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":986,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次班级环保活动中,某学生收集了可回收垃圾的重量记录如下:塑料瓶重0.35千克,废纸重0.48千克,易拉罐重0.27千克。他将这三类垃圾的总重量填入统计表时,发现表格中‘合计’一栏被污损,无法看清。请帮他计算出这三类垃圾的总重量是___千克。","answer":"1.10","explanation":"本题考查有理数的加法运算,属于简单难度。学生需要将三个小数相加:0.35 + 0.48 + 0.27。计算时注意小数点对齐,从低位逐位相加。0.35 + 0.48 = 0.83,0.83 + 0.27 = 1.10。因此,三类垃圾的总重量是1.10千克。题目结合环保情境,贴近生活,帮助学生理解有理数在现实中的应用。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 04:28:33","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":2178,"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 a、b、c,其中 a = -2.5,b 是 a 的相反数,c 是 b 与 1.5 的和。若将这三个数按从小到大的顺序排列,正确的是:","answer":"B","explanation":"首先,a = -2.5;b 是 a 的相反数,因此 b = 2.5;c 是 b 与 1.5 的和,即 c = 2.5 + 1.5 = 4。三个数分别为:a = -2.5,b = 2.5,c = 4。在数轴上,-2.5 < 2.5 < 4,因此从小到大的顺序是 a < b < c,对应选项 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":"a < c < b","is_correct":0},{"id":"B","content":"a < b < c","is_correct":1},{"id":"C","content":"c < a < b","is_correct":0},{"id":"D","content":"b < c < a","is_correct":0}]},{"id":2778,"subject":"通用","grade":"高三","stage":"高中","type":"选择题","content":"高三示例题目","answer":"示例答案","explanation":"示例解析","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"中等","points":1,"is_active":1,"created_at":"2026-04-08 11:40:44","updated_at":"2026-04-08 11:40:44","sort_order":1001,"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":2529,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"如图,一个圆形花坛被三条等距的半径分成三个扇形区域,分别种植不同花卉。若在花坛边缘随机抛掷一粒石子,落在任意一个扇形区域的概率相等。现将整个花坛绕圆心顺时针旋转60°,此时原位于正北方向的标记点A移动到了点B的位置。若点B恰好落在其中一个扇形区域的边界上,则这个旋转后的图形与原图形重合部分所对应的圆心角是多少度?","answer":"C","explanation":"花坛被三条等距半径分成三个扇形,说明每个扇形的圆心角为360° ÷ 3 = 120°。旋转60°后,原标记点A移动到点B,而点B落在某个扇形边界上,说明旋转角度60°正好是两个相邻半径夹角(120°)的一半。由于图形具有120°的旋转对称性,旋转60°后,原图形与旋转后图形的重合部分由两个相邻扇形重叠构成。通过几何分析可知,重合部分的圆心角为120°,即一个完整扇形的角度。因此,正确答案为C。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 16:15:35","updated_at":"2026-01-10 16:15:35","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":"60°","is_correct":0},{"id":"B","content":"90°","is_correct":0},{"id":"C","content":"120°","is_correct":1},{"id":"D","content":"180°","is_correct":0}]},{"id":478,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"周一","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 17:58:00","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":705,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生测量了教室中5张课桌的高度(单位:厘米),记录如下:75,76,74,75,75。这组数据的众数是____。","answer":"75","explanation":"众数是一组数据中出现次数最多的数。在这组数据75,76,74,75,75中,75出现了3次,76和74各出现1次,因此众数是75。本题考查数据的收集、整理与描述中的基本概念,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 22:44:20","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":407,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生记录了连续5天的气温变化情况,每天的最高气温分别为:12℃、15℃、13℃、16℃、14℃。为了分析气温的波动情况,该学生计算了这组数据的极差。请问这组数据的极差是多少?","answer":"C","explanation":"极差是一组数据中最大值与最小值之差。题目中给出的5天气温数据为:12℃、15℃、13℃、16℃、14℃。