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

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==Problem ==
 
==Problem ==
What is the reciprocal of <math>\frac{1}{2}+\frac{2}{3}</math>?
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What is the [[reciprocal]] of <math>\frac{1}{2}+\frac{2}{3}</math>?
  
 
<math>\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} </math>
 
<math>\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} </math>

Revision as of 23:51, 20 February 2008

Problem

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

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

Solution

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

See Also

2008 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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