[{"id":2212,"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，将比零度高记为正，比零度低记为负。已知周一的气温变化为上升3度，周二为下降5度，周三为上升2度，周四为下降4度。若这四天的气温变化总和为负数，则这个总和是____度。","answer":"-4","explanation":"根据题意，将每天的气温变化用正负数表示：周一为+3，周二为-5，周三为+2，周四为-4。将这些数相加：+3 + (-5) + (+2) + (-4) = (3 + 2) + (-5 - 4) = 5 - 9 = -4。因此，这四天的气温变化总和为-4度，符合题目中‘总和为负数’的条件。","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":2291,"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-3，点B与点A的距离为7个单位长度，且点B在原点右侧。点C是线段AB的中点，点D与点C的距离为4个单位长度，且点D在点C的左侧。那么点D表示的数是___。","answer":"-3.5","explanation":"点A表示-3，点B在原点右侧且与A相距7个单位，因此点B表示的数为-3 + 7 = 4。点C是AB的中点，坐标为(-3 + 4) ÷ 2 = 0.5。点D在点C左侧4个单位，因此点D表示的数为0.5 - 4 = -3.5。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-09 16:44:29","updated_at":"2026-01-09 16:44:29","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":553,"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（单位：小时）时，记录了以下5个数据：2.5，3，3.5，4，4.5。如果他想用这组数据制作频数分布表，并将数据分为两组：3小时以下（不含3小时）和3小时及以上，那么这两组的频数分别是多少？","answer":"A","explanation":"首先明确分组标准：第一组是“3小时以下（不含3小时）”，即小于3；第二组是“3小时及以上”，即大于或等于3。原始数据为：2.5，3，3.5，4，4.5。其中，只有2.5小于3，属于第一组，频数为1；其余数据3、3.5、4、4.5均大于或等于3，属于第二组，共4个数据，频数为4。因此，两组的频数分别是1和4，正确答案为A。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 19:11:16","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":"A","content":"1和4","is_correct":1},{"id":"B","content":"2和3","is_correct":0},{"id":"C","content":"3和2","is_correct":0},{"id":"D","content":"4和1","is_correct":0}]},{"id":804,"subject":"数学","grade":"初一","stage":"小学","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30分钟到60分钟之间的学生人数占总调查人数的40%。如果总调查人数为50人，那么阅读时间不在这个区间内的学生有___人。","answer":"30","explanation":"总调查人数为50人，阅读时间在30到60分钟之间的占40%，即50 × 40% = 20人。因此，不在这个区间内的学生人数为50 - 20 = 30人。本题考查数据的收集与整理，涉及百分比的实际应用，属于简单难度。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 00:21:14","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":1384,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市为优化公交线路，对一条主干道上的乘客流量进行了为期7天的调查。调查数据显示，每天早高峰时段（7:00-9:00）的乘客人数分别为：120人、135人、150人、165人、180人、195人、210人。调查发现，乘客人数每天以固定数值递增。公交公司计划根据这7天的平均乘客人数，安排每辆公交车的载客量。已知每辆公交车最多可载客45人，且要求每趟车的载客率不低于80%。若公交公司希望用最少数量的公交车完成运输任务，且每辆车每天只运行一趟，问：该公司至少需要安排多少辆公交车？请通过计算说明理由。","answer":"第一步：计算7天乘客人数的总和。\n120 + 135 + 150 + 165 + 180 + 195 + 210 = 1155（人）\n\n第二步：计算平均每天的乘客人数。\n1155 ÷ 7 = 165（人）\n\n第三步：确定每辆公交车的最低有效载客量（载客率不低于80%）。\n每辆车最多可载45人，80%载客量为：\n45 × 0.8 = 36（人）\n即每辆车每天至少运送36人才能满足载客率要求。\n\n第四步：计算满足平均每天165人运输所需的最少车辆数。\n设需要x辆车，则每辆车平均载客量为165 ÷ x。\n要求：165 ÷ x ≥ 36\n解不等式：\n165 ≥ 36x\nx ≤ 165 ÷ 36 ≈ 4.583\n由于x必须为整数，且要满足每辆车载客量不低于36人，因此x最大可取4，但需验证是否可行。