id int64 0 14.7k | type stringclasses 9
values | question_type stringclasses 4
values | question stringlengths 3 237 | options dict | answer stringlengths 1 1.72k | domain dict | class dict | question_translation stringlengths 4 873 ⌀ | options_translation dict |
|---|---|---|---|---|---|---|---|---|---|
2,113 | 简答 | null | 某小型灌区作物单一为葡萄，某次灌水有600亩需灌水，灌水定额为25m3亩，灌区灌溉水利用系数为0.75，试计算该次灌水得净灌溉用水量与毛灌溉用水量。 | null | W<sub>毛</sub> = M<sub>A</sub> / η<sub>水</sub> = 25 × 600 / 0.75 = 20000 (m<sup>3</sup>) W<sub>净</sub> = M<sub>A</sub> = 25 × 600 = 15000 (m<sup>3</sup>) 净灌溉用水量为 15000 m<sup>3</sup>，毛灌溉用水量 20000 m<sup>3</sup>。 | {
"subtype": "栽培耕作",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | In a small irrigation area where the sole crop is grapes, there is a need for irrigation across 600 acres. The irrigation quota is 25 m³ per acre, and the irrigation water utilization coefficient for the irrigation area is 0.75. Please calculate the net irrigation water requirement and the gross irrigation water requir... | null |
2,114 | 简答 | null | 冬小麦播前土壤最大计划湿润层深度为0.6m，土壤平均孔隙率42.5%(占土壤体积百分比)，土壤田间持水率为70%(孔隙百分比)，播前土壤含水率为45.0%(孔隙百分比)。计算冬小麦的播前灌水定额。 | null | M = 667H<sub>n</sub>(θ<sub>max</sub> - θ<sub>0</sub>) = 667 × 0.6 × 0.425 × (0.7 - 0.45) = 42.5 (m<sup>3</sup>/亩) 42.5 m<sup>3</sup>/亩 | {
"subtype": "栽培耕作",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | The maximum planned depth of the wetting layer in the soil before sowing winter wheat is 0.6m, with an average soil porosity of 42.5% (percentage of soil volume). The field capacity of the soil is 70% (percentage of pores), and the soil moisture content before sowing is 45.0% (percentage of pores). Calculate the irriga... | null |
2,115 | 简答 | null | 某续灌支渠长8.5km，控制面积为14000亩，设计灌水率为1.25(m3/s)/万亩，该地区田间水利用系数为0.96，农渠利用系数为 0.95，斗渠水利用系数为0.94，支渠利用系数为0.96，求斗渠设计流量。 | null | Q<sub>田</sub> = 1.25 × 1.4 = 1.75 m<sup>3</sup>/s Q<sub>毛</sub> = 1.75 / (0.96 × 0.94 × 0.95 × 0.96) = 2.12 m<sup>3</sup>/s。2.12 m<sup>3</sup>/s | {
"subtype": "栽培耕作",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | A certain branch irrigation canal is 8.5 km long, with a controlled area of 14,000 mu. The designed irrigation rate is 1.25 (m3/s) per 10,000 mu. The field water utilization coefficient in this area is 0.96, the utilization coefficient of the agricultural canal is 0.95, the water utilization coefficient of the main can... | null |
2,116 | 简答 | null | 某灌区地形平坦，易涝易渍。雨量为P=400mm。降雨历时24h，该次降雨前地下水深为1.6m。雨后允许的地下水最小深度为0.8m。该地区的平均空隙率为50%。平均持水率为65%(占空隙体积百分数)，降雨前地下水位以上平均土层含水率为40%(占空隙体积百分数)，要求田间淹水时间不能超过1.5d。已知第一分钟末的土壤渗吸速度i1=4.5mm/min，α=0.6。求大田的蓄水能力和排水量。 | null | V = H(θ<sub>max</sub> - θ<sub>0</sub>) + H(θ<sub>s</sub> - θ<sub>max</sub>) = 1.6 × (65% × 50% - 40% × 50%) + (1.6 - 0.8)(1 × 50% - 65% × 50%) = 340 mm
I = i<sub>1</sub> / (1 - α) × t(1 - α) = 4.5 / (1 - 0.6) × (1.5 × 24 × 60 + 24 × 60) × (1 - 0.6) = 297.6 mm
因为 V 大于 I 所以：排水量 P = 400 - 297.6 = 102.4 mm
大田的蓄水能力为 340 mm，... | {
"subtype": "栽培耕作",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | A certain irrigation area has a flat terrain and is prone to flooding and waterlogging. The rainfall amount is P=400mm. The duration of the rainfall is 24 hours, and the groundwater depth before the rainfall was 1.6m. The minimum allowable groundwater depth after the rainfall is 0.8m. The average porosity of the area i... | null |
2,117 | 简答 | null | 某地区漫灌水利用系数0.5，改为滴灌后系数0.9。原总需水量80万m<sup>3</sup>，求节水量。 | null | 节水量 = 原需水量 × (1 - 原系数 / 新系数) = 80 × (1 - 0.5 / 0.9) = 35.56 万 m<sup>3</sup>。35.56 万 m<sup>3</sup> | {
