[{"id":1093,"subject":"数学","grade":"七年级","stage":"小学","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和玻璃瓶，其中塑料瓶的数量比玻璃瓶的3倍多5个。若设玻璃瓶的数量为x个，则塑料瓶的数量可表示为______。","answer":"3x + 5","explanation":"根据题意，塑料瓶的数量比玻璃瓶的3倍多5个。玻璃瓶的数量为x，那么它的3倍就是3x，再加上5个，就是塑料瓶的数量，因此表达式为3x + 5。这是整式加减中的基本概念，考查学生将文字语言转化为代数表达式的能力。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-06 08:55:58","updated_at":"2026-01-06 08:55:58","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":1695,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市为改善交通状况，计划在一条主干道上设置若干个智能公交站。已知该道路在平面直角坐标系中沿x轴方向延伸，起点坐标为(0, 0)，终点坐标为(12, 0)。规划部门决定在这些站点中设置A、B、C三类站点，其中A类站点每2千米设一个，B类站点每3千米设一个，C类站点每4千米设一个，均从起点开始设置（即起点处同时设有A、B、C三类站点）。若某学生从起点出发，沿道路步行，每经过一个站点就记录一次，问：该学生在到达终点前，共会经过多少个不同的站点？（注：若某位置同时设有多个类型的站点，只算作一个站点）","answer":"1. 确定各类站点的位置：\n - A类站点：每2千米一个，位置为 x = 0, 2, 4, 6, 8, 10, 12\n 共 7 个位置\n - B类站点：每3千米一个，位置为 x = 0, 3, 6, 9, 12\n 共 5 个位置\n - C类站点：每4千米一个，位置为 x = 0, 4, 8, 12\n 共 4 个位置\n\n2. 列出所有站点坐标并去重：\n 合并三类站点的所有x坐标：\n {0, 2, 3, 4, 6, 8, 9, 10, 12}\n 注意：6出现在A和B类，4和12出现在A和C类，0出现在三类中，但每个坐标只算一次\n\n3. 统计不同站点的总数：\n 上述集合中共有 9 个不同的x坐标值\n\n4. 因此，该学生从起点到终点（含起点和终点），共经过 9 个不同的站点\n\n答：该学生共会经过 9 个不同的站点。","explanation":"本题综合考查了平面直角坐标系、有理数（坐标值）、数据的收集与整理（分类统计、去重）以及实际应用建模能力。解题关键在于理解‘不同站点’的含义——即使多个类型站点位于同一位置，也只计为一个物理站点。因此需要分别列出A、B、C三类站点的所有位置，然后合并并去除重复的坐标点。这涉及集合思想的应用，虽然七年级尚未系统学习集合，但通过列表和观察可以实现去重操作。题目背景新颖，结合了城市规划与数学建模，避免了传统行程问题的套路，强调对‘位置唯一性’的理解和数据处理能力，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 13:39:12","updated_at":"2026-01-06 13:39:12","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":1059,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次环保知识竞赛中，某班级共收集了120份有效问卷。统计结果显示，其中表示‘经常进行垃圾分类’的学生人数是表示‘偶尔进行垃圾分类’人数的2倍少10人。如果设‘偶尔进行垃圾分类’的学生人数为x人，则根据题意可列出一元一次方程：________。","answer":"x + (2x - 10) = 120","explanation":"题目中已知总人数为120人，分为两类：‘偶尔进行垃圾分类’的人数为x人，‘经常进行垃圾分类’的人数比这个数的2倍少10人，即(2x - 10)人。根据总人数等于两部分人数之和，可列出方程：x + (2x - 10) = 120。此方程符合一元一次方程的形式，且基于实际问题建立，考查了学生将文字信息转化为数学表达式的能力。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 06:46:46","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":422,"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，发现一周内每天阅读时间（单位：分钟）分别为：25，30，28，35，32，27，33。为了分析阅读时间的分布情况，该学生计算了这组数据的平均数。请问这组数据的平均数是多少？","answer":"C","explanation":"要计算这组数据的平均数，需要将所有数据相加，然后除以数据的个数。具体步骤如下：首先，将每天的阅读时间相加：25 + 30 + 28 + 35 + 32 + 27 + 33 = 210（分钟）。然后，用总和除以天数（7天）：210 ÷ 7 = 30（分钟）。因此，这组数据的平均数是30分钟，正确答案是C。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:32:46","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":"28分钟","is_correct":0},{"id":"B","content":"29分钟","is_correct":0},{"id":"C","content":"30分钟","is_correct":1},{"id":"D","content":"31分钟","is_correct":0}]},{"id":999,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次环保主题活动中，某学生记录了连续5天每天收集的废旧电池数量（单位：节），分别为：-3，5，0，-2，7。这里规定：收集到电池记为正数，丢失或损坏电池记为负数。这5天该学生实际收集的电池总数为___节。","answer":"7","explanation":"题目中给出的数据是有理数，包含正数、负数和零。根据题意，正数表示收集到的电池数量，负数表示丢失或损坏的数量，因此需要将所有数值相加得到净收集量。计算过程为：(-3) + 5 + 0 + (-2) + 7 = (5 + 7) + (-3 - 2) + 0 = 12 - 5 = 7。所以这5天实际收集的电池总数为7节。