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

Revision as of 14:43, 17 February 2008

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

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

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

2008 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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All AMC 12 Problems and Solutions

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