Difference between revisions of "2008 AMC 12A Problems/Problem 10"
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Since rate times time gives output, <math>\left(\frac{1}{5}+\frac{1}{7}\right)\left(t-1\right)=1 \Rightarrow \mathrm{(D)}</math> | Since rate times time gives output, <math>\left(\frac{1}{5}+\frac{1}{7}\right)\left(t-1\right)=1 \Rightarrow \mathrm{(D)}</math> | ||
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=== Solution 2 === | === Solution 2 === |
Revision as of 20:35, 3 July 2013
- The following problem is from both the 2008 AMC 12A #10 and 2008 AMC 10A #13, so both problems redirect to this page.
Problem
Doug can paint a room in hours. Dave can paint the same room in hours. Doug and Dave paint the room together and take a one-hour break for lunch. Let be the total time, in hours, required for them to complete the job working together, including lunch. Which of the following equations is satisfied by ?
Solution
Solution 1
Doug can paint of a room per hour, Dave can paint of a room in an hour, and the time they spend working together is .
Since rate times time gives output, The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
Solution 2
If one person does a job in hours and another person does a job in hours, the time it takes to do the job together is hours.
Since Doug paints a room in 5 hours and Dave paints a room in 7 hours, they both paint in hours. They also take 1 hour for lunch, so the total time hours.
Looking at the answer choices, is the only one satisfied by .
See Also
2008 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 9 |
Followed by Problem 11 |
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 |
2008 AMC 10A (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 10 Problems and Solutions |