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

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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>.  
 
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>.  
 
Plugging into the second equation:
 
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>
 
 
<cmath>(6.2-L)p=\frac{27}{200}</cmath>
 
<cmath>(6.2-L)p=\frac{27}{200}</cmath>
  
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<cmath>5p=.26-\frac{27}{200}</cmath>
 
<cmath>5p=.26-\frac{27}{200}</cmath>
 
 
<math>p=\frac{1}{40}</math>.  
 
<math>p=\frac{1}{40}</math>.  
Plugging <math>p</math> into our third equation gives:
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Plugging <math>p</math> into our third equation gives: <cmath>L=\frac{4}{5}</cmath>
 
 
<cmath>L=\frac{4}{5}</cmath>
 
  
Since we let <math>L</math> be in hours, the lunch break then takes <math>\frac{4}{5}\cdot 60 = 48</math> minutes, which leads us to the correct answer of <math>\boxed{\textbf{(D)}\ 48}</math>.
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Converting <math>L</math> from hours to minutes gives <math>L=48</math> minutes, which is <math>\boxed{\textbf{(D)}\ 48}</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 22:13, 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:

\[(8-L)(p+h)=.50\]

\[(6.2-L)h=.24\]

\[(11.2-L)p=.26\]

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:

\[(6.2-L)\frac{16}{9}p=.24\] \[(6.2-L)p=\frac{27}{200}\]

We can then subtract this from the third equation:

\[5p=.26-\frac{27}{200}\] $p=\frac{1}{40}$. Plugging $p$ into our third equation gives: \[L=\frac{4}{5}\]

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

See Also

2012 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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