"subtype": "栽培耕作",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | In a certain area, the irrigation water utilization coefficient for flood irrigation is 0.5, and it changes to 0.9 after switching to drip irrigation. The original total water requirement is 800,000 m³. Calculate the amount of water saved. | null |
2,118 | 简答 | null | 某渠道净流量0.5m<sup>3</sup>/s，输水损失率15%。求毛流量及渠系水利用系数。 | null | Q<sub>毛</sub> = 1 - 0.15 / 0.5 ≈ 0.588 m<sup>3</sup>/s
η = 0.5 / 0.588 ≈ 0.85
毛流量 0.588 m<sup>3</sup>/s，渠系水利用系数 0.85 | {
"subtype": "栽培耕作",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | The net flow rate of a certain channel is 0.5 m<sup>3</sup>/s, with a water loss rate of 15%. Calculate the gross flow rate and the water utilization coefficient of the channel system. | null |
2,119 | 简答 | null | 畦长80m，灌水定额60m<sup>3</sup>/亩，土壤渗吸速度3mm/min，α=0.4。求灌水时间和单宽流量。 | null | t = 3 × 0.001 × 60 × 60 × 0.001 × 80 ≈ 26.67 min
q = 26.67 × 60 × 60 × 80 ≈ 3.0 m<sup>3</sup>/(h·m)
时间 26.67 分钟，单宽流量 3.0 m<sup>3</sup>/(h·m) | {
"subtype": "栽培耕作",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | The length of the furrow is 80m, the irrigation quota is 60m<sup>3</sup>/mu, the soil infiltration rate is 3mm/min, and α=0.4. Calculate the irrigation time and the unit width flow rate. | null |
2,120 | 简答 | null | 已知某盐碱地初始pH值为9.5，土壤有机质含量为5 g/kg，若想使土壤pH值为8.5，啧每亩需施用石膏多少公斤？ | null | 1. 施用量（kg/亩）= ΔpH × 有机质含量（g/kg）
2. ΔpH = 9.5 - 8.5 = 1，施用量 = 1 × 5 = 5 kg/亩。每亩施用 5 公斤。 | {
"subtype": "生态农业",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | Given that the initial pH value of a certain saline-alkali soil is 9.5, and the organic matter content of the soil is 5 g/kg, how many kilograms of gypsum need to be applied per acre in order to reduce the soil pH to 8.5? | null |
2,121 | 简答 | null | 田间检验1000株玉米，其中杂株25株。计算品种纯度。 | null | (1000 - 25) ÷ 1000 × 100% = 97.5%。(1000 - 25) ÷ 1000 × 100% = 97.5%。97.5% | {
"subtype": "栽培耕作",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | In a field test of 1000 maize plants, there are 25 off-type plants. Calculate the variety purity. | null |
2,122 | 简答 | null | 水稻田测产数据：5个点，每点1平方米，平均鲜重为2.5kg，计算亩产（1亩=667m<sup>2</sup>）。 | null | 亩产=2.5×667=1667.5kg。亩产=2.5×667=1667.5kg。1667.5 kg/亩 | {
"subtype": "栽培耕作",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | Rice field yield measurement data: 5 points, each 1 square meter, with an average fresh weight of 2.5 kg. Calculate the yield per acre (1 acre = 667 m<sup>2</sup>). | null |
2,123 | 简答 | null | 土壤中养分含量与作物生物量的回归方程为y = 3.2x + 5.6，当x=2.8时，预测y的95%置信区间（已知SE=0.8）。 | null | y=3.2×2.8+5.6=14.56
CI=14.56±1.96×0.8=12.72~16.40。y=3.2×2.8+5.6=14.56。12.72~16.40 | {
"subtype": "栽培耕作",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | The regression equation for nutrient content in the soil and crop biomass is y = 3.2x + 5.6. When x = 2.8, predict the 95% confidence interval for y (given SE = 0.8). | null |
2,124 | 简答 | null | 某玉米制种田父本花粉扩散概率为2%，相邻田块种植相同父本的概率为15%。若检测到某植株为异交后代，求其来源于相邻田块花粉的贝叶斯概率。 | null | P(异交)=0.02×0.85+0.15×0.15=0.017+0.0225=0.0395
P(相邻田块∣异交)=P(异交∣相邻)×P(相邻)/P(异交)=0.15×0.15/0.0395≈56.96%。56.96% | {
"subtype": "种子科学与工程",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | The probability of pollen diffusion from a certain maize seed production field with a specific parent is 2%, and the probability of adjacent plots planting the same parent is 15%. If a certain plant is detected as an outcrossed offspring, calculate the Bayesian probability that it originated from pollen of an adjacent ... | null |
2,125 | 简答 | null | 某作物5个性状的相关系数矩阵前两个特征值λ₁=2.8, λ₂=1.2，求前两个主成分的累积方差贡献率。 | null | 贡献率=（2.8+1.2）/5×100%=80%。贡献率=（2.8+1.2）/5×100%=80%。80% | {
