A tank has two pipes entering it.when operating together,they fill the tank in 40 minutes.operating independently,one of the pipes fills the tank 60 minutes faster than the other one does. Calculate the time taken by each pipe to fill the tank on its own.
rate of slower --- x units/min
rate of faster ---- y units/min
combined rate = x+y units/min
let the volume of the whole thing have 1 unit
1/(x+y) = 40
x+y = 1/40
1/x - 1/y = 60
(y - x)/xy = 60
60xy = y - x
60xy + x = y
x(60y+1) = y
x = y/(60y+1)
then in x+y = 1/40
y/(60y+1) + y = 1/40
multiply by 40(60y+1)
40y + 40y(60y+1) = 60y+1
40y + 240y^2 + 40y = 60y+1
2400y^2 + 20y - 1 = 0
60y - 1)(40y + 1) = -
y = 1/60 or y is negative, which is not possible
so if y = 1/60, then x = (1/60)/(2) = 1/120
so time of slower = 1/(1/120) = 120 minutes
time for faster = 1/(1/60) = 60 minutes
check:
combined rate = 1/60 + 1/20
= 1/40
time at combined rate = 1/(1/40) = 40 minutes, as given