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Aptitude::Pipes & Cistern

Home > Quantitative Aptitude > Pipes & Cistern > Subjective Solved Examples

Example 1 / 10

Two pipes A and B can fill a tank in 36 hours and 40 hours respectively. If both the pipes are opened simultaneously, how much time will be taken to fill the tank?
Time taken by pipe A alone to fill the tank= 36 hours
So, part filled by pipe A in 1 hour= (1/36)

Time taken by pipe B alone to fill the tank= 40 hours
So, part filled by pipe B in 1 hour= (1/40)

Now, if both the pipes are opened, then part filled by both in 1 hour= (1/36 + 1/40) = (19/360)

Therefore, time taken by A and B together to fill the tank= (360/19) hours = 18 hours 57 min. Ans.

Simply we have to calculate the part filled by a particular tap in unit time. This unit time can be in hour, minutes and day. After getting do LCM for both the taps and then calculate the total time.
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Example 2 / 10

Two pipes can fill a tank in 10 hours and 15 hours respectively while a third pipe empties the full tank in 25 hours. If all the three pipes operate simultaneously, in how much time will the tank be fulfilled?
Let us name the three pipes as pipe A, B and C respectively.

Time taken by pipe A alone to fill the tank= 10 hours
So, part filled by pipe A in 1 hour= (1/10)

Time taken by pipe B alone to fill the tank= 15 hours
So, part filled by pipe B in 1 hour= (1/15)

Time taken by pipe C alone to empty the tank= 25 hours
So, part emptied by pipe C in 1 hour= (1/25)

Now, if all the three pipes operate at the same time, part filled by A, B and C in 1 hour = (1/10 + 1/15 - 1/25) = (19/150)
Therefore, Time taken to fill the tank if all three pipes operate = (150/19) hours 7 hrs. 53 min.Ans.
We have done negative because the third tap is going to decrease the content of the tank. Rest as type 1
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Example 3 / 10

If two pipes function simultaneously, the reservoir will be filled in 12 hours. One pipe fills the reservoir 10 hours faster than the other. How many hours does it take the second pipe to fill the reservoir?
Let the reservoir is filled by first pipe in x hours.
Then, second pipe will fill it in (x+10) hours
Then 1/x +1/(x+10) = 1/12 ----- (eq.)
By Solving above equation, we will get x=20

So, the second pipe will take 20+10=30 hrs to fill the reservoir.Ans.


It is similar to type 1 only one difference is there that here we know the combined work of both the pipes and one of the pipe. We only need to calculate the work done by another pipe.
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Example 4 / 10

A cistern has two taps which fills it in 12 minutes and 16 minutes respectively. There is also a waste pipe in the cistern .When all the three are opened, the empty cistern is full in 20 minutes. How long will the waste pipe take to empty the full cistern?
Let us name the two taps which are filling the cistern as A and B, and the waste pipe which is emptying the cistern as C.

Time taken by A alone to fill the tank= 12 hours
So, part filled by A in 1 hour = (1/12)

Time taken by B alone to fill the tank = 16 hours
So, part filled by B in 1 hour = (1/16)

Time taken to fill the cistern when all three operates = 20 hours
Part filled in 1 hour when all three operate = (1/20)

Now, if all the three operate at the same time then we can write,
(1/A)+(1/B)-(1/C) = (1/20)
Or
(1/12)+(1/16)-(1/C) = (1/20)

By, Solving the above equation we get,
C = (240/23) hours

Therefore, the waste pipe C takes (240/23) hours to empty the cistern alone.Ans.

We just have to calculate the work done by the waste pipe in unit time. We will get negative value because it is decreasing the contents of the cistern hence negative.
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Work done by waste pipe in 1 minute = 1/20 - (1/12+1/16) = -23/240
Hence the waste pipe will empty the full cistern = 240/23 min. = 10.43 minutes Ans.
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Example 5 / 10

An electric pump can fill a tank in 4 hours. Because of a leak in the tank, it took 4.5 hours to fill the tank. If the tank is full, how much time will the leak take to empty it?
Let the time taken by the leak to empty the tank be 'L' hours.

