Aptitude - Pipes and Cistern - Discussion

How come 7/3 hours to fill 1 part for the leakage?

Their total tank divided in to 7 parts i.e 20 min to fill 1/3min to fill 1 part

How cum 7/3 hours to fill 1 part for the leakage?

Very well explaination.

I understand very well.

Nice explanation by Paushali Mukhoty.

Paushali Mukhoty
Good explanation thank you.

I'm confused, please check my logic
1. It takes 2 hours for the pump to fill the tank without the leak.
2. The leak adds 1/3hrs onto that time time, which is an increase in time of 1/6th.
3. Then the amount leaked = 1/6 of the amount pumped.
4. When the pumped is turned off the leak will drain the tank at 1/6 of the speed at which the pump fills it
5. If it takes the pump 2hrs to fill it then it should take 12 hours for the leak to drain the tank

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 (1x14/1)hours= 14 hours
So the leak can drain all the water of the tank in 14 hours.

First, let's find the rate of filling up with and without leakage.

Assume that the volume of the tank is 2 liters (for easy calculation).
Then the rate of filling without leak = 1 liter/hour.
Rate of filling with leakage = 2 liters/ (7/3) hours = 6/7 liters per hour.
Now we get the leakage rate ie = 1 - 6/7 = 1/7 liters/hour.

So, time taken = total volume/rate of leakage = 2/ (1/7) = 14 hours.

First, it takes 120 min for leakage it lake (120+1/3*60=120+20)=140 min then LCM(120 and 140 it. It give 840min. 840/60 = 14hours.