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

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<math> \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} </math>.
 
<math> \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} </math>.
  
−
Thus our answer is <math>\textbf{(D)}\ \frac{15}{8}</math>.
+
Thus our answer is <math>\boxed{\textbf{(D)}\ \frac{15}{8}}</math>.
  
 
==See Also==
 
==See Also==
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:07, 9 November 2013

Problem

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

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

Solution

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 $\boxed{\textbf{(D)}\ \frac{15}{8}}$.

See Also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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