初中
数学
中等
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知识点: 初中数学
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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":660,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次班级环保活动中,某学生收集了若干节废旧电池。若每3节电池可兑换1个环保积分,该学生共获得了8个环保积分,则他收集的电池总数为____节。","answer":"24","explanation":"根据题意,每3节电池兑换1个环保积分,获得8个积分说明兑换了8组,每组3节电池。因此总电池数为 8 × 3 = 24 节。本题考查一元一次方程的实际应用,学生可通过简单的乘法运算得出结果,符合七年级‘一元一次方程’知识点的简单难度要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 22:15:10","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":630,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"某学校组织七年级学生参加环保知识竞赛,参赛学生的成绩被分为五个等级:优秀、良好、中等、及格、不及格。为了统计各等级人数,老师将数据整理成如下表格:\n\n| 等级 | 人数 |\n|--------|------|\n| 优秀 | 12 |\n| 良好 | 18 |\n| 中等 | 15 |\n| 及格 | 10 |\n| 不及格 | 5 |\n\n如果老师想用扇形统计图表示各等级人数所占百分比,那么表示“良好”等级的扇形圆心角的度数是多少?","answer":"B","explanation":"首先计算总人数:12 + 18 + 15 + 10 + 5 = 60人。\n“良好”等级人数为18人,占总人数的比例为18 ÷ 60 = 0.3,即30%。\n扇形统计图中,整个圆为360度,因此“良好”等级对应的圆心角为:360 × 0.3 = 108度。\n所以正确答案是B选项。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 21:55:35","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":"90度","is_correct":0},{"id":"B","content":"108度","is_correct":1},{"id":"C","content":"120度","is_correct":0},{"id":"D","content":"144度","is_correct":0}]},{"id":2174,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"已知有理数 a 和 b 满足 a > 0,b < 0,且 |a| < |b|。某学生计算 a + b 的结果,并比较其与 a 和 b 的大小关系。以下结论中正确的是:","answer":"D","explanation":"根据题意,a 是正数,b 是负数,且 |a| < |b|,说明 b 的绝对值更大。因此 a + b 的结果为负数,但比 b 更接近 0。例如,若 a = 2,b = -5,则 a + b = -3。此时有 -5 < -3 < 2,即 b < a + b < a。选项 D 正确描述了这一大小关系。选项 A 错误,因为 a + b < a;选项 B 错误,因为 a + b > b;选项 C 错误,因为 a + b < 0。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-09 14:12:20","updated_at":"2026-01-09 14:12:20","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 + b > a","is_correct":0},{"id":"B","content":"a + b < b","is_correct":0},{"id":"C","content":"a + b > 0","is_correct":0},{"id":"D","content":"b < a + b < a","is_correct":1}]},{"id":2258,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"在数轴上,点A表示的数是-3,点B与点A之间的距离是5个单位长度,且点B在原点右侧。那么点B表示的数是___。","answer":"B","explanation":"点A在数轴上表示-3,点B与点A相距5个单位长度。由于点B在原点右侧,说明点B表示的数大于0。从-3向右移动5个单位,即-3 + 5 = 2,因此点B表示的数是2。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 16:03:06","updated_at":"2026-01-09 16:03:06","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","is_correct":0},{"id":"B","content":"2","is_correct":1},{"id":"C","content":"8","is_correct":0},{"id":"D","content":"-2","is_correct":0}]},{"id":1941,"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)和点B(8, 7)是某矩形的两个对角顶点,且该矩形的边分别平行于坐标轴。若该矩形的周长是面积的$\\frac{1}{2}$,则这个矩形的另外两个顶点坐标的横坐标之和为____。","answer":"10","explanation":"由A、B为对角顶点且边平行坐标轴,可得另两点为(2,7)和(8,3)。设长为6,宽为4,周长=20,面积=24。验证20 = 24×½成立。另两点横坐标为2和8,和为10。