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(1) |x| = y = positive, so x can be positive or negative
NOT SUFFICIENT

(2) Squaring both sides,
(x-y)^2 > (x+y)^2
x^2 + y^2 -2xy > x^2 + y^2 +2xy
-4xy>0
Since y is positive, x must be negative
SUFFICIENT

FINAL ANSWER IS (B)

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Quote:
If x and y are integers with y > 0, is x positive ?


(1) |x| - y = 0

(2) |x - y| > |x + y|

(1) insufic

|x|-y=0…(y>0)…|x|=y
|x|>0: x>0 or x<0

(2) sufic

|x - y| > |x + y|
if x=0 then y = y (invalid)
if x>0 then x-y < x+y (invalid)
if x<0 then -x-y > -x+y (valid)

Ans (B)
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given y>0
#1
|x| - y = 0
possible when x >y and since value os x is lxl so x can be either + or -ve so insufficient
#2

|x - y| > |x + y|

only possible when value of x is -ve integer
IMO B ; sufficient


If x and y are integers with y > 0, is x positive ?


(1) |x| - y = 0

(2) |x - y| > |x + y|
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If x and y are integers with y > 0, is x positive ?


(1) |x| - y = 0

(2) |x - y| > |x + y|

Statement 1:

(1) |x| - y = 0

If we consider X as positive X= 6, y=6

6-6=0 or if X is negative X=-6 the modulus will make it 6. So X can be positive or negative

(2) |x - y| > |x + y|

Lets consider X as positive X=6

|6-2|>|6+2| Therefore X cannot be positive

|-6-2|>|-6+2| 8>4 Therefore X is not positive

And X is not positive hence statement B is sufficient
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