习题练习

[{"id":470,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生调查了班级同学每天用于课外阅读的时间（单位：分钟），并将数据整理如下：15, 20, 25, 30, 35, 40, 45。如果再加入一个数据后，这组数据的中位数变为30，那么加入的这个数据可能是多少？","answer":"B","explanation":"原数据有7个数，按从小到大排列为：15, 20, 25, 30, 35, 40, 45。中位数是第4个数，即30。加入一个数据后，总共有8个数，中位数是第4个和第5个数的平均数。要使中位数为30，则第4个和第5个数的平均数必须为30。若加入30，则新数据为：15, 20, 25, 30, 30, 35, 40, 45，此时第4个数是30，第5个数也是30，中位数为(30+30)÷2=30，符合条件。其他选项加入后，中位数均不等于30。因此正确答案是B。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:53:54","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":"25","is_correct":0},{"id":"B","content":"30","is_correct":1},{"id":"C","content":"35","is_correct":0},{"id":"D","content":"40","is_correct":0}]},{"id":751,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次校园环保活动中，某学生收集了若干千克废纸。若每千克废纸可生产再生纸0.8千克，则该学生收集的废纸共可生产再生纸____千克。已知他最终生产出的再生纸比收集的废纸少6千克，则他最初收集的废纸是____千克。","answer":"0.8x, 30","explanation":"设该学生收集的废纸为x千克。根据题意，每千克废纸可生产0.8千克再生纸，因此可生产的再生纸为0.8x千克。又知再生纸比废纸少6千克，即x - 0.8x = 6，解得0.2x = 6，x = 30。因此，第一空填0.8x（表示再生纸质量与废纸质量的关系），第二空填30（表示收集的废纸质量）。本题综合考查了一元一次方程的建立与求解，以及有理数的运算，符合七年级数学课程要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 23:24:25","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":416,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"2","answer":"待完善","explanation":"解析待完善","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:31:06","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":2390,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某工程队计划在一条笔直的道路旁修建一个等腰三角形花坛，设计要求花坛的底边长为6米，两腰相等且与底边的夹角均为60°。施工过程中，一名学生提出：若将该花坛沿底边的垂直平分线对折，则两个部分完全重合。现测得花坛的高为h米，面积为S平方米。下列说法正确的是：","answer":"A","explanation":"根据题意，花坛为等腰三角形，底边为6米，两腰与底边的夹角均为60°。在等腰三角形中，若底角均为60°，则顶角也为60°（因为三角形内角和为180°），因此该三角形三个角都是60°，是等边三角形。等边三角形三边相等，故腰长也为6米。作底边的高h，将底边分为两段各3米，在直角三角形中，由勾股定理得：h = √(6² - 3²) = √(36 - 9) = √27 = 3√3。面积为S = (底 × 高)\/2 = (6 × 3√3)\/2 = 9√3。同时，等边三角形是轴对称图形，对称轴为底边的垂直平分线，对折后两部分完全重合。因此选项A正确。选项B错误，因为不是直角三角形；选项C的高计算错误；选项D错误，因为等边三角形是轴对称图形。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 11:51:13","updated_at":"2026-01-10 11:51:13","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":"该花坛是等边三角形，h = 3√3，S = 9√3","is_correct":1},{"id":"B","content":"该花坛是等腰直角三角形，h = 3，S = 9","is_correct":0},{"id":"C","content":"该花坛的高h = √39，S = 3√39","is_correct":0},{"id":"D","content":"该花坛不是轴对称图形，无法沿任何直线对折重合","is_correct":0}]},{"id":393,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"159.5","answer":"答案待完善","explanation":"解析待完善","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:14:17","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":932,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次班级环保活动中，某学生记录了5个小组每周回收的废纸重量（单位：千克），分别为：3.5、4.2、3.8、4.0、4.5。为了计算平均每个小组回收的废纸重量，需要先求出总重量，再除以小组数量。那么这5个小组平均每周回收废纸____千克。","answer":"4.0","explanation":"首先将5个小组回收的废纸重量相加：3.5 + 4.2 + 3.8 + 4.0 + 4.5 = 20.0（千克）。然后将总重量除以小组数量5：20.0 ÷ 5 = 4.0（千克）。因此，平均每个小组每周回收废纸4.0千克。本题考查数据的收集与整理中的平均数计算，属于简单难度的基础运算。