习题练习

[{"id":2301,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"在一次班级数学测验中，某学生记录了5名同学完成一次函数应用题所用的时间（单位：分钟），分别为：8、10、12、10、15。这组数据的中位数和众数分别是多少？","answer":"A","explanation":"首先将数据从小到大排列：8、10、10、12、15。共有5个数据，奇数个，因此中位数是中间的那个数，即第3个数，为10。众数是出现次数最多的数，10出现了两次，其余数各出现一次，因此众数是10。所以正确答案是A。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 10:44:00","updated_at":"2026-01-10 10:44:00","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":"A","content":"中位数是10，众数是10","is_correct":1},{"id":"B","content":"中位数是12，众数是10","is_correct":0},{"id":"C","content":"中位数是10，众数是12","is_correct":0},{"id":"D","content":"中位数是11，众数是10","is_correct":0}]},{"id":793,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"某学生测量了教室里5个不同位置的气温，分别为-2℃、3℃、0℃、-5℃和4℃，这些气温的平均值是___℃。","answer":"待完善","explanation":"首先将所有气温相加：-2 + 3 + 0 + (-5) + 4 = 0。然后将总和除以数据的个数5，得到平均值为0 ÷ 5 = 0。因此，这些气温的平均值是0℃。本题考查有理数的加减运算及平均数的计算方法，属于数据的收集、整理与描述知识点，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 00:09:30","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":692,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次班级图书角整理活动中，某学生统计了同学们捐赠的图书类型，其中故事书有15本，科普书比故事书少6本，漫画书是科普书的2倍。那么漫画书有___本。","answer":"18","explanation":"首先根据题意，故事书有15本，科普书比故事书少6本，因此科普书数量为15 - 6 = 9本。漫画书是科普书的2倍，即9 × 2 = 18本。因此漫画书有18本。本题考查的是有理数的基本运算在实际问题中的应用，属于数据的收集、整理与描述知识点范畴，计算过程简单明了，适合七年级学生。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 22:37:16","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":2391,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某学生测量了一块三角形金属片的三个内角，发现其中两个角分别为55°和65°。若该金属片被一条垂直于最长边的直线从顶点垂直平分，形成两个全等的小三角形，则这条平分线将原三角形分成的两个小三角形中，每个小三角形的周长与原三角形周长的比值最接近以下哪个选项？（假设原三角形三边长度分别为a、b、c，且c为最长边）","answer":"D","explanation":"首先，根据三角形内角和为180°，可求得第三个角为180° - 55° - 65° = 60°。因此三个角分别为55°、60°、65°，对应最长边为对角65°的边。题目中提到‘一条垂直于最长边的直线从顶点垂直平分’，此处表述存在歧义：若指从对角顶点向最长边作高，则不一定平分该边，除非是等腰三角形；但本题三角形三内角均不相等，故不是等腰三角形，高不会平分底边。因此，无法保证分出的两个小三角形全等。题目条件自相矛盾——在非等腰三角形中，从顶点到对边的高不可能同时满足‘垂直’和‘平分’并形成两个全等三角形。因此，题设条件不成立，无法确定具体周长比值。正确选项为D。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 11:51:55","updated_at":"2026-01-10 11:51:55","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":"1:2","is_correct":0},{"id":"B","content":"√2:2","is_correct":0},{"id":"C","content":"(1+√3):4","is_correct":0},{"id":"D","content":"无法确定具体比值","is_correct":1}]},{"id":194,"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":"题目中设每支铅笔的价格为x元，因为每本笔记本比每支铅笔贵3元，所以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。根据总花费为18元，可列出方程：3x + 2(x + 3) = 18。因此，正确选项是A。其他选项错误地将笔记本价格设为比铅笔便宜，或混淆了数量与单价的关系。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 14:03:39","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":749,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次班级大扫除中，某学生负责统计清洁工具的分配情况。已知每2名学生共用1把扫帚，每3名学生共用1个拖把，每4名学生共用1个水桶。如果总共使用了26件清洁工具，那么参加大扫除的学生人数是___人。","answer":"24","explanation":"设参加大扫除的学生人数为x。根据题意，扫帚的数量为x\/2，拖把的数量为x\/3，水桶的数量为x\/4。总工具数为26件，因此可列方程：x\/2 + x\/3 + x\/4 = 26。通分后得(6x + 4x + 3x)\/12 = 26，即13x\/12 = 26。两边同乘以12，得13x = 312，解得x = 24。因此，参加大扫除的学生人数是24人。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 23:22:54","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":672,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次班级大扫除中，某学生负责统计擦窗户和拖地的人数。已知擦窗户的人数比拖地人数的2倍少3人，而两项工作总共有27人参与。