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Is |x| > |y|? (1) x^2 > y^2 (2) x > y 13 May 2012, 05:22
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A
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73% (00:55) correct 27% (01:07) wrong based on 2348 sessions

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Is |x| > |y|?

(1) x^2 > y^2
(2) x > y
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Is |x| > |y|? (1) x^2 > y^2 (2) x > y 10 Feb 2013, 05:50
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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.
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Is |x| > |y|? (1) x^2 > y^2 (2) x > y 30 Jun 2013, 10:51
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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)
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Is |x| > |y|? (1) x^2 > y^2 (2) x > y 13 May 2012, 07:00
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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.
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Is |x| > |y|? (1) x^2 > y^2 (2) x > y 15 May 2014, 13:36
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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
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Is |x| > |y|? (1) x^2 > y^2 (2) x > y Updated on: 06 May 2020, 09:20
Originally posted by jabhatta2 on 06 May 2020, 08:20.
Last edited by jabhatta2 on 06 May 2020, 09:20, edited 2 times in total.
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Bunuel
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.

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 ? [ x^2 > y^2 -- implies |x| > |y| ]

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

Thank you !
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Bunuel
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.

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 ? [ x^2 > y^2 -- implies |x| > |y| ]

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|.
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Is |x| > |y|? (1) x^2 > y^2 (2) x > y 06 May 2020, 08:56
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dvinoth86
Is |x| > |y|?

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

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

Posted from my mobile device
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Bunuel
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.

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 ? [ x^2 > y^2 -- implies |x| > |y| ]

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|.

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.
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Is |x| > |y|? (1) x^2 > y^2 (2) x > y 21 Dec 2021, 03:37
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Bunuel
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.

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 ? [ x^2 > y^2 -- implies |x| > |y| ]

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|.

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.[/quote]

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.
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dvinoth86
Is |x| > |y|?

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

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
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dvinoth86
Is |x| > |y|?

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

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

(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 and x<-y. This means x<y<0<-y. This means x is farther away from 0 as compared to y. |x|>|y|
Sufficient
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