take 1) alone and add z both the side
=> z+ z > x + y + z +1
=> 2z > 1
=> z > .5 so not sufficient as z = .7 then no and if z =1.2 then yes
take 2) alone and add z both the side
=> x + y + z +1 < z
=> 1 < z thus sufficient.
So answer is B
variables in DS
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smackmartine
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IMO B
Given x+y+z> 0
Asked : whether z>1?
1) z> x+y+1
if x = 3, y =5 , z> 3+5+1 => z> 9 ,So z>1
if x=-3 , y= -5, z> -3-5+1 => z> -7 ,So z can be -6,-5......1, 2... so z may or may not be greater than 1
Insufficient.
2) x + y + 1 < 0
also , x+y+z> 0 (given)
adding 1 on both side , we get
x+y+z+1> 1
(x + y + 1) + z >1
as (x + y + 1)<0 , for above inequality to be true z >1
because , say (x + y + 1) = -1
so (x + y + 1) + z >1 can be written as -1+z >1 => z>2 , Sufficient
So B
Given x+y+z> 0
Asked : whether z>1?
1) z> x+y+1
if x = 3, y =5 , z> 3+5+1 => z> 9 ,So z>1
if x=-3 , y= -5, z> -3-5+1 => z> -7 ,So z can be -6,-5......1, 2... so z may or may not be greater than 1
Insufficient.
2) x + y + 1 < 0
also , x+y+z> 0 (given)
adding 1 on both side , we get
x+y+z+1> 1
(x + y + 1) + z >1
as (x + y + 1)<0 , for above inequality to be true z >1
because , say (x + y + 1) = -1
so (x + y + 1) + z >1 can be written as -1+z >1 => z>2 , Sufficient
So B
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GMATGuruNY
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One approach is to link the inequalities by rephrasing them in terms of x+y.amar66 wrote:If x + y + z > 0, is z > 1?
(1) z > x + y +1
(2) x + y + 1 < 0
Please explain the methodology to solve these type of questions.
This approach will yield an inequality in which z is the only remaining variable.
Given information: x+y+z > 0.
Isolating x+y, we get:
-z < x+y.
Statement 1: z > x + y + 1.
Isolating x+y, we get:
x+y < z-1.
Linking together -z < x+y and x+y < z-1, we get:
-z < x+y < z-1
-z < z-1
1 < 2z
z > .5.
Thus, it is possible that z<1, that z=1, or that z>1.
Insufficient.
Statement 2: x+y+1 < 0.
Isolating x+y, we get:
x+y < -1.
Linking together -z < x+y and x+y < -1, we get:
-z < x+y < -1.
-z < -1.
z > 1.
Sufficient.
The correct answer is B.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
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SoCan
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This boils down to the same method irock and GMATGuru showed, but is a slightly different way to look at it. I like to add the inequalities where possible. Just make sure the sign is facing the same way. It would be even clearer if the formatting let things line up properly, but you should still be able to see it
1)
x+y+z>0
-x-y+z>1
========
2z>1, or z>.5
2)
x+y+z>0
-x-y+0>1
========
z>1, sufficient
Again, this is basically the same thing the other two approaches took, but just presents it differently.
1)
x+y+z>0
-x-y+z>1
========
2z>1, or z>.5
2)
x+y+z>0
-x-y+0>1
========
z>1, sufficient
Again, this is basically the same thing the other two approaches took, but just presents it differently.
SoCan,
Your way seems pretty straightforward when trying to solve this problem but I had one question pertaining to your reasoning on statement 2. Looking at statement 2 it is safe to assume when you have a 0 solely on the other side of the inequality we must transfer a value to the other side so we are truly comparing two different values?
Your way seems pretty straightforward when trying to solve this problem but I had one question pertaining to your reasoning on statement 2. Looking at statement 2 it is safe to assume when you have a 0 solely on the other side of the inequality we must transfer a value to the other side so we are truly comparing two different values?
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SoCan
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I don't know if I quite understand your question, so let me know if I'm not answering it.factor26 wrote:SoCan,
Your way seems pretty straightforward when trying to solve this problem but I had one question pertaining to your reasoning on statement 2. Looking at statement 2 it is safe to assume when you have a 0 solely on the other side of the inequality we must transfer a value to the other side so we are truly comparing two different values?
You could leave the zero on the side and still add the two expressions. You'd just get
z-1>0
and you'd end up adding 1 to both sides anyway.
If you're asking about the 0 in
-x-y+0>1
I just put the zero there to emphasize that there's no z in that inequality.
. Anyway thanks again!!
