Question

A tank can be filled in 10h but owing to a leakage in its bottom it requires 5h more to fill it. If the cistern is full, in what time can the leak empty it?​

Answer 1

Answer:

your answer is 60 hr

Step-by-step explanation:

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