Difference between revisions of "2008 AMC 12A Problems/Problem 10"
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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>? | ||
Revision as of 22:35, 18 February 2008
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 ?
$\textbf{(A)}\ \left( \frac{1}{5}+\frac{1}{7}\right)\left( t+1 \right)=1 \qquad \textbf{(B)}\ \left( \frac{1}{5}+\frac{1}{7}\right)t+1=1 \qquad \textbf{(C)}\left( \frac{1}{5}+\frac{1}{7}\right)t=1 \\
\textbf{(D)}\ \left( \frac{1}{5}+\frac{1}{7}\right)\left(t-1\right)=1 \qquad \textbf{(E)}\ \left(5+7\right)t=1$ (Error compiling LaTeX. ! Missing $ inserted.)
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 |