Difference between revisions of "1981 AHSME Problems/Problem 3"

Revision as of 15:24, 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 $\left(D\right)$ $\frac{11}{6x}$ .

Revision as of 15:24, 21 November 2018 (view source) Hong2021 (talk \| contribs) (→‎Solution) ← Older edit		Revision as of 15:24, 21 November 2018 (view source) Hong2021 (talk \| contribs) (→‎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 <~~cmath~~>\left(D\right)</~~cmath~~> <math>\frac{11}{6x}</math>.	+	The answer is <math>\left(D\right)</math> <math>\frac{11}{6x}</math>.