1
/ 10
中等
第 1 题
已完成 0 题
练习设置
题目内容
数学
初中
选择题
请选择一道题目开始练习
我的笔记
答案解析
请选择一道题目查看解析
练习完成!
恭喜您完成了本次练习
0
总题数
0
正确
0
错误
答题卡
[{"id":205,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生计算一个数的相反数时,将 -5 写成了 5,那么他计算的是 ___ 的相反数。","answer":"-5","explanation":"相反数的定义是:一个数 a 的相反数是 -a。题目中说某学生将 -5 写成了 5,说明他实际上是把原数的相反数算成了 5。也就是说,-a = 5,那么 a = -5。因此,他计算的是 -5 的相反数。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 14:39:31","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":748,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次班级环保活动中,某学生收集了若干千克废纸,第一天卖出了总量的三分之一,第二天又卖出了2千克,此时还剩下5千克。该学生最初收集的废纸共有___千克。","answer":"10.5","explanation":"设该学生最初收集的废纸为x千克。根据题意,第一天卖出了x的三分之一,即(1\/3)x千克,第二天卖出了2千克,剩下5千克。可以列出方程:x - (1\/3)x - 2 = 5。化简得:(2\/3)x = 7。两边同时乘以3\/2,得到x = 7 × (3\/2) = 10.5。因此,该学生最初收集的废纸共有10.5千克。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 23:22:42","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":2190,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在一条东西方向的直线上做实验,规定向东为正方向。他从原点出发,先向东走了5米,记作+5米,然后向西走了8米。此时他所在的位置应记作多少米?","answer":"D","explanation":"该学生从原点出发,向东走5米到达+5米的位置,再向西走8米,相当于从+5米减去8米,即5 - 8 = -3米。因此,他最终位置应记作-3米。选项D正确。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 14:25:31","updated_at":"2026-01-09 14:25:31","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":"+13米","is_correct":0},{"id":"B","content":"+3米","is_correct":0},{"id":"C","content":"-3米","is_correct":0},{"id":"D","content":"-13米","is_correct":1}]},{"id":499,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某班级在一次数学测验中,收集了30名学生的成绩,并制作了频数分布表。已知成绩在80~89分这一组的学生有8人,占总人数的百分比最接近以下哪个选项?","answer":"C","explanation":"题目考查的是数据的收集、整理与描述中的百分比计算。总人数为30人,80~89分的学生有8人。计算该组所占百分比:8 ÷ 30 ≈ 0.2667,即约26.67%。比较选项,26.67%最接近27%,因此正确答案是C。此题帮助学生理解频数与百分比之间的关系,属于简单难度的基础统计题。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 18:09:41","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":"A","content":"20%","is_correct":0},{"id":"B","content":"25%","is_correct":0},{"id":"C","content":"27%","is_correct":1},{"id":"D","content":"30%","is_correct":0}]},{"id":1931,"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每日运动时间数据时,发现若将数据按从小到大的顺序排列,第8个和第9个数据分别为25分钟和27分钟。已知这组数据共有15个,且唯一众数为20分钟,出现4次。若去掉一个最大值和一个最小值后,剩余13个数据的平均数恰好比原平均数多1分钟,则原数据中的最大值是____分钟。","answer":"40","explanation":"中位数为(25+27)\/2=26。设原平均数为x,则新平均数为x+1。总和关系:15x - (最小值+最大值) = 13(x+1),化简得最大值+最小值=2x-13。结合众数、中位数和整数约束,推得最大值为40。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-07 14:10:11","updated_at":"2026-01-07 14:10:11","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":1102,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"某学生测量了教室中一张长方形桌子的长和宽,发现长比宽多0.6米,且周长为3.6米。设桌子的宽为x米,则可列出一元一次方程为:2(x + ___) = 3.6","answer":"x + 0.6","explanation":"根据题意,桌子的长比宽多0.6米,宽为x米,则长为x + 0.6米。长方形的周长公式为2(长 + 宽),代入得2(x + (x + 0.6)) = 3.6,化简括号内为2(2x + 0.6) = 3.6,但题目要求填写的是方程中的空白部分,即长与宽之和的表达式,因此应为x + (x + 0.6)中的第二部分,即x + 0.6。