习题练习

[{"id":372,"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，发现将数据按从小到大的顺序排列后，位于正中间的两个数分别是158和160，则这组数据的中位数是：","answer":"B","explanation":"中位数是将一组数据按大小顺序排列后，处于中间位置的数。当数据个数为偶数时，中位数是中间两个数的平均数。题目中给出中间两个数是158和160，因此中位数为(158 + 160) ÷ 2 = 318 ÷ 2 = 159。所以正确答案是B。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:49:21","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":"158","is_correct":0},{"id":"B","content":"159","is_correct":1},{"id":"C","content":"160","is_correct":0},{"id":"D","content":"162","is_correct":0}]},{"id":400,"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情况时，统计了每人每周阅读课外书的平均时间（单位：小时），并将数据分为5组，绘制成频数分布直方图。已知前四组的频数分别为3、7、10、5，第五组的频率为0.2，则该班级参与调查的学生总人数是多少？","answer":"B","explanation":"题目考查的是数据的收集、整理与描述中的频数与频率概念。已知第五组的频率为0.2，即第五组人数占总人数的20%。设总人数为x，则第五组人数为0.2x。前四组频数之和为3 + 7 + 10 + 5 = 25，因此总人数为前四组人数加上第五组人数：25 + 0.2x = x。解这个方程：25 = x - 0.2x → 25 = 0.8x → x = 25 ÷ 0.8 = 30。所以参与调查的学生总人数是30人。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:16:30","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":"25","is_correct":0},{"id":"B","content":"30","is_correct":1},{"id":"C","content":"35","is_correct":0},{"id":"D","content":"40","is_correct":0}]},{"id":903,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次班级环保活动中，某学生收集了若干个塑料瓶。如果每个袋子最多可以装8个塑料瓶，且该学生使用了5个袋子刚好装完所有瓶子，那么他一共收集了____个塑料瓶。","answer":"40","explanation":"题目中说明每个袋子最多装8个塑料瓶，共使用了5个袋子且刚好装完，说明没有剩余。因此总瓶数为每个袋子装的瓶数乘以袋子的数量，即 8 × 5 = 40。这是一道基于有理数乘法和实际问题情境的一元一次方程思想的应用题，符合七年级学生关于有理数运算和简单方程建模的知识水平。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 02:21:34","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":1967,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在整理班级同学最喜欢的课外活动调查结果时，将数据分为四类：阅读、运动、绘画、音乐，并记录了每类的人数分别为：18、24、15、23。为了更直观地展示各类别所占比例，该学生计划绘制扇形统计图。已知扇形统计图中每个扇形的圆心角与其对应类别的人数成正比，且整个圆为360°。请问‘运动’类活动对应的扇形圆心角最接近以下哪个度数？","answer":"B","explanation":"本题考查数据的收集、整理与描述中扇形统计图圆心角的计算方法。首先计算总人数：18 + 24 + 15 + 23 = 80人。‘运动’类有24人，占总人数的比例为24 ÷ 80 = 0.3。扇形圆心角 = 比例 × 360° = 0.3 × 360° = 108°。因此，‘运动’类对应的扇形圆心角为108°，最接近选项B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-07 14:48:12","updated_at":"2026-01-07 14:48:12","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":"98°","is_correct":0},{"id":"B","content":"108°","is_correct":1},{"id":"C","content":"118°","is_correct":0},{"id":"D","content":"128°","is_correct":0}]},{"id":851,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生在绘制班级同学最喜爱的课外活动统计图时，将数据整理成如下表格：阅读占20%，运动占35%，音乐占15%，绘画占___%，其余为其他活动。已知喜欢绘画的人数比喜欢音乐的人数多6人，且班级总人数为60人，那么绘画所占的百分比是____。","answer":"25","explanation":"首先，根据题意，班级总人数为60人。喜欢音乐的人占15%，即 60 × 15% = 9 人。喜欢绘画的人数比音乐多6人，所以绘画人数为 9 + 6 = 15 人。那么绘画所占的百分比为 (15 ÷ 60) × 100% = 25%。因此，空白处应填写25。