Time and work. Two men and a woman working. > Two men and a woman are entrusted with a task. The second man needs three hours more to cope with the job than the second man and the woman would need working together. The first man, working alone, would need as much time as the second man and the woman working together.The first man working alone, would spend eight hours less than double the period of time the second man would spend working alone. How much time would the two men and the woman need to complete the task if they all worked together? What I have tried: Let the efficiencies of A,B,C(two man and a woman) be a work/h, b work/h, c work/h. Therefore: $$b.(t_1+3)=(a+c)*t_1$$ $$a.t_2=(b+c)t_2$$ $$a.(2t_3-8)=b.t_3$$

Let the efficiencies of the first man be $a$, the second be $b$ and that of the woman, $c$.

It is given that: $$a = b + c \tag{1}$$ $$\frac{1}{b} = \frac{1}{b+c} + 3 \tag{2}$$ $$\frac{1}{a}=\frac{2}{b}-8 \tag{3}$$

Solving this, can you get $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$?