Question

Taps A & B can fill a cistern in 20 & 30hr respectively. Both pipes are open to fill the tank but when the tank is 1/4 full a leak develop im the bottom of the tank, through which 1/5 of the water supply by both pipes leak out. Find how much time the tank will full​

Answer 1

Solution :-

Given : Taps A & B can fill a cistern in 20 & 30hr respectively.

(A + B)'s one hour work = (1/20 + 1/30)

= (3 + 2)/60

= (5/60) = 1/12

Given : Both pipes are open to fill the tank but when the tank is 1/4 full a leak develop in the bottom of the tank, through which 1/5 of the water supply by both pipes leak out.

1 work completed by A & B = 12 hours

1/4 work completed by A & B = 12/4 = 3 hours

Remaining work = 1 - 1/4

= (4 - 1)/4 = 3/4

1/5 (A + B)'s work = (1/20) × (1/5) = 1/100

1 work completed by 1/5 (A + B) = 100 hours

3/4 work completed by 1/5 (A + B) = (3 × 100)/4 = 75 hours

Total time taken by A & B to complete the work together = (3 + 75) hours = 78 hours.

Hence,

A & B can fill the tank in 78 hours.