1966 AHSME Problems/Problem 37


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 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

See also

1966 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 36
Followed by
Problem 38
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