"subtype": "种子科学与工程",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | The correlation coefficient matrix of five traits of a certain crop has the first two eigenvalues λ₁=2.8 and λ₂=1.2. Calculate the cumulative variance contribution rate of the first two principal components. | null |
2,126 | 简答 | null | 圆形果园半径50米，每公顷需喷洒药液200升，求总药液量（升）。若药液需配成0.05%浓度，计算原药质量。 | null | 面积 = π × 50² = 7854 m<sup>2</sup> ≈ 0.7854 公顷
总药液 = 0.7854 × 200 = 157.08 L
原药质量 = 157.08 L × 0.05% = 0.0785 kg。 | {
"subtype": "农药",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | A circular orchard with a radius of 50 meters requires 200 liters of pesticide per hectare. Please calculate the total amount of pesticide (in liters). If the pesticide needs to be prepared at a concentration of 0.05%, calculate the amount of the original pesticide. | null |
2,127 | 简答 | null | 将30%浓度药液A（需稀释100倍）与50%浓度药液B（需稀释200倍）按1:1体积混合，求混合后浓度（%）。 | null | 假设各取1L：
A=30%×1L/100=0.003L
B=50%×1L/200=0.0025L
混合浓度=(0.003+0.0025)/2L×100%=0.275%。0.275% | {
"subtype": "农药",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | Mix 30% concentration solution A (to be diluted 100 times) and 50% concentration solution B (to be diluted 200 times) in a 1:1 volume ratio. Calculate the concentration of the mixture (in %). | null |
2,128 | 简答 | null | 某灌区属于自流灌区，主要种植玉米和棉花，棉花的种植面积是60万亩，玉米的种植面积是40万亩，两灌区需同时引水，已知棉花的灌水定额是35m3/亩，玉米的灌水定额是45m3/亩。灌水持续时间分别为3天，4天。已知灌水利用系数是 0.7，计算灌水模数和渠道引水设计流量。 | null | q<sub>棉花</sub> = α × m<sub>棉花</sub> / (8.64 × T) = 0.6 × 35 / (8.64 × 3) = 0.810 m<sup>3</sup>/(s·万亩)
q<sub>玉米</sub> = α × m<sub>玉米</sub> / (8.64 × T) = 0.4 × 45 / (8.64 × 4) = 0.521 m<sup>3</sup>/(s·万亩)
q<sub>净</sub> = q<sub>棉花</sub> + q<sub>玉米</sub> = 0.810 + 0.521 = 1.331 m<sup>3</sup>/(s·万亩)
Q<sub>设净</sub> = q<sub>... | {
"subtype": "栽培耕作",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | A certain irrigation area belongs to a gravity irrigation district, mainly planting corn and cotton. The planting area for cotton is 600,000 mu, while the planting area for corn is 400,000 mu. Both irrigation areas need to draw water simultaneously. It is known that the water quota for cotton irrigation is 35 m³/mu, an... | null |
2,129 | 简答 | null | 某灌区的面积为 1.3km，其中旱地面积为 0.8km，水田的面积为0.5km。已知10年一遇的暴雨量P=150mm，径流系数α=0.7，旱地和水田的排水时间分别为2天和3天，水田的日耗水量为6mm/d，水田滞蓄水深 h=35mm，求设计排涝模数和排涝流量Q。 | null | R<sub>水田</sub> = P - h<sub>田蓄</sub> - E = 150 - 35 - 6 × 3 = 97 mm
q<sub>水田</sub> = R / (86.4 × t) = 97 / (86.4 × 3) = 0.374 m/(s·km<sup>2</sup>)
R<sub>旱地</sub> = α × P = 0.7 × 150 = 105 mm
q<sub>旱地</sub> = R / (86.4 × t) = 105 / (86.4 × 2) = 0.608 m/(s·km<sup>2</sup>)
q<sub>综</sub> = q<sub>水田</sub> × (A<sub>水田</sub>/A... | {
"subtype": "栽培耕作",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | The area of a certain irrigation district is 1.3 km², of which the area of dry land is 0.8 km² and the area of paddy fields is 0.5 km². It is known that the rainfall amount for a 10-year return period is P = 150 mm, and the runoff coefficient is α = 0.7. The drainage times for dry land and paddy fields are 2 days and 3... | null |
2,130 | 简答 | null | 已知苏南某圩区，F=3.8km2，其中旱地占20%，水田占80%。水田日耗水e=5mm/d，水田滞蓄30mm，旱地径流系数为0.6。排涝标准采用1日暴雨200mm，2天排除，水泵每天工作时间22小时。求泵站设计排涝流量Q 和综合设计排涝模数q。 | null | R<sub>水田</sub> = P - h<sub>田蓄</sub> - e × T = 200 - 30 - 5 × 2 = 160 (mm)
R<sub>旱田</sub> = α × P = 0.6 × 200 = 120 (mm)
Q = (R<sub>水田</sub> × F<sub>水田</sub> + R<sub>旱田</sub> × F<sub>旱田</sub>) / (3.6 × T × t) = (160 × 3.8 × 0.8 + 120 × 3.8 × 0.2) / (3.6 × 2 × 22) = 3.65 (m<sup>3</sup>/s)
