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If xy > 0 and yz < 0, which of the following must be negativ [#permalink]
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27 Feb 2014, 07:02
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The Official Guide For GMAT® Quantitative Review, 2ND EditionIf xy > 0 and yz < 0, which of the following must be negative? (A) \(xyz\) (B) \(x*y*z^2\) (C) \(x*y^2*z\) (D) \(x*y^2*z^2\) (E) \(x^2*y^2*z^2\) Problem Solving Question: 117 Category: Arithmetic Properties of numbers Page: 76 Difficulty: 600 GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition  Quantitative Questions ProjectEach 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. Thank you!
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Re: If xy > 0 and yz < 0, which of the following must be negativ [#permalink]
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27 Feb 2014, 07:02
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Re: If xy > 0 and yz < 0, which of the following must be negativ [#permalink]
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01 Mar 2014, 02:19
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1. X . Y > 0
Here 2 Cases are possible. Either Both X,Y are Positive or Both are Negative X= +ve & Y=+ve X= ve & Y=ve
2. Y.Z < 0
following from the above case When: Y= +ve then Z= ve When: Y= ve then Z= +ve
After combining, In all we have 2 cases: CASE 1 > X: +ve, Y: +ve, Z: ve CASE2 > X: ve, Y: ve, Z: +ve
Checking for Options:
(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



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Re: If xy > 0 and yz < 0, which of the following must be negativ [#permalink]
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01 Mar 2014, 04:50



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Re: If xy > 0 and yz < 0, which of the following must be negativ [#permalink]
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22 Mar 2014, 05:13
Bunuel wrote: SOLUTION
If xy > 0 and yz < 0, which of the following must be negative?
(A) xyz (B) xyz^2 (C) xy^2z (D) xy^2z^2 (E) x^2y^2z^2
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



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Re: If xy > 0 and yz < 0, which of the following must be negativ [#permalink]
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22 Mar 2014, 05:17



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Re: If xy > 0 and yz < 0, which of the following must be negativ [#permalink]
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03 Jun 2014, 05:07
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
5) Don't even need to check : all will be +ve .



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Re: If xy > 0 and yz < 0, which of the following must be negativ [#permalink]
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22 Oct 2014, 03:36
Please add some brackets in the answer choices. At first, I got mislead by some of the answer choices in the heat of the moment.



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Re: If xy > 0 and yz < 0, which of the following must be negativ [#permalink]
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22 Oct 2014, 03:39



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Re: If xy > 0 and yz < 0, which of the following must be negativ [#permalink]
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23 Jun 2015, 10:30
xy>0 yz<0 means x y z + +  Negativ   + Positiv Just test the answer choices C is the correct answer. x y^2 z + +  Negativ  + + Negativ
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If xy > 0 and yz < 0, which of the following must be negativ [#permalink]
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06 Dec 2015, 13:46
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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
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If xy > 0 and yz < 0, which of the following must be negativ [#permalink]
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06 Dec 2015, 13:56
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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. Hope this helps.



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Re: If xy > 0 and yz < 0, which of the following must be negativ [#permalink]
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07 Mar 2017, 17:45
Quote: If xy > 0 and yz < 0, which of the following must be negative?
(A) \(xyz\) (B) \(x*y*z^2\) (C) \(x*y^2*z\) (D) \(x*y^2*z^2\) (E) \(x^2*y^2*z^2\)
Since xy > 0 and yz < 0, we see the following: When xy > 0: x = positive and y = positive or x = negative and y = negative When yz < 0: y = negative and z = positive or y = positive and z = negative Putting our two statements together: 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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Re: If xy > 0 and yz < 0, which of the following must be negativ [#permalink]
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08 Mar 2017, 03:32
xy is positive yz is negative xy.yz will be negative or x. y^2. z will be negative
Option C



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If xy > 0 and yz < 0, which of the following must be negativ [#permalink]
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16 Apr 2017, 06:35
Bunuel wrote: The Official Guide For GMAT® Quantitative Review, 2ND EditionIf xy > 0 and yz < 0, which of the following must be negative? (A) \(xyz\) (B) \(x*y*z^2\) (C) \(x*y^2*z\) (D) \(x*y^2*z^2\) (E) \(x^2*y^2*z^2\) Problem Solving Question: 117 Category: Arithmetic Properties of numbers Page: 76 Difficulty: 600 GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition  Quantitative Questions ProjectEach 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. 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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If xy > 0 and yz < 0, which of the following must be negativ
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