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Bunuel
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Is x > y? (1) 1/x < 1/y (2) 1/x > 1 21 Nov 2016, 02:42
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C
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Difficulty:

55% (hard)

Question Stats:

58% (01:42) correct 42% (01:32) wrong based on 196 sessions

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Is x > y?

(1) 1/x < 1/y

(2) 1/x > 1
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C
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Is x > y? (1) 1/x < 1/y (2) 1/x > 1 21 Nov 2016, 11:56
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(1) is not sufficient as we cannot decide from the inequality whether x and y are positive or not.

(2) gives x to be positive and less than 1. But is insufficient to answer the question.

(1,2) gives 0<x<1. Let's take x to be 1/2
=> 1/y > 2
=> y has to be positive, and y < 1/2, or y < x.
Therefore together they are sufficient.

Answer is C.


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Is x > y? (1) 1/x < 1/y (2) 1/x > 1 21 Nov 2016, 12:47
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Statement (1) would be sufficient when we know if x and y are positive or negative.

Statement (2) gives 0<x<1 but nothing about y.

Together we can get the soultion
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Is x > y? (1) 1/x < 1/y (2) 1/x > 1 22 Feb 2017, 22:15
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Bunuel
Is \(x> y\)?

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

(2) \(\frac{1}{x} > 1\)
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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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Is x > y? (1) 1/x < 1/y (2) 1/x > 1 20 Jul 2017, 06:44
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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
Show Spoiler
C
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.
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Is x > y? (1) 1/x < 1/y (2) 1/x > 1 20 Jul 2017, 06:57
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natesalzman
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
Show Spoiler
C
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.
Show more

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.
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Is x > y? (1) 1/x < 1/y (2) 1/x > 1 30 Jul 2017, 17:43
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Bunuel
Is x > y?

(1) 1/x < 1/y

(2) 1/x > 1
Show more

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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Is x > y? (1) 1/x < 1/y (2) 1/x > 1 07 Mar 2024, 04:44
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