2008 AMC 12A Problems/Problem 2

Problem

What is the reciprocal of $\frac{1}{2}+\frac{2}{3}$?

$\mathrm{(A)}\ \frac{6}{7}\qquad\mathrm{(B)}\ \frac{7}{6}\qquad\mathrm{(C)}\ \frac{5}{3}\qquad\mathrm{(D)}\ 3\qquad\mathrm{(E)}\ \frac{7}{2}$

Solution

Solution 1

Here's a cheapshot: Obviously, $\frac{1}{2}+\frac{2}{3}$ is greater than $1$. Therefore, its reciprocal is less than $1$, and the answer must be $\boxed{\frac{6}{7}}$.

Solution 2

$\left(\frac{1}{2}+\frac{2}{3}\right)^{-1}=\left(\frac{3}{6}+\frac{4}{6}\right)^{-1}=\left(\frac{7}{6}\right)^{-1}=\frac{6}{7}\Rightarrow A$.

See Also

2008 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AMC 12 Problems and Solutions

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