Difference between revisions of "2008 AMC 12A Problems/Problem 10"
| Line 12: | Line 12: | ||
Since rate multiplied by time gives output: | Since rate multiplied by time gives output: | ||
<math>\left(\frac{1}{5}+\frac{1}{7}\right)\left(t-1\right)=1 \Rightarrow D</math> | <math>\left(\frac{1}{5}+\frac{1}{7}\right)\left(t-1\right)=1 \Rightarrow D</math> | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC12 box|year=2008|ab=A|num-b=9|num-a=11}} | ||
Revision as of 21:33, 18 February 2008
Problem 10
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 ?
$\textbf{(A)}\ \left( \frac{1}{5}+\frac{1}{7}\right)\left( t+1 \right)=1 \qquad \textbf{(B)}\ \left( \frac{1}{5}+\frac{1}{7}\right)t+1=1 \qquad \textbf{(C)}\left( \frac{1}{5}+\frac{1}{7}\right)t=1 \\
\textbf{(D)}\ \left( \frac{1}{5}+\frac{1}{7}\right)\left(t-1\right)=1 \qquad \textbf{(E)}\ \left(5+7\right)t=1$ (Error compiling LaTeX. Unknown error_msg)
Solution
Doug can paint 
of a room per hour. Dave can paint 
of a room in an hour. The time that they spend working together is 
.
Since rate multiplied by 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 | |
