Top 130 Pipe and cistern questions for Competitive Exams [ 100% FREE ]

Introducing an important aspect of the quantitative aptitude section: the Pipe and cistern questions. A common challenge candidates face in various competitive exams, be it banks, SSC, RRB, Insurance and more, is problems related to pipes and cisterns. In this comprehensive article, we delve deep into “Pipe and cistern questions“, ensuring you get the most effective preparation material.

For many, the term “Pipe and cistern questions” can seem daunting, but with the right approach and understanding of the “pipes and cisterns formula“, it becomes manageable. This is why we’re emphasizing on the crucial examples like “two pipes p and q together can fill a cistern” and “two pipes a and b can fill a cistern“. Both these examples will give you a clear perspective on how to approach these problems.

We also recognize the importance of language adaptability in preparation. Therefore, we have also incorporated “pipe and cistern questions in Hindi” and insights on “pipe and cistern in Hindi“. This ensures that you not only understand the crux of the “pipes and cisterns problems” but also can relate to them in a language you’re comfortable with.

Word problems, especially around the “pipes and cisterns formula“, have been a mainstay in competitive exams. While “two pipes p and q together can fill a cistern” or “two pipes a and b can fill a cistern” might seem similar, they can carry nuances that are essential for aspirants to grasp.

Pipes and Cisterns Formula :

Given below are aimportant formulas which shall help you solve the pipes and cistern based questions quicker and more efficiently:

  • If x hours are required to fill up a tank, then part filled in 1 hr =1/x
  • If y hours are required to empty the tank, then part emptied in 1 hour = 1/y
  • If a pipe can fill a tank in x hours and can empty the same tank in y hours. When both the pipes are opened at the same time, then the net part of the tank filled in 1 hr = {(xy) / (y-x)}, provided y>x
  • If a pipe can fill a tank in x hours and can empty the same tank in y hours. When both the pipes are opened at the same time, then the net part of the tank filled in 1 hr = {(xy) / (x-y)}, provided x>y
  • Net work done = (Sum of work done by Inlets) – (Sum of work done by Outlets)
  • One inlet can fill the tank in x hr and the other inlet can fill the same tank in y hrs, if both the inlets are opened at the same time, the time taken to fill the whole tank = {(xy) / (y+x)}
  • If two pipes take x and y hours respectively to fill a tank of water and a third pipe is opened which takes z hours to empty the tank, then the time taken to fill the tank = {1 / (1/x)+(1/y)+(1/z)} and the net part of the tank filled in 1 hr = (1/x)+(1/y)-(1/z)

The syllabus for competitive exams is vast, as we all know. But every topic, including “Pipe and cistern questions“, holds its weight. Therefore, giving due time to understand and practice these “pipes and cisterns problems” is paramount. Especially when considering the intense competition and the challenge to crack these exams.

In conclusion, this article offers an in-depth understanding of “Pipe and cistern questions“, covering essential examples like “two pipes p and q together can fill a cistern” and “two pipes a and b can fill a cistern“, not forgetting the importance of language with “pipe and cistern questions in Hindi” and “pipe and cistern in Hindi“.

Whether you’re just starting out or brushing up your skills, this article is your guide to excel. So, gear up and dive into the world of “Pipe and cistern questions” and ace your upcoming Government sector exams!

Top 130 Pipe and cistern questions for Competitive Exams :

31. Pipe A can fill a tank in 15 minutes and the tap at bottom of a tank can drain water at a rate of 4 liters per minute. If both the pipe and tap are opened together then they can fill the tank in 20 minutes. Find the capacity of the tank. 

पाइप A एक टंकी को 15 मिनट में भर सकता है और टंकी के नीचे एक नल 4 लीटर प्रति मिनट की दर से पानी निकाल सकता है। यदि पाइप और नल दोनों को एक साथ खोला जाता है, तो वे टंकी को 20 मिनट में भर सकते हैं। टंकी की क्षमता ज्ञात कीजिए।


Option “A” is correct.

