习题练习

[{"id":126,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"若一个数的相反数是 -7，则这个数是 ____。","answer":"7","explanation":"本题考查有理数中相反数的概念。相反数的定义是：只有符号不同的两个数互为相反数。也就是说，一个数 a 的相反数是 -a。题目中给出一个数的相反数是 -7，设这个数为 x，则有 -x = -7。两边同时乘以 -1，得到 x = 7。因此，这个数是 7。","solution_steps":"Array","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-24 08:54:36","updated_at":"2025-12-24 08:54:36","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":2435,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"在一次校园绿化项目中，工人师傅用四块相同的等腰直角三角形地砖拼接成一个轴对称图形，拼接方式如图所示（每块地砖的直角边长为√2米）。若拼接后的大图形是一个正方形，且内部形成一个较小的空白正方形区域，则该空白正方形的面积是多少？","answer":"B","explanation":"每块等腰直角三角形地砖的直角边长为√2米，因此每条直角边对应的斜边（即等腰直角三角形的斜边）长度为：√[(√2)² + (√2)²] = √(2 + 2) = √4 = 2（米）。四块这样的三角形地砖以斜边朝外、直角顶点朝内拼接，可形成一个大正方形，其边长等于原三角形斜边的长度，即2米，故大正方形面积为 2 × 2 = 4 平方米。每块三角形面积为 (1\/2) × √2 × √2 = (1\/2) × 2 = 1 平方米，四块总面积为 4 × 1 = 4 平方米。由于大正方形总面积也为4平方米，说明拼接紧密，但中间空白区域实际由四个直角顶点围成。观察可知，四个直角顶点位于大正方形的中心区域，彼此间距构成一个小正方形，其边长等于两个直角边在水平和垂直方向上的投影差。通过坐标法或几何分析可得，空白正方形边长为√2米，因此面积为 (√2)² = 2 平方米。故正确答案为 B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 13:07:22","updated_at":"2026-01-10 13:07:22","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 平方米","is_correct":0},{"id":"B","content":"2 平方米","is_correct":1},{"id":"C","content":"√2 平方米","is_correct":0},{"id":"D","content":"4 平方米","is_correct":0}]},{"id":509,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某班级组织了一次环保活动，收集废旧纸张。第一周收集了总量的40%，第二周收集了30千克，此时已收集的与未收集的质量比为3:2。问这批废旧纸张的总质量是多少千克？","answer":"D","explanation":"设这批废旧纸张的总质量为x千克。第一周收集了40%即0.4x千克，第二周收集了30千克，因此已收集的总量为0.4x + 30千克。未收集的部分为x - (0.4x + 30) = 0.6x - 30千克。根据题意，已收集与未收集的质量比为3:2，可列方程：(0.4x + 30) \/ (0.6x - 30) = 3 \/ 2。交叉相乘得：2(0.4x + 30) = 3(0.6x - 30)，即0.8x + 60 = 1.8x - 90。移项整理得：60 + 90 = 1.8x - 0.8x，即150 = x。因此总质量为150千克，正确答案为D。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 18:14:45","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":"75千克","is_correct":0},{"id":"B","content":"100千克","is_correct":0},{"id":"C","content":"120千克","is_correct":0},{"id":"D","content":"150千克","is_correct":1}]},{"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":494,"subject":"数学","grade":"初一","stage":"小学","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。根据表格信息，成绩在80分及以上的人数占总人数的百分比最接近以下哪个选项？\n\n| 分数段（分） | 人数 |\n|--------------|------|\n| 60以下 | 5 |\n| 60—69 | 8 |\n| 70—79 | 12 |\n| 80—89 | 15 |\n| 90—100 | 10 |","answer":"C","explanation":"首先计算总人数：5 + 8 + 12 + 15 + 10 = 50（人）。\n成绩在80分及以上的人数包括80—89和90—100两个分数段，共15 + 10 = 25（人）。\n所求百分比为：25 ÷ 50 × 100% = 50%。\n因此，正确答案是C选项。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 18:06:22","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":"40%","is_correct":0},{"id":"C","content":"50%","is_correct":1},{"id":"D","content":"60%","is_correct":0}]},{"id":291,"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情况时，收集了10名同学每周课外阅读时间（单位：小时），数据如下：3，5，4，6，5，7，5，4，6，5。这组数据的众数和中位数分别是多少？","answer":"A","explanation":"首先将数据从小到大排序：3，4，4，5，5，5，5，6，6，7。众数是出现次数最多的数，其中5出现了4次，次数最多，因此众数是5。中位数是数据按顺序排列后位于中间位置的数。由于共有10个数据（偶数个），中位数为第5个和第6个数的平均数，即(5 + 5) ÷ 2 = 5。因此，众数是5，中位数是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 15:32:36","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":"众数是5，中位数是5","is_correct":1},{"id":"B","content":"众数是4，中位数是5","is_correct":0},{"id":"C","content":"众数是5，中位数是4.5","is_correct":0},{"id":"D","content":"众数是6，中位数是5.5","is_correct":0}]},{"id":1892,"subject":"数学","grade":"七年级","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(0, 0)、B(4, 0)、C(5, 3)，且四边形ABCD是一个平行四边形。