Is x > y? (1) 1/x < 1/y (2) 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.

Is x>y?

(1) \(\frac{1}{x} < \frac{1}{y}\)
(2) \(\frac{1}{x} > 1\)



Having a problem with the OA. The answer provided is
But they don't consider the possibility of y being a non-integer... I feel like I am probably missing something obvious but it has always been my understanding that one could not assume anything with DS questions.

To answer your question, please check alternative solution below.

Is x>y?

(1) \(\frac{1}{x} < \frac{1}{y}\). If x and y have the same sign, then when cross-multiplying we'll get y < x but if x and y have the opposite signs, when cross-multiplying we'll get y > x. Not sufficient.

(2) \(\frac{1}{x} > 1\). Clearly insufficient because we know nothing about y. Though from this statement we can get that x must be positive.

(1)+(2) Since from (2) x is positive, then from (1) we'll get that (positive) < 1/y, which would mean that y must also be positive. Therefore, when cross-multiplying \(\frac{1}{x} < \frac{1}{y}\), we'll get y < x. Sufficient.

Answer: C.

We need to determine whether x is greater than y.

Statement One Alone:

1/x < 1/y

Statement one alone is not sufficient to answer the question. We see that if x = -2 and y = 4, then x is less than y. On the other hand, if x = 1/2 and y = 1/4, then x is greater than y.

Statement Two Alone:

1/x > 1

Since we don’t know anything about y, statement two is not sufficient to answer the question.

Statements One and Two Together:

Using the information from statement two, we see that x must be positive. Thus, we can reciprocate both sides of the inequality and switch the inequality sign:

1/x > 1

x < 1

Using the information from statement one, we see that y has to be positive if x is positive. Thus, we can reciprocate both sides of the inequality and switch the inequality sign:

1/x < 1/y

x > y

We see that x is indeed greater than y.

Answer: C

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