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":2140,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在解方程时,将方程 2(x - 3) = 4 的两边同时除以2,得到 x - 3 = 2,然后解得 x = 5。这一解法的依据是等式的哪一条性质?","answer":"D","explanation":"该学生在解方程时,将方程两边同时除以2,这是运用了等式的基本性质:等式两边同时除以同一个不为零的数,等式仍然成立。这一步骤是解一元一次方程的常用方法,符合七年级数学课程中关于等式性质的教学内容。","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":"等式两边同时加上同一个数,等式仍然成立","is_correct":0},{"id":"B","content":"等式两边同时减去同一个数,等式仍然成立","is_correct":0},{"id":"C","content":"等式两边同时乘以同一个数,等式仍然成立","is_correct":0},{"id":"D","content":"等式两边同时除以同一个不为零的数,等式仍然成立","is_correct":1}]},{"id":1384,"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,对一条主干道上的乘客流量进行了为期7天的调查。调查数据显示,每天早高峰时段(7:00-9:00)的乘客人数分别为:120人、135人、150人、165人、180人、195人、210人。调查发现,乘客人数每天以固定数值递增。公交公司计划根据这7天的平均乘客人数,安排每辆公交车的载客量。已知每辆公交车最多可载客45人,且要求每趟车的载客率不低于80%。若公交公司希望用最少数量的公交车完成运输任务,且每辆车每天只运行一趟,问:该公司至少需要安排多少辆公交车?请通过计算说明理由。","answer":"第一步:计算7天乘客人数的总和。\n120 + 135 + 150 + 165 + 180 + 195 + 210 = 1155(人)\n\n第二步:计算平均每天的乘客人数。\n1155 ÷ 7 = 165(人)\n\n第三步:确定每辆公交车的最低有效载客量(载客率不低于80%)。\n每辆车最多可载45人,80%载客量为:\n45 × 0.8 = 36(人)\n即每辆车每天至少运送36人才能满足载客率要求。\n\n第四步:计算满足平均每天165人运输所需的最少车辆数。\n设需要x辆车,则每辆车平均载客量为165 ÷ x。\n要求:165 ÷ x ≥ 36\n解不等式:\n165 ≥ 36x\nx ≤ 165 ÷ 36 ≈ 4.583\n由于x必须为整数,且要满足每辆车载客量不低于36人,因此x最大可取4,但需验证是否可行。\n\n若x = 4,则每辆车平均载客量为165 ÷ 4 = 41.25人,满足≥36人,且41.25 ≤ 45,未超载。\n因此4辆车可行。\n\n但题目要求“用最少数量的公交车”,我们需确认是否可以更少。\n若x = 3,则每辆车平均载客量为165 ÷ 3 = 55人 > 45人,超载,不可行。\n\n因此,最少需要4辆公交车。\n\n答案:至少需要安排4辆公交车。","explanation":"本题综合考查了数据的收集、整理与描述(计算平均数)、有理数的运算(加减与除法)、不等式与不等式组(建立并求解不等式)以及实际应用问题的建模能力。解题关键在于理解“载客率不低于80%”转化为数学条件为每辆车平均载客量不低于36人,并结合最大载客量限制,通过不等式分析确定最小车辆数。同时需验证解的合理性,排除超载情况,体现数学思维的严谨性。题目情境新颖,贴近生活,考查学生从数据中提取信息、建立数学模型并解决实际问题的能力,符合七年级数学课程标准要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 11:17:21","updated_at":"2026-01-06 11:17:21","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":2321,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某学生在整理班级同学最喜欢的运动项目数据时,制作了如下频数分布表。已知喜欢跳绳的人数是喜欢踢毽子的2倍,且喜欢跳绳和踢毽子的总人数为36人。如果喜欢打篮球的人数比喜欢踢毽子的多6人,那么喜欢打篮球的有多少人?","answer":"A","explanation":"设喜欢踢毽子的人数为x,则喜欢跳绳的人数为2x。根据题意,跳绳和踢毽子的总人数为36人,可得方程:x + 2x = 36,解得x = 12。因此,喜欢踢毽子的有12人,喜欢跳绳的有24人。又知喜欢打篮球的人数比喜欢踢毽子的多6人,即12 + 6 = 18人。故喜欢打篮球的有18人,正确答案为A。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 10:50:27","updated_at":"2026-01-10 10:50:27","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":"18人","is_correct":1},{"id":"B","content":"20人","is_correct":0},{"id":"C","content":"24人","is_correct":0},{"id":"D","content":"30人","is_correct":0}]},{"id":947,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在某次班级环保活动中,学生们收集废纸进行回收。