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Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation.

(A) xyz : For Case1: Result= -ve For Case2: Result= +ve SO INCORRECT

(B) xyz^2 For Case1: Result= +ve ( For the answer we need MUST be Negative for both the cases & after checking for the 1st Case we are getting a positive Value. So, we don't need to check the 2nd case here ) For Case2: Result= +ve SO INCORRECT

(C) xy^2z For Case1: Result= -ve For Case2: Result= -ve CORRECT

(D) xy^2z^2 For Case1: Result= +ve For Case2: Result= -ve SO INCORRECT

(E) x^2y^2z^2

(At first look we can see it will give only positive result, So can be ignored straight away ) For Case1: Result= +ve For Case2: Result= +ve SO INCORRECT

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

Answer: C.

how we rule out D? xy^2 z^2 = x (yz)^2

(D) \(xy^2z^2=x*positive*positive=x*positive\) --> if x is also positive, then \(x*positive=positive\), thus this option is not necessarily negative. For example, consider x=positive, y=positive, and z=negative.

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

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03 Jun 2014, 04:07

It took me a while but i broke it down to the following: xy is +ve and yz is -ve so we know: 1- x and y have the same sign . Either both +ve or both -ve. 2- yz have different signs. Either y +ve and z -ve OR y -ve and z +ve. What matters here then is the sign of y because it determines the signs of the rest .

Lets test each case: 1)xyz: If y is -ve : (-ve)(-ve)(+ve) = +ve If y is +ve: (+ve)(+ve)(-ve) = -ve

2)xyz^2: If y is -ve : (-ve)(-ve)(+ve) = +ve If y is +ve: (+ve)(+ve)(+ve) = +ve

3) xy^2z: If y is -ve : (-ve)(+ve)(+ve) = -ve If y is +ve: (+ve)(+ve)(-ve) = -ve -----> Correct (both negative)

4)xy^2z^2: If y is -ve : (-ve)(+ve)(+ve) = -ve If y is +ve: (+ve)(+ve)(+ve) = +ve

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

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06 Dec 2015, 12:56

2

This post received KUDOS

MensaNumber wrote:

Bunuel: could the right answer choice also be- x^p*y^q*z^r where p&r are odd and q is an even integer? Thanks

Let me try to answer your question.

As per your conditions, \(x^p*y^q*z^r\) , p,r are odd and q=even. So in order to look for the combination that MUST be negative, neglect 'y'.

Also from yz<0 and xy>0 you know that x and z will have opposite signs. As both of them are raised to odd powers, the "nature" will remain the same (positive --> positive and negative ---> negative). Thus, 1 of the 2, x and z will be positive and the other will be negative ---> \(x^p*y^q*z^r\) always < 0 for all values of x,y,z.

1) When x is positive, y is positive and z is negative

2) When x is negative, y is negative and z is positive.

Using those two statements, let’s determine which answer choice must be true.

A) Is xyz negative?

Using statement two, we see that xyz does not have to be negative.

B) Is (x)(y)(z^2) negative?

In order for answer choice B to be true, xy must be negative. However, both statements one and two prove that xy is not negative.

C) Is (x)(y^2)(z) negative?

In order for answer choice C to be true, xz must be negative. Looking at both statements one and two, we see that in either statement xz is negative. Thus, answer choice C is true. We can stop here.

Answer: C
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Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

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Thank you!

can be only two possibilities case 1 -,-,+=x,y,z Case 2 +,+,-=x,y,z only option C qualifies that x*y^2*z since when we put x^2 in both cases result will be negative
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