[{"id":495,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生在整理班级同学的课外阅读时间数据时，发现一周内每天阅读时间（单位：分钟）分别为：30、45、0、60、35、50、40。如果去掉一个最低值和一个最高值后，剩余数据的平均数是多少？","answer":"C","explanation":"首先将数据按从小到大排列：0、30、35、40、45、50、60。最低值是0，最高值是60，去掉这两个值后，剩余数据为：30、35、40、45、50。计算这五个数的平均数：(30 + 35 + 40 + 45 + 50) ÷ 5 = 200 ÷ 5 = 42。因此，正确答案是C。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 18:06:51","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":"40","is_correct":0},{"id":"B","content":"41","is_correct":0},{"id":"C","content":"42","is_correct":1},{"id":"D","content":"43","is_correct":0}]},{"id":213,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"某学生计算一个数的相反数时，将原数 5 写成了 -5，那么他得到的结果与原数的正确相反数相比，相差____。","answer":"10","explanation":"原数是 5，它的正确相反数是 -5。某学生误将原数当作 -5，计算其相反数得到 -(-5) = 5。正确结果是 -5，而学生得到的是 5，两者相差 5 - (-5) = 10。因此答案是 10。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 14:40:00","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":2310,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某学生在研究轴对称图形时，发现一个等腰三角形的顶角为80°，底边长为6 cm。若将该三角形沿其对称轴对折，则对折后两部分完全重合。请问这个等腰三角形的腰长最接近下列哪个值？（结果保留一位小数）","answer":"A","explanation":"该题考查轴对称与等腰三角形性质的综合应用。已知等腰三角形顶角为80°，则每个底角为(180°−80°)÷2=50°。作底边的高（即对称轴），将底边分为两段，每段长3 cm，并构成两个全等的直角三角形。在其中一个直角三角形中，已知一个锐角为50°，邻边（底边一半）为3 cm，要求斜边（即腰长）。利用余弦函数：cos(50°) = 邻边 \/ 斜边 = 3 \/ 腰长，得腰长 = 3 \/ cos(50°)。查表或计算器得cos(50°)≈0.6428，因此腰长≈3 ÷ 0.6428 ≈ 4.667 cm，保留一位小数约为4.7 cm，最接近选项A的4.6 cm。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 10:45:32","updated_at":"2026-01-10 10:45:32","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":"4.6 cm","is_correct":1},{"id":"B","content":"5.2 cm","is_correct":0},{"id":"C","content":"6.8 cm","is_correct":0},{"id":"D","content":"7.4 cm","is_correct":0}]},{"id":927,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"某学生测量了一个角的度数为75度，这个角的补角是___度。","answer":"105","explanation":"补角是指两个角的和为180度。已知一个角是75度，设其补角为x度，则有方程：75 + x = 180。解这个一元一次方程得：x = 180 - 75 = 105。因此，这个角的补角是105度。本题考查补角的概念及简单的一元一次方程应用，属于几何图形初步与一元一次方程的结合知识点。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 02:49:57","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":1524,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某校七年级开展‘校园植物分布调查’项目，学生需记录不同区域植物种类数量，并进行数据分析。调查区域被划分为A、B、C三个区域，分别位于平面直角坐标系中的矩形范围内：A区为点(0,0)到(4,3)，B区为点(4,0)到(8,3)，C区为点(0,3)到(8,6)。已知A区每平方米有2种植物，B区每平方米有3种植物，C区每平方米有1.5种植物。调查过程中发现，B区实际记录的植物种类总数比理论值少6种，而C区比理论值多4种。若三个区域总记录植物种类为86种，求A区的实际面积（单位：平方米）。注：所有区域均为矩形，面积单位为平方米，植物种类数为整数或一位小数。","answer":"解：\n\n第一步：计算各区域的面积。\n\nA区：从(0,0)到(4,3)，长为4，宽为3，面积为 4 × 3 = 12（平方米）\nB区：从(4,0)到(8,3)，长为4，宽为3，面积为 4 × 3 = 12（平方米）\nC区：从(0,3)到(8,6)，长为8，宽为3，面积为 8 × 3 = 24（平方米）\n\n第二步：计算各区域理论植物种类数。\n\nA区理论种类：12 × 2 = 24（种）\nB区理论种类：12 × 3 = 36（种）\nC区理论种类：24 × 1.5 = 36（种）\n\n第三步：设A区实际记录的植物种类为A_actual。\n\n根据题意：\nB区实际 = 36 - 6 = 30（种）\nC区实际 = 36 + 4 = 40（种）\n\n三个区域总记录种类为86种，因此：\nA_actual + 30 + 40 = 86\nA_actual = 86 - 70 = 16（种）\n\n第四步：设A区实际面积为x平方米。\n\n已知A区每平方米有2种植物，因此实际种类数为 2x。\n所以有方程：\n2x = 16\n解得：x = 8\n\n答：A区的实际面积为8平方米。","explanation":"本题综合考查了平面直角坐标系中矩形面积的确定、实数运算、一元一次方程的建立与求解，以及数据的整理与分析能力。解题关键在于理解‘理论值’与‘实际值’的差异，并通过总数量反推未知量。首先利用坐标确定各区域几何尺寸并计算面积，再结合单位面积植物密度求出理论种类数；接着根据题设调整B、C两区的实际记录数，利用总和求出A区实际记录种类；最后设A区实际面积为未知数，建立一元一次方程求解。题目融合了坐标、面积、密度、方程与数据分析，逻辑链条完整，难度较高，适合训练学生综合应用能力。