AMC10 2012 A

AMC10 2012 A · Q19

AMC10 2012 A · Q19. It mainly tests Systems of equations, Work problems.

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?

画家Paula和她的两名助手以恒定但不同的速率作画。他们总是从上午8:00开始，三人午饭时间相同。周一三人画了房子50%，下午4:00停止。周二Paula不在，两名助手只画了24%，下午2:12停止。周三Paula独自工作，到下午7:12完成房子。每日的午饭休息时间有多少分钟？

(A) 30 30

(B) 36 36

(D) 48 48

(E) 60 60

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Let the length of the lunch break be $m$ minutes. Then the three painters each worked $480-m$ minutes on Monday, the two helpers worked $372-m$ minutes on Tuesday, and Paula worked $672-m$ minutes on Wednesday. If Paula paints $p\%$ of the house per minute and her helpers paint a total of $h\%$ of the house per minute, then $$(p+h)(480-m)=50,$$ $$h(372-m)=24,\ \text{and}$$ $$p(672-m)=26.$$ Adding the last two equations gives $672p+372h-mp-mh=50$, and subtracting this equation from the first one gives $108h-192p=0$, so $h=\frac{16p}{9}$. Substitution into the first equation then leads to the system $$\frac{25p}{9}(480-m)=50,$$ $$p(672-m)=26.$$ The solution of this system is $p=\frac{1}{24}$ and $m=48$. Note that $h=\frac{2}{27}$.

答案（D）：设午休时间为 $m$ 分钟。则三位油漆工在周一各工作 $480-m$ 分钟，两位助手在周二工作 $372-m$ 分钟，Paula 在周三工作 $672-m$ 分钟。若 Paula 每分钟刷完房子的 $p\%$，她的助手合计每分钟刷完房子的 $h\%$，则 $$(p+h)(480-m)=50,$$ $$h(372-m)=24,\ \text{且}$$ $$p(672-m)=26.$$ 将后两式相加得 $672p+372h-mp-mh=50$，再用第一式减去该式得 $108h-192p=0$，所以 $h=\frac{16p}{9}$。代入第一式得到方程组 $$\frac{25p}{9}(480-m)=50,$$ $$p(672-m)=26.$$ 该方程组解为 $p=\frac{1}{24}$，$m=48$。注意 $h=\frac{2}{27}$。

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