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1981 AHSME Problems/Problem 3

Revision as of 15:24, 21 November 2018 by Hong2021 (talk | contribs) (→‎Solution)

Solution

The least common multiple of ${\frac{1}{x}}$, $\frac{1}{2x}$, and $\frac{1}{3x}$ is $\frac{1}{6x}$.

$\frac{1}{x}$ = $\frac{6}{6x}$, $\frac{1}{2x}$ = $\frac{3}{6x}$, $\frac{1}{3x}$ = $\frac{2}{6x}$.

$\frac{6}{6x}$ + $\frac{3}{6x}$ + $\frac{2}{6x}$ = $\frac{11}{6x}$

The answer is \[\left(D\right)\] $\frac{11}{6x}$.

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