[{"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":[]},{"id":2155,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在数轴上从原点出发，先向右移动3.5个单位长度，再向左移动5.2个单位长度，最后向右移动1.7个单位长度。此时该学生所在位置表示的有理数是多少？","answer":"B","explanation":"该学生从原点0出发，第一次向右移动3.5，到达+3.5；第二次向左移动5.2，即3.5 - 5.2 = -1.7；第三次向右移动1.7，即-1.7 + 1.7 = 0。因此最终位置表示的有理数是0。本题结合数轴与有理数加减的实际情境，考查学生对有理数运算的理解，符合七年级课程要求。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 13:07:43","updated_at":"2026-01-09 13:07:43","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":"-0.5","is_correct":0},{"id":"B","content":"0","is_correct":1},{"id":"C","content":"0.5","is_correct":0},{"id":"D","content":"1","is_correct":0}]},{"id":1490,"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‘校园绿化角’项目，计划在矩形花坛中种植不同种类的植物。花坛的长比宽多4米，若将长减少2米，宽增加3米，则新花坛的面积比原来增加18平方米。现需在花坛四周铺设宽度相同的步行道，使得整个区域（花坛+步行道）的外轮廓仍为一个矩形，且其周长为60米。已知步行道的铺设成本为每平方米80元，求铺设步行道的总费用。","answer":"设原花坛的宽为x米，则长为(x + 4)米。\n\n根据题意，原面积为：x(x + 4) = x² + 4x（平方米）\n\n长减少2米，变为(x + 4 - 2) = (x + 2)米；\n宽增加3米，变为(x + 3)米；\n新面积为：(x + 2)(x + 3) = x² + 5x + 6（平方米）\n\n由题意得：新面积比原面积多18平方米，列方程：\n(x² + 5x + 6) - (x² + 4x) = 18\n化简得：x + 6 = 18\n解得：x = 12\n\n因此，原花坛宽为12米，长为16米。\n\n设步行道的宽度为y米，则整个区域（含步行道）的长为(16 + 2y)米，宽为(12 + 2y)米。\n\n整个区域的周长为60米，列方程：\n2[(16 + 2y) + (12 + 2y)] = 60\n化简：2(28 + 4y) = 60 → 56 + 8y = 60 → 8y = 4 → y = 0.5\n\n步行道宽度为0.5米。\n\n整个区域面积：(16 + 2×0.5)(12 + 2×0.5) = 17 × 13 = 221（平方米）\n原花坛面积：16 × 12 = 192（平方米）\n步行道面积：221 - 192 = 29（平方米）\n\n铺设费用：29 × 80 = 2320（元）\n\n答：铺设步行道的总费用为2320元。","explanation":"本题综合考查了一元一次方程、整式的加减、几何图形初步及实际问题建模能力。首先通过设未知数表示花坛的长和宽，利用面积变化建立一元一次方程，求出原花坛尺寸。接着引入步行道宽度作为新未知数，结合矩形周长公式建立第二个方程，解出步行道宽度。最后通过面积差计算步行道面积，并结合单价求总费用。题目融合了代数运算与几何图形分析，要求学生具备较强的逻辑推理和综合应用能力，属于困难难度。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 12:00:17","updated_at":"2026-01-06 12:00:17","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":968,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生测量了一个长方形花坛的长和宽，记录数据时不小心把单位弄混了。已知花坛的实际周长是 12 米，长比宽多 2 米。如果设宽为 x 米，则根据题意可列出一元一次方程：______ = 12。","answer":"2(x + x + 2)","explanation":"根据题意，设宽为 x 米，则长为 (x + 2) 米。长方形的周长公式为：周长 = 2 × (长 + 宽)。代入得：2 × (x + x + 2) = 12。因此空白处应填写 2(x + x + 2)。该题考查一元一次方程的实际应用，符合七年级数学课程要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 04:04:02","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":1699,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市地铁系统在某一周内每日客流量（单位：万人次）记录如下：周一为 a，周二比周一多 2，周三比周二少 1，周四是周三的 2 倍，周五比周四少 3，周六是周五的一半，周日比周六多 1。已知这一周的平均每日客流量为 8 万人次，且该周总客流量为整数。若 a 为有理数，求 a 的值，并验证该周每日客流量是否均为正数。","answer":"设周一客流量为 a 万人次。\n\n根据题意，逐日表示客流量：\n- 周一：a\n- 周二：a + 2\n- 周三：(a + 2) - 1 = a + 1\n- 周四：2 × (a + 1) = 2a + 2\n- 周五：(2a + 2) - 3 = 2a - 1\n- 周六：(2a - 1) ÷ 2 = a - 0.5\n- 周日：(a - 0.5) + 1 = a + 0.5\n\n一周总客流量为七天之和：\na + (a + 2) + (a + 1) + (2a + 2) + (2a - 1) + (a - 0.5) + (a + 0.5)\n\n合并同类项：\n= a + a + 2 + a + 1 + 2a + 2 + 2a - 1 + a - 0.5 + a + 0.5\n= (a + a + a + 2a + 2a + a + a) + (2 + 1 + 2 - 1 - 0.5 + 0.5)\n= 9a + 4\n\n已知平均每日客流量为 8 万人次，则总客流量为：\n7 × 8 = 56（万人次）\n\n列方程：\n9a + 4 = 56\n\n解方程：\n9a = 56 - 4 = 52\na = 52 ÷ 9 = 52\/9\n\n所以 a = 52\/9\n\n验证每日客流量是否为正数：\n- 周一：52\/9 ≈ 5.78 > 0\n- 周二：52\/9 + 2 = 52\/9 + 18\/9 = 70\/9 ≈ 7.78 > 0\n- 周三：52\/9 + 1 = 52\/9 + 9\/9 = 61\/9 ≈ 6.78 > 0\n- 周四：2 × 61\/9 = 122\/9 ≈ 13.56 > 0\n- 周五：2 × 52\/9 - 1 = 104\/9 - 9\/9 = 95\/9 ≈ 10.56 > 0\n- 周六：95\/9 ÷ 2 = 95\/18 ≈ 5.28 > 0\n- 周日：95\/18 + 1 = 95\/18 + 18\/18 = 113\/18 ≈ 6.28 > 0\n\n所有日客流量均为正数，符合实际意义。