[{"id":1419,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某校组织七年级学生开展‘校园绿化区域规划’项目活动。在平面直角坐标系中，校园内一块矩形绿化区域ABCD的顶点坐标分别为A(0, 0)、B(8, 0)、C(8, 6)、D(0, 6)（单位：米）。现计划在矩形内部修建一条宽度为1米的L形步道，步道由两条互相垂直且宽度均为1米的路径组成：一条从点E(2, 0)垂直向上延伸至点F(2, 4)，另一条从点F(2, 4)水平向右延伸至点G(7, 4)。步道所占区域需从绿化面积中扣除。此外，为美化环境，将在剩余绿化区域中种植花卉，每平方米种植成本为30元。若学校预算为5000元，问：该预算是否足够支付花卉种植费用？若不够，最多还能增加多少平方米的种植面积？（精确到0.1平方米）","answer":"第一步：计算矩形绿化区域ABCD的总面积。\n矩形长 = 8 - 0 = 8 米，宽 = 6 - 0 = 6 米，\n面积 = 8 × 6 = 48 平方米。\n\n第二步：计算L形步道的面积。\n步道由两部分组成：\n（1）竖直部分：从E(2,0)到F(2,4)，长度为4米，宽度为1米，\n面积为 4 × 1 = 4 平方米。\n（2）水平部分：从F(2,4)到G(7,4)，长度为5米，宽度为1米，\n面积为 5 × 1 = 5 平方米。\n注意：两部分在F点重叠一个1×1的正方形区域，不能重复计算。\n因此，步道总面积 = 4 + 5 - 1 = 8 平方米。\n\n第三步：计算剩余绿化面积。\n剩余面积 = 48 - 8 = 40 平方米。\n\n第四步：计算花卉种植总成本。\n每平方米30元，总成本 = 40 × 30 = 1200 元。\n\n第五步：比较预算与实际费用。\n学校预算为5000元，1200 < 5000，因此预算足够。\n\n第六步：计算在预算范围内最多还能增加多少种植面积。\n剩余预算 = 5000 - 1200 = 3800 元。\n每平方米30元，可增加的面积 = 3800 ÷ 30 ≈ 126.666... 平方米。\n精确到0.1平方米，最多可增加 126.7 平方米。\n\n答：该预算足够支付花卉种植费用；最多还能增加126.7平方米的种植面积。","explanation":"本题综合考查平面直角坐标系中图形位置的确定、矩形面积计算、重叠区域的处理以及一元一次方程与不等式的实际应用。解题关键在于准确理解L形步道的几何结构，识别出竖直与水平路径在交点F处存在1平方米的重叠区域，避免重复计算。通过分步计算总面积、扣除步道面积、核算成本，并最终利用预算差额反推可增加面积，体现了数学建模与实际问题解决能力。题目融合了几何图形初步、平面直角坐标系、有理数运算和一元一次方程的应用，难度较高，适合能力较强的七年级学生挑战。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 11:30:52","updated_at":"2026-01-06 11:30:52","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":2524,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"如图，一个圆形花坛的半径为6米，某学生从花坛边缘的点A出发，沿直线走到花坛中心O，再从O沿另一条直线走到边缘的点B，且∠AOB = 60°。则该学生从A经O到B所走的总路程为多少米？","answer":"A","explanation":"该学生从点A走到圆心O，再从O走到点B。由于A和B都在圆周上，OA和OB都是圆的半径，长度为6米。因此，AO = 6米，OB = 6米。总路程为AO + OB = 6 + 6 = 12米。虽然∠AOB = 60°，但题目问的是沿AO和OB走的路径长度，不是弦AB的长度，因此角度信息是干扰项，不影响路程计算。故正确答案为A。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 16:04:19","updated_at":"2026-01-10 16:04:19","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":"12","is_correct":1},{"id":"B","content":"12 + 2√3","is_correct":0},{"id":"C","content":"12 + 6√3","is_correct":0},{"id":"D","content":"18","is_correct":0}]},{"id":2169,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在数轴上标记了三个有理数点A、B、C，其中点A表示的数是-3.5，点B位于点A右侧4.2个单位长度处，点C位于点B左侧2.8个单位长度处。若将这三个点所表示的数按从小到大的顺序排列，正确的顺序是？","answer":"B","explanation":"首先确定各点表示的有理数：点A为-3.5；点B在A右侧4.2个单位，即-3.5 + 4.2 = 0.7；点C在B左侧2.8个单位，即0.7 - 2.8 = -2.1。因此三个数分别为：A=-3.5，B=0.7，C=-2.1。比较大小：-3.5 < -2.1 < 0.7，即A < C < B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-09 13:53:54","updated_at":"2026-01-09 13:53:54","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":"A < B < C","is_correct":0},{"id":"B","content":"A < C < B","is_correct":1},{"id":"C","content":"C < A < B","is_correct":0},{"id":"D","content":"B < C < A","is_correct":0}]},{"id":2245,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某学生在研究温度变化时，记录了连续7天的每日最低气温（单位：℃），这些数据分别为：-3，2，-5，0，-1，4，-2。该学生想计算这7天中，气温低于零度的天数占总天数的几分之几，并进一步求出这些负温度的绝对值的平均数。请完成以下两个任务：(1) 求出气温低于零度的天数占总天数的几分之几（结果用最简分数表示）；(2) 求出所有负温度的绝对值的平均数（结果保留一位小数）。","answer":"(1) 4\/7；(2) 2.8","explanation":"本题综合考查了正数、负数的识别，绝对值的概念，以及分数和平均数的计算。七年级学生已掌握负数的意义、绝对值的求法以及基本统计量的计算。题目通过真实情境（气温记录）引导学生分析数据，区分正负数，并进行多步运算，体现了数学在实际生活中的应用，难度较高，符合困难级别要求。","solution_steps":"第一步：确定气温低于零度的天数。观察数据：-3，2，-5，0，-1，4，-2。其中小于0的数有：-3，-5，-1，-2，共4天。总天数为7天，因此所求分数为4\/7，已是最简分数。第二步：找出所有负温度：-3，-5，-1，-2。求它们的绝对值：| -3 | = 3，| -5 | = 5，| -1 | = 1，| -2 | = 2。