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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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New post Updated on: 19 Mar 2013, 11:54
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A
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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)

Originally posted by GMAThirst on 19 Mar 2013, 11:42.
Last edited by Bunuel on 19 Mar 2013, 11:54, edited 2 times in total.
Renamed the topic, edited the question and the tags.
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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 19 Mar 2013, 11:57
14
4
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)


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

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 19 Mar 2013, 17:05
1
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
++ (or)
--
yz < 0
+- (or)
-+

Remember Square of any number is always +ve. so we can remove all squares

A. xyz there is a chance for --+ so may be +
B. xy ++ or --
C. xz if x is + then z should be - and vice versa
D. x may be + or -
E. +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 20 Mar 2013, 05:39
Hi, sorry to bother again, but I still didn't understand either of the explanations.....

I saw a long way in the book to solve it but I didn't get how both of you were able to solve it quick in that manner.

Thanks.
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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 20 Mar 2013, 05:43
1
GMAThirst wrote:
Hi, sorry to bother again, but I still didn't understand either of the explanations.....

I saw a long way in the book to solve it but I didn't get how both of you were able to solve it quick in that manner.

Thanks.


xy > 0 means that xy is positive.
yz < 0 means that yz is negative.

x(y^2)z = (xy)(yz) = positive*negative = negative. Thus option C is always 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 20 Mar 2013, 05:48
Hey Banuel,

Thanks it's much clearer now. Also, I'll make sure to follow the GMAT Rules for posting
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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 25 Jul 2013, 20:20
I follow the tables for these types of questions

XY>0 => Both must be -ve or both must be +ve.

YZ<0 =>Both must be of opposite signs.

X | Y | Z

+ + -
- - +

Evaluate all the options by this you will arrive at (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 29 Aug 2013, 04:36
2
I also follow the table approach for these types of questions, as per the attached file. I set up the columns for x, y, and z, and check which combination of positive and negative values I need to satisfy the conditions given (xy>0 and yz<0), then I check each answer choice in turn. When actually doing these questions I wouldn't fill in the whole table (since the answer is C there is no point checking D and E), but for the sake of completeness I added them. Hope it helps a bit, I'm more visual so didn't either consider the algebraic approach given above!
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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 05 Jan 2014, 07:14
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)


If 'x' and 'y' have the same sign, and 'y' and 'z' have different signs, then it must follow that 'x' and 'z' have different sign

Hence y^2 will always be positive and x*z will always be negative

Thus C gives the correct answer

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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 15 Apr 2015, 02:59
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)



from stem, x and y have same sign and z is just opposite to x and y .
clear 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 05 May 2016, 17:30
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Attached is a visual that should help.
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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 12 May 2017, 02:20
since xy>0 (pos) and yz<0 (neg) --> xy (neg) * yz (pos) must be negative

rewrite xy (neg) * yz (pos) --> x(y^2)z
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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 Aug 2017, 19:13
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)



I enjoyed solving this problem. After solving it, It was also a pleasure seeing it categorized as Medium and a 600 question.

---

Starting with the givens:

Given XY > 0
this given should be interpreted as Either both X and Y are positive or Negative. For the product of two like signs always yields a positive.
Given YZ < 0
this given should be interpreted as either Y or Z as positive or negative. For the product of two different signs always yields a negative.

Considering the implications of the givens, this problem branches off in two directions: the first is Y is negative; therefore, X is negative and Z is positive. And the second, Both X and Y are positive, while Z is negative

This problem asks for a MUST solution, so our two branches need to overlap the same answer.

If X and Y = -
Z = +
then (A) (B) (E) is eliminated, leaving us with (C) and (D) as possible Answers

If X and Y = +
Z = -
then (B)(D)(E) is eliminated, leaving us with (A) and (C) as possible answers.

Since the only overlapping solution to our two branches is (C), (C) must be the answer
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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 Aug 2017, 21:15
Clearly C is the Answer.

See the pic, solution is attached,
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If xy > 0 and yz < 0, which of the following must be negativ  [#permalink]

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New post 27 Aug 2017, 17:11
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 negativ  [#permalink]

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New post 31 Aug 2017, 09:24
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 negativ  [#permalink]

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Re: If xy > 0 and yz < 0, which of the following must be negativ &nbs [#permalink] 16 Sep 2018, 03:37
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