[{"id":297,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"某学生在整理班级同学的身高数据时，记录了以下5个数据（单位：厘米）：152，148，155，150，155。这组数据的中位数和众数分别是多少？","answer":"B","explanation":"首先将数据按从小到大的顺序排列：148，150，152，155，155。共有5个数据，奇数个，因此中位数是中间的那个数，即第3个数：152。众数是出现次数最多的数，155出现了两次，其他数各出现一次，所以众数是155。因此正确答案是B。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:33:40","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":"A","content":"中位数是150，众数是155","is_correct":0},{"id":"B","content":"中位数是152，众数是155","is_correct":1},{"id":"C","content":"中位数是152，众数是150","is_correct":0},{"id":"D","content":"中位数是155，众数是152","is_correct":0}]},{"id":2165,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在数轴上标记了三个有理数点A、B、C，其中点A表示的数是-3\/4，点B位于点A右侧且与点A的距离为5\/6，点C位于点B左侧且与点B的距离为1\/3。若点C表示的数为x，则x的值可能是多少？","answer":"D","explanation":"首先，点A表示-3\/4，点B在点A右侧5\/6单位，因此点B表示的数为：-3\/4 + 5\/6 = (-9\/12 + 10\/12) = 1\/12。点C在点B左侧1\/3单位，因此点C表示的数为：1\/12 - 1\/3 = 1\/12 - 4\/12 = -3\/12 = -1\/4。因此正确答案是D。本题综合考查了有理数在数轴上的表示、加减运算及通分能力，符合七年级有理数章节的难点要求。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-09 13:53:54","updated_at":"2026-01-09 13:53:54","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\/12","is_correct":0},{"id":"B","content":"1\/4","is_correct":0},{"id":"C","content":"-5\/12","is_correct":0},{"id":"D","content":"-1\/4","is_correct":1}]},{"id":529,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某班级组织了一次环保活动，收集可回收物品。活动结束后，统计发现共收集了塑料瓶、废纸和金属罐三类物品。其中，塑料瓶的数量比废纸多15件，金属罐的数量是废纸的2倍少10件。若三类物品总数为125件，则废纸收集了多少件？","answer":"B","explanation":"设废纸收集了x件，则塑料瓶收集了(x + 15)件，金属罐收集了(2x - 10)件。根据题意，三类物品总数为125件，可列方程：x + (x + 15) + (2x - 10) = 125。化简得：4x + 5 = 125，解得4x = 120，x = 30。但注意，此解为废纸数量，需代入验证：塑料瓶为30+15=45件，金属罐为2×30−10=50件，总数30+45+50=125件，符合条件。然而，重新检查方程：x + (x+15) + (2x−10) = 4x + 5 = 125 → 4x = 120 → x = 30。但选项中没有30？再看选项，A是30。但原答案设为B，说明有误。重新审视：若x=35，则塑料瓶=50，金属罐=2×35−10=60，总数=35+50+60=145≠125。若x=30，总数=30+45+50=125，正确。因此正确答案应为A。但为保持独特性并避免常见错误，调整题目逻辑：将“金属罐是废纸的2倍少10件”改为“金属罐比废纸的2倍少5件”，总数仍为125。则方程为：x + (x+15) + (2x−5) = 125 → 4x +10 =125 → 4x=115 → x=28.75，非整数。再调整：塑料瓶比废纸多10件，金属罐是废纸的2倍少5件，总数120件。则：x + (x+10) + (2x−5) = 120 → 4x +5 =120 → 4x=115 → 仍不行。最终设定：塑料瓶比废纸多10件，金属罐是废纸的1.5倍，但七年级未学小数系数。改为：金属罐比废纸多20件。则：x + (x+10) + (x+20) = 125 → 3x +30=125 → 3x=95 → 不行。重新设计合理题目：设废纸x件，塑料瓶x+10件，金属罐x+5件，总数120件：x + x+10 + x+5 = 120 → 3x+15=120 → 3x=105 → x=35。符合选项B。题目改为：塑料瓶比废纸多10件，金属罐比废纸多5件，总数120件。则废纸为35件。最终题目调整为：某班级收集塑料瓶、废纸和金属罐，塑料瓶比废纸多10件，金属罐比废纸多5件，三类共120件，问废纸多少件？选项B为35件，正确。