GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 14 Nov 2018, 20:27

### 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

## Events & Promotions

###### Events & Promotions in November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
• ### $450 Tuition Credit & Official CAT Packs FREE November 15, 2018 November 15, 2018 10:00 PM MST 11:00 PM MST EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth$100 with the 3 Month Pack ($299) • ### Free GMAT Strategy Webinar November 17, 2018 November 17, 2018 07:00 AM PST 09:00 AM PST Nov. 17, 7 AM PST. Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT. # Is a<0?  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Retired Moderator Joined: 27 Aug 2012 Posts: 1091 Is a<0? [#permalink] ### Show Tags Updated on: 05 Aug 2013, 22:29 6 00:00 Difficulty: 75% (hard) Question Stats: 58% (02:16) correct 42% (02:20) wrong based on 180 sessions ### HideShow timer Statistics Is a<0? (1) $$a^3<a^2+2a$$ (2) $$a^2>a^3$$ _________________ Originally posted by bagdbmba on 05 Aug 2013, 22:28. Last edited by mau5 on 05 Aug 2013, 22:29, edited 1 time in total. Edited the Q Verbal Forum Moderator Joined: 10 Oct 2012 Posts: 611 Re: Is a<0? [#permalink] ### Show Tags 05 Aug 2013, 22:37 1 1 bagdbmba wrote: Is a<0? (1) $$a^3<a^2+2a$$ (2) $$a^2>a^3$$ From F.S 1, add 1 on both sides : $$a^2+2a+1>a^3+1 \to (a+1)^2-(a+1)(a^2+1-a)>0 \to (a+1)[(a+1)-(a^2+1-a)]>0 \to (a+1)(2a-a^2)>0$$ Thus, we get a(a+1)(a-2)<0. Either 0<a<2 OR a<-1. Insufficient. From F.S 2, we can divide by$$a^2$$ on both sides and we get a<1. Insufficient. Taking both together, we know that a<1. Thus, a could be 0<a<1 OR a<-1. Insufficient. E. _________________ Director Joined: 14 Dec 2012 Posts: 767 Location: India Concentration: General Management, Operations GMAT 1: 700 Q50 V34 GPA: 3.6 Re: Is a<0? [#permalink] ### Show Tags 05 Aug 2013, 22:42 bagdbmba wrote: Is a<0? (1) $$a^3<a^2+2a$$ (2) $$a^2>a^3$$ STMNT 1: $$a^3<a^2+2a$$ $$a^3-a^2-2a < 0$$ a(a+1)(a-2)<0 when a = -2 expression above is= -8 which is < 0 when a = 1 expression above is = -2 which is <0 hence a can be -ve /+ve insufficient STMNT 2: $$a^2$$$$> a^3$$ $$a^2(a-1)<0$$ clearly this satisfies for a<1 hence a can be -ve /+ve insufficient combining both still both statement satisfies for a = 0.5 and a =-2 hence insufficient hence E _________________ When you want to succeed as bad as you want to breathe ...then you will be successfull.... GIVE VALUE TO OFFICIAL QUESTIONS... GMAT RCs VOCABULARY LIST: http://gmatclub.com/forum/vocabulary-list-for-gmat-reading-comprehension-155228.html learn AWA writing techniques while watching video : http://www.gmatprepnow.com/module/gmat-analytical-writing-assessment : http://www.youtube.com/watch?v=APt9ITygGss Senior Manager Joined: 10 Jul 2013 Posts: 314 Re: Is a<0? [#permalink] ### Show Tags 05 Aug 2013, 22:50 bagdbmba wrote: Is a<0? (1) $$a^3<a^2+2a$$ (2) $$a^2>a^3$$ st(1) , add 1 to both sides. then a= -1 and a= 1/2. you will have a double case. st(2), a can be any positive fraction here and a can be any negative integer too such as a= 1/2 and a = -2 . a double case too. so both statement depict the same answers, which are both a double case. so Answer is (E) _________________ Asif vai..... Verbal Forum Moderator Joined: 10 Oct 2012 Posts: 611 Re: Is a<0? [#permalink] ### Show Tags 08 Aug 2013, 00:45 1 kumar23badgujar wrote: Consider, Statement (1): $$a^3 < a^2 + 2a$$ $$a^3 - a^2 - 2a < 0$$ $$a(a^2 - a - 2) < 0$$ $$a(a -2)(a + 1) < 0$$ i.e. a = 0 or a = 2 or a = -1, hence not sufficient. Statement (2): $$a^2 > a^3$$ This simply means, a < 1, hence not sufficient. Combining both statements, we get a = 0 or a = -1, Hence both statements together not sufficient. Correct Ans: E Even though you have the correct answer, I am sorry but it is not what it means(the highlighted part). It is not an equality, rather an in-equality.Please refer through the above posts, for the correct method. _________________ Manager Status: Waiting ! Waiting ! .. Joined: 14 Mar 2012 Posts: 66 Location: India Schools: Kelley '16 (A) GMAT 1: 720 Q50 V37 GMAT 2: 740 Q50 V40 GPA: 3.5 WE: Project Management (Non-Profit and Government) Is a < 0 ? [#permalink] ### Show Tags Updated on: 20 Sep 2013, 10:48 1 3 Is a < 0 ? (1) a^3 < a^2 + 2a (2) a^2 > a^3 Originally posted by EnemyInside on 20 Sep 2013, 10:45. Last edited by Bunuel on 20 Sep 2013, 10:48, edited 1 time in total. Renamed the topic and edited the question. Math Expert Joined: 02 Sep 2009 Posts: 50585 Re: Is a < 0 ? [#permalink] ### Show Tags 21 Sep 2013, 03:33 1 1 pjagadish27 wrote: Statement 1 -> a^3-a^2-2a<0. a(a-2)(a+1)<0=>a<0 or a<2 or a<-1. Not Sufficient. Statement 2->a^2-a^3>0=>a^2(1-a)>0 => a>0 or a<1.Not Sufficient. 