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

1

This post received KUDOS

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.

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

(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.

Re: Is IxI > IyI ? [#permalink]
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.........

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|

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|

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

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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