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[{"id":1509,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某学生在研究平面直角坐标系中的点运动规律时,发现一个动点P从原点O(0, 0)出发,按照以下规则移动:第1次向右移动1个单位,第2次向上移动2个单位,第3次向左移动3个单位,第4次向下移动4个单位,第5次再向右移动5个单位,第6次再向上移动6个单位,依此类推,每次移动方向按右、上、左、下循环,移动步长为当前次数的数值。设第n次移动后点P的坐标为(x_n, y_n)。已知该学生记录了前k次移动后点P的横坐标与纵坐标的绝对值之和为S_k = |x_k| + |y_k|,且发现当k = 2024时,S_k = 1012。请判断这一结论是否正确,并通过计算说明理由。","answer":"我们分析动点P的移动规律:\n\n移动方向按周期为4的循环进行:右(+x)、上(+y)、左(-x)、下(-y),对应第1、2、3、4次,然后第5次又回到右,依此类推。\n\n将移动分为每4次一组,称为一个完整周期。\n\n在一个周期内(如第4m+1到第4m+4次):\n- 第4m+1次:向右移动 (4m+1) 单位 → x 增加 (4m+1)\n- 第4m+2次:向上移动 (4m+2) 单位 → y 增加 (4m+2)\n- 第4m+3次:向左移动 (4m+3) 单位 → x 减少 (4m+3)\n- 第4m+4次:向下移动 (4m+4) 单位 → y 减少 (4m+4)\n\n计算一个周期内x和y的净变化:\nΔx = (4m+1) - (4m+3) = -2\nΔy = (4m+2) - (4m+4) = -2\n\n即每完成一个完整的4次移动,x减少2,y减少2。\n\n现在考虑k = 2024次移动。\n\n2024 ÷ 4 = 506,即恰好完成506个完整周期,无剩余移动。\n\n初始位置为(0, 0),经过506个周期后:\nx = 0 + 506 × (-2) = -1012\ny = 0 + 506 × (-2) = -1012\n\n因此,S_k = |x| + |y| = |-1012| + |-1012| = 1012 + 1012 = 2024\n\n但题目中说S_k = 1012,这与计算结果2024不符。\n\n因此,该学生的结论是错误的。\n\n正确答案是:S_{2024} = 2024,而不是1012。","explanation":"本题综合考查了平面直角坐标系中点的坐标变化规律、周期性运动分析、整式运算以及绝对值的计算。解题关键在于识别移动模式的周期性(每4次为一个周期),并计算每个周期内坐标的净变化。通过分组求和,将2024次移动划分为506个完整周期,利用整式加减计算总位移。由于每个周期使x和y各减少2,因此总位移为(-1012, -1012),进而求得绝对值之和为2024。题目设置的陷阱在于学生可能误认为每次移动后坐标绝对值之和呈线性增长或忽略方向变化,导致错误判断。本题需要较强的逻辑推理能力和模式识别能力,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 12:08:01","updated_at":"2026-01-06 12:08:01","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":689,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生在绘制平面直角坐标系中的点时,先向右移动4个单位,再向上移动3个单位,最后向左移动1个单位,此时他所在位置的坐标是(___,3)。","answer":"3","explanation":"该学生从原点出发,先向右移动4个单位,横坐标变为4;再向上移动3个单位,纵坐标变为3;最后向左移动1个单位,横坐标减少1,变为4 - 1 = 3。因此,最终位置的横坐标是3,纵坐标是3,题目中已给出纵坐标为3,所以空格应填3。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 22:36:07","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":851,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生在绘制班级同学最喜爱的课外活动统计图时,将数据整理成如下表格:阅读占20%,运动占35%,音乐占15%,绘画占___%,其余为其他活动。已知喜欢绘画的人数比喜欢音乐的人数多6人,且班级总人数为60人,那么绘画所占的百分比是____。","answer":"25","explanation":"首先,根据题意,班级总人数为60人。喜欢音乐的人占15%,即 60 × 15% = 9 人。喜欢绘画的人数比音乐多6人,所以绘画人数为 9 + 6 = 15 人。那么绘画所占的百分比为 (15 ÷ 60) × 100% = 25%。因此,空白处应填写25。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 01:04:52","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":358,"subject":"数学","grade":"初一","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名同学的数学成绩(单位:分)分别为:82,76,90,88,84。为了分析整体情况,该学生需要计算这组数据的平均数。请问这组数据的平均数是多少?","answer":"B","explanation":"计算平均数的方法是将所有数据相加,然后除以数据的个数。首先将5个成绩相加:82 + 76 + 90 + 88 + 84 = 420。然后将总和除以人数5:420 ÷ 5 = 84。因此,这组数据的平均数是84分,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 15:44:21","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":"82分","is_correct":0},{"id":"B","content":"84分","is_correct":1},{"id":"C","content":"86分","is_correct":0},{"id":"D","content":"88分","is_correct":0}]},{"id":675,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"某学生测量了一个矩形花坛的长和宽,发现长比宽多2米。若花坛的周长为20米,则花坛的宽是___米。","answer":"4","explanation":"设花坛的宽为x米,则长为(x + 2)米。根据矩形周长公式:周长 = 2 × (长 + 宽),代入得:2 × (x + x + 2) = 20。化简得:2 × (2x + 2) = 20,即4x + 4 = 20。