Two taps running together can fill a tank in

Let one tap fills the tank in $x$ hours.

Therefore, the other tap will fill the tank in $(x+3)$ hours.

Time taken by both taps, running together, to fill the tank $=3 \frac{1}{13}$ hours $=\frac{40}{13}$ hours

Part filled by one tap in 1 hour $=\frac{1}{x}$

Part filled by the other tap in 1 hour $=\frac{1}{x+3}$

Part filled by both taps, running together, in 1 hour $=\frac{1}{x}+\frac{1}{x+3}$

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

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

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

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

$\Rightarrow 13 x^{2}-41 x-120=0$

$\Rightarrow 13 x^{2}-(65-24) x-120=0$

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

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

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

$\Rightarrow x-5=0$ or $13 x+24=0$

$\Rightarrow x=5$ or $x=\frac{-24}{13}$

$\Rightarrow x=5 \quad(\because$ time cannot be a negative fraction $)$

Thus, one pipe will take 5 hour and the other will take $\{(5+3)=8\}$ hour s to fill the tank.