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This is an excellent question on rates with a few trick elements thrown in. This means that you can expect a question of this sort when you are in the 650 – 700 range.

The question is based on independent work, specifically related to Pipes & Cisterns.

Attachment:
15th May 2019 - Reply 1.JPG
15th May 2019 - Reply 1.JPG [ 20.71 KiB | Viewed 19367 times ]

Let us try to solve this question by taking the total capacity of the tank ( which represents the work to be done by the pipes ) to be the LCM of 60 and 90. The LCM of 60 and 90 is 180.
But, we also have a leak present at 3/4th the height of the tank. It is given that it takes 36 minutes to empty TILL 3/4th the height of the tank. This is where you need to be careful in interpreting the data given.

For leaks/emptying pipes, the reference is always a completely filled tank. Similarly, for taps/filling pipes, the reference is always an empty tank.

So when the question says that the leak takes 36 minutes to empty TILL 3/4th height of the tank, it means that it takes 36 minutes to empty the TOP quarter of the tank. The entire solution of this question hinges on whether you interpreted this data right or not.

If a leak takes 36 minutes to empty 1/4th of the tank, it will take 144 minutes (36 * 4) to empty the whole tank. So, now we have to take the LCM of 60, 90 and 144. The LCM of these values is 720.

So, let the capacity of the tank be 720 litres.
720 litres is filled by pipe A in 60 minutes; this means pipe A fills at the rate of 12 litres per minute.
720 litres is filled by pipe B in 90 minutes; this means pipe B fills at the rate of 8 litres per minute.
If the leak was at the bottom of the tank, it could have emptied the whole tank i.e. 720 litres in 144 minutes; this means, the leak empties at the rate of 5 litres per minute.

Beyond this stage, we will have to work this problem out in two parts.

Part 1 – Till 3/4th of the tank is filled, only pipe A and pipe B are working. When both of them work together, they can fill at the rate of 20 litres per minute. Therefore, they will take 27 minutes to fill 3/4th of the volume i.e. 540 litres.

Part 2 – Once 3/4th of the tank is filled, along with pipe A and pipe B, the leak will also start functioning. Therefore, the effective work done by all the three together will be (12+8 – 5) i.e. 15 litres per minute. At this rate, it will take 12 minutes to fill the remaining 180 litres.
Therefore, the total time taken = 27 + 12 = 39 minutes. So, the correct answer option is A.

The language in this problem could prove to be a barrier to understanding it in the right way. Once that is taken care of, the solution is rather simple. It pays, therefore, to spend time on reading the question carefully before deciding on your approach.

Hope this helps! :)
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ArvindCrackVerbal

Thank you for putting detailed explanation to my same solution posted above.

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Two pipes A and B can fill a tank in 60 min and 90 min respectively. A leak is present at \(\frac{3}{4}\) of the height. The leak alone takes 36 min to empty till \(\frac{3}{4}\) height of the tank. Find the time taken to fill the tank when all of the taps and the leak are opened simultaneously.

A. 39 min
B. 27 min
C. 37 min
D. 33 min
E. 45 min


A+B together takes 36 mins to fill the tank

1/60 + 1/90 = 180/5 = 36 mins

To fill the tank upto 3/4 height , pipes will take 3/4(36) = 27 mins.

so they will take 9 mins to fill the last 1/4 of the tank.

For remaining 9 mins => 1/9 - 1/36 (leak) = 12 mins [I can not understand this part]

Total = 27 + 12 = 39 mins.
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DisciplinedPrep
Two pipes A and B can fill a tank in 60 min and 90 min respectively. A leak is present at \(\frac{3}{4}\) of the height. The leak alone takes 36 min to empty till \(\frac{3}{4}\) height of the tank. Find the time taken to fill the tank when all of the taps and the leak are opened simultaneously.

A. 39 min
B. 27 min
C. 37 min
D. 33 min
E. 45 min

If Pipe A and B alone are working, time taken to fill completely is Reci(1/60 + 1/90) = Reci (5/180) = 180/5 = 36 min

Leak takes 36 min to empty 1/4th tank. So it takes 144 min to empty entire tank.
So, If Pipe A and B and Leak are working together, time taken will be to completely fill tank (assuming leak at the bottom of tank) Reci(1/60 + 1/90 - 1/144) = Reci((12 + 8 - 5)/720) = Reci(15/720) = 48 min.

