习题练习

[{"id":1719,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市为了优化公交线路，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的车辆通过数量（单位：辆），数据如下：120，135，128，142，130，138，145。交通部门计划根据这些数据调整红绿灯时长，并设定一个‘高峰阈值’——若某时段车流量超过该阈值，则启动延长绿灯时间的方案。已知该阈值为这7天数据的平均数向上取整后的值。同时，为评估调整效果，工作人员在实施新方案后又连续观测了5天，得到新的车流量数据：148，152，146，150，154。现要求：\n\n（1）计算原始7天数据的平均数，并确定‘高峰阈值’；\n（2）将原始7天数据与新观测的5天数据合并，求这12天车流量的中位数；\n（3）若规定‘车流量超过高峰阈值的天数占比超过50%’，则认为交通压力显著增大。请判断实施新方案后是否出现这一情况，并说明理由；\n（4）假设每辆车平均占用道路长度为6米，道路有效通行长度为800米，利用不等式估算在高峰阈值下，道路上的车辆是否会发生拥堵（即车辆总长度是否超过道路有效长度），并给出结论。","answer":"（1）原始7天数据之和为：120 + 135 + 128 + 142 + 130 + 138 + 145 = 938。\n平均数为：938 ÷ 7 = 134。\n向上取整后，高峰阈值为135。\n\n（2）合并12天数据并按从小到大排序：\n120，128，130，135，138，142，145，146，148，150，152，154。\n共有12个数据，中位数为第6和第7个数据的平均数：(142 + 145) ÷ 2 = 143.5。\n\n（3）高峰阈值为135。在原始7天中，超过135的数据有：138，142，145（共3天），占比3\/7 ≈ 42.9%，未超过50%。\n在新观测的5天中，所有数据均大于135（148，152，146，150，154），即5天全部超过阈值，占比5\/5 = 100%。\n但题目要求判断的是‘实施新方案后’是否出现‘车流量超过高峰阈值的天数占比超过50%’，应仅针对新观测的5天数据判断。\n由于5天中有5天超过阈值，占比100% > 50%，因此交通压力显著增大。\n\n（4）高峰阈值为135辆，即每小时最多135辆车通过。\n每辆车平均占用6米，则135辆车总长度为：135 × 6 = 810（米）。\n道路有效通行长度为800米。\n因为810 > 800，所以车辆总长度超过道路有效长度，会发生拥堵。\n结论：在高峰阈值下，道路会发生拥堵。","explanation":"本题综合考查了数据的收集、整理与描述中的平均数、中位数、百分比比较，以及有理数运算、不等式在实际问题中的应用。第（1）问考察平均数计算和取整规则；第（2）问要求对12个数据排序并求中位数，注意偶数个数据时取中间两数平均值；第（3）问强调对‘实施新方案后’这一时间范围的准确理解，避免误将全部12天数据纳入判断，体现数据分析的严谨性；第（4）问将实际问题转化为不等式模型，通过比较总长度与道路容量判断是否拥堵，体现数学建模能力。题目情境真实，逻辑层层递进，难度较高，符合困难等级要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 14:12:28","updated_at":"2026-01-06 14:12:28","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":2231,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"某学生在数轴上从原点出发，先向右移动5个单位长度，再向左移动8个单位长度，接着又向右移动3个单位长度，最后向左移动4个单位长度。此时该学生所在位置对应的数是___。","answer":"-4","explanation":"根据正负数在数轴上的表示，向右移动为正，向左移动为负。因此，该学生的移动过程可表示为：+5 - 8 + 3 - 4。计算过程为：5 - 8 = -3；-3 + 3 = 0；0 - 4 = -4。最终位置对应的数是-4。此题综合考查了正负数的加减运算及在数轴上的实际意义，符合七年级学生对有理数运算的理解要求。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-09 14:39:22","updated_at":"2026-01-09 14:39:22","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":2205,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生记录了连续五天的气温变化情况（单位：℃），其中正数表示比前一天升温，负数表示比前一天降温：+3，-2，+1，-4，+2。这五天中，气温变化幅度最大的一天是第几天？","answer":"D","explanation":"气温变化幅度是指变化的绝对值大小，不考虑正负。计算各天变化的绝对值：|+3|=3，|-2|=2，|+1|=1，|-4|=4，|+2|=2。