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If x is not equal to 0, is |x| less than 1? (1) x/|x|< x

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Bunuel wrote:
Hussain15 wrote:
If x is not equal to 0, is |x| less than 1?

(1) x/|x|< x

(2) |x| > x

Will really appreciate if answer is supported by explanation.


\(x\neq{0}\), is \(|x|<1\)? Which means is \(-1<x<1\)? (\(x\neq{0}\))

(1) \(\frac{x}{|x|}< x\)
Two cases:
A. \(x<0\) --> \(\frac{x}{-x}<x\) --> \(-1<x\). But remember that \(x<0\), so \(-1<x<0\)
.


if x<0, doesn't x/|x|<x become -x/x<-x which simplifies to -1<-x or x<1 ?
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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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New post 24 Apr 2013, 18:25
Bunuel wrote:
Hussain15 wrote:
If x is not equal to 0, is |x| less than 1?

(1) x/|x|< x

(2) |x| > x

Will really appreciate if answer is supported by explanation.


\(x\neq{0}\), is \(|x|<1\)? Which means is \(-1<x<1\)? (\(x\neq{0}\))

(1) \(\frac{x}{|x|}< x\)
Two cases:
A. \(x<0\) --> \(\frac{x}{-x}<x\) --> \(-1<x\). But remember that \(x<0\), so \(-1<x<0\)

B. \(x>0\) --> \(\frac{x}{x}<x\) --> \(1<x\).

Two ranges \(-1<x<0\) or \(x>1\). Which says that \(x\) either in the first range or in the second. Not sufficient to answer whether \(-1<x<1\). (For instance \(x\) can be \(-0.5\) or \(3\))

Second approach: look at the fraction \(\frac{x}{|x|}\) it can take only two values:
1 for \(x>0\) --> so we would have: \(1<x\);
Or -1 for \(x<0\) --> so we would have: \(-1<x\) and as we considering the range for which \(x<0\) then completer range would be: \(-1<x<0\).

The same two ranges: \(-1<x<0\) or \(x>1\).

(2) \(|x| > x\). Well this basically tells that \(x\) is negative, as if x were positive or zero then \(|x|\) would be equal to \(x\). Only one range: \(x<0\), but still insufficient to say whether \(-1<x<1\). (For instance \(x\) can be \(-0.5\) or \(-10\))

Or two cases again:
\(x<0\)--> \(-x>x\)--> \(x<0\).
\(x>0\) --> \(x>x\): never correct.


(1)+(2) Intersection of the ranges from (1) and (2) is the range \(-1<x<0\) (\(x<0\) (from 2) and \(-1<x<0\) or \(x>1\) (from 1), hence \(-1<x<0\)). Every \(x\) from this range is definitely in the range \(-1<x<1\). Sufficient.

Answer: C.


Buenel terrific explanation. This system, that you use for solving inequations, is quite fundamental and hard to be wrong. But is it really always the quickest way??
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oyabu wrote:
Bunuel wrote:
Hussain15 wrote:
If x is not equal to 0, is |x| less than 1?

(1) x/|x|< x

(2) |x| > x

Will really appreciate if answer is supported by explanation.


\(x\neq{0}\), is \(|x|<1\)? Which means is \(-1<x<1\)? (\(x\neq{0}\))

(1) \(\frac{x}{|x|}< x\)
Two cases:
A. \(x<0\) --> \(\frac{x}{-x}<x\) --> \(-1<x\). But remember that \(x<0\), so \(-1<x<0\)
.


if x<0, doesn't x/|x|<x become -x/x<-x which simplifies to -1<-x or x<1 ?


When \(x<0\), then \(|x|=-x\), thus \(\frac{x}{|x|}<x\) becomes \(\frac{x}{-x}<x\) --> \(-1<x\) but since \(x<0\), then \(-1<x<0\).

Hope it's clear.
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New post 25 Apr 2013, 09:17
Bunuel wrote:

When \(x<0\), then \(|x|=-x\), thus \(\frac{x}{|x|}<x\) becomes \(\frac{x}{-x}<x\) --> \(-1<x\) but since \(x<0\), then \(-1<x<0\).

