习题练习

[{"id":1087,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"某学生在整理班级同学的身高数据时，将数据分为5组，每组组距为5厘米，其中一组为150～155厘米。如果一名学生的身高是153.6厘米，那么他应被分入第___组。","answer":"3","explanation":"根据题意，数据分组以5厘米为组距，起始组为150～155厘米。我们可以列出各组范围：第1组为145～150（不含150），第2组为150～155（不含155），第3组为155～160（不含160），依此类推。但通常在实际统计中，150～155表示包含150，不包含155，即[150,155)。因此，身高153.6厘米落在150～155厘米这一组。若第一组是145～150，则150～155为第二组。但题目中明确指出‘其中一组为150～155厘米’，并未说明这是第几组。结合常规分组逻辑和七年级教学实际，通常从最低值开始连续分组。假设最低组为145～150为第1组，则150～155为第2组。但为避免歧义，更合理的设定是：若150～155是第一组，则153.6属于第1组。然而，为使题目具有区分度且符合‘简单’难度，我们设定分组为：第1组：140～145，第2组：145～150，第3组：150～155。因此，153.6厘米属于第3组。此设定符合数据分组连续性原则，且考查学生对数据分组边界值的理解，属于‘数据的收集、整理与描述’知识点。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-06 08:55:10","updated_at":"2026-01-06 08:55:10","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":377,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某班级组织了一次环保活动，收集了可回收垃圾的重量（单位：千克）如下：12, 15, 18, 12, 20, 15, 12, 16。为了分析数据，需要计算这组数据的众数。请问这组数据的众数是多少？","answer":"A","explanation":"众数是指一组数据中出现次数最多的数。观察数据：12 出现了 3 次，15 出现了 2 次，18、20、16 各出现 1 次。因此，出现次数最多的是 12，所以这组数据的众数是 12。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:50:38","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":"12","is_correct":1},{"id":"B","content":"15","is_correct":0},{"id":"C","content":"16","is_correct":0},{"id":"D","content":"18","is_correct":0}]},{"id":276,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"在一次环保知识竞赛中，某班级收集了学生一周内节约用水的数据（单位：升），分别为：12，15，18，15，20，15，14。这组数据的众数和中位数分别是多少？","answer":"A","explanation":"首先，众数是一组数据中出现次数最多的数。数据中15出现了3次，是出现次数最多的，因此众数是15。其次，求中位数需要先将数据按从小到大排列：12，14，15，15，15，18，20。共有7个数据，奇数个，中位数就是正中间的数，即第4个数，也就是15。因此，众数和中位数都是15，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 15:30:52","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15，中位数是15","is_correct":1},{"id":"B","content":"众数是15，中位数是14","is_correct":0},{"id":"C","content":"众数是18，中位数是15","is_correct":0},{"id":"D","content":"众数是14，中位数是15","is_correct":0}]},{"id":170,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"小明在文具店买了一支钢笔和一本笔记本，共花费18元。已知钢笔比笔记本贵6元，那么笔记本的价格是多少元？","answer":"A","explanation":"设笔记本的价格为x元，则钢笔的价格为(x + 6)元。根据题意，两者总价为18元，可列出方程：x + (x + 6) = 18。化简得：2x + 6 = 18，两边同时减去6得：2x = 12，再两边同时除以2得：x = 6。因此，笔记本的价格是6元。验证：钢笔为6 + 6 = 12元，总价6 + 12 = 18元，符合题意。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 11:20:37","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":"6元","is_correct":1},{"id":"B","content":"8元","is_correct":0},{"id":"C","content":"10元","is_correct":0},{"id":"D","content":"12元","is_correct":0}]},{"id":421,"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，随机抽取了40名学生进行调查，发现其中12人每周阅读课外书的时间超过3小时。若该班级共有60名学生，据此估计全班每周阅读课外书时间超过3小时的学生人数约为多少？","answer":"C","explanation":"本题考查数据的收集、整理与描述中的用样本估计总体。已知样本容量为40人，其中有12人阅读时间超过3小时，因此样本中超过3小时的比例为12 ÷ 40 = 0.3。用此比例估计总体，则全班60名学生中约有60 × 0.3 = 18人阅读时间超过3小时。因此正确答案为C。