Difference between revisions of "1961 AHSME Problems/Problem 1"

(Solution to Problem 1)
m (Problem 1)
 
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== Problem 1==
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== Problem==
  
 
When simplified, <math>(-\frac{1}{125})^{-2/3}</math> becomes:
 
When simplified, <math>(-\frac{1}{125})^{-2/3}</math> becomes:
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\textbf{(C)}\ 25\qquad
 
\textbf{(C)}\ 25\qquad
 
\textbf{(D)}\ -25\qquad
 
\textbf{(D)}\ -25\qquad
\textbf{(E)}\ 25\sqrt{-1}</math>  
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\textbf{(E)}\ 25\sqrt{-1}</math>
  
 
==Solution==
 
==Solution==

Latest revision as of 12:04, 17 May 2018

Problem

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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
All AHSME Problems and Solutions


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