习题练习

[{"id":623,"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，参赛学生分为若干小组。统计结果显示，若每3人一组，则多出2人；若每5人一组，则正好分完。已知参赛人数在30到50之间，请问参赛学生共有多少人？","answer":"B","explanation":"题目要求找出一个在30到50之间的整数，满足两个条件：除以3余2，且能被5整除。我们逐个验证选项：A选项30除以3余0，不符合‘多出2人’；B选项35除以3得11余2，符合第一个条件，且35能被5整除，符合第二个条件；C选项40除以3余1，不符合；D选项45除以3余0，也不符合。因此，只有35同时满足两个条件。本题考查的是有理数中的整除与余数概念，结合一元一次方程的思想（可设人数为x，则x ≡ 2 (mod 3)，x ≡ 0 (mod 5)），适合七年级学生理解。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 21:50:15","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":"30","is_correct":0},{"id":"B","content":"35","is_correct":1},{"id":"C","content":"40","is_correct":0},{"id":"D","content":"45","is_correct":0}]},{"id":2467,"subject":"数学","grade":"八年级","stage":"初中","type":"解答题","content":"如图，在平面直角坐标系中，点A(0, 4)，点B(6, 0)，点C在x轴正半轴上，且△ABC是以∠ACB为直角的直角三角形。点D是线段AB上一点，过点D作DE⊥AC于点E，DF⊥BC于点F，使得四边形DECF为矩形。已知矩形DECF的面积S与点D的横坐标x满足关系式：S = -x² + 6x。若点P是该矩形对角线交点，求当点P到原点的距离最小时，点P的坐标。","answer":"待完善","explanation":"解析待完善","solution_steps":"待完善","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 14:29:26","updated_at":"2026-01-10 14:29:26","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":406,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生调查了班级同学每周用于课外阅读的时间（单位：小时），并将数据整理如下表。已知这组数据的平均数为5，且所有数据均为正整数。若其中五个数据分别是3、4、5、6、7，那么第六个数据可能是多少？","answer":"B","explanation":"题目考查数据的收集、整理与描述中的平均数概念。已知6个数据的平均数是5，因此总和为6 × 5 = 30。已知的五个数据之和为3 + 4 + 5 + 6 + 7 = 25。设第六个数据为x，则25 + x = 30，解得x = 5。又因题目说明所有数据均为正整数，5符合条件。因此第六个数据是5，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 17:26:55","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":"A","content":"4","is_correct":0},{"id":"B","content":"5","is_correct":1},{"id":"C","content":"6","is_correct":0},{"id":"D","content":"7","is_correct":0}]},{"id":2402,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"在一次校园科技节活动中，某学生设计了一个由两个全等直角三角形拼接而成的轴对称图形，如图所示（图形描述：两个直角边分别为3和4的直角三角形沿斜边上的高对称拼接，形成一个四边形）。若该图形的周长为20，则其面积的最大可能值为多少？","answer":"A","explanation":"本题综合考查勾股定理、全等三角形、轴对称及一次函数最值思想。已知两个全等直角三角形直角边为3和4，则斜边为5（由勾股定理得√(3²+4²)=5）。每个三角形面积为(1\/2)×3×4=6，两个总面积为12。拼接方式沿斜边上的高对称，形成轴对称四边形。斜边上的高h可由面积法求得：(1\/2)×5×h=6 ⇒ h=12\/5=2.4。拼接后图形的周长由四条边组成：两条直角边（3和4）各出现两次，但拼接时部分边重合。实际外周长包括两个直角边和一个对称轴两侧的边。但题目给出周长为20，需验证合理性。实际上，若两个三角形沿斜边上的高对称拼接，形成的四边形有两条边为3，两条为4，总周长为2×(3+4)=14，与题设20不符，说明拼接方式并非简单并列。重新理解题意：可能是将两个三角形以不同方式组合，使整体呈轴对称且周长为20。但无论拼接方式如何，总面积恒为两个三角形面积之和，即2×6=12。因此，面积最大可能值即为12，无法更大。选项中A为12，符合逻辑。题目通过设定周长条件制造干扰，实则考查学生对面积守恒的理解。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 12:08:13","updated_at":"2026-01-10 12:08:13","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":"15","is_correct":0},{"id":"C","content":"18","is_correct":0},{"id":"D","content":"24","is_correct":0}]},{"id":560,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"102千克","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 19:22:34","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":12,"subject":"语文","grade":"初一","stage":"初中","type":"选择题","content":"《朝花夕拾》的作者是？","answer":"A","explanation":"《朝花夕拾》是鲁迅创作的回忆性散文集。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-08-29 16:33:04","updated_at":"2025-08-29 16:33:04","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":"A","content":"鲁迅","is_correct":1},{"id":"B","content":"郭沫若","is_correct":0},{"id":"C","content":"茅盾","is_correct":0},{"id":"D","content":"老舍","is_correct":0}]},{"id":1473,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市为了优化公交线路，对一条主干道的车流量进行了为期7天的观测，记录每天上午7:00至9:00的车辆通过数量（单位：百辆），数据如下：12, 15, 18, 14, 16, 20, 17。