上，量得A地与B地的距离是 $6 cm$ 。甲、乙两辆汽车分别从 $A$ 地和 $B$ 地同时出发, 相向而行, 2 小时后相遇。已知甲、乙两车的速度之比是 $2: 3$ , 甲车每小时行驶多少千米?",
+    "options": [],
+    "answer": "60千米",
+    "solution": "【分析】先根据实际距离 $=$ 图上距离 $\\div$ 比例尺, 求出两地的实际距离, 再根据速度和 $=$ 路程 $\\div$ 时间, 求出甲、乙的速度和, 再由 '甲车与乙车速度的比是 $2: 3$ , ' 利用按比例分配的方法列式解答即可。本题主要应用的知识点是: 实际距离 = 图上距离 $\\div$ 比例尺, 速度和 $\\times$ 相遇时间 = 路程及利用按比例分配的方法解决问题。\n【详解】 $6 \\div \\frac{1}{5000000} = 30000000 (cm)$\n$30000000cm = 300km$\n$300 \\div 2 \\times \\frac {2}{2 + 3} \\\\ =150 \\times \\frac {2}{5} \\\\ = 60 \\text {(千 米)}$\n答: 甲车每小时行驶 60 千米。",
+    "knowledge_points": ["比例尺应用"],
+    "source": "20-21六年级下·全国·单元测试",
+    "merged_id": "unit4_296",
+    "original_id": "unit4_solve_296"
+  },
+  {
+    "id": "unit4_solve_297",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "天天服装厂4天加工240套童装。照这样计算，要生产720套这样的童装，需要多少天？（用比例解）",
+    "options": [],
+    "answer": "12天",
+    "solution": "【分析】根据题意可知, 每天加工的童装套数是一定的,即加工的童装套数与天数比值一定, 据此解答。此题考查了用比例解决问题，先确定两个量之间是比值一定还是乘积一定，再列比例解答。\n【详解】解：设需要 $x$ 天。\n$240: 4 = 720: x \\\\240 x= 2880\\\\x = 12$\n答: 需要 12 天。",
+    "knowledge_points": ["比例的应用"],
+    "source": "20-21六年级下·全国·单元测试",
+    "merged_id": "unit4_297",
+    "original_id": "unit4_solve_297"
+  },
+  {
+    "id": "unit4_solve_298",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "在比例尺为1:7500000的地图上，甲、乙两地的距离是 $6 cm$ 。现有一辆客车和一辆货车分别从两地同时出发，相向而行，经过 3 小时相遇。客车与货车行驶的路程比是 8:7，货车每小时行驶多少千米？",
+    "options": [],
+    "answer": "$70 \\mathrm{~km}$",
+    "solution": "【分析】本题可根据'实际距离 $=$ 图上距离 $\\div$ 比例尺'求出实际距离 $6 \\div \\frac{1}{7500000} = 45000000(cm)$ , $45000000 cm = 450 \\mathrm{~km}$ 。两车3小时共行驶 $450 \\mathrm{~km}$ , 则每小时行驶路程之和为 $450 \\div 3 = 150(\\mathrm{~km})$ ,客车与货车行驶的路程比是 8:7, 时间相同, 速度比等于路程比, 所以客车与货车的速度比是 8:7, 货车每小时行驶 $150 \\times \\frac{7}{8 + 7} = 70(\\mathrm{~km})$ 。据此解答。理解掌握'实际距离 = 图上距离 ÷ 比例尺'及'速度比等于路程比'是解答本题的关键。\n【详解】 $6 \\div \\frac{1}{7500000} = 45000000(cm)$\n$45000000 \\mathrm {c m} = 450 \\mathrm {k m} \\\\ 450 \\div 3 = 150 (\\mathrm {k m}) \\\\ 150 \\times \\frac {7}{8 + 7} = 70 (\\mathrm {k m}) $\n答: 货车每小时行驶 $70 \\mathrm{~km}$ 。",
+    "knowledge_points": ["比例尺应用"],
+    "source": "20-21 六年级下·全国·课后作业",
+    "merged_id": "unit4_298",
+    "original_id": "unit4_solve_298"
+  },
+  {
+    "id": "unit4_solve_299",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "中等",
+    "question": "在长 28 厘米, 宽 18 厘米的纸上, 画学校的平面图。校园东西长 520 米, 南北宽320 米。用多大的比例尺比较合适？运动场长 150 米，在图上应画多长？",
+    "options": [],
+    "answer": "(答案不唯一)1:20; (答案不唯一)7.5 厘米",
+    "solution": "【分析】根据图上距离：实际距离 $=$ 比例尺，确定比例尺；根据图上距离 $=$ 实际距离 $\\times$ 比例尺，确定图上长度。关键是理解比例尺的意义，比例尺没有单位名称，为了方便，通常把比例尺的前项化作 1（图上距离大于实际距离的，常把后项化为 1）。\n【详解】 $25: 550 = 1: 22\\\\20: 320 = 1: 16$比例尺用 $1: 20$ 即可\n$150 \\times \\frac{1}{20} = 7.5$ (厘米)答: 比例尺用 1:20 比较合适; 运动场长 150 米, 在图上应画 7.5 厘米长。",
+    "knowledge_points": ["图上距离与实际距离的换算", "比例尺的意义"],
+    "source": "2019·浙江杭州·小升初真题",
+    "merged_id": "unit4_299",
+    "original_id": "unit4_solve_299"
+  },
+  {
+    "id": "unit4_solve_300",
+    "unit": "第四单元：比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "在一幅比例尺为 $\\frac{1}{5000000}$ 的地图上, 量得 $A 、 B$ 两地之间的公路长度是 $6 cm$ , 甲、乙两车同时从 $A 、 B$ 两地出发, 相向而行。已知乙车速度与甲车速度之比是 $2:3$ 。两车相遇时, 甲车比乙车多行驶了多少千米?",
+    "options": [],
+    "answer": "60千米",
+    "solution": "【分析】两地的实际距离 $=$ 图上距离 $\\div$ 比例尺, 据此求出 $A 、 B$ 两地之间的公路长度, 因为两车行驶的时间相同, 两车的速度比就是两车行驶的路程之比, 按比例分配, 即可求出甲车比乙车多行驶的路程。此题考查了比例尺与按比例分配的综合应用，先求出两地的实际距离，明确两车速度之比等于行驶的路程之比是解题关键。\n【详解】 $6 \\div \\frac{1}{5000000} = 30000000$ （厘米） 30000000 厘米 = 300 千米\n$300 \\div (2 + 3) \\times (3 - 2) \\\\ = 300 \\div 5 \\times 1 \\\\ = 60 (\\text {千 米}) $\n答: 甲车比乙车多行驶了 60 千米。",
+    "knowledge_points": ["比例尺应用", "按比分配问题"],
+    "source": "18-19 六年级下·浙江·期末",
+    "merged_id": "unit4_300",
+    "original_id": "unit4_solve_300"
+  },  {
+    "id": "unit4_choice_301",
+    "unit": "第四单元：比例",
+    "chapter": "比例尺的应用",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "把线段比例尺![](images/e70e294d3cd6cfc7e421baf81425cb28.png)改写成数值比例尺是（）。",
+    "options": [
+      "A. 1:50",
+      "B. 1:200",
+      "C. 1:5000000"
+    ],
+    "answer": "C",
+    "solution": "【分析】把线段比例尺改写成数值比例尺的方法: 写出 1 厘米和 1 厘米所代表的实际距离的比, 统一单位后再化成最简比的形式即可。把线段比例尺改写成数值比例尺时，一定要先进行单位换算。\n【详解】图上距离：实际距离=1厘米：50千米 =1 厘米：5000000 厘米 = 1:5000000；故答案为: C。",
+    "knowledge_points": [
+      "比例尺的意义"
+    ],
+    "source": "19-20 六年级下·江苏·单元测试",
+    "merged_id": "unit4_301",
+    "original_id": "unit4_choice_301"
+  },
+  {
+    "id": "unit4_choice_302",
+    "unit": "第四单元：比例",
+    "chapter": "比例的应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "如图, 阴影部分占大圆的 $\\frac{1}{5}$, 占小圆的 $\\frac{1}{4}$, 小圆面积与大圆面积的比是 ()。\n![](images/4c13e26df60ada52ba046c1a70d77b7d9e71fe3cf04118cadf549de25c5d134e.jpg)",
+    "options": [
+      "A. $\\frac{1}{4}: \\frac{1}{5}$",
+      "B. $5: 4$",
+      "C. 4:5"
+    ],
+    "answer": "C",
+    "solution": "【分析】分析题目, 大圆面积的 $\\frac{1}{5}$ 等于阴影部分的面积, 小圆面积的 $\\frac{1}{4}$ 也等于阴影部分的面积, 即大圆面积 $\\times \\frac{1}{5} =$ 小圆面积 $\\times \\frac{1}{4}$, 据此结合比例的基本性质写出小圆和大圆面积的比, 最后根据比的基本性质把结果化成最简整数比即可。\n【详解】大圆面积 $\\times \\frac{1}{5} =$ 小圆面积 $\\times \\frac{1}{4}$小圆面积：大圆面积 $= \\frac{1}{5}: \\frac{1}{4}$ $\\frac {1}{5}: \\frac {1}{4} = (\\frac {1}{5} \\times 2 0): (\\frac {1}{4} \\times 2 0) = 4: 5$ 阴影部分占大圆的 $\\frac{1}{5}$, 占小圆的 $\\frac{1}{4}$, 小圆面积与大圆面积的比是 $4: 5$ 。故答案为: C",
+    "knowledge_points": [
+      "比例的基本性质",
+      "比的意义",
+      "比的化简"
+    ],
+    "source": "24-25 六年级下·海南海口·单元测试",
+    "merged_id": "unit4_302",
+    "original_id": "unit4_choice_302"
+  },
+  {
+    "id": "unit4_choice_303",
+    "unit": "第四单元：比例",
+    "chapter": "比例的应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "如图是由两个三角形重叠而成的, 重叠部分的面积占三角形 A 的 $\\frac{1}{9}$, 占三角形 B 的 $\\frac{1}{5}$, 则三角形 A 的面积: 三角形 B 的面积是 ( )。\n![](images/6e84adc7b0ba856b07822013756483ef2714c7ebca1df17ded2782a10e946288.jpg)",
+    "options": [
+      "A. $9: 5$",
+      "B. $5: 9$",
+      "C. $5: 3$"
+    ],
+    "answer": "A",
+    "solution": "【分析】根据题意, 重叠部分的面积占三角形 A 的 $\\frac{1}{9}$, 占三角形 B 的 $\\frac{1}{5}$, 根据分数乘法的意义可得出: 三角形 A 的面积 $\\times \\frac{1}{9} =$ 三角形 B 的面积 $\\times \\frac{1}{5}$, 根据比例的基本性质改写成比例式为三角形 A 的面积: 三角形 B 的面积 $= \\frac{1}{5}: \\frac{1}{9}$, 再化简比即可。\n【详解】三角形A的面积 $\\times \\frac{1}{9} =$ 三角形B的面积 $\\times \\frac{1}{5}$三角形 A 的面积：三角形 B 的面积 $= \\frac{1}{5}: \\frac{1}{9} = (\\frac {1}{5} \\times 4 5): (\\frac {1}{9} \\times 4 5) = 9: 5$ 则三角形 A 的面积：三角形 B 的面积是 $9: 5$ 。故答案为: A",
+    "knowledge_points": [
+      "比例的基本性质",
+      "求一个数的几分之几的问题",
+      "比的化简"
+    ],
+    "source": "24-25 六年级下·全国·单元测试",
+    "merged_id": "unit4_303",
+    "original_id": "unit4_choice_303"
+  },
+  {
+    "id": "unit4_choice_304",
+    "unit": "第四单元：比例",
+    "chapter": "正比例与反比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "收音机刻度盘上的波长和频率分别是用米（m）和千赫兹（kHz）为单位标刻的。下面是一些对应的数值，那么波长和频率的比例关系是什么？（）\n   | 波长/m   | 300  | 500  | 600  | 1000 |\n   | -------- | ---- | ---- | ---- | ---- |\n   | 频率/kHz | 1000 | 600  | 500  | 300  |",
+    "options": [
+      "A. 成正比例关系",
+      "B. 成反比例关系",
+      "C. 不成比例关系"
+    ],
+    "answer": "B",
+    "solution": "【分析】判断两个相关联的量之间成什么比例，就看这两个量是对应的比值一定，还是对应的乘积一定，如果是比值一定，就成正比例，如果是乘积一定，则成反比例。通过计算判断。\n【详解】 $300 \\div 1000 = 0.3$ $5 0 0 \\div 6 0 0 \\approx 0. 8 3$ $6 0 0 \\div 5 0 0 = 1. 2$ $1 0 0 0 \\div 3 0 0 \\approx 3. 3 3$ 可知，波长和频率的比值不一定，不成正比例关系。$ 3 0 0 \\times 1 0 0 0 = 3 0 0 0 0 0$ $ 5 0 0 \\times 6 0 0 = 3 0 0 0 0 0$ $ 6 0 0 \\times 5 0 0 = 3 0 0 0 0 0$ $ 1 0 0 0 \\times 3 0 0 = 3 0 0 0 0 0$ 波长 $\\times$ 频率 $= 300000$ （一定），所以波长和频率的比例关系是反比例关系。故答案为: B",
+    "knowledge_points": [
+      "正比例的意义及辨识",
+      "反比例的意义及辨识"
+    ],
+    "source": "24-25 六年级下·全国·单元测试",
+    "merged_id": "unit4_304",
+    "original_id": "unit4_choice_304"
+  },
+  {
+    "id": "unit4_choice_305",
+    "unit": "第四单元：比例",
+    "chapter": "正比例与反比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "有两个相关联的量,它们的关系如图所示。这两个量可能是 ( )。\n![](images/71cf23d61651af8e199bb6c169d5cd52a2dea87e66ffaf3b81ce93e81e09a3f4.jpg)",
+    "options": [
+      "A. 货物总量一定, 每天运送的吨数和所需天数",
+      "B. 数量一定，总价和单价",
+      "C. 修一条路, 已修的路和未修的路",
+      "D. 平行四边形的面积一定, 它的底和高"
+    ],
+    "answer": "B",
+    "solution": "【分析】两种相关联的量，如果这两种量中相对应的两个数比值一定，这两种量就成正比例；如果这两种量中相对应的两个数乘积一定，这两种量就成反比例，根据图象可知，图形中两种相关联的量是正比例，逐项分析各选项，进行解答。\n【详解】A．因为每天运送的吨数 $\\times$ 所需天数 $=$ 货物总量（一定），每天运送的吨数和所需天数的乘积一定，则每天运送的吨数和所需天数成反比例，不符合题意；B. 数量 (一定) = 总价 $\\div$ 单价, 所以比值一定, 单价和总价成正比例, 符合题意; C. 已修的路 + 未修的路 = 这段路的总长（一定），是对应的两个量的和一定，所以修一段路，已经修的与未修的不成比例，不符合题意； D. 底 $\\times$ 高 = 平行四边形的面积（一定），乘积一定，所以它的底和高成反比例关系，不符合题意。故答案为: B",
+    "knowledge_points": [
+      "正比例的意义及辨识",
+      "反比例的意义及辨识",
+      "平行四边形面积的计算",
+      "经济问题"
+    ],
+    "source": "23-24 六年级下·河北保定·期末",
+    "merged_id": "unit4_305",
+    "original_id": "unit4_choice_305"
+  },
+  {
+    "id": "unit4_choice_306",
+    "unit": "第四单元：比例",
+    "chapter": "比例尺的应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "在一幅图上用 5 厘米的线段正好表示实际距离 300 千米, 这幅图的比例尺是 ( )。\n![](images/bc653ed69716a09b3b3ee313e9543503.png)",
+    "options": [
+      "A. ![](images/bc653ed69716a09b3b3ee313e9543503.png)",
+      "B. ![](images/1f74538b9b49aa781324596075a3a66a.png)",
+      "C. ![](images/62a300bb0b9765b177a399f558b40d8f.png)",
+      "D. ![](images/1c5436adb7cb785de04e6b2227628fbe.png)"
+    ],
+    "answer": "C",
+    "solution": "【分析】图上距离：实际距离 = 比例尺，据此写出图上距离与实际距离的比，化简即可。\n【详解】$5厘米：300千米=5 厘米：30000000 厘米 = 1:6000000$ 6000000 厘米 = 60 千米所以在一幅图上用 5 厘米的线段正好表示实际距离 300 千米,这幅图的比例尺是 $0.60$ 千米。故答案为: C",
+    "knowledge_points": [
+      "比例尺的意义"
+    ],
+    "source": "2024·山东临沂·小升初真题",
+    "merged_id": "unit4_306",
+    "original_id": "unit4_choice_306"
+  },
+  {
+    "id": "unit4_choice_307",
+    "unit": "第四单元：比例",
+    "chapter": "正比例与反比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "x 和 y 是两种相关联的量, 它们的关系可以用下面的图象表示。那么, 这个图象可能表示的是 ( ) 的关系。\n![](images/cf8543fceaa8516928c96bff7b9bedc7b6a8097784ae71df0bad8807bd3c8a51.jpg)",
+    "options": [
+      "A. 看一本书，看了的页数和没看的页数",
+      "B. 正方形的面积和边长",
+      "C. 圆柱的高一定, 体积和底面积",
+      "D. 平行四边形的面积一定, 底和高"
+    ],
+    "answer": "C",
+    "solution": "【分析】图象是一条直线，说明 $x$ 和 $y$ 成正比例关系。乘积一定的两个量成反比例关系，比值（或商）一定的两个量成正比例关系。由此分析各个选项中的两个量，从而解题。\n【详解】A．看了的页数+没看的页数=总页数，所以看了的页数和没看的页数不成比例； B. 正方形面积 $\\div$ 边长 $= \\text{边}$ 长, 边长和面积不固定, 所以正方形的面积和边长数不成比例;C. 圆柱体积÷底面积=高（一定），那么圆柱的高一定，体积和底面积成正比例关系； D. 底 $\\times$ 高 = 平行四边形的面积（一定），那么平行四边形的面积一定，底和高成反比例关系。所以，这个图象可能表示的是圆柱的高一定时，圆柱体积和底面积的关系。故答案为: C",
+    "knowledge_points": [
+      "正比例的意义及辨识",
+      "反比例的意义及辨识",
+      "平行四边形面积的计算",
+      "圆柱的体积"
+    ],
+    "source": "23-24 六年级下·北京西城·期末",
+    "merged_id": "unit4_307",
+    "original_id": "unit4_choice_307"
+  },
+  {
+    "id": "unit4_choice_308",
+    "unit": "第四单元：比例",
+    "chapter": "正比例与反比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "小明去超市购买鸡蛋，已知鸡蛋的价格为10元/千克。下面哪幅图正确表示了总价和数量之间的关系？（）\nA.\n![](images/44f83f3a844f0af099ec4ffcf6b4c9a27b3a05e24179f388db6f686e4d71bc1b.jpg)\nB.\n![](images/5e2e2ed229a63d8fd99d502b08069c965d4cfcfc2fab60c909dce01a794d4ed8.jpg)\nC.\n![](images/1d95b6bcf730d36e4399c81f10b686865427b9190fa1c76c4704a545f3e1fbcf.jpg)\nD.\n![](images/23a343109bf325167c82093ee58afd665f129352b25f5a1dbaa5704bb301f34d.jpg)",
+    "options": [
+      "A. ![](images/44f83f3a844f0af099ec4ffcf6b4c9a27b3a05e24179f388db6f686e4d71bc1b.jpg)",
+      "B. ![](images/5e2e2ed229a63d8fd99d502b08069c965d4cfcfc2fab60c909dce01a794d4ed8.jpg)",
+      "C. ![](images/1d95b6bcf730d36e4399c81f10b686865427b9190fa1c76c4704a545f3e1fbcf.jpg)",
+      "D. ![](images/23a343109bf325167c82093ee58afd665f129352b25f5a1dbaa5704bb301f34d.jpg)"
+    ],
+    "answer": "C",
+    "solution": "【分析】已知鸡蛋的价格为 10 元/千克, 根据总价 $\\div$ 数量 = 单价 (一定), 商一定, 那么总价和数量成正比例关系,据此解答。两种相关联的量，一种量变化，另一种量也随着变化，如果这两种量中相对应的两个数的比值一定（也就是商一定），这两种量成正比例关系。\n【详解】A．数量在变，总价不变，不符合题意；B. 数量不变, 总价在变, 不符合题意;C. $20 \\div 2 = 10$ (元) $40 \\div 4 = 10$ （元） $60 \\div 6 = 10$ (元) $80 \\div 8 = 10$ （元） 随着总价和数量的变化，鸡蛋的单价一定，都是10元，符合题意； D. 买 1 千克鸡蛋大于 10 元, 不符合题意。 故答案为: C",
+    "knowledge_points": [
+      "正比例图象的认识",
+      "正比例的意义及辨识",
+      "经济问题"
+    ],
+    "source": "23-24六年级下·河南三门峡·期中",
+    "merged_id": "unit4_308",
