习题练习

[{"id":1911,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在整理班级同学最喜欢的运动项目数据时，制作了如下频数分布表。已知喜欢篮球的人数占总调查人数的30%，且总人数为40人，那么喜欢篮球的学生有多少人？","answer":"B","explanation":"题目考查的是数据的收集、整理与描述中的百分比计算。已知总人数为40人，喜欢篮球的人数占30%，即求40的30%是多少。计算过程为：40 × 30% = 40 × 0.3 = 12（人）。因此，喜欢篮球的学生有12人，正确答案为B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-07 13:11:55","updated_at":"2026-01-07 13:11:55","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":"10人","is_correct":0},{"id":"B","content":"12人","is_correct":1},{"id":"C","content":"15人","is_correct":0},{"id":"D","content":"18人","is_correct":0}]},{"id":321,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某班级组织了一次环保知识竞赛，共收集了50份有效问卷。根据统计结果，喜欢‘垃圾分类’主题的有28人，喜欢‘节约用水’主题的有25人，同时喜欢两个主题的有12人。那么，只喜欢其中一个主题的学生共有多少人？","answer":"A","explanation":"本题考查数据的收集、整理与描述中的集合思想。设喜欢‘垃圾分类’的人数为A = 28，喜欢‘节约用水’的人数为B = 25，两者都喜欢的人数为A ∩ B = 12。只喜欢‘垃圾分类’的人数为28 - 12 = 16人，只喜欢‘节约用水’的人数为25 - 12 = 13人。因此，只喜欢其中一个主题的学生总数为16 + 13 = 29人。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:37:48","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":"29","is_correct":1},{"id":"B","content":"30","is_correct":0},{"id":"C","content":"31","is_correct":0},{"id":"D","content":"32","is_correct":0}]},{"id":225,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"一个三角形的内角和是____度。","answer":"180","explanation":"根据三角形内角和定理，任意一个三角形的三个内角之和恒等于180度。这是七年级几何中的基本知识点，适用于所有类型的三角形，无论其形状或大小如何。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 14:40:43","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":519,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"在一次环保主题活动中，某学校七年级学生收集了可回收垃圾的重量数据（单位：千克），整理如下表所示。若将数据按从小到大的顺序排列，则中位数是多少？\n\n| 班级 | 垃圾重量（千克） |\n|------|------------------|\n| 七（1）班 | 12 |\n| 七（2）班 | 8 |\n| 七（3）班 | 15 |\n| 七（4）班 | 10 |\n| 七（5）班 | 13 |\n| 七（6）班 | 9 |","answer":"B","explanation":"首先将所有班级的垃圾重量按从小到大的顺序排列：8, 9, 10, 12, 13, 15。共有6个数据，是偶数个，因此中位数是第3个和第4个数的平均数。第3个数是10，第4个数是12，所以中位数为 (10 + 12) ÷ 2 = 22 ÷ 2 = 11。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 18:20:52","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":"10.5","is_correct":0},{"id":"B","content":"11","is_correct":1},{"id":"C","content":"11.5","is_correct":0},{"id":"D","content":"12","is_correct":0}]},{"id":1330,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市地铁线路规划部门正在设计一条新线路，需要在平面直角坐标系中确定两个站点A和B的位置。已知站点A位于点(2, 3)，站点B位于第一象限，且满足以下条件：\n\n1. 站点B到x轴的距离是到y轴距离的2倍；\n2. 线段AB的长度为√58；\n3. 在站点A和B之间需要设置一个临时中转站C，使得C是线段AB的中点；\n4. 规划部门还要求中转站C的纵坐标必须大于4。\n\n请根据以上条件，求出站点B的坐标，并验证中转站C是否满足规划要求。若存在多个可能的B点，请说明理由并给出所有符合条件的解。","answer":"设站点B的坐标为(x, y)，其中x > 0，y > 0（因为B在第一象限）。\n\n根据条件1：站点B到x轴的距离是|y|，到y轴的距离是|x|。由于在第一象限，x > 0，y > 0，所以有：\n y = 2x （1）\n\n根据条件2：AB的距离为√58，A(2, 3)，B(x, y)，由两点间距离公式得：\n √[(x - 2)² + (y - 3)²] = √58\n两边平方得：\n (x - 2)² + (y - 3)² = 58 （2）\n\n将（1）代入（2）：\n (x - 2)² + (2x - 3)² = 58\n展开：\n (x² - 4x + 4) + (4x² - 12x + 9) = 58\n合并同类项：\n 5x² - 16x + 13 = 58\n移项：\n 5x² - 16x - 45 = 0\n\n解这个一元二次方程：\n 判别式 Δ = (-16)² - 4×5×(-45) = 256 + 900 = 1156 = 34²\n x = [16 ± 34] \/ (2×5)\n x₁ = (16 + 34)\/10 = 50\/10 = 5\n x₂ = (16 - 34)\/10 = -18\/10 = -1.8\n\n由于B在第一象限，x > 0，故舍去x = -1.8，取x = 5\n代入（1）得：y = 2×5 = 10\n所以B点坐标为(5, 10)\n\n求中点C的坐标：\n C = ((2 + 5)\/2, (3 + 10)\/2) = (7\/2, 13\/2) = (3.5, 6.5)\n\n验证条件4：C的纵坐标为6.5 > 4，满足要求。