在一次班级环保活动中,某学生收集了可回收垃圾的重量为 3.5 千克,比另一名同学多收集了 1.2 千克。设另一名同学收集的垃圾重量为 x 千克,则可列出一元一次方程为:_3.5 = x + 1.2_,解得 x = _2.3_。
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众数是一组数据中出现次数最多的数。首先统计每个数出现的次数:3出现1次,4出现3次,5出现3次,6出现2次,7出现1次。可以看出,4和5都出现了3次,是出现次数最多的数,因此这组数据的众数是4和5。当一组数据中有两个数出现次数相同且最多时,这两个数都是众数。所以正确答案是C。
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[{"id":2020,"content":"在一次校园绿化活动中,某学生用一根长度为12米的篱笆围成一个一边靠墙的矩形花圃(靠墙的一边不需要篱笆)。为了使花圃的面积最大,该学生应如何设计长和宽?设垂直于墙的一边长度为x米,则花圃面积S与x的函数关系为S = x(12 - 2x)。当x取何值时,面积S取得最大值?","type":"选择题","subject":"数学","grade":"八年级","stage":"初中","difficulty":"简单","answer":"B","explanation":"题目给出面积函数 S = x(12 - 2x),可展开为 S = -2x² + 12x。这是一个开口向下的二次函数,其最大值出现在顶点处。顶点横坐标公式为 x = -b\/(2a),其中 a = -2,b = 12。代入得 x = -12 \/ (2 × (-2)) = 3。因此当 x = 3 米时,面积最大。此时平行于墙的一边为 12 - 2×3 = 6 米,面积为 3×6 = 18 平方米。本题考查一次函数与二次函数在实际问题中的应用,结合几何情境,难度适中,符合八年级学生认知水平。","options":[{"id":"A","content":"x = 2"},{"id":"B","content":"x = 3"},{"id":"C","content":"x = 4"},{"id":"D","content":"x = 6"}]},{"id":419,"content":"某学生在整理班级同学的课外阅读时间时,记录了5名同学每周阅读的小时数分别为:3、5、4、6、2。如果再加入一名同学的阅读时间后,这组数据的平均数变为4小时,那么这名同学的阅读时间是多少小时?","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"A","explanation":"首先计算原来5名同学的阅读总时间:3 + 5 + 4 + 6 + 2 = 20(小时)。设新加入的同学阅读时间为x小时,则6名同学的总阅读时间为20 + x。根据题意,平均数为4小时,因此有方程:(20 + x) ÷ 6 = 4。两边同时乘以6得:20 + x = 24,解得x = 4。所以这名同学的阅读时间是4小时,正确答案是A。","options":[{"id":"A","content":"4"},{"id":"B","content":"5"},{"id":"C","content":"3"},{"id":"D","content":"6"}]},{"id":444,"content":"在一次班级大扫除中,某学生负责统计同学们带来的清洁工具数量。他记录了抹布、扫帚和拖把的总数为28件。已知抹布比扫帚多4件,拖把比扫帚少2件。问扫帚有多少件?","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"B","explanation":"设扫帚有x件,则抹布有(x + 4)件,拖把有(x - 2)件。根据题意,三种工具的总数为28件,可列方程:x + (x + 4) + (x - 2) = 28。化简得:3x + 2 = 28,解得3x = 26,x = 10。因此,扫帚有10件。此题考查一元一次方程的实际应用,通过设未知数、列方程、解方程的过程,帮助学生理解如何将生活问题转化为数学问题并求解。","options":[{"id":"A","content":"8件"},{"id":"B","content":"10件"},{"id":"C","content":"12件"},{"id":"D","content":"14件"}]},{"id":182,"content":"小明在文具店买了一支钢笔和一本笔记本,共花费18元。已知钢笔的价格是笔记本价格的2倍,那么笔记本的价格是多少元?","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"A","explanation":"设笔记本的价格为x元,则钢笔的价格为2x元。根据题意,钢笔和笔记本共花费18元,可列出方程:x + 2x = 18,即3x = 18。解得x = 6。因此,笔记本的价格是6元。