 It is currently 30 Nov 2020, 02:41 ### GMAT 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

every week, we’ll send you an estimated GMAT 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. # x < 0  Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:
Founder  Joined: 18 Apr 2015
Posts: 13918
GRE 1: Q160 V160 Followers: 315

Kudos [?]: 3687 , given: 12942

Expert's post 00:00

Question Stats: 52% (00:23) correct 47% (00:27) wrong based on 40 sessions

Last edited by Carcass on 27 Dec 2018, 09:47, edited 2 times in total.
Edited the OA
Manager Joined: 29 Nov 2017
Posts: 190
Location: United States
GRE 1: Q142 V146 WE: Information Technology (Computer Software)
Followers: 1

Kudos [?]: 94 , given: 99

if X is negative then if we suppose entity A will be over all positive then A is greater than B.

Last edited by IshanGre on 16 May 2018, 05:40, edited 1 time in total. Retired Moderator Joined: 07 Jun 2014
Posts: 4803
GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
Followers: 175

Kudos [?]: 3037  , given: 394

1
KUDOS
Expert's post
IshanGre wrote:
if X is negative the if we suppose entity A will be over all positive then A is greater than B.

This is actually a tricky question

Given that x<0

Quantity A: $$x^2-x^5 = x^2(1-x^3)$$ now $$x^2$$ is always positve as $$x \neq 0$$ but $$1-x^3$$ may or maynot be positive.

If $$|x| > 0$$ then $$1-x^3$$ is negative.

Try plugging values: $$x=-0.5$$

Quanity A: 0.21875 here quantity A is greater.

$$x=-1.5$$

Quanity A: -5.34375 here quantity B is greater.

Hence the correct option is D.
_________________

Sandy
If you found this post useful, please let me know by pressing the Kudos Button

Try our free Online GRE Test Intern Joined: 08 Apr 2018
Posts: 44
Followers: 0

Kudos [?]: 17  , given: 15

1
KUDOS
sandy wrote:
IshanGre wrote:
if X is negative the if we suppose entity A will be over all positive then A is greater than B.

This is actually a tricky question

Given that x<0

Quantity A: $$x^2-x^5 = x^2(1-x^3)$$ now $$x^2$$ is always positve as $$x \neq 0$$ but $$1-x^3$$ may or maynot be positive.

If $$|x| > 0$$ then $$1-x^3$$ is negative.

Try plugging values: $$x=-0.5$$

Quanity A: 0.21875 here quantity A is greater.

$$x=-1.5$$

Quanity A: -5.34375 here quantity B is greater.

Hence the correct option is D.

Hi Sandy, if the correct answer is D, can you please update the Reveal Answer to D? It is still A.
Intern Joined: 15 May 2018
Posts: 38
Followers: 0

Kudos [?]: 6 , given: 1

Carcass wrote:
$$x < 0$$

 Quantity A Quantity B $$x^2 - x^5$$ $$0$$

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

answer must be a in all cases Intern Joined: 25 May 2018
Posts: 3
Followers: 0

Kudos [?]: 3  , given: 0

1
KUDOS
Founder  Joined: 18 Apr 2015
Posts: 13918
GRE 1: Q160 V160 Followers: 315

Kudos [?]: 3687 , given: 12942

Expert's post Re: x < 0   [#permalink] 04 Jul 2018, 00:59
Display posts from previous: Sort by

# x < 0  Question banks Downloads My Bookmarks Reviews Important topics  Powered by phpBB © phpBB Group Kindly note that the GRE® test is a registered trademark of the Educational Testing Service®, and this site has neither been reviewed nor endorsed by ETS®.