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(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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