[{"id":688,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次环保活动中，某学生记录了连续5天每天回收的塑料瓶数量（单位：个）分别为：18、22、20、25、15。若将这5天的数据按从小到大的顺序排列，则位于中间的那个数是____。","answer":"20","explanation":"题目考查的是数据的收集与整理中的中位数概念。首先将数据从小到大排序：15、18、20、22、25。由于共有5个数据（奇数个），中位数就是正中间的那个数，即第3个数，为20。因此答案是20。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 22:34:23","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":2480,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"某学生用一块半径为6 cm的圆形纸板制作一个圆锥形帽子，他将圆形纸板剪去一个扇形后，将剩余部分沿半径粘合形成圆锥的侧面。若圆锥底面圆的周长恰好为4π cm，则被剪去的扇形的圆心角是多少度？","answer":"C","explanation":"本题考查圆的周长与扇形圆心角的关系，属于圆的相关知识，难度为简单。\n\n解题思路如下：\n\n1. 原圆形纸板半径为6 cm，即圆锥的母线长为6 cm。\n2. 圆锥底面周长为4π cm，根据圆周长公式 C = 2πr，可得底面半径 r = (4π) \/ (2π) = 2 cm。\n3. 圆锥侧面展开图是一个扇形，其弧长等于底面圆的周长，即弧长为4π cm。\n4. 扇形所在圆的半径为6 cm，整个圆的周长为 2π × 6 = 12π cm。\n5. 扇形的圆心角 θ 满足比例关系：θ \/ 360° = 弧长 \/ 圆周长 = 4π \/ 12π = 1\/3。\n6. 因此，θ = 360° × (1\/3) = 120°，这是剩余扇形的圆心角。\n7. 被剪去的扇形圆心角 = 360° - 120° = 240°。\n\n故正确答案为 C。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 15:08:32","updated_at":"2026-01-10 15:08:32","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":"60°","is_correct":0},{"id":"B","content":"120°","is_correct":0},{"id":"C","content":"240°","is_correct":1},{"id":"D","content":"300°","is_correct":0}]},{"id":513,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"在一次班级环保活动中，某学生收集了废旧纸张和塑料瓶两类可回收物。已知他收集的塑料瓶数量比废旧纸张数量的2倍少3个，总共收集了27个物品。设废旧纸张的数量为x个，则根据题意可列出一元一次方程，求出x的值是：","answer":"A","explanation":"设废旧纸张的数量为x个，则塑料瓶的数量为2x - 3个。根据题意，总数量为27个，因此可列方程：x + (2x - 3) = 27。化简得：3x - 3 = 27，移项得：3x = 30，解得：x = 10。因此，废旧纸张的数量为10个，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 18:17:07","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":"12","is_correct":0},{"id":"C","content":"15","is_correct":0},{"id":"D","content":"18","is_correct":0}]},{"id":2134,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在解方程时，将方程 3x + 5 = 20 的第一步写为 3x = 15。请问该学生在这一步中运用了等式的哪一条基本性质？","answer":"B","explanation":"该学生将方程 3x + 5 = 20 变形为 3x = 15，是将等式两边同时减去了 5，从而消去左边的常数项。这一操作依据的是等式的基本性质：等式两边同时减去同一个数，等式仍然成立。因此正确答案是 B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 12:56:39","updated_at":"2026-01-09 12:56:39","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":"等式两边同时加上同一个数，等式仍然成立","is_correct":0},{"id":"B","content":"等式两边同时减去同一个数，等式仍然成立","is_correct":1},{"id":"C","content":"等式两边同时乘以同一个数，等式仍然成立","is_correct":0},{"id":"D","content":"等式两边同时除以同一个数，等式仍然成立","is_correct":0}]},{"id":2289,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"在数轴上，点A表示的数是-3，点B与点A的距离为7个单位长度，且点B在原点右侧。若点C位于点A和点B之间，且AC:CB = 2:5，则点C所表示的数为____。","answer":"-1","explanation":"首先，点A表示-3，点B在A右侧且距离为7，因此点B表示的数为-3 + 7 = 4。点C在A和B之间，且AC:CB = 2:5，说明将AB线段分成2+5=7等份，AC占2份。AB总长为7，每份为1单位长度，因此AC = 2。从点A（-3）向右移动2个单位，得到点C的坐标为-3 + 2 = -1。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-09 16:44:29","updated_at":"2026-01-09 16:44:29","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":846,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次环保活动中，某班级学生收集废旧纸张和塑料瓶两类物品。已知收集废旧纸张的人数比收集塑料瓶的人数多5人，两类活动共有31人参加，且每人只参加一类。若设收集塑料瓶的人数为x，则根据题意可列出一元一次方程：_x + (x + 5) = 31_，解得x = _13_，因此收集废旧纸张的人数是_18_人。","answer":"x + (x + 5) = 31；13；18","explanation":"设收集塑料瓶的人数为x，则收集废旧纸张的人数为x + 5。根据总人数为31人，可列方程：x + (x + 5) = 31。化简得2x + 5 = 31，解得2x = 26，x = 13。因此收集塑料瓶的有13人，收集废旧纸张的有13 + 5 = 18人。本题考查一元一次方程的实际应用，结合生活情境，帮助学生理解方程建模的基本方法。