$\underline{\mathbf{Answer: A}}$
$\underline{\mathbf{Explanation:}}$
The discriminant of the equation $\mathrm{x^2 + px + q = 0}$ is $\Delta_1 = \mathrm{p^2-4q}$
and discriminant of $\mathrm{x^2 = rx = s = 0}$ is given by $\Delta_2 = \mathrm{r^2 - 4s}$
Now, $\Delta_1 + \Delta_2 = \mathrm{p^2 - 4q + r^2 -4s = p^2 + r^2 -4(q+s) = p^2 + r^2 -2pr = (p-r)^2} \;[\text{if}\;\mathrm{pr = 2(q+s)}]$
$\therefore \Delta_1 + \Delta_2 = \mathrm{(p-r)^2} \ge 0$
$\because$ Sum of these two $\mathbf{Discriminant}$ is $\mathbf{positive}$.
$\therefore $ At least one of them has to be $\mathbf{positive}$.
$\therefore$ At least one of the equations between $\mathrm{x^2+px+q = 0}$ and $\mathrm{x^2 + rx + s = 0}$ must have real roots.
$\therefore \mathbf A$ is the correct option.