# 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 $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$?

• Are you getting 1 h as the answer or 2 h as the answer? Dec 21, 2017 at 15:12
• @Soumee The Answer is 1 hr, right?
– user371838
Dec 21, 2017 at 15:13
• Even I got 1 hour. Book says 2 hr. Dec 21, 2017 at 15:17
• @Soumee No problem, we both got the same answer, so high probability the book is wrong.
– user371838
Dec 21, 2017 at 15:18
• right...Can you show the shortcut method for the question which was duplicate..... Dec 21, 2017 at 15:19

Hint:

Let us start from another angle.

Let the first man, the second man, the woman individually can do the job in $m_1,m_2,w$ hours respectively.

So, in one hour the first man, the second man, the woman individually can do $1/m_1,1/m_2,1/w$ part respectively of the job .

So, if the second man & the woman work together all along, in one hour they can do $1/m_1+1/m_2=?$ parts.

So, they will require $\dfrac{m_1+m_2}{m_1m_2}$ hours $\implies m_2-\dfrac{m_1+m_2}{m_1m_2}=3$

Similarly form other equations.

Can you take it from here?