1966 AHSME Problems/Problem 37
Contents
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 
be the number of hours needed by Alpha and Beta, working together, to do the job. Then equals:
Solution
Solution 2
Let 
,
,
denote the number of hours needed by Alpha, Beta, Gamma, respectively. We also have their respective efficiency 
, 
, and 
. Thus we get the equations
![\[\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{1}{A-6}\]](http://latex.artofproblemsolving.com/e/2/6/e26c40b89eeabe776619cbb25ba5c46c570e35b7.png)
![\[\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{1}{B-1}\]](http://latex.artofproblemsolving.com/f/3/c/f3c35188ae70a4151610aa17bef358284e15aa27.png)
![\[\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{2}{C}\Longrightarrow \frac{1}{A}+\frac{1}{B}=\frac{1}{C}\]](http://latex.artofproblemsolving.com/5/f/5/5f5070052534001d5d3275be6187807bde3f2227.png)
Equating the first 
equations gets
![\[\frac{1}{A-6}=\frac{1}{B-1}\Longrightarrow A=B-5\]](http://latex.artofproblemsolving.com/d/e/2/de20913d0661970005db06bd3af2bed5a3acfc5d.png)
Substituting the new relation along with the third equation into the first equation gets
![\[\frac{2}{A}+\frac{2}{A-5}=\frac{1}{A-6}\]](http://latex.artofproblemsolving.com/5/5/d/55d8542d0e43ec1c29da6109de68f6bd013ed73f.png)
Solving the quadratic gets 
B=A-5>0
A=\frac{20}{3}$is the only legit solution.
Thus$ (Error compiling LaTeX. Unknown error_msg)B=\frac{5}{2}
h=\frac{1}{\frac{1}{A}+\frac{1}{B}}=\frac{4}{3}$.
See also
| 1966 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 36 |
Followed by Problem 38 | |
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