Hope it's clear.


Thanks. One more follow-up for clarification - Isn't the absolute value of a negative value positive? So if x<0 (x is a negative number), then the absolute value of x should be positive? i.e. |-x|=x?
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oyabu wrote:
Bunuel wrote:

When \(x<0\), then \(|x|=-x\), thus \(\frac{x}{|x|}<x\) becomes \(\frac{x}{-x}<x\) --> \(-1<x\) but since \(x<0\), then \(-1<x<0\).

Hope it's clear.


Thanks. One more follow-up for clarification - Isn't the absolute value of a negative value positive? So if x<0 (x is a negative number), then the absolute value of x should be positive? i.e. |-x|=x?


First of all \(|-x|=|x|\). Next, if \(x<0\), then \(|x|=|negative|=-x=-negative=positive\).

It seems that you need to go through basics: math-absolute-value-modulus-86462.html
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New post 23 Jun 2013, 01:36
One question:

if you pug in numbers, for the first statement is the statement not only true for number between -1 and 0? I just do not get it completely!
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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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New post 11 Jul 2013, 14:43
If x is not equal to 0, is |x| less than 1?

|x|<1
Is -1<x<1

(1) x/|x|< x

x/|x|< x
Two cases: x>0, x<0

x>0
x/|x|< x
x/x<x
1<x

If x>0 and x>1 then x>1
If x>1 then |x| is NOT less than 1

x<0
x/|x|< x
-x/|-x|<x
-1<x

-1<x<0
if -1<x<0 then |x| IS less than 1
INSUFFICIENT

(2) |x| > x

If the absolute value of x is greater than x than x must be negative. However, |-x| could be less than 1 or greater than 1 depending on the value of x.
INSUFFICIENT

1+2 x must be negative and x from #1 x is either >1 or between -1 and 0. Therefore, we know -1<x<0 which means that |x| is always less than 1.
SUFFICIENT

(C)
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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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New post 06 Aug 2013, 08:42
Bunuel wrote:
Hussain15 wrote:
If x is not equal to 0, is |x| less than 1?

(1) x/|x|< x

(2) |x| > x

Will really appreciate if answer is supported by explanation.


\(x\neq{0}\), is \(|x|<1\)? Which means is \(-1<x<1\)? (\(x\neq{0}\))

(1) \(\frac{x}{|x|}< x\)
Two cases:
A. \(x<0\) --> \(\frac{x}{-x}<x\) --> \(-1<x\). But remember that \(x<0\), so \(-1<x<0\)

B. \(x>0\) --> \(\frac{x}{x}<x\) --> \(1<x\).

Two ranges \(-1<x<0\) or \(x>1\). Which says that \(x\) either in the first range or in the second. Not sufficient to answer whether \(-1<x<1\). (For instance \(x\) can be \(-0.5\) or \(3\))

Second approach: look at the fraction \(\frac{x}{|x|}\) it can take only two values:
1 for \(x>0\) --> so we would have: \(1<x\);
Or -1 for \(x<0\) --> so we would have: \(-1<x\) and as we considering the range for which \(x<0\) then completer range would be: \(-1<x<0\).

The same two ranges: \(-1<x<0\) or \(x>1\).

(2) \(|x| > x\). Well this basically tells that \(x\) is negative, as if x were positive or zero then \(|x|\) would be equal to \(x\). Only one range: \(x<0\), but still insufficient to say whether \(-1<x<1\). (For instance \(x\) can be \(-0.5\) or \(-10\))

Or two cases again:
\(x<0\)--> \(-x>x\)--> \(x<0\).
\(x>0\) --> \(x>x\): never correct.


(1)+(2) Intersection of the ranges from (1) and (2) is the range \(-1<x<0\) (\(x<0\) (from 2) and \(-1<x<0\) or \(x>1\) (from 1), hence \(-1<x<0\)). Every \(x\) from this range is definitely in the range \(-1<x<1\). Sufficient.

Answer: C.