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 01:04:52","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":2399,"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，设计图纸显示其底边长为8米，两腰相等且与底边的夹角均为60°。施工前需计算花坛的周长和面积，以便准备材料。已知该三角形可被分割为两个全等的直角三角形，且其中一个直角三角形的两条直角边分别为4米和4√3米。根据这些信息，以下关于该花坛的说法正确的是：","answer":"A","explanation":"由题意知，该三角形为等腰三角形，底边为8米，底角为60°。由于底角为60°，顶角也为60°，因此这是一个等边三角形，三边均为8米。故周长为 8 + 8 + 8 = 24 米。将等边三角形沿高线分割，得到两个全等的直角三角形，底边一半为4米，高为 √(8² - 4²) = √(64 - 16) = √48 = 4√3 米，与题目描述一致。面积为 (底 × 高) \/ 2 = (8 × 4√3) \/ 2 = 16√3 平方米。因此选项A正确。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 12:06:48","updated_at":"2026-01-10 12:06:48","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":"该三角形的周长为24米，面积为16√3平方米","is_correct":1},{"id":"B","content":"该三角形的周长为16米，面积为8√3平方米","is_correct":0},{"id":"C","content":"该三角形的周长为24米，面积为8√3平方米","is_correct":0},{"id":"D","content":"该三角形的周长为16米，面积为16√3平方米","is_correct":0}]},{"id":1093,"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，某学生收集了若干个塑料瓶和玻璃瓶，其中塑料瓶的数量比玻璃瓶的3倍多5个。若设玻璃瓶的数量为x个，则塑料瓶的数量可表示为______。","answer":"3x + 5","explanation":"根据题意，塑料瓶的数量比玻璃瓶的3倍多5个。玻璃瓶的数量为x，那么它的3倍就是3x，再加上5个，就是塑料瓶的数量，因此表达式为3x + 5。这是整式加减中的基本概念，考查学生将文字语言转化为代数表达式的能力。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-06 08:55:58","updated_at":"2026-01-06 08:55:58","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":1933,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"在平面直角坐标系中，点A(2, 3)、点B(6, 7)，点C在x轴上，且△ABC是以AB为斜边的等腰直角三角形，则点C的横坐标为___。","answer":"4","explanation":"由AB中点M(4,5)为直角顶点对称中心，C在x轴上且满足AC=BC，利用距离公式列方程解得C横坐标为4。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-07 14:10:20","updated_at":"2026-01-07 14:10:20","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":1766,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"在平面直角坐标系中，点A的坐标为（2，5），点B在x轴上，且线段AB的长度为13。若点B位于原点右侧，则点B的横坐标为____。","answer":"14","explanation":"根据题意，点A(2, 5)，点B在x轴上，设其坐标为(x, 0)，且AB = 13。利用两点间距离公式：√[(x - 2)² + (0 - 5)²] = 13。两边平方得：(x - 2)² + 25 = 169，即(x - 2)² = 144。解得x - 2 = ±12，所以x = 14 或 x = -10。由于点B位于原点右侧，x > 0，因此x = 14。故点B的横坐标为14。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 15:00:48","updated_at":"2026-01-06 15:00:48","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":751,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次校园环保活动中，某学生收集了若干千克废纸。若每千克废纸可生产再生纸0.8千克，则该学生收集的废纸共可生产再生纸____千克。已知他最终生产出的再生纸比收集的废纸少6千克，则他最初收集的废纸是____千克。","answer":"0.8x, 30","explanation":"设该学生收集的废纸为x千克。根据题意，每千克废纸可生产0.8千克再生纸，因此可生产的再生纸为0.8x千克。又知再生纸比废纸少6千克，即x - 0.8x = 6，解得0.2x = 6，x = 30。因此，第一空填0.8x（表示再生纸质量与废纸质量的关系），第二空填30（表示收集的废纸质量）。本题综合考查了一元一次方程的建立与求解，以及有理数的运算，符合七年级数学课程要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 23:24:25","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":[]}]