Difference between revisions of "2008 AMC 12A Problems/Problem 10"

(Problem 10)
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==Problem 10==
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==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>?
  

Revision as of 22:35, 18 February 2008

Problem

Doug can paint a room in $5$ hours. Dave can paint the same room in $7$ hours. Doug and Dave paint the room together and take a one-hour break for lunch. Let $t$ 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 $t$?

$\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 $\frac{1}{5}$ of a room per hour. Dave can paint $\frac{1}{7}$ of a room in an hour. The time that they spend working together is $t-1$.

Since rate multiplied by time gives output: $\left(\frac{1}{5}+\frac{1}{7}\right)\left(t-1\right)=1 \Rightarrow D$

See Also

2008 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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All AMC 12 Problems and Solutions
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