Difference between revisions of "1958 AHSME Problems/Problem 1"

Latest revision as of 05:10, 3 October 2014

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 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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

@@ Line 1: / Line 1: @@
 ==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}}
+{{AHSME 50p box|year=1958|before=First question|num-a=2}}
+{{MAA Notice}}