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[{"id":360,"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身高数据时,记录了10名同学的身高(单位:厘米)如下:152, 148, 155, 160, 158, 153, 149, 157, 161, 154。如果将这些数据按从小到大的顺序排列,则中位数是多少?","answer":"B","explanation":"首先将数据按从小到大的顺序排列:148, 149, 152, 153, 154, 155, 157, 158, 160, 161。共有10个数据,为偶数个,因此中位数是第5个和第6个数据的平均数。第5个数是154,第6个数是155,所以中位数为 (154 + 155) ÷ 2 = 154.5。因此正确答案是B。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:45:12","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":"154","is_correct":0},{"id":"B","content":"154.5","is_correct":1},{"id":"C","content":"155","is_correct":0},{"id":"D","content":"155.5","is_correct":0}]},{"id":332,"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,随机抽取了10名学生的成绩(单位:分)如下:78, 85, 88, 92, 76, 85, 90, 85, 82, 87。这组数据的众数是多少?","answer":"B","explanation":"众数是一组数据中出现次数最多的数。观察给出的数据:78, 85, 88, 92, 76, 85, 90, 85, 82, 87。统计每个数出现的次数:76出现1次,78出现1次,82出现1次,85出现3次,87出现1次,88出现1次,90出现1次,92出现1次。其中85出现的次数最多,共3次,因此这组数据的众数是85。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:39:30","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":"76","is_correct":0},{"id":"B","content":"85","is_correct":1},{"id":"C","content":"87","is_correct":0},{"id":"D","content":"90","is_correct":0}]},{"id":707,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生在调查班级同学最喜欢的运动项目时,共收集了30份有效问卷,其中喜欢篮球的有12人,喜欢足球的有8人,喜欢跳绳的有5人,其余同学喜欢乒乓球。那么喜欢乒乓球的同学占全班人数的____(填最简分数)。","answer":"1\/6","explanation":"总人数为30人,喜欢篮球、足球和跳绳的人数分别为12人、8人和5人,合计为12 + 8 + 5 = 25人。因此喜欢乒乓球的人数为30 - 25 = 5人。喜欢乒乓球的人数占全班人数的比例为5\/30,约分后得到最简分数1\/6。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 22:46:42","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":559,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"18","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 19:22:23","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":1444,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某校七年级组织学生参加数学实践活动,要求每名学生从A、B、C三个任务中至少选择一个完成。已知共有120名学生参与,其中选择A任务的有78人,选择B任务的有65人,选择C任务的有52人。同时,恰好选择两个任务的学生人数是恰好选择三个任务学生人数的3倍,且没有学生一个任务都不选。问:恰好选择三个任务的学生有多少人?","answer":"设恰好选择三个任务的学生人数为x人。\n\n根据题意,恰好选择两个任务的学生人数是3x人。\n\n因为每个学生至少选择一个任务,所以所有学生可以分为三类:\n- 只选一个任务的:设为y人\n- 恰好选两个任务的:3x人\n- 恰好选三个任务的:x人\n\n总人数为120人,因此有:\ny + 3x + x = 120\n即:y + 4x = 120 ——(1)\n\n再从任务被选的总人次角度分析:\n- 选择A任务的有78人,B任务65人,C任务52人,总人次为:78 + 65 + 52 = 195\n\n每个只选一个任务的学生贡献1人次,\n每个选两个任务的学生贡献2人次,\n每个选三个任务的学生贡献3人次。\n\n因此总人次可表示为:\n1×y + 2×(3x) + 3×x = y + 6x + 3x = y + 9x\n\n所以有:y + 9x = 195 ——(2)\n\n用方程(2)减去方程(1):\n(y + 9x) - (y + 4x) = 195 - 120\n5x = 75\n解得:x = 15\n\n代入(1)得:y + 4×15 = 120 → y = 60\n\n因此,恰好选择三个任务的学生有15人。\n\n答:恰好选择三个任务的学生有15人。","explanation":"本题考查数据的收集、整理与描述中的集合思想与方程建模能力,结合一元一次方程和二元一次方程组的解法。