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 22:03, 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:
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:
We can then subtract this from the third equation:
Plugging into our third equation gives:
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.