## Aptitude-pipes & Cistern Problem

1. A leak in the bottom of a tank can empty the full tank in 8 hours. An inlet pipe fills water at the rate of 6 liters a minute. When the tank is full, the inlet is opened and due to the leak, the tank is empty in 12 hours. How many liters does the cistern hold?

Explanation:

Work done by the inlet in 1 hour = ( 1 / 8 ) – ( 1 / 12 )

= ( 3 – 2 ) / 24

= 1 / 24

Work done by the inlet in 1 min = ( 1 / 24 ) * ( 1 / 60 )

= 1 / 1440

Given, volume of ( 1 / 1440 ) part = 6 liters

Volume of whole = 6 * 1440

= 8640 liters

2. A cistern can be filled by a tap in 4 hours while it can be emptied by another tap in 9 hours. If both the taps are opened simultaneously, then after how much time will the cistern get filled?

Explanation:

Net part filled in 1 hour = ( 1 / 4 ) – ( 1 / 9 )

= ( 9 – 4 ) / 36

= ( 5 / 36 )

The cistern will be filled in 36/5 hours.

= 7.2 hours

3. Three pipes A, B and C can fill a tank from empty to full in 30 minutes, 20 minutes and 10 minutes respectively. When the tank is empty, all the three pipes are opened. A, B and C discharge chemical solutions P, Q and R respectively. What is the proportion of solution R in the liquid in the tank after 3 minutes?

Explanation:

Part filled by ( A + B + C ) = (( 1 / 30 ) + ( 1 / 20 ) + ( 1 / 10 ))

= 11 / 20

Part filled by C in 3 minutes = 3 / 10

Required ratio = ( 3 / 10 ) * ( 20 / 11 )

= ( 3 * 2 ) / 11

= 6 / 11

4. Two pipes A and B can fill a cistern in 12 minutes and 15 minutes respectively while a third pipe C can empty the full tank in 6 minutes. A and B are kept open for 5 minutes in the beginning and then C is also opened. In what time is the cistern emptied?

Explanation:

Part filled in 5 min = 5 * (( 1 / 12 ) + ( 1 / 15 ))

= 5 * (( 5 + 4 ) / 60 )

= 5 * ( 9 / 60 )

= 3 / 4

Part emptied in 1 min when all the pipes are opened = ( 1 / 6 ) – (( 1 / 12 ) + ( 1 / 15 ))

= 1 / 60

Now, ( 1 / 60 ) part is emptied in 1 min.

( 3 / 4 ) part will be emptied in 60 * ( 3 / 4 ) = 15 * 3

= 45 min