Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 25 Nov 2015, 00:28

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 June
Open Detailed Calendar

Is a>0? (1) a^3-a<0 (2) 1-a^2<0

Author Message
TAGS:
Manager
Joined: 02 Oct 2009
Posts: 93
Followers: 1

Kudos [?]: 18 [0], given: 2

Is a>0? (1) a^3-a<0 (2) 1-a^2<0 [#permalink]  12 Nov 2009, 06:48
4
This post was
BOOKMARKED
00:00

Difficulty:

65% (hard)

Question Stats:

57% (02:41) correct 43% (01:44) wrong based on 138 sessions
Is a>0?

(1) a^3-a<0
(2) 1-a^2<0
[Reveal] Spoiler: OA
Math Expert
Joined: 02 Sep 2009
Posts: 30362
Followers: 5079

Kudos [?]: 57126 [1] , given: 8806

Re: Inequality DS [#permalink]  12 Nov 2009, 07:17
1
KUDOS
Expert's post
2
This post was
BOOKMARKED
KocharRohit wrote:
Is a>0?
________________________________________
Is a>0?

1. a^3-a<0
2. 1-a^2<0

(1) $$a^3-a<0$$ --> $$a*(a^2-1)<0$$

Two cases:
A. $$a<0$$ and $$a^2-1>0$$, (which means $$a<-1$$ or $$a>1$$). As $$a<0$$ then the range would be $$a<-1$$.

B. $$a>0$$ and $$a^2-1<0$$, (which means $$-1<a<1$$). As $$a>0$$ then the range would be $$0<a<1$$.

So from this statement we have that $$a<-1$$ OR $$0<a<1$$. Not sufficient.

(2) $$1-a^2<0$$, which is the same as $$a^2-1>0$$ --> $$a<-1$$ or $$a>1$$. Again two possible answers to the question is $$a$$ positive. Not sufficient.

(1)+(2) Intersection of the ranges form (1) and (2) gives unique range $$a<-1$$ (From (1): $$a<-1$$ OR $$0<a<1$$ AND from (2): $$a<-1$$ or $$a>1$$). So the answer to the question is a positive is NO. sufficient.

_________________
VP
Joined: 05 Mar 2008
Posts: 1473
Followers: 11

Kudos [?]: 230 [0], given: 31

Re: Inequality DS [#permalink]  12 Nov 2009, 07:34
Bunuel wrote:
KocharRohit wrote:
Is a>0?
________________________________________
Is a>0?

1. a^3-a<0
2. 1-a^2<0

(1) $$a^3-a<0$$ --> $$a*(a^2-1)<0$$

Two cases:
A. $$a<0$$ and $$a^2-1>0$$, (which means $$a<-1$$ or $$a>1$$). As $$a<0$$ then the range would be $$a<-1$$.

B. $$a>0$$ and $$a^2-1<0$$, (which means $$-1<a<1$$). As $$a>0$$ then the range would be $$0<a<1$$.

So from this statement we have that $$a<-1$$ OR $$0<a<1$$. Not sufficient.

(2) $$1-a^2<0$$, which is the same as $$a^2-1>0$$ --> $$a<-1$$ or $$a>1$$. Again two possible answers to the question is $$a$$ positive. Not sufficient.

(1)+(2) Intersection of the ranges form (1) and (2) gives unique range $$a<-1$$ (From (1): $$a<-1$$ OR $$0<a<1$$ AND from (2): $$a<-1$$ or $$a>1$$). So the answer to the question is a positive is NO. sufficient.

Question:

on statement 1, why couldn't you say:
a^3-a <0
=a^3 < a
=a^2<1
= -1<a<1
Math Expert
Joined: 02 Sep 2009
Posts: 30362
Followers: 5079

Kudos [?]: 57126 [1] , given: 8806

Re: Inequality DS [#permalink]  12 Nov 2009, 07:47
1
KUDOS
Expert's post
lagomez wrote:
Question:

on statement 1, why couldn't you say:
a^3-a <0
=a^3 < a
=a^2<1
= -1<a<1

This is not correct as when you get $$a^2<1$$ from $$a^3 < a$$, you are reducing (dividing) by a. We can not divide the inequality by the variable (or by any expression) sign of which is unknown.

