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Is x > y? (1) x^2 > y^2 (2) x  y > 0 [#permalink]
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14 Jun 2012, 01:07
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84% (00:30) correct 16% (00:38) wrong based on 146 sessions
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Is x > y? (1) x^2 > y^2 (2) x  y > 0
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Re: Is x > y? [#permalink]
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14 Jun 2012, 01:11



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Re: Is x > y? [#permalink]
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14 Jun 2012, 01:16
Bunuel wrote: Is x > y?
(1) x^2 > y^2 > \(x>y\) > so, \(x\) may or may not be greater than \(y\). Consider: \(x=2\) and \(y=1\) for a NO answer and \(x=2\) and \(y=1\) for an YES answer. Not sufficient.
(2) x  y > 0 > \(x>y\). Directly answers the question. Sufficient.
Answer: B.
Hope it's clear. thanks for the help



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Re: Is x > y? (1) x^2 > y^2 (2) x  y > 0 [#permalink]
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14 Jun 2012, 01:30
sarb wrote: Is x > y?
(1) x^2 > y^2 (2) x  y > 0 Hi, x>y? Using (1), x^2 > y^2 or x^2  y^2>0 or (xy)(x+y)>0 so, either xy>0 & x+y>0 or xy<0 & x+y<0 so, x may or may not be less than y. Insufficient. Using (2), xy>0 or x>y, so answer to question is Yes. Sufficient Thus, answer is (B) Regards,



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Re: Is x > y? (1) x^2 > y^2 (2) x  y > 0 [#permalink]
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14 Jun 2012, 01:33
cyberjadugar wrote: sarb wrote: Is x > y?
(1) x^2 > y^2 (2) x  y > 0 Hi, x>y? Using (1), x^2 > y^2 or x^2  y^2>0 or (xy)(x+y)>0 so, either xy>0 & x+y>0 or xy<0 & x+y<0 so, x may or may not be less than y. Insufficient. Using (2), xy>0 or x>y, so answer to question is Yes. Sufficient Thus, answer is (B) Regards, thanks a lot for helping me



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Concept of Squaring a variable,taking roots and interference of signs? [#permalink]
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15 Jan 2015, 01:07
As the title suggests,i can't figure out when to be wary of the "could be positive or negative,so we can't say" rule in algebra.Please help in understanding this. Acc. to the books,if x^2 is lets say 25,x could be positive or negative both so we cant say what the result could be when we use x in any other calculation.That is fine.Great. Next is √x^2= x Coz we take square of x and x becomes +ve and then ROOT INSTRUCTS US TO TAKE ONLY THE POSITIVE VALUE OF X. Is x > y? (1) x^2> y^2 So,here if we take the root on both sides of the equation,will we not get x>y???Please correct me if i am wrong or missing something here. (2) x  y > 0 Answer is B. Bunuel , insights please!!



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Re: Is x > y? (1) x^2 > y^2 (2) x  y > 0 [#permalink]
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15 Jan 2015, 21:48
Hi Ralphcuisak, You bring up some important examples/math concepts that are worth remembering for Test Day. However, they both involve an EQUALS sign, while this question involves an INEQUALITY. It's a different situation, so you have use different rules. Since "squaring" a positive has the same effect as "squaring" a negative (the result is POSITIVE in both options), you have to think about the "relative" value of X and Y when you're told.... X^2 > Y^2 X and/or Y COULD be positive OR negative (or even 0).....this really means the same thing as X > Y IF.... X = 2 Y = 1 (2)^2 > (1)^1 2 > 1 2 > 1 IF.... X = 2 Y = 1 (2)^2 > (1)^2 2 is NOT > 1 2 > 1 GMAT assassins aren't born, they're made, Rich
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Re: Is x > y? (1) x^2 > y^2 (2) x  y > 0 [#permalink]
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17 Jan 2015, 01:06
It's clear now..Thanks EMPOWERgmatRichC..



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Re: Is x > y? (1) x^2 > y^2 (2) x  y > 0 [#permalink]
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17 Jan 2015, 18:15
Hi Ralphcuisak, I'm happy to help. Remember that math rules don't exist on their own  they have to "work" with real numbers. That means there's always a practical way to "test out" your ideas to make sure that they're correct: TEST VALUES. For example, what happens when you square a negative fraction? You don't have to theorize over this (and rubbing your chin and staring at the ceiling is a waste of time)  choose a negative fraction, write it down, then square it. (1/2)^2 = (1/2)(1/2) = + 1/4 Now you have PROOF of the rule. Keep this idea in mind anytime you feel stuck on a question or "your answer" doesn't match any of the 5 given answers. You should be able to prove any rule that you're using. GMAT assassins aren't born, they're made, Rich
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Re: Is x > y? (1) x^2 > y^2 (2) x  y > 0 [#permalink]
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