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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 \(-1<x<1\).

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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GMAT is not a game for losers , and the moment u decide to appear for it u are no more a loser........ITS A BRAIN GAME

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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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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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 \(-1<x<1\).

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 \(-1<x<1\).

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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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

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