[{"id":273,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"在一次班级调查中，某学生记录了10名同学的身高（单位：厘米）：150，152，155，155，158，160，162，165，168，170。这组数据的中位数是多少？","answer":"C","explanation":"中位数是将一组数据从小到大排列后，处于中间位置的数。本题共有10个数据，是偶数个，因此中位数是第5个和第6个数据的平均数。数据已按顺序排列：150，152，155，155，158，160，162，165，168，170。第5个数是158，第6个数是160。中位数为（158 + 160）÷ 2 = 318 ÷ 2 = 159。因此正确答案是C。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:30:20","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":"155","is_correct":0},{"id":"B","content":"158","is_correct":0},{"id":"C","content":"159","is_correct":1},{"id":"D","content":"160","is_correct":0}]},{"id":1025,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"某学生调查了班级同学最喜欢的运动项目，收集数据后发现：喜欢篮球的人数是喜欢跳绳人数的2倍，喜欢跳绳的人数比喜欢踢毽子的人数多3人，而喜欢踢毽子的人数是4人。那么，喜欢篮球的人数是____人。","answer":"14","explanation":"根据题意，喜欢踢毽子的人数是4人。喜欢跳绳的人数比踢毽子多3人，因此跳绳人数为 4 + 3 = 7 人。喜欢篮球的人数是跳绳人数的2倍，所以篮球人数为 7 × 2 = 14 人。本题考查数据的收集与整理，结合有理数运算，通过逐步推理得出结果。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 05:42:40","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":13,"subject":"语文","grade":"初二","stage":"初中","type":"填空题","content":"《桃花源记》的作者是______，他是______（朝代）的诗人。","answer":"陶渊明, 东晋","explanation":"《桃花源记》是东晋诗人陶渊明的作品。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":2,"is_active":1,"created_at":"2025-08-29 16:33:04","updated_at":"2025-08-29 16:33:04","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":667,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次环保活动中，某学生收集了若干个废旧电池，其中可回收电池比不可回收电池多8个。如果可回收电池的数量是15个，那么不可回收电池有___个。","answer":"7","explanation":"题目中已知可回收电池比不可回收电池多8个，且可回收电池为15个。设不可回收电池的数量为x，根据题意可得方程：15 = x + 8。解这个一元一次方程，两边同时减去8，得到x = 7。因此，不可回收电池有7个。本题考查了一元一次方程的实际应用，属于七年级数学课程中的重点内容。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 22:19:52","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":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":326,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生调查了班级同学最喜欢的运动项目，并将数据整理成如下表格。已知喜欢篮球的人数比喜欢足球的多6人，喜欢乒乓球的人数是喜欢羽毛球的2倍，且总人数为40人。如果喜欢足球的有8人，那么喜欢羽毛球的有多少人？","answer":"B","explanation":"根据题意，喜欢足球的有8人，喜欢篮球的比足球多6人，所以喜欢篮球的有 8 + 6 = 14 人。设喜欢羽毛球的有 x 人，则喜欢乒乓球的有 2x 人。总人数为40人，因此可以列出方程：足球人数 + 篮球人数 + 羽毛球人数 + 乒乓球人数 = 总人数，即 8 + 14 + x + 2x = 40。化简得 22 + 3x = 40，解得 3x = 18，x = 6。