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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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Is |x| > |y|?

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

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Re: Is |x| > |y|? [#permalink]

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

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New post 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|? [#permalink]

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

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New post 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.

The answer is A.

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Re: Is IxI > IyI ? [#permalink]

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

Please provide explanations. Thanks!


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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New post 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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New post 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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New post 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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New post 08 Jul 2017, 08:03
Is |x| > |y|?

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

I used this approach:
1) + - x> + - Y ---> case 1) x> y case 2) - x >- y which means x < y. Two answers, so not sufficient
2) x> y --> sufficient
What is wrong in this approach? Would appreciate a reply!

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Re: Is |x| > |y|?   [#permalink] 08 Jul 2017, 08:03
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