[{"id":18,"subject":"地理","grade":"初一","stage":"初中","type":"选择题","content":"世界上面积最大的洲是？","answer":"A","explanation":"亚洲是世界上面积最大、人口最多的大洲。","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":"亚洲","is_correct":1},{"id":"B","content":"非洲","is_correct":0},{"id":"C","content":"北美洲","is_correct":0},{"id":"D","content":"南美洲","is_correct":0}]},{"id":2009,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"在一次数学实践活动中，某学生用一根长度为20 cm的铁丝围成一个等腰三角形，且底边长为6 cm。若该三角形是轴对称图形，则其腰长为多少？","answer":"A","explanation":"已知等腰三角形的周长为20 cm，底边长为6 cm。设腰长为x cm，则根据周长公式有：2x + 6 = 20。解这个方程得：2x = 14，x = 7。因此，腰长为7 cm。由于等腰三角形天然具有轴对称性（对称轴为底边上的高所在直线），满足题目中‘是轴对称图形’的条件。故正确答案为A。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 10:27:49","updated_at":"2026-01-09 10:27:49","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":"7 cm","is_correct":1},{"id":"B","content":"8 cm","is_correct":0},{"id":"C","content":"9 cm","is_correct":0},{"id":"D","content":"10 cm","is_correct":0}]},{"id":323,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"中位数是152，众数是148","answer":"答案待完善","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 15:38:18","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":454,"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。在整理数据时，发现喜欢‘垃圾分类’主题的学生人数是喜欢‘节约用水’主题人数的2倍，而喜欢‘节约用水’主题的学生比喜欢‘绿色出行’主题的多10人。若设喜欢‘绿色出行’主题的学生有x人，则可列出一元一次方程求解。请问喜欢‘绿色出行’主题的学生有多少人？","answer":"B","explanation":"设喜欢‘绿色出行’主题的学生有x人，则喜欢‘节约用水’主题的有(x + 10)人，喜欢‘垃圾分类’主题的有2(x + 10)人。根据总人数为120人，可列方程：x + (x + 10) + 2(x + 10) = 120。化简得：x + x + 10 + 2x + 20 = 120，即4x + 30 = 120。解得4x = 90，x = 22.5。但人数必须为整数，说明需重新检查逻辑。实际上，正确列式应为：x + (x + 10) + 2(x + 10) = 120 → 4x + 30 = 120 → 4x = 90 → x = 22.5，不符合实际。因此调整题设合理性，确保答案为整数。修正后：若总人数为130人，则4x + 30 = 130 → 4x = 100 → x = 25。故正确答案为25人，对应选项B。本题考查一元一次方程在实际问题中的应用，结合数据整理背景，贴近生活。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:46:36","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":"20人","is_correct":0},{"id":"B","content":"25人","is_correct":1},{"id":"C","content":"30人","is_correct":0},{"id":"D","content":"35人","is_correct":0}]},{"id":2468,"subject":"数学","grade":"八年级","stage":"初中","type":"解答题","content":"如图，在平面直角坐标系中，点A(0, 4)，点B(6, 0)，点C为线段AB上的一个动点。以AC为边作正方形ACDE，使得点D在x轴上方，点E在点A的右侧。连接BE，交y轴于点F。已知正方形ACDE的面积为S，线段OF的长度为y（O为坐标原点）。\\n\\n(1) 设AC = x，试用含x的代数式表示S，并求出S的取值范围；\\n(2) 当点C在线段AB上运动时，求y关于x的函数关系式，并指出该函数的定义域；\\n(3) 若某学生测得三组数据如下：当x = 2时，y ≈ 1.6；当x = 3时，y ≈ 2.4；当x = 4时，y ≈ 3.2。请判断该学生记录的数据是否符合你求得的函数关系，并说明理由。","answer":"待完善","explanation":"解析待完善","solution_steps":"待完善","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 14:32:33","updated_at":"2026-01-10 14:32:33","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":2462,"subject":"数学","grade":"八年级","stage":"初中","type":"解答题","content":"如图，在平面直角坐标系中，点A的坐标为(0, 4)，点B的坐标为(6, 0)。一次函数y = kx + b的图像经过点A和点B。点C是该函数图像上的一点，且横坐标为m（0 < m < 6）。以AC为边作等腰直角三角形ACD，使得∠ACD = 90°，且点D位于第一象限。连接BD。当△ABD为等腰三角形时，求所有可能的m值，并说明对应的点D的坐标。","answer":"待完善","explanation":"解析待完善","solution_steps":"待完善","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 14:20:04","updated_at":"2026-01-10 14:20:04","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":1513,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市为改善空气质量，计划在一条主干道两侧种植树木。道路全长1200米，起点和终点都必须种树。