\n\n因此，a 的值为 52\/9。","explanation":"本题综合考查有理数运算、整式加减、一元一次方程的建立与求解，以及数据的整理与合理性分析。解题关键在于根据文字描述准确列出每日客流量的代数表达式，利用平均数求出总客流量，建立方程求解未知数 a。同时需注意 a 为有理数，且结果需符合实际情境（客流量为正数）。通过分步推导和验证，确保答案的科学性和合理性。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 13:41:29","updated_at":"2026-01-06 13:41:29","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":2547,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"如图，在平面直角坐标系中，抛物线 y = x² - 4x + 3 与反比例函数 y = k\/x 的图像在第一象限内有一个公共点 P，且点 P 到 x 轴的距离为 1。若将该抛物线绕其顶点旋转 180°，得到新的抛物线，则新抛物线与反比例函数图像的交点个数为多少？","answer":"B","explanation":"首先，求原抛物线 y = x² - 4x + 3 的顶点：配方得 y = (x - 2)² - 1，顶点为 (2, -1)。点 P 在第一象限且在抛物线上，且到 x 轴距离为 1，即纵坐标为 1。代入抛物线方程：1 = x² - 4x + 3，解得 x² - 4x + 2 = 0，解得 x = 2 ± √2。因在第一象限，取 x = 2 + √2，故 P(2 + √2, 1)。又 P 在反比例函数 y = k\/x 上，代入得 k = x·y = (2 + √2)·1 = 2 + √2，故反比例函数为 y = (2 + √2)\/x。将原抛物线绕顶点 (2, -1) 旋转 180°，其开口方向反向，形状不变，新抛物线方程为 y = -(x - 2)² - 1 = -x² + 4x - 5。联立新抛物线与反比例函数：-x² + 4x - 5 = (2 + √2)\/x，两边乘以 x（x ≠ 0）得：-x³ + 4x² - 5x = 2 + √2，即 -x³ + 4x² - 5x - (2 + √2) = 0。此三次方程在实数范围内分析图像趋势：当 x → 0⁺ 时，左边 → -∞；当 x → +∞ 时，-x³ 主导，→ -∞；在 x = 2 附近函数值变化分析可知，函数图像仅穿过 x 轴一次，故仅有一个实数解。因此，新抛物线与反比例函数图像有 1 个交点。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 17:02:12","updated_at":"2026-01-10 17:02:12","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":"0 个","is_correct":0},{"id":"B","content":"1 个","is_correct":1},{"id":"C","content":"2 个","is_correct":0},{"id":"D","content":"3 个","is_correct":0}]},{"id":550,"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120份有效问卷。统计结果显示，其中喜欢数学题的学生有45人，喜欢语文题的有38人，既喜欢数学题又喜欢语文题的有15人。问只喜欢数学题的学生有多少人？","answer":"A","explanation":"根据题意，喜欢数学题的学生共有45人，其中包括了既喜欢数学又喜欢语文的15人。因此，只喜欢数学题的学生人数为：45 - 15 = 30人。本题考查的是数据的收集与整理中的集合基本概念，属于简单难度的应用题，符合七年级‘数据的收集、整理与描述’知识点要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 19:09:10","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":"A","content":"30人","is_correct":1},{"id":"B","content":"33人","is_correct":0},{"id":"C","content":"45人","is_correct":0},{"id":"D","content":"60人","is_correct":0}]},{"id":539,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"在一次班级环保活动中，某学生收集了若干节废旧电池。他将这些电池按每5节装一盒，发现最后剩下2节；如果改为每7节装一盒，则刚好装完，没有剩余。已知他收集的电池总数在30到50之间，那么他一共收集了多少节电池？","answer":"C","explanation":"设该学生收集的电池总数为x节。根据题意：\n1. 每5节装一盒，剩下2节，说明 x 除以5余2，即 x ≡ 2 (mod 5)；\n2. 每7节装一盒，刚好装完，说明 x 能被7整除，即 x ≡ 0 (mod 7)；\n3. 且 30 < x < 50。\n\n在30到50之间，7的倍数有：35、42、49。\n- 35 ÷ 5 = 7 余 0 → 不符合“余2”的条件；\n- 42 ÷ 5 = 8 余 2 → 符合余2的条件；\n- 49 ÷ 5 = 9 余 4 → 不符合。\n\n因此，只有42同时满足被7整除、被5除余2，并且在30到50之间。\n故正确答案是C。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 18:51:32","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":"35","is_correct":0},{"id":"B","content":"37","is_correct":0},{"id":"C","content":"42","is_correct":1},{"id":"D","content":"47","is_correct":0}]},{"id":620,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"72度","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 21:45:21","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":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":[]}]