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# |x|y > x|y|

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

54% (01:20) correct 45% (00:39) wrong based on 37 sessions
$$|x| y > x |y|$$

 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]  10 Aug 2018, 14:09
1
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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]  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]  28 Aug 2018, 00:35
What about when both are negative
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Re: |x|y > x|y| [#permalink]  28 Aug 2018, 06:02
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Expert's post
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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