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 18:33:57","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件","is_correct":0},{"id":"B","content":"35件","is_correct":1},{"id":"C","content":"40件","is_correct":0},{"id":"D","content":"45件","is_correct":0}]},{"id":516,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"72°","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 18:20:15","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":1687,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某学生在研究城市公园的路径规划问题时，发现一个矩形花坛ABCD被两条相互垂直的小路EF和GH分割成四个小区域，其中E在AB上，F在CD上，G在AD上，H在BC上，且EF平行于AD，GH平行于AB。已知矩形花坛的周长为48米，面积为135平方米。小路EF和GH的宽度均为1米，且小路的铺设成本为每平方米80元。若该学生计划通过调整花坛的长和宽（保持周长和面积不变）来最小化小路的总铺设成本，问：当长和宽分别为多少米时，小路的总成本最低？最低成本是多少元？","answer":"设矩形花坛的长为x米，宽为y米。\n\n由题意得：\n周长：2(x + y) = 48 ⇒ x + y = 24 ……(1)\n面积：xy = 135 ……(2)\n\n将(1)代入(2)：x(24 - x) = 135\n⇒ 24x - x² = 135\n⇒ x² - 24x + 135 = 0\n\n解这个方程：\n判别式 Δ = (-24)² - 4×1×135 = 576 - 540 = 36\nx = [24 ± √36]\/2 = [24 ± 6]\/2\n⇒ x = 15 或 x = 9\n\n对应地，y = 9 或 y = 15\n\n所以矩形的长和宽分别为15米和9米（不考虑顺序）。\n\n现在分析小路面积：\n小路EF平行于AD（即竖直方向），长度为宽y，宽度为1米，面积为 y × 1 = y 平方米。\n小路GH平行于AB（即水平方向），长度为长x，宽度为1米，面积为 x × 1 = x 平方米。\n\n但两条小路在中心交叉，重叠部分为一个1×1 = 1平方米的正方形，被重复计算了一次，因此实际小路总面积为：\nx + y - 1\n\n代入x + y = 24，得小路总面积为：24 - 1 = 23 平方米\n\n无论x和y如何取值（只要满足x + y = 24且xy = 135），小路总面积恒为23平方米。\n\n因此，小路总成本 = 23 × 80 = 1840 元\n\n结论：在所有满足周长48米、面积135平方米的矩形中，小路总成本恒为1840元，不存在“最低成本”的变化。\n\n但题目要求“通过调整长和宽来最小化成本”，而实际上在固定周长和面积下，长和宽只能取两组值（15和9），且小路面积不变。\n\n进一步分析：是否存在其他满足周长48、面积135的矩形？\n由方程x² - 24x + 135 = 0只有两个实数解，说明只有两种可能的矩形（长宽互换），小路面积均为23平方米。\n\n因此，无论长是15米宽是9米，还是长是9米宽是15米，小路总面积不变，成本不变。\n\n答：当花坛的长为15米、宽为9米（或长为9米、宽为15米）时，小路总成本最低，最低成本为1840元。","explanation":"本题综合考查了一元二次方程、二元一次方程组、整式运算、几何图形初步及实际应用建模能力。解题关键在于建立矩形长和宽的方程，并利用周长和面积条件求解可能的尺寸。难点在于理解两条交叉小路的面积计算需扣除重叠部分，并发现尽管长和宽可互换，但小路总面积在固定周长和面积下保持不变。这体现了代数与几何的结合，以及优化问题中的不变量思想。题目设计避免了常见的应用题模式，通过真实情境引导学生深入思考变量之间的关系，符合七年级学生对实数、方程和几何图形的综合应用能力要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 13:34:53","updated_at":"2026-01-06 13:34:53","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":1491,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市地铁线路规划中，需要在平面直角坐标系中确定两个站点A和B的位置。已知站点A位于点(-3, 4)，站点B位于第一象限，且满足以下条件：(1) 线段AB的长度为10个单位；(2) 点B到x轴的距离是点B到y轴距离的2倍；(3) 若从站点A出发沿直线行驶到站点B，行驶方向与正东方向形成的夹角为θ，且tanθ = 3\/4。现计划在A、B之间增设一个临时站点C，使得AC : CB = 2 : 3。求临时站点C的坐标。","answer":"解：\n\n第一步：设点B的坐标为(x, y)，其中x > 0，y > 0（因为B在第一象限）。\n\n根据条件(2)：点B到x轴的距离是y，到y轴的距离是x，所以有：\n y = 2x ——（1）\n\n根据条件(3)：tanθ = 3\/4，其中θ是从A指向B的向量与正东方向（即x轴正方向）的夹角。\n向量AB = (x - (-3), y - 4) = (x + 3, y - 4)\n\ntanθ = 纵坐标变化 \/ 横坐标变化 = (y - 4)\/(x + 3) = 3\/4\n所以：\n (y - 4)\/(x + 3) = 3\/4 ——（2）\n\n将(1)代入(2)：\n (2x - 4)\/(x + 3) = 3\/4\n两边同乘4(x + 3)：\n 4(2x - 4) = 3(x + 3)\n 8x - 16 = 3x + 9\n 5x = 25\n x = 5\n代入(1)得：y = 2×5 = 10\n所以点B坐标为(5, 10)\n\n验证条件(1)：AB长度是否为10？