习题练习

[{"id":486,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"某学生在整理班级同学的课外阅读时间数据时，记录了5位同学一周内每天阅读的分钟数（均为整数），并计算出这组数据的平均数为30分钟。如果其中4位同学的阅读时间分别是28分钟、32分钟、25分钟和35分钟，那么第五位同学的阅读时间是多少分钟？","answer":"B","explanation":"已知5位同学阅读时间的平均数是30分钟，因此5人总阅读时间为 5 × 30 = 150 分钟。已知4位同学的阅读时间分别为28、32、25和35分钟，它们的和为 28 + 32 + 25 + 35 = 120 分钟。那么第五位同学的阅读时间为 150 - 120 = 30 分钟。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 18:00:47","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":"28","is_correct":0},{"id":"B","content":"30","is_correct":1},{"id":"C","content":"32","is_correct":0},{"id":"D","content":"34","is_correct":0}]},{"id":1009,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生在整理班级同学的课外阅读时间时，将一周内每天阅读超过30分钟的人数记录如下：周一5人，周二7人，周三6人，周四8人，周五4人，周六9人，周日10人。若该学生想计算这周平均每天有多少人阅读超过30分钟，则计算结果为___人。","answer":"7","explanation":"本题考查数据的收集、整理与描述中的平均数计算。首先将每天的人数相加：5 + 7 + 6 + 8 + 4 + 9 + 10 = 49，共有7天，因此平均每天人数为49 ÷ 7 = 7（人）。计算过程简单，符合七年级学生对平均数概念的理解和应用能力。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 05:14:13","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":767,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次班级环保活动中，某学生收集了可回收垃圾的重量为 3.5 千克，比另一名同学多收集了 1.2 千克。设另一名同学收集的垃圾重量为 x 千克，则可列出一元一次方程为：_3.5 = x + 1.2_，解得 x = _2.3_。","answer":"3.5 = x + 1.2；2.3","explanation":"根据题意，某学生收集的 3.5 千克比另一名同学多 1.2 千克，说明另一名同学的收集量加上 1.2 千克等于 3.5 千克，因此可列方程 3.5 = x + 1.2。解这个方程，两边同时减去 1.2，得到 x = 3.5 - 1.2 = 2.3。本题考查一元一次方程的建立与求解，属于简单难度，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 23:43:56","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":2326,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某学生在研究一次函数的图像时，发现函数 y = 2x - 4 的图像与 x 轴、y 轴分别交于点 A 和点 B。若将该图像沿直线 x = 1 作轴对称变换，得到新的图像，则新图像与坐标轴围成的三角形面积是原图像与坐标轴围成三角形面积的多少倍？","answer":"A","explanation":"首先求原函数 y = 2x - 4 与坐标轴的交点：令 x = 0，得 y = -4，即点 B(0, -4)；令 y = 0，得 2x - 4 = 0，解得 x = 2，即点 A(2, 0)。原图像与坐标轴围成的三角形是以原点 O(0,0)、A(2,0)、B(0,-4) 为顶点的直角三角形，面积为 (1\/2) × 2 × 4 = 4。\n\n将该图像沿直线 x = 1 作轴对称变换。点 A(2,0) 关于 x = 1 的对称点为 A'(0,0)，点 B(0,-4) 关于 x = 1 的对称点为 B'(2,-4)。新图像经过 A' 和 B'，其解析式可通过两点确定：斜率 k = (-4 - 0)\/(2 - 0) = -2，截距为 0，故新函数为 y = -2x。\n\n新图像与坐标轴交于原点 O(0,0) 和点 (0,0)（重合），但实际与 x 轴交于原点，与 y 轴也交于原点，因此需重新分析：实际上，y = -2x 过原点，与两轴仅交于原点，但结合对称变换后的几何意义，新三角形应由对称后的线段与坐标轴形成。更准确地说，原三角形 OAB 经对称后变为三角形 OA'B'，其中 O'(2,0) 并非原点。正确做法是：原三角形顶点为 O(0,0)、A(2,0)、B(0,-4)，对称后对应点为 O'(2,0)、A'(0,0)、B'(2,-4)。新三角形为 A'O'B'，即顶点为 (0,0)、(2,0)、(2,-4)，仍是直角三角形，底为 2，高为 4，面积仍为 (1\/2)×2×4=4。因此面积不变，是原面积的 1 倍。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 10:51:34","updated_at":"2026-01-10 10:51:34","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倍","is_correct":1},{"id":"B","content":"2倍","is_correct":0},{"id":"C","content":"3倍","is_correct":0},{"id":"D","content":"4倍","is_correct":0}]},{"id":2758,"subject":"历史","grade":"七年级","stage":"初中","type":"选择题","content":"考古学家在河南安阳发现了一处大型商代遗址，出土了大量刻有文字的龟甲和兽骨。这些文字主要用于记录商王占卜的内容，对研究商朝历史具有重要价值。这种文字被称为：","answer":"A","explanation":"题干中提到‘刻有文字的龟甲和兽骨’以及‘用于记录商王占卜的内容’，这是甲骨文的典型特征。甲骨文是商朝时期刻在龟甲和兽骨上的文字，主要用于占卜记事，是中国已发现的古代文字中时代最早、体系较为完整的文字。金文主要铸刻在青铜器上，盛行于西周；小篆是秦朝统一后的标准字体；隶书则流行于汉代。因此，根据出土文物的材质和用途，正确答案为A。