Pipes and cistern questions are mostly the trickiest questions to solve in an examination. To handle the questions in the best possible manner, you can start practising them today. Practise makes a man perfect and more accurate in solving problems.

These types of questions generally appear in the **banking exams**, SSC, RRB, insurance, Olympiads, and others. To secure your place in this competitive examination, practise the questions today.

## Formulas To Solve Pies and Cistern Questions

here are some basic formulas that will help you to solve the Pipes and cistern questions with them. Remmember these formulas may not be enough for each question, you have to practice more questions to bring flow in solving questions.

- If it will take x hours to fill a tank, then half tank will be filled in
**1 hr =1/x** - In the event that y hours are expected to discharge the tank, part exhausted in
**1 hour = 1/y** - If a line can fill a tank in x hours and can discharge similar tank in y hours. At the point when both the lines are opened simultaneously, then, at that point, the net piece of the tank filled in
**1 hr = {(xy)/(y-x)}, gave y>x** - If a line can fill a tank in x hours and can exhaust similar tank in y hours. At the point when both the lines are opened simultaneously, then, at that point, the net piece of the tank filled in
**1 hr = {(xy)/(x-y)}, gave x>y** - Net work done = (Amount of work done by Gulfs) – (Amount of work done by Outlets)
- One gulf can fill the tank in x hr and the other channel can fill the similar tank in y hrs, assuming both the bays are opened simultaneously, the time taken to fill the entire tank
**= {(xy)/(y+x)}** - If two lines take x and y hours separately to fill a tank of water and a third line is opened which takes z hours to exhaust the tank, then, at that point, the time taken to fill the tank =
**{1/(1/x)+(1/y)+(1/z)} and the net piece of the tank filled in 1 hr = (1/x)+(1/y)- (1/z).**

## Pipes and Cistern Questions and Answers

Below you can find different questions and answers related to Pipes and Cisterns:

**Q1:Two pipes A and B can separately fill a cistern in 60 min and 75 min respectively. There is a third pipe in the bottom of the cistern to empty it. If all the three pipes are simultaneously opened, then the cistern is full in 50 min. In how much time, the third pipe alone can empty the cistern ?**

85 Min.

95 Min.

105 Min.

105 Min.

Answer: D) 100 min

Explanation:

Work done by the third pipe in 1 min = 1/50 – (1/60 + 1/75) = – 1/100.

[-ve sign means emptying]

The third pipe alone can empty the cistern in 100 min.

**Q2: A water tank is two-fifth full.Pipe A can fill a tank in 10 minutes and pipe B can empty it in 6 minutes.If both the pipes are open,how long will it take to empty or fill the tank completely?**

6 min.to empty

9 min. to empty

10min. to fill

6 min. to fill

Answer: A

Explanation:

Clearly,pipe B is faster than pipe A and so,the tank will be emptied.

part to be emptied = 2/5

part emptied by (A+B) in 1 minute = (16−110)=11516-110=115

115:25::1:x115:25::1:

25*15=6mins25*15=6

so, the tank will be emptied in 6 minutes`

**Q3: A tap can fill a tank in 6 hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the tank completely?**

2hrs 45 min

3hrs 25 min

1hr 20 min

2hr 20min

Answer: B

Explanation:

Time taken by one tap to fill **half of the tank** = 3 hrs.

Part filled by the four taps in 1 hour =4*1/6 =2/3

Remaining part =(1−12)=121-12=12

23:12::1:x23:12::1

=> x = (12*1*32)=3412*1*32=34

So, total time taken = 3 hrs. 45 mins.

**Q4: A pump can fill a tank with water in 2 hours. Because of a leak, it took 213213 hours to fill the tank. The leak can drain all the water of the tank in:**

11 Hours

10 Hours

9 Hours

14 Hours

Answer: D

Explanation:

If the total area of pump=1 part

The pumop take 2 hrs to fill 1 part

The pumop take1 hour to fill 1/2 portion

Due to lickage

The pumop take 7/3 hrs to fill 1 part

The pumop take1 hour to fill 3/7 portion

Now the difference of area = (1/2-3/7)=1/14

This 1/14 part of water drains in 1 hour

Total area=1 part of water drains in (1×14/1)hours= 14 hours

So the leak can drain all the water of the tank in 14 hours.

Leak will empty the tank in 14 hrs

**Q5: A cistern is normally filled in 8 hours but takes two hours longer to fill because of a leak in its bottom. If the cistern is full, the leak will empty it in ?**

20 Hours

10 Hours

9 Hours

21 Hours

Answer: A) 20 hrs

Explanation:

1/8 – 1/x = 1/10

=> x = 40 hrs

**Q6: 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:**

23

43

14

18

Answer: C

Explanation:

Part filled in 2 hours = 2/6=1/3

Remaining part =(1−13)=231-13=23

(A + B)’s 7 hour’s work = 2/3

(A + B)’s 1 hour’s work = 2/21

C’s 1 hour’s work = { (A + B + C)’s 1 hour’s work } – { (A + B)’s 1 hour’s work }

=1/6-2/21 = 1/14

C alone can fill the tank in 14 hours.

