Disclaimer: All the Information provided in this post are prepared & compiled by A. Praveen Kumar, SPM, Papannapet SO502303, Telangana State for in good faith of Postal Assistant Exam Aspirants. Author of blog does not accepts any responsibility in relation to the accuracy, completeness, usefulness or otherwise, of contents.
Pipes and Cisterns:
It’s based on the Time and work model.
Terms which are used in these problems are
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.
Properties:
§ If a pipe can fill a tank in x hours, then:
part filled in 1 hour = 1/x
§ If a pipe can empty a tank in y hours, then:
part filled in 1 hour = 1/y
§ If a pipe can fill a tank in x hours and another pipe can empty the tank the full tank in y hours ( where y>x), there on opening both the pipes, the net part filled in 1 hour = { 1/x – 1/y }
§ If a pipe can fill a tank in x hours and another pipe can empty the tank the full tank in y hours ( where y
E represents Total time taken to empty the tank
F represents Total time taken to fill the tank
L represents the L.C.M
e represents time taken to empty the tank
f represents time taken to fill the tank
§ Time for emptying, (emptying pipe is bigger in size.)
E = (f * e)/(f – e)
§ Time for filling, (Filling pipe is bigger in size.)
F = (e * f)/(e – f)
§ Pipes ‘A’ & ‘B’ can fill a tank in f1 hrs & f2 hrs respectively. Another pipe ‘C’ can empty the full tank in ‘e’ hrs. If the three pipes are opened simultaneously then the tank is filled in ,
F = L/[(L/f1) + (L/f2)  (L/e)]
§ Two taps ‘A’ & ‘B’ can fill a tank in ‘t1′ & ‘t2′ hrs respectively. Another pipe ‘C’ can empty the full tank in ‘e’ hrs. If the tank is full & all the three pipes are opened simultaneously. Then the tank will be emptied in,
E = L/[(L/e)  (L/f1)  (L/f2)]
§ Capacity of the tank is , F = (f * e)/(e – f)
§ A filling tap can fill a tank in ‘f’ hrs. But it takes ‘e’ hrs longer due to a leak at the bottom. The leak will empty the full tank in ,
E = [ t(f + e) * tf ] / [ t(f + e) – tf ]
Example:
1.If a pipe can fill the tank in 6 hrs but unfortunately there was a leak in the tank due to which it took 30 more minutes .Now if the tank was full how much time will it take to get emptied through the leak?
Sol: By last property,
t(f+e) = 6+0.5 =6.5hrs
tf = 6 hrs
E = 6.5*6 / (6.5 – 6)
= 78 hrs .
2.A leak in the bottom of a tank can empty the full tank in 6 hours. An inlet pipe fills water at the rate of 4 litres a minute. When the tank is full, the inlet is opened and due to the leak the tank is empty in 8 hours. The capacity of the tank (in litres) is
Sol: Work done by the inlet in 1 hour = (1/6)  (1/8) = 1/24
Work done by the inlet in 1 min = (1/24) * (1/60) = 1/1440
Volume of 1/1440 part = 4 liters.
Volume of whole = (1440 * 4) litres = 5760 litres.
Work done by the inlet in 1 min = (1/24) * (1/60) = 1/1440
Volume of 1/1440 part = 4 liters.
Volume of whole = (1440 * 4) litres = 5760 litres.
3.12 buckets of water fill a tank when the capacity of each bucket is 13.5 litres. How many buckets will be needed to fill the same tank, if the capacity of each bucket is 9 litres?
Sol: Capacity of the tank = (12 * 13.5) litres = 162 litres.
Capacity of each bucket = 9 litres
Number of buckets needed = (162/9) = 18.
Capacity of each bucket = 9 litres
Number of buckets needed = (162/9) = 18.
4.One tap can fill a cistern in 2 hours and another tap can empty the cistern in 3 hours. How long will they take to fill the cistern if both the taps are opened ?
Sol: Net part filled in 1 hour = (1/2)  (1/3) = 1/6
Cistern will be full in 6 hours 
5.A cistern can be filled in 9 hours but it takes 10 hours due to in its bottom. If the cistern is full, then the time that the leak will take to empty it, is:
Sol: Work done by the leak in 1 hour = (1/9  1/10) = 1/90.
Leak will empty the full cistern in 90 hrs
Leak will empty the full cistern in 90 hrs
6.A leak in the bottom of a tank can empty the full tank in 8 hours. An inlet pipe fills water at the rate of 6 litres 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 hod?
Sol: Work done by the inlet in 1 hour = (1/8)  (1/12) = 1/24
Work done by the inlet in 1 min = (1/24) * (1/60) = 1/1440
Volume of 1/1440 part = 6 litres
Volume of whole = (1440 x 6) litres 8640 litres
Work done by the inlet in 1 min = (1/24) * (1/60) = 1/1440
Volume of 1/1440 part = 6 litres
Volume of whole = (1440 x 6) litres 8640 litres
7.An electric pump can fill a tank in 3 hours. Because of a leak in the tank, it took 3 hours 30 min to fill the tank. The leak can drain out all the water of the tank in :.
Sol: Work done by the leak in 1 hour = (1/3)  (2/7) = 1/21 .
Leak will mpty the tank in 21 hours.
Leak will mpty the tank in 21 hours.
