1961 AHSME Problems/Problem 1

Revision as of 12:04, 17 May 2018 by Rockmanex3 (talk | contribs) (Solution to Problem 1)

Problem 1

When simplified, $(-\frac{1}{125})^{-2/3}$ becomes:

$\textbf{(A)}\ \frac{1}{25} \qquad \textbf{(B)}\ -\frac{1}{25} \qquad \textbf{(C)}\ 25\qquad \textbf{(D)}\ -25\qquad \textbf{(E)}\ 25\sqrt{-1}$

Solution

To remove the negative exponent, flip the fraction of the base. This results in $(-125)^{2/3}$.

Then, cube root $-125$ and and square the result to get the answer of $25$, or answer choice $\boxed{\textbf{(C)}}$.

See Also

1961 AHSC (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
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All AHSME Problems and Solutions


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