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":361,"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身高数据时,发现一组数据的最小值是148厘米,最大值是172厘米。若将这组数据分为5组,则每组的组距最接近多少厘米?","answer":"B","explanation":"首先计算极差:最大值减去最小值,即172 - 148 = 24厘米。要将数据分为5组,则组距 = 极差 ÷ 组数 = 24 ÷ 5 = 4.8厘米。由于组距通常取整数,且要覆盖整个数据范围,因此应向上取整为5厘米。若取4厘米,则5组只能覆盖20厘米(5×4),不足以包含24厘米的极差;而5厘米可以覆盖25厘米,满足要求。因此最接近且合理的组距是5厘米。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:45:34","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":"4厘米","is_correct":0},{"id":"B","content":"5厘米","is_correct":1},{"id":"C","content":"6厘米","is_correct":0},{"id":"D","content":"7厘米","is_correct":0}]},{"id":805,"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情况时,收集了每位同学每月阅读的书籍数量,并将数据按从小到大的顺序排列。已知这组数据的中位数是4,且数据个数为奇数。如果去掉最大的一个数据后,新的中位数变为3.5,那么原数据中最少有多少个数据?____","answer":"7","explanation":"设原数据有n个,且n为奇数。中位数为第(n+1)\/2个数,已知为4。去掉最大的一个数据后,剩下n-1个数据(偶数个),中位数为中间两个数的平均数,即第(n-1)\/2个和第(n+1)\/2个数据的平均值为3.5。由于原数据有序,去掉最大值后,中间两个数应分别为3和4(因为(3+4)\/2=3.5)。为了使这种情况成立,原数据中第(n+1)\/2个数必须是4,且其前一个数为3。当n=7时,原数据第4个数为4,去掉最大值后剩下6个数,第3和第4个数分别为3和4,满足新中位数为3.5。若n<7(如n=5),则无法满足去掉最大值后中间两数为3和4的条件。因此原数据最少有7个。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 00:22:41","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":2229,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"某学生记录了连续三天的气温变化:第一天上升了5℃,第二天下降了3℃,第三天又下降了4℃。如果这三天的气温变化用正数和负数表示,则这三天的气温变化总和为____℃。","answer":"-2","explanation":"根据正负数的意义,气温上升用正数表示,下降用负数表示。因此,三天的气温变化分别为:+5℃、-3℃、-4℃。将它们相加:5 + (-3) + (-4) = 5 - 3 - 4 = -2。所以总和为-2℃。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 14:27:19","updated_at":"2026-01-09 14:27:19","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":475,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"某学生测量了班级10名同学的身高(单位:厘米),数据如下:152, 155, 148, 160, 158, 153, 157, 150, 156, 154。这组数据的众数是多少?","answer":"D","explanation":"众数是指一组数据中出现次数最多的数。观察给出的数据:152, 155, 148, 160, 158, 153, 157, 150, 156, 154,每个数值都只出现了一次,没有任何一个数重复出现。因此,这组数据中没有众数。正确答案是D。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:57:08","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":"152","is_correct":0},{"id":"B","content":"154","is_correct":0},{"id":"C","content":"155","is_correct":0},{"id":"D","content":"没有众数","is_correct":1}]},{"id":2047,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某公园计划修建一个菱形花坛,设计图纸上标注了两条对角线的长度分别为6米和8米。施工过程中,工人需要在外围铺设一圈装饰灯带,灯带必须沿着菱形的四条边铺设。已知每米灯带的成本为15元,则铺设完整圈灯带的总成本是多少元?","answer":"D","explanation":"本题考查菱形的性质与勾股定理的应用。菱形的两条对角线互相垂直且平分,因此可以将菱形分成四个全等的直角三角形。每条对角线的一半分别为3米和4米,根据勾股定理,菱形边长为√(3² + 4²) = √(9 + 16) = √25 = 5米。菱形周长为4 × 5 = 20米。每米灯带15元,总成本为20 × 15 = 300元。因此正确答案为D。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-09 10:49:58","updated_at":"2026-01-09 10:49:58","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":"120元","is_correct":0},{"id":"B","content":"150元","is_correct":0},{"id":"C","content":"180元","is_correct":0},{"id":"D","content":"300元","is_correct":1}]},{"id":1810,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某公园计划修建一个等腰三角形花坛,设计要求其底边长为6米,两腰相等且每腰长为5米。