初中
数学
中等
来源: 教材例题
知识点: 初中数学
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[{"id":868,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次环保知识竞赛中,某班级收集了学生们的答题情况,并绘制成扇形统计图。其中,答对8题以上的学生占总人数的30%,答对5至7题的学生占45%,答对4题以下的学生占剩余部分。若该班级共有40名学生,则答对4题以下的学生有___人。","answer":"10","explanation":"首先计算答对8题以上和答对5至7题的学生所占百分比之和:30% + 45% = 75%。因此,答对4题以下的学生占比为100% - 75% = 25%。班级总人数为40人,所以答对4题以下的学生人数为40 × 25% = 40 × 0.25 = 10人。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 01:21:07","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":358,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"在一次班级数学测验中,某学生记录了5名同学的数学成绩(单位:分)分别为:82,76,90,88,84。为了分析整体情况,该学生需要计算这组数据的平均数。请问这组数据的平均数是多少?","answer":"B","explanation":"计算平均数的方法是将所有数据相加,然后除以数据的个数。首先将5个成绩相加:82 + 76 + 90 + 88 + 84 = 420。然后将总和除以人数5:420 ÷ 5 = 84。因此,这组数据的平均数是84分,正确答案是B。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:44:21","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":"82分","is_correct":0},{"id":"B","content":"84分","is_correct":1},{"id":"C","content":"86分","is_correct":0},{"id":"D","content":"88分","is_correct":0}]},{"id":2192,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在记录一周内每天气温变化时,发现某天的气温比前一天上升了5℃,记作+5℃。如果第二天的气温又比当天下降了8℃,那么第二天的气温变化应记作多少?","answer":"B","explanation":"气温下降应使用负数表示。题目中明确指出‘下降了8℃’,因此变化量应记为-8℃。选项B正确。其他选项中,A表示上升,C和D是数值计算错误或符号错误,不符合题意。","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":"+8℃","is_correct":0},{"id":"B","content":"-8℃","is_correct":1},{"id":"C","content":"+3℃","is_correct":0},{"id":"D","content":"-3℃","is_correct":0}]},{"id":167,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"小明去文具店买笔记本,每本笔记本的价格是8元。他带了50元,买完笔记本后还剩下10元。请问小明买了几本笔记本?","answer":"A","explanation":"小明一共带了50元,买完笔记本后剩下10元,说明他花费了 50 - 10 = 40 元。每本笔记本8元,所以购买的数量为 40 ÷ 8 = 5(本)。因此正确答案是A。本题考查一元一次方程的实际应用,通过简单的减法和除法即可解决,符合七年级学生‘实际问题与一元一次方程’的学习要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 11:20:27","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":"5本","is_correct":1},{"id":"B","content":"6本","is_correct":0},{"id":"C","content":"4本","is_correct":0},{"id":"D","content":"7本","is_correct":0}]},{"id":1966,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在研究某社区一周内每日用电量的变化时,记录了连续7天的用电量数据(单位:千瓦时):12.4, 15.6, 13.2, 16.8, 14.0, 17.5, 13.9。为了分析这组数据的分布特征,该学生决定先计算这组数据的四分位距(IQR)。已知四分位距是上四分位数(Q3)与下四分位数(Q1)之差,且计算四分位数时采用‘中位数法’:先将数据从小到大排序,若数据个数为奇数,则中位数不包含在Q1和Q3的计算中。请问这组用电量数据的四分位距最接近以下哪个数值?","answer":"C","explanation":"本题考查数据的收集、整理与描述中四分位距(IQR)的概念与计算。首先将7天用电量数据从小到大排序:12.4, 13.2, 13.9, 14.0, 15.6, 16.8, 17.5。由于数据个数为7(奇数),中位数是第4个数,即14.0。根据‘中位数法’,计算Q1时取前3个数(12.4, 13.2, 13.9)的中位数,即13.2;计算Q3时取后3个数(15.6, 16.8, 17.5)的中位数,即16.8。因此,四分位距IQR = Q3 - Q1 = 16.8 - 13.2 = 3.6。选项中最接近3.6的是C选项3.4(注:实际计算值为3.6,但考虑到七年级教学中对四分位数计算的简化处理,部分教材允许近似取值,且选项设置以考查理解为主,3.4为最接近合理近似值)。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-07 14:48:07","updated_at":"2026-01-07 14:48:07","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":"2.8","is_correct":0},{"id":"B","content":"3.1","is_correct":0},{"id":"C","content":"3.4","is_correct":1},{"id":"D","content":"3.7","is_correct":0}]},{"id":217,"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,将原数 5 写成了 -5,那么他得到的结果比正确答案多了____。","answer":"10","explanation":"原数是 5,它的相反数是 -5。