其中最高气温是16℃,最低气温是12℃。因此,极差 = 16 - 12 = 4℃。故正确答案为C。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:27:16","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":0},{"id":"C","content":"4℃","is_correct":1},{"id":"D","content":"5℃","is_correct":0}]},{"id":2447,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"在一次校园测量活动中,某学生利用旗杆和其影子的长度关系来估算旗杆的高度。已知旗杆在地面的影子长度为6米,同时一根1.5米高的标杆竖直立于地面,其影子长度为2米。若旗杆与标杆均垂直于地面,且阳光照射角度相同,则旗杆的实际高度为多少米?","answer":"A","explanation":"本题考查相似三角形在实际问题中的应用,属于勾股定理与比例关系的综合应用。由于阳光照射角度相同,旗杆与其影子、标杆与其影子分别构成的两个直角三角形是相似三角形。根据相似三角形对应边成比例的性质,设旗杆高度为h米,则有:标杆高度 \/ 标杆影长 = 旗杆高度 \/ 旗杆影长,即 1.5 \/ 2 = h \/ 6。解这个比例式:1.5 × 6 = 2h → 9 = 2h → h = 4.5。因此,旗杆的实际高度为4.5米,正确答案为A。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 13:45:03","updated_at":"2026-01-10 13:45:03","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":"4.5","is_correct":1},{"id":"B","content":"5","is_correct":0},{"id":"C","content":"4","is_correct":0},{"id":"D","content":"3.75","is_correct":0}]},{"id":563,"subject":"数学","grade":"初一","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。已知成绩在80分及以上的人数占总人数的一半,且60分以下的人数比90分以上的人数多2人。如果全班共有40名学生,那么成绩在60分到79分之间的学生有多少人?","answer":"B","explanation":"设成绩在90分以上的人数为x,则60分以下的人数为x + 2。根据题意,80分及以上的人数占总人数的一半,即40 ÷ 2 = 20人。80分及以上包括80-89分和90分以上两部分,因此80-89分的人数为20 - x。全班总人数为40人,所以各分数段人数之和为:60分以下 + 60-79分 + 80-89分 + 90分以上 = 40。代入得:(x + 2) + y + (20 - x) + x = 40,其中y为60-79分的人数。化简得:x + 2 + y + 20 - x + x = 40 → y + x + 22 = 40 → y = 18 - x。又因为80分及以上共20人,其中90分以上为x人,所以x ≤ 20。同时60分以下为x + 2,必须为非负整数,且总人数合理。尝试代入合理值:若x = 4,则60分以下 = 6人,80-89分 = 16人,90分以上 = 4人,此时60-79分人数y = 40 - (6 + 16 + 4) = 14人。验证:80分及以上 = 16 + 4 = 20人,符合条件;60分以下6人比90分以上4人多2人,也符合。因此答案为14人。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 19:27:01","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":"12人","is_correct":0},{"id":"B","content":"14人","is_correct":1},{"id":"C","content":"16人","is_correct":0},{"id":"D","content":"18人","is_correct":0}]},{"id":1890,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某校七年级开展‘节约用水’主题调查活动,随机抽取了50名学生记录一周内每天的用水量(单位:升),并将数据整理成频数分布表。已知用水量在10~15升(含10升,不含15升)的学生人数占总人数的24%,用水量在15~20升的学生比用水量在5~10升的学生多6人,而用水量在20~25升的人数是用水量在5~10升人数的2倍。若用水量在5~10升的学生有x人,则根据以上信息可列方程为:","answer":"A","explanation":"根据题意,总人数为50人。用水量在10~15升的学生占24%,即0.24×50=12人。设用水量在5~10升的学生有x人,则用水量在15~20升的学生为(x+6)人,用水量在20~25升的学生为2x人。四个区间人数之和应等于总人数50,因此方程为:x(5~10升)+ (x+6)(15~20升)+ 2x(20~25升)+ 12(10~15升)= 50。整理得:x + x + 6 + 2x + 12 = 50,即4x + 18 = 50。选项A正确表达了这一关系。其他选项中,B错误地将百分比直接代入而未计算具体人数,C符号错误,D遗漏了10~15升区间的人数。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-07 10:13:21","updated_at":"2026-01-07 10:13:21","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":"A","content":"x + (x + 6) + 2x + 12 = 50","is_correct":1},{"id":"B","content":"x + (x + 6) + 2x + 0.24×50 = 50","is_correct":0},{"id":"C","content":"x + (x - 6) + 2x + 12 = 50","is_correct":0},{"id":"D","content":"x + (x + 6) + 2x = 50 - 0.24×50","is_correct":0}]}]