+    "original_id": "unit4_choice_308"
+  },
+  {
+    "id": "unit4_choice_309",
+    "unit": "第四单元：比例",
+    "chapter": "比例尺的应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "这幅图上的 1 厘米表示实际距离是 ( )。\n![](images/d85e6711957058bdac26a9d5e2ead44c24af0f92cc5fd714f7a4fa4814668db0.jpg)",
+    "options": [
+      "A. 100千米",
+      "B. 400千米",
+      "C. 200千米",
+      "D. 300千米"
+    ],
+    "answer": "A",
+    "solution": "【分析】比例尺的表现形式分为数值比例尺与线段比例尺。从线段比例尺中可知, 图上 1 厘米的距离等于 100 千米, 据此解答。\n【详解】由分析可得: 这幅图上的 1 厘米表示实际距离是 100 千米。故答案为: A",
+    "knowledge_points": [
+      "比例尺的意义"
+    ],
+    "source": "22-23 六年级下·福建厦门·期中",
+    "merged_id": "unit4_309",
+    "original_id": "unit4_choice_309"
+  },
+  {
+    "id": "unit4_choice_310",
+    "unit": "第四单元：比例",
+    "chapter": "比例尺的应用",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "把线段比例尺 0.4080千米 改写成数值比例尺是（）。",
+    "options": [
+      "A. $\\frac{1}{40}$",
+      "B. $\\frac{1}{4000}$",
+      "C. $\\frac{1}{4000000}$"
+    ],
+    "answer": "C",
+    "solution": "【分析】根据线段比例尺可知, 1 厘米表示 40 千米, 先将 40 千米化为 4000000 厘米, 再根据数值比例尺 $=$ $\\frac{\\text { 图上距离 }}{\\text { 实际距离}}$ 求出数值比例尺。本题考查了线段比例尺和数值比例尺的认识和应用。\n【详解】40千米 $= 4000000$ 厘米数值比例尺是 $\\frac{1}{4000000}$ 。故答案为: C",
+    "knowledge_points": [
+      "比例尺的意义"
+    ],
+    "source": "22-23 六年级下·全国·单元测试",
+    "merged_id": "unit4_310",
+    "original_id": "unit4_choice_310"
+  },
+  {
+    "id": "unit4_choice_311",
+    "unit": "第四单元：比例",
+    "chapter": "正比例与反比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "表中, 如果 a 和 b 成反比例, 括号里应填 ( )。\n| a    | 4    | 8    |\n| ---- | ---- | ---- |\n| b    | ()   | 12   |",
+    "options": [
+      "A. 2",
+      "B. 8",
+      "C. 24"
+    ],
+    "answer": "C",
+    "solution": "【分析】两种相关联的量，一种量变化，另一种量也随着变化，如果这两种量中相对应的乘积一定，这两种量就叫做成反比例的量，它们的关系叫做反比例关系。因为 a 和 b成反比例, 所以它们的乘积一定, 可先求出其中一组数值的乘积, 再除以 4 , 就是所求。抓住反比例的概念来判断, 利用反比例关系灵活地解决问题, 是解题关键。\n【详解】因为 a 和 b 成反比例, 所以 $ab = 8 \\times 12 = 96$ (一定)。 $96 \\div 4 = 24$故答案为: C",
+    "knowledge_points": [
+      "反比例的应用"
+    ],
+    "source": "22-23 六年级下·全国·单元测试",
+    "merged_id": "unit4_311",
+    "original_id": "unit4_choice_311"
+  },
+  {
+    "id": "unit4_choice_312",
+    "unit": "第四单元：比例",
+    "chapter": "比例尺的应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "有四张图，比例尺分别如下，图中5厘米长的线段表示的实际距离最长的是（）。",
+    "options": [
+      "A. $1: 3000000$",
+      "B. $\\frac{1}{1500000}$",
+      "C. ![](images/7b04308545574b96d34670e1419201626c450dc4cdf181ef51683e3c8ada243c.jpg)",
+      "D. $50: 1$"
+    ],
+    "answer": "A",
+    "solution": "【分析】根据公式：实际距离 $=$ 图上距离 $\\div$ 比例尺，用图中5 厘米分别计算出选项中比例尺的实际距离, 比较后找出最长, 据此解答。此题考查了比例尺的运用，关键是熟记计算公式以及单位换算的方法。\n【详解】A. $5 \\div \\frac{1}{3000000} = 15000000$ （厘米），15000000 厘米 $= 150$ 千米；B. $5 \\div \\frac{1}{150000} = 750000$ (厘米), 750000 厘米 = 75 千米;C. 表示图上的 1 厘米是实际距离 20 千米,$5 \\times 20 = 100$ （千米）；D. $5 \\div 50 = 0.1$ 厘米;因为 150 千米 $>100$ 千米 $>75$ 千米 $>0.1$ 厘米故答案为: A",
+    "knowledge_points": [
+      "比例尺应用",
+      "图上距离与实际距离的换算"
+    ],
+    "source": "2021·浙江温州·小升初真题",
+    "merged_id": "unit4_312",
+    "original_id": "unit4_choice_312"
+  },
+  {
+    "id": "unit4_choice_313",
+    "unit": "第四单元：比例",
+    "chapter": "比例的基本性质",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "根据下图中的信息判断, 下列等式不成立的是 ( )。\n![](images/32dd64c2e8690f8f668ac5e13d129c0479b1cbc5b82d3307b76338e11f5017b5.jpg)",
+    "options": [
+      "A. $a: c = d: b$",
+      "B. $\\frac{a}{d} = \\frac{c}{b}$",
+      "C. $a: c = b: d$",
+      "D. $\\frac{b}{c} = \\frac{d}{a}$"
+    ],
+    "answer": "C",
+    "solution": "【分析】根据平行四边形的面积公式： $S = ab = cd$ ，然后根据比例的基本性质，内项积等于外项积，把各项的比例式化为乘积式，然后与平行四边形的面积公式对比即可。本题考查比例的基本性质，明确内项积等于外项积是解题的关键。\n【详解】A．因为 $a:c = d:b$ ，所以 $\\mathrm{ab} = \\mathrm{cd}$ ，符合题意；B. 因为 $\\frac{a}{d} = \\frac{c}{b}$, 所以 $ab = cd$, 符合题意; C. 因为 $a: c = b: d$, 所以 $\\mathrm{ad} = \\mathrm{bc}$, 不符合题意; D. 因为 $\\frac{b}{c} = \\frac{d}{a}$, 所以 $ab = cd$, 符合题意。故答案为: C",
+    "knowledge_points": [
+      "比例的基本性质",
+      "平行四边形的面积"
+    ],
+    "source": "21-22 六年级下·山东济宁·期末",
+    "merged_id": "unit4_313",
+    "original_id": "unit4_choice_313"
+  },
+  {
+    "id": "unit4_choice_314",
+    "unit": "第四单元：比例",
+    "chapter": "正比例图象的认识",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "下面图 ( ) 表示的是成正比例关系的图像。\nA.![](images/a945ce1823754fb05b62eed2528418f161fd2ceb706180c4f13a8832ae9fbe3b.jpg)\nB.![](images/a05b715ec49807ed807678432e8773da22e27ca0d0a5ef151ed446d213553696.jpg)\nC.![](images/dbf9ef402bfdc98ab68588f881e9e73eef58f9877d4b3df64fde53478eace4b8.jpg)",
+    "options": [
+      "A. ![](images/a945ce1823754fb05b62eed2528418f161fd2ceb706180c4f13a8832ae9fbe3b.jpg)",
+      "B. ![](images/a05b715ec49807ed807678432e8773da22e27ca0d0a5ef151ed446d213553696.jpg)",
+      "C. ![](images/dbf9ef402bfdc98ab68588f881e9e73eef58f9877d4b3df64fde53478eace4b8.jpg)"
+    ],
+    "answer": "B",
+    "solution": "【分析】首先要知道成正比例关系的图象特点是一条递增的直线，再者相关联的两个量应是比值或商一定，而不是和或差一定。本题是成正比例关系知识的拓展, 是把数和形结合起来, 研究两个相关联的量之间的关系。\n【详解】A．如图虽然是直线，但是它是两个量的和一定，即：售出个数+剩下个数=总个数（一定），不是比值或商一定，所以不正确； B. 工作总量 $\\div$ 工作人员 = 每人的工作量（一定），是成正比例关系的图像； C. 成正比例关系的图象特点是一条递增的直线, 所以不正确。故答案为: B",
+    "knowledge_points": [
+      "正比例的意义及辨识",
+      "单式折线统计图"
+    ],
+    "source": "16-17 六年级下·全国·单元测试",
+    "merged_id": "unit4_314",
+    "original_id": "unit4_choice_314"
+  },
+  {
+    "id": "unit4_choice_315",
+    "unit": "第四单元：比例",
+    "chapter": "正比例图象的认识",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "图中表示斑马和长颈鹿的奔跑情况。下面叙述错误的是 ( )。\n![](images/dc699a1f682856b0a552a35d03c52fb4fb9fa47bcfde5a09307d9607e378687a.jpg)",
+    "options": [
+      "A. 上图是一个正比例关系的图象",
+      "B. 在 20 分钟时,它们路程相差 $8km$",
+      "C. 斑马和长颈鹿的速度之比是 $2: 3$",
+      "D. 在 15 分钟时,长颈鹿跑了 $12km$"
+    ],
+    "answer": "C",
+    "solution": "【分析】由图可知：斑马和长颈鹿的奔跑的路程和时间的商一定，也就是速度一定，可以判断图象是正比例图象；20 分钟时, 长颈鹿奔跑了 $16km$, 斑马奔跑了 $24km$, 两者相减, 可得出它们的路程差; 分别计算出 20 分钟时斑马和长颈鹿的速度，再进行比的运算可知斑马和长颈鹿的速度比；15分钟时，长颈鹿奔跑的路程是 $12km$ 。本题考查了对成正比例关系图象的认识。根据图象, 按要求找出有用的信息进行计算是解答本题的关键。\n【详解】A. 斑马和长颈鹿的奔跑的路程和时间的商一定,奔跑的路程和时间成正比例。B. 20 分钟时, 长颈鹿奔跑了 $16km$, 斑马奔跑了 $24km$ 。$2 4 - 1 6 = 8 (km)$C. $24 \\div 20 = 1.2 \\text{ (km/分)}$ $1 6 \\div 2 0 = 0. 8 (km / \\text {分})$$1.2: 0. 8 = 5: 4$ D. 在 15 分钟时, 长颈鹿跑了 $12km$ 。 故答案为: C。",
+    "knowledge_points": [
+      "正比例图象的认识"
+    ],
+    "source": "20-21 六年级下·广东广州·期末",
+    "merged_id": "unit4_315",
+    "original_id": "unit4_choice_315"
+  },
+  {
+    "id": "unit4_choice_316",
+    "unit": "第四单元：比例",
+    "chapter": "图形的放大与缩小",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "下列图形是按一定比例缩小的, $x =$ ( )。\n    ![](images/945f1d9c42374165322abf1a19d6350b.png)",
+    "options": [
+      "A. 10",
+      "B. 9.5",
+      "C. 8",
+      "D. 7.5"
+    ],
+    "answer": "D",
+    "solution": "【分析】根据图形放大与缩小的方法以及比例的意义可知:把这个长方形按一定比例缩小后, 则原来的长与缩小后的长的比值等于原来的宽与缩小后的宽的比值; 据此可列比例 $12: x = 8: 5$, 解比例即可。利用放大与缩小的方法列出比例式，进而进行解答。\n【详解】 $12: x = 8: 5\\\\8 x = 1 2 \\times 5\\\\8 x = 6 0\\\\x = 6 0 \\div 8\\\\x = 7. 5$ 故答案为: D",
+    "knowledge_points": [
+      "图形的放大与缩小",
+      "正比例的应用"
+    ],
+    "source": "17-18 六年级下·全国·课后作业",
+    "merged_id": "unit4_316",
+    "original_id": "unit4_choice_316"
+  },
+  {
+    "id": "unit4_choice_317",
+    "unit": "第四单元：比例",
+    "chapter": "比例的基本性质",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "如图：平行四边形a边上的高为b，c边上的高为d，根据这些信息，下列式子中（）不成立。\n![](images/2756777a0b7b8644de617cff0044da604ee565c7a2e0584c30ffd3687747d58f.jpg)",
+    "options": [
+      "A. a:c=d:b",
+      "B. a:c=b:d",
+      "C. $\\frac{a}{d} = \\frac{c}{b}$",
+      "D. $\\frac{b}{c}= \\frac {d}{a}$"
+    ],
+    "answer": "B",
+    "solution": "【分析】根据平行四边形的面积 $= \\text{底} \\times$ 高, 得出 $ab = cd$, 再利用比例的基本性质 (在比例里, 两个外项的积等于两个内项的积) 对给出的选项逐一分析, 做出选择。本题主要是利用平行四边形的面积公式与比例的基本性质解决问题。\n【详解】因为平行四边形a边上的高为b，c边上的高为d，所以ab=cd,A. a:c=d:b, cd=ab, 所以符合题意, 此选项正确; B. a:c=b:d, ad=bc, 与题意不符, 此选项错误; C. $\\frac{a}{d} = \\frac{c}{b}, ab = cd$, 符合题意, 此选项正确; D. $\\frac{b}{c} = \\frac{d}{a}, ab = dc$, 符合题意, 此选项正确;故答案为: B",
+    "knowledge_points": [
+      "比例的基本性质",
+      "平行四边形的面积"
+    ],
+    "source": "2017·重庆·小升初真题",
+    "merged_id": "unit4_317",
+    "original_id": "unit4_choice_317"
+  },
+  {
+    "id": "unit4_choice_318",
+    "unit": "第四单元：比例",
+    "chapter": "比例尺的应用",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "把线段比例尺![](images/162c5d8ca687de1414bfcdc40aeef497.png)改成数值比例尺，正确的是（）。",
+    "options": [
+      "A. $1: 600000$",
+      "B. $1: 200000$",
+      "C. $1: 20000$"
+    ],
+    "answer": "B",
+    "solution": "【分析】图上 1 厘米表示实际 2 千米, 把 2 千米换算成厘米, 写出图上距离与实际距离的比就是数值比例尺。熟练掌握线段比例尺和数值比例尺的转化方法是解答本题的关键。\n【详解】2千米 $= 200000$ 厘米比例尺为 $1: 200000$ 故答案为：B。",
+    "knowledge_points": [
+      "比例尺的意义"
+    ],
+    "source": "20-21六年级下·全国·单元测试",
+    "merged_id": "unit4_318",
+    "original_id": "unit4_choice_318"
+  },
+  {
+    "id": "unit4_choice_319",
+    "unit": "第四单元：比例",
+    "chapter": "比例的应用",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "乐乐沿着直尺的方向拉橡皮筋（如下图）。如果点 $A$ 的位置固定不变，将橡皮筋拉长，使点 $C$ 的位置到 $12 \\mathrm{~cm}$ 处，那么此时点 $B$ 的位置在（）cm处。\n![](images/772670612707ffe8474e14ebf077646a372c0f84f5837a0bbe1798b1926b7e68.jpg)",
+    "options": [
+      "A. 8",
+      "B. 9",
+      "C. 10",
+      "D. 11"
+    ],
+    "answer": "A",
+    "solution": "【分析】根据题意可知, BC 和 AC 长度之间的比值一定,据此列比例解答。此题考查了用比例解决实际问题, 找准比例关系,认真解答即可。\n【详解】解：设此时B点的位置在 $x$ 厘米处。$\\frac {x}{1 2} = \\frac {6}{9}\\\\9 x = 1 2 \\times 6\\\\x = 8$那么此时点 $B$ 的位置在 8 厘米处。故选择：A",
+    "knowledge_points": [
+      "比例的应用"
+    ],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_319",
+    "original_id": "unit4_choice_319"
+  },
+  {
+    "id": "unit4_choice_320",
+    "unit": "第四单元：比例",
+    "chapter": "正比例与反比例",
+    "question_type": "选择题",
+    "difficulty": "中等",
+    "question": "下面表示比例关系的图象是（）。\nA. ![](images/1a22196a444d195d2738856e63126418ee5632e1b5dc3ca894611de415647194.jpg)\nB.![](images/3dac187e42b9ee7c1d540195d76d4efe96a9a8fa1e31f63560690765ccebf82b.jpg)\nC. ![](images/7286d01ac367d612918c752dd8d0f43685e632f3fe07a257f031123228254c99.jpg)\nD.![](images/80d523c63c5304c2cc4f862cbbb3c728905dafb101f45215b584f37757cb56bd.jpg)",
+    "options": [
+      "A. ![](images/1a22196a444d195d2738856e63126418ee5632e1b5dc3ca894611de415647194.jpg)",
+      "B. ![](images/3dac187e42b9ee7c1d540195d76d4efe96a9a8fa1e31f63560690765ccebf82b.jpg)",
+      "C. ![](images/7286d01ac367d612918c752dd8d0f43685e632f3fe07a257f031123228254c99.jpg)",
+      "D. ![](images/80d523c63c5304c2cc4f862cbbb3c728905dafb101f45215b584f37757cb56bd.jpg)"
+    ],
+    "answer": "C",
+    "solution": "【分析】两种相关联的量，一种量变化，另一种量也随着变化，如果这两种量相对应的两个数的比值一定，那么这两种量就是正比例关系，如果这两种量相对应的两个数的乘积一定，这两种量就是反比例关系，据此解答。此题考查了辨别两个量之间是否存在比例关系，主要看它们对应的数是否比值或乘积一定。\n【详解】A.![](images/e4d9b5e6d905a1c420886bf7a9c572e5af85395bcae7055de96e52c1fec928d0.jpg)，当时间超过20时，路程不再变化，不成比例。B.![](images/d9267708a4e447d4964b340880ae4f3e8109b56ed86e18d18a0f7fed693bdb26.jpg), 60 分前后, 温度与时间的比值不同，不成比例。C.![](images/6a61738b463a9ec66bcbab173b4ca8d87ed2bb59bf9087bb79428ebbb840b4ec.jpg) 数量/件，数量随时间的变化而变化,并且比值一定，成正比例关系。D.![](images/2d35f68312c56936ef4e608e7385b77005169b03d839c5beaff83d172af3a755.jpg) ，售出的数量与剩下的数量的和一定，不成比例。故选择：C",
+    "knowledge_points": [
+      "正比例的意义及辨识",
+      "反比例的意义及辨识"
+    ],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_320",
+    "original_id": "unit4_choice_320"
+  },
+  {
+    "id": "unit4_choice_321",
+    "unit": "第四单元：比例",
+    "chapter": "比例尺的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一幅高铁地图的比例尺为 $0-250km$，改写成数值比例尺是( )，量得北京到上海的图上距离是 $5.3 \\mathrm{~cm}$, 则实际距离是( )km。",
+    "options": [],
+    "answer": "$1: 25000000$; 1325",
+    "solution": "【分析】根据比例尺的意义可把线段比例尺改写成数值比例尺。求北京到上海的实际距离，根据'图上距离÷比例尺 = 实际距离'，代入数据计算即可。此题主要考查学生对比例尺的认识与灵活应用。\n【详解】（1）根据线段比例尺可知，图上 1 厘米表示实际距离 $250km, 250km = 25000000 \\mathrm{~cm}$，写成比例尺：1:25000000;(2) $5.3 \\div \\frac{1}{25000000} = 5.3 \\times 25000000 = 132500000 (\\mathrm{~cm}) = 1325km$",
+    "knowledge_points": [
+      "比例尺的意义"
+    ],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_321",
+    "original_id": "unit4_choice_321"
+  },
+  {
+    "id": "unit4_choice_322",
+    "unit": "第四单元：比例",
