If xy 0 and yz

Question

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

Bunuel · Accepted Answer

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.

niyantg · Answer

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

PPhoenix77 · Answer

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

Bunuel · Answer

(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.

Henriano · Answer

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.

Bunuel · Answer

No brackets are needed there, all is correct. Still formatted to avoid confusion.

NoHalfMeasures · Answer

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

ENGRTOMBA2018 · Answer

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.

mcelroytutoring · Answer

Attached is a visual that should help.

JeffTargetTestPrep · Answer

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

sashiim20 · Answer

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)...

JeffTargetTestPrep · Answer

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

Abhishek009 · Answer

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)

satish_sahoo · Answer

Hi  Bunuel ,

I got the right answer but it took me 2 mins here. In your solution when you say notice, do you went and checked all the options from A to C or had a glance at all the options first and then noticed that option C can be re-written in (positive)*(negative) form and then derived at your answer.

I want to learn how to notice such things coz this will save huge amount of time.

Thanks.

Bunuel · Answer

So, we are given information about xy and yz. I quickly glanced through the options to see if any of them could be broken down into these known terms. Option C was the only one that fit, making it easy to spot the solution.

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

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

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