解题关键在于理解“人次”与“人数”的区别,并合理设未知数,建立两个不同角度的等量关系:一是总人数,二是任务被选的总人次。通过设恰好选三个任务的人数为x,利用“恰好选两个任务的人数是其3倍”建立联系,再结合总人数和总人次列出方程组,最终求解。本题综合性强,需要学生具备较强的逻辑分析和方程建模能力,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 11:41:23","updated_at":"2026-01-06 11:41:23","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":1865,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市地铁1号线在平面直角坐标系中沿直线铺设,已知A站坐标为(-3, 2),B站坐标为(5, -6)。现计划在AB之间增设一个临时站点C,使得从A到C的距离与从C到B的距离之比为2:3。同时,为方便乘客换乘,需在C点正东方向4个单位处设置一个公交接驳点D。若一名学生从A站出发,先乘地铁到C站,再步行到D点,求该学生行走的总路程(精确到0.1)。","answer":"1. 设C点坐标为(x, y)。由于C在AB线段上,且AC:CB = 2:3,使用定比分点公式:\n x = (3×(-3) + 2×5)\/(2+3) = (-9 + 10)\/5 = 1\/5 = 0.2\n y = (3×2 + 2×(-6))\/5 = (6 - 12)\/5 = -6\/5 = -1.2\n 所以C点坐标为(0.2, -1.2)\n\n2. D点在C点正东方向4个单位,即横坐标加4,纵坐标不变:\n D点坐标为(0.2 + 4, -1.2) = (4.2, -1.2)\n\n3. 计算AC距离:\n AC = √[(0.2 - (-3))² + (-1.2 - 2)²] = √[(3.2)² + (-3.2)²] = √[10.24 + 10.24] = √20.48 ≈ 4.5\n\n4. 计算CD距离:\n CD = 4(正东方向水平距离)\n\n5. 总路程 = AC + CD ≈ 4.5 + 4 = 8.5\n\n答:该学生行走的总路程约为8.5个单位长度。","explanation":"本题综合考查平面直角坐标系中的定比分点、两点间距离公式及坐标变换。关键步骤是运用定比分点公式确定C点坐标,再根据方向确定D点坐标,最后分段计算距离并求和。难点在于比例关系的坐标化处理和精确计算带小数的平方根。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-07 09:40:17","updated_at":"2026-01-07 09:40:17","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":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":196,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"小明在超市买了3支铅笔,每支铅笔2元,又买了1个笔记本,价格是5元。他付给收银员20元,应找回多少钱?","answer":"A","explanation":"首先计算小明购买物品的总花费:3支铅笔,每支2元,共花费 3 × 2 = 6 元;加上1个笔记本5元,总花费为 6 + 5 = 11 元。他付了20元,所以应找回 20 - 11 = 9 元。因此正确答案是A。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 14:04:07","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":"A","content":"9元","is_correct":1},{"id":"B","content":"11元","is_correct":0},{"id":"C","content":"13元","is_correct":0},{"id":"D","content":"15元","is_correct":0}]},{"id":1027,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"某班级组织了一次环保知识竞赛,共收集了50份有效问卷。统计结果显示,有32名学生表示经常进行垃圾分类,有25名学生表示每天步行或骑自行车上学。已知每位学生至少符合其中一项环保行为,那么同时做到垃圾分类和绿色出行的学生至少有___人。","answer":"7","explanation":"根据容斥原理,设同时做到两项的学生人数为x。总人数 = 垃圾分类人数 + 绿色出行人数 - 同时做到两项的人数。即:50 = 32 + 25 - x,解得x = 7。因此,同时做到两项的学生至少有7人。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 05:45:38","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":803,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次环保活动中,某学校七年级学生共收集了120千克废旧纸张。已知男生收集的纸张比女生多20千克,设女生收集的纸张为x千克,则可列出一元一次方程:_x + (x + 20) = 120_,解得女生收集了___千克。","answer":"50","explanation":"根据题意,女生收集x千克,男生比女生多20千克,即男生收集(x + 20)千克。总重量为120千克,因此方程为x + (x + 20) = 120。解这个方程:2x + 20 = 120 → 2x = 100 → x = 50。所以女生收集了50千克。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 00:20:18","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":[]}]