This is one of the most frequently used catch from GMAT.

NEVER EVER divide inequality by the variable (or expression with variable) with unknown sign.
_________________
VP
Joined: 05 Mar 2008
Posts: 1473
Followers: 11

Kudos [?]: 230 [0], given: 31

Re: Inequality DS [#permalink]  12 Nov 2009, 07:52
Bunuel wrote:
lagomez wrote:
Question:

on statement 1, why couldn't you say:
a^3-a <0
=a^3 < a
=a^2<1
= -1<a<1

This is not correct as when you get $$a^2<1$$ from $$a^3 < a$$, you are reducing (dividing) by a. We can not divide the inequality by the variable (or by any expression) sign of which is unknown.

This is one of the most frequently used catch from GMAT.

NEVER EVER divide inequality by the variable (or expression with variable) with unknown sign.

Yes, good answer. that gets me every time

what if we are told a>0 in the question
then we can divide?
Math Expert
Joined: 02 Sep 2009
Posts: 30362
Followers: 5079

Kudos [?]: 57126 [0], given: 8806

Re: Inequality DS [#permalink]  12 Nov 2009, 07:56
Expert's post
lagomez wrote:

Yes, good answer. that gets me every time

what if we are told a>0 in the question
then we can divide?

Not only if we were told that a is positive, but even if we were told that a is negative we could divide. But in this case the question wouldn't make any sense as the question exactly asks if a is positive.
_________________
Intern
Joined: 07 May 2011
Posts: 42
GMAT 1: Q V
GMAT 2: Q V
Followers: 0

Kudos [?]: 14 [0], given: 11

Re: Is a>0? 1. a^3-a<0 2. 1-a^2<0 [#permalink]  29 May 2012, 15:31
I wanted to offer my thoughts in case my line of reasoning resonates with someone and helps them better understand.
Ok, so we need to know if a>0.
given,
1. a^3-a<0
i could factor out a from both terms but that may make it complicated, but I ma told that the left hand side is negative, so whatever the value of a, whether positive or negative, I ought to be able to say that a is bigger than a^3, otherwise a^3-a would not be less than 0.
so, a^3<a.
if something is cubed and it is still less than the original something, then something is funky, and not your regular positive whole numbers. it could be fraction, positive or negative, i don't know yet, it could be a negative whole number too i suppose, but i would need to test that. let me do that real quick. clearly, one of the possibilities has to be a positive fraction between 0 and 1 because then it would be say (1/2)^3<1/2. that statement holds. so a can be positive. i just need to check if it can be negative too. what if a is a negative fraction? a=-1/2 say. then -1/8<-1/2. that's not right, so a is not a negative fraction, could it be a negative whole number? instantly, i can see -2 cubed would be -8 is less than -2. so a could be a negative whole or a positive fraction. that doesn't conclusively answer the question. so crossing out statement 1, noting that a could be a positive fraction or negative whole. actually, i am gonna draw it on the number line. ok. circle the 0 to 1 portion and less than -1 portion. moving on.
2.1-a^2>0
ok, instantly i can see that 1 has to be greater than a^2 for their difference to be positive. so without worrying about what sign mumbojumbo, i can say a^2<1. well, i have learned from mgmat advanced quant book that anytime i see a^2<1, i can simply write it as |a|<1 and that i can write it as -1<a<1. so a is between -1 and 1. so a could be positive or negative and doesn't conclusively answer the question. so insufficient.
BUT drawing this statement 2 relation on a number line and comparing it to the earlier number line i drew for statement 1, i can see that there is a common region from 0 to 1. so taken together, i can see that a lies between 0 and 1 which are all positive. and it sufficiently answers the original question, Yes, a>0.
Hope my line of reasoning is not a wrong way that still got the right answer, and I hope it at least makes someone follow and understand the solution to this problem.
Manager
Joined: 29 Mar 2010
Posts: 142
Location: United States
GMAT 1: 590 Q28 V38
GPA: 2.54
WE: Accounting (Hospitality and Tourism)
Followers: 1

