初中
数学
中等
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知识点: 初中数学
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[{"id":428,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"86.2分","answer":"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 17:34:37","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":480,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某班级在一次数学测验中,随机抽取了10名学生的成绩(单位:分)如下:78,82,85,88,90,90,92,94,96,98。关于这组数据的描述,以下哪一项是正确的?","answer":"B","explanation":"首先将数据按从小到大排列:78,82,85,88,90,90,92,94,96,98。数据个数为10,是偶数,因此中位数为第5和第6个数的平均数,即(90 + 90) ÷ 2 = 90。众数是出现次数最多的数,90出现了两次,其余数均出现一次,因此众数是90。平均数为所有数据之和除以个数:(78 + 82 + 85 + 88 + 90 + 90 + 92 + 94 + 96 + 98) ÷ 10 = 893 ÷ 10 = 89.3。极差是最大值减最小值:98 - 78 = 20。因此,选项B中‘平均数是89.3,极差是20’是正确的。选项A中位数正确但表述不完整(虽正确但不是最全面判断),选项C中位数错误,选项D极差和平均数均错误。综合分析,只有B完全正确。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:58:18","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":"这组数据的众数是90,中位数是90","is_correct":0},{"id":"B","content":"这组数据的平均数是89.3,极差是20","is_correct":1},{"id":"C","content":"这组数据的中位数是89,众数是90","is_correct":0},{"id":"D","content":"这组数据的极差是18,平均数是90","is_correct":0}]},{"id":575,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","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记录如下:纸类3.5千克,塑料2.8千克,金属1.7千克。如果每千克可回收垃圾可获得0.6元环保积分,那么该学生一共可以获得多少元环保积分?","answer":"A","explanation":"首先计算该学生收集的可回收垃圾总重量:3.5 + 2.8 + 1.7 = 8.0(千克)。然后根据每千克可获得0.6元积分,计算总积分:8.0 × 0.6 = 4.8(元)。因此,该学生一共可以获得4.8元环保积分,正确答案是A。本题考查有理数的加减与乘法运算在实际生活中的应用,符合七年级数学课程中‘有理数’知识点的教学目标。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 19:58:21","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":"4.8元","is_correct":1},{"id":"B","content":"5.2元","is_correct":0},{"id":"C","content":"4.2元","is_correct":0},{"id":"D","content":"5.0元","is_correct":0}]},{"id":523,"subject":"数学","grade":"初一","stage":"初中","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名同学每周的阅读时间(单位:小时)分别为:3,5,4,6,2。如果他想用条形统计图来展示这些数据,并希望每个条形的高度与对应数值成正比,那么当阅读时间为4小时的同学对应的条形高度为8厘米时,阅读时间为6小时的同学对应的条形高度应为多少厘米?","answer":"B","explanation":"题目考查的是数据的收集、整理与描述中的比例关系应用。已知阅读时间与条形高度成正比,即高度 = k × 时间。根据条件,当时间为4小时时,高度为8厘米,可求出比例系数 k = 8 ÷ 4 = 2(厘米\/小时)。因此,当时间为6小时时,高度 = 2 × 6 = 12厘米。故正确答案为B。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 18:25:28","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厘米","is_correct":0},{"id":"B","content":"12厘米","is_correct":1},{"id":"C","content":"14厘米","is_correct":0},{"id":"D","content":"16厘米","is_correct":0}]},{"id":176,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"已知函数 $ y = ax^2 + bx + c $ 的图像经过点 $ (1, 0) $、$ (3, 0) $ 和 $ (0, 3) $,且该函数在区间 $ [2, 4] $ 上的最大值为 $ M $,最小值为 $ m $。若 $ M - m = k $,则 $ k $ 的值为多少?","answer":"D","explanation":"首先,由题意知二次函数 $ y = ax^2 + bx + c $ 经过三点:$ (1, 0) $、$ (3, 0) $、$ (0, 3) $。\n\n因为函数过 $ (1, 0) $ 和 $ (3, 0) $,说明 $ x = 1 $ 和 $ x = 3 $ 是方程的两个根,因此可设函数为:\n$$\ny = a(x - 1)(x - 3)\n$$\n又因为函数过点 $ (0, 3) $,代入得:\n$$\n3 = a(0 - 1)(0 - 3) = a \\cdot (-1) \\cdot (-3) = 3a \\Rightarrow a = 1\n$$\n所以函数表达式为:\n$$\ny = (x - 1)(x - 3) = x^2 - 4x + 3\n$$\n\n接下来求该函数在区间 $ [2, 4] $ 上的最大值 $ M $ 和最小值 $ m $。\n\n二次函数 $ y = x^2 - 4x + 3 $ 的对称轴为:\n$$\nx = \\frac{-(-4)}{2 \\cdot 1} = 2\n$$\n开口向上,因此在区间 $ [2, 4] $ 上,最小值出现在顶点 $ x = 2 $ 处,最大值出现在离对称轴最远的端点 $ x = 4 $ 处。\n\n计算函数值:\n- 当 $ x = 2 $ 时,$ y = (2)^2 - 4 \\cdot 2 + 3 = 4 - 8 + 3 = -1 $,即 $ m = -1 $\n- 当 $ x = 4 $ 时,$ y = (4)^2 - 4 \\cdot 4 + 3 = 16 - 16 + 3 = 3 $,即 $ M = 3 $\n\n所以 $ k = M - m = 3 - (-1) = 4 $\n\n因此正确答案是 D。