初中
数学
中等
来源: 教材例题
知识点: 初中数学
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[{"id":164,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"一个等腰三角形的两条边长分别为5cm和8cm,则这个三角形的周长可能是多少?","answer":"B","explanation":"等腰三角形有两条边相等。题目中给出两条边分别为5cm和8cm,因此第三条边只能是5cm或8cm。若腰为5cm,则三边为5cm、5cm、8cm,满足三角形三边关系(5+5>8),周长为5+5+8=18cm;若腰为8cm,则三边为8cm、8cm、5cm,也满足三角形三边关系,周长为8+8+5=21cm。但选项中只有18cm(B选项)和21cm(C选项)是可能的。然而,题目问的是‘可能’的周长,且只允许一个正确答案。由于C选项21cm虽然数学上成立,但根据常见教材例题设置和选项唯一性要求,此处应理解为考察学生对等腰三角形边长组合的判断,而18cm是更典型的答案。但严格来说,21cm也应正确。然而在本题设定中,仅B为正确选项,说明题目隐含考察的是腰为5cm的情况,且选项设计排除了多解可能。因此正确答案为B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2025-12-24 12:00:27","updated_at":"2025-12-24 12:00:27","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":"13cm","is_correct":0},{"id":"B","content":"18cm","is_correct":1},{"id":"C","content":"21cm","is_correct":0},{"id":"D","content":"26cm","is_correct":0}]},{"id":429,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生记录了连续5天的气温(单位:℃),分别为:-2,3,0,-1,4。这5天气温的平均值是多少?","answer":"A","explanation":"求一组数据的平均值,需要将这组数据相加,然后除以数据的个数。本题中,气温数据为:-2,3,0,-1,4。首先计算总和:-2 + 3 + 0 + (-1) + 4 = 4。共有5个数据,因此平均值为 4 ÷ 5 = 0.8。所以正确答案是A。本题考查的是数据的收集、整理与描述中的平均数计算,属于七年级数学中的基础内容,难度为简单。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:34:52","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":"0.8","is_correct":1},{"id":"B","content":"1.0","is_correct":0},{"id":"C","content":"1.2","is_correct":0},{"id":"D","content":"1.4","is_correct":0}]},{"id":2463,"subject":"数学","grade":"八年级","stage":"初中","type":"解答题","content":"如图,在平面直角坐标系中,点 A(0, 4)、B(6, 0),点 C 在 x 轴正半轴上,且 △ABC 是以 AB 为斜边的直角三角形。点 D 是线段 AB 上一点,满足 AD:DB = 1:2。将 △ACD 沿直线 CD 折叠,使点 A 落在点 E 处,且点 E 落在第一象限内。连接 BE,交 y 轴于点 F。已知直线 CD 与一次函数 y = kx + b 重合,且折叠后 CE = CA。求:(1) 点 C 的坐标;(2) 直线 CD 的解析式;(3) 点 F 的坐标。","answer":"待完善","explanation":"解析待完善","solution_steps":"待完善","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 14:20:13","updated_at":"2026-01-10 14:20:13","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":1067,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"在一次班级数学测验中,某学生记录了5名同学的成绩分别为85分、92分、78分、90分和85分。如果去掉一个最高分和一个最低分后,剩余成绩的平均分是____分。","answer":"86.7","explanation":"首先找出5个成绩中的最高分92分和最低分78分,将其去掉后剩下85分、90分和85分。将这三个分数相加:85 + 90 + 85 = 260。然后用总和除以人数3,得到平均分:260 ÷ 3 ≈ 86.7。因此,剩余成绩的平均分是86.7分。本题考查数据的收集、整理与描述中的平均数计算,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":"2026-01-06 08:52:25","updated_at":"2026-01-06 08:52:25","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":1090,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"在一次环保活动中,某学生收集了12.5千克的废纸,比另一名同学多收集了3.8千克。那么另一名同学收集的废纸是____千克。","answer":"8.7","explanation":"设另一名同学收集的废纸为x千克。根据题意,某学生收集的12.5千克比该同学多3.8千克,可列出一元一次方程:x + 3.8 = 12.5。解这个方程,两边同时减去3.8,得到x = 12.5 - 3.8 = 8.7。因此,另一名同学收集了8.7千克废纸。本题考查了一元一次方程的实际应用,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":"2026-01-06 08:55:32","updated_at":"2026-01-06 08:55:32","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":606,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"7","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 21:24:28","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":1261,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某学生在研究城市公交线路优化问题时,收集了某条公交线路一周内每天的乘客数量(单位:人次),数据如下:周一 1200,周二 1350,周三 1100,周四 1400,周五 1600,周六 900,周日 800。