若每5千克废纸可兑换1个环保积分,某小组共收集了37千克废纸,最多可以兑换___个环保积分。","answer":"7","explanation":"根据题意,每5千克废纸兑换1个环保积分。将总重量37千克除以5,得到37 ÷ 5 = 7.4。由于只能兑换完整的积分,不能兑换部分积分,因此取商的整数部分,即最多可以兑换7个环保积分。本题考查的是有理数中的除法运算及实际问题中的取整应用,属于简单难度,符合七年级学生对有理数运算的理解水平。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 03:27:53","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":[]},{"id":2513,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"某学生在观察一个由三个相同正方体堆叠而成的立体图形时,从正面、上面和左面分别看到了不同的平面图形。已知从正面看到的图形是一个高为3个单位、宽为1个单位的长方形,从上面看到的图形是一个边长为1个单位的正方形,那么从左面看到的图形最可能是什么形状?","answer":"A","explanation":"该立体图形由三个相同的正方体竖直堆叠而成,形成一个高度为3个单位、底面为1×1的正方柱。从正面观察时,看到的是三个正方体垂直排列形成的3×1长方形;从上面观察时,只能看到最上面一个正方体的顶面,即1×1的正方形。由于该立体图形在左右方向上没有延伸(宽度始终为1个单位),因此从左面观察时,看到的仍然是三个正方体竖直堆叠的侧面,形状与正面视图相同,即高为3个单位、宽为1个单位的长方形。因此正确答案为A。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 15:42:00","updated_at":"2026-01-10 15:42:00","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":"一个高为3个单位、宽为1个单位的长方形","is_correct":1},{"id":"B","content":"一个边长为1个单位的正方形","is_correct":0},{"id":"C","content":"一个高为2个单位、宽为1个单位的长方形","is_correct":0},{"id":"D","content":"一个高为1个单位、宽为3个单位的长方形","is_correct":0}]},{"id":636,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"在一次环保活动中,某学校七年级学生共收集了120千克废纸。其中,A班收集的废纸比B班多10千克,且两班收集的废纸总量正好是全年级收集量的一半。设B班收集的废纸为x千克,则根据题意可列方程为:","answer":"A","explanation":"题目中说明A班比B班多收集10千克,B班收集了x千克,则A班收集了(x + 10)千克。两班共收集的废纸是全年级的一半,全年级共收集120千克,因此两班共收集120 ÷ 2 = 60千克。所以可列方程:x + (x + 10) = 60。选项A正确。选项B错误地将总量设为120;选项C错误地将A班的收集量表示为10x;选项D虽然表达式正确,但等式右边应为60而非120。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 22:01:35","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 + (x + 10) = 60","is_correct":1},{"id":"B","content":"x + (x - 10) = 120","is_correct":0},{"id":"C","content":"x + 10x = 60","is_correct":0},{"id":"D","content":"x + (x + 10) = 120","is_correct":0}]},{"id":1971,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在研究某次学校科技节中各参赛小组完成项目所用时间时,记录了八个小组的数据(单位:分钟):28.5, 32.1, 26.8, 30.4, 29.7, 33.6, 27.9, 31.2。为了分析这组数据的集中趋势和离散程度,该学生先计算了平均数,再计算了各数据与平均数之差的绝对值,并求出这些绝对值的平均数(即平均绝对偏差,MAD)。请问这组数据的平均绝对偏差最接近以下哪个数值?","answer":"B","explanation":"本题考查数据的收集、整理与描述中平均绝对偏差(MAD)的概念与计算。首先计算八个小组所用时间的平均数:(28.5 + 32.1 + 26.8 + 30.4 + 29.7 + 33.6 + 27.9 + 31.2) ÷ 8 = 240.2 ÷ 8 = 30.025。然后计算每个数据与平均数之差的绝对值:|28.5−30.025|=1.525,|32.1−30.025|=2.075,|26.8−30.025|=3.225,|30.4−30.025|=0.375,|29.7−30.025|=0.325,|33.6−30.025|=3.575,|27.9−30.025|=2.125,|31.2−30.025|=1.175。将这些绝对值相加:1.525 + 2.075 + 3.225 + 0.375 + 0.325 + 3.575 + 2.125 + 1.175 = 14.4。