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# If x 0, is x^2/|x| < 1?

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

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10 Oct 2013, 07:15
1
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: If x 0, is x^2/|x| < 1?  [#permalink]

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28 Feb 2014, 17:58
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.
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Re: If x 0, is x^2/|x| < 1?  [#permalink]

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01 Mar 2014, 03:46
2
prsnt11 wrote:
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.
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Re: If x 0, is x^2/|x| < 1?  [#permalink]

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01 Mar 2014, 12:21
1
Oh yes...so silly of me! If \sqrt{(x^2)} = |x|, x^2 = |x|*|x|
Thanks a lot Bunuel for your prompt help!:)
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Re: If x 0, is x^2/|x| < 1?  [#permalink]

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01 Mar 2014, 12:49
1
msand wrote:
If x ≠ 0, is x^2/|x| < 1?

(1) x < 1
(2) x > -1

Another way i would look at this is : x^2 is always positive, |x| is always positive too .. so a square value divided by the same number would only be less than 1 if the number was a proper fraction instead of an integer .. so the question is asking is x between -1 and 1 or |x| < 1 ?
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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

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

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13 Mar 2016, 06:05
Here from the question statement
x^2/|x| >1 => the question is actually asking us that does x lies in the (-1,1) range
as you can well see that the combination statement works.
Hence C
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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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Re: If x ≠ 0, is x^2/|x| < 1?  [#permalink]

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Updated on: 26 Mar 2016, 15:13

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?

Originally posted by iliavko on 26 Mar 2016, 14:57.
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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Re: 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]

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21 Jun 2017, 14:19
Is x^2/|x| < 1?

Which means - is lxl < 1?

or -1 < x < 1?

(1) x < 1 ======> Not Sufficient
(2) x > -1 ======> Not Sufficient

Combining we get: -1 < x < 1

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If x 0, is x^2/|x| < 1?  [#permalink]

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17 Sep 2018, 10:28
Hi,

here are my two cents for this question.

This looks long but actually i did solve in 45 secs, but took another 15 secs because of the range that combined statement 1 and 2 gave.

If you are not comfortable by reducing the inequality , though Bunuel has superbly explained it we can also think in terms of this approach.

we are given that x$$\neq{0}$$

IS $$\frac{x^2}{|x|} <1$$ ?

lets try to understand what we are being asked by trying to rephrase the question
we can write this as
$$\frac{x^2}{|x|} -1 <0$$
now
$$\frac{x^2-|x|} {|x|}<0$$

Since |x| is positive we can multiply both sides by |x| to get rid of it in denominator
$${x^2-|x|}$$ <0 ------(a)

now for |x|= x if x $$\geq{0}$$
|x|= - x if x <0
so above equation (a) for x$$\geq{0}$$ becomes
$${x^2-x}$$ <0

if we plot it on number line we will have
0<x<1

and for x<0 the equation (a) becomes
$${x^2 +x}$$ <0

if we plot it on number line we will have
-1<x<0

combining we get that inequality hold if
-1<x<1 except for x$$\neq{0}$$

So question is actually asking this does x lie in this range for this inequality

Now if we see statement 1 and 2 we can say they are clearly insufficient . Combining 1 and 2 we get
-1<x<1

But since we are given that x$$\neq{0}$$ so in this range ) is not considered as apart of the values of x for which the inequality holds.

PS : There can be several ways to solve the question, but if you follow one process for all questions there are chances you master that technique and solve quick enough to mark the answer confidently. No point in knowing multiple techniques and mastering none.

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If x 0, is x^2/|x| < 1?   [#permalink] 17 Sep 2018, 10:28

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