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# if x and y are both integers and x>y, is x positive? 1)

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Senior Manager
Joined: 17 Jul 2007
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if x and y are both integers and x>y, is x positive? 1) [#permalink]

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27 Jul 2007, 17:56
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if x and y are both integers and x>y, is x positive?

1) y<0
2) 2x+y+4>x+y+4

I get how 1 is Insufficient, but not grasping 2.

Last edited by zakk on 28 Jul 2007, 07:40, edited 2 times in total.

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Senior Manager
Joined: 04 Jun 2007
Posts: 344

Kudos [?]: 36 [0], given: 0

Re: DS: Easy problem [#permalink]

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27 Jul 2007, 22:44
zakk wrote:
if x and y are both integers and x>y, is x positive?

1) y<0>x+y+4

I get how 1 is Insufficient, but not grasping 2.

Problem with HTML I think. Please repost after disabling HTML.

Kudos [?]: 36 [0], given: 0

Senior Manager
Joined: 04 Jun 2007
Posts: 344

Kudos [?]: 36 [0], given: 0

Re: DS: Easy problem [#permalink]

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28 Jul 2007, 08:01
zakk wrote:
if x and y are both integers and x>y, is x positive?

1) y<0
2) 2x+y+4>x+y+4

I get how 1 is Insufficient, but not grasping 2.

I think statement 2 is also insufficient.
2x+y+4 > x+y+4
can be written as
x + (x+y+4) > (x+y+4)

If (x+y+4) > 0, then x > 0.
But, if (x+y+4) < 0, then x > 0 or x <0.

Kudos [?]: 36 [0], given: 0

Director
Joined: 26 Feb 2006
Posts: 900

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Re: DS: Easy problem [#permalink]

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28 Jul 2007, 08:12
sumande wrote:
zakk wrote:
if x and y are both integers and x>y, is x positive?

1) y<0
2) 2x+y+4>x+y+4

I get how 1 is Insufficient, but not grasping 2.

I think statement 2 is also insufficient.
2x+y+4 > x+y+4
can be written as
x + (x+y+4) > (x+y+4)

If (x+y+4) > 0, then x > 0.
But, if (x+y+4) < 0, then x > 0 or x <0.

B. st 2 should be sufficient:

2x + y + 4 > x + y + 4
2x > x
2x - x > 0
x > 0

Kudos [?]: 165 [0], given: 0

Senior Manager
Joined: 04 Jun 2007
Posts: 344

Kudos [?]: 36 [0], given: 0

Re: DS: Easy problem [#permalink]

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28 Jul 2007, 09:31
Himalayan wrote:
sumande wrote:
zakk wrote:
if x and y are both integers and x>y, is x positive?

1) y<0
2) 2x+y+4>x+y+4

I get how 1 is Insufficient, but not grasping 2.

I think statement 2 is also insufficient.
2x+y+4 > x+y+4
can be written as
x + (x+y+4) > (x+y+4)

If (x+y+4) > 0, then x > 0.
But, if (x+y+4) < 0, then x > 0 or x <0.

B. st 2 should be sufficient:

2x + y + 4 > x + y + 4
2x > x
2x - x > 0
x > 0

You are right. I don't know what I was thinking.

Kudos [?]: 36 [0], given: 0

Re: DS: Easy problem   [#permalink] 28 Jul 2007, 09:31
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# if x and y are both integers and x>y, is x positive? 1)

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