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# Is x > 0 ?

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Re: Is x > 0 ? [#permalink]

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15 Jan 2013, 06:22
rxs0005 wrote:
Is x > 0 ?
(1) x < x^2
(2) x < x^3

1. x < x^2
Test x = 2: 2 < 4 OK!
Test x = -1: -1 < 1 OK!
Test x = -1/4: -1/4 < -1/16 OK!
INSUFFICIENT!

2. x < x^3
Test x = 2: 2 < 8 OK!
Test x = -1: -1 = -1 NOT OK!
Test x = -1/4: -1/4 < -1/64 OK!
INSUFFICIENT!

Together: When x is negative fraction or x is greater than 2, both inequalities work!

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Re: Is x > 0 ? [#permalink]

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08 Jul 2014, 00:22
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12 Nov 2015, 00:45
Hello from the GMAT Club BumpBot!

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03 Dec 2015, 18:08
Is x > 0?

(1) $$x < x^2$$

(2) $$x < x^3$$

[Reveal] Spoiler:
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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03 Dec 2015, 18:17
gmatser1 wrote:
Is x > 0?

(1) $$x < x^2$$

(2) $$x < x^3$$

[Reveal] Spoiler:
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).

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.

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Re: Is x > 0? [#permalink]

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03 Dec 2015, 21:37
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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03 Dec 2015, 22:43
gmatser1 wrote:
Is x > 0?

(1) $$x < x^2$$

(2) $$x < x^3$$

[Reveal] Spoiler:
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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04 Dec 2015, 03:39
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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21 Dec 2016, 06:43
Hello from the GMAT Club BumpBot!

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Re: Is x > 0 ?   [#permalink] 21 Dec 2016, 06:43

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