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xyz > 0 and y>0
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14 Feb 2022, 06:00
14 Feb 2022, 06:00
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71% (00:58) correct
28% (00:38) wrong based on 67 sessions
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\(xyz > 0\) and \(y > 0\)
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
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
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14 Feb 2022, 06:30
14 Feb 2022, 06:30
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Carcass wrote:
\(xyz > 0\) and \(y > 0\)
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
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
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14 Feb 2022, 09:49
14 Feb 2022, 09:49
1
Carcass wrote:
\(xyz > 0\) and \(y > 0\)
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
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
