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[{"id":1921,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某班级组织了一次环保知识竞赛,共收集了120份有效问卷。在整理数据时,一名学生将各分数段人数绘制成扇形统计图。已知得分在80~100分的人数占总人数的35%,则该分数段对应的扇形圆心角的度数是多少?","answer":"B","explanation":"扇形统计图中,每个扇形的圆心角度数 = 该部分所占百分比 × 360°。题目中80~100分的人数占35%,因此对应的圆心角为:35% × 360° = 0.35 × 360° = 126°。故正确答案为B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-07 13:14:46","updated_at":"2026-01-07 13:14:46","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":"105°","is_correct":0},{"id":"B","content":"126°","is_correct":1},{"id":"C","content":"140°","is_correct":0},{"id":"D","content":"150°","is_correct":0}]},{"id":471,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"在一次环保知识竞赛中,某班级共收集了120份有效问卷。统计结果显示,喜欢垃圾分类宣传活动的学生人数是喜欢节水宣传活动的2倍,而喜欢节水宣传活动的学生比喜欢低碳出行宣传活动的多10人。设喜欢低碳出行宣传活动的学生有x人,则根据题意可列出一元一次方程为:","answer":"A","explanation":"设喜欢低碳出行宣传活动的学生有x人。根据题意,喜欢节水宣传活动的学生比喜欢低碳出行的多10人,因此为(x + 10)人;喜欢垃圾分类宣传活动的学生是喜欢节水宣传的2倍,即为2(x + 10)人。三类人数之和等于总有效问卷数120,因此方程为:x + (x + 10) + 2(x + 10) = 120。选项A正确列出了该方程。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:54:03","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":"x + (x + 10) + 2(x + 10) = 120","is_correct":1},{"id":"B","content":"x + (x - 10) + 2x = 120","is_correct":0},{"id":"C","content":"x + 2x + (x + 10) = 120","is_correct":0},{"id":"D","content":"x + (x + 10) + 2x = 120","is_correct":0}]},{"id":974,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生测量了学校花坛一周的温度变化,记录了连续5天的最高温度分别为:23℃、25℃、24℃、26℃、22℃。这5天最高温度的平均值是______℃。","answer":"24","explanation":"求平均数的方法是将所有数据相加,再除以数据的个数。计算过程为:(23 + 25 + 24 + 26 + 22) ÷ 5 = 120 ÷ 5 = 24。因此,这5天最高温度的平均值是24℃。本题考查的是数据的收集、整理与描述中的平均数计算,属于七年级数学简单难度内容。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 04:11:53","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":2497,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"某学生在学习投影与视图时,观察一个底面为正方形的直棱柱。已知该棱柱的高为6 cm,底面边长为4 cm。若将该棱柱沿一条侧棱方向正投影到与其底面垂直的平面上,则投影图形的面积是多少?","answer":"A","explanation":"该直棱柱底面为正方形,边长为4 cm,高为6 cm。当沿一条侧棱方向进行正投影,且投影平面与底面垂直时,投影图形为一个矩形。这个矩形的一条边是底面正方形的边长4 cm,另一条边是棱柱的高6 cm。因为投影方向沿着侧棱(即高度方向),所以高度方向在投影中保持不变,而底面的另一条边在投影中也被保留(因投影面与底面垂直,底面的一条边与投影方向垂直,故投影后长度不变)。因此,投影图形是一个长为6 cm、宽为4 cm的矩形,面积为 6 × 4 = 24 cm²。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 15:18:49","updated_at":"2026-01-10 15:18:49","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":"24 cm²","is_correct":1},{"id":"B","content":"32 cm²","is_correct":0},{"id":"C","content":"48 cm²","is_correct":0},{"id":"D","content":"16 cm²","is_correct":0}]},{"id":507,"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,制作了如下频数分布表。已知身高在150~155cm(含150cm,不含155cm)的人数为8人,155~160cm的人数为12人,160~165cm的人数为15人,165~170cm的人数为10人。若该班共有50名学生,且没有其他身高段的学生,那么身高不低于160cm的学生占总人数的百分比是多少?","answer":"A","explanation":"题目要求计算身高不低于160cm的学生占总人数的百分比。根据频数分布表,身高不低于160cm包括两个区间:160~165cm(15人)和165~170cm(10人),共15 + 10 = 25人。班级总人数为50人,因此百分比为(25 ÷ 50) × 100% = 50%。故正确答案为A。本题考查数据的整理与描述中的频数统计与百分比计算,属于简单难度,符合七年级数学课程内容。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 18:13:20","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":"50%","is_correct":1},{"id":"B","content":"60%","is_correct":0},{"id":"C","content":"70%","is_correct":0},{"id":"D","content":"80%","is_correct":0}]},{"id":513,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"在一次班级环保活动中,某学生收集了废旧纸张和塑料瓶两类可回收物。