初中
数学
中等
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[{"id":2191,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在记录一周内每天气温变化时,发现某天的气温比前一天上升了3℃,记作+3℃。如果第二天的气温比第一天下降了5℃,那么第二天的气温变化应记作多少?","answer":"D","explanation":"气温下降应使用负数表示。题目中明确指出气温比第一天下降了5℃,因此变化量应记为-5℃。正数表示上升,负数表示下降,符合七年级正负数在现实情境中的应用知识点。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 14:25:31","updated_at":"2026-01-09 14:25:31","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":"-3℃","is_correct":0},{"id":"C","content":"+2℃","is_correct":0},{"id":"D","content":"-5℃","is_correct":1}]},{"id":1878,"subject":"语文","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在整理班级同学的数学测验成绩时,制作了如下频数分布表:\n\n| 成绩区间(分) | 频数(人) |\n|----------------|-----------|\n| 60 ≤ x < 70 | 4 |\n| 70 ≤ x < 80 | 8 |\n| 80 ≤ x < 90 | 12 |\n| 90 ≤ x ≤ 100 | 6 |\n\n已知全班平均成绩为81分,若将每位学生的成绩都加上5分后重新计算平均分,并绘制新的频数分布直方图,则下列说法正确的是:\n\nA. 新数据的平均数为86分,各组频数保持不变,但组中值整体增加5\nB. 新数据的平均数为86分,各组频数按比例增加,组距变为原来的1.05倍\nC. 新数据的平均数仍为81分,因为数据分布形状未变,仅位置平移\nD. 新数据的平均数为86分,但90 ≤ x ≤ 100这一组的频数会减少,因为部分学生超过100分","answer":"A","explanation":"本题考查数据的收集、整理与描述中对数据变换的理解。当所有原始数据统一加上一个常数(此处为5)时,平均数也会相应增加该常数,因此新平均数为81 + 5 = 86分。频数反映的是落在各区间内的人数,由于每个数据点都加5,原属于某一区间的数据整体平移到更高区间,但人数不变,故各组频数保持不变。例如,原60≤x<70区间变为65≤x<75,依此类推。组中值(如65、75、85、95)也相应增加5。选项B错误,因为频数不按比例变化;C错误,平均数会变;D错误,虽然理论上成绩可能超过100,但题目未说明有上限限制,且即使超过,也只是进入新区间,不会导致原组频数‘减少’,而是重新归类。因此,A最准确描述了数据变换后的统计特征。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-07 09:54:35","updated_at":"2026-01-07 09:54:35","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":"新数据的平均数为86分,各组频数保持不变,但组中值整体增加5","is_correct":1},{"id":"B","content":"新数据的平均数为86分,各组频数按比例增加,组距变为原来的1.05倍","is_correct":0},{"id":"C","content":"新数据的平均数仍为81分,因为数据分布形状未变,仅位置平移","is_correct":0},{"id":"D","content":"新数据的平均数为86分,但90 ≤ x ≤ 100这一组的频数会减少,因为部分学生超过100分","is_correct":0}]},{"id":1429,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市地铁系统正在进行客流量数据分析。已知某条线路在早高峰期间(7:00—9:00)的乘客到达情况如下:每5分钟为一个统计时段,共24个时段。统计发现,前12个时段的平均客流量比后12个时段少180人,且整个早高峰期间总客流量为12960人。若设前12个时段的平均客流量为x人,后12个时段的平均客流量为y人。\n\n(1)根据题意列出关于x和y的二元一次方程组;\n(2)解该方程组,求出x和y的值;\n(3)若地铁公司规定,当某时段客流量超过600人时,需增派工作人员。问:后12个时段中有多少个时段需要增派工作人员?(假设每个时段的客流量等于该时段的平均客流量)\n(4)为进一步优化调度,地铁公司计划将总客流量按每100人一组进行分组统计。请计算共可分成多少组?余下多少人?","answer":"(1)根据题意,前12个时段的平均客流量为x人,后12个时段为y人。\n前12个时段总客流量为12x,后12个时段为12y。\n整个早高峰共24个时段,总客流量为12960人,因此有:\n12x + 12y = 12960\n又已知前12个时段的平均客流量比后12个时段少180人,即:\nx = y - 180\n所以方程组为:\n12x + 12y = 12960\nx = y - 180\n\n(2)将第二个方程代入第一个方程:\n12(y - 180) + 12y = 12960\n12y - 2160 + 12y = 12960\n24y - 2160 = 12960\n24y = 12960 + 2160 = 15120\ny = 15120 ÷ 24 = 630\n代入x = y - 180得:\nx = 630 - 180 = 450\n所以,x = 450,y = 630\n\n(3)后12个时段的平均客流量为630人,每个时段客流量为630人。\n规定超过600人需增派工作人员,630 > 600,因此每个后12个时段都需要增派。\n共12个时段需要增派工作人员。\n\n(4)总客流量为12960人,按每100人一组分组:\n12960 ÷ 100 = 129 余 60\n所以可分成129组,余下60人。","explanation":"本题综合考查二元一次方程组、有理数运算、不等式判断及数据整理能力。第(1)问要求学生从实际问题中抽象出数学模型,建立方程组;第(2)问考查代入法解方程组的基本技能;第(3)问结合不等关系进行逻辑判断,体现数学应用意识;第(4)问涉及带余除法在实际数据分组中的应用,强化数据处理能力。题目背景新颖,贴近现实,考查点多维,逻辑链条完整,符合困难难度要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 11:35:59","updated_at":"2026-01-06 11:35:59","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":14,"subject":"英语","grade":"初一","stage":"初中","type":"选择题","content":"What is the plural form of \"child\"?","answer":"B","explanation":"\"child\"的复数形式是\"children\"。