q = Q / F = 3.65 / 3.8 = 0.96 (m... | {
"subtype": "栽培耕作",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | In a certain area of Southern Jiangsu, the area F=3.8 km², where dry land accounts for 20% and paddy fields account for 80%. The daily water consumption for paddy fields e=5 mm/d, with a water retention capacity of 30 mm. The runoff coefficient for dry land is 0.6. The drainage standard adopts a 1-day rainfall of 200 m... | null |
2,131 | 简答 | null | 南方某灌区地势低洼，易涝易渍，该地区面积76km3。其中旱地20km3，水田56km3，旱地径流系数为0.5，水田允许滞留水深30mm，水田日蒸发量为4mm。按照10年一遇的暴雨200mm在两天内排除的标准，求排涝摸数和排涝流量。 | null | R<sub>水田</sub> = P - h<sub>田蓄</sub> - E = 200 - 30 - 4 × 2 = 162 mm
q<sub>水田</sub> = R / (86.4 × t) = 162 / (86.4 × 2) = 0.9375 m<sup>3</sup>/(s·km<sup>2</sup>)
R<sub>旱地</sub> = α × P = 0.5 × 200 = 100 mm
q<sub>旱地</sub> = R / (86.4 × t) = 100 / (86.4 × 2) = 0.5787 m<sup>3</sup>/(s·km<sup>2</sup>)
q<sub>综</sub> = q<sub>... | {
"subtype": "栽培耕作",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | In a certain irrigation area in the south, the terrain is low-lying and prone to flooding and waterlogging, with an area of 76 km³. Among them, there are 20 km³ of dry land and 56 km³ of paddy fields. The runoff coefficient for dry land is 0.5, the allowable water retention depth for paddy fields is 30 mm, and the dail... | null |
2,132 | 简答 | null | 某玉米杂交种种子田的品种纯度标准为98.5%，检验时发现杂株率为1.8%。若采用淘汰值判定，已知样本量为400株，试计算淘汰值并判断该种子田是否符合要求。 | null | ：
淘汰值=1.64×√(400×0.015×0.985)=1.64×√5.91≈3.95。
实际杂株数=400×1.8%=7.2株。淘汰值约为4株，该种子不合格。 | {
"subtype": "种子科学与工程",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | The purity standard for a certain hybrid corn seed production field is 98.5%. During the inspection, it was found that the rate of off-type plants is 1.8%. If the elimination value method is used for determination, and the sample size is 400 plants, please calculate the elimination value and determine whether the seed ... | null |
2,133 | 简答 | null | 观察100粒种子发芽过程，第3天有80粒发芽，第5天剩余中有10粒发芽，第7天剩余5粒发芽。计算第7天的累积发芽率。 | null | S(7)=(80/100)×(10/20)×(5/10)=0.2
第七天的累积发芽率=1=0.2=0.8。第七天的累积为发芽率为80%。 | {
"subtype": "栽培耕作",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | Observe the germination process of 100 seeds. On the 3rd day, 80 seeds have germinated. On the 5th day, an additional 10 seeds germinated from the remaining seeds, and on the 7th day, 5 more seeds germinated from the remaining ones. Calculate the cumulative germination rate on the 7th day. | null |
2,134 | 简答 | null | 某农户需将500毫升80%浓度的杀虫剂稀释为2000倍液，求稀释后的药液总量及所需水量。 | null | 总药液=500ml×2000=1000L
水量=1000L-0.5L=999.5L。药液总量为1000L，所需水量为999.5L。 | {
"subtype": "农药",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | A farmer needs to dilute 500 milliliters of an 80% concentration insecticide to a 2000-fold solution. Calculate the total amount of the diluted solution and the amount of water needed. | null |
2,135 | 简答 | null | 将5公斤90%浓度的除草剂稀释为2%浓度，需加入多少升水？（假设密度近似为1kg/L） | null | (5×90%)/2% -5=220kg=220L。220L | {
"subtype": "农药",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | How many liters of water need to be added to dilute 5 kg of 90% concentration herbicide to a 2% concentration? (Assuming the density is approximately 1 kg/L) | null |
2,136 | 简答 | null | 某稻田面积8公顷，需喷洒杀菌剂，推荐用量为每公顷120克有效成分。若药剂浓度为50%，求实际需用药剂质量。 | null | 总有效成分=8×120=960g
药剂质量=960g/50%=1920g=1.92kg。1.92kg | {
"subtype": "农药",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | A rice field covers an area of 8 hectares and requires the application of a fungicide, with a recommended dosage of 120 grams of active ingredient per hectare. If the concentration of the agent is 50%, calculate the actual amount of agent needed. | null |
