Difference between revisions of "2012 AMC 10A Problems/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? | ||
| − | <math> \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)}\ </math> |
==Solution== | ==Solution== | ||
| − | Let Paula work at a rate of < | + | 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 < | + | 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 < | + | 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 < | + | Plugging <math>p</math> into our third equation gives: <cmath>L=\frac{4}{5}</cmath> |
| − | Converting < | + | Converting <math>L</math> from hours to minutes gives us <math>L=48</math> minutes, which is <math>\boxed{\textbf{(D)}\ 48}</math>. |
== See Also == | == See Also == | ||
Revision as of 21:03, 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?
Solution
Let Paula work at a rate of 
, the two helpers work at a combined rate of 
, and the time it takes to eat lunch be 
, where and 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:
![\[(8-L)(p+h)=50\]](http://latex.artofproblemsolving.com/a/6/b/a6b426a40f74b220a311ec1c88a811b4140cbbec.png)
![\[(6.2-L)h=24\]](http://latex.artofproblemsolving.com/c/a/0/ca0bb9613fe2541315ddae223bb083c7ce71c150.png)
![\[(11.2-L)p=26\]](http://latex.artofproblemsolving.com/c/5/4/c54ea3b1829b133d3a8c50813b201e554b787f2c.png)
With three equations and three variables, we need to find the value of .
Adding the second and third equations together gives us 
. Subtracting the first equation from this new one gives us 
, so we get 
.
Plugging into the second equation:
![\[(6.2-L)\frac{16}{9}p=24\]](http://latex.artofproblemsolving.com/0/4/3/0432514c9d8a2a0ebb087ffbae80de16d08ff8c1.png)
![\[(6.2-L)p=\frac{27}{2}\]](http://latex.artofproblemsolving.com/e/4/3/e4345c90982e72d78f0fb4e7b05f92af7c323bfc.png)
We can then subtract this from the third equation:
![\[5p=26-\frac{27}{2}\]](http://latex.artofproblemsolving.com/8/d/6/8d69c9bb94b0f7f1f858955aa67404de7277fe40.png)
![\[p=\frac{5}{2}\]](http://latex.artofproblemsolving.com/5/9/5/595d0f14b12bcd573f3b83d5026c5599072df59d.png)
Plugging into our third equation gives: ![\[L=\frac{4}{5}\]](http://latex.artofproblemsolving.com/b/7/5/b75a1c103aa62e4c825d52dcb8bcdd4351285201.png)
Converting from hours to minutes gives us 
minutes, which is 
.
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. 