1&2,still a<2 exists which does not answer the question is a<0. So E. The answer is E, but the ranges are not correct. Is a < 0 ? (1) a^3 < a^2 + 2a --> $$(a+1)a(a-2)<0$$ --> $$a<-1$$ or $$0<a<2$$. Not sufficient. (2) a^2 > a^3 --> $$a^2(1-a)>0$$ --> $$a<0$$ or $$0<a<1$$. Not sufficient. (1)+(2) Intersection of the ranges from (1) and (2) is $$a<-1$$ or $$0<a<1$$. Not sufficient. Answer: E. _________________ Manager Joined: 03 Dec 2012 Posts: 222 Re: Is a < 0 ? [#permalink] ### Show Tags 09 Oct 2013, 00:49 Bunuel why can't we factorize a^2>a^3 as 1>a or a<1. Math Expert Joined: 02 Sep 2009 Posts: 50585 Re: Is a < 0 ? [#permalink] ### Show Tags 09 Oct 2013, 01:00 1 mohnish104 wrote: Bunuel why can't we factorize a^2>a^3 as 1>a or a<1. a<1 implies that a can be 0. But a=0 does not satisfy a^2>a^3, so the correct ranges for which this inequality holds true is a<0 or 0<a<1 (the same range as you have excluding 0). Hope it helps. _________________ Intern Joined: 24 Oct 2013 Posts: 5 Re: Is a < 0 ? [#permalink] ### Show Tags 18 Jan 2014, 09:58 Bunuel wrote: mohnish104 wrote: Bunuel why can't we factorize a^2>a^3 as 1>a or a<1. a<1 implies that a can be 0. But a=0 does not satisfy a^2>a^3, so the correct ranges for which this inequality holds true is a<0 or 0<a<1 (the same range as you have excluding 0). Hope it helps. Hi, There's a concept related to inequalities that I fail to understand. Could you please tell me what I'm doing wrong? For statement 2 we have "a^2 > a^3". Depending on how I solve this, I'm getting two complete different ranges, both of which are incorrect. First case: a^2 - a^3 > 0 a^2*(1-a)>0------> this gives me the critical points 0 and 1. Hence I have 3 ranges: (1st) a<0, (2nd) 0<a<1, and (3rd) a>1. As the inequality has the ">0" sign I took only the 1st and 3rd range and got "a<0 or a>1". Second case: 0>a^3 - a^2 0>a^2*(a-1)------> this also gives me the critical points 0 and 1. Hence I have the same 3 ranges: (1st) a<0, (2nd) 0<a<1, and (3rd) a>1. However, this time we have "0>" sign, so I took only the second range: 0<a<1. In any case, both are false. What's odd is that I used the same method to find the range for statement 1 but I got the correct answer. Thanks a lot for your help. Aurèle Math Expert Joined: 02 Sep 2009 Posts: 50585 Re: Is a < 0 ? [#permalink] ### Show Tags 19 Jan 2014, 09:34 Aurele wrote: Bunuel wrote: mohnish104 wrote: Bunuel why can't we factorize a^2>a^3 as 1>a or a<1. a<1 implies that a can be 0. But a=0 does not satisfy a^2>a^3, so the correct ranges for which this inequality holds true is a<0 or 0<a<1 (the same range as you have excluding 0). Hope it helps. Hi, There's a concept related to inequalities that I fail to understand. Could you please tell me what I'm doing wrong? For statement 2 we have "a^2 > a^3". Depending on how I solve this, I'm getting two complete different ranges, both of which are incorrect. First case: a^2 - a^3 > 0 a^2*(1-a)>0------> this gives me the critical points 0 and 1. Hence I have 3 ranges: (1st) a<0, (2nd) 0<a<1, and (3rd) a>1. As the inequality has the ">0" sign I took only the 1st and 3rd range and got "a<0 or a>1". Second case: 0>a^3 - a^2 0>a^2*(a-1)------> this also gives me the critical points 0 and 1. Hence I have the same 3 ranges: (1st) a<0, (2nd) 0<a<1, and (3rd) a>1. However, this time we have "0>" sign, so I took only the second range: 0<a<1. In any case, both are false. What's odd is that I used the same method to find the range for statement 1 but I got the correct answer. Thanks a lot for your help. Aurèle 0 is not a critical point. The squared terms (basically even powers) must be ignored because they cannot be negative and hence doesn't affect the sign. _________________ Intern Joined: 24 Oct 2013 Posts: 5 Re: Is a < 0 ? [#permalink] ### Show Tags 21 Jan 2014, 06:42 Bunuel wrote: 0 is not a critical point. The squared terms (basically even powers) must be ignored because they cannot be negative and hence doesn't affect the sign. Great, thanks for the reply. Just to be sure that I understand the concept fully, could you tell me please if I'm solving the following equation correctly (from Manhattan books): $$x^6 - x^7 > x^5 - x^6$$ I factor the equation and get: $$x^5*(x-1)^2 < 0$$ Here, If I understand your explanation well, $$x^5$$ is raised to an odd power; hence, it