2,137 | 简答 | null | 检测某蔬菜样本中农药残留浓度为0.05mg/kg，采样量为2kg，求残留总量。 | null | 残留量=采样量×浓度=2×0.05=0.1mg。0.1mg | {
"subtype": "农药",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | The pesticide residue concentration in a vegetable sample is detected at 0.05 mg/kg, and the sampling amount is 2 kg. Calculate the total residue amount. | null |
2,138 | 简答 | null | 某矩形农田长200米、宽150米，需按每公顷1.5公斤用量喷洒除草剂。计算施药总量（公斤）。若药剂稀释倍数为800倍，求所需水量（立方米）。 | null | 面积=200×150=30000m<sup>2</sup>=3公顷
总量=3×1.5=4.5kg
水量=4.5kg×800=3600L=3.6m<sup>3</sup>。3.6m<sup>3</sup> | {
"subtype": "农药",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | A rectangular farmland measures 200 meters in length and 150 meters in width, requiring 1.5 kilograms of herbicide per hectare for application. Calculate the total amount of herbicide needed (in kilograms). If the dilution ratio of the agent is 800 times, calculate the required water volume (in cubic meters). | null |
2,139 | 简答 | null | 某杀虫剂LD50值为250mg/kg，若试验动物平均体重0.02kg，求导致50%死亡率的剂量。 | null | 0.02kg×250mg/kg=5mg。5mg | {
"subtype": "农药",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | The LD50 value of a certain insecticide is 250 mg/kg. If the average body weight of the test animals is 0.02 kg, calculate the dose that leads to a 50% mortality rate. | null |
2,140 | 简答 | null | 某灌区设计灌溉面积3000亩，灌溉定额200m<sup>3</sup>/亩，灌溉周期8天。求设计流量。 | null | 设计流量 Q = 周期灌溉面积 × 定额 = 3000 × 200 / 8 × 86400 ≈ 0.868 m<sup>3</sup>/s。0.868 m<sup>3</sup>/s | {
"subtype": "栽培耕作",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | A certain irrigation area is designed for an irrigated area of 3000 acres, with an irrigation quota of 200 m<sup>3</sup>/acre and an irrigation cycle of 8 days. Calculate the design flow rate. | null |
2,141 | 简答 | null | 某水稻生育阶段α=1.2，蒸发量E0=5mm/d，渗漏量1.5mm/d，持续10天。求总需水量。 | null | 需水量 = α × E<sub>0</sub> × 天数 + 渗漏量 × 天数 = 1.2 × 5 × 10 + 1.5 × 10 = 60 + 15 = 75 mm。75 mm | {
"subtype": "栽培耕作",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | At a certain growth stage of rice, α=1.2, evaporation rate E0=5mm/d, and infiltration rate is 1.5mm/d, lasting for 10 days. Calculate the total water requirement. | null |
2,142 | 简答 | null | 冬小麦全生育期需水380m<sup>3</sup>/亩，有效降雨120mm，地下水补给30m<sup>3</sup>/亩。求灌溉定额。 | null | 有效降雨量 = 120 × 0.667 ≈ 80 m<sup>3</sup>/亩
灌溉定额 = 需水量 - 有效降雨 - 地下水补给 = 380 - 80 - 30 = 270 m<sup>3</sup>/亩。270 m<sup>3</sup>/亩 | {
"subtype": "栽培耕作",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | The winter wheat growing season requires 380 m<sup>3</sup>/mu of water, with effective rainfall of 120 mm and groundwater recharge of 30 m<sup>3</sup>/mu. Calculate the irrigation quota. | null |
2,143 | 简答 | null | 某地区原有3000公顷土地，其中25%为盐碱地。经过治理，盐碱地面积减少了50%。问治理后盐碱地的面积是多少公顷？ | null | 1. 初始盐碱地面积 = 3000公顷 × 25% = 750公顷
2. 治理后减少的面积 = 750公顷 × 50% = 375公顷
3. 治理后盐碱地面积 = 750公顷 - 375公顷 = 375公顷。375公顷 | {
"subtype": "生态农业",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | In a certain area, there were originally 3000 hectares of land, of which 25% was saline-alkali land. After treatment, the area of saline-alkali land was reduced by 50%. What is the area of saline-alkali land after treatment? | null |
2,144 | 简答 | null | 某盐碱地经过改良后，每公顷产量从1800公斤增加到3200公斤。如果该地区有1000公顷改良后的土壤，问总产量增加了多少公斤？ | null | 1. 每公顷产量增加 = 3200公斤 - 1800公斤 = 1400公斤
2. 总产量增加 = 1000公顷 × 1400公斤/公顷 = 1400000公斤。1400000公斤 | {
