A and B can together do a piece of work in 10 days. If A works with twice of his efficiency and B works with an efficiency 1/3rd less than his efficiency, then the work gets completed in 6 days. In how many days can A and B do the work alone respectively?

Let total work to be done = 10 units

A and B can together do a piece of work in 10 days, => (A+B)'s efficiency = $$\frac{10}{10}=1$$ unit/day

Let A's efficiency = $$x$$ units/day and B's efficiency = $$y$$ units/day

=> $$x+y=1$$ -----------(i)

If A works with twice of his efficiency, => A's new efficiency = $$2x$$ units/day

Similarly, B's new efficiency = $$y-\frac{y}{3}=\frac{2y}{3}$$ units/day

Now, the work is completed in 6 days, => $$(2x+\frac{2y}{3})\times6=10$$

=> $$6x+2y=\frac{10}{2}=5$$ -------------(ii)

Solving equations (i) and (ii), => $$x=\frac{3}{4}$$ and $$y=\frac{1}{4}$$

$$\therefore$$ Time taken by A alone to complete the work = $$\frac{10}{\frac{3}{4}}=\frac{40}{3}$$ days

and time taken by B = $$\frac{10}{\frac{1}{4}}=40$$ days

=> Ans - (A)