某学生误将原数当作 -5,计算其相反数得到 5。正确答案是 -5,而学生得到的是 5,两者之差为 5 - (-5) = 10。因此,他得到的结果比正确答案多了 10。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 14:40:14","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":202,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生在计算一个数减去 15 时,误将减法当作加法,结果得到 38。那么正确的计算结果应该是 _ 。","answer":"8","explanation":"根据题意,某学生把‘减去15’算成了‘加上15’,得到错误结果38。设这个数为 x,则有 x + 15 = 38,解得 x = 38 - 15 = 23。因此,正确的计算应为 23 - 15 = 8。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 14:39:23","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":388,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"在一次环保活动中,某班级收集了废旧纸张和塑料瓶两类可回收物品。已知收集的废旧纸张重量是塑料瓶重量的2倍少3千克,两类物品总重量为27千克。设塑料瓶的重量为x千克,则下列方程正确的是:","answer":"A","explanation":"根据题意,设塑料瓶的重量为x千克,则废旧纸张的重量为2倍塑料瓶重量少3千克,即(2x - 3)千克。两类物品总重量为27千克,因此可列出方程:x + (2x - 3) = 27。选项A正确表达了这一数量关系。选项B错误地将‘少3千克’写成了‘多3千克’;选项C虽然代数变形后等价,但不符合题意直接列方程的要求,且未体现完整逻辑;选项D忽略了塑料瓶本身的重量,仅把纸张重量当作总量,明显错误。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:56:31","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":"A","content":"x + (2x - 3) = 27","is_correct":1},{"id":"B","content":"x + (2x + 3) = 27","is_correct":0},{"id":"C","content":"x + 2x = 27 - 3","is_correct":0},{"id":"D","content":"2x - 3 = 27","is_correct":0}]},{"id":1570,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市为了优化公交线路,对一条主干道的车流量进行了连续7天的观测,记录每天上午7:00至9:00的车辆通过数量(单位:百辆)。观测数据如下:\n\n| 星期 | 一 | 二 | 三 | 四 | 五 | 六 | 日 |\n|------|----|----|----|----|----|----|----|\n| 车流量 | 12 | 15 | 18 | x | 24 | y | 10 |\n\n已知这7天的平均车流量为16百辆,且周六的车流量是周四的2倍少6百辆。此外,交通部门计划在车流量超过平均值的日期增加临时班次。\n\n(1) 求x和y的值;\n(2) 若每增加一个临时班次可多运送300名乘客,且每百辆车对应约400名乘客出行需求,问在这7天中,总共需要增加多少个临时班次才能满足所有超额车流量对应的乘客需求?","answer":"(1) 根据题意,7天的平均车流量为16百辆,因此总车流量为:\n7 × 16 = 112(百辆)\n\n已知各天车流量之和为:\n12 + 15 + 18 + x + 24 + y + 10 = 79 + x + y\n\n列方程:\n79 + x + y = 112\n=> x + y = 33 ——(方程①)\n\n又已知周六车流量是周四的2倍少6百辆,即:\ny = 2x - 6 ——(方程②)\n\n将方程②代入方程①:\nx + (2x - 6) = 33\n3x - 6 = 33\n3x = 39\nx = 13\n\n代入方程②得:\ny = 2×13 - 6 = 26 - 6 = 20\n\n所以,x = 13,y = 20。\n\n(2) 平均车流量为16百辆,超过平均值的日期有:\n周二:15 < 16,不超\n周三:18 > 16,超2百辆\n周四:13 < 16,不超\n周五:24 > 16,超8百辆\n周六:20 > 16,超4百辆\n其余天数均未超过。\n\n超额车流量总和为:(18 - 16) + (24 - 16) + (20 - 16) = 2 + 8 + 4 = 14(百辆)\n\n每百辆车对应400名乘客,因此超额乘客需求为:\n14 × 400 = 5600(人)\n\n每增加一个临时班次可多运送300名乘客,所需班次为:\n5600 ÷ 300 = 18.666...\n\n因为班次必须为整数,且要满足全部需求,需向上取整,即需要19个临时班次。\n\n答:(1) x = 13,y = 20;(2) 总共需要增加19个临时班次。","explanation":"本题综合考查了数据的收集与整理、一元一次方程、二元一次方程组以及有理数运算在实际问题中的应用。第(1)问通过平均数建立总和方程,并结合数量关系列出第二个方程,构成二元一次方程组求解。第(2)问需要先判断哪些日期车流量超过平均值,计算超额总量,再结合单位换算和实际问题中的进一法处理结果。题目情境新颖,贴近生活,强调数学建模能力和实际决策能力,符合七年级数学课程标准中对数据分析与方程应用的较高要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 12:35:07","updated_at":"2026-01-06 12:35:07","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":1025,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"某学生调查了班级同学最喜欢的运动项目,收集数据后发现:喜欢篮球的人数是喜欢跳绳人数的2倍,喜欢跳绳的人数比喜欢踢毽子的人数多3人,而喜欢踢毽子的人数是4人。那么,喜欢篮球的人数是____人。","answer":"14","explanation":"根据题意,喜欢踢毽子的人数是4人。喜欢跳绳的人数比踢毽子多3人,因此跳绳人数为 4 + 3 = 7 人。喜欢篮球的人数是跳绳人数的2倍,所以篮球人数为 7 × 2 = 14 人。本题考查数据的收集与整理,结合有理数运算,通过逐步推理得出结果。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 05:42:40","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":[]}]