设拖地的人数为x，则根据题意可列出一元一次方程：___。","answer":"x + (2x - 3) = 27","explanation":"设拖地的人数为x，则擦窗户的人数为2x - 3（因为比拖地人数的2倍少3人）。两项工作总人数为27人，因此拖地人数加上擦窗户人数等于27，即x + (2x - 3) = 27。该方程正确反映了题目中的数量关系，属于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":"2025-12-29 22:22:30","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":592,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"某班级进行了一次数学测验，成绩分布如下表所示。根据统计表，该班级成绩在80分到89分之间的人数占总人数的百分比是多少？\n\n| 分数段 | 人数 |\n|--------|------|\n| 90-100 | 8 |\n| 80-89 | 12 |\n| 70-79 | 10 |\n| 60-69 | 5 |\n| 60以下 | 3 |","answer":"B","explanation":"首先计算总人数：8 + 12 + 10 + 5 + 3 = 38（人）。成绩在80-89分之间的人数为12人。所求百分比为 (12 ÷ 38) × 100% ≈ 31.58%，四舍五入后最接近的选项是30%。因此正确答案是B。本题考查数据的收集、整理与描述中的百分比计算，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 20:35:39","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":"24%","is_correct":0},{"id":"B","content":"30%","is_correct":1},{"id":"C","content":"36%","is_correct":0},{"id":"D","content":"40%","is_correct":0}]},{"id":539,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"在一次班级环保活动中，某学生收集了若干节废旧电池。他将这些电池按每5节装一盒，发现最后剩下2节；如果改为每7节装一盒，则刚好装完，没有剩余。已知他收集的电池总数在30到50之间，那么他一共收集了多少节电池？","answer":"C","explanation":"设该学生收集的电池总数为x节。根据题意：\n1. 每5节装一盒，剩下2节，说明 x 除以5余2，即 x ≡ 2 (mod 5)；\n2. 每7节装一盒，刚好装完，说明 x 能被7整除，即 x ≡ 0 (mod 7)；\n3. 且 30 < x < 50。\n\n在30到50之间，7的倍数有：35、42、49。\n- 35 ÷ 5 = 7 余 0 → 不符合“余2”的条件；\n- 42 ÷ 5 = 8 余 2 → 符合余2的条件；\n- 49 ÷ 5 = 9 余 4 → 不符合。\n\n因此，只有42同时满足被7整除、被5除余2，并且在30到50之间。\n故正确答案是C。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 18:51:32","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":"37","is_correct":0},{"id":"C","content":"42","is_correct":1},{"id":"D","content":"47","is_correct":0}]},{"id":1426,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某学校组织七年级学生参加数学实践活动，要求学生利用平面直角坐标系设计一个‘校园寻宝’路线。已知校园平面图上以正门为原点O(0,0)，向东为x轴正方向，向北为y轴正方向。第一个藏宝点A位于(3,4)，第二个藏宝点B位于(-2,6)，第三个藏宝点C位于(5,-3)。一名学生从正门出发，依次经过A、B、C三个点后返回正门。若该学生每走1个单位长度需要消耗2分钟，且在每个藏宝点停留整理数据的时间为5分钟。已知该学生总共用时不超过150分钟，问：该学生是否能在规定时间内完成整个寻宝任务？如果不能，最多可以跳过几个藏宝点（只能跳过B或C，不能跳过A），才能确保总时间不超过150分钟？请通过计算说明。","answer":"首先计算从原点O(0,0)到A(3,4)的距离：\n距离OA = √[(3-0)² + (4-0)²] = √(9+16) = √25 = 5\n\n从A(3,4)到B(-2,6)的距离：\n距离AB = √[(-2-3)² + (6-4)²] = √[(-5)² + 2²] = √(25+4) = √29 ≈ 5.385\n\n从B(-2,6)到C(5,-3)的距离：\n距离BC = √[(5+2)² + (-3-6)²] = √[7² + (-9)²] = √(49+81) = √130 ≈ 11.402\n\n从C(5,-3)返回原点O(0,0)的距离：\n距离CO = √[(5-0)² + (-3-0)²] = √(25+9) = √34 ≈ 5.831\n\n总行走距离 = OA + AB + BC + CO ≈ 5 + 5.385 + 11.402 + 5.831 = 27.618（单位长度）\n\n行走时间 = 27.618 × 2 ≈ 55.236（分钟）\n\n停留时间：共3个藏宝点，每个停留5分钟，总停留时间 = 3 × 5 = 15（分钟）\n\n总用时 ≈ 55.236 + 15 = 70.236（分钟）\n\n由于70.236 < 150，因此该学生能在规定时间内完成整个寻宝任务。\n\n但题目要求判断“是否能在规定时间内完成”，并进一步问“如果不能，最多可以跳过几个点”。然而根据计算，实际用时远小于150分钟，因此无需跳过任何点。\n\n但为严谨起见，我们验证是否存在理解偏差：题目中“总共用时不超过150分钟”是上限，而实际仅需约70分钟，远低于限制。\n\n因此结论是：该学生能在规定时间内完成整个寻宝任务，不需要跳过任何藏宝点。\n\n答案：能完成，不需要跳过任何点。","explanation":"本题综合考查了平面直角坐标系中两点间距离公式、实数的运算、近似计算以及实际问题的建模能力。解题关键在于正确运用距离公式√[(x₂−x₁)²+(y₂−y₁)²]计算各段路径长度，再结合时间与距离的关系（每单位2分钟）和停留时间进行总时间估算。虽然题目设置了‘是否超时’和‘跳过点’的复杂情境，但通过精确计算发现实际耗时远低于限制，体现了数学建模中数据验证的重要性。本题难度较高，因其融合了多个知识点并要求学生进行多步推理和实际判断，符合困难级别要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 11:34:57","updated_at":"2026-01-06 11:34:57","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":[]}]