小明去文具店买笔记本,每本笔记本的价格是5元。他买了3本,付给收银员20元,应找回多少钱?
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设收集的纸张数量为x张,则塑料瓶数量为(x + 10)个。根据题意,纸张每5张换1积分,可得纸张积分为x/5;塑料瓶每3个换1积分,可得塑料瓶积分为(x + 10)/3。总积分为12,因此方程为x/5 + (x + 10)/3 = 12。解这个方程:两边同乘15得3x + 5(x + 10) = 180,即3x + 5x + 50 = 180,8x = 130,x = 25。故答案依次为x/5、(x+10)/3、25。
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[{"id":868,"content":"在一次环保知识竞赛中,某班级收集了学生们的答题情况,并绘制成扇形统计图。其中,答对8题以上的学生占总人数的30%,答对5至7题的学生占45%,答对4题以下的学生占剩余部分。若该班级共有40名学生,则答对4题以下的学生有___人。","type":"填空题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"10","explanation":"首先计算答对8题以上和答对5至7题的学生所占百分比之和:30% + 45% = 75%。因此,答对4题以下的学生占比为100% - 75% = 25%。班级总人数为40人,所以答对4题以下的学生人数为40 × 25% = 40 × 0.25 = 10人。","options":[]},{"id":245,"content":"某学生计算一个数的相反数时,将该数加上3,结果得到8,那么这个数的相反数是____。","type":"填空题","subject":"数学","grade":"初一","stage":"小学","difficulty":"简单","answer":"-5","explanation":"设这个数为x。根据题意,某学生将这个数加上3得到8,即x + 3 = 8,解得x = 5。那么这个数的相反数是-5。题目考查的是相反数的概念和一元一次方程的简单应用,符合七年级数学课程内容。","options":[]},{"id":2018,"content":"在一次校园绿化项目中,工人在一块矩形空地的四周铺设了宽度相同的步行道。已知原空地的长为 8 米,宽为 6 米,铺设步行道后整个区域(包括步行道)的面积为 120 平方米。若设步行道的宽度为 x 米,则可列方程为:","type":"选择题","subject":"数学","grade":"八年级","stage":"初中","difficulty":"简单","answer":"B","explanation":"步行道围绕矩形空地四周铺设,宽度为 x 米,因此整个区域的长和宽都要在原来的基础上各增加 2x 米(左右各 x 米,上下各 x 米)。原空地长 8 米,宽 6 米,所以铺设后整个区域的长为 (8 + 2x) 米,宽为 (6 + 2x) 米。根据题意,整个区域的面积为 120 平方米,因此可列方程为:(8 + 2x)(6 + 2x) = 120。选项 B 正确。","options":[{"id":"A","content":"(8 + x)(6 + x) = 120"},{"id":"B","content":"(8 + 2x)(6 + 2x) = 120"},{"id":"C","content":"8x + 6x = 120"},{"id":"D","content":"(8 - 2x)(6 - 2x) = 120"}]},{"id":592,"content":"某班级进行了一次数学测验,成绩分布如下表所示。根据统计表,该班级成绩在80分到89分之间的人数占总人数的百分比是多少?\n\n| 分数段 | 人数 |\n|--------|------|\n| 90-100 | 8 |\n| 80-89 | 12 |\n| 70-79 | 10 |\n| 60-69 | 5 |\n| 60以下 | 3 |","type":"选择题","subject":"数学","grade":"初一","stage":"小学","difficulty":"简单","answer":"B","explanation":"首先计算总人数:8 + 12 + 10 + 5 + 3 = 38(人)。成绩在80-89分之间的人数为12人。所求百分比为 (12 ÷ 38) × 100% ≈ 31.58%,四舍五入后最接近的选项是30%。因此正确答案是B。本题考查数据的收集、整理与描述中的百分比计算,属于简单难度。","options":[{"id":"A","content":"24%"},{"id":"B","content":"30%"},{"id":"C","content":"36%"},{"id":"D","content":"40%"}]},{"id":2300,"content":"某公园内有一块平行四边形形状的草坪,已知其相邻两边的长分别为√12米和√27米,且其中一条对角线恰好等于这两边之和。