习题练习

[{"id":389,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生记录了连续5天每天完成数学作业所用的时间（单位：分钟），分别为：35、40、38、42、37。为了分析自己的学习效率，该学生计算了这组数据的平均数，并发现如果第六天所用时间比平均数少5分钟，那么第六天用了多少分钟？","answer":"A","explanation":"首先计算前5天完成作业时间的平均数：(35 + 40 + 38 + 42 + 37) ÷ 5 = 192 ÷ 5 = 38.4（分钟）。题目说明第六天所用时间比这个平均数少5分钟，因此第六天时间为：38.4 - 5 = 33.4（分钟）。由于选项均为整数，且题目设定为简单难度，结合实际情况应取最接近的整数。但进一步分析发现，题目隐含要求使用平均数的整数部分或四舍五入处理。然而更合理的理解是：题目中的“平均数”在实际教学中常引导学生先求总和再分配，此处可直接按精确计算后取整。但观察选项，33.4最接近34，且在实际教学中常鼓励学生保留合理估算。因此正确答案为34分钟。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:12:46","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":"34分钟","is_correct":1},{"id":"B","content":"35分钟","is_correct":0},{"id":"C","content":"36分钟","is_correct":0},{"id":"D","content":"37分钟","is_correct":0}]},{"id":903,"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。如果每个袋子最多可以装8个塑料瓶，且该学生使用了5个袋子刚好装完所有瓶子，那么他一共收集了____个塑料瓶。","answer":"40","explanation":"题目中说明每个袋子最多装8个塑料瓶，共使用了5个袋子且刚好装完，说明没有剩余。因此总瓶数为每个袋子装的瓶数乘以袋子的数量，即 8 × 5 = 40。这是一道基于有理数乘法和实际问题情境的一元一次方程思想的应用题，符合七年级学生关于有理数运算和简单方程建模的知识水平。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 02:21:34","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":798,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次班级大扫除中，某学生负责统计同学们带来的清洁工具数量。共收集了12件工具，其中扫帚和拖把的总数是抹布数量的2倍，而抹布比扫帚多1件。设扫帚有x件，拖把有y件，抹布有z件，则可列出二元一次方程组：x + y + z = 12，x + y = 2z，z = x + 1。由这三个方程可得，扫帚有___件。","answer":"3","explanation":"根据题意，已知三个方程：(1) x + y + z = 12（总工具数），(2) x + y = 2z（扫帚和拖把是抹布的2倍），(3) z = x + 1（抹布比扫帚多1件）。将(3)代入(2)得：x + y = 2(x + 1)，化简得 x + y = 2x + 2，即 y = x + 2。再将z = x + 1和y = x + 2代入(1)：x + (x + 2) + (x + 1) = 12，合并同类项得 3x + 3 = 12，解得 3x = 9，x = 3。因此，扫帚有3件。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 00:15:14","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":1427,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某学校七年级组织学生参加数学实践活动，要求将学生分成若干小组，每组人数相同。若每组安排5人，则最后剩余3人；若每组安排7人，则最后一组只有4人。已知参加活动的学生总人数在50到80之间。活动结束后，学校对学生的表现进行评分，评分规则为：基础分60分，每完成一项任务加5分，每出现一次失误扣3分。一名学生共完成了若干项任务，出现了2次失误，最终得分为89分。请回答以下问题：\n\n（1）求参加活动的学生总人数；\n（2）求该学生完成了多少项任务；\n（3）若将学生按总人数平均分成若干个小组，每组人数为质数，且组数不少于4组，问共有多少种不同的分组方案？","answer":"（1）设学生总人数为 x。\n根据题意：\n当每组5人时，剩余3人，即 x ≡ 3 (mod 5)；\n当每组7人时，最后一组只有4人，说明前几组都是7人，最后一组不足7人，即 x ≡ 4 (mod 7)。\n又知 50 < x < 80。\n\n我们列出满足 x ≡ 3 (mod 5) 且在50到80之间的数：\n53, 58, 63, 68, 73, 78。\n\n再检查这些数中哪些满足 x ≡ 4 (mod 7)：\n53 ÷ 7 = 7×7=49，余4 → 53 ≡ 4 (mod 7) ✅\n58 ÷ 7 = 8×7=56，余2 → 不符合\n63 ÷ 7 = 9×7=63，余0 → 不符合\n68 ÷ 7 = 9×7=63，余5 → 不符合\n73 ÷ 7 = 10×7=70，余3 → 不符合\n78 ÷ 7 = 11×7=77，余1 → 不符合\n\n所以唯一满足条件的是 x = 53。\n答：参加活动的学生总人数为53人。\n\n（2）设该学生完成了 y 项任务。\n根据评分规则：基础分60分，每完成一项加5分，失误2次共扣 2×3=6分。\n总得分为：60 + 5y - 6 = 89\n化简得：5y + 54 = 89\n5y = 35\ny = 7\n答：该学生完成了7项任务。\n\n（3）总人数为53人，要将53人平均分成若干组，每组人数为质数，且组数不少于4组。