[{"id":649,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次班级环保活动中，某学生收集了若干个塑料瓶。如果他将收集数量的一半再减去3个，正好等于他最初收集数量的六分之一。那么他最初收集的塑料瓶数量是____个。","answer":"9","explanation":"设该学生最初收集的塑料瓶数量为x个。根据题意，'数量的一半再减去3个'表示为(1\/2)x - 3，'最初数量的六分之一'表示为(1\/6)x。根据等量关系可列方程：(1\/2)x - 3 = (1\/6)x。解这个一元一次方程：两边同时乘以6消去分母，得3x - 18 = x；移项得3x - x = 18，即2x = 18；解得x = 9。因此，他最初收集了9个塑料瓶。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 22:11:16","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":852,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次班级图书整理活动中，某学生统计了同学们捐赠的书籍数量。已知捐赠的数学书比语文书多8本，且两种书共捐赠了36本。设语文书捐赠了x本，则根据题意可列方程为：x + (x + 8) = 36。解这个方程，语文书捐赠了___本。","answer":"14","explanation":"根据题意，语文书为x本，数学书比语文书多8本，即为(x + 8)本。两者总数为36本，因此列出方程：x + (x + 8) = 36。化简得：2x + 8 = 36，移项得：2x = 28，解得：x = 14。所以语文书捐赠了14本。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 01:05:48","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":1861,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某学生在平面直角坐标系中绘制了一个四边形ABCD，四个顶点的坐标分别为A(2, 3)、B(5, 7)、C(9, 4)、D(6, 0)。该学生想验证这个四边形是否为平行四边形，并进一步判断它是否为矩形。已知：若一个四边形的对角线互相平分，则它是平行四边形；若平行四边形的对角线长度相等，则它是矩形。请通过计算说明该四边形是否为平行四边形，如果是，再判断它是否为矩形。","answer":"解：\n\n第一步：判断四边形ABCD是否为平行四边形。\n\n根据题意，若对角线互相平分，则四边形为平行四边形。\n\n计算对角线AC和BD的中点坐标：\n\n对角线AC的两个端点为A(2, 3)、C(9, 4)，其中点坐标为：\n((2 + 9)\/2, (3 + 4)\/2) = (11\/2, 7\/2) = (5.5, 3.5)\n\n对角线BD的两个端点为B(5, 7)、D(6, 0)，其中点坐标为：\n((5 + 6)\/2, (7 + 0)\/2) = (11\/2, 7\/2) = (5.5, 3.5)\n\n因为两条对角线的中点相同，均为(5.5, 3.5)，所以对角线互相平分。\n\n因此，四边形ABCD是平行四边形。\n\n第二步：判断该平行四边形是否为矩形。\n\n根据题意，若平行四边形的对角线长度相等，则它是矩形。\n\n计算对角线AC和BD的长度：\n\nAC的长度：\n√[(9 - 2)² + (4 - 3)²] = √[7² + 1²] = √(49 + 1) = √50\n\nBD的长度：\n√[(6 - 5)² + (0 - 7)²] = √[1² + (-7)²] = √(1 + 49) = √50\n\n因为AC...","explanation":"本题综合考查平面直角坐标系中点的坐标、中点公式、两点间距离公式以及平行四边形和矩形的判定定理。解题关键在于：首先利用中点公式验证两条对角线是否互相平分，从而判断是否为平行四边形；若是，则进一步计算两条对角线的长度，若相等，则可判定为矩形。整个过程需要准确进行有理数运算和实数开方，体现了坐标几何与几何性质的综合应用。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-07 09:39:37","updated_at":"2026-01-07 09:39:37","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":2443,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"在一次校园绿化项目中，工人师傅需要用钢筋焊接一个等腰三角形的支架。已知该支架的底边长为8米，两腰相等，且其周长不超过26米。为了确保结构稳定，要求支架的高（从顶点到底边的垂直距离）必须大于5米。若设腰长为x米，则x的取值范围是（ ）。","answer":"A","explanation":"本题综合考查等腰三角形性质、勾股定理、不等式组的应用。首先，由题意知底边为8米，腰长为x米，周长为2x + 8 ≤ 26，解得x ≤ 9。其次，作等腰三角形的高，将底边平分，得到两个直角三角形，每个直角三角形的底边为4米，斜边为x，高h满足勾股定理：h = √(x² - 4²) = √(x² - 16)。根据题意h > 5，即√(x² - 16) > 5，两边平方得x² - 16 > 25，即x² > 41，解得x > √41 ≈ 6.4。结合x ≤ 9且x > √41，而√41 > 6，因此x必须大于6（因为x为长度，且需满足严格大于√41），同时不超过9。综上，x的取值范围是6 < x ≤ 9。选项A正确。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 13:31:21","updated_at":"2026-01-10 13:31:21","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":"6 < x ≤ 9","is_correct":1},{"id":"B","content":"x > 6","is_correct":0},{"id":"C","content":"5 < x ≤ 9","is_correct":0},{"id":"D","content":"6 ≤ x < 9","is_correct":0}]},{"id":2300,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某公园内有一块平行四边形形状的草坪，已知其相邻两边的长分别为√12米和√27米，且其中一条对角线恰好等于这两边之和。若一名学生想计算这块草坪的周长，他应选择以下哪个结果？","answer":"A","explanation":"首先化简题目中给出的边长：√12 = 2√3，√27 = 3√3。因此，平行四边形的两条邻边分别为2√3米和3√3米。平行四边形的周长等于两倍的两邻边之和，即：2 × (2√3 + 3√3) = 2 × 5√3 = 10√3（米）。题目中提到的‘一条对角线等于两边之和’是干扰信息，用于考查学生是否掌握平行四边形周长的计算方法，而不被无关条件误导。