## 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**

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 = | 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 | = | |

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 = | |

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 = | = | = | = | ||||

3 | 4 | 12 | 12 |

*By Formula,*

Here, x = 3 hours and y = 4 hoours

∴ Time taken to fill the tank = | |

x + y |

= | |

(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,

= | - | = | |||

3 | B | 3.5 |

or, | - | = | |||

3 | 3.5 | B |

or, | - | = | |||

21 | 21 | B |

or, | = | ||

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 = | 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 ) = | + | + | |||

4 | 5 | 6 |

Part filled by ( A + B + C ) = | |

60 |

= | |

60 |

∴ Time taken to fill the tank = | |

60 |

= 1 | |

27 |

*By Formula, *

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

∴ Time taken to fill the tank = | |

(yz + xz + xy) |

= | |

(5 × 6 + 4 × 6 + 4 × 5) |

= | |

(30 + 24 + 20) |

= | |

37 |

= 1 | = 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 = | |

(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 = | + | ||

8 | 32 |

= | |

32 |

= | min. | |

32 |

So, time taken by both pipes to fill the tank working together = | min. | |

32 |

*By Formula,*Here, x = 4 and t = 24

Time taken to fill the tank by both pipes together = | min. | |

(x ² - 1) |

= | |

(4^{2} - 1) |

= 32/5 min

Capacity of tank

= | - y | = | ||||

x | z |

or, | - | = y | ||

x | z |

or, c | - | = y | ||

x | z |

or, c | z - | = y | |||

xz |

∴ c = | |

(z - x) |

y = admit water

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,

= | - (10 × 60) = | ||

4 | 12 |

or, | - | = (10 × 60) | ||

4 | 12 |

or, 3c - | = 600 | |

12 |

or, | = 600 | |

12 |

or, | = 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 = | |

(z - x) |

= | |

(12 - 4) |

= | |

8 |

**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

= | + | = 1 | ||

12 | 16 |

or, | = 1 | |

48 |

or, 7x = 48 + 16

or, 7x = 64

or, x = 64/7

∴ Time required to fill the tank = | 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 = | × 2 = | ||

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 = | × 2 = | ||

6 | 3 |

Part filled by both ( A + B ) in 2 hour = | + | ||||

4 | 3 |

= | |

12 |

= 7/12

Remaining part = 1 - | = | ||

12 | 12 |

= | ||||

12 | ||||

6 |

= | |

12 |

= 2.5 hour