本题考查有理数的加法运算，结合生活情境，帮助学生理解有理数在实际问题中的意义。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 04:51:06","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":1006,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"某班级组织植树活动，若每名学生种3棵树，则还剩10棵树没人种；若每名学生种4棵树，则最后一名学生只需种2棵树。这个班级共有___名学生。","answer":"12","explanation":"设这个班级共有x名学生。根据题意，树的总数不变。第一种情况：每名学生种3棵，还剩10棵，所以总树数为3x + 10。第二种情况：前(x - 1)名学生每人种4棵，最后一名学生种2棵，总树数为4(x - 1) + 2 = 4x - 4 + 2 = 4x - 2。因为树的数量相同，列方程：3x + 10 = 4x - 2。解这个一元一次方程：移项得10 + 2 = 4x - 3x，即12 = x。所以这个班级共有12名学生。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 05:03:53","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":2280,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"在数轴上，点A表示的数是-5，点B与点A的距离是8个单位长度，且点B在原点右侧。若点C是点A和点B之间的一个点，且AC:CB = 3:1，则点C表示的数是___。","answer":"1","explanation":"首先，点A表示-5，点B在原点右侧且与A距离为8，因此点B表示的数是-5 + 8 = 3。点C在A和B之间，且AC:CB = 3:1，说明点C将线段AB按3:1的比例内分。根据内分点公式，点C的坐标为：(1×(-5) + 3×3) ÷ (3+1) = (-5 + 9) ÷ 4 = 4 ÷ 4 = 1。因此，点C表示的数是1。此题综合考查了数轴上的距离、位置关系以及线段的按比例分割，符合七年级数轴与有理数运算的综合应用要求。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-09 16:27:13","updated_at":"2026-01-09 16:27:13","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":400,"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，统计了每人每周阅读课外书的平均时间（单位：小时），并将数据分为5组，绘制成频数分布直方图。已知前四组的频数分别为3、7、10、5，第五组的频率为0.2，则该班级参与调查的学生总人数是多少？","answer":"B","explanation":"题目考查的是数据的收集、整理与描述中的频数与频率概念。已知第五组的频率为0.2，即第五组人数占总人数的20%。设总人数为x，则第五组人数为0.2x。前四组频数之和为3 + 7 + 10 + 5 = 25，因此总人数为前四组人数加上第五组人数：25 + 0.2x = x。解这个方程：25 = x - 0.2x → 25 = 0.8x → x = 25 ÷ 0.8 = 30。所以参与调查的学生总人数是30人。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:16:30","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":"25","is_correct":0},{"id":"B","content":"30","is_correct":1},{"id":"C","content":"35","is_correct":0},{"id":"D","content":"40","is_correct":0}]},{"id":1931,"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每日运动时间数据时，发现若将数据按从小到大的顺序排列，第8个和第9个数据分别为25分钟和27分钟。已知这组数据共有15个，且唯一众数为20分钟，出现4次。若去掉一个最大值和一个最小值后，剩余13个数据的平均数恰好比原平均数多1分钟，则原数据中的最大值是____分钟。","answer":"40","explanation":"中位数为(25+27)\/2=26。设原平均数为x，则新平均数为x+1。总和关系：15x - (最小值+最大值) = 13(x+1)，化简得最大值+最小值=2x-13。结合众数、中位数和整数约束，推得最大值为40。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-07 14:10:11","updated_at":"2026-01-07 14:10:11","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":1926,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某班级为了了解学生最喜欢的课外活动，随机抽取了40名学生进行调查，并将结果整理成如下频数分布表：\n\n| 活动类型 | 频数 |\n|----------|------|\n| 阅读 | 8 |\n| 运动 | 15 |\n| 绘画 | 6 |\n| 音乐 | 11 |\n\n若该班级共有200名学生，估计喜欢运动的学生人数最接近以下哪个数值？","answer":"C","explanation":"根据频数分布表，40名学生中有15人最喜欢运动，所占比例为 15 ÷ 40 = 0.375。用此比例估计整个班级200名学生中喜欢运动的人数：200 × 0.375 = 75。因此，估计喜欢运动的学生人数最接近75人，正确答案为C。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-07 13:16:48","updated_at":"2026-01-07 13:16:48","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":"A","content":"50","is_correct":0},{"id":"B","content":"65","is_correct":0},{"id":"C","content":"75","is_correct":1},{"id":"D","content":"85","is_correct":0}]}]