A, B and C can all together do a piece of work in 10 days, in which B takes 3 times as long as A and C together to do the work. In how many days can B alone do the work?

40 days

If B takes 3 times time taken by A and C

Then efficiency of (A + C) = 3B (where B = efficiency of B) 

Total efficiency of the group(A + B + C) = A + C + B = 4B

Total work = efficiency x no. of days 

                = 4B x 10 = 40B

Time taken by B alone = \(\frac{total\; work}{efficiency}\)

                                 = \(\frac{40B}{B}\) = 40 days