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Is xy + xy < xy ?

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Is xy + xy < xy ?  [#permalink]

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Updated on: 18 Jun 2014, 01:04
9
00:00

Difficulty:

85% (hard)

Question Stats:

53% (02:15) correct 47% (02:08) wrong based on 132 sessions

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Is xy + xy < xy ?

(1) x^2/y < 0

(2) x^9*(y^3)^3 < (x^2)^4*y^8

Originally posted by nitin1negi on 17 Jun 2014, 19:45.
Last edited by Bunuel on 18 Jun 2014, 01:04, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to DS forum.
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Re: Is xy + xy < xy ?  [#permalink]

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18 Jun 2014, 01:17
nitin1negi wrote:
Is xy + xy < xy ?

(1) x^2/y < 0

(2) x^9*(y^3)^3 < (x^2)^4*y^8

Is xy+xy<xy?

Is $$xy+xy<xy$$? --> cancel xy in both sides: is $$xy<0$$? So, the question basically asks whether x and y have the opposite signs.

(1) x^2/y<0. This statement implies that y is negative and x can be anything but zero. Not sufficient.

(2) x^9*(y^3)^3 < (x^2)^4*y^8 --> $$(xy)^9<(xy)^8$$. Since $$(xy)^8>0$$ (from the inequality we can deduce that neither of the unknowns is zero, thus even power of xy will ensure that it's positive), then we can safely reduce by it: $$xy<1$$ ($$xy\neq{0}$$). So, we cannot say whether $$xy<0$$. Not sufficient.

(1)+(2) $$y<0$$ and $$xy<1$$. If $$x=-\frac{1}{2}$$ and $$y=-1$$, then the answer is NO but if $$x=1$$ and $$y=-1$$, then the answer is YES. Not sufficient.

Answer: E.

Hope it's clear.

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html pay attention to rules, 2, 3, 5 and 7. Thank you.

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Is xy + xy < xy?  [#permalink]

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Updated on: 20 Mar 2017, 22:38
Is xy + xy < xy?

(1) x^2/y<0

(2) x^9(y^3)^3 < (x^2)^4*y^8

Originally posted by harshal123 on 20 Mar 2017, 14:15.
Last edited by Bunuel on 20 Mar 2017, 22:38, edited 2 times in total.
Renamed the topic and edited the question.
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Re: Is xy + xy < xy?  [#permalink]

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20 Mar 2017, 16:48
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harshal123 wrote:
is xy+xy <xy?

(1) x^2/y<0

(2) x^9(y^3)^3 < (x^2)^4 y^8

Dear harshal123,

I'm happy to respond.

I will say that this Jamboree question looks remarkably like this MGMAT question. At least these are how the sources have been cited on GMAT Club. If these two companies really are the authors of these respective questions, I would say that something is suspicious here. I will let MGMAT and Jamboree sort that out.

Here are a couple blog articles you may find relevant:
Exponent Properties on the GMAT
Exponent Powers on the GMAT

First, let's decipher the prompt:
xy + xy < xy
subtract xy
xy < 0
That's what the question is really asking: is the product of those two variables negative?

Statement #1: $$\frac{x^2}{y} < 0$$
We know that neither variable can equal zero. Anything non-zero squared is positive, so x^2 is positive. Divide both sides by this.

$$\frac{1}{y} < 0$$

We know that y is negative. We know nothing about x. Thus, this statement, alone and by itself, is insufficient.

Statement #2: $$x^9(y^3)^3 < (x^2)^4 y^8$$

This can be simplified to
$$(xy)^9 < (xy)^8$$

Again, we know that neither variable can be zero. Thus, xy is a non-zero product, and any non-zero number raised to an even power is positive. We know $$(xy)^8$$ is a positive number, so we can divide both sides of the inequality by this.

xy < 1

We know that the product is less than 1. It could be less than zero, but it doesn't have to be. Thus, this statement, alone and by itself, is insufficient.

Combined:
From S#1, we know y < 0
From S#2, we know xy < 1

Example #1: $$x = +2$$, $$y = -3$$. This satisfies both statements and gives a "yes" answer to the prompt.
Example #2: $$x = -2$$, $$y = -\frac{1}{3}$$. This satisfies both statements and gives a "no" answer to the prompt.

Two different answers, nothing is sufficient. Combined, the statements are not sufficient.

OA = (E)

Does all this make sense?
Mike
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Mike McGarry
Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)
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Re: Is xy + xy < xy ?  [#permalink]

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20 Mar 2017, 22:40
harshal123 wrote:
Is xy + xy < xy?

(1) x^2/y<0

(2) x^9(y^3)^3 < (x^2)^4*y^8

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Merging topics.
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Re: Is xy + xy < xy ?  [#permalink]

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24 Jan 2019, 22:54
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Re: Is xy + xy < xy ?   [#permalink] 24 Jan 2019, 22:54
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