+    "chapter": "比例尺的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一幅地图的比例尺是![](images/82bdf78d4c14d3e8dbe2362cd8a66d77.png) 把它改写成数值比例尺是( )。",
+    "options": [],
+    "answer": "1:3000000",
+    "solution": "【分析】观察线段比例尺可知, 图上 1 厘米表示实际 30 千米, 根据图上距离: 实际距离 = 比例尺, 转化成数值比例尺即可。要注意把 30 千米转化为以厘米为单位。\n【详解】1 厘米 : 30 千米 = 1 厘米 : 3000000 厘米 = 1:3000000一幅地图的比例尺是 ![](images/82bdf78d4c14d3e8dbe2362cd8a66d77.png) 把它改写成数值比例尺是 $1: 3000000$。",
+    "knowledge_points": [
+      "比例尺的意义"
+    ],
+    "source": "23-24 六年级下·湖北十堰·期末",
+    "merged_id": "unit4_322",
+    "original_id": "unit4_choice_322"
+  },
+  {
+    "id": "unit4_choice_323",
+    "unit": "第四单元：比例",
+    "chapter": "比例尺的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "比例尺![](images/cda7f3388d31ece439f0ec78cf3c1702.png)表示图上() 厘米的距离相当于实际距离的() 千米，把它化成数值比例尺是() 。如果量得 A、B 两地的距离是 8.5 厘米，那么 A、B 两地的实际距离是() 千米。",
+    "options": [],
+    "answer": "1;50 ;1:5000000或 $\\frac{1}{5000000}$; 425",
+    "solution": "【分析】根据线段比例尺的意义可知，比例尺![](images/cda7f3388d31ece439f0ec78cf3c1702.png) 表示图上1厘米相当于实际距离50千米;根据'图上距离：实际距离 $=$ 比例尺'把线段比例尺化成数值比例尺;已知量得 A、B 两地的距离是 8.5 厘米, 根据 '实际距离 =图上距离÷比例尺'求出 A、B 两地的实际距离。注意单位的换算：1千米 $= 100000$ 厘米。\n【详解】$1厘米：50千米=1 厘米：(50×100000) 厘米\\\\= 1: 5 0 0 0 0 0$$8.5 \\div \\frac {1}{5 0 0 0 0 0} \\\\ = 8. 5 \\times 5 0 0 0 0 0 0 \\\\ = 4 2 5 0 0 0 0 0 \\text {(厘 米)}$42500000 厘米 = 425 千米比例尺 0 50 100 150千米 表示图上1厘米的距离相当于实际距离的50千米，把它化成数值比例尺是1:5000000。如果量得 A、B 两地的距离是 8.5 厘米, 那么A、B 两地的实际距离是 425 千米。",
+    "knowledge_points": [
+      "图上距离与实际距离的换算",
+      "比例尺的意义"
+    ],
+    "source": "24-25 六年级下·海南海口·单元测试",
+    "merged_id": "unit4_323",
+    "original_id": "unit4_choice_323"
+  },
+  {
+    "id": "unit4_choice_324",
+    "unit": "第四单元：比例",
+    "chapter": "比例尺的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "比例尺 ![](images/d708cb4f389f065d057c808c77f16540.png) 表示图上 1 厘米相当于地面实际距离( )米, 把它改成数值比例尺是( )。如果量得 A、B 两地的距离上 4.5 厘米, 则实际距离是( )米。",
+    "options": [],
+    "answer": "50; $1: 5000 或 \\frac{1}{5000}$; 225",
+    "solution": "【分析】线段比例尺![](images/d708cb4f389f065d057c808c77f16540.png) 表示图上 1 厘米相当于地面实际距离 50 米; 再根据 '图上距离: 实际距离 = 比例尺' 把它改写成数值比例尺。已知 A、B 两地的距离上 4.5 厘米, 根据 '实际距离 = 图上距离÷比例尺'求出A、B两地的实际距离。注意单位的换算：1米 $= 100$ 厘米。\n【详解】$1 厘米：50 米=1 厘米：(50×100) 厘米\\\\ = 1: 5 0 0 0$ $4. 5 \\div \\frac {1}{5 0 0 0} \\\\ = 4. 5 \\times 5 0 0 0 \\\\ = 2 2 5 0 0 \\text {(厘 米)}$22500 厘米 = 225 米比例尺![](images/d708cb4f389f065d057c808c77f16540.png) 表示图上 1 厘米相当于地面实际距离（50）米，把它改成数值比例尺是（1:5000）。如果量得 A、B 两地的距离上 4.5 厘米, 则实际距离是 (225) 米。",
+    "knowledge_points": [
+      "图上距离与实际距离的换算",
+      "比例尺的意义",
+      "厘米和米之间的进率与换算"
+    ],
+    "source": "24-25 六年级下·海南海口·期中",
+    "merged_id": "unit4_324",
+    "original_id": "unit4_choice_324"
+  },
+  {
+    "id": "unit4_choice_325",
+    "unit": "第四单元：比例",
+    "chapter": "正比例的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "A 和 B 是自行车上的两个齿轮。A 转 2 圈, B 转 5 圈。如果 A 转了 150 圈, B 转( )圈; 如果 B 转了 90 圈, A 转( )圈。\n![](images/ac4fc87df13cf3ca879aaec6e574f49c0aa4712b06cd7676979088500e1de460.jpg)",
+    "options": [],
+    "answer": "375 ;36",
+    "solution": "【分析】两种相关联的量，一种量变化，另一种量也随着变化，如果这两种量中相对应的两个数的比值一定，这两种量就叫作成正比例的量，它们的关系叫作正比例关系，A转2圈，B转5圈，A和B转的圈数比2:5是一定的，那么A转的圈数和B转的圈数成正比例，据此解答。\n【详解】解：设如果 A 转了 150 圈，B 转 x 圈。$1 5 0: x = 2: 5\\\\2 x = 1 5 0 \\times 5\\\\ 2 x = 7 5 0 \\\\ x = 7 5 0 \\div 2 \\\\ x = 3 7 5$所以，如果 A 转了 150 圈，B 转 375 圈。解：设如果 B 转了 90 圈，A 转 y 圈。$y: 9 0 = 2: 5 \\\\ 5 y = 9 0 \\times 2 \\\\ 5 y = 1 8 0 \\\\ y = 1 8 0 \\div 5 \\\\ y = 3 6$所以，如果 B 转了 90 圈，A 转 36 圈。",
+    "knowledge_points": [
+      "正比例的应用"
+    ],
+    "source": "24-25 六年级下·海南海口·单元测试",
+    "merged_id": "unit4_325",
+    "original_id": "unit4_choice_325"
+  },
+  {
+    "id": "unit4_choice_326",
+    "unit": "第四单元：比例",
+    "chapter": "反比例的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "如果支架右侧第3个孔挂4个同样大的正方体，则支架左侧第2个孔应该挂_这样的正方体才保持平衡。\n![](images/afdbbc2bb53c98dac0c22b7810e2ae617370310315c666c3349b46b2d9b06183.jpg)",
+    "options": [],
+    "answer": "6",
+    "solution": "【分析】根据题意可知，孔数和挂正方体的数量成反比例，即左边的孔数 $\\times$ 挂的正方体数量 $=$ 右边的孔数 $\\times$ 挂的正方体数量，设左侧第 2 个孔应挂 $x$ 个正方体，列方程： $2 x = 3 \\times 4$，解方程，即可解答。\n【详解】解：设支架左侧第2个孔挂 $\\mathbf{x}$ 个正方体。$ 2 x = 3 \\times 4 \\\\ 2 x = 1 2 \\\\ x = 1 2 \\div 2 \\\\ x = 6$如果支架右侧第 3 个孔挂 4 个同样大的正方体, 则支架左侧第 2 个孔应该挂 6 这样的正方体才保持平衡。",
+    "knowledge_points": [
+      "列方程解含一个未知数的问题",
+      "反比例的应用"
+    ],
+    "source": "24-25 六年级下·海南海口·单元测试",
+    "merged_id": "unit4_326",
+    "original_id": "unit4_choice_326"
+  },
+  {
+    "id": "unit4_choice_327",
+    "unit": "第四单元：比例",
+    "chapter": "正比例与反比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "如图, 如果 a 与 b 成正比例，可以填( )；如果a与b成反比例，可以填( )。\n| a    | 3    | 5    |\n| ---- | ---- | ---- |\n| b    | 45   | ()   |",
+    "options": [],
+    "answer": "75; 27",
+    "solution": "【分析】两种相关联的量，一种量变化，另一种量也随着变化，如果这两种量中相对应的两个数的比值一定（也就是商一定），这两种量成正比例关系。如果a与b成正比例，则 $a:b = 3:45$；把 $a = 5$ 代入比例式，解比例求出b的值。两种相关联的量，一种量变化，另一种量也随着变化，如果这两种量中相对应的两个数的乘积一定，这两种量成反比例关系。如果a与b成反比例，则 $ab = 3 \\times 45$；把 $a = 5$ 代入比例式，解比例求出b的值。\n【详解】如果a与b成正比例，则 $a:b = 3:45$； 当 $a = 5$ 时 $5:b = 3:45$ 解： $3b = 5 \\times 45$ $3b = 225$ $b = 225 \\div 3$ $b = 75$ 如果a与b成反比例，则 $ab = 3 \\times 45$； 当 $a = 5$ 时 $5b = 3 \\times 45$ 解： $5b = 135$ $b = 135 \\div 5$ $b = 27$ 填空如下： 如果a与b成正比例，可以填（75）；如果a与b成反比例，可以填（27）。",
+    "knowledge_points": [
+      "正比例的应用",
+      "解比例",
+      "反比例的应用"
+    ],
+    "source": "2014 六年级·全国·课后作业",
+    "merged_id": "unit4_327",
+    "original_id": "unit4_choice_327"
+  },
+  {
+    "id": "unit4_choice_328",
+    "unit": "第四单元：比例",
+    "chapter": "正比例与反比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "如表中 $m$ 和 $n$ 是两个相关联的量。\n(1)当 $x = \\frac{1}{2}$ 时， $\\mathrm{m}$ 与 $\\mathrm{n}$ 成（）比例关系。\n| m    | 6    | x    |\n| ---- | ---- | ---- |\n| n    | 3    | 1/4  |\n(2)当 $x = (\\quad)$ 时， $m$ 与 $n$ 成反比例关系。",
+    "options": [],
+    "answer": "(1)正 (2)72",
+    "solution": "【分析】判断两个相关联的量之间成什么比例，就看这两个量是对应的比值一定，还是对应的乘积一定；如果是比值一定，就成正比例；如果是乘积一定，则成反比例。\n【详解】 (1) 当 $x = \\frac{1}{2}$ 时, $6 \\div 3 = 2$ $\\frac {1}{2} \\div \\frac {1}{4} = \\frac {1}{2} \\times 4 = 2$ $m \\div n = 2$ (一定), 比值一定, $m$ 与 $n$ 成正比例。所以，当 $x = \\frac{1}{2}$ 时， $\\mathrm{m}$ 与 $\\mathrm{n}$ 成正比例。(2) $\\mathrm{m}$ 与 $\\mathrm{n}$ 成反比例, 那么 $\\mathrm{m}$ 与 $\\mathrm{n}$ 的乘积一定,$\\frac {1}{4} x = 6 \\times 3$解: $\\frac{1}{4} x = 18\\\\ x = 1 8 \\div \\frac {1}{4} \\\\ x = 1 8 \\times 4 \\\\ x = 7 2$当 $x = 72$ 时, $m$ 与 $n$ 成反比例关系。",
+    "knowledge_points": [
+      "正比例的意义及辨识",
+      "反比例的意义及辨识",
+      "用字母表示数",
+      "数量关系"
+    ],
+    "source": "23-24 六年级下·贵州黔西·期末",
+    "merged_id": "unit4_328",
+    "original_id": "unit4_choice_328"
+  },
+  {
+    "id": "unit4_choice_329",
+    "unit": "第四单元：比例",
+    "chapter": "正比例的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "宁宁发现学校有一个漏水的水龙头, 他记录了这个水龙头的滴水情况。\n| 滴水量/毫升 | 15   | 30   | 45   | 60   | 75   | …    |\n| ----------- | ---- | ---- | ---- | ---- | ---- | ---- |\n| 时间/分     | 1    | 2    | 3    | 4    | 5    | …    |\n(1)滴水量和时间成( )比例。\n(2)宁宁在这个水龙头的下方放一个水桶, 1 小时可以接水 ( )升。",
+    "options": [],
+    "answer": "(1)正 (2) $0.9 \\mathrm{~g}$",
+    "solution": "【分析】（1）判断两个相关联的量之间成什么比例，就看这两个量是对应的比值一定，还是对应的乘积一定，如果是比值一定，就成正比例，如果是乘积一定，则成反比例。(2) 1 小时 $= 60$ 分钟, 用 60 乘 1 分钟的滴水量即可得解。\n【详解】 (1) 滴水量: 时间 $= 15: 1 = 30: 2 = 45: 3 = 60: 4 =$ 75:5=15所以滴水量和时间是比值一定，滴水量和时间成正比例。(2) 1时 $= 60$ 分 $6 0 \\times 1 5 = 9 0 0 (\\text {毫 升})$ $9 0 0 \\text {毫 升} = 0. 9 \\text {升}$ 1 小时可以接水 0.9 升。",
+    "knowledge_points": [
+      "正比例的应用",
+      "正比例的意义及辨识"
+    ],
+    "source": "23-24 六年级下·河南洛阳·期末",
+    "merged_id": "unit4_329",
+    "original_id": "unit4_choice_329"
+  },
+  {
+    "id": "unit4_choice_330",
+    "unit": "第四单元：比例",
+    "chapter": "正比例的意义及辨识",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "如图表示两辆汽车行驶的路程与相应的时间关系的图象, 完成填空。\n![](images/0e7cc6d40ff953416e07c8361f73dc15e0b0bb45ea455444b45db21fa425c111.jpg)\n(1)两辆汽车行驶的路程和时间都成( )比例关系。\n(2)从昆明到普洱大约有 320 千米, A 车从昆明到普洱大约要( )小时。\n(3)从图像上看，( )车的速度比较快。",
+    "options": [],
+    "answer": "(1)正 (2)4 (3) A",
+    "solution": "【分析】 (1) $x \\div y = k$ （一定）， $x$ 和 $y$ 成正比例关系；正比例图像是一条经过原点的直线，据此分析。(2) 看图可知, A 车 1 小时行驶 80 千米, 根据路程 $\\div$ 速度 = 时间, 列式计算即可。 (3) 观察图像, A 车 4 小时行驶 320 千米, B 车 6 小时行驶 320 千米, 用时越少速度越快。\n【详解】（1）观察图像，是一条经过原点的直线，两辆汽车行驶的路程和时间都成正比例关系。(2) $320 \\div 80 = 4$ (小时)从昆明到普洱大约有 320 千米, A 车从昆明到普洱大约要 4 小时。(3) 4 小时 $< 6$ 小时, 从图像上看, A 车的速度比较快。",
+    "knowledge_points": [
+      "正比例的意义及辨识"
+    ],
+    "source": "22-23 六年级下·云南·期中",
+    "merged_id": "unit4_330",
+    "original_id": "unit4_choice_330"
+  },
+  {
+    "id": "unit4_choice_331",
+    "unit": "第四单元：比例",
+    "chapter": "正比例图象的认识",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "下图是出租车所行的路程和时间图象。图中反映了当( )一定时，所行路程和时间成( )比例。照这样计算，这辆汽车8小时行( )千米。\n![](images/01dbb5a6691b4dfd6cb4daf9ea3de9ddd3adbeb0461ccd091fbe99b560ea3fb5.jpg)",
+    "options": [],
+    "answer": "速度 ;正; 640",
+    "solution": "【分析】根据正比例图象可知，两种相关联的量如果成正比例，则两种量会是一条经过原点的直线，则表示两个相关联的量成正比例关系，因此行驶的路程与时间成正比例，即速度一定； $80 \\div 1 = 80$ （千米）， $160 \\div 2 = 80$ （千米）， $240 \\div 3 = 80$ （千米），因此这辆汽车每小时行驶的速度是 80 千米；根据路程 $=$ 时间 $\\times$ 速度，即可求出这辆汽车 8 小时行驶的路程。明确正比例图象是一条经过原点的直线是解题的关键。\n【详解】这辆汽车每小时行驶的速度是 80 千米, 速度一定,它行驶的路程和时间成正比例;8小时行驶的路程为： $80 \\times 8 = 640$ 千米因此这辆汽车8小时行驶640千米。",
+    "knowledge_points": [
+      "正比例图象的认识",
+      "基础行程问题"
+    ],
+    "source": "22-23六年级下·吉林·期末",
+    "merged_id": "unit4_331",
+    "original_id": "unit4_choice_331"
+  },
+  {
+    "id": "unit4_choice_332",
+    "unit": "第四单元：比例",
+    "chapter": "比例尺的意义",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "用 表示图上 $1\\mathrm{cm}$ 相当于实际距离 $\\_\\mathrm{m}$，把它改写成数值比例尺是",
+    "options": [],
+    "answer": "20 ;1:2000",
+    "solution": "【分析】比例尺的种类：按表现形式分为数值比例尺与线段比例尺![](images/759ce1c72a311c024de0288c70949a7e5f02c04638651d443379fd7cbb2ea0b0.jpg)表示图上 $1 \\mathrm{~cm}$ 代表实际距离 $20 \\mathrm{~m}$, 再根据比例尺的意义, 比例尺 = 图上距离: 实际距离, 统一单位后再把数据代入, 即可求改写成数值比例尺。此题的解题关键是理解掌握比例尺的意义。\n【详解】0 20m用 表示图上1cm相当于实际距离 $20\\mathrm{m}$。$ 1 \\mathrm {c m}: 2 0 \\mathrm {m} \\\\ = 1 \\mathrm {c m}: 2 0 0 0 \\mathrm {c m} \\\\ = 1: 2 0 0 0$即把它改写成数值比例尺是 $1: 2000$。",
+    "knowledge_points": [
+      "比例尺的意义"
+    ],
+    "source": "21-22六年级下·北京·期中",
+    "merged_id": "unit4_332",
+    "original_id": "unit4_choice_332"
+  },
+  {
+    "id": "unit4_choice_333",
+    "unit": "第四单元：比例",
+    "chapter": "正比例与反比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "下表中, 如果 $x$ 与 $y$ 成正比例, 那么 $\\triangle = (\\quad)$; 如果 $x$ 与 $y$ 成反比例, 那么 $\\triangle = (\\quad)$。\n| x    | 9    | 4.5         |\n| ---- | ---- | ----------- |\n| y    | 3    | $\\triangle$ |",
+    "options": [],
+    "answer": "1.5 ; 6",
+    "solution": "【分析】两种相关联的量，一种量变化，另一种量也随着变化，如果这两种量中相对应的两个数的比值一定，这两种量就叫做成正比例的量，它们的关系叫做正比例关系；两种相关联的量，一种量变化，另一种量也随着变化，如果这两种量中相对应的两个数的乘积一定，这两种量就叫做成反比例的量，它们的关系叫做反比例关系；据此解答。掌握正、反比例关系的意义是解答题目的关键。\n【详解】如果 $x$ 与 $y$ 成正比例，那么 $x$ 与 $y$ 的比值一定。$4.5: \\triangle = 9: 3\\\\9 \\triangle = 4. 5 \\times 3\\\\9 \\triangle = 1 3. 5\\\\9 \\triangle \\div 9 = 1 3. 5 \\div 9\\\\\\triangle = 1. 5$如果 $x$ 与 $y$ 成反比例, 那么 $x$ 与 $y$ 的乘积一定。$4.5 \\times \\Delta = 9 \\times 3\\\\4.5 \\triangle = 27\\\\ \\\\4.5 \\triangle \\div 4.5 = 27 \\div 4.5\\\\\\triangle = 6$",
+    "knowledge_points": [
+      "正比例的应用",
+      "反比例的应用"
+    ],
+    "source": "22-23 六年级下·全国·单元测试",
+    "merged_id": "unit4_333",
+    "original_id": "unit4_choice_333"
+  },{
+    "id": "unit4_choice_334",
+    "unit": "第四单元：比例",
+    "chapter": "正比例图象的认识",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "如图的图象表示一个水龙头打开后的时间和出水量的关系。\n(1)看图填表。![](images/93512662f67ee2cd3ff0b47b949ccda976c94b96cd2aba4b600dcecdd5a14ed3.jpg)\n| 时间/秒   | 40   | ()   |\n| --------- | ---- | ---- |\n| 出水量/升 | ()   | 9    |\n(2)这个水龙头打开的时间与出水量成( )比例。\n(3)如果自来水管的内直径是 2 厘米, 水管内水的流速是每秒 8 厘米。一位同学去洗手, 走时忘记关掉水龙头, 5 分钟浪费( )升水。",
+    "options": [],
+    "answer": "(1) 45; 8 (2)正 (3) $7.536$",
