A1 alone can do a work in 12 days and A2 can do the same work in 15 days. With help of A3, they finish the work in 5 days. In how many days A3 alone can do the same work?

20 days

The total work = LCM of 12, 15 and 5

Efficiency of A3 = (A1+A2+A3) - A1 - A2 = 12 - 5 - 4 

                        = 3

Time taken by A3 alone to do the work = $\frac{total\;work}{efficiency\;of\;A3}$

                            = $\frac{60}{3}$  = 20 days