[{"id":1687,"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路径规划问题时，发现一个矩形花坛ABCD被两条相互垂直的小路EF和GH分割成四个小区域，其中E在AB上，F在CD上，G在AD上，H在BC上，且EF平行于AD，GH平行于AB。已知矩形花坛的周长为48米，面积为135平方米。小路EF和GH的宽度均为1米，且小路的铺设成本为每平方米80元。若该学生计划通过调整花坛的长和宽（保持周长和面积不变）来最小化小路的总铺设成本，问：当长和宽分别为多少米时，小路的总成本最低？最低成本是多少元？","answer":"设矩形花坛的长为x米，宽为y米。\n\n由题意得：\n周长：2(x + y) = 48 ⇒ x + y = 24 ……(1)\n面积：xy = 135 ……(2)\n\n将(1)代入(2)：x(24 - x) = 135\n⇒ 24x - x² = 135\n⇒ x² - 24x + 135 = 0\n\n解这个方程：\n判别式 Δ = (-24)² - 4×1×135 = 576 - 540 = 36\nx = [24 ± √36]\/2 = [24 ± 6]\/2\n⇒ x = 15 或 x = 9\n\n对应地，y = 9 或 y = 15\n\n所以矩形的长和宽分别为15米和9米（不考虑顺序）。\n\n现在分析小路面积：\n小路EF平行于AD（即竖直方向），长度为宽y，宽度为1米，面积为 y × 1 = y 平方米。\n小路GH平行于AB（即水平方向），长度为长x，宽度为1米，面积为 x × 1 = x 平方米。\n\n但两条小路在中心交叉，重叠部分为一个1×1 = 1平方米的正方形，被重复计算了一次，因此实际小路总面积为：\nx + y - 1\n\n代入x + y = 24，得小路总面积为：24 - 1 = 23 平方米\n\n无论x和y如何取值（只要满足x + y = 24且xy = 135），小路总面积恒为23平方米。\n\n因此，小路总成本 = 23 × 80 = 1840 元\n\n结论：在所有满足周长48米、面积135平方米的矩形中，小路总成本恒为1840元，不存在“最低成本”的变化。\n\n但题目要求“通过调整长和宽来最小化成本”，而实际上在固定周长和面积下，长和宽只能取两组值（15和9），且小路面积不变。\n\n进一步分析：是否存在其他满足周长48、面积135的矩形？\n由方程x² - 24x + 135 = 0只有两个实数解，说明只有两种可能的矩形（长宽互换），小路面积均为23平方米。\n\n因此，无论长是15米宽是9米，还是长是9米宽是15米，小路总面积不变，成本不变。\n\n答：当花坛的长为15米、宽为9米（或长为9米、宽为15米）时，小路总成本最低，最低成本为1840元。","explanation":"本题综合考查了一元二次方程、二元一次方程组、整式运算、几何图形初步及实际应用建模能力。解题关键在于建立矩形长和宽的方程，并利用周长和面积条件求解可能的尺寸。难点在于理解两条交叉小路的面积计算需扣除重叠部分，并发现尽管长和宽可互换，但小路总面积在固定周长和面积下保持不变。这体现了代数与几何的结合，以及优化问题中的不变量思想。题目设计避免了常见的应用题模式，通过真实情境引导学生深入思考变量之间的关系，符合七年级学生对实数、方程和几何图形的综合应用能力要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 13:34:53","updated_at":"2026-01-06 13:34:53","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":1513,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市为改善空气质量，计划在一条主干道两侧种植树木。道路全长1200米，起点和终点都必须种树。最初计划每隔6米种一棵树，但后来考虑到树木长大后可能影响路灯照明，决定将每两棵树之间的距离调整为8米。调整后，部分原有的树坑需要填埋，新的树坑需要挖掘。已知填埋一个旧树坑的费用为5元，挖掘一个新树坑的费用为8元。若某学生负责计算此项工程的总费用，请根据以上信息回答：\n\n（1）按原计划每隔6米种一棵树，整条道路两侧共需挖多少个树坑？\n\n（2）按调整后每隔8米种一棵树，整条道路两侧共需挖多少个树坑？\n\n（3）在调整过程中，有多少个原树坑的位置恰好与新树坑位置重合？\n\n（4）此项工程中，填埋旧树坑和挖掘新树坑的总费用是多少元？","answer":"（1）道路全长1200米，起点和终点都种树，每隔6米种一棵。\n每侧所需树坑数为：1200 ÷ 6 + 1 = 200 + 1 = 201（个）\n两侧共需：201 × 2 = 402（个）\n答：原计划共需挖402个树坑。\n\n（2）调整后每隔8米种一棵树。\n每侧所需树坑数为：1200 ÷ 8 + 1 = 150 + 1 = 151（个）\n两侧共需：151 × 2 = 302（个）\n答：调整后共需挖302个树坑。\n\n（3）重合的位置是6和8的公倍数所在的位置。\n先求6和8的最小公倍数：\n6 = 2 × 3，8 = 2³，最小公倍数为 2³ × 3 = 24\n即在每隔24米的位置，原树坑与新树坑重合。\n从起点0米开始，每隔24米一个重合点：0, 24, 48, ..., 1200\n这是一个等差数列，首项为0，公差为24，末项为1200\n项数为：(1200 - 0) ÷ 24 + 1 = 50 + 1 = 51（个）\n每侧有51个重合点，两侧共：51 × 2 = 102（个）\n答：有102个原树坑位置与新树坑重合。