Kudos [?]: 68 [0], given: 16

Re: Is a>0? (1) a^3-a<0 (2) 1-a^2<0 [#permalink]  30 May 2012, 01:18
For statement 1 can't you just factor out an a and make it a(a^2-1) where a has to equal zero making it not sufficient?
_________________

4/28 GMATPrep 42Q 36V 640

Manager
Joined: 08 Apr 2012
Posts: 129
Followers: 11

Kudos [?]: 81 [0], given: 14

Re: Is a>0? (1) a^3-a<0 (2) 1-a^2<0 [#permalink]  30 May 2012, 04:07
hfbamafan wrote:
For statement 1 can't you just factor out an a and make it a(a^2-1) where a has to equal zero making it not sufficient?

Hi,

My 2 cents:

Bunuel's explanation is great. Hats off.
To put all confusion to rest note that we cannot divide both sides both sides of the statement (1) by a, as we don't know whether a is positive or negative.

Hence, bunuel's approach is the logical one.

Regards,

Shouvik.
_________________

Shouvik
http://www.Edvento.com

Intern
Joined: 19 Mar 2012
Posts: 20
Followers: 1

Kudos [?]: 4 [0], given: 0

Re: Is a>0? (1) a^3-a<0 (2) 1-a^2<0 [#permalink]  07 Jun 2012, 09:09
I'm confused... Maybe someone can show me where I'm going wrong.
This how I approached (I actually came up with A).

a^3-a<0
a(a^2-1)<0
factors out to a(a-1)(a+1)<0 - so it seems here we have factors that are consecutive, (a-1)(a)(a+1)<0
The only product that would be less than 0 would require all 3 to be negative, so I determined a as negative, which would answer the question "no."

Where am I going wrong?
Math Expert
Joined: 02 Sep 2009
Posts: 30362
Followers: 5079

Kudos [?]: 57126 [0], given: 8806

Re: Is a>0? (1) a^3-a<0 (2) 1-a^2<0 [#permalink]  07 Jun 2012, 09:19
Expert's post
joshhowatt wrote:
I'm confused... Maybe someone can show me where I'm going wrong.
This how I approached (I actually came up with A).

a^3-a<0
a(a^2-1)<0
factors out to a(a-1)(a+1)<0 - so it seems here we have factors that are consecutive, (a-1)(a)(a+1)<0
The only product that would be less than 0 would require all 3 to be negative, so I determined a as negative, which would answer the question "no."

Where am I going wrong?

The product of three multiples to be negative either all three must be negative or one must be negative and other two must be positive.

For more on how to solve such kind of inequalities check:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.
_________________
Intern
Joined: 19 Mar 2012
Posts: 20
Followers: 1

Kudos [?]: 4 [0], given: 0

Re: Is a>0? (1) a^3-a<0 (2) 1-a^2<0 [#permalink]  07 Jun 2012, 09:28
Bunuel wrote:
joshhowatt wrote:
I'm confused... Maybe someone can show me where I'm going wrong.
This how I approached (I actually came up with A).

a^3-a<0
a(a^2-1)<0
factors out to a(a-1)(a+1)<0 - so it seems here we have factors that are consecutive, (a-1)(a)(a+1)<0
The only product that would be less than 0 would require all 3 to be negative, so I determined a as negative, which would answer the question "no."

Where am I going wrong?

The product of three multiples to be negative either all three must be negative or one must be negative and other two must be positive.

For more on how to solve such kind of inequalities check:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.

Exactly...
So if they are consecutive numbers we have a couple of options:

Say the numbers are -1,0,1. This won't work because one is zero, therefore the answer would not be less than 0.

