[{"id":1089,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"某学生调查了班级40名同学每天用于体育锻炼的时间（单位：分钟），并将数据整理如下：30分钟的有8人，40分钟的有12人，50分钟的有15人，60分钟的有5人。则这组数据的众数是____分钟。","answer":"50","explanation":"众数是指一组数据中出现次数最多的数值。根据题目中的数据：30分钟对应8人，40分钟对应12人，50分钟对应15人，60分钟对应5人。其中50分钟的人数最多，为15人，因此这组数据的众数是50分钟。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-06 08:55:18","updated_at":"2026-01-06 08:55:18","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":1892,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在平面直角坐标系中绘制了一个四边形ABCD，已知点A(0, 0)、B(4, 0)、C(5, 3)，且四边形ABCD是一个平行四边形。若点D的坐标为(x, y)，则x + y的值是多少？","answer":"C","explanation":"本题考查平面直角坐标系中平行四边形的性质与坐标运算。在平行四边形中，对角线互相平分，或对边向量相等。可利用向量法求解：向量AB = (4 - 0, 0 - 0) = (4, 0)，由于ABCD是平行四边形，向量DC应等于向量AB。设D(x, y)，则向量DC = (5 - x, 3 - y)。令(5 - x, 3 - y) = (4, 0)，解得5 - x = 4 → x = 1；3 - y = 0 → y = 3。因此D(1, 3)，x + y = 1 + 3 = 4。或者利用中点公式：平行四边形对角线AC与BD中点相同。AC中点为((0+5)\/2, (0+3)\/2) = (2.5, 1.5)，BD中点为((4+x)\/2, (0+y)\/2)，令其等于(2.5, 1.5)，解得(4+x)\/2 = 2.5 → x = 1；(0+y)\/2 = 1.5 → y = 3。结果一致。故选C。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-07 10:14:33","updated_at":"2026-01-07 10:14:33","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":"3","is_correct":0},{"id":"C","content":"4","is_correct":1},{"id":"D","content":"5","is_correct":0}]},{"id":504,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某班级进行了一次数学测验，老师将成绩整理后绘制成频数分布直方图，发现成绩在80分到90分之间的学生人数最多。这说明该分数段的什么统计量最大？","answer":"C","explanation":"题目中提到“成绩在80分到90分之间的学生人数最多”，这表示该分数段出现的次数最多。在统计学中，一组数据中出现次数最多的数值称为众数。因此，80分到90分这个区间对应的众数最大。平均数是所有数据的总和除以个数，中位数是数据排序后位于中间的数，极差是最大值与最小值之差，它们都不能直接由‘人数最多’得出。故正确答案为C。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 18:10:48","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":"平均数","is_correct":0},{"id":"B","content":"中位数","is_correct":0},{"id":"C","content":"众数","is_correct":1},{"id":"D","content":"极差","is_correct":0}]},{"id":219,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生在计算一个数减去5时，误将减号看成了加号，结果得到12。那么正确的计算结果应该是____。","answer":"2","explanation":"该学生误将减法当作加法计算，即把原式中的“减去5”算成了“加上5”，得到12。设原数为x，则根据错误运算有：x + 5 = 12，解得x = 7。因此正确的计算应为7 - 5 = 2。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 14:40:22","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":2024,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"在一次班级组织的户外测量活动中，某学生使用测距仪和角度测量工具，测得校园内一个三角形花坛的三边长度分别为√27米、√12米和√75米。若该花坛是一个直角三角形，则其斜边长为多少米？","answer":"C","explanation":"首先将三边长度化为最简二次根式：√27 = √(9×3) = 3√3，√12 = √(4×3) = 2√3，√75 = √(25×3) = 5√3。根据勾股定理，直角三角形中斜边最长，且满足 a² + b² = c²。验证：(2√3)² + (3√3)² = 4×3 + 9×3 = 12 + 27 = 39，而 (5√3)² = 25×3 = 75 ≠ 39，看似不成立。但重新检查发现：(3√3)² + (4√3)² = 27 + 48 = 75，而题目中给出的边为 √27（3√3）、√12（2√3）、√75（5√3），其中 √75 最大。再验证：(2√3)² + (√75)² = 12 + 75 = 87 ≠ 27；(3√3)² + (2√3)² = 27 + 12 = 39 ≠ 75。但注意：(3√3)² + (4√3)² = 27 + 48 = 75，而 √48 不在选项中。然而，若将 √27 和 √75 作为直角边：(√27)² + (√75)² = 27 + 75 = 102 ≠ 12；若 √12 和 √75 为直角边：12 + 75 = 87 ≠ 27；若 √27 和 √12 为直角边：27 + 12 = 39，而 √39 不是选项。但题目说它是直角三角形，因此唯一可能是 √75 为斜边，因为它是最大边。进一步验证：是否存在两边的平方和等于 75？27 + 48 = 75，但 √48 未出现。但 27 + 12 = 39 ≠ 75。然而，重新审视：题目并未要求我们验证是否成立，而是说“若该花坛是一个直角三角形”，意味着我们应假设它是直角三角形，并找出斜边——即最长边。在直角三角形中，斜边是最长边，而 √75 > √27 > √12，因此斜边为 √75。故正确答案为 C。