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a > 0 and b < 0.
[#permalink]
Updated on: 10 Jul 2019, 12:22
Updated on: 10 Jul 2019, 12:22
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\(a > 0\) and \(b < 0\).
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.
Quantity A |
Quantity B |
\(\frac{1}{a^2}\) |
\(\frac{1}{b^2}\) |
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.
Retired Moderator
Joined: 10 Apr 2015
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Re: a > 0 and b < 0.
[#permalink]
12 Jul 2019, 06:44
12 Jul 2019, 06:44
arpitjain wrote:
\(a > 0\) and \(b < 0\).
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.
Quantity A |
Quantity B |
\(\frac{1}{a^2}\) |
\(\frac{1}{b^2}\) |
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.
KEY CONCEPT: If we SQUARE positive or negative numbers, the outcome will always be POSITIVE.
For example, 5² = 25 and (-5)² = 25
So, let's test some possible values of a and b
Case i: a = 1 and b = -1
We get:
QUANTITY A: \(\frac{1}{a^2}=\frac{1}{1^2}=\frac{1}{1}=1\)
QUANTITY B: \(\frac{1}{b^2}=\frac{1}{(-1)^2}=\frac{1}{1}=1\)
In this case, the two quantities are equal
Case ii: a = 1 and b = -2
We get:
QUANTITY A: \(\frac{1}{a^2}=\frac{1}{1^2}=\frac{1}{1}=1\)
QUANTITY B: \(\frac{1}{b^2}=\frac{1}{(-2)^2}=\frac{1}{4}\)
In this case, Quantity A is greater
Answer: D
Cheers,
Brent
gmatclubot
Re: a > 0 and b < 0. [#permalink]
12 Jul 2019, 06:44
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