[{"id":431,"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时间数据时，记录了5名同学每周的阅读时间（单位：小时）分别为：3，5，4，6，2。如果他想用条形统计图表示这些数据，那么纵轴上表示频数的刻度最小应设置为多少才能完整显示所有数据？","answer":"B","explanation":"题目中给出的5个数据分别是3、5、4、6、2，其中最大的数值是6。在绘制条形统计图时，纵轴表示频数（即阅读时间），为了完整显示最高的条形，纵轴的刻度必须大于或等于最大值6。通常为了图形美观和留有余地，刻度会设置为比最大值稍大的整数。选项中比6大的最小整数是7，因此纵轴刻度最小应设置为7。选项B正确。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:35:50","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":"6","is_correct":0},{"id":"B","content":"7","is_correct":1},{"id":"C","content":"5","is_correct":0},{"id":"D","content":"8","is_correct":0}]},{"id":345,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"在平面直角坐标系中，点 A 的坐标是 (3, 4)，点 B 的坐标是 (3, -2)。这两点之间的距离是多少？","answer":"A","explanation":"点 A 和点 B 的横坐标相同，都是 3，说明它们位于同一条竖直线上。两点之间的距离等于它们纵坐标之差的绝对值。计算：|4 - (-2)| = |4 + 2| = |6| = 6。因此，两点之间的距离是 6。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:41:02","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":"6","is_correct":1},{"id":"B","content":"5","is_correct":0},{"id":"C","content":"7","is_correct":0},{"id":"D","content":"8","is_correct":0}]},{"id":1371,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某学校组织七年级学生开展‘校园植物多样性调查’活动。调查小组在校园内选取了5个不同区域进行植物种类统计，并将数据整理如下表。已知每个区域的植物种类数均为正整数，且满足以下条件：\n\n1. 区域A的植物种类数比区域B多3种；\n2. 区域C的植物种类数是区域D的2倍；\n3. 区域E的植物种类数比区域A少5种；\n4. 五个区域植物种类总数为67种；\n5. 区域D的植物种类数比区域B少2种；\n6. 所有区域的植物种类数都不超过20种。\n\n请根据以上信息，求出每个区域的植物种类数。","answer":"设区域B的植物种类数为 x 种。\n\n根据条件1：区域A = x + 3\n根据条件5：区域D = x - 2\n根据条件2：区域C = 2 × (x - 2) = 2x - 4\n根据条件3：区域E = (x + 3) - 5 = x - 2\n\n根据条件4，五个区域总数为67：\nA + B + C + D + E = 67\n代入表达式：\n(x + 3) + x + (2x - 4) + (x - 2) + (x - 2) = 67\n合并同类项：\nx + 3 + x + 2x - 4 + x - 2 + x - 2 = 67\n( x + x + 2x + x + x ) + (3 - 4 - 2 - 2) = 67\n6x - 5 = 67\n6x = 72\nx = 12\n\n代回各区域：\n区域B：x = 12 种\n区域A：x + 3 = 15 种\n区域D：x - 2 = 10 种\n区域C：2x - 4 = 2×12 - 4 = 20 种\n区域E：x - 2 = 10 种\n\n验证总数：15 + 12 + 20 + 10 + 10 = 67，正确。\n验证条件6：所有数值均 ≤ 20，满足。\n\n答：区域A有15种，区域B有12种，区域C有20种，区域D有10种，区域E有10种植物。","explanation":"本题综合考查了二元一次方程组的思想（虽未显式列出两个方程，但通过多个等量关系建立一元一次方程）、整式的加减运算、有理数的四则运算以及数据的整理与分析能力。解题关键在于合理设元，将多个文字条件转化为代数表达式，再通过列方程求解。题目设置了多个约束条件，包括总数限制和范围限制（不超过20种），要求学生在解出答案后进行验证，体现了数学建模与逻辑推理的结合。情境贴近生活，考查学生从实际问题中抽象出数学模型的能力，属于综合性较强的困难题。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 11:12:47","updated_at":"2026-01-06 11:12:47","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":2044,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某公园计划修建一个等腰三角形花坛，设计要求花坛的两条等边长度均为√50米，底边为整数米，且整个花坛的周长不超过30米。若从美观和结构稳定性考虑，要求该等腰三角形的高尽可能大，则底边的长度应为多少米？","answer":"A","explanation":"本题综合考查勾股定理、二次根式化简、三角形三边关系及最值分析。已知等腰三角形两腰长为√50 = 5√2 ≈ 7.07米，设底边为x米（x为整数），则周长为2×5√2 + x ≈ 14.14 + x ≤ 30，得x ≤ 15.86，即x ≤ 15。