# Problem on Pipes and Cisterns

**Question 1: ** 9 pumps working 8 hours a day can empty a reservoir in 20 days. How many such pumps needed to empty the same reservoir working 6 hours a day in 16 days.

**Solution: ** Apply formula ** P _{1}D_{1}H_{1} /W_{1}= P_{2}D_{2}H_{2}/W_{2}**

9 x 20 x 8 = P2 x 16 x 6

P

_{2}=

**15 pumps**required

**Question 2: ** A leak can empty a completely filled tank in 10 hours. If a tap is opened in completely filled tank which admits 4 liters of water per minute, then leak takes 15 hours to empty the tank. How many litres of water does the tank hold?

**Solution: ** Take LCM (10, 15) = 30

Let leak pipe is A and A’s efficiency = 30/10 = 3

Let inlet pipe B and B’s efficiency= 30/15 = 2

Pipe A is emptying at 3 units/hour and Pipe B is filling using then emptying rate down to 2 units/hour.

So, filling pipe efficiency is 3 – 2 = 1 unit/ hour.

Pipe B will fill tank in 30/1=30 hours

Filling rate = 4 litre/minute

It will fill 4 x 60 = 240 litre/hour.

Total capacity= 240 x 30 = **7200 litres**

**Question 3: ** A, B and C can fill a tank in 6 hours together. After working for 2 hours, C is closed and A and B fill it in 7 hours more. Find the time taken by C alone to fill the tank?

**Solution: ** Let the total capacity is 42 units.

(A + B + C) per hour work = 42/6 = 7 units.

They all worked for 2 hour.

Total water filled = 7 x 2 = 14 units.

Remaining capacity= 42 – 14 =28 units.

(A+B)’s efficiency= 28/7 = 4 units/hr

C’s efficiency = 7 – 4 =3 units/hr

C alone can fill the tank in 42/3 = **14 hours**

**Question 4: ** A tap drives at a rate of one drop/second and 800 drops = 100 ml. The number of litres water wasted in 30 days of a month is ?.

**Solution: ** 1 sec = 1 drop

Number of seconds in 30 days= 30_{days} x 24_{hrs} x 60_{min} x 60_{sec}

Number of milli-litres wasted = (100 x 30_{days} x 24_{hrs} x 60_{min} x 60_{sec})/800

= 324000 ml

Number of litre= 324000/1000

=**324 litres** water wasted.

**Question 5: ** Two pipes A and B independently can fill a tank in 20 hours and 25 hours. Both are opened together for 5 hours after which the second pipe is turned off. What is the time taken by first pipe alone to fill the remaining portion of the tank?

**Solution: ** Total unit water = LCM(20, 25) = 100 unit

A’s efficiency = 100/20 = 5 unit/hour

B’s efficiency =100/25 = 4 unit/hour

After 5 hour the water filled by A and B together = 5 x 9 =45 unit

Remaining unit = 100 – 45 = 55 unit

Time taken by A alone = 55/5 = **11 hours**

## Recommended Posts:

- Pipes and Cisterns
- Problem on HCF and LCM
- Problem on Numbers
- Problem on Trains, Boat and streams
- Problem on Time Speed and Distance
- Age of AI-based recruitment... What to expect?
- Placement 100 Course Walk-Through
- AMCAT Mock Paper | Verbal Aptitude 2
- AMCAT Mock Paper | Quantitative Aptitude 4
- AMCAT Mock Paper | Quantitative Aptitude 5
- AMCAT Mock Paper | Verbal Aptitude 1
- AMCAT Mock Paper | Verbal Aptitude 5
- AMCAT Mock Paper | Quantitative Aptitude 1
- AMCAT Mock Paper | Logical Aptitude 5