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If xy > 0 and yz < 0, which of the following must be negativ [#permalink]
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19 Mar 2013, 11:42
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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)
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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]
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19 Mar 2013, 11:57
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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, 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]
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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]
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20 Mar 2013, 05:43



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



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Re: If xy > 0 and yz < 0, which of the following must be negativ [#permalink]
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14 Apr 2015, 23:21
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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, 02:59
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, 17:30
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Attached is a visual that should help.
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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, 02:20
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]
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23 Aug 2017, 19:13
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) I enjoyed solving this problem. After solving it, It was also a pleasure seeing it categorized as Medium and a 600 question.  Starting with the givens: Given XY > 0 this given should be interpreted as Either both X and Y are positive or Negative. For the product of two like signs always yields a positive. Given YZ < 0 this given should be interpreted as either Y or Z as positive or negative. For the product of two different signs always yields a negative.Considering the implications of the givens, this problem branches off in two directions: the first is Y is negative; therefore, X is negative and Z is positive. And the second, Both X and Y are positive, while Z is negativeThis problem asks for a MUST solution, so our two branches need to overlap the same answer. If X and Y =  Z = + then (A) (B) (E) is eliminated, leaving us with (C) and (D) as possible Answers If X and Y = + Z =  then (B)(D)(E) is eliminated, leaving us with (A) and (C) as possible answers. Since the only overlapping solution to our two branches is (C), (C) must be the answer



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Re: If xy > 0 and yz < 0, which of the following must be negativ [#permalink]
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23 Aug 2017, 21:15
Clearly C is the Answer. See the pic, solution is attached,
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If xy > 0 and yz < 0, which of the following must be negativ [#permalink]
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27 Aug 2017, 17:11
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) \(xy>0\)  (Both \(x\) and \(y\) are either positive or negative. ie; \(x\) and \(y\) have same signs) \(yz<0\)  (\(y\) and \(z\) have different signs) Lets check the options. Case one : \(x\) and \(y\) are positive and \(z\) is negative. Square of negative is positive. A. \(xyz = (positive)(positive)(negative) = negative\) B. \(xy(z^2) = (positive)(positive)(negative^2) = positive\) C. \(x(y^2)z = (positive)(positive^2)(negative) = negative\) D. \(x(y^2)(z^2) = (positive)(positive^2)(negative^2) = positive\) E. \((x^2)(y^2)(z^2) = (positive^2)(positive^2)(negative^2) = positive\) Case two: \(x\) and \(y\) are negative and \(z\) is positive. Square of negative is positive. A. \(xyz = (negative)(negative)(positive) = positive\) B. \(xy(z^2) = (negative)(negative)(positive^2) = positive\) C. \(x(y^2)z = (negative)(negative^2)(positive) = negative\) D. \(x(y^2)(z^2) = (negative)(negative^2)(positive^2) = negative\) E. \((x^2)(y^2)(z^2) = (negative^2)(negative^2)(positive^2) = positive\) Question is asking which of the following "must be" negative. From both the cases, we get, C is negative. Hence Answer C. Answer (C)...



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Re: If xy > 0 and yz < 0, which of the following must be negativ [#permalink]
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31 Aug 2017, 09:24
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) SInce xy > 0: x = pos and y = pos OR x = neg and y = neg SInce yz < 0: y = neg and z = pos OR y = pos and z = neg Thus, we have two scenarios: 1) x = pos, y = pos, and z = neg 2) x = neg, y = neg, and z = pos Since y^2 is always positive regardless of whether y is positive or negative, we see that the product of x and z will always be negative, and thus x(y^2)z will always be negative. Answer: C
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