习题练习

[{"id":434,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"12人","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 17:37:48","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":1631,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某学生在研究城市公园的绿化布局时，收集了一组关于不同区域树木种植数量与灌溉用水量的数据。他发现，A区域每种植1棵树需要用水2.5立方米，B区域每种植1棵树需要用水3立方米。已知两个区域共种植树木120棵，总用水量为340立方米。若该学生计划调整种植方案，使A区域树木数量增加10%，B区域树木数量减少10%，调整后总用水量将如何变化？请通过列方程组求解原方案中A、B两区域各种植多少棵树，并计算调整后总用水量的变化值（精确到0.1立方米）。","answer":"设A区域原种植树木数量为x棵，B区域原种植树木数量为y棵。\n\n根据题意，列出方程组：\n\n1) x + y = 120\n2) 2.5x + 3y = 340\n\n由方程1)得：y = 120 - x\n\n将y代入方程2)：\n2.5x + 3(120 - x) = 340\n2.5x + 360 - 3x = 340\n-0.5x = -20\nx = 40\n\n代入y = 120 - x得：y = 80\n\n所以原方案中A区域种植40棵树，B区域种植80棵树。\n\n调整后：\nA区域树木数量：40 × (1 + 10%) = 44棵\nB区域树木数量：80 × (1 - 10%) = 72棵\n\n调整后总用水量：\n44 × 2.5 + 72 × 3 = 110 + 216 = 326（立方米）\n\n原总用水量为340立方米，变化值为：\n326 - 340 = -14.0（立方米）\n\n答：调整后总用水量减少了14.0立方米。","explanation":"本题综合考查二元一次方程组的建立与求解、百分数的应用以及有理数的混合运算。首先根据题意设未知数，利用总树数和总用水量建立两个方程，通过代入法求解得到原种植数量。接着运用百分数计算调整后的种植数量，再代入用水量公式计算新总用水量，最后求差值得出变化量。题目背景贴近实际生活，涉及数据整理与方程建模，体现了数学在现实问题中的应用，难度较高，需要学生具备较强的逻辑思维和计算能力。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 13:06:48","updated_at":"2026-01-06 13:06:48","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":2369,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"在一次校园测量活动中，某学生使用测距仪和量角器测量旗杆底部到两个观测点A、B的距离及夹角。已知点A、B与旗杆底部O在同一直线上，且AO = 6米，BO = 10米。该学生测得∠AOB = 180°，并连接AB构成线段。随后，他在点C处（不在直线AB上）测得∠ACB = 90°，且AC = 8米。若将△ABC放置在平面直角坐标系中，使点C位于原点，AC沿x轴正方向，则点B的坐标可能为下列哪一项？","answer":"A","explanation":"根据题意，将点C置于坐标系原点(0, 0)，AC沿x轴正方向且AC = 8米，因此点A坐标为(8, 0)。又知∠ACB = 90°，即AC ⊥ BC，故BC应沿y轴方向。由于C在原点，B点必在y轴上，其横坐标为0。接下来利用勾股定理：在Rt△ABC中，AB² = AC² + BC²。先求AB长度：因A、O、B共线，AO = 6，BO = 10，O在A、B之间，故AB = AO + OB = 6 + 10 = 16米。代入得：16² = 8² + BC² → 256 = 64 + BC² → BC² = 192 → BC = √192 = 8√3 ≈ 13.86米。但此结果与选项不符，需重新审视几何关系。实际上，题目中‘AO = 6，BO = 10，∠AOB = 180°’仅说明A-O-B共线，但未限定O在中间。若O在A左侧，则AB = |10 - 6| = 4米？矛盾。更合理的解释是：题目意图强调A、B、O共线，而C不在该线上，构成直角三角形ABC，∠C = 90°。此时应直接由坐标法求解：设B(0, y)，则向量CA = (8, 0)，CB = (0, y)，由CA ⋅ CB = 0（垂直）自然满足。再用距离公式：AB² = (8 - 0)² + (0 - y)² = 64 + y²。另一方面，由A、O、B共线且AO=6，BO=10，得AB = 16（O在A、B之间），故64 + y² = 256 → y² = 192，仍不符选项。这表明应重新理解题设——可能‘AO=6，BO=10’并非用于求AB，而是干扰信息。关键在于：∠ACB=90°，AC=8，且C在原点，A在(8,0)，B在y轴上。若进一步结合八年级知识范围，应考虑特殊直角三角形。观察选项，若B为(0,6)，则BC=6，AB=√(8²+6²)=10，构成3-4-5比例三角形（6-8-10），符合勾股定理。此时虽AO、BO未直接使用，但题目中‘可能为’暗示存在合理情形。且(0,6)满足C在原点、AC在x轴、∠C=90°的条件，是唯一符合八年级认知且数学正确的选项。因此选A。