+    "solution": "【分析】（1）分别找出 40 秒对应的出水量和 9 升对应的时间；观察统计图可知，这个水龙头 40 秒出水 8 升，出 9 升，水需要 45 秒据此填表；(2) 因为 $8 \\div 40 = 0.2$ (升), $9 \\div 45 = 0.2$ (升), 出水量 $\\div$ 打开时间 = 每秒的出水量 (一定), 出水量和打开时间的商一定, 则这个水龙头的出水量和打开时间成正比例; (3) 先根据圆柱的体积公式 $V = \\pi r^{2} h$ 算出每秒流的水所占的体积, 然后算出 5 分钟浪费了多少升水。\n【详解】 (1)\n| 时间/秒   | 40   | 60   |\n| --------- | ---- | ---- |\n| 出水量/升 | 6    | 9    |\n(2) 根据分析, 这个水龙头打开的时间与出水量成正比例;\n(3) $3.14 \\times (2 \\div 2)^{2} \\times 8\\\\ = 3. 1 4 \\times 1 ^ {2} \\times 8 \\\\ = 3. 1 4 \\times 8 \\\\=25.12 (立方厘米)$\n5分钟=300秒 $25.12 \\times 300 = 7536$ （立方厘米） 7536 立方厘米=7.536 升 所以5分钟浪费7.536升水。",
+    "knowledge_points": [
+      "正比例图象的认识",
+      "圆柱的体积"
+    ],
+    "source": "21-22 六年级下·福建福州·期中",
+    "merged_id": "unit4_334",
+    "original_id": "unit4_choice_334"
+  },
+  {
+    "id": "unit4_choice_335",
+    "unit": "第四单元：比例",
+    "chapter": "比例尺的意义",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一张地图的比例尺为 ![](images/b7252ff946e99caca392ca3452b12796.png), 把这个线段比例尺改写为数值比例尺是( )。",
+    "options": [],
+    "answer": "1:4000000",
+    "solution": "【分析】观察线段比例尺, 图上 $1 \\mathrm{~cm}$ 对应 $40km$, 根据图上距离: 实际距离 = 比例尺, 写出图上距离与实际距离的比,化简即可。比例尺没有单位名称。为了方便, 通常把比例尺的前项化作 1 (图上距离大于实际距离的, 常把后项化为 1)。\n【详解】 $1 \\mathrm{~cm}: 40km = 1 \\mathrm{~cm}: 4000000 \\mathrm{~cm} = 1: 4000000$ ，这个线段比例尺改写为数值比例尺是 $1: 4000000$ 。",
+    "knowledge_points": [
+      "比例尺的意义"
+    ],
+    "source": "20-21 六年级下·湖北十堰·期中",
+    "merged_id": "unit4_335",
+    "original_id": "unit4_choice_335"
+  },
+  {
+    "id": "unit4_choice_336",
+    "unit": "第四单元：比例",
+    "chapter": "反比例的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一辆自行车前齿轮有 36 齿, 后齿轮有 18 齿, 前后齿轮齿数比是( ); 若前齿轮转 2 圈, 后齿轮转( )圈。\n![](images/20c3d1f432fef215aa284cf1921d3a00017d4e1e1c8fa953ce23158499af8396.jpg)",
+    "options": [],
+    "answer": "2:1 ;4",
+    "solution": "【分析】由题意可知，前轮齿数：后轮齿数 $= 36:18$ ，利用比的基本性质把 36:18 化为最简整数比；前后齿轮行驶的路程相等，齿轮齿数 $\\times$ 转的圈数 = 行驶的路程，等量关系式：后轮齿数 $\\times$ 后轮转的圈数 = 前轮齿数 $\\times$ 前轮转的圈数，据此解答。本题主要考查应用比例解决实际问题, 理解前后轮行驶的路程相等并找出等量关系式是解答题目的关键。\n【详解】前轮齿数：后轮齿数 $= 36:18 =\\\\(3 6 \\div 1 8) \\quad : \\quad (1 8 \\div 1 8) \\\\= 2: 1$\n解：设前齿轮转 2 圈，后齿轮转 x 圈。\n$1 8 x = 3 6 \\times 2\\\\1 8 x = 7 2\\\\ x = 7 2 \\div 1 8 \\\\ x = 4$\n所以，若前齿轮转 2 圈，后齿轮转 4 圈。",
+    "knowledge_points": [
+      "反比例的应用",
+      "比的意义"
+    ],
+    "source": "21-22 六年级下·河北沧州·期末",
+    "merged_id": "unit4_336",
+    "original_id": "unit4_choice_336"
+  },
+  {
+    "id": "unit4_choice_337",
+    "unit": "第四单元：比例",
+    "chapter": "正比例的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "如图是一列动车行驶情况统计图。这列动车每分钟行驶( )km；这列动车行驶的路程和时间成比例。按照这样的速度，要行驶480km,需要( )小时。\n![](images/65a52efd94b77862e0791009f157feaeca371196c8c75ec2781de007bafe50f8.jpg)",
+    "options": [],
+    "answer": "5; 1.6",
+    "solution": "【分析】第一个空，观察统计图，找到1分钟对应的路程即可；第二个空，用路程÷速度=时间，直接计算即可。从图像上看, 成正比例关系的图像是一条经过原点的直线。\n【详解】 $480 \\div 5 = 96$ （分钟） $= 1.6$ （小时）这列动车每分钟行驶 $5km$; 这列动车行驶的路程和时间成比例。按照这样的速度, 要行驶 $480km$, 需要 1.6 小时。",
+    "knowledge_points": [
+      "正比例的应用"
+    ],
+    "source": "20-21六年级下·广东梅州·期末",
+    "merged_id": "unit4_337",
+    "original_id": "unit4_choice_337"
+  },
+  {
+    "id": "unit4_choice_338",
+    "unit": "第四单元：比例",
+    "chapter": "比例尺的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "下图是一张地图上的线段比例尺, 如果将它转化成数值比例尺是____; 在这张地图上量得两城市之间的距离为 $5.5 \\mathrm{~cm}$, 则两城市之间的实际距离是_____km。\n![](images/5c31813355a41391436c7141e9b19d834cf025f151a69eb604c4648b6cbcb848.jpg)",
+    "options": [],
+    "answer": "$1: 3000000$ ;165",
+    "solution": "【分析】线段比例尺 1 厘米表示实际距离 30 千米, 把 30 千米换算成厘米后求出图上距离与实际距离的比即可化成数值比例尺; 实际距离 = 图上距离 $\\times$ 比例尺; 据此解答。明确数值比例尺和线段比例尺互化的方法以及实际距离、图上距离和比例尺之间的关系是解答本题的关键。\n【详解】 $30km = 3000000 \\mathrm{~cm}$, 则数值比例尺是 $1: 3000000$ 实际距离: $5.5 \\times 30 = 165 \\text{ (km)}$",
+    "knowledge_points": [
+      "图上距离与实际距离的换算",
+      "比例尺的意义"
+    ],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_338",
+    "original_id": "unit4_choice_338"
+  },
+  {
+    "id": "unit4_choice_339",
+    "unit": "第四单元：比例",
+    "chapter": "比例尺的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "一幅图的比例尺是![](images/ecd53fc5b77efe66357eb505316c324d.png)，那么图上的1厘米表示实际距离()；实际距离 50 千米在图上要画()厘米, 把这个线段比例尺改写成数值比例尺是()。",
+    "options": [],
+    "answer": "5千米 ;10; 1:500000",
+    "solution": "【分析】此题主要考查了比例尺的认识，图上距离：实际距离 = 比例尺，观察线段比例尺可知，图上的 1 厘米表示实际距离 5 千米；已知比例尺与实际距离，要求图上距离，实际距离 $\\times$ 比例尺 = 图上距离，据此计算；把线段比例尺化成数值比例尺时，要将单位化统一，据此解答。认识比例尺和数值比例尺, 知道图上距离: 实际距离 = 比例尺是解决此题的关键。\n【详解】一幅图的比例尺是![](images/ecd53fc5b77efe66357eb505316c324d.png)，那么图上的1 厘米表示实际距离 5 千米; 实际距离 50 千米在图上要画 10 厘米, 把这个线段比例尺改写成数值比例尺是 $1: 500000$。",
+    "knowledge_points": [
+      "图上距离与实际距离的换算",
+      "比例尺的意义"
+    ],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_339",
+    "original_id": "unit4_choice_339"
+  },
+  {
+    "id": "unit4_choice_340",
+    "unit": "第四单元：比例",
+    "chapter": "正比例的意义及辨识",
+    "question_type": "填空题",
+    "difficulty": "容易",
+    "question": "下图是某造纸厂今年 5 月上旬的生产情况统计图，这个造纸厂 4 天的生产量是（）；生产 640 吨纸需要（）天；这家造纸厂的生产量与时间成( )比例。\n![](images/17372f9abb14a8c224f9e0054fb5a51186ee03761c0cde4ce739ef7430231249.jpg)",
+    "options": [],
+    "answer": "320; 8; 正",
+    "solution": "【分析】如果两个数成正比例关系, 即两个数的比值一定,如果两个数成反比例关系, 则两个数的乘积一定; 根据统计图可知, $\\frac{80}{1} = \\frac{160}{2} =$ ……, 比值一定, 所以生产量与时间成正比例, 由此解答即可。此题考查了正比例关系的运用，也可运用比的性质解答。\n【详解】这个造纸厂 4 天的生产量是 320 ; 生产 640 吨纸需要 8 天; 这家造纸厂的生产量与时间成正比例。",
+    "knowledge_points": [
+      "正比例的意义及辨识"
+    ],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_340",
+    "original_id": "unit4_choice_340"
+  },
+  {
+    "id": "unit4_choice_341",
+    "unit": "第四单元：比例",
+    "chapter": "平行四边形的面积",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "如图, 已知平行四边形 $A B C D$ 中, 线段 $A B$ 长 $30 \\mathrm{~cm}$, 那么线段 $A D$ 长() cm。\n![](images/88f06e1b8835f24527b13937f1a52eec0db845f0aa70ade682f4be8adef6668a.jpg)",
+    "options": [],
+    "answer": "20",
+    "solution": "【分析】根据平行四边形的面积 $= \\text{底} \\times$ 高, 先求出平行四边形的面积, 再除以底边对应的高即可。此题考查了平行四边形的面积计算, 注意底与高的对应关系。\n【详解】 $18 \\times 30 \\div 27\\\\= 5 4 0 \\div 2 7 \\\\ = 2 0$\n线段 $AD$ 长 20 厘米。",
+    "knowledge_points": [
+      "平行四边形的面积"
+    ],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_341",
+    "original_id": "unit4_choice_341"
+  },
+  {
+    "id": "unit4_choice_342",
+    "unit": "第四单元：比例",
+    "chapter": "图形的放大与缩小",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "如果把圆的半径按 1:2 缩小，那么缩小后的圆与原来的圆的面积比是( )。",
+    "options": [],
+    "answer": "1:4",
+    "solution": "【分析】可以设原来圆的半径是 1 , 则缩小后的半径为 $\\frac{1}{2}$ ,根据圆的面积公式分别表示出缩小后的圆和原来的圆的面积, 再相比即可。一个图形放大或缩小的倍数是指对应边放大或缩小的倍数，面积放大或缩小这个倍数的平方倍。\n【详解】解：设原来圆的半径为 1，则缩小后的半径为 $\\frac{1}{2}$。缩小后的圆和原来圆的面积比：\n$ (\\pi \\times \\frac {1}{2} \\times \\frac {1}{2}) \\quad : \\quad (\\pi \\times 1 \\times 1) \\\\ = \\frac {1}{4} \\pi : \\pi \\\\ = 1: 4$",
+    "knowledge_points": [
+      "图形的放大与缩小"
+    ],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_342",
+    "original_id": "unit4_choice_342"
+  },
+  {
+    "id": "unit4_choice_343",
+    "unit": "第四单元：比例",
+    "chapter": "比例的应用",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "笑笑用 $100 \\mathrm{~g}$ 水和 $20 \\mathrm{~g}$糖制成一杯糖水。如果多放 $10 \\mathrm{~g}$ 糖, 要保证这杯糖水与原来一样甜, 应再加水( ) $\\mathrm{g}$。",
+    "options": [],
+    "answer": "50",
+    "solution": "【分析】根据题意, 将 $20 \\mathrm{~g}$ 糖放入 $100 \\mathrm{~g}$ 水中, 此时糖和水的比为 $20: 100$; 多放入 $10 \\mathrm{~g}$ 糖, 要保证这杯糖水和原来一样甜, 即糖水中糖和水的比不变, 设应再加入 $\\mathrm{xg}$ 水, 根据糖和水的比不变, 列出比例解答即可。此题主要考查学生解比例的实际应用。\n【详解】解：设应再加入 $\\mathrm{xg}$ 水。\n$2 0: 1 0 0 = 3 0: (1 0 0 + x)\\\\2 0 \\times (1 0 0 + x) = 3 0 \\times 1 0 0\\\\2 0 0 0 + 2 0 x = 3 0 0 0\\\\2 0 x = 1 0 0 0\\\\x = 5 0$",
+    "knowledge_points": [
+      "比例的应用"
+    ],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_343",
+    "original_id": "unit4_choice_343"
+  },
+  {
+    "id": "unit4_choice_344",
+    "unit": "第四单元：比例",
+    "chapter": "正比例与反比例",
+    "question_type": "填空题",
+    "difficulty": "中等",
+    "question": "x, y 均不为 0 , 如果 $\\frac{4}{x} = 2 y$, 那么 x 与 y 成( )比例关系; 如果 $y = \\frac{5}{8} x$, 那么 x 与 y 成( )比例关系。",
+    "options": [],
+    "answer": "反;正",
+    "solution": "【分析】 $\\frac{4}{x} = 2y$ 经计算得: $xy = \\frac{4}{2} = 2; y = \\frac{5}{8} x$ 经计算得: $x:y = 8:5 = \\frac{8}{5}$ 。再根据正比例、反比例的概念进行解答。本题考查了了正、反比例的相关知识。明确正、反比例的概念，是解答本题的关键。\n【详解】因为 $\\frac{4}{x} = 2 y$, 所以 $x y = \\frac{4}{2} = 2$ (一定), 所以 x 与 y 成反比例关系; 因为 $y = \\frac{5}{8} x$, 所以 $x: y = 8: 5 = \\frac{8}{5}$ (一定), 所以 x 与 y 成正比例关系。",
+    "knowledge_points": [
+      "正比例的意义及辨识",
+      "反比例的意义及辨识"
+    ],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_344",
+    "original_id": "unit4_choice_344"
+  },
+  {
+    "id": "unit4_choice_345",
+    "unit": "第四单元：比例",
+    "chapter": "比例尺应用",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "甲、乙两城市之间的距离是 120 千米, 把它画在比例尺为 $0 \\sim 50$ 千米的地图上应画 2.4 厘米。( )",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】根据线段比例尺可知, 图上距离 1 厘米表示实际距离 50 千米, 用实际距离除以 1 厘米所代表的距离, 即可求出应画的图上距离。明确线段比例尺的意义是解答本题的关键。\n【详解】 $120 \\div 50 = 2.4$ （厘米）图上距离为 2.4 厘米, 原题说法正确;故答案为: √",
+    "knowledge_points": [
+      "比例尺应用"
+    ],
+    "source": "19-20 六年级下·重庆万州·期中",
+    "merged_id": "unit4_345",
+    "original_id": "unit4_choice_345"
+  },
+  {
+    "id": "unit4_choice_346",
+    "unit": "第四单元：比例",
+    "chapter": "正比例与反比例",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "X 与 Y 是两种相关联的量, $X - Y = 0$, 那么 X 与 Y 不成比例。( )",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】成正比例的量：两种相关联的量，若一种量变化，另一种也随着变化，且这两种量中相对应的两个数的比值（商）一定，这两种量就叫做成正比例的量；成反比例的量：两种相关联的量，若一种量变化，另一种也随着变化，如果这两种中相对应的两个数的积一定，这两种量就叫做成反比例的量。在本题中，相关联的两种量 X 和 Y，它们的数量关系可以转化为 $X = Y$, 这样 X 和 Y 就成正比例。解题时，对原题提供的数量关系的形式所做的变形是很重要的，变形恰当，就能够及时发现其存在的比例关系。\n【详解】由分析得:X与Y是两种相关联的量， $\\mathrm{X} - \\mathrm{Y} = 0$ ，那么X与Y成正比例。原说法是错误的。故答案为: ×",
+    "knowledge_points": [
+      "正比例的意义及辨识",
+      "反比例的意义及辨识"
+    ],
+    "source": "2021 六年级下·全国·专题练习",
+    "merged_id": "unit4_346",
+    "original_id": "unit4_choice_346"
+  },
+  {
+    "id": "unit4_choice_347",
+    "unit": "第四单元：比例",
+    "chapter": "比例尺应用",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "在美术本上画了一栋50 米高的房子，比较合适的比例尺是 $1: 50$。（）",
+    "options": [],
+    "answer": "×",
+    "solution": "【分析】因为图上距离：实际距离 $=$ 比例尺，所以图上距离 $=$ 实际距离 $\\times$ 比例尺，50 米 $\\times \\frac{1}{50} = 1$ （米），即画在纸上的房子的高度是 1 米，这显然不正确。仅从比例尺的数据不容易判断它是否合适，但是当求出图上距离后，一切豁然开朗，能够准确地做出判断。\n【详解】由分析得：在美术本上画了一栋50米高的房子，比较合适的比例尺是1:50。原题说法错误。故答案为×。",
+    "knowledge_points": [
+      "比例尺应用"
+    ],
+    "source": "2021 六年级下·全国·专题练习",
+    "merged_id": "unit4_347",
+    "original_id": "unit4_choice_347"
+  },
+  {
+    "id": "unit4_choice_348",
+    "unit": "第四单元：比例",
+    "chapter": "比例尺的意义",
+    "question_type": "判断题",
+    "difficulty": "中等",
+    "question": "比例尺![](images/2d325958d7fdeed3de07f81e03941845.png)倍。（ ）",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】观察线段比例尺可知, 图上 1 厘米表示 10 千米,统一单位后，用实际距离÷图上距离即可。\n【详解】10千米 $= 1000000$ 厘米 $1 0 0 0 0 0 0 \\div 1 = 1 0 0 0 0 0$比例尺![](images/2d325958d7fdeed3de07f81e03941845.png) 表示实际距离是图上距离的 1000000 倍, 说法正确。故答案为: √",
+    "knowledge_points": [
+      "比例尺的意义"
+    ],
+    "source": "22-23六年级下·安徽马鞍山·期末",
+    "merged_id": "unit4_348",
+    "original_id": "unit4_choice_348"
+  },
+  {
+    "id": "unit4_choice_349",
+    "unit": "第四单元：比例",
+    "chapter": "正比例图象的认识",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "从下面图象中可以知道斑马、长颈鹿各自的奔跑路程与奔跑时间都成正比例。（ ）\n![](images/4a987a34767da90b084f9c7b55d28150d6c5480ada3aba18ce49c1da7f446d4f.jpg)",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】两种相关联的量，一种量变化，另一种量也随着变化，如果这两种量中相对应的两个数的比值一定（也就是商一定），这两种量成正比例关系。正比例关系的图象是一条直线，据此判断。\n【详解】斑马、长颈鹿各自的奔跑路程与奔跑时间的关系图象是一条直线，表示斑马、长颈鹿各自的奔跑速度一定，即路程与时间的比值一定，所以从图象中可以知道斑马、长颈鹿各自的奔跑路程与奔跑时间都成正比例。\n原题说法正确。故答案为: √",
+    "knowledge_points": [
+      "正比例图象的认识",
+      "正比例的意义及辨识"
+    ],
+    "source": "23-24 六年级下·湖北武汉·期末",
+    "merged_id": "unit4_349",
+    "original_id": "unit4_choice_349"
+  },
+  {
+    "id": "unit4_choice_350",
+    "unit": "第四单元：比例",
+    "chapter": "比例尺的意义",
+    "question_type": "判断题",
+    "difficulty": "容易",
+    "question": "把线段比例尺![](images/3c2e3714a8c97ca4a37cbe89e5f107b7ae2ff5ffcc034d75b60924c04a12cc89.jpg)改写成数值比例是 $1:8000000$。（每一小段有 1 厘米) ( )",
+    "options": [],
+    "answer": "√",
