Pipes A, B and C can fill a tank in 15, 30 and 40 hours, respectively. Pipes A, B and C are opened at 6 a.m., 8 a.m. and 10 a.m., respectively, on the same day. When will the tank be full?

3:20 p.m.

Capacity of the tank = LCM of 15, 30, 40

From 6a.m. to 8a.m., tap A works for 2hrs

Capacity of tank after two hrs = time x efficiency 

                                            = 2 x 8 = 16

From 8a.m. to 10a.m. A and B work together,

Capacity at 10a.m. = 16 units(from 6-8) + time(2hrs) x eff.(A+B)

                             = 16 + (2 x 12) = 16 + 24 = 40

Capacity to be filled after this = 120 - 40 = 80

This will be done by A, B, C together in $\frac{80}{eff,\;of\;A\;+\;B\;+\;C}$ 

                                = $\frac{80}{15}$ hrs = 5$\frac{5}{15}$ hrs

Thus, the tank will be filled at 3:20p.m.