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2025-12-29 12:32:35","updated_at":"2025-12-29 12:32:35","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":"1","is_correct":0},{"id":"B","content":"2","is_correct":0},{"id":"C","content":"3","is_correct":0},{"id":"D","content":"4","is_correct":1}]},{"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":493,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"30人","answer":"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 18:05:14","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":299,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生在平面直角坐标系中画了一个点,该点的横坐标是-3,纵坐标是5。这个点位于第几象限?","answer":"B","explanation":"在平面直角坐标系中,四个象限的划分如下:第一象限横纵坐标均为正,第二象限横坐标为负、纵坐标为正,第三象限横纵坐标均为负,第四象限横坐标为正、纵坐标为负。题目中给出的点横坐标是-3(负),纵坐标是5(正),因此该点位于第二象限。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:34:00","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":"第一象限","is_correct":0},{"id":"B","content":"第二象限","is_correct":1},{"id":"C","content":"第三象限","is_correct":0},{"id":"D","content":"第四象限","is_correct":0}]},{"id":1534,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某学校组织七年级学生开展‘城市绿地规划’数学实践活动。活动要求学生在平面直角坐标系中设计一个矩形绿化区域,其四个顶点坐标均为整数,且满足以下条件:\n\n1. 矩形的一组对边平行于x轴,另一组对边平行于y轴;\n2. 矩形的周长为20个单位长度;\n3. 矩形的面积不小于24个单位面积;\n4. 矩形完全位于第一象限,且其左下角顶点位于原点(0, 0);\n5. 设矩形的右上角顶点坐标为(x, y),其中x和y均为正整数。\n\n现从所有满足上述条件的矩形中随机选取一个,求该矩形的面积恰好为24的概率。","answer":"解:\n\n由题意,矩形左下角顶点为(0, 0),右上角顶点为(x, y),其中x > 0,y > 0,且x、y均为正整数。\n\n因为矩形对边分别平行于坐标轴,所以其长为x,宽为y。\n\n根据条件2:周长为20,\n即:2(x + y) = 20 \n⇒ x + y = 10 \n(方程①)\n\n根据条件3:面积不小于24,\n即:xy ≥ 24 \n(不等式②)\n\n又x、y为正整数,且x + y = 10,我们可以列出所有满足方程①的正整数解:\n\n(x, y) 的可能组合为:\n(1,9), (2,8), (3,7), (4,6), (5,5), (6,4), (7,3), (8,2), (9,1)\n\n计算每种组合的面积xy:\n1×9 = 9 < 24 → 不满足\n2×8 = 16 < 24 → 不满足\n3×7 = 21 < 24 → 不满足\n4×6 = 24 ≥ 24 → 满足\n5×5 = 25 ≥ 24 → 满足\n6×4 = 24 ≥ 24 → 满足\n7×3 = 21 < 24 → 不满足\n8×2 = 16 < 24 → 不满足\n9×1 = 9 < 24 → 不满足\n\n因此,满足所有条件的(x, y)组合有:\n(4,6), (5,5), (6,4)\n共3种。\n\n其中,面积恰好为24的有:(4,6) 和 (6,4),共2种。\n\n注意:虽然(4,6)和(6,4)表示不同的矩形(长宽不同),但在坐标系中它们是不同的图形,应视为两个不同的矩形。\n\n因此,所求概率为:\n满足条件的矩形总数:3\n面积恰好为24的矩形数:2\n\n概率 = 2 \/ 3\n\n答:该矩形的面积恰好为24的概率是 2\/3。","explanation":"本题综合考查了平面直角坐标系、二元一次方程组、不等式与不等式组以及数据的整理与描述等知识点。解题关键在于:\n\n1. 利用矩形顶点坐标与边长的关系,将几何问题转化为代数问题;\n2. 由周长条件建立方程 x + y = 10;\n3. 由面积条件建立不等式 xy ≥ 24;\n4. 枚举所有满足方程的正整数解,并结合不等式筛选出符合条件的解;\n5. 在满足所有条件的样本空间中,计算目标事件(面积为24)发生的概率。\n\n本题难度较高,体现在需要综合运用多个知识点,并进行分类讨论与逻辑推理。同时,题目情境新颖,避免了传统应用题的套路,强调数学建模与数据分析能力,符合七年级数学课程的综合应用要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 12:17:55","updated_at":"2026-01-06 12:17:55","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":352,"subject":"数学","grade":"初一","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调查数据时,制作了如下频数分布表。已知喜欢篮球的人数占总人数的40%,总人数为50人,那么喜欢足球的人数是多少?\n\n| 运动项目 | 人数 |\n|----------|------|\n| 篮球 | ? |\n| 足球 | ? |\n| 乒乓球 | 12 |\n| 羽毛球 | 8 |\n\nA. 10\nB. 15\nC. 20\nD. 25","answer":"A","explanation":"首先根据题意,总人数为50人,喜欢篮球的人数占40%,因此喜欢篮球的人数为:50 × 40% = 20人。\n\n已知喜欢乒乓球的人数为12人,喜欢羽毛球的人数为8人,因此这三类运动的总人数为:20(篮球)+ 12(乒乓球)+ 8(羽毛球)= 40人。\n\n总人数为50人,所以喜欢足球的人数为:50 - 40 = 10人。\n\n因此正确答案是A。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:42:55","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","is_correct":1},{"id":"B","content":"15","is_correct":0},{"id":"C","content":"20","is_correct":0},{"id":"D","content":"25","is_correct":0}]}]