AMC12 2012 A

AMC12 2012 A · Q13

AMC12 2012 A · Q13. It mainly tests Systems of equations, Rates (speed).

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 P.M. How long, in minutes, was each day's lunch break?

画家 Paula 和她的两个助手各自以恒定但不同的速率作画。他们总是在上午 8:00 开始，而且三个人每天吃午饭所用的时间都相同。周一三人画完了一栋房子的 50%，并在下午 4:00 停工。周二 Paula 不在时，两位助手只画了房子的 24%，并在下午 2:12 停工。周三 Paula 独自工作，一直干到晚上 7:12 才完成整栋房子。问每天的午饭休息时间是多少分钟？

(A) 30 30

(B) 36 36

(D) 48 48

(E) 60 60

Answer

Correct choice: (D)

正确答案：（D）

Solution

Let Paula work at a rate of $p$, the two helpers work at a combined rate of $h$, and the time it takes to eat lunch be $L$, where $p$ and $h$ are in house/hours and L is in hours. Then the labor on Monday, Tuesday, and Wednesday can be represented by the three following equations: \[(8-L)(p+h)=50\] \[(6.2-L)h=24\] \[(11.2-L)p=26\] With three equations and three variables, we need to find the value of $L$. Adding the second and third equations together gives us $6.2h+11.2p-L(p+h)=50$. Subtracting the first equation from this new one gives us $-1.8h+3.2p=0$, so we get $h=\frac{16}{9}p$. Plugging into the second equation: \[(6.2-L)\frac{16}{9}p=24\] \[(6.2-L)p=\frac{27}{2}\] We can then subtract this from the third equation: \[5p=26-\frac{27}{2}\] \[p=\frac{5}{2}\] Plugging $p$ into our third equation gives: \[L=\frac{4}{5}\] Converting $L$ from hours to minutes gives us $L=48$ minutes, which is $\boxed{\textbf{(D)}\ 48}$.

设 Paula 的工作速率为 $p$，两位助手合计的工作速率为 $h$，吃午饭所用时间为 $L$，其中 $p$ 与 $h$ 的单位为（房子/小时），$L$ 的单位为小时。则周一、周二、周三的工作量可分别表示为以下三个方程： \[(8-L)(p+h)=50\] \[(6.2-L)h=24\] \[(11.2-L)p=26\] 有三个方程和三个未知量，我们需要求出 $L$。将第二个与第三个方程相加得到 $6.2h+11.2p-L(p+h)=50$。用这个新方程减去第一个方程得到 $-1.8h+3.2p=0$，因此 $h=\frac{16}{9}p$。代入第二个方程： \[(6.2-L)\frac{16}{9}p=24\] \[(6.2-L)p=\frac{27}{2}\] 再用第三个方程减去它： \[5p=26-\frac{27}{2}\] \[p=\frac{5}{2}\] 将 $p$ 代入第三个方程得：\[L=\frac{4}{5}\] 把 $L$ 从小时换算成分钟得到 $L=48$ 分钟，即 $\boxed{\textbf{(D)}\ 48}$。

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