2012 AMC 10A Problems/Problem 19
- The following problem is from both the 2012 AMC 12A #13 and 2012 AMC 10A #19, so both problems redirect to this page.
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[hide]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 .
Solution 2 (Modular Arithmetic)
Because Paula worked from to , she worked for 11 hours and 12 minutes = 672 minutes. Since there is % of the house left, we get the equation . Because is mod , looking at our answer choices, the only answer that is is . So the answer 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 |
Solution 3 (GCD)
We factor out the equations to be , where n is the number of hours for the break, p is the time Paula requires, and h is the time her helpers require. We find that when we select , we have them being and , which correspond to being multiples of 13 and 6. Checking, we find that this satisfies the first equation, so multiplying The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.