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Can someone solve this D.S. problem for me...

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Can someone solve this D.S. problem for me... [#permalink] New post 03 Nov 2005, 15:08
This is my first post, please help. I don't understand the answer and will appreciate good explanation. Thanks in advance

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

(1) z>x+y+z
(2) x+y+1<0
Manager
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 [#permalink] New post 03 Nov 2005, 15:29
I got D!

Here's my try..

Given x+y+z>0
or z > -(x+y)

Q: Is z>1?
Since we know z > -(x+y), it will suffice to know if -(x+y) > 1 that way we know z>1
In other words, treat -(x+y) as a constant K. If such constant K is >1 then z will be >1 because we already know that z > K

-(x+y) > 1 simplifies as x+y <1?

(1) z>x+y+z
Simplifies as x+y <0 => x+y can be -0.1, -1, -6,..
Therefore x+y <1
SUFF => AD

(2) x+y+1<0
Simplifies as x+y <-1 => x+y can be -1.1, -2, -6,..
This means x+y <1
SUFF => D
Director
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 [#permalink] New post 03 Nov 2005, 15:30
B

A) Insuff.
If x+y+z=0.5,z=0.6 -> Z>1 false
If x+y+z=9,z=10 -> Z>1 True
B) suff.
x+y<-1, Given x+y+z>0
x+y = -1.1, -2, -3.3...
To make x+y+z>0, z has to be >= |x+y| => z can be 1.1, 2, 3.3 ...
Hence z is always > 1
Manager
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 [#permalink] New post 03 Nov 2005, 15:40
Changing my answer to B. Here's why..

Given x+y+z>0
or z > -(x+y)

Q: Is z>1?

(1) z>x+y+z
Simplifies as x+y <0 => x+y can be -0.1, -1, -6,..
Therefore -(x+y) will be 0.1, 1, 6,..
Since z > -(x+y), z can be <, =, or > 1
NOT SUFF => BCE

(2) x+y+1<0
Simplifies as x+y <-1 => x+y can be -1.1, -2, -6,..
Therefore -(x+y) will be 1.1, 2, 6,..
Since z > -(x+y), z>1 all the time
SUFF => B
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 [#permalink] New post 03 Nov 2005, 23:28
(1) Insufficient

(2) Sufficient: combine x+y+z > 0 and x + y + 1 < 0 and you get z > 1

Answer is B.
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Re: Can someone solve this D.S. problem for me... [#permalink] New post 04 Nov 2005, 00:57
norae wrote:
This is my first post, please help. I don't understand the answer and will appreciate good explanation. Thanks in advance

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

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


It is B.
From 1, we get z > 0 so it may or may not be greater than 1, so insufficient.

From 2 we get (X+Y)< 1, so z > 1 to make x+y+z > 0.

Hence statement 2 is sufficient.
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Thanks for all your help but..... [#permalink] New post 04 Nov 2005, 07:55
I just realised I made a mistake in the options I provided
The question is correct but the options should have been

(1) Z>X+Y+1
(2) X+Y+1<0
[/size]

Any help you can give me is appreciated.
Manager
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 [#permalink] New post 04 Nov 2005, 08:44
norae wrote:
I just realised I made a mistake in the options I provided
The question is correct but the options should have been

(1) Z>X+Y+1
(2) X+Y+1<0
[/size]

Any help you can give me is appreciated.


I would still say the answer as B. B is sufficient as per the very elaborate explanations given before.

I find A insufficient because:-

Z>X+Y+1
Z-1>X+Y
Considering Z-1=X+Y and substituing it in X+Y+Z>0 ,we get Z-1+Z>0
2Z-1>0;Z>1/2. IF it was a number greater than 1, we could have said A is sufficient. But in this case when the sign is equality then we find Z>1/2 .So the real Z should be greater than 1/2. It still can be 3/4 or 1.2. So I would say A is insufficient.
  [#permalink] 04 Nov 2005, 08:44
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