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":2376,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某学生用一张矩形纸片制作一个无盖长方体盒子,纸片的长为 24 cm,宽为 18 cm。从四个角各剪去一个边长为 x cm 的正方形,然后将四边折起形成盒子。若要求盒子的容积为 400 cm³,则 x 的值应满足的方程是:","answer":"A","explanation":"制作无盖长方体盒子时,从矩形纸片的四个角各剪去一个边长为 x 的正方形后,折起四边形成盒子。此时,盒子的高为 x cm,底面的长为 (24 - 2x) cm,宽为 (18 - 2x) cm。容积 = 长 × 宽 × 高,即 V = x(24 - 2x)(18 - 2x)。题目给出容积为 400 cm³,因此方程为 x(24 - 2x)(18 - 2x) = 400。选项 A 正确。选项 B 错误,因为未考虑两边都剪去 x;选项 C 缺少高度项 x;选项 D 错误地将 x 平方,不符合实际几何意义。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 11:27:47","updated_at":"2026-01-10 11:27:47","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(24 - 2x)(18 - 2x) = 400","is_correct":1},{"id":"B","content":"x(24 - x)(18 - x) = 400","is_correct":0},{"id":"C","content":"(24 - x)(18 - x) = 400","is_correct":0},{"id":"D","content":"x²(24 - 2x)(18 - 2x) = 400","is_correct":0}]},{"id":983,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"某班级组织了一次环保知识竞赛,共收集了30名学生的成绩。为了分析数据,老师将成绩按10分为一段进行分组,得到如下频数分布表:90~100分有5人,80~89分有12人,70~79分有8人,60~69分有4人,60分以下有1人。则这次竞赛成绩的中位数落在_______分数段内。","answer":"80~89","explanation":"中位数是将一组数据从小到大排列后,处于中间位置的数。本题共有30名学生,因此中位数是第15个和第16个数据的平均数。根据频数累计:60分以下1人,60~69分4人(累计5人),70~79分8人(累计13人),80~89分12人(累计25人)。第15和第16个数据均落在80~89分区间内,因此中位数落在80~89分数段。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 04:23:16","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":414,"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。若要将这些数据整理成频数分布直方图,则80~89分这一组的频数是多少?\n\n| 分数段 | 人数 |\n|--------|------|\n| 60~69 | 4 |\n| 70~79 | 8 |\n| 80~89 | ? |\n| 90~100| 6 |\n\n已知全班共有30名学生参加测验。","answer":"C","explanation":"题目考查的是数据的收集、整理与描述中的频数计算。已知全班总人数为30人,其他分数段的人数分别为:60~69分有4人,70~79分有8人,90~100分有6人。因此,80~89分这一组的人数为:30 - 4 - 8 - 6 = 12(人)。所以80~89分这一组的频数是12,正确答案是C。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:30:28","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":"10","is_correct":0},{"id":"B","content":"11","is_correct":0},{"id":"C","content":"12","is_correct":1},{"id":"D","content":"13","is_correct":0}]},{"id":2775,"subject":"历史","grade":"七年级","stage":"初中","type":"选择题","content":"下列哪一项是唐朝对外友好交往的典型事例,体现了当时中外文化交流的繁荣?","answer":"B","explanation":"本题考查唐朝时期中外交流的史实。A项张骞出使西域发生在西汉时期,不属于唐朝;C项郑和下西洋是明朝的事件;D项玄奘西行虽为唐朝中外交流的重要事件,但其主要目的是求取佛经,而鉴真东渡日本则是主动将唐朝的佛教、建筑、医学等文化传播到日本,是唐朝对外友好交往和文化输出的典型代表,更符合‘对外友好交往’和‘文化交流繁荣’的题意。