Pipes A, B and C can fill a tank in 30 h, 40 h and 60 h respectively. Pipes A, B and C are opened at 7 a.m., 8 a.m., and 10 a.m., respectively on the same day. When will the tank be full?

Let the total work be 120 units.|
($$\because$$ LCM of 30, 40 and 60 be 120.)
Efficiency of pipe A = 120/30 = 4 units/hr
Efficiency of pipe B = 120/40 = 3 units/hr
Efficiency of pipe C = 120/60 = 2 units/hr
Work done by pipe A in 3 hr(7 am to 10 am) = 4 $$\times$$ 3 = 12
Work done by pipe B in 2 hr(8 am to 10 am) = 3 $$\times$$ 2 = 6
remaining work = 120 - 12 - 6 = 102 units
At 10 am, all pipes are opened. 
So, time taken by all pipes to complete the remaining work = $$\frac{112}{4 + 3 + 2} = \frac{102}{9} = 11 frac{3}{9} = 11hr 20 min
Time when tank will be filled = 10 am + 11 hr 20 min = 9 .20 p.m.
$$\therefore$$ The correct answer is option C.