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(x^2 - 9)/3

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(x^2 - 9)/3 [#permalink] New post 26 Jun 2017, 01:44
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Question Stats:

50% (00:07) correct 50% (01:06) wrong based on 2 sessions


GIVEN X>4

Quantity A
Quantity B
\large{[\frac{(x^2 - 9)/3}{(x+3)/8}]\large}^{-1}
\frac{3}{8}



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.
[Reveal] Spoiler: OA

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Re: (x^2 - 9)/3 [#permalink] New post 26 Jul 2017, 15:28
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Can anyone help me with this question?
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Re: (x^2 - 9)/3 [#permalink] New post 21 Sep 2017, 09:32
As stated the expression above should be equal to [\frac{x^2-9}{3}*\frac{8}{x+3}]^{-1} = [\frac{(x-3)(x+3)}{3}*\frac{8}{x+3}]^{-1} = [\frac{(x-3)}{3}*8]^{-1} = \frac{3}{8}*\frac{1}{x-3}. Thus, the answer is uncertain since \frac{1}{x-3} can be 1 making the two expression equal but also greater or lower than 1. How can the answer be B?
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Re: (x^2 - 9)/3 [#permalink] New post 21 Sep 2017, 20:54
Expert's post
\large[\frac{\frac{x+3}{8}}{\frac{(x^2 - 9)}{3}}\large]^{1}

The first quantity is raised to -1, then switch it in the reverse form.

\frac{3(x+3)}{8(x+3)(x-3)}

(x+3) cancel out

\frac{3}{8(x-3)}

Since x > 4, the denominator in Quantity A must be greater than 8; since the numerators in Quantity A and Quantity B are the same and the denominator in Quantity A is larger, Quantity B must be greater.

Hope now is clear
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Re: (x^2 - 9)/3 [#permalink] New post 22 Sep 2017, 19:20
how X>4; there is statement given that x>4. Ans is D
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Re: (x^2 - 9)/3 [#permalink] New post 22 Sep 2017, 21:40
Pria wrote:
how X>4; there is statement given that x>4. Ans is D



Question re aranged as per Source and X>4 (statement was missing) , Now corrected

Hope it helps

Explanation already given by Carcass
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Re: (x^2 - 9)/3   [#permalink] 22 Sep 2017, 21:40
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