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

Revision as of 23:42, 25 April 2008

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

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

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

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

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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>\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}</math>
 ==Solution==