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|x|y > x|y|

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|x|y > x|y| [#permalink] New post 10 Aug 2018, 10:20
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Question Stats:

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\(|x| y > x |y|\)

Quantity A
Quantity B
\((x + y)^2\)
\((x – y)^2\)


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

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[Reveal] Spoiler: OA

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Re: |x|y > x|y| [#permalink] New post 10 Aug 2018, 14:09
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Carcass wrote:
\(|x|y > x|y|\)

Quantity A
Quantity B
\((x + y)^2\)
\((x – y)^2\)


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

Kudos for R.A.E



\(|x|y > x|y|\)

x and y are on both sides. then why |x|y is greater than x|y|. x is negative ans y is positive.

The best answer is B. Plug values for x and y in order to prove.
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Re: |x|y > x|y| [#permalink] New post 10 Aug 2018, 15:04
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here it is given |x|y > x|y|
meaning if x is negative and y is positive.
so from we get (-x+y)2 and (-x-y)2
clearly (-x-y)2 is greater. So B is greater.
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Re: |x|y > x|y| [#permalink] New post 28 Aug 2018, 00:35
What about when both are negative
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Re: |x|y > x|y| [#permalink] New post 28 Aug 2018, 06:02
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Vivek13791 wrote:
What about when both are negative


If both x and y are negative then the inequality \(|x|y>x|y|\) is not true as for x<0 and y<0 we can write \(|x|y>x|y|\) as \(-xy> - xy\) which is clearly not possible.


The best way to solve this is to rewrite the inequality:

\(|x|y>x|y|\) as

\(\frac{y}{|y|}>\frac{x}{|x|}\) since \(|x|\) and \(|y|\) are always positive numbers it wont effect the inequality.

Also \(\frac{n}{|n|}\) can be either +1 or -1.

Hence for this inequality to hold y is positive and x is negative.


Once you have the fact that x is negative you can substitute in the equations given in the quantity A and B to calculate which is greater!

For example put \(x =-n\) where n is a positive number and y is positive as we have seen before!

Qty A: \((x+y)^2=(y-n)^2\)

Qty B: \((x-y)^2=(y+n)^2\)

Quantity B is greater!
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Re: |x|y > x|y|   [#permalink] 28 Aug 2018, 06:02
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