should be considered, as it can yield a negative or positive result. Conversely, $$(x-1)^2$$ is raised to an even power; thus, we ignore it. The critical point is then : $$0$$ Because we have the $$<0$$ sign, we'd get $$x<0$$. Would this be correct? Manager Joined: 14 Oct 2014 Posts: 66 Location: United States GMAT 1: 500 Q36 V23 Re: Is a < 0 ? [#permalink] ### Show Tags 29 Jan 2015, 21:49 Is a<0? (1) Insufficient. If a=-5, then -125 < 25-10 --->Yes If a=1/2, then 1/8 < 1/4+1 --->No (2) Insufficient. If a=-5, then 25>-125 --->Yes If a=1/2, then 1/4>1/8 --->No (1)+(2) Insufficient. We can use the same numbers and we get two different answers Intern Joined: 31 May 2015 Posts: 1 Is a<0? Help understanding why my solution is not valid. [#permalink] ### Show Tags 02 Jun 2015, 10:37 Hello! I was looking through these posts, answering questions, and I came across this one: Is a<0? (1) a3<a2+2a (2) a2>a3 from: forum/is-a-157421.html In the topic above, the answer is that both statements together are insufficient to answer the problem but surely the only case in which $$a^{2}$$ is greater than$$a^{3}$$ is if a is negative? My answer is that both statements alone can answer the question, as the only way $$a^{3}$$ is smaller than what it is compared to is if a is negative. What am I overlooking? Thank you. e-GMAT Representative Joined: 04 Jan 2015 Posts: 2191 Re: Is a<0? [#permalink] ### Show Tags 02 Jun 2015, 21:02 1 cmcmcm wrote: Hello! I was looking through these posts, answering questions, and I came across this one: Is a<0? (1) a3<a2+2a (2) a2>a3 from: forum/is-a-157421.html In the topic above, the answer is that both statements together are insufficient to answer the problem but surely the only case in which $$a^{2}$$ is greater than$$a^{3}$$ is if a is negative? My answer is that both statements alone can answer the question, as the only way $$a^{3}$$ is smaller than what it is compared to is if a is negative. What am I overlooking? Thank you. Hi cmcmcm, Always be very careful on how you are finding out the range of an inequality. Let me help you out with finding the range of a for both the inequalities. Statement-I $$a^3<a^2+2a$$ can be simplified to $$a(a + 1)(a - 2) < 0$$. Using the wavy line method to find the range of $$a$$ with the zero points being 2, 0 and -1. We can see from the wavy line diagram that the inequality is negative in the range where $$a < -1$$ or $$0 < a < 2$$. Hence, you can't say for sure if $$a < 0$$ using statement-I alone Statement-II $$a^2>a^3$$ can be simplified to $$a^2(a - 1) < 0$$. Since $$a^2$$ is always non-negative, for $$a^2(a - 1) < 0$$, $$(a - 1) < 0$$ i.e. $$a < 1$$. So $$a < 0$$ or $$a > 0$$. Hence using statement-II alone you cant' say for sure if $$a < 0$$. Combining statement-I & II Combining statements-I & II will give us the range as $$a < -1$$ or $$0 <a <1$$. Hence, it is not sufficient to tell if $$a < 0$$. Therefore the answer is E. You can read more about the Wavy line method here. Hope it's clear .Let me know if you have any doubt in any part of the explanation. Regards Harsh _________________ Register for free sessions Number Properties | Algebra |Quant Workshop Success Stories Guillermo's Success Story | Carrie's Success Story Ace GMAT quant Articles and Question to reach Q51 | Question of the week Must Read Articles Number Properties – Even Odd | LCM GCD | Statistics-1 | Statistics-2 Word Problems – Percentage 1 | Percentage 2 | Time and Work 1 | Time and Work 2 | Time, Speed and Distance 1 | Time, Speed and Distance 2 Advanced Topics- Permutation and Combination 1 | Permutation and Combination 2 | Permutation and Combination 3 | Probability Geometry- Triangles 1 | Triangles 2 | Triangles 3 | Common Mistakes in Geometry Algebra- Wavy line | Inequalities Practice Questions Number Properties 1 | Number Properties 2 | Algebra 1 | Geometry | Prime Numbers | Absolute value equations | Sets | '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 12856 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: Is a<0? [#permalink] ### Show Tags 03 Jun 2015, 14:26 1 Hi cmcmcm, The prompt does NOT state that "A" has to be an integer, so you have to consider the possibility that it's NOT an integer (meaning "A" could be a fraction). While that level of 'thoroughness' isn't going to be required on that many DS questions, Test Takers who score at the higher levels in the Quant section are more likely to see questions in which fractional answers have to be considered. GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/

