初中
数学
中等
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知识点: 初中数学
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[{"id":800,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"某学生在整理班级同学的课外阅读情况时,随机抽取了30名同学,统计他们每月阅读课外书的数量。其中,阅读2本书的有8人,阅读3本书的有12人,阅读4本书的有6人,其余同学阅读5本书。那么这30名同学每月平均阅读课外书的数量是___本。","answer":"3.2","explanation":"首先计算阅读5本书的人数:30 - 8 - 12 - 6 = 4人。然后计算总阅读量:2×8 + 3×12 + 4×6 + 5×4 = 16 + 36 + 24 + 20 = 96本。最后求平均数:96 ÷ 30 = 3.2本。因此,平均每月阅读3.2本书。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 00:15:45","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":796,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次班级图书角整理活动中,某学生统计了上周同学们借阅的图书数量,发现科技类图书比文学类图书多借出8本,两类图书共借出46本。设文学类图书借出x本,则科技类图书借出___本,根据题意可列方程为___。","answer":"x + 8;x + (x + 8) = 46","explanation":"题目中明确指出科技类图书比文学类多8本,若文学类借出x本,则科技类为x + 8本。两类图书共借出46本,因此可列出方程:x + (x + 8) = 46。本题考查用字母表示数量关系及建立一元一次方程的能力,属于‘一元一次方程’知识点,符合七年级教学要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 00:14:51","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":1070,"subject":"数学","grade":"七年级","stage":"小学","type":"填空题","content":"在一次班级环保活动中,某学生收集了若干个废旧电池,若每3个旧电池可兑换1个新电池,该学生最终共获得了12个新电池,则他最初收集的废旧电池至少有___个。","answer":"36","explanation":"根据题意,每3个旧电池可兑换1个新电池,要获得12个新电池,则需要 12 × 3 = 36 个旧电池。由于兑换过程是整组进行的(不能兑换部分电池),且题目问的是‘至少’需要多少个,因此不需要考虑额外余数或多次兑换的情况。直接计算即可得出最少需要36个废旧电池。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-06 08:52:49","updated_at":"2026-01-06 08:52:49","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":2269,"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的距离为5个单位长度,且位于点A的右侧。点C与点B关于原点对称。那么点C表示的数是___","answer":"D","explanation":"点A表示-3,点B在点A右侧且距离为5,因此点B表示的数是-3 + 5 = 2。点C与点B关于原点对称,即点C是点B的相反数,所以点C表示的数是-2的相反数,即8。因此正确答案是D。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-09 16:09:15","updated_at":"2026-01-09 16:09:15","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":"-8","is_correct":0},{"id":"B","content":"2","is_correct":0},{"id":"C","content":"-2","is_correct":0},{"id":"D","content":"8","is_correct":1}]},{"id":292,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"众数是85,中位数是85","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 15:32:46","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":2760,"subject":"历史","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在参观博物馆时,看到一件出土于河南安阳的青铜器,器身刻有‘司母戊’三字,形制庄重,纹饰精美。这件文物最有可能属于哪个历史时期?","answer":"B","explanation":"司母戊鼎是中国目前已发现的最大、最重的青铜礼器,出土于河南安阳殷墟,而殷墟是商朝后期的都城遗址。‘司母戊’三字表明这是商王为祭祀母亲戊而铸造的青铜器,属于商朝晚期典型器物。夏朝尚未发现成熟青铜铭文,西周青铜器铭文较长且风格不同,春秋时期青铜器风格趋于轻巧,与此鼎特征不符。因此,正确答案为B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-12 10:39:47","updated_at":"2026-01-12 10:39:47","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":"夏朝","is_correct":0},{"id":"B","content":"商朝","is_correct":1},{"id":"C","content":"西周","is_correct":0},{"id":"D","content":"春秋时期","is_correct":0}]},{"id":2267,"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之间的距离为7个单位长度,且点B位于点A的右侧。现在将点B向左移动4个单位长度到达点C,再将点C向右移动2个单位长度到达点D。那么点D表示的数是多少?","answer":"B","explanation":"首先,点A表示-3,点B在点A右侧且距离为7,因此点B表示的数是-3 + 7 = 4。将点B向左移动4个单位,到达点C,即4 - 4 = 0。再将点C向右移动2个单位,到达点D,即0 + 2 = 2。因此点D表示的数是2,正确答案是B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-09 16:09:15","updated_at":"2026-01-09 16:09:15","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":"-2","is_correct":0},{"id":"B","content":"2","is_correct":1},{"id":"C","content":"4","is_correct":0},{"id":"D","content":"6","is_correct":0}]},{"id":2477,"subject":"数学","grade":"八年级","stage":"初中","type":"解答题","content":"如图,在平面直角坐标系中,点 A(0, 4),点 B(6, 0),点 C 在 x 轴正半轴上,且 △ABC 是等腰三角形,AB = AC。过点 A 作直线 l 垂直于 BC,垂足为点 D。点 E 是线段 AD 上一点(不与 A、D 重合),连接 BE 并延长交 y 轴于点 F。已知直线 BE 的解析式为 y = kx + b,且满足 k = -\\\\frac{1}{2}。若四边形 AOFC 的面积为 15,其中 O 为坐标原点,求点 C 的横坐标。","answer":"待完善","explanation":"解析待完善","solution_steps":"待完善","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 15:05:28","updated_at":"2026-01-10 15:05:28","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":850,"subject":"数学","grade":"初一","stage":"初中","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,并计算出最小值为148cm,最大值为172cm。若该学生想用组距为5cm进行分组,则最多可以分成___组。","answer":"5","explanation":"首先计算数据的全距:172 - 148 = 24(cm)。然后用全距除以组距:24 ÷ 5 = 4.8。由于分组数必须为整数,且要覆盖所有数据,因此需要向上取整,得到5组。例如,可分组为:148-152,153-157,158-162,163-167,168-172(注意实际分组时边界处理可微调,但组数确定为5)。因此最多可以分成5组。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 01:04:37","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":1977,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"某学生在纸上画了一个矩形,其长为8 cm,宽为6 cm。若以该矩形的一个顶点为旋转中心,将矩形绕此点顺时针旋转90°,则旋转后原对角线所扫过的区域面积最接近以下哪个值?(π取3.14)","answer":"A","explanation":"本题考查旋转与圆的综合应用。矩形对角线长度为√(8² + 6²) = √(64 + 36) = √100 = 10 cm。以某一顶点为旋转中心旋转90°,对角线的另一端点将绕该中心作半径为10 cm的圆弧运动,扫过的区域是一个半径为10 cm、圆心角为90°的扇形。扇形面积为 (90°\/360°) × π × 10² = (1\/4) × 3.14 × 100 = 78.5 cm²。因此,对角线扫过的区域面积最接近78.5 cm²。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-07 15:00:36","updated_at":"2026-01-07 15:00:36","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":"A","content":"78.5 cm²","is_correct":1},{"id":"B","content":"50.2 cm²","is_correct":0},{"id":"C","content":"113.0 cm²","is_correct":0},{"id":"D","content":"25.1 cm²","is_correct":0}]}]