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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]

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The Official Guide For GMAT® Quantitative Review, 2ND Edition

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) \(x^2*y^2*z^2\)

Problem Solving
Question: 117
Category: Arithmetic Properties of numbers
Page: 76
Difficulty: 600


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

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

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

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New post 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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New post 22 Mar 2014, 05:17
DestinyGMAT wrote:
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


(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.

Hope it's clear.
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Re: If xy > 0 and yz < 0, which of the following must be negativ [#permalink]

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

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

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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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New post 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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New post 16 Apr 2017, 06:35
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

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) \(x^2*y^2*z^2\)

Problem Solving
Question: 117
Category: Arithmetic Properties of numbers
Page: 76
Difficulty: 600


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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   [#permalink] 16 Apr 2017, 06:35
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