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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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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 05 Feb 2011, 09:33
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Question Stats:

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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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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 05 Feb 2011, 09:45
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SOLUTION

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.

Adding/subtracting/multiplying/dividing inequalities: http://gmatclub.com/forum/help-with-add ... 55290.html
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Re: 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 05 Feb 2011, 09:52
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2
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: 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 05 Feb 2011, 09:56
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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: 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 05 Feb 2011, 09:58
Thanks guys ! these explanations really clarify things :)
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Re: 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 07 Feb 2011, 07:43
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thanks for the expalanation bunuel
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Re: 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 16 Feb 2011, 19:41
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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: 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 19 Jan 2012, 09:02
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 + 1 < 0  [#permalink]

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New post 20 Jan 2012, 05:54
Thanks Bunuel, makes the questions clear and easy to solve!
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Re: 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 12 Feb 2014, 12:02
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1
If x + y + z > 0, is z > 1?

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

Sol: st1 says z>x+y+1 and x+y+z>0

If x=-2, y=-3,z=6 then z>1
But if x=-0.1,y=-0.1 thus z>0.8 so it may or may not be greater than 1
Not sufficient. A and D ruled out

St 2 says x+y+1>0
Again x=-1/2,y=1/2 then z can take any value greater than or less than 1 but if x=1000,y=1 then again z can take any value so clearly
Not sufficient

On combining we get that z>0 but z may or may not be greater than1

Ans E

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Re: 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 Updated on: 02 Mar 2018, 15:37
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Top Contributor
RisingForceX wrote:
If x + y + z > 0, is z > 1?

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


Target question: Is z > 1

Given: x + y + z > 0

Statement 1: z > x + y +1
Let's create a similar inequality to x + y + z > 0
Take z > x + y +1 and subtract x and y from both sides to get: z - x - y > 1
We now have two inequalities with the inequality signs facing the same direction.
z - x - y > 1
x + y + z > 0
ADD them to get: 2z > 1
Divide both sides by 2 to get: z > 1/2
So, z COULD equal 2, in which case z > 1
Or z COULD equal 3/4, in which case z < 1
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: x + y + 1 < 0
Let's use the same strategy.
This time, let's multiply both sides by -1 to get: -x - y - 1 > 0
We now have two inequalities with the inequality signs facing the same direction.
-x - y - 1 > 0
x + y + z > 0
ADD them to get: z - 1 > 0
Add 1 to both sides to get z > 1
Perfect!!!
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: B

Cheers,
Brent
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Originally posted by GMATPrepNow on 17 Jul 2016, 07:45.
Last edited by GMATPrepNow on 02 Mar 2018, 15:37, edited 1 time in total.
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Re: If x + y + z > 0, is z > 1? (1) z > x + y +1 (2) x + y + 1 < 0  [#permalink]

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Re: If x + y + z > 0, is z > 1? (1) z > x + y +1 (2) x + y + 1 < 0   [#permalink] 21 Sep 2019, 17:36
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