Given:

Time to fill a tank by Pipe A = 15 minutes

Tap drains water rate = 4 liters/minute

Time to fill the tank Pipe A and Tap opened together = 20 minutes

Formula used:

Work done by both pipes in one minute = ((1/T) – (1/t))

Where,

T = Time taken by pipe 1 to fill the tank

t = Time taken by pipe 2 to empty the tank

Calculation:

⇒ Work done = ((1/15) – (1/t))

⇒ Work done = 1/20

⇒ 1/20 = ((1/15) – (1/t))

⇒ 1/t = (1/15) – (1/20)

⇒ 1/t = 1/60

⇒ t = 60 minutes

60 minutes is draining time of water

Now, tap drains water at rate of 4 liters per minute and work done by tap to drain water is 1/60

Hence, Volume of tank = 4 × 60

∴ Volume of tank is 240 liters

32. Pipe A can fill a tank in 6 hours. Pipe B can fill the same tank in 8 hours. Pipe A, B and C together can fill the same tank in 12 hours. Then which of the following statements is true for pipe C?

पाइप A एक टैंक को 6 घंटे में भर सकता है। पाइप B उसी टैंक को 8 घंटे में भर सकता है। पाइप A, B और C मिलकर उसी टंकी को 12 घंटे में भर सकते हैं। तो पाइप C के लिए निम्नलिखित में से कौन सा कथन सत्य है?


Option “C” is correct.

Given:

Time taken by A to fill tank = 6 hours

Time taken by B to fill tank = 8 hours

Time taken by A, B and, C together to fill the tank = 12 hours

Concept used:

Total work = time × efficiency 

Calculation:

Let the capacity of the tank ( work to be done) be 24x units (LCM of 6, 8, 12)

⇒ The efficiency of pipe A = 24x/6 = 4x units/day

⇒ Efficiency of pipe B = 24x/8 = 3x units/day

⇒ Efficiency of pipe (A + B + C) = 24x/12 = 2x units/day

⇒ Efficiency of pipe C = efficiency of (A + B = C) – efficiency of (A + B)

Efficiency of pipe C = 2x – (4x + 3x) = – 5x units/day

Negative efficiency implies that pipe C is emptying pipe.

⇒ Time taken by pipe C to empty the filled tank = 24x/5x

= 4.8 hours or 4 hrs 48 min

∴ The pipe C will empty the tank in 4 hrs 48 mins.

33. Two taps can separately fill a cistern in 20 minutes and 25 minutes. Both taps are open for 10 minutes after which the slower one is closed. How long will it take to fill the remaining portion by the other tap alone?

दो नल अलग-अलग एक टंकी को 20 मिनट और 25 मिनट में भर सकते हैं। दोनों नलों को 10 मिनट के लिए खोला जाता है, जिसके बाद धीमी गति वाले नल को बंद कर दिया जाता है। दूसरे नल द्वारा अकेले टंकी के शेष भाग को भरने में कितना समय लगेगा?


Option “C” is correct.

Total work = 100 units

⇒ Work done by both pipe in 10 min = (5 + 4) × 10 = 90 units

⇒ Remaining work = 100 – 90 = 10 units

∴ Faster tap fill the remaining tank in = 10/5 = 2 min

34. A water cooler is filled when it is half empty with the help of a pipe which can fill the whole cooler in 50/3 minutes. There is an outlet tap at the bottom of the cooler which can empty the tank in 25 minutes. When the inlet pipe is started, a person comes and takes out the water from that tap for 1 minute then after a gap of 2 minutes another person comes and takes out the water for 1 minute and this process goes on. In how much time cooler will be completely full?

एक पाइप, जो पूरे कूलर को 50/3 मिनट में भर सकता है, की सहायता से एक वाटर कूलर आधा खाली होने पर भरा जाता है। कूलर के नीचे एक आउटलेट नल है जो 25 मिनट में टैंक को खाली कर सकता है। जब इनलेट पाइप चालू होता है तो एक व्यक्ति आता है और उस नल से 1 मिनट के लिए पानी निकालता है फिर 2 मिनट के अंतराल के बाद दूसरा व्यक्ति आता है और 1 मिनट के लिए पानी निकालता है और यह प्रक्रिया चलती रहती है। कूलर कितने समय में पूरी तरह से भर जाएगा?