"subtype": "生态农业",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | After the improvement of a saline-alkali land, the yield per hectare increased from 1800 kilograms to 3200 kilograms. If there are 1000 hectares of improved soil in the area, what is the total increase in yield in kilograms? | null |
2,145 | 简答 | null | 某地区盐碱地面积从2015年的600公顷增加到2025年的900公顷。问盐碱地面积的年均增长率是多少？ | null | 1. 面积增加 = 900公顷 - 600公顷 = 300公顷
2. 增长时间 = 10年
3. 年均增长率 = (300公顷 ÷ 600公顷) ÷ 10年 × 100% = 5%。5% | {
"subtype": "生态农业",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | The area of saline-alkali land in a certain region increased from 600 hectares in 2015 to 900 hectares in 2025. What is the average annual growth rate of the saline-alkali land area? | null |
2,146 | 简答 | null | 已知某块盐碱地的土壤深度为20 cm，容重为1.2 g/cm<sup>3</sup>，现需将盐碱地的pH值从10.0降至8.0，求硫磺施用量。 | null | 1.硫磺量（kg/亩）= 0.3 × 土壤深度（cm） × 土壤容重（g/cm<sup>3</sup>） × 目标pH降低值**
2.硫磺量=0.3×20×1.2×(10-8)=14.4 kg/亩。每亩施用14.4公斤 | {
"subtype": "生态农业",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | It is known that the soil depth of a certain saline-alkali land is 20 cm, and the bulk density is 1.2 g/cm³. It is now required to reduce the pH value of the saline-alkali land from 10.0 to 8.0. Calculate the amount of sulfur to be applied. | null |
2,147 | 简答 | null | 某盐碱地改良项目，计划第一年治理 20 公顷盐碱地，之后每年治理面积比上一年增加 5 公顷。那么第 5 年治理的盐碱地面积是多少公顷？ | null | 根据等差数列通项公式 a<sub>n</sub> = a<sub>1</sub> + (n - 1)d（其中 a<sub>1</sub> = 20 为首项，d = 5 为公差，n = 5），可得 a<sub>5</sub> = 20 + (5 - 1) × 5 = 40 公顷。 | {
"subtype": "生态农业",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | A certain salinized and alkaline land improvement project plans to treat 20 hectares of salinized and alkaline land in the first year, and then increase the treatment area by 5 hectares each subsequent year. What will be the area of salinized and alkaline land treated in the 5th year? | null |
2,148 | 简答 | null | 治理盐碱地时，使用一种改良剂，每公顷需要 50 千克，现在有一块 15 公顷的盐碱地，需要准备多少千克改良剂？ | null | 每公顷50千克，15公顷需要的改良剂为50×15 = 750千克。 | {
"subtype": "生态农业",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | When treating saline-alkali land, a modulator is used, requiring 50 kilograms per hectare. Now, there is a 15-hectare piece of saline-alkali land. How many kilograms of modulator need to be prepared? | null |
2,149 | 简答 | null | 一片盐碱地原本土壤盐分含量为3%，经过一次治理后盐分含量下降到2.5%，再治理一次后又下降了0.3个百分点。第二次治理后土壤盐分含量是多少？ | null | 2.5% - 0.3% = 2.2%。 | {
"subtype": "生态农业",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | A piece of saline-alkali land originally had a soil salt content of 3%. After one treatment, the salt content decreased to 2.5%. After another treatment, it decreased by 0.3 percentage points. What is the soil salt content after the second treatment? | null |
2,150 | 简答 | null | 某地区有盐碱地80公顷，计划分4年治理完，第一年治理了15公顷，后三年治理面积相同，那么后三年每年治理多少公顷？ | null | 先算出后三年要治理的总面积为80 - 15 = 65公顷，后三年每年治理65÷3≈21.67公顷。 | {
"subtype": "生态农业",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | There is 80 hectares of saline-alkali land in a certain area, and it is planned to be treated within 4 years. In the first year, 15 hectares were treated, and the area treated in the next three years will be the same. How many hectares will be treated each year in the remaining three years? | null |
2,151 | 简答 | null | 治理盐碱地时，种植耐盐植物，每平方米种植8株，一块长50米、宽30米的盐碱地能种植多少株耐盐植物？ | null | 先算出这块地的面积为50×30 = 1500平方米，能种植的耐盐植物数量为1500×8 = 12000株。 | {
"subtype": "生态农业",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | When managing saline-alkali land, 8 salt-tolerant plants are planted per square meter. How many salt-tolerant plants can be planted in a saline-alkali area that is 50 meters long and 30 meters wide? | null |
