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

(Problem 19)
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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?
 
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)}\$
+
<math> \textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\$
  
 
==Solution==
 
==Solution==
Let Paula work at a rate of <math>p</math>, the two helpers work at a combined rate of <math>h</math>, and the time it takes to eat lunch be <math>L</math>, where <math>p</math> and <math>h</math> 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:
+
Let Paula work at a rate of </math>p<math>, the two helpers work at a combined rate of </math>h<math>, and the time it takes to eat lunch be </math>L<math>, where </math>p<math> and </math>h<math> 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>(8-L)(p+h)=50</cmath>
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<cmath>(11.2-L)p=26</cmath>
 
<cmath>(11.2-L)p=26</cmath>
  
With three equations and three variables, we need to find the value of <math>L</math>.
+
With three equations and three variables, we need to find the value of </math>L<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>.  
+
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:
  
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<cmath>5p=26-\frac{27}{2}</cmath>
 
<cmath>5p=26-\frac{27}{2}</cmath>
 
<cmath>p=\frac{5}{2}</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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Plugging </math>p<math> into our third equation gives: <cmath>L=\frac{4}{5}</cmath>
  
Converting <math>L</math> from hours to minutes gives us <math>L=48</math> minutes, which is <math>\boxed{\textbf{(D)}\ 48}</math>.
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Converting </math>L<math> from hours to minutes gives us </math>L=48<math> minutes, which is </math>\boxed{\textbf{(D)}\ 48}$.
  
 
== See Also ==
 
== See Also ==

Revision as of 00:33, 21 January 2018

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. ! Missing $ inserted.)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:

<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. ! Missing $ inserted.)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:

<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. ! Missing $ inserted.)p$into our third equation gives: <cmath>L=\frac{4}{5}</cmath>

Converting$ (Error compiling LaTeX. ! Missing $ inserted.)L$from hours to minutes gives us$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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