Dear Reader, Below are three practice problems on Pipes and Cisterns.

Question 1

Fieasta Fountain Square is one of the largest housing projects in Bangalore. It has over 300 flats, a swimming pool, play area, community hall, walking lawns etc. In one of the blocks named Arjun, there are two water tanks A and B. A is much smaller than B. Water fills at the rate of one litre every hour in A. At the end of first hour tank B gets filled with 10 litres of water. At the end of second hour the capacity raises to 20 litres. At the end of third hour the capacity raises to 40 litres and so on. If 1/16th of the total capacity of tank B is filled after 24 hours , what is the total time required for tank B to get completely filled?

(a) 26 hours (b) 27 hours (c) 29 hours (d) 28 hours

Answer : d) 28 hours

Solution:

Though two tanks namely A and B are discussed, the question relates to Tank B only. Hence any data about Tank A can be ignored.

Based on data given, the capacity of tank B that is getting filled is doubled compared to the previous hour.

1/16 of the tank gets filled in 24 hours. Therefore double of 1/16 i.e 2(1/16) or 1/8th of the tank will be filled in 25 hours.

Similarly 1/4th of the tank will get filled in 26 hours.
1/2th of the tank will get filled in 27 hours
Finally the full capacity will be reached in 28 hours.

Question 2

Arunodaya Enclave has 20 flats. Water from three different sources are fed into the cistern. Three taps can fill the cistern in 10mins, 15mins and 18mins respectively. The cistern being empty, all the three taps are kept open by the watchman of the enclave. After 3mins, the watchman closes the third tap. After third tap is closed, how many minutes would be required by the first two taps to fill the cistern completely.

(a)1 min (b) 2 min (c) 3 min (d) 4 min

Answer : b) 2 min

Solution:

In 1 minute the first tap can fill 1/10th of the cistern
In 1 minute the second tap can fill 1/15th of the cistern
In 1 minute the third tap can filll 1/18th of the cistern
Therefore, the capacity that can be filled by all the three taps simultaneously in 1 minute = 1/10 + 1/15 + 1/18 = 2/9th of the total capacity of cistern
For 3 minutes, the capacity filled by three taps operating simultaneously = 3 x 2/9 = 2/3rd of the total capacity.
At the end of the 3rd minute, 3rd tap is closed.
Remaining capacity that is to be filled by the first and second taps alone = 1 - 2/3 = 1/3rd of the total capacity.
the capacity that can be filled by first two taps simultaneously in 1 minute = 1/10 + 1/15 = 5/30 = 1/6th of the total capacity.

Capacity Filled            Time Required
1/6                            1 minute
1/3                             ?

1/3rd capacity of the cistern (remaining capacity after third tap is closed) can be filled by first two taps in 1/3/1/6 = 2 minutes.
So option (b) is correct.

Question 3

Srivari Riveria is a big housing complex in Coimbatore. Gaint tanks are placed in every complex building to cater to the needs of the residents. In a block named Annapurna, three taps P, Q and R can fill a tank in 12, 15 and 30 hours respectively. The caretaker of the complex has instructions to keep tap P open all the time and Q and R are to be opened for one hour alternately. When will the tank in Annapurna become full?

a) 5 hrs b) 6 hrs c) 7 hrs d) 8 hrs

Answer : c) 7 hrs

Solution:

In 1 hour P can fill 1/12th of the tank
In 1 hour Q can fill 1/15th of the tank
In 1 hour R can fill 1/30th of the tank

For the first two hours P is kept open for the entire duration of 2 hours while Q and R are opened for 1 hour each.

Therefore, capacity that will be filled in 2 hours by all the pipes

= Capacity filled by first pipe in 2 hours + Capacity filled by second tap in 1 hour + Capacity filled by third tap in 1 hour
= 2(1/12) + 1/15 + 1/30
= 1/6 +1/15+1/30 = 4/15th of the tank.

The process is repeated and hence in 6 hours (3 x 2 hours), 4/15x3 = 4/5th of the tank will be filled. In the seventh hour, the remaining 1/5th of the tank will be filled. The entire tank thus will be full in 7 hours.