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# If (x+y)/z >0, is x<0? 1.) x<y 2.) z<0

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Manager
Joined: 24 Dec 2005
Posts: 94

Kudos [?]: 9 [0], given: 0

If (x+y)/z >0, is x<0? 1.) x<y 2.) z<0 [#permalink]

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26 Feb 2006, 10:29
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If (x+y)/z >0, is x<0?

1.) x<y
2.) z<0

Kudos [?]: 9 [0], given: 0

Manager
Joined: 14 Dec 2005
Posts: 73

Kudos [?]: 5 [0], given: 0

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26 Feb 2006, 10:52
If (x+y)/z >0, is x<0?

1.) x<y
2.) z<0

1) is insufficient...pick numbers y = 2, x = 1 or -1 (two cases), z = 3.

2) if Z < 0 , the only way x+y/z >0 if (x+y) <0

x+ y < 0 if x<0, y<0 or x >0 y<0, but x is small positive number and y is a big negative number..

different conditions implies that this is not sufficient as well.

3) 1 and 2 doesn't change the scenario...

my answer is E..what is OA?

Kudos [?]: 5 [0], given: 0

GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5036

Kudos [?]: 436 [0], given: 0

Location: Singapore

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26 Feb 2006, 19:16
1) Could be positive or negative depending on y and z.
E.g x = -1, y =10, z = 5 Then (x+y)/z = -1+10/5 = 9/5 > 0
E.g x = 1, y = 10, z = 5 Then (x+y)/z = 11/5 > 0
E.g x = -10, y = -1, z=-5 Then (x+y)/z = 11/5 > 0

Insufficient.

2) z<0. (x+y) must be negative. (x+y) < 0 ---> x < -y (x must be a negative number smaller than -y).

Sufficient.

Ans B

Kudos [?]: 436 [0], given: 0

VP
Joined: 20 Sep 2005
Posts: 1016

Kudos [?]: 38 [0], given: 0

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27 Feb 2006, 01:37
ywilfred,

For II, x can be greater than 0 and still x+y < 0. e.g. x =5, y = -10.

ywilfred wrote:
1) Could be positive or negative depending on y and z.
E.g x = -1, y =10, z = 5 Then (x+y)/z = -1+10/5 = 9/5 > 0
E.g x = 1, y = 10, z = 5 Then (x+y)/z = 11/5 > 0
E.g x = -10, y = -1, z=-5 Then (x+y)/z = 11/5 > 0

Insufficient.

2) z<0. (x+y) must be negative. (x+y) < 0 ---> x < -y (x must be a negative number smaller than -y).

Sufficient.

Ans B

Kudos [?]: 38 [0], given: 0

VP
Joined: 20 Sep 2005
Posts: 1016

Kudos [?]: 38 [0], given: 0

Re: OG -DS Inequality [#permalink]

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27 Feb 2006, 01:40
C.

I. x<y.....INSUFF x=1,y=3 and x=0, y=3....don't know sign of z.

II. z<0 => x+y < 0..still INSUFF. x=5,y=-10...x > 0 and x=-5,y-10, x < 0.

I+II gives z<0 , x+y< 0 and x<y...only possible if x is negative.

Gordon wrote:
If (x+y)/z >0, is x<0?

1.) x<y
2.) z<0

Kudos [?]: 38 [0], given: 0

Manager
Joined: 24 Dec 2005
Posts: 94

Kudos [?]: 9 [0], given: 0

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27 Feb 2006, 04:07
OA is C. nice explanation lhotseface. Thanks!!

Kudos [?]: 9 [0], given: 0

27 Feb 2006, 04:07
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# If (x+y)/z >0, is x<0? 1.) x<y 2.) z<0

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