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Is |x| > |y|? [#permalink] New post 10 Feb 2013, 05:37
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Is |x| > |y|?

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

Last edited by Bunuel on 10 Feb 2013, 05:41, edited 1 time in total.
Edited the question.
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Re: Is |x| > |y|? [#permalink] New post 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 IxI > IyI ? [#permalink] New post 26 Feb 2013, 03:45
Is IxI > IyI ?
(1) x^2 > y^2
(2) x > y

Please provide explanations. Thanks!
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Re: Is IxI > IyI ? [#permalink] New post 26 Feb 2013, 03:48
fozzzy wrote:
Is IxI > IyI ?
(1) x^2 > y^2
(2) x > y

Please provide explanations. Thanks!


Merging similar topics. Please refer to the solution above.
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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!!

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!


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Re: Is IxI > IyI ? [#permalink] New post 26 Feb 2013, 03:59
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

Consider kudos if my post helps

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Re: Is IxI > IyI ?   [#permalink] 26 Feb 2013, 03:59
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