初中
数学
中等
来源: 教材例题
知识点: 初中数学
答案预览
点击下方'查看答案'按钮查看详细解析并跳转到题目详情页
直接前往详情页
练习完成!
恭喜您完成了本次练习,继续加油提升自己的知识水平!
学习建议
您在一元一次方程的应用方面掌握良好,但仍有提升空间。建议重点复习方程求解步骤和实际应用问题。
[{"id":727,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在某次班级大扫除中,学生们被分成若干小组清理教室。如果每组安排5人,则多出3人;如果每组安排6人,则最后一组只有4人。这个班级共有___名学生。","answer":"28","explanation":"设班级共有x名学生。根据题意,当每组5人时,多出3人,说明x除以5余3,即x = 5a + 3(a为组数)。当每组6人时,最后一组只有4人,说明x除以6余4,即x = 6b + 4(b为组数)。寻找同时满足这两个条件的最小正整数。尝试代入:当x=28时,28 ÷ 5 = 5组余3,符合第一种情况;28 ÷ 6 = 4组余4,也符合第二种情况。因此,班级共有28名学生。本题考查一元一次方程的实际应用与整数解问题,属于简单难度。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 23:02:09","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":136,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"一个长方形的长比宽多3厘米,若其周长为26厘米,则这个长方形的宽是____厘米。","answer":"5","explanation":"设长方形的宽为x厘米,则长为(x + 3)厘米。根据长方形周长公式:周长 = 2 × (长 + 宽),代入得:2 × (x + x + 3) = 26,化简为2 × (2x + 3) = 26,即4x + 6 = 26。解得4x = 20,x = 5。因此,宽为5厘米。本题考查一元一次方程在几何问题中的简单应用,符合初一学生对方程和几何基础的学习要求。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-24 09:40:59","updated_at":"2025-12-24 09:40:59","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":1875,"subject":"语文","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在整理班级同学的课外阅读情况时,收集了以下数据:全班40人,每人每周阅读时间(单位:小时)分布在区间[1, 10]内,且均为整数。他将数据分为5组,每组8人,并计算出每组的平均阅读时间分别为:3.5、4.25、5.0、6.75、8.0。若该学生想用这些数据绘制一个频数分布直方图,并发现其中某一组的实际总阅读时间比按平均数估算的总时间多出2小时,则该组最可能是哪一组?","answer":"C","explanation":"本题考查数据的收集、整理与描述,以及对平均数与总和关系的理解。每组有8人,因此按平均数估算的总阅读时间 = 平均数 × 8。实际总时间比估算多出2小时,说明该组的实际总和 = 平均数 × 8 + 2。由于每人阅读时间为整数,总时间也必为整数。我们逐项分析:A组:3.5 × 8 = 28,+2 = 30(整数,可能);B组:4.25 × 8 = 34,+2 = 36(整数,可能);C组:6.75 × 8 = 54,+2 = 56(整数,可能);D组:8.0 × 8 = 64,+2 = 66(整数,可能)。但关键在于“平均数为6.75”意味着总和为54,而54 ÷ 8 = 6.75,说明原始数据总和为54。若实际多出2小时,则总和为56,平均为7.0。但题目说“按平均数估算”是基于报告的6.75,而实际更高,说明原始分组数据可能被低估。然而,6.75 = 27\/4,说明总和54是3的倍数,而56不是8的倍数导致平均变为7,这在整数数据中是可能的。但更关键的是,6.75是唯一一个非半整数的平均数(3.5、4.25、5.0、8.0均为0.25的倍数,但6.75也符合),但结合“多出2小时”这一异常,最可能出现在中间偏高组,因为极端组(如3.5或8.0)数据分布受限,而6.75组处于中间偏上,数据波动空间大,更容易出现统计偏差。综合分析,C组最可能因数据分布不均导致估算偏差,故选C。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-07 09:54:14","updated_at":"2026-01-07 09:54:14","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":"平均阅读时间为3.5小时的一组","is_correct":0},{"id":"B","content":"平均阅读时间为4.25小时的一组","is_correct":0},{"id":"C","content":"平均阅读时间为6.75小时的一组","is_correct":1},{"id":"D","content":"平均阅读时间为8.0小时的一组","is_correct":0}]},{"id":1249,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某学生在研究平面直角坐标系中的几何问题时,发现一个有趣的规律:若将一个点P(x, y)先向右平移3个单位,再向上平移2个单位,得到点P';然后将点P'绕原点逆时针旋转90°,得到点P''。