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If x 0, is x^2 / |x| < 1?

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If x 0, is x^2 / |x| < 1? [#permalink] New post 14 Dec 2009, 01:03
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If x ≠ 0, is x^2 / |x| < 1?

(1) x < 1
(2) x > −1
[Reveal] Spoiler: OA
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Re: Modulus question from GMAT tests [#permalink] New post 14 Dec 2009, 01:47
zaarathelab wrote:
If x ≠ 0, is x^2 / |x| < 1?
(1) x < 1
(2) x > −1


We can safely reduce LHS of inequality \frac{x^2}{|x|}<1 by |x|. We'll get: is |x|<1? Basically question asks whether x is in the range -1<x<1.

Statements 1 and 2 are not sufficient, but together they are defining the range for x as -1<x<1.

Answer: C.
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Re: Modulus question from GMAT tests [#permalink] New post 14 Dec 2009, 09:22
IMO c

the expression x^2/lxl<1 will always be +ve as the N is a sqaure and mod x is always +ve

the expression is true only for fractional value

s1) tells us that x can be a +ve or -ve fractional and also can be a -int hence insuff.
s2) tells us that x can be a +ve or -ve fractional and also can be a +int hence insuff....

combinig x is only a fraction and hence suff...
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Re: Modulus question from GMAT tests [#permalink] New post 16 Dec 2009, 20:57
Bunuel wrote:
zaarathelab wrote:
If x ≠ 0, is x^2 / |x| < 1?
(1) x < 1
(2) x > −1


We can safely reduce inequality \frac{x^2}{|x|}<1 by |x|. We'll get: is |x|<1? Basically question asks whether x is in the range -1<x<1.

Statements 1 and 2 are not sufficient, but together they are defining the range for x as -1<x<1.

Answer: C.


Bunuel - please clarify this divison -\frac{x^2}{|x|}<1 by |x|. We'll get: is |x|<1? can be done because |x| is always +ve or is there are any other reason ?
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Re: Modulus question from GMAT tests [#permalink] New post 17 Dec 2009, 04:04
ichha148 wrote:
Bunuel - please clarify this divison -\frac{x^2}{|x|}<1 by |x|. We'll get: is |x|<1? can be done because |x| is always +ve or is there are any other reason ?


We should never multiply (or reduce) inequality by variable (or expression with variable) if we don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero.

But in this case we are not reducing the inequality we are reducing only one part of it. So, it's safe to do so.

For example if we had: \frac{x^4}{x^3}<0 we can reduce LHS by x^3 and write x<0.
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Re: Mod [#permalink] New post 29 Dec 2009, 06:10
If x ≠ 0, is x^2/|x|< 1?

1: x<1 - Insufficient. x can be -2 or 1/2
2:x>-1 - Insufficient. x can be 2 or 1/2

Together, sufficient. x must be a fraction, and thus the answer is C.
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If x≠0, is x^2/abs(x)<1? [#permalink] New post 08 Jan 2013, 04:12
If x ≠ 0, is x^2/abs(x)<1?
(1) x < 1
(2) x > −1

Question is came from GWD. My approach is time comsuming. So suggest me a quicker approch. Thanks
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Re: If x≠0, is x^2/abs(x)<1? [#permalink] New post 08 Jan 2013, 04:30
Re: If x≠0, is x^2/abs(x)<1?   [#permalink] 08 Jan 2013, 04:30
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