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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

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).

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

Re: Is x between 0 and 1? (1) - x < x^3 (2) x < x^2 [#permalink]

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17 Jul 2012, 12:29

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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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.

(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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