又由三角形三边关系，底边x必须满足：0 < x < 2×5√2 ≈ 14.14，所以x ≤ 14。因此x的可能取值为1到14之间的整数。\n\n要求高尽可能大，即面积尽可能大。等腰三角形的高h可由勾股定理求得：h = √[(5√2)² - (x\/2)²] = √[50 - x²\/4]。要使h最大，即要使50 - x²\/4最大，也就是x²\/4最小，即x最小。但x不能太小，否则不满足实际结构需求，但数学上在允许范围内x越小，高越大。\n\n然而，题目隐含要求是“在满足周长不超过30米且底边为整数的条件下，使高最大”，因此应在x ≤ 14的整数中找使h最大的x。由于h = √(50 - x²\/4)是关于x的减函数，x越小，h越大。但还需验证三角形是否存在：当x=14时，x\/2=7，h=√(50-49)=√1=1；当x=12时，h=√(50-36)=√14≈3.74；x=10时，h=√(50-25)=√25=5；x=8时，h=√(50-16)=√34≈5.83；x=6时，h=√(50-9)=√41≈6.40；x=4时，h=√(50-4)=√46≈6.78；x=2时，h=√(50-1)=√49=7。但x=2或4时，虽然高更大，但周长分别为14.14+2=16.14和18.14，虽满足≤30，但题目强调“美观和结构稳定性”，过小的底边会导致三角形过于尖锐，不符合实际工程要求。\n\n但题目明确要求“高尽可能大”，在数学上应取使h最大的合法x。然而，进一步分析发现：当x减小时，高增大，但题目选项只给出6、8、10、12。在这四个选项中，x=6时，h=√(50 - 9)=√41≈6.40；x=8时，h=√(50-16)=√34≈5.83；x=10时，h=5；x=12时，h≈3.74。显然x=6时高最大。同时验证周长：2×5√2 + 6 ≈ 14.14 + 6 = 20.14 < 30，满足条件。因此，在给定选项中，底边为6米时高最大，符合题意。故选A。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-09 10:49:03","updated_at":"2026-01-09 10:49:03","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","is_correct":1},{"id":"B","content":"8","is_correct":0},{"id":"C","content":"10","is_correct":0},{"id":"D","content":"12","is_correct":0}]},{"id":1323,"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兴趣小组活动，活动分为A、B、C三个项目。已知报名参加A项目的人数比B项目多10人，C项目的人数是A项目与B项目人数之和的一半。后来由于场地限制，学校决定对报名人数进行调整：从A项目中调出5人到B项目，从C项目中调出3人到A项目。调整后，三个项目的人数恰好构成一个等差数列，且总人数不变。若调整后B项目的人数不少于15人，求原来报名参加A、B、C三个项目的人数各是多少？","answer":"设原来报名参加B项目的人数为x人，则A项目人数为(x + 10)人。\n\n根据题意，C项目人数是A与B人数之和的一半，即：\nC = (A + B) \/ 2 = ((x + 10) + x) \/ 2 = (2x + 10) \/ 2 = x + 5\n\n所以原来三个项目人数分别为：\nA：x + 10\nB：x\nC：x + 5\n\n总人数为：(x + 10) + x + (x + 5) = 3x + 15\n\n调整后：\n- A项目调出5人，调入3人 → A' = (x + 10) - 5 + 3 = x + 8\n- B项目调入5人 → B' = x + 5\n- C项目调出3人 → C' = (x + 5) - 3 = x + 2\n\n调整后三个项目人数为：A' = x + 8，B' = x + 5，C' = x + 2\n\n题目说明这三个数构成一个等差数列。观察发现：\n(x + 2), (x + 5), (x + 8) 是公差为3的等差数列，顺序为C', B', A'\n\n因此，只要满足这个顺序，就构成等差数列。\n\n同时题目给出条件：调整后B项目人数不少于15人，即：\nB' = x + 5 ≥ 15\n→ x ≥ 10\n\n由于x代表人数，必须为正整数，且所有人数均为非负整数，因此x ≥ 10即可。\n\n但我们还需验证是否还有其他限制。目前没有其他约束，因此最小的合理解为x = 10。\n\n代入得：\n原来B项目人数：x = 10人\nA项目人数：x + 10 = 20人\nC项目人数：x + 5 = 15人\n\n验证调整后人数：\nA' = 20 - 5 + 3 = 18\nB' = 10 + 5 = 15\nC' = 15 - 3 = 12\n\n检查是否构成等差数列：12, 15, 18 → 是，公差为3\nB' = 15 ≥ 15，满足条件\n总人数：20 + 10 + 15 = 45；调整后：18 + 15 + 12 = 45，守恒\n\n因此，原来报名参加A、B、C项目的人数分别为20人、10人、15人。","explanation":"本题综合考查了一元一次方程、不等式与不等式组、数据的整理与逻辑推理能力。解题关键在于合理设未知数，准确表达各项目原有人数，并根据调动规则计算调整后人数。通过分析‘构成等差数列’这一条件，发现调整后人数自然形成等差关系，从而简化问题。最后结合‘B项目不少于15人’的不等式条件，确定最小合理整数值。整个过程涉及代数表达、等差数列性质、不等式和实际问题的建模，属于综合性强、思维层次高的困难题。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 10:55:14","updated_at":"2026-01-06 10:55:14","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":2226,"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，发现某天的气温比前一天上升了5℃，记作+5℃。