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

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Is x > y? (1) 1/x < 1/y (2) 1/x > 1  [#permalink]

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21 Nov 2016, 02:42
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55% (hard)

Question Stats:

58% (01:50) correct 42% (01:32) wrong based on 120 sessions

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Is x > y?

(1) 1/x < 1/y

(2) 1/x > 1

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Joined: 21 Sep 2016
Posts: 4
Re: Is x > y? (1) 1/x < 1/y (2) 1/x > 1  [#permalink]

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21 Nov 2016, 11:56
(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.

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Re: Is x > y? (1) 1/x < 1/y (2) 1/x > 1  [#permalink]

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21 Nov 2016, 12:47
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  [#permalink]

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22 Feb 2017, 22:15
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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20 Jul 2017, 06:44
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.
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Posts: 54401
Re: Is x > y? (1) 1/x < 1/y (2) 1/x > 1  [#permalink]

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20 Jul 2017, 06:57
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natesalzman wrote:
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.

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.

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Re: Is x > y? (1) 1/x < 1/y (2) 1/x > 1  [#permalink]

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30 Jul 2017, 17:43
Bunuel wrote:
Is x > y?

(1) 1/x < 1/y

(2) 1/x > 1

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.

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Re: Is x > y? (1) 1/x < 1/y (2) 1/x > 1   [#permalink] 30 Jul 2017, 17:43
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# Is x > y? (1) 1/x < 1/y (2) 1/x > 1

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