If x^2-10 a x-11 b=0 have roots c and d and x^2-10 c x-11 d=0 have...

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 c²-10 a c-11 b=0 Similarly, $a$ is the root of $$\begin{aligned} \Rightarrow& x²-10 c x+11 d=0 \\ & a²-10 c a-11 d=0 \end{aligned}$$On subtracting Eq. (v) from Eq. (iv), we get \begin{aligned}& \left(c^2-a²\right)=11(b-d) \\ & \therefore \quad(c+a)(c-a)=11 \times11(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}