2012 AMC 10A Problems/Problem 19

The following problem is from both the 2012 AMC 12A #13 and 2012 AMC 10A #19, so both problems redirect to this page.

Problem 19

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?

$\textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 60$

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:

$(8-L)(p+h)=.50$

$(6.2-L)h=.24$

$(11.2-L)p=.26$

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: $(6.2-L)\frac{16}{9}p=.24$

$(6.2-L)p=\frac{9}{16}\cdot \frac{24}{100}=\frac{27}{200}$

We can then subtract this from the third equation:

$5p=.26-\frac{27}{200}=\frac{25}{200}=\frac{1}{8}$

Therefore $p=\frac{1}{40}$ . Plugging this into our third equation gives:

$(11.2-L)\frac{1}{40}=\frac{26}{100}$

$L=\frac{4}{5}$

Since we let $L$ be in hours, the lunch break then takes $\frac{4}{5}\cdot 60 = 48$ minutes, which leads us to the correct answer of $\boxed{\textbf{(D)}\ 48}$ .

2012 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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All AMC 10 Problems and Solutions

2012 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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All AMC 12 Problems and Solutions

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2012 AMC 10A Problems/Problem 19

Problem 19

Solution

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