Time taken by the electric pump to fill the tank= 4 hours
So, part filled by the electric pump in 1 hour = (1/4)

Time taken by the leak to empty the empty the tank = L hours
So, part emptied in 1 hour = (1/L)

Therefore,
(1/4) - (1/L) = 1/4.5
Solving the above equation we get,
L= 36 hours

Hence, time taken by the leak to empty the tank = 36 hours (Ans)
The catch of the question is the extra time which is taken is that amount of work done by the leak.
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Work done by the leak in in the tank in 1 hour = [1/4 - 1/4.5] = 1/36
Hence, the leak in tank will empty the tank completely = 36 hrs.(Ans)
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Example 6 / 10

Two pipes can fill a cistern in 10 and 16 hours respectively. The pipes are opened simultaneously and it is found that due to leakage in the bottom it took 28 minutes more to fill the cistern. When the cistern is full, in what time will the leak empty it?
Work done by the 2 pipes in 1 hour = 1/10+1/16 = 13/80
∴ Time taken by these pipes to fill the tank = 80/13 =6 hr 2minutes
Due to leakage, time taken = 6 hr 2 mins +28 mins= 6 hr 30 minutes
∴ Work done by (2 pipes + leak) in 1 hr =1/6.5
Work done by the leak in 1 hour = 13/80 - 1/6.5 = 9/1040
∴ Leak will empty the pipe in 115.55 hrs (Ans.)

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Example 7 / 10

Two pipes A and B can fill a tank in 40 and 45 minutes respectively. A water pipe C can empty the tank in 50 minutes. First A and B are opened. After 7 minutes, C is also opened. In how much time, the tank is full?
Time taken by pipe A to fill the tank alone = 40 mins
So, part filled by pipe A in 1 hour = (1/40)

Similarly, time taken by pipe B to fill the tank alone = 45 mins
So, part filled by pipe B in 1 hour = (1/45)

And, time taken by pipe C to empty the tank alone = 50 mins
So, part emptied by pipe C in 1 hour = (1/50)

Now, part filled by pipe A and B together = (1/40) + (1/45) = 17/360

So, part filled by both in 7 minutes = 7 x (17/360) = 119/360

Remaining part = 1-(119/360) = 241/360

Now, part filled in 1 min when A, B and C are operating at the same time = (1/40) + (1/45) - (1/50) = 49/1800

But, time taken to fill (49/1800)th part of the tank = 1 min
Or, time taken to fill 1 part of the tank = (1/(49/1800)) min = (1800/49) mins
Therefore, time taken to fill the remaining (241/360)th part of the tank = (1800/49) x (241/360)
=(1205/49) mins = 24.59 mins

Therefore, total time taken = (24.59 + 7) mins = 31.59 mins Ans.
First approach with the amount of work done by the 2 pipes in 7 minutes and after that all the three pipes will work.
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Example 8 / 10

Two pipes A and B can fill a tank in 20 min and 30 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?
Let us assume that pipe B be closed after 'x' mins.

Part filled by A and B together in 1 min = (1/20) + (1/30) = 1/12
So, part filled by A and B in 'x' mins = x/12

Now, part filled by A alone in 1 min = 1/20
So, part filled by A alone in (18-x) mins = (18-x)/20

According to the question:
(x/12) + (18-x)/20 = 1 [Since, total work done is equal to 1]

By, solving the above equation we get:
=> x = 3 mins

Therefore, pipe B should be closed after 3 mins. Ans.
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Example 9 / 10

A tap supplies 8 liters of water per minute into a cistern. A leak at the bottom of the cistern can empty the cistern in 10 hours. A full tank with the tap open is emptied by the leak in 15 hours. What is the capacity of the tank?
Tank can be emptied by leak in 1 hour = 1/10
or
Time required to emptied 1/10 part of tank = 1 hour
Given full tank can be emptied by leak = 15 hours
or
Time need to emptied 1/15 part of tank by leak = 1 hours
so (1/10 - 1/15) part can be filled by tap = 1 hours
=> 1/30 part can be filled by tap = 1 hours or
Time needed to fill the cistern by tap = 30 hours = 1800 minutes
So, Capacity of cistern = 8 * 1800 = 14400 liters (Ans.)
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Example 10 / 10

Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the tank is:
Time taken by A, B and C to fill the tank together = 6 hours

So, part filled by A, B and C in 1 hour = 1/6
Or
part filled by A, B and C in 2 hours = 2 x (1/6) = 1/3

Remaining part = 1 - (1/3) = 2/3

Now, part of the tank filled by A and B in 7 hours = 2/3
Or
part of the tank filled by A and B In 1 hour = (2/3) / 7 = 2/21


[Part filled by (A+B+C) in 1 hour] - [Part filled by (A+B) in 1 hour] = C's 1 hour work
Or
C's 1-hour work = (1/6) - (2/21) = 1/14

Therefore, time taken by C alone to fill the tank = 14 hours Ans.
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