习题练习

[{"id":968,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生测量了一个长方形花坛的长和宽，记录数据时不小心把单位弄混了。已知花坛的实际周长是 12 米，长比宽多 2 米。如果设宽为 x 米，则根据题意可列出一元一次方程：______ = 12。","answer":"2(x + x + 2)","explanation":"根据题意，设宽为 x 米，则长为 (x + 2) 米。长方形的周长公式为：周长 = 2 × (长 + 宽)。代入得：2 × (x + x + 2) = 12。因此空白处应填写 2(x + x + 2)。该题考查一元一次方程的实际应用，符合七年级数学课程要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 04:04:02","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":1715,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某校七年级组织学生参加环保知识竞赛，参赛学生需完成两项任务：任务一为线上答题，任务二为实地调查。竞赛结束后，统计发现：若每名参与任务一的学生得分为正整数，且得分不低于5分；参与任务二的学生得分也为正整数，且得分不低于3分。已知共有30名学生参与竞赛，其中同时参与两项任务的学生有8人。若只参与任务一的学生平均得分为7分，只参与任务二的学生平均得分为5分，同时参与两项任务的学生在任务一和任务二中分别平均得分为6分和4分。现定义总得分为所有学生在各自参与任务中的得分之和（例如，同时参与两项的学生，其得分计入两次）。若总得分不超过500分，求同时参与两项任务的学生人数是否可能为8人？若可能，求此时总得分的最小值；若不可能，说明理由。","answer":"设只参与任务一的学生人数为x，只参与任务二的学生人数为y，同时参与两项任务的学生人数为z。\n\n根据题意，z = 8（题目给定），总人数为30人，因此有：\nx + y + z = 30\n代入z = 8，得：\nx + y = 22 （1）\n\n计算总得分：\n- 只参与任务一的学生总得分：7x\n- 只参与任务二的学生总得分：5y\n- 同时参与两项任务的学生在任务一中的总得分：6 × 8 = 48\n- 同时参与两项任务的学生在任务二中的总得分：4 × 8 = 32\n\n因此，总得分S为：\nS = 7x + 5y + 48 + 32 = 7x + 5y + 80\n\n由（1）得 y = 22 - x，代入上式：\nS = 7x + 5(22 - x) + 80\n = 7x + 110 - 5x + 80\n = 2x + 190\n\n要求总得分不超过500分，即：\n2x + 190 ≤ 500\n2x ≤ 310\nx ≤ 155\n\n但x为只参与任务一的人数，且x ≥ 0，y = 22 - x ≥ 0，故x ≤ 22。\n因此x的取值范围是 0 ≤ x ≤ 22，且x为整数。\n\n此时S = 2x + 190，当x取最小值0时，S最小：\nS_min = 2×0 + 190 = 190\n\n验证是否满足所有条件：\n- 只参与任务一：0人，平均7分 → 合理（无人参与，无矛盾）\n- 只参与任务二：22人，平均5分 → 总得分110\n- 同时参与两项：8人，任务一总得分48，任务二总得分32\n- 总得分：0 + 110 + 48 + 32 = 190 ≤ 500，满足\n\n因此，同时参与两项任务的学生人数为8人是可能的。\n此时总得分的最小值为190分。","explanation":"本题综合考查了二元一次方程组、不等式与不等式组、数据的收集与整理等知识点。解题关键在于正确理解“总得分”是各任务得分的累加，包括重复计算同时参与两项的学生得分。通过设定变量，建立人数关系式，再表达总得分函数，并结合不等式约束进行分析。难点在于识别“总得分”的定义方式以及合理处理平均分与总人数之间的关系。通过代数建模，将实际问题转化为数学表达式，最终通过最小化目标函数得到结果。题目情境新颖，融合环保主题与数据统计，考查学生综合应用能力。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 14:10:12","updated_at":"2026-01-06 14:10:12","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":2207,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在一条东西走向的直线上做标记，规定向东为正方向。他从原点出发，先向东走了5米，记作+5米，接着又向西走了8米。此时他的位置相对于原点的方向和距离应如何表示？","answer":"B","explanation":"向东走5米记作+5，向西走8米记作-8。总位移为+5 + (-8) = -3，表示最终位于原点西侧3米处，应记作-3米。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":"向东3米，记作+3米","is_correct":0},{"id":"B","content":"向西3米，记作-3米","is_correct":1},{"id":"C","content":"向东13米，记作+13米","is_correct":0},{"id":"D","content":"向西13米，记作-13米","is_correct":0}]},{"id":782,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在某次班级大扫除中，某学生负责统计清洁工具的数量。他发现扫帚的数量比拖把多5把，而两种工具的总数是17把。如果设拖把的数量为x把，那么根据题意可以列出方程：x + (x + 5) = 17。解这个方程可得x = ___。","answer":"6","explanation":"根据题意，拖把数量为x，则扫帚数量为x + 5。两者总数为17，因此方程为x + (x + 5) = 17。化简得2x + 5 = 17，移项得2x = 12，解得x = 6。所以拖把有6把。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 23:59:20","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":357,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"在一次环保活动中，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：12、15、18、14、16。为了分析数据，他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":"A","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。具体计算如下：12 + 15 + 18 + 14 + 16 = 75，共有5天，所以平均数为75 ÷ 5 = 15。因此，正确答案是A。本题考查的是数据的收集、整理与描述中的平均数计算，属于七年级数学的基础知识。