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

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New post 08 Dec 2008, 14:27
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A
B
C
D
E

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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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Re: DS: OG suppl. 85 [#permalink]

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New post 08 Dec 2008, 15:49
From 2) we get -z+1 < 0 I.e. z>1 ==> 2) sufficient

1) is not sufficient:
for example, consider (x,y,z)=(-0.5,-0.2,1): x+y+z>0 and z>x+y+1
and for (x,y,z)=(-0.5,-0.2,3): also both inequalities are verified


answer is B

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Re: DS: OG suppl. 85 [#permalink]

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New post 08 Dec 2008, 16:05
I would go for E.

I didn't understand how statement 2 is sufficient as stated by echizen

according to statement 2 , x+y+1 < 0 , so x+y < -1 , so say x+y = -2

from qs, x+y+z>0 , -2+z>0 , z>2 , but that does not necessarily mean z>1 ??
Hence statement 2 is also insufficient.

Whats the QA ? ( pls post it along with explanation if given )

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Re: DS: OG suppl. 85 [#permalink]

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New post 08 Dec 2008, 16:16
botirvoy wrote:
If x+y+z>0, is z>1?
1) z>x+y+1
2)x+y+1<0


B

x+y+z>0
x+y+1+z>1

stmt1 : from z>x+y+1 and x+y+z+1>1 -> z could be 2/3 or >1 - Not Suff
stmt2: since x+y+1<0, z >1 - Suff

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Re: DS: OG suppl. 85 [#permalink]

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New post 08 Dec 2008, 16:17
gameCode wrote:
I would go for E.

I didn't understand how statement 2 is sufficient as stated by echizen

according to statement 2 , x+y+1 < 0 , so x+y < -1 , so say x+y = -2

from qs, x+y+z>0 , -2+z>0 , z>2 , but that does not necessarily mean z>1 ??
Hence statement 2 is also insufficient.

Whats the QA ? ( pls post it along with explanation if given )


Q: x+y+z>0 => x + y > - z
(2): x+y+1<0 => x + y < -1

=> -z < -1 => z > 1 => suff

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Re: DS: OG suppl. 85 [#permalink]

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New post 09 Dec 2008, 17:41
gameCode wrote:
I would go for E.

I didn't understand how statement 2 is sufficient as stated by echizen

according to statement 2 , x+y+1 < 0 , so x+y < -1 , so say x+y = -2

from qs, x+y+z>0 , -2+z>0 , z>2 , but that does not necessarily mean z>1 ??
Hence statement 2 is also insufficient.

Whats the QA ? ( pls post it along with explanation if given )


The two inequalities say x+y<-1, and x+y+z>0, therefore, z will need to be greater than 1 (and in your example, greater than 2), otherwise x+y(which is <-1)+z will be less than 0.

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Re: DS: OG suppl. 85 [#permalink]

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New post 09 Dec 2008, 21:48
botirvoy wrote:
If x+y+z>0, is z>1?
1) z>x+y+1
2)x+y+1<0



1) z>x+y+1
--> z+z >(x+y+z) +1
--> 2z>1 --> z>1/2
Not sufficient

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

B
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Re: DS: OG suppl. 85 [#permalink]

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New post 09 Dec 2008, 22:01
gameCode wrote:
I would go for E.

I didn't understand how statement 2 is sufficient as stated by echizen

according to statement 2 , x+y+1 < 0 , so x+y < -1 , so say x+y = -2

from qs, x+y+z>0 , -2+z>0 , z>2 , but that does not necessarily mean z>1 ??
Hence statement 2 is also insufficient.

Whats the QA ? ( pls post it along with explanation if given )


If Z>2 then it is definitely greater than 1.

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Re: DS: OG suppl. 85 [#permalink]

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New post 09 Dec 2008, 22:26
yeah i realize my mistake in that one. I get confused in this kind of DS. Thanks everyone.

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Director
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Re: DS: OG suppl. 85 [#permalink]

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New post 22 Dec 2008, 06:15
OA is indeed B. Thanks everyone!

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Re: DS: OG suppl. 85 [#permalink]

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New post 28 Dec 2008, 13:46
x2suresh wrote:
botirvoy wrote:
If x+y+z>0, is z>1?
1) z>x+y+1
2)x+y+1<0



1) z>x+y+1
--> z+z >(x+y+z) +1
--> 2z>1 --> z>1/2
Not sufficient

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

B


Nice way of doing it by adding Z on both sides.

Got a B

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Re: DS: OG suppl. 85   [#permalink] 28 Dec 2008, 13:46
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