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

Revision as of 11:04, 17 May 2018

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

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

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 $\boxed{\textbf{(C)}}$ .

1961 AHSC (Problems • Answer Key • Resources)
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

@@ Line 1: / Line 1: @@
-When simplified <math>(-\frac{1}{125})^{-\frac{2}{3}}</math> becomes:
+== Problem 1==
+When simplified, <math>(-\frac{1}{125})^{-2/3}</math> becomes:
+<math>\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}</math>
+==Solution==
+To remove the negative exponent, flip the fraction of the base.  This results in <math>(-125)^{2/3}</math>.
+Then, cube root <math>-125</math> and and square the result to get the answer of <math>25</math>, or answer choice <math>\boxed{\textbf{(C)}}</math>.
+==See Also==
+{{AHSME 40p box|year=1961|before=First Question|num-a=2}}
 {{MAA Notice}}