第三步：计算这些绝对值的和：3 + 5 + 1 + 2 = 11。第四步：求平均数：11 ÷ 4 = 2.75，保留一位小数为2.8。","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-09 14:44:04","updated_at":"2026-01-09 14:44:04","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":291,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"某学生在整理班级同学的课外阅读情况时，收集了10名同学每周课外阅读时间（单位：小时），数据如下：3，5，4，6，5，7，5，4，6，5。这组数据的众数和中位数分别是多少？","answer":"A","explanation":"首先将数据从小到大排序：3，4，4，5，5，5，5，6，6，7。众数是出现次数最多的数，其中5出现了4次，次数最多，因此众数是5。中位数是数据按顺序排列后位于中间位置的数。由于共有10个数据（偶数个），中位数为第5个和第6个数的平均数，即(5 + 5) ÷ 2 = 5。因此，众数是5，中位数是5，正确答案是A。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:32:36","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":"众数是5，中位数是5","is_correct":1},{"id":"B","content":"众数是4，中位数是5","is_correct":0},{"id":"C","content":"众数是5，中位数是4.5","is_correct":0},{"id":"D","content":"众数是6，中位数是5.5","is_correct":0}]},{"id":503,"subject":"数学","grade":"初一","stage":"初中","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。根据表格，喜欢阅读的人数占总调查人数的百分比是多少？\n\n| 活动类型 | 人数 |\n|----------|------|\n| 阅读 | 12 |\n| 运动 | 18 |\n| 音乐 | 10 |\n| 绘画 | 10 |","answer":"B","explanation":"首先计算总调查人数：12 + 18 + 10 + 10 = 50（人）。喜欢阅读的人数为12人，因此所占百分比为 (12 ÷ 50) × 100% = 24%。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 18:10:26","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":"20%","is_correct":0},{"id":"B","content":"24%","is_correct":1},{"id":"C","content":"30%","is_correct":0},{"id":"D","content":"36%","is_correct":0}]},{"id":737,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生调查了班级同学每天用于课外阅读的时间（单位：分钟），将数据整理后绘制成频数分布直方图。已知阅读时间在30～40分钟这一组的频数是8，频率是0.2，则该学生所在班级的总人数是____。","answer":"40","explanation":"根据频率的定义，频率 = 频数 ÷ 总人数。题目中给出频数为8，频率为0.2，因此总人数 = 频数 ÷ 频率 = 8 ÷ 0.2 = 40。该题考查数据的收集、整理与描述中的频率与频数关系，属于简单计算题。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 23:09:43","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":713,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生测量了教室里5盏灯的功率，分别为40瓦、60瓦、40瓦、100瓦和40瓦。这组数据的中位数是____瓦。","answer":"40","explanation":"首先将这组数据按从小到大的顺序排列：40、40、40、60、100。共有5个数据，是奇数个，因此中位数是正中间的那个数，即第3个数，为40瓦。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 22:49:55","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":2265,"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-3，点B与点A的距离为7个单位长度，且点B在原点右侧。若点C是点B关于原点的对称点，则点C表示的数是___。","answer":"B","explanation":"点A表示-3，点B与点A的距离为7个单位长度，且点B在原点右侧。因此点B可能在-3的右侧7个单位，即-3 + 7 = 4，所以点B表示4。点C是点B关于原点的对称点，即与4到原点距离相等但方向相反，因此点C表示-4。故正确答案为B。","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":"4","is_correct":0},{"id":"B","content":"-4","is_correct":1},{"id":"C","content":"10","is_correct":0},{"id":"D","content":"-10","is_correct":0}]},{"id":2503,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"某学生观察一个由两个相似直角三角形组成的几何图形，其中较小三角形的斜边长为5 cm，较大三角形的对应斜边长为15 cm。若较小三角形的一条直角边为3 cm，则较大三角形中对应的直角边长度为多少？","answer":"B","explanation":"由于两个三角形相似，对应边的长度成比例。较小三角形与较大三角形的斜边之比为 5:15 = 1:3，因此相似比为 1:3。较小三角形中一条直角边为 3 cm，则较大三角形中对应的直角边应为 3 × 3 = 9 cm。故正确答案为 B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 15:26:45","updated_at":"2026-01-10 15:26:45","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 cm","is_correct":0},{"id":"B","content":"9 cm","is_correct":1},{"id":"C","content":"12 cm","is_correct":0},{"id":"D","content":"15 cm","is_correct":0}]}]