We know that diameter in a circle is half the radius, so the radius of the circle having diameter X is $$\frac{X}{2}$$


Now, the circumference of the circle having diameter $$X$$ would be $$2 \pi r=2 \times \pi \times \frac{X}{2}=X \pi$$

The area of the circle having diameter X would be $$\pi r^2=\pi \times (\frac{X}{2})^2=\frac{X^2 \pi}{4}$$

To compare $$X \pi$$ with $$\frac{X^2 \pi}{4}$$ is same as comparing $$X$$ with $$\frac{X^2}{4}$$

Since nothing is said about the value of X , a unique comparison cannot be formed. For example taking $$\mathrm{x}=2$$, we get $$\mathrm{X}=2 \& \frac{\mathrm{X}^2}{4}=\frac{2^2}{4}=1$$, so column $$A$$ gets higher quantity.

But if we take $$X=8$$, we get $$X=8 \& \frac{X^2}{4}=\frac{8^2}{4}=16$$, so column $B$ gets higher quantity.
Hence the answer is (D).