1958 AHSME Problems/Problem 1: Difference between revisions

Revision as of 11:10, 3 June 2011

The value of $[2 - 3(2 - 3)^{-1}]^{-1}$ is:

$\textbf{(A)}\ 5\qquad \textbf{(B)}\ -5\qquad \textbf{(C)}\ \frac{1}{5}\qquad \textbf{(D)}\ -\frac{1}{5}\qquad \textbf{(E)}\ \frac{5}{3}$

$\frac{1}{2-3(-1)}=\frac{1}{5}$ , so the answer is $\boxed{\text{C}}$ .

1958 AHSME (Problems • Answer Key • Resources)
Preceded by First question	Followed by Problem 2
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All AHSME Problems and Solutions

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 ==Problem==
-The value of <math> [2 \minus{} 3(2 \minus{} 3)^{\minus{}1}]^{\minus{}1}</math> is:
+The value of <math> [2 - 3(2 - 3)^{-1}]^{-1}</math> is:
 <math> \textbf{(A)}\ 5\qquad
-\textbf{(B)}\ \minus{}5\qquad
+\textbf{(B)}\ -5\qquad
 \textbf{(C)}\ \frac{1}{5}\qquad
-\textbf{(D)}\ \minus{}\frac{1}{5}\qquad
+\textbf{(D)}\ -\frac{1}{5}\qquad
 \textbf{(E)}\ \frac{5}{3}</math>
 ==Solution==
-<math> \frac{1}{2\minus{}3(\minus{}1)}=\frac{1}{5}</math>, so the answer is <math>\boxed{\text{C}}</math>.
+<math> \frac{1}{2-3(-1)}=\frac{1}{5}</math>, so the answer is <math>\boxed{\text{C}}</math>.
 ==See also==
 {{AHSME box|year=1958|before=First question|num-a=2}}