其中第四天的变化绝对值为4，是五天中最大的，因此气温变化幅度最大的是第四天。","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":"第一天","is_correct":0},{"id":"B","content":"第二天","is_correct":0},{"id":"C","content":"第三天","is_correct":0},{"id":"D","content":"第四天","is_correct":1}]},{"id":516,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"72°","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 18:20:15","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":470,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生调查了班级同学每天用于课外阅读的时间（单位：分钟），并将数据整理如下：15, 20, 25, 30, 35, 40, 45。如果再加入一个数据后，这组数据的中位数变为30，那么加入的这个数据可能是多少？","answer":"B","explanation":"原数据有7个数，按从小到大排列为：15, 20, 25, 30, 35, 40, 45。中位数是第4个数，即30。加入一个数据后，总共有8个数，中位数是第4个和第5个数的平均数。要使中位数为30，则第4个和第5个数的平均数必须为30。若加入30，则新数据为：15, 20, 25, 30, 30, 35, 40, 45，此时第4个数是30，第5个数也是30，中位数为(30+30)÷2=30，符合条件。其他选项加入后，中位数均不等于30。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 17:53:54","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":"25","is_correct":0},{"id":"B","content":"30","is_correct":1},{"id":"C","content":"35","is_correct":0},{"id":"D","content":"40","is_correct":0}]},{"id":177,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"已知函数 $ f(x) = |x - 2| + |x + 3| $，若关于 $ x $ 的不等式 $ f(x) < a $ 有解，则实数 $ a $ 的取值范围是（ ）","answer":"A","explanation":"本题考查绝对值函数的性质与不等式有解问题。函数 $ f(x) = |x - 2| + |x + 3| $ 表示数轴上点 $ x $ 到点 2 和点 -3 的距离之和。根据绝对值几何意义，当 $ x $ 在区间 $[-3, 2]$ 内时，该距离和最小，最小值为 $ |2 - (-3)| = 5 $。因此，$ f(x) $ 的最小值为 5，即 $ f(x) \\geq 5 $ 对所有实数 $ x $ 成立。要使不等式 $ f(x) < a $ 有解，必须存在某个 $ x $ 使得 $ f(x) < a $，这就要求 $ a $ 必须大于 $ f(x) $ 的最小值 5。若 $ a = 5 $，则 $ f(x) < 5 $ 无解，因为 $ f(x) \\geq 5 $；只有当 $ a > 5 $ 时，才能找到某些 $ x $ 使得 $ f(x) < a $。因此，实数 $ a $ 的取值范围是 $ a > 5 $。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2025-12-29 12:32:47","updated_at":"2025-12-29 12:32:47","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"$ a > 5 $","is_correct":1},{"id":"B","content":"$ a \\geq 5 $","is_correct":0},{"id":"C","content":"$ a > 0 $","is_correct":0},{"id":"D","content":"$ a \\geq 0 $","is_correct":0}]},{"id":2772,"subject":"历史","grade":"七年级","stage":"初中","type":"选择题","content":"在隋唐时期，中国与外部世界的交流日益频繁。某学生在查阅资料时发现，唐朝都城长安是当时世界上规模最大的城市之一，吸引了来自不同国家的人在此居住和经商。以下哪一项最能体现唐朝对外交流的开放性和包容性？","answer":"A","explanation":"本题考查学生对唐朝对外交流特点的理解。唐朝是中国历史上对外开放程度较高的朝代，长安作为国际大都市，汇聚了来自中亚、西亚乃至欧洲的人员和商品。鸿胪寺是唐朝负责接待外宾的官方机构，而波斯（今伊朗）、大食（阿拉伯帝国）商人活跃于长安，正体现了唐朝对外来文化的接纳与包容。选项B、C、D所述内容均与史实不符：唐朝并未限制外国人活动，反而鼓励通商；佛教在唐朝得到广泛传播和发展；唐朝也与多国保持友好往来，如与日本的遣唐使交流频繁。因此，正确答案为A。