Difference between revisions of "1962 AHSME Problems/Problem 1"

Revision as of 22:34, 10 November 2013

The expression $\frac{1^{4y-1}}{5^{-1}+3^{-1}}$ is equal to:

$\textbf{(A)}\ \frac{4y-1}{8}\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ \frac{15}{2}\qquad\textbf{(D)}\ \frac{15}{8}\qquad\textbf{(E)}\ \frac{1}{8}$

We simplify the expression to yield:

$\dfrac{1^{4y-1}}{5^{-1}+3^{-1}}=\dfrac{1}{5^{-1}+3^{-1}}=\dfrac{1}{\dfrac{1}{5}+\dfrac{1}{3}}=\dfrac{1}{\dfrac{8}{15}}=\dfrac{15}{8}$ .

Thus our answer is $\boxed{\textbf{(D)}\ \frac{15}{8}}$ .

1962 AHSC (Problems • Answer Key • Resources)
Preceded by First Question	Followed by Problem 2
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All AHSME Problems and Solutions

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 ==See Also==
+{{AHSME 50p box|year=1962|before=First Question|num-a=2}}
 [[Category:Introductory Algebra Problems]]
 {{MAA Notice}}