# Difference between revisions of "1962 AHSME Problems/Problem 24"

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− | {{ | + | 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 23:01, 16 April 2014

## Problem

Three machines working together, can do a job in hours. When working alone, needs an additional hours to do the job; , one additional hour; and , additional hours. The value of is:

## Solution

Machine P takes hours, machine Q takes hours, and machine R takes hours. We also know that all three working together take 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, Obviously, the number of hours is positive, so the answer is .