初中
数学
中等
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知识点: 初中数学
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[{"id":423,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"在一次环保知识竞赛中,某班级收集了学生家庭一周内节约用水的数据(单位:升),整理后发现:有3个家庭节约了15升,5个家庭节约了20升,2个家庭节约了25升。请问该班级学生家庭平均每周节约用水多少升?","answer":"B","explanation":"要计算平均节约用水量,需先求总节水量,再除以家庭总数。总节水量 = 3×15 + 5×20 + 2×25 = 45 + 100 + 50 = 195(升)。家庭总数 = 3 + 5 + 2 = 10(个)。平均节水量 = 195 ÷ 10 = 19(升)。因此,正确答案是B。本题考查数据的收集、整理与描述中的平均数计算,属于简单难度的基础应用。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:32:50","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":"18升","is_correct":0},{"id":"B","content":"19升","is_correct":1},{"id":"C","content":"20升","is_correct":0},{"id":"D","content":"21升","is_correct":0}]},{"id":567,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"平均数是5.2,中位数是5,众数是5","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 19:35:05","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":2146,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在解方程时,将方程 2x + 3 = 9 的解题步骤写为:第一步,两边同时减去3,得到 2x = 6;第二步,两边同时除以2,得到 x = 3。这名学生使用的解方程依据是___。","answer":"B","explanation":"该学生在解方程过程中,第一步使用了等式的基本性质:两边同时减去3,保持等式成立;第二步两边同时除以2(不为0),也符合等式的基本性质。因此正确依据是选项B所描述的内容。选项C和D虽然也是方程变形中的方法,但不是本题中直接体现的依据。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 13:00:46","updated_at":"2026-01-09 13:00:46","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":"等式两边同时加上同一个数,等式仍然成立","is_correct":0},{"id":"B","content":"等式两边同时减去同一个数,等式仍然成立,且等式两边同时除以同一个不为0的数,等式仍然成立","is_correct":1},{"id":"C","content":"移项时符号要改变","is_correct":0},{"id":"D","content":"合并同类项法则","is_correct":0}]},{"id":499,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某班级在一次数学测验中,收集了30名学生的成绩,并制作了频数分布表。已知成绩在80~89分这一组的学生有8人,占总人数的百分比最接近以下哪个选项?","answer":"C","explanation":"题目考查的是数据的收集、整理与描述中的百分比计算。总人数为30人,80~89分的学生有8人。计算该组所占百分比:8 ÷ 30 ≈ 0.2667,即约26.67%。比较选项,26.67%最接近27%,因此正确答案是C。此题帮助学生理解频数与百分比之间的关系,属于简单难度的基础统计题。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 18:09:41","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%","is_correct":0},{"id":"B","content":"25%","is_correct":0},{"id":"C","content":"27%","is_correct":1},{"id":"D","content":"30%","is_correct":0}]},{"id":1613,"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‘校园植物分布调查’项目,要求学生在平面直角坐标系中标记校园内不同区域植物的种类与数量。已知校园主干道为一条直线,其方程为 y = 2x + 1,花坛区域是一个以点 A(1, 3) 为圆心、半径为 √5 的圆形区域。调查发现,在花坛内及边界上的植物共有 15 种,其中喜阴植物占总数的 40%,其余为喜阳植物。另有一条灌溉水渠从点 B(0, -1) 出发,与主干道垂直相交于点 P。若每种植一株喜阳植物需要 0.5 升水,每种植一株喜阴植物需要 0.3 升水,且水渠每分钟供水 2 升。问:要完成花坛区域内所有植物的首次灌溉,至少需要多少分钟?(结果保留一位小数)","answer":"解题步骤如下:\n\n第一步:确定花坛区域与主干道的几何关系。\n花坛是以 A(1, 3) 为圆心、半径为 √5 的圆,其方程为 (x - 1)² + (y - 3)² = 5。\n主干道方程为 y = 2x + 1。\n\n第二步:求水渠与主干道的交点 P。\n水渠与主干道垂直,主干道斜率为 2,因此水渠斜率为 -1\/2。\n水渠过点 B(0, -1),其方程为 y + 1 = (-1\/2)(x - 0),即 y = -½x - 1。\n联立主干道与水渠方程:\n2x + 1 = -½x - 1\n两边同乘 2 得:4x + 2 = -x - 2\n5x = -4 → x = -0.8\n代入 y = 2x + 1 得:y = 2×(-0.8) + 1 = -1.6 + 1 = -0.6\n所以交点 P 坐标为 (-0.8, -0.6)\n\n第三步:计算植物种类与需水量。\n花坛内共有 15 种植物。\n喜阴植物占 40%:15 × 0.4 = 6 种\n喜阳植物:15 - 6 = 9 种\n(注:题目中‘种’理解为‘株’,因涉及单株用水量)\n每株喜阳植物需水 0.5 升,总需水:9 × 0.5 = 4.5 升\n每株喜阴植物需水 0.3 升,总需水:6 × 0.3 = 1.8 升\n总需水量:4.5 + 1.8 = 6.3 升\n\n第四步:计算灌溉所需时间。