We know that:

\(\frac{y^2}{x^{12}}\) is an integer

\(=> \frac{y^2}{x^{12}} ≥ 1\)

\(=> y^2 ≥ x^12\)

\(=> y ≥ x^6\) (since x and y are both greater than 1, i.e. they are both positive, we can take the square root without changing the signs)

Taking square root on both sides (ignoring the negative part since x and y are positive):

\(\sqrt{y}\) \(≥ x^3\)

However, since \(x > 1\), we must have: \(x^3 > x^2\)


Thus, we can conclude that:

\(\sqrt{y} ≥ x^3 > x^2\)

\(=> \sqrt{y} > x^2\)

Thus, Quantity B is greater than Quantity A

Answer B