因此,正确答案为B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-12 10:42:55","updated_at":"2026-01-12 10:42:55","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":"A","content":"张骞出使西域,开辟丝绸之路","is_correct":0},{"id":"B","content":"鉴真东渡日本,传播唐朝文化与佛教","is_correct":1},{"id":"C","content":"郑和下西洋,访问亚非多个国家","is_correct":0},{"id":"D","content":"玄奘西行天竺,取回大量佛经并翻译","is_correct":0}]},{"id":1719,"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为期7天的观测,记录每天上午8:00至9:00的车辆通过数量(单位:辆),数据如下:120,135,128,142,130,138,145。交通部门计划根据这些数据调整红绿灯时长,并设定一个‘高峰阈值’——若某时段车流量超过该阈值,则启动延长绿灯时间的方案。已知该阈值为这7天数据的平均数向上取整后的值。同时,为评估调整效果,工作人员在实施新方案后又连续观测了5天,得到新的车流量数据:148,152,146,150,154。现要求:\n\n(1)计算原始7天数据的平均数,并确定‘高峰阈值’;\n(2)将原始7天数据与新观测的5天数据合并,求这12天车流量的中位数;\n(3)若规定‘车流量超过高峰阈值的天数占比超过50%’,则认为交通压力显著增大。请判断实施新方案后是否出现这一情况,并说明理由;\n(4)假设每辆车平均占用道路长度为6米,道路有效通行长度为800米,利用不等式估算在高峰阈值下,道路上的车辆是否会发生拥堵(即车辆总长度是否超过道路有效长度),并给出结论。","answer":"(1)原始7天数据之和为:120 + 135 + 128 + 142 + 130 + 138 + 145 = 938。\n平均数为:938 ÷ 7 = 134。\n向上取整后,高峰阈值为135。\n\n(2)合并12天数据并按从小到大排序:\n120,128,130,135,138,142,145,146,148,150,152,154。\n共有12个数据,中位数为第6和第7个数据的平均数:(142 + 145) ÷ 2 = 143.5。\n\n(3)高峰阈值为135。在原始7天中,超过135的数据有:138,142,145(共3天),占比3\/7 ≈ 42.9%,未超过50%。\n在新观测的5天中,所有数据均大于135(148,152,146,150,154),即5天全部超过阈值,占比5\/5 = 100%。\n但题目要求判断的是‘实施新方案后’是否出现‘车流量超过高峰阈值的天数占比超过50%’,应仅针对新观测的5天数据判断。\n由于5天中有5天超过阈值,占比100% > 50%,因此交通压力显著增大。\n\n(4)高峰阈值为135辆,即每小时最多135辆车通过。\n每辆车平均占用6米,则135辆车总长度为:135 × 6 = 810(米)。\n道路有效通行长度为800米。\n因为810 > 800,所以车辆总长度超过道路有效长度,会发生拥堵。\n结论:在高峰阈值下,道路会发生拥堵。","explanation":"本题综合考查了数据的收集、整理与描述中的平均数、中位数、百分比比较,以及有理数运算、不等式在实际问题中的应用。第(1)问考察平均数计算和取整规则;第(2)问要求对12个数据排序并求中位数,注意偶数个数据时取中间两数平均值;第(3)问强调对‘实施新方案后’这一时间范围的准确理解,避免误将全部12天数据纳入判断,体现数据分析的严谨性;第(4)问将实际问题转化为不等式模型,通过比较总长度与道路容量判断是否拥堵,体现数学建模能力。题目情境真实,逻辑层层递进,难度较高,符合困难等级要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 14:12:28","updated_at":"2026-01-06 14:12:28","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":2150,"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 2x + 5 = 13 的两边同时减去5,得到 2x = 8,然后再将两边同时除以2,得到 x = 4。这名学生使用的解题方法体现了等式的哪一条基本性质?","answer":"D","explanation":"该学生先对等式两边同时减去5,再同时除以2,整个过程体现了对等式两边进行相同运算时,等式依然成立这一基本性质。虽然选项B和C分别描述了其中一步所依据的性质,但整个解题过程综合体现了等式的基本性质:等式两边同时进行相同的运算(加、减、乘、除同一个数,除数不为零),等式仍然成立。因此,最全面且准确的答案是D。","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":"等式两边同时乘或除以同一个不为零的数,等式仍然成立","is_correct":0},{"id":"D","content":"等式两边同时进行相同的运算,等式仍然成立","is_correct":1}]},{"id":403,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生在平面直角坐标系中绘制了一个四边形ABCD,已知点A的坐标为(1, 2),点B的坐标为(4, 2),点C的坐标为(4, 5),点D的坐标为(1, 5)。该学生想判断这个四边形的形状,以下说法正确的是:","answer":"B","explanation":"首先根据坐标确定四边形各边的位置:AB从(1,2)到(4,2),是水平线段,长度为3;BC从(4,2)到(4,5),是垂直线段,长度为3;CD从(4,5)到(1,5),是水平线段,长度为3;DA从(1,5)到(1,2),是垂直线段,长度为3。因此四条边长度均为3,且相邻边互相垂直,说明四个角都是直角。虽然四条边相等且角为直角,看似是正方形,但进一步观察发现,正方形是特殊的矩形,而题目中并未强调‘邻边相等’这一正方形的关键特征是否被学生验证。然而,根据坐标可直接看出:对边平行(AB∥CD,AD∥BC),且四个角均为90度,符合矩形的定义。