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

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If xy > 0 and yz < 0, which of the following must be negativ [#permalink] New post 19 Mar 2013, 11:42
00:00
A
B
C
D
E

Difficulty:

  25% (low)

Question Stats:

71% (01:00) correct 28% (00:42) wrong based on 103 sessions
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, 11:54, edited 2 times in total.
Renamed the topic, edited the question and the tags.
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Re: If xy > 0 and yz < 0, which of the following must be negativ [#permalink] New post 19 Mar 2013, 11:57
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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.

Answer: C.

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Re: If xy > 0 and yz < 0, which of the following must be negativ [#permalink] New post 19 Mar 2013, 17:05
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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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Re: If xy > 0 and yz < 0, which of the following must be negativ [#permalink] New post 20 Mar 2013, 05:39
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.
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Re: If xy > 0 and yz < 0, which of the following must be negativ [#permalink] New post 20 Mar 2013, 05:43
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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Re: If xy > 0 and yz < 0, which of the following must be negativ [#permalink] New post 20 Mar 2013, 05:48
Hey Banuel,

Thanks it's much clearer now. Also, I'll make sure to follow the GMAT Rules for posting
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Re: If xy > 0 and yz < 0, which of the following must be negativ [#permalink] New post 25 Jul 2013, 20:20
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).

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Re: If xy > 0 and yz < 0, which of the following must be negativ [#permalink] New post 29 Aug 2013, 04:36
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!
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Table.png
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Re: If xy > 0 and yz < 0, which of the following must be negativ [#permalink] New post 05 Jan 2014, 07:14
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 :)
Re: If xy > 0 and yz < 0, which of the following must be negativ   [#permalink] 05 Jan 2014, 07:14
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