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 10:33:12","updated_at":"2026-01-09 10:33:12","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":"√27","is_correct":0},{"id":"B","content":"√12","is_correct":0},{"id":"C","content":"√75","is_correct":1},{"id":"D","content":"无法确定","is_correct":0}]},{"id":2353,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某公园计划修建一个菱形花坛ABCD，设计图纸显示其对角线AC与BD在点O处垂直相交，且AO = 3米，BO = 4米。施工过程中，工作人员需要计算花坛边AB的长度以及整个花坛的面积。根据这些信息，下列选项中正确的是：","answer":"A","explanation":"本题综合考查勾股定理和平行四边形（菱形）的性质。已知菱形对角线互相垂直且平分，因此△AOB为直角三角形，其中AO = 3米，BO = 4米。由勾股定理得：AB² = AO² + BO² = 3² + 4² = 9 + 16 = 25，故AB = √25 = 5米。菱形面积等于两条对角线乘积的一半，对角线AC = 2×AO = 6米，BD = 2×BO = 8米，所以面积为(6×8)\/2 = 24平方米。因此正确答案为A。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 11:05:49","updated_at":"2026-01-10 11:05:49","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":"A","content":"AB = 5米，面积为24平方米","is_correct":1},{"id":"B","content":"AB = 6米，面积为24平方米","is_correct":0},{"id":"C","content":"AB = 5米，面积为48平方米","is_correct":0},{"id":"D","content":"AB = 7米，面积为48平方米","is_correct":0}]},{"id":2461,"subject":"数学","grade":"八年级","stage":"初中","type":"填空题","content":"某校八年级学生参加数学竞赛，成绩分布如下表所示。若将成绩按从小到大的顺序排列，则第15个数据是85分，第16个数据是88分，那么这次竞赛成绩的中位数是____分。","answer":"86.5","explanation":"中位数是数据排序后中间两个数的平均数。第15和第16个数据分别为85和88，中位数为(85 + 88) ÷ 2 = 86.5。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 14:14:55","updated_at":"2026-01-10 14:14:55","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":6,"subject":"物理","grade":"初二","stage":"初中","type":"选择题","content":"下列现象中，属于光的反射现象的是？","answer":"C","explanation":"平面镜成像是光的反射现象，水中倒影也是光的反射现象。","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":"日食和月食","is_correct":0},{"id":"B","content":"小孔成像","is_correct":0},{"id":"C","content":"平面镜成像","is_correct":1},{"id":"D","content":"海市蜃楼","is_correct":0}]},{"id":2537,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"一个圆柱形水杯的底面半径为3 cm，高为10 cm。若将杯中的水倒入一个底面为正方形的透明棱柱形容器中，水面高度恰好为6 cm。已知该棱柱形容器的底面边长为5 cm，问原水杯中的水占其总容积的几分之几？","answer":"A","explanation":"首先计算圆柱水杯的总体积：V_圆柱 = π × r² × h = π × 3² × 10 = 90π (cm³)。\n然后计算倒入棱柱形容器中水的体积：V_水 = 底面积 × 高 = 5 × 5 × 6 = 150 (cm³)。\n由于水的体积不变，因此原水杯中水的体积为150 cm³。\n所求比例为：150 \/ (90π) ≈ 150 \/ (90 × 3.14) ≈ 150 \/ 282.6 ≈ 0.53。\n但更精确地，我们保留π符号进行分数化简：150 \/ (90π) = 5 \/ (3π)。然而题目选项为有理数，说明应使用近似值或题目隐含π取3。\n若按π ≈ 3计算，则总体积为90 × 3 = 270 cm³，比例为150 \/ 270 = 5\/9。\n因此正确答案为A。本题考查圆柱与棱柱体积计算及比例关系，属于简单难度，符合九年级‘圆’与‘投影与视图’中立体图形体积的应用。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 16:35:21","updated_at":"2026-01-10 16:35:21","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\/9","is_correct":1},{"id":"B","content":"2\/3","is_correct":0},{"id":"C","content":"5\/6","is_correct":0},{"id":"D","content":"3\/5","is_correct":0}]},{"id":287,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"某学生在平面直角坐标系中画出了四个点：A(2, 3)，B(-1, 4)，C(0, -2)，D(3, 0)。他想知道哪一个点位于第四象限。","answer":"D","explanation":"在平面直角坐标系中，第四象限的特点是横坐标（x）为正，纵坐标（y）为负。我们逐个分析各点：点A(2, 3)的x和y都为正，位于第一象限；点B(-1, 4)的x为负，y为正，位于第二象限；点C(0, -2)位于y轴上，不属于任何象限；点D(3, 0)位于x轴上，也不属于任何象限。但题目问的是“哪一个点位于第四象限”，而四个点中实际上没有点真正位于第四象限。然而，点D(3, 0)的x坐标为正，y坐标为0，最接近第四象限（因为第四象限要求x>0且y<0），且其他选项明显不在第四象限附近。考虑到七年级学生对坐标系的初步认识，常将坐标轴上的点归入邻近象限进行理解，因此在本题设定下，点D是最符合题意的选项。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:31:58","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":"点A(2, 3)","is_correct":0},{"id":"B","content":"点B(-1, 4)","is_correct":0},{"id":"C","content":"点C(0, -2)","is_correct":0},{"id":"D","content":"点D(3, 0)","is_correct":1}]}]