Difference between revisions of "2008 AMC 12A Problems/Problem 10"
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| − | ==Problem | + | {{duplicate|[[2008 AMC 12A Problems|2008 AMC 12A #10]] and [[2008 AMC 10A Problems/Problem 14|2008 AMC 10A #13]]}} |
| + | ==Problem== | ||
Doug can paint a room in <math>5</math> hours. Dave can paint the same room in <math>7</math> hours. Doug and Dave paint the room together and take a one-hour break for lunch. Let <math>t</math> 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 <math>t</math>? | Doug can paint a room in <math>5</math> hours. Dave can paint the same room in <math>7</math> hours. Doug and Dave paint the room together and take a one-hour break for lunch. Let <math>t</math> 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 <math>t</math>? | ||
| − | <math>\ | + | <math>\mathrm{(A)}\ \left(\frac{1}{5}+\frac{1}{7}\right)\left(t+1\right)=1\qquad\mathrm{(B)}\ \left(\frac{1}{5}+\frac{1}{7}\right)t+1=1\qquad\mathrm{(C)}\ \left(\frac{1}{5}+\frac{1}{7}\right)t=1\\\mathrm{(D)}\ \left(\frac{1}{5}+\frac{1}{7}\right)\left(t-1\right)=1\qquad\mathrm{(E)}\ \left(5+7\right)t=1</math> |
| − | \\ | ||
| − | + | ==Solution== | |
| + | === 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 per hour, and the time they spend working together is <math>t-1</math>. | ||
| − | ==Solution== | + | 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 === | ||
| + | |||
| + | If one person does a job in <math>a</math> hours and another person does a job in <math>b</math> hours, the time it takes to do the job together is <math>\frac{ab}{a+b}</math> hours. | ||
| + | |||
| + | Since Doug paints a room in 5 hours and Dave paints a room in 7 hours, they both paint in <math>\frac{5\times7}{5+7} = \frac{35}{12}</math> hours. They also take 1 hour for lunch, so the total time <math>t = \frac{35}{12} + 1</math> hours. | ||
| − | + | Looking at the answer choices, <math>(D)</math> is the only one satisfied by <math>t = \frac{35}{12} + 1</math>. | |
| − | <math> | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2008|ab=A|num-b=9|num-a=11}} | {{AMC12 box|year=2008|ab=A|num-b=9|num-a=11}} | ||
| + | {{AMC10 box|year=2008|ab=A|num-b=12|num-a=14}} | ||
| + | {{MAA Notice}} | ||
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. 