8.TapsA and B can fill a bucket in 12 minutes and 15 minutes respectively. If both are opened and A is closed after 3 minutes, how much further time would it take for B to fill the bucket
Sol: Part filled in 3min = 3[(1/12) + (1/15)] = 3 * (9/60) = 9/20
Remaining part = 1  (9/20) = 11/20
Part filled by B in 1 min = 1/15
(1/15) : (11/20) = 1 : x or x = (11/20) * 1 * (15/1) = 8 min 15 sec
Remaining part is filled by B in 8 ruin. 15 sec.
Remaining part = 1  (9/20) = 11/20
Part filled by B in 1 min = 1/15
(1/15) : (11/20) = 1 : x or x = (11/20) * 1 * (15/1) = 8 min 15 sec
Remaining part is filled by B in 8 ruin. 15 sec.
9. Two pipes A and B can fill a cistern in 12 minutes and 16 minutes respectively. If both the pipes are opened together, then after how much time B should be closed so that the tank is full in 9 minutes ?
Sol: Let B be closed after x minutes. Then,
Part filled by (A + B) in x min. + Part filled by A in (9 — x) min, = 1
x[(1/12) + (1/16)] + (9  x)(1/12) = 1 or (7x/48) + (9x)/12 = 1
or7x + 36 — 4x = 48 or x=4.
So, B must be closed after 4 minutes
Part filled by (A + B) in x min. + Part filled by A in (9 — x) min, = 1
x[(1/12) + (1/16)] + (9  x)(1/12) = 1 or (7x/48) + (9x)/12 = 1
or7x + 36 — 4x = 48 or x=4.
So, B must be closed after 4 minutes
10.Bucket P has thrice the capacity as bucket Q It takes 60 turns for bucket P to fill the empty drum. How many turns it wifi take for both the buckets P and Q, having each turn together to fill the empty drum
Sol: Let capacity of P be y litres. Then, capacity of Q = y/3 litres. Capacity of the drum = 60y litres.
Required number of turns = 60y /(y + (y/3)) = (60 * (3/4y) = 45
Required number of turns = 60y /(y + (y/3)) = (60 * (3/4y) = 45
11.To fill a cistern, pipes A, B and C take 20 minutes, 15 minutes and 12 minutes respectively. The time in minutes that the three pipes together will take to fill the cistern, is : .
Sol; Part filled by (A +B+ c) in 1 min. = (1/20) +(1/15) + (1/12) = 12/60 = 1/5
All the three pipes together will fill the tank in 5 min.
All the three pipes together will fill the tank in 5 min.
12. A tap can fill a tank in 16 minutes and another can empty it in8 minutes. If the tank is already half full and both the tanks are oped together, the tank will be:
Sol; Rate of waste pipe being more, the tank will be emptied when both the pipes are opened.
Net emptying work done by both in 1 min = (1/8)  (1/16) = 1/16
Now, full tank will be emptied by them in 16 min.
Half full tank will be emptied in 8 min.
Net emptying work done by both in 1 min = (1/8)  (1/16) = 1/16
Now, full tank will be emptied by them in 16 min.
Half full tank will be emptied in 8 min.
13.A tank can be filled by a tap in 20 minutes and by another tap in 6O minutes. Both the taps are kept open for 10 minutes and then the first tap is shut off. After this, the tank will be completely filled in
Sol; Part filled in 10 min = 10[(1/20) + (1/60)] = 10 * (4/60) = 2/3
Remaining part = (1  (2/3)) = 1/3
Part filled by second tap in 1 min = 1/60
(1/60) : (1/3) ∷ 1 : x
Hence, the remaining part will be filled in 20 min
Remaining part = (1  (2/3)) = 1/3
Part filled by second tap in 1 min = 1/60
(1/60) : (1/3) ∷ 1 : x
Hence, the remaining part will be filled in 20 min
14.Two taps A and B can fill a tank in 10 hours and 15 hours respectively. If both the taps are opened together, the tank will be full in:
Sol: As hours work=1/10, Bs 1 hours work = 1/15,
(A+B)s 1 hours work = (1/10) + (1/15) = 5/30 = 1/6
Both the taps can fill the tank in 6 hours.
(A+B)s 1 hours work = (1/10) + (1/15) = 5/30 = 1/6
Both the taps can fill the tank in 6 hours.
15.Two pipes can fill a tank in 10 hours 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 the tank will 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.
The tank will be full in 15/2 hrs = 7 hrs 30 min.
16.Two pipes A and B can fill a tank in 20 and 30 minutes respectively. If both the pipes are used together, then how long will it take to fill the tank?
 
 
 
17.Pipes A and B can fill a tank in 5 and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in:
 
 
18.One pipe ca fill a tank three times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes, then the slower pipe alone will be able to fill the tank in:
Sol:
 
Let the slower pipe alone fill the tank in x minutes.
 
 
 
19.A cistern can be filled by a tap in 4 hours while it can be emptied by another tap in 9 hours. If both the taps are opened simultaneously, then after how much time will the cistern get filled?
 
 

20.A tank is filled in 5 hours by three pipes A, B and C. The pipe C is twice as fast as B and B is twice as fast as A. How much time will pipe A alone take to fill the tank?
Suppose pipe A alone takes x hours to fill the tank.
 
 
 
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