若一名学生想计算这块草坪的周长,他应选择以下哪个结果?","type":"选择题","subject":"数学","grade":"八年级","stage":"初中","difficulty":"简单","answer":"A","explanation":"首先化简题目中给出的边长:√12 = 2√3,√27 = 3√3。因此,平行四边形的两条邻边分别为2√3米和3√3米。平行四边形的周长等于两倍的两邻边之和,即:2 × (2√3 + 3√3) = 2 × 5√3 = 10√3(米)。题目中提到的‘一条对角线等于两边之和’是干扰信息,用于考查学生是否掌握平行四边形周长的计算方法,而不被无关条件误导。因此,正确答案为A。","options":[{"id":"A","content":"10√3 米"},{"id":"B","content":"12√3 米"},{"id":"C","content":"14√3 米"},{"id":"D","content":"16√3 米"}]},{"id":2327,"content":"某学生在研究轴对称图形时,发现一个四边形ABCD关于直线MN对称,其中点A与点C对称,点B与点D对称。若∠ABC = 70°,则∠ADC的度数为多少?","type":"选择题","subject":"数学","grade":"八年级","stage":"初中","difficulty":"简单","answer":"A","explanation":"由于四边形ABCD关于直线MN轴对称,且点A与点C对称,点B与点D对称,说明图形在对称轴两侧完全重合。因此,对应角相等。∠ABC与∠ADC是关于对称轴对应的角,故∠ADC = ∠ABC = 70°。本题考查轴对称图形的性质:对称点所连线段被对称轴垂直平分,且对称图形中对应角、对应边相等。","options":[{"id":"A","content":"70°"},{"id":"B","content":"110°"},{"id":"C","content":"90°"},{"id":"D","content":"140°"}]},{"id":721,"content":"在一次班级图书整理活动中,某学生统计了同学们捐赠的图书数量,发现捐赠数量最多的比最少的多8本,而最多的数量是最少的3倍。如果最少捐赠了___本书,那么最多捐赠了___本书。","type":"填空题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"4, 12","explanation":"设最少捐赠了x本书,则最多捐赠了3x本书。根据题意,最多比最少多8本,可列方程:3x - x = 8,解得2x = 8,x = 4。因此最少捐赠了4本,最多捐赠了3 × 4 = 12本。本题考查一元一次方程的实际应用,属于简单难度。","options":[]},{"id":319,"content":"8人","type":"选择题","subject":"数学","grade":"初一","stage":"小学","difficulty":"简单","answer":"答案待完善","explanation":"解析待完善","options":[]},{"id":2210,"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5℃,记作+5℃;而另一天的气温比前一天下降了3℃,应记作____℃。","type":"填空题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"简单","answer":"-3","explanation":"根据正数和负数表示相反意义的量的知识点,气温上升用正数表示,下降则用负数表示。题目中气温下降3℃,因此应记作-3℃。这符合七年级学生对正负数在实际生活中应用的理解要求。","options":[]},{"id":2506,"content":"如图,一个圆形花坛被两条互相垂直的小路分成四个面积相等的扇形区域,其中一条小路的长度为8米。若要在花坛边缘安装一圈LED灯带,则所需灯带的最短长度为多少米?","type":"选择题","subject":"数学","grade":"九年级","stage":"初中","difficulty":"简单","answer":"A","explanation":"题目中描述两条互相垂直的小路将圆形花坛分成四个面积相等的扇形,说明这两条小路是圆的两条互相垂直的直径。已知其中一条小路的长度为8米,即圆的直径为8米,因此半径r = 4米。要在花坛边缘安装灯带,即求圆的周长。圆的周长公式为C = 2πr = 2π × 4 = 8π(米)。因此,所需灯带的最短长度为8π米,对应选项A。","options":[{"id":"A","content":"8π"},{"id":"B","content":"16π"},{"id":"C","content":"4π"},{"id":"D","content":"32π"}]}]