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Relevant wiki: Vieta's Formula Problem Solving - Intermediate

Using Vieta's on first and second equation: $c+d=10a~,~a+b=10c$

Adding we get:- $a+b+c+d=10(a+c)..(1)$ Subtracting we get:- $b-d=11(c-a)..(2)$

Since $c$ is a root of first equation: $\implies c^2-10ac-11b=0$ and since $a$ is a root of second equation: $\implies a^2-10ac-11d=0$

Subtracting these two we get:- $c^2-a^2=11(b-d)=121(c-a)~$ (Using $2$ ) $\implies c+a=121$ Substituting in $1$ we get:-

$a+b+c+d=10(121)=\boxed{1210}$