1981 AHSME Problems/Problem 3: Difference between revisions

Revision as of 16:03, 21 November 2018

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 $\boxed\left(D\right)$ (Error compiling LaTeX. Unknown error_msg) $\frac{11}{6x}$ .

Revision as of 16:02, 21 November 2018 view source Hong2021 (talk \| contribs) 35 edits →Solution ← Older edit		Revision as of 16:03, 21 November 2018 view source Hong2021 (talk \| contribs) 35 edits →Solution Newer edit →
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	<math>\frac{6}{6x}</math> + <math>\frac{3}{6x}</math> + <math>\frac{2}{6x}</math> = <math>\frac{11}{6x}</math>		<math>\frac{6}{6x}</math> + <math>\frac{3}{6x}</math> + <math>\frac{2}{6x}</math> = <math>\frac{11}{6x}</math>

	The answer is <math>\boxed{\left(D\right)</math> <math>\frac{11}{6x}}</math>.		The answer is <math>\boxed\left(D\right)</math> <math>\frac{11}{6x}</math>.