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Intern  Joined: 21 Dec 2010
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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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52 00:00

Difficulty:   65% (hard)

Question Stats: 61% (01:55) correct 39% (02:06) wrong based on 883 sessions

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

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

DS18602.01
Math Expert V
Joined: 02 Sep 2009
Posts: 58340
If x + y + z > 0, is z > 1? (1) z > x + y +1 (2) x + y + 1 < 0  [#permalink]

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49
35
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.

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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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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 _________________
General Discussion
Retired Moderator Joined: 20 Dec 2010
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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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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: If x + y + z > 0, is z > 1? (1) z > x + y +1 (2) x + y + 1 < 0  [#permalink]

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Thanks guys ! these explanations really clarify things Intern  Joined: 27 Jun 2008
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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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1
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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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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Intern  Joined: 14 Dec 2010
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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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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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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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1
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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Posts: 4003
Re: If x + y + z > 0, is z > 1? (1) z > x + y +1 (2) x + y + 1 < 0  [#permalink]

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7
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

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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