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:44:04","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":"15","is_correct":1},{"id":"B","content":"14","is_correct":0},{"id":"C","content":"16","is_correct":0},{"id":"D","content":"13","is_correct":0}]},{"id":1322,"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，对一条主干道的车流量进行了为期7天的观测，记录每天上午8:00至9:00的车辆通行数量（单位：辆）如下：320，345，332，358，340，367，350。交通部门计划根据这组数据制定新的公交发车间隔方案。已知公交车的平均载客量为40人，每辆车每小时最多运行2个单程，且每辆公交车每天最多工作8小时。若要求在任何观测时段内，公交车运力至少能满足该时段车流量的15%（假设每辆车平均载客1.2人），同时总运营成本不能超过每日120个‘车次’（一个车次指一辆车完成一个单程）。问：为满足上述条件，该线路每日至少需要安排多少辆公交车？并说明如何安排发车班次才能使运力覆盖最紧张的一天，且总车次不超过限制。","answer":"第一步：计算7天中最大车流量\n观测数据中最大值为367辆（第6天）。\n\n第二步：计算该时段所需最小运力\n每辆车平均载客1.2人，因此367辆车对应乘客数约为：\n367 × 1.2 = 440.4 ≈ 441人\n要求公交运力至少满足15%，即：\n441 × 15% = 66.15 ≈ 67人\n\n第三步：计算每小时所需最少公交车运力\n每辆公交车每小时可运行2个单程，每个单程载客40人，因此一辆车每小时最大运力为：\n2 × 40 = 80人\n要满足67人的运力需求，至少需要：\n67 ÷ 80 = 0.8375 → 向上取整为1辆车（每小时）\n\n第四步：考虑全天工作安排\n每辆车每天最多工作8小时，每小时最多贡献80人运力，因此一辆车每天最多提供：\n8 × 80 = 640人运力\n但高峰时段（8:00–9:00）只需67人运力，因此从运力角度看，1辆车即可满足高峰需求。\n\n第五步：分析车次限制\n总车次上限为每日120个单程。\n若安排n辆车，每辆车每天最多运行8小时 × 2单程\/小时 = 16个单程，\n则总车次最多为16n。\n要求16n ≤ 120 → n ≤ 7.5 → 最多可用7辆车。\n\n第六步：验证最少车辆数是否可行\n虽然1辆车可满足高峰运力，但需确保其在8:00–9:00运行。\n假设安排1辆车专门在高峰时段运行，其余时间可调度。\n该辆车在高峰1小时内可运行2个单程，提供80人运力 > 67人，满足要求。\n总车次使用2个，远低于120限制。\n\n第七步：结论\n因此，每日至少需要安排1辆公交车即可满足运力要求和车次限制。\n安排方式：该辆车在8:00–9:00运行2个单程（如8:00发车，8:30返回；8:30再发车），其余时间可灵活调度或停运，确保总车次不超过120。\n\n最终答案：每日至少需要安排1辆公交车。","explanation":"本题综合考查数据的收集与整理（分析7天车流量）、有理数运算（乘法、百分数计算）、不等式思想（车次限制）、实际应用建模（运力与车辆调度）以及最优化思维（最少车辆数）。解题关键在于识别‘最紧张的一天’作为约束条件，将实际问题转化为数学不等式与整数规划问题。通过计算高峰时段所需最小运力，并结合车辆运行能力与车次上限，逐步推理得出最小车辆数。题目情境新颖，融合交通规划与数学建模，体现数学在现实决策中的应用，符合七年级学生已学的实数运算、一元一次不等式、数据统计等知识点，难度较高，需多步逻辑推理与综合分析。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 10:54:43","updated_at":"2026-01-06 10:54:43","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":2142,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在解方程 3(x - 2) = 2x + 1 时，第一步将方程两边同时展开，得到 3x - 6 = 2x + 1。接下来，他应该进行的正确步骤是：","answer":"B","explanation":"解一元一次方程时，展开后应通过移项将含未知数的项移到等式一边，常数项移到另一边。选项 B 正确地将 2x 移到左边变为 -2x，将 -6 移到右边变为 +6，符合等式性质，是标准解法步骤。其他选项或错误合并项，或不当操作，不符合解方程的基本规则。","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":"将 3x 和 2x 相加，得到 5x - 6 = 1","is_correct":0},{"id":"B","content":"将 2x 移到左边，-6 移到右边，得到 3x - 2x = 1 + 6","is_correct":1},{"id":"C","content":"将方程两边同时除以 3，得到 x - 2 = (2x + 1)\/3","is_correct":0},{"id":"D","content":"将 -6 和 +1 相加，得到 3x = 2x - 5","is_correct":0}]},{"id":2843,"subject":"政治","grade":"高三","stage":"高中","type":"选择题","content":"驾驶人在使用驾驶自动化系统时，确实避免了一些车祸的发生，但也存在发生交通事故的情形。如交通事故纯粹因系统缺陷导致，关于汽车生产厂商责任，说法正确的是（ ）","answer":"B","explanation":"产品责任实行无过错责任原则，只要产品存在缺陷造成损害，生产者就应承担责任。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"中等","points":1,"is_active":1,"created_at":"2026-04-08 20:01:23","updated_at":"2026-04-08 20:01:23","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":"A. 只有生产汽车时有过错，厂商才承担责任","is_correct":0},{"id":"B","content":"B. 不管生产汽车时有没有过错，厂商都应承担责任","is_correct":1},{"id":"C","content":"C. 如能证明生产汽车时没有过错，厂商就无须承担责任","is_correct":0},{"id":"D","content":"D. 不管生产汽车时有没有过错，厂商都应与驾驶人共同承担责任","is_correct":0}]},{"id":511,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"4题","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 18:16:19","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":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}]}]