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 01:02:32","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":1974,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"某学生在操场上竖立了一根高度为2米的旗杆，正午时太阳光线与地面形成的仰角为30°。若此时旗杆在地面上的影长为a米，则a的值最接近以下哪个选项？（已知√3≈1.732）","answer":"C","explanation":"本题考查锐角三角函数中正切函数的应用。旗杆垂直于地面，影长与旗杆构成一个直角三角形，其中旗杆为对边，影长为邻边，太阳光线与地面的夹角为30°。根据正切定义：tan(30°) = 对边 \/ 邻边 = 2 \/ a。又因为 tan(30°) = 1\/√3 ≈ 0.577，所以有 2 \/ a = 1\/√3，解得 a = 2√3 ≈ 2 × 1.732 = 3.464。因此，影长a最接近3.46米，正确答案为C。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-07 14:59:07","updated_at":"2026-01-07 14:59:07","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":"1.15","is_correct":0},{"id":"B","content":"2.00","is_correct":0},{"id":"C","content":"3.46","is_correct":1},{"id":"D","content":"4.62","is_correct":0}]},{"id":2521,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"某学生观察一个由三个全等的等边三角形拼接而成的轴对称图形（如图，未展示），若将该图形绕其对称中心旋转一定角度后能与原图形完全重合，则这个旋转角度最小为多少？","answer":"C","explanation":"该图形由三个全等的等边三角形拼接而成，且具有轴对称性。由于等边三角形的每个内角为60°，三个三角形围绕中心拼接时，中心点周围的角度总和为360°，因此每个三角形占据120°的扇形区域。要使图形绕对称中心旋转后与自身重合，最小的旋转角度应等于其旋转对称的最小单位角度。因为图形具有三重旋转对称性（即每转120°重合一次），所以最小旋转角度为360° ÷ 3 = 120°。选项C正确。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 15:56:56","updated_at":"2026-01-10 15:56:56","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":"60°","is_correct":0},{"id":"B","content":"90°","is_correct":0},{"id":"C","content":"120°","is_correct":1},{"id":"D","content":"180°","is_correct":0}]},{"id":1906,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某班级组织了一次环保知识竞赛，参赛学生需完成一份包含10道选择题的试卷。每答对一题得5分，答错或不答扣2分。一名学生最终得分为29分，请问这名学生答对了多少道题？","answer":"B","explanation":"设这名学生答对了x道题，则答错或不答的题目数为(10 - x)道。根据得分规则：每答对一题得5分，答错或不答扣2分，总得分为29分，可列出一元一次方程：5x - 2(10 - x) = 29。展开并化简：5x - 20 + 2x = 29 → 7x = 49 → x = 7。因此，这名学生答对了7道题。验证：7×5 = 35分，答错3题扣3×2 = 6分，35 - 6 = 29分，符合题意。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-07 13:10:44","updated_at":"2026-01-07 13:10:44","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":"6道","is_correct":0},{"id":"B","content":"7道","is_correct":1},{"id":"C","content":"8道","is_correct":0},{"id":"D","content":"9道","is_correct":0}]},{"id":1347,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某学生在研究平面直角坐标系中的图形变换时，发现一个有趣的规律：将点 A(a, b) 先向右平移 3 个单位，再向下平移 2 个单位，得到点 A'；然后将点 A' 关于 x 轴对称，得到点 A''。已知点 A'' 的坐标为 (5, -4)。同时，该学生还发现，若将原点 O(0, 0) 按照同样的变换步骤（先向右平移 3 个单位，再向下平移 2 个单位，最后关于 x 轴对称），得到的新点与原点之间的距离是一个无理数。求：(1) 点 A 的原始坐标 (a, b)；(2) 原点 O 经过上述变换后得到的点与原点之间的距离（保留根号形式）。","answer":"(1) 设点 A 的原始坐标为 (a, b)。\n第一步：向右平移 3 个单位，得到点 (a + 3, b)；\n第二步：向下平移 2 个单位，得到点 (a + 3, b - 2)；\n第三步：关于 x 轴对称，横坐标不变，纵坐标变为相反数，得到点 (a + 3, -(b - 2)) = (a + 3, -b + 2)。\n根据题意，该点即为 A''(5, -4)，所以有：\n a + 3 = 5\n -b + 2 = -4\n解第一个方程：a = 5 - 3 = 2\n解第二个方程：-b = -6 ⇒ b = 6\n因此，点 A 的原始坐标为 (2, 6)。\n\n(2) 对原点 O(0, 0) 进行相同变换：\n第一步：向右平移 3 个单位 → (0 + 3, 0) = (3, 0)\n第二步：向下平移 2 个单位 → (3, 0 - 2) = (3, -2)\n第三步：关于 x 轴对称 → (3, -(-2)) = (3, 2)\n得到的新点为 P(3, 2)。\n计算点 P 与原点 O(0, 0) 之间的距离：\n距离 = √[(3 - 0)² + (2 - 0)²] = √(9 + 4) = √13\n因此，距离为 √13。","explanation":"本题综合考查了平面直角坐标系中的坐标变换（平移与对称）、坐标运算以及两点间距离公式。第一问通过逆向推理，从最终坐标反推出原始坐标，需要学生理解每一步变换对坐标的影响，并建立方程求解。第二问则要求学生正确执行变换步骤，并运用勾股定理计算距离，涉及实数中的无理数概念。题目设计避免了常见的生活情境，以数学探究为背景，强调逻辑推理与多步骤操作能力，符合困难难度要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 11:03:37","updated_at":"2026-01-06 11:03:37","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":[]}]