Two pipes running together can fill a tank in

Let the first pipe takes $x$ minutes to fill the tank. Then the second pipe will takes $=(x+5)$ minutes to fill the tank.

Since, the first pipe takes $x$ minutes to fill the tank.

Therefore, portion of the tank filled by the first pipe in one minutes $=\frac{1}{x}$

So, portion of the tank filled by the first pipe in $11 \frac{1}{9}$ minutes $=\frac{100}{9 x}$

Similarly,

Portion of the tank filled by the second pipe in $11 \frac{1}{9}$ minutes $=\frac{100}{9(x+5)}$

It is given that the tank is filled in $11 \frac{1}{9}$ minutes.

So,

$\frac{100}{9 x}+\frac{100}{9(x+5)}=1$

$\frac{100(x+5)+100 x}{9 x(x+5)}=1$

$100 x+500+100 x=9 x^{2}+45 x$

$9 x^{2}+45 x-200 x-500=0$

$9 x^{2}-155 x-500=0$

$9 x^{2}-180 x+25 x-500=0$

$9 x(x-20)+25(x-20)=0$

$(x-20)(9 x+25)=0$

But, $x$ cannot be negative.

Therefore, when $x=20$ then

$(x+5)=20+5=25$

Hence, the first water tape will takes 20 min to fill the tank, and the second water tape will take 25 min to fill the tank.