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Math Expert V
Joined: 02 Sep 2009
Posts: 56239
If xy > 0 and yz < 0, which of the following must be negativ  [#permalink]

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2
19 00:00

Difficulty:   45% (medium)

Question Stats: 68% (01:49) correct 32% (01:32) wrong based on 1162 sessions

### HideShow timer Statistics 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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Math Expert V
Joined: 02 Sep 2009
Posts: 56239
Re: If xy > 0 and yz < 0, which of the following must be negativ  [#permalink]

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8
15
SOLUTION

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) $$x^2*y^2*z^2$$

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

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Posts: 35
Re: If xy > 0 and yz < 0, which of the following must be negativ  [#permalink]

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5
3
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
##### General Discussion
Math Expert V
Joined: 02 Sep 2009
Posts: 56239
Re: If xy > 0 and yz < 0, which of the following must be negativ  [#permalink]

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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 $$x(y^2)z = (xy)(yz) = positive*negative = negative$$.

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

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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 $$x(y^2)z = (xy)(yz) = positive*negative = negative$$.

how we rule out D?
xy^2 z^2 = x (yz)^2
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Math Expert V
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Posts: 56239
Re: If xy > 0 and yz < 0, which of the following must be negativ  [#permalink]

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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 $$x(y^2)z = (xy)(yz) = positive*negative = negative$$.

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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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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Math Expert V
Joined: 02 Sep 2009
Posts: 56239
Re: If xy > 0 and yz < 0, which of the following must be negativ  [#permalink]

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Henriano wrote:

No brackets are needed there, all is correct. Still formatted to avoid confusion.
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Re: If xy > 0 and yz < 0, which of the following must be negativ  [#permalink]

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xy>0 yz<0 means

x y z
+ + - Negativ
- - + Positiv

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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1
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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2
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

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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1
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.

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

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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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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

GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition - Quantitative Questions Project

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

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Bunuel, they dont call you genius for no reason, epic solution.
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Re: If xy > 0 and yz < 0, which of the following must be negativ  [#permalink]

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Solved it in 02:50. Tricky if you dont identify the wrong answers right away. Somehow I ended up leaving a) and c) last because I was unsure which one it was. However, spent an extra 30 s figuering out the answer must be c)  Re: If xy > 0 and yz < 0, which of the following must be negativ   [#permalink] 08 Aug 2018, 13:13
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