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

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If x+y+z > 0, is z > 1 ? [#permalink] New post 05 Feb 2011, 08:33
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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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Re: is z > 1 ? [#permalink] New post 05 Feb 2011, 08:45
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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


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.

Answer: B.

Hope it's clear.
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Re: is z > 1 ? [#permalink] New post 05 Feb 2011, 08:52
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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] New post 05 Feb 2011, 08:56
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] New post 05 Feb 2011, 08:58
Thanks guys ! these explanations really clarify things :)
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Re: is z > 1 ? [#permalink] New post 07 Feb 2011, 06:43
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thanks for the expalanation bunuel
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Re: is z > 1 ? [#permalink] New post 16 Feb 2011, 18:41
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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


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.

Answer (B)
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Re: is z > 1 ? [#permalink] New post 16 Feb 2011, 23:36
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.

Answer B.
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Re: is z > 1 ? [#permalink] New post 03 Mar 2011, 06:13
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.

Answer: B.

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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Re: DS inequalities [#permalink] New post 11 Jan 2012, 23:44
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:
adding z to either sides
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
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Re: If x+y+z > 0, is z > 1 ? 1) z > x + y + 1 2) x + y [#permalink] New post 19 Jan 2012, 08:02
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] New post 20 Jan 2012, 04:54
Thanks Bunuel, makes the questions clear and easy to solve!
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Re: If x+y+z > 0, is z > 1 ? [#permalink] New post 22 Jul 2013, 21:26
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Re: If x+y+z > 0, is z > 1 ? [#permalink] New post 22 Jul 2014, 15:42
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] 22 Jul 2014, 15:42
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