Difference between revisions of "2008 AMC 12A Problems/Problem 10"
(→Problem 10) |
I like pie (talk | contribs) (Duplicate, AMC 10 box, style, answer choices) |
||
| Line 1: | Line 1: | ||
| + | {{duplicate|[[2008 AMC 12A Problems|2008 AMC 12A #10]] and [[2008 AMC 10A Problems/Problem 14|2008 AMC 10A #13]]}} | ||
==Problem== | ==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== | ||
| − | Doug can paint <math>\frac{1}{5}</math> of a room per hour | + | Doug can paint <math>\frac{1}{5}</math> of a room per hour, Dave can paint <math>\frac{1}{7}</math> of a room in an hour, and the time they spend working together is <math>t-1</math>. |
| − | Since rate | + | Since rate times time gives output, <math>\left(\frac{1}{5}+\frac{1}{7}\right)\left(t-1\right)=1 \Rightarrow \mathrm{(D)}</math> |
| − | <math>\left(\frac{1}{5}+\frac{1}{7}\right)\left(t-1\right)=1 \Rightarrow D</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}} | ||
Revision as of 00:05, 26 April 2008
- 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
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, 
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 | ||
