Question

A tank can be filled up by two taps in 6 hours. The smaller tap alone takes 5 hours more than the bigger tap alone. Find the time required by each tap to fill the tank separately.

Answer 1

Let the bigger tap alone take x hours to fill the tank.

Then the smaller tap alone takes (x + 5) hours to fill the tank.

The bigger tap fills `1/x` part of the tank in 1 hour, and the smaller tap fills `1/(x+5)` part of the tank in 1 hour.

∴ both the taps together fill `(1/x + 1/(x + 5))` part of the tank in 1 hour. Both the taps together fill the tank in 6 hours. (Given)

∴ Both the taps together fill `1/6` part of the tank in 1 hour.

∴ `1/x+1/(x+5)=1/6`

∴ `(x+5+x)/(x(x+5))=1/6`

∴ `(2x+5)/(x^2+5x)=1/6`

∴ 6(2x + 5) = x² + 5x

∴ 12x + 30 = x² + 5x

∴ x² + 5x − 12x - 30 = 0

∴ x² − 7x − 30 = 0

∴ x² − 10x + 3x − 30 = 0

∴ x(x − 10) + 3(x − 10) = 0 

∴ (x − 10) (x + 3) = 0

∴ x − 10 = 0 or x + 3 = 0

∴ x = 10 or x = −3

But the time cannot be negative.

∴ x = −3 is unacceptable.

∴ x = 10 and x + 5 = 10 + 5 = 15.

The bigger tap alone fills the tank in 10 hours and the smaller tap alone in 15 hours.