习题练习

[{"id":432,"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课外阅读时间数据时，记录了5位同学每周阅读课外书的小时数分别为：3，5，4，6，7。如果他想用这组数据制作一个频数分布表，并将数据按“3-4小时”“5-6小时”“7小时及以上”进行分组，那么“5-6小时”这一组的频数是多少？","answer":"B","explanation":"首先，列出原始数据：3，5，4，6，7。按照分组标准，“3-4小时”包括3和4，对应数据中的3和4，共2个；“5-6小时”包括5和6，对应数据中的5和6，共2个；“7小时及以上”只有7，共1个。因此，“5-6小时”这一组的频数是2。本题考查数据的整理与分组，属于‘数据的收集、整理与描述’知识点，难度简单，适合七年级学生。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:36:00","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":"1","is_correct":0},{"id":"B","content":"2","is_correct":1},{"id":"C","content":"3","is_correct":0},{"id":"D","content":"4","is_correct":0}]},{"id":564,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"在一次班级数学测验中，某学生记录了5名同学的数学成绩（单位：分）分别为：85，90，78，92，85。如果老师决定将每位同学的成绩都增加5分，那么这组数据的中位数会如何变化？","answer":"A","explanation":"首先将原始数据从小到大排列：78，85，85，90，92。共有5个数据，中位数是中间的那个数，即第3个数，为85分。当每位同学的成绩都增加5分后，新的数据为：83，90，90，95，97。重新排序后为：83，90，90，95，97，中位数是第3个数，即90分。90 - 85 = 5，因此中位数增加了5分。所以正确答案是A。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 19:31:22","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":"增加5分","is_correct":1},{"id":"B","content":"增加10分","is_correct":0},{"id":"C","content":"不变","is_correct":0},{"id":"D","content":"无法确定","is_correct":0}]},{"id":517,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"在一次环保活动中，某班级收集了废旧纸张共120千克。第一周收集了总量的1\/3，第二周收集了剩余部分的1\/2。请问第二周收集了多少千克废旧纸张？","answer":"C","explanation":"首先，第一周收集的废旧纸张为总量的1\/3，即120 × 1\/3 = 40千克。剩余部分为120 - 40 = 80千克。第二周收集了剩余部分的1\/2，即80 × 1\/2 = 40千克。因此，第二周收集了40千克，正确答案是C。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 18:20:33","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":"20千克","is_correct":0},{"id":"B","content":"30千克","is_correct":0},{"id":"C","content":"40千克","is_correct":1},{"id":"D","content":"60千克","is_correct":0}]},{"id":2451,"subject":"数学","grade":"八年级","stage":"初中","type":"填空题","content":"某学生在研究一个轴对称图形时，发现其对称轴是直线x=3，且图形上一点A的坐标为(5, 4)。若点A关于该对称轴的对称点为B，则点B的横坐标为___。","answer":"1","explanation":"对称轴为x=3，点A(5,4)到对称轴的距离为5−3=2，对称点B在对称轴另一侧相同距离处，故横坐标为3−2=1。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 13:54:59","updated_at":"2026-01-10 13:54:59","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":2242,"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，再向左移动8个单位，然后向右移动3个单位，最后向左移动6个单位。此时该学生所在位置表示的数是___。","answer":"-6","explanation":"根据正负数在数轴上的移动规则，向右为正，向左为负。起始位置为0，第一次移动+5，第二次移动-8，第三次移动+3，第四次移动-6。计算过程为：0 + 5 - 8 + 3 - 6 = (5 + 3) - (8 + 6) = 8 - 14 = -6。因此最终位置表示的数是-6。","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":1599,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某市为了解七年级学生数学学习负担情况，随机抽取了若干名学生进行问卷调查。调查结果显示，学生每天完成数学作业的时间（单位：分钟）分布如下：30分钟以下占10%，30到60分钟占40%，60到90分钟占35%，90分钟以上占15%。已知被调查学生中，完成作业时间在60分钟以上的学生共有200人。现从这些学生中按分层抽样的方法抽取50人进行深度访谈，其中‘90分钟以上’组应抽取多少人？若该市共有12000名七年级学生，请估算全市每天完成数学作业超过90分钟的学生人数。","answer":"第一步：设被调查学生总人数为x人。\n根据题意，完成作业时间在60分钟以上的学生包括‘60到90分钟’和‘90分钟以上’两组，占比为35% + 15% = 50%。\n因此有：\n50% × x = 200\n即：\n0.5x = 200\n解得：x = 400\n所以被调查学生总人数为400人。\n\n第二步：计算‘90分钟以上’组的人数。\n该组占比15%，人数为：\n15% × 400 = 0.15 × 400 = 60（人）\n\n第三步：进行分层抽样，总样本为50人。\n分层抽样要求各组抽取人数比例与原群体一致。\n因此‘90分钟以上’组应抽取人数为：\n(60 \/ 400) × 50 = (3\/20) × 50 = 7.5\n由于人数必须为整数，且分层抽样通常四舍五入处理，但此处需保持总人数为50，应合理分配。