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-07 14:11:56","updated_at":"2026-01-07 14:11:56","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":2261,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"在数轴上,点A表示的数是-3,点B与点A的距离是5个单位长度,且点B在原点右侧。一名学生认为点B表示的数可能是2或-8,那么该学生的说法是否正确?","answer":"B","explanation":"点A表示-3,与点B的距离是5个单位长度,数学上确实有两个可能的位置:-3 + 5 = 2,或-3 - 5 = -8。但题目明确指出点B在原点右侧,即表示的数必须大于0,因此点B只能是2。该学生忽略了位置限制,错误地认为-8也符合条件,所以其说法不正确。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 16:03:06","updated_at":"2026-01-09 16:03:06","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":"正确,因为-3加5等于2,减5等于-8","is_correct":0},{"id":"B","content":"不正确,因为点B在原点右侧,只能表示正数,所以只能是2","is_correct":1},{"id":"C","content":"正确,因为距离为5的点有两个,分别是2和-8","is_correct":0},{"id":"D","content":"不正确,因为点B应该在-3的左侧,所以只能是-8","is_correct":0}]},{"id":1910,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某班级组织植树活动,计划将一批树苗平均分给若干小组。如果每组分配5棵树苗,则剩余3棵;如果每组分配6棵树苗,则最后一组不足3棵但至少有1棵。已知小组数量为整数,且树苗总数不超过50棵,则该班级最多可能有多少个小组?","answer":"B","explanation":"设小组数量为x(x为正整数),树苗总数为y。根据题意:\n\n1. 每组5棵,剩3棵:y = 5x + 3;\n2. 每组6棵时,最后一组不足3棵但至少有1棵,说明前(x−1)组每组6棵,最后一组有1、2棵,即:\n 6(x−1) + 1 ≤ y < 6(x−1) + 3\n 化简得:6x − 5 ≤ y < 6x − 3\n\n将y = 5x + 3代入不等式:\n6x − 5 ≤ 5x + 3 < 6x − 3\n\n解左边:6x − 5 ≤ 5x + 3 → x ≤ 8\n解右边:5x + 3 < 6x − 3 → 3 + 3 < x → x > 6\n\n所以x的取值范围是:6 < x ≤ 8,即x = 7 或 8\n\n又因为树苗总数不超过50棵:y = 5x + 3 ≤ 50 → 5x ≤ 47 → x ≤ 9.4,满足x=7和x=8\n\n当x=8时,y = 5×8 + 3 = 43\n验证第二种分法:前7组每组6棵,共42棵,最后一组43−42=1棵,符合“不足3棵但至少有1棵”\n\n当x=9时,y=48,但6×8 + 3 = 51 > 48,不满足y < 6x−3(即48 < 51成立),但检查分配:前8组48棵,最后一组0棵,不符合“至少有1棵”,故x=9不成立\n\n因此,满足所有条件的最大x为8。\n\n故正确答案为B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-07 13:11:51","updated_at":"2026-01-07 13:11:51","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":"7个","is_correct":0},{"id":"B","content":"8个","is_correct":1},{"id":"C","content":"9个","is_correct":0},{"id":"D","content":"10个","is_correct":0}]},{"id":573,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生测量了一个长方形花坛的长和宽,发现长比宽多2米,且周长为20米。若设花坛的宽为x米,则根据题意可列出一元一次方程,求出花坛的面积是多少平方米?","answer":"D","explanation":"设花坛的宽为x米,则长为(x + 2)米。根据长方形周长公式:周长 = 2 × (长 + 宽),代入已知条件得:2 × (x + x + 2) = 20。化简得:2 × (2x + 2) = 20 → 4x + 4 = 20 → 4x = 16 → x = 4。因此,宽为4米,长为6米。面积为长 × 宽 = 4 × 6 = 24平方米。故正确答案为D。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 19:52:38","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":"12","is_correct":0},{"id":"B","content":"16","is_correct":0},{"id":"C","content":"20","is_correct":0},{"id":"D","content":"24","is_correct":1}]},{"id":649,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次班级环保活动中,某学生收集了若干个塑料瓶。如果他将收集数量的一半再减去3个,正好等于他最初收集数量的六分之一。那么他最初收集的塑料瓶数量是____个。","answer":"9","explanation":"设该学生最初收集的塑料瓶数量为x个。根据题意,'数量的一半再减去3个'表示为(1\/2)x - 3,'最初数量的六分之一'表示为(1\/6)x。根据等量关系可列方程:(1\/2)x - 3 = (1\/6)x。解这个一元一次方程:两边同时乘以6消去分母,得3x - 18 = x;移项得3x - x = 18,即2x = 18;解得x = 9。因此,他最初收集了9个塑料瓶。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 22:11:16","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":[]}]