Two pipes A and B can fill a tank independently in 20 and 30 minutes respectively. Both the pipes are opened simultaneously and after 6 minutes tap B is closed. Then the total time taken, in minutes,to fill the tank, is (assume that the tank is emptyinitially)

Solution:

Let total work be 60 units [LCM of 20,30]

A does 60 units in 20 minutes. Hence in 1 minute A does (60/20)units = 3 units.

B does 60 units in 30 minutes. Hence in 1 minute B does (60/30)units =2 units.

Total work in 1 minute = 5 units

Now, for the given scenario,

First,

A and B work for 6 minutes together @ 5 units/min

Work done = (5 x 6) units = 30

Work remaining =Total units of work - work done = 60-30 =30 UNITS .

Then,

A work alone @ 3 units/min.

Time taken = (Work remaining / speed) = 30/3 mins. = 10 minutes

Total time = 6  +10 = 16  minutes