# Pipes and Cistern Questions

**FACTS AND FORMULAE FOR PIPES AND CISTERN QUESTIONS**

**1. Inlet :** A pipe connected with a tank or a cistern or a reservoir, that fills it, is known as an inlet.

**2. ****Outlet :** A pipe connected with a tank or cistern or reservoir, emptying it, is known as an outlet.

**(i)** If a pipe can fill a tank in x hours, then:

part filled in 1 hour = 1/x

**(ii)** If a pipe can empty a tank in y hours, then:

part emptied in 1 hour = 1/y

**(iii)** If a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours (where y > x), then on opening both the pipes, then the net part filled in 1 hour = $\left(\frac{1}{x}-\frac{1}{y}\right)$

**(iv)** If a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours (where x > y), then on opening both the pipes, the net part emptied in 1 hour = $\left(\frac{1}{y}-\frac{1}{x}\right)$

A) 10 min. 20 sec. | B) 11 min. 45 sec. |

C) 12 min. 30 sec. | D) 14 min. 40 sec. |

Explanation:

Part filled in 4 minutes =4(1/15+1/20) = 7/15

Remaining part =(1-7/15) = 8/15

Part filled by B in 1 minute =1/20 : 8/15 :: 1:x

x = (8/15*1*20) = $10\frac{2}{3}min=10min40sec$

The tank will be full in (4 min. + 10 min. + 40 sec.) = 14 min. 40 sec

A) 6 min.to empty | B) 6 min.to fill |

C) 9 min.to empty | D) 9 min.to fill |

Explanation:

Clearly,pipe B is faster than pipe A and so,the tank will be emptied.

part to be emptied = 2/5

part emptied by (A+B) in 1 minute=$\left(\frac{1}{6}-\frac{1}{10}\right)=\frac{1}{15}$

$\frac{1}{15}:\frac{2}{5}::1:x$

$\frac{2}{5}*15=6mins$

so, the tank will be emptied in 6 min

A) 3 hrs 15 min | B) 3 hrs 45 min |

C) 4 hrs 15 min | D) 4 hrs 1 |

Explanation:

Time taken by one tap to fill **half of the tank** = 3 hrs.

Part filled by the four taps in 1 hour =4*1/6 =2/3

Remaining part =$\left(1-\frac{1}{2}\right)=\frac{1}{2}$

$\frac{2}{3}:\frac{1}{2}::1:x$

=> x = $\left(\frac{1}{2}*1*\frac{3}{2}\right)=\frac{3}{4}$

So, total time taken = 3 hrs. 45 mins.

A) 7 hours | B) 8 hours |

C) 12 hours | D) 14 hours |

Explanation:

If the total area of pump=1 part

The pumop take 2 hrs to fill 1 part

The pumop take1 hour to fill 1/2 portion

Due to lickage

The pumop take 7/3 hrs to fill 1 part

The pumop take1 hour to fill 3/7 portion

Now the difference of area = (1/2-3/7)=1/14

This 1/14 part of water drains in 1 hour

Total area=1 part of water drains in (1x14/1)hours= 14 hours

So the leak can drain all the water of the tank in 14 hours.

Leak will empty the tank in 14 hrs

A) 20 hrs | B) 28 hrs |

C) 36 hrs | D) 40 hrs |

A) 10 | B) 12 |

C) 14 | D) 16 |

Explanation:

Part filled in 2 hours = 2/6=1/3

Remaining part =$\left(1-\frac{1}{3}\right)=\frac{2}{3}$

(A + B)'s 7 hour's work = 2/3

(A + B)'s 1 hour's work = 2/21

C's 1 hour's work = { (A + B + C)'s 1 hour's work } - { (A + B)'s 1 hour's work }

=1/6-2/21 = 1/14

C alone can fill the tank in 14 hours.

A) 6 hrs | B) 20/3 hrs |

C) 7 hrs | D) 15/2 hrs |

Explanation:

$\left(A+B\right)\text{'}s1hourwork=\left(\frac{1}{12}+\frac{1}{15}\right)=\frac{9}{60}=\frac{3}{20}$

$\left(A+C\right)\text{'}s1hourwork=\left(\frac{1}{12}+\frac{1}{20}\right)=\frac{8}{60}=\frac{2}{15}$

$partfilledin2hrs=\left(\frac{3}{20}+\frac{2}{15}\right)=\frac{17}{60}$

$partfilledin6hrs=\left(3*\frac{17}{60}\right)=\frac{17}{20}$

$remainingpart=\left(1-\frac{17}{20}\right)=\frac{3}{20}$

Now, it is the turn of A and B (3/20) part is filled by A and B in 1 hour.

Therefore, Total time taken to fill the tank =(6+1)hrs= 7 hrs

A) 50 m^3/min | B) 60 m^3/min |

C) 72 m^3/min | D) None of these |

Explanation:

Let the filling capacity of the pump be x ${m}^{3}$/min.

Then, emptying capacity of the pump=(x+10)${m}^{3}$/min.

so,$\frac{2400}{x}-\frac{{\displaystyle 2400}}{{\displaystyle x+10}}=8\iff {x}^{2}+10x-3000=0$

$\Rightarrow \left(x-50\right)+\left(x+60\right)=0\iff x=50$