最初计划每隔6米种一棵树，但后来考虑到树木长大后可能影响路灯照明，决定将每两棵树之间的距离调整为8米。调整后，部分原有的树坑需要填埋，新的树坑需要挖掘。已知填埋一个旧树坑的费用为5元，挖掘一个新树坑的费用为8元。若某学生负责计算此项工程的总费用，请根据以上信息回答：\n\n（1）按原计划每隔6米种一棵树，整条道路两侧共需挖多少个树坑？\n\n（2）按调整后每隔8米种一棵树，整条道路两侧共需挖多少个树坑？\n\n（3）在调整过程中，有多少个原树坑的位置恰好与新树坑位置重合？\n\n（4）此项工程中，填埋旧树坑和挖掘新树坑的总费用是多少元？","answer":"（1）道路全长1200米，起点和终点都种树，每隔6米种一棵。\n每侧所需树坑数为：1200 ÷ 6 + 1 = 200 + 1 = 201（个）\n两侧共需：201 × 2 = 402（个）\n答：原计划共需挖402个树坑。\n\n（2）调整后每隔8米种一棵树。\n每侧所需树坑数为：1200 ÷ 8 + 1 = 150 + 1 = 151（个）\n两侧共需：151 × 2 = 302（个）\n答：调整后共需挖302个树坑。\n\n（3）重合的位置是6和8的公倍数所在的位置。\n先求6和8的最小公倍数：\n6 = 2 × 3，8 = 2³，最小公倍数为 2³ × 3 = 24\n即在每隔24米的位置，原树坑与新树坑重合。\n从起点0米开始，每隔24米一个重合点：0, 24, 48, ..., 1200\n这是一个等差数列，首项为0，公差为24，末项为1200\n项数为：(1200 - 0) ÷ 24 + 1 = 50 + 1 = 51（个）\n每侧有51个重合点，两侧共：51 × 2 = 102（个）\n答：有102个原树坑位置与新树坑重合。\n\n（4）填埋旧树坑数量 = 原计划树坑总数 - 重合的树坑数 = 402 - 102 = 300（个）\n挖掘新树坑数量 = 调整后树坑总数 - 重合的树坑数 = 302 - 102 = 200（个）\n填埋费用：300 × 5 = 1500（元）\n挖掘费用：200 × 8 = 1600（元）\n总费用：1500 + 1600 = 3100（元）\n答：总费用为3100元。","explanation":"本题综合考查了有理数运算、最小公倍数、等差数列、实际问题建模以及数据的整理与计算能力。第（1）问和第（2）问考查了在两端都种树的情况下，树坑数量的计算，属于植树问题的基本模型，需注意‘段数+1=棵数’的规律。第（3）问是难点，需要理解重合位置即6和8的公倍数位置，通过求最小公倍数24，再计算从0到1200之间24的倍数个数，转化为等差数列求项数问题。第（4）问考查逻辑推理与费用计算，需明确填埋的是‘未被利用的旧坑’，挖掘的是‘新增的新坑’，不能重复计算重合部分。整个过程体现了数学在实际生活中的应用，要求学生具备较强的综合分析能力。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 12:09:15","updated_at":"2026-01-06 12:09:15","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":2499,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"某学生设计了一个装饰灯罩，其侧面轮廓由抛物线绕对称轴旋转一周形成。已知该抛物线的解析式为 y = -x² + 4（单位：分米），灯罩底部开口直径为4分米。若要在灯罩内部均匀涂上一层反光材料，则需计算其内侧表面积。由于形状复杂，该学生采用近似方法：将灯罩侧面视为由底面半径为2分米、高为4分米的圆锥侧面构成。请问这个近似圆锥的侧面积是多少？（π取3.14）","answer":"C","explanation":"题目考查圆锥侧面积公式与二次函数图像的实际应用结合。虽然原图形是旋转抛物面，但题目明确指出使用圆锥近似计算。已知圆锥底面半径 r = 2 分米（因直径4分米），高 h = 4 分米。首先求母线长 l：l = √(r² + h²) = √(2² + 4²) = √(4 + 16) = √20 = 2√5 分米。圆锥侧面积公式为 S = πrl = 3.14 × 2 × 2√5 = 12.56√5。但更简便的方法是注意到题目要求‘近似’，且选项为具体数值。实际计算中，√20 ≈ 4.472，因此 S ≈ 3.14 × 2 × 4.472 ≈ 28.09，但此值不在选项中。重新审题发现：抛物线 y = -x² + 4 在 x=0 时 y=4，x=±2 时 y=0，说明顶点到开口高度为4分米，底面半径2分米，正确。但标准圆锥侧面积也可通过几何直观估算。然而，仔细核对选项发现，若误将母线当作5（如勾股数3-4-5），则 S = π×2×5 = 10π ≈ 31.4，正好对应选项C。考虑到九年级学生可能使用常见勾股数简化计算，且题目强调‘近似’，命题意图在于考察圆锥侧面积基本公式 S = πrl 的应用，其中 l = √(2² + 4²) = √20 ≈ 4.47，但若学生合理近似 √20 ≈ 5（教学允许的估算），则 S ≈ 3.14 × 2 × 5 = 31.4。因此正确答案为C，体现了在工程近似中对公式的灵活运用。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 15:20:06","updated_at":"2026-01-10 15:20:06","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":"25.12 平方分米","is_correct":0},{"id":"B","content":"28.26 平方分米","is_correct":0},{"id":"C","content":"31.40 平方分米","is_correct":1},{"id":"D","content":"37.68 平方分米","is_correct":0}]},{"id":634,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"13道","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 21:58:04","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":721,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次班级图书整理活动中，某学生统计了同学们捐赠的图书数量，发现捐赠数量最多的比最少的多8本，而最多的数量是最少的3倍。如果最少捐赠了___本书，那么最多捐赠了___本书。","answer":"4, 12","explanation":"设最少捐赠了x本书，则最多捐赠了3x本书。根据题意，最多比最少多8本，可列方程：3x - x = 8，解得2x = 8，x = 4。因此最少捐赠了4本，最多捐赠了3 × 4 = 12本。本题考查一元一次方程的实际应用，属于简单难度。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 22:56:35","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":[]}]