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

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<cmath>(11.2-L)p=.26</cmath>
 
<cmath>(11.2-L)p=.26</cmath>
  
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Adding the second and third equations together gives us <math>6.2h+11.2p-L(p+h)=.50</math>. Subtracting the first equation from this new one gives us <math>-1.8h+3.2p=0</math>, and solving for <math>h</math> in terms of <math>p</math> gives <math>h=\frac{16}{9}p</math>. We can plug this into the second equation:
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Adding the second and third equations together gives us <math>6.2h+11.2p-L(p+h)=.50</math>. Subtracting the first equation from this new one gives us <math>-1.8h+3.2p=0</math>, so we get <math>h=\frac{16}{9}p</math>.  
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+
Plugging into the second equation:
 
<cmath>(6.2-L)\frac{16}{9}p=.24</cmath>
 
<cmath>(6.2-L)\frac{16}{9}p=.24</cmath>
  
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<cmath>(11.2-L)\frac{1}{40}=\frac{26}{100}</cmath>
 
<cmath>(11.2-L)\frac{1}{40}=\frac{26}{100}</cmath>
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<cmath>L=\frac{4}{5}</cmath>
 
<cmath>L=\frac{4}{5}</cmath>

Revision as of 21:05, 11 September 2013

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 $\boxed{\textbf{(D)}\ 48}$.

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

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