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Is x > y? (1) x^2 > y^2 (2) x > y
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13 May 2012, 05:22
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68% (00:37) correct 32% (00:43) wrong based on 766 sessions
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Is x > y? (1) x^2 > y^2 (2) x > y
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Re: Is x > y?
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10 Feb 2013, 05:50




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Re: Is x > y? (1) x^2 > y^2 (2) x > y
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13 May 2012, 07:00
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. A it is.
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Re: Is x > y? (1) x^2 > y^2 (2) x > y
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27 Jun 2012, 11:19
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.
The answer is A.



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Re: Is IxI > IyI ?
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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 Archit



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Re: Is x > y? (1) x^2 > y^2 (2) x > y
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16 May 2013, 09:36
dvinoth86 wrote: Is x > y?
(1) x^2 > y^2 (2) x > y Stmt1: is sufficients as taking root can give us mod stmt 2: can be flawed as below: 1. 4 , 5 and 4, 5 substitute for answer.



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Re: Is x > y?
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30 Jun 2013, 10:51
Is x > y?
Is x>y? OR Is x>y?
We can also square both sides as we know that x, y are >=0
x > y is x^2 > y^2? Is x^2 > y^2?
(1) x^2 > y^2
This tells us directly that x^2 is greater than y^2 SUFFICIENT
(2) x > y
5>4 5 > 4 5 > 4 (Valid) Or 3>8 3 > 8 3 > 8 (Invalid) INSUFFICIENT
(A)



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Re: Is x > y?
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15 May 2014, 13:36
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 sufficientOption 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 SO, 2nd option is insufficientcorrect option is Kudos is the best form of appreciation



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Re: Is x > y?
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15 May 2014, 22:53
dhirajx wrote: Is \(x\) > \(y\)? (1) \(x^2\) > \(y^2\) (2) \(x\) > \(y\) Sol: We need to know whether x>y St 1 tells us that x^2>y^2 or \(\sqrt{x^2}\) > \(\sqrt{y^2}\) Also x=\(\sqrt{x^2}\) So we have x>y St 1 is clearly sufficient St 2 says x>y if x=5,y=3 then x>y but if x=3 and y=5 then y>x Ans is A
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Re: Is x > y?
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08 Jul 2017, 09:03
Is x > y?
(1) x^2 > y^2 (2) x > y
I used this approach: 1) +  x> +  Y > case 1) x> y case 2)  x > y which means x < y. Two answers, so not sufficient 2) x> y > sufficient What is wrong in this approach? Would appreciate a reply!



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Re: Is x > y?
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19 Jul 2018, 12:25
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Re: Is x > y? &nbs
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