1988 AJHSME Problems/Problem 15: Difference between revisions

Latest revision as of 22:55, 4 July 2013

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 $\frac{6}{5}\rightarrow \boxed{\text{C}}$ .

1988 AJHSME (Problems • Answer Key • Resources)
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

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 {{AJHSME box|year=1988|num-b=14|num-a=16}}
 [[Category:Introductory Algebra Problems]]
+{{MAA Notice}}

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1988 AJHSME Problems/Problem 15: Difference between revisions

Latest revision as of 22:55, 4 July 2013

Problem

Solution

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