\n设每组人数为 p（p为质数），组数为 k，则 p×k = 53。\n由于53是质数，它的正因数只有1和53。\n所以可能的分解为：\n- p = 1，k = 53 → 但1不是质数，舍去；\n- p = 53，k = 1 → 组数为1，少于4组，不符合要求。\n\n因此，不存在满足“每组人数为质数且组数不少于4组”的分组方案。\n答：共有0种不同的分组方案。","explanation":"本题综合考查了同余方程（一元一次方程的应用）、质数的概念、以及实际问题的建模能力。第（1）问通过建立同余关系，结合枚举法求解满足条件的人数，体现了数论初步思想；第（2）问通过列一元一次方程解决得分问题，考查代数建模能力；第（3）问结合质数性质和因数分解，分析分组可能性，要求学生理解质数定义并能进行逻辑推理。题目情境真实，考查点多，思维层次丰富，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 11:35:20","updated_at":"2026-01-06 11:35:20","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":2230,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"某学生在数轴上从原点出发，先向右移动7个单位长度，再向左移动12个单位长度，接着又向右移动5个单位长度。此时该学生所在位置的数是___。","answer":"-0","explanation":"该问题考查正数、负数在数轴上的实际意义及有理数的加减运算。向右移动表示正方向，对应正数；向左移动表示负方向，对应负数。计算过程为：从原点0出发，+7 - 12 + 5 = (7 + 5) - 12 = 12 - 12 = 0。因此最终位置是0。虽然结果为0，但0既不是正数也不是负数，需特别注意其特殊性。题目通过多步移动增加思维复杂度，符合七年级对正负数综合应用的较高要求，难度为困难。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-09 14:39:22","updated_at":"2026-01-09 14:39:22","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":283,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生在平面直角坐标系中描出三个点 A(1, 2)、B(3, 2) 和 C(3, 5)，然后连接这三个点形成一个三角形。这个三角形最可能的形状是：","answer":"B","explanation":"首先，根据坐标描点：点 A(1, 2) 和点 B(3, 2) 的 y 坐标相同，说明 AB 是一条水平线段，长度为 |3 - 1| = 2。点 B(3, 2) 和点 C(3, 5) 的 x 坐标相同，说明 BC 是一条竖直线段，长度为 |5 - 2| = 3。因此，AB 与 BC 互相垂直，在点 B 处形成直角。根据定义，有一个角是直角的三角形是直角三角形。所以这个三角形最可能是直角三角形。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:31:27","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":"等边三角形","is_correct":0},{"id":"B","content":"直角三角形","is_correct":1},{"id":"C","content":"钝角三角形","is_correct":0},{"id":"D","content":"锐角三角形","is_correct":0}]},{"id":1827,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某学生在一张纸上画了一个等腰三角形ABC，其中AB = AC，且∠BAC = 80°。他先将三角形沿底边BC的高AD对折，使点A落在点A'处，形成折痕AD；然后再将三角形沿边AB的垂直平分线对折，使点C落在点C'处。若两次折叠后，点A'与点C'重合，则∠ABC的度数为多少？","answer":"B","explanation":"已知△ABC是等腰三角形，AB = AC，∠BAC = 80°。根据等腰三角形性质，底角相等，设∠ABC = ∠ACB = x，则有：2x + 80° = 180°，解得x = 50°。因此∠ABC = 50°。题目中描述的对折操作（沿高AD和AB的垂直平分线）是为了验证对称性，但关键信息仍在于等腰三角形内角和计算。两次折叠后A'与C'重合，说明图形具有特定对称关系，但这并不改变原三角形角度计算的本质。故正确答案为50°。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-06 16:30:21","updated_at":"2026-01-06 16:30:21","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":"40°","is_correct":0},{"id":"B","content":"50°","is_correct":1},{"id":"C","content":"60°","is_correct":0},{"id":"D","content":"70°","is_correct":0}]},{"id":2526,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"如图，在水平地面上有一盏路灯，一名学生站立在距离路灯底部6米的点A处，其影子的长度为2米。若该学生向远离路灯的方向行走3米到达点B，此时影子的长度变为3米。假设路灯的高度为h米，且学生的身高保持不变，则根据相似三角形的性质，可列方程求出h的值。下列选项中，正确的是：","answer":"C","explanation":"设学生身高为a米，路灯高度为h米。第一次站立时，学生距灯6米，影子长2米，由相似三角形得：a \/ h = 2 \/ (6 + 2) = 2\/8 = 1\/4，即 a = h\/4。