B) 2ay³;г) -6p2q;

$$\frac{2ay^3}{-4a^2b} = -\frac{y^3}{2ab}$$

$$\frac{-6p^2q}{-2q^3} = \frac{3p^2}{q^2}$$

Ответ: $$\frac{3p^2}{q^2}$$