Question

Two taps running together can fill a tank in `3 1/13` hours. If one tap takes 3 hours more than the other to fill the tank, then how much time will each tap take to fill the tank?

Answer 1

Let one pipe fills the cistern is x hours.

Then the other pipe will fill the cistern is (x + 3) hours.
Given:

Time taken by both pipes, running together, to fill the cistern = `3 1/13 h = 40/13 h`

Part of the cistern filled by one pipe in 1 h = `1/x`

Part of the cistern filled by other pipe in 1 `h = 1/(x+3)`

So, part of the cistern filled by both pipes, running together, in 1 h = `1/x + 1/(x + 3)`

`:. 1/x + 1/(x + 3) = 13/40`

`=> (2x + 3)/(x^2 + 3x) = 13/40`

⇒ 13x² + 39x = 80x + 120

⇒ 13x² − 41x − 120 = 0

⇒ 13x² − 65x + 24x − 120 = 0

⇒ 13x(x − 5) + 24(x − 5) = 0

⇒ (x − 5)(13x + 24) = 0

⇒x − 5 = 0 or 13x + 24 = 0

`=> x = 5 or x  = -24/13`

Since time cannot be negative, so x = 5.

∴ Time taken by one pipe to fill the cistern = 5 hours

Time taken by the other pipe to fill the cistern = 5 + 3 = 8 hours