2008 AMC 10A Problem 13

Below is the professionally curated solution for Problem 13 of the 2008 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2008 AMC 10A solutions, or check the answer key.

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Concepts:ratelinear equation

Difficulty rating: 1280

13.

Doug can paint a room in 55 hours. Dave can paint the same room in 77 hours. Doug and Dave paint the room together and take a one-hour break for lunch. Let tt 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?t?

(15+17)(t+1)=1\left(\dfrac{1}{5} + \dfrac{1}{7}\right)(t + 1) = 1

(15+17)t+1=1\left(\dfrac{1}{5} + \dfrac{1}{7}\right)t + 1 = 1

(15+17)t=1\left(\dfrac{1}{5} + \dfrac{1}{7}\right)t = 1

(15+17)(t1)=1\left(\dfrac{1}{5} + \dfrac{1}{7}\right)(t - 1) = 1

(5+7)t=1(5 + 7)t = 1

Solution:

Working together, Doug and Dave paint 15+17\dfrac{1}{5} + \dfrac{1}{7} of the room per hour.

Because they break for one hour, they work for only t1t - 1 hours, and this must complete the whole room:

(15+17)(t1)=1. \left(\dfrac{1}{5} + \dfrac{1}{7}\right)(t - 1) = 1.

Thus, the correct answer is D.

Problem 13 in Other Years