[{"id":986,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次班级环保活动中，某学生收集了可回收垃圾的重量记录如下：塑料瓶重0.35千克，废纸重0.48千克，易拉罐重0.27千克。他将这三类垃圾的总重量填入统计表时，发现表格中‘合计’一栏被污损，无法看清。请帮他计算出这三类垃圾的总重量是___千克。","answer":"1.10","explanation":"本题考查有理数的加法运算，属于简单难度。学生需要将三个小数相加：0.35 + 0.48 + 0.27。计算时注意小数点对齐，从低位逐位相加。0.35 + 0.48 = 0.83，0.83 + 0.27 = 1.10。因此，三类垃圾的总重量是1.10千克。题目结合环保情境，贴近生活，帮助学生理解有理数在现实中的应用。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 04:28:33","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":1420,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市为改善交通状况，计划在一条主干道上设置若干个公交站点。经调查，若每两个相邻站点之间的距离相等，且总站点数为n（n ≥ 3），则整条线路的总长度为L = 100(n - 1) 米。现因城市规划调整，要求总长度L必须满足 500 ≤ L ≤ 1200，同时站点数量n必须为整数。此外，为便于管理，站点数n还需满足不等式组：\n\n2n + 3 > 15\n3n - 5 ≤ 2n + 7\n\n请回答以下问题：\n（1）求满足上述所有条件的站点数n的所有可能取值；\n（2）若每增加一个站点，运营成本增加800元，而每米线路的维护费用为0.5元\/年，求在满足条件的所有方案中，年总成本最低的站点数量及对应的最低年总成本。","answer":"（1）首先根据题意，总长度L = 100(n - 1)，且满足 500 ≤ L ≤ 1200。\n\n代入得：\n500 ≤ 100(n - 1) ≤ 1200\n两边同时除以100：\n5 ≤ n - 1 ≤ 12\n加1得：\n6 ≤ n ≤ 13\n\n再解不等式组：\n① 2n + 3 > 15 → 2n > 12 → n > 6\n② 3n - 5 ≤ 2n + 7 → 3n - 2n ≤ 7 + 5 → n ≤ 12\n\n综合得：n > 6 且 n ≤ 12，即 7 ≤ n ≤ 12\n\n结合前面的 6 ≤ n ≤ 13，取交集得：7 ≤ n ≤ 12\n\n又n为整数，所以n的可能取值为：7, 8, 9, 10, 11, 12\n\n（2）年总成本 = 站点运营成本 + 线路维护成本\n站点运营成本 = 800n 元\n线路长度L = 100(n - 1) 米，维护费用 = 0.5 × 100(n - 1) = 50(n - 1) 元\n\n所以年总成本 C = 800n + 50(n - 1) = 800n + 50n - 50 = 850n - 50\n\n这是一个关于n的一次函数，且系数850 > 0，因此C随n的增大而增大。\n要使C最小，应取n的最小可能值，即n = 7\n\n当n = 7时：\nC = 850 × 7 - 50 = 5950 - 50 = 5900（元）\n\n答：（1）n的可能取值为7, 8, 9, 10, 11, 12；（2）当年总成本最低时，站点数量为7个，最低年总成本为5900元。","explanation":"本题综合考查了一元一次不等式组的解法、代数式的建立与最值分析。第（1）问需将实际问题转化为数学不等式，通过解多个不等式并求交集得到整数解范围，体现了数学建模能力。第（2）问要求建立成本函数，理解一次函数的单调性，并应用于优化决策，考查了函数思想在实际问题中的应用。题目融合了不等式组、代数式、函数最值等多个七年级核心知识点，情境新颖，逻辑层次清晰，难度较高，适合用于选拔性评价。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 11:31:15","updated_at":"2026-01-06 11:31:15","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":2552,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"某圆形花坛的半径为6米，现计划在花坛中心安装一个旋转喷头，其喷洒范围为一个扇形区域，该扇形的圆心角为120°。若喷头每分钟旋转一周，且喷洒半径可在3米到8米之间调节，问：当喷洒半径为多少米时，喷头在旋转过程中恰好能完全覆盖整个花坛，但不会超出花坛边缘？","answer":"A","explanation":"要使喷头在旋转过程中恰好完全覆盖整个圆形花坛且不超出边缘，喷洒范围必须恰好等于花坛的面积。花坛是半径为6米的圆，因此其覆盖范围的最大半径不能超过6米，否则会超出花坛。同时，由于喷头每分钟旋转一周，且喷洒区域为120°的扇形，意味着每转一圈，喷头会分三次（每次120°）喷洒不同方向，从而在连续旋转中覆盖整个圆周。只要喷洒半径等于花坛半径6米，就能在旋转过程中逐步覆盖整个花坛，而不会越界。若半径大于6米（如7米或8米），则会超出花坛边缘，不符合“不超出”的要求。因此，正确答案是6米。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 17:07:42","updated_at":"2026-01-10 17:07:42","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米","is_correct":1},{"id":"B","content":"7米","is_correct":0},{"id":"C","content":"8米","is_correct":0},{"id":"D","content":"无法完全覆盖","is_correct":0}]},{"id":801,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次班级环保活动中，某学生收集废旧电池的数量比另一名学生的3倍少5节。如果两人一共收集了27节电池，那么收集较少的学生收集了___节电池。","answer":"8","explanation":"设收集较少的学生收集了x节电池，则另一名学生收集了(3x - 5)节。根据题意，两人共收集27节，列出方程：x + (3x - 5) = 27。化简得4x - 5 = 27，解得4x = 32，x = 8。因此，收集较少的学生收集了8节电池。本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 00:16:45","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":2133,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在解方程 3(x - 2) = 2x + 1 时，第一步将等式两边同时展开，得到 3x - 6 = 2x + 1。