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1962 AHSME Problems/Problem 24: Difference between revisions

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==Solution==
==Solution==
{{solution}}
Machine P takes <math>x+6</math> hours, machine Q takes <math>x+1</math> hours, and machine R takes <math>2x</math> hours.
We also know that all three working together take <math>x</math> hours.
Now the time it takes for all three machines to complete the job is the harmonic mean of the times the three machines take individually; that is,
<cmath>x=\frac1{\frac1{x+6}+\frac1{x+1}+\frac1{2x}}</cmath>
<cmath>\frac{x}{x+6}+\frac{x}{x+1}+\frac{x}{2x}=1</cmath>
<cmath>\frac{x}{x+6}+\frac{x}{x+1}=\frac12</cmath>
<cmath>2x(x+1)+2x(x+6)=(x+1)(x+6)</cmath>
<cmath>2x^2+2x+2x^2+12x=x^2+7x+6</cmath>
<cmath>3x^2+7x-6=0</cmath>
<cmath>(3x-2)(x+3)=0</cmath>
<cmath>x\in\{\frac23, -3\}</cmath>
Obviously, the number of hours is positive, so the answer is
<math>\boxed{\frac23 \textbf{ (A)}}</math>.

Latest revision as of 22:01, 16 April 2014

Problem

Three machines $\text{P, Q, and R,}$ working together, can do a job in $x$ hours. When working alone, $\text{P}$ needs an additional $6$ hours to do the job; $\text{Q}$, one additional hour; and $R$, $x$ additional hours. The value of $x$ is:

$\textbf{(A)}\ \frac{2}3\qquad\textbf{(B)}\ \frac{11}{12}\qquad\textbf{(C)}\ \frac{3}2\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3$

Solution

Machine P takes $x+6$ hours, machine Q takes $x+1$ hours, and machine R takes $2x$ hours. We also know that all three working together take $x$ hours. Now the time it takes for all three machines to complete the job is the harmonic mean of the times the three machines take individually; that is, \[x=\frac1{\frac1{x+6}+\frac1{x+1}+\frac1{2x}}\] \[\frac{x}{x+6}+\frac{x}{x+1}+\frac{x}{2x}=1\] \[\frac{x}{x+6}+\frac{x}{x+1}=\frac12\] \[2x(x+1)+2x(x+6)=(x+1)(x+6)\] \[2x^2+2x+2x^2+12x=x^2+7x+6\] \[3x^2+7x-6=0\] \[(3x-2)(x+3)=0\] \[x\in\{\frac23, -3\}\] Obviously, the number of hours is positive, so the answer is $\boxed{\frac23 \textbf{ (A)}}$.

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