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

Revision as of 21:15, 9 November 2013

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

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

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 The expression <math>\frac{1^{4y-1}}{5^{-1}+3^{-1}}</math> is equal to:
-<math>\textbf{(A)}\ \frac{4y-1}{8} \qquad</math>
+<math> \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} </math>
-<math>\textbf{(B)}\ 8 \qquad</math>
-<math>\textbf{(C)}\ \frac{15}{2} \qquad</math>
-<math>\textbf{(D)}\ \frac{15}{8} \qquad</math>
-<math>\textbf{(E)}\ \frac{1}{8} \qquad</math>