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

Revision as of 16:24, 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 $\left(D\right)$ $\frac{11}{6x}$ .

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 ==Solution==
-The least common multiple of <math>\displaystyle{\frac{1}{x}}</math>, <math>\frac{1}{2x}</math>, and <math>\frac{1}{3x}</math> is <math>\frac{1}{6x}</math>.
+The least common multiple of <math>{\frac{1}{x}}</math>, <math>\frac{1}{2x}</math>, and <math>\frac{1}{3x}</math> is <math>\frac{1}{6x}</math>.
 <math>\frac{1}{x}</math> = <math>\frac{6}{6x}</math>, <math>\frac{1}{2x}</math> = <math>\frac{3}{6x}</math>, <math>\frac{1}{3x}</math> = <math>\frac{2}{6x}</math>.
@@ Line 7: / Line 7: @@
 <math>\frac{6}{6x}</math> + <math>\frac{3}{6x}</math> + <math>\frac{2}{6x}</math> = <math>\frac{11}{6x}</math>
-The answer is (D) <math>\frac{11}{6x}</math>.
+The answer is <cmath>\left(D\right)</cmath> <math>\frac{11}{6x}</math>.