+    "solution": "【分析】根据比例尺的意义可知，题目中的线段比例尺表示图上1厘米代表实际距离80千米，根据比例尺=图上距离：实际距离，单位换算后代入数据即可求出数值比例尺。此题的解题关键是理解比例尺的意义以及掌握比例尺不同的种类。\n【详解】$1 厘米：80 千米＝1厘米∶8000000厘米＝1∶8000000$改写成数值比例尺是 $1: 8000000$。故答案为: √\n=1 厘米：8000000 厘米\n$= 1:8000000$",
+    "knowledge_points": [
+      "比例尺的意义"
+    ],
+    "source": "2022·湖南湘西·小升初真题",
+    "merged_id": "unit4_350",
+    "original_id": "unit4_choice_350"
+  },
+  {
+    "id": "unit4_choice_351",
+    "unit": "第四单元：比例",
+    "chapter": "图形的放大与缩小",
+    "question_type": "作图题",
+    "difficulty": "中等",
+    "question": "画出图形①按 $2:1$ 放大后的图形, 画出图形②按 $1:3$ 缩小后的图形。\n![](images/54a905034bd1f3274b010d4c7348645c6466f1711fe393128a7f08aa45ec970d.jpg)",
+    "options": [],
+    "answer": "见详解",
+    "solution": "【分析】把图形按照 2:1 放大, 就是将图形的每一条边放大到原来的 2 倍; 把图形按照 1:3 缩小, 就是将图形的每一条边缩小到原来的 $\\frac{1}{3}$ 。据此先确定放大或缩小后的边长, 然后作图即可。本题考查图形按比例放大或缩小的相关知识。缩小后或放大后的图形与原图形相比，形状相同，大小不同。\n【详解】\n![](images/fc64fc7ec10f2a62c39facf7d775f872e9369a475dc89b2db2ff886075f9fbb1.jpg)",
+    "knowledge_points": [
+      "图形的放大与缩小"
+    ],
+    "source": "20-21 六年级下·全国·单元测试",
+    "merged_id": "unit4_351",
+    "original_id": "unit4_choice_351"
+  },
+  {
+    "id": "unit4_choice_352",
+    "unit": "第四单元：比例",
+    "chapter": "图形的放大与缩小",
+    "question_type": "作图题",
+    "difficulty": "中等",
+    "question": "画一画。\n![](images/e99d437b88b99806baf5687387e9c0d60751d2eadf7046a8d3a64dcb21889ef3.jpg)\n(1) 画出图形 A 按 $2:1$ 放大后的图形。\n(2) 把图形 B 绕点 O 顺时针旋转 $90^{\\circ}$。\n(3) 把图形 C 先向左平移 5 格, 再向上平移 6 格, 画出最终得到的图形。\n(4) 画出图形 D 的另一半, 使它成为一个轴对称图形。",
+    "options": [],
+    "answer": "见详解",
+    "solution": "【分析】（1）根据图形放大的方法，先计算按 $2:1$ 放大后的平行四边形的底和高各是多少，再作图即可。\n(2) 根据图形旋转的性质, 以 $\\mathrm{O}$ 点为旋转中心, 找出构成图形 $\\mathrm{B}$ 的关键点, 分别作出各关键点绕 $\\mathrm{O}$ 点顺时针旋转 $90^{\\circ}$ 的对应点, 顺次连接旋转后的关键点即可。\n(3) 根据图形平移的性质, 找出构成图形的关键点, 过关键点沿平移方向画出平行线, 由平移的距离确定关键点平移后对应点的位置, 最后连接对应点即可。\n(4) 根据轴对称图形的性质, 各对称点到对称轴的距离相等, 据此先描出各对称点的位置, 然后顺连接各点画出轴对称图形的另一半。\n【详解】 (1) $3 \\times 2 = 6$ $2 \\times 2 = 4$ (1) (2) (3) (4) 作图如下:\n![](images/feaae272769a61397c91b7a1a72e0c027941f4fc556c45977dcc02cac7d58146.jpg)",
+    "knowledge_points": [
+      "图形的放大与缩小",
+      "作旋转后的图形",
+      "补全轴对称图形",
+      "作平移后的图形"
+    ],
+    "source": "2015·山东·小升初真题",
+    "merged_id": "unit4_352",
+    "original_id": "unit4_choice_352"
+  },
+  {
+    "id": "unit4_choice_353",
+    "unit": "第四单元：比例",
+    "chapter": "应用比例尺画图",
+    "question_type": "作图题",
+    "difficulty": "中等",
+    "question": "黄晴家在学校正西方向300米，李海家在黄晴家北偏东 $45^{\\circ}$ 方向100米处，王欣家在李海家正东方向 400 米处，先确定比例尺，再画出他们三家和学校的位置平面图。我选的线段比例尺是（），数值比例尺是（）。\n![](images/c7100ab780841493c7dd85dab912922f3c774c0d81995fc791842ea51f7f2bef.jpg)",
+    "options": [],
+    "answer": "$100 \\mathrm{~m}$ ；1:10000 见详解",
+    "solution": "【分析】因为三家距离各自观测点的距离都是几百米，所以比例尺选择图上距离1厘米代表实际距离100米，画出线段比例尺，数值比例尺是1厘米：100米=1厘米：10000厘米=1:10000，据此画出图即可。\n【详解】线段比例尺： $100\\mathrm{m}$ 数值比例尺：1:10000图上：黄晴家在学校正西方向 $300 \\div 100 = 3$ （厘米）处；李海家在黄晴家北偏东 $45^{\\circ}$ 方向 $100 \\div 100 = 1$ （厘米）处；王欣家在李海家正东方向 $400 \\div 100 = 4$ （厘米）处；\n![](images/04777e8bfe0f6a4711e4a1862852a16d598750353835ed3699d81da4b42f7df9.jpg)",
+    "knowledge_points": [
+      "根据方向、角度和距离确定物体的位置",
+      "应用比例尺画图"
+    ],
+    "source": "24-25 六年级下·全国·单元测试",
+    "merged_id": "unit4_353",
+    "original_id": "unit4_choice_353"
+  },
+  {
+    "id": "unit4_choice_354",
+    "unit": "第四单元：比例",
+    "chapter": "图形的放大与缩小",
+    "question_type": "作图题",
+    "difficulty": "中等",
+    "question": "操作。\n![](images/a993415dcc700561edb08b3e72f98ff13068361e64c7cd3f3bbe4de1b390cff3.jpg)\n(1) 按 $1:3$ 画出长方形缩小后的图形。\n(2) 按 $2: 1$ 画出梯形放大后的图形。",
+    "options": [],
+    "answer": "见详解",
+    "solution": "【分析】（1）图中长方形按 $1:3$ 缩小，则原来长方形的长和宽都除以 3，求出缩小后长方形的长和宽，据此画出缩小后的长方形。(2) 图中梯形按 $2: 1$ 放大, 则原来梯形的上底、下底和高都乘 2 , 求出放大后梯形的上底、下底和高, 据此画出放大后的梯形。\n【详解】 (1) 缩小后长方形的长: $6 \\div 3 = 2$缩小后长方形的宽: $3 \\div 3 = 1$缩小后的长方形如下图。(2) 放大后梯形的上底: $4 \\times 2 = 8$放大后梯形的下底: $2 \\times 2 = 4$放大后梯形的高: $2 \\times 2 = 4$放大后的梯形如下图。\n![](images/f5cc1ee73b655d8a071212b90395ba2727f4da35439a2f84fbe7ae2b3675d859.jpg)",
+    "knowledge_points": [
+      "图形的放大与缩小"
+    ],
+    "source": "24-25 六年级下·全国·单元测试",
+    "merged_id": "unit4_354",
+    "original_id": "unit4_choice_354"
+  },
+  {
+    "id": "unit4_choice_355",
+    "unit": "第四单元：比例",
+    "chapter": "应用比例尺画图",
+    "question_type": "作图题",
+    "difficulty": "中等",
+    "question": "小明家正北方向300米是科技馆，科技馆正东方向600米是动物园，动物园南偏西 $30^{\\circ}$ 方向 400 米是书店。画出上述地点的平面图。(比例尺是 $1: 10000$ )\n![](images/8dae9040c83694e067bbdbf8aed86c8f.png)",
+    "options": [],
+    "answer": "见详解",
+    "solution": "【分析】以图上的 '上北下南, 左西右东' 为准, 比例尺是 $1: 10000$ 表示图上 1 厘米相当于实际距离 100 米。在小明家正北方向上画 $300 \\div 100 = 3$ 厘米长的线段即是科技馆;在科技馆正东方向上画 $600 \\div 100 = 6$ 厘米长的线段即是动物园;在动物园南偏西 $30^{\\circ}$ 方向上画 $400 \\div 100 = 4$ 厘米长的线段即是书店。\n【详解】10000 厘米 = 100 米比例尺 1:10000 表示图上 1 厘米相当于实际距离 100 米。$300 \\div 100 = 3 (厘米) 600\\div100＝6（厘米);400 \\div 100 = 4 （厘米）$\n$600 \\div 100 = 6$ （厘米）\n如图:\n![](images/7559d5af0f22ba89c3ca585cb5c4e20fe70e3163471d64fcff8b66b60bbc946a.jpg)",
+    "knowledge_points": [
+      "根据方向、角度和距离确定物体的位置",
+      "应用比例尺画图"
+    ],
+    "source": "23-24六年级下·内蒙古通辽·期中",
+    "merged_id": "unit4_355",
+    "original_id": "unit4_choice_355"
+  },
+  {
+    "id": "unit4_choice_356",
+    "unit": "第四单元：比例",
+    "chapter": "图形的放大与缩小",
+    "question_type": "作图题",
+    "difficulty": "中等",
+    "question": "(1) 画出三角形先绕O点逆时针方向旋转 $90^{\\circ}$, 再向左平移两格后的图形。\n(2) 画出原来三角形按 $2:1$ 放大后的图形。\n![](images/d9f01f124ab3f1f3c20e511186cf4917ff922d64a177356250b4b59adbef3921.jpg)",
+    "options": [],
+    "answer": "见详解",
+    "solution": "【分析】（1）根据旋转的特征，三角形绕点O逆时针旋转 $90^{\\circ}$，先把点O所在的两条直角边分别绕此点按逆时针方向旋转 $90^{\\circ}$，再画出旋转后的三角形的斜边，得到旋转后的三角形；把旋转后的三角形的三个顶点都向左平移两格得到它们的对应点，再把对应点依次连接起来组成三角形，据此画图。(2) 原来三角形按 2:1 放大, 也就是把对应的底和高都扩大到原来的 2 倍, 已知三角形的底有 3 格, 高有 2 格, 则放大后的三角形的底是 $3 \\times 2 = 6$ (格), 高是 $2 \\times 2 = 4$ (格),据此画图。\n![](images/ec46b20d242f9f4ac888c0cc16260812f93a7201d99d542f9aa5521962ef87d1.jpg)\n【详解】![](images/7f8ebcb9364afe4cf09f3805fc3ff838.png)",
+    "knowledge_points": [
+      "图形的放大与缩小",
+      "作旋转后的图形",
+      "作平移后的图形"
+    ],
+    "source": "23-24 六年级下·广东韶关·期末",
+    "merged_id": "unit4_356",
+    "original_id": "unit4_choice_356"
+  },
+  {
+    "id": "unit4_choice_357",
+    "unit": "第四单元：比例",
+    "chapter": "应用比例尺画图",
+    "question_type": "作图题",
+    "difficulty": "中等",
+    "question": "在图中按要求画出指定的位置。\n![](images/cc812281248b20378cab52d7d3b50b158715a23efd243c29128c41451f97e471.jpg)\n(1) 医院在学校正西方向 2000 米处。\n(2) 图书馆在学校北偏东 $60^{\\circ}$ 大约 1500 米处。\n(3) 菜市场在图书馆南偏东 $45^{\\circ}$ 大约 1000 米处。",
+    "options": [],
+    "answer": "见详解",
+    "solution": "【分析】以图上的 '上北下南, 左西右东' 为准, 比例尺 $1: 50000$ 表示图上 1 厘米相当于实际距离 500 米。(1) 在学校正西方向上画 $2000 \\div 500 = 4$ 厘米长的线段, 即是医院; (2) 在学校北偏东 $60^{\\circ}$ 上画 $1500 \\div 500 = 3$ 厘米长的线段,即是图书馆; (3) 在图书馆南偏东 $45^{\\circ}$ 上画 $1000 \\div 500 = 2$ 厘米长的线段,即是菜市场。\n【详解】50000 厘米 = 500 米比例尺 1:50000 表示图上 1 厘米相当于实际距离 500 米。$2000 \\div 500 = 4$ （厘米）$1500 \\div 500 = 3$ （厘米） $1000 \\div 500 = 2$ （厘米）\n如图:\n![](images/390578038f28afe1ff19d329c21e8455429700ba7e428b861719c67a02310bb5.jpg)",
+    "knowledge_points": [
+      "根据方向、角度和距离确定物体的位置",
+      "应用比例尺画图",
+      "比例尺的意义"
+    ],
+    "source": "23-24六年级下·安徽黄山·期末",
+    "merged_id": "unit4_357",
+    "original_id": "unit4_choice_357"
+  },
+  {
+    "id": "unit4_choice_358",
+    "unit": "第四单元：比例",
+    "chapter": "图形的放大与缩小",
+    "question_type": "作图题",
+    "difficulty": "容易",
+    "question": "分别按3:1和1:2的比画出长方形放大和缩小后的图形。\n![](images/0470787b40cba4b5fa74c0af4fa6fa11a8622ff9c10d4cf53573c315b5f74ce9.jpg)",
+    "options": [],
+    "answer": "见详解",
+    "solution": "【分析】根据题意, 把长为 4、宽为 2 的长方形按 $3: 1$ 放大,那么长方形的长、宽都要乘 3 , 据此画出放大后的长方形;把长为 4、宽为 2 的长方形按 $1: 2$ 缩小, 那么长方形的长、宽都要除以 2 , 据此画出缩小后的长方形。\n【详解】放大后长方形的长: $4 \\times 3 = 12$\n放大后长方形的宽: $2 \\times 3 = 6$\n缩小后长方形的长: $4 \\div 2 = 2$\n缩小后长方形的宽: $2 \\div 2 = 1$\n如图:\n![](images/d8e27800dbc746615f03cbd5f542c72d57498df76f85b426a7bf4a4cf87912ab.jpg)",
+    "knowledge_points": [
+      "图形的放大与缩小"
+    ],
+    "source": "14-15六年级·全国·期末",
+    "merged_id": "unit4_358",
+    "original_id": "unit4_choice_358"
+  },
+  {
+    "id": "unit4_choice_359",
+    "unit": "第四单元：比例",
+    "chapter": "图形的放大与缩小",
+    "question_type": "作图题",
+    "difficulty": "中等",
+    "question": "作图题。\n![](images/ba07034f6cf1722e0dbc26ba990f1d4aba8205d85dcd041df8a5a2f01ec7a741.jpg)\n(1) 按 $1: 2$ 的比画出三角形缩小后的图形。缩小后的三角的面积与原三角形面积的比是 ( )。\n(2) 按 $2: 1$ 的比画出平行四边形放大后的图形。放大后的平行四边形的面积与原来平行四边形面积的比是 ( )。",
+    "options": [],
+    "answer": "（1）作图见详解；1:4 (2) 作图见详解; $4: 1$",
+    "solution": "【分析】（1）把图形按照 $1: n$ 缩小, 就是将图形的每一条边缩小到原来的 $\\frac{1}{n}$, 缩小后图形与原图形对应边长的比是 $1: n$ 。据此画出三角形缩小后的图形, 根据三角形面积 = 底$x$ 高 $\\div 2$, 两数相除又叫两个数的比, 分别求出三角形缩小前后的面积, 写出缩小后的三角的面积与原三角形面积的比,化简即可;(2) 把图形按照 n:1 放大, 就是将图形的每一条边放大到原来的 n 倍, 放大后图形与原图形对应边长的比是 n:1。据此画出平行四边形放大后的图形, 平行四边形面积 = 底 $\\times$ 高,分别计算出放大前后平行四边形的面积, 根据比的意义,写出放大后的平行四边形的面积与原来平行四边形面积的比, 化简即可。\n【详解】\n![](images/e641472b564136b44d591397f6c7fb0178b98597913047c7630562d168d34e79.jpg)\n(1) $(2 \\times 1 \\div 2): (4 \\times 2 \\div 2) = 1: 4$ 缩小后的三角的面积与原三角形面积的比是 $1: 4$ 。(2) $(6 \\times 4): (3 \\times 2) = 24: 6 = (24 \\div 6): (6 \\div 6) = 4: 1$ 放大后的平行四边形的面积与原来平行四边形面积的比是 $4: 1$ 。",
+    "knowledge_points": [
+      "图形的放大与缩小",
+      "三角形面积的计算",
+      "平行四边形面积的计算",
+      "比的意义"
+    ],
+    "source": "24-25 六年级下·海南海口·单元测试",
+    "merged_id": "unit4_359",
+    "original_id": "unit4_choice_359"
+  },
+  {
+    "id": "unit4_choice_360",
+    "unit": "第四单元：比例",
+    "chapter": "图形的放大与缩小",
+    "question_type": "作图题",
+    "difficulty": "中等",
+    "question": "操作题。画出图 A 向右平移 4 格后的图形。将图 B 按 2:1 放大, 画出放大后的图形。\n![](images/a28baebcf6911a23d1cbd81419264e33888bb0f979343932de227da53810e707.jpg)",
+    "options": [],
+    "answer": "图见详解",
+    "solution": "【分析】作平移后的图形步骤：找点一找出构成图形的关键点；定方向、距离一确定平移方向和平移距离；画线一过关键点沿平移方向画出平行线；定点一由平移的距离确定关键点平移后的对应点的位置；连点一连接对应点。\n把图形按照 n:1 放大, 就是将图形的每一条边放大到原来的 n 倍, 放大后图形与原图形对应边长的比是 n:1; 据此解答。\n【详解】作图如下：![](images/5c51f6c80bd573fb688efdca62d586d8857db806274a394323d1361ed9a7f56b.jpg)",
+    "knowledge_points": [
+      "图形的放大与缩小",
+      "作平移后的图形"
+    ],
+    "source": "24-25 六年级下·海南海口·单元测试",
+    "merged_id": "unit4_360",
+    "original_id": "unit4_choice_360"
+  },
+  {
+    "id": "unit4_choice_361",
+    "unit": "第四单元：比例",
+    "chapter": "图形的放大与缩小",
+    "question_type": "作图题",
+    "difficulty": "中等",
+    "question": "(1) 画出图中梯形向右平移 6 格后的图形。(2) 画出三角形绕 C 点按顺时针方向旋转 $90^{\\circ}$ 后的图形。 (3) 每个小方格是边长为 1 厘米的正方形。请按 1:2 的比例画出长方形缩小后的图形。缩小后长方形的面积是 ( ) 平方分米。\n![](images/7710999d34217085eccb8accadde7d4875731213a00c652bee4e88e7287f1921.jpg)",
+    "options": [],
+    "answer": "(1) 见详解(2) 见详解 (3) 见详解; 0.08",
+    "solution": "【分析】（1）根据平移的特征，将梯形的各顶点分别向右平移6格，依次连接即可得到平移后的图形。\n(2) 根据旋转的特征, 将三角形绕 C 点按顺时针方向旋转 $90^{\\circ}$, 点 C 位置不变, 其余各部分均绕此点按相同方向旋转相同度数, 即可画出旋转后的图形。 (3) 图中长方形的长是 8 厘米, 宽是 4 厘米, 按 $1:2$ 缩小,原来长方形的长和宽都除以 2 ，即是缩小后长方形的长和宽,据此画出缩小后的长方形；再根据长方形的面积 $=$ 长 $\\times$ 宽,求出缩小后长方形的面积, 并根据进率 '1 平方分米 = 100平方厘米' 换算单位。\n【详解】（1）梯形向右平移6格后的图形见下图。(2) 三角形绕 C 点按顺时针方向旋转 $90^{\\circ}$ 后的图形见下图。 (3) 缩小后长方形的长: $8 \\div 2 = 4$ (厘米)缩小后长方形的宽: $4 \\div 2 = 2$ (厘米)缩小后长方形的面积: $4 \\times 2 = 8$ (平方厘米)8平方厘米=0.08平方分米缩小后长方形的面积是 0.08 平方分米。如图:\n![](images/a1780202729862d95c6df432485dbc1194af448a2cdac892bcd80bea98ae55fd.jpg)",
+    "knowledge_points": [
+      "图形的放大与缩小",
+      "作旋转后的图形",
+      "作平移后的图形",
+      "长方形的面积"
+    ],
+    "source": "24-25 六年级下·海南海口·期末",
+    "merged_id": "unit4_361",
+    "original_id": "unit4_choice_361"
+  },
+  {
+    "id": "unit4_choice_362",
+    "unit": "第四单元：比例",
+    "chapter": "图形的放大与缩小",
+    "question_type": "作图题",
+    "difficulty": "中等",
+    "question": "按要求在方格中作图。\n(1) 根据给定的对称轴画出图形的另一半。\n(2) 画出将这个轴对称图形按 $2:1$ 放大后的图形。\n![](images/e06cc6b087bba4306230eb9c70d578789d6eb739e18e09b1b0b0e3ff0df9cb44.jpg)",
+    "options": [],
+    "answer": "见详解",
+    "solution": "【分析】（1）画出轴对称图形的另一半，根据轴对称图形的特征可知，沿着对称轴对折，另一半与已知图形完全重合，可以找到左边图形的几个关键点，再通过数格子的形式在右边找出离对称轴距离相同的关键点，再把这些关键点连接起来即可。(2) 把这个轴对称图形按 $2:1$ 放大, 即把原来的图形各个边都放大到原来的 2 倍, 据此即可画出此图。\n【详解】 (1) (2) 如下图所示:\n![](images/6a19676ad7f2a9d17ba61916879871cbba725ee90b261c14d79023a5560dc4bc.jpg)",
