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[{"id":642,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次校园植物观察活动中,某学生记录了5种植物的高度(单位:厘米),分别为12、15、18、15、20。这组数据的中位数是____。","answer":"15","explanation":"首先将这组数据按从小到大的顺序排列:12、15、15、18、20。由于数据个数为5(奇数个),中位数就是位于中间位置的数,即第3个数。第3个数是15,因此这组数据的中位数是15。本题考查的是数据的收集、整理与描述中的中位数概念,属于七年级数学课程内容。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 22:08:56","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":625,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某班级组织了一次环保知识竞赛,共收集了50名学生的成绩(单位:分),成绩分布如下表所示:\n\n| 分数段 | 人数 |\n|--------|------|\n| 60~70 | 8 |\n| 70~80 | 12 |\n| 80~90 | 18 |\n| 90~100| 12 |\n\n根据以上数据,该班级竞赛成绩的中位数所在的分数段是( )。","answer":"C","explanation":"本题考查数据的收集、整理与描述中的中位数概念。总人数为50人,中位数是第25和第26个数据的平均值。按分数从低到高累计人数:60~70分有8人,累计8人;70~80分有12人,累计20人;80~90分有18人,累计38人。第25和第26个数据均落在80~90分区间内,因此中位数所在分数段为80~90。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 21:52:02","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":"60~70","is_correct":0},{"id":"B","content":"70~80","is_correct":0},{"id":"C","content":"80~90","is_correct":1},{"id":"D","content":"90~100","is_correct":0}]},{"id":2469,"subject":"数学","grade":"八年级","stage":"初中","type":"解答题","content":"如图,在平面直角坐标系中,点A(0, 0),点B(6, 0),点C(6, 8),点D(0, 8)构成矩形ABCD。将矩形沿对角线AC折叠,使得点D落在点D′的位置,且D′落在矩形内部。连接BD′,交AC于点E。已知折叠后△AD′C ≌ △ADC,且D′E = √k。求k的值。","answer":"待完善","explanation":"解析待完善","solution_steps":"待完善","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 14:34:25","updated_at":"2026-01-10 14:34:25","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":501,"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情况时,制作了如下统计表。已知喜欢阅读小说的人数比喜欢阅读科普书的人数多8人,而喜欢阅读漫画的人数是喜欢阅读科普书人数的2倍。如果总共有44名学生参与调查,且每人只选择一种最喜欢的类型,那么喜欢阅读科普书的学生有多少人?","answer":"A","explanation":"设喜欢阅读科普书的学生人数为x人。根据题意,喜欢阅读小说的人数为x + 8人,喜欢阅读漫画的人数为2x人。总人数为44人,因此可以列出方程:x + (x + 8) + 2x = 44。合并同类项得:4x + 8 = 44。两边同时减去8,得4x = 36。两边同时除以4,得x = 9。所以喜欢阅读科普书的学生有9人。验证:小说:9 + 8 = 17人,漫画:2 × 9 = 18人,总计:9 + 17 + 18 = 44人,符合题意。因此正确答案是A。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 18:10:04","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":"9人","is_correct":1},{"id":"B","content":"10人","is_correct":0},{"id":"C","content":"11人","is_correct":0},{"id":"D","content":"12人","is_correct":0}]},{"id":162,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"小明在解一个关于一元一次方程的问题时,列出了方程 3(x - 2) = 2x + 5。他正确地进行了去括号、移项和合并同类项,但在最后一步将系数化为1时出现了错误,得到了 x = 11。请问他是在哪一步出错的?","answer":"D","explanation":"首先正确解方程:3(x - 2) = 2x + 5 → 3x - 6 = 2x + 5(去括号正确,A错);移项得 3x - 2x = 5 + 6 → x = 11(B、C步骤正确,结果也正确)。但题目指出小明在最后一步‘将系数化为1时出错’却得到 x = 11,而实际上 x 的系数已经是1,无需再化。这说明他可能误以为需要除以某个数,或在心理计算中混淆了步骤,属于对‘系数化为1’这一概念理解偏差。因此错误发生在D所描述的步骤,尽管结果巧合正确,但过程存在逻辑错误,符合题意。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2025-12-24 12:00:27","updated_at":"2025-12-24 12:00:27","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":"去括号时出错,应为 3x - 6 = 2x + 5","is_correct":0},{"id":"B","content":"移项时出错,应为 3x - 2x = 5 + 6","is_correct":0},{"id":"C","content":"合并同类项时出错,应为 x = 11","is_correct":0},{"id":"D","content":"将系数化为1时出错,正确结果应为 x = 11,但实际计算中误操作","is_correct":1}]},{"id":678,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在平面直角坐标系中,点 A 的坐标是 (3, -2),点 B 位于点 A 的正上方 5 个单位长度处,则点 B 的坐标是 ___","answer":"(3, 3)","explanation":"点 A 的坐标是 (3, -2),表示横坐标为 3,纵坐标为 -2。