Is |X+Y| >|X-Y|? (1) |X| > |Y| (2) |X-Y| < |X|

[richachampion]

Is |X+Y| >|X-Y|?

(1) |X| > |Y|
(2) |X-Y| < |X|

[richachampion]

Is |X+Y| >|X-Y|?

(1) |X| > |Y|
(2) |X-Y| < |X|

In order to evaluate such question. We need to know their sign more than the modulus.
Statement 1 doesn't tell us the signs of X and Y. We won't be able to evaluate this question based on the information provided.

Statement 2→
We will have to consider two cases for this statement.
|X-Y| < |X|

Since the left-hand side is less than the modulus of X that means something is literally subtracted from X.
Case #1
X → +VE
Y→ -VE
With the case#1 we can solve the question.

Case #2 → Second possibility
X → -VE
Y → -VE

In this case
|X+Y| =|X| + |Y|
|X+Y| =|X| - |Y|
HENCE
Is |X+Y| >|X-Y|?
NO
Thus statement 2 can solve the question.

[richachampion]

OA: B

[GMATGuruNY]

Is |x+y|>|x-y|?

1) |x|>|y|

2) |x-y|<|x|

Is |x+y| > |x-y|?
When there is absolute value notation on each side, we can square the inequality.

(x+y)² > (x-y)²
x² + 2xy + y² > x² - 2xy + y²
4xy > 0
xy > 0.

Question rephrased: Do x and y have the same sign?

Statement 1: |x| > |y|
Here, x and y could have the same sign or different signs.
INSUFFICIENT.

Statement 2: |x-y| < |x|
Squaring both sides, we get:
(x-y)² < x²
x² - 2xy + y² < x²
y² < 2xy
xy > y²/2.
Since the square of a value cannot be negative, y²/2 cannot be negative.
Thus, xy>0, implying that x and y have the same sign.
SUFFICIENT.

The correct answer is B.