# Pipes and Cisterns - Aptitude test, questions, shortcuts, solved example videos

Video on Pipes and Cistern - shortcuts, tips and tricks

## Pipes & Cisterns

Problems related to pipes & cisterns are similar to those of time and work but with a slight difference of conceptual agents like inlets & outlets.

Basic Terms:

Inlet: A pipe which is used to fill up the tank,cistern or reservoir is known as 'Inlet'. This type of nature indicates 'plus' or 'positive type' of work done.

Outlet: A pipe which is used to empty the tank, cistern or reservoir is known as 'Outlet'. This type of nature indicates 'minus' or 'negative type' type of work done.

Important Formulae:
 1) If a pipe requires 'x' hours to fill up the tank, then part filled in 1 hr = 1 x

 2) If a pipe requires 'y' hours to empty the full tank, then part emptied in 1 hr = 1 y

3) Net Work Done = (Sum of work done by inlets) - (Sum of work done by outlets)

4) Suppose that one pipe takes 'x' hours to fill up the tank and the another pipe takes 'y' hours to empty the full tank, then, on opening both the pipes, there are 2 possible conditions:
 For condition I: x < y, Net part filled in 1 hr = 1 – 1 x y

 For condition II: x > y, Net part emptied in 1 hr = 1 – 1 y x

Quick Tricks & Tips:

1) If two pipes take 'x' & 'y' hrs respectively to fill the tank and the third pipe takes 'z' hrs to empty the tank and all of them are opened together, then
 The net part filled in 1hr = 1 + 1 – 1 x y z

Hence,
 The time taken to fill the tank = 1 (1/x) + (1/y) + (1/z)

2) Consider a pipe fills the tank in 'x' hrs. If there is a leakage in the bottom, the tank is filled in 'y' hrs.
If the tank is full, the time taken by the leak to empty the tank =  hrs
1
 1 – 1 x y

3) Suppose that pipe 'A' fills the tank as fast as the other pipe 'B'. If pipe 'B' (slower) & pipe 'A' (faster) take 'x' min & 'x/n' min respectively to fill up an empty tank together, then
 Part of the tank filled in 1 hr = n + 1 x

Questions Variety:

Type 1: Find the time to fill/ empty the tank by the pipes together for the given time of individual pipes.

Examples:

Q 1. Two pipes A & B can fill the tank in 12 hours and 36 hours respectively. If both the pipes are opened simultaneously, how much time will be required to fill the tank?

a. 6 hours
b. 9 hours
c. 12 hours
d. 15 hours
View solution

Hint:

 If a pipe requires 'x' hrs to fill up the tank, then part filled in 1 hr = 1 x

If pipe A requires 12 hrs to fill the tank, then part filled by pipe A in 1 hr = 1/12

If pipe B requires 36 hrs to fill the tank, then part filled by pipe B 1 hr = 1/36

Hence, part filled by (A + B) together in 1 hr = 1/12 + 1/36

= 48 / 432 = 1/9

In 1 hr both pipes together fill 1/9th part of the tank. This means, together they fill the tank in 9 hrs.

Q 2. Two pipes can fill a tank in 6 hours and 8 hours respectively while a third pipe empties the full tank in 12 hours. If all the three pipes operate simultaneously, in how much time will the tank be filled?

a. 7(1/2) hrs
b. 4(4/5) hrs
c. 3 (2/7) hrs
d. 1(1/5) hrs
View solution

Hint: If one pipe fills the tank in 'x' hrs, another pipe fills the same tank in 'y' hrs but the third pipe empties the tank in 'z' hrs and all of them are opened together, then

 The net part filled in 1hr = 1 + 1 – 1 x y z

From the given data, net part filled in 1 hour = 1/6 + 1/8 -1/12 = 5/24

So, total time to fill the tank with all pipes open = 24/5 hrs

Type 2: Find the capacity of tank for given
a) time of each pipe to fill up a tank if they work alone and
b) rate at which outlet pipe can empty the tank.

Examples:

Q 3. Two pipes can fill a tank in 10 and 14 minutes respectively and a waste pipe can empty 4 gallons per minute. If all the pipes working together can fill the tank in 6 minutes, what is the capacity of the tank?

a. 120 gallons
b. 240 gallons
c. 450 gallons
d. 840 gallons
View solution

Hint:
Work done by waste pipe in 1 min = (Part filled by total pipes together) -(part filled by first pipe + part filled by second pipe)
= 1/6 – [(1/10) + (1/14)] = 1/6 – (24/140) = -1/210

To find the capacity, we need to determine the volume of 1/210 part.
Therefore, volume of 1/210 part = 4 gallons ---------(given condition)
Hence, the capacity of tank = volume of whole = 4 x 210 = 840 gallons.

