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# is xy >0 1) x-y > -2 2 ) x-2y < -6

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Manager
Joined: 20 Mar 2005
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is xy >0 1) x-y > -2 2 ) x-2y < -6 [#permalink]

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06 Aug 2007, 00:06
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is xy >0

1) x-y > -2
2 ) x-2y < -6

Kudos [?]: 104 [0], given: 0

Intern
Joined: 06 Aug 2007
Posts: 15

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07 Aug 2007, 11:07
Here is what I am thinking:

Question is asking if x and y are both positive or both negative.

stmt 1: insuff b/c it says nothing about the sign (+/-) for x and y
stmt 2: insuff b/c it says nothing about the sign for x and y

stmt 1 & 2:
stmt 1 can be rewritten as: x+2>y
stmt 2 can be rewritten as: x+6<2y which is also: x/2 + 3 < y
combine the stmts to get: x/2 + 3 < y < x+2
therefore, if x is positive, y must be positive
or if x is negative, then y must be negative

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VP
Joined: 09 Jul 2007
Posts: 1098

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Location: London

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07 Aug 2007, 11:18
trancing wrote:
Here is what I am thinking:

Question is asking if x and y are both positive or both negative.

stmt 1: insuff b/c it says nothing about the sign (+/-) for x and y
stmt 2: insuff b/c it says nothing about the sign for x and y

stmt 1 & 2:
stmt 1 can be rewritten as: x+2>y
stmt 2 can be rewritten as: x+6<2y which is also: x/2 + 3 < y
combine the stmts to get: x/2 + 3 < y < x+2
therefore, if x is positive, y must be positive
or if x is negative, then y must be negative

let's say X=Y so the sign should be +or- in that case, only statement A alone is sufficient, no need for additional calculation
any suggestions

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Intern
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07 Aug 2007, 11:34
but if x = 1 and y = -1 then A is still true.

A alone is not sufficient.

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Intern
Joined: 06 Aug 2007
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07 Aug 2007, 12:14

If I plug in values for X, say -8, -1, etc... the output doesn't look convincing,
Say X=-8, then -1<Y<-6??

Am I overlooking something here?

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Intern
Joined: 06 Aug 2007
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07 Aug 2007, 12:33
hmm....good point. But I think it just means that Y is not defined with X = -8

For example, let X = -8 and plugging into the original equations in the question you will get y > -1 and y < -6
which means that Y is not defined when X = -8

let me know if you have another way to approach this problem.

Balvinder, can you give us the answer?

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07 Aug 2007, 12:33
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