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Is |x−z−y| > x−z+y?.......Veritas question

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Source: — Data Sufficiency |

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by Jay@ManhattanReview » Fri Apr 14, 2017 4:40 am
Mo2men wrote:Is |x−z−y| > x−z+y?

(1) 0<x<z<y
(2) (x-z-y) is negative

OA: A

Source: Veritas
Hi Mo2men,

Statement 1: 0<x<z<y

0<x<z<y implies that each of x, y, and z is positive.

|x-z-y| must be a negative number since x < (y+z).

Thus, we have: Is -(x-z-y) > x-z+y?

=> -x+z+y > x-z+y

=> 2z > 2x; y cancells

=> z > x--> This is a fact given in the statement. Sufficient.

Statement 2: (x-z-y) is negative

x-z-y < 0

=> x < (y+z)

|x-z-y| must be a negative number since x < (y+z), Thus, we have: Is |a negative number| > x-z+y?

Now we have the same situation as was in statement 1.

So, we reach at: Is x < z? We do not know this. This fact belonged to statement 1 and not to statement 2. Insufficient.

The correct answer: A

Hope this helps!

Relevant book: Manhattan Review GMAT Number Properties Guide

-Jay
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Last edited by Jay@ManhattanReview on Fri Apr 14, 2017 6:12 am, edited 1 time in total.
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by GMATGuruNY » Fri Apr 14, 2017 5:04 am
Mo2men wrote:Is |x−z−y| > x−z+y?

(1) 0<x<z<y
(2) (x-z-y) is negative
Statement 1: 0<x<z<y
Since x<z, x-z < 0.
Inequalities in which the <> faces the SAME DIRECTION can be ADDED TOGETHER.
Adding together x-z < 0 and 0 < y, we get:
(x-z) + 0 < 0 + y
x-z-y < 0.

Since x-z-y < 0, |x-z-y| = -(x-z-y).
Substituting |x-z-y| = -(x-z-y) into the question stem, we get:
Is -(x-z-y) > x-z+y?

Simplifying the question stem, we get:
-(x-z-y) > x-z+y ?
-x+z+y > x-z+y ?
2z > 2x ?
z > x ?

Since Statement 1 indicates that z>x, the answer to the question stem is YES.
SUFFICIENT.

Statement 2: (x-z-y) is negative
As shown in the blue portion above, x-z-y < 0 enables us to rephrase the question stem as follows:
z > x ?
x-z-y < 0 implies that z > x-y.
No way to determine whether z>x.
INSUFFICIENT.

The correct answer is A.
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by Mo2men » Fri Apr 14, 2017 5:53 am
Jay@ManhattanReview wrote:
Mo2men wrote:Is |x−z−y| > x−z+y?

(1) 0<x<z<y
(2) (x-z-y) is negative

OA: A

Source: Veritas
Hi Mo2men,

Statement 1: 0<x<z<y

0<x<z<y implies that each of x, y, and z is positive.

|x-z-y| must be a negative number since x < (y+z).

Thus, we have: Is -(x-z-y) < x-z+y?; mind the reversal of inequality sign

=> -x+z+y < x-z+y

=> 2x < 2z; y cancells

=> x < z--> This is a fact given in the statement. Sufficient.
I feel something wrong here in red.Following your final step, it gives:

-x+z+y < x-z+y

2z < 2x.
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by Jay@ManhattanReview » Fri Apr 14, 2017 6:11 am
Mo2men wrote:
Jay@ManhattanReview wrote:
Mo2men wrote:Is |x−z−y| > x−z+y?

(1) 0<x<z<y
(2) (x-z-y) is negative

OA: A

Source: Veritas
Hi Mo2men,

Statement 1: 0<x<z<y

0<x<z<y implies that each of x, y, and z is positive.

|x-z-y| must be a negative number since x < (y+z).

Thus, we have: Is -(x-z-y) < x-z+y?; mind the reversal of inequality sign

=> -x+z+y < x-z+y

=> 2x < 2z; y cancells

=> x < z--> This is a fact given in the statement. Sufficient.
I feel something wrong here in red.Following your final step, it gives:

-x+z+y < x-z+y

2z < 2x.
Hi Mo2men,

Thanks for pointing it out. It's incorrect, I edit my response.

-Jay
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