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Difference between revisions of "1954 AHSME Problems/Problem 6"

(Created page with "== Problem 6== The value of <math>\frac{1}{16}a^0+\left (\frac{1}{16a} \right )^0- \left (64^{-\frac{1}{2}} \right )- (-32)^{-\frac{4}{5}}</math> is: <math>\textbf{(A)}\ 1 ...")
 
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== Solution ==
 
== Solution ==
−
<math>\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)^{\frac{1}{5}})^4\implies 1-\frac{1}{16}-(-2)^4\implies 1-16-\frac{1}{16}\implies -15-\frac{1}{16}\implies\frac{-241}{16}</math>, which is not one of the answer options
+
<math>\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}}</math>.

Revision as of 14:59, 23 June 2018

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

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

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