# Pipes and Cisterns

Posted on 4:11 AM in

PIPES & CISTERNS

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.

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

(i) Part filled in 1 hour = 1/x
(ii) If a pipe can empty a full tank in y hours, then: Part emptied in 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].

SOLVED PROBLEMS

Ex. 1. Two pipes A and B can fill a tank in 36 hours and 45 hours respectively. If both the pipes are opened simultaneously, how much 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. A pipe can fill a tank in 16 hours. Due to a leak in the bottom,it is filled in 24 hours. If the tank is full, how much time will the leak take to empty it ?
Sol. Work done by the leak in 1 hour = [ 1/16 – 1/24 ]=1/48
:. Leak will empty the full cistern in '48 hours.

Ex. 3. A cistern is filled by pipe A in 10 hours and the full cistern can be leaked out by an exhaust pipe B in 12 hours. If both the pipes are opened, in what time the cistern is full?
Sol. Work done by the inlet in 1 hour: = 1/10
Work done by the outlet in 1 hour = 1/12
Net part filled in 1 hour = [ 1/10 – 1/12 ] = 1/60
The cistern will be full in 60 hours.

Ex. 4. 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, 32 minutes extra are taken for the cistern to be filled up. 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 hours
= (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. 5. Pipes A and B can fill a tank in 20 hours and 30 hours respectively and pipe C can empty the full tank in 40 hours. If all the pipes are opened together, how much time will be needed to make the tank full ?

Sol. Net part filled m 1 hour = [1/20 + 1/30 – 1 /40] = 7/120
The tank will be full in 120/7 = 17 ½

Ex. 6. Two pipes A and B can frll 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 minutes?
Sol. Let B be closed after x minutes. Then,
part filled by (A + B) in x min. + part filled by A in (18 - x) min. = 1.
X {1/24 + 1/32] + (18 – x) X 1/24 = 1 or 7x/96 + (18 – x)/24 = 1
7x + 4(18 –x) = 96 or x=8
Hence, B must be closed after 8 minutes

Ex. 7. Two pipes A and B can fill a tank in 36 min. and 4S min. respectively. A water pipe C can empty the tank in 30 min. First A and B are opened. After 7 minutes, C is 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 1 min. When A,B and C are opened

= [ 1/36 + 1/45 + 1/30 ] = 1/60
Now, 1/60 part is filled in 1 min.

13/20 part is filled in [60x13/20] = 39 min
Total time taken to fill the tank = (39 + 7) min. = 46 min.