Difference between revisions of "2012 AMC 10A Problems/Problem 19"

Revision as of 00:57, 11 February 2017

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:

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

$(6.2-L)h=24$

$(11.2-L)p=26$

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:

$5p=26-\frac{27}{200}$ $p=\frac{5}{2}$ Plugging $p$ into our third equation gives: $L=\frac{4}{5}$

Converting $L$ from hours to minutes gives us $L=48$ minutes, which is $\boxed{\textbf{(D)}\ 48}$ .

2012 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

2012 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 26: / Line 26: @@
 <cmath>5p=26-\frac{27}{200}</cmath>
-<cmath>p=\frac{1}{40}</cmath>
+<cmath>p=\frac{5}{2}</cmath>
 Plugging <math>p</math> into our third equation gives: <cmath>L=\frac{4}{5}</cmath>

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Difference between revisions of "2012 AMC 10A Problems/Problem 19"

Revision as of 00:57, 11 February 2017

Problem 19

Solution

See Also