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My understtanding: All statement 1 tells us for sure is x is not 0 Statement 2 we can break into two staments

x>y OR x>-y

What I dont get is only one of those statements being true makes B the correct choice...so how is B the correct choice when there are two possibilities for a solution?

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We appreciate your kudos'

Last edited by Bunuel on 13 Feb 2013, 08:50, edited 1 time in total.

|y| is always positive, but y may or may not be positive. If y < 0, then |y| = -y. Also, x > |y| means x is > a positive number, so x is always positive.

|y| is always positive, but y may or may not be positive. If y < 0, then |y| = -y. Also, x > |y| means x is > a positive number, so x is always positive.

|y| is always positive, but y may or may not be positive. If y < 0, then |y| = -y. Also, x > |y| means x is > a positive number, so x is always positive.

My understtanding: All statement 1 tells us for sure is x is not 0 Statement 2 we can break into two staments

x>y OR x>-y

What I dont get is only one of those statements being true makes B the correct choice...so how is B the correct choice when there are two possibilities for a solution?

Yes u got it right way! but u gotta complete it buddy

ya definitely S1 is not sufficent

But S2 is a trap. x>y OR x>-y

assume numbers for x and y in this case let x=4 and y=3 so 4>3 like wise x=4 and y=-3 this also 4>-3

So x>|y| always will be x>y

Hope i helped u
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GMAT - Practice, Patience, Persistence Kudos if u like

From F.S 1, we can have x=-5,y=-3. Thus , x^2>y, but x<y. Hence the answer is NO. Now, for x=5,y=3, x^2>y and x>y. Here the answer is a YES. Thus not sufficient.

Fro F.S 2, x-mod(y)>0. Thus, x>mod(y). Whatever be the sign of y, x>y(always). Sufficient.

My understtanding: All statement 1 tells us for sure is x is not 0 Statement 2 we can break into two staments

x>y OR x>-y

What I dont get is only one of those statements being true makes B the correct choice...so how is B the correct choice when there are two possibilities for a solution?

From 2, we get that x - |y| is positive.

Since |y| is postive, this means that x is also positive.

So, if "y" is negative, it automatically means that x is greater than y.

If "y" is positive,

x - y still gives a postive number. Hence x is again greater than y.

Answer is hence B.
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