1988 AJHSME Problems/Problem 15

Revision as of 06:24, 3 June 2009 by 5849206328x (talk | contribs)

Problem

The reciprocal of $\left( \frac{1}{2}+\frac{1}{3}\right)$ is

$\text{(A)}\ \frac{1}{6} \qquad \text{(B)}\ \frac{2}{5} \qquad \text{(C)}\ \frac{6}{5} \qquad \text{(D)}\ \frac{5}{2} \qquad \text{(E)}\ 5$

Solution

The reciprocal for a fraction $\frac{a}{b}$ turns out to be $\frac{b}{a}$, so if we can express the expression as a single fraction, we're basically done.

The expression is equal to $\frac{3}{6}+\frac{2}{6}=\frac{5}{6}$, so the reciprocal is $\frac{6}{5}\rightarrow \boxed{\text{C}}$.

See Also

1988 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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 AJHSME/AMC 8 Problems and Solutions