\n\n（4）填埋旧树坑数量 = 原计划树坑总数 - 重合的树坑数 = 402 - 102 = 300（个）\n挖掘新树坑数量 = 调整后树坑总数 - 重合的树坑数 = 302 - 102 = 200（个）\n填埋费用：300 × 5 = 1500（元）\n挖掘费用：200 × 8 = 1600（元）\n总费用：1500 + 1600 = 3100（元）\n答：总费用为3100元。","explanation":"本题综合考查了有理数运算、最小公倍数、等差数列、实际问题建模以及数据的整理与计算能力。第（1）问和第（2）问考查了在两端都种树的情况下，树坑数量的计算，属于植树问题的基本模型，需注意‘段数+1=棵数’的规律。第（3）问是难点，需要理解重合位置即6和8的公倍数位置，通过求最小公倍数24，再计算从0到1200之间24的倍数个数，转化为等差数列求项数问题。第（4）问考查逻辑推理与费用计算，需明确填埋的是‘未被利用的旧坑’，挖掘的是‘新增的新坑’，不能重复计算重合部分。整个过程体现了数学在实际生活中的应用，要求学生具备较强的综合分析能力。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 12:09:15","updated_at":"2026-01-06 12:09:15","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":424,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"在一次班级数学测验中，老师收集了10名学生的成绩（单位：分）如下：85，78，92，88，76，90，84，89，81，87。如果老师想用一个统计量来代表这次测验的整体水平，并且希望这个值能反映大多数学生的成绩情况，那么最合适的统计量是：","answer":"B","explanation":"题目要求选择一个能代表整体水平并反映大多数学生成绩情况的统计量。首先观察数据：85，78，92，88，76，90，84，89，81，87。这些数据分布较为均匀，没有明显的极端值（如特别高或特别低的分数），但也没有重复出现的数值，因此众数不存在或无法体现‘大多数’。最大值（92）仅代表最高分，不能反映整体。平均数虽然能反映整体平均水平，但容易受极端值影响；而中位数是将数据按大小顺序排列后位于中间的值，能较好地代表中间水平，避免极端值干扰。将数据从小到大排列：76，78，81，84，85，87，88，89，90，92。共有10个数据，中位数为第5和第6个数的平均数，即(85 + 87) ÷ 2 = 86。这个值位于数据中间位置，能较好地反映大多数学生的成绩集中趋势，因此最合适。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:32:56","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":"平均数","is_correct":0},{"id":"B","content":"中位数","is_correct":1},{"id":"C","content":"众数","is_correct":0},{"id":"D","content":"最大值","is_correct":0}]},{"id":2271,"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-4，点B表示的数是6。某学生在数轴上标出了点C，使得点C到点A的距离是点C到点B的距离的2倍。那么点C表示的数可能是多少？","answer":"D","explanation":"设点C表示的数为x。根据题意，点C到点A的距离为|x + 4|，点C到点B的距离为|x - 6|。由条件得：|x + 4| = 2|x - 6|。分情况讨论：当x ≥ 6时，x + 4 = 2(x - 6)，解得x = 16；当-4 ≤ x < 6时，x + 4 = 2(6 - x)，解得x = 16\/3；当x < -4时，-(x + 4) = 2(6 - x)，解得x = -16。经检验，x = -16和x = 16\/3均满足原方程，因此点C表示的数可能是-16或16\/3。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-09 16:09:15","updated_at":"2026-01-09 16:09:15","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":"-16","is_correct":0},{"id":"B","content":"8\/3","is_correct":0},{"id":"C","content":"16","is_correct":0},{"id":"D","content":"-16或16\/3","is_correct":1}]},{"id":889,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在某次班级大扫除中，一组学生负责擦窗户。如果每扇窗户需要2分钟擦干净，他们一共用了24分钟完成任务，那么这组学生一共擦了____扇窗户。","answer":"12","explanation":"题目中给出每扇窗户需要2分钟，总用时24分钟。要求擦了多少扇窗户，可以用总时间除以每扇窗户所需时间：24 ÷ 2 = 12。这是一道简单的一元一次方程应用题，设擦了x扇窗户，则2x = 24，解得x = 12。考查学生将实际问题转化为方程并求解的能力，属于一元一次方程知识点，难度简单。