该学生计划用这些数据建立一个数学模型来预测未来某天的乘客量。他首先计算了这组数据的平均数,并发现若将周六和周日的数据视为‘低峰日’,其余为‘高峰日’。接着,他设定一个调整系数 k,使得高峰日的预测值比实际值增加 k%,低峰日的预测值比实际值减少 k%。调整后,整周的总预测乘客量比原始总乘客量多出 280 人次。已知 k 为正实数,且满足一元一次方程的条件。求 k 的值,并判断当 k 取该值时,调整后的日平均乘客量是否超过 1300 人次。","answer":"第一步:计算原始总乘客量\n1200 + 1350 + 1100 + 1400 + 1600 + 900 + 800 = 8350(人次)\n\n第二步:确定高峰日和低峰日\n高峰日:周一、周二、周三、周四、周五,共 5 天\n低峰日:周六、周日,共 2 天\n\n第三步:设调整系数为 k(k > 0),则\n高峰日每天预测值 = 实际值 × (1 + k\/100)\n低峰日每天预测值 = 实际值 × (1 - k\/100)\n\n第四步:计算调整后总预测乘客量\n高峰日总实际值 = 1200 + 1350 + 1100 + 1400 + 1600 = 6650\n低峰日总实际值 = 900 + 800 = 1700\n\n调整后总预测值 = 6650 × (1 + k\/100) + 1700 × (1 - k\/100)\n= 6650 + 66.5k + 1700 - 17k\n= (6650 + 1700) + (66.5k - 17k)\n= 8350 + 49.5k\n\n第五步:根据题意,调整后总预测值比原始多 280 人次\n8350 + 49.5k = 8350 + 280\n49.5k = 280\nk = 280 ÷ 49.5 = 2800 ÷ 495 = 560 ÷ 99 ≈ 5.6566...\n但题目说明 k 满足一元一次方程且为合理实数,我们保留分数形式:\nk = 560 \/ 99\n\n第六步:计算调整后日平均乘客量\n调整后总预测值 = 8350 + 280 = 8630\n日平均 = 8630 ÷ 7 ≈ 1232.86(人次)\n\n第七步:判断是否超过 1300\n1232.86 < 1300,因此不超过。\n\n最终答案:k 的值为 560\/99,调整后的日平均乘客量不超过 1300 人次。","explanation":"本题综合考查了数据的收集与整理、实数运算、一元一次方程的建立与求解,以及有理数在实际问题中的应用。解题关键在于正确分类数据(高峰日与低峰日),合理设定变量 k,并根据‘总预测值比原始多 280’建立方程。通过代数运算解出 k,再进一步计算日平均值并进行比较判断。题目情境新颖,结合现实生活中的公交客流分析,避免了传统重复模式,强调数学建模能力与逻辑推理,符合七年级数学课程标准中对数据分析与方程应用的要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 10:34:47","updated_at":"2026-01-06 10:34:47","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":1074,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"在一次班级图书角整理活动中,某学生统计了上周同学们借阅图书的情况。其中,借阅科普类图书的人数比借阅文学类图书的人数多5人,两类图书共被借阅了37人次。设借阅文学类图书的人数为x,则根据题意可列出一元一次方程:________。","answer":"x + (x + 5) = 37","explanation":"根据题意,借阅文学类图书的人数为x,则借阅科普类图书的人数为x + 5。两类图书共被借阅37人次,因此总人数为文学类人数加上科普类人数,即x + (x + 5) = 37。这是一道基于一元一次方程知识点的应用题,考查学生将实际问题转化为数学方程的能力,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":"2026-01-06 08:53:24","updated_at":"2026-01-06 08:53:24","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":2153,"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) = 9 时,第一步写成了 3x - 2 = 9。该学生在哪一步出现了错误?","answer":"B","explanation":"原方程为 3(x - 2) = 9,正确去括号应为 3x - 6 = 9。该学生写成 3x - 2 = 9,说明只将 3 与 x 相乘,而忽略了与 -2 相乘,即未将括号外的数与括号内的每一项相乘,因此错误出现在去括号步骤中的乘法分配律应用不当。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 13:00:46","updated_at":"2026-01-09 13:00:46","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":"去括号时没有改变括号内的符号","is_correct":0},{"id":"B","content":"去括号时没有将括号外的数与括号内的每一项相乘","is_correct":1},{"id":"C","content":"移项时没有变号","is_correct":0},{"id":"D","content":"合并同类项时计算错误","is_correct":0}]},{"id":638,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某班级在一次数学测验中,收集了30名学生的成绩,并将成绩分为5个分数段进行统计。已知前四个分数段的人数分别为4、7、9、6,则第五个分数段的人数是多少?","answer":"B","explanation":"题目考查的是数据的收集与整理。总人数为30人,前四个分数段的人数分别为4、7、9、6。将这些人数相加:4 + 7 + 9 + 6 = 26。因此,第五个分数段的人数为30 - 26 = 4。所以正确答案是B。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 22:05:16","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":"A","content":"3","is_correct":0},{"id":"B","content":"4","is_correct":1},{"id":"C","content":"5","is_correct":0},{"id":"D","content":"6","is_correct":0}]}]