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MGMAT CAT 1: DS Problem

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Source: — Data Sufficiency |
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by theCodeToGMAT » Mon Nov 11, 2013 9:42 am
To find: (x)(y + z) > 0

Statement 1:
|x + y| = |x| + |y|
this means that x & y have same signs..
x= +ve, y= +ve, z = +ve & < y
(+ve) (+ve + +ve) ==> YES

x= +ve, y= +ve, z = -ve & > y
(+ve) (+ve + -ve) ==> (+ve) (-ve) < 0 NO
INSUFFICIENT

Statement 2:
|z + y| = |y| + |z|
this means that z & y have same signs..
z = +ve, y = +ve, x = +ve
(+ve) (+ve + +ve) ==> > 0 YES

z = +ve, y = +ve, x = +ve
(-ve) (+ve + +ve) ==> < 0 NO
INSUFFICIENT

Combining...
x, y & z have same sign...
If positive
(+ve)(+ve + +ve) > 0 YES

If Negative
(-ve) (-ve + -ve) ==> (-VE)(-VE) > 0 YES
SUFFICIENT

Answer [spoiler]{C}[/spoiler]
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by GMATGuruNY » Mon Nov 11, 2013 11:15 am
josh80 wrote:If x, y, and z are nonzero numbers, is (x)(y + z) > 0?

(1) |x + y| = |x| + |y|

(2) |z + y| = |y| + |z|
Test the following combinations for (x,y) and (y,z):
Case 1: 1, 1
Case 2: 1, -1
Case 3: -1, 1
Case 4: -1, -1

Statement 1: |x+y| = |x| + |y|
Case 1: |1+1| = |1| + |1|
Case 2: |1 + (-1)| = |1| + |1|
Case 3: |(-1) + 1| = |-1| + |1|
Case 4: |(-1) + (-1)| = |-1| + |-1|

Only Cases 1 and 4 work.
Implication:
x and y have the SAME SIGN.

If x=1, y=1, and z=1, then (x)(y+z) > 0.
If x=1, y=1, and z=-1, then (x)(y+z) = 0.
INSUFFICIENT.

Statement 2: |z+y| = |y| + |z|
Implication:
y and z have the SAME SIGN (just as x and y have the same sign in statement 1).

If x=1, y=1, and z=1, then (x)(y+z) > 0.
If x=1, y=-1 and z=-1, then (x)(y+z) < 0.
INSUFFICIENT.

Statements combined:
Since x and y have the same sign, and y and z have the same sign, ALL 3 VARIABLES -- x, y, and z -- have the SAME SIGN.

If x>0, y>0 and z>0, then (x)(y+z) = (positive)(positive) = positive.
If x<0, y<0, and z<0, then (x)(y+z) = (negative)(negative) = positive.
Thus, (x)(y+z) > 0.
SUFFICIENT.

The correct answer is C.
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