\n更精确做法是按比例分配：\n各组人数分别为：\n- 30分钟以下：10% × 400 = 40人 → 抽取 (40\/400)×50 = 5人\n- 30到60分钟：40% × 400 = 160人 → 抽取 (160\/400)×50 = 20人\n- 60到90分钟：35% × 400 = 140人 → 抽取 (140\/400)×50 = 17.5人\n- 90分钟以上：60人 → 抽取 (60\/400)×50 = 7.5人\n将小数部分调整：17.5和7.5分别取18和7，或17和8。为使总和为50，可取：\n5 + 20 + 17 + 8 = 50\n因此‘90分钟以上’组应抽取8人。\n\n第四步：估算全市超过90分钟的学生人数。\n样本中‘90分钟以上’占比为15%，以此估计全市：\n12000 × 15% = 12000 × 0.15 = 1800（人）\n\n答：分层抽样中‘90分钟以上’组应抽取8人；全市估计有1800名学生每天完成数学作业超过90分钟。","explanation":"本题综合考查数据的收集、整理与描述中的百分比计算、分层抽样原理及用样本估计总体的统计思想。解题关键在于先通过已知部分人数反推总样本量，再根据各层比例进行分层抽样人数分配，注意实际抽样中人数必须为整数，需合理调整。最后利用样本比例推断总体数量，体现统计推断的基本方法。题目情境贴近学生实际，数据真实合理，考查学生综合运用统计知识解决实际问题的能力，难度较高。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 12:50:16","updated_at":"2026-01-06 12:50:16","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":177,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"已知函数 $ f(x) = |x - 2| + |x + 3| $，若关于 $ x $ 的不等式 $ f(x) < a $ 有解，则实数 $ a $ 的取值范围是（ ）","answer":"A","explanation":"本题考查绝对值函数的性质与不等式有解问题。函数 $ f(x) = |x - 2| + |x + 3| $ 表示数轴上点 $ x $ 到点 2 和点 -3 的距离之和。根据绝对值几何意义，当 $ x $ 在区间 $[-3, 2]$ 内时，该距离和最小，最小值为 $ |2 - (-3)| = 5 $。因此，$ f(x) $ 的最小值为 5，即 $ f(x) \\geq 5 $ 对所有实数 $ x $ 成立。要使不等式 $ f(x) < a $ 有解，必须存在某个 $ x $ 使得 $ f(x) < a $，这就要求 $ a $ 必须大于 $ f(x) $ 的最小值 5。若 $ a = 5 $，则 $ f(x) < 5 $ 无解，因为 $ f(x) \\geq 5 $；只有当 $ a > 5 $ 时，才能找到某些 $ x $ 使得 $ f(x) < a $。因此，实数 $ a $ 的取值范围是 $ a > 5 $。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2025-12-29 12:32:47","updated_at":"2025-12-29 12:32:47","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"$ a > 5 $","is_correct":1},{"id":"B","content":"$ a \\geq 5 $","is_correct":0},{"id":"C","content":"$ a > 0 $","is_correct":0},{"id":"D","content":"$ a \\geq 0 $","is_correct":0}]},{"id":361,"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，发现一组数据的最小值是148厘米，最大值是172厘米。若将这组数据分为5组，则每组的组距最接近多少厘米？","answer":"B","explanation":"首先计算极差：最大值减去最小值，即172 - 148 = 24厘米。要将数据分为5组，则组距 = 极差 ÷ 组数 = 24 ÷ 5 = 4.8厘米。由于组距通常取整数，且要覆盖整个数据范围，因此应向上取整为5厘米。若取4厘米，则5组只能覆盖20厘米（5×4），不足以包含24厘米的极差；而5厘米可以覆盖25厘米，满足要求。因此最接近且合理的组距是5厘米。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:45:34","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":"A","content":"4厘米","is_correct":0},{"id":"B","content":"5厘米","is_correct":1},{"id":"C","content":"6厘米","is_correct":0},{"id":"D","content":"7厘米","is_correct":0}]},{"id":1010,"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完成数学作业所用的时间（单位：分钟），整理数据后发现，时间在30到40分钟之间的学生人数最多，共有12人；时间在40到50分钟之间的有8人；时间在20到30分钟之间的有5人；时间在50到60分钟之间的有3人。那么，完成作业时间在___分钟范围内的人数最多。","answer":"30到40","explanation":"题目中给出了不同时间段内完成数学作业的学生人数：30到40分钟有12人，40到50分钟有8人，20到30分钟有5人，50到60分钟有3人。比较各组人数可知，12人是最大值，对应的时间范围是30到40分钟。因此，完成作业时间在30到40分钟范围内的人数最多。本题考查数据的收集与整理，要求学生能从分组数据中识别频数最高的组，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":"2025-12-30 05:15:24","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":983,"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30名学生的成绩。为了分析数据，老师将成绩按10分为一段进行分组，得到如下频数分布表：90～100分有5人，80～89分有12人，70～79分有8人，60～69分有4人，60分以下有1人。则这次竞赛成绩的中位数落在_______分数段内。","answer":"80～89","explanation":"中位数是将一组数据从小到大排列后，处于中间位置的数。本题共有30名学生，因此中位数是第15个和第16个数据的平均数。根据频数累计：60分以下1人，60～69分4人（累计5人），70～79分8人（累计13人），80～89分12人（累计25人）。第15和第16个数据均落在80～89分区间内，因此中位数落在80～89分数段。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 04:23:16","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":[]}]