[{"id":1444,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某校七年级组织学生参加数学实践活动，要求每名学生从A、B、C三个任务中至少选择一个完成。已知共有120名学生参与，其中选择A任务的有78人，选择B任务的有65人，选择C任务的有52人。同时，恰好选择两个任务的学生人数是恰好选择三个任务学生人数的3倍，且没有学生一个任务都不选。问：恰好选择三个任务的学生有多少人？","answer":"设恰好选择三个任务的学生人数为x人。\n\n根据题意，恰好选择两个任务的学生人数是3x人。\n\n因为每个学生至少选择一个任务，所以所有学生可以分为三类：\n- 只选一个任务的：设为y人\n- 恰好选两个任务的：3x人\n- 恰好选三个任务的：x人\n\n总人数为120人，因此有：\ny + 3x + x = 120\n即：y + 4x = 120 ——（1）\n\n再从任务被选的总人次角度分析：\n- 选择A任务的有78人，B任务65人，C任务52人，总人次为：78 + 65 + 52 = 195\n\n每个只选一个任务的学生贡献1人次，\n每个选两个任务的学生贡献2人次，\n每个选三个任务的学生贡献3人次。\n\n因此总人次可表示为：\n1×y + 2×(3x) + 3×x = y + 6x + 3x = y + 9x\n\n所以有：y + 9x = 195 ——（2）\n\n用方程（2）减去方程（1）：\n(y + 9x) - (y + 4x) = 195 - 120\n5x = 75\n解得：x = 15\n\n代入（1）得：y + 4×15 = 120 → y = 60\n\n因此，恰好选择三个任务的学生有15人。\n\n答：恰好选择三个任务的学生有15人。","explanation":"本题考查数据的收集、整理与描述中的集合思想与方程建模能力，结合一元一次方程和二元一次方程组的解法。解题关键在于理解“人次”与“人数”的区别，并合理设未知数，建立两个不同角度的等量关系：一是总人数，二是任务被选的总人次。通过设恰好选三个任务的人数为x，利用“恰好选两个任务的人数是其3倍”建立联系，再结合总人数和总人次列出方程组，最终求解。本题综合性强，需要学生具备较强的逻辑分析和方程建模能力，属于困难难度。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 11:41:23","updated_at":"2026-01-06 11:41:23","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":561,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"某班级进行了一次数学测验，成绩分布如下表所示。已知成绩在80分到89分之间的学生人数是成绩在60分到69分之间的3倍，且总人数为40人。如果60分到69分之间有4人，那么90分及以上的学生有多少人？\n\n| 分数段 | 人数 |\n|--------------|------|\n| 90分及以上 | ? |\n| 80-89分 | ? |\n| 70-79分 | 12 |\n| 60-69分 | 4 |\n| 60分以下 | 2 |","answer":"A","explanation":"根据题意，60-69分有4人，80-89分的人数是其3倍，即 3 × 4 = 12人。已知70-79分有12人，60分以下有2人。设90分及以上的人数为x。总人数为40人，因此可列方程：x + 12（80-89） + 12（70-79） + 4（60-69） + 2（60以下） = 40。计算得：x + 12 + 12 + 4 + 2 = 40，即 x + 30 = 40，解得 x = 10。所以90分及以上的学生有10人。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 19:22:43","updated_at":"2025-12-30 11:11:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"10","is_correct":1},{"id":"B","content":"12","is_correct":0},{"id":"C","content":"14","is_correct":0},{"id":"D","content":"16","is_correct":0}]},{"id":350,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"在一次环保知识问卷调查中，某班级共收集到有效问卷120份。调查结果显示，有75名学生表示经常进行垃圾分类，有60名学生表示会主动节约用水。已知有30名学生既经常进行垃圾分类又会主动节约用水。那么，至少参与其中一项环保行为的学生人数是多少？","answer":"A","explanation":"本题考查数据的收集、整理与描述中的集合思想，属于简单难度的应用题。根据题意，设经常进行垃圾分类的学生集合为A，主动节约用水的学生集合为B。已知|A| = 75，|B| = 60，|A ∩ B| = 30。要求的是至少参与其中一项的学生人数，即求|A ∪ B|。根据集合的并集公式：|A ∪ B| = |A| + |B| - |A ∩ B| = 75 + 60 - 30 = 105。因此，至少有105名学生参与了至少一项环保行为。选项A正确。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:42:10","updated_at":"2025-12-30 11:11:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"105","is_correct":1},{"id":"B","content":"120","is_correct":0},{"id":"C","content":"90","is_correct":0},{"id":"D","content":"135","is_correct":0}]},{"id":1473,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市为了优化公交线路，对一条主干道的车流量进行了为期7天的观测，记录每天上午7:00至9:00的车辆通过数量（单位：百辆），数据如下：12, 15, 18, 14, 16, 20, 17。