Difference between revisions of "1991 AHSME Problems/Problem 3"

Latest revision as of 16:14, 23 February 2018

$(4^{-1}-3^{-1})^{-1}=$

(A) $-12$ (B) $-1$ (C) $\frac{1}{12}$ (D) $1$ (E) $12$

$\fbox{A}$ This is $\frac{1}{\frac{1}{4} - \frac{1}{3}} = \frac{1}{\frac{-1}{12}} = -12.$

1991 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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
All AHSME Problems and Solutions

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 == Solution ==
-<math>\fbox{}</math>
+<math>\fbox{A}</math> This is <math>\frac{1}{\frac{1}{4} - \frac{1}{3}} = \frac{1}{\frac{-1}{12}} = -12.</math>
 == See also ==