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-06 08:57:55","updated_at":"2026-01-06 08:57:55","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":686,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次班级环保活动中,某学生收集了若干千克废旧纸张。如果将这些纸张平均分给5个小组,每组可得12千克;后来又有3个小组加入,现在要将这些纸张重新平均分给所有小组,那么每个小组分到的纸张是___千克。","answer":"7.5","explanation":"首先根据题意,原来有5个小组,每组12千克,所以总纸张重量为 5 × 12 = 60 千克。后来增加了3个小组,总小组数变为 5 + 3 = 8 个。将60千克纸张平均分给8个小组,每个小组分到 60 ÷ 8 = 7.5 千克。本题考查了一元一次方程的实际应用和整数的除法运算,属于简单难度,符合七年级学生对有理数和一元一次方程知识点的掌握水平。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 22:33:25","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":575,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","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:纸类3.5千克,塑料2.8千克,金属1.7千克。如果每千克可回收垃圾可获得0.6元环保积分,那么该学生一共可以获得多少元环保积分?","answer":"A","explanation":"首先计算该学生收集的可回收垃圾总重量:3.5 + 2.8 + 1.7 = 8.0(千克)。然后根据每千克可获得0.6元积分,计算总积分:8.0 × 0.6 = 4.8(元)。因此,该学生一共可以获得4.8元环保积分,正确答案是A。本题考查有理数的加减与乘法运算在实际生活中的应用,符合七年级数学课程中‘有理数’知识点的教学目标。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 19:58:21","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":"4.8元","is_correct":1},{"id":"B","content":"5.2元","is_correct":0},{"id":"C","content":"4.2元","is_correct":0},{"id":"D","content":"5.0元","is_correct":0}]},{"id":15,"subject":"英语","grade":"初二","stage":"初中","type":"填空题","content":"Fill in the blank: I have _____ (go) to school every day.","answer":"to go","explanation":"\"have to\"表示\"必须,不得不\",后接动词原形。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"中等","points":1,"is_active":1,"created_at":"2025-08-29 16:33:04","updated_at":"2025-08-29 16:33:04","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":1748,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市为优化公交线路,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通过的公交车数量。观测数据如下(单位:辆):12, 15, 18, 15, 20, 15, 17。交通部门计划根据这些数据调整发车间隔,并规定:若某天的车流量超过平均车流量的1.2倍,则当天需增加临时班次。同时,为满足环保要求,临时班次的增加数量必须满足不等式 2x + 3 ≤ 11,其中x为增加的临时班次数量(x为非负整数)。已知每增加一个临时班次,运营成本增加200元。现需确定:在这7天中,有多少天需要增加临时班次?在这些需要增加班次的天数里,最多可以安排多少个临时班次,使得总成本不超过1000元?","answer":"第一步:计算7天的平均车流量。\n数据总和:12 + 15 + 18 + 15 + 20 + 15 + 17 = 112\n平均车流量:112 ÷ 7 = 16(辆)\n\n第二步:计算触发临时班次的阈值。\n1.2 × 16 = 19.2\n因此,只有当某天车流量 > 19.2 时,才需增加临时班次。\n查看数据:只有第5天的20辆 > 19.2,其余均 ≤ 19.2。\n所以,只有1天需要增加临时班次。\n\n第三步:解不等式确定最多可增加的临时班次数量。\n给定不等式:2x + 3 ≤ 11\n解:2x ≤ 8 → x ≤ 4\n又x为非负整数,所以x可取0,1,2,3,4。\n即每天最多可增加4个临时班次。\n\n第四步:计算在成本限制下的最大可安排班次总数。\n每天最多增加4个班次,共1天需要增加,因此最多可安排4个临时班次。\n每个班次成本200元,总成本为:4 × 200 = 800元 ≤ 1000元,满足条件。\n若尝试增加更多,但只有1天需要增加,且每天最多4个,故无法超过4个。\n\n最终答案:\n有1天需要增加临时班次;在这些天数里,最多可以安排4个临时班次,总成本800元,不超过1000元。","explanation":"本题综合考查了数据的收集、整理与描述(计算平均数)、有理数运算、一元一次不等式的求解以及实际应用中的最优化决策。首先通过求平均数确定基准值,再结合倍数关系判断哪些天需要干预;接着利用不等式约束确定单日最大增班数;最后结合成本限制验证可行性。题目设置了真实情境,要求学生在多步骤推理中整合多个知识点,体现数据分析与数学建模能力,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 14:29:25","updated_at":"2026-01-06 14:29:25","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":[]}]