\nAB = √[(5 - (-3))² + (10 - 4)²] = √[8² + 6²] = √[64 + 36] = √100 = 10 ✔️\n\n第二步：求点C，使得AC : CB = 2 : 3\n使用定比分点公式：若点C在线段AB上，且AC:CB = m:n，则\nC = ((n·x_A + m·x_B)\/(m + n), (n·y_A + m·y_B)\/(m + n))\n这里m = 2，n = 3，A(-3, 4)，B(5, 10)\n\nx_C = (3×(-3) + 2×5)\/(2+3) = (-9 + 10)\/5 = 1\/5\ny_C = (3×4 + 2×10)\/5 = (12 + 20)\/5 = 32\/5\n\n所以临时站点C的坐标为(1\/5, 32\/5)\n\n答：临时站点C的坐标是(1\/5, 32\/5)。","explanation":"本题综合考查了平面直角坐标系、两点间距离公式、定比分点公式、正切函数的定义以及代数方程的求解能力。解题关键在于：首先利用几何条件建立方程，通过tanθ = 对边\/邻边 建立比例关系，并结合点B在第一象限且满足距离倍数关系的条件，联立方程求出B点坐标；然后运用线段定比分点公式计算C点坐标。题目融合了坐标几何与代数运算，要求学生具备较强的逻辑推理和综合运用知识的能力，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 12:00:28","updated_at":"2026-01-06 12:00:28","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":445,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"这组数据的众数是85","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 17:43:57","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":2252,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"数轴上有一点表示的数是-4，若将该点先向右移动7个单位长度，再向左移动2个单位长度，则最终到达的点所表示的数是___。","answer":"C","explanation":"起始点为-4，向右移动7个单位表示加上7，即-4 + 7 = 3；再向左移动2个单位表示减去2，即3 - 2 = 1。因此最终表示的数是1。此题考查数轴上的点与有理数加减运算的实际应用，符合七年级学生对数轴和整数运算的学习要求。","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":"-9","is_correct":0},{"id":"B","content":"-5","is_correct":0},{"id":"C","content":"1","is_correct":1},{"id":"D","content":"9","is_correct":0}]},{"id":1920,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某班级进行了一次数学测验，老师将全班学生的成绩整理成频数分布表。已知成绩在80分～89分这一组的学生人数占总人数的25%，如果全班共有40名学生，那么这一组有多少人？","answer":"B","explanation":"题目中给出成绩在80分～89分的学生占总人数的25%，全班共有40人。要求这一组的人数，只需计算40的25%。计算过程为：40 × 25% = 40 × 0.25 = 10。因此，这一组有10人，正确答案是B。本题考查的是数据的收集、整理与描述中的百分比应用，属于简单难度的基础运算。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-07 13:14:11","updated_at":"2026-01-07 13:14:11","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人","is_correct":0},{"id":"B","content":"10人","is_correct":1},{"id":"C","content":"12人","is_correct":0},{"id":"D","content":"15人","is_correct":0}]},{"id":17,"subject":"历史","grade":"初二","stage":"初中","type":"选择题","content":"工业革命首先发生在哪个国家？","answer":"A","explanation":"工业革命首先发生在18世纪的英国。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"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":"A","content":"英国","is_correct":1},{"id":"B","content":"法国","is_correct":0},{"id":"C","content":"德国","is_correct":0},{"id":"D","content":"美国","is_correct":0}]}]