2008 AMC 10A Problem 13

1. Start with how fast each person works. Doug paints the whole room in $5$ hours, so in one hour he paints $\frac{1}{5}$ of the room. Dave paints $\frac{1}{7}$ of the room in one hour.
2. Working together, in one hour they paint
$$\frac{1}{5} + \frac{1}{7} \text{ of the room}.$$
3. Now the time. The total time $t$ includes a one-hour lunch, and nobody paints during lunch. So the time they actually paint is $t - 1$ hours.
4. Work done = (amount per hour) × (hours of painting). They finish exactly one room, so
$$\left(\frac{1}{5}+\frac{1}{7}\right)(t - 1) = 1.$$
5. Quick check: $\frac15 + \frac17 = \frac{12}{35}$, so the painting time is $\frac{35}{12}$, a little under $3$ hours. Adding lunch, $t$ is a little under $4$ hours, which makes sense.

The answer is $\boxed{\textbf{(D)}\ \left(\frac{1}{5}+\frac{1}{7}\right)\left(t-1\right)=1}$.

In work problems, think "fraction of the job per hour". Those amounts can be added when people work together. Then multiply by the hours spent working, which here is the total time minus the break.

Doug paints $\frac15$ of the room per hour and Dave paints $\frac17$ per hour, so together they paint $\frac15 + \frac17$ of the room per hour.

The total time $t$ includes a one-hour lunch during which no painting happens, so the actual painting time is $t - 1$ hours. Painting one whole room means
$$\left(\frac{1}{5}+\frac{1}{7}\right)(t - 1) = 1.$$

The answer is $\boxed{\textbf{(D)}\ \left(\frac{1}{5}+\frac{1}{7}\right)\left(t-1\right)=1}$.