已知点P''的坐标为(-5, 4),求原点P的坐标(x, y)。此外,若该点P满足不等式组:2x - y > 1 且 x + 3y ≤ 10,请验证所求得的点P是否满足该不等式组。","answer":"解:\n\n第一步:设原点P的坐标为(x, y)。\n\n根据题意,点P先向右平移3个单位,再向上平移2个单位,得到点P'。\n平移变换规则:向右平移a个单位,横坐标加a;向上平移b个单位,纵坐标加b。\n因此,P'的坐标为:\n P' = (x + 3, y + 2)\n\n第二步:将点P'绕原点逆时针旋转90°,得到点P''。\n旋转90°逆时针的坐标变换公式为:\n 若点A(a, b)绕原点逆时针旋转90°,则新坐标为(-b, a)\n\n对P'(x + 3, y + 2)应用该公式:\nP'' = (-(y + 2), x + 3) = (-y - 2, x + 3)\n\n题目已知P''的坐标为(-5, 4),因此列出方程组:\n -y - 2 = -5\n x + 3 = 4\n\n解第一个方程:\n -y - 2 = -5\n → -y = -3\n → y = 3\n\n解第二个方程:\n x + 3 = 4\n → x = 1\n\n所以,原点P的坐标为(1, 3)。\n\n第三步:验证点P(1, 3)是否满足不等式组:\n 2x - y > 1\n x + 3y ≤ 10\n\n代入x = 1,y = 3:\n\n第一式:2(1) - 3 = 2 - 3 = -1\n -1 > 1? 不成立。\n\n第二式:1 + 3×3 = 1 + 9 = 10\n 10 ≤ 10? 成立。\n\n由于第一式不满足,因此点P(1, 3)不满足整个不等式组。\n\n最终答案:\n点P的坐标为(1, 3),但该点不满足给定的不等式组。","explanation":"本题综合考查了平面直角坐标系中的平移变换、旋转变换、二元一次方程组的建立与求解,以及不等式组的验证。解题关键在于掌握坐标变换的代数表示:平移是坐标的加减,旋转90°逆时针的公式为(a, b) → (-b, a)。通过逆向推理,从P''的坐标反推出P',再反推出P。最后将所得坐标代入不等式组进行验证,体现了数形结合与逻辑推理能力。题目设计新颖,融合了多个知识点,要求学生具备较强的综合运用能力,符合困难难度要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 10:31:09","updated_at":"2026-01-06 10:31:09","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":446,"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,收集了10名同学每天阅读的分钟数:25,30,35,30,40,35,30,45,35,30。如果将这些数据按从小到大的顺序排列,那么位于中间两个数的平均数是多少?","answer":"B","explanation":"首先将数据从小到大排序:25,30,30,30,30,35,35,35,40,45。共有10个数据(偶数个),因此中位数是中间两个数的平均数,即第5个和第6个数的平均值。第5个数是30,第6个数是35,所以中位数为 (30 + 35) ÷ 2 = 65 ÷ 2 = 32.5。本题考查数据的整理与描述中的中位数概念,属于简单难度,符合七年级数学课程标准要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:44:01","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":"30","is_correct":0},{"id":"B","content":"32.5","is_correct":1},{"id":"C","content":"35","is_correct":0},{"id":"D","content":"37.5","is_correct":0}]},{"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":172,"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两本笔记本,共花费18元。已知每本笔记本的价格是5元,那么这支钢笔的价格是多少元?","answer":"A","explanation":"根据题意,两本笔记本的总价为:2 × 5 = 10(元)。小明总共花费18元,因此钢笔的价格为:18 - 10 = 8(元)。所以正确答案是A选项。