## Pipes and Cistern

Problems related to Pipes and Cistern are the same as those of time and work.

Inlet pipe:- A pipe which fills up the tank is known as inlet.
Outlet pipe:- A pipe which empties a tank is known as outlet.

Important Points
1.  If a pipe can fill a tank in x hours, then the part of tank filled in 1 hour = 1 x

Example:

 A pipe can fill the tank in 5 hours, then the tank filled in 1 hour = 1 5

2. If a pipe can fill 1/x part in 1 hour, then it can fill the whole tank in x hours.

Example: If a pipe can fill 1/3 part of a tank in 1 h, then it can fill the whole tank in 3 h.
3. Time taken to fill a tank is taken positive and time taken to empty a tank is taken negative.

### Important Formulas

Rule 1: If a pipe can fill a tank in x hours and an another pipe can fill the same tank in y hours, then the

 time taken to fill the tank when both pipes are opened  = xy hours X + y

Example: Two pipes A and B can fill a tank in 9 and 18 hours, respectively. If both the pipes are opened simultaneously, how much time will be taken to fill the tank ?
Solution: Part filled by A in 1 hour = 1/9
Part filled by B in 1 hour = 1/18

 Part filled by ( A + B ) in 1 hour = 1 + 1 = (2 + 1) = 3 = 1 9 18 18 18 6
 Hence, both the pipes together will fill the tank = 1 = 6 hours 1 6

By Formula,
Here, x = 9 hours and y = 18 hours

 ∴ Time taken to fill the tank = xy x + y

 = (9 × 18) = 9 × 18 = 6 hours 9 + 18 27

Rule 2: If a pipe fill a tank in x hours, and an another pipe can empty in y hours, then the

 time taken to fill the tank when both pipes are opened  = xy x + y

Example: A pipe can fill a tank in 3 hours, while another pipe can empty it in 4 hours. If both the pipes are opened simultaneously, how much time will be taken to fill the tank ?

Solution:
Part filled by 1st pipe in 1 hours = 1/3
Part filled by 2nd pipe in 1 hours = 1/4
 Part filled in 1 h by both pipes = 1 = 1 = (4 - 3) = 1 3 4 12 12
Hence, the tank will be filled completely in = 12 hours

By Formula,

Here, x = 3 hours and y = 4 hoours
 ∴ Time taken to fill the tank = xy x + y

 = (3 × 4) (4 - 3)

Example: A pipe can fill a tank in 3 hours and because of a leakage it is taking 3.5 hours to fill the tank. Find the time in which the leakage can drain all the water when it is full.
Solution: Let the lick drain all the water in B min.
Part filled by A in 1 min = 1/3
Part filled by ( A + B ) in 1 min = 1/3.5
Part drain by lick in 1 min = 1/B
According to the question,

 = 1 - 1 = 1 3 B 3.5

 or, 1 - 1 = 1 3 3.5 B

 or, 7 - 6 = 1 21 21 B
 or, 1 = 1 21 B

∴ B = 21 min
So, the leakage will drain all water in 21 min.

Rule 3: If three pipes can fill a tank separately in x, y and z hours respectively, then part of tank filled in 1 hour by all three pipes 1/x + 1/y + 1/z and total

 time taken to fill the tank is = xyz hours (x + y + z)

Example: Three pipes A, B and C can fill a tank separately in 4 hours, 5 hours and 6 hours respectively. Find the time taken by all the three pipes to fill the tank when the pipes are opened together.
Solution:-
Part filled by A in 1 hour = 1/4
Part filled by B in 1 hour = 1/5
Part filled by C in 1 hour = 1/6

 Part filled by ( A + B + C ) = 1 + 1 + 1 4 5 6
 Part filled by ( A + B + C ) = ( 15 + 12 + 10 ) 60

 = 37 60
 ∴ Time taken to fill the tank = 37 60
 =  1 23 27

By Formula,

Here, x = 4 hours , y = 5 hours and z = 6 hours

 ∴ Time taken to fill the tank = xyz (yz + xz + xy)

