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

Let time taken by $1^{st}$ tap to fill the tank = x hrs

Time taken by $2^{nd}$ tap = $(x+3)$ hrs

Together they fill in 3$\frac{1}{13}=\frac{40}{13}$hrs

Now,

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

$\Rightarrow \frac{x+3+x}{\left(x\right)(x+3)}=\frac{13}{40}$

$\Rightarrow \frac{2x+3}{x^2+3x}=\frac{13}{40}$

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

$\Rightarrow 13x^2-41x-120=0$ $\Rightarrow 13x^2-65x+24x-120=0$

$\Rightarrow 13x(x-5)+24(x-5)=0$

$\Rightarrow (x-5)(13x+24)=0$

Either $x-5=0or13x+24=0$

$x=5,x=-\frac{24}{13}\text{[Rejected]}$
So, $1^{st}$ tap fill the tank in 5 hrs

and $2^{nd}$ tap fill the tank in $5+3=8$ hrs