Option “D” is correct.

Suppose the capacity of the tank = 50 units (LCM of 50/3 and 25)

∴ Efficiency of Inlet pipe = 50/(50/3) = 3

Efficiency of Outlet pipe = 50/25 = 2

Since the cooler is half empty

∴ Quantity of water to be filled = 50/2 = 25 units

A person takes the water out for 1 minute and then 2 minute gap that means in 3 minutes, outlet tap is working for only 1 minute.

After 9 minutes∶

Quantity of water filled = 3 × 9 – 2 × 3 = 21 units

10th minute∶ Both inlet and outlet will be working;

∴ Quantity of water filled = 3 – 2 = 1 unit

11th minute∶ Only inlet working;

∴ Quantity of water filled = 3 units

∴ Quantity of water filled upto 11 minutes = 21 + 1 + 3 = 25 units

That means cooler is full in 11 minutes.

35. Pipes A and B are emptying pipes and can empty a tank in 6 hours and 16 hours, respectively. C is a filling pipe. All three pipes were opened together. They took 80 minutes to empty 5/18th of the tank. Pipe C alone can fill the tank in:

पाइप A और B खाली करने वाले पाइप हैं और क्रमशः 6 घंटे और 16 घंटे में एक टैंक खाली  कर सकते हैं। C एक भरने वाला पाइप है। तीनों पाइप एक साथ खोले गए। उन्हें टैंक का 5/18वां हिस्सा खाली करने में 80 मिनट का समय लगा। पाइप C अकेले कितने समय में टैंक को भर सकता है?


Option “C” is correct.

Given:

A pipe = 6 hours

B pipe = 16 hours

Time = 80 minutes to empty 5/18 of the tank

Formula used:

Capacity = time × efficiency

Calculation:

Let the efficiency of Pipe C be x, then

(8 + 3 – x) × 80/60 = 48 × 5/18

(11 – x) × 4/3 = 40/3

(11 – x) = 10

x = 11 – 10

x = 1

Pipe C alone can empty the tank in 48/1 = 48 hrs

∴ The correct answer is 48 hrs.

36. A pipe can fill a tank in 10 minutes while another pipe can empty it in 12 minutes. If the pipes are opened alternately each for 1 minute, beginning with the first pipe. the tank will be full after (in minutes):

एक पाइप 10 मिनट में एक टैंक भर सकता है जबकि दूसरा पाइप 12 मिनट में इसे खाली कर सकता है। यदि शुरुआत पहले पाइप से करके, पाइपों को 1 मिनट के लिए बारी-बारी से खोला जाए। टैंक कितनी देर में पूरा भर जाएगा (मिनट में)?


Option “B” is correct.

Given:

A pipe can fill a tank in 10 minutes

Another pipe can empty it in 12 minutes

They are opened alternately

Beginning with the first pipe

Calculation:

A → +10, B → -12 (lcm of 10,12 is 60) than 

efficiency of A is +6 and B is -5

They opened alternative so A fill 6 litres in 1 minute and B empty 5 litres in 1 minute

They fill together 1 litre in 2 minutes 

They fill 1 × 54 is 54 litre in 2 × 54 is 108 minutes 

(∵ total = 60 litre)

Remaining 6 litre fill by the first pipe in 1 minute because his efficiency is 6litre/min

∴ the tank will full after 108 + 1 is 109 minutes

Note:

This type of question always filling pipe per unit work or capacity subtracted.

⇒ 60 – 6 = 54 litre, 

Then clear the multiply 54 in (2 min → 1 litre)

⇒ 108 min → 54 litre

Add the 1 min, previous subtracted

⇒ 108 + 1 min → 54 + 6

⇒ 109 min → 60 litres. 

Approach this method, definitely your answer is correct.

