Question

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 1

Two taps can fill a tank in $3\frac{1}{13}$ hours. One tap takes $3$ hours more than the other tap to fill the tank. Let, tap $1$ takes "x hours" of time, the tap $2$ takes "$x+3$ hours" of time.

In one hour tap $1$ does $\frac{1}{x}$ of work, i.e., fill $1_{/x}$ of tank. In one hour tap $2$, fills $\frac{1}{x+3}$ of tank.

$(\frac{1}{x}+\frac{1}{x+3})\left(3\frac{1}{13}\right)=1$

$\frac{2x+3}{x^2+3x}\left(\frac{40}{13}\right)=1$

$80x+120=13x^2+39x$

$13x^2-41x-120=0$.

$x=10$ hours.

So, one tap takes $10$ hours, and other takes $13$ hours each to fill the bank.