GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 13 Dec 2019, 18:52 ### 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

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x

Author Message
TAGS:

### Hide Tags

Manager  B
Status: May The Force Be With Me (D-DAY 15 May 2012)
Joined: 06 Jan 2012
Posts: 191
Location: India
Concentration: General Management, Entrepreneurship
If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

### Show Tags

13
84 00:00

Difficulty:   55% (hard)

Question Stats: 62% (01:42) correct 38% (01:48) wrong based on 1511 sessions

### HideShow timer Statistics

If x ≠ 0, is |x| < 1 ?

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

Originally posted by boomtangboy on 08 Apr 2012, 09:43.
Last edited by Bunuel on 09 Jun 2019, 22:05, edited 3 times in total.
Renamed the topic and edited the question.
Math Expert V
Joined: 02 Sep 2009
Posts: 59721
Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

### Show Tags

29
28
boomtangboy wrote:
If x≠0, is |x| <1?
(1) x2<1
(2) |x| < 1/x

If $$x\neq{0}$$, is $$|x| <1$$?

Is $$|x| <1$$? --> is -$$1<x<1$$ ($$x\neq{0}$$)?

(1) $$x^2<1$$ --> $$-1<x<1$$. Sufficient.

(2) $$|x| < \frac{1}{x}$$ --> since LHS (|x|) is an absolute value which is always non-negative then RHS (1/x), must be positive (as $$|x| < \frac{1}{x}$$), so $$\frac{1}{x}>0$$ --> $$x>0$$.

Now, if $$x>0$$ then $$|x|=x$$ and we have: $$x<\frac{1}{x}$$ --> since $$x>0$$ then we can safely multiply both parts by it: $$x^2<1$$ --> $$-1<x<1$$, but as $$x>0$$, the final range is $$0<x<1$$. Sufficient.

_________________
##### General Discussion
Manager  Joined: 14 Feb 2012
Posts: 89
Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

### Show Tags

1
Explanation->
|x|<1 means x should be between -1 and 1.

from (1)
x2-1<0 => x between -1 and 1. sufficient

from (2)
if x +ve then same as 1 and sufficient
if x -ve then |x|=-x
-x2<1 or x2 > -1 not sufficient

so and A

Math Expert V
Joined: 02 Sep 2009
Posts: 59721
Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

### Show Tags

2
shikhar wrote:
If x≠0, is |x| <1?

(1) x2<1
(2) |x| < 1/x

Merging similar topics.

shikhar wrote:
Explanation->
|x|<1 means x should be between -1 and 1.

from (1)
x2-1<0 => x between -1 and 1. sufficient

from (2)
if x +ve then same as 1 and sufficient
if x -ve then |x|=-x
-x2<1 or x2 > -1 not sufficient

so and A

$$x$$ cannot be negative. Refer to the solution above.

Also if $$x<0$$ then we have $$-x<\frac{1}{x}$$ and now if we cross multiply by negative $$x$$ then we should flip the sign: $$-x^2>1$$ --> $$x^2+1<0$$ which cannot be true for any real value of $$x$$ (the sum of two positive value cannot be less than zero).
_________________
Intern  Joined: 01 Mar 2012
Posts: 16
Concentration: Operations, Finance
GMAT 1: 740 Q49 V41 GPA: 3.3
WE: Engineering (Manufacturing)
Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

### Show Tags

Bunuel wrote:
shikhar wrote:
If x≠0, is |x| <1?

(1) x2<1
(2) |x| < 1/x

Merging similar topics.

shikhar wrote:
Explanation->
|x|<1 means x should be between -1 and 1.

from (1)
x2-1<0 => x between -1 and 1. sufficient

from (2)
if x +ve then same as 1 and sufficient
if x -ve then |x|=-x
-x2<1 or x2 > -1 not sufficient

so and A

$$x$$ cannot be negative. Refer to the solution above.

Also if $$x<0$$ then we have $$-x<\frac{1}{x}$$ and now if we cross mulitply by negative $$x$$ then we should flip the sign: $$-x^2>1$$ --> $$x^2+1<0$$ which cannot be true for any real value of $$x$$ (the sum of two positive value cannot be less than zero).

Well Explained Bunuel
Director  Joined: 25 Apr 2012
Posts: 651
Location: India
GPA: 3.21
Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

### Show Tags

1
boomtangboy wrote:
If x≠0, is |x| < 1 ?

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

Given question stem asks if |x|<1------> Is -1<x<1

from St 1 we have x^2<1 ------> -1<x<1 So Sufficient

from St2 we have

|x|<1/x

Notice that |x| is a positive value and for any Integer value |x|> 1/x -----This implies X is a fraction. In order to satisfy the above equation let us take some fractional value of x and check what happens to the above equation

x= -1/2 so we have 1/2<-2 ------No
x=3/4 so we have 3/4< 4/3 ------Yes
x=4/3 so we have 4/3 <3/4 ------- no

We see that when fraction is between 0<x<1 then the above equation holds true and hence x is between -1 and 1
Ans should be D....

