Three men, Alpha, Beta, 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?

Through the use of some algerbaic gymnastics, I arrived at some solutions.

\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{2}{c}
\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}

Solving this conglomeration, I arrived at:

a=\frac{20}{3}, \;\ b=\frac{5}{3}, \;\ c=\frac{4}{3}

Since a and b work together, a+b must be less than a or b working alone.

Therefore, \frac{3}{20}+\frac{3}{5}=\frac{3}{4}

I am sure there is an easier method, but the main thing is... does anyone concur?