[{"id":661,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次班级环保活动中，某学生收集了若干节废旧电池。如果他将收集到的电池数量增加5节后，总数恰好是原来数量的2倍。那么他原来收集了___节电池。","answer":"5","explanation":"设该学生原来收集了x节电池。根据题意，增加5节后总数为x + 5，而这个数量等于原来数量的2倍，即2x。因此可以列出方程：x + 5 = 2x。解这个一元一次方程，将x移到右边得5 = 2x - x，即5 = x。所以原来收集了5节电池。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 22:15:41","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":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":286,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生在平面直角坐标系中画了一个点，该点的横坐标是3，纵坐标是-2。若将该点先向右平移4个单位，再向下平移3个单位，则平移后的点的坐标是？","answer":"A","explanation":"原点的坐标为(3, -2)。向右平移4个单位，横坐标增加4，即3 + 4 = 7；再向下平移3个单位，纵坐标减少3，即-2 - 3 = -5。因此，平移后的点的坐标是(7, -5)。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 15:31:54","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":"A","content":"(7, -5)","is_correct":1},{"id":"B","content":"(7, 1)","is_correct":0},{"id":"C","content":"(-1, -5)","is_correct":0},{"id":"D","content":"(-1, 1)","is_correct":0}]},{"id":2329,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"在一次校园植物观察活动中，某学生测量了四块三角形花坛的三边长度（单位：米），并记录了以下数据。根据勾股定理，可以判断为直角三角形的是哪一块？","answer":"B","explanation":"根据勾股定理，若一个三角形是直角三角形，则其两直角边的平方和等于斜边的平方，即满足 a² + b² = c²，其中 c 为最长边。逐一验证各选项：\n\nA：3² + 4² = 9 + 16 = 25 ≠ 6² = 36，不满足；\nB：5² + 12² = 25 + 144 = 169 = 13²，满足勾股定理，是直角三角形；\nC：7² + 8² = 49 + 64 = 113 ≠ 9² = 81，不满足；\nD：6² + 7² = 36 + 49 = 85 ≠ 8² = 64，不满足。\n\n因此，只有选项 B 满足勾股定理，正确答案为 B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 10:52:33","updated_at":"2026-01-10 10:52:33","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":"三边分别为 3，4，6","is_correct":0},{"id":"B","content":"三边分别为 5，12，13","is_correct":1},{"id":"C","content":"三边分别为 7，8，9","is_correct":0},{"id":"D","content":"三边分别为 6，7，8","is_correct":0}]},{"id":2221,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","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5℃，记作+5℃；第二天又下降了3℃，记作-3℃。如果这两天的温度变化总和用正负数表示，那么这两天的总变化是___℃。","answer":"2","explanation":"根据正负数表示相反意义的量，温度上升记为正，下降记为负。两天的变化分别为+5℃和-3℃，总变化为+5 + (-3) = 2℃，因此答案是2。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 14:27:19","updated_at":"2026-01-09 14:27:19","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":202,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生在计算一个数减去 15 时，误将减法当作加法，结果得到 38。那么正确的计算结果应该是 _ 。","answer":"8","explanation":"根据题意，某学生把‘减去15’算成了‘加上15’，得到错误结果38。设这个数为 x，则有 x + 15 = 38，解得 x = 38 - 15 = 23。因此，正确的计算应为 23 - 15 = 8。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 14:39:23","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":2151,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"在一次数学测验中，某学生解答一道关于一元一次方程的题目时，列出了方程：3x + 5 = 20。该方程的解表示的意义是：某数的三倍加上5等于20，那么这个数是多少？解这个方程得到的正确结果是：","answer":"B","explanation":"解方程 3x + 5 = 20，首先两边同时减去5，得到 3x = 15，然后两边同时除以3，得到 x = 5。因此，这个数是5，对应选项B。该题考查一元一次方程的基本解法，符合七年级数学课程内容。