If the 7-digit number x468y05is divisible by 11, then whatis the value of (x + y)?

Given numbers $$x 4 68y05 $$

For divisibility by 11 difference of the sum of digits at the alternate place is taken and made divisible by 11

then $$(x +6+y+5) - (4+8+0)$$

$$\Rightarrow (x+y+11)-12$$

$$\Rightarrow \dfrac (x+y-1){11}$$

Now (x+y) should be such number that on subtracting 1 from it we should get a number divisible by 11 so the probable 

the answer would be  12 is $$(x=y)= 12 

As $$ \dfrac{(x+y)-1}{11} $$

$$\Rightarrow \dfrac {12-1}{11}$$

$$\Rightarrow \dfrac {11}{11}= 1 $$

Ans = 12