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

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Current Student
Joined: 11 May 2008
Posts: 555

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

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05 Sep 2008, 20:00
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

Kudos [?]: 222 [0], given: 0

VP
Joined: 17 Jun 2008
Posts: 1374

Kudos [?]: 406 [0], given: 0

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05 Sep 2008, 20:07
arjtryarjtry wrote:
If x + y + z > 0, is z > 1?
(1) z > x + y +1
(2) x + y + 1 < 0

IMO B

(1) z> x+y+1 => z-1>x+y => z-1+z>x+y+z =>2z-1>0 (from given conditions) => z>1/2 => z can be < 1 or >1 =>INSUFFI
(2)x+y+1<0 => x+y+1<x+y+z (From given conditions)=> z>1 SUFFI
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Manager
Joined: 03 Jun 2008
Posts: 133

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Schools: ISB, Tuck, Michigan (Ross), Darden, MBS

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06 Sep 2008, 05:45
spriya wrote:
arjtryarjtry wrote:
If x + y + z > 0, is z > 1?
(1) z > x + y +1
(2) x + y + 1 < 0

IMO B

(1) z> x+y+1 => z-1>x+y => z-1+z>x+y+z =>2z-1>0 (from given conditions) => z>1/2 => z can be < 1 or >1 =>INSUFFI
(2)x+y+1<0 => x+y+1<x+y+z (From given conditions)=> z>1 SUFFI

IMO : spriya's right
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Senior Manager
Joined: 04 Jan 2006
Posts: 276

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06 Sep 2008, 07:13
arjtryarjtry wrote:
If x + y + z > 0, is z > 1?
(1) z > x + y +1
(2) x + y + 1 < 0

IMO, B.

Let's simplify the question
Assume x + y = a
If a + z > 0 ---------- [1]
is z > 1?

(1) z > a + 1
Or z - a > 1 ---------- [2]
[1] + [2]; 2z > 1
z > 1/2
z could be less or more than 1 INSUFF

(2) a + 1 < 0
-a > 1 ---------- [3]
[1] + [3]; z > 1
SUFF

Kudos [?]: 43 [0], given: 0

Re: is z>0?   [#permalink] 06 Sep 2008, 07:13
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