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x<0 and y<1
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13 Aug 2022, 08:55
13 Aug 2022, 08:55
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Question Stats:
59% (01:07) correct
40% (01:06) wrong based on 22 sessions
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x<0 and y<1
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Quantity A |
Quantity B |
\(xy[x+z]\) |
\(x^2+xyz\) |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Part of the project: GRE QCQ & PS 3 Questions Every One DAILY Challenge - (2022)
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GRE Math Essentials (2022)
GRE Geometry Formulas
GRE - Math Book
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
GRE Math Essentials (2022)
GRE Geometry Formulas
GRE - Math Book
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Re: x<0 and y<1
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16 Aug 2022, 06:33
16 Aug 2022, 06:33
1
Carcass wrote:
x<0 and y<1
Quantity A |
Quantity B |
\(xy[x+z]\) |
\(x^2+xyz\) |
We can solve this question using matching operations (see the video lesson below )
Given:
Quantity A: \(xy(x+z)\)
Quantity B: \(x^2+xyz\)
Expand Quantity A:
Quantity A: \(x^2y + xyz\)
Quantity B: \(x^2+xyz\)
Subtract \(xyz\) from both quantities:
Quantity A: \(x^2y\)
Quantity B: \(x^2\)
Since \(x < 0\), we can be certain that \(x^2\) is positive, which means we can safely divide both quantities by \(x^2\) to get:
Quantity A: \(y\)
Quantity B: \(1\)
Since we are told \(y < 1\), it must be the case that Quantity B is greater.
Answer: B
RELATED VIDEO
gmatclubot
Re: x<0 and y<1 [#permalink]
16 Aug 2022, 06:33
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