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# if x+y+z>0 is z>1? 1) z>x+y+1 2) x+y+1<0 I said

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Intern
Joined: 02 Aug 2003
Posts: 8
Location: FL
Followers: 0

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

if x+y+z>0 is z>1? 1) z>x+y+1 2) x+y+1<0 I said [#permalink]  07 Sep 2003, 15:38
if x+y+z>0 is z>1?

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

I said "d" but the answer says "B" Can someone explain why?
Manager
Joined: 26 Aug 2003
Posts: 233
Location: United States
Followers: 1

Kudos [?]: 2 [0], given: 0

The first choice doesn't give enough information about x and y (or x + y). Either one of them can be 0 and other positive or both could be positive... many possiblities. BUT the second choice leads you to see that x + y is less than -1. So for x + y + z > 0 to be true, 'z' has to be greater than 1.
SVP
Joined: 30 Oct 2003
Posts: 1794
Location: NewJersey USA
Followers: 5

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

From A)
x = 1 y = -2 z = 0.5
Z > x+y+1 because 0.5 > 0 and also x+y+z > 0 but z < 1
x = 1 y = -2 z = 2
Z > x+y+1 because 0.5 > 0 and also x+y+z > 0 but z > 1
So we cannot say for sure

From B)
x+y+1 < 0
or X+y < -1
since x+y < -1 then to make x+y+z > 0 z has to be greater than 1

So B is sufficient.
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