Difference between revisions of "2008 AMC 12A Problems/Problem 2"

Revision as of 20:34, 3 July 2013

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

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

2008 AMC 12A (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 ==See Also==
 {{AMC12 box|year=2008|ab=A|num-b=1|num-a=3}}
+{{MAA Notice}}

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