 = (4 × 5 × 6) (5 × 6 + 4 × 6 + 4 × 5)
 = 120 (30 + 24 + 20)

 = 60 37
 =  1 23 = hours 37

Rule 4: If two taps A and B, which can fill a tank, such that efficiency of A is x times of B and takes t min less/more than B to fill the tank, then

 Time taken to fill the tank by both pipes together = xt (x² - 1)

Example: If tap A can fill a tank in 4 times faster than tap B and takes 24 min less than tap B to fill the tank. If both the taps are opened simultaneously, then find the time taken to fill the tank.
Solution: Let the time taken by pipe A to fill the tank be x min. Then, time taken by pipe B to fill the tank be 4x min.
According to the question,
4x - x = 24
or, 3x = 24 or, x = 24/3 = 8
∴ Time taken by pipe A = 8 min
Time taken by pipe B = 4 × 8 = 32 min
Time taken to fill the tank by pipe A in 1 min = 1/8
Time taken to fill the tank by pipe B in 1 min = 1/32

 Time taken to fill the tank by pipes in 1 min = 1 + 1 8 32
 = (4 + 1) 32
 = 5 min. 32
 So, time taken by both pipes to fill the tank working together = 5 min. 32
By Formula,
Here, x = 4 and t = 24
 Time taken to fill the tank by both pipes together = xt min. (x ² - 1)
 = ( 4 × 24 ) (42 - 1)
= 96/15
= 32/5 min

### Capacity of tank

 = c -   y = c x z
 or, c - c =   y x z
 or, c 1 - 1 =   y x z
 or, c z - x =   y xz
 ∴ c = xyz (z - x)
Where, c = capacity of tank

x = time required to empty
z = new time required to empty

Example: A cistern had leakage which can empty in 4 hours. A pipe which admit 10 L of water per minute into the cistern is turn on and now the cistern is empty in 12 h, Find the capacity of tank the tank.
Solution:- Let the capacity of the tank be c litres.
According to the question,

 = c - (10 × 60) = c 4 12
 or, c - c =  (10 × 60) 4 12
 or, 3c - c =  600 12
 or, 2c =  600 12
 or, c =  600 6

or, c = 600 × 6
∴ Capacity = 3600 L

By Formula,
Here, x = 4 hours , y = 10 L per minute = 10 × 60 = 600 L per hour and z = 12 hours

 ∴ c   = xyz (z - x)
 = 4 × 600 × 12 (12 - 4)
 = 4 × 600 × 12 8
= 3600 L

Example: A pipe can fill a tank in 12 min and B can fill the same tank in 16 min. Both pipes are open together. But 4 min before the tank is full, one pipe A is closed. How much time will it take to fill the tank.
Solution: Let pipe B open for x min.
then, pipe A open for ( x - 4 ) min.
Part filled by pipe A in 1 min = 1/12 min
Part filled by pipe A in ( x - 4 ) min = ( x - 4 )/12
Part filled by Pipe B in 1 min = 1/16
Part filled by pipe B in x min = x/16

 = (x - 4) + x = 1 12 16
 or, 4 ( x - 4 ) + 3x = 1 48
or, 4x - 16 + 3x 48
or, 7x = 48 + 16
or, 7x = 64
or, x = 64/7
 ∴ Time required to fill the tank = 67 min. 7

Example: Pipe A can fill a tank in 8 hours , and pipe B in 6 hours. If both the pipes are open and after 2 hours , pipe A is closed then how much time is required to fill the tank by pipe B?
Solution: Time taken by pipe A = 8 hours
Part filled by pipe A in 1 hour = 1/8

 Part filled by pipe A in 2 hour = 1 × 2 = 1 8 4

Time taken by pipe B = 6 hours
Part filled by pipe B in 1 hour = 1/6
 Part filled by pipe B in 2 hour = 1 × 2 = 1 6 3
 Part filled by both ( A + B ) in 2 hour = 1 + 1 4 3
 = ( 3 + 4 ) 12

= 7/12
 Remaining part = 1 - 7 = 5 12 12
 = 1 12 1 6
 = (5 × 6) 12
= 5/2
= 2.5 hour