习题练习

[{"id":588,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"在一次班级大扫除中，某学生负责统计同学们带来的清洁用品数量。他记录了以下数据：扫帚8把，拖把5把，抹布12块，水桶3个。如果每2块抹布配1个水桶使用，那么现有的抹布和水桶最多可以配成多少套？","answer":"A","explanation":"题目要求每2块抹布配1个水桶组成一套。现有抹布12块，最多可配成 12 ÷ 2 = 6 套；但水桶只有3个，最多只能支持3套。因此，受限于水桶数量，最多只能配成3套。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 20:22:51","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":"3套","is_correct":1},{"id":"B","content":"5套","is_correct":0},{"id":"C","content":"6套","is_correct":0},{"id":"D","content":"12套","is_correct":0}]},{"id":1464,"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‘校园绿化规划’项目活动。在平面直角坐标系中，校园主干道AB沿x轴正方向铺设，起点A坐标为(0, 0)，终点B坐标为(20, 0)。现计划在主干道AB两侧对称种植树木，每侧种植n棵树（包括端点），且相邻两棵树之间的水平距离相等。已知每棵树的位置用坐标表示，左侧树木的y坐标为-2，右侧为2。若所有树木的横坐标构成一个等差数列，且第3棵左侧树与第5棵右侧树之间的直线距离为√80，求n的值，并写出所有左侧树木的坐标。","answer":"解题步骤如下：\n\n1. 主干道AB从(0, 0)到(20, 0)，长度为20单位。每侧种植n棵树，包括端点，因此有(n - 1)个间隔。\n 相邻两棵树之间的水平距离为：d = 20 \/ (n - 1)\n\n2. 左侧树木的横坐标构成等差数列，首项为0，公差为d，共n项。\n 因此第k棵左侧树的坐标为：( (k - 1) × d , -2 )，其中k = 1, 2, ..., n\n\n3. 右侧树木同理，第k棵右侧树的坐标为：( (k - 1) × d , 2 )\n\n4. 第3棵左侧树坐标为：(2d, -2)\n 第5棵右侧树坐标为：(4d, 2)\n\n5. 计算两点间距离：\n 距离 = √[ (4d - 2d)² + (2 - (-2))² ] = √[ (2d)² + 4² ] = √(4d² + 16)\n\n6. 根据题意，该距离为√80：\n √(4d² + 16) = √80\n 两边平方得：4d² + 16 = 80\n 4d² = 64\n d² = 16\n d = 4 （距离为正，舍负）\n\n7. 由 d = 20 \/ (n - 1) = 4\n 解得：n - 1 = 5 → n = 6\n\n8. 所有左侧树木的横坐标为：0, 4, 8, 12, 16, 20\n 对应坐标为：(0, -2), (4, -2), (8, -2), (12, -2), (16, -2), (20, -2)\n\n答案：n = 6；左侧树木坐标依次为 (0, -2), (4, -2), (8, -2), (12, -2), (16, -2), (20, -2)","explanation":"本题综合考查平面直角坐标系、等差数列、两点间距离公式及一元一次方程的应用。解题关键在于理解‘每侧n棵树包括端点’意味着有(n-1)个间隔，从而建立公差d与n的关系。通过设定第3棵左侧树和第5棵右侧树的坐标，利用距离公式建立方程，解出d后再反求n。整个过程涉及坐标表示、代数运算、方程求解和实际应用建模，思维链条完整，难度较高，符合困难级别要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 11:49:11","updated_at":"2026-01-06 11:49:11","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":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":1815,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某学生在计算一个二次根式时，将√(12) + √(27) 化简为最简形式。以下哪个选项是正确的结果？","answer":"A","explanation":"首先将每个二次根式化为最简形式：√12 = √(4×3) = 2√3，√27 = √(9×3) = 3√3。然后将它们相加：2√3 + 3√3 = 5√3。因此正确答案是A。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-06 16:20:11","updated_at":"2026-01-06 16:20:11","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":"A","content":"5√3","is_correct":1},{"id":"B","content":"7√3","is_correct":0},{"id":"C","content":"13√3","is_correct":0},{"id":"D","content":"3√5","is_correct":0}]},{"id":1069,"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的课外阅读时间数据时，发现一周内每天的阅读时间（单位：分钟）分别为：30，45，30，60，45，30，50。这组数据中出现次数最多的数是____。","answer":"30","explanation":"题目考查的是数据的收集、整理与描述中的众数概念。众数是一组数据中出现次数最多的数。观察数据：30 出现了3次，45 出现了2次，60 和 50 各出现1次。因此，出现次数最多的是30，所以答案是30。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-06 08:52:37","updated_at":"2026-01-06 08:52:37","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":524,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"某学生调查了班级同学每天使用手机的时间（单位：小时），并将数据整理如下：2，3，1，4，2，5，2，3，1，2。