所以喜欢羽毛球的有6人，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 15:38:49","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":"5人","is_correct":0},{"id":"B","content":"6人","is_correct":1},{"id":"C","content":"7人","is_correct":0},{"id":"D","content":"8人","is_correct":0}]},{"id":1216,"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参加数学实践活动，要求学生测量校园内一个不规则花坛的边界，并用数学方法估算其面积。花坛的边界由五条线段组成，形成一个凸五边形ABCDE。学生们在平面直角坐标系中建立了模型，测得五个顶点的坐标分别为：A(0, 0)，B(4, 0)，C(6, 3)，D(3, 6)，E(0, 4)。为了估算面积，一名学生提出将五边形分割为三个三角形：△ABC、△ACD和△ADE。请根据该学生的分割方法，利用坐标几何知识，计算该五边形的面积。（提示：可使用向量叉积法或坐标法中的‘鞋带公式’，但需通过三角形面积公式逐步计算）","answer":"解：\n\n我们将五边形ABCDE分割为三个三角形：△ABC、△ACD和△ADE。利用平面直角坐标系中三角形面积的坐标公式：\n\n对于顶点为 (x₁, y₁)，(x₂, y₂)，(x₃, y₃) 的三角形，其面积为：\n\n面积 = ½ |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|\n\n第一步：计算△ABC的面积\nA(0, 0)，B(4, 0)，C(6, 3)\n\nS₁ = ½ |0×(0 - 3) + 4×(3 - 0) + 6×(0 - 0)|\n  = ½ |0 + 4×3 + 0| = ½ × 12 = 6\n\n第二步：计算△ACD的面积\nA(0, 0)，C(6, 3)，D(3, 6)\n\nS₂ = ½ |0×(3 - 6) + 6×(6 - 0) + 3×(0 - 3)|\n  = ½ |0 + 6×6 + 3×(-3)| = ½ |36 - 9| = ½ × 27 = 13.5\n\n第三步：计算△ADE的面积\nA(0, 0)，D(3, 6)，E(0, 4)\n\nS₃ = ½ |0×(6 - 4) + 3×(4 - 0) + 0×(0 - 6)|\n  = ½ |0 + 3×4 + 0| = ½ × 12 = 6\n\n第四步：求总面积\nS = S₁ + S₂ + S₃ = 6 + 13.5 + 6 = 25.5\n\n答：该五边形的面积为25.5平方单位。","explanation":"本题考查平面直角坐标系中多边形面积的坐标计算方法，属于几何与代数综合应用题。解题关键在于将不规则多边形合理分割为若干三角形，并运用坐标法中的三角形面积公式进行逐项计算。题目要求不使用直接套用鞋带公式，而是通过三角形分割的方式，训练学生的图形分析能力和坐标运算能力。该方法不仅巩固了平面直角坐标系的知识，还融合了整式运算（含绝对值与代数式化简），体现了数形结合的思想。难度较高，因涉及多个坐标点的代入、符号处理及多步运算，适合能力较强的七年级学生挑战。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 10:23:18","updated_at":"2026-01-06 10:23:18","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":1282,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某校七年级组织学生参加数学实践活动，调查校园内不同区域的植物种类分布情况。调查结果显示，校园被划分为A、B、C三个区域，每个区域的植物种类数量满足以下条件：A区域的植物种类比B区域多2种；C区域的植物种类是A区域与B区域种类数之和的一半；三个区域植物种类总数为18种。若将A区域的植物种类数设为x，B区域为y，C区域为z，请建立方程组并求解各区域的植物种类数。此外，若学校计划在植物种类最少的区域增加种植，使得该区域种类数增加后，三个区域植物种类数的平均数变为7种，求该区域需要增加多少种植物？","answer":"设A区域的植物种类数为x，B区域为y，C区域为z。\n\n根据题意，列出以下三个方程：\n\n1. A区域比B区域多2种：x = y + 2\n2. C区域是A与B之和的一半：z = (x + y) \/ 2\n3. 三个区域总数为18种：x + y + z = 18\n\n将第1个方程代入第2个方程：\nz = ((y + 2) + y) \/ 2 = (2y + 2) \/ 2 = y + 1\n\n再将x = y + 2 和 z = y + 1 代入第3个方程：\n(y + 2) + y + (y + 1) = 18\n3y + 3 = 18\n3y = 15\ny = 5\n\n代入得：x = 5 + 2 = 7，z = 5 + 1 = 6\n\n所以，A区域有7种，B区域有5种，C区域有6种。