**Q7: A cistern has a leak which would empty the cistern in 20 minutes. A tap is turned on which admits 4 liters a minute into the cistern, and it is emptied in 24 minutes. How many liters does the cistern hold ?**

400 lit

480 lit

500 lit

580 lit

Answer: B

Explanation:

1/k – 1/20 = -1/24

k = 120

120 x 4 = 480

Therefore, the capacity of the cistern is 480 liters.

**Q8: Taps X and Y can fill a tank in 30 and 40 minutes respectively.Tap Z can empty the filled tank in 60 minutes.If all the three taps are kept open for one minute each,how much time will the taps take to fill the tank?**

15min.

10min.

24min.

20min.

Answer: C

Explanation:

Given taps X and Y can fill the tank in 30 and 40 minutes respectively. Therefore,

part filled by tap X in 1 minute = 1/30

part filled by tap Y in 1 minute = 1/40

Tap Z can empty the tank in 60 minutes. Therefore,

part emptied by tap Z in 1 minute = 1/60

Net part filled by Pipes X,Y,Z together in 1 minute = [1/30 +1/40 – 1/60] = 5/120 = 1/24

i.e., the tank can be filled in 24 minutes.

**Q9: A large tanker can be filled by two pipes A and B in 60 and 40 minutes respectively. How many minutes will it take to fill the tanker from empty state if B is used for half the time and A and B fill it together for the other half ?**

50 min.

45 min.

60 min.

30 min.

Answer: D

Explanation:

Part filled by (A + B) in 1 minute = (1/60 + 1/40) = 1/24

Suppose the tank is filled in x minutes.

Then, x/2(1/24 + 1/40) = 1

(x/2) * (1/15) = 1 => x = 30 min.

**Q10: 12 buckets of water fill a tank when the capacity of each tank is 13.5 liters. How many buckets will be needed to fill the same tank,if the capacity of each bucket is 9 liters?**

10

12

30

18

Answer: D

Explanation:

Capacity of the tank =(12 x 13.5) liters =162 liters.

Capacity of each bucket =9 liters

Number of buckets needed = 162/9 =18.

**Q11: Water flows into a tank which is 200 m long and 150 m wide, through a pipe of cross-section (0.3m x 0.2m) at 20 km/h. In what time will the water level be 12m ?**

200 hours

100 hours

300 hours

400 hours

Answer: C

Explanation:

Volume of water collected in the tank in 1 hour

⇒ (0.3 × 0.2 × 20km × 1000mts) = 1200 m cubic

If after t hours, the water is at height of 12m,

1200t=200×150×12

⇒ t = 300 Hours.

**Q12:**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?7(1/2) hrs

4(4/5) hrs

3 (2/7) hrs

1(1/5) hrs

Answer: B

Explanation: Speed of the man (upstream) = (4 / 40 * 60) km/hr

= 6 km/hr

Speed of the man (downstream) = ( 4 / 36 * 60) km/hr

= (60 / 9)km/hr

= 20/3 km/h

Rate of the current = ½[downstream speed – upstream speed]

= (1/2) [ 20 / 3 – 6] km/hr

Rate of the current = 1/3 km/hr

**Q13:**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?120 gallons

240 gallons

450 gallons

840 gallons

Answer:D**Explanation:**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

Here, negative (-) sign indicates emptying of tank.

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.

**Q14:**A booster pump can be used to fill as to empty the tank. The capacity of the tank is 1200 m^{3}. The emptying capacity of the tank is 10 m^{3}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.25

50

75

100

Answer : B**Explanation:**Assume that the filling capacity of the pump = x m

^{3}/min

Given condition : Emptying capacity of the tank is 10 m

^{3}per minute higher than its filling capacity. This means that the emptying capacity of the pump is = x + 10 m

^{3}/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, x

^{2}+ 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 m

^{3}/min.

**Q 15.**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?8 hrs

16 hrs

21 hrs

36 hrs

Answer: D**Explanation:**

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 =

xyhrs / y – x

**Q16.**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?35 hrs

70 hrs

180 hrs

300 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

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.

**Q17.**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?12 hrs

30 hrs

44 hrs

60 hrs

Answer: D**Explanation:**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

x

^{2}– 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.

**Q18.**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?6 hrs

7 hrs

8 hrs

14 hrs

Answer: B**Explanation:**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 = ⅛

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

**Q19.**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?3 hrs

7 hrs

12 hrs

20 hrs

Answer: C**Explanation:**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

**Q20.**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?4 min

6 min

14 min

20 min

Answer: B**Explanation:**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

## FAQs

**How do you solve pipe problems in aptitude?**

Read the question multiple times and try to answer the query with the application of simplest formulas and using common sense.

**In which exam does Pipes and Cistern Questions come?**

There are various exams like bank exams, SSC, RRB, Insurance, Olympiads and others in which these types of questions actually comes in.

**Is this tuff to solve these types of problems?**

No by using common sense and correct formulas you can easily solve these types of questions.