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Intern  Joined: 21 Dec 2010
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If x+y+z > 0, is z > 1 ?  [#permalink]

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Difficulty:   55% (hard)

Question Stats: 63% (01:53) correct 37% (02:04) wrong based on 1060 sessions

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

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

This is a question from OG 10, could someone please explain this. The explanation in the OG is not clear enough.

Thanks,
Eddy
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Joined: 02 Sep 2009
Posts: 55681
Re: is z > 1 ?  [#permalink]

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47
31
eddyed911 wrote:
If x+y+z > 0, is z > 1 ?

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

this is a question from OG 10, could someone please explain this. The explanation in the OG is not clear enough.

Thanks,
Eddy

Welcome to Gmat Club Eddy!

Note that:
You can only add inequalities when their signs are in the same direction:

If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.

You can only apply subtraction when their signs are in the opposite directions:

If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.

Back to the original question:

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

(1) z > x + y + 1 --> as the signs are in the same direction we can add two inequalities: $$(x+y+z)+z>x+y+1$$ --> $$2z>1$$ --> $$z>\frac{1}{2}$$, so z may or may not be more than 1. Not sufficient.

(2) x + y + 1 < 0 --> as the signs are in the opposite direction we can subtract two inequalities: $$(x+y+z)-(x+y+1)>0$$ --> $$z-1>0$$ --> $$z>1$$. Sufficient.

Hope it's clear.
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Re: is z > 1 ?  [#permalink]

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6
1
Statement 1:
Given:
z>x+y+1 <---- Equ. 1
&
x+y+z>0

Now add z to both sides of Equ. 1
2z>x+y+z+1
Since x+y+z > 0
2z>GT0+1 (GT0 means a quantity greater than 0)
2z>GT1
z>GT(1/2)
So we cannot conclude that z>1
Therefore, insufficient!

Statement 2:
x+y+1<0
Adding z to both sides of the Equation
x+y+z+1<Z
GT0+1<z
or
GT1<z
Therefore, z>1
Sufficient!

Ans: 'B'

I hope the explanation is clear enough _________________
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Re: is z > 1 ?  [#permalink]

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3
If x+y+z > 0, is z > 1 ?

1) z > x + y + 1

z - 1 > x+ y
and
x+y > -z

z-1 > -z
2z>1

z>1/2

There are infinite real numbers between 1/2 and 1.

Not sufficient.

2)
x + y + 1 < 0
x+y < -1

To make x+y+z>0; z should be greater than 1 because x+y is less than -1.

Sufficient.

Ans: "B"
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Re: is z > 1 ?  [#permalink]

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Thanks guys ! these explanations really clarify things Intern  Joined: 27 Jun 2008
Posts: 47
Location: United States (AL)
Concentration: General Management, Technology
GMAT 1: 660 Q48 V34 WE: Consulting (Computer Software)
Re: is z > 1 ?  [#permalink]

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1
thanks for the expalanation bunuel
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Joined: 16 Oct 2010
Posts: 9331
Location: Pune, India
Re: is z > 1 ?  [#permalink]

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6
2
eddyed911 wrote:
If x+y+z > 0, is z > 1 ?

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

this is a question from OG 10, could someone please explain this. The explanation in the OG is not clear enough.

Thanks,
Eddy

In case you get confused with addition/subtraction of inequalities, you can stick with addition only, if you like. Just rewrite one inequality to give them the same sign. Remember that you can add inequalities only when they have the same sign.

Ques: Is z > 1?
Given: x+y+z > 0 .......... (I)

Stmnt 1: z > x + y + 1 .........(II)
(I) and (II) both have the same inequality sign '>' so we can add them.
x + y + 2z > x + y + 1
We get, z > (1/2)
z may or may not be greater than 1. Not sufficient.

Stmnt 2: x + y + 1 < 0
We can re-write this as 0 > x + y + 1 .......(III)
Now, (I) and (III) have the same sign '>' so you can add them.
x + y + z > x + y + 1
We get z > 1. Sufficient.