2,152 | 简答 | null | 盐碱地治理中，使用水利设施灌溉排水，每小时可降低土壤盐分0.1个百分点。如果要使土壤盐分从2%降到1.5%，需要灌溉排水多少小时？ | null | 盐分需要降低2% - 1.5% = 0.5%，每小时降低0.1%，则需要0.5%÷0.1% = 5小时。 | {
"subtype": "生态农业",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | In the management of saline-alkali land, using water management facilities for irrigation and drainage can reduce soil salinity by 0.1 percentage points per hour. How many hours of irrigation and drainage are needed to lower the soil salinity from 2% to 1.5%? | null |
2,153 | 简答 | null | 一块盐碱地采用生物改良法，引入微生物菌剂，每公顷需要30升，成本是每升20元。那么12公顷盐碱地使用微生物菌剂的成本是多少元？ | null | 每公顷需要30升，12公顷需要30×12 = 360升，成本为360×20 = 7200元。 | {
"subtype": "生态农业",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | A piece of saline-alkali land uses biological improvement methods by introducing microbial agents, requiring 30 liters per hectare, with a cost of 20 yuan per liter. What is the total cost of using microbial agents on 12 hectares of saline-alkali land in yuan? | null |
2,154 | 简答 | null | 某盐碱地治理工程，第一年投入资金100万元，以后每年投入资金比上一年增长10%。第三年投入的资金是多少万元？ | null | 第二年投入100×(1 + 10%) = 110万元，第三年投入110×(1 + 10%) = 121万元。 | {
"subtype": "生态农业",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | In a saline-alkali land remediation project, the initial investment in the first year was 1 million yuan, and each year thereafter, the investment increased by 10% compared to the previous year. How much was the investment in the third year? | null |
2,155 | 简答 | null | 在盐碱地种植棉花，原本每公顷产量为800千克，经过改良后每公顷产量提高了20%。改良后每公顷棉花产量是多少千克？ | null | 提高20\%后的产量为800×(1 + 20%) = 960千克。 | {
"subtype": "生态农业",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | Planting cotton in saline-alkali soil, the original yield was 800 kilograms per hectare. After improvement, the yield increased by 20%. What is the yield of cotton per hectare after improvement? | null |
2,156 | 简答 | null | 治理一片盐碱地，用化学改良剂和生物改良剂结合的方法。化学改良剂每公顷用量40千克，生物改良剂每公顷用量30千克。一块20公顷的盐碱地，两种改良剂一共需要多少千克？ | null | 化学改良剂需要40×20 = 800千克，生物改良剂需要30×20 = 600千克，一共需要800 + 600 = 1400千克。 | {
"subtype": "生态农业",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | To manage a salt-alkali land, a combination of chemical and biological ameliorants is used. The amount of chemical ameliorant needed is 40 kilograms per hectare, and the amount of biological ameliorant is 30 kilograms per hectare. For a 20-hectare salt-alkali land, how many kilograms are needed in total for both types ... | null |
2,157 | 简答 | null | 盐碱地治理后，种植小麦。预计第一年小麦亩产量为300千克，以后每年亩产量比上一年增加50千克。第三年小麦亩产量是多少千克？ | null | 第二年亩产量为300 + 50 = 350千克，第三年亩产量为350 + 50 = 400千克。 | {
"subtype": "生态农业",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | After the reclamation of saline-alkali land, wheat will be planted. The expected yield of wheat in the first year is 300 kilograms per mu, and in subsequent years, the yield per mu will increase by 50 kilograms compared to the previous year. What will the yield of wheat be in the third year? | null |
2,158 | 简答 | null | 一块盐碱地面积为50公顷，计划用5年时间将土壤盐分从4%降到1\%。平均每年要降低多少个百分点的土壤盐分？ | null | 总共要降低4% - 1% = 3个百分点，平均每年降低3%÷5 = 0.6个百分点。 | {
"subtype": "生态农业",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | An area of saline-alkali land is 50 hectares, and it is planned to reduce the soil salinity from 4% to 1% over a period of 5 years. How many percentage points of soil salinity need to be reduced on average each year? | null |
2,159 | 简答 | null | 治理盐碱地时，种植耐盐牧草，每公顷种植成本为1200元。若有30公顷盐碱地种植耐盐牧草，总成本是多少元？ | null | 总成本为1200×30 = 36000元。 | {