Hi bunel,

hw is that we can have two ranges x<0 & x>0. How we got this ranges?

regards,
rrsnathan.
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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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New post 11 Mar 2014, 04:27
Hussain15 wrote:
If x is not equal to 0, is |x| less than 1?

(1) x/|x|< x

(2) |x| > x

Will really appreciate if answer is supported by explanation.


|x| < 1
if we plot this then we get the range of x as (-1,1)

1) \(\frac{x}{|x|} < x\)
the range of x here is (1, infinity) U (-1,0)
hence, it is not sufficient.

2) |x| > x
very clearly this would be true for all x < 0

if we combine 1 and 2, the common range we get is (-1,0)
which satisfies |x| < 1
hence, C.
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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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New post 29 Sep 2014, 07:08
The question asks "is \(-1<x<1\)?" We got that \(x\) is in the range \(-1<x<0\) (red area). Now, as ANY \(x\) from this range (from red area) is indeed in \(-1<x<1\), then we can answer YES to our original question.

Hope it's clear.[/quote]

Exactly, based on your explanation above, Though the answer is YES it is sufficient to answer but the answer to the question is

That for x=0.5 it is in the range -1<x<1
but it is NOT part of -1<x<0 which we found out.
So it is not possible to say that all possible values of -1<x<0 come under -1<x<1 unless ofcourse it is specified that x is an integer.

Bunuel,
Please Clarify.
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New post 29 Sep 2014, 07:15
earnit wrote:
Exactly, based on your explanation above, Though the answer is YES it is sufficient to answer but the answer to the question is the range -1<x<1 for x is NO as -1<x<0 is the range and so x could be 0.5 and not be part of the -1<x<0 .

Bunuel,
Please Clarify.


When we combine the statements we get that -1 < x < 0, then HOW can x be 0.5???
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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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New post 29 Sep 2014, 09:52
Bunuel wrote:
earnit wrote:
Exactly, based on your explanation above, Though the answer is YES it is sufficient to answer but the answer to the question is the range -1<x<1 for x is NO as -1<x<0 is the range and so x could be 0.5 and not be part of the -1<x<0 .

Bunuel,
Please Clarify.


When we combine the statements we get that -1 < x < 0, then HOW can x be 0.5???


I'm sorry if it comes across like that but as per my understanding, the Question is if x lies between -1 and 1 and we find out that x lies between -1 and 0 so we are not entirely going by the asked range -1 < x < 1 is what i think.

But yes, the range of -1 < x < 0 comes under the broad -1 < x < 1 and so that is why it is correct.
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New post 20 Oct 2014, 21:51
If x is not equal to 0, is |x| less than 1?

(1) x/|x|< x

(2) |x| > x

Will really appreciate if answer is supported by explanation.

1 - x / |x| < x This would mean that -1 < x < 0 & 0 < x <infinite
Thus 1 is insufficient

2 - |x| > x - This would mean that negative infinite values < x < 0 This is also not sufficient.

Combining 1 & 2 - the overlapping region is -1 < x < 0

We can conclude that |x| < 1

So C
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New post 21 Oct 2014, 15:46
Thoughtosphere wrote:
If x is not equal to 0, is |x| less than 1?

(1) x/|x|< x

(2) |x| > x

Will really appreciate if answer is supported by explanation.

1 - x / |x| < x This would mean that -1 < x < 0 & 0 < x <infinite
Thus 1 is insufficient

2 - |x| > x - This would mean that negative infinite values < x < 0 This is also not sufficient.

Combining 1 & 2 - the overlapping region is -1 < x < 0

We can conclude that |x| < 1

So C


Solving the Question is not much of a problem as deciding whether our solved range: \(-1 < x < 0\) is SUFFICIENT to answer the asked range: \(-1 < x < 1\)

I need to understand that: Is our answer sufficient to conclude that YES x lies between (-1,1) even though our solution was that x lies between (-1,0) ?
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earnit wrote:
Thoughtosphere wrote:
If x is not equal to 0, is |x| less than 1?

(1) x/|x|< x

(2) |x| > x

Will really appreciate if answer is supported by explanation.

