1954 AHSME Problems/Problem 6

Problem 6

The value of $\frac{1}{16}a^0+\left (\frac{1}{16a} \right )^0- \left (64^{-\frac{1}{2}} \right )- (-32)^{-\frac{4}{5}}$ is:

$\textbf{(A)}\ 1 \frac{13}{16} \qquad \textbf{(B)}\ 1 \frac{3}{16} \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ \frac{7}{8}\qquad\textbf{(E)}\ \frac{1}{16}$

Solution

$\frac{1}{16}a^0+\left (\frac{1}{16a} \right )^0- \left (64^{-\frac{1}{2}} \right )- (-32)^{-\frac{4}{5}}\implies \frac{1}{16}+1-\frac{1}{8}-((-32)^4)^\frac{1}{5}\implies 1-\frac{1}{16}-\frac{1}{16}\implies1-\frac{1}{8}\implies\boxed{\textbf{(D) }\frac{7}{8}}$ .

Page

Toolbox

Search

1954 AHSME Problems/Problem 6

Problem 6

Solution