Bunuel wrote:
Is \(x> y\)?
(1) \(\frac{1}{x} < \frac{1}{y}\)
(2) \(\frac{1}{x} > 1\)
Official solution from Veritas Prep.
This inequality problem fairly clearly includes the concept of reciprocals, but less conspicuously invokes the concept of positive/negative number properties. With statement 1, the "obvious" cases seem to support that \(x>y.\) If you take the statement that \(\frac{1}{x}<\frac{1}{y}\) and plug in \(x=3\) and \(y=2\), that holds with the statement \(\frac{1}{3}<\frac{1}{2}\) and means that \(x>y\). And if both variables are negative, the same can hold. For \(\frac{1}{x} <\frac{1}{y}\) to hold where x and y are negative, then \(\frac{1}{x}\) has to be farther from zero. So that might look like \(x=−2\) and \(y=−3\), where x is still greater than y. But think of a case where the x term will always be smaller than the y term: where x is negative and y is positive. Then the process of taking reciprocals doesn't matter: \(\frac{1}{x}\) is going to still be negative, and \(\frac{1}{y}\) is still going to be positive. So in this case, you get the answer "no," that x is not greater than y. So statement 1 is not sufficient, and that is because of the fact that x could be negative while y is positive.
Statement 2 alone is certainly not sufficient, as it does not provide enough information about y. You know from this statement that \(0<x<1\), because the reciprocal of x is greater than 1. But y still has infinite possibilities.However, when you take the statements together, statement 2 rules out the "exception" for statement 1. It guarantees that x is positive, meaning that both x and y are positive. And that then allows you to simply cross-multiply without changing the sign (since there are no negatives in that multiplication). \(\frac{1}{x} <\frac{1}{y}\) then becomes \(y<x\), proving that the answer is "yes" and making C the correct response.
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