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2008 AMC 12A Problems/Problem 10

Revision as of 00:05, 26 April 2008 by I like pie (talk | contribs) (Duplicate, AMC 10 box, style, answer choices)
The following problem is from both the 2008 AMC 12A #10 and 2008 AMC 10A #13, so both problems redirect to this page.

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 $t$?

$\mathrm{(A)}\ \left(\frac{1}{5}+\frac{1}{7}\right)\left(t+1\right)=1\qquad\mathrm{(B)}\ \left(\frac{1}{5}+\frac{1}{7}\right)t+1=1\qquad\mathrm{(C)}\ \left(\frac{1}{5}+\frac{1}{7}\right)t=1\\\mathrm{(D)}\ \left(\frac{1}{5}+\frac{1}{7}\right)\left(t-1\right)=1\qquad\mathrm{(E)}\ \left(5+7\right)t=1$

Solution

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

Since rate times time gives output, $\left(\frac{1}{5}+\frac{1}{7}\right)\left(t-1\right)=1 \Rightarrow \mathrm{(D)}$

See Also

2008 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions
2008 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
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