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

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==Solution==
 
==Solution==
  
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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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The least common multiple of <math>x</math>, <math>2x</math>, and <math>3x</math> is <math>6x</math>, so <math>\frac{1}{x} + \frac{1}{2x} + \frac{1}{3x} = \frac{6}{6x}</math> + <math>\frac{3}{6x}</math> + <math>\frac{2}{6x} = \boxed{\left(\mathbf{D}\right) \frac{11}{6x}}</math>.
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<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>.
 
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<math>\frac{6}{6x}</math> + <math>\frac{3}{6x}</math> + <math>\frac{2}{6x}</math> = <math>\frac{11}{6x}</math>
 
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The answer is <math>\boxed{\left(D\right) \frac{11}{6x}}</math>.
 
  
 
== See Also ==
 
== See Also ==
 
{{AHSME box|year=1981|num-b=2|num-a=4}}
 
{{AHSME box|year=1981|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 13:59, 29 June 2025

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 $x$, $2x$, and $3x$ is $6x$, so $\frac{1}{x} + \frac{1}{2x} + \frac{1}{3x} = \frac{6}{6x}$ + $\frac{3}{6x}$ + $\frac{2}{6x} = \boxed{\left(\mathbf{D}\right) \frac{11}{6x}}$.

See Also

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

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