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(x^2 - 9)/3

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Carcass
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(x^2 - 9)/3 [#permalink] New post   Updated on: 27 Dec 2018, 05:43
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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.
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B

Originally posted by Carcass on 26 Jun 2017, 01:44.
Last edited by Carcass on 27 Dec 2018, 05:43, edited 1 time in total.
Edited by Carcass
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(x^2 - 9)/3 [#permalink] New post   25 Mar 2021, 22:48
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Carcass wrote:

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.


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

Col. A: \(\left[ \begin{array}{cc|r}{\frac{(x+3)(x-3)}{3}\frac{8}{(x+3)} \end{array} \right] ^{-1}\)
Col. B: \(\frac{3}{8}\)

Col. A: \(\frac{3}{8(x-3)}\)
Col. B: \(\frac{3}{8}\)

Multiplying both sides by \(\frac{8}{3}\);

Col. A: \(\frac{1}{(x-3)}\)
Col. B: \(1\)

Since, \(x > 4, (x-3) > 0\)
Multiplying both sides by \((x-3)\);

Col. A: \(1\)
Col. B: \((x-3)\)

Adding \(3\) to both sides;

Col. A: \(4\)
Col. B: \(x\)

Hence, option B
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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
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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
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\(\large[\frac{\frac{x+3}{8}}{\frac{(x^2 - 9)}{3}}\large]^{1}\)

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
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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

Explanation already given by Carcass
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Re: (x^2 - 9)/3 [#permalink] New post   20 Dec 2018, 04:48
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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] New post   21 Dec 2018, 10:51
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do not be tricked by putting x=4
keep in mind that x > 4
therefore adjust the end conclusion accordingly
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Re: (x^2 - 9)/3 [#permalink] New post   26 Mar 2021, 03:14
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KarunMendiratta wrote:
Carcass wrote:

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.


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

Col. A: \(\left[ \begin{array}{cc|r}{\frac{(x+3)(x-3)}{3}\frac{8}{(x+3)} \end{array} \right] ^{-1}\)
Col. B: \(\frac{3}{8}\)

Col. A: \(\frac{3}{8(x-3)}\)
Col. B: \(\frac{3}{8}\)

Multiplying both sides by \(\frac{8}{3}\);

Col. A: \(\frac{1}{(x-3)}\)
Col. B: \(1\)

Since, \(x > 4, (x-3) > 0\)
Multiplying both sides by \((x-3)\);

Col. A: \(1\)
Col. B: \((x-3)\)

Adding \(3\) to both sides;

Col. A: \(4\)
Col. B: \(x\)

Hence, option B


The latex code you have implemented is perfect

You are 5 stars asset for the forum and the students......until last :)

:thumbsup:
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(x^2 - 9)/3 [#permalink] New post   26 Mar 2021, 07:22
Carcass

Thanks :please:
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Re: (x^2 - 9)/3 [#permalink] New post   16 Oct 2022, 07:58
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