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 13:00:46","updated_at":"2026-01-09 13:00:46","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":"4","is_correct":0},{"id":"B","content":"5","is_correct":1},{"id":"C","content":"6","is_correct":0},{"id":"D","content":"7","is_correct":0}]},{"id":628,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某次环保活动中，某班学生收集废旧纸张和塑料瓶进行回收。已知每3千克废旧纸张和每2千克塑料瓶可兑换15元环保基金。如果该班共收集了9千克废旧纸张和6千克塑料瓶，那么他们可以兑换多少元环保基金？","answer":"B","explanation":"根据题意，每3千克废旧纸张和2千克塑料瓶可兑换15元。观察所收集的数量：9千克废旧纸张是3千克的3倍，6千克塑料瓶是2千克的3倍，说明收集的总量正好是基本兑换单位的3倍。因此，兑换金额为15元 × 3 = 45元。本题考查学生对比例关系的理解与简单整数倍的应用，属于有理数在实际问题中的简单运用。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 21:54:40","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":"30元","is_correct":0},{"id":"B","content":"45元","is_correct":1},{"id":"C","content":"60元","is_correct":0},{"id":"D","content":"75元","is_correct":0}]},{"id":2260,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"在数轴上，点P表示的数是-3，点Q与点P之间的距离是5个单位长度，且点Q在原点的右侧。那么点Q表示的数是___","answer":"B","explanation":"点P表示的数是-3，点Q与点P相距5个单位长度，因此点Q可能在-3的左边或右边。若在左边，则为-3 - 5 = -8；若在右边，则为-3 + 5 = 2。题目中明确指出点Q在原点的右侧，即表示的数大于0，因此点Q表示的数是2。选项B正确。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 16:03:06","updated_at":"2026-01-09 16:03:06","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":"-8","is_correct":0},{"id":"B","content":"2","is_correct":1},{"id":"C","content":"8","is_correct":0},{"id":"D","content":"-2","is_correct":0}]},{"id":1706,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某学校组织七年级学生开展‘校园植物分布调查’活动，要求将校园划分为若干区域，并在平面直角坐标系中记录每种植物的位置。已知校园被划分为四个象限，某学生在第一象限内发现一种植物，其位置坐标为 (a, b)，其中 a 和 b 是正实数，且满足以下条件：\n\n① a 和 b 是方程组\n 2x + y = 8\n x - y = -2\n 的解；\n\n② 该点到原点的距离为 d，且 d² 是一个整数；\n\n③ 若将该点绕原点逆时针旋转 90°，得到新点 P'，求点 P' 的坐标；\n\n④ 若以原点、点 P 和点 P' 为三个顶点构成三角形，判断该三角形的形状（按边和角分类），并说明理由。\n\n请依次解答上述四个问题。","answer":"① 解方程组：\n 2x + y = 8 （1）\n x - y = -2 （2）\n\n 将（2）式变形得：x = y - 2，代入（1）式：\n 2(y - 2) + y = 8\n 2y - 4 + y = 8\n 3y = 12\n y = 4\n 代入 x = y - 2 得：x = 4 - 2 = 2\n 所以 a = 2，b = 4，点 P 坐标为 (2, 4)\n\n② 计算到原点的距离 d：\n d² = 2² + 4² = 4 + 16 = 20\n 20 是整数，满足条件。\n\n③ 将点 P(2, 4) 绕原点逆时针旋转 90°，旋转公式为：\n (x, y) → (-y, x)\n 所以 P' 坐标为 (-4, 2)\n\n④ 三点坐标：O(0, 0)，P(2, 4)，P'(-4, 2)\n\n 计算三边长度：\n OP = √(2² + 4²) = √20\n OP' = √((-4)² + 2²) = √(16 + 4) = √20\n PP' = √[(2 - (-4))² + (4 - 2)²] = √(6² + 2²) = √(36 + 4) = √40\n\n 因为 OP = OP'，所以是等腰三角形。\n\n 再判断是否为直角三角形：\n 检查是否满足勾股定理：\n OP² + OP'² = 20 + 20 = 40 = PP'²\n 所以 ∠POP' = 90°，是直角三角形。\n\n 综上，该三角形是等腰直角三角形。","explanation":"本题综合考查了二元一次方程组的解法、实数运算、平面直角坐标系中的坐标变换（旋转变换）、两点间距离公式以及三角形形状的判定。解题关键在于：\n\n1. 通过代入法准确求解方程组，得到点的坐标；\n2. 利用勾股定理计算点到原点的距离平方，并验证其为整数；\n3. 掌握绕原点逆时针旋转 90° 的坐标变换规则：(x, y) → (-y, x)；\n4. 利用坐标计算三角形三边长度，通过边长关系判断三角形类型：两边相等说明是等腰三角形，三边满足勾股定理说明是直角三角形，因此是等腰直角三角形。\n\n本题融合了代数与几何知识，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 13:44:30","updated_at":"2026-01-06 13:44:30","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":[]}]