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Is x^2>y^2?

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Is x^2>y^2?  [#permalink]

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New post 18 May 2017, 01:07
1
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A
B
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Difficulty:

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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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Re: Is x^2>y^2?  [#permalink]

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New post 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
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Re: Is x^2>y^2?  [#permalink]

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New post 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
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Re: Is x^2>y^2?  [#permalink]

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New post 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

Hence answer is B
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Re: Is x^2>y^2?  [#permalink]

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New post 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 \(x^2>y^2\)
Since unique answer, therefore SUFFICIENT

Thus Option "B"

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Re: Is x^2>y^2?  [#permalink]

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New post 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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