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Difficulty: 605-655 Level,    Inequalities,                                     
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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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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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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]
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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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]
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]
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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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

Posted from my mobile device
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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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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

Originally posted by BrentGMATPrepNow on 17 Jul 2016, 07:45.
Last edited by BrentGMATPrepNow 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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eddyed911 wrote:
If \(x + y + z > 0\), is \(z > 1\) ?

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


DS18602.01



1) Add inequalities: x + y + 2z > x + y + 1
z > 1/2
Not sufficient
2) Add inequalities: x + y + z > x + y + 1
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]
I have a question regarding statement 1. If I know that z>x+y+1, why can´t I say 0>x+y+1-z, then I would know that z is larger than 1 since x+y+z>0m if I add one but substract Z and it gives me a negative number, z>1. What am I doing wrong here?
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Re: If x + y + z > 0, is z > 1? (1) z > x + y +1 (2) x + y + 1 < 0 [#permalink]
If \(x + y + z > 0\), is \(z > 1\) ?

(1) \(z > x + y + 1\)

Add the statement into the given information:

\(x + y + 2z > x + y + 1\)
\(2z > 1\)
\(z > \frac{1}{2}\)

INSUFFICIENT.

(2) \(x + y + 1 < 0\)

Subtract the statement into the given information:

\(Z - 1 > 0\)
\(Z > 1\)

SUFFICIENT.

Answer is 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]
I don't understand when I'm supposed to add the inequalities. How did you know you were supposed to add them instead of just plugging in numbers or trying to solve for a single variable?
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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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eddyed911 wrote:
If \(x + y + z > 0\), is \(z > 1\) ?

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

DS18602.01

We need to keep in mind that inequalities with the same sign can be added.

Statement-I (Insufficient)
1. \(x + y + z > 0\) can be rewritten as \(z > - x - y\)
2. \(z > x + y + 1\) can be added to \(z > - x - y\) to yield \(2z > 1\) -> \(z > 0.5\)
3. Hence we cannot conclude if \(z > 1\)

Statement-II (Sufficient)
1. \(x + y + z > 0\) can be rewritten as \(x + y > - z\)
2. Multiplying \(x + y + 1 < 0\) with -1 we get \(- x - y > 1\)
3. Adding the above two inequalities we get \(0 > - z + 1\) -> \(-1 > -z\) -> \(z > 1\)

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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howareyu wrote:
I don't understand when I'm supposed to add the inequalities. How did you know you were supposed to add them instead of just plugging in numbers or trying to solve for a single variable?



Honestly, for inequalities, make it a habit to try and avoid plugging values as much as possible. This should be your last resort because that's one rabbit hole deep enough to sink your whole exam. another reason why GMAT loves this topic (it's one of the most important topic).

try & make inferences about signs of the variables involved before you start solving that question, eg here:

\(z >x+y+1\)

\(z-1> x+y \)(now this is where GMAT wants you to try all those versions of numbers , we don't even know if x,y,z are integers, so the possibilities are just so many, enough to make you waste 10-15 min.)


take a moment, and see:- even at its best, \(x + y\) can take maximum value equal to \(z-1\) (even tho we know \(x+y <z\), but let's take this for the sake of the argument. )

so let,\( x+y= z-1 \)

now we're given in stem,\( x+y+z>0\) --------- (A)

put x+y = z-1 in equation (A) ------> z-1+z>0 -------> 2z-1>0 ---------> 2z>1 -------> z>1/2--------> z>0.5. so, z COULD be >1 but it could also be 0.51 and still satisfy our inequality while remaining less than 1.
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If x + y + z > 0, is z > 1? (1) z > x + y +1 (2) x + y + 1 < 0 [#permalink]
ueh55406

I solved this question, but I am wondering: what's the 'trigger/sign' that we should use to know whether we should add or subtract inequalities. I did so here only based on intuition more than any real practical insight.

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

(1) z>x+y+1
Add both inequalities and we get z > 1/2
Insufficient
(2) x+y+1<0
Subtract both inequalities and we get z > 1
Sufficient.

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]
If x+y+z>0 then x+y > -1

(1) z>x+y+1
-> z > -1 + 1 = 0: Insuff

(2) x+y+1<0
-> x+y < -1
-> -1 + z > 0
-> z > 1: Suff
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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eddyed911 wrote:
If \(x + y + z > 0\), is \(z > 1\) ?

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


DS18602.01

Solution:

We need to determine whether z > 1 given that x + y + z > 0.

Statement One Alone:

If x + y = 0, then z > 1. However, if x + y = -0.5 and z = 0.6, then z is not greater than 1. Statement one alone is not sufficient.

Statement Two Alone:

Since x + y < -1, we see that z must be greater than 1 so that x + y + z > 0. Statement two alone is sufficient.

Answer: B
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