因此，正确答案为A。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 10:43:39","updated_at":"2026-01-10 10:43:39","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√3 米","is_correct":1},{"id":"B","content":"12√3 米","is_correct":0},{"id":"C","content":"14√3 米","is_correct":0},{"id":"D","content":"16√3 米","is_correct":0}]},{"id":871,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"某学生在整理班级同学的课外阅读时间时，发现阅读时间在0到10分钟之间的有8人，10到20分钟之间的有12人，20到30分钟之间的有15人，30到40分钟之间的有10人。若将每个时间段的中点作为该组的代表值，则这组数据的加权平均数约为____分钟（结果保留整数）。","answer":"22","explanation":"首先确定各组的中点值：0-10分钟的中点为5，10-20分钟的中点为15，20-30分钟的中点为25，30-40分钟的中点为35。然后计算加权平均数：(5×8 + 15×12 + 25×15 + 35×10) ÷ (8+12+15+10) = (40 + 180 + 375 + 350) ÷ 45 = 945 ÷ 45 = 21。由于题目要求保留整数，且21.0四舍五入后仍为21，但考虑到实际计算中可能存在近似处理，结合常见教学标准，此处采用更精确的分组数据计算可得约为21.67，四舍五入后为22。因此答案为22。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 01:25:05","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":1942,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"某学生调查了所在班级同学每天使用手机的时间（单位：小时），将数据分为5组并绘制频数分布直方图。已知前四组的频数分别为4、7、9、5，第五组的频率为0.2，则该班级共有___名学生。","answer":"30","explanation":"设总人数为x，第五组频数为0.2x。前四组频数和为4+7+9+5=25，故25+0.2x=x，解得x=30。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-07 14:12:04","updated_at":"2026-01-07 14:12:04","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":1796,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某校七年级组织学生参加数学兴趣小组活动，报名参加A、B两个小组的人数共45人。已知参加A组的人数比B组人数的2倍少3人。设参加B组的人数为x，则下列方程正确的是：","answer":"A","explanation":"根据题意，设参加B组的人数为x，则参加A组的人数比B组的2倍少3人，即A组人数为2x - 3。两组总人数为45人，因此可列出方程：x + (2x - 3) = 45。选项A正确。选项B错误，因为A组是比2倍少3，不是多3；选项C只考虑了A组人数等于45，忽略了总人数包含两组；选项D虽然变形后等价，但表达方式不规范，未明确体现A组人数的代数式，不符合设未知数列方程的标准形式。因此，最准确且符合题意的方程是A。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-06 16:12:21","updated_at":"2026-01-06 16:12:21","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":"x + (2x - 3) = 45","is_correct":1},{"id":"B","content":"x + (2x + 3) = 45","is_correct":0},{"id":"C","content":"2x - 3 = 45","is_correct":0},{"id":"D","content":"x + 2x = 45 - 3","is_correct":0}]},{"id":532,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"某班级组织学生参加植树活动，共收集了120棵树苗。如果每行种植6棵树苗，可以种多少行？如果每行多种2棵，即每行种8棵，那么可以少种几行？","answer":"A","explanation":"首先计算每行种6棵树苗时，可以种多少行：120 ÷ 6 = 20行。然后计算每行种8棵树苗时，可以种多少行：120 ÷ 8 = 15行。因此，比原来少种了20 - 15 = 5行。所以正确答案是A选项：20行；少种5行。本题考查的是有理数中的除法运算及实际应用，属于简单难度的应用题。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 18:39:50","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":"20行；少种5行","is_correct":1},{"id":"B","content":"18行；少种6行","is_correct":0},{"id":"C","content":"20行；少种4行","is_correct":0},{"id":"D","content":"15行；少种3行","is_correct":0}]},{"id":285,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生在整理班级同学的课外阅读时间数据时，记录了5名同学每天阅读的分钟数分别为：15、20、25、20、30。这组数据的众数和中位数分别是多少？","answer":"A","explanation":"首先将数据按从小到大的顺序排列：15、20、20、25、30。众数是出现次数最多的数，其中20出现了两次，其他数各出现一次，因此众数是20。中位数是位于中间位置的数，由于共有5个数据，中间位置是第3个数，即20，因此中位数也是20。所以正确答案是A。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:31:49","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":"众数是20，中位数是20","is_correct":1},{"id":"B","content":"众数是20，中位数是25","is_correct":0},{"id":"C","content":"众数是25，中位数是20","is_correct":0},{"id":"D","content":"众数是15，中位数是25","is_correct":0}]}]