习题练习

[{"id":704,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次班级大扫除中，某学生负责统计各小组的垃圾重量（单位：千克），记录如下：第一组 3.5，第二组 4.2，第三组 3.8，第四组 4.5。如果学校规定每班平均垃圾重量不超过 4 千克为合格，那么该班四个小组的平均垃圾重量是 ___ 千克，因此该班 ___（填“合格”或“不合格”）。","answer":"4.0，合格","explanation":"首先计算四个小组垃圾重量的总和：3.5 + 4.2 + 3.8 + 4.5 = 16.0（千克）。然后用总重量除以小组数 4，得到平均重量：16.0 ÷ 4 = 4.0（千克）。由于 4.0 千克等于学校规定的上限 4 千克，因此该班达到合格标准，应填“合格”。本题考查数据的收集、整理与描述中的平均数计算及简单比较，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 22:43:57","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":1329,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某学生在研究城市公交线路优化问题时，收集了A、B两条公交线路在一天中不同时段的乘客数量数据，并绘制成如下表格。已知A线路每辆公交车最多可载客40人，B线路每辆最多可载客35人。若要求每条线路在每个时段运行的公交车数量必须为整数，且总运行车辆数最少，同时确保所有乘客都能被运送（不允许超载），请根据以下数据建立数学模型并求解：\n\n| 时段 | A线路乘客数 | B线路乘客数 |\n|------|---------------|---------------|\n| 早高峰（7:00-9:00） | 320 | 280 |\n| 平峰（9:00-17:00） | 160 | 140 |\n| 晚高峰（17:00-19:00） | 360 | 315 |\n\n假设每条线路在每个时段独立安排车辆，不考虑车辆跨时段调度。请分别求出A、B两条线路在三个时段各自所需的最少公交车数量，并计算全天两条线路总共需要的最少公交车班次（即各时段车辆数之和）。","answer":"解：\n\n我们分别计算每条线路在每个时段所需的最少公交车数量。由于每辆车有最大载客限制，且车辆数必须为整数，因此需要使用“向上取整”的方法。\n\n**第一步：计算A线路各时段所需车辆数**\n\n- 早高峰：320 ÷ 40 = 8（恰好整除），需8辆车\n- 平峰：160 ÷ 40 = 4（恰好整除），需4辆车\n- 晚高峰：360 ÷ 40 = 9（恰好整除），需9辆车\n\n**第二步：计算B线路各时段所需车辆数**\n\n- 早高峰：280 ÷ 35 = 8（恰好整除），需8辆车\n- 平峰：140 ÷ 35 = 4（恰好整除），需4辆车\n- 晚高峰：315 ÷ 35 = 9（恰好整除），需9辆车\n\n**第三步：计算全天总班次**\n\nA线路总班次：8 + 4 + 9 = 21（班次）\nB线路总班次：8 + 4 + 9 = 21（班次）\n\n全天两条线路总共需要的最少公交车班次为：21 + 21 = 42（班次）\n\n答：A线路在早高峰、平峰、晚高峰分别需要8、4、9辆车；B线路分别需要8、4、9辆车；全天总共需要最少42个公交车班次。","explanation":"本题综合考查了有理数的除法运算、实际问题中的整数解处理（向上取整思想）、数据的收集与整理，以及优化思想（最小化资源使用）。虽然计算本身不复杂，但难点在于理解‘不允许超载’意味着必须向上取整，即使除法结果接近整数也不能向下舍入。同时，题目设置了真实情境——城市公交调度，要求学生从数据中提取信息，建立数学模型（即每个时段的车辆数 = 乘客数 ÷ 每车载客量，结果向上取整），并进行多步推理与汇总。尽管所有除法结果恰好为整数，避免了余数处理，但情境复杂、信息量大，且要求系统性分析，符合‘困难’难度标准。此外，题目未使用常见人名，情境新颖，考查角度独特，避免了传统应用题的重复模式。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 10:56:38","updated_at":"2026-01-06 10:56:38","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":1231,"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从原点O(0, 0)出发，沿直线y = x向右上方移动。同时，另一个动点Q从点A(6, 0)出发，沿x轴向负方向以每秒1个单位的速度匀速运动。已知点P的运动速度是每秒√2个单位。设运动时间为t秒（t ≥ 0），当t为何值时，线段PQ的长度最短？并求出这个最短长度。","answer":"解：\n\n设运动时间为t秒。\n\n点P从原点O(0, 0)出发，沿直线y = x运动，速度为每秒√2个单位。\n由于直线y = x的方向向量为(1, 1)，其模长为√(1² + 1²) = √2，\n因此点P在t秒后的坐标为：\n x_P = t × (1) = t\n y_P = t × (1) = t\n即 P(t, t)\n\n点Q从A(6, 0)出发，沿x轴向负方向以每秒1个单位速度运动，\n因此Q的坐标为：\n x_Q = 6 - t\n y_Q = 0\n即 Q(6 - t, 0)\n\n线段PQ的长度为：\n|PQ| = √[(t - (6 - t))² + (t - 0)²]\n = √[(2t - 6)² + t²]\n = √[4t² - 24t + 36 + t²]\n = √[5t² - 24t + 36]\n\n令函数 f(t) = 5t² - 24t + 36，则 |PQ| = √f(t)\n由于平方根函数在定义域内单调递增，因此当f(t)最小时，|PQ|最小。\n\nf(t) 是一个开口向上的二次函数，其最小值出现在顶点处：\n t = -b\/(2a) = 24\/(2×5) = 24\/10 = 2.4\n\n因此，当 t = 2.4 秒时，PQ长度最短。\n\n最短长度为：\n|PQ| = √[5×(2.4)² - 24×2.4 + 36]\n = √[5×5.76 - 57.6 + 36]\n = √[28.8 - 57.6 + 36]\n = √[7.2]\n = √(72\/10) = √(36×2 \/ 10) = 6√2 \/ √10 = (6√20)\/10 = (6×2√5)\/10 = (12√5)\/10 = (6√5)\/5\n\n或者直接保留为 √7.2，但更规范地化简：\n7.2 = 72\/10 = 36\/5\n所以 √(36\/5) = 6\/√5 = (6√5)\/5\n\n答：当 t = 2.4 秒时，线段PQ的长度最短，最短长度为 (6√5)\/5 个单位。","explanation":"本题综合考查了平面直角坐标系、函数思想、二次函数最值以及两点间距离公式，属于跨知识点综合应用题。解题关键在于：\n1. 根据运动方向和速度，正确写出两个动点的坐标表达式；\n2. 利用两点间距离公式建立关于时间t的距离函数；\n3. 将距离的平方视为二次函数，利用顶点公式求最小值对应的t值；\n4. 注意距离是平方根形式，但由于根号单调递增，最小值点一致；\n5. 