1 - x / |x| < x This would mean that -1 < x < 0 & 0 < x <infinite
Thus 1 is insufficient

2 - |x| > x - This would mean that negative infinite values < x < 0 This is also not sufficient.

Combining 1 & 2 - the overlapping region is -1 < x < 0

We can conclude that |x| < 1

So C


Solving the Question is not much of a problem as deciding whether our solved range: \(-1 < x < 0\) is SUFFICIENT to answer the asked range: \(-1 < x < 1\)

I need to understand that: Is our answer sufficient to conclude that YES x lies between (-1,1) even though our solution was that x lies between (-1,0) ?


The question asks whether -1 < x< 1. We got that -1 < x < 0. So, the answer is YES: x is definitely from the range (-1, 1).
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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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New post 24 Nov 2014, 11:26
The question is asking if
-1 < x < 1

Statement 1) x/|x| < x Now if x + ve, x/|x| = 1
and if x is -ve
x / |x| = -1
So essentially, the equation becomes +1 < x or -1 < -x or 1 > x (multiplying both sides by -ve)
In two cases, one case says x > 1 and the other one says x < 1, hence Not sufficient.

Statement 2. |x| > x , this will only happen when x is -ve
id x is -ve |x| can be > 1 or |x| can be < 1 or equal to 1. - Not sufficient.

Combining the two statements:
x is -ve
1 > x hence sufficient. C)
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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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New post 26 Nov 2014, 14:05
Bunuel wrote:
Hussain15 wrote:
If x is not equal to 0, is |x| less than 1?

(1) x/|x|< x

(2) |x| > x

Will really appreciate if answer is supported by explanation.


\(x\neq{0}\), is \(|x|<1\)? Which means is \(-1<x<1\)? (\(x\neq{0}\))

(1) \(\frac{x}{|x|}< x\)
Two cases:
A. \(x<0\) --> \(\frac{x}{-x}<x\) --> \(-1<x\). But remember that \(x<0\), so \(-1<x<0\)

B. \(x>0\) --> \(\frac{x}{x}<x\) --> \(1<x\).

Two ranges \(-1<x<0\) or \(x>1\). Which says that \(x\) either in the first range or in the second. Not sufficient to answer whether \(-1<x<1\). (For instance \(x\) can be \(-0.5\) or \(3\))

Second approach: look at the fraction \(\frac{x}{|x|}\) it can take only two values:
1 for \(x>0\) --> so we would have: \(1<x\);
Or -1 for \(x<0\) --> so we would have: \(-1<x\) and as we considering the range for which \(x<0\) then completer range would be: \(-1<x<0\).

The same two ranges: \(-1<x<0\) or \(x>1\).

(2) \(|x| > x\). Well this basically tells that \(x\) is negative, as if x were positive or zero then \(|x|\) would be equal to \(x\). Only one range: \(x<0\), but still insufficient to say whether \(-1<x<1\). (For instance \(x\) can be \(-0.5\) or \(-10\))

Or two cases again:
\(x<0\)--> \(-x>x\)--> \(x<0\).
\(x>0\) --> \(x>x\): never correct.


(1)+(2) Intersection of the ranges from (1) and (2) is the range \(-1<x<0\) (\(x<0\) (from 2) and \(-1<x<0\) or \(x>1\) (from 1), hence \(-1<x<0\)). Every \(x\) from this range is definitely in the range \(-1<x<1\). Sufficient.

Answer: C.



Bunuel
I'm very unclear about this. if x<0, then -x/|x| <-x (since x is negative in our scenario 1)...so where do I go from here?
if i cut both -x's, I get 1/|x| > 1....so x>1....I can't understand how we got x>-1...
If you could help, that'll be great. I tried plugging in for this question and got in right, but I've been trying to teach myself your conceptual way as I believe it is much better than plugging in.
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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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New post 14 Dec 2014, 12:04
i want some more questions of this type, Bunuel sir can u please help me.
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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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New post 15 Dec 2014, 08:04
Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x   [#permalink] 15 Dec 2014, 08:04

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