如果第二天的气温又比这一天下降了8℃，那么第二天的气温变化应记作____℃。","answer":"-8","explanation":"根据正负数表示相反意义的量的知识点，气温上升用正数表示，下降则用负数表示。题目中气温下降了8℃，因此应记作-8℃。本题考查学生对正负数在实际情境中应用的理解，符合七年级课程标准要求。","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":[]},{"id":1927,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某班级组织一次环保活动，收集可回收垃圾。第一周收集了x千克废纸，第二周收集的比第一周的2倍少3千克。已知两周共收集了17千克废纸，则第一周收集了多少千克？","answer":"C","explanation":"设第一周收集废纸x千克，则第二周收集了(2x - 3)千克。根据题意，两周共收集17千克，可列方程：x + (2x - 3) = 17。化简得3x - 3 = 17，移项得3x = 20，解得x = 7。因此第一周收集了7千克废纸。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-07 13:18:07","updated_at":"2026-01-07 13:18:07","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":"5","is_correct":0},{"id":"B","content":"6","is_correct":0},{"id":"C","content":"7","is_correct":1},{"id":"D","content":"8","is_correct":0}]},{"id":5,"subject":"数学","grade":"初三","stage":"初中","type":"选择题","content":"二次函数y = x² - 4x + 3的对称轴是？","answer":"B","explanation":"二次函数y = ax² + bx + c的对称轴为x = -b\/(2a)，这里a = 1, b = -4，所以对称轴为x = -(-4)\/(2*1) = 2。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"中等","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":"x = 1","is_correct":0},{"id":"B","content":"x = 2","is_correct":1},{"id":"C","content":"x = 3","is_correct":0},{"id":"D","content":"x = 4","is_correct":0}]},{"id":506,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"在一次班级组织的环保活动中，某学生收集了若干个塑料瓶和废纸。已知每个塑料瓶可兑换0.3元，每公斤废纸可兑换1.2元。该学生总共收集了20个物品（包括塑料瓶和废纸），共获得兑换金额9.6元。若设塑料瓶的数量为x个，则根据题意可列出一元一次方程为：","answer":"A","explanation":"设塑料瓶数量为x个，则废纸的数量为(20 - x)公斤（因为总共有20个物品）。每个塑料瓶兑换0.3元，所以塑料瓶总价值为0.3x元；每公斤废纸兑换1.2元，所以废纸总价值为1.2(20 - x)元。根据题意，总兑换金额为9.6元，因此可列方程：0.3x + 1.2(20 - x) = 9.6。选项A正确。选项B错误地将废纸数量也设为x；选项C颠倒了塑料瓶和废纸的系数关系；选项D使用了减法，不符合实际兑换逻辑。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 18:13:16","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":"0.3x + 1.2(20 - x) = 9.6","is_correct":1},{"id":"B","content":"0.3x + 1.2x = 9.6","is_correct":0},{"id":"C","content":"0.3(20 - x) + 1.2x = 9.6","is_correct":0},{"id":"D","content":"0.3x - 1.2(20 - x) = 9.6","is_correct":0}]},{"id":1077,"subject":"数学","grade":"七年级","stage":"小学","type":"填空题","content":"在一次班级环保活动中，某学生收集了若干节废旧电池。若每5节电池装一盒，则最后剩下3节；若每7节电池装一盒，则刚好装完。该学生至少收集了___节废旧电池。","answer":"28","explanation":"设该学生收集的电池总数为x节。根据题意，x除以5余3，即x ≡ 3 (mod 5)；同时x能被7整除，即x ≡ 0 (mod 7)。我们寻找满足这两个条件的最小正整数。列出7的倍数：7, 14, 21, 28, 35…，检查哪些数除以5余3。7÷5=1余2，14÷5=2余4，21÷5=4余1，28÷5=5余3，满足条件。因此最小的x是28。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-06 08:53:45","updated_at":"2026-01-06 08:53:45","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":[]}]