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GRE Prep Club Legend  Joined: 07 Jun 2014
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x2 > y2 and x > –|y| [#permalink]
Expert's post 00:00

Question Stats: 55% (00:44) correct 44% (00:30) wrong based on 43 sessions
$$x^2 > y^2$$ and $$x > -|y|$$

 Quantity A Quantity B x y

A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

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Sandy
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Re: x2 > y2 and x > –|y| [#permalink]
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sandy wrote:
$$x^2 > y^2$$ and $$x > -|y|$$

 Quantity A Quantity B x y

A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.

Given $$x^2 > y^2$$ and $$x > -|y|$$

From $$x^2 > y^2$$ => $$x^2$$ - $$y^2$$ > 0 => ( x + y ) ( x - y ) > 0.

Here ( x + y ) ( x - y ) > 0 either both should be +ve or both -ve to be greater than 0.

But from $$x > -|y|$$ we get ( x + y ) > 0 ( Note here |y| is converted to y )

Since we came to know that ( x + y ) > 0 then must be ( x - y ) > 0

So ( x - y ) > 0 => x > y.

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Re: x2 > y2 and x > –|y| [#permalink]
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if $$x^2>y^2$$, then $$|x|>|y|$$

if x>0 y>0, then x>y since $$|x|>|y|$$;

if x>0 y<0, then x>y. Because $$x>-|y|$$ and $$-|y|=-(-y)=y$$;

x<0 y>0 is not possible. Because $$x>-|y|$$ and $$-|y|=-y$$, so $$x>-y$$, which means $$|x|<|y|$$, but we already know $$|x|>|y|$$, so this is a contradiction;

x<0 y<0 is also not possbile, Because $$x>-|y|$$ and $$-|y|=-(-y)=y$$ therefore x>y but then $$|x|<|y|$$, and we already know $$|x|>|y|$$, so this is a contradiction;
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Re: x2 > y2 and x > –|y| [#permalink]
Yuan wrote:
if $$x^2>y^2$$, then $$|x|>|y|$$

if x>0 y>0, then x>y since $$|x|>|y|$$;

if x>0 y<0, then x>y. Because $$x>-|y|$$ and $$-|y|=-(-y)=y$$;

x<0 y>0 is not possible. Because $$x>-|y|$$ and $$-|y|=-y$$, so $$x>-y$$, which means $$|x|<|y|$$, but we already know $$|x|>|y|$$, so this is a contradiction;

x<0 y<0 is also not possbile, Because $$x>-|y|$$ and $$-|y|=-(-y)=y$$ therefore x>y but then $$|x|<|y|$$, and we already know $$|x|>|y|$$, so this is a contradiction;

$$x>-y$$, which means $$|x|<|y|$$
How please explain?
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Re: x2 > y2 and x > –|y| [#permalink]
Expert's post
sandy wrote:
$$x^2 > y^2$$ and $$x > -|y|$$

 Quantity A Quantity B x y

A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.

$$x^2>y^2$$ means $$|x|>|y|$$..
This means x is farther away from 0 as compared to y..
So there can be four ways...
x....y....0
X.............0....y
0....y....x
y....0.............x

So if x>0 then x>y , but if x<0, then x<y...
So the relationship depend on the SIGN of x..

Now next we are given x>-|y|
Let us check the four cases
x....y....0. => If x<0 then x>-|y| is not true as -|y| >0.
X..........0....y => If x<0 then x>-|y| is not true as -|y| <0, but -|y| will be closer to 0.
0....y....x => If x>0 then x>-|y| is possible.
y....0..........x => If x<0 then x>-|y| is again possible..

In both the cases that are possible, x>y..

Hence A
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Some useful Theory.
1. Arithmetic and Geometric progressions : https://greprepclub.com/forum/progressions-arithmetic-geometric-and-harmonic-11574.html#p27048
2. Effect of Arithmetic Operations on fraction : https://greprepclub.com/forum/effects-of-arithmetic-operations-on-fractions-11573.html?sid=d570445335a783891cd4d48a17db9825
3. Remainders : https://greprepclub.com/forum/remainders-what-you-should-know-11524.html
4. Number properties : https://greprepclub.com/forum/number-property-all-you-require-11518.html
5. Absolute Modulus and Inequalities : https://greprepclub.com/forum/absolute-modulus-a-better-understanding-11281.html Re: x2 > y2 and x > –|y|   [#permalink] 02 Jan 2019, 19:14
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