Last visit was: 05 Nov 2024, 05:35 It is currently 05 Nov 2024, 05:35

Close

GRE Prep Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GRE score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel
Verbal Expert
Joined: 18 Apr 2015
Posts: 29891
Own Kudos [?]: 36119 [2]
Given Kudos: 25919
Send PM
Most Helpful Community Reply
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Own Kudos [?]: 3208 [2]
Given Kudos: 172
Location: India
WE:Education (Education)
Send PM
General Discussion
avatar
Intern
Intern
Joined: 01 May 2017
Posts: 3
Own Kudos [?]: 5 [0]
Given Kudos: 0
Send PM
avatar
Director
Director
Joined: 03 Sep 2017
Posts: 518
Own Kudos [?]: 701 [0]
Given Kudos: 0
Send PM
Re: (x^2 - 9)/3 [#permalink]
2
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?
Verbal Expert
Joined: 18 Apr 2015
Posts: 29891
Own Kudos [?]: 36119 [2]
Given Kudos: 25919
Send PM
Re: (x^2 - 9)/3 [#permalink]
1
Expert Reply
1
Bookmarks
\(\large[\frac{\frac{x+3}{8}}{\frac{(x^2 - 9)}{3}}\large]^{1}\)

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
avatar
Intern
Intern
Joined: 28 Feb 2017
Posts: 6
Own Kudos [?]: 2 [0]
Given Kudos: 0
Send PM
Re: (x^2 - 9)/3 [#permalink]
how X>4; there is statement given that x>4. Ans is D
avatar
Retired Moderator
Joined: 20 Apr 2016
Posts: 1307
Own Kudos [?]: 2272 [0]
Given Kudos: 251
WE:Engineering (Energy and Utilities)
Send PM
Re: (x^2 - 9)/3 [#permalink]
1
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
Verbal Expert
Joined: 18 Apr 2015
Posts: 29891
Own Kudos [?]: 36119 [0]
Given Kudos: 25919
Send PM
Re: (x^2 - 9)/3 [#permalink]
Expert Reply
Question improved in its format. Now is very clear to read and perhaps to solve.

Hard question.

Regards
avatar
Intern
Intern
Joined: 21 Dec 2018
Posts: 20
Own Kudos [?]: 8 [0]
Given Kudos: 0
Send PM
Re: (x^2 - 9)/3 [#permalink]
1
do not be tricked by putting x=4
keep in mind that x > 4
therefore adjust the end conclusion accordingly
Verbal Expert
Joined: 18 Apr 2015
Posts: 29891
Own Kudos [?]: 36119 [0]
Given Kudos: 25919
Send PM
Re: (x^2 - 9)/3 [#permalink]
Expert Reply
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:
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Own Kudos [?]: 3208 [0]
Given Kudos: 172
Location: India
WE:Education (Education)
Send PM
(x^2 - 9)/3 [#permalink]
Carcass

Thanks :please:
User avatar
GRE Prep Club Legend
GRE Prep Club Legend
Joined: 07 Jan 2021
Posts: 5006
Own Kudos [?]: 74 [0]
Given Kudos: 0
Send PM
Re: (x^2 - 9)/3 [#permalink]
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Prep Club for GRE Bot
Re: (x^2 - 9)/3 [#permalink]
Moderators:
GRE Instructor
77 posts
GRE Forum Moderator
37 posts
Moderator
1111 posts
GRE Instructor
228 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne