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.
