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munchkin123
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If x is not equal to 0, is |x| less than 1? (1) x/|x|< x (2) |x| > x 29 Mar 2018, 06:33
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I did it this way.

Is |x| < 1?

Statement 1

x/|x| < x

For, x>1
Divide by x on both sides, without changing the sign.
1/|x| < 1
|x| > 1

For x<1
Divide by x on both sides, and change the sign.
1/|x| > 1
|x| < 1

Insufficient, as we have both |x| > 1 (x is positive) and |x| < 1 (x is negative)

Statement 2

|x| > x
This is possible only when x < 0. (x is negative)

This is still insufficient to prove the given question.

Taking them both,

From (2), we get (x is negative) and from (1), the corresponding answer is |x| < 1.

Hence, C.


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

(1) x/|x|< x

(2) |x| > x

PLEASE READ THE WHOLE DISCUSSION
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If x is not equal to 0, is |x| less than 1? (1) x/|x|< x (2) |x| > x 02 May 2018, 07:19
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Bunuel
If x is not equal to 0, is |x| less than 1?

Is \(|x|<1\)?
Is \(-1<x<1\)? (\(x\neq{0}\))
So, the question asks whether x is in the range shown below:



(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 consider 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 \(-1<x<1\). So, we have a definite YES answer to the question. Sufficient.


Answer: C.


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Hi Bunuel

Surely i am missing something here..my basic doubt is in scenario when we take X<0

If we assume, x<0, then shouldn't the statement x/|x|<x = -x/-x <-x (my thinking here is if we considering x<0, then x should be negative throughout )
So, the equation becomes 1<-x = -1>x, we can keep this option as we have initially assumed X<0
Then if we take any values of x less then -1, |x| will always be greater than 1

if we assume X>0, then x/|x|<x = x/x<x.....which is 1<x, we can keep this option as initially we have assumed X>0
Then we take any values of x greater than 1 then |x| will always be greater than 1.

So statement 1 in either case is sufficient..

Thanks in advance for your help..
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If x is not equal to 0, is |x| less than 1? (1) x/|x|< x (2) |x| > x 02 May 2018, 09:24
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Bunuel
If x is not equal to 0, is |x| less than 1?

Is \(|x|<1\)?
Is \(-1<x<1\)? (\(x\neq{0}\))
So, the question asks whether x is in the range shown below:



(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 consider 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 \(-1<x<1\). So, we have a definite YES answer to the question. Sufficient.


Answer: C.


Show Spoiler
Attachment:
MSP340910303aagbh2c14gf0000420i437if9d12bhg.gif
Attachment:
MSP2380151735322d2i2d790000280i6giebh17fdf3.gif
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MSP8506173e0f30207c4a2700005eg6g59c48d51i5b.gif
Attachment:
MSP111401047g43cgaehag920000640367fgf148e668.gif

Hi Bunuel

Surely i am missing something here..my basic doubt is in scenario when we take X<0

If we assume, x<0, then shouldn't the statement x/|x|<x = -x/-x <-x (my thinking here is if we considering x<0, then x should be negative throughout )
So, the equation becomes 1<-x = -1>x, we can keep this option as we have initially assumed X<0
Then if we take any values of x less then -1, |x| will always be greater than 1

if we assume X>0, then x/|x|<x = x/x<x.....which is 1<x, we can keep this option as initially we have assumed X>0
Then we take any values of x greater than 1 then |x| will always be greater than 1.

So statement 1 in either case is sufficient..

Thanks in advance for your help..
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Negative x does not mean that you should replace x with -x. x just represents a negative number. You should replace |x| with -x, though because if x < 0, then |x| = -x.
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If x is not equal to 0, is |x| less than 1? (1) x/|x|< x (2) |x| > x 03 May 2018, 08:37
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Bunuel
If x is not equal to 0, is |x| less than 1?

Is \(|x|<1\)?
Is \(-1<x<1\)? (\(x\neq{0}\))
So, the question asks whether x is in the range shown below:



(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 consider 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 \(-1<x<1\). So, we have a definite YES answer to the question. Sufficient.


Answer: C.


Show Spoiler
Attachment:
MSP340910303aagbh2c14gf0000420i437if9d12bhg.gif
Attachment:
MSP2380151735322d2i2d790000280i6giebh17fdf3.gif
Attachment:
MSP8506173e0f30207c4a2700005eg6g59c48d51i5b.gif
Attachment:
MSP111401047g43cgaehag920000640367fgf148e668.gif

Hi Bunuel

Surely i am missing something here..my basic doubt is in scenario when we take X<0

If we assume, x<0, then shouldn't the statement x/|x|<x = -x/-x <-x (my thinking here is if we considering x<0, then x should be negative throughout )
So, the equation becomes 1<-x = -1>x, we can keep this option as we have initially assumed X<0
Then if we take any values of x less then -1, |x| will always be greater than 1

if we assume X>0, then x/|x|<x = x/x<x.....which is 1<x, we can keep this option as initially we have assumed X>0
Then we take any values of x greater than 1 then |x| will always be greater than 1.