\n\n若x = 4，则每辆车平均载客量为165 ÷ 4 = 41.25人，满足≥36人，且41.25 ≤ 45，未超载。\n因此4辆车可行。\n\n但题目要求“用最少数量的公交车”，我们需确认是否可以更少。\n若x = 3，则每辆车平均载客量为165 ÷ 3 = 55人 > 45人，超载，不可行。\n\n因此，最少需要4辆公交车。\n\n答案：至少需要安排4辆公交车。","explanation":"本题综合考查了数据的收集、整理与描述（计算平均数）、有理数的运算（加减与除法）、不等式与不等式组（建立并求解不等式）以及实际应用问题的建模能力。解题关键在于理解“载客率不低于80%”转化为数学条件为每辆车平均载客量不低于36人，并结合最大载客量限制，通过不等式分析确定最小车辆数。同时需验证解的合理性，排除超载情况，体现数学思维的严谨性。题目情境新颖，贴近生活，考查学生从数据中提取信息、建立数学模型并解决实际问题的能力，符合七年级数学课程标准要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 11:17:21","updated_at":"2026-01-06 11:17:21","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":1013,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次环保活动中，某班级收集了可回收垃圾共120千克，其中纸张占总重量的五分之三，塑料占总重量的四分之一，其余为金属。那么金属的重量是___千克。","answer":"18","explanation":"首先计算纸张的重量：120 × 3\/5 = 72 千克；然后计算塑料的重量：120 × 1\/4 = 30 千克；纸张和塑料共重 72 + 30 = 102 千克；因此金属的重量为 120 - 102 = 18 千克。本题考查有理数中的分数乘法与加减运算在实际问题中的应用。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 05:24:02","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":1230,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某学生在研究平面直角坐标系中的几何问题时，发现一个动点P(x, y)始终满足以下两个条件：(1) 点P到点A(3, 0)的距离与到点B(-3, 0)的距离之和恒为10；(2) 点P的纵坐标y满足不等式 2y + 4 < 3y - 1。已知该动点P的轨迹与x轴围成一个封闭图形，求该图形的面积，并判断是否存在这样的点P同时满足上述两个条件。","answer":"解：\n\n第一步：分析条件(1)\n点P(x, y)到A(3, 0)和B(-3, 0)的距离之和为10，即：\n√[(x - 3)² + y²] + √[(x + 3)² + y²] = 10\n这是椭圆的定义：到两个定点（焦点）距离之和为定值（大于两焦点间距离）的点的轨迹。\n两焦点A(3,0)、B(-3,0)之间的距离为6，而定值为10 > 6，符合条件。\n因此，点P的轨迹是以A、B为焦点，长轴长为10的椭圆。\n\n椭圆标准形式：中心在原点，焦点在x轴上。\n焦距2c = 6 ⇒ c = 3\n长轴2a = 10 ⇒ a = 5\n由椭圆关系：b² = a² - c² = 25 - 9 = 16 ⇒ b = 4\n所以椭圆方程为：x²\/25 + y²\/16 = 1\n\n该椭圆与x轴围成的封闭图形即为椭圆本身，其面积为：\nS = πab = π × 5 × 4 = 20π\n\n第二步：分析条件(2)\n解不等式：2y + 4 < 3y - 1\n移项得：4 + 1 < 3y - 2y ⇒ 5 < y ⇒ y > 5\n\n第三步：判断是否存在同时满足两个条件的点P\n由椭圆方程 x²\/25 + y²\/16 = 1，可知y的取值范围为：\n-4 ≤ y ≤ 4（因为y²\/16 ≤ 1 ⇒ |y| ≤ 4）\n但条件(2)要求 y > 5，而5 > 4，因此y > 5不在椭圆的y取值范围内。\n\n结论：不存在同时满足两个条件的点P。\n\n最终答案：\n该封闭图形的面积为20π；不存在同时满足两个条件的点P。","explanation":"本题综合考查了平面直角坐标系、椭圆的几何定义、实数运算、不等式求解以及逻辑推理能力。首先利用椭圆的定义将距离和转化为标准椭圆方程，进而求出面积；然后通过解不等式得到y的范围；最后通过比较椭圆的y值范围与不等式解集，判断是否存在公共解。题目融合了代数与几何，要求学生具备较强的综合分析能力，属于困难难度。解题关键在于理解椭圆的定义及其几何性质，并准确进行不等式的求解与范围比较。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 10:26:43","updated_at":"2026-01-06 10:26:43","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":788,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次环保活动中，某学校七年级学生共收集了120千克废纸。如果每5千克废纸可以生产3千克再生纸，那么这些废纸一共可以生产____千克再生纸。","answer":"72","explanation":"根据题意，每5千克废纸可生产3千克再生纸。先求出120千克废纸中有多少个5千克：120 ÷ 5 = 24。