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Is x > 0 ?

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Is x > 0?  [#permalink]

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New post 03 Dec 2015, 19:08
Is x > 0?

(1) \(x < x^2\)

(2) \(x < x^3\)

Explanation: Statement (1) is insufficient. If \(x < x^3\), x could be negative: \(x^2\) will always be positive, so for any negative value of x, \(x^2\) will be greater. x could also be positive and greater than 1: for any such value, \(x^2\) will be larger. If x is positive and less than 1, \(x^2\) is smaller. Long story short: according to (1), x must be negative or greater than 1. That isn’t enough information.

Statement (2) is also insufficient. x must be between –1 and 0, or greater than 1. If x is a negative fraction, \(x^3\) is a larger number: a negative fraction closer to zero. If x is a positive number greater than 1, \(x^3\) is greater still.

Taken together, it’s still not enough information to determine whether x is positive. x could be between –1 and 0, or it could be greater than 1. The correct choice is (E).
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Re: Is x > 0?  [#permalink]

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New post 03 Dec 2015, 19:17
gmatser1 wrote:
Is x > 0?

(1) \(x < x^2\)

(2) \(x < x^3\)

Explanation: Statement (1) is insufficient. If \(x < x^3\), x could be negative: \(x^2\) will always be positive, so for any negative value of x, \(x^2\) will be greater. x could also be positive and greater than 1: for any such value, \(x^2\) will be larger. If x is positive and less than 1, \(x^2\) is smaller. Long story short: according to (1), x must be negative or greater than 1. That isn’t enough information.

Statement (2) is also insufficient. x must be between –1 and 0, or greater than 1. If x is a negative fraction, \(x^3\) is a larger number: a negative fraction closer to zero. If x is a positive number greater than 1, \(x^3\) is greater still.

Taken together, it’s still not enough information to determine whether x is positive. x could be between –1 and 0, or it could be greater than 1. The correct choice is (E).


Question asks if x>0 ?

Per statement 1, x^2>x --> x^2-x>0 ---> x(x-1)>0 ---> either x<0 or x>1. Not sufficient to answer the question asked.

Per statement 2, x^3>x ---> x^3-x>0 ---> x(x+1)(x-1)>0 ---> either -1<x<0 or x> 1. Again, not sufficient to answer the question asked.

Combining the 2 statements, you get the ranges for x as -1<x<0 or x>1. Again, still not a definitive answer to the question asked.

E is the correct answer.
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Re: Is x > 0?  [#permalink]

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New post 03 Dec 2015, 22:37
Engr2012 wrote:
gmatser1 wrote:
Is x > 0?

either x<0


what is the rationale for inverting the inequality sign?
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Re: Is x > 0 ?  [#permalink]

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New post 03 Dec 2015, 23:43
gmatser1 wrote:
Is x > 0?

(1) \(x < x^2\)

(2) \(x < x^3\)

Explanation: Statement (1) is insufficient. If \(x < x^3\), x could be negative: \(x^2\) will always be positive, so for any negative value of x, \(x^2\) will be greater. x could also be positive and greater than 1: for any such value, \(x^2\) will be larger. If x is positive and less than 1, \(x^2\) is smaller. Long story short: according to (1), x must be negative or greater than 1. That isn’t enough information.

Statement (2) is also insufficient. x must be between –1 and 0, or greater than 1. If x is a negative fraction, \(x^3\) is a larger number: a negative fraction closer to zero. If x is a positive number greater than 1, \(x^3\) is greater still.

Taken together, it’s still not enough information to determine whether x is positive. x could be between –1 and 0, or it could be greater than 1. The correct choice is (E).


Merging topics.

Please refer to the discussion on previous 2 pages.
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Re: Is x > 0 ?  [#permalink]

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New post 04 Dec 2015, 04:39
AS7x wrote:
Engr2012 wrote:
gmatser1 wrote:
Is x > 0?

either x<0


what is the rationale for inverting the inequality sign?


What step in particular are you talking about ? I don't see any inversion of the inequality sign!
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Re: Is x > 0 ?  [#permalink]

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New post 05 Dec 2015, 13:03
1
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is x > 0 ?

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

If the range of the question covers that of the condition, the condition is sufficient
There is one variable (x) in the original condition and 2 equations are given by the conditions, so there is high chance (D) will be the answer
For condition 1, x<x^2, 0<x^2-x, 0<x(x-1)--> x<0, 1<x, and this is insufficient as this range is not included in the question,
For condition 2, x<x^3, 0<x^3-x, 0<x(x-1)(x+1) --> -1<x<0, 1<x is also not sufficient as this is also not included in the question
Looking at the conditions together, 1<x<0, 1<x is also insufficient, as the range of the question does not include that of the conditions
Hence, the answer becomes (E).

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Re: Is x > 0 ?  [#permalink]

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New post 21 Aug 2018, 06:13
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Re: Is x > 0 ? &nbs [#permalink] 21 Aug 2018, 06:13

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