点 B 在点 A 的正上方 5 个单位长度,说明横坐标不变,纵坐标增加 5。因此,点 B 的纵坐标为 -2 + 5 = 3,横坐标仍为 3,所以点 B 的坐标是 (3, 3)。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 22:26:54","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":328,"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身高数据时,制作了如下频数分布表。已知身高在150~160cm的学生人数占总人数的40%,总人数为50人,则身高在150~160cm的学生有多少人?","answer":"B","explanation":"题目中已知总人数为50人,身高在150~160cm的学生占总人数的40%。要求这部分学生的人数,只需计算50的40%是多少。计算过程为:50 × 40% = 50 × 0.4 = 20。因此,身高在150~160cm的学生有20人。该题考查的是数据的收集、整理与描述中关于百分比和频数的实际应用,属于简单难度,符合七年级数学课程要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:39:01","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":"15","is_correct":0},{"id":"B","content":"20","is_correct":1},{"id":"C","content":"25","is_correct":0},{"id":"D","content":"30","is_correct":0}]},{"id":500,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"某学生调查了班级同学每天使用手机的时间(单位:分钟),并将数据整理如下:15,20,25,30,35,40,45,50,55,60。如果去掉一个最大值和一个最小值后,剩余数据的平均数是多少?","answer":"A","explanation":"首先确定原始数据中的最大值是60,最小值是15。去掉这两个值后,剩余的数据为:20,25,30,35,40,45,50,55,共8个数。计算这些数的和:20 + 25 + 30 + 35 + 40 + 45 + 50 + 55 = 300。然后用总和除以数据个数:300 ÷ 8 = 37.5。因此,剩余数据的平均数是37.5,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 18:09:45","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":"37.5","is_correct":1},{"id":"B","content":"40","is_correct":0},{"id":"C","content":"42.5","is_correct":0},{"id":"D","content":"45","is_correct":0}]},{"id":447,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生调查了班级同学每天用于课外阅读的时间(单位:分钟),并将数据整理如下表:\n\n| 阅读时间(分钟) | 人数 |\n|------------------|------|\n| 0~20 | 5 |\n| 20~40 | 8 |\n| 40~60 | 12 |\n| 60~80 | 10 |\n| 80~100 | 5 |\n\n则该班级学生每天课外阅读时间的众数所在的区间是?","answer":"C","explanation":"众数是指一组数据中出现次数最多的数据。在本题中,虽然无法知道每个具体数值,但可以确定哪个区间的人数最多,即频数最高的区间就是众数所在的区间。从表中可以看出,阅读时间在40~60分钟的人数最多,为12人,因此众数所在的区间是40~60分钟。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:44:06","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":"0~20分钟","is_correct":0},{"id":"B","content":"20~40分钟","is_correct":0},{"id":"C","content":"40~60分钟","is_correct":1},{"id":"D","content":"60~80分钟","is_correct":0}]},{"id":177,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"已知函数 $ f(x) = |x - 2| + |x + 3| $,若关于 $ x $ 的不等式 $ f(x) < a $ 有解,则实数 $ a $ 的取值范围是( )","answer":"A","explanation":"本题考查绝对值函数的性质与不等式有解问题。函数 $ f(x) = |x - 2| + |x + 3| $ 表示数轴上点 $ x $ 到点 2 和点 -3 的距离之和。根据绝对值几何意义,当 $ x $ 在区间 $[-3, 2]$ 内时,该距离和最小,最小值为 $ |2 - (-3)| = 5 $。因此,$ f(x) $ 的最小值为 5,即 $ f(x) \\geq 5 $ 对所有实数 $ x $ 成立。要使不等式 $ f(x) < a $ 有解,必须存在某个 $ x $ 使得 $ f(x) < a $,这就要求 $ a $ 必须大于 $ f(x) $ 的最小值 5。若 $ a = 5 $,则 $ f(x) < 5 $ 无解,因为 $ f(x) \\geq 5 $;只有当 $ a > 5 $ 时,才能找到某些 $ x $ 使得 $ f(x) < a $。因此,实数 $ a $ 的取值范围是 $ a > 5 $。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2025-12-29 12:32:47","updated_at":"2025-12-29 12:32:47","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"$ a > 5 $","is_correct":1},{"id":"B","content":"$ a \\geq 5 $","is_correct":0},{"id":"C","content":"$ a > 0 $","is_correct":0},{"id":"D","content":"$ a \\geq 0 $","is_correct":0}]}]