1981 AHSME Problems/Problem 3

Problem

For $x\neq0$ , $\dfrac{1}{x}+\dfrac{1}{2x}+\dfrac{1}{3x}$ equals

$\textbf{(A)}\ \dfrac{1}{2x}\qquad\textbf{(B)}\ \dfrac{1}{6x}\qquad\textbf{(C)}\ \dfrac{5}{6x}\qquad\textbf{(D)}\ \dfrac{11}{6x}\qquad\textbf{(E)}\ \dfrac{1}{6x^3}$

Solution

The least common multiple of $x$ , $2x$ , and $3x$ is $6x$ , so $\frac{1}{x} + \frac{1}{2x} + \frac{1}{3x} = \frac{6}{6x}$ + $\frac{3}{6x}$ + $\frac{2}{6x} = \boxed{\left(\mathbf{D}\right) \frac{11}{6x}}$ .

1981 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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All AHSME Problems and Solutions

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1981 AHSME Problems/Problem 3

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