同时,由于所有边长也相等,它实际上是一个正方形,但选项中D的描述虽然正确,但‘正方形’属于更特殊的分类,而题目要求选择‘正确’的说法,B和D都看似合理。但考虑到七年级学生对图形的初步认识,通常先掌握矩形定义(直角+对边相等),且题目中坐标明确显示水平与垂直边构成直角,最直接、稳妥的判断是矩形。此外,选项D虽数学上正确,但‘正方形’需额外验证邻边相等,而题目未突出这一点。综合教学重点和选项表述,B为最符合七年级认知水平的正确答案。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:17:13","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":"这是一个平行四边形,因为有两组对边分别平行","is_correct":0},{"id":"B","content":"这是一个矩形,因为四个角都是直角且对边相等","is_correct":1},{"id":"C","content":"这是一个菱形,因为四条边长度都相等","is_correct":0},{"id":"D","content":"这是一个正方形,因为四条边相等且四个角都是直角","is_correct":0}]},{"id":850,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"某学生在整理班级同学的身高数据时,将数据按从小到大的顺序排列,并计算出最小值为148cm,最大值为172cm。若该学生想用组距为5cm进行分组,则最多可以分成___组。","answer":"5","explanation":"首先计算数据的全距:172 - 148 = 24(cm)。然后用全距除以组距:24 ÷ 5 = 4.8。由于分组数必须为整数,且要覆盖所有数据,因此需要向上取整,得到5组。例如,可分组为:148-152,153-157,158-162,163-167,168-172(注意实际分组时边界处理可微调,但组数确定为5)。因此最多可以分成5组。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 01:04:37","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":1638,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市为了解七年级学生每日完成数学作业所用时间,随机抽取了100名学生进行调查,并将数据整理如下:作业时间在30分钟以下的有15人,30~60分钟的有40人,60~90分钟的有30人,90分钟以上的有15人。现计划从这100名学生中按比例抽取20人进行深入访谈。已知被抽中的学生中,作业时间在60分钟以上的学生人数为m,且该城市共有5000名七年级学生。若用样本中作业时间在90分钟以上的学生频率估计总体,求该城市七年级学生中作业时间在90分钟以上的人数;并求m的值。","answer":"第一步:根据样本数据,作业时间在90分钟以上的学生有15人,总样本为100人,因此频率为15 ÷ 100 = 0.15。\n\n第二步:用样本频率估计总体,该城市共有5000名七年级学生,因此作业时间在90分钟以上的人数约为:\n5000 × 0.15 = 750(人)。\n\n第三步:从100名学生中按比例抽取20人,抽样比例为20 ÷ 100 = 1\/5。\n\n第四步:原样本中作业时间在60分钟以上的学生包括60~90分钟和90分钟以上两部分,共30 + 15 = 45人。\n\n第五步:按比例抽取,则被抽中的学生中作业时间在60分钟以上的人数为:\n45 × (1\/5) = 9(人),即m = 9。\n\n最终答案:该城市七年级学生中作业时间在90分钟以上的人数约为750人,m的值为9。","explanation":"本题综合考查了数据的收集、整理与描述中的频数、频率、样本估计总体以及按比例抽样等核心概念。解题关键在于理解频率的定义(频数 ÷ 总数),并能将其应用于总体估计;同时掌握按比例抽样的方法,即各组抽取人数 = 原组人数 × 抽样比例。题目设置了真实情境,要求学生从多个数据组中提取信息并进行两步计算,体现了数据分析在实际问题中的应用,难度较高,适合考查学生的综合数据处理能力。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 13:08:48","updated_at":"2026-01-06 13:08:48","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":537,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"某学生在整理班级同学最喜欢的课外活动调查数据时,将收集到的信息绘制成扇形统计图。已知喜欢阅读的同学所占的圆心角为72度,那么喜欢阅读的同学占全班人数的百分比是多少?","answer":"C","explanation":"扇形统计图中,整个圆的圆心角为360度,代表全班100%的人数。喜欢阅读的同学对应的圆心角是72度,因此所占百分比为:72 ÷ 360 × 100% = 0.2 × 100% = 20%。所以正确答案是C。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 18:49:05","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":"10%","is_correct":0},{"id":"B","content":"15%","is_correct":0},{"id":"C","content":"20%","is_correct":1},{"id":"D","content":"25%","is_correct":0}]}]