选项A正确。","options":[{"id":"A","content":"6元"},{"id":"B","content":"9元"},{"id":"C","content":"12元"},{"id":"D","content":"3元"}]},{"id":2238,"content":"某学生在数轴上从原点出发,先向右移动5个单位长度,再向左移动8个单位长度,接着向右移动3个单位长度,最后向左移动6个单位长度。该学生最终所在位置的数与其起始位置(原点)的距离是___。","type":"填空题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"困难","answer":"6","explanation":"该学生从原点0出发,按照顺序移动:+5 → -8 → +3 → -6。计算总位移:5 - 8 + 3 - 6 = -6。最终位置是-6,与原点0的距离是|−6| = 6。题目考查正负数在数轴上的实际应用及绝对值的理解,要求学生掌握连续正负数运算和距离的非负性,属于综合应用型难题。","options":[]},{"id":2228,"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℃;而另一天的气温比前一天下降了2℃,应记作____℃。","type":"填空题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"简单","answer":"-2","explanation":"根据正数和负数表示相反意义的量的规则,气温上升用正数表示,气温下降则用负数表示。下降2℃应记作-2℃,符合七年级正负数在实际生活中的应用知识点。","options":[]},{"id":200,"content":"一个长方形的长是8厘米,宽是5厘米,它的周长是______厘米。","type":"填空题","subject":"数学","grade":"初一","stage":"小学","difficulty":"简单","answer":"26","explanation":"长方形的周长计算公式是:周长 = 2 × (长 + 宽)。将已知的长8厘米和宽5厘米代入公式,得到:2 × (8 + 5) = 2 × 13 = 26(厘米)。因此,这个长方形的周长是26厘米。","options":[]},{"id":450,"content":"某学生在整理班级同学的课外阅读时间数据时,记录了10名学生每周的阅读时间(单位:小时)如下:3, 5, 4, 6, 4, 7, 5, 4, 6, 5。为了分析数据,他计算了这组数据的众数。请问这组数据的众数是多少?","type":"选择题","subject":"数学","grade":"初一","stage":"小学","difficulty":"简单","answer":"C","explanation":"众数是一组数据中出现次数最多的数。首先统计每个数出现的次数:3出现1次,4出现3次,5出现3次,6出现2次,7出现1次。可以看出,4和5都出现了3次,是出现次数最多的数,因此这组数据的众数是4和5。当一组数据中有两个数出现次数相同且最多时,这两个数都是众数。所以正确答案是C。","options":[{"id":"A","content":"4"},{"id":"B","content":"5"},{"id":"C","content":"4和5"},{"id":"D","content":"6"}]},{"id":939,"content":"在一次环保知识竞赛中,某班级学生共收集了120条有效答题记录。经统计,其中答对一题得3分,答错或不答扣1分。若该班级总得分为280分,则他们答对了____道题。","type":"填空题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"100","explanation":"设答对的题数为x,则答错或不答的题数为(120 - x)。根据得分规则,总得分为3x - 1×(120 - x) = 280。化简方程得:3x - 120 + x = 280,即4x = 400,解得x = 100。因此,他们答对了100道题。本题考查一元一次方程的实际应用,属于简单难度。","options":[]},{"id":1080,"content":"在一次环保活动中,某学生收集了可回收垃圾和不可回收垃圾共12千克,其中可回收垃圾比不可回收垃圾多4千克。设不可回收垃圾为x千克,则可列出一元一次方程为:______。","type":"填空题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"简单","answer":"x + (x + 4) = 12","explanation":"设不可回收垃圾为x千克,根据题意,可回收垃圾比不可回收垃圾多4千克,因此可回收垃圾为(x + 4)千克。两者总重量为12千克,所以方程为x + (x + 4) = 12。该题考查一元一次方程的实际建模能力,属于简单难度。","options":[]}]