交通部门计划根据这些数据调整红绿灯时长，并设定一个‘高峰阈值’，若某天的车流量超过该阈值，则启动延长绿灯时间的应急方案。已知该阈值设定为这组数据的中位数与平均数的较大者。同时，为评估调整效果，工程师在平面直角坐标系中绘制了车流量与绿灯延长时间的函数关系图，其中绿灯延长时间 y（单位：秒）与车流量 x（单位：百辆）满足一次函数关系，且当 x = 15 时 y = 10，当 x = 20 时 y = 20。若某天观测到车流量为 19 百辆，且该天启动了应急方案，求该天绿灯延长时间的理论值，并判断该天车流量是否确实超过了设定的高峰阈值。","answer":"第一步：计算7天车流量的平均数。\n数据：12, 15, 18, 14, 16, 20, 17\n总和 = 12 + 15 + 18 + 14 + 16 + 20 + 17 = 112\n平均数 = 112 ÷ 7 = 16（百辆）\n\n第二步：求中位数。\n将数据从小到大排列：12, 14, 15, 16, 17, 18, 20\n共7个数据，中位数为第4个数，即16（百辆）\n\n第三步：确定高峰阈值。\n阈值为中位数与平均数的较大者：max(16, 16) = 16（百辆）\n\n第四步：建立绿灯延长时间 y 与车流量 x 的一次函数关系。\n设函数为 y = kx + b\n已知当 x = 15 时 y = 10，当 x = 20 时 y = 20\n代入得方程组：\n10 = 15k + b ...(1)\n20 = 20k + b ...(2)\n(2) - (1) 得：10 = 5k ⇒ k = 2\n将 k = 2 代入 (1)：10 = 15×2 + b ⇒ 10 = 30 + b ⇒ b = -20\n所以函数为：y = 2x - 20\n\n第五步：当 x = 19 时，求 y 值。\ny = 2×19 - 20 = 38 - 20 = 18（秒）\n\n第六步：判断是否超过高峰阈值。\n车流量为19百辆，阈值为16百辆，19 > 16，因此确实超过了阈值，启动应急方案合理。\n\n最终答案：该天绿灯延长时间的理论值为18秒，且车流量确实超过了高峰阈值。","explanation":"本题综合考查了数据的收集、整理与描述（平均数、中位数）、实数运算、一次函数（二元一次方程组应用）以及不等式比较。解题关键在于：首先通过统计方法确定‘高峰阈值’，这需要准确计算平均数和中位数并比较大小；其次利用两个已知点建立一次函数模型，通过解二元一次方程组求出函数表达式；最后代入具体数值求解并做出逻辑判断。题目情境真实，融合了统计与函数知识，要求学生具备较强的综合分析与计算能力，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 11:52:51","updated_at":"2026-01-06 11:52:51","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":2470,"subject":"数学","grade":"八年级","stage":"初中","type":"解答题","content":"如图，在平面直角坐标系中，点A(0, 6)，点B(8, 0)，点C为线段AB上的动点。以AC为边作正方形ACDE，使得点D在x轴正半轴上，点E在第一象限。连接BE，交y轴于点F。已知正方形ACDE的边长为a，且满足a² = 4x + 12，其中x为点C的横坐标。求当△BEF的面积最大时，点C的坐标及此时△BEF的面积。","answer":"待完善","explanation":"解析待完善","solution_steps":"待完善","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 14:39:17","updated_at":"2026-01-10 14:39:17","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":1859,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市地铁线路规划图中，两条平行轨道AB和CD被一条斜向联络线EF所截，形成多个角。已知∠1与∠2是同旁内角，且∠1的度数是∠2的2倍少30°。同时，在平面直角坐标系中，点A的坐标为(2, 3)，点B在x轴正方向上，且AB的长度为5个单位。若将线段AB向右平移3个单位，再向下平移2个单位得到线段A'B'，求：(1) ∠1和∠2的度数；(2) 点A'的坐标；(3) 若点C是线段A'B'的中点，求点C的坐标。","answer":"(1) 设∠2的度数为x°，则∠1 = (2x - 30)°。\n因为AB∥CD，EF为截线，∠1与∠2是同旁内角，所以∠1 + ∠2 = 180°。\n列方程：(2x - 30) + x = 180\n3x - 30 = 180\n3x = 210\nx = 70\n所以∠2 = 70°，∠1 = 2×70 - 30 = 110°。\n\n(2) 点A(2, 3)向右平移3个单位，横坐标加3，得(5, 3)；再向下平移2个单位，纵坐标减2，得(5, 1)。\n所以点A'的坐标为(5, 1)。\n\n(3) 点B在x轴正方向上，且AB = 5，A(2, 3)，设B(x, 0)，由距离公式：\n√[(x - 2)² + (0 - 3)²] = 5\n(x - 2)² + 9 = 25\n(x - 2)² = 16\nx - 2 = ±4\nx = 6 或 x = -2\n因为B在x轴正方向上，且从A向右延伸更合理（结合平移方向），取x = 6，即B(6, 0)。\n将B(6, 0)同样平移：向右3单位得(9, 0)，向下2单位得(9, -2)，即B'(9, -2)。\n点C是A'B'的中点，A'(5, 1)，B'(9,...","explanation":"本题综合考查平行线性质、一元一次方程、平面直角坐标系中的平移与坐标计算、中点公式。第(1)问利用同旁内角互补建立方程求解角度；第(2)问考查图形平移对坐标的影响；第(3)问需先通过距离公式确定点B坐标，再经平移得B'，最后用中点公式求C。关键步骤是正确理解几何关系与坐标变换规则，并准确进行代数运算。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-07 09:39:21","updated_at":"2026-01-07 09:39:21","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":2223,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"某学生在记录一周气温变化时，发现某地周一的气温比标准气温低3℃，记作-3℃；周三的气温比标准气温高5℃，记作+5℃。如果标准气温为0℃，那么周一和周三的气温相差___℃。","answer":"8","explanation":"周一气温为-3℃，周三气温为+5℃。求两天气温的差值，即计算5 - (-3) = 5 + 3 = 8。因此，两天气温相差8℃。本题考查正负数在实际情境中的意义及简单运算，符合七年级学生对正负数应用的理解水平。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 14:27:19","updated_at":"2026-01-09 14:27:19","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":[]}]