Is |x| |y|? (1) x^2 y^2 (2) x y

Question

Is |x| > |y|?

(1) x^2 > y^2
(2) x > y

GyanOne · Answer

Statement 1: x^2 > y^2 => |x| > |y|. Sufficient. Statement 2: x>y. If x>0 and y>0 then x>y implies |x|>|y|. If x<0 and y<0 then x>y implies |x|<|y|. Insufficient. A it is.

Bunuel · Answer

Is |x| > |y|?

(1) x^2 > y^2. Since both sides are non-negative, then we can safely take the square root: |x| > |y|. Sufficient.

Or: "is |x| > |y|?" can be rewritten as: is x^2 > y^2? (we can safely square the whole inequality since both sides are non-negative). This statement directly answers the question. Sufficient.

(2) x > y. Clearly insufficient: consider x=1 and y=0 for an YES answer and x=1 and y=-2 for a NO answer. Not sufficient.

Answer: A.

WholeLottaLove · Answer

Is |x| > |y|?

Is x>y?
OR
Is x>-y?

We can also square both sides as we know that x, y are >=0

|x| > |y|
is |x|^2 > |y|^2?
Is x^2 > y^2?

(1) x^2 > y^2

This tells us directly that x^2 is greater than y^2
SUFFICIENT

(2) x > y

5>4
|5| > |4|
5 > 4 (Valid)
Or
-3>-8
|-3| > |-8|
3 > 8 (Invalid)
INSUFFICIENT

(A)

gmatacequants · Answer

To find out the sufficiency for the problem statement,

Option 1:
x^2 > y^2

Quick (and dirty) method : to look for the cases where the option would lead to contradictory or insufficient conclusions to the problem statement

Looking at modulus function, one can verify by checking the inequality scenario in positive and negative domains.
Using values
 i) x= 5,y=4 
 ii) x=-5,y=4
 iii) x=-5,y=-4
 iv) x= 5,y=-4 ,
all such cases would lead to a definitive conclusion on inequality |x|>|y|

Otherwise also, x^2 > y^2 
 => |x|*|x|>|y|*|y|
 => |x| > |y| .. taking sq. root of both sides(which are positive)

SO, 1st option is sufficient

Option 2 :
x > y
Quick (and dirty) method

Using values 
i) x = 5, y = -6
ii) x = 5 , y = 4 ,
both give different conclusions on inequality |x|>|y|

SO, 2nd option is insufficient

correct option is  (B)

Kudos is the best form of appreciation

jabhatta2 · Answer

Hi Bunuel -- 
For S1 -- can you please elaborate the theory re: why

if x^2 > y^2 -- that implies |x| > |y|

I can memorize this but i would like to understand logically the theory behind something like. How would one make this assumption logically without memorizing this information ?

I have tested a few test cases and this is true -- but logically or on a number line -- how does one make sense of this theory ?

For example -- why does |x| > |y| NOT imply x^3 > y^3 but x^2 > y^2

Thank you !

Bunuel · Answer

Since both side of the inequality are non-negative, then we can safely take the square root to get  \sqrt{x^2} > \sqrt{y^2} . Next, since  \sqrt{a^2}=|a| , then  \sqrt{x^2} > \sqrt{y^2}  gives |x| > |y|.

Kinshook · Answer

Asked: Is |x| > |y|?

(1) x^2 > y^2
|x| > |y|
SUFFICIENT

(2) x > y
If x = 3; y = - 4; |x| = 3 < |y| = 4
But if x = 4; y = 3; |x| = 4 > |y| = 3
NOT SUFFICIENT

IMO A

jabhatta2 · Answer

karishma   Bunuel  - I was trying to do this DS problem via testing cases as this is a Yes / No question

I attempted S2 first as its easier

Question on S1

Does it make sense for a test taker to try to use the same test cases once again that they used in the other statement ? (assuming the restrictions in this test case allow for it )

I think it might because in case both S1 and S2 are both INS

I think you have a ready-made case which is applicable for S1 and S2 when testing for C or E specifically

but just wanted your thoughts !

Thank you !

jabhatta2 · Answer

Bunuel  .. Can you answer my question when you have a moment

Posted from my mobile device

Lazybum · Answer

does anyone know why this is "not negative" as bunuel pointed out? oh actually nvm, its a fundamental identity that if you square a number, it always is greater than or equal to 0. i tried to solve this question differently but it didnt work, does anyone understand why? I used the "critical method" cited by bunuel, here: https://gmatclub.com/forum/absolute-mod ... 10849.html i listed out all the scenarios x > y when x>0 and y>0 -x > y when x<0 and y>0 x > -y when x>0 and y<0 -x > -y when x<0 and y<0 this just doesn't work and i'm confused why.

BLTN · Answer

does anyone know why this is "not negative" as bunuel pointed out? oh actually nvm, its a fundamental identity that if you square a number, it always is greater than or equal to 0. i tried to solve this question differently but it didnt work, does anyone understand why? I used the "critical method" cited by bunuel, here: https://gmatclub.com/forum/absolute-mod ... 10849.html i listed out all the scenarios x > y when x>0 and y>0 -x > y when x<0 and y>0 x > -y when x>0 and y<0 -x > -y when x<0 and y<0 this just doesn't work and i'm confused why. Dear Lazybum the values are non-negative because you cannot take square root from the negative value. Even if the X = -2 then the (-2)^2= 4 and \sqrt{4} = 2 not the - 2 The question is asking is the absolute value of X grate then the absolute value of Y 2d statement gives you variables X and Y in lieu of Absolute variables.

PriyamRathor · Answer

Hello  chetan2u   gmatphobia  ,

I did S1 in some other way.

S1  x^2 >y^2

x^2-y^2>0 
(x-y) (x+y)>0
Hence,

x>y and x>-y

Lets take x>-y

Say 1>-2 -------------> Now when we put this in Question Stem |1| > |-2| which implies |1| >|2| ------  Not possible

Hence S1 is not sufficient.

Please guide.

Regards

chetan2u · Answer

(x-y)(x+y)>0 means both x-y and x+y have same sign. 1) Both positive: x>y and x>-y. This means x>|y| 2) Both negative: x|y| Sufficient

bumpbot · Answer

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Is |x| > |y|? (1) x^2 > y^2 (2) x > y

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