In two days A, B and C together can finish $$\frac{1}{2}$$ of a work and in another 2 days B and C together can finish $$\frac{3}{10}$$ part of the work. Then A alone can complete the whole work in

Let the total work to be done = L.C.M. (2,10) = 10 units

Let efficiencies of A, B and C be $$a,b$$ and $$c$$ units/day respectively.

A, B and C together can finish $$\frac{1}{2}$$ of a work (i.e. 5 units) in 2 days, => $$(a+b+c)=\frac{5}{2}=2.5$$ units/day -------------(i)

Similarly, $$(b+c)=\frac{3}{2}=1.5$$ units/day ---------(ii)

Subtracting equation (ii) from (i), we get :

=> $$a=2.5-1.5=1$$ unit/day

$$\therefore$$ Time taken by A alone to complete the work = $$\frac{10}{1}=10$$ days

=> Ans - (D)