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If X+Y+Z >0, is Z>1? 1)Z>X+Y+1 2)X+Y+1<0

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If X+Y+Z >0, is Z>1? 1)Z>X+Y+1 2)X+Y+1<0 [#permalink] New post 23 Oct 2007, 18:27
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If X+Y+Z >0, is Z>1?

1)Z>X+Y+1

2)X+Y+1<0


Please expalin, thanks
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Re: OG 10 DS: #84 [#permalink] New post 23 Oct 2007, 18:43
[quote="garyxb"]If X+Y+Z >0, is Z>1?

1)Z>X+Y+1

2)X+Y+1<0>0 --------(a)

(1) says that Z > X+Y+1

add (1) and (a)

we get X+Y+2Z > X+Y+1

which means Z >1/2 INSUFF

from(2)
X+Y+1<0> X+Y+1 -------(II)

add (a) and (II)

you get z>1
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 [#permalink] New post 23 Oct 2007, 19:08
St1:
z > x+y+1
If x+y = -2, then z > -1. So z could be less than 1 or more than 1. Insufficient.

St2:
x+y+1 < 0
x+y < -1

So z > 1 in order for x+y+z > 0. Sufficient.

Ans B

Last edited by ywilfred on 23 Oct 2007, 19:09, edited 2 times in total.
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 [#permalink] New post 23 Oct 2007, 19:08
Given:
x + y + z > 0
z > -x - y
z > -1(x+y)

z > 1?

1. z > x + y + 1
add: z > -x - y
--------------------
2z > 1

z > 1?
z > 1/2 :INSUFF

2. x + y + 1 < 0
rewrite: x + y < -1
originally given: z > -1(x+y)
original question: z > 1?
one possible solution: z > -1(x+y) > 1?

(-1)(x+y) > 1?
x + y < -1? Yes! :SUFF

Ans: B
  [#permalink] 23 Oct 2007, 19:08
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