习题练习

[{"id":583,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"9","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 20:11:43","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":175,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"小明买了3支铅笔和2本笔记本，共花费18元。已知每支铅笔的价格是2元，那么每本笔记本的价格是多少元？","answer":"A","explanation":"首先计算3支铅笔的总价格：3 × 2 = 6（元）。小明总共花费18元，因此2本笔记本的价格为：18 - 6 = 12（元）。那么每本笔记本的价格是：12 ÷ 2 = 6（元）。所以正确答案是A。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 12:29:21","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":"6元","is_correct":1},{"id":"B","content":"5元","is_correct":0},{"id":"C","content":"4元","is_correct":0},{"id":"D","content":"3元","is_correct":0}]},{"id":1910,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某班级组织植树活动，计划将一批树苗平均分给若干小组。如果每组分配5棵树苗，则剩余3棵；如果每组分配6棵树苗，则最后一组不足3棵但至少有1棵。已知小组数量为整数，且树苗总数不超过50棵，则该班级最多可能有多少个小组？","answer":"B","explanation":"设小组数量为x（x为正整数），树苗总数为y。根据题意：\n\n1. 每组5棵，剩3棵：y = 5x + 3；\n2. 每组6棵时，最后一组不足3棵但至少有1棵，说明前(x−1)组每组6棵，最后一组有1、2棵，即：\n 6(x−1) + 1 ≤ y < 6(x−1) + 3\n 化简得：6x − 5 ≤ y < 6x − 3\n\n将y = 5x + 3代入不等式：\n6x − 5 ≤ 5x + 3 < 6x − 3\n\n解左边：6x − 5 ≤ 5x + 3 → x ≤ 8\n解右边：5x + 3 < 6x − 3 → 3 + 3 < x → x > 6\n\n所以x的取值范围是：6 < x ≤ 8，即x = 7 或 8\n\n又因为树苗总数不超过50棵：y = 5x + 3 ≤ 50 → 5x ≤ 47 → x ≤ 9.4，满足x=7和x=8\n\n当x=8时，y = 5×8 + 3 = 43\n验证第二种分法：前7组每组6棵，共42棵，最后一组43−42=1棵，符合“不足3棵但至少有1棵”\n\n当x=9时，y=48，但6×8 + 3 = 51 > 48，不满足y < 6x−3（即48 < 51成立），但检查分配：前8组48棵，最后一组0棵，不符合“至少有1棵”，故x=9不成立\n\n因此，满足所有条件的最大x为8。\n\n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 13:11:51","updated_at":"2026-01-07 13:11:51","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":"7个","is_correct":0},{"id":"B","content":"8个","is_correct":1},{"id":"C","content":"9个","is_correct":0},{"id":"D","content":"10个","is_correct":0}]},{"id":610,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生在整理班级同学的课外阅读时间时，记录了5位同学每周阅读的小时数分别为：3，5，4，6，2。如果老师要求每位同学的阅读时间都增加相同的整数小时，使得新的数据中位数变为5，那么每位同学至少需要增加多少小时？","answer":"A","explanation":"原始数据为：3，5，4，6，2。先将数据从小到大排序：2，3，4，5，6。当前中位数是中间的数，即4。设每位同学增加x小时（x为正整数），则新数据为：2+x，3+x，4+x，5+x，6+x。排序后仍为：2+x，3+x，4+x，5+x，6+x，中位数是4+x。要求中位数为5，即4 + x = 5，解得x = 1。因此，每位同学至少需要增加1小时。验证：增加1小时后数据为3，4，5，6，7，排序后中位数为5，符合条件。故正确答案为A。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 21:36:13","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","is_correct":1},{"id":"B","content":"2","is_correct":0},{"id":"C","content":"3","is_correct":0},{"id":"D","content":"4","is_correct":0}]},{"id":820,"subject":"数学","grade":"初一","stage":"小学","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和不可回收垃圾共30袋。已知可回收垃圾每袋重2千克，不可回收垃圾每袋重1.5千克，这些垃圾总重量为54千克。设可回收垃圾有x袋，则根据题意可列出一元一次方程：2x + 1.5(______) = 54。","answer":"30 - x","explanation":"题目中已知垃圾总袋数为30袋，可回收垃圾有x袋，则不可回收垃圾的袋数就是总袋数减去可回收袋数，即30 - x袋。因此，在列方程时，不可回收垃圾的总重量应为1.5乘以(30 - x)。所以空白处应填30 - x。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 00:37:52","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":795,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生在整理班级同学的课外阅读书籍数量时，制作了频数分布表。已知阅读书籍数量为3本的学生有5人，4本的有8人，5本的有7人，其余学生均阅读2本。若全班共有30名学生，则阅读2本书的学生有___人。","answer":"10","explanation":"根据题意，全班共有30名学生。已知阅读3本、4本、5本书的学生人数分别为5人、8人、7人，合计为5 + 8 + 7 = 20人。因此，阅读2本书的学生人数为总人数减去已知人数：30 - 20 = 10人。本题考查数据的收集与整理，属于简单难度的计算题。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 00:14:15","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":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":1084,"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最喜欢的运动项目调查数据时，共收集了60份有效问卷。其中喜欢篮球的人数占总人数的$\\frac{1}{3}$，喜欢足球的人数是喜欢篮球人数的$\\frac{1}{2}$，其余同学喜欢羽毛球。那么喜欢羽毛球的同学有___人。","answer":"30","explanation":"总人数为60人。喜欢篮球的人数为60 × $\\frac{1}{3}$ = 20人。喜欢足球的人数是篮球人数的$\\frac{1}{2}$，即20 × $\\frac{1}{2}$ = 10人。因此，喜欢羽毛球的人数为60 - 20 - 10 = 30人。本题考查了数据的收集与整理，以及有理数的乘法与加减运算，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":"2026-01-06 08:54:27","updated_at":"2026-01-06 08:54:27","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":2475,"subject":"数学","grade":"八年级","stage":"初中","type":"解答题","content":"如图，在平面直角坐标系中，点 A(0, 4)、B(3, 0)、C(0, 0) 构成直角三角形 ABC，∠C = 90°。将 △ABC 沿直线 l 折叠，使得点 A 落在 x 轴上的点 A′ 处，且 A′ 位于点 B 的右侧。已知折叠后的折痕 l 与边 AB 相交于点 D，与边 AC 相交于点 E。若折痕 l 是线段 AA′ 的垂直平分线，且四边形 ADEC 的面积为 6，求折痕 l 的长度。","answer":"待完善","explanation":"解析待完善","solution_steps":"待完善","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 14:53:41","updated_at":"2026-01-10 14:53:41","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":966,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"某学生测量了学校花坛中5株向日葵的高度（单位：厘米），分别为：82，75，90，78，_85_。如果这5株向日葵的平均高度是82厘米，那么被遮盖的那个数据应该是多少？","answer":"85","explanation":"已知5株向日葵的平均高度是82厘米，因此总高度为 5 × 82 = 410 厘米。已知的四个高度分别是82、75、90、78，它们的和为 82 + 75 + 90 + 78 = 325 厘米。所以被遮盖的数据为 410 - 325 = 85 厘米。本题考查数据的收集与整理中的平均数计算，属于简单难度，符合七年级‘数据的收集、整理与描述’知识点。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 04:03:03","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":[]}]