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

## 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 $\boxed{\frac23 \textbf{ (A)}}$.