Shortcut Trick for Problems based on Pipe and Cistern Questions

Pipe and Cistern problems are similar to time and work problems. A pipe is used to fill or empty the tank or cistern.

1. Inlet Pipe : A pipe used to fill the tank or cistern is known as Inlet Pipe.

2. Outlet Pipe : A pipe used to empty the tank or cistern is known as Outlet Pipe.


3. Pipe completed Work = Time x  Efficiency .



4. Take inlet pipe efficiency positive sign and  outlet pipe efficiency negative sign.


5. A pipe fills Tank in " n " hours , B pipe fills Tank in " m " hours.


                             
For efficiency take A and B, L.C.M. = mn

 "A" efficiency = mn/n = m

"B" efficiency = mn/m= n


Here A separately completed work = time x efficiency = mn

B separately Completed Work.         = time x efficiency = mn


6. Work done by individuals separately is always equal .





 Examples :



Ques 1.

Two taps A and B can fill a tank in 20 hours and 30 hours respectively. If both the taps are opened together, the tank will be full in ?

Solution  :

Time taken by " A" = 20 hours

Time taken by "B" = 30 hours

L.C.M of 20 and 30 = 60

A efficiency = 60/20= 3

B efficiency = 60/30 = 2


A and B working Together , so efficiency of two pipes = 3 + 2= 5


Work = 60


A and B together completed work in time of = work/efficiency

= 60/5 = 12 hours


Ques 2.

To fill a cistern, pipes A, B and C take 20 minutes, 15 minute and 12 minutes respectively. The time in minutes that the three pipes together will take to fill the cistern is


Solution  :

Time taken by " A" = 20 hours

Time taken by "B" = 15 hours

Time taken by "C" = 12 hours


L.C.M of 20,15 and 12 = 60

A efficiency = 60/20= 3

B efficiency = 60/15 = 2

C efficiency = 60/12 =5

A,B and C together fill tank in time of = work/(A+B+C Efficiency)


= 60/(3+4+5) = 60/12 = 5 min



Ques 3.

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


Solution  :


Time taken by " A" = 10 hours

Time taken by "B" = 12 hours

Time taken by "C" = 20 hours


L.C.M 10,12 and 20 is = 60

A efficiency = 60/10= 6

B efficiency = 60/12 = 5

C efficiency = 60/20 = 3


Here " C " pipe is outlet pipe ,take it's efficiency in negative = (-3)


A,B and C operating simultaneously , total efficiency = 6+5-3= 8


Time take to fill tank = 60/8 = 7 1/2 hours


Ques 4.

A cistern can be filled in 9 hours but it takes 10 hours due to a leak in its bottom. If the cistern is full, then the time that the leak will take to empty it is


Solution  :
Time taken by A = 9 hours

With leakage time taken by A = 10 hours

Take L.C.M of 9 and 10 = 90

A efficiency = 90/9 = 10

A and Leakage efficiency = 90/10 =9

A efficiency  - Leak Efficiency = 9

==>  10 - Leak Efficiency.          = 9

==> Leak Efficiency = 10-9= 1


Time taken by Leak to empty the full tank = 90/1 = 90 hours


Ques 5.

A leak in the bottom of a tank can empty the full tank in 8 hours. An inlet pipe fills water are 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 litres does the cistern hold ?


Solution  :

 Time taken by Leak = 8 hours
                                   
Time taken by Leak+Inlet = 12


L.C.M of 8 and 12 = 24

Leak Efficiency = 24/8= 3

Leak and inlet efficiency = 24/2 = 2

Work = 24

Here in two cases tank is empty so take efficiencies negative sign

Outlet + Inlet = -2

==>    -3+ Inlet = -2

==> Inlet efficiency. = -2+3= 1

Time taken by inlet to full tank = 24/1=24

Inlet delivers 6 litres in a minute

Inlet delivers in 24 hours litres = 24x60x6 = 8640 litres


Ques 6.

Two pipes fills a cistern in 15 hrs and 20 hrs respectively. The pipes are opened simultaneously and it is observed that it took 26 min more to fill the cistern because of leakage at the bottom. If the cistern is full, then in what time will the leak empty it?


Solution  :
A and B are inlet pipes , C is outlet pipe


Time taken by A = 15 hours

Time taken by B = 20 hours

A and B pipes fill tank in hours = 60/7 hours = 8 hours 34 min

According to problem A,B and  C open simultaneously its take   26 minutes more Than A and B working together

==> Time taken by A,B and C together = 8 hours + 26 min = 9 hours

work done by leak in 1 hr = work done by two pipes - 1/9

= 7 /60 – 1/9 = 3/ 540 = 1/180

Therefore, leak will empty the full cistern in 180 hours.



Note : By using same procedure you can solve " Time and Work " problems also.


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