Let $a$ and $b$ be the roots of the equation $x^2 -10cx - 11d=0$ and those of $x^2-10ax-11b=0$ are $c, d$ find the value of $a+b+c+d$ when $a\ne b\ne c\ne d.$

Given $a, b$ are the roots of $x^2-10cx-11d=0$ and $c, d$ are the roots of $x^2-10ax-11b=0.$

Thus, $a+b=10c,c+d=10a$

and $ab=-11d,cd=-11b$

So, $a+b+c+d=10(c+a) ...\left(i\right)$

and $(a+b)-(c+d)=10(c-a)$

$\Rightarrow (b-d)=11(c-a) ...\left(ii\right)$

Also, $a$ is a root of $x^2-10cx-11d=0$ and $c$ is a root of $x^2-10ax-11b=0.$

Thus, $c^2-10ac=11b,a^2-10ca=11d$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{c}^2-a^2=11(b-d)\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{c}^2-a^2=11(b-d)=11\times 11(c-d)\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}c+a=11\times 11=121\end{array}$

From Eq. (i), we get, 

$a+b+c+d=121\times 10=1210.$