Two pipes P and O can fill a tank in 12 minutes and 15 minutes respectively. If both are opened together and P is closed after 3 minutes then the time required to fill the tank (in minutes) is

Solution:

Let total work be 60 units [LCM of 12,15]

P does 60 units in 12 minutes. Hence in 1 minute P does (60/12)units = 5 units.

O does 60 units in 15 minutes. Hence in 1 minute O does (60/15)units = 4 units. 

Total work in 1 minute = 9 units

Now, for the given scenario,

First, 

P and O work for 3minutes together @ 9 units/min

Work done = (9 x 3) units = 27

Work remaining =Total units of work - work done = 60-27 =33 UNITS .

Then,

O work alone @ 4 units/min.

Time taken = (Work remaining / speed) = 33/4 mins. = 8.25 minutes

Total time = 3+8.25 = 11.25 minutes or 11$$\frac{1}{4}$$ minutes