Rahul,

Here's my approach, is it correct?


X^2-Y^2>0 meaning X+Y and X-Y have to have the same sign. I'm just going to look at Statement 2 here:

2) |X|>Y means X>Y or X-Y>0 OR -X>Y meaning X+Y<0 (Is this correct re-statement?)

Since the signs have to be the same when X-Y then X+Y is also positive, that step is meaningless because it already answer our question of is X-Y>0.

The second condition of X+Y<0 means X-Y also has to be less than Zero. Now we have two scenarios that will satisfy X^2-Y^2 but gives different result of X-Y so insufficient.

gmatusa2010 wrote:If X^2>Y^2 is X>y?

1) X>|Y|
2) |X|>Y


Let's see some different approaches....

simplifying problem => |x|>|y| => x>y and x<y

st(1) x>y and x>-y => x>y only if -x<-y, however st(1) defines only -x<y, Not sufficient
st(2) x>y and -x>y => x>y only if -x<-y, however st(2) defines only x<-y, Not sufficient

Combining st(1&2) => x>-y and -x>y we get |x|<|y| from where only possible x<y Satisfies - Answer No x<y

IOM C

yes Rahul, I see my mistakes. I approached mods mechanically, better was plug-in to see that A is correct.

Rahul,

I noticed that when u did your cases there were 3. X is positive, X is negative and Y negative, and X is negative and Y is positive. I got the statement correctly but am I missing things with just two cases?

statement 1:

x > |y| ==> x > 0 ---(1)

now lets see whether x > y :

case 1) y > 0

==> x >|y| = x > y (if y > 0 |y| = y) -- yes x > y

case 2) y< 0

x > 0 from inequality (1) & y < 0 ...==> yes x > y

Hence Statement 1 is Sufficient


Statement 2:

|x| > y lets assume |x| = K (positive real no) , so now we can rewrite statement 2 as

y < K (a positive real no)

above inequality can be represented on number line as given below:



means y can be positive or negative

with this knowledge lets check whether x > y

case 1 : x > 0

then
|x| > y = x > y ==> yes x > y

case 2 : x < 0

to understand x > y under this condition lets take some e.g

example 1: x = -4 => |x| = 4 => y < 4 => y = (....-5,-4, -3 , -2, -1, 0, 1, 2, 3)

=> for some cases x > y and for some cases x < y

hence Statement 2 is Insufficient

Answer A


Please let me know if my answer is correct or not.
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X^2 > y^2 => |x| > |y|

Is x > y ?

Let us represent the relationship (|x| > |y|) on a number line :


--------------X'-----Y'-------0--------Y----X------------------


Statement 1:

x > |y| , Since |y| is always non-negative, that means x = X' in the diagram above
Hence, clearly X > Y ---- Sufficient


Statement 2 :

|x| > y
which is true for both X and X'
When x = X, then X > Y,
but when x = X' then x < Y

hence - Insufficient

Answer is A.

Interesting approach... how did you get x^2=y^2?

anshumishra wrote:X^2 = y^2 => |x| > |y|

Is x > y ?

Let us represent the relationship (|x| > |y|) on a number line :


--------------X'-----Y'-------0--------Y----X------------------


Statement 1:

x > |y| , Since |y| is always non-negative, that means x = X' in the diagram above
Hence, clearly X > Y ---- Sufficient


Statement 2 :

|x| > y
which is true for both X and X'
When x = X, then X > Y,
but when x = X' then x < Y

hence - Insufficient

Answer is A.

gmatusa2010 wrote:Interesting approach... how did you get x^2=y^2?

anshumishra wrote:X^2 = y^2 => |x| > |y|

Is x > y ?

Let us represent the relationship (|x| > |y|) on a number line :


--------------X'-----Y'-------0--------Y----X------------------


Statement 1:

x > |y| , Since |y| is always non-negative, that means x = X' in the diagram above
Hence, clearly X > Y ---- Sufficient


Statement 2 :

|x| > y
which is true for both X and X'
When x = X, then X > Y,
but when x = X' then x < Y

hence - Insufficient

Answer is A.

Hey gmatusa2010, that was a typo, I have fixed it.

It is in the given question :

If X^2>Y^2 is X>y?

1) X>|Y|
2) |X|>Y