We know that the number x belongs to a set of prime numbers less than 10 which implies that x can take any value out of $\((2,3,5,7)\)$ and the number $y$ belongs to a set of prime numbers greater than 10 , so $\(y\)$ can take any value out of $\((11,13,17,19 \&\)$ so on)

It is clear that the value of $\(y\)$ must be an odd number whereas $\(x\)$ can be either even or odd.

If we consider $\(\mathrm{x}=$ even prime $=2 \& \mathrm{y}\)$ any odd prime number, we get column A quantity as $\((-1)^{(x+y)}=(-1)^{(\text {even+odd })}=(-1)^{(\text {odd })}=-1 \& \quad\)$ column $\(\quad \mathrm{B}\)$ quantity comes out to be $\((-1)^{(\mathrm{xy})}=(-1)^{(\mathrm{cven} \mathrm{\times odd})}=(-1)^{(\mathrm{cven})}=1\)$ which implies column B has higher quantity.

Next if we take $x$ as well as $y$ as odd prime numbers we get $\((-1)^{(\mathrm{x}+\mathrm{y})}=(-1)^{(\text {odd }+ \text { odd })}=(-1)^{(\text {even })}=1 \&(-1)^{(\mathrm{xy})}=(-1)^{(\text {odd } \times \text { odd })}=(-1)^{(\text {odd })}=-1\)$ which gives column A higher than column B

As a unique relation cannot be formed between column A \& column B quantities, the answer is (D).