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":1877,"subject":"语文","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在研究一组数据的分布特征时,绘制了频数分布直方图,并记录了以下信息:数据最小值为12,最大值为48,组距为6。若该学生将数据分为若干组,且最后一组的上限恰好为48,则这组数据被分成了多少组?若该学生进一步发现,其中一个组的频数为0,但该组仍被保留在直方图中,这说明该统计图遵循了哪项基本原则?","answer":"D","explanation":"首先计算分组数:数据范围 = 最大值 - 最小值 = 48 - 12 = 36,组距为6,因此理论组数 = 36 ÷ 6 = 6。由于最后一组上限恰好为48,说明分组从12开始,依次为[12,18)、[18,24)、[24,30)、[30,36)、[36,42)、[42,48],共6组(注意最后一组包含48,为闭区间)。因此分组数为6。其次,频数为0的组仍被保留,说明统计图完整呈现了所有预设区间,即使某区间无数据也不删除,这体现了‘频数为零的组也应保留以反映真实分布’的原则,避免误导数据连续性。选项D正确。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-07 09:54:27","updated_at":"2026-01-07 09:54:27","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":"分了5组;遵循了组间不重叠原则","is_correct":0},{"id":"B","content":"分了6组;遵循了等距分组原则","is_correct":0},{"id":"C","content":"分了7组;遵循了组限明确且不遗漏数据原则","is_correct":0},{"id":"D","content":"分了6组;遵循了频数为零的组也应保留以反映真实分布的原则","is_correct":1}]},{"id":1920,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某班级进行了一次数学测验,老师将全班学生的成绩整理成频数分布表。已知成绩在80分~89分这一组的学生人数占总人数的25%,如果全班共有40名学生,那么这一组有多少人?","answer":"B","explanation":"题目中给出成绩在80分~89分的学生占总人数的25%,全班共有40人。要求这一组的人数,只需计算40的25%。计算过程为:40 × 25% = 40 × 0.25 = 10。因此,这一组有10人,正确答案是B。本题考查的是数据的收集、整理与描述中的百分比应用,属于简单难度的基础运算。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-07 13:14:11","updated_at":"2026-01-07 13:14:11","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":"A","content":"8人","is_correct":0},{"id":"B","content":"10人","is_correct":1},{"id":"C","content":"12人","is_correct":0},{"id":"D","content":"15人","is_correct":0}]},{"id":2429,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某学生在一张方格纸上画了一个四边形ABCD,其顶点坐标分别为A(0, 0)、B(4, 0)、C(5, 2)、D(1, 2)。该学生声称这个四边形是平行四边形,并尝试通过计算对边长度和斜率来验证。若只根据坐标信息判断,以下哪个结论最能支持该四边形是平行四边形?","answer":"D","explanation":"判断一个四边形是否为平行四边形,有多种方法。在坐标系中,最直接且可靠的方法之一是验证对角线是否互相平分,即两条对角线的中点是否重合。计算对角线AC的中点:A(0,0)、C(5,2),中点为((0+5)\/2, (0+2)\/2) = (2.5, 1);对角线BD的中点:B(4,0)、D(1,2),中点为((4+1)\/2, (0+2)\/2) = (2.5, 1)。两者中点相同,说明对角线互相平分,因此四边形ABCD是平行四边形。选项D正确。其他选项虽部分正确(如A、B、C中提到的边长或斜率关系),但单独使用可能存在反例(如等腰梯形满足某些边等长或斜率相同但不是平行四边形),而中点重合是平行四边形的充要条件之一,更具说服力。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 12:52:54","updated_at":"2026-01-10 12:52:54","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":"AB与CD的长度相等,且AD与BC的斜率相同","is_correct":0},{"id":"B","content":"AB与CD的斜率相同,且AD与BC的长度相等","is_correct":0},{"id":"C","content":"AB与CD的斜率相同,且AD与BC的斜率也相同","is_correct":0},{"id":"D","content":"对角线AC和BD的中点坐标相同","is_correct":1}]},{"id":2020,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"在一次校园绿化活动中,某学生用一根长度为12米的篱笆围成一个一边靠墙的矩形花圃(靠墙的一边不需要篱笆)。