交通部门计划根据这些数据调整红绿灯时长，并设定一个‘高峰阈值’，若某天的车流量超过该阈值，则启动延长绿灯时间的应急方案。已知该阈值设定为这组数据的中位数与平均数的较大者。同时，为评估调整效果，工程师在平面直角坐标系中绘制了车流量与绿灯延长时间的函数关系图，其中绿灯延长时间 y（单位：秒）与车流量 x（单位：百辆）满足一次函数关系，且当 x = 15 时 y = 10，当 x = 20 时 y = 20。若某天观测到车流量为 19 百辆，且该天启动了应急方案，求该天绿灯延长时间的理论值，并判断该天车流量是否确实超过了设定的高峰阈值。","answer":"第一步：计算7天车流量的平均数。\n数据：12, 15, 18, 14, 16, 20, 17\n总和 = 12 + 15 + 18 + 14 + 16 + 20 + 17 = 112\n平均数 = 112 ÷ 7 = 16（百辆）\n\n第二步：求中位数。\n将数据从小到大排列：12, 14, 15, 16, 17, 18, 20\n共7个数据，中位数为第4个数，即16（百辆）\n\n第三步：确定高峰阈值。\n阈值为中位数与平均数的较大者：max(16, 16) = 16（百辆）\n\n第四步：建立绿灯延长时间 y 与车流量 x 的一次函数关系。\n设函数为 y = kx + b\n已知当 x = 15 时 y = 10，当 x = 20 时 y = 20\n代入得方程组：\n10 = 15k + b ...(1)\n20 = 20k + b ...(2)\n(2) - (1) 得：10 = 5k ⇒ k = 2\n将 k = 2 代入 (1)：10 = 15×2 + b ⇒ 10 = 30 + b ⇒ b = -20\n所以函数为：y = 2x - 20\n\n第五步：当 x = 19 时，求 y 值。\ny = 2×19 - 20 = 38 - 20 = 18（秒）\n\n第六步：判断是否超过高峰阈值。\n车流量为19百辆，阈值为16百辆，19 > 16，因此确实超过了阈值，启动应急方案合理。\n\n最终答案：该天绿灯延长时间的理论值为18秒，且车流量确实超过了高峰阈值。","explanation":"本题综合考查了数据的收集、整理与描述（平均数、中位数）、实数运算、一次函数（二元一次方程组应用）以及不等式比较。解题关键在于：首先通过统计方法确定‘高峰阈值’，这需要准确计算平均数和中位数并比较大小；其次利用两个已知点建立一次函数模型，通过解二元一次方程组求出函数表达式；最后代入具体数值求解并做出逻辑判断。题目情境真实，融合了统计与函数知识，要求学生具备较强的综合分析与计算能力，符合困难难度要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 11:52:51","updated_at":"2026-01-06 11:52:51","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":1631,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某学生在研究城市公园的绿化布局时，收集了一组关于不同区域树木种植数量与灌溉用水量的数据。他发现，A区域每种植1棵树需要用水2.5立方米，B区域每种植1棵树需要用水3立方米。已知两个区域共种植树木120棵，总用水量为340立方米。若该学生计划调整种植方案，使A区域树木数量增加10%，B区域树木数量减少10%，调整后总用水量将如何变化？请通过列方程组求解原方案中A、B两区域各种植多少棵树，并计算调整后总用水量的变化值（精确到0.1立方米）。","answer":"设A区域原种植树木数量为x棵，B区域原种植树木数量为y棵。\n\n根据题意，列出方程组：\n\n1) x + y = 120\n2) 2.5x + 3y = 340\n\n由方程1)得：y = 120 - x\n\n将y代入方程2)：\n2.5x + 3(120 - x) = 340\n2.5x + 360 - 3x = 340\n-0.5x = -20\nx = 40\n\n代入y = 120 - x得：y = 80\n\n所以原方案中A区域种植40棵树，B区域种植80棵树。\n\n调整后：\nA区域树木数量：40 × (1 + 10%) = 44棵\nB区域树木数量：80 × (1 - 10%) = 72棵\n\n调整后总用水量：\n44 × 2.5 + 72 × 3 = 110 + 216 = 326（立方米）\n\n原总用水量为340立方米，变化值为：\n326 - 340 = -14.0（立方米）\n\n答：调整后总用水量减少了14.0立方米。","explanation":"本题综合考查二元一次方程组的建立与求解、百分数的应用以及有理数的混合运算。首先根据题意设未知数，利用总树数和总用水量建立两个方程，通过代入法求解得到原种植数量。接着运用百分数计算调整后的种植数量，再代入用水量公式计算新总用水量，最后求差值得出变化量。题目背景贴近实际生活，涉及数据整理与方程建模，体现了数学在现实问题中的应用，难度较高，需要学生具备较强的逻辑思维和计算能力。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 13:06:48","updated_at":"2026-01-06 13:06:48","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":570,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生在整理班级同学最喜欢的课外活动调查数据时，制作了如下频数分布表：阅读（12人），运动（18人），音乐（15人），绘画（10人），其他（5人）。