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?
Answer & explanation
Correct answer: option 1
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.