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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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08 Sep 2010, 11:51
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If $$x \neq 0$$, is $$\frac{x^2}{|x|}< 1$$?

(1) x < 1
(2) x > −1
[Reveal] Spoiler: OA
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If x#0, is x^2/|x| < 1? [#permalink]

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08 Sep 2010, 11:57
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udaymathapati wrote:
If $$x \neq 0$$, is $$\frac{x^2}{|x|}< 1$$?

(1) x < 1
(2) x > −1

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.

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Re: If x#0, is x^2/|x| < 1? [#permalink]

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08 Sep 2010, 13:11
2
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udaymathapati wrote:
[img]
Attachment:
Maths1.JPG
[/img]

(1) x < 1
(2) x > −1

x^2/abs(x) <1 ?

Another way to look at it...

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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Re: If x#0, is x^2/|x| < 1? [#permalink]

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09 Sep 2010, 04:38
Bunuel wrote:
udaymathapati wrote:
[img]
Attachment:
Maths1.JPG
[/img]

(1) x < 1
(2) x > −1

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.

Bunuel,
Can you explain how it reduce it to $$|x|$$ from the expression?
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Re: If x#0, is x^2/|x| < 1? [#permalink]

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09 Sep 2010, 10:12
1
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Expert's post
udaymathapati wrote:
Bunuel wrote:
udaymathapati wrote:
[img]
Attachment:
Maths1.JPG
[/img]

(1) x < 1
(2) x > −1

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.

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

So $$-1<x<1$$.
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Re: If x#0, is x^2/|x| < 1? [#permalink]

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14 Oct 2010, 14:44
tatane90 wrote:
If x ≠ 0, is x^2/|x| < 1?
(1) x < 1
(2) x > −1

$$\frac{x^2}{|x|} \lt 1$$
If x>0, then this implies x<1
If x<0, then this implies x>-1
So it is only true if either 0<x<1 or -1<x<0

(1) Not sufficient clealry, as x is not bounded on lower side
(2) Not sufficient clealrly, as x is not bounded on upper side

(1+2) Exactly defined the range for which the inequality holds. Sufficient

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Re: If x#0, is x^2/|x| < 1? [#permalink]

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16 Oct 2010, 04:49
Bunuel, did anybody tell you that you are a genius? Stay away from scientists, they might start researching on your brain

+1
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http://greatlakes.edu.in

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Re: If x#0, is x^2/|x| < 1? [#permalink]

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04 Jul 2013, 01:24
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on Abolute Values: math-absolute-value-modulus-86462.html

DS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=37
PS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=58

Hard set on Abolute Values: inequality-and-absolute-value-questions-from-my-collection-86939.html

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Re: If x#0, is x^2/|x| < 1? [#permalink]

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04 Jul 2013, 07:12
1
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$$x^2/|x|$$ 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).
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Re: If x#0, is x^2/|x| < 1? [#permalink]

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04 Jul 2013, 12:15
1
KUDOS
x^2/|x| < 1
(x^2 = |x|*|x|)
SO
(|x|*|x|)/|x| < 1
Is |x|<1?
is x<1 or is x>-1
is -1<x<1?

(1) x < 1
The issue here is depending on what x is, x may not be in the range of -1<x<1.
INSUFFICIENT

(2) x > −1
The same problem that applied to a) applies to b).
INSUFFICIENT

a+b) this gives us a range of -1<x<1 which is what the question is looking for.
SUFFICIENT

(C)
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Re: If x#0, is x^2/|x| < 1? [#permalink]

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24 Sep 2013, 01:22
1
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udaymathapati wrote:
If x#0, is |x|/x<1?

(1) x < 1
(2) x > −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.
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Re: If x#0, is x^2/|x| < 1? [#permalink]

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24 Sep 2013, 01:38
stne wrote:
udaymathapati wrote:
If x#0, is |x|/x<1?

(1) x < 1
(2) x > −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.
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Re: If x#0, is x^2/|x| < 1? [#permalink]

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24 Sep 2013, 02:35
Bunuel wrote:
stne wrote:
udaymathapati wrote:
If x#0, is |x|/x<1?

(1) x < 1
(2) x > −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.

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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Re: If x#0, is x^2/|x| < 1? [#permalink]

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09 Sep 2015, 07:14
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Re: If x#0, is x^2/|x| < 1? [#permalink]

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26 Mar 2016, 11:32
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?
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Re: If x#0, is x^2/|x| < 1? [#permalink]

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26 Mar 2016, 11:39
iliavko wrote:
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.
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If x#0, is x^2/|x| < 1? [#permalink]

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26 Mar 2016, 14:57

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, 15:13, edited 1 time in total.
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Re: If x#0, is x^2/|x| < 1? [#permalink]

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26 Mar 2016, 15:06
iliavko wrote:

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.

Hope it's clear.
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If x#0, is x^2/|x| < 1? [#permalink]

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26 Mar 2016, 15:24
Thank you for the replies!

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?
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Re: If x#0, is x^2/|x| < 1? [#permalink]

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26 Mar 2016, 15:39
iliavko wrote:
Thank you for the replies!

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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Re: If x#0, is x^2/|x| < 1?   [#permalink] 26 Mar 2016, 15:39
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