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, $a+b+c+d=10(c+a)$   ... (i)

and $(a+b)-(c+d)=10(c-a)$

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

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

Thus, $c^2-10ac=11b,a^2-10ca=11d$

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

From Eq. (i), we get

a+b+c+d= 121 x 10 = 1210