Question

A cistern has two pipes. One pipe can fill it with water in 8 hours and the other can empty it in 5 hours. In how many hours will be emptied if both the pipes are opened together when 3/4th of the cistern is alrrady full of water ?​

Answer 1

Answer:

Part of the tank filled by the first pipe in 1 hour = 1/8

Part of the tank emptied by the second pipe in 1 hour = 1/5

Therefore, part of the tank emptied when both the pipes are opened together = 1/5 – 1/8 = 3/40

Therefore, 3/40 part of the tank is emptied in 1 hour

Therefore, 3/4th of the tank will be emptied in

$( \frac{40}{3} \times \frac{3}{4} ) \\ = 10 \: \: hours$

Answer 2

Answer:

Take Lcm of 5&8= 40( this is the capacity of the cistern)

40*3/4= 30 part is filled with water

capacity of both the pipes is 8-5=3

part of the cistern to be emptied is 30 we will decide it by 3

30/3 hence it will take 10 hour to completely empty the cistern.