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-12 10:39:39","updated_at":"2026-01-12 10:39:39","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":"甲骨文","is_correct":1},{"id":"B","content":"金文","is_correct":0},{"id":"C","content":"小篆","is_correct":0},{"id":"D","content":"隶书","is_correct":0}]},{"id":2343,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某公园计划修建一个等腰三角形花坛，设计要求其周长为24米，且其中一条边长为9米。已知该三角形为轴对称图形，且满足三角形三边关系。若设底边为x米，两腰各为y米，则下列哪组方程能正确描述该三角形的设计条件？","answer":"D","explanation":"本题考查等腰三角形的性质、周长计算及三角形三边关系。已知花坛为等腰三角形，周长为24米，设底边为x，两腰为y，则周长公式为 x + 2y = 24。又因三角形任意两边之和大于第三边，任意两边之差小于第三边，即 |y - y| < x < y + y 可简化为 0 < x < 2y；同时需满足 |x - y| < y < x + y。由于 y > 0 且 x > 0，最关键的约束是两边之差小于第三边：|x - y| < y，即 -y < x - y < y，化简得 0 < x < 2y，这与三角形不等式一致。选项D中的 |x - y| < y < x + y 正确表达了以y为一边时，其余两边x与y需满足的不等关系，且结合 x + 2y = 24 可完整描述设计条件。其他选项要么逻辑错误（如A中|y−y|=0，表述冗余），要么不等式方向混乱。因此正确答案为D。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 11:00:01","updated_at":"2026-01-10 11:00:01","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 + 2y = 24 且 |y - y| < x < y + y","is_correct":0},{"id":"B","content":"x + 2y = 24 且 |y - x| < y < y + x","is_correct":0},{"id":"C","content":"x + 2y = 24 且 |y - y| < x < 2y","is_correct":0},{"id":"D","content":"x + 2y = 24 且 |x - y| < y < x + y","is_correct":1}]},{"id":500,"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使用手机的时间（单位：分钟），并将数据整理如下：15，20，25，30，35，40，45，50，55，60。如果去掉一个最大值和一个最小值后，剩余数据的平均数是多少？","answer":"A","explanation":"首先确定原始数据中的最大值是60，最小值是15。去掉这两个值后，剩余的数据为：20，25，30，35，40，45，50，55，共8个数。计算这些数的和：20 + 25 + 30 + 35 + 40 + 45 + 50 + 55 = 300。然后用总和除以数据个数：300 ÷ 8 = 37.5。因此，剩余数据的平均数是37.5，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 18:09:45","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":"37.5","is_correct":1},{"id":"B","content":"40","is_correct":0},{"id":"C","content":"42.5","is_correct":0},{"id":"D","content":"45","is_correct":0}]},{"id":2315,"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身高数据时，记录了5名同学的身高（单位：cm）分别为：158, 162, 160, 165, 155。若再加入一名同学的身高后，这组数据的平均数恰好为160 cm，则这名同学的身高是多少？","answer":"A","explanation":"首先计算原有5名同学身高的总和：158 + 162 + 160 + 165 + 155 = 800（cm）。设新加入同学的身高为x cm，则6名同学的总身高为(800 + x) cm。根据题意，平均数为160 cm，因此有方程：(800 + x) ÷ 6 = 160。解这个方程：800 + x = 960，得x = 160。所以这名同学的身高是160 cm，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 10:47:15","updated_at":"2026-01-10 10:47:15","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":"160 cm","is_correct":1},{"id":"B","content":"158 cm","is_correct":0},{"id":"C","content":"162 cm","is_correct":0},{"id":"D","content":"164 cm","is_correct":0}]},{"id":1091,"subject":"数学","grade":"七年级","stage":"小学","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，发现最矮的同学身高为148厘米，最高的同学身高为165厘米。如果将所有同学的身高都增加3厘米，则新的数据中，最高身高与最矮身高的差是___厘米。","answer":"17","explanation":"原数据中最高身高为165厘米，最矮为148厘米，两者相差165 - 148 = 17厘米。当所有数据都增加相同的数值（3厘米）时，数据的分布形状不变，极差（最大值与最小值之差）保持不变。因此，新的最高身高为165 + 3 = 168厘米，新的最矮身高为148 + 3 = 151厘米，差值为168 - 151 = 17厘米。所以答案是17。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-06 08:55:35","updated_at":"2026-01-06 08:55:35","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":[]}]