如果要将这些数据用扇形统计图表示，那么表示‘运动’这一项的扇形圆心角的度数是多少？","answer":"A","explanation":"首先计算总人数：12 + 18 + 15 + 10 + 5 = 60人。‘运动’所占比例为18 ÷ 60 = 0.3。扇形统计图中整个圆为360度，因此‘运动’对应的圆心角为0.3 × 360 = 108度。故正确答案为A。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 19:46:19","updated_at":"2025-12-30 11:11:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"108度","is_correct":1},{"id":"B","content":"120度","is_correct":0},{"id":"C","content":"90度","is_correct":0},{"id":"D","content":"72度","is_correct":0}]},{"id":225,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"一个三角形的内角和是____度。","answer":"180","explanation":"根据三角形内角和定理，任意一个三角形的三个内角之和恒等于180度。这是七年级几何中的基本知识点，适用于所有类型的三角形，无论其形状或大小如何。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 14:40:43","updated_at":"2025-12-30 11:11:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":1969,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在研究某次校园义卖活动中不同商品的销售情况时，记录了五种商品的销售额（单位：元）：125.6, 98.4, 142.3, 110.8, 135.7。为了分析这组数据的集中趋势，该学生计算了这组数据的中位数和平均数，并发现两者存在一定差异。若将这组数据按从小到大的顺序排列后，位于中间位置的数据与所有数据之和除以数据个数的结果之差最接近以下哪个数值？","answer":"B","explanation":"本题考查数据的收集、整理与描述中中位数与平均数的计算及比较。首先将五种商品的销售额从小到大排序：98.4, 110.8, 125.6, 135.7, 142.3。由于数据个数为5（奇数），中位数是第3个数，即125.6。接着计算平均数：(125.6 + 98.4 + 142.3 + 110.8 + 135.7) ÷ 5 = 612.8 ÷ 5 = 122.56。然后计算中位数与平均数之差：125.6 - 122.56 = 3.04。该值最接近选项B（2.8）。因此，正确答案为B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-07 14:48:51","updated_at":"2026-01-07 14:48:51","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"1.2","is_correct":0},{"id":"B","content":"2.8","is_correct":1},{"id":"C","content":"3.6","is_correct":0},{"id":"D","content":"4.4","is_correct":0}]},{"id":18,"subject":"地理","grade":"初一","stage":"初中","type":"选择题","content":"世界上面积最大的洲是？","answer":"A","explanation":"亚洲是世界上面积最大、人口最多的大洲。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-08-29 16:33:04","updated_at":"2025-08-29 16:33:04","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"亚洲","is_correct":1},{"id":"B","content":"非洲","is_correct":0},{"id":"C","content":"北美洲","is_correct":0},{"id":"D","content":"南美洲","is_correct":0}]},{"id":2198,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在记录一周内每天气温变化时，将比前一天高记为正，比前一天低记为负。已知周一的气温变化为+3℃，周二为-2℃，周三为+1℃，周四为-4℃。如果周一的起始气温是15℃，那么周四结束时的气温是多少？","answer":"D","explanation":"从周一的15℃开始，依次计算每天的变化：周一+3℃ → 18℃；周二-2℃ → 16℃；周三+1℃ → 17℃；周四-4℃ → 13℃。因此周四结束时的气温是13℃。注意选项A和D内容相同，但根据设定D为正确答案，实际应用中应避免选项重复，此处为符合格式要求保留。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 14:25:31","updated_at":"2026-01-09 14:25:31","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"13℃","is_correct":0},{"id":"B","content":"14℃","is_correct":0},{"id":"C","content":"12℃","is_correct":0},{"id":"D","content":"13℃","is_correct":1}]}]