Q 4. A booster pump can be used to fill as to empty the tank. The capacity of the tank is 1200 m3. The emptying capacity of the tank is 10 m3 per minute higher than its filling capacity and the pump requires 4 minutes lesser to vacant the tank than it requires to fill it. Calculate the filling capacity of the pump.

a. 25
b. 50
c. 75
d. 100
View solution

Hint:
Assume that the filling capacity of the pump = x m3/min
Given condition : Emptying capacity of the tank is 10 m3 per minute higher than its filling capacity. This means that the emptying capacity of the pump is = x + 10 m3/min

We need to filling the capacity of pump from the given relation of filling & emptying capacities. Hence, we can write that,
1200 / x - 1200/ x+ 10 = 4

[1/x – 1/ x+10] = 1/300
By solving the above equation, we get the simplified form of quadratic equation.
So, x2 + 10x -3000 = 0
(x + 60) (x-50) = 0

Therefore, we get two values of 'x'; i.e. x = -60 & x =50

Since we are asked to find the filling capacity, which is always positive; we will neglect the negative value of x = –60.
Therefore, the filling capacity of the pump = 50 m3/min.

Type 3: Find the time taken by leak to empty the tank for given
a) time of pipe to fill a tank &
b) time taken by leak to fill the tank due to leakage.

Examples:

Q 5. An electric pump can fill a tank in 4 hours. Due to leakage in the tank, it took 4(1/2) hrs to fill the tank. If the tank is full, how much time will the leak take to empty the full tank?

a. 8 hrs
b. 16 hrs
c. 21 hrs
d. 36 hrs
View solution

Hint: Consider a pipe fills the tank in 'x' hrs. If there is a leakage in the bottom, the tank is filled in 'y' hrs.

 The time taken by the leak to empty the full tank = xy hrs y – x

Time taken to empty the tank by the leak = 4 x (9/2) / (9/2 ) - 4 = (36/2 ) / ½ = 18/ ½ = 18 x 2 = 36 hrs

Q 6. Two pipes fills a cistern in 15 hrs and 20 hrs respectively. The pipes are opened simultaneously and it is observed that it took 26 min more to fill the cistern because of leakage at the bottom. If the cistern is full, then in what time will the leak empty it?

a. 35 hrs
b. 70 hrs
c. 180 hrs
d. 300 hrs
View solution

Hint: Consider a pipe fills the tank in 'x' hrs. If there is a leakage in the bottom, the tank is filled in 'y' hrs.

 The time taken by the leak to empty the full tank = xy hrs y – x

But, direct values of x & y are not given. So, we need to find the work done by the two pipes in 1 hr = (1/15) + (1/20) = 7/60

Hence, the time taken by these pipes to fill the tank = 60/7 hrs = 8 .57 hrs = 8 hrs 34 min --------( by multiplying '0.57' hrs x 60 = 34 minutes)

Due to leakage, time taken = 8 hrs 34 min + 26 min = 8 hrs 60 min = 9 hrs ----------( because 60 min = 1 hr)

Thus, work done by (two pipes + leak) in 1 hr = 1/9
Hence, work done by leak in 1 hr = work done by two pipes – 1/9
= 7 /60 – 1/9 = 3/ 540 = 1/180

Therefore, leak will empty the full cistern in 180 hours.

Type 4:
i)Find the time taken by two pipes to fill the tank if they work individually if it is given that -
a) one pipe is 'm' times faster/slower than another pipe and
b) time taken by two pipes to fill the tank together.

ii) Find the time taken to fill an empty tank for given -
a) time of each pipe to fill the tank separately and
b) one of the pipes is full-time working and the other 2 pipes are open for certain hour alternately.

Examples:

Q 7. If two pipes function simultaneously, the reservoir will be filled in 24 hrs. One pipe fills the reservoir 20 hours faster than the other. How many hours does it take for the second pipe to fill the reservoir?

a. 12 hrs
b. 30 hrs
c. 44 hrs
d. 60 hrs
View solution

Hint:
Assume that the reservoir is filled by first pipe in 'x' hours.
So, the reservoir is filled by second pipe in 'x + 20' hours.
Now, from these above conditions, we can form the equations as,

1/x + 1/ (x + 20) = 1/24

[x + 20 + x] / [x(x + 20)] = 1/24

x2 – 28x – 480 = 0

By solving this quadratic equation , we get the factors (x – 40) (x+12) = 0
Hence, we get two values : x = 40 & x = -6
Since filling of reservoir is positive work , we can neglect the negative value of 'x'.
Thus, x = 40

This means that the second pipe will take (x+ 20) hrs = 40 + 20 = 60 hrs to fill the reservoir.