已知他收集的塑料瓶数量比废旧纸张数量的2倍少3个,总共收集了27个物品。设废旧纸张的数量为x个,则根据题意可列出一元一次方程,求出x的值是:","answer":"A","explanation":"设废旧纸张的数量为x个,则塑料瓶的数量为2x - 3个。根据题意,总数量为27个,因此可列方程:x + (2x - 3) = 27。化简得:3x - 3 = 27,移项得:3x = 30,解得:x = 10。因此,废旧纸张的数量为10个,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 18:17:07","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":"10","is_correct":1},{"id":"B","content":"12","is_correct":0},{"id":"C","content":"15","is_correct":0},{"id":"D","content":"18","is_correct":0}]},{"id":814,"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课外活动时,收集了以下数据:阅读、运动、绘画、音乐。他将这些数据整理成扇形统计图,其中表示‘运动’的扇形圆心角为108度。如果全班共有40名学生,那么喜欢‘运动’的学生人数是___人。","answer":"12","explanation":"扇形统计图中,每个部分的圆心角占整个圆(360度)的比例等于该部分数据占总数据的比例。‘运动’对应的圆心角是108度,因此喜欢运动的学生所占比例为108 ÷ 360 = 0.3。全班共有40名学生,所以喜欢运动的学生人数为40 × 0.3 = 12人。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 00:30:51","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":2423,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某校八年级组织学生参加户外测量活动,一名学生使用测角仪和卷尺测量操场旁一座旗杆的高度。他在距离旗杆底部8米的点A处测得旗杆顶端的仰角为60°,然后向旗杆方向前进4米到达点B,再次测得旗杆顶端的仰角为θ。若该学生眼睛离地面高度忽略不计,且地面为水平面,则根据勾股定理和三角函数关系,旗杆的高度最接近下列哪个值?","answer":"A","explanation":"设旗杆高度为h米。在点A(距旗杆底部8米)测得仰角为60°,根据正切函数定义:tan(60°) = h \/ 8,而tan(60°) = √3,因此 h = 8√3 米。虽然题目中提到前进到点B并测得新仰角θ,但实际只需利用第一次测量数据即可直接求出旗杆高度,因为已知距离和仰角,且地面水平、观测点与旗杆底部共线。该题结合生活情境考查勾股定理与三角函数的初步应用,重点在于识别直角三角形中的边角关系。计算得 h = 8 × √3 ≈ 13.856 米,最接近选项A。其他选项分别为:B(12)、C(约10.392)、D(约6.928),均小于正确值,故选A。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 12:36:19","updated_at":"2026-01-10 12:36:19","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":"A","content":"8√3 米","is_correct":1},{"id":"B","content":"12 米","is_correct":0},{"id":"C","content":"6√3 米","is_correct":0},{"id":"D","content":"4√3 米","is_correct":0}]},{"id":280,"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,随机抽取了30名学生的阅读时间(单位:小时\/周),并将数据整理如下:5, 6, 7, 5, 8, 6, 7, 9, 5, 6, 7, 8, 6, 5, 7, 8, 9, 6, 7, 5, 8, 7, 6, 5, 7, 8, 6, 7, 5, 6。为了分析这组数据的集中趋势,该学生想求出这组数据的中位数。请问这组数据的中位数是多少?","answer":"B","explanation":"首先将30个数据按从小到大的顺序排列:5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 9, 9, 9。由于数据个数为30(偶数),中位数是第15个和第16个数据的平均数。第15个数是7,第16个数也是7,因此中位数为(7 + 7) ÷ 2 = 7。但仔细核对排序后发现:实际排序中第15个是6,第16个是7。正确排序后前14个为5和6,第15个是6,第16个是7,因此中位数为(6 + 7) ÷ 2 = 6.5。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 15:31:13","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":"6.5","is_correct":1},{"id":"C","content":"7","is_correct":0},{"id":"D","content":"7.5","is_correct":0}]},{"id":1817,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某学生在研究一次函数图像时,发现函数 y = 2x - 4 的图像与 x 轴和 y 轴分别交于点 A 和点 B。若以原点 O 为顶点,△OAB 为直角三角形,则该三角形的面积为多少?","answer":"A","explanation":"首先求一次函数 y = 2x - 4 与坐标轴的交点。令 y = 0,得 0 = 2x - 4,解得 x = 2,所以点 A 为 (2, 0)。令 x = 0,得 y = -4,所以点 B 为 (0, -4)。原点 O 为 (0, 0)。△OAB 是以 OA 和 OB 为直角边的直角三角形,其中 OA = 2(x 轴上的长度),OB = 4(y 轴上的长度,取绝对值)。直角三角形面积公式为 (1\/2) × 底 × 高,因此面积为 (1\/2) × 2 × 4 = 4。故正确答案为 A。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-06 16:20:47","updated_at":"2026-01-06 16:20:47","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":"4","is_correct":1},{"id":"B","content":"6","is_correct":0},{"id":"C","content":"8","is_correct":0},{"id":"D","content":"10","is_correct":0}]}]