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Concentration: Entrepreneurship, International Business

GMAT 1: 440 Q33 V13

GPA: 3

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

This post received KUDOS

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 _________________

GMAT - Practice, Patience, Persistence Kudos if u like

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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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!

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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23 Aug 2017, 20:13

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 negative

This 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

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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