Difference between revisions of "2012 AMC 10A Problems/Problem 19"
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==Solution 3 (GCD)== | ==Solution 3 (GCD)== | ||
− | We factor out the equations to be <math>(8-n)\left(\frac{1}{p}+\frac{1}{h}\right)=\frac{1}{2},\left(\frac{31}{5}-n\right)\left(\frac{1}{h}\right)=\frac{6}{25} \text{and } (\frac{56}{5}-n)(\frac{1}{p}=\frac{13}{50}</math>, 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 <math>\mathbf{D}</math>, we have them being <math>10.4</math> and <math>5.4</math>, which correspond to being multiples of 13 and 6. Checking, we find that this satisfies the first equation, so multiplying <math>0.8\cdot 60= \boxed{48.}</math> | + | We factor out the equations to be <math>(8-n)\left(\frac{1}{p}+\frac{1}{h}\right)=\frac{1}{2},\left(\frac{31}{5}-n\right)\left(\frac{1}{h}\right)=\frac{6}{25} \text{and } \left(\frac{56}{5}-n\right)\left(\frac{1}{p}=\frac{13}{50}\right)</math>, 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 <math>\mathbf{D}</math>, we have them being <math>10.4</math> and <math>5.4</math>, which correspond to being multiples of 13 and 6. Checking, we find that this satisfies the first equation, so multiplying <math>0.8\cdot 60= \boxed{48.}</math> |
== See Also == | == See Also == |
Revision as of 18:32, 8 November 2024
- The following problem is from both the 2012 AMC 12A #13 and 2012 AMC 10A #19, so both problems redirect to this page.
Contents
[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 .
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
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.