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).