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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]

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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]

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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 $$-1<x<1$$.

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Re: Modulus question from GMAT tests [#permalink]

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14 Dec 2009, 09:22
2
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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]

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16 Dec 2009, 20:57
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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 $$-1<x<1$$.

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]

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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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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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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]

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08 Jan 2013, 04:30
curtis0063 wrote:
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

Merging similar topics. Please refer to the solutions above.
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Re: Modulus question from GMAT tests [#permalink]

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10 Oct 2013, 06:48
Bunuel wrote:
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 $$-1<x<1$$.

How do you get that $$|x|<1$$?
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Re: Modulus question from GMAT tests [#permalink]

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10 Oct 2013, 07:15
waltiebikkiebal wrote:
Bunuel wrote:
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 $$-1<x<1$$.

How do you get that $$|x|<1$$?

$$\frac{x^2}{|x|}<1$$;

$$\frac{|x|*|x|}{|x|}<1$$;

$$|x|<1$$.

Does this make sense?
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Re: Modulus question from GMAT tests [#permalink]

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10 Oct 2013, 07:17
yes, thanks a lot.
It is clear now.
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Re: If x 0, is x^2 / |x| < 1? [#permalink]

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09 Jun 2015, 10:19
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Re: If x 0, is x^2 / |x| < 1? [#permalink]

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09 Jun 2015, 23:13
Hi All,

This question is loaded with Number Property rules. If you know the rules, then you can make relatively quick work of this question; you can also solve it by TESTing VALUES...

We're told that X ≠ 0. We're asked if (X^2) /(|X|) < 1. This is a YES/NO question.

Before dealing with the two Facts, I want to point out a couple of Number Properties in the question stem:

1) X^2 will either be 0 or positive.
2) |X| will either be 0 or positive.
3) For the fraction in the question to be LESS than 1, X^2 must be LESS than |X|.

Fact 1: X < 1

IF...
X = 1/2
(1/4)/|1/2| = 1/2 and the answer to the question is YES

IF...
X = -1
(1)/|-1| = 1 and the answer to the question is NO
Fact 1 is INSUFFICIENT

Fact 2: X > −1

IF...
X = 1/2
(1/4)/|1/2| = 1/2 and the answer to the question is YES

IF...
X = 1
(1)/|1| = 1 and the answer to the question is NO
Fact 2 is INSUFFICIENT

Combined, we know...
-1 < X < 1

Since X cannot equal 0, X must be a FRACTION (either negative or positive). In ALL cases, X^2 will be LESS than |X|, so the fraction will ALWAYS be less than 1 and the answer to the question is ALWAYS YES.
Combined, SUFFICIENT

[Reveal] Spoiler:
C

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Re: If x 0, is x^2 / |x| < 1?   [#permalink] 09 Jun 2015, 23:13
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