Question

油漆工 Paula 和她的两名助手各自以恒定但不同的速度刷漆。他们总是在上午八点开始工作，并且三人每天午餐都花同样长的时间。星期一，三人一起粉刷了房子的 $50\%$ ，在下午四点停工。星期二 Paula 不在，两名助手只粉刷了房子的 $24\%$ ，并在下午两点十二分停工。星期三 Paula 独自工作，一直到晚上七点十二分才完成整栋房子。每天的午餐休息时间是多少分钟？

Paula the painter and her two helpers each paint at constant, but different, rates. They always start at 8:00 AM and all three always take the same amount of time to eat lunch. On Monday the three of them painted $50\%$ of a house, quitting at 4:00 PM. On Tuesday, when Paula wasn't there, the two helpers painted only $24\%$ of the house and quit at 2:12 PM. On Wednesday Paula worked by herself and finished the house by working until 7:12 PM. How long, in minutes, was each day's lunch break?

$30$

$36$

$42$

$48$

$60$

答案：D

知识点：速率方程组

难度评级：1810

解答：

设午餐时长为 $m$ 分钟。星期一三人工作了 $480 - m$ 分钟，星期二助手工作了 $372 - m$ 分钟，星期三 Paula 工作了 $672 - m$ 分钟。

若 Paula 每分钟刷 $p\%$ ，两名助手合计每分钟刷 $h\%$ ，则 $\begin{aligned} (p+h)(480-m) &= 50, \\ h(372-m) &= 24, \\ p(672-m) &= 26. \end{aligned}$

将后两个方程相加，再从第一个方程中减去，得到 $108h - 192p = 0$ ，所以 $h = \tfrac{16}{9}p$ 。解这个方程组得到 $p = \tfrac{1}{24}$ 且 $m = 48$ 。

因此，正确答案是 D。

Let $m$ be the lunch length in minutes. The three worked $480 - m$ minutes Monday, the helpers $372 - m$ minutes Tuesday, and Paula $672 - m$ minutes Wednesday.

If Paula paints $p\%$ per minute and the helpers together paint $h\%$ per minute, then $\begin{aligned} (p+h)(480-m) &= 50, \\ h(372-m) &= 24, \\ p(672-m) &= 26. \end{aligned}$

Adding the last two equations and subtracting from the first gives $108h - 192p = 0,$ so $h = \tfrac{16}{9}p.$ Solving the system gives $p = \tfrac{1}{24}$ and $m = 48.$

Thus, the correct answer is D.

2012 AMC 12A 第 13 题

答案：D

难度评级：1810

解答：

其他年份的第 13 题