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x^2 y > 0

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x^2 y > 0 [#permalink] New post 12 Dec 2015, 07:26
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Question Stats:

89% (00:27) correct 10% (00:23) wrong based on 19 sessions
\(x^2y\) > 0

\(xy^2\) < 0

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.




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Question: 4
Page: 151
Difficulty: medium
[Reveal] Spoiler: OA

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Last edited by Carcass on 09 Jun 2019, 13:21, edited 5 times in total.
Edited the question
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Re: x^2 y > 0 [#permalink] New post 12 Dec 2015, 07:47
Expert's post
Solution

Whereas one possible strategy to attack the question should be pick numbers and test what's going on.

Instead, notice that no matter what in our quantities \(x^2\) and \(y^2\) are always positive, regardless the sign.

But in the first one to be > 0 y must be positive. In the second X must be negative to have > 0

So \(y > x\)

The answer is \(B\)
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Re: x^2 y > 0 [#permalink] New post 14 Jul 2016, 08:09
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I don't understand this solution. Why X must be negative? And what is the [?] symbol?
Carcass wrote:
Solution

Whereas one possible strategy to attack the question should be pick numbers and test what's going on.

Instead, notice that no matter what in our quantities \(x^2\) and \(y^2\) are always positive, regardless the sign.

But in the first one to be > 0 y must be positive. In the second X must be negative to have > 0

So \(y > x\)

The answer is \(B\)
1 KUDOS received
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Re: x^2 y > 0 [#permalink] New post 15 Jul 2016, 02:05
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Sorry for the inconvenience. :( Apologize

In the inequality \(XY^2\) is NOT > 0 but < 0. For this reason there was the symbol [?] because a mismatch .

Thank you
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Re: x^2 y > 0 [#permalink] New post 16 Jul 2016, 01:30
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Thanks. I thought I missed something :crazy:

Carcass wrote:
Sorry for the inconvenience. :( Apologize

In the inequality \(XY^2\) is NOT > 0 but < 0. For this reason there was the symbol [?] because a mismatch .

Thank you
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Re: x^2 y > 0 [#permalink] New post 16 Jul 2016, 01:40
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Thank you to pointed out. The bb code or else was the problem..........
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Re: x^2 y > 0 [#permalink] New post 21 Jul 2016, 09:45
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Carcass wrote:
x²y > 0

xy² < 0

Quantity A
Quantity B
x
y




Given: x²y > 0
Since x² must be POSITIVE, we can divide both sides of this inequality by x² to get: y > 0
In other words y is POSITIVE

Given: xy² < 0
Since y² must be POSITIVE, we can divide both sides of this inequality by y² to get: x < 0
In other words x is NEGATIVE

Answer:
[Reveal] Spoiler:
B


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Re: x^2 y > 0   [#permalink] 21 Jul 2016, 09:45
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