\n\n植物种类最少的是B区域（5种）。\n\n设B区域增加k种植物后，三个区域总数为：7 + (5 + k) + 6 = 18 + k\n\n此时平均数为7，即：(18 + k) \/ 3 = 7\n18 + k = 21\nk = 3\n\n答：A区域有7种植物，B区域有5种，C区域有6种；B区域需要增加3种植物，才能使平均数变为7种。","explanation":"本题综合考查二元一次方程组和一元一次方程的应用，结合数据的收集与整理背景，贴近实际生活。首先根据文字描述建立三元一次方程组，通过代入法逐步消元，转化为一元一次方程求解。解题关键在于准确理解‘C区域是A与B之和的一半’这一条件，并将其转化为代数表达式。求得各区域种类数后，进一步分析最小值，并利用平均数的概念建立新方程求解增加量。整个过程涉及方程建模、代数运算和逻辑推理，符合七年级学生对二元一次方程组和数据分析的学习要求，难度较高。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 10:40:35","updated_at":"2026-01-06 10:40:35","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":947,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在某次班级环保活动中，学生们收集废纸进行回收。若每5千克废纸可兑换1个环保积分，某小组共收集了37千克废纸，最多可以兑换___个环保积分。","answer":"7","explanation":"根据题意，每5千克废纸兑换1个环保积分。将总重量37千克除以5，得到37 ÷ 5 = 7.4。由于只能兑换完整的积分，不能兑换部分积分，因此取商的整数部分，即最多可以兑换7个环保积分。本题考查的是有理数中的除法运算及实际问题中的取整应用，属于简单难度，符合七年级学生对有理数运算的理解水平。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 03:27:53","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":1414,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市为改善交通状况，计划在一条主干道旁修建一条自行车专用道。该专用道由两段组成：第一段为直线段，第二段为半圆形弯道，连接直线段的终点并使其与另一条平行道路平滑衔接。已知直线段长度为120米，半圆形弯道的直径与直线段垂直，且整个自行车道的总长度为(120 + 15π)米。现需在该自行车道旁每隔6米安装一盏路灯，起点和终点都必须安装。若每盏路灯的安装成本为80元，且预算中还包含一次性施工费500元，问：该自行车道照明系统的总造价是多少元？请通过计算说明。","answer":"1. 计算半圆形弯道的长度：\n 设半圆形弯道的半径为r米，则其周长为πr（半圆）。\n 根据题意，整个自行车道总长度为：120 + πr = 120 + 15π\n 解得：πr = 15π → r = 15（米）\n\n2. 计算自行车道总长度：\n 直线段：120米\n 半圆段：π × 15 = 15π ≈ 47.1米\n 总长度 = 120 + 15π 米（保留π形式更精确）\n\n3. 计算路灯数量：\n 每隔6米安装一盏，起点和终点都必须安装。\n 路灯数量 = 总长度 ÷ 间隔 + 1\n 但需注意：由于是闭合路径的一部分（非环形），直接按线段处理。\n 总长度为 (120 + 15π) 米，约为 120 + 47.1 = 167.1 米\n 167.1 ÷ 6 ≈ 27.85，说明可以完整安装27个间隔，共28盏灯。\n 验证：27个间隔 × 6米 = 162米 < 167.1米，第28盏灯在终点，符合要求。\n 因此，路灯数量为28盏。\n\n4. 计算总造价：\n 路灯费用：28 × 80 = 2240（元）\n 施工费：500（元）\n 总造价 = 2240 + 500 = 2740（元）\n\n答：该自行车道照明系统的总造价是2740元。","explanation":"本题综合考查了实数运算、一元一次方程、几何图形初步（半圆周长）、有理数运算以及实际应用建模能力。解题关键在于：首先通过总长度表达式建立方程求出半径；其次理解‘每隔6米安装一盏，起点终点都装’意味着路灯数为总长除以间隔后向上取整再加1，但因总长略大于整数倍，需判断最后一个间隔是否足够容纳一盏灯；最后结合有理数乘法与加法完成造价计算。题目情境新颖，融合工程背景，要求学生具备较强的阅读理解与数学建模能力，属于困难级别。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 11:29:31","updated_at":"2026-01-06 11:29:31","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":[]}]