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GMAT 1: 710 Q48 V40 GMAT 2: 740 Q49 V42 Re: is z > 1 ?  [#permalink]

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eddyed911 wrote:
If x+y+z > 0, is z > 1 ?

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

this is a question from OG 10, could someone please explain this. The explanation in the OG is not clear enough.

Thanks,
Eddy

Statement 1: just tells us z-1>x+y. Tells us nothing about the signs of any variable. Insufficient.
Statement 2: tells us x+y<-1. Now for x+y+z>0, minimum quantity we have to add has to be greater than one. Test numbers. Sufficient.

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Re: is z > 1 ?  [#permalink]

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Bunuel wrote:
eddyed911 wrote:
If x+y+z > 0, is z > 1 ?

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

this is a question from OG 10, could someone please explain this. The explanation in the OG is not clear enough.

Thanks,
Eddy

Welcome to Gmat Club Eddy!

Note that:
You can only add inequalities when their signs are in the same direction:

If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.

Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.

You can only apply subtraction when their signs are in the opposite directions:

If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.

Back to the original question:

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

(1) z > x + y + 1 --> as the signs are in the same direction we can add two inequalities: $$(x+y+z)+z>x+y+1$$ --> $$2z>1$$ --> $$z>\frac{1}{2}$$, so z may or may not be more than 1. Not sufficient.

(2) x + y + 1 < 0 --> as the signs are in the opposite direction we can subtract two inequalities: $$(x+y+z)-(x+y+1)>0$$ --> $$z-1>0$$ --> $$z>1$$. Sufficient.

Hope it's clear.

is the option (1) is like : x+y+z+z>0+x+y+1 (as a>b+c>d= a+c>b+d), and option 2
(2) x+y+z -(x+y+z)>0-0 (as a>b -c<d = a-b>c-d)
Help me for my understanding.
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Smita04 wrote:
If x + y + z > 0, is z > 1?

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

stmnt1: adding z to both sides of the equation

2z >x+y+z+1
2z>1 or z> 0.5. this can give us any value for z (0.6,1,1.5).

taking numeric values
assume
x= - 0.2
y= - 0.3
z= + 0.6

z> x+y+1 but z<1

for x= 1, y= 2 and z= 5

we have x+y+z>0

and z> x+y+1 and z>1. Hence insuff

stmnt2:
x+y+1+z<0+z
or z>1

also given x+y+1<0 or x+y < -1
we have x+y+z>0 or -1 + z >0 or z>1

hence suff

so B
Intern  Joined: 14 Dec 2010
Posts: 41
Re: If x+y+z > 0, is z > 1 ? 1) z > x + y + 1 2) x + y  [#permalink]

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Thanks Bunuel...bcoz of the property u told this Q took 30 seconds to solve...
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Re: If x+y+z > 0, is z > 1 ? 1) z > x + y + 1 2) x + y  [#permalink]

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Thanks Bunuel, makes the questions clear and easy to solve!
Manager  Joined: 22 Feb 2009
Posts: 160
Re: If x+y+z > 0, is z > 1 ?  [#permalink]

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Entwistle wrote:
Statement 1:
Given:
z>x+y+1 <---- Equ. 1
&
x+y+z>0

Now add z to both sides of Equ. 1
2z>x+y+z+1
Since x+y+z > 0
2z>GT0+1 (GT0 means a quantity greater than 0)
2z>GT1
z>GT(1/2)
So we cannot conclude that z>1
Therefore, insufficient!

Statement 2:
x+y+1<0
Adding z to both sides of the Equation
x+y+z+1<Z
GT0+1<z
or
GT1<z
Therefore, z>1
Sufficient!

Ans: 'B'

I hope the explanation is clear enough Yeah, Add z to the both sides of the equation is the fastest way _________________
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Re: If x+y+z > 0, is z > 1 ?  [#permalink]

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_________________ Re: If x+y+z > 0, is z > 1 ?   [#permalink] 16 Oct 2018, 04:03
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