**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 tankRest 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 = 1/6B’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**