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[{"id":1643,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市为优化公交线路,对一条主干道的车流量进行了为期一周的观测,记录每天上午7:00至9:00的车辆通过数量(单位:辆),数据如下:周一 1200,周二 1350,周三 1420,周四 1380,周五 1500,周六 900,周日 750。交通部门计划根据这些数据调整发车间隔,并设定以下规则:若某日平均车流量超过1300辆,则工作日(周一至周五)发车间隔为4分钟;否则为6分钟。周末发车间隔固定为8分钟。已知每辆公交车单程运行时间为40分钟,且每辆车每天最多运行6个单程。现需在平面直角坐标系中绘制该周车流量的折线图,并计算满足运营需求所需的最少公交车数量。假设所有公交车均从总站出发,且发车间隔必须严格保持。","answer":"第一步:整理数据并判断每日发车间隔\n周一:1200 ≤ 1300 → 发车间隔6分钟\n周二:1350 > 1300 → 发车间隔4分钟\n周三:1420 > 1300 → 发车间隔4分钟\n周四:1380 > 1300 → 发车间隔4分钟\n周五:1500 > 1300 → 发车间隔4分钟\n周六:900 ≤ 1300,但为周末 → 发车间隔8分钟\n周日:750 ≤ 1300,但为周末 → 发车间隔8分钟\n\n第二步:计算每天需要的发车班次\n每天运营时间:7:00–9:00,共2小时 = 120分钟\n发车班次 = 120 ÷ 发车间隔(向上取整)\n周一:120 ÷ 6 = 20 班\n周二至周五:120 ÷ 4 = 30 班\n周六、周日:120 ÷ 8 = 15 班\n\n第三步:计算每天所需公交车数量\n每辆车每天最多运行6个单程,即最多参与6个班次(假设每个班次为单程)\n所需车辆数 = 总班次数 ÷ 6(向上取整)\n周一:20 ÷ 6 ≈ 3.33 → 需4辆车\n周二至周五:30 ÷ 6 = 5 → 需5辆车\n周六、周日:15 ÷ 6 = 2.5 → 需3辆车\n\n第四步:确定整周所需最少公交车数量\n由于车辆可重复使用,需找出单日最大需求量\n最大需求出现在周二至周五,每天需5辆车\n因此,整周至少需要5辆公交车才能满足高峰日需求\n\n第五步:在平面直角坐标系中绘制折线图(描述性说明)\n横轴:星期(周一至周日),共7个点\n纵轴:车流量(单位:辆),范围建议0–1600\n依次标出点:(1,1200), (2,1350), (3,1420), (4,1380), (5,1500), (6,900), (7,750)\n用线段连接各点,形成折线图,标注坐标轴名称和单位\n\n最终答案:满足运营需求所需的最少公交车数量为5辆。","explanation":"本题综合考查数据的收集与整理、有理数运算、不等式判断、一元一次方程思想(发车班次计算)、平面直角坐标系绘图以及实际应用中的最优化问题。解题关键在于理解发车间隔与车流量的关系,并通过不等式判断每日调度策略;再结合时间、班次与车辆运行能力,建立数学模型计算最少车辆数。折线图的绘制要求学生掌握坐标系的基本使用方法。题目情境贴近现实,逻辑链条较长,需分步分析,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 13:11:11","updated_at":"2026-01-06 13:11:11","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":2157,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在数轴上标出两个有理数,其中一个数位于原点左侧3个单位长度处,另一个数位于原点右侧5个单位长度处。这两个有理数的和是多少?","answer":"B","explanation":"位于原点左侧3个单位长度的有理数是-3,位于原点右侧5个单位长度的有理数是5。根据有理数加法法则,-3 + 5 = 2。因此正确答案是B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 13:07:43","updated_at":"2026-01-09 13:07:43","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":"-8","is_correct":0},{"id":"B","content":"2","is_correct":1},{"id":"C","content":"8","is_correct":0},{"id":"D","content":"-2","is_correct":0}]},{"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":2520,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"某学生设计了一个圆形花坛,其边缘由一段圆弧和两条半径围成,形成一个扇形区域。已知该扇形的圆心角为60°,面积为6π平方米。若在该扇形区域内接一个最大的等边三角形(三个顶点均在扇形边界上),则这个等边三角形的边长是多少?","answer":"A","explanation":"首先,根据扇形面积公式 S = (θ\/360°) × πr²,其中θ = 60°,S = 6π,代入得:6π = (60\/360) × πr² → 6π = (1\/6)πr² → r² = 36 → r = 6米。因此扇形半径为6米。由于圆心角为60°,若将扇形的两条半径端点与圆心连接,可构成一个边长为6米的等边三角形(因为两边为半径,夹角60°,三边相等)。此三角形完全位于扇形内,且是内接于该扇形中最大的等边三角形(任何其他构造都会导致边长更短或超出边界)。故该等边三角形边长为6米。