Two pipes together can fill a tank in 12 hours.

Let the time required to fill the tank by second pipe be x hours.
Then, the time required to fill the tank by first pipe is (x − 10) hours.

Given:
Two pipes together can fill a tank in 12 hours.

According to the question,

$\frac{1}{x}+\frac{1}{x-10}=\frac{1}{12}$

$\Rightarrow \frac{(x-10)+(x)}{(x)(x-10)}=\frac{1}{12}$

$\Rightarrow \frac{2 x-10}{x^{2}-10 x}=\frac{1}{12}$

$\Rightarrow 24 x-120=x^{2}-10 x$

$\Rightarrow x^{2}-10 x-24 x+120=0$

$\Rightarrow x^{2}-34 x+120=0$

$\Rightarrow x^{2}-30 x-4 x+120=0$

$\Rightarrow x(x-30)-4(x-30)=0$

$\Rightarrow(x-30)(x-4)=0$

$\Rightarrow x=30,4$

But $(x-10)$ is the time required by the first pipe to fill the tank, which is always positive.

Thus, $x=30$ and $x \neq 4$

Hence, the second pipe will take 30 hours to fill the tank.