Two different numbers are taken from the set {0, I ,2,3,4,5,6,7,8, 9, 10).The probability that their sum and  positive difference, are both multiple of 4, is $\frac{x}{55}$ then x equals

$\text{ Let the two numbers are ' }a\text{ ' and ' }b\text{ ' }$

$\text{ According to the question }a+b=4p\text{ and }a-b=4q\text{ where }p\text{, }q\in I$

$\begin{array}{l}\Rightarrow 2a=4(p+q)\text{ and }2b=4(p-q)\\ \Rightarrow a=2I_1\text{ and }b=2I_2\end{array}$

Hence both a and b must be even. 

$\text{ Hence }n\left(S\right){=}^{11}{C}_2$

$\begin{array}{l}n\left(A\right)=(0,4),(0,8),(2,6),(2,10),(4,8),(6,10)=6\\ \therefore P\left(A\right)=\frac{6}{{ }^{11}{C}_2}=\frac{6}{55}\end{array}$