## Quantitative Aptitude Notes: Time and Work (Part - 4) (PIPES AND CISTERNS)

### Quantitative Aptitude Notes: Time and Work (Part - 4) (PIPES AND CISTERNS)

Inlet: A pipe connected with a tank or cistern or a reservoir, that fills it, is known as an inlet.
Outlet: A pipe connected with a tank or cistern or reservoir, emptying it is known as an outlet.
Important Concepts
a) If a pipe can fill a tank in x hours, part filled in 1 hour = 1/x.
b) If a pipe can fill a tank in x hours and another pipe in y hours, part of tank filled in 1 hour when both the pipes are opened simultaneously = (1/x + 1/y) = ( x+y)/xy
Therefore, Time taken to fill the tank by both the pipes when opened simultaneously = xy/(x+y)
c) If a pipe can empty a tank in "y" hours, then tank emptied in 1 hour = 1/y
d) If a pipe can empty a tank in y hours and another pipe in x hours, part of tank emptied in 1 hour when both the pipes are opened simultaneously = (1/x + 1/y) = (x+y)/xy
Therefore, Time taken to empty the tank by both the pipes when opened simultaneously = xy/(x+y)
e) 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 = (y - x)/xy.
Therefore, When both the pipes are opened simultaneously, time taken to fill the tank fully = xy/(y - x) hours.
f) If a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours (where x > y), then on opening both the pipes, the net part emptied in 1 hour = 1/y - 1/x = (x - y)/xy
Therefore, When both the pipes are open simultaneously, time taken to empty the tank fully = xy/(x - y) hours.
Memory Based Example Problems based on various types
1. Two pipes A and B can fill a tank in 24 hours and 30 hours separately. In the event that both the channel are opened all the while in the void tank, the amount of the truth will surface eventually taken by them to fill it?
Solution:
Part filled by A in 1 hour = 1/24, part filled by B in 1 hour = 1/30
Part filled by (A+B) in 1 hour = (1/24+ 1/30) = 9/120 = 3/40
Time taken by both to fill the tank = 40/3 hrs = 13 hrs 20 min.
2. Funnels A and B can fill a tank in 6 hours and 9 hours individually and channel C can purge it in 12 hours. On the off chance that every one of the funnels is opened together in the vacant tank in what amount of the reality of the situation will become obvious eventually be full?
Solution:
Net part filled in 1 hour = (1/6+1/9+1/12) = 7/36
Thus, the tank will be full in 36/7 hrs.
3. Two pipes that pipe 1 and pipe 2 can fill a water reservoir in 36 hours and 45 hours respectively. If both the pipes are simultaneously, how much time will be taken to fill the water reservoir?
Solution:
Here both pipes are filled so we can easily say it is a positive.
Step 1: So at first of we calculate fill pipe in 1 hour time taken
Pipe 1 + Pipe 2 together filled a water reservoir in 1 hour = (1 / 36 + 1 / 45) = 9 / 180 = 1 / 20
Step 2: So in 1 hour it fill with 1 / 20
Hence time taken both the pipe 1 and pipe 2 will fill the water reservoir in 20 hours.
4. There is two types Pipes, Pipe 1 and Pipe 2 can fill a water reservoir in 3 and 6 hours respectively. Pipe 3 can empty it in 12 hours. If all the three pipes are opened together, then the water reservoir will be filled in:
Solution:
Net part filled in 1 hour = (1/3 + 1/6 – 1/12) = 5/12
The water reservoir will be full in 12/15 hours
5. Two funnels can fill a reservoir in 14 hours and 16 hours individually. The channels are opened all the while and it is found that because of spillage in the base it took 32 minutes more to fill the reservoir. Once the reservoir is full, in what time, spillage will discharge the full reservoir?
Solution:
Work done by the two funnels in 1 hour= (1/14+ 1/16) = 15/112
Time taken by these funnels to fill the tank = 112/15 hrs = 7 hrs 28 min
Because of spillage, time taken = 7 hrs. 28 min + 32 min = 8 hrs
Therefore, Work done by (two funnels + spill) in 1 hour = 1/8
Work done by the break in 1 hour = (15/112-1/8) = 1/112
Break will discharge the full reservoir in 8 hours.

6. A and B can fill a cistern in 7.5 minutes and 5 minutes respectively and C can carry off 14litres per minute. If the cistern is already full and all the three pipes are opened, then it is emptied in 1hour. How many litres can it hold?
Solution:
If the capacity is L litres,
Water filled in 1 hour = L/7.5*60 + L/5*60 = 20L
Water removed in 1 hour = 14*60
Since the tank was full initially, hence L + 20L – 14 * 60 = 0 => L = 40 litres.
7. Tap A can fill a tank in 20 hour, tap B in 25 hour but tap C can empty a full tank in 30 hour. Starting with A, followed by B and C, each tap is opened alternatively for 1 hour period till the tank get filled up completely. In how many hours will the tank be filled up completely?
Solution:
LCM (20,25,30)=300
Suppose capacity of the tank is 300 litre.
Quantity filled by tap A in 1 hour = 300/20 = 15 litre.
Quantity filled by tap B in 1 hour = =300/25 = 12 litre.
Quantity emptied by tap C in 1 hour =300/30 = 10 litre.
In every three hour, 15+12−10=17 litre is filled.
In 45 hours, 17×15=255 litre is filled.
Remaining quantity to be filled =300−255=45 litre.
In next 3 hour, 17 more litre is filled.
Remaining quantity to be filled =45−17=28 litre.
In next 2 hour hour, 15+12=27 more litre is filled.
Remaining quantity to be filled =28−27=1 litre.
But in next 1 hour, 10 litre is emptied.
Remaining quantity to be filled =1+10=11 litre.
Pipe A fills this remaining quantity in 11/15 hour (i.e, in 44 minutes).
Therefore, the tank is filled in 51 hour 44 minutes
8. A cistern has two inlet pipes – A & B and one outlet pipe – C. Pipe A and B can fill the cistern in 20 and 25 hours respectively, while pipe C can empty it in 10 hours. First, pipe A was opened, after 1 hour pipe B was opened and after another 1 hour pipe C was opened. Find the time required to empty the cistern completely.
Solution:
Efficiency of pipe A = 5%
Efficiency of pipe B = 4%
Efficiency of pipe c = 10%
In the first hour, work done = 5% [Pipe A only], in the 2nd hour, work done = 5 + 4 = 9% [Pipes A and B both]
Total work done in 2 hours = 14% [i.e., 9% + 5%]
Combined efficiency of 3 pipes = 5 + 4 – 10 = -1%
(i.e.,) net effect is of emptying the tank;
So, 1% of tank is emptied in 1 hour.
Hence, 14% of tank would be emptied in 14 hours.
9. A tank can be filled in 2 (2 / 5) hours, when all the five pipes (including both inlet and outlet pipes) are turned on simultaneously. Also it is given that each inlet pipe can fill the tank in 4 hours when working separately and each outlet pipe can empty the tank in 6 hours, when working separately. How many outlet pipes are attached to the tank?
Solution:
Work done in 1 hour by all five pipes = 1 / (12 / 5) = 5 / 12
Let there be ‘n’ inlet pipes, hence no. of outlet pipes = 5 – n
Now, (1 / 4) (n) – (1 / 6) (5 – n) = 5 / 12
(n / 4) – (5 / 6) + (n / 6) = 5 / 12
5n / 12 = (5 / 12) + (5 / 6) or 15 / 12
n = (12 × 15) / (12 × 5) = 3
Hence, there are 3 inlet and 2 outlet pipes.