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 12:13:23","updated_at":"2026-01-06 12:13:23","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":2247,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某学生在一次数学实践活动中，记录了一周内某城市每日的气温变化情况。规定：气温上升记为正，下降记为负。已知这七天的气温变化依次为：+3℃，-2℃，+5℃，-4℃，+1℃，-6℃，+2℃。若第一天的起始气温为-1℃，请回答以下问题：经过这七天的连续变化后，最终气温是多少摄氏度？并判断最终气温比起始气温是升高了还是降低了，变化了多少摄氏度？","answer":"最终气温是-2℃，比起始气温降低了1℃。","explanation":"本题综合考查正负数在连续变化中的加减运算，要求学生理解正负数表示相反意义的量，并能进行多步有理数加法运算。题目设置了真实情境（气温变化），避免机械计算，强调过程推理。通过逐日累加变化量，最终得出结果，并比较起始与结束状态的差异，体现了正负数在实际问题中的应用，符合七年级课程标准中‘有理数运算’与‘实际问题建模’的要求。","solution_steps":"1. 起始气温为-1℃。\n2. 第一天变化：-1 + (+3) = 2℃\n3. 第二天变化：2 + (-2) = 0℃\n4. 第三天变化：0 + (+5) = 5℃\n5. 第四天变化：5 + (-4) = 1℃\n6. 第五天变化：1 + (+1) = 2℃\n7. 第六天变化：2 + (-6) = -4℃\n8. 第七天变化：-4 + (+2) = -2℃\n9. 最终气温为-2℃。\n10. 比起始气温-1℃的变化量：-2 - (-1) = -1℃，即降低了1℃。","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-09 14:44:04","updated_at":"2026-01-09 14:44:04","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":691,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"某学生测量了家中客厅地面的长和宽，发现长为 4.5 米，宽为 3.2 米。若用边长为 0.3 米的正方形地砖铺满整个地面（不考虑损耗），则至少需要 ___ 块地砖。","answer":"160","explanation":"首先计算客厅地面的面积：4.5 × 3.2 = 14.4（平方米）。然后计算每块地砖的面积：0.3 × 0.3 = 0.09（平方米）。最后用总面积除以单块地砖面积：14.4 ÷ 0.09 = 160。因为题目要求‘至少需要’且‘铺满’，所以结果为整数 160 块。本题综合考查了有理数的乘除运算和实际问题中的面积计算，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 22:37:03","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":1638,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市为了解七年级学生每日完成数学作业所用时间，随机抽取了100名学生进行调查，并将数据整理如下：作业时间在30分钟以下的有15人，30～60分钟的有40人，60～90分钟的有30人，90分钟以上的有15人。现计划从这100名学生中按比例抽取20人进行深入访谈。已知被抽中的学生中，作业时间在60分钟以上的学生人数为m，且该城市共有5000名七年级学生。若用样本中作业时间在90分钟以上的学生频率估计总体，求该城市七年级学生中作业时间在90分钟以上的人数；并求m的值。","answer":"第一步：根据样本数据，作业时间在90分钟以上的学生有15人，总样本为100人，因此频率为15 ÷ 100 = 0.15。\n\n第二步：用样本频率估计总体，该城市共有5000名七年级学生，因此作业时间在90分钟以上的人数约为：\n5000 × 0.15 = 750（人）。\n\n第三步：从100名学生中按比例抽取20人，抽样比例为20 ÷ 100 = 1\/5。\n\n第四步：原样本中作业时间在60分钟以上的学生包括60～90分钟和90分钟以上两部分，共30 + 15 = 45人。\n\n第五步：按比例抽取，则被抽中的学生中作业时间在60分钟以上的人数为：\n45 × (1\/5) = 9（人），即m = 9。\n\n最终答案：该城市七年级学生中作业时间在90分钟以上的人数约为750人，m的值为9。","explanation":"本题综合考查了数据的收集、整理与描述中的频数、频率、样本估计总体以及按比例抽样等核心概念。解题关键在于理解频率的定义（频数 ÷ 总数），并能将其应用于总体估计；同时掌握按比例抽样的方法，即各组抽取人数 = 原组人数 × 抽样比例。题目设置了真实情境，要求学生从多个数据组中提取信息并进行两步计算，体现了数据分析在实际问题中的应用，难度较高，适合考查学生的综合数据处理能力。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 13:08:48","updated_at":"2026-01-06 13:08:48","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":1929,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"在平面直角坐标系中，点A(2, 3)、点B(5, y)、点C(x, 7)共线，且线段AC的中点在直线y = 2x - 1上，则x + y的值为____。","answer":"11","explanation":"利用三点共线斜率相等得(y-3)\/3 = (7-y)\/(x-5)，中点((2+x)\/2, 5)代入直线方程得5 = 2·((2+x)\/2) -1，解得x=6，代入得y=5，故x+y=11。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-07 14:10:00","updated_at":"2026-01-07 14:10:00","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":346,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生在整理班级同学的课外阅读时间数据时，记录了10名同学每周阅读的小时数分别为：3，5，4，6，3，7，5，4，5，6。这组数据的众数是多少？","answer":"C","explanation":"众数是指一组数据中出现次数最多的数。将数据从小到大排列为：3，3，4，4，5，5，5，6，6，7。其中，3出现2次，4出现2次，5出现3次，6出现2次，7出现1次。因此，出现次数最多的数是5，所以这组数据的众数是5。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:41:14","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":"3","is_correct":0},{"id":"B","content":"4","is_correct":0},{"id":"C","content":"5","is_correct":1},{"id":"D","content":"6","is_correct":0}]}]