37. Pipe A is a filling pipe, while B and C are emptying pipes. Pipe A alone can fill a tank in 10 hours and pipe C alone can empty the full tank in 24 hours. If all three pipes are opened together, the tank is completely filled in 40 hours. In how many hours can pipe B alone empty two-third part of the tank?

पाइप A एक भरने वाला पाइप है, जबकि B और C खाली करने वाले पाइप हैं। अकेले पाइप A एक टंकी को 10 घंटे में भर सकता है और अकेले पाइप C भरी टंकी को 24 घंटे में खाली कर सकता है। यदि तीनों पाइपों को एक साथ खोल दिया जाए, तो टंकी 40 घंटे में पूरी तरह से भर जाती है। पाइप B अकेले टंकी के दो-तिहाई भाग को कितने घंटे में खाली कर सकता है?


Option “B” is correct.

Given:

Pipe A fills in 10 hours

Pipe C empties in 24 hours

Pipes A, B, and C together fill in 40 hours

Concept Used: 

Capacity of a tank is LCM of time taken by all pipes to fill and empty the tank.

Efficiency of Pipe = Capacity of a tank/ Time taken to fill the tank

Calculation:

Capacity of a tank is the LCM of 10, 24, and 40

⇒ Capacity of a tank = 120 liters.

Efficiency of Pipe A = 120/10

⇒ Efficiency of Pipe A = 12 liters/hour

Efficiency of Pipe C = 120/24

⇒ Efficiency of Pipe C = 5 liters/hour

Efficiency of Pipe A – B – C(– because B and C empties the tank) = 120/40

⇒ Efficiency of Pipe A – B – C = 30 liters/hour

Putting the values of result (I) and (II), in above equation

⇒ 12 – B – 5 = 3

⇒ 7 – B = 30

⇒ – B = 3 – 7 = – 4

⇒ B = 4 liters/hour  

Now B needs to empty 2/3rd of a tank,

⇒ 2/3 × 120

⇒ 80

Time taken by Pipe B,

⇒ 80/4 = 20 

∴ Time taken by Pipe B to empty the 2/3rd tank is 20 hours.

38. Pipes A, B and C can fill a tank in 15, 30 and 40 hours, respectively. Pipes A, B and C are opened at 6 a.m., 8 a.m. and 10 a.m., respectively, on the same day. When will the tank be full?
पाइप A, B और C क्रमशः 15, 30 और 40 घंटे में एक टैंक भर सकते हैं। पाइप A , B और C उसी दिन क्रमशः सुबह 6 बजे, 8 बजे और 10 बजे खोले जाते हैं। टंकी कब भर जाएगी?


Option “B” is correct.

Given:

A tank is filled by pipes A, B, C in 15, 30, and 40 hours respectively

A, B, C are opened at 6 am , 8 am and 10 am respectively

Calculation:

Pipe A fill the tank in 15 hours

Pipe A fill the tank in 1 hours = 1/15

Pipe B fill the tank in 30 hours

Pipe B fill the tank in 1 hour = 1/30

Pipe C fill the tank in 40 hour

Pipe C fill the tank in 1 hour = 1/40

Pipe A work done since it is opened at 6 am to 10 am i.e for 4 hours

Then , pipe A works in 4 hours = 1/15 × 4 = 4/15

Similarly, Pipe B works done since it is opened to 8 am to 10 am i.e for 2 hours

Then, pipe A works for 2 hours = 1/30 × 2 = 2/30

Total work done from 6 am to 10 am = 4/15 + 2/30

⇒ (8 + 2)/30 = 1/3

Now, Remaining work = 1 – 1/3 = 2/3

Now , 2/3 work done by  (A + B + C) together after 10 am

Work done by (A + B + C)’s in 1 day = 1/15 + 1/30 + 1/40

⇒ (8 + 4 + 3)/120 = 15/120 = 1/8

Now , 1/8 part of work done by (A + B + C)’s in 1 hour 

1 part of work done by  (A + B + C)’s = 8 hours

2/3 part of work done by (A + B + C)’s = 8 × 2/3 = 16/3 hours = 5 hours 20 minutes

Then , Time taken to fill the tank = 10 am + 5 hours 20 min = 3 : 20 pm

The tank will full at 3:20 pm

39. If two pipes function simultaneously, a tank is filled in 12 hours. One pipe fills the tank 10 hours faster than the other. How many hours does the faster pipe alone take to fill the tank?