So statement 1 in either case is sufficient..

Thanks in advance for your help..

Negative x does not mean that you should replace x with -x. x just represents a negative number. You should replace |x| with -x, though because if x < 0, then |x| = -x.
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Hi Bunuel

Thanks for your reply. But i am still not able to get why we shouldn't consider negative x's throughout the equation.
If we assuming |x|= -x, we are assuming x<0, which means we are assuming the variable "X" as negative . So, we consider it negative only for Mode but keep it positive in other parts of the equation it seems inconsistent

Thanks again..
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If x is not equal to 0, is |x| less than 1? (1) x/|x|< x (2) |x| > x 03 May 2018, 09:53
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Hi Bunuel

Thanks for your reply. But i am still not able to get why we shouldn't consider negative x's throughout the equation.
If we assuming |x|= -x, we are assuming x<0, which means we are assuming the variable "X" as negative . So, we consider it negative only for Mode but keep it positive in other parts of the equation it seems inconsistent

Thanks again..
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Consider simple example: x = -1. So, we know that x is negative. Do you replace x there by -x and write -x = -1?
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If x is not equal to 0, is |x| less than 1? (1) x/|x|< x (2) |x| > x 10 May 2020, 10:33
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If x is not equal to 0, is |x| less than 1?

(1) x/|x|< x
(2) |x| > x

_________________________________________

from the statement 1,
--> x < x|x|
--> 0 < x|x| - x
--> 0 < x(|x| - 1)
so, it is either
x < 0 and |x| < 1
or
x > 0 and |x| > 1
we do not know which is correct.
Not sufficient.

from the statement (2),
--> |x| > x
we know that x < 0. but this is clearly not sufficient.

(1) + (2)
from (2), we know that x <0. therefore we know that |x| < 1
sufficient.

the answer is C
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If x is not equal to 0, is |x| less than 1? (1) x/|x|< x (2) |x| > x 10 Nov 2022, 06:27
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Given, X is not equal to 0
Mod of x can either by positive or 0. Since the given info states that x is not equal to 0,
mod of x can only be positive.

(1): x /|x|<x

x /|x|<x
x <|x|*x
x *|x|- x>0
x (|x|-1)>0

if x is positive, then (|x|-1) is positive for the whole expression to be greater than 0.
i.e |x|>1
Back to the question, is |x|<1? NO!

if x is negative, then (|x|-1) is negative for the whole expression to be greater than 0.
i.e |x|<1
Back to the question, is |x|<1? YES!
No definite answers.
So options A & D are out.

(2)|x| > x
since |x| is always positive in this case, x has to be negative for |x| > x to be true.
But this does not answer the question |x|<1
So option B is out.

We are left with options C and E.

C) From both statements, we establish that when x is negative, |x|<1.

From (1)if x is negative, |x|<1 is true.
(2)x has to be negative for |x| > x to be true.

Therefore, X is negative and hence |x|<1 is true.

C is the answer.
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If x is not equal to 0, is |x| less than 1? (1) x/|x|< x (2) |x| > x 27 Jul 2023, 15:30
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Bunuel
If x is not equal to 0, is |x| less than 1?

Is \(|x|<1\)?
Is \(-1<x<1\)? (\(x\neq{0}\))
So, the question asks whether x is in the range shown below:



(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 consider 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 \(-1<x<1\). So, we have a definite YES answer to the question. Sufficient.


Answer: C.


Show Spoiler
Attachment:
MSP340910303aagbh2c14gf0000420i437if9d12bhg.gif
Attachment:
MSP2380151735322d2i2d790000280i6giebh17fdf3.gif
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MSP8506173e0f30207c4a2700005eg6g59c48d51i5b.gif
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MSP111401047g43cgaehag920000640367fgf148e668.gif
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Hi Bunuel,

Really thank you for all your support.

I have a small confusion which is really troubling me.

How do we open the mode when if suppose that x<0. Do we just add a -ve on the RHS or do we also change the sign?

For say in this question x/|x| < x
if x < 0,Then, X/-X > X ----> -1>X ?

I see that you have not changed the sign, but what I have understood till now is that we need to change the sign as well.

Like say |X-1| < 3 then if x<0 then it will be X-1> -3 ---> X> -2. And if we don't change the sign then it will be X<-2 which will be wrong if I plot the equation on the no. line.

Please help me on this. I know I'm getting confused with the concept.

Posted from my mobile device
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If x is not equal to 0, is |x| less than 1? (1) x/|x|< x (2) |x| > x 27 Jul 2023, 23:48
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Bunuel
If x is not equal to 0, is |x| less than 1?

Is \(|x|<1\)?
Is \(-1<x<1\)? (\(x\neq{0}\))
So, the question asks whether x is in the range shown below:



(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 consider 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 \(-1<x<1\). So, we have a definite YES answer to the question. Sufficient.


Answer: C.


Show Spoiler
Attachment:
MSP340910303aagbh2c14gf0000420i437if9d12bhg.gif
Attachment:
MSP2380151735322d2i2d790000280i6giebh17fdf3.gif
Attachment:
MSP8506173e0f30207c4a2700005eg6g59c48d51i5b.gif
Attachment:
MSP111401047g43cgaehag920000640367fgf148e668.gif


Hi Bunuel,

Really thank you for all your support.

