It is currently 17 Mar 2018, 11:32

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Intern
Joined: 26 Feb 2017
Posts: 5
Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

### Show Tags

17 Mar 2017, 01:48
Hi,
I'm new to this and my head is spining... so many details in all the posts, can't even begin to read them!

I thought in S2, x<0.
So we know all the signs in S1 and therefore we can solve it like a simple inequality...and we find |x|<1. S1 &S2 together suff.

But this is too simple!? Am i wrong in my approach even if the result is right?... I m a newbie with no GMAT confidence...help
Manager
Joined: 23 Jan 2016
Posts: 224
Location: India
GPA: 3.2
Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

### Show Tags

13 May 2017, 08:25
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.

Should the answer not be A? because in option A, we can simply take the mod to the other side - x<|x|*x - and the sign will remain the same since mod can only be positive; further on. because x multiplied by mod x is is larger than x, it tells us that x must be positive, because were it negative, x multiplied by mod x would have been smaller than x.

Would really appreciate some light on this basic concept. Thanks.
Math Expert
Joined: 02 Sep 2009
Posts: 44289
Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

### Show Tags

13 May 2017, 08:31
OreoShake 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 $$-1<x<0$$

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.

Should the answer not be A? because in option A, we can simply take the mod to the other side - x<|x|*x - and the sign will remain the same since mod can only be positive; further on. because x multiplied by mod x is is larger than x, it tells us that x must be positive, because were it negative, x multiplied by mod x would have been smaller than x.

Would really appreciate some light on this basic concept. Thanks.

_________________
Intern
Joined: 18 Mar 2017
Posts: 3
Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

### Show Tags

15 Oct 2017, 14:02
For (1), wouldn't x<0 give you:

-x/x < -x
-1 < -x
1 > x?

When I plug in numbers for that, it doesn't work, but I don't see how we get x/-x < x when plugging in x<0?

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.

Math Expert
Joined: 02 Sep 2009
Posts: 44289
Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

### Show Tags

15 Oct 2017, 21:17
cgarmestani wrote:
For (1), wouldn't x<0 give you:

-x/x < -x
-1 < -x
1 > x?

When I plug in numbers for that, it doesn't work, but I don't see how we get x/-x < x when plugging in x<0?

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.

If x < 0, then |x| = -x.

Substitute |x| by -x in x/|x|< x to get x/(-x) < x and then to get -1 < x. Since we consider the range when x < 0, then -1 < x < 0.
_________________
Intern
Joined: 14 May 2017
Posts: 1
Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

### Show Tags

29 Nov 2017, 07:09
Bunuel wrote:
udaymathapati wrote:
Hi Bunuel,
Thanks for detail explanation. I am finding it difficult only last intersection part. Can you explain it further. My doubut is...If we combine "-1<x<0" or "x>1" these two inequalities, how come range for x fall betweeen -1<x<1 since x>1 is area which will not fit into this equation. Can you explain?

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.

the red zone indicates that the range is -1<x<0 , how do we arrive at -1<x<1?
Math Expert
Joined: 02 Sep 2009
Posts: 44289
Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

### Show Tags

29 Nov 2017, 07:22
yousufkhan wrote:
Bunuel wrote:
udaymathapati wrote:
Hi Bunuel,
Thanks for detail explanation. I am finding it difficult only last intersection part. Can you explain it further. My doubut is...If we combine "-1<x<0" or "x>1" these two inequalities, how come range for x fall betweeen -1<x<1 since x>1 is area which will not fit into this equation. Can you explain?

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.

the red zone indicates that the range is -1<x<0 , how do we arrive at -1<x<1?

The question asks whether $$-1<x<1$$ is true. We got that $$-1<x<0$$. Any, x from $$-1<x<0$$ IS in the range from -1 to 1, so we have an YES answer to the question.
_________________
Senior Manager
Joined: 31 Jul 2017
Posts: 317
Location: Malaysia
WE: Consulting (Energy and Utilities)
Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

### Show Tags

29 Nov 2017, 08:52
Hussain15 wrote:
If x is not equal to 0, is |x| less than 1?

(1) x/|x|< x

(2) |x| > x

Stmnt 2: |x| - x >0, which means x < 0... So, Insufficient
Stmnt 1: x (1-1/|x|)>0, which means x < 0 , 1-1/|x| < 0 or x > 0, 1-1/|x| >0.. Insufficeint

Combining 1 +2,

We know x <0, 1-1/|x| < 0.. which is |x| < 1.

Hope it helps.
_________________

If my Post helps you in Gaining Knowledge, Help me with KUDOS.. !!

Retired Moderator
Status: The best is yet to come.....
Joined: 10 Mar 2013
Posts: 528
Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

### Show Tags

27 Dec 2017, 03:15
Similar topic: https://gmatclub.com/forum/if-x-is-diff ... 26715.html
_________________

Hasan Mahmud

Intern
Joined: 03 Sep 2017
Posts: 15
Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

### Show Tags

17 Feb 2018, 11:51
(1) We dont have to remove the abs(x)...we can treat it as a variable that is always positive.

So if we multiply both sides by abs(x),we dont have to change the sign of the inequality and we get

abs(x)*x> x

if x>0 ---> abs(x) >1
if x<0----> abs(x) <1
Not suf.

(2) abs(x) < x
x must be <0, but we cant say that abs (x)<1, because this will be true for any x<0.

(1)(2): We are restricted to the inference:
if x<0----> abs(x) <1
so we its suf.
Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x   [#permalink] 17 Feb 2018, 11:51

Go to page   Previous    1   2   3   4   5   [ 90 posts ]

Display posts from previous: Sort by

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

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.