\n水渠供水速度为每分钟 2 升。\n所需时间 = 总需水量 ÷ 供水速度 = 6.3 ÷ 2 = 3.15 分钟\n保留一位小数:3.2 分钟\n\n答:至少需要 3.2 分钟。","explanation":"本题综合考查平面直角坐标系中直线的垂直关系、圆的方程、百分比计算、有理数运算及实际问题建模能力。解题关键在于理解‘垂直’意味着斜率乘积为 -1,从而求出水渠方程,并与主干道联立求交点。虽然交点 P 的坐标在本题中不影响最终灌溉时间(因供水速度恒定),但其计算过程体现了坐标系中几何关系的综合运用。植物种类按比例分配后,结合单位需水量计算总需水量,再根据供水速率求时间,涉及小数乘除和有理数运算。题目情境新颖,融合数据统计、几何与代数,难度较高,适合考查学生综合应用能力。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 12:57:33","updated_at":"2026-01-06 12:57:33","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":504,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某班级进行了一次数学测验,老师将成绩整理后绘制成频数分布直方图,发现成绩在80分到90分之间的学生人数最多。这说明该分数段的什么统计量最大?","answer":"C","explanation":"题目中提到“成绩在80分到90分之间的学生人数最多”,这表示该分数段出现的次数最多。在统计学中,一组数据中出现次数最多的数值称为众数。因此,80分到90分这个区间对应的众数最大。平均数是所有数据的总和除以个数,中位数是数据排序后位于中间的数,极差是最大值与最小值之差,它们都不能直接由‘人数最多’得出。故正确答案为C。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 18:10:48","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":"平均数","is_correct":0},{"id":"B","content":"中位数","is_correct":0},{"id":"C","content":"众数","is_correct":1},{"id":"D","content":"极差","is_correct":0}]},{"id":544,"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,将数据按从小到大的顺序排列,并制作了频数分布表。他发现身高在150cm到160cm之间的学生人数占总人数的40%,而身高在160cm到170cm之间的学生人数比前者多10人。如果全班共有50名学生,那么身高在160cm到170cm之间的学生有多少人?","answer":"C","explanation":"首先,根据题意,全班共有50名学生。身高在150cm到160cm之间的学生占40%,即 50 × 40% = 20人。题目说明身高在160cm到170cm之间的学生比前者多10人,因此该区间人数为 20 + 10 = 30人。故正确答案为C。本题考查数据的收集、整理与描述中的百分比计算和简单推理,符合七年级数学知识点要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 19:01:30","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":"20人","is_correct":0},{"id":"B","content":"25人","is_correct":0},{"id":"C","content":"30人","is_correct":1},{"id":"D","content":"35人","is_correct":0}]},{"id":244,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"一个长方形的长是 8 厘米,宽是 5 厘米,它的面积是 _ 平方厘米。","answer":"40","explanation":"长方形的面积计算公式是:面积 = 长 × 宽。题目中给出的长是 8 厘米,宽是 5 厘米,因此面积为 8 × 5 = 40 平方厘米。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 14:42:06","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":610,"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时,记录了5位同学每周阅读的小时数分别为:3,5,4,6,2。如果老师要求每位同学的阅读时间都增加相同的整数小时,使得新的数据中位数变为5,那么每位同学至少需要增加多少小时?","answer":"A","explanation":"原始数据为:3,5,4,6,2。先将数据从小到大排序:2,3,4,5,6。当前中位数是中间的数,即4。设每位同学增加x小时(x为正整数),则新数据为:2+x,3+x,4+x,5+x,6+x。排序后仍为:2+x,3+x,4+x,5+x,6+x,中位数是4+x。要求中位数为5,即4 + x = 5,解得x = 1。因此,每位同学至少需要增加1小时。验证:增加1小时后数据为3,4,5,6,7,排序后中位数为5,符合条件。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":"2025-12-29 21:36:13","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":"1","is_correct":1},{"id":"B","content":"2","is_correct":0},{"id":"C","content":"3","is_correct":0},{"id":"D","content":"4","is_correct":0}]},{"id":734,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生测量了教室里一盏灯到地面的垂直距离为2.8米,灯正下方地面上有一张课桌,课桌的高度为0.75米,那么灯到课桌桌面的垂直距离是______米。","answer":"2.05","explanation":"灯到地面的距离是2.8米,课桌高度为0.75米,课桌桌面距离地面0.75米。因此灯到桌面的垂直距离为2.8减去0.75,即2.8 - 0.75 = 2.05(米)。本题考查有理数的减法在实际生活中的应用,属于简单难度的计算题。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 23:06:44","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":[]}]