Question

Two pipes can fill a tank in 10 hours and 12
hours while a third pipe empties the full
tank in 20 hours. If all the three pipes
operate simultaneously, in how much
time will the tank be filled? *​

Answer 1

Answer:

7 hours and 30 minutes

Step-by-step explanation:

Let u be the capacity of tank.

and x be the time in hours required to fill the tank if all three pipes operate simultaneously.

Capacity of water filled by tank A in 10 hours = u

Capacity of water filled by tank A in 1 hours = u/10

Capacity of water filled by tank A in x hours = (x*u)/10

Capacity of water filled by tank B in 12 hours = u

Capacity of water filled by tank B in 1 hours= u/12

Capacity of water filled by tank B in x hours = (x*u)/12

Capacity of water emptied by tank C in 20 hours =u

Capacity of water emptied by tank C in 1 hours =u/20

Capacity of water emptied by tank C in x hours = (x*u)/20

According to question:

(x*u)/10 + (x*u)/12 - (x*u)/20 = u

x*u(1/10 + 1/12 - 1/20)=u

on solving we get

x= 7.5 hours