\n\n因此，唯一符合条件的站点B的坐标为(5, 10)，中转站C(3.5, 6.5)满足规划要求。","explanation":"本题综合考查了平面直角坐标系、两点间距离公式、一元二次方程的解法以及不等式判断。解题关键在于将几何条件转化为代数方程：利用‘到坐标轴距离’的关系建立y = 2x；利用距离公式建立二次方程；通过解方程并结合第一象限的限制筛选有效解；最后计算中点坐标并验证纵坐标是否大于4。虽然方程有两个解，但负值解因不符合第一象限被排除，体现了数学建模中的实际意义检验。整个过程涉及多个知识点的融合应用，逻辑链条完整，属于困难级别的综合解答题。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 10:57:14","updated_at":"2026-01-06 10:57:14","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":910,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"某学生记录了连续5天的气温变化情况，以20℃为标准，超出部分记为正，不足部分记为负，记录如下：+3，-2，0，+5，-1。这5天的平均气温比标准气温高____℃。","answer":"1","explanation":"首先将每天的温差相加：(+3) + (-2) + 0 + (+5) + (-1) = 3 - 2 + 0 + 5 - 1 = 5。然后将总温差除以天数5，得到平均温差：5 ÷ 5 = 1。因此，这5天的平均气温比标准气温高1℃。本题考查有理数的加减运算及平均数计算，属于有理数与数据整理的综合应用。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 02:30:07","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":234,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生在计算一个数减去3.5时，误将减号看成了加号，结果得到8.2。那么正确的计算结果应该是____。","answer":"1.2","explanation":"该学生误将减法算成加法，即他计算的是：原数 + 3.5 = 8.2。由此可求出原数为：8.2 - 3.5 = 4.7。那么正确的计算应为：4.7 - 3.5 = 1.2。因此正确答案是1.2。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 14:41:11","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":2413,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"在一次数学实践活动中，某学生测量了一个等腰三角形的底边和腰长，发现底边长为8 cm，腰长为5 cm。随后，该学生将这个三角形沿其对称轴折叠，使两个腰完全重合。若将折叠后的图形展开，并在三角形内部作一条平行于底边的线段，使得这条线段将三角形的面积分为相等的两部分，则这条线段的长度是多少？","answer":"A","explanation":"首先，已知等腰三角形底边为8 cm，腰长为5 cm。利用勾股定理可求出高：从顶点向底边作高，将底边平分，得到两个直角三角形，直角边分别为4 cm和h，斜边为5 cm。由勾股定理得 h² + 4² = 5²，解得 h = 3 cm，因此三角形面积为 (1\/2)×8×3 = 12 cm²。要求作一条平行于底边的线段，将面积分为相等的两部分，即上方小三角形面积为6 cm²。由于小三角形与原三角形相似，面积比为1:2，因此边长比为 √(1\/2) = 1\/√2。原底边为8 cm，故所求线段长度为 8 × (1\/√2) = 8\/√2 = 4√2 cm。因此正确答案为A。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 12:26:35","updated_at":"2026-01-10 12:26:35","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":"A","content":"4√2 cm","is_correct":1},{"id":"B","content":"4 cm","is_correct":0},{"id":"C","content":"2√6 cm","is_correct":0},{"id":"D","content":"3√3 cm","is_correct":0}]},{"id":1836,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"如图，在平面直角坐标系中，点A(0, 4)、B(3, 0)、C(-3, 0)构成△ABC。若点D是线段BC上的一点，且△ABD与△ACD的周长相等，则点D的横坐标为多少？","answer":"B","explanation":"由题意，点B(3,0)、C(-3,0)，所以线段BC在x轴上，中点为原点O(0,0)。因为△ABD与△ACD的周长相等，即AB + BD + AD = AC + CD + AD。两边同时减去AD，得AB + BD = AC + CD。计算AB和AC的长度：AB = √[(3-0)² + (0-4)²] = √(9+16) = 5；AC = √[(-3-0)² + (0-4)²] = √(9+16) = 5。所以AB = AC，代入得BD = CD。因此D是BC的中点，坐标为(0,0)，横坐标为0。故选B。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-06 16:49:42","updated_at":"2026-01-06 16:49:42","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","is_correct":0},{"id":"B","content":"0","is_correct":1},{"id":"C","content":"1","is_correct":0},{"id":"D","content":"2","is_correct":0}]},{"id":1943,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"某学生记录了一周内每天使用手机的时间（单位：分钟）：45，60，_，75，90，105，120。已知这组数据的中位数与平均数相等，则缺失的数据是____。","answer":"82.5","explanation":"设缺失数据为x，按顺序排列后中位数为第四个数。若x在第三或第四位，中位数为(75+x)\/2或(75+90)\/2。通过计算平均数并令其等于中位数，解得x=82.5。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-07 14:12:08","updated_at":"2026-01-07 14:12:08","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":[]}]