Two pipes x and y can fill an empty tank in 16 hours and 20 hours respectively. Pipe z alone can empty the completely filled tank in 25 hours. Firstly both pipes x and y are opened and after 6 hours pipe z is also opened. What will be the total time (in hours) taken to completely fill the tank?
Answers
Answer:
10.48 hours ≅ 10.5 hours
Step-by-step explanation:
Let the total capacity of the tank = C
x can fill the tank of C in = 16 hours
The quantity that x can fill in 1 hour = C/16
y can fill the tank of C in = 20 hours
The quantity that y can fill in 1 hour = C/20
z can empty the tank of C in = 25 hours
The quantity that z can empty in 1 hour = C/25
The quantity that x and y can fill the tank of C in 1 hour = (C/16 + C/20) = 9C/80
The quantity that x and y can fill the tank of C in 6 hours = 6(C/16 + C/20) = 54C/80
z is open after 6 hours, so the quantity of x, y and z can fill the tank in 1 hour will change.
So, the quantity that x, y and z can fill the tank in 1 hour = (C/16 + C/20 - C/25)
= 29C/400
Let us convert the calculations into equal units
6 hours past and already filled part of the tank = 54C/80 = (54C/80) x (5/5)
= 270C/400
x, y and z are open
Remaining parts of the tank to be filled = (400 - 270) = 130C/400
For 1 hour -- 29C/400
? how many hours -- 130C/400
Number of hours to fill the remaining part (14C/400) = [(130C/400) x 1] / (29C/400)
= 130/29 = 4.48 hours
Therefore, the total number of hours required for the tank of capacity C = 6 + 4.48 = 10.48 ≅ 10.5 hours