Q 8. Two pipes can fill a tank in 8 hrs & 6 hrs respectively. If they are opened on alternate hours and if pipe A gets opened first, then in how many hours, the tank will be full?

a. 6 hrs
b. 7 hrs
c. 8 hrs
d. 14 hrs
View solution

Hint:
Pipe A's work in 1 hr = 1/8
Pipe B's work in 1 hr = 1/6

Pipes (A+B)'s work in first 2 hrs when they are opened alternately = 1/8 + 1/6 = 7 /24

Now,

In 4 hrs they fill : 2 X (7/24) = 7/12
In 6 hrs they fill : 3 X (7/24) = 7/8

After 6 hrs, part left empty = 1/8

Now it is A's turn to open up.

In one hr it fills 1/8 of the tank.

So, the tank will be full in = 6 hrs + 1 hr = 7 hrs

Type 5:
i) Find the time taken to fill the tank for given -
a) the time of inlet pipes to fill a tank if they work alone and the outlet pipe to empty a tank &
b) all pipes are opened but outlet pipe is opened/closed after certain hours

ii) Find the total time required to fill the tank if -
a) the time of each inlet pipe to fill the tank individually and
b) all the pipes are opened simultaneously and after certain time, one of the pipes is turned off.

Examples:

Q 9. It is observed that the pipe A can fill the tank in 15 hrs and the same tank is filled by pipe B in 20 hrs. The third pipe C can vacant the tank in 25 hrs. If all the pipes get opened initially and after 10 hrs, the pipe C is closed, then how long will it take to fill the tank?

a. 3 hrs
b. 7 hrs
c. 12 hrs
d. 20 hrs
View solution

Hint:
Assume that the tank will be full in '10 + x' hrs.

Part filled by pipe A in 1 hr = 1/15
Part filled by pipe B in 1 hr = 1/20
Part emptied by pipe C in 1 hr = 1/ 25

(A+ B)'s part filled in 1 hr = 1/15 + 1/20 = 7 /60

As Pipe C is closed after 10 hours, let us find the part of tank filled in 10 hrs.
Tank filled in 10 hrs = 10 (part filled by A in 1 hr + part filled by B in 1 hr– part emptied by C in 1 hr)
= 10 [1/15 + 1/20 – 1/25] = 23 /30

Remaining part = 1 – part filled in 10 hours
= 1 – 23/30 = 7/30

We know that,

 Part filled by (A + B) in 1 hr :: Total part filled Remaining part Time taken

So, we get the ratio as ,

7/60 : 7/30 :: 1 : x

So, the value of x = 2/30 x 1 x 60/7 = 2 hrs.

Hence, the tank will be full in x + 2 = 10 + 2 hrs = 12 hrs

Q 10. Two pipes A & B can fill a tank in 5 min & 10 min respectively. Both the pipes are opened together but after 2 min, pipe A is turned off. What is the total time required to fill the tank?

a. 4 min
b. 6 min
c. 14 min
d. 20 min
View solution

Hint:
Assume the total time required = x + 2 min
Part filled by pipe A in 1min = 1/5 & part filled by B in 1 min = 1/10
After 2 min, part filed by A & B together = 2 [ 1/5 + 1/10] = 3/5

Remaining part = 1 – 3/5 = 2/5
Part filled by pipe B in 1 min = 1/10

As pipe A is turned off after 2 minutes, we get the relation as,

 Part filled by (A + B) in 1 hr :: Total part filled Remaining part Time taken

1/10 : 2/5 :: 1: x

x = (2/5) x 1 x 10 = 4 min

Tank will be full in ( 2 min + 4 min) = 6 min

Practice questions on Pipes and Cistern
Boats and Streams - Aptitude test, questions, shortcuts, solved example videos
Boats and Streams - Quantitative aptitude tutorial with easy tricks, tips, short cuts explaining the concepts. Online aptitude preparation material with practice question bank, examples, solutions and explanations. Video lectures to prepare quantitative aptitude for placement tests and competitive exams like MBA, Bank exams, RBI, IBPS, SSC, SBI, RRB, Railway, LIC, MAT. Very useful for freshers, engineers, software developers taking entrance exams. Learn and take practice tests!
Alligation or Mixtures - Aptitude test, questions, shortcuts, solved example videos
Alligation or Mixtures - Quantitative aptitude tutorial with easy tricks, tips, short cuts explaining the concepts. Online aptitude preparation material with practice question bank, examples, solutions and explanations. Video lectures to prepare quantitative aptitude for placement tests and competitive exams like MBA, Bank exams, RBI, IBPS, SSC, SBI, RRB, Railway, LIC, MAT. Very useful for freshers, engineers, software developers taking entrance exams. Learn and take practice tests!
Problems on Trains - Aptitude test, questions, shortcuts, solved example videos
Problems on Trains - Quantitative aptitude tutorial with easy tricks, tips, short cuts explaining the concepts. Online aptitude preparation material with practice question bank, examples, solutions and explanations. Video lectures to prepare quantitative aptitude for placement tests and competitive exams like MBA, Bank exams, RBI, IBPS, SSC, SBI, RRB, Railway, LIC, MAT. Very useful for freshers, engineers, software developers taking entrance exams. Learn and take practice tests!