A cistern can be filled by three pipes A, B and C alone 12 hrs, 24hrs and 48 hrs respectively. There is an opening D in the cistern that empties the cistern at the rate of 6m/hr. If the cistern is 96m deep then, in how much time will it be filled upto 72hrs of its depth if all the pipes are opened together at the start but B is closed after an hour?

Tank filled by A alone in 1 hr = 1/12
Tank filled by B alone in 1 hr = 1/24
Tank filled by C alone in 1 hr = 1/48
D empty the tank at the rate of 6m/hr

So,
Tank empty by D in 1hr = 6/96 = 1/16
Now, tan is to be filled up to 72m i.e., 72/96 =3/4 of tank
So,
Let the 3/4^th of the tank to be filled in 't' hours time

For 1 hr all are opened then B closed

So, for (t - 1) hr A, C and D opened
(1/12 + 1/24+ 1/48 - 1/16) + (t - 1) (1/12 + 1/48 - 1/16) = 3/4
⇒ (1/12) + (t - 1) (1/24) = 3/4
⇒ (t - 1)/24 = (3/4) - (1/12)
⇒ (t - 1)/24 = 2/3
⇒ 3t - 3 = 48
⇒ 3t = 51
⇒ t = 17 hours.