施工前需要计算该花坛的高,以便准备支撑材料。请问这个等腰三角形花坛的高是多少米?","answer":"B","explanation":"此题考查勾股定理在等腰三角形中的应用。等腰三角形底边上的高将底边平分为两段,每段长度为3米。由此可构造一个直角三角形,其中一条直角边为3米(底边的一半),斜边为5米(腰长),所求高为另一条直角边。根据勾股定理:高² = 5² - 3² = 25 - 9 = 16,因此高 = √16 = 4米。故正确答案为B。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-06 16:18:43","updated_at":"2026-01-06 16:18:43","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":"A","content":"3米","is_correct":0},{"id":"B","content":"4米","is_correct":1},{"id":"C","content":"5米","is_correct":0},{"id":"D","content":"6米","is_correct":0}]},{"id":131,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"一个长方形的长比宽多5厘米,若其周长为30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 + 5)厘米。根据长方形周长公式:周长 = 2 × (长 + 宽),代入得:2 × (x + x + 5) = 30,即2 × (2x + 5) = 30,化简得4x + 10 = 30,解得4x = 20,x = 5。因此,宽为5厘米。本题结合代数设未知数与一元一次方程求解,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 08:59:12","updated_at":"2025-12-24 08:59:12","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":2154,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在解一个关于一元一次方程的问题时,列出了方程 3(x - 2) = 2x + 1。该方程的解是下列哪一个?","answer":"B","explanation":"解方程 3(x - 2) = 2x + 1:首先去括号得 3x - 6 = 2x + 1,移项得 3x - 2x = 1 + 6,合并同类项得 x = 7。因此正确答案是 B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 13:00:46","updated_at":"2026-01-09 13:00:46","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":"x = 5","is_correct":0},{"id":"B","content":"x = 7","is_correct":1},{"id":"C","content":"x = -5","is_correct":0},{"id":"D","content":"x = -7","is_correct":0}]},{"id":640,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某班级组织了一次环保活动,收集废纸和塑料瓶。已知每千克废纸可兑换0.8元,每千克塑料瓶可兑换1.2元。一名学生共收集了15千克废品,兑换后获得16元。若设该学生收集的废纸为x千克,则根据题意可列出一元一次方程为:","answer":"A","explanation":"设收集的废纸为x千克,则塑料瓶为(15 - x)千克。废纸每千克兑换0.8元,总价值为0.8x元;塑料瓶每千克兑换1.2元,总价值为1.2(15 - x)元。两者之和等于16元,因此方程为0.8x + 1.2(15 - x) = 16。选项A正确。选项B错误地将两种废品都设为x千克;选项C颠倒了废纸和塑料瓶的对应关系;选项D使用了减法,不符合实际兑换逻辑。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 22:07:00","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":"0.8x + 1.2(15 - x) = 16","is_correct":1},{"id":"B","content":"0.8x + 1.2x = 16","is_correct":0},{"id":"C","content":"0.8(15 - x) + 1.2x = 16","is_correct":0},{"id":"D","content":"0.8x - 1.2(15 - x) = 16","is_correct":0}]},{"id":520,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某班级组织了一次环保知识竞赛,共收集了50份有效问卷。统计结果显示,有28人答对了第一题,有25人答对了第二题,有15人两道题都答对了。那么,两道题都没有答对的人数是多少?","answer":"A","explanation":"本题考查数据的收集、整理与描述中的集合思想应用。已知总人数为50人,答对第一题的有28人,答对第二题的有25人,两道题都答对的有15人。根据容斥原理,至少答对一道题的人数为:28 + 25 - 15 = 38人。因此,两道题都没有答对的人数为:50 - 38 = 12人。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 18:24:43","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":"A","content":"12","is_correct":1},{"id":"B","content":"13","is_correct":0},{"id":"C","content":"14","is_correct":0},{"id":"D","content":"15","is_correct":0}]}]