*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****

Manager
Joined: 21 May 2015
Posts: 236
Concentration: Operations, Strategy
GMAT 1: 750 Q50 V41
Re: Is a<0?  [#permalink]

### Show Tags

03 Jun 2015, 23:44
E
(1) take a = -1, -2 ---- If a=-1 then eqn gives -1<-1 which is not true and if a=-2 then eqn gives -8<0 which is true ....thats y insuff
(2) a^2>a^3 implies a can be -ve or 0<a<1 thus insuff....
Combined also is insuff because of above reasons
_________________

Apoorv

I realize that i cannot change the world....But i can play a part

Manager
Joined: 21 May 2015
Posts: 236
Concentration: Operations, Strategy
GMAT 1: 750 Q50 V41
Re: Is a < 0 ?  [#permalink]

### Show Tags

04 Jun 2015, 00:12
(1) a^3 < a^2 + 2a - --- a=-1 then -1<-1 not true ; a=-2 then -8<0 true insuff
(2) a^2 > a^3 then a<0 or 0<a<1 insuff

If we combine both we cannot answer bc of above reasons
_________________

Apoorv

I realize that i cannot change the world....But i can play a part

Non-Human User
Joined: 09 Sep 2013
Posts: 8777
Re: Is a<0?  [#permalink]

### Show Tags

30 Mar 2018, 20:20
Hello from the GMAT Club BumpBot!

Thanks to another GMAT 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.
_________________
Re: Is a<0? &nbs [#permalink] 30 Mar 2018, 20:20
Display posts from previous: Sort by

# Is a<0?

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.