I have a small confusion which is really troubling me.

How do we open the mode when if suppose that x<0. Do we just add a -ve on the RHS or do we also change the sign?

For say in this question x/|x| < x
if x < 0,Then, X/-X > X ----> -1>X ?

I see that you have not changed the sign, but what I have understood till now is that we need to change the sign as well.

Like say |X-1| < 3 then if x<0 then it will be X-1> -3 ---> X> -2. And if we don't change the sign then it will be X<-2 which will be wrong if I plot the equation on the no. line.

Please help me on this. I know I'm getting confused with the concept.

Posted from my mobile device
Show more

I think you'd benefit from going through some theory:


10. Absolute Value



For more check Ultimate GMAT Quantitative Megathread



Hope it helps.
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If x is not equal to 0, is |x| less than 1? (1) x/|x|< x (2) |x| > x 28 Jul 2023, 00:53
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Bunuel
If x is not equal to 0, is |x| less than 1?

Is \(|x|<1\)?
Is \(-1<x<1\)? (\(x\neq{0}\))
So, the question asks whether x is in the range shown below:



(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 consider 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 \(-1<x<1\). So, we have a definite YES answer to the question. Sufficient.


Answer: C.


Show Spoiler
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Hi Bunuel,

Really thank you for all your support.

I have a small confusion which is really troubling me.

How do we open the mode when if suppose that x<0. Do we just add a -ve on the RHS or do we also change the sign?

For say in this question x/|x| < x
if x < 0,Then, X/-X > X ----> -1>X ?

I see that you have not changed the sign, but what I have understood till now is that we need to change the sign as well.

Like say |X-1| < 3 then if x<0 then it will be X-1> -3 ---> X> -2. And if we don't change the sign then it will be X<-2 which will be wrong if I plot the equation on the no. line.

Please help me on this. I know I'm getting confused with the concept.

Posted from my mobile device

I think you'd benefit from going through some theory:


10. Absolute Value



For more check Ultimate GMAT Quantitative Megathread



Hope it helps.
Show more


Hi Bunuel,

Thank you for the reply. I have gone through the entire theory.

I think I have understood something please correct me if I'm wrong.

So as for this question x/|x| < x will be X/-X < X if X< 0 bec the value of the entire mode is -ve, and we do not need to change the sign.

But here |X-1| < 3, if X< 0 then the value inside the mode will be -ve so we need to make it positive by multiplying LHS or RHS with -ve and to keep the inequality the same as before we also change the sign.

Please correct me if I'm wong.

Also if there are any more such questions, please share them.

Thank you.
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If x is not equal to 0, is |x| less than 1? (1) x/|x|< x (2) |x| > x 01 Aug 2023, 08:00
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Bunuel
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Bunuel
If x is not equal to 0, is |x| less than 1?

Is \(|x|<1\)?
Is \(-1<x<1\)? (\(x\neq{0}\))
So, the question asks whether x is in the range shown below:



(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 consider 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 \(-1<x<1\). So, we have a definite YES answer to the question. Sufficient.


Answer: C.


Show Spoiler
Attachment:
MSP340910303aagbh2c14gf0000420i437if9d12bhg.gif
Attachment:
MSP2380151735322d2i2d790000280i6giebh17fdf3.gif
Attachment:
MSP8506173e0f30207c4a2700005eg6g59c48d51i5b.gif
Attachment:
MSP111401047g43cgaehag920000640367fgf148e668.gif


Hi Bunuel,

Really thank you for all your support.

I have a small confusion which is really troubling me.

How do we open the mode when if suppose that x<0. Do we just add a -ve on the RHS or do we also change the sign?

For say in this question x/|x| < x
if x < 0,Then, X/-X > X ----> -1>X ?

I see that you have not changed the sign, but what I have understood till now is that we need to change the sign as well.

Like say |X-1| < 3 then if x<0 then it will be X-1> -3 ---> X> -2. And if we don't change the sign then it will be X<-2 which will be wrong if I plot the equation on the no. line.

Please help me on this. I know I'm getting confused with the concept.

Posted from my mobile device


Hi Bunuel,

Thank you for the reply. I have gone through the entire theory.

I think I have understood something please correct me if I'm wrong.

So as for this question x/|x| < x will be X/-X < X if X< 0 bec the value of the entire mode is -ve, and we do not need to change the sign.

But here |X-1| < 3, if X< 0 then the value inside the mode will be -ve so we need to make it positive by multiplying LHS or RHS with -ve and to keep the inequality the same as before we also change the sign.

Please correct me if I'm wong.

Also if there are any more such questions, please share them.

Thank you.
Show more

Hi Bunuel,

I have my exam tomorrow, was hoping if you can help me with this.

Thank you in advance :)
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If x is not equal to 0, is |x| less than 1? (1) x/|x|< x (2) |x| > x 25 Sep 2025, 05:12
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