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

## 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: $\left(\frac{1}{5}+\frac{1}{7}\right)\left(t-1\right)=1 \Rightarrow D$