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a < b and a < 0
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22 Nov 2023, 03:03
22 Nov 2023, 03:03
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a < −b and a < 0
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 |
\(a^2\) |
\(b^2\) |
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.
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Free Materials for the GRE General Exam - Where to get it!!
GRE Geometry Formulas
GRE - Math Book
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Re: a < b and a < 0
[#permalink]
01 Dec 2023, 17:30
01 Dec 2023, 17:30
2
Key point is that b could be either a negative or a positive number.
Case 1: b is a positive number
E.g., b = 1. Given that a < -b, say a = -2. Plug it into the columns A and B.
\(1^2 = 1\) and \((-2)^2 = 4\)
So when b is positive, \(b^2 > a^2\).
Case 2: b is a negative number
E.g., b = -1 so a < 1. However, we must still satisfy the rule that a < 0. Let's say a = -2.
\((-2)^2 =4\) and \((-1)^2 = 1\).
Thus, \(a^2 > b^2\).
Since there are values of a and b that satisfy both the rules but have differing inequalities, the answer is D.
Case 1: b is a positive number
E.g., b = 1. Given that a < -b, say a = -2. Plug it into the columns A and B.
\(1^2 = 1\) and \((-2)^2 = 4\)
So when b is positive, \(b^2 > a^2\).
Case 2: b is a negative number
E.g., b = -1 so a < 1. However, we must still satisfy the rule that a < 0. Let's say a = -2.
\((-2)^2 =4\) and \((-1)^2 = 1\).
Thus, \(a^2 > b^2\).
Since there are values of a and b that satisfy both the rules but have differing inequalities, the answer is D.
gmatclubot
Re: a < b and a < 0 [#permalink]
01 Dec 2023, 17:30
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