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Let $x$ and $y$ be the two numbers.

$ x^{2} + y^{2} = 97 \quad \longrightarrow (1) $

The geometric mean cannot exceed the arithmetic mean. $ \boxed{ \text{AM} \geqslant \text{GM}} $

$ \Rightarrow \boxed{\frac{a_{1} + a_{2} + \dots + a_{n} } {n} \geqslant \sqrt[n]{a_{1} a_{2} \dots a_{n}}} $

Now$, \frac{x^{2} + y^{2}}{2} \geqslant \sqrt{x^{2} \cdot y^{2}} $

$ \Rightarrow \frac{x^{2} + y^{2}}{2} \geqslant \sqrt{(xy)^{2}} $

$ \Rightarrow \frac{x^{2} + y^{2}}{2} \geqslant xy $

$ \Rightarrow x^{2} + y^{2} \geqslant 2xy $

$ \Rightarrow 97 \geqslant 2xy \quad [\because \text{From equation (1)}]$

$ \Rightarrow 2xy \leqslant 97 $

$ \Rightarrow xy \leqslant \frac{97}{2} $

$ \Rightarrow \boxed{xy \leqslant 48. 5}$

So$,xy$ cannot be more than $48. 5.$

$\therefore$ Only option $\text{(C)}$ not possible.

Correct Answer $: \text{C}$

$ x^{2} + y^{2} = 97 \quad \longrightarrow (1) $

The geometric mean cannot exceed the arithmetic mean. $ \boxed{ \text{AM} \geqslant \text{GM}} $

$ \Rightarrow \boxed{\frac{a_{1} + a_{2} + \dots + a_{n} } {n} \geqslant \sqrt[n]{a_{1} a_{2} \dots a_{n}}} $

Now$, \frac{x^{2} + y^{2}}{2} \geqslant \sqrt{x^{2} \cdot y^{2}} $

$ \Rightarrow \frac{x^{2} + y^{2}}{2} \geqslant \sqrt{(xy)^{2}} $

$ \Rightarrow \frac{x^{2} + y^{2}}{2} \geqslant xy $

$ \Rightarrow x^{2} + y^{2} \geqslant 2xy $

$ \Rightarrow 97 \geqslant 2xy \quad [\because \text{From equation (1)}]$

$ \Rightarrow 2xy \leqslant 97 $

$ \Rightarrow xy \leqslant \frac{97}{2} $

$ \Rightarrow \boxed{xy \leqslant 48. 5}$

So$,xy$ cannot be more than $48. 5.$

$\therefore$ Only option $\text{(C)}$ not possible.

Correct Answer $: \text{C}$