为了使花圃的面积最大,该学生应如何设计长和宽?设垂直于墙的一边长度为x米,则花圃面积S与x的函数关系为S = x(12 - 2x)。当x取何值时,面积S取得最大值?","answer":"B","explanation":"题目给出面积函数 S = x(12 - 2x),可展开为 S = -2x² + 12x。这是一个开口向下的二次函数,其最大值出现在顶点处。顶点横坐标公式为 x = -b\/(2a),其中 a = -2,b = 12。代入得 x = -12 \/ (2 × (-2)) = 3。因此当 x = 3 米时,面积最大。此时平行于墙的一边为 12 - 2×3 = 6 米,面积为 3×6 = 18 平方米。本题考查一次函数与二次函数在实际问题中的应用,结合几何情境,难度适中,符合八年级学生认知水平。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 10:31:29","updated_at":"2026-01-09 10:31:29","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 = 2","is_correct":0},{"id":"B","content":"x = 3","is_correct":1},{"id":"C","content":"x = 4","is_correct":0},{"id":"D","content":"x = 6","is_correct":0}]},{"id":2209,"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的温度变化时,以20℃为标准,高于20℃的部分记为正数,低于20℃的部分记为负数。已知周三的温度变化记为-3℃,周五的温度变化记为+5℃。那么周三和周五的实际温度相差多少摄氏度?","answer":"D","explanation":"周三的温度变化为-3℃,表示实际温度是20 - 3 = 17℃;周五的温度变化为+5℃,表示实际温度是20 + 5 = 25℃。两者相差25 - 17 = 8℃。因此正确答案是D。","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":"2℃","is_correct":0},{"id":"B","content":"3℃","is_correct":0},{"id":"C","content":"5℃","is_correct":0},{"id":"D","content":"8℃","is_correct":1}]},{"id":2168,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在数轴上标出三个有理数 a、b、c,已知 a < b < c,且 |a| = |c|,b 是 a 与 c 的算术平均数。若 a + c = -8,则下列说法正确的是:","answer":"B","explanation":"由已知 a + c = -8,且 b 是 a 与 c 的算术平均数,得 b = (a + c) \/ 2 = -8 \/ 2 = -4,因此选项 B 正确。又因为 |a| = |c|,说明 a 和 c 到原点的距离相等,但 a + c = -8 ≠ 0,所以 a 和 c 不互为相反数(相反数之和为 0),排除 A。由于 |a| = |c|,C 错误。a 与 c 不相等(因 a < b < c),距离不可能为 0,D 错误。本题综合考查有理数在数轴上的表示、绝对值、相反数及平均数概念,需多步推理,符合七年级困难题要求。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-09 13:53:54","updated_at":"2026-01-09 13:53:54","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 和 c 互为相反数","is_correct":0},{"id":"B","content":"b 的值为 -4","is_correct":1},{"id":"C","content":"c 的绝对值小于 a 的绝对值","is_correct":0},{"id":"D","content":"a 与 c 之间的距离为 0","is_correct":0}]},{"id":473,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"某学生调查了班级同学每天完成数学作业所用的时间(单位:分钟),并将数据整理如下:30, 35, 40, 40, 45, 50, 55, 60, 60, 60。如果他想用一个统计量来代表大多数同学完成作业的时间,最合适的统计量是:","answer":"C","explanation":"题目中给出的数据是:30, 35, 40, 40, 45, 50, 55, 60, 60, 60。观察数据发现,60分钟出现了3次,是出现次数最多的数据,因此众数是60。题目要求用一个统计量来代表‘大多数’同学的时间,而‘众数’正是反映数据集中趋势、体现出现频率最高的值,最适合描述‘大多数’的情况。虽然平均数和中位数也能反映集中趋势,但它们不一定对应实际出现最多的数值;极差只反映数据范围,不能代表典型情况。因此最合适的统计量是众数。