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","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":"childs","is_correct":0},{"id":"B","content":"children","is_correct":1},{"id":"C","content":"childes","is_correct":0},{"id":"D","content":"childies","is_correct":0}]},{"id":12,"subject":"语文","grade":"初一","stage":"初中","type":"选择题","content":"《朝花夕拾》的作者是?","answer":"A","explanation":"《朝花夕拾》是鲁迅创作的回忆性散文集。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","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":"鲁迅","is_correct":1},{"id":"B","content":"郭沫若","is_correct":0},{"id":"C","content":"茅盾","is_correct":0},{"id":"D","content":"老舍","is_correct":0}]},{"id":1797,"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的课外阅读时间数据时,随机抽取了30名学生,记录了他们每周课外阅读的时间(单位:小时),并将数据整理如下:5人每周阅读2小时,8人每周阅读3小时,10人每周阅读4小时,4人每周阅读5小时,3人每周阅读6小时。若该学生想用这组数据估计全班50名同学每周课外阅读的总时间,那么估算结果最接近以下哪个数值?","answer":"B","explanation":"首先计算样本中30名学生的总阅读时间:5×2 + 8×3 + 10×4 + 4×5 + 3×6 = 10 + 24 + 40 + 20 + 18 = 112小时。然后求出样本平均阅读时间:112 ÷ 30 ≈ 3.73小时\/人。用此平均值估算全班50人的总阅读时间:3.73 × 50 ≈ 186.5小时。最接近的选项是190小时,因此选B。本题考查数据的收集、整理与描述中的样本估计总体思想,以及有理数的乘除运算,符合七年级数学课程标准要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-06 16:12:41","updated_at":"2026-01-06 16:12:41","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":"180小时","is_correct":0},{"id":"B","content":"190小时","is_correct":1},{"id":"C","content":"200小时","is_correct":0},{"id":"D","content":"210小时","is_correct":0}]},{"id":2278,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"在数轴上,点A表示的数是-3,点B与点A的距离为7个单位长度,且点B位于点A的右侧;点C与点B的距离为4个单位长度,且点C位于点B的左侧。那么点C表示的数是___。","answer":"0","explanation":"首先,点A表示-3,点B在点A右侧且距离为7,因此点B表示的数是-3 + 7 = 4。接着,点C在点B左侧且距离为4,因此点C表示的数是4 - 4 = 0。本题综合考查了数轴上点的位置关系与有理数加减运算,要求学生理解‘右侧’表示加法,‘左侧’表示减法,并能分步推理,属于较难题型。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-09 16:27:13","updated_at":"2026-01-09 16:27:13","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":10,"subject":"生物","grade":"初一","stage":"初中","type":"选择题","content":"植物细胞和动物细胞都具有的结构是?","answer":"D","explanation":"植物细胞和动物细胞都具有细胞膜、细胞质和细胞核,但植物细胞还具有细胞壁、液泡和叶绿体。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-08-29 16:33:04","updated_at":"2025-11-17 16:31:46","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":"细胞壁、细胞膜、细胞质","is_correct":0},{"id":"B","content":"细胞壁、液泡、叶绿体","is_correct":0},{"id":"C","content":"细胞膜、液泡、叶绿体","is_correct":0},{"id":"D","content":"细胞膜、细胞质、细胞核","is_correct":1}]},{"id":1081,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","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共30件。已知每张废旧纸张可兑换0.2元,每个塑料瓶可兑换0.5元,该学生共获得10.8元。若设废旧纸张有x张,则可列出一元一次方程为:____ + 0.5(30 - x) = 10.8","answer":"0.2x","explanation":"题目中已知废旧纸张和塑料瓶总数为30件,设废旧纸张有x张,则塑料瓶有(30 - x)个。每张废旧纸张兑换0.2元,因此x张可兑换0.2x元;每个塑料瓶兑换0.5元,(30 - x)个可兑换0.5(30 - x)元。总金额为10.8元,所以方程为:0.2x + 0.5(30 - x) = 10.8。空白处应填写的是废旧纸张兑换的金额部分,即0.2x。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-06 08:54:09","updated_at":"2026-01-06 08:54:09","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":702,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次班级图书角的统计中,某学生记录了本周同学们借阅科普类图书的次数,数据如下:3次、5次、4次、6次、4次、3次、5次。这组数据的中位数是____。","answer":"4","explanation":"首先将这组数据按从小到大的顺序排列:3, 3, 4, 4, 5, 5, 6。共有7个数据,是奇数个,因此中位数是正中间的那个数,即第4个数。第4个数是4,所以这组数据的中位数是4。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 22:43:34","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":[]}]