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?
$\textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}$
==Solution== Let Paula work at a rate of$ (Error compiling LaTeX. Unknown error_msg)phLph$are in house/hours and L is in hours. Then the labor on Monday, Tuesday, and Wednesday can be represented by the three following equations:
<cmath>(8-L)(p+h)=50</cmath>
<cmath>(6.2-L)h=24</cmath>
<cmath>(11.2-L)p=26</cmath>
With three equations and three variables, we need to find the value of$ (Error compiling LaTeX. Unknown error_msg)L6.2h+11.2p-L(p+h)=50-1.8h+3.2p=0h=\frac{16}{9}p$. Plugging into the second equation:
<cmath>(6.2-L)\frac{16}{9}p=24</cmath> <cmath>(6.2-L)p=\frac{27}{2}</cmath>
We can then subtract this from the third equation:
<cmath>5p=26-\frac{27}{2}</cmath> <cmath>p=\frac{5}{2}</cmath> Plugging$ (Error compiling LaTeX. Unknown error_msg)p$into our third equation gives: <cmath>L=\frac{4}{5}</cmath>
Converting$ (Error compiling LaTeX. Unknown error_msg)LL=48\boxed{\textbf{(D)}\ 48}$.
See Also
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.