Now, Pipe A and B work till 3/4th height, So time to fill till 3/4th height = 3/4x36 = 27 min
and Pipe A, B and leak work together for 1/4th height, So time taken to fill remaining 1/4th of the height = 1/4x48 = 12 min

So, Total time = 27 + 12 = 39 min
Option A
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MofeBhatia
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Two pipes A and B can fill a tank in 60 min and 90 min respectively. A leak is present at \(\frac{3}{4}\) of the height. The leak alone takes 36 min to empty till \(\frac{3}{4}\) height of the tank. Find the time taken to fill the tank when all of the taps and the leak are opened simultaneously.

A. 39 min
B. 27 min
C. 37 min
D. 33 min
E. 45 min


A+B together takes 36 mins to fill the tank

1/60 + 1/90 = 180/5 = 36 mins

To fill the tank upto 3/4 height , pipes will take 3/4(36) = 27 mins.

so they will take 9 mins to fill the last 1/4 of the tank.

For remaining 9 mins => 1/9 - 1/36 (leak) = 12 mins [I can not understand this part]

Total = 27 + 12 = 39 mins.

When both A and B are working , they can fill the tank in 36 mins. So, lets just say , it's 36 units of work

Now in 27 mins , they are completing 3/4W of the work i.e. 27 units

Remaining work (36-27) 9 units. This work is completed in 12 mins because of the leakage.

=> 1/9 - 1/36 => 12 mins

3/4 of work is completed in 27 mins and 1/4 of work is completed in 12 mins ( 3 mins more than required) because of the leakage.
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DisciplinedPrep
Two pipes A and B can fill a tank in 60 min and 90 min respectively. A leak is present at \(\frac{3}{4}\) of the height. The leak alone takes 36 min to empty till \(\frac{3}{4}\) height of the tank. Find the time taken to fill the tank when all of the taps and the leak are opened simultaneously.

A. 39 min
B. 27 min
C. 37 min
D. 33 min
E. 45 min

Say the capacity of the tank is 180 units. So 1/4 of the tank will be 45 units and 3/4 of the tank will be 135 units.

So pipe A fills 3 units per min, pipe B fills 2 units per min.
Leak removes 45 units in 36 mins i.e. 5/4 units per min.

To fill 135 units (3/4 of the tank), pipe A and B together take 135/(3+2) mins = 27 mins
To fill rest of the 45 units, both pipes and leak together take 45/(3+2 - 5/4) = 12 mins

Total time taken = 39 mins

Answer (A)
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DisciplinedPrep
Two pipes A and B can fill a tank in 60 min and 90 min respectively. A leak is present at \(\frac{3}{4}\) of the height. The leak alone takes 36 min to empty till \(\frac{3}{4}\) height of the tank. Find the time taken to fill the tank when all of the taps and the leak are opened simultaneously.

A. 39 min
B. 27 min
C. 37 min
D. 33 min
E. 45 min

rate of A =1/60
rate of B=1/90
total time to fill tank = 5/180 ; 36 mins
let total tank vol ; 4
so its rate ; 4/36 ; 1/9 ; 9 mins to fill
36-9 = 27 mins for 3/4 vol
since 1/36 is filling and 1/9 ; 1/9-1/36 ; 3/36 = 1/12 ; 12 mins this total vol-inflow = outflow
so 27+12 ; 39 mins total time
IMO A
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DisciplinedPrep
Two pipes A and B can fill a tank in 60 min and 90 min respectively. A leak is present at \(\frac{3}{4}\) of the height. The leak alone takes 36 min to empty till \(\frac{3}{4}\) height of the tank. Find the time taken to fill the tank when all of the taps and the leak are opened simultaneously.

A. 39 min
B. 27 min
C. 37 min
D. 33 min
E. 45 min

Case 1: The leak is at 3/4th height from the bottom of the tank, this means, till 3/4th height, A and B together will fill the tank without any water leaks.
Case 2: After that, water will leak for the remaining 1/4th height, and consecutively, A and B will keep filling too.

For case 1, A and B's efficiency 1/60 + 1/90 = 1/36 i.e. 36 mins. So to fill 3/4th of the height they will take 36*3/4=27 mins.
Now for case 2, for 1/4th of the height, A and B will take 36*1/4=9 mins and the leak will take 36 mins to release that water. Therefore, 1/9 - 1/36 = 1/12 i.e. 12 mins.
Therefore total time taken in 27+12=39 mins(A).
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