+    "knowledge_points": [
+      "图形的放大与缩小",
+      "补全轴对称图形"
+    ],
+    "source": "23-24 六年级下·浙江杭州·期末",
+    "merged_id": "unit4_362",
+    "original_id": "unit4_choice_362"
+  },
+  {
+    "id": "unit4_choice_363",
+    "unit": "第四单元：比例",
+    "chapter": "图形的放大与缩小",
+    "question_type": "作图题",
+    "difficulty": "中等",
+    "question": "按 3:1 的比画出长方形放大的图形, 再按 1:2 的比画出三角形缩小后的图形。\n![](images/9161a959943e6d1bdc30e05ae4baf0265eb5a169a4053da802f9f1849b30192f.jpg)",
+    "options": [],
+    "answer": "图见详解",
+    "solution": "【分析】根据放大和缩小的特征：长方形的长是 3 格，宽是 1 格；按 3:1 放大，放大后的长是： $3 \\times 3 = 9$ 格；宽是： $1 \\times 3= 3$ （格）；画出放大后的长方形图形；三角形的直角边分别是 4 格和 6 格，按 $1:2$ 缩小，缩小后的直角边分别是： $4 \\div 2 = 2$ 格； $6 \\div 2 = 3$ 格，据此画出缩小后的三角形。\n【详解】如图：\n![](images/f0180245af99b386658a4cfd60dd4070ed301c6c8e0b733776c547388b298bdb.jpg)",
+    "knowledge_points": [
+      "图形的放大与缩小"
+    ],
+    "source": "22-23 六年级下·安徽马鞍山·期末",
+    "merged_id": "unit4_363",
+    "original_id": "unit4_choice_363"
+  },
+  {
+    "id": "unit4_choice_364",
+    "unit": "第四单元：比例",
+    "chapter": "应用比例尺画图",
+    "question_type": "作图题",
+    "difficulty": "中等",
+    "question": "根据信息画一画。\n(比例尺 1:50000)\n![](images/4d8ca7b0b79ef8d009d7be55aa9a4cc19d32ad58876da6ca621175d29caf113d.jpg)\n(1) 新华书店在广场北偏西 $60^{\\circ}$ 方向 1500 米处, 请在图上标出新华书店的位置。(2) 若要从新华书店修一条小路到新开路，画出新华书店到新开路最近的路线。",
+    "options": [],
+    "answer": "(1) (2) 图见详解",
+    "solution": "【分析】（1）根据图上距离 $=$ 实际距离 $\\times$ 比例尺，将新华书店与广场的实际距离按 1:50000 换算成图上距离。再根据地图上方向的规定'上北下南，左西右东'，以广场为观测点，确定新华书店的位置即可。\n(2) 直线外一点与这条直线所有点的连线中, 垂线段最短,所以只要做出新华书店到新开路的垂线段即可。\n【详解】 (1) $1: 50000 = \\frac{1}{50000} \\quad 1500$ 米 $= 150000$ 厘米 $150000 \\times \\frac{1}{50000} = 3$ （厘米）新华书店与广场的图上距离是 3 厘米。作图如下:(2) 作出新华书店到新开路的垂线段就是最短路线, 作图如下:\n![](images/72afede3346cfbe6e21003c047696cf222d46541fd22955709f7cdc8e49850ba.jpg)",
+    "knowledge_points": [
+      "画垂线",
+      "根据方向、角度和距离确定物体的位置",
+      "应用比例尺画图",
+      "图上距离与实际距离的换算"
+    ],
+    "source": "23-24六年级下·河北邢台·期末",
+    "merged_id": "unit4_364",
+    "original_id": "unit4_choice_364"
+  },
+  {
+    "id": "unit4_choice_365",
+    "unit": "第四单元：比例",
+    "chapter": "图形的放大与缩小",
+    "question_type": "作图题",
+    "difficulty": "中等",
+    "question": "把平行四边形按 $1:2$ 缩小，把圆形按 $2:1$ 扩大。\n![](images/a7801e541a642bb1f11126fd1e21468e4b9d7c21cfad4b9ab095890ae6c593db.jpg)",
+    "options": [],
+    "answer": "见详解",
+    "solution": "【分析】图中平行四边形的底是 8 , 高是 4 , 按 $1:2$ 缩小后,缩小后的平行四边形的底和高都除以 2 , 据此画出缩小后的平行四边形;图中圆的半径是 1 , 按 2:1 扩大, 扩大后的圆的半径乘 2 ,据此画出扩大后的圆。掌握作放大和缩小后图形的作图方法是解题的关键, 明确放大或缩小图形, 只改变图形的大小, 不改变图形的形状。\n【详解】缩小后平行四边形的底： $8 \\div 2 = 4$ 缩小后平行四边形的高: $4 \\div 2 = 2$ 扩大后圆的半径: $1 \\times 2 = 2$如图:\n![](images/4bf4293dffd1cc1456ed617f793c7096556a15c2754899fd9113d5e264171892.jpg)",
+    "knowledge_points": [
+      "图形的放大与缩小"
+    ],
+    "source": "2023·贵州黔西·小升初真题",
+    "merged_id": "unit4_365",
+    "original_id": "unit4_choice_365"
+  },
+  {
+    "id": "unit4_choice_366",
+    "unit": "第四单元：比例",
+    "chapter": "图形的放大与缩小",
+    "question_type": "作图题",
+    "difficulty": "容易",
+    "question": "把三角形 A 向右平移5格，得到三角形B，再将三角形A按2:1扩大，得到三角形C。\n![](images/f411959a63c9ace56513ab1f13be8191487d79c8adf77bfb561a98faa5fccd4d.jpg)",
+    "options": [],
+    "answer": "见详解",
+    "solution": "【分析】根据平移的特征, 将三角形的各顶点分别向右平移 5 格, 依次连接即可得到平移后的图形三角形 $\\mathrm{B}$ 。\n把三角形 A 按 2:1 扩大, 即三角形 A 的每一条边扩大到原来的 2 倍, 原三角形 A 的底和高分别乘 2 , 得出扩大后三角形 A 的底和高, 据此画出扩大后的图形三角形 C。此题主要考查图形的平移和图形的放大与缩小,掌握其作图方法是解答题目的关键。\n【详解】作图如下：\n![](images/22996d953636d3bbe522c52bc974dab20d093983d2fb708397e4c9099a905902.jpg)",
+    "knowledge_points": [
+      "图形的放大与缩小",
+      "作平移后的图形"
+    ],
+    "source": "2022·新疆吐鲁番·小升初真题",
+    "merged_id": "unit4_366",
+    "original_id": "unit4_choice_366"
+  },{
+    "id": "unit4_choice_367",
+    "unit": "第四单元:比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "小红家在公园正东方向, 距离公园 400m; 小英家在公园北偏东 60° 方向距离是 300m; 小青家在小英家正西方向 200m 处。在下图中画出他们三家和公园的位置平面图(比例尺是 1:10000)。\n\n![](images/ec3a4c91e1fa7901c4361caaad66d89cb04a4dfb41722241f5219f48ce11805f.jpg)",
+    "options": [],
+    "answer": "见详解",
+    "solution": "【分析】根据图上距离 = 实际距离 × 比例尺, 先分别求出图上距离。弄清要标示的物体在哪个方位上, 有多少度, 按要求的方位和度数准确画图; 注意各场所离中心点的距离,根据要求的比例画出相应的长度。将方向和距离结合起来描述位置时,要注意三个要素:一是观测点,二是方向,三是距离。\n\n【详解】400m = 40000cm\n300m = 30000cm\n200m = 20000cm\n40000 ÷ 10000 = 4 (cm)\n30000 ÷ 10000 = 3 (cm)\n20000 ÷ 10000 = 2 (cm)\n\n![](images/50b71051dba03316898216d54d1b90e6a7042dbd56bc4d811bff01b85019a637.jpg)",
+    "knowledge_points": ["根据方向、角度和距离确定物体的位置", "图上距离与实际距离的换算"],
+    "source": "(21-22 六年级下·贵州铜仁·期中)",
+    "merged_id": "unit4_367",
+    "original_id": "unit4_choice_367"
+  },
+  {
+    "id": "unit4_choice_368",
+    "unit": "第四单元:比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "按要求作图\n\n![](images/571f5b3a1931ffe2ed1b441d17d4151be9bdfed5331b8428c2a32809826e1ea1.jpg)\n\n(1) 过 C 点作 AB 边上的垂线, 垂足为 D, 点 D 的位置用数对表示是 ( , )。\n(2) 画出三角形 ABC 绕 B 点逆时针旋转 90 度后的图形。\n(3) 在合适的位置画出三角形 ABC 按 2:1 的比放大后的图形。",
+    "options": [],
+    "answer": "(1) 画图见详解; (1, 6) (2) 和 (3) 见详解",
+    "solution": "【分析】(1)延长BA,再过C点向延长线上做垂线即可;数对中第一个数字表示列,第二个数字表示行,由此写出点D的位置即可;(2) 根据旋转的方法, 将三角形与点 B 相连的两条边绕点 B 逆时针旋转 90 度, 再将其它边连起来即可; (3) 将三角形 ABC 按 2:1 的比放大, 则放大后三角形的边长是原三角形的 2 倍, 据此画图即可。本题综合性较强,熟练掌握数对表示位置的特点,图形旋转的方法以及图形放大与缩小的方法是解答本题的关键。\n\n【详解】\n\n![](images/76396eab55e8d91b61a8b6ff35ab32f340e9a1539fffb07495fa8d4cf575c46b.jpg)\n\n点 D 的位置用数对表示是 (1, 6)。",
+    "knowledge_points": ["图形的放大与缩小", "画垂线", "作旋转后的图形", "用数对表示位置"],
+    "source": "(2021·河南信阳·小升初真题)",
+    "merged_id": "unit4_368",
+    "original_id": "unit4_choice_368"
+  },
+  {
+    "id": "unit4_choice_369",
+    "unit": "第四单元:比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "按 2:1 画出下面长方形放大后的图形。\n\n![](images/31fd0d752f701936455b23baf5e6d4e5656cab09d6dfc095646a47f454722bfb.jpg)",
+    "options": [],
+    "answer": "见详解",
+    "solution": "【分析】将长方形的长和宽都扩大到原来的2倍即可。图形放大或缩小时, 图形形状不变, 每条边的边长扩大或缩小。\n\n【详解】如图:\n\n![](images/d972e971236ee726692e023cfbb6c529cf5c237fadbe9a18c26041d8941bdc68.jpg)",
+    "knowledge_points": ["图形的放大与缩小"],
+    "source": "(18-19 六年级下·广东阳江·期中)",
+    "merged_id": "unit4_369",
+    "original_id": "unit4_choice_369"
+  },
+  {
+    "id": "unit4_choice_370",
+    "unit": "第四单元:比例",
+    "chapter": "比例",
+    "question_type": "选择题",
+    "difficulty": "容易",
+    "question": "画出下面图形按 3:1 放大后的图形。\n\n![](images/72881d95c6b739ed51aaa15fed1b1ef7fe96eab248c9b4a66ce652e25bb38094.jpg)",
+    "options": [],
+    "answer": "见详解",
+    "solution": "【分析】把图形按照 n:1 放大, 就是将图形的每一条边放大到原来的 n 倍, 放大后图形与原图形对应边长的比是n:1。图形放大或缩小的倍数是指对应边放大或缩小的倍数。\n\n【详解】作图如下:\n\n![](images/8f8049747d717da6af31bc93eec602df32e329787a49ddebd01305f6ca755294.jpg)",
+    "knowledge_points": ["图形的放大与缩小"],
+    "source": "(20-21 五年级下·全国·期中)",
+    "merged_id": "unit4_370",
+    "original_id": "unit4_choice_370"
+  },
+  {
+    "id": "unit4_choice_371",
+    "unit": "第四单元:比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "一张设计图纸的比例尺是 1:600, 图中的一个长方形大厅长4厘米,宽2.5厘米。这个大厅的实际面积是多少平方米?",
+    "options": [],
+    "answer": "360 平方米",
+    "solution": "【分析】比例尺 = 图上距离 : 实际距离, 所以实际距离 = 图上距离 ÷ 比例尺, 分别计算出长方形的实际长和实际宽,再根据长方形的面积 = 长 × 宽计算即可, 注意单位转化。明确图上距离、实际距离和比例尺之间的关系,需要分别算出长和宽再计算其面积。\n\n【详解】实际长: 4 ÷ (1:600) = 2400 厘米 = 24 米\n实际宽: 2.5 ÷ (1:600) = 1500 厘米 = 15 米\n实际面积: 24 × 15 = 360 (平方米)\n答:这个大厅的实际面积是360平方米。",
+    "knowledge_points": ["图上距离与实际距离的换算"],
+    "source": "(20-21六年级下·辽宁·单元测试)",
+    "merged_id": "unit4_371",
+    "original_id": "unit4_choice_371"
+  },
+  {
+    "id": "unit4_choice_372",
+    "unit": "第四单元:比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "若蜡烛每分钟燃烧的长度一定,则蜡烛未点燃前是多少厘米?\n\n![](images/a2042a5a539d3bb1993b544afb42b361f62f67b53b57b66c8904a7b534b9ad76.jpg)",
+    "options": [],
+    "answer": "20cm",
+    "solution": "【分析】由题意可知: 蜡烛燃烧 (20 - 10) 分钟, 燃烧的长度是 (14 - 8)cm, 根据蜡烛每分钟燃烧的长度一定, 蜡烛燃烧的长度和燃烧时间成正比例关系, 设蜡烛 20 分钟燃烧 xcm, 列出比例 (20 - 10):(14 - 8) = 20:x, 解出 x = 12, 所以蜡烛未点燃前是 12 + 8 = 20(cm)。本题考查正比例的应用, 理解'蜡烛每分钟燃烧的长度一定, 蜡烛燃烧的长度和燃烧时间成正比例关系'是解题的关键。\n\n【详解】解:设蜡烛20分钟燃烧 xcm。\n\n(20 - 10): (14 - 8) = 20: x\n10x = 120\nx = 12\n\n12 + 8 = 20 (cm)\n答:蜡烛未点燃前是 20cm。",
+    "knowledge_points": ["正比例的应用"],
+    "source": "(20-21 六年级下·全国·单元测试)",
+    "merged_id": "unit4_372",
+    "original_id": "unit4_choice_372"
+  },
+  {
+    "id": "unit4_choice_373",
+    "unit": "第四单元:比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "一辆货车从甲地到乙地,计划每小时行驶 50km,9 小时到达。实际速度比计划慢,这辆货车从甲地到乙地实际行驶了多少小时?\n\n![](images/c1a5c98c71adddc1bb57160c0fab69928d9442e3d23c693383cb2c15012ff502.jpg)\n\n提示:实际2.5小时只行驶了112.5 km",
+    "options": [],
+    "answer": "10小时",
+    "solution": "【分析】甲地到乙地的路程一定,货车的行驶速度与行驶时间成反比例关系。由此可设这辆货车从甲地到乙地实际行驶了 x 小时,列出比例 50 × 9 = (112.5 ÷ 2.5) x。本题考查正比例的应用, 理解'甲地到乙地的路程一定, 货车的行驶速度与行驶时间成反比例关系'是解题的关键。\n\n【详解】解:设这辆货车从甲地到乙地实际行驶了 x 小时。\n\n50 × 9 = (112.5 ÷ 2.5) x\n45x = 450\nx = 10\n\n答: 这辆货车从甲地到乙地实际行驶了 10 小时。",
+    "knowledge_points": ["反比例的应用"],
+    "source": "(20-21 六年级下·全国·单元测试)",
+    "merged_id": "unit4_373",
+    "original_id": "unit4_choice_373"
+  },
+  {
+    "id": "unit4_choice_374",
+    "unit": "第四单元:比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "读图与作图。\n\n![](images/107db2ab0f647bce17f35d5d0f90674c60c6b686457e99a5fdda60554b95dc6b.jpg)\n\n(1) 新冠肺炎疫情期间,为与新冠病毒竞速,武汉市火速建设了雷神山、火神山医院,用以集中收治肺炎患者。根据位置示意图,雷神山医院位于火神山医院()偏()()°方向()km处。\n(2) 火神山医院的建成,极大缓解了金银潭医院的就诊压力,金银潭医院位于火神山医院北偏东 50° 方向 25km 处,请在图中画出金银潭医院的位置。",
+    "options": [],
+    "answer": "(1) 东; 南; 30; 40;(答案不唯一) (2)见详解",
+    "solution": "【分析】(1)由题图中的线段比例尺可知,图中 1cm 的表示实际距离 10km。图中雷神山医院位于火神山院东偏南 30° 方向 4cm 处,所以实际上,雷神山医院位于火神山医院东偏南 30° 方向 40km 处;(2) 金银潭医院与火神山医院的实际距离是 25km, 图上距离就是 2.5cm, 也就是在图上, 金银潭医院位于火神山医院北偏东 50° 方向 2.5cm 处, 由此画出金银潭医院在图上的位置即可。此题主要考查学生利用方向、角度和距离描述物体位置和比例尺的实际应用。\n\n【详解】(1)根据分析可知,雷神山医院位于火神山医院东偏南 30° 方向 40km 处;\n\n(2) 如图:\n\n![](images/acaa472631382365091b5ee78679bf9a094e972900a0958787ed2aaf37c6a489.jpg)",
+    "knowledge_points": ["根据方向、角度和距离确定物体的位置", "应用比例尺画图"],
+    "source": "(20-21 六年级下·全国·单元测试)",
+    "merged_id": "unit4_374",
+    "original_id": "unit4_choice_374"
+  },
+  {
+    "id": "unit4_choice_375",
+    "unit": "第四单元:比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "李老师应该付给邮局多少钱?\n\n![](images/7d067e77e8520020bf004a138c3bcdcd4520d93de999289393f742c228eb660a.jpg)",
+    "options": [],
+    "answer": "468元",
+    "solution": "【分析】表示两个比相等的式子,叫做比例。根据总价÷数量=单价,可以设李老师应该付给邮局x元,则312元:8份 = x: 12份, 据此列出比例方程, 并求解。\n\n【详解】解:李老师应该付给邮局 x 元,\n\n312: 8 = x: 12\n8x = 312 × 12\n8x = 3744\n8x ÷ 8 = 3744 ÷ 8\nx = 468\n\n答: 李老师应该付给邮局 468 元。",
+    "knowledge_points": ["比例的应用"],
+    "source": "(24-25六年级下·全国·单元测试)",
+    "merged_id": "unit4_375",
+    "original_id": "unit4_choice_375"
+  },
+  {
+    "id": "unit4_choice_376",
+    "unit": "第四单元:比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "如下图, 甲城和乙城两点间的距离是 ( ) 厘米。实际距离是()千米?\n\n![](images/371d27594296ca936e917b39142a65909d76c5e84bc0675e7ad55f6c6082e2f5.jpg)",
+    "options": [],
+    "answer": "2.5 ;125千米",
+    "solution": "【分析】分析题目,先用直尺量出甲城和乙城两点间的距离;再根据图中给出的比例尺可知图上的 1 厘米表示实际的 50 千米,据此用量出的图上距离乘 50 即可得出实际距离是多少千米。\n\n【详解】量得甲城和乙城两点间的距离是 2.5 厘米。2.5 × 50 = 125 (千米)\n答: 甲城和乙城两点间的距离是 2.5 厘米。实际距离是 125 千米。",
+    "knowledge_points": ["比例尺应用", "长度的测量方法"],
+    "source": "(24-25 六年级下·海南海口·单元测试)",
+    "merged_id": "unit4_376",
+    "original_id": "unit4_choice_376"
+  },
+  {
+    "id": "unit4_choice_377",
+    "unit": "第四单元:比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "这只蚂蚁实际长 5 毫米, 量一量图中蚂蚁的长度, 求出这幅图的比例尺。\n\n![](images/a561b91208d770d424eece01a43209c11f29706c97851b32c67d6eb9bc03a603.jpg)",
+    "options": [],
+    "answer": "6:1",
+    "solution": "【分析】分析题目,先用直尺量出蚂蚁的图上长度,再根据 1 厘米 = 10 毫米把单位换算统一, 最后根据比例尺 = 图上距离: 实际距离求出比例尺即可。\n\n【详解】量得蚂蚁的图上距离是 3 厘米。\n\n图上距离:实际距离 =3 厘米:5毫米 = (3×10) 毫米:5毫米 =30:5= (30 ÷ 5) : (5 ÷ 5)= 6:1\n答: 这幅图的比例尺是 6: 1。",
+    "knowledge_points": ["长度的测量方法", "比例尺的意义"],
+    "source": "(24-25 六年级下·海南海口·单元测试)",
+    "merged_id": "unit4_377",
+    "original_id": "unit4_choice_377"
+  },
+  {
+    "id": "unit4_choice_378",
