# RS Aggarwal Quantitative Aptitude PDF Free Download: PIPES AND CISTERNS

**PIPES AND CISTERNS**

__IMPORTANT FACTS ____AND FORMULAE__

__IMPORTANT FACTS__

__AND FORMULAE__

__ __

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

**Outlet:** A pipe connected with a tank or a cistern or a 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 full 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, the net part filled in 1 hour = (1/x)-(1/y)

(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 = (1/y)-(1/x)

# SOLVED EXAMPLES

__ __

** Ex. 1:Two pipes ****A and B can fill a tank ****in 36 bours and 46 bours respectively. ****If both the pipes are opened simultaneously, bow mucb time will be taken to fill the**

**tank?**

** **

**Sol:** Part filled by A in 1 hour = (1/36);

Part filled by B in 1 hour = (1/45);

Part filled by (A + B) In 1 hour =(1/36)+(1/45)=(9/180)=(1/20)

Hence, both the pipes together will fill the tank in 20 hours.

**Ex. 2: Two pipes can fill a tank in 10hours and 12 hours respectively while a third, pipe empties the full tank in 20 hours. If all the three pipes operate simultaneously, in how much time will the tank be filled?**

**Sol:** Net part filled In 1 hour =(1/10)+(1/12)-(1/20)=(8/60)=(2/15).

The tank will be full in __15/2 __hrs = 7 hrs 30 min.

**Ex. 3: If two pipes function simultaneously, tbe reservoir will ****be filled in 12 hours. One pipe fills the reservoir 10 hours faster than tbe otber. How many hours ****does it take the second pipe to fill the reservoir?**

**Sol**:let the reservoir be filled by first pipe in x hours.

Then ,second pipe fill it in (x+10)hrs.

Therefore (1/x)+(1/x+10)=(1/12) ó(x+10+x)/(x(x+10))=(1/12).

ó x^2 –14x-120=0 ó (x-20)(x+6)=0

óx=20 [neglecting the negative value of x]

so, the second pipe will take (20+10)hrs. (i.e) 30 hours to fill the reservoir

Ex. 4: A cistern has two taps which fill it in 12 minutes and 15minutes respectively. There is also a waste pipe in the cistern. When all the 3 are opened , the empty cistern is full in 20 minutes. How long will the waste pipe take to empty the full cistern?

**Sol**: Workdone by the waste pipe in 1min

=(1/20)-(1/12)+(1/15) = -1/10 [negative sign means emptying]

therefore the waste pipe will empty the full cistern in 10min

**Ex. 5: An electric pump can fill a tank in 3 hours. Because ****of a leak in ,the tank it took 3(1/2) hours to fill the tank. If the tank is full, how much time will the leak take **

**to empty it ?**

**Sol**: work done by the leak in 1 hour=(1/3)-(1/(7/2))=(1/3)-(2/7)=(1/21).

The leak will empty .the tank in 21 hours.

**Ex. 6. Two pipes can fill a cistern in 14 hours and 16 hours respectively. The pipes **

**are opened simultaneously and it is found that due to leakage in the ****bottom it tooki 32 minutes more to fill the cistern.When the cistern is full, in what time will the leak empty it?**

** **

**Sol:** Work done by the two pipes in 1 hour =(1/14)+(1/16)=(15/112).

Time taken by these pipes to fill the tank = (112/15) hrs = 7 hrs 28 min.

Due to leakage, time taken = 7 hrs 28 min + 32 min = 8 hrs

Work done by (two pipes + leak) in 1 hour = (1/8).

Work done by the leak m 1 hour =(15/112)-(1/8)=(1/112).

Leak will empty the full cistern in 112 hours.

** **

Ex. 7: Two pipes A and B can fill a tank in 36 min. and 45 min. respectively. A water pipe C can empty the tank in 30 min. First A and B are opened. after 7 min,C is also opened. In how much time, the tank is full?

**Sol:**Part filled in 7 min. = 7*((1/36)+(1/45))=(7/20).

Remaining part=(1-(7/20))=(13/20).

Net part filled in 1min. when A,B and C are opened=(1/36)+(1/45)-(1/30)=(1/60).

Now,(1/60) part is filled in one minute.

(13/20) part is filled in (60*(13/20))=39 minutes.

Ex.8: Two pipes A,B can fill a tank in 24 min. and 32 min. respectively. If both the pipes are opened simultaneously, after how much time B should be closed so that the tank is full in 18 min.?

**Sol**: let B be closed after x min. then ,

Part filled by (A+B) in x min. +part filled by A in (18-x)min.=1

Therefore x*((1/24)+(1/32))+(18-x)*(1/24)=1 ó (7x/96) + ((18-x)/24)=1.

ó 7x +4*(18-x)=96.

Hence, be must be closed after 8 min.