यदि दो पाइप एक साथ कार्य करते हैं, तो एक टंकी 12 घंटे में भर जाता है। एक पाइप टंकी को दूसरे की तुलना में 10 घंटे तेजी से भरता है। टंकी को भरने के लिए तेज पाइप को अकेले कितने घंटे लगते हैं?


Option “B” is correct.

Given:

Two pipes working simultaneously can fill a tank in 12 hours.

One pipe fills the tank 10 hours faster than the other.

Calculation:

Let the slower pipe can fill the tank in x hours,

Then faster pipe can fill the tank in (x – 10) hours.

Tank filled by slower pipe in 1 hour = 1/x

Tank filled by faster pipe in 1 hour = 1/(x – 10)

Part filled by both the pipes in 1 hour = 1/12

According to the question

1/x + 1/(x – 10) = 1/12

⇒ [(x – 10) + x]/x(x – 10) = 1/12

⇒ 12(x – 10) + 12x = x(x – 10)

⇒ 12x – 120 + 12x = x2 – 10x

⇒ x2 – 34x + 120 = 0

⇒ x2 – 30x – 4x + 120 = 0

⇒ x(x – 30) – 4(x – 30) = 0

⇒ (x – 30)(x – 4) = 0

⇒ x = 30, 4

But x = 4 is not possible since (x – 10) i.e. (4 – 10) will give a negative value

Hence slower pipe can fill the tank in 30 hours and

Faster pipe can fill the tank  in = 30 – 10 hours = 20 hours

∴ The faster pipe alone take to fill the tank in 20 hours

40. Pipes A and B can fill a tank in 43.2 minutes and 108 minutes, respectively. Pipe C can empty it at 3 litres/minute. When all the three pipes are opened together, they fill the tank in 54 minutes. The capacity (in litres) of the tank is:

पाइप A और B क्रमशः 43.2 मिनट और 108 मिनट में एक टंकी भर सकते हैं। पाइप C इसे 3 लीटर/मिनट में खाली कर सकता है। जब तीनों पाइप एक साथ खोले जाते हैं, तो वे 54 मिनट में टंकी को भर देते हैं। टंकी की क्षमता (लीटर में) है:


Option “B” is correct.

Given : 

Pipe A can fill the tank in 43.2 minutes 

Pipe B can fill the tank in 108 minutes 

Pipe C can empty 3 litres/minute 

All of them take 54 minutes to fill the tank

Formula used :

Pipe A is filling pipe and take time a to fill the tank  

Pipe B is filling pipe and take time b to fill the tank

Pipe C is a drain pipe and takes time c to empty the tank 

If all of them take total ‘t’ time to fill the tank then 

1/t = (1/a) + (1/b) – (1/c) 

Calculation:

Let the time taken by pipe C be ‘c’ minutes 

According to the question 

1/54 = (1/43.2) + (1/108) – (1/c) 

⇒ c = 72 minutes 

The capacity of the tank = water drain per minutes by pipe C × Total time 

⇒ 3 × 72 

⇒ 216 litres 

∴ The capacity of the tank will be 216 litres 

Alternate Method

Formula used:

Efficiency = Total work/Time

Calculation:

LCM of 43.2, 108 and 54 = 432 units = Total work

Efficiency of A = 432/43.2 = 10 units/min

Efficiency of B = 432/108 = 4 units/min

Efficiency of (A + B + C) = 432/54 = 8 units/min

Efficiency of C = (10 + 4 – 8) units/min

⇒ 6 units/min

Capacity of tank = (432/6) × 3 litres

⇒ (72 × 3) litres

⇒ 216 litres

∴ The capacity of tank is 216 litres

 

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