最后求平均绝对偏差:14.4 ÷ 8 = 1.8。1.8 最接近选项 B 的 1.7,因此答案为 B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-07 14:49:19","updated_at":"2026-01-07 14:49:19","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":"A","content":"1.5","is_correct":0},{"id":"B","content":"1.7","is_correct":1},{"id":"C","content":"1.9","is_correct":0},{"id":"D","content":"2.1","is_correct":0}]},{"id":712,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次环保活动中,某学生记录了连续5天每天回收的塑料瓶数量,分别为:12个、15个、_个、18个、20个。已知这5天回收数量的平均数是16个,那么第三天回收的塑料瓶数量是___个。","answer":"15","explanation":"根据平均数的定义,5天回收总数的平均数是16个,因此5天的总回收数量为 5 × 16 = 80 个。已知第1天到第5天中,第1、2、4、5天分别回收了12、15、18、20个,合计为 12 + 15 + 18 + 20 = 65 个。所以第三天回收的数量为 80 - 65 = 15 个。本题考查数据的收集与整理中的平均数应用,属于简单难度的实际问题建模。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 22:49:38","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":810,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次班级图书捐赠活动中,某学生第一天捐了若干本书,第二天比第一天多捐了5本,两天一共捐了23本。设第一天捐了___本书。","answer":"9","explanation":"设第一天捐了x本书,则第二天捐了(x + 5)本。根据题意,两天共捐书数量为:x + (x + 5) = 23。解这个一元一次方程:2x + 5 = 23,移项得2x = 18,解得x = 9。因此,第一天捐了9本书。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 00:25:12","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":1476,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某校七年级组织学生参加环保知识竞赛,竞赛成绩以百分制记录。为分析成绩分布情况,某学生随机抽取了50名参赛学生的成绩,整理后得到如下信息:成绩在60分以下的有5人,60~69分的有8人,70~79分的有12人,80~89分的有15人,90~100分的有10人。已知所有被抽取学生的平均成绩为78.6分,且90~100分这一组中,最低分为92分,最高分为100分,该组平均分为96分。若将80~89分这一组的所有成绩都提高5分,同时将60~69分这一组的所有成绩都降低3分,其余组数据不变,求调整后这50名学生的平均成绩(精确到0.1分)。","answer":"解题步骤如下:\n\n第一步:计算原始总分。\n已知平均成绩为78.6分,总人数为50人,\n所以原始总分 = 78.6 × 50 = 3930(分)。\n\n第二步:计算90~100分组原始总分。\n该组有10人,平均分为96分,\n所以该组原始总分 = 96 × 10 = 960(分)。\n\n第三步:计算其余四组的原始总分。\n其余四组总人数 = 50 - 10 = 40人,\n其余四组原始总分 = 3930 - 960 = 2970(分)。\n\n第四步:分析调整情况。\n- 60~69分组:8人,每人成绩降低3分,总分减少 8 × 3 = 24(分)。\n- 80~89分组:15人,每人成绩提高5分,总分增加 15 × 5 = 75(分)。\n- 其他组(60分以下、70~79分、90~100分)成绩不变,总分不变。\n\n第五步:计算调整后总分。\n调整后总分 = 原始总分 - 24 + 75 = 3930 + 51 = 3981(分)。\n\n第六步:计算调整后平均成绩。\n调整后平均成绩 = 3981 ÷ 50 = 79.62(分)。\n精确到0.1分,结果为79.6分。\n\n答:调整后这50名学生的平均成绩为79.6分。","explanation":"本题综合考查了数据的收集、整理与描述中的频数分布、平均数计算,以及有理数的混合运算和一元一次方程思想的应用(虽未显式列方程,但总分与平均数的关系本质上是线性关系)。解题关键在于理解平均数与总分之间的转换,并能准确计算各组调整对总分的影响。题目设置了真实情境,要求学生在多组数据中识别变化部分,排除干扰信息(如90~100分组的详细数据仅用于验证,实际解题中只需其总分),体现了数据分析能力和逻辑推理能力。难度较高,因涉及多步运算、信息筛选和精确计算,符合困难级别要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 11:53:43","updated_at":"2026-01-06 11:53:43","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":[]}]