So all three must be negative no?
Math Expert
Joined: 02 Sep 2009
Posts: 30362
Followers: 5079

Kudos [?]: 57126 [0], given: 8806

Re: Is a>0? (1) a^3-a<0 (2) 1-a^2<0 [#permalink]  07 Jun 2012, 09:35
Expert's post
joshhowatt wrote:
Bunuel wrote:
joshhowatt wrote:
I'm confused... Maybe someone can show me where I'm going wrong.
This how I approached (I actually came up with A).

a^3-a<0
a(a^2-1)<0
factors out to a(a-1)(a+1)<0 - so it seems here we have factors that are consecutive, (a-1)(a)(a+1)<0
The only product that would be less than 0 would require all 3 to be negative, so I determined a as negative, which would answer the question "no."

Where am I going wrong?

The product of three multiples to be negative either all three must be negative or one must be negative and other two must be positive.

For more on how to solve such kind of inequalities check:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.

Exactly...
So if they are consecutive numbers we have a couple of options:

Say the numbers are -1,0,1. This won't work because one is zero, therefore the answer would not be less than 0.

So all three must be negative no?

We are not told that the $$a$$ is an integer so $$a-1$$, $$a$$ and $$a+1$$ are not necessarily consecutive integers. Please read the solutions above and plug some numbers from the correct ranges to check whether inequality holds true for them.

For example if $$a=\frac{1}{2}$$ then $$a^2-a=\frac{1}{8}-\frac{1}{2}=-\frac{3}{8}<0$$.

_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 7252
Followers: 383

Kudos [?]: 93 [0], given: 0

Re: Is a>0? (1) a^3-a<0 (2) 1-a^2<0 [#permalink]  07 Aug 2014, 05:57
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.
_________________
Intern
Joined: 12 Feb 2013
Posts: 10
Location: India
Concentration: Finance, Marketing
Followers: 1

Kudos [?]: 7 [0], given: 61

Re: Is a>0? (1) a^3-a<0 (2) 1-a^2<0 [#permalink]  07 Aug 2014, 08:31
Statement 1 : a^3-a<0 => a(a^2-1)<0, which can be possible in two cases:
Case 1 : a<0 , a^2 - 1 > 0 Case 2 : a>0 , a^2 - 1 < 0 . Hence, we can't say whether a > 0 .
Statement 2 : 1 - a^2 < 0 => a^2 > 1 , which is again possible in two cases
case 1 : a>0 Case 2: a<0 . Hence, we can't say whether a > 0.

Taking Staement 1 and 2 together : From statement 2 we got to know a^2 > 1 from which using Statement 1 [case 1] we can confirm a<0

Hence Both statements are needed to conclude this.
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 7252
Followers: 383

Kudos [?]: 93 [0], given: 0

Re: Is a>0? (1) a^3-a<0 (2) 1-a^2<0 [#permalink]  27 Aug 2015, 04:48
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.
_________________
Senior Manager
Joined: 23 Jan 2013
Posts: 463
Schools: Cambridge'16
Followers: 2

Kudos [?]: 34 [0], given: 35

Re: Is a>0? (1) a^3-a<0 (2) 1-a^2<0 [#permalink]  30 Sep 2015, 23:32
Is a>0? Y/N

St1. a^3-a<0 means a(a^2-1)<0

so two options: a>0 & a^2<1 OR a<0 & a^2>1. INSUFF

St2. 1-a^2<0 means a^2>1, so can be a<>0. INSUFF

St1+St2 means that a<0 & a^2>1. SUFF

C
Re: Is a>0? (1) a^3-a<0 (2) 1-a^2<0   [#permalink] 30 Sep 2015, 23:32
Similar topics Replies Last post
Similar
Topics:
1 If abc ≠ 0, is a > 0? (1) 3*a/b > 0 (2) b/c^2 < 0 3 16 Sep 2015, 02:51
Is X^2Y^3Z^2>0 ? (1) XY>0 (2) YZ<0 3 17 Aug 2015, 13:59
1 x+y)/Z>0 is x<0? (1) x<y (2) z<0 2 14 Jun 2011, 06:43
1 Is m^n >= 0? 1. m< 0 and n is an integer . 2. n<0 1 20 Dec 2010, 09:21
36 Is (x^7)(y^2)(z^3) > 0 ? (1) yz < 0 (2) xz > 0 27 10 Jun 2010, 11:52
Display posts from previous: Sort by