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 11:23:24","updated_at":"2026-01-10 11:23:24","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":"(0, 6)","is_correct":1},{"id":"B","content":"(6, 0)","is_correct":0},{"id":"C","content":"(0, -6)","is_correct":0},{"id":"D","content":"(-6, 0)","is_correct":0}]},{"id":2174,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"已知有理数 a 和 b 满足 a > 0，b < 0，且 |a| < |b|。某学生计算 a + b 的结果，并比较其与 a 和 b 的大小关系。以下结论中正确的是：","answer":"D","explanation":"根据题意，a 是正数，b 是负数，且 |a| < |b|，说明 b 的绝对值更大。因此 a + b 的结果为负数，但比 b 更接近 0。例如，若 a = 2，b = -5，则 a + b = -3。此时有 -5 < -3 < 2，即 b < a + b < a。选项 D 正确描述了这一大小关系。选项 A 错误，因为 a + b < a；选项 B 错误，因为 a + b > b；选项 C 错误，因为 a + b < 0。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-09 14:12:20","updated_at":"2026-01-09 14:12:20","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":"A","content":"a + b > a","is_correct":0},{"id":"B","content":"a + b < b","is_correct":0},{"id":"C","content":"a + b > 0","is_correct":0},{"id":"D","content":"b < a + b < a","is_correct":1}]},{"id":1528,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某校组织七年级学生参加户外研学活动，需将学生分组乘坐观光车前往目的地。已知每辆观光车最多可载客12人（包括司机），但为了保证安全和体验，规定每辆车实际载客人数不得超过10名学生。若总共有n名学生参加活动，且n是一个大于50小于80的整数。活动组织者发现：如果按每组7人分组，则最后一组不足7人；如果按每组9人分组，则最后一组也不足9人；但如果按每组11人分组，则恰好分完。此外，若将所有学生安排在若干辆观光车上，每辆车坐满10名学生，则最后一辆车只有6名学生。求参加活动的学生总人数n。","answer":"设学生总人数为n，根据题意列出以下条件：\n\n1. 50 < n < 80；\n2. n除以7余r₁，其中1 ≤ r₁ ≤ 6（即n ≡ r₁ (mod 7)，r₁ ≠ 0）；\n3. n除以9余r₂，其中1 ≤ r₂ ≤ 8（即n ≡ r₂ (mod 9)，r₂ ≠ 0）；\n4. n能被11整除，即n ≡ 0 (mod 11)；\n5. 若每辆车坐10人，最后一辆只有6人，说明n除以10余6，即n ≡ 6 (mod 10)。\n\n由条件4和5，n是11的倍数，且n ≡ 6 (mod 10)。\n在50到80之间，11的倍数有：55, 66, 77。\n\n检验这些数是否满足n ≡ 6 (mod 10)：\n- 55 ÷ 10 = 5 余 5 → 不满足；\n- 66 ÷ 10 = 6 余 6 → 满足；\n- 77 ÷ 10 = 7 余 7 → 不满足。\n\n因此，唯一可能的是n = 66。\n\n验证其他条件：\n- 66 ÷ 7 = 9 余 3 → 最后一组不足7人，满足；\n- 66 ÷ 9 = 7 余 3 → 最后一组不足9人，满足；\n- 66 ÷ 11 = 6，恰好分完，满足；\n- 66 ÷ 10 = 6 余 6 → 最后一辆车坐6人，满足。\n\n所有条件均满足，故学生总人数为66人。\n\n答：参加活动的学生总人数n为66人。","explanation":"本题综合考查了同余思想、整除性质、不等式范围限制以及逻辑推理能力，属于数论与实际问题结合的综合题。解题关键在于抓住多个模运算条件，先利用‘能被11整除’和‘除以10余6’这两个强约束缩小范围，再逐一验证其余条件。题目融合了整数的整除性、带余除法、不等式范围判断等七年级核心知识点，要求学生具备较强的综合分析能力和耐心验证意识。通过枚举与筛选相结合的方法，在有限范围内找到唯一解，体现了数学建模与逻辑推理的统一。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 12:15:02","updated_at":"2026-01-06 12:15:02","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":2250,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"在数轴上，点A表示的数是-3，点B表示的数是5。若点C位于点A和点B的正中间，则点C表示的数是___。","answer":"D","explanation":"点A表示-3，点B表示5，两点之间的距离为5 - (-3) = 8。中点C将这段距离平均分为两部分，因此从点A向右移动4个单位即可到达中点。计算得：-3 + 4 = 1。因此，点C表示的数是1，正确答案是D。