# Pipe and Cisterns -02

Question 1

Two pipes A and B can fill a tank in 15 minutes and 20 minutes respectively. Both the pipes are opened together but after 4 minutes, pipe A is turned off. What is the total time required to fill the tank?

Sol:
Pipe A can fill the tank in 15 minutes
Part of the tank filled by Pipe A in 1 minute =1/15

Pipe B can fill the tank in 20 minutes
Part of the tank filled by Pipe A in 1 minute =1/20

Part filled in 4 minutes = 4 (1/5 + 1/20) = 7/15
Remaining part = 1 – 7/15 = 8/15

Part filled by pipe B in 1 minute = 1/20

Therefore, time taken by Pipe B to fill remaining part
1/20 * x =8/15
x = 8/15 × 1 × 20 = 10 minutes = 10 min 40 sec

Thus, the tank will be full in = 4 min + 10 min + 40 sec = 14 min 40 sec

Question 02

Two pipes A and B can fill a tank in 15 hours and 20 hours respectively while a third pipe C can empty the full tank in 25 hours. All the three pipes are opened in the beginning. After 10 hours, C is closed. In how much time, will the tank be full?

Pipe A can fill the tank in 15 hours
Part of the tank filled by Pipe A in 1 hour =1/15

Pipe B can fill the tank in 20 hours
Part of the tank filled by Pipe B in 1 hour =1/20

Pipe C can empty the tank in 25 hours
Part of the tank emptied by Pipe C in 1 hour =1/25

Part of the tank filled when all the three pipes are opened for 10 hours = 10 × ( 1/15 + 1/20 – 1/25) = 230/300 = 23/30
Remaining empty part = 1 – 23/30 = 7/30

Part of the tank filled in 1 hour by pipes A and B = 1/15 + 1/20 = 7/60
Since, 7/60 part of the tank is filled in 1 hour.

Therefore, 7/30 part is filled in 2 hours.

Thus, total time required to fill the tank = ( 10 + 2 ) = 12 hours

Question 03

A large tanker can be filled by two pipes A and B in 60 minutes and 40 minutes respectively. How many minutes will it take to fill the tanker from empty state if B is used for half the time and A and B fill it together for the other half?

Pipe A can fill the tank in 60 minutes
Part of the tank filled by Pipe A in 1 minute =1/60

Pipe B can fill the tank in 40 minutes
Part of the tank filled by Pipe B in 1 minute =1/40

Part filled by (A+B) in 1 minute = (1/60 + 1/40) = 1/24

Suppose the tank is filled in x minutes.
For x/2 time –> Both Pipe A & B is filling the tank
Rest x/2 time –> Only Pipe B is filling the tank

So, , x/2 (1/24 + 1/40) = 1

x/2 × 1/15 = 1

x = 30 minutes

Question 04

Two pipes A and B can fill a tank in 6 hours and 4 hours respectively. If they are opened on alternate hours and if pipe A is opened first, in how many hours, the tank shall be full?

A’s work in 1 hour = 1/6
B’s work in 1 hour = ¼

(A+B)’s 2 hour’s work when opened alternately = 1/6 + ¼ = 5/12

(A+B)’s 4 hour’s work when opened alternately = 10/12 = 5/6

Thus, remaining part = 1 – 5/6 = 1/6

Now, it’s A’s turn and 1/6 part is filled by A in 1 hour.

Therefore, Total time taken to fill the tank = (4+1) = 5 hours