If the square of a two-digit number is reduced by the square of the number formed by reversing the digits of the number, the final result is :

Answer: Option D

Let the two-digit number be 10x + y
Then, number formed by reversing the digits = 10y + x
Difference of square of the numbers :
$$={\left(10x+y\right)}^2-{\left(10y+x\right)}^2$$
$$=\left(100x^2+y^2+20xy\right)-$$     $$\left(100y^2+x^2+20xy\right)$$
$$=99\left(x^2-y^2\right)$$     which is divisible by both 9 and 11