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 12:29:10","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":"8元","is_correct":1},{"id":"B","content":"10元","is_correct":0},{"id":"C","content":"13元","is_correct":0},{"id":"D","content":"15元","is_correct":0}]},{"id":2404,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某校八年级开展了一次数学实践活动,要求学生测量校园内一个不规则四边形花坛ABCD的边长与角度。已知AB = 5 m,BC = 12 m,CD = 9 m,DA = 8 m,且对角线AC将四边形分成两个直角三角形△ABC和△ADC,其中∠ABC = 90°,∠ADC = 90°。若一名学生想计算该花坛的面积,以下哪个选项是正确的?","answer":"A","explanation":"题目中给出四边形ABCD被对角线AC分成两个直角三角形:△ABC和△ADC,且∠ABC = 90°,∠ADC = 90°。因此,可以分别计算两个直角三角形的面积,再相加得到整个四边形的面积。\n\n在△ABC中,AB = 5 m,BC = 12 m,∠ABC = 90°,所以面积为:\n(1\/2) × AB × BC = (1\/2) × 5 × 12 = 30 m²。\n\n在△ADC中,AD = 8 m,DC = 9 m,∠ADC = 90°,所以面积为:\n(1\/2) × AD × DC = (1\/2) × 8 × 9 = 36 m²。\n\n因此,花坛总面积为:30 + 36 = 66 m²。\n\n本题综合考查了勾股定理的应用背景(直角三角形识别)、三角形面积计算以及实际问题中的几何建模能力,符合八年级学生知识水平。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 12:09:17","updated_at":"2026-01-10 12:09:17","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":"A","content":"66 m²","is_correct":1},{"id":"B","content":"72 m²","is_correct":0},{"id":"C","content":"78 m²","is_correct":0},{"id":"D","content":"84 m²","is_correct":0}]},{"id":2498,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"某学生设计了一个圆形花坛,其外围铺设了一条宽度均匀的环形步道。已知花坛的半径为3米,整个花坛与步道合起来的总面积为25π平方米。若设步道宽度为x米,则可列出一元二次方程求解x。根据题意,下列方程正确的是:","answer":"A","explanation":"花坛半径为3米,步道宽度为x米,且步道均匀围绕花坛一周,因此整个结构(花坛+步道)的外圆半径为3 + x米。整个区域的总面积为外圆面积,即π(3 + x)²。题目给出总面积为25π平方米,因此可列出方程:π(3 + x)² = 25π。两边同时除以π,得(3 + x)² = 25,解得x = 2(舍去负值)。选项A正确反映了这一关系。选项B错误地将直径增加当作半径增加;选项C是展开后的形式但未体现几何意义;选项D表示半径减小,与题意不符。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 15:19:44","updated_at":"2026-01-10 15:19:44","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":"π(3 + x)² = 25π","is_correct":1},{"id":"B","content":"π(3 + 2x)² = 25π","is_correct":0},{"id":"C","content":"πx² + 6πx = 25π","is_correct":0},{"id":"D","content":"π(3 - x)² = 25π","is_correct":0}]},{"id":974,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生测量了学校花坛一周的温度变化,记录了连续5天的最高温度分别为:23℃、25℃、24℃、26℃、22℃。这5天最高温度的平均值是______℃。","answer":"24","explanation":"求平均数的方法是将所有数据相加,再除以数据的个数。计算过程为:(23 + 25 + 24 + 26 + 22) ÷ 5 = 120 ÷ 5 = 24。因此,这5天最高温度的平均值是24℃。本题考查的是数据的收集、整理与描述中的平均数计算,属于七年级数学简单难度内容。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 04:11:53","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":[]}]