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# If xy > 0 and yz < 0, which of the following must be negativ

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Intern
Joined: 14 Jul 2012
Posts: 8
Concentration: Marketing, Entrepreneurship
WE: Sales (Other)
If xy > 0 and yz < 0, which of the following must be negativ [#permalink]

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19 Mar 2013, 12:42
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Question Stats:

66% (02:06) correct 34% (00:51) wrong based on 413 sessions

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If xy > 0 and yz < 0, which of the following must be negative:

A. xyz
B. xy(z^2)
C. x(y^2)z
D. x(y^2)(z^2)
E. (x^2)(y^2)(z^2)
[Reveal] Spoiler: OA

Last edited by Bunuel on 19 Mar 2013, 12:54, edited 2 times in total.
Renamed the topic, edited the question and the tags.
Math Expert
Joined: 02 Sep 2009
Posts: 39702
Re: If xy > 0 and yz < 0, which of the following must be negativ [#permalink]

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19 Mar 2013, 12:57
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GMAThirst wrote:
If xy > 0 and yz < 0, which of the following must be negative:

A. xyz
B. xy(z^2)
C. x(y^2)z
D. x(y^2)(z^2)
E. (x^2)(y^2)(z^2)

Notice that option C is x(y^2)z = (xy)(yz) = positive*negative = negative.

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Re: If xy > 0 and yz < 0, which of the following must be negativ [#permalink]

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19 Mar 2013, 18:05
1
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GMAThirst wrote:
If xy > 0 and yz < 0, which of the following must be negative:

A. xyz
B. xy(z^2)
C. x(y^2)z
D. x(y^2)(z^2)
E. (x^2)(y^2)(z^2)

xy > 0
++ (or)
--
yz < 0
+- (or)
-+

Remember Square of any number is always +ve. so we can remove all squares

A. xyz there is a chance for --+ so may be +
B. xy ++ or --
C. xz if x is + then z should be - and vice versa
D. x may be + or -
E. +ve
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Intern
Joined: 14 Jul 2012
Posts: 8
Concentration: Marketing, Entrepreneurship
WE: Sales (Other)
Re: If xy > 0 and yz < 0, which of the following must be negativ [#permalink]

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20 Mar 2013, 06:39
Hi, sorry to bother again, but I still didn't understand either of the explanations.....

I saw a long way in the book to solve it but I didn't get how both of you were able to solve it quick in that manner.

Thanks.
Math Expert
Joined: 02 Sep 2009
Posts: 39702
Re: If xy > 0 and yz < 0, which of the following must be negativ [#permalink]

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20 Mar 2013, 06:43
1
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Expert's post
GMAThirst wrote:
Hi, sorry to bother again, but I still didn't understand either of the explanations.....

I saw a long way in the book to solve it but I didn't get how both of you were able to solve it quick in that manner.

Thanks.

xy > 0 means that xy is positive.
yz < 0 means that yz is negative.

x(y^2)z = (xy)(yz) = positive*negative = negative. Thus option C is always negative.

Hope it's clear.
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Intern
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Posts: 8
Concentration: Marketing, Entrepreneurship
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Re: If xy > 0 and yz < 0, which of the following must be negativ [#permalink]

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20 Mar 2013, 06:48
Hey Banuel,

Thanks it's much clearer now. Also, I'll make sure to follow the GMAT Rules for posting
Director
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GMAT 1: 630 Q47 V29
GMAT 2: 680 Q50 V32
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Re: If xy > 0 and yz < 0, which of the following must be negativ [#permalink]

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25 Jul 2013, 21:20
I follow the tables for these types of questions

XY>0 => Both must be -ve or both must be +ve.

YZ<0 =>Both must be of opposite signs.

X | Y | Z

+ + -
- - +

Evaluate all the options by this you will arrive at (C).

Rgds,
TGC !
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Intern
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Re: If xy > 0 and yz < 0, which of the following must be negativ [#permalink]

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29 Aug 2013, 05:36
I also follow the table approach for these types of questions, as per the attached file. I set up the columns for x, y, and z, and check which combination of positive and negative values I need to satisfy the conditions given (xy>0 and yz<0), then I check each answer choice in turn. When actually doing these questions I wouldn't fill in the whole table (since the answer is C there is no point checking D and E), but for the sake of completeness I added them. Hope it helps a bit, I'm more visual so didn't either consider the algebraic approach given above!
Attachments

File comment: Solution table

Table.png [ 3.87 KiB | Viewed 5508 times ]

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Re: If xy > 0 and yz < 0, which of the following must be negativ [#permalink]

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05 Jan 2014, 08:14
GMAThirst wrote:
If xy > 0 and yz < 0, which of the following must be negative:

A. xyz
B. xy(z^2)
C. x(y^2)z
D. x(y^2)(z^2)
E. (x^2)(y^2)(z^2)

If 'x' and 'y' have the same sign, and 'y' and 'z' have different signs, then it must follow that 'x' and 'z' have different sign

Hence y^2 will always be positive and x*z will always be negative

Thus C gives the correct answer

Cheers!
J
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Posts: 15990
Re: If xy > 0 and yz < 0, which of the following must be negativ [#permalink]

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15 Apr 2015, 00:21
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Re: If xy > 0 and yz < 0, which of the following must be negativ [#permalink]

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15 Apr 2015, 03:59
GMAThirst wrote:
If xy > 0 and yz < 0, which of the following must be negative:

A. xyz
B. xy(z^2)
C. x(y^2)z
D. x(y^2)(z^2)
E. (x^2)(y^2)(z^2)

from stem, x and y have same sign and z is just opposite to x and y .
clear C .
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If xy > 0 and yz < 0, which of the following must be negativ [#permalink]

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05 May 2016, 18:30
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Attached is a visual that should help.
Attachments

Screen Shot 2016-05-05 at 6.44.19 PM.png [ 141.98 KiB | Viewed 2113 times ]

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Manager
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Re: If xy > 0 and yz < 0, which of the following must be negativ [#permalink]

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12 May 2017, 03:20
since xy>0 (pos) and yz<0 (neg) --> xy (neg) * yz (pos) must be negative

rewrite xy (neg) * yz (pos) --> x(y^2)z
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Re: If xy > 0 and yz < 0, which of the following must be negativ   [#permalink] 12 May 2017, 03:20
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