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 ${\frac{1}{x}}$ , $\frac{1}{2x}$ , 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 $\boxed{\left(D\right) \frac{11}{6x}}$ .

1981 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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

Page

Toolbox

Search

1981 AHSME Problems/Problem 3

Problem

Solution

See Also