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

1966 AHSME (Problems • Answer Key • Resources)
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1966 AHSME Problems/Problem 37

Contents

Problem

Solution

Solution 2

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