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# Is x^2>y^2?

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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GMAT 1: 760 Q51 V42
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18 May 2017, 01:07
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13
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Difficulty:

65% (hard)

Question Stats:

52% (01:35) correct 48% (01:27) wrong based on 161 sessions

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Is $$x^2>y^2$$?

1) |x|>y
2) x>|y|

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MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" CEO Status: GMATINSIGHT Tutor Joined: 08 Jul 2010 Posts: 2978 Location: India GMAT: INSIGHT Schools: Darden '21 WE: Education (Education) Re: Is x^2>y^2? [#permalink] ### Show Tags 18 May 2017, 07:12 MathRevolution wrote: Is $$x^2>y^2$$? 1) |x|>y 2) x>|y| Question : Is $$x^2>y^2$$? Statement : 1) |x|>y @y = -4, x=2, $$x^2<y^2$$ @y = -4, x=5, $$x^2>y^2$$ NOT SUFFICIENT Statement : 2) x>|y| i.e. value of x is positive and absolute value of x is greater than absolute value of y. hence, squaring both sides of the inequation we get $$x^2>y^2$$ SUFFICIENT Answer: Option B _________________ Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772 Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8033 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Is x^2>y^2? [#permalink] ### Show Tags 21 May 2017, 17:58 ==> In the original condition, there are 2 variables (x,y), and in order to match the number of variables to the number of equations, there must be 2 variables. Since there is 1 for con 1) and 1 for con 2), C is most likely to be the answer. By solving con 1) and con 2), from x>|y|≥0, it is x>0, so from x>|y|, it is |x|>|y|, which becomes $$x^2>y^2$$, hence yes, it is sufficient. The answer is C. However, this is an absolute value question, one of the key questions, so you apply CMT 4 (A: if you get C too easily, consider A or B). For con 1), you get (x,y)=(2,1) yes (2,-3) no, hence not sufficient. For con 2), from x>|y|≥0, you get x>0, which becomes |x|>|y|, then $$x^2>y^2$$, it is always yes and sufficient. Therefore, the answer is B. Answer: B _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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21 May 2017, 23:05
x^2 > y^2 if and only if |x|>|y|

Statement 1. |x|>y.
So absolute value of x is greater than y. But we don't know whether |x|>|y| or not.

Eg, take x = 4, y = 3. Here |x| > y and |x| is also > |y|. So x^2 > y^2
Now take x= -4, y= -5. Here |x| > y But |x| < |y|. Here x^2 < y^2
So Insufficient.

Statement 2. x > |y|
If x > |y| then obviously |x| will also be > |y|.
(Because absolute value of x must be either greater than x if x is negative, or equal to x if x is non-negative)

Thus definitely x^2 > y^2. Statement is Sufficient

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Joined: 27 Apr 2015
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19 Feb 2018, 06:56
1
MathRevolution wrote:
Is $$x^2>y^2$$?

1) |x|>y
2) x>|y|

To Find Is $$x^2>y^2$$ ?
=> OR $$x^2-y^2>0$$
=> OR $$(x-y)(x+y)>0$$
Thus $$x^2>y^2$$ is possible if both $$(x-y)(x+y)$$ are of same sign

PL. NOTE Now IF $$a>b$$----(1) and
=> Case 1 IF $$(a+b)>0$$ then SQUARING both the side of the inequality (1) the inequality sign (here '>') remains SAME i.e (2) becomes $$a^2>b^2$$
=> Case 2 IF $$(a+b)<0$$ then SQUARING both the side of the inequality (1) the inequality sign (here '>') REVERSE. i.e (2) becomes $$a^2<b^2$$
One can verify the above OUTCOMES by plugging values

Stat 1 $$|x|>y$$ ---(2)
So from (2)we have LHS '|x|' is ALWAYS +ve and RHS 'y' is less than |x|
=> Therefore 'y'can be +ve OR -ve and SO $$(|x|+y)$$ can be >0 OR <0. Thus SQUARING both the side of the inequality (2) will give 2 OUTCOMES i.e
=> Case 1 $$(|x|+y)>0$$ then $$|x|^2>y^2$$ or $$x^2>y^2$$
=> Case 2 $$(|x|+y)<0$$ then $$|x|^2<y^2$$ or $$x^2<y^2$$
Since no unique answer, therefore NOT SUFFICIENT

Stat 2 $$x>|y|$$
Since RHS '|y| is ALWAYS +ve and LHS 'x' is greater than |y|
=> Therefore 'x' is ALSO +ve and SO $$(x+|y|)$$ is >0. Thus SQUARING both the side of the inequality (2) will give ONLY 1 OUTCOMES
=> Case 1 $$(x+|y|)>0$$ then $$x^2>|y|^2$$ or $$x^2>y^2$$

Thus Option "B"

Regards
Dinesh
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21 Jul 2019, 04:35
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Re: Is x^2>y^2?   [#permalink] 21 Jul 2019, 04:35
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