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1. if you set x= positive decimal you get the original value which is <1 now you can try x = negative integer(-5) which results in the positive version which is >1 so INSUFF 2. this is INSUFF since x could be a huge positive number which makes it >1 OR it could be a small decimal number which makes it <1

combining you see -1< X <1 which means X is a +/- decimal which also means it will be <1 so C
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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?

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

Consider this: \(\frac{x^2}{|x|}=\frac{|x|*|x|}{|x|}=|x|\). It's basically the same as if it were \(\frac{x^2}{x}\) --> we could reduce this fraction by \(x\) and we would get \(x\), and when \(x\) is positive, result is positive and when \(x\) is negative, result is negative. Now, \(\frac{x^2}{|x|}\) is the ratio of two positive values and the result can not be negative, so we can not get \(x\), we should get \(|x|\) to guarantee that the result is positive.

OR: \(x<0\)--> then \(|x|=-x\) --> \(\frac{x^2}{|x|}=\frac{x^2}{-x}=-x<1\) --> \(x>-1\);

\(x>0\)--> then \(|x|=x\) --> \(\frac{x^2}{|x|}=\frac{x^2}{x}=x<1\);

\(x^2/|x|\) reduces to |x||x|/|x| which reduces the qn to is |x| <1 ? This again reduces to -1< x <1. Only combining (1) and two answers this qn. Hence answer is (C).

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.
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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.

as you can see the question has now been corrected

originally udaymathapati had changed x^2 to |x| and posted the question

where is my kudo for pointing this out ?
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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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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?

Last edited by iliavko on 26 Mar 2016, 14:13, edited 1 time in total.

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?

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.

Sorry, but something is still not 100% clear to me, so the meaning of "undefined" is not like "can't possibly be done, so don't even consider it" but more like a valid answer? Isn't undefined a sort of a dead end? Or is it just "correct" to write X#0 with no particular practical reason?

Sorry, but something is still not 100% clear to me, so the meaning of "undefined" is not like "can't possibly be done, so don't even consider it" but more like a valid answer? Isn't undefined a sort of a dead end? Or is it just "correct" to write X#0 with no particular practical reason?

The question asks: is x^2/|x| < 1 if -1 < x < 1. Now, if x is any number but 0 from -1 to 1, then the answer is YES. But if x is 0, then we cannot answer the question, because if x = 0 , then x^2/|x| is undefined.
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