tough one

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tough one

by buoyant » Fri Nov 14, 2014 12:47 pm
|x+2|=|y+2| what is the value of x+y?
(1) xy<0
(2) x>2 y<2

[spoiler]OA: D[/spoiler]

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by GMATGuruNY » Fri Nov 14, 2014 1:16 pm
buoyant wrote:|x+2|=|y+2| what is the value of x+y?
(1) xy<0
(2) x>2 y<2

[spoiler]OA: D[/spoiler]
|x+2|=|y+2| implies TWO CASES:

Case 1: No signs changed
x+2 = y+2
x=y.

Case 2: Signs changed on ONE SIDE
x+2 = -(y+2)
x+2 = -y-2
x+y = -4.

Thus, |x+2|=|y+2| implies that x=y and/or that x+y=-4.

Statement 1: xy<0
Implication:
It is not possible that x=y.
Thus, x+y = -4.
SUFFICIENT.

Statement 2: x>2, y<2
Implication:
It is not possible that x=y.
Thus, x+y = -4.
SUFFICIENT.

The correct answer is D.
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by buoyant » Fri Nov 14, 2014 2:16 pm
Mitch,

in each statement even though case 1 doesn't hold good, how do we know that case 2 will apply. the information may be insufficient right?

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by GMATGuruNY » Fri Nov 14, 2014 5:21 pm
buoyant wrote:Mitch,

in each statement even though case 1 doesn't hold good, how do we know that case 2 will apply. the information may be insufficient right?
The prompt indicates that |x+2| = |y+2|.
This equation REQUIRES that Case 1 be true (x=y) or that Case 2 be true (x+y = -4).
Thus, if x≠y, then Case 2 must be true, implying that x+y = -4.
Since each statement indicates that x≠y, we know in each statement that x+y = -4.
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