习题练习

[{"id":1917,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某班级进行了一次数学测验，成绩分布如下表所示。若将成绩分为“优秀”（90分及以上）、“良好”（75～89分）、“及格”（60～74分）和“不及格”（60分以下）四个等级，则成绩为“良好”的学生人数占总人数的百分比是多少？\n\n| 分数段 | 人数 |\n|--------------|------|\n| 90～100 | 8 |\n| 75～89 | 12 |\n| 60～74 | 6 |\n| 60以下 | 4 |","answer":"B","explanation":"首先计算总人数：8 + 12 + 6 + 4 = 30（人）。成绩为“良好”（75～89分）的学生有12人。因此，“良好”等级所占百分比为：(12 ÷ 30) × 100% = 40%。故正确答案为B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-07 13:13:10","updated_at":"2026-01-07 13:13:10","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":"30%","is_correct":0},{"id":"B","content":"40%","is_correct":1},{"id":"C","content":"50%","is_correct":0},{"id":"D","content":"60%","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":[]},{"id":655,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次环保主题活动中，某学生记录了连续5天每天节约用水的升数，分别为：3.5升、4.2升、3.8升、4.0升、3.6升。这5天平均每天节约用水______升。","answer":"3.82","explanation":"要计算平均每天节约用水的升数，需将5天的用水量相加后除以天数。计算过程为：(3.5 + 4.2 + 3.8 + 4.0 + 3.6) ÷ 5 = 19.1 ÷ 5 = 3.82（升）。本题考查的是数据的收集、整理与描述中的平均数计算，属于七年级数学中数据处理的基础知识，难度为简单。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 22:13:04","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":2021,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某学生在整理班级数学测验成绩时，发现一组数据的平均数为85分，后来发现漏记了一个成绩90分。将这个成绩加入后，新的平均数变为85.5分。请问原来这组数据共有多少个成绩？","answer":"A","explanation":"设原来有n个成绩，则原来总分是85n。加入90分后，总人数变为n+1，总分变为85n + 90，新的平均数为85.5。根据平均数公式列出方程：(85n + 90) \/ (n + 1) = 85.5。两边同乘(n + 1)得：85n + 90 = 85.5(n + 1) = 85.5n + 85.5。移项整理：85n - 85.5n = 85.5 - 90 → -0.5n = -4.5 → n = 9。因此原来有9个成绩，正确答案是A。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 10:31:38","updated_at":"2026-01-09 10:31:38","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":"A","content":"9","is_correct":1},{"id":"B","content":"10","is_correct":0},{"id":"C","content":"11","is_correct":0},{"id":"D","content":"12","is_correct":0}]},{"id":1065,"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，某学生收集了可回收垃圾的重量为 (3x - 2) 千克，其他同学共收集了 (x + 5) 千克。若全班总共收集了 20 千克可回收垃圾，则 x 的值是___。","answer":"17\/4","explanation":"根据题意，某学生收集的垃圾重量为 (3x - 2) 千克，其他同学收集了 (x + 5) 千克，全班总重量为 20 千克。可列方程：(3x - 2) + (x + 5) = 20。合并同类项得：4x + 3 = 20。移项得：4x = 17，解得 x = 17\/4。该题考查整式的加减与一元一次方程的综合应用，符合七年级数学知识范围。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-06 08:52:17","updated_at":"2026-01-06 08:52:17","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":890,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次班级环保活动中，某学生收集了12.5千克废纸，另一名同学收集的废纸比这名学生多3.7千克，两人一共收集了___千克废纸。","answer":"28.7","explanation":"首先，第二名同学收集的废纸重量为12.5 + 3.7 = 16.2千克。然后将两人收集的废纸重量相加：12.5 + 16.2 = 28.7千克。