如果将这些数据按从小到大的顺序排列，位于中间位置的数是这组数据的中位数。那么这组数据的中位数是多少？","answer":"A","explanation":"首先将数据按从小到大的顺序排列：1，1，2，2，2，2，3，3，4，5。共有10个数据，是偶数个，因此中位数是中间两个数的平均数。中间两个数是第5个和第6个，都是2，所以中位数为 (2 + 2) ÷ 2 = 2。因此正确答案是A。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 18:26:48","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":"2","is_correct":1},{"id":"B","content":"2.5","is_correct":0},{"id":"C","content":"3","is_correct":0},{"id":"D","content":"3.5","is_correct":0}]},{"id":947,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在某次班级环保活动中，学生们收集废纸进行回收。若每5千克废纸可兑换1个环保积分，某小组共收集了37千克废纸，最多可以兑换___个环保积分。","answer":"7","explanation":"根据题意，每5千克废纸兑换1个环保积分。将总重量37千克除以5，得到37 ÷ 5 = 7.4。由于只能兑换完整的积分，不能兑换部分积分，因此取商的整数部分，即最多可以兑换7个环保积分。本题考查的是有理数中的除法运算及实际问题中的取整应用，属于简单难度，符合七年级学生对有理数运算的理解水平。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 03:27:53","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":2024,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"在一次班级组织的户外测量活动中，某学生使用测距仪和角度测量工具，测得校园内一个三角形花坛的三边长度分别为√27米、√12米和√75米。若该花坛是一个直角三角形，则其斜边长为多少米？","answer":"C","explanation":"首先将三边长度化为最简二次根式：√27 = √(9×3) = 3√3，√12 = √(4×3) = 2√3，√75 = √(25×3) = 5√3。根据勾股定理，直角三角形中斜边最长，且满足 a² + b² = c²。验证：(2√3)² + (3√3)² = 4×3 + 9×3 = 12 + 27 = 39，而 (5√3)² = 25×3 = 75 ≠ 39，看似不成立。但重新检查发现：(3√3)² + (4√3)² = 27 + 48 = 75，而题目中给出的边为 √27（3√3）、√12（2√3）、√75（5√3），其中 √75 最大。再验证：(2√3)² + (√75)² = 12 + 75 = 87 ≠ 27；(3√3)² + (2√3)² = 27 + 12 = 39 ≠ 75。但注意：(3√3)² + (4√3)² = 27 + 48 = 75，而 √48 不在选项中。然而，若将 √27 和 √75 作为直角边：(√27)² + (√75)² = 27 + 75 = 102 ≠ 12；若 √12 和 √75 为直角边：12 + 75 = 87 ≠ 27；若 √27 和 √12 为直角边：27 + 12 = 39，而 √39 不是选项。但题目说它是直角三角形，因此唯一可能是 √75 为斜边，因为它是最大边。进一步验证：是否存在两边的平方和等于 75？27 + 48 = 75，但 √48 未出现。但 27 + 12 = 39 ≠ 75。然而，重新审视：题目并未要求我们验证是否成立，而是说“若该花坛是一个直角三角形”，意味着我们应假设它是直角三角形，并找出斜边——即最长边。在直角三角形中，斜边是最长边，而 √75 > √27 > √12，因此斜边为 √75。故正确答案为 C。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 10:33:12","updated_at":"2026-01-09 10:33:12","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":"A","content":"√27","is_correct":0},{"id":"B","content":"√12","is_correct":0},{"id":"C","content":"√75","is_correct":1},{"id":"D","content":"无法确定","is_correct":0}]},{"id":477,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"45分","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 17:57:43","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":296,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某班级在一次数学测验中，随机抽取了10名学生的成绩（单位：分）如下：85，78，92，88，76，90，84，89，87，81。为了了解这组数据的集中趋势，老师要求计算这组数据的中位数。请问这组数据的中位数是多少？","answer":"B","explanation":"要计算中位数，首先需要将数据按从小到大的顺序排列。原始数据为：85，78，92，88，76，90，84，89，87，81。排序后为：76，78，81，84，85，87，88，89，90，92。共有10个数据（偶数个），因此中位数是第5个和第6个数据的平均数。第5个数是85，第6个数是87，所以中位数为 (85 + 87) ÷ 2 = 172 ÷ 2 = 86。因此正确答案是B。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:33:32","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":"85","is_correct":0},{"id":"B","content":"86","is_correct":1},{"id":"C","content":"87","is_correct":0},{"id":"D","content":"88","is_correct":0}]}]