2008 AMC 12A Problems/Problem 10: Difference between revisions
New page: 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. The time that they spend working together is <math>t-1</math>. Si... |
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==Problem 10== | |||
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>\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</math> | |||
==Solution== | |||
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. The time that they spend working together is <math>t-1</math>. | 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. The time that they spend working together is <math>t-1</math>. | ||
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> | ||
Revision as of 21:31, 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:
