1991 AHSME Problems/Problem 3

Problem

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

(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
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