Pipe And Cisterns

Pipes and Cisterns problems are almost the same as those of Time and Work Problems. Thus, if a pipe fills a tank in 6 hours, then the pipe fills 1/6th of the tank in 1 hour. The only difference with pipes and cisterns problems is that there are outlets as well as inlets. Thus, there agents(the outlets) which perform negative work too. The rest of the process is almost similar.

A pipe is connected to a cistern or tank. It is used to fill or empty the cistern; accordingly, it is called an inlet or an outlet.

Inlet

A pipe connected with full tank (or cistern or a reservoir) is called an inlet.

Outlet

A pipe which is connected to empty a tank is known as an outlet, if it empties it.

Problems on pipes and cisterns or tank are similar to problems on time and work. In pipes and cistern problems, the amount of work done is the part of the tank of filled or emptied. The time taken to do a piece of work is the time take to fill or empty a tank completely or to the desired level.

Point to Remember

• If an inlet connected to a tank fills it in x hours, part of the tank filled in one hour is = 1x
• If an outlet connected to a tank empties it in y hours, part of the tank emptied in one hour is = 1y
• An inlet can fill a tank in x hours and an outlet can empty the same tank in y hours. If both the pipes are opened at the same time and y>x, the net part of the tank filled in one hour is given by,
= (1x1y)
Therefore,
when both the pipes are open the time taken to fill the whole tank is given by,
= xyy-x Hours
If X is greater than Y, more water is flowing out of the tank than flowing into the tank. The net part of the tank emptied in one hour is given by,
= (1y1x)
Therefore, when both the pipes are open the time taken to empty the full tank is given by,
= yxx – y Hours
• An inlet can fill a tank in x hours and another inlet can fill the same tank in y hours. If both the inlets are opened at the same time, the net part of the tank filled in one hour is given by,
= (1x + 1y)
Therefore, the time taken to fill the whole tank is given by,
= yxx + y Hours Hours
In a similar way, If an outlet can empty a tank in x hours and another outlet can empty the same tank in y hours, the part of the tank emptied in one hour when both the pipes start working together is given by,
= (1x + 1y)
• If a pipe fills a tank in x hrs and another fills the same tank in y hrs, but a third one empties the full tank in z hrs, and all of them are opened together, the net part filled in 1 hr = [1x + 1y1z]
∴ Time taken to fill the tank = xyz(yz + xz – xy) hrs.

Problems related to this topic

(1). A water tank has three taps X, Y and Z. X fills four buckets in 24 minutes, Y fills 8 buckets in 1 hour and Z fills 2 buckets in 20 minutes. If all the taps are opened together, a full tank is emptied in 2 hours. If a bucket can hold 5 liters of water, what is the capacity of the tank?
Solution:
X fills 4 buckets in 24 minutes.
Thus, X fills 1 bucket in 244 = 6 minutes
Similarly,
Y fills 8 buckets in 1 hour.
Thus Y fills 1 bucket in 608 minutes
Similarly,
Z fills one bucket in 202 = 10 minutes
Thus Y fills 1 bucket in 2 hours,
No. of buckets filled by X will be = 1206 = 20 buckets
No. of buckets filled by Y will be = 120/ (60/8) = (120 * 8) / 60 = 16 buckets
No. of buckets filled by Z will be = 12010 = 12 buckets
Total number of buckets filled = (20 + 16 + 12) = 48 buckets
Total amount of water coming out of the tank = capacity of the tank = 48 * 5 liters = 240 liters


(2). Two taps X and Y can separately fill a cistern in 30 and 20 minutes respectively. They started to fill a cistern together but tap P is turned off after few minutes and tap Q fills the rest part of cistern in 5 minutes. After how many minutes was tap X turned-off?
Solution:
Let X was turned off after x min. Then, cistern filled by X in x min + cistern
filled by Y in (x+5) min = 1
x30 + (x+5)20 = 1
5x+15=60 , x = 9 min.


(3). Three fill pipes P, Q and R can fill separately a cistern in 12, 16 and 20 minutes respectively. P was opened first. After 2 minutes, Q was opened and after 2 minutes from the start of Q, R was also opened. Find the time when the cistern will be full after the opening of R?
Solution:
Let cistern will be full in x min.
Then part filled by P in x min + part filled by Q in (x-2) min + part filled by R in (x-4) min = 1
= x12 + (x-2)16 + (x-4)20 = 1
47x – 78 = 240
x = 16247 = 32147 min


(4). There is a leak in the bottom of the cistern. This leak can empty a full tank in 8 hours. When the tank is full, a tap is opened into the tank which admit 6 liters per hour and the tank is now emptied in 12 hours. What is the capacity of the cistern?
Solution:
The leak can empty the cistern in 8 hours,
In 1 hour, part of the tank emptied by the leak = 18
Also, after opening the tap, in one hour, part of the cistern emptied = 112
Let the tap can fill the cistern in x hours. Therefore,
In 1 hour, part of the cistern filled by the tap = 1x
By the question, 1x18 = 112
Or x = 24
The tap admits 6 liters of water per hour, it will admit (6 x 24) =144 liters of water in 24 hours, which should be the capacity of the tank.


(5). Three taps A,B and C can fill a tank in 10,20 and 30hours respectively. If A is open all the time and B and C are open for one hour each alternately, then the tank will be full in?
Solution:
(A+B)’s 1 hour’s work = (110 + 120) = 320
(A+C)’s 1 hour’s work = (110 + 130) = 215
Part filled in 2 hrs = (320 + 215) = 1760
Part filled in 6 hrs = (3 × 1760) = 1720
Remaining Part = (1 – 1720) = 320
Now, it is the turn of A and B and 320 part is filled by A and B in 1 hour.
Therefore,
Total time taken to fill the tank = (6+1) hrs = 7 hrs

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