若点D的坐标为(x, y)，则x + y的值是多少？","answer":"C","explanation":"本题考查平面直角坐标系中平行四边形的性质与坐标运算。在平行四边形中，对角线互相平分，或对边向量相等。可利用向量法求解：向量AB = (4 - 0, 0 - 0) = (4, 0)，由于ABCD是平行四边形，向量DC应等于向量AB。设D(x, y)，则向量DC = (5 - x, 3 - y)。令(5 - x, 3 - y) = (4, 0)，解得5 - x = 4 → x = 1；3 - y = 0 → y = 3。因此D(1, 3)，x + y = 1 + 3 = 4。或者利用中点公式：平行四边形对角线AC与BD中点相同。AC中点为((0+5)\/2, (0+3)\/2) = (2.5, 1.5)，BD中点为((4+x)\/2, (0+y)\/2)，令其等于(2.5, 1.5)，解得(4+x)\/2 = 2.5 → x = 1；(0+y)\/2 = 1.5 → y = 3。结果一致。故选C。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-07 10:14:33","updated_at":"2026-01-07 10:14:33","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":"4","is_correct":1},{"id":"D","content":"5","is_correct":0}]},{"id":845,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次班级大扫除中，某学生负责统计各小组收集的废旧纸张重量（单位：千克）。记录如下：第一组收集3.5千克，第二组收集4.2千克，第三组收集2.8千克，第四组收集5.1千克。若全班平均每组收集4千克，则第五组应收集___千克才能达到平均标准。","answer":"4.4","explanation":"要使五组的平均重量为4千克，则总重量应为 5 × 4 = 20 千克。前四组共收集 3.5 + 4.2 + 2.8 + 5.1 = 15.6 千克。因此第五组需要收集 20 - 15.6 = 4.4 千克。本题考查数据的收集与整理中的平均数计算，属于简单难度的应用题。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 01:01:56","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":1355,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某校组织七年级学生参加环保主题研学活动，活动分为A、B两组，每组人数不同。已知A组人数比B组多8人，若从A组调2人到B组，则A组人数恰好是B组人数的2倍。活动结束后，学校对两组学生收集的可回收垃圾重量进行了统计，发现A组平均每人收集垃圾重量比B组多0.5千克，且两组共收集了120千克垃圾。若设B组原有人数为x人，A组原有人数为y人，A组平均每人收集垃圾重量为z千克。请根据以上信息：(1) 列出关于x、y的二元一次方程组，并求出A、B两组原有的人数；(2) 用含z的代数式表示B组平均每人收集的垃圾重量，并建立关于z的一元一次方程，求出z的值；(3) 若学校规定每人至少收集3千克垃圾才能获得‘环保小卫士’称号，请判断A、B两组中哪些组的所有学生都能获得该称号，并说明理由。","answer":"(1) 根据题意，A组人数比B组多8人，可得方程：y = x + 8。\n若从A组调2人到B组，则A组变为(y - 2)人，B组变为(x + 2)人，此时A组人数是B组的2倍，得方程：y - 2 = 2(x + 2)。\n将第一个方程代入第二个方程：\n(x + 8) - 2 = 2(x + 2)\nx + 6 = 2x + 4\n6 - 4 = 2x - x\nx = 2\n代入y = x + 8，得y = 10。\n所以，B组原有2人，A组原有10人。\n\n(2) A组平均每人收集z千克，则A组共收集10z千克。\nB组平均每人收集垃圾重量为：(120 - 10z) \/ 2 = 60 - 5z（千克）。\n根据题意，A组平均比B组多0.5千克，得方程：\nz = (60 - 5z) + 0.5\nz = 60.5 - 5z\nz + 5z = 60.5\n6z = 60.5\nz = 60.5 ÷ 6 = 121\/12 ≈ 10.083（千克）\n所以，z = 121\/12 千克。\n\n(3) A组平均每人收集121\/12 ≈ 10.083千克 > 3千克，满足条件，因此A组所有学生都能获得称号。\nB组平均每人收集60 - 5z = 60 - 5×(121\/12) = 60 - 605\/12 = (720 - 605)\/12 = 115\/12 ≈ 9.583千克 > 3千克，也满足条件。\n因此，A、B两组的所有学生都能获得‘环保小卫士’称号。","explanation":"本题综合考查二元一次方程组、一元一次方程、整式运算及实际问题的建模能力。第(1)问通过人数变化建立方程组，考查学生对等量关系的理解与解方程组的能力；第(2)问引入平均数概念，结合总重量建立代数表达式并求解，涉及有理数运算与方程应用；第(3)问结合不等式思想（隐含比较），判断是否满足最低标准，体现数学在生活中的应用。题目情境新颖，融合环保主题，考查知识点全面，逻辑层次清晰，难度递进，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 11:06:01","updated_at":"2026-01-06 11:06:01","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":592,"subject":"数学","grade":"初一","stage":"小学","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。根据统计表，该班级成绩在80分到89分之间的人数占总人数的百分比是多少？\n\n| 分数段 | 人数 |\n|--------|------|\n| 90-100 | 8 |\n| 80-89 | 12 |\n| 70-79 | 10 |\n| 60-69 | 5 |\n| 60以下 | 3 |","answer":"B","explanation":"首先计算总人数：8 + 12 + 10 + 5 + 3 = 38（人）。成绩在80-89分之间的人数为12人。所求百分比为 (12 ÷ 38) × 100% ≈ 31.58%，四舍五入后最接近的选项是30%。因此正确答案是B。本题考查数据的收集、整理与描述中的百分比计算，属于简单难度。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 20:35:39","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":"24%","is_correct":0},{"id":"B","content":"30%","is_correct":1},{"id":"C","content":"36%","is_correct":0},{"id":"D","content":"40%","is_correct":0}]}]