M03-14

Official Solution:


Is \(x < y\)?

(1) \(x > y^2\).

By testing various numbers, we can easily find instances that result in a NO answer for the question. Select a large positive value for \(x\) and a small positive value for \(y\). For example, if \(x=10\) and \(y=1\), the answer is NO. To obtain a YES answer, we must identify a value for \(y\) that is greater than \(x\) but, when squared, becomes smaller than \(x\). This leads us to consider positive proper fractions (values between 0 and 1). If we set \(x=\frac{1}{3}\) and \(y=\frac{1}{2}\), the answer is YES. As such, the given information is insufficient.

(2) \(x\) and \(y\) are positive integers.

This statement is clearly not sufficient on its own.

(1)+(2) Since from (2) \(y\) is a positive integer, then \(y^2 \geq y\) (with equality when \(y = 1\)). Therefore, from (1), \(x > y^2 \geq y\), which means that \(x > y\). So, the answer to the question "Is \(x < y\)?" is NO. Sufficient.


Answer: C
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Bunuel

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Thanks in advance.

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