Croatia Mathematical Competitions

From Vieta's formulae we have $a + b = 10c$ and $c + d = 10a$. By adding these equations we get
$$a + b + c + d = 10(a + c).$$
Since $a$ is a solution of the equation $x^2 - 10c x - 11d = 0$, and $d = 10a - c$, it follows that
$$0 = a^2 - 10a c - 11d = a^2 - 10a c - 11(10a - c) = a^2 - 110a + 11c - 10a c.$$
Analogously, we get
$$c^2 - 110c + 11a - 10a c = 0.$$
By subtracting these equations it follows that
$$(a - c)(a + c - 121) = 0.$$
Since $a \neq c$, we get $a + c = 121$. Therefore, $a + b + c + d = 10 \cdot 121 = 1210$.