+    "unit": "第四单元:比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "在一幅比例尺是 1:250000 的地图上,展览馆到文化馆是 8 厘米,文化馆到小新家是 4 厘米。已知出租车在 3 千米以内 (含 3 千米)按起步价10元计算,以后每增加1千米车费就增加1.2元。小新从家坐出租车到展览馆一共需要多少元?\n\n![](images/36f62f29288b120d029c37af6ff4d65b94ac1b70d582022c2a3fc971648cb40c.jpg)",
+    "options": [],
+    "answer": "42.4元",
+    "solution": "【分析】比例尺 = 图上距离 : 实际距离, 则实际距离 = 图上距离 ÷ 比例尺, 先计算出展览馆到小新家的距离, 计算可知展览馆距离小新家 30 千米, 其中 3 千米按照起步价 10 元收费, 超出的 (30-3) 千米按照每千米 1.2 元收费, 根据'总价 = 单价 × 数量'求出超出部分需要付的钱数, 最后加上起步价, 据此解答。\n\n【详解】8 ÷ 1/250000 + 4 ÷ 1/250000\n= 8 × 250000 + 4 × 250000\n= (8 + 4) × 250000\n= 12 × 250000\n= 3000000 (厘米)\n\n3000000 厘米 = 30 千米\n\n(30 - 3) × 1.2 + 10\n= 27 × 1.2 + 10\n= 32.4 + 10\n= 42.4 (元)\n\n答:小新从家坐出租车到展览馆一共需要42.4元。",
+    "knowledge_points": ["分段计费问题(小数乘法)", "图上距离与实际距离的换算", "经济问题"],
+    "source": "(24-25六年级下·海南海口·单元测试)",
+    "merged_id": "unit4_378",
+    "original_id": "unit4_choice_378"
+  },
+  {
+    "id": "unit4_choice_379",
+    "unit": "第四单元:比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "按要求作图并填空。\n\n![](images/78cb5014b7adf58a2f71255617333a2fd67055e2794f1b86cd9cf2d8b5b3693e.jpg)\n\n(1) 画出图形①绕点A逆时针方向旋转 180° 后的图形②。\n(2) 如果点 B 的位置是 (4, 3) 那么旋转后点 B 的对应点 B' 的位置是 ( )。\n(3) 如果把图形①按 2: 1 放大, 请画出放大后的图形③。\n(4) 图形③和图形①的面积比是 ( )。",
+    "options": [],
+    "answer": "(1)见详解 (2) (8, 5); (3) 见详解 (4) 4: 1",
+    "solution": "【分析】(1)根据题意,注意图形的形状和大小在旋转过程中保持不变,绕点A画出逆时针方向旋转 180° 的图形即可。(2) 将画出的图形点 B', 在方格中读出其数对位置即可。第一个数字表示列, 第二个数字表示行。(3) 按 2: 1 放大意味着将图形的每条边的长度都 ×2。在放大过程中,要注意保持图形的形状特征不变,角度不变,各部分之间的比例关系也不变。(4) 观察画好后的图形, 按 2: 1 放大后的图形与原图形的面积比是 4: 1。\n\n【详解】(1) (3)\n\n![](images/d471580eb7b9d66d916d5e16bbc21da6097ba61020a06cfb3209d8c627af8bf9.jpg)\n\n(2) 旋转后点 B 的对应点 B' 的位置是 (8, 5);\n\n(4) (6 × 2 ÷ 2): (3 × 1 ÷ 2)\n= 6: 1.5\n= 4: 1\n\n图形③和图形①的面积比是 4: 1。",
+    "knowledge_points": ["图形的放大与缩小", "作旋转后的图形", "用数对表示位置", "比的意义"],
+    "source": "(2024·四川巴中·小升初真题)",
+    "merged_id": "unit4_379",
+    "original_id": "unit4_choice_379"
+  },
+  {
+    "id": "unit4_choice_380",
+    "unit": "第四单元:比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "按要求完成三幅图,并填空。\n\n![](images/ea30b0455c3cf5dbc886183b531a5aaae563240b4142dffaebf14f4268c72333.jpg)\n\n(1) 画出圆向右平移 3 格, 再向上平移 4 格后的图形, 移动后圆心 O 的位置用数对表示为 ( , )。\n(2) 画出平行四边形绕 A 点顺时针旋转 90° 后的图形。\n(3)画出正方形按 2:1 扩大后的图形, 扩大后的正方形面积是原来的 ( ) 倍。",
+    "options": [],
+    "answer": "见详解",
+    "solution": "【分析】(1)将圆心先向右平移3格,再向上平移4格,确定好圆心的位置,以半径为2个格子画圆;依据用数对表示位置的方法, 括号内先写列再写行, 中间用','隔开表示圆心的位置;(2) 将各个关键点按顺时针方旋转 90° 作出它们的对应点, 再将对应点顺次连接即可; (3) 将正方形按照 2:1 放大, 就是将正方形的边放大到原来的 2 倍; 正方形的面积 = 边长 × 边长, 用放大后的面积除以放大前的面积, 求出扩大后的正方形面积是原来的几倍。\n\n【详解】(1)移动后圆心0的位置用数对表示为(6,7);\n(3) 扩大后的边长 = 2 × 2 = 4\n扩大前的面积 = 2 × 2 = 4\n扩大后的面积 = 4 × 4 = 16\n16 ÷ 4 = 4\n扩大后的正方形面积是原来的 4 倍;\n(1)(2)(3) 图:\n\n![](images/9098d49c14b627fddff320cd36f83578b77928c1b7bbcc849082711ee772fc50.jpg)",
+    "knowledge_points": ["图形的放大与缩小", "作旋转后的图形", "用数对表示位置", "作平移后的图形"],
+    "source": "(23-24 六年级下·辽宁鞍山·期末)",
+    "merged_id": "unit4_380",
+    "original_id": "unit4_choice_380"
+  },
+  {
+    "id": "unit4_choice_381",
+    "unit": "第四单元:比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "王阿姨一家四口打算从长沙来南昌游玩。如果乘高铁来南昌,1.2小时到达。如果选择自驾游来南昌,需要5小时才能到达。王阿姨自驾游时的平均车速是多少?\n\n| 出行方式 | 平均速度    | 时间     |\n| -------- | ----------- | -------- |\n| 乘高铁   | 300 千米/时 | 1.2 小时 |\n| 自驾游   | () 千米/时  | 5 小时   |",
+    "options": [],
+    "answer": "72 千米/时",
+    "solution": "【分析】长沙到南昌的路程是不变的,速度与时间的乘积是一定的,即速度与时间这两种量成反比例,由此设王阿姨自驾游时的平均车速是 x 千米/时,列出比例式解答即可。\n\n【详解】解:设王阿姨自驾游时的平均车速是 x 千米/时。\n\n5x = 300 × 1.2\n5x = 360\nx = 360 ÷ 5\nx = 72\n\n答: 王阿姨自驾游时的平均车速是 72 千米/时。",
+    "knowledge_points": ["列方程解含一个未知数的问题", "反比例的应用"],
+    "source": "(23-24六年级下·江西南昌·期末)",
+    "merged_id": "unit4_381",
+    "original_id": "unit4_choice_381"
+  },
+  {
+    "id": "unit4_choice_382",
+    "unit": "第四单元:比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "学校要给一间功能教室铺地砖,每块地砖的面积与所需地砖的数量如表。\n\n| 每块地砖的面积/平方米 | 0.2  | 0.3  | 0.4  | 0.6  | 0.8  | ...  |\n| --------------------- | ---- | ---- | ---- | ---- | ---- | ---- |\n| 所需地砖的数量/块     | 600  | 400  | 300  | 200  | 150  | ...  |\n\n(1) 每块地砖的面积和所需地砖的数量成 ( ) 比例关系。\n(2) 如果铺这一地面用了 500 块地砖, 所用的地砖每块面积是多大?(用比例解答)",
+    "options": [],
+    "answer": "(1) 反 (2) 0.24 平方米",
+    "solution": "【分析】(1)判断两个相关联的量之间成什么比例,就看这两个量是对应的比值一定,还是对应的乘积一定;如果是比值一定,就成正比例;如果是乘积一定,则成反比例。据此解答即可。\n(2) 可假设所用的地砖每块面积是 x 平方米, 根据每块地砖的面积和所需地砖的数量成反比例关系, 据此即可列比例求解。\n\n【详解】(1) 0.2 × 600 = 0.3 × 400 = 0.4 × 300 = 0.6 × 200 =0.8 × 150 = 120 (平方米)\n因为每块地砖的面积 × 需要的块数 = 铺地面积(一定),所以每块地砖的面积和所需地砖的数量成反比例关系。\n\n(2) 解:设所用地砖的面积为 x 平方米。\n\n500x = 0.2 × 600\n500x = 120\n500x ÷ 500 = 120 ÷ 500\nx = 0.24\n\n答: 所用的地砖每块面积是 0.24 平方米。",
+    "knowledge_points": ["反比例的意义及辨识", "反比例的应用"],
+    "source": "(23-24六年级下·浙江宁波·期末)",
+    "merged_id": "unit4_382",
+    "original_id": "unit4_choice_382"
+  },{
+    "id": "unit4_choice_383",
+    "unit": "第四单元:比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "中国首颗太阳探测卫星'羲和号',可以连续对太阳进行观测。其搭载的成像光谱仪通过对太阳的日面扫描,获得光谱的数量(单位:条)与对应的时长(单位:秒)关系如下:\n\n| 时长/秒           | 0    | 5    | 10   | 15   | 20   | 25   | 30   | ...  |\n| ----------------- | ---- | ---- | ---- | ---- | ---- | ---- | ---- | ---- |\n| 获得光谱的数量/条 | 0    | 50   | 100  | 150  | 200  | 250  | 300  | ...  |\n\n(1) 根据表中数据,写出一个正确的比例式:;\n(2) 根据表中数据,在下图中描出表示获得光谱的数量与对应的时长的点,并把这些点顺次连接起来。\n\n![](images/7fdde9125a606d24154f04d4b908d34d0a4fa0737dcd9958248338053eabebbf.jpg)\n\n(3) 如果连续观测 5 分钟, 获得光谱的数量大约是 条。",
+    "options": [],
+    "answer": "(1) 500: 5 = 1000: 10 (2)见详解 (3) 30000",
+    "solution": "【分析】(1)观察表格中数据,会发现每一组时长和对应的获得光谱的数量的比值是相等的,5秒对应500条,10秒对应1000条等,它们的比值都是 1/100,所以比例式可以为:5:500 = 10:1000, 20:2000 = 25:2500 等。只要满足这样的比例关系即可。(2) 按照表格中的数据,在坐标图上找到对应的点,然后把这些点依次连接起来即可。(3) 先把 5 分钟换算成秒, 即 5 × 60 = 300 秒, 30 秒时是 3000 条, 300 秒是 30 秒的 10 倍, 所以获得光谱的数量大约是 3000 × 10 = 30000 条。\n\n【详解】(1)根据表中数据,写出一个正确的比例式:5: 500 = 10: 1000。(答案不唯一)(2) 根据表格中提供的数据分别在对应位置描上点, 然后将点连接起来。\n\n![](images/0f7319dc072e9a09e236964aed86a13424aa556a43ddf7e978a99e069c187118.jpg)\n\n(3) 因为 1 分钟 = 60 秒, 5 分钟 = (5 × 60) 秒 = 300 秒,从表格中可以看出 30 秒时获得 3000 条光谱, 300 秒大约是30秒的10倍,所以获得光谱的数量是 3000 × 10 = 30000 条。如果连续观测 5 分钟, 获得光谱的数量大约是 30000 条。",
+    "knowledge_points": ["比例的应用", "单式折线统计图"],
+    "source": "(23-24 六年级下·北京房山·期末)",
+    "merged_id": "unit4_383",
+    "original_id": "unit4_choice_383"
+  },
+  {
+    "id": "unit4_choice_384",
+    "unit": "第四单元:比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "你知道吗? 远古时候,'小山羊'在人们的生命中起着'钱'的作用,64只羊可以换多少把斧头?请用列比例的方法解决问题。\n\n![](images/75b96584bd89b77b67cba83a014566637e66bf75fe7dbc516c6f73d95b88a8a5.jpg)",
+    "options": [],
+    "answer": "96 把",
+    "solution": "【分析】根据'4只羊可以换6把斧头'可知,羊的只数:斧头的把数 = 4:6,据此列出比例方程,并求解。\n\n【详解】解:设64只羊可以换 x 把斧头。\n\n64: x = 4: 6\n4x = 64 × 6\n4x = 384\nx = 384 ÷ 4\nx = 96\n\n答: 64 只羊可以换 96 把斧头。",
+    "knowledge_points": ["解比例", "比例的应用"],
+    "source": "(23-24 六年级下·江西九江·期末)",
+    "merged_id": "unit4_384",
+    "original_id": "unit4_choice_384"
+  },
+  {
+    "id": "unit4_choice_385",
+    "unit": "第四单元:比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "订阅《小学数学报》。\n\n小红:我们班订了35份,用了630元。小明:我们班订了25份,需要多少钱?\n\n(1) 小红在解答这题时, 分两步计算:第一步: 630 ÷ 35 = 18 (元)她是先算出();第二步应该这样列式:()。(2) 小明说: 我可以用比例来解答。请你也用比例来解答。",
+    "options": [],
+    "answer": "(1) 《小学数学报》的单价; 18 × 25 = 450 (元) (2) 450 元",
+    "solution": "【分析】(1)根据总价÷数量 = 单价,用的总钱数 ÷ 订的份数 = 每份钱数,每份钱数 × 小明班订的份数 = 小明班需要的钱数,进行分析。(2) 用比例解决问题只要比例两边的比统一即可, 设小明班需要 x 元钱, 根据小明班需要的钱数: 小明班订的份数 = 小红班用的钱数: 小红班订的份数, 列出比例解答即可。\n\n【详解】(1)小红在解答这题时,分两步计算:第一步: 630 ÷ 35 = 18 (元)她是先算出《小学数学报》的单价;第二步应该这样列式: 18 × 25 = 450 (元)。\n\n(2) 解:设小明班需要 x 元钱。\n\nx: 25 = 630: 35\n35x = 25 × 630\n35x = 15750\n35x ÷ 35 = 15750 ÷ 35\nx = 450\n\n答: 小明班需要 450 元钱。",
+    "knowledge_points": ["归一问题", "比例的应用"],
+    "source": "(23-24 六年级下·浙江台州·期末)",
+    "merged_id": "unit4_385",
+    "original_id": "unit4_choice_385"
+  },
+  {
+    "id": "unit4_choice_386",
+    "unit": "第四单元:比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "一列动车匀速行驶,请将下表补充完成。\n\n| 时间(分) | 1    | 2    | 3    | 4    | 5    | 6    |\n| ---------- | ---- | ---- | ---- | ---- | ---- | ---- |\n| 路程 (km)  | 5    | 10   | ()   | ()   | () | () |\n\n(1) 把上表中的路程和时间所对应的点描在方格纸上, 再顺次连接起来。\n\n![](images/8cf42657f671db7bd2e1e90bf0d903a80281b262adf1659b5a1548997903dfb8.jpg)\n\n(2) 路程和时间成正比例吗? 为什么?\n(3) 从图可以看出, 这列动车的速度是多少? 照这样计算, 12 分可以行驶多少千米?",
+    "options": [],
+    "answer": "15; 20; 25; 30\n(1) 图见详解(2) 成正比例;见详解 (3) 5千米/分; 60千米",
+    "solution": "【分析】根据统计表,动车速度是每分钟 (5 ÷ 1) 千米,根据关系式:速度 × 时间 = 路程,由此完成上表。(1) 根据统计表所提供的数据,在图中描出表示时间及所对应的路程的点,依次连接即可; (2) 要求是否成正比例, 先求出列车的速度, 如果速度一定, 就成正比例; (3) 从统计图可以看出, 动车每分钟行驶 5 千米, 根据路程 = 速度 × 时间, 用 12 分钟乘动车每分钟行驶的路程即可解答。\n\n【详解】填空如下:\n\n| 时间(分) | 1    | 2    | 3    | 4    | 5    | 6    |\n| ---------- | ---- | ---- | ---- | ---- | ---- | ---- |\n| 路程 (km)  | 5    | 10   | 15   | 20   | 25   | 30   |\n\n(1) 如图:![](images/52cdd3434fbe4b8c4e2354dd01fba21aaf87e77edbd51042e066af48e4d62d7c.jpg)\n\n(2) 5 ÷ 1 = 5 (千米) 10 ÷ 2 = 5 (千米) 路程÷时间 = 速度(一定),因为路程和时间的比值一定,所以路程和时间成正比例。(3) 5 × 12 = 60 (千米)从图可以看出,这列动车的速度是5千米/分,照这样计算,12分可以行驶60千米。",
+    "knowledge_points": ["正比例的应用", "正比例的意义及辨识", "基础行程问题", "单式折线统计图"],
+    "source": "(23-24 六年级下·河南周口·期中)",
+    "merged_id": "unit4_386",
+    "original_id": "unit4_choice_386"
+  },
+  {
+    "id": "unit4_choice_387",
+    "unit": "第四单元:比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "困难",
+    "question": "黄金分割是指将整体一分为二,较大部分与整体的比值等于较小部分与较大部分的比值,其比值约为0.618,这个比例被公认为是最美的美感,如果一个人的肚脐以上的高度与肚脐以下的高度比符合黄金比,这个人的身材比较好看。如图中点C是线段AB的黄金分割点,AC>BC。\n\n![](images\\4b4d95152c3742b997d88dad3ee07e12.png)\n\n(1) 根据上面的线段写出一个比例: ( ) : ( ) = ( ) : ( )。\n(2) 一个参加空姐选拔活动的选手, 其肚脐以上部分长 65 厘米, 以下部分长 100 厘米。为了显得更好看一些 (精确到 1 厘米), 她应该穿多少厘米高的鞋子?",
+    "options": [],
+    "answer": "(1)(答案不唯一) AC; AB; BC; AC; (2) 5 厘米",
+    "solution": "【分析】(1)由题意可知,点C是线段AB的黄金分割点,较长的线段:整条线段=较短的线段:较长的线段;(2) 当选手肚脐以上的高度与肚脐以下的高度比符合黄金比时, 肚脐以上的高度: 肚脐以下的高度 = 0.618, 则肚脐以下的高度 = 肚脐以上的高度 ÷ 0.618, 最后减去这个选手肚脐以下部分的长度就是鞋子的高度, 据此解答。本题主要考查解决与比相关的问题的能力,理解黄金比例的意义是解答题目的关键。\n\n【详解】(1)分析可知,根据线段写出的比例为:AC:AB=BC:AC。(答案不唯一)(2) 65 ÷ 0.618 ≈ 105 (厘米) 105 - 100 = 5 (厘米)答: 她应该穿 5 厘米高的鞋子。",
+    "knowledge_points": ["比与分数、除法的关系", "比的意义", "比例的意义"],
+    "source": "(22-23六年级下·山西晋中·期中)",
+    "merged_id": "unit4_387",
+    "original_id": "unit4_choice_387"
+  },
+  {
+    "id": "unit4_choice_388",
+    "unit": "第四单元:比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "下图是某社区平面图,已知快乐书店到广场的实际距离是300米。\n\n![](images/0dc38181b69fce8ea99d0bb75057ff1c2f8d9262b44cde24fe7f85b01288e1c6.jpg)\n\n(1) 求出这幅平面图的比例尺。\n(2) 肯德基到广场的实际距离是多少米?",
+    "options": [],
+    "answer": "(1) 1: 10000; (2) 500 米",
+    "solution": "【分析】先分别量出快乐书店到广场的图上距离,再据'比例尺 = 图上距离:实际距离',即可求出这幅图的比例尺;进而依据'实际距离 = 图上距离 ÷ 比例尺'即可求解。此题主要考查比例尺的意义以及图上距离、实际距离和比例尺的关系。\n\n【详解】(1)快乐书店到广场的图上距离是3厘米。300 米=30000 厘米 3:30000 = 1:10000答: 这幅平面图的比例尺是 1: 10000。\n\n(2) 肯德基到广场的图上距离是 5 厘米。5 × 10000 = 50000 (厘米) 50000 厘米 = 500 米答: 肯德基到广场的实际距离是 500 米。",
+    "knowledge_points": ["比例尺应用", "图上距离与实际距离的换算"],
+    "source": "(21-22 六年级下·江苏·单元测试)",
+    "merged_id": "unit4_388",
+    "original_id": "unit4_choice_388"
+  },
+  {
+    "id": "unit4_choice_389",
+    "unit": "第四单元:比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "一列动车在高速铁路上行驶的时间和路程如图。\n\n![](images/7cafaca52a8980135bdb461152ef37f1b0ede3d1e9cffe63137732dec110daf6.jpg)\n\n(1) 看图填写下表。\n\n| 时间(时)   | 3    | ()   |\n| ------------ | ---- | ---- |\n| 路程(千米) | ()   | 800  |\n\n(2) 这列动车行驶的时间和路程成()比例。(填'正'或'反')\n(3) 照这样的速度, 这列动车行驶 1800 千米需要多少时?",
+    "options": [],
+    "answer": "(1) 4;600(2) 正; (3) 9时",
+    "solution": "【分析】(1)先在横轴上找到时间是 3 时的点,再沿着此点所在的纵向格线与图象的交点水平向左,便可以找到与时间 3 时相对应的纵轴上的数据 600 千米。所以 3 时行驶 600 千米;先在纵轴上找到路程是 800 千米的点,再沿着此点所在的横向格线与图象的交点竖直向下,便可以找到与路程 800 千米相对应的横轴上的数据 4 时。所以 800 千米需要行驶 4 时;(2) 两种相关联的量相对应的两个数的比值一定, 这两种量是成正比例的量; 两种相关联的量相对应的两个数的乘积一定, 这两种量是成反比例的量。据此来判断; (3) 先根据图象中的条件求出这列动车的速度; 再根据'路程÷速度 = 时间'求出这列动车行驶 1800 千米所用的时间。正比例关系图象是一条从(0,0)出发的无限延伸的射线。据此也可判断两个相关联的变量是成正比例关系。\n\n【详解】(1)通过观察上图,可知:\n\n| 时间(时)   | 3    | 16   |\n| ------------ | ---- | ---- |\n| 路程(千米) | 150  | 800  |\n\n(2) 因为 200/1 = 400/2 = 600/3 = 800/4 = 200 (一定), 即速度是 200 千米/时。当速度一定时, 路程和时间成正比例关系。(3) 200 ÷ 1 = 200 (千米/时)1800 ÷ 200 = 9 (时)答: 这列动车行驶 1800 千米需要 9 时。",
