A takes 10 days than the time taken by B to finish a piece of work. If both A and B together can finish the work in 12 days,

Solution:

Let B takes $x$ days to complete the work.

Therefore, A will take $(x-10)$ days.

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

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

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

$\Rightarrow x^{2}-10 x=12(2 x-10)$

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

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

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

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

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

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

$\Rightarrow x=30$ or $x=4$

Number of days to complete the work by B cannot be less than that by A; therefore, we get:

$x=30$

Thus, B completes the work in 30 days.