Is |x| > |y|? : GMAT Data Sufficiency (DS)
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# Is |x| > |y|?

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Is |x| > |y|? (1) x^2 > y^2 (2) x > y [#permalink]

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13 May 2012, 04:22
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Is |x| > |y|?

(1) x^2 > y^2
(2) x > y
[Reveal] Spoiler: OA

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Re: Is |x| > |y|? (1) x^2 > y^2 (2) x > y [#permalink]

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13 May 2012, 06:00
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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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Re: Is |x| > |y|? (1) x^2 > y^2 (2) x > y [#permalink]

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27 Jun 2012, 10:19
Is |x| > |y|?
(1) x^2 > y^2
(2) x > y

1) This means |x|>|y|. Sufficient.
2) We do not know signs of x and y. If both were positive, then the statement would be true. If both were negative, then the statement would be false. If they had different signs, we would then need to know the vaue of x and y. INSUFFICIENT.

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10 Feb 2013, 04:37
Is |x| > |y|?

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

Last edited by Bunuel on 10 Feb 2013, 04:41, edited 1 time in total.
Edited the question.
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Re: Is |x| > |y|? [#permalink]

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10 Feb 2013, 04:50
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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.

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Re: Is IxI > IyI ? [#permalink]

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26 Feb 2013, 02:59
fozzzy wrote:
Is IxI > IyI ?
(1) x^2 > y^2
(2) x > y

Hi fozzy

when IxI = IyI that implies x^2 = Y^2
hence clearly statement 1 is sufficient

But for statement 2 substitute -ve values for x and y to satisfy the inequality....for -vel values you will get an answer to the question at hand ....but for +ve value it will be a definite yes......hence statement 2 is insufficient.
Moreover when IxI = I yI ,than either x = -y or y = -x.........

Hope that helps

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Re: Is |x| > |y|? (1) x^2 > y^2 (2) x > y [#permalink]

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16 May 2013, 08:36
dvinoth86 wrote:
Is |x| > |y|?

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

Stmt1: is sufficients as taking root can give us mod

stmt 2: can be flawed as below:
1. -4 , -5 and 4, 5 substitute for answer.
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Re: Is |x| > |y|? [#permalink]

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30 Jun 2013, 09:51
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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Re: Is |x| > |y|? [#permalink]

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15 May 2014, 12:36
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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
[Reveal] Spoiler:
(B)

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Re: Is |x| > |y|? [#permalink]

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15 May 2014, 14:45
gmatacequants wrote:
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
[Reveal] Spoiler:
(B)

Kudos is the best form of appreciation

why no consideration is given to decimal values of x and y ?
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Re: Is |x| > |y|? [#permalink]

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15 May 2014, 21:53
dhirajx wrote:
Is $$|x|$$ > $$|y|$$?
(1) $$x^2$$ > $$y^2$$
(2) $$x$$ > $$y$$

Sol:

We need to know whether |x|>|y|

St 1 tells us that x^2>y^2 or $$\sqrt{x^2}$$ > $$\sqrt{y^2}$$
Also |x|=$$\sqrt{x^2}$$
So we have |x|>|y| St 1 is clearly sufficient

St 2 says x>y if x=5,y=3 then |x|>|y|
but if x=-3 and y=-5 then |y|>|x|

Ans is A
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Re: Is |x| > |y|? [#permalink]

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14 Aug 2015, 07:24
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Re: Is |x| > |y|? (1) x^2 > y^2 (2) x > y [#permalink]

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20 Sep 2015, 01:50
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Re: Is |x| > |y|? (1) x^2 > y^2 (2) x > y   [#permalink] 20 Sep 2015, 01:50
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