Here, $a+b=10 c$
$$
c+d=10 a
$$
On subtracting Eq. (ii) from Eq. (i), we get, $\quad(a-c)+(b-d)=10(c-a)$ $\Rightarrow \quad(b-d)=11(c-a)$
Since, $c$ is the root of $x^2-10 a x+11 b=0$
$$
\Rightarrow \quad c^2-10 a c-11 b=0
$$
Similarly, $a$ is the root of
$$
\begin{aligned}
\Rightarrow & x^2-10 c x+11 d=0 \\
& a^2-10 c a-11 d=0
\end{aligned}
$$
On subtracting Eq. (v) from Eq. (iv), we get
$$
\begin{aligned}
& \left(c^2-a^2\right)=11(b-d) \\
& \therefore \quad(c+a)(c-a)=11 \times 11(c-a) \quad\{\text { from (i) \& (ii) } \\
& \Rightarrow \quad c+a=121 . \\
& \therefore \quad a+b+c+d=10 c+10 a \\
& =10(c+a)=1210 \\
&
\end{aligned}
$$