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Director  Joined: 07 Jun 2004
Posts: 583
Location: PA
Is x between 0 and 1? (1) - x < x^3 (2) x < x^2  [#permalink]

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3
2 00:00

Difficulty:   75% (hard)

Question Stats: 52% (01:45) correct 48% (01:59) wrong based on 247 sessions

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Is x between 0 and 1?

(1) - x < x^3
(2) x < x^2
Math Expert V
Joined: 02 Sep 2009
Posts: 55226
Re: DS Algebra  [#permalink]

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1
1
rxs0005 wrote:
Is x between 0 and 1?

(1) - x < x^3
(2) x < x^2

Is $$0<x<1$$?

(1) -x<x^3 --> $$x^3+x>0$$ --> $$x(x^2+1)>0$$ --> $$x>0$$ (as x^2+1 is always positive). Not sufficient.

(2) x<x^2 --> $$x^2-x>0$$ --> $$x(x-1)>0$$ --> either $$x>1$$ or $$x<0$$, so the answer is NO. Sufficient.

Answer: B.

Check this for more: x2-4x-94661.html#p731476
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Director  Joined: 07 Jun 2004
Posts: 583
Location: PA
Re: DS Algebra  [#permalink]

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Bunuel

thanks , I had a Q , your approach to these DS algebra has always been using algebra itself to get the answers will this ( should ) work most of the times right ?

I prefer your approach as its clear cut rather than plug in values and check more time consuming

do share your thoughts on this

thanks

rxs0005
Math Expert V
Joined: 02 Sep 2009
Posts: 55226
Re: DS Algebra  [#permalink]

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rxs0005 wrote:
Bunuel

thanks , I had a Q , your approach to these DS algebra has always been using algebra itself to get the answers will this ( should ) work most of the times right ?

I prefer your approach as its clear cut rather than plug in values and check more time consuming

do share your thoughts on this

thanks

rxs0005

Yes, algebraic approach will work most of the times, though sometimes other approaches might be faster or/and simpler, also there are certain GMAT questions which are pretty much only solvable with plug-in or trial and error methods (well at leas in 2-3 minutes).

Check this for more on this issue:
positive-integers-85437.html#p831985
gmat-prep-exponent-101358.html#p797434
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Manager  Joined: 19 Dec 2010
Posts: 95
Re: DS Algebra  [#permalink]

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pick a positive integer, positive fraction. Negative integer, negative fraction approach to this problem...
statement 1 says that x is > 0 (integer or fraction doesn't matter...it satisfies the condition)
statement 2 says that x is <0 and >1 hence it never satisfies the original problem. NO is an acceptable answer.
Solved. B
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Re: DS Algebra  [#permalink]

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(1)

-x < x^3 , this is correct for any +ve fraction, but also for a number > 1, not sufficient.

(2)

x < x^2

This is correct for any number(-ve integer or -ve fraction), but also for a number > 1

x< 0 or x > 1, so answer is B
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Manager  Joined: 16 Feb 2012
Posts: 172
Concentration: Finance, Economics
Re: Is x between 0 and 1? (1) - x < x^3 (2) x < x^2  [#permalink]

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If someone could elaborate the first statement in more detail. From the statement I know that x(x^2 + 1)> 0, so x>0 and x^2 + 1> 1 isn't it?
Furthermore, I can conclude that x could be x>0 and x>1, so because I don't know the exact value of x the statement is insufficient.
Is that correct?
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Math Expert V
Joined: 02 Sep 2009
Posts: 55226
Re: Is x between 0 and 1? (1) - x < x^3 (2) x < x^2  [#permalink]

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Stiv wrote:
If someone could elaborate the first statement in more detail. From the statement I know that x(x^2 + 1)> 0, so x>0 and x^2 + 1> 1 isn't it?
Furthermore, I can conclude that x could be x>0 and x>1, so because I don't know the exact value of x the statement is insufficient.
Is that correct?

Not quite.

The question asks: is $$0<x<1$$?

The first statement says: $$x(x^2+1)>0$$. So, we have that the product of two multiples, $$x$$ and $$x^2+1$$, is positive. Now, since the second multiple is always positive ($$x^2+1= nonnegative +positive=positive$$), then the first multiple must also be positive in order the product to be positive, therefore $$x>0$$. So, from this statement, we cannot say whether $$0<x<1$$ is true.

Hope it's clear.
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Re: DS Algebra  [#permalink]

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Bunuel wrote:
rxs0005 wrote:
Is x between 0 and 1?

(1) - x < x^3
(2) x < x^2

Is $$0<x<1$$?

(1) -x<x^3 --> $$x^3+x>0$$ --> $$x(x^2+1)>0$$ --> $$x>0$$ (as x^2+1 is always positive). Not sufficient.

(2) x<x^2 --> $$x^2-x>0$$ --> $$x(x-1)>0$$ --> either $$x>1$$ or $$x<0$$, so the answer is NO. Sufficient.

Answer: B.

Dear Bunuel, if x is a negative integer, even then we can conclude that x < x^2

from B, how can we know that x cannot be a negative integer??
Math Expert V
Joined: 02 Sep 2009
Posts: 55226
Re: DS Algebra  [#permalink]

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asifibaju wrote:
Bunuel wrote:
rxs0005 wrote:
Is x between 0 and 1?

(1) - x < x^3
(2) x < x^2

Is $$0<x<1$$?

(1) -x<x^3 --> $$x^3+x>0$$ --> $$x(x^2+1)>0$$ --> $$x>0$$ (as x^2+1 is always positive). Not sufficient.

(2) x<x^2 --> $$x^2-x>0$$ --> $$x(x-1)>0$$ --> either $$x>1$$ or $$x<0$$, so the answer is NO. Sufficient.

Answer: B.

Dear Bunuel, if x is a negative integer, even then we can conclude that x < x^2

from B, how can we know that x cannot be a negative integer??

Not quite sure understand your question.

We are asked: is $$0<x<1$$? From (2) we have that $$x>1$$ or $$x<0$$, which means that $$x$$ could be any negative number (including integers) as well as any number greater than 1 . Hence the answer to the question whether $$0<x<1$$ is NO.
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Re: DS Algebra  [#permalink]

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oops !! i am really sorry !! Got it now.. I was out of my mind and asked such a silly question.. effect of a hectic day.. sorry again Intern  B
Joined: 01 Jun 2015
Posts: 1
Re: Is x between 0 and 1? (1) - x < x^3 (2) x < x^2  [#permalink]

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dear Bunuel

could you please explain the statement 2 solution in more detail. i followed till the eq x(x-1) > 0..after it how are we moving to x<0 or X>1

Thanks
Math Expert V
Joined: 02 Sep 2009
Posts: 55226
Re: Is x between 0 and 1? (1) - x < x^3 (2) x < x^2  [#permalink]

### Show Tags Re: Is x between 0 and 1? (1) - x < x^3 (2) x < x^2   [#permalink] 19 Jun 2017, 03:59
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# Is x between 0 and 1? (1) - x < x^3 (2) x < x^2

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