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1981 AHSME Problems/Problem 3

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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 (D) $\frac{11}{6x}$.

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