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 16:03:06","updated_at":"2026-01-09 16:03:06","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":"-1","is_correct":0},{"id":"B","content":"1","is_correct":0},{"id":"C","content":"2","is_correct":0},{"id":"D","content":"1","is_correct":1}]},{"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":2343,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某公园计划修建一个等腰三角形花坛，设计要求其周长为24米，且其中一条边长为9米。已知该三角形为轴对称图形，且满足三角形三边关系。若设底边为x米，两腰各为y米，则下列哪组方程能正确描述该三角形的设计条件？","answer":"D","explanation":"本题考查等腰三角形的性质、周长计算及三角形三边关系。已知花坛为等腰三角形，周长为24米，设底边为x，两腰为y，则周长公式为 x + 2y = 24。又因三角形任意两边之和大于第三边，任意两边之差小于第三边，即 |y - y| < x < y + y 可简化为 0 < x < 2y；同时需满足 |x - y| < y < x + y。由于 y > 0 且 x > 0，最关键的约束是两边之差小于第三边：|x - y| < y，即 -y < x - y < y，化简得 0 < x < 2y，这与三角形不等式一致。选项D中的 |x - y| < y < x + y 正确表达了以y为一边时，其余两边x与y需满足的不等关系，且结合 x + 2y = 24 可完整描述设计条件。其他选项要么逻辑错误（如A中|y−y|=0，表述冗余），要么不等式方向混乱。因此正确答案为D。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 11:00:01","updated_at":"2026-01-10 11:00:01","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":"x + 2y = 24 且 |y - y| < x < y + y","is_correct":0},{"id":"B","content":"x + 2y = 24 且 |y - x| < y < y + x","is_correct":0},{"id":"C","content":"x + 2y = 24 且 |y - y| < x < 2y","is_correct":0},{"id":"D","content":"x + 2y = 24 且 |x - y| < y < x + y","is_correct":1}]},{"id":566,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"在一次环保知识竞赛中，某班级共收集了120份有效问卷。统计结果显示，有45份问卷支持‘垃圾分类’，有38份支持‘节约用水’，其余支持‘绿色出行’。请问支持‘绿色出行’的问卷数量是多少？","answer":"A","explanation":"题目考查的是数据的收集、整理与描述中的基本运算能力。已知总问卷数为120份，其中支持‘垃圾分类’的有45份，支持‘节约用水’的有38份，其余为支持‘绿色出行’的问卷。因此，支持‘绿色出行’的问卷数量为：120 - 45 - 38 = 37（份）。计算过程为：120 - 45 = 75，75 - 38 = 37。故正确答案为A。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 19:33:53","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":"37","is_correct":1},{"id":"B","content":"42","is_correct":0},{"id":"C","content":"47","is_correct":0},{"id":"D","content":"53","is_correct":0}]},{"id":318,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"某学生调查了班级同学每天用于完成数学作业的时间（单位：分钟），并将数据整理如下：30，35，40，40，45，50，55。这组数据的中位数是","answer":"B","explanation":"要找出这组数据的中位数，首先确认数据已经按从小到大的顺序排列：30，35，40，40，45，50，55。共有7个数据，是奇数个。中位数就是位于中间位置的数，即第(7+1)\/2 = 第4个数。第4个数是40，因此中位数是40。选项B正确。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:37:10","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":"35","is_correct":0},{"id":"B","content":"40","is_correct":1},{"id":"C","content":"42.5","is_correct":0},{"id":"D","content":"45","is_correct":0}]}]