Two taps running together can fill a tank in 3$\frac{1}{13}$ hours

Two taps running together can fill a tank in 3$\frac{1}{13}$ 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

Let one tap fill the tank in x hrs. Therefore, other tap fills the tank in (x + 3) hrs.

Work done by both the taps in one hour is

$$ \frac{1}{x} +\frac{1}{x+3} = \frac{13}{40} $$

(2x + 3) 40 = 13(x² + 3x)

13x² - 41x - 120 = 0

(13x + 24)(x - 5) = 0

x = 5 

(rejecting the negative value)

Hence one tap takes 5 hrs and another 8 hrs separately to fill the tank.