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Re: If xy > 0 and yz < 0, which of the following must be negative? [#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 \(x(y^2)z = (xy)(yz) = positive*negative = negative\).

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

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 negative? [#permalink]
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 negative? [#permalink]
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Henriano wrote:
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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Re: If xy > 0 and yz < 0, which of the following must be negative? [#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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Re: If xy > 0 and yz < 0, which of the following must be negative? [#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 negative? [#permalink]
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Attached is a visual that should help.
Attachments

Screen Shot 2016-05-05 at 6.44.19 PM.png
Screen Shot 2016-05-05 at 6.44.19 PM.png [ 141.98 KiB | Viewed 31366 times ]

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

A. xyz
B. xy(z^2)
C. x(y^2)z
D. x(y^2)(z^2)
E. (x^2)(y^2)(z^2)


\(xy>0\) ------- (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)...
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Re: If xy > 0 and yz < 0, which of the following must be negative? [#permalink]
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GMAThirst wrote:
If xy > 0 and yz < 0, which of the following must be negative:

A. xyz
B. xy(z^2)
C. x(y^2)z
D. x(y^2)(z^2)
E. (x^2)(y^2)(z^2)


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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Re: If xy > 0 and yz < 0, which of the following must be negative? [#permalink]
Bunuel wrote:
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) \(x^2*y^2*z^2\)


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