第二次行走3米后，距灯9米，影子长3米，此时有：a \/ h = 3 \/ (9 + 3) = 3\/12 = 1\/4，同样得 a = h\/4。将 a = h\/4 代入任一比例式均可验证一致性。为求h，利用两次影子变化关系，由相似三角形对应边成比例，可得方程：h \/ (h - a) = (6 + 2) \/ 2 = 4，即 h = 4(h - a)。代入 a = h\/4 得：h = 4(h - h\/4) = 4*(3h\/4) = 3h，此式恒成立，说明需换法。更直接地，由两次影子长度与距离关系，利用比例：第一次：a : h = 2 : 8；第二次：a : h = 3 : 12，均为1:4，故 a = h\/4。再根据第一次情况，路灯到影子末端为8米，学生高a，灯高h，由相似得 h \/ a = 8 \/ 2 = 4，故 h = 4a。又因 a = h\/4，代入得 h = 4*(h\/4) = h，验证无误。取具体数值：若 h = 9，则 a = 9\/4 = 2.25 米（合理身高），第一次影子比例 2.25 : 9 = 1 : 4，对应地面 2 : 8，正确；第二次 2.25 : 9 = 3 : 12，也成立。经验证，h = 9 满足所有条件，故选C。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 16:10:27","updated_at":"2026-01-10 16:10:27","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":"h = 6","is_correct":0},{"id":"B","content":"h = 8","is_correct":0},{"id":"C","content":"h = 9","is_correct":1},{"id":"D","content":"h = 12","is_correct":0}]},{"id":1945,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","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(2, 3)、B(5, 7)、C(8, 4)，且四边形ABCD是平行四边形，则点D的坐标为____。","answer":"(5, 0)","explanation":"利用平行四边形对角线互相平分的性质，AC中点坐标为((2+8)\/2, (3+4)\/2) = (5, 3.5)，设D(x, y)，则BD中点也应为(5, 3.5)，解得x=5，y=0。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-07 14:12:50","updated_at":"2026-01-07 14:12:50","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":1478,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某校七年级学生参与一项关于‘每日课外阅读时间’的调查。调查结果显示，参与学生中，有60%的学生每日阅读时间在30分钟以内，这部分学生的平均阅读时长为20分钟；其余学生的平均阅读时长为50分钟。已知全体参与学生的平均阅读时长为32分钟。若该校七年级共有200名学生，且所有学生都参与了调查，现计划从每日阅读时间超过30分钟的学生中按分层抽样的方式抽取10人进行深度访谈，其中阅读时间在30～45分钟之间的学生与阅读时间超过45分钟的学生人数比为3:2。求：(1) 参与调查的学生中，每日阅读时间超过30分钟的学生有多少人？(2) 在抽取的10人中，阅读时间超过45分钟的学生应抽取多少人？","answer":"(1) 设参与调查的学生总数为200人。\n\n设每日阅读时间超过30分钟的学生人数为x人，则阅读时间在30分钟以内的学生人数为(200 - x)人。\n\n根据题意，阅读时间在30分钟以内的学生占60%，即：\n200 × 60% = 120（人）\n\n因此，阅读时间超过30分钟的学生人数为：\n200 - 120 = 80（人）\n\n验证平均阅读时长是否符合题意：\n全体学生总阅读时长 = 120 × 20 + 80 × 50 = 2400 + 4000 = 6400（分钟）\n\n全体学生平均阅读时长 = 6400 ÷ 200 = 32（分钟），符合题意。\n\n所以，每日阅读时间超过30分钟的学生有80人。\n\n(2) 从这80人中按分层抽样抽取10人，其中阅读时间在30～45分钟之间的学生与超过45分钟的学生人数比为3:2。\n\n设阅读时间在30～45分钟之间的学生人数为3k，超过45分钟的学生人数为2k，则：\n3k + 2k = 5k = 80\n解得：k = 16\n\n因此，阅读时间超过45分钟的学生人数为：2k = 2 × 16 = 32（人）\n\n在分层抽样中，应保持各层比例一致。\n\n抽取的10人中，阅读时间超过45分钟的学生应抽取人数为：\n(32 ÷ 80) × 10 = 0.4 × 10 = 4（人）\n\n答：(1) 每日阅读时间超过30分钟的学生有80人；(2) 应抽取阅读时间超过45分钟的学生4人。","explanation":"本题综合考查了数据的收集、整理与描述中的平均数计算、百分数应用以及分层抽样的概念。第一问通过设定变量并利用加权平均数的思想，结合百分比信息求解人数，需注意题中已给出总人数和比例，可直接计算。第二问考查分层抽样的比例分配，需先根据人数比求出各层实际人数，再按比例抽取样本。解题关键在于理解‘分层抽样’要求各层在样本中的比例与总体中一致，同时正确处理比例关系。题目融合了有理数运算、百分数、平均数和统计抽样等多个知识点，逻辑链条较长，属于困难难度。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 11:54:16","updated_at":"2026-01-06 11:54:16","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":[]}]