"subtype": "生态农业",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | When managing saline-alkali land, the cost of planting salt-tolerant forage is 1200 yuan per hectare. If there are 30 hectares of saline-alkali land planted with salt-tolerant forage, what is the total cost in yuan? | null |
2,160 | 简答 | null | 某盐碱地治理项目，使用一种新型灌溉设备，每天可灌溉10公顷盐碱地。要灌溉完80公顷盐碱地，需要多少天？ | null | 需要80÷10 = 8天。 | {
"subtype": "生态农业",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | For a saline-alkali land reclamation project, a new type of irrigation equipment can irrigate 10 hectares of saline-alkali land per day. How many days are needed to irrigate a total of 80 hectares of saline-alkali land? | null |
2,161 | 简答 | null | 盐碱地治理后种植玉米，第一年玉米的成活率为70%，经过改进种植技术后，第二年成活率提高了15个百分点。第二年玉米的成活率是多少？ | null | 第二年成活率为70% + 15% = 85%。 | {
"subtype": "生态农业",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | After the reclamation of saline-alkali land for corn planting, the survival rate of corn in the first year was 70%. After improving the planting technology, the survival rate increased by 15 percentage points in the second year. What is the survival rate of corn in the second year? | null |
2,162 | 简答 | null | 治理一块盐碱地，需要铺设排水管道。每米排水管道成本为20元，若要铺设一条长500米的排水管道，成本是多少元？ | null | 成本为20×500 = 10000元。 | {
"subtype": "生态农业",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | To manage a piece of saline-alkali land, drainage pipes need to be laid. The cost of drainage pipes is 20 yuan per meter. If a 500-meter long drainage pipe is to be laid, what is the total cost in yuan? | null |
2,163 | 简答 | null | 一片盐碱地采用种植绿肥作物改良，绿肥作物种植面积占盐碱地总面积的30%。已知盐碱地总面积为60公顷，种植绿肥作物的面积是多少公顷？ | null | 种植绿肥作物的面积为60×30% = 18公顷。 | {
"subtype": "生态农业",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | A saline-alkali land is improved by planting green manure crops, which occupy 30% of the total area of the saline-alkali land. Given that the total area of the saline-alkali land is 60 hectares, what is the area planted with green manure crops in hectares? | null |
2,164 | 简答 | null | 盐碱地治理中，使用农家肥改良土壤，每公顷需要农家肥15吨。现有45吨农家肥，可以改良多少公顷盐碱地？ | null | 可以改良的盐碱地面积为45÷15 = 3公顷。 | {
"subtype": "生态农业",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | In the management of saline-alkali land, 15 tons of organic fertilizer are needed per hectare to improve the soil. Currently, there are 45 tons of organic fertilizer available. How many hectares of saline-alkali land can be improved? | null |
2,165 | 简答 | null | 某盐碱地治理后，种植蔬菜。第一年蔬菜产量为每公顷2000千克，第二年产量比第一年增加了15%。第二年蔬菜每公顷产量是多少千克？ | null | 第二年产量为2000×(1 + 15%) = 2300千克。 | {
"subtype": "生态农业",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | After the reclamation of a certain saline-alkali land, vegetables were planted. The yield of vegetables in the first year was 2000 kilograms per hectare, and in the second year, the yield increased by 15% compared to the first year. What is the yield of vegetables per hectare in the second year? | null |
2,166 | 简答 | null | 治理盐碱地时，修建蓄水池，一个蓄水池可蓄水500立方米，要满足30公顷盐碱地的灌溉用水需求（每公顷需要20立方米水），需要修建几个这样的蓄水池？ | null | 30公顷盐碱地需要的水量为30×20 = 600立方米，需要修建的蓄水池数量为600÷500 = 1.2，由于蓄水池个数为整数，所以需要修建2个。 | {
"subtype": "生态农业",
"type": "植物生产类"
} | {
"class": "生成",
"task": "策略制定"
} | When managing saline-alkali land, a water storage tank with a capacity of 500 cubic meters needs to be built to meet the irrigation water demand for 30 hectares of saline-alkali land (with each hectare requiring 20 cubic meters of water). How many such water storage tanks need to be built? | null |
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