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

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

Latest revision as of 23:16, 9 February 2020

Problem

What is the least common multiple of ${\frac{1}{x}}$, $\frac{1}{2x}$, and $\frac{1}{3x}$ is $\frac{1}{6x}$?

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