接下来，他应该进行的正确步骤是：","answer":"B","explanation":"解一元一次方程时，通常采用移项的方法，将含未知数的项移到等式一边，常数项移到另一边。由 3x - 6 = 2x + 1，正确的移项应为：3x - 2x = 1 + 6，即选项 B 所述。选项 A 移项时符号错误，选项 C 过早除以系数不符合常规步骤，选项 D 虽可接受但不是最直接的移项方式，而题目问的是‘接下来应该进行的正确步骤’，B 是最标准且合理的操作。","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":"将 3x 移到右边，得到 -6 = -x + 1","is_correct":0},{"id":"B","content":"将 2x 移到左边，-6 移到右边，得到 3x - 2x = 1 + 6","is_correct":1},{"id":"C","content":"两边同时除以 3，得到 x - 2 = (2x + 1)\/3","is_correct":0},{"id":"D","content":"将等式两边同时加 6，得到 3x = 2x + 7","is_correct":0}]},{"id":195,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"小明买了3支铅笔和2本笔记本，共花费18元。已知每本笔记本比每支铅笔贵3元，设每支铅笔的价格为x元，则下列方程正确的是（ ）。","answer":"A","explanation":"设每支铅笔的价格为x元，根据题意，每本笔记本比每支铅笔贵3元，因此每本笔记本的价格为(x + 3)元。小明买了3支铅笔，总价为3x元；买了2本笔记本，总价为2(x + 3)元。两者相加等于总花费18元，因此方程为：3x + 2(x + 3) = 18。选项A正确。其他选项中，B错误地将笔记本价格设为比铅笔便宜，C和D则颠倒了铅笔和笔记本的数量与单价对应关系，均不符合题意。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 14:04:01","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":"3x + 2(x + 3) = 18","is_correct":1},{"id":"B","content":"3x + 2(x - 3) = 18","is_correct":0},{"id":"C","content":"3(x + 3) + 2x = 18","is_correct":0},{"id":"D","content":"3(x - 3) + 2x = 18","is_correct":0}]},{"id":2775,"subject":"历史","grade":"七年级","stage":"初中","type":"选择题","content":"下列哪一项是唐朝对外友好交往的典型事例，体现了当时中外文化交流的繁荣？","answer":"B","explanation":"本题考查唐朝时期中外交流的史实。A项张骞出使西域发生在西汉时期，不属于唐朝；C项郑和下西洋是明朝的事件；D项玄奘西行虽为唐朝中外交流的重要事件，但其主要目的是求取佛经，而鉴真东渡日本则是主动将唐朝的佛教、建筑、医学等文化传播到日本，是唐朝对外友好交往和文化输出的典型代表，更符合‘对外友好交往’和‘文化交流繁荣’的题意。因此，正确答案为B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-12 10:42:55","updated_at":"2026-01-12 10:42:55","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":"A","content":"张骞出使西域，开辟丝绸之路","is_correct":0},{"id":"B","content":"鉴真东渡日本，传播唐朝文化与佛教","is_correct":1},{"id":"C","content":"郑和下西洋，访问亚非多个国家","is_correct":0},{"id":"D","content":"玄奘西行天竺，取回大量佛经并翻译","is_correct":0}]},{"id":868,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次环保知识竞赛中，某班级收集了学生们的答题情况，并绘制成扇形统计图。其中，答对8题以上的学生占总人数的30%，答对5至7题的学生占45%，答对4题以下的学生占剩余部分。若该班级共有40名学生，则答对4题以下的学生有___人。","answer":"10","explanation":"首先计算答对8题以上和答对5至7题的学生所占百分比之和：30% + 45% = 75%。因此，答对4题以下的学生占比为100% - 75% = 25%。班级总人数为40人，所以答对4题以下的学生人数为40 × 25% = 40 × 0.25 = 10人。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 01:21:07","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":2458,"subject":"数学","grade":"八年级","stage":"初中","type":"填空题","content":"在一次数学实践活动中，某学生测量了一块等腰三角形花坛的两条腰长均为5米，底边上的高为4米。若要在花坛中铺设一条从顶点到底边中点的装饰带，则这条装饰带的长度为____米。","answer":"4","explanation":"等腰三角形底边上的高、中线、顶角平分线三线合一，因此从顶点到底边中点的线段就是高，长度为4米。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 14:05:26","updated_at":"2026-01-10 14:05:26","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":653,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次班级环保活动中，某学生收集了若干个塑料瓶和玻璃瓶，其中塑料瓶的数量比玻璃瓶多8个。若两种瓶子一共有36个，那么玻璃瓶有___个。","answer":"14","explanation":"设玻璃瓶的数量为x个，则塑料瓶的数量为x + 8个。根据题意，两种瓶子总数为36个，可列方程：x + (x + 8) = 36。化简得2x + 8 = 36，解得2x = 28，x = 14。因此，玻璃瓶有14个。本题考查一元一次方程的实际应用，属于七年级数学课程内容。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 22:11:43","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":[]}]