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

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Is a>0? (1) a^3-a<0 (2) 1-a^2<0 [#permalink]

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12 Nov 2009, 07:48
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Is a>0?

(1) a^3-a<0
(2) 1-a^2<0
[Reveal] Spoiler: OA

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12 Nov 2009, 08:17
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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.

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12 Nov 2009, 08:34
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

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Is a>0? (1) a^3-a<0 (2) 1-a^2<0 [#permalink]

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12 Nov 2009, 08:47
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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 cannot 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.
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12 Nov 2009, 08:52
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?

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12 Nov 2009, 08:56
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.
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Re: Is a>0? 1. a^3-a<0 2. 1-a^2<0 [#permalink]

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29 May 2012, 16:31
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.

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Re: Is a>0? (1) a^3-a<0 (2) 1-a^2<0 [#permalink]

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30 May 2012, 02:18
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?
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Re: Is a>0? (1) a^3-a<0 (2) 1-a^2<0 [#permalink]

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30 May 2012, 05:07
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.
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Re: Is a>0? (1) a^3-a<0 (2) 1-a^2<0 [#permalink]

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07 Jun 2012, 10:09
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?

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Re: Is a>0? (1) a^3-a<0 (2) 1-a^2<0 [#permalink]

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07 Jun 2012, 10:19
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.
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Re: Is a>0? (1) a^3-a<0 (2) 1-a^2<0 [#permalink]

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07 Jun 2012, 10:28
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?

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Re: Is a>0? (1) a^3-a<0 (2) 1-a^2<0 [#permalink]

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07 Jun 2012, 10:35
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 $$a^2-a=\frac{1}{8}-\frac{1}{2}=-\frac{3}{8}<0$$.

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Re: Is a>0? (1) a^3-a<0 (2) 1-a^2<0 [#permalink]

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07 Aug 2014, 06:57
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Re: Is a>0? (1) a^3-a<0 (2) 1-a^2<0 [#permalink]

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07 Aug 2014, 09:31
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.

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Re: Is a>0? (1) a^3-a<0 (2) 1-a^2<0 [#permalink]

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27 Aug 2015, 05:48
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Re: Is a>0? (1) a^3-a<0 (2) 1-a^2<0 [#permalink]

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01 Oct 2015, 00:32
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

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Re: Is a>0? (1) a^3-a<0 (2) 1-a^2<0 [#permalink]

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Is a > 0 ? [#permalink]

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05 Jan 2017, 20:00
Is a > 0 ?

1, $$a^{3} - a < 0$$
2, $$1 - a^{2} > 0$$
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Re: Is a > 0 ? [#permalink]

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05 Jan 2017, 20:09
Is a positive?
Statement 1:
a^3-a<0
a^3<a
So either a<-1 or 0<a<1
NOT SUFFICIENT

Statement 2:
1-a^2>0
1>a^2
So -1<a<1
NOT SUFFICIENT

Statement 1 & 2 together :

0<a<1
SUFFICIENT

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Re: Is a > 0 ?   [#permalink] 05 Jan 2017, 20:09

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