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

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 \(x<0\).

If \(x\neq{0}\), is \(\frac{x^2}{|x|}<1\)? --> reduce by \(|x|\) --> is \(|x|<1\)? or is \(-1<x<1\)?

Two statements together give us the sufficient info.

Answer: C.

Bunuel,
Can you explain how it reduce it to \(|x|\) from the expression?

Bunuel, did anybody tell you that you are a genius? Stay away from scientists, they might start researching on your brain

+1

I think\(\frac{|x|}{x} <1\)

(1) x < 1
(2) x > −1

Here the answer should be E

x= 1/2 satisfies both the statements and answer to the stem is no, 1 is not less 1

X= - 1/2 satisfies both the statements and answer to the stem is yes , -1<1


but for question \(\frac{x^2}{x} <1\)

(1) x < 1
(2) x > −1


here the answer is C as shown above

Please do correct if I am missing something
thanks.

If it were:
If x#0, is |x|/x<1?

(1) x < 1
(2) x > −1

Then the answer is E. The question basically asks whether x is negative and we cannot answer that even when we combine the statements given.

If it were:
If x#0, is x^2/x<1?

(1) x < 1
(2) x > −1

Then the answer is C. The question basically asks whether x<0 or 0<x<1. When we combine the statements, we get that -1<x<1 (x#0). So, the answer to the question is YES.

How do you get that \(|x|<1\)?

\(\frac{x^2}{|x|}<1\);

\(\frac{|x|*|x|}{|x|}<1\);

\(|x|<1\).

Does this make sense?

Hey Bunuel,
Can you please elaborate how you reduced the left-hand side to |x| < 1? I'm getting confused because of the absolute value sign.

\(\frac{x^2}{|x|}=\frac{|x|*|x|}{|x|}=|x|\).

Hope it helps.

Oh yes...so silly of me! If \sqrt{(x^2)} = |x|, x^2 = |x|*|x|
Thanks a lot Bunuel for your prompt help!:)

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

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Why can't we manipulate the question to get is x^2<|x| ? can't we pass the |x| to the right? It says in the stem that it's #0. Am i wrong here?

We can do this but not because |x| is not 0, but because |x| > 0 and we can multiply both sides by it.

thank you for the reply!

Does this mean that the notation x#0 in the question stem doesn't tell anything new? it's already known from the rules that the divisor can't be zero. So if you look at the stem wihtout the x#0 information, the question doesn't change.

Anyways, because whe know that divisor is not zero, we know that x<0 or x>0 and since it's an absolute value, it must be x>0. Is this correct?

Ps. does it mean that in a question where it is mentioned that X#0 but the denominator is not a module, we still must somewhow be sure of the sign of the denominator to be able to multiply like in the case of say, x^2\x so X in both, numerator and denomiator?

Division by 0 is not allowed so \(x \neq 0\) rules out this case. If we were not told that, then when considering the two statements together we were not be able to tell whether \(\frac{x^2}{|x|}<1\) because if x= 0 then \(\frac{x^2}{|x|}\) is undefined not less than 1.

Hope it's clear.