Question

Q) Two taps, which can fill a tank in 12 hours and 36 hours, are opened
simultaneously. When the tank was supposed to be full, it was found that
only (5/6)th of it was full due to a leak at the bottom. Find the time in which
the remaining part of the tank would be filled (in hours).​

Answer 1

Answer:

Cistern filled by both pipes in one hour = 1/12+1/20=2/15th

Therefore both pipes filled the cistern in 15/2hrs.

Now, due to leakage both pipes filled the cistern in 15/2+30/60=8hrs.

Therefore Due to leakage, filled part in one hour = 1/8

Therefore part of cistern emptied, due to leakage in one hour = 2/15-1/8= 1/120th

In 120 hrs, the leak would empty the cistern

Answer 2

Answer:

1.8 hours

Step-by-step explanation:

Lets say the two taps are A and B and lets call the leak as L

Now, the time taken by A&B is 12 and 36 hours, so lets assume the capacity of the tank to be 36litres.

So, now the rate of A&B is 3 and 1 respectively.

Combined rate of A&B is 4

Only 5/6th of tank is full, which means 5/6*36=30litres is filled instead of 36litres, that is because of a leak(L)

So we need to find the combined rate of A,B and L, lets say x

If 30litres is filled instead of 36litres, then the rate would be x instead of 4

Therefore x=4*(36/30)=10/3, i.e. x=10/3litres/hour

Therefore rate of leak L is 10/3 -(3+1)= -2/3/hour

For A,B and L to fill the 36 litres together would take t=work/combined rate(x), i.e. 36litres/(10/3)litres/hour=10.8 hours

It would take total 10.8 hour to fill all of the tank, but till now only 5/6th of the tank is full in 9 hours, therefore time taken for the remaining 1/6th of tank to fill is 10.8-9=1.8 hours.