If xy > 0 and yz < 0, which of the following must be negative?

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

Answer: C.
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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

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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Attached is a visual that should help.

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

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

\(xy>0\) ------- (Both \(x\) and \(y\) are either positive or negative. ie; \(x\) and \(y\) have same signs)

\(yz<0\) ------- (\(y\) and \(z\) have different signs)

Lets check the options.

Case one : \(x\) and \(y\) are positive and \(z\) is negative. Square of negative is positive.

A. \(xyz = (positive)(positive)(negative) = negative\)
B. \(xy(z^2) = (positive)(positive)(negative^2) = positive\)
C. \(x(y^2)z = (positive)(positive^2)(negative) = negative\)
D. \(x(y^2)(z^2) = (positive)(positive^2)(negative^2) = positive\)
E. \((x^2)(y^2)(z^2) = (positive^2)(positive^2)(negative^2) = positive\)

Case two: \(x\) and \(y\) are negative and \(z\) is positive. Square of negative is positive.

A. \(xyz = (negative)(negative)(positive) = positive\)
B. \(xy(z^2) = (negative)(negative)(positive^2) = positive\)
C. \(x(y^2)z = (negative)(negative^2)(positive) = negative\)
D. \(x(y^2)(z^2) = (negative)(negative^2)(positive^2) = negative\)
E. \((x^2)(y^2)(z^2) = (negative^2)(negative^2)(positive^2) = positive\)

Question is asking which of the following "must be" negative. From both the cases, we get, C is negative. Hence Answer C.

Answer (C)...

(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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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.

No brackets are needed there, all is correct. Still formatted to avoid confusion.
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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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Comparing \(xy > 0\) and \(yz < 0\) we can certainly say that -

\(y < 0\) , ie y must be -ve as \(yz < 0\) , further since \(xy > 0\) ie +ve , \(x\) must be -ve as well....

Hence we have, \(x = y =\) -ve and \(z\) is +ve

(A) \(-*-*+ = +\)

(B) \(-*-*+^2 = +\)

(C) \(-*-^2*+ = -\)

(D) \(-*-^2*+^2 = +\)

(E) \(-^2*-^2*+^2 = +\)

Clearly, Answer will be (C)
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