Difference between revisions of "2012 AMC 10A Problems/Problem 19"
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<cmath>5p=.26-\frac{27}{200}=\frac{25}{200}=\frac{1}{8}</cmath> | <cmath>5p=.26-\frac{27}{200}=\frac{25}{200}=\frac{1}{8}</cmath> | ||
− | Therefore <math>p=\frac{1}{40}</math>. Plugging this into our third equation | + | Therefore <math>p=\frac{1}{40}</math>. Plugging this into our third equation gives: |
<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 22:03, 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?
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:
Adding the second and third equations together gives us . Subtracting the first equation from this new one gives us , and solving for in terms of gives . We can plug this into the second equation:
We can then subtract this from the third equation:
Therefore . Plugging this into our third equation gives:
Since we let be in hours, the lunch break then takes minutes, which leads us to the correct answer of .
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.