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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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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 04 Jul 2026, 13:20
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hey, quick question here, below is what I tried to do but not sure if its correct.

From statement 1, I took 2 cases. X as 1/2 and -1/2. I get a yes and no answer. Using statement 2 I know that x is negative. Hence only -1/2 is possible which gives a yes answer to the question.

Is this a correct approach?
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If x is not equal to 0, is |x| less than 1? (1) x/|x|< x (2) |x| > x 05 Jul 2026, 01:07
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Is the level of question really 805?

Bunuel


Range from (1): -----(-1)----(0)----(1)---- \(-1<x<0\) or \(x>1\), green area;

Range from (2): -----(-1)----(0)----(1)---- \(x<0\), blue area;

From (1) and (2): ----(-1)----(0)----(1)---- \(-1<x<0\), common range of \(x\) from (1) and (2) (intersection of ranges from (1) and (2)), red area.

Hope it's clear.
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If x is not equal to 0, is |x| less than 1? (1) x/|x|< x (2) |x| > x 05 Jul 2026, 01:32
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Is the level of question really 805?


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The difficulty level of a question on the site, after sufficient attempts, is determined automatically based on various parameters collected from users' attempts via timer, such as the percentage of correct answers and the time taken to answer the question. So, this is an 805+ (Hard) Level question based on our statistics.
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If x is not equal to 0, is |x| less than 1? (1) x/|x|< x (2) |x| > x 06 Jul 2026, 09:34
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Hi architkap,

Good instinct: plugging in numbers is a completely valid way to attack a DS question. But there's one rule that makes or breaks the method, and it's exactly where your test slipped.

Every value you plug in must actually satisfy the statement you're testing. Let's check your two picks against Statement (1), which says x/|x| < x:

- x = -1/2: (-1/2)/|-1/2| = -1, and is -1 < -1/2? Yes. This value is legal, and it gives a YES to "is |x| < 1?"
- x = 1/2: (1/2)/|1/2| = 1, and is 1 < 1/2? No. So x = 1/2 does not obey Statement (1) at all.

That's the catch. Because x = 1/2 breaks the statement, it can't be used to show anything about Statement (1). You reached the right final answer, but the "insufficiency" of Statement (1) wasn't really demonstrated - you got lucky that your surviving case still pointed to C.

The valid pair for Statement (1)

To prove Statement (1) alone is not sufficient, you need two values that both satisfy x/|x| < x yet answer the question differently:

- x = -1/2: obeys (1), and |x| < 1 - YES
- x = 2: 2/|2| = 1 < 2 obeys (1), and |x| < 1 - NO

Same statement, two answers - not sufficient.

Now bring in Statement (2), |x| > x, which forces x < 0. That kills the x = 2 case, leaving only negative fractions like -1/2 - always YES. So together they're sufficient - C.

Takeaway: in DS, before a test value can "count," confirm it satisfies the statement. Then look for two legal values that disagree.

Answer: C

architkap
hey, quick question here, below is what I tried to do but not sure if its correct.

From statement 1, I took 2 cases. X as 1/2 and -1/2. I get a yes and no answer. Using statement 2 I know that x is negative. Hence only -1/2 is possible which gives a yes answer to the question.

Is this a correct approach?
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