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 15:56:03","updated_at":"2026-01-10 15:56:03","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":"6米","is_correct":1},{"id":"B","content":"3√3米","is_correct":0},{"id":"C","content":"4√3米","is_correct":0},{"id":"D","content":"2√6米","is_correct":0}]},{"id":224,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生在计算一个数减去5时,误将减号看成了加号,结果得到20。那么这个数正确的计算结果应该是____。","answer":"10","explanation":"根据题意,某学生把'减去5'误看成'加上5',得到结果是20。设这个数为x,则有 x + 5 = 20,解得 x = 15。那么正确的计算应是 15 - 5 = 10。因此正确答案是10。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 14:40:41","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":1093,"subject":"数学","grade":"七年级","stage":"小学","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和玻璃瓶,其中塑料瓶的数量比玻璃瓶的3倍多5个。若设玻璃瓶的数量为x个,则塑料瓶的数量可表示为______。","answer":"3x + 5","explanation":"根据题意,塑料瓶的数量比玻璃瓶的3倍多5个。玻璃瓶的数量为x,那么它的3倍就是3x,再加上5个,就是塑料瓶的数量,因此表达式为3x + 5。这是整式加减中的基本概念,考查学生将文字语言转化为代数表达式的能力。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-06 08:55:58","updated_at":"2026-01-06 08:55:58","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":216,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"一个长方形的长是8厘米,宽是5厘米,它的周长是_空白处_厘米。","answer":"26","explanation":"长方形的周长计算公式是:周长 = 2 × (长 + 宽)。将已知的长8厘米和宽5厘米代入公式:2 × (8 + 5) = 2 × 13 = 26。因此,这个长方形的周长是26厘米。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 14:40:10","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":1046,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次班级环保活动中,某学生收集了若干千克可回收垃圾。若他再收集3.5千克,则总量将达到8千克。请问他最初收集了___千克垃圾。","answer":"4.5","explanation":"设该学生最初收集的垃圾为x千克。根据题意,可列出方程:x + 3.5 = 8。解这个一元一次方程,两边同时减去3.5,得到x = 8 - 3.5 = 4.5。因此,他最初收集了4.5千克垃圾。本题考查了一元一次方程在实际生活中的应用,属于七年级数学课程中的基础题型。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 06:24:45","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":490,"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情况时,绘制了如下条形统计图(描述性文字替代图像):横轴表示阅读书籍数量(单位:本),纵轴表示对应人数。图中显示:阅读0本的有2人,1本的有5人,2本的有8人,3本的有4人,4本的有1人。请问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值。根据题目描述,阅读0本的有2人,1本的有5人,2本的有8人,3本的有4人,4本的有1人。其中,阅读2本的人数最多,为8人,因此这组数据的众数是2。选项C正确。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 18:03:49","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","is_correct":0},{"id":"B","content":"1","is_correct":0},{"id":"C","content":"2","is_correct":1},{"id":"D","content":"3","is_correct":0}]},{"id":571,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"某学生调查了班级同学最喜欢的课外活动,并将数据整理成如下表格。如果喜欢阅读的人数占总调查人数的20%,且总共有50人参与调查,那么喜欢阅读的同学有多少人?","answer":"B","explanation":"题目中给出总调查人数为50人,喜欢阅读的人数占20%。要计算喜欢阅读的人数,只需将总人数乘以百分比:50 × 20% = 50 × 0.2 = 10(人)。因此,喜欢阅读的同学有10人,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 19:47:25","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":"5人","is_correct":0},{"id":"B","content":"10人","is_correct":1},{"id":"C","content":"15人","is_correct":0},{"id":"D","content":"20人","is_correct":0}]}]