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:55:53","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":"A","content":"平均数","is_correct":0},{"id":"B","content":"中位数","is_correct":0},{"id":"C","content":"众数","is_correct":1},{"id":"D","content":"极差","is_correct":0}]},{"id":1801,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"在平面直角坐标系中,点A(2, 3)、B(6, 7),线段AB的中点为M。若点P(x, y)满足PM = 5且x + y = 10,则点P的横坐标x的可能值为___。","answer":"4或8","explanation":"先求中点M(4,5),设P(x,10−x),利用距离公式列方程(x−4)²+(5−x)²=25,化简得x²−12x+32=0,解得x=4或8。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 16:15:51","updated_at":"2026-01-06 16:15:51","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":551,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"某学生记录了一周内每天完成的数学练习题数量,分别为:8道、10道、7道、9道、11道、6道、12道。这组数据的众数是多少?","answer":"D","explanation":"众数是一组数据中出现次数最多的数。观察数据:6、7、8、9、10、11、12,每个数都只出现了一次,没有任何一个数重复出现。因此,这组数据中没有众数。正确答案是D。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 19:09:31","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":"6","is_correct":0},{"id":"B","content":"8","is_correct":0},{"id":"C","content":"10","is_correct":0},{"id":"D","content":"没有众数","is_correct":1}]},{"id":1329,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某学生在研究城市公交线路优化问题时,收集了A、B两条公交线路在一天中不同时段的乘客数量数据,并绘制成如下表格。已知A线路每辆公交车最多可载客40人,B线路每辆最多可载客35人。若要求每条线路在每个时段运行的公交车数量必须为整数,且总运行车辆数最少,同时确保所有乘客都能被运送(不允许超载),请根据以下数据建立数学模型并求解:\n\n| 时段 | A线路乘客数 | B线路乘客数 |\n|------|---------------|---------------|\n| 早高峰(7:00-9:00) | 320 | 280 |\n| 平峰(9:00-17:00) | 160 | 140 |\n| 晚高峰(17:00-19:00) | 360 | 315 |\n\n假设每条线路在每个时段独立安排车辆,不考虑车辆跨时段调度。请分别求出A、B两条线路在三个时段各自所需的最少公交车数量,并计算全天两条线路总共需要的最少公交车班次(即各时段车辆数之和)。","answer":"解:\n\n我们分别计算每条线路在每个时段所需的最少公交车数量。由于每辆车有最大载客限制,且车辆数必须为整数,因此需要使用“向上取整”的方法。\n\n**第一步:计算A线路各时段所需车辆数**\n\n- 早高峰:320 ÷ 40 = 8(恰好整除),需8辆车\n- 平峰:160 ÷ 40 = 4(恰好整除),需4辆车\n- 晚高峰:360 ÷ 40 = 9(恰好整除),需9辆车\n\n**第二步:计算B线路各时段所需车辆数**\n\n- 早高峰:280 ÷ 35 = 8(恰好整除),需8辆车\n- 平峰:140 ÷ 35 = 4(恰好整除),需4辆车\n- 晚高峰:315 ÷ 35 = 9(恰好整除),需9辆车\n\n**第三步:计算全天总班次**\n\nA线路总班次:8 + 4 + 9 = 21(班次)\nB线路总班次:8 + 4 + 9 = 21(班次)\n\n全天两条线路总共需要的最少公交车班次为:21 + 21 = 42(班次)\n\n答:A线路在早高峰、平峰、晚高峰分别需要8、4、9辆车;B线路分别需要8、4、9辆车;全天总共需要最少42个公交车班次。","explanation":"本题综合考查了有理数的除法运算、实际问题中的整数解处理(向上取整思想)、数据的收集与整理,以及优化思想(最小化资源使用)。虽然计算本身不复杂,但难点在于理解‘不允许超载’意味着必须向上取整,即使除法结果接近整数也不能向下舍入。同时,题目设置了真实情境——城市公交调度,要求学生从数据中提取信息,建立数学模型(即每个时段的车辆数 = 乘客数 ÷ 每车载客量,结果向上取整),并进行多步推理与汇总。尽管所有除法结果恰好为整数,避免了余数处理,但情境复杂、信息量大,且要求系统性分析,符合‘困难’难度标准。此外,题目未使用常见人名,情境新颖,考查角度独特,避免了传统应用题的重复模式。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 10:56:38","updated_at":"2026-01-06 10:56:38","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":[]}]