One 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.

Explanation

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 smalller 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)=x^(2)+5x`
`:.12x+30=x^(2)+5x`
`:.x^(2)+5x-12x-30=0`
`:.x^(2)-7x-30=0`
`:.x^(2)-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`.
Ans. The bigger tap alone fills the tank in 10 hours and the smaller tap alone in 15 hours.