Is The Information Enough?

Since $a$ and $b$ are the roots of the quadratic equation $x^2 - 10cx - 11d$ , we have the following equations:

$a + b = 10c$ ...... $(1)$
$ab = -11d$ ......... $(2)$
$a^2 - 10ac -11d =0$ ......... $(3)$

Similarly, $c$ and $d$ are the roots of the quadratic equation $x^2 - 10ax - 11b$ , we have the following equations:

$c + d = 10a$ ...... $(4)$
$cd = -11b$ ......... $(5)$
$c^2 - 10ac -11b =0$ ......... $(6)$

Multiplying equation 2 and 5, we get,
$abcd = 121bd$ $\Rightarrow ac=121$ ...... $(7)$

Subtracting equation 4 from 1, we get,
$b - d = 11(c - a)$ ...... $(8)$

Subtracting equation 6 from 3, we get,
$a^2 - c^2 - 11d + 11b = 0$
$\Rightarrow (a+c)(a-c) = 11(d-b)$
$\Rightarrow (a+c)(a-c) = 121(a-c)$ (Using equation 8)
Since all of them are distinct, $(a+c) = 121$ .... $(9)$

Adding equation 1 and 4, we get,
$a+b+c+d = 10(a+c)$ ........ $(10)$

Combining equation 9 and 10, we get
$\boxed{a+b+c+d = 1210}$

my solution is same as that of Mr.sudeep

by the basic knowledge of quadratic eqn, a+b =10c and ab=-11c c+d =10a and cd=-11b clearly , abcd=121bd implies that ac=121......................................(1) now ab+cd=-11(b+d) ......................................(2) Also a+b+c+d =10(a+c) implies that b+d=9(a+c).........(3) now from (1) and (2) and (3) a^2 + c^2 -99(a+c)-20ac=0 (a+c)^2 -22*121-99(a+c)=0 from this, a+c=121 or a+c=-22 But if a+c = -22 then a=c=-11 and a,c are distinct So , a+c=121 and b+d =9(a+c) so a+b+c+d=10(a+c)=1210

a+b = 10c ---(i) ab = -11d ---(ii) c+d = 10a ---(iii) cd = -11b ---(iv)

(i) + (iii) gives a+b+c+d = 10(a+c) ---(v)

or, b+d = 9(a+c) ---(vi)

Multiply (i) by a gives: a2-ab = 10ac or, a2-11d = 10ac ---(vii)

Multiply (iii) by c gives: c2-cd = 10ac or, c2-11b = 10ac ---(viii)

(vii)+(viii) gives a2+c2 -11(b+d) = 20ac

Using (vi) we get a2+2ac +c2 -99(a+c) = 22ac or, (a+c)2 -99(a+c) = 22ac ---(ix)

Using (ii) and (iv) ab = -11d or, ab = -11(-11*b/c) or, ac = 121

Replacing the value of ac in (ix) (a+c)2 -99(a+c) = 22*121

Solving this we get a+c = 121

So a+b+c+d = 10(a+c) = 10*121 = 1210