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 02:01:51","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":469,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"在一次环保知识问卷调查中，某班级共发放了60份问卷，回收有效问卷54份。请问该问卷的有效回收率是多少？","answer":"B","explanation":"有效回收率的计算公式为：有效回收率 = (有效问卷数量 ÷ 发放问卷总数) × 100%。根据题意，有效问卷为54份，发放总数为60份，因此有效回收率为 (54 ÷ 60) × 100% = 0.9 × 100% = 90%。故正确答案为B。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:53:49","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":"85%","is_correct":0},{"id":"B","content":"90%","is_correct":1},{"id":"C","content":"95%","is_correct":0},{"id":"D","content":"100%","is_correct":0}]},{"id":2237,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"某学生在数轴上从原点出发，先向右移动5个单位长度，再向左移动8个单位长度，接着又向右移动3个单位长度，最后向左移动6个单位长度。此时该学生所在位置的数与它相反数的和是___。","answer":"0","explanation":"该学生从原点0出发，依次进行移动：+5（右移5），-8（左移8），+3（右移3），-6（左移6）。计算最终位置：0 + 5 - 8 + 3 - 6 = -6。设该位置的数为-6，其相反数为6，两者之和为-6 + 6 = 0。根据相反数的性质，任何数与其相反数之和恒为0，因此无论最终位置为何，该和始终为0。本题综合考查数轴上的正负数运算及相反数的概念，需多步推理，难度较高。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-09 14:39:22","updated_at":"2026-01-09 14:39:22","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":2450,"subject":"数学","grade":"八年级","stage":"初中","type":"填空题","content":"在一次函数 y = kx + b 的图像中，已知该函数与 x 轴交于点 (4, 0)，与 y 轴交于点 (0, -6)，则 k 的值为___。","answer":"3\/2","explanation":"由 y 轴交点得 b = -6，代入 x 轴交点 (4, 0) 得 0 = 4k - 6，解得 k = 3\/2。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 13:54:56","updated_at":"2026-01-10 13:54:56","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":686,"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废旧纸张。如果将这些纸张平均分给5个小组，每组可得12千克；后来又有3个小组加入，现在要将这些纸张重新平均分给所有小组，那么每个小组分到的纸张是___千克。","answer":"7.5","explanation":"首先根据题意，原来有5个小组，每组12千克，所以总纸张重量为 5 × 12 = 60 千克。后来增加了3个小组，总小组数变为 5 + 3 = 8 个。将60千克纸张平均分给8个小组，每个小组分到 60 ÷ 8 = 7.5 千克。本题考查了一元一次方程的实际应用和整数的除法运算，属于简单难度，符合七年级学生对有理数和一元一次方程知识点的掌握水平。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 22:33:25","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":727,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在某次班级大扫除中，学生们被分成若干小组清理教室。如果每组安排5人，则多出3人；如果每组安排6人，则最后一组只有4人。这个班级共有___名学生。","answer":"28","explanation":"设班级共有x名学生。根据题意，当每组5人时，多出3人，说明x除以5余3，即x = 5a + 3（a为组数）。当每组6人时，最后一组只有4人，说明x除以6余4，即x = 6b + 4（b为组数）。寻找同时满足这两个条件的最小正整数。尝试代入：当x=28时，28 ÷ 5 = 5组余3，符合第一种情况；28 ÷ 6 = 4组余4，也符合第二种情况。因此，班级共有28名学生。本题考查一元一次方程的实际应用与整数解问题，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 23:02:09","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":[]}]