最后代入求最短距离，并进行合理的根式化简。\n本题难度较高，要求学生具备较强的建模能力和代数运算技巧，同时理解函数最值在实际问题中的应用。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 10:27:01","updated_at":"2026-01-06 10:27:01","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":478,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"周一","answer":"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 17:58:00","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":397,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"在一次校园植物观察活动中，某学生记录了五种植物一周内每天的生长高度（单位：厘米），并将数据整理如下表。已知这五种植物的平均每日生长高度为1.2厘米，其中四种植物的生长高度分别为0.8、1.0、1.5和1.3厘米，那么第五种植物的每日生长高度是多少？","answer":"C","explanation":"题目考查数据的收集、整理与描述中的平均数计算。已知五种植物的平均每日生长高度为1.2厘米，因此总生长高度为 5 × 1.2 = 6.0 厘米。已知四种植物的生长高度分别为0.8、1.0、1.5和1.3厘米，它们的和为 0.8 + 1.0 + 1.5 + 1.3 = 4.6 厘米。因此第五种植物的生长高度为 6.0 - 4.6 = 1.4 厘米。故正确答案为C。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:15:08","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":"1.1厘米","is_correct":0},{"id":"B","content":"1.2厘米","is_correct":0},{"id":"C","content":"1.4厘米","is_correct":1},{"id":"D","content":"1.6厘米","is_correct":0}]},{"id":2458,"subject":"数学","grade":"八年级","stage":"初中","type":"填空题","content":"在一次数学实践活动中，某学生测量了一块等腰三角形花坛的两条腰长均为5米，底边上的高为4米。若要在花坛中铺设一条从顶点到底边中点的装饰带，则这条装饰带的长度为____米。","answer":"4","explanation":"等腰三角形底边上的高、中线、顶角平分线三线合一，因此从顶点到底边中点的线段就是高，长度为4米。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 14:05:26","updated_at":"2026-01-10 14:05:26","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":2019,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"在一次校园绿化设计中，工人师傅需要在一块矩形空地的对角线上铺设一条石板路。已知这块空地的长为12米，宽为5米。为了估算材料用量，一名学生想计算这条对角线的长度。请问该对角线的长度是多少？","answer":"A","explanation":"本题考查勾股定理的应用。矩形空地可看作一个长方形，其对角线将长方形分成两个直角三角形。根据勾股定理，对角线长度 c 满足 c² = a² + b²，其中 a = 12 米，b = 5 米。计算得：c² = 12² + 5² = 144 + 25 = 169，因此 c = √169 = 13 米。选项A正确。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 10:31:24","updated_at":"2026-01-09 10:31:24","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":"13米","is_correct":1},{"id":"B","content":"15米","is_correct":0},{"id":"C","content":"17米","is_correct":0},{"id":"D","content":"√119米","is_correct":0}]},{"id":452,"subject":"数学","grade":"初一","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50名学生进行调查，并将结果绘制成扇形统计图。其中，喜欢阅读的学生所占的圆心角为72度。那么，喜欢阅读的学生人数是多少？","answer":"A","explanation":"扇形统计图中，每个部分的圆心角占整个圆（360度）的比例等于该部分人数占总人数的比例。已知喜欢阅读的学生对应的圆心角是72度，总调查人数为50人。计算方法是：(72 ÷ 360) × 50 = 0.2 × 50 = 10。因此，喜欢阅读的学生有10人。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:44:55","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":"10人","is_correct":1},{"id":"B","content":"12人","is_correct":0},{"id":"C","content":"15人","is_correct":0},{"id":"D","content":"20人","is_correct":0}]},{"id":1060,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"在一次班级环保活动中，某学生收集了废旧纸张和塑料瓶共12件，其中废旧纸张比塑料瓶多4件。设塑料瓶的数量为x件，则根据题意可列出一元一次方程：_x + (x + 4) = 12_，解得x = _4_，因此塑料瓶有_4_件，废旧纸张有_8_件。","answer":"x + (x + 4) = 12；4；4；8","explanation":"设塑料瓶数量为x件，则废旧纸张数量为x + 4件。根据总数量为12件，可列方程x + (x + 4) = 12。解这个方程：2x + 4 = 12 → 2x = 8 → x = 4。因此塑料瓶有4件，废旧纸张有4 + 4 = 8件。本题考查一元一次方程的建立与求解，符合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":"2026-01-06 08:51:55","updated_at":"2026-01-06 08:51:55","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":[]}]