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If x>y, then is 1/x < 1/y?

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If x>y, then is 1/x < 1/y?  [#permalink]

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New post 28 Nov 2018, 10:01
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Question Stats:

66% (01:41) correct 34% (01:23) wrong based on 64 sessions

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If x>y, then is 1/x < 1/y?

A. x is negative

B. y is negative

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Re: If x>y, then is 1/x < 1/y?  [#permalink]

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New post 28 Nov 2018, 21:30
A) x is negative.
Case 1: x = neg y = neg, eg.x = -3, y = -5. Answer would be YES.
Case 2: x = neg y = pos ....but wait if this happens then x cannot be greater than y. Thus, discard case 2.

A is sufficient.

B) case 1: similar like that in A)..yes
Case 2: x = pos y = neg
x = 3 y = -5
This doesn't satisfy our condition. So, No for this case. Therefore, not sufficient.

Hence, A.

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Re: If x>y, then is 1/x < 1/y?  [#permalink]

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New post 29 Nov 2018, 17:52
Top Contributor
fitzpratik wrote:
If x>y, then is 1/x < 1/y?

A. x is negative
B. y is negative


Here's another approach....

Given: y < x

Target question: Is 1/x < 1/y?

Statement 1: x is negative
Since we already know that y < x, we can conclude that y is also NEGATIVE
If x and y are both NEGATIVE, then the product xy is POSITIVE
Since xy is POSITIVE, we can safely take the target question,Is 1/x < 1/y?, and multiply both sides by xy.
When we do so, we get the NEW target question: Is y < x?
Since it's given that y < x, the answer to the NEW target question is YES, y IS less than x
Since we can answer the NEW target question with certainty, statement 1 is SUFFICIENT

Statement 2: y is negative
There are several values of x and y that satisfy statement 2. Here are two:
Case a: x = -1/3 and y = -1/2. In this case, 1/x = -3 and 1/y = 2. So, the answer to the target question is YES, 1/x is NOT less than 1/y
Case b: x = 1/2 and y = -1/2. In this case, 1/x = 2 and 1/y = -2. So, the answer to the target question is NO, 1/x is NOT less than 1/y
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

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Re: If x>y, then is 1/x < 1/y?  [#permalink]

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New post 17 Apr 2019, 12:22
From the question stem
\(\frac{1}{x}-\frac{1}{y}<0\) -> \(\frac{y-x}{xy}<0\)?
since \(x>y\) -> \((y-x)<0\) then \(\frac{negative}{xy}<0\)

Two cases

\(\frac{negative}{positive}<0\)
\(\frac{negative}{negative}\) becomes positive


Simplified question stem is xy>0 (or simpler question stem does x & y have same signs?)

a) x is negative,
y must also be negative as x>y so sufficient

b) y is negative
since x>y
x can be positive, then NO
x can be negative then Yes
Not sufficient.

Answer A
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Re: If x>y, then is 1/x < 1/y?   [#permalink] 17 Apr 2019, 12:22
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