因此，两人一共收集了28.7千克废纸。本题考查的是有理数的加法运算，属于简单难度，符合七年级有理数知识点要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 02:08:03","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":1977,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"某学生在纸上画了一个矩形，其长为8 cm，宽为6 cm。若以该矩形的一个顶点为旋转中心，将矩形绕此点顺时针旋转90°，则旋转后原对角线所扫过的区域面积最接近以下哪个值？（π取3.14）","answer":"A","explanation":"本题考查旋转与圆的综合应用。矩形对角线长度为√(8² + 6²) = √(64 + 36) = √100 = 10 cm。以某一顶点为旋转中心旋转90°，对角线的另一端点将绕该中心作半径为10 cm的圆弧运动，扫过的区域是一个半径为10 cm、圆心角为90°的扇形。扇形面积为 (90°\/360°) × π × 10² = (1\/4) × 3.14 × 100 = 78.5 cm²。因此，对角线扫过的区域面积最接近78.5 cm²。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-07 15:00:36","updated_at":"2026-01-07 15:00:36","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":"78.5 cm²","is_correct":1},{"id":"B","content":"50.2 cm²","is_correct":0},{"id":"C","content":"113.0 cm²","is_correct":0},{"id":"D","content":"25.1 cm²","is_correct":0}]},{"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":1099,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"在一次班级图书捐赠活动中，某学生捐出的图书数量比班级平均每人捐书数量的2倍还多3本。如果班级共有30名学生，总共捐了150本书，那么这名学生捐了___本书。","answer":"13","explanation":"首先根据题意，班级共有30名学生，总共捐了150本书，因此平均每人捐书数量为150 ÷ 30 = 5本。题目中说某学生捐出的图书数量比平均每人捐书数量的2倍还多3本，即2 × 5 + 3 = 10 + 3 = 13本。因此，这名学生捐了13本书。本题考查了有理数的四则运算和一元一次方程的基本思想，通过平均数建立数量关系，适合七年级学生水平。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-06 08:57:21","updated_at":"2026-01-06 08:57:21","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":1368,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市地铁线路规划中，需确定两个站点A和B之间的最短运行时间。已知列车在平直轨道上的平均速度为每小时60千米，但在弯道处需减速至每小时40千米。线路设计图显示，从A站到B站的总轨道长度为12千米，其中包含一段弯道。若列车全程运行时间不超过15分钟，且弯道长度至少为2千米，试求弯道长度的可能取值范围。假设列车在直道和弯道上均以恒定速度行驶，且不考虑停站和加减速时间。","answer":"解：\n设弯道长度为x千米，则直道长度为(12 - x)千米。\n根据题意，弯道长度至少为2千米，即：\nx ≥ 2。\n列车在弯道上的速度为40千米\/小时，行驶时间为：\n弯道时间 = x \/ 40 小时。\n列车在直道上的速度为60千米\/小时，行驶时间为：\n直道时间 = (12 - x) \/ 60 小时。\n总运行时间为两者之和，且不超过15分钟，即15\/60 = 0.25小时。\n因此，建立不等式：\nx \/ 40 + (12 - x) \/ 60 ≤ 0.25。\n为消去分母，两边同乘以120（40和60的最小公倍数）：\n120 × (x \/ 40) + 120 × ((12 - x) \/ 60) ≤ 120 × 0.25\n3x + 2(12 - x) ≤ 30\n3x + 24 - 2x ≤ 30\nx + 24 ≤ 30\nx ≤ 6\n结合弯道长度至少为2千米的条件，得：\n2 ≤ x ≤ 6\n因此，弯道长度的可能取值范围是大于等于2千米且小于等于6千米。\n答：弯道长度的取值范围是2千米到6千米（含端点）。","explanation":"本题综合考查了一元一次不等式的建立与求解，以及实际问题的数学建模能力。首先根据题意设定未知数x表示弯道长度，利用速度、时间与路程的关系分别表示直道和弯道的行驶时间，再根据总时间不超过15分钟（即0.25小时）建立不等式。通过通分消去分母，化简不等式得到x ≤ 6，再结合题设中弯道长度至少为2千米的条件，最终确定x的取值范围为2 ≤ x ≤ 6。解题过程中需注意单位统一（时间换算为小时），并合理运用不等式的性质进行变形。本题背景新颖，贴近现实，考查学生将实际问题转化为数学表达式的能力，属于困难难度的综合性应用题。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 11:11:20","updated_at":"2026-01-06 11:11:20","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":[]}]