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27x^-2/9x^-4
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02 Mar 2021, 08:59
02 Mar 2021, 08:59
Expert Reply
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Question Stats:
70% (01:04) correct
29% (01:26) wrong based on 54 sessions
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Quantity A |
Quantity B |
\(\frac{27x^{-2}}{9x^{-4}}\) |
\(\frac{12^{-1}x^3}{36^{-1}x^2}\) |
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.
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
Re: 27x^-2/9x^-4
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02 Mar 2021, 09:04
02 Mar 2021, 09:04
1
Let's simplify both A & B
Qt A
\(\frac{27x^{-2}}{9x^{-4}}\) = \(\frac{27 * x^4}{9 * x^2}\) = 3\(x^2\)
Qt B
\(\frac{12^{-1}x^3}{36^{-1}x^2}\) = \(\frac{36 * x^3}{12 * x^2}\) = 3x
Now if x = 1 ; Qt A = Qt B
if x = 2 ; Qt A > Qt B
Answer D
Qt A
\(\frac{27x^{-2}}{9x^{-4}}\) = \(\frac{27 * x^4}{9 * x^2}\) = 3\(x^2\)
Qt B
\(\frac{12^{-1}x^3}{36^{-1}x^2}\) = \(\frac{36 * x^3}{12 * x^2}\) = 3x
Now if x = 1 ; Qt A = Qt B
if x = 2 ; Qt A > Qt B
Answer D
Re: 27x^-2/9x^-4
[#permalink]
28 Apr 2021, 06:58
28 Apr 2021, 06:58
1
It is important to read the question carefully; note that there are no boundary conditions for x.
Therefore, x could be 0, negative, positive, or a fraction.
x=0, C
x=-1, A
x=1/2, B
Therefore, D.
Therefore, x could be 0, negative, positive, or a fraction.
x=0, C
x=-1, A
x=1/2, B
Therefore, D.
