OG DS Question - Disagree with OE

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OG DS Question - Disagree with OE [#permalink]

03 Sep 2004, 11:18

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Hi,

Went through this OG question for DS. Didn't quite agree with the explanation. Wonder if anyone has a better idea.

If x+y+z > 0, is z > 1?

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

I'll post the OE after some discussion.

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03 Sep 2004, 11:49

B is sufficient because if X+Y<-1 then Z must be more that 1 in order for X+Y+Z to be more than 1. My impression is that A is correct as well, but I think it might be a trap answer. So I will guess is D but i have feeling I am making a mistake.

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03 Sep 2004, 12:00

Yeah, I also came to choice D

st 1.
I didn't use numbers but:
z > x + y + 1
z - x - y > 1

x + y + z > 0
z > -x - y

st 2.
I used numbers and came to all that z > 1

I am honest, and did not look that careful to this question.
(6 hours GMAT today cant be that healthy...time for a break)

Please correct me if I am wrong.

Regards,

Alex

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03 Sep 2004, 12:10

Here's what the OE did, which I wasn't entirely convinced.

For statement 1: z > x+y+1

It says statement 1 is not sufficient. What it did was to use x + y + z > 0, so x+y > -z, so that using (1), z > x+y+1>-z+1, impllies z > 0.5. What happens here is it took x+y to be equal to -z and substituted it inside the inequality. But x+y is not equals -z, it is greater than -z. So x+y can be anything, say, -z+1, or -z+20 etc. So if x+y=-z+20, then z > -z+20+1, 2z > 21, z > 21/2 !

For statement 2: x+y<-1, so z>1 for x+y+z>0.

What do you think of the OE for statement (1) ?

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OG is right [#permalink]

03 Sep 2004, 13:05

Statement 1 is insufficient while statement 2 is sufficient.
1) is insufficient cause -
We have x + y + z > 0 given and by 1) z> x + y + 1.

Take x = -.75, y = 0.25 and z = 0.75. We satisfy both the equations.
Yet z<1.
Take x = 1, y = 1 and z =4. We again satisy both equations but z>1

So insufficient.

Mathematical proof.
Add both inequatities - Addition does not change signs.
we get x + y + 2z > x + y + 1
Which is 2z > 1 or z > 1/2

Thus if the question were - is z > 1/2 then answer would be D.
Hope this ends the debate....
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03 Sep 2004, 15:40

My 1st draft :

1) x+y+z>0 & z>x+y+1

id est 2z>x+y+z+1>1 which means that z>1/2 No

2) x+y+z>0 <=> x+y>-z

id est -z+1 <x+y+1<0 => z>1 OK

I would choose answer B

[#permalink] 03 Sep 2004, 15:40

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OG DS Question - Disagree with OE

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