+    "knowledge_points": ["正比例的意义及辨识", "反比例的意义及辨识", "基础行程问题"],
+    "source": "(21-22 六年级下·四川广安·期末)",
+    "merged_id": "unit4_389",
+    "original_id": "unit4_choice_389"
+  },
+  {
+    "id": "unit4_choice_390",
+    "unit": "第四单元:比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "按计划,中国空间站将于 2022 年底完成建造,可支持 3 名航天员长期在轨驻留,据测算,空间站在轨飞行时间与路程的关系如下表。\n\n| 时间/秒   | 1    | 2    | 3    | 4    | … |\n| --------- | ---- | ---- | ---- | ---- | ---- |\n| 路程/千米 | 7.8  | 15.6 | 23.4 | 31.2 | … |\n\n![](images/f423207bccdb2a2d4033b6902ba8b603e545cbcdddf19d1972011332a64849fb.jpg)\n\n(1) 根据表中的数据填空并描点,再顺次连接。\n(2) 中国空间站的在轨飞行路程与时间()成()比例,写出你判断的理由。",
+    "options": [],
+    "answer": "(1) 见详解; (2) 是; 正; 见详解",
+    "solution": "【分析】(1)折线统计图的绘制方法是:整理数据;用一个长度单位表示一定的数量; 根据统计表中数量的多少描出各点, 再把各点用线段顺次连接起来。(2) 判断两个相关联的量之间成什么比例, 就看这两个量是对应的比值一定, 还是对应的乘积一定; 如果是比值一定, 就成正比例; 如果是乘积一定, 则成反比例。因为路程÷时间 = 速度, 所以先计算出飞行路程与时间的比值, 然后判断这两个相关联的量成什么比例即可。此题的解题关键是掌握折线统计图的画法及特征, 另外辨识成正、反比例的量, 就看这两个量是对应的比值一定,还是对应的乘积一定,再做判断。\n\n【详解】(1) 作图如下:\n\n![](images/d375d0f25f0354c9635e9a0e217f914aa9f6253d5e2efc5ee301863a8e3c84aa.jpg)\n\n(2) 7.8 ÷ 1 = 7.8 (千米/秒) 15.6 ÷ 2 = 7.8 (千米/秒) 23.4 ÷ 3 = 7.8 (千米/秒) 31.2 ÷ 4 = 7.8 (千米/秒)可得 7.8/1 = 15.6/2 = 23.4/3 = 31.2/4 = 7.8 (一定),比值一定,符合正比例的意义,所以中国空间站的在轨飞行路程与时间是成正比例。",
+    "knowledge_points": ["正比例的意义及辨识", "基础行程问题", "单式折线统计图"],
+    "source": "(2022·河南郑州·小升初真题)",
+    "merged_id": "unit4_390",
+    "original_id": "unit4_choice_390"
+  },
+  {
+    "id": "unit4_choice_391",
+    "unit": "第四单元:比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "根据图中提供的信息,完成下面各题。\n\n![](images/c1cd3dc2cd737c1feafb656d5537ad0122131cbd18ac119884356618f6655925.jpg)\n\n(1) 陈晨家与汽车站的图上距离是 ( ) 厘米, 实际距离是 ( ) 米。(2) 学校在陈晨家正西方向 1000 米处, 学校与陈晨家的图上距离是 ( ) 厘米。(3) 请在图中画出学校的位置,并标出名称。",
+    "options": [],
+    "answer": "(1) 2.5, 625; (2) 4; (3) 见详解",
+    "solution": "【分析】(1)量得陈晨家与汽车站的图上距离是 2.5 厘米,根据比例尺, 即可求出陈晨家到汽车站的实际距离;(2) 利用图上距离 = 实际距离 × 比例尺, 即可求出学校与陈晨家的图上距离; (3) 以陈晨家为中心, 向正西方向测量描出学校与陈晨家的图上距离, 即可得出学校的位置。解答本题的关键是掌握图上距离、实际距离和比例尺三者之间的关系,同时要掌握利用方向和距离确定物体位置的方法。\n\n【详解】(1)经过测量可知,陈晨家与汽车站的图上距离是2.5厘米。2.5 × 250 = 625 (米)(2) 1000 ÷ 250 = 4 (厘米) (3) 作图如下:\n\n![](images/dd49de3669edf8501c19932170ae38e78b7e6923fe0f33b6be05f01eef5011b8.jpg)",
+    "knowledge_points": ["根据方向、角度和距离确定物体的位置", "图上距离与实际距离的换算"],
+    "source": "(21-22 六年级下·山东济宁·期末)",
+    "merged_id": "unit4_391",
+    "original_id": "unit4_choice_391"
+  },
+  {
+    "id": "unit4_choice_392",
+    "unit": "第四单元:比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "文具店甲的作业本总价与数量如下表:\n\n| 总价/元 | 1.5  | 3    | 4.5  | ()   | … |\n| ------- | ---- | ---- | ---- | ---- | ---- |\n| 数量/本 | 1    | 2    | ()   | 4    | … |\n\n![](images/12c39d5357449155d3036baf0efec72de1a68f6d3fb59af2918f5de7a8d9fb09.jpg)\n\n(1) 请根据表格中的数量关系,将表格补充完整。\n(2) 把总价与数量所对应的点在图中描出来,并连线。\n(3) 根据图象, 判断总价与数量成 ( ) 比例关系, 并说明理由。",
+    "options": [],
+    "answer": "(1) (2) 见详解 (3) 正",
+    "solution": "【分析】(1)根据'单价 × 数量 = 总价''数量 = 总价 ÷ 单价'进行解答;(2) 根据数据, 找出各点, 然后把点依次连接即可;(3) 因为总价 ÷ 数量 = 单价 (一定), 所以总价和数量成正比例。此题考查了判断成正比例的量的方法及根据数据画出正比例图象,应注意知识的灵活运用。\n\n【详解】(1) 4.5 ÷ 1.5 = 3 (本) 1.5 × 4 = 6 (元)\n\n| 总价/元 | 1.5  | 3    | 4.5  | 6    | … |\n| ------- | ---- | ---- | ---- | ---- | ---- |\n| 数量/本 | 1    | 2    | 3    | 4    | … |\n\n(2) 如图:![](images/3e19120f0f1e62814b8bc76616a8253ce5deba0411e326ab3f36b02abcb5204d.jpg)\n\n(3) 根据图象、判断总价与数量成正比例关系。因为总价÷数量 = 单价(一定),所以总价和数量成正比例。",
+    "knowledge_points": ["正比例图象的认识", "正比例的意义及辨识", "经济问题", "单式折线统计图"],
+    "source": "(2022·云南昆明·小升初真题)",
+    "merged_id": "unit4_392",
+    "original_id": "unit4_choice_392"
+  },
+  {
+    "id": "unit4_choice_393",
+    "unit": "第四单元:比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "同一时间、同一地点测得树高和影长的数据如表:\n\n| 树高/m | 2    | 3    | 4    | 6    |\n| ------ | ---- | ---- | ---- | ---- |\n| 影长/m | 1.6  | 2.4  | 3.2  | 4.8  |\n\n(1) 在下图中描出表示树高和对应影长的点, 然后把它们连起来。\n\n![](images/a1a764974f604947f455da21637da84d6af75886798f8c7ba0a8bd1fb721e908.jpg)\n\n(2) 连线以后观察, 树高和影长成 ( ) 关系。\n(3) 利用图象判断, 树高 8 米时, 影长 ( ) 米。",
+    "options": [],
+    "answer": "(1) 见详解;(2) 正比例; (3) 6.4",
+    "solution": "【分析】(1)根据统计表,在图中描出各点,再连起来即可;(2) 正比例图象的特点是过原点的一条直线, 据此解答即可; (3) 观察图象可知, 树高 8 米时, 影长为 6.4 米, 据此解答即可。熟练掌握正比例的意义及图象的特点是解答本题的关键。\n\n【详解】(1) 如图:\n\n![](images/0647dea0eab38089d98758a48d026d33c7c1adefdac1e4aa286ce0bfa60a85a1.jpg)\n\n(2) 连线以后观察, 树高和影长成正比例关系; (3) 利用图象判断, 树高 8 米时, 影长 6.4 米。",
+    "knowledge_points": ["正比例图象的认识"],
+    "source": "(19-20六年级下·福建厦门·期末)",
+    "merged_id": "unit4_393",
+    "original_id": "unit4_choice_393"
+  },
+  {
+    "id": "unit4_choice_394",
+    "unit": "第四单元:比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "'精心操作'我最行。(1) 按 1: 2 画出图形①缩小后的图形; 若每个小正方形边长为 1cm, 则缩小后的图形面积是 ( )。(2) 画出图形②绕点C逆时针旋转 90° 后的图形,旋转后点B的位置用数对表示为(__,__)。\n\n![](images/ad282cb4362dec54f44dba8e9ef30beb0735f32d08b76fbe38a114455bfbe947.jpg)",
+    "options": [],
+    "answer": "(1) 见详解; 3cm² (2) 见详解; (13, 1)",
+    "solution": "【分析】(1)图①是底为 6 格,高为 2 格的平行四边形,根据图形放大与缩小的意义,按 1:2 缩小后的平行四边形是对应底为 3 格,高为 1 格,对应角大小不变的平行四边形。根据平行四边形面积计算公式'S = ah'即可计算出缩小后图形的面积。(2) 根据旋转的特征, 图形②绕点 C 逆时针旋转 90°, 点 C 的位置不动, 其余各部分均绕此点按相同方向旋转相同的度数即可画出旋转后的图形。根据旋转后点 B 所在的列与行及用数对表示点的位置的方法, 第一个数字表示列数,第二个数字表示行数, 即可用数对表示出点 B 的位置。本题考查图形的放大与缩小、平行四边形的面积的计算、作旋转一定度数后的图形、数对与位置,解答本题的关键是掌握图形的放大与缩小的方法。\n\n【详解】(1)按 1: 2 画出图形①缩小后的图形(图中红色平行四边形);若每个小正方形边长为 1cm,则缩小后的图形面积是 3 × 1 = 3 (cm²)。(2) 画出图形②绕点C逆时针旋转 90° 后的图形(三角形 A'B'C),旋转后点B的位置用数对表示为(13,1)。![](images/d30e9c7443eab17f70422a6b031f4f3381963ed407d5bc455c44848b8b55f0e1.jpg)",
+    "knowledge_points": ["图形的放大与缩小", "作旋转后的图形", "用数对表示位置", "三角形的面积"],
+    "source": "(20-21 六年级下·江苏·单元测试)",
+    "merged_id": "unit4_394",
+    "original_id": "unit4_choice_394"
+  },
+  {
+    "id": "unit4_choice_395",
+    "unit": "第四单元:比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "一根弹簧挂上物体(质量不超过 20 千克)后长度会伸长。下图表示一个物体的质量和弹簧伸长的长度之间的关系。\n\n![](images/d9bb4b6f34f73b686a4c4cb69b02aa02df1feab1e900e52db370bc0fce41be2c.jpg)\n\n(1) 物体的质量与弹簧伸长的长度成______比例。\n(2) 如果挂上 7 千克的物体, 那么弹簧应伸长多少厘米?\n(3) 要使弹簧伸长 4.5 厘米, 应挂上多少千克的物体?",
+    "options": [],
+    "answer": "(1) 正 (2) 7/4 厘米 (3) 18 千克",
+    "solution": "【分析】(1)判断两个相关联的量之间成什么比例,就看这两个量是对应的比值一定,还是对应的乘积一定;如果是比值一定, 就成正比例; 如果是乘积一定, 则成反比例, 1/4 = 2/8, 由此即可知道弹簧伸长的长度: 物体的质量 = 1/4 (一定), 即弹簧伸长的长度和物体的质量成正比例关系。(2) 根据第一问可知, 伸长的长度: 物体的质量 = 1/4, 伸长的长度 = 1/4 × 物体的质量, 把物体的质量等于 7 代入公式,即 1/4 × 7 即可求出此时弹簧伸长的长度; (3) 物体的质量 = 伸长的长度 ÷ 1/4, 把数代入公式即可求解。此题属于辨识成正、反比例的量以及正比例的应用题;两个量是对应的比值一定,这两个量成正比例关系;两个量的乘积一定,则这两个量成反比例关系。\n\n【详解】(1)由分析可知,弹簧伸长的长度和物体的质量比值一定,所以是正比例关系。(2) 1/4 × 7 = 7/4 (厘米)\n\n答: 如果挂上 7 千克的物体, 那么弹簧应伸长 7/4 厘米。(3) 4.5 ÷ 1/4 = 18 (千克)答: 应挂上 18 千克的物体。",
+    "knowledge_points": ["正比例的应用", "正比例的意义及辨识", "反比例的意义及辨识"],
+    "source": "(20-21 六年级下·辽宁·单元测试)",
+    "merged_id": "unit4_395",
+    "original_id": "unit4_choice_395"
+  },
+  {
+    "id": "unit4_choice_396",
+    "unit": "第四单元:比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "在一张长方形彩纸上摆满小正方形,每个小正方形面积与所需小正方形的数量如表:\n\n| 每个小正方形的面积/cm² | 4    | 9    | 16   |\n| ---------------------- | ---- | ---- | ---- |\n| 所需小正方形的数量/个  | 216  | 96   | 54   |\n\n(1) 每个小正方形的面积与所需小正方形的数量成 ( )比例关系。\n(2) 如果采用面积是 36cm² 的小正方形来摆满这张长方形彩纸, 需要多少个小正方形? (用比例方法解答)",
+    "options": [],
+    "answer": "(1) 反;(2) 24个",
+    "solution": "【分析】(1)每个小正方形的面积 × 小正方形的数量 = 长方形彩纸的面积;长方形彩纸的面积一定,每个小正方形的面积与所需小正方形的数量成反比例关系。(2) 长方形彩纸的面积 = 36 × 需要小正方形个数, 由此解答即可。解决此题关键是根据比值一定或乘积一定,先列出比例,进而根据比例的性质先把比例式转化为乘积式来解比例得解;注意等号要对齐。\n\n【详解】(1)长方形彩纸的面积一定,每个小正方形的面积与所需小正方形的数量成反比例关系;(2) 解: 设需要多 x 个小正方形;\n\n36x = 216 × 4\n36x ÷ 36 = 216 × 4 ÷ 36\nx = 24\n\n答: 需要 24 个小正方形。",
+    "knowledge_points": ["反比例的应用"],
+    "source": "(20-21 六年级下·全国·单元测试)",
+    "merged_id": "unit4_396",
+    "original_id": "unit4_choice_396"
+  },
+  {
+    "id": "unit4_choice_397",
+    "unit": "第四单元:比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "下边的图象表示汽车在公路上行驶的路程与耗油量的关系。\n\n① 请你用学过的数学知识描述这辆汽车行驶的路程和耗油量的关系,并讲明理由。\n(2) 根据图象, 这辆汽车行驶 75km 耗 6 升。计算这辆汽车行驶 180km 耗油多少升?\n\n![](images/2d569aaebb2cfa5c1e8de641bb84ee998672b3cea0a06c28fd4781931cce5bde.jpg)",
+    "options": [],
+    "answer": "(1)汽车行驶路程与耗油量是正比例关系;因为 50: 4 = 100: 8 = 150: 12 = … = 12.5 (一定);汽车行驶路程与耗油量的比值一定,所以汽车行驶路程与耗油量是正比例关系。(2) 14.4 升",
+    "solution": "【分析】(1)表中有两种相关联的量,行驶的路程和耗油量,耗油量随着行驶的路程变化而变化,且行驶路程和耗油量的比值是一定的, 50:4 = 100:8 = 150:12…,符合正比例关系式 x:y = k(一定),所以汽车行驶路程与耗油量是正比例关系; (2) 因为行驶的路程和耗油量成正比例, 设这辆汽车行驶 180km 耗油 x 升, 据此列比例解答。此题主要考查从图象中获得信息的能力,以及正比例的意义的实际应用。\n\n【详解】(1)汽车行驶路程与耗油量是正比例关系;因为 50:4 = 100:8 = 150:12 = … = 12.5 (一定);汽车行驶路程与耗油量的比值一定,所以汽车行驶路程与耗油量是正比例关系;(2) 解: 设这辆汽车行驶 180km 耗油 x 升;\n\n75/6 = 180/x\n75x = 6 × 180\nx = (6 × 180)/75\nx = 14.4;\n\n答: 辆汽车行驶 180km 耗油 14.4 升。",
+    "knowledge_points": ["正比例的应用", "正比例的意义及辨识"],
+    "source": "(18-19六年级下·河北秦皇岛·期末)",
+    "merged_id": "unit4_397",
+    "original_id": "unit4_choice_397"
+  },
+  {
+    "id": "unit4_choice_398",
+    "unit": "第四单元:比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "(1)按1:2的比画出三角形缩小后的图形。\n(2) 如果每个小方格的边长表示 1 厘米。缩小后三角形的面积是 ( ) 平方厘米。\n\n![](images/17fca04639b4fb3e2f96e249bc970e984915857607db911f74a6420e5125d6cb.jpg)",
+    "options": [],
+    "answer": "(1)\n\n![](images/40e8098704318aa2916ef8c16e3ec5adb227dcdc6ed0232c670fdcdd81fff433.jpg)\n\n(2) 3",
+    "solution": "【分析】(1)通过观察图形可知,这个三角形的底是4厘米,高是6厘米,首先求出缩小后三角形的底和高各是多少厘米,据此画除缩小后的三角形。(2) 根据三角形的面积公式: S = ah ÷ 2, 把数据代入公式解答。此题考查的目的是理解掌握图形缩小的方法及应用,以及三角形面积公式的灵活运用,关键熟记公式。\n\n【详解】(1) 4 ÷ 2 = 2 (厘米)6 ÷ 2 = 3 (厘米) 作图如下:\n\n![](images/426eb085c884504c38c12442e697eb8067b5dc79188c36072dc8f30d68238ce7.jpg)\n\n(2) 2 × 3 ÷ 2 = 3 (平方厘米)\n\n答: 缩小后三角形的面积是 3 平方厘米。",
+    "knowledge_points": ["图形的放大与缩小", "三角形的面积"],
+    "source": "(2020·贵州铜仁·小升初真题)",
+    "merged_id": "unit4_398",
+    "original_id": "unit4_choice_398"
+  },
+  {
+    "id": "unit4_choice_399",
+    "unit": "第四单元:比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "杨叔叔驾驶电动汽车外出游玩,请你根据下表中的相关数据,回答下列问题。\n\n| 行驶路程/km   | 50   | 100  | 150  | 200  | ...  |\n| ------------- | ---- | ---- | ---- | ---- | ---- |\n| 耗电量/千瓦时 | 7.5  | 15   | 22.5 | 30   | ...  |\n\n(1) 耗电量和行驶路程成 ( ) 比例关系, 因为 ( )。\n(2) 在图中描出表示行驶路程和相对应耗电量的点, 然后把它们按顺序连起来。观察图象, 你有什么发现?\n\n![](images/21a2d761de24b13c1dfae665c70057da3c1357511fe501280276e2ac8da3c548.jpg)\n\n(3) 汽车电池充满电后有 45 千瓦时的电, 根据图象推测可行驶 ( ) km。",
+    "options": [],
+    "answer": "(1)正;耗电量与行驶路程的比值一定 (2) 见详解 (3) 300",
+    "solution": "【分析】(1)根据表格中的耗电量和行驶路程,如果比值一定,就是正比例关系,如果乘积一定,就是反比例关系; (2) 根据表格中的数据, 描点连线即可。认真观察图象找出它的特点即可。(3) 根据比例关系, 计算耗电量是 45 千瓦时, 对应的行驶路程。此题考查了有关正比例的应用,明确如果相关联的两个量对应的数据的比值一定,它们就成正比例关系。\n\n【详解】(1)耗电量和行驶路程成正比例关系,因为耗电量与行驶路程的比值一定。(2) 耗电量/千瓦时\n\n![](images/88d9f243b3d00fe99df39a9ea44dbc4bfcd5853a121b3f6d47533024ba5cadfd.jpg)\n\n耗电量/千瓦时示例:我发现图象是一条以 0 点为端点的射线。\n\n(3) 解:设可行驶 x 千米。\n\n7.5/50 = 45/x\n7.5x = 225\nx = 300\n\n答:根据图象推测可行驶 300 千米。",
+    "knowledge_points": ["正比例的应用", "反比例的意义及辨识"],
+    "source": "(20-21 六年级下·全国·单元测试)",
+    "merged_id": "unit4_399",
+    "original_id": "unit4_choice_399"
+  },
+  {
+    "id": "unit4_choice_400",
+    "unit": "第四单元:比例",
+    "chapter": "比例",
+    "question_type": "解答题",
+    "difficulty": "容易",
+    "question": "动手实践, 操作应用。\n\n![](images/0c31a68560eb28866b85c13c76a5bd73b34fd7e5d98334ae2b1cf463bd3d207f.jpg)\n\n(1) ( ) 号长方形是①号长方形缩小后的图形, 它是按 ( ) : ( ) 缩小的。\n(2) ( ) 号长方形是①号长方形放大后的图形, 它是按 ( ) : ( ) 放大的。\n(3) 在方格图中画出②号平行四边形按 2: 1 放大后的图形。\n(4) 在方格图中画出④号长方形按 1:3 缩小后的图形。\n(5) 用①号长方形长和宽的比(不化简)与⑤号长方形长和宽的比(不化简)组成一个比例是()。",
+    "options": [],
+    "answer": "(1) ③; 1; 2 (2) ⑤; 3; 2 (3) (4) 见详解 (5) 6: 2 = 9: 3",
+    "solution": "【分析】(1)找①号长方形缩小后的图形,只要看那个图形比它小的长方形,而且长与宽的比是否相等;进而求出缩小的比;(2) 找①号长方形放大后的图形, 只要看那个图形比它大的长方形, 而且长与宽的比是否相等; 进而求出放大的比; (3) 在方格图中画出②号平行四边形按放大后的图形, 把②号平行四边形的各边扩大 2 倍; (4) 在方格图中画出④号长方形按 1: 3 缩小后的图形, 把④号长方形的各边缩小 3 倍; (5) 用①号长方形长和宽的比与⑤号长方形长和宽的比组成一个比例即可。图形的放大与缩小:形状相同,大小不同。\n\n【详解】(1) (3) 号长方形是①号长方形缩小后的图形,它是按(1):(2)缩小的。(2) (5) 号长方形是①号长方形放大后的图形, 它是按 (3) : (2) 放大的。(3) (4) 题如下图:\n\n![](images/2214ad112eea4228864277c969f51de99ed7787729236452df855f336825b0fe.jpg)(5) 6: 2 = 9: 3。",
+    "knowledge_points": ["图形的放大与缩小", "比与比例的区别与联系", "比例的意义"],
+    "source": "(20-21 六年级下·全国·单元测试)",
+    "merged_id": "unit4_400",
+    "original_id": "unit4_choice_400"
+  }
+]
+