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 03:03:40","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":[]},{"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":2248,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某学生在研究温度变化时，记录了一周内每天中午12点的气温（单位：摄氏度），其中正数表示高于0℃，负数表示低于0℃。已知这七天的气温分别为：+3，-2，+5，-4，+1，-3，+2。该学生发现，若将其中某一天的气温值取相反数后，整周气温的总和恰好变为0。请问：是哪一天的气温被取了相反数？并说明理由。","answer":"被取相反数的是第四天的气温，即-4℃。理由如下：原始七天气温总和为+2℃，要使总和变为0，需减少2℃。将-4变为+4，相当于总和增加8℃，但实际只需调整使总和减少2℃。重新计算发现，只有将+2变为-2（即第七天的气温取相反数），总和才会减少4℃，不符合。进一步分析发现，原始总和为+2，若将+2变为-2，总和变为-2；若将-2变为+2，总和变为+6；若将+3变为-3，总和变为-4；若将-3变为+3，总和变为+8；若将+5变为-5，总和变为-8；若将-4变为+4，总和变为+10；若将+1变为-1，总和变为0。因此，只有将第一天的+3变为-3，或第七天的+2变为-2，或第五天的+1变为-1，才可能影响总和。但经逐一验证，只有将第五天的+1变为-1时，总和从+2变为0。故正确答案是第五天的气温+1被取了相反数。","explanation":"本题综合考查正负数的加减运算、相反数的概念以及代数方程的建立与求解能力。题目通过真实情境（气温记录）引入，要求学生在理解总和变化机制的基础上，建立数学模型（变化量 = -2 × 原值），并解出符合条件的具体数值。解题关键在于理解‘取相反数’对总和的影响是两倍于原数的变化量，从而将问题转化为解简单的一元一次方程。此题难度较高，因其需要学生从现象中抽象出数学关系，并进行逻辑推理和验证，符合七年级学生对正负数应用的深化要求。","solution_steps":"1. 计算原始七天气温的总和：+3 + (-2) + (+5) + (-4) + (+1) + (-3) + (+2) = (3 - 2 + 5 - 4 + 1 - 3 + 2) = 2。\n2. 设第i天的气温为a_i，若将其取相反数，则总和变化量为：-2 × a_i（因为原来加a_i，现在加-a_i，差值为-2a_i）。\n3. 要使新总和为0，需满足：原总和 + 变化量 = 0，即 2 + (-2 × a_i) = 0。\n4. 解方程：2 - 2a_i = 0 → 2a_i = 2 → a_i = 1。\n5. 在原始数据中，只有第五天的气温为+1，因此是将第五天的气温+1取相反数变为-1。\n6. 验证：新气温序列为+3，-2，+5，-4，-1，-3，+2，总和为3 - 2 + 5 - 4 - 1 - 3 + 2 = 0，符合条件。","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-09 14:44:04","updated_at":"2026-01-09 14:44:04","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":513,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"在一次班级环保活动中，某学生收集了废旧纸张和塑料瓶两类可回收物。已知他收集的塑料瓶数量比废旧纸张数量的2倍少3个，总共收集了27个物品。设废旧纸张的数量为x个，则根据题意可列出一元一次方程，求出x的值是：","answer":"A","explanation":"设废旧纸张的数量为x个，则塑料瓶的数量为2x - 3个。根据题意，总数量为27个，因此可列方程：x + (2x - 3) = 27。化简得：3x - 3 = 27，移项得：3x = 30，解得：x = 10。因此，废旧纸张的数量为10个，正确答案是A。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 18:17:07","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":"A","content":"10","is_correct":1},{"id":"B","content":"12","is_correct":0},{"id":"C","content":"15","is_correct":0},{"id":"D","content":"18","is_correct":0}]},{"id":1969,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在研究某次校园义卖活动中不同商品的销售情况时，记录了五种商品的销售额（单位：元）：125.6, 98.4, 142.3, 110.8, 135.7。为了分析这组数据的集中趋势，该学生计算了这组数据的中位数和平均数，并发现两者存在一定差异。若将这组数据按从小到大的顺序排列后，位于中间位置的数据与所有数据之和除以数据个数的结果之差最接近以下哪个数值？","answer":"B","explanation":"本题考查数据的收集、整理与描述中中位数与平均数的计算及比较。首先将五种商品的销售额从小到大排序：98.4, 110.8, 125.6, 135.7, 142.3。由于数据个数为5（奇数），中位数是第3个数，即125.6。接着计算平均数：(125.6 + 98.4 + 142.3 + 110.8 + 135.7) ÷ 5 = 612.8 ÷ 5 = 122.56。然后计算中位数与平均数之差：125.6 - 122.56 = 3.04。该值最接近选项B（2.8）。因此，正确答案为B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-07 14:48:51","updated_at":"2026-01-07 14:48:51","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":"A","content":"1.2","is_correct":0},{"id":"B","content":"2.8","is_correct":1},{"id":"C","content":"3.6","is_correct":0},{"id":"D","content":"4.4","is_correct":0}]}]