Question

A pipe can fill a cistern in 3 hours
due to leak in the bottom it is filled in four hours
when the cistern is full in how much time will it be emptied by the leak​

Answer 1

Answer:

Let V = volume of cistern, gal

RateP = volumetric rate of fill pipe, gal/min

tP = 10 hrs (time to fill V with no leaks)

V = RateP × tP

or RateP = V/tP

also,

V = RateL × t

where: RateL = volumetric leak rate, gal/min

t = time to drain V alone = ?

and RateL = V/t

Combined balance with both fill and leak pipes open:

V = RateP×(12 hr) - RateL×(12 hr)

V = (V/tP)×(12 hr) - (V/t)×(12 hr)

and solve for t:

1/(12 hr) = 1/tP - 1/t

1/t = 1/tP - 1/(12 hr) = 1/(10 hr) - (1/12 hr)

t = 1/( 1/(10 hr) - 1/(12 hr) ) = 60 hr

Answer 2

Question :-

A pipe can fill a cistern in 3 hours due to leak in the bottom it is filled in 4 hours when the cistern is full in how much time will it be emptied by the leak?

When there is no leakage :-

Time taken by Pipe to fill the cistern = 3hours.

Work done by Pipe in 1 hours = 1/3

When there is no leakage.

Time taken by Pipe to fill the Tank = 4

Work done by Pipe in 1 hours = -1/4

Work done by the leak in 1 hours = 1/3- 1/4 = 4-3/12 = 1/12

Hence the cistern will be emptied by leakage in 12 hours.