解这个一元一次方程:4x = 16,x = 4。因此,花坛的宽是4米。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 22:25:10","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":567,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"平均数是5.2,中位数是5,众数是5","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 19:35:05","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":494,"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。根据表格信息,成绩在80分及以上的人数占总人数的百分比最接近以下哪个选项?\n\n| 分数段(分) | 人数 |\n|--------------|------|\n| 60以下 | 5 |\n| 60—69 | 8 |\n| 70—79 | 12 |\n| 80—89 | 15 |\n| 90—100 | 10 |","answer":"C","explanation":"首先计算总人数:5 + 8 + 12 + 15 + 10 = 50(人)。\n成绩在80分及以上的人数包括80—89和90—100两个分数段,共15 + 10 = 25(人)。\n所求百分比为:25 ÷ 50 × 100% = 50%。\n因此,正确答案是C选项。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 18:06:22","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":"25%","is_correct":0},{"id":"B","content":"40%","is_correct":0},{"id":"C","content":"50%","is_correct":1},{"id":"D","content":"60%","is_correct":0}]},{"id":1869,"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连续7天的观测,记录每天上午8:00至9:00的车辆通行数量(单位:辆),数据如下:312,298,305,310,307,299,304。交通部门计划根据这组数据预测未来某周的车流量,并设定一个合理的通行能力标准。已知该道路的设计通行能力为每天平均车流量的1.2倍,且要求实际车流量不超过设计通行能力的90%才算安全运行。若未来某周的车流量服从本次观测的平均水平,请通过计算判断该道路在未来是否满足安全运行要求。若不能满足,则至少需要将设计通行能力提升到当前观测平均车流量的多少倍(精确到0.01)才能满足安全要求?","answer":"解:\n\n第一步:计算7天观测数据的平均车流量。\n\n平均车流量 = (312 + 298 + 305 + 310 + 307 + 299 + 304) ÷ 7\n= (2135) ÷ 7\n= 305(辆)\n\n第二步:计算当前设计通行能力。\n\n设计通行能力 = 平均车流量 × 1.2 = 305 × 1.2 = 366(辆)\n\n第三步:计算安全运行上限(即设计通行能力的90%)。\n\n安全上限 = 366 × 90% = 366 × 0.9 = 329.4(辆)\n\n第四步:比较实际平均车流量与安全上限。\n\n实际平均车流量为305辆,小于329.4辆,因此当前道路满足安全运行要求。\n\n但题目要求判断“若不能满足”的情况下的处理方式,因此需进一步分析假设情形。\n\n然而根据计算,305 < 329.4,满足安全要求,故当前无需提升。\n\n但为完整解答问题,假设未来车流量上升至等于安全上限临界值,我们反向求解所需的设计通行能力倍数。\n\n设所需设计通行能力为平均车流量的k倍,则:\n\n安全上限 = k × 305 × 0.9 ≥ 305(因实际车流量为305)\n\n即:k × 305 × 0.9 ≥ 3...","explanation":"本题综合考查数据的收集与整理(计算平均数)、有理数运算、一元一次不等式的应用。解题关键在于理解‘安全运行’的定义:实际车流量 ≤ 设计通行能力 × 90%。先通过平均数反映典型车流量,再建立不等式模型求解最小安全倍数。难点在于将实际问题转化为数学不等式,并理解倍数关系的逻辑链条。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-07 09:41:09","updated_at":"2026-01-07 09:41:09","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":928,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次环保活动中,某学生记录了班级同学一周内回收的废纸重量(单位:千克),数据如下:2.5,3.0,2.8,3.2,2.5,3.1,2.9。这组数据的众数是___。","answer":"2.5","explanation":"众数是一组数据中出现次数最多的数。观察数据:2.5 出现了两次,3.0、2.8、3.2、3.1、2.9 各出现一次。因此,2.5 是出现次数最多的数,即这组数据的众数是 2.5。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 02:51:57","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":371,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某班级组织了一次环保知识竞赛,共20道题,答对一题得5分,答错或不答扣2分。一名学生最终得分为65分,请问他答对了多少道题?","answer":"A","explanation":"设这名学生答对了x道题,则答错或不答的题数为(20 - x)道。根据题意,答对一题得5分,答错或不答扣2分,总得分为65分,可列出一元一次方程:5x - 2(20 - x) = 65。展开并化简:5x - 40 + 2x = 65,合并同类项得7x - 40 = 65,移项得7x = 105,解得x = 15。因此,该学生答对了15道题。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:49:17","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":"15","is_correct":1},{"id":"B","content":"14","is_correct":0},{"id":"C","content":"13","is_correct":0},{"id":"D","content":"12","is_correct":0}]}]