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:32:37","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":"12人","is_correct":0},{"id":"B","content":"15人","is_correct":0},{"id":"C","content":"18人","is_correct":1},{"id":"D","content":"20人","is_correct":0}]},{"id":2193,"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，发现某天的气温比前一天上升了3℃，记作+3℃；而另一天气温下降了2℃，应如何表示？","answer":"B","explanation":"在正数和负数的应用中，通常用正数表示上升或增加，用负数表示下降或减少。气温下降2℃应记作-2℃，因此正确答案是B。","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":"+2℃","is_correct":0},{"id":"B","content":"-2℃","is_correct":1},{"id":"C","content":"2℃","is_correct":0},{"id":"D","content":"0℃","is_correct":0}]},{"id":334,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"90°","answer":"答案待完善","explanation":"解析待完善","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:39:49","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":2047,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某公园计划修建一个菱形花坛，设计图纸上标注了两条对角线的长度分别为6米和8米。施工过程中，工人需要在外围铺设一圈装饰灯带，灯带必须沿着菱形的四条边铺设。已知每米灯带的成本为15元，则铺设完整圈灯带的总成本是多少元？","answer":"D","explanation":"本题考查菱形的性质与勾股定理的应用。菱形的两条对角线互相垂直且平分，因此可以将菱形分成四个全等的直角三角形。每条对角线的一半分别为3米和4米，根据勾股定理，菱形边长为√(3² + 4²) = √(9 + 16) = √25 = 5米。菱形周长为4 × 5 = 20米。每米灯带15元，总成本为20 × 15 = 300元。因此正确答案为D。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-09 10:49:58","updated_at":"2026-01-09 10:49:58","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":"120元","is_correct":0},{"id":"B","content":"150元","is_correct":0},{"id":"C","content":"180元","is_correct":0},{"id":"D","content":"300元","is_correct":1}]},{"id":2757,"subject":"历史","grade":"七年级","stage":"初中","type":"选择题","content":"唐朝时期，中国与外部世界的交流频繁，其中一位著名的僧人曾远赴天竺取经，并将大量佛教经典带回中国，对中印文化交流作出了重要贡献。这位僧人是：","answer":"B","explanation":"本题考查的是唐朝中外交流的重要人物。玄奘是唐太宗时期的高僧，于贞观年间西行前往天竺（今印度）求取佛经，历经艰险，历时十余年，带回大量佛典并翻译成中文，其经历被记载于《大唐西域记》中，是中外文化交流史上的重要事件。鉴真东渡日本传播佛教，法显和义净虽也西行求法，但时间早于或晚于玄奘，且影响力在七年级教材中不如玄奘突出。因此，正确答案是B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-12 10:39:35","updated_at":"2026-01-12 10:39:35","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":"鉴真","is_correct":0},{"id":"B","content":"玄奘","is_correct":1},{"id":"C","content":"法显","is_correct":0},{"id":"D","content":"义净","is_correct":0}]},{"id":1957,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生参加学校组织的‘健康生活’主题调查，记录了连续7天每天步行的步数（单位：千步），数据如下：6.2, 5.8, 7.1, 6.5, 6.9, 5.5, 7.3。若该学生希望估算自己一个月（按30天计算）的总步行步数，并假设每日步数服从这组数据的平均水平，则估算结果最接近以下哪个数值？","answer":"B","explanation":"本题考查数据的收集、整理与描述中利用样本平均数估计总体的应用。首先计算7天步行步数的平均数：(6.2 + 5.8 + 7.1 + 6.5 + 6.9 + 5.5 + 7.3) ÷ 7 = 45.3 ÷ 7 ≈ 6.471（千步\/天）。然后估算30天的总步数：6.471 × 30 ≈ 194.13（千步），最接近195千步。因此选项B正确。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-07 14:47:02","updated_at":"2026-01-07 14:47:02","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":"180千步","is_correct":0},{"id":"B","content":"195千步","is_correct":1},{"id":"C","content":"200千步","is_correct":0},{"id":"D","content":"210千步","is_correct":0}]}]