# 1966 AHSME Problems/Problem 37

## Problem

Three men, Alpha, Beta, and Gamma, working together, do a job in 6 hours less time than Alpha alone, in 1 hour less time than Beta alone, and in one-half the time needed by Gamma when working alone. Let $h$ be the number of hours needed by Alpha and Beta, working together, to do the job. Then $h$ equals:

$\text{(A) } \frac{5}{2} \quad \text{(B) } \frac{3}{2} \quad \text{(C) } \frac{4}{3} \quad \text{(D) } \frac{5}{4} \quad \text{(E) } \frac{3}{4}$

## Solution

$\fbox{C}$

## Solution 2

Let $A$,$B$,$C$ denote the number of hours needed by Alpha, Beta, Gamma, respectively. We also have their respective efficiency $\frac{1}{A}$, $\frac{1}{B}$, and $\frac{1}{C}$. Thus we get the equations $$\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{1}{A-6}$$ $$\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{1}{B-1}$$ $$\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{2}{C}\Longrightarrow \frac{1}{A}+\frac{1}{B}=\frac{1}{C}$$ Equating the first $2$ equations gets $$\frac{1}{A-6}=\frac{1}{B-1}\Longrightarrow A=B+5$$ Substituting the new relation along with the third equation into the first equation gets $$\frac{2}{A}+\frac{2}{A-5}=\frac{1}{A-6}$$ Solving the quadratic gets $A=3$ or $\frac{20}{3}$. Since $B=A-5>0$, $A=\frac{20}{3}$ is the only legit solution.

Thus $B=\frac{5}{3}$ and $h=\frac{1}{\frac{1}{A}+\frac{1}{B}}=\frac{4}{3}$, $\boxed{C}$.

~ Nafer