每个5千克对应3千克再生纸，因此总共可生产 24 × 3 = 72 千克再生纸。本题考查有理数的乘除运算在实际问题中的应用，属于简单难度。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 00:06:44","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":2483,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"一个圆形花坛被均匀划分为6个扇形区域，分别种植不同颜色的花。若将整个花坛绕其中心顺时针旋转60°，则每个扇形区域会与原来相邻的下一个区域重合。现在随机选择一个点落在花坛上，该点落在红色区域的概率是1\/6。若花坛旋转两次（每次60°），则该点最终落在红色区域的概率是多少？","answer":"A","explanation":"由于花坛被均匀分为6个扇形，每个区域占1\/6的面积，且旋转是绕中心进行的刚体变换，不改变区域的面积和分布。每次顺时针旋转60°，相当于将整个图案向右移动一个扇形位置。旋转两次共120°，即移动两个位置，但整个图案的结构保持不变，每个颜色区域仍然占据1\/6的面积。因此，无论旋转多少次（只要旋转角度是60°的整数倍），每个颜色区域在整体中所占比例不变。所以，随机点落在红色区域的概率始终是1\/6。本题考查的是旋转对称性与概率初步的结合，强调几何变换不改变面积比例这一核心思想。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 15:10:16","updated_at":"2026-01-10 15:10:16","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":"1\/6","is_correct":1},{"id":"B","content":"1\/3","is_correct":0},{"id":"C","content":"1\/2","is_correct":0},{"id":"D","content":"选项D","is_correct":0}]},{"id":1534,"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‘城市绿地规划’数学实践活动。活动要求学生在平面直角坐标系中设计一个矩形绿化区域，其四个顶点坐标均为整数，且满足以下条件：\n\n1. 矩形的一组对边平行于x轴，另一组对边平行于y轴；\n2. 矩形的周长为20个单位长度；\n3. 矩形的面积不小于24个单位面积；\n4. 矩形完全位于第一象限，且其左下角顶点位于原点(0, 0)；\n5. 设矩形的右上角顶点坐标为(x, y)，其中x和y均为正整数。\n\n现从所有满足上述条件的矩形中随机选取一个，求该矩形的面积恰好为24的概率。","answer":"解：\n\n由题意，矩形左下角顶点为(0, 0)，右上角顶点为(x, y)，其中x > 0，y > 0，且x、y均为正整数。\n\n因为矩形对边分别平行于坐标轴，所以其长为x，宽为y。\n\n根据条件2：周长为20，\n即：2(x + y) = 20 \n⇒ x + y = 10 \n（方程①）\n\n根据条件3：面积不小于24，\n即：xy ≥ 24 \n（不等式②）\n\n又x、y为正整数，且x + y = 10，我们可以列出所有满足方程①的正整数解：\n\n(x, y) 的可能组合为：\n(1,9), (2,8), (3,7), (4,6), (5,5), (6,4), (7,3), (8,2), (9,1)\n\n计算每种组合的面积xy：\n1×9 = 9 < 24 → 不满足\n2×8 = 16 < 24 → 不满足\n3×7 = 21 < 24 → 不满足\n4×6 = 24 ≥ 24 → 满足\n5×5 = 25 ≥ 24 → 满足\n6×4 = 24 ≥ 24 → 满足\n7×3 = 21 < 24 → 不满足\n8×2 = 16 < 24 → 不满足\n9×1 = 9 < 24 → 不满足\n\n因此，满足所有条件的(x, y)组合有：\n(4,6), (5,5), (6,4)\n共3种。\n\n其中，面积恰好为24的有：(4,6) 和 (6,4)，共2种。\n\n注意：虽然(4,6)和(6,4)表示不同的矩形（长宽不同），但在坐标系中它们是不同的图形，应视为两个不同的矩形。\n\n因此，所求概率为：\n满足条件的矩形总数：3\n面积恰好为24的矩形数：2\n\n概率 = 2 \/ 3\n\n答：该矩形的面积恰好为24的概率是 2\/3。","explanation":"本题综合考查了平面直角坐标系、二元一次方程组、不等式与不等式组以及数据的整理与描述等知识点。解题关键在于：\n\n1. 利用矩形顶点坐标与边长的关系，将几何问题转化为代数问题；\n2. 由周长条件建立方程 x + y = 10；\n3. 由面积条件建立不等式 xy ≥ 24；\n4. 枚举所有满足方程的正整数解，并结合不等式筛选出符合条件的解；\n5. 在满足所有条件的样本空间中，计算目标事件（面积为24）发生的概率。\n\n本题难度较高，体现在需要综合运用多个知识点，并进行分类讨论与逻辑推理。同时，题目情境新颖，避免了传统应用题的套路，强调数学建模与数据分析能力，符合七年级数学课程的综合应用要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 12:17:55","updated_at":"2026-01-06 12:17:55","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":[]}]