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# (x^2 - 9)/3

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(x^2 - 9)/3 [#permalink]  26 Jun 2017, 01:44
Expert's post
00:00

Question Stats:

72% (00:51) correct 27% (01:41) wrong based on 74 sessions

Given $$x >4$$

 Quantity A Quantity B $$\left[ \begin{array}{cc|r} \frac{\frac{(x^2 - 9)}{3}}{\frac{(x+3)}{8} \end{array} \right] ^{-1}$$ $$\frac{3}{8}$$

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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Last edited by Carcass on 27 Dec 2018, 05:43, edited 1 time in total.
Edited by Carcass
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Re: (x^2 - 9)/3 [#permalink]  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]  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]  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]  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]  22 Sep 2017, 21:40
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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

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Founder
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Re: (x^2 - 9)/3 [#permalink]  20 Dec 2018, 04:48
Expert's post
Question improved in its format. Now is very clear to read and perhaps to solve.

Hard question.

Regards
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Re: (x^2 - 9)/3 [#permalink]  21 Dec 2018, 10:51
do not be tricked by putting x=4
keep in mind that x > 4
therefore adjust the end conclusion accordingly
Re: (x^2 - 9)/3   [#permalink] 21 Dec 2018, 10:51
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