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-12 10:41:20","updated_at":"2026-01-12 10:41:20","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":"长安城内设有专门接待外国使节的鸿胪寺，并有来自波斯、大食等地的商人开设店铺","is_correct":1},{"id":"B","content":"唐朝政府严格限制外国人在中国境内活动，只允许他们在边境进行贸易","is_correct":0},{"id":"C","content":"唐朝禁止佛教传播，以维护本土文化的纯粹性","is_correct":0},{"id":"D","content":"唐朝实行闭关锁国政策，拒绝与任何外国建立外交关系","is_correct":0}]},{"id":324,"subject":"数学","grade":"初一","stage":"小学","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，记录了以下数据：2，3，5，3，4，3，6。这组数据的众数是多少？","answer":"B","explanation":"众数是一组数据中出现次数最多的数。观察数据：2，3，5，3，4，3，6。其中数字2出现1次，3出现3次，4出现1次，5出现1次，6出现1次。因此，出现次数最多的是3，共出现3次。所以这组数据的众数是3。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:38:21","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":0},{"id":"B","content":"3","is_correct":1},{"id":"C","content":"4","is_correct":0},{"id":"D","content":"5","is_correct":0}]},{"id":2247,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某学生在一次数学实践活动中，记录了一周内某城市每日的气温变化情况。规定：气温上升记为正，下降记为负。已知这七天的气温变化依次为：+3℃，-2℃，+5℃，-4℃，+1℃，-6℃，+2℃。若第一天的起始气温为-1℃，请回答以下问题：经过这七天的连续变化后，最终气温是多少摄氏度？并判断最终气温比起始气温是升高了还是降低了，变化了多少摄氏度？","answer":"最终气温是-2℃，比起始气温降低了1℃。","explanation":"本题综合考查正负数在连续变化中的加减运算，要求学生理解正负数表示相反意义的量，并能进行多步有理数加法运算。题目设置了真实情境（气温变化），避免机械计算，强调过程推理。通过逐日累加变化量，最终得出结果，并比较起始与结束状态的差异，体现了正负数在实际问题中的应用，符合七年级课程标准中‘有理数运算’与‘实际问题建模’的要求。","solution_steps":"1. 起始气温为-1℃。\n2. 第一天变化：-1 + (+3) = 2℃\n3. 第二天变化：2 + (-2) = 0℃\n4. 第三天变化：0 + (+5) = 5℃\n5. 第四天变化：5 + (-4) = 1℃\n6. 第五天变化：1 + (+1) = 2℃\n7. 第六天变化：2 + (-6) = -4℃\n8. 第七天变化：-4 + (+2) = -2℃\n9. 最终气温为-2℃。\n10. 比起始气温-1℃的变化量：-2 - (-1) = -1℃，即降低了1℃。","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-09 14:44:04","updated_at":"2026-01-09 14:44:04","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":2187,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在数轴上标记了三个有理数：-2.5、1 和 0.75。若将这三个数按从小到大的顺序排列，并计算相邻两个数之间的差值之和（即最大数减中间数，加上中间数减最小数），最终结果是多少？","answer":"B","explanation":"首先将三个有理数按从小到大的顺序排列：-2.5 < 0.75 < 1。计算相邻两个数之间的差值之和：(0.75 - (-2.5)) + (1 - 0.75) = (0.75 + 2.5) + 0.25 = 3.25 + 0.25 = 3.5。因此正确答案是B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-09 14:21:04","updated_at":"2026-01-09 14:21:04","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":"3.25","is_correct":0},{"id":"B","content":"3.5","is_correct":1},{"id":"C","content":"3.75","is_correct":0},{"id":"D","content":"4.0","is_correct":0}]}]