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2008 AMC 12A Problems/Problem 10

Revision as of 21:30, 18 February 2008 by 123456789 (talk | contribs) (New page: Doug can paint <math>\frac{1}{5}</math> of a room per hour. Dave can paint <math>\frac{1}{7}</math> of a room in an hour. The time that they spend working together is <math>t-1</math>. Si...)
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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: $\left(\frac{1}{5}+\frac{1}{7}\right)\left(t-1\right)=1 \Rightarrow D$

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