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xyz > 0 and y>0

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Carcass
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xyz > 0 and y>0 [#permalink] New post   14 Feb 2022, 06:00
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\(xyz > 0\) and \(y > 0\)

Quantity A
Quantity B
\(xy^2z^3 \)
\(0\)




A) The quantity in Column A is greater.
B) The quantity in Column B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.



Answer: A
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xyz > 0 and y>0 [#permalink] New post   14 Feb 2022, 06:30
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Carcass wrote:
\(xyz > 0\) and \(y > 0\)

Quantity A
Quantity B
\(xy^2z^3\)
\(0\)




A) The quantity in Column A is greater.
B) The quantity in Column B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Answer: A


We can solve this question using matching operations (my favorite Quantitative Comparison strategy)
Given:
Quantity A: \(xy^2z^3\)
Quantity B: \(0\)

Since we are told \(y\) is POSITIVE, we can safely divide both quantities by \(y\) to get:
Quantity A: \(xyz^3\)
Quantity B: \(0\)

Since \(z^2\) must be POSITIVE, we can safely divide both quantities by \(z^2\) to get:
Quantity A: \(xyz\)
Quantity B: \(0\)

Since it's given that \(xyz > 0\), the correct answer is A

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Re: xyz > 0 and y>0 [#permalink] New post   14 Feb 2022, 09:49
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Carcass wrote:
\(xyz > 0\) and \(y > 0\)

Quantity A
Quantity B
\(xy^2z^3 0\)
\(0\)




A) The quantity in Column A is greater.
B) The quantity in Column B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.



Answer: A


Since y>0, xz must be >0 for statement to hold true. Now x & z can either be +ve or -ve together.

Case 1: When both x & z are +v : Ans is A
Cash 2: When booth x & z are -ve: In xy^2z^3 0, we can ignore y as it's +v and also squared. Z^3 will always be -ve as it's raised to odd power, so all in all both -ve will multiply and whole no. will become +ve
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Re: xyz > 0 and y>0 [#permalink] New post   21 Apr 2022, 21:27
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There is a typo in the question.

Zero should not be a part of the A Term.
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Re: xyz > 0 and y>0 [#permalink] New post   21 Apr 2022, 21:45
Expert Reply
avnishkanungo wrote:
There is a typo in the question.

Zero should not be a part of the A Term.


Fixed the typo

Thak you
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steveharvi
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Re: xyz > 0 and y>0 [#permalink] New post   18 Apr 2024, 05:02
what if both x and z is negative, xyz would still be positive, and the above expression can be negative...,
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xyz > 0 and y>0 [#permalink] New post   18 Apr 2024, 05:45
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\(xyz > 0\) and \(y > 0\)

QA \(xy^2z^3 \)

QB zero

Now both in the stem and in QA y is always positive so we can disregard y for now whatever is the value of y

we left with x and z that we do know >0

Now to have xz>0 both are positive or both are negative

if both negative and y^3 will be alwayspositive but multiplied negative x and negative z they will become positive ANDDDD y is positive and QA is positive

if xz>0 are both positive, then y is always positive and QA will be positive

A>B

A is the answer
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Re: xyz > 0 and y>0 [#permalink] New post   18 Apr 2024, 09:26
ahh yes, I missed that part, thank you for the explanation :-)
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Re: xyz > 0 and y>0 [#permalink]
18 Apr 2024, 09:26
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