Difference between revisions of "2008 AMC 12A Problems/Problem 10"
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==Solution== | ==Solution== | ||
=== Solution 1 === | === Solution 1 === | ||
| − | Doug can paint <math>\frac{1}{5}</math> of a room per hour, Dave can paint <math>\frac{1}{7}</math> of a room | + | Doug can paint <math>\frac{1}{5}</math> of a room per hour, Dave can paint <math>\frac{1}{7}</math> of a room per hour, and the time they spend working together is <math>t-1</math>. |
| − | Since rate | + | Since rate multiplied by time gives output, <math>\left(\frac{1}{5}+\frac{1}{7}\right)\left(t-1\right)=1 \Rightarrow \mathrm{(D)}</math> |
=== Solution 2 === | === Solution 2 === | ||
Latest revision as of 17:51, 4 June 2021
- 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 per hour, and the time they spend working together is 
.
Since rate multiplied by time gives output, 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
