Question

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 $t?$

$\left(\dfrac{1}{5} + \dfrac{1}{7}\right)(t + 1) = 1$

$\left(\dfrac{1}{5} + \dfrac{1}{7}\right)t + 1 = 1$

$\left(\dfrac{1}{5} + \dfrac{1}{7}\right)t = 1$

$\left(\dfrac{1}{5} + \dfrac{1}{7}\right)(t - 1) = 1$

$(5 + 7)t = 1$

Answer: D

Concepts:rate linear equation

Difficulty rating: 1380

Solution:

In one hour Doug paints $\tfrac{1}{5}$ of the room and Dave paints $\tfrac{1}{7},$ so together they paint $\tfrac{1}{5} + \tfrac{1}{7}$ of the room per hour.

Of the total time $t,$ one hour is spent at lunch, so they work for $t - 1$ hours. The fraction painted must equal $1,$ giving $\left(\dfrac{1}{5} + \dfrac{1}{7}\right)(t - 1) = 1.$

Thus, D is the correct answer.

2008 AMC 12A Problem 10

Answer: D

Difficulty rating: 1380

Solution:

Problem 10 in Other Years