Bunuel's solution is superb. Saves time
_________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”
Senior Manager  Joined: 13 May 2013
Posts: 396
Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

### Show Tags

1
If x≠0, is |x| < 1 ?

is x<1
OR
is -x<1
x>-1

Is -1<x<1?

(1) x^2 < 1
x^2<1
|x|<1

This tells us exactly what the stem looks for.
SUFFICIENT

(2) |x| < 1/x
is x < 1/x
OR
is -x < 1/x
is x > -1/x
SO
-1/x < x < 1/x
-1 < x^2 < 1
SUFFICIENT

(I am going to use Bunuel's method only because I think it makes more sense and I believe mine is wrong anyways)

|x|<1/x
if |x|<1/x then 1/x MUST be positive as it is greater than an absolute value. If 1/x is positive then x must also be positive and therefore |x|<1/x is actually equal to x<1/x. Because we know that x is positive we can multiply both sides by x to simplify.

(x)* x < 1/x *(x)
x^2 < 1
|x|<1
x<1 or x>-1
-1<x<1

This tells us exactly what the stem is looking for
SUFFICIENT
(D)

(also, could someone explain to me how to solve by the method I originally chose in #2...the one where I get the positive and negative case for |X|)

Thanks!
Verbal Forum Moderator B
Joined: 10 Oct 2012
Posts: 583
Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

### Show Tags

1
WholeLottaLove wrote:
If x≠0, is |x| < 1 ?

(also, could someone explain to me how to solve by the method I originally chose in #2...the one where I get the positive and negative case for |X|)

Thanks!

I.x>0

$$x<\frac{1}{x}$$ As x>0, we can safely cross-multiply $$\to x^2<1 \to |x|<1.$$

II.x<0

$$-x<\frac{1}{x}$$ multiply both sides by$$x^2$$, which is a positive quantity $$\to -x^3<x$$$$[x\neq{0}]$$

or $$x(1+x^2)>0 \to$$ $$(1+x^2)$$ and x have same sign and as$$(1+x^2)$$ is always positive, thus x>0. However, this goes against our assumption. Thus, x is not negative.

Sufficient.
_________________
Senior Manager  Joined: 13 May 2013
Posts: 396
Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

### Show Tags

If x≠0, is |x| < 1 ?

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

Valid X is a number that is greater than -1 (i.e. -3/4, -1/2) and less than 1 (i.e. 1/2, 3/4) either way the absolute value of any of those numbers will be less than 1.
SUFFICIENT

(2) |x| < 1/x
Multiply both sides by x
x^2<1
Same solution as above.
SUFFICIENT

(D)

Is it valid to multiply the absolute value of x on one side (like in number two) to simplify as long as x is positive?
Math Expert V
Joined: 02 Sep 2009
Posts: 59721
Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

### Show Tags

1
1
WholeLottaLove wrote:
If x≠0, is |x| < 1 ?

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

Valid X is a number that is greater than -1 (i.e. -3/4, -1/2) and less than 1 (i.e. 1/2, 3/4) either way the absolute value of any of those numbers will be less than 1.
SUFFICIENT

(2) |x| < 1/x
Multiply both sides by x
x^2<1
Same solution as above.
SUFFICIENT

(D)

Is it valid to multiply the absolute value of x on one side (like in number two) to simplify as long as x is positive?

As, since from |x| < 1/x we concluded that x is positive, then yes we can do that: x < 1/x --> x^2 < 1.

Hope it helps.
_________________
Manager  B
Joined: 14 Jan 2013
Posts: 127
Concentration: Strategy, Technology
GMAT Date: 08-01-2013
GPA: 3.7
WE: Consulting (Consulting)
Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

### Show Tags

Bunuel wrote:
shikhar wrote:
If x≠0, is |x| <1?

(1) x2<1
(2) |x| < 1/x

Merging similar topics.

shikhar wrote:
Explanation->
|x|<1 means x should be between -1 and 1.

from (1)
x2-1<0 => x between -1 and 1. sufficient

from (2)
if x +ve then same as 1 and sufficient
if x -ve then |x|=-x
-x2<1 or x2 > -1 not sufficient

so and A

$$x$$ cannot be negative. Refer to the solution above.

Also if $$x<0$$ then we have $$-x<\frac{1}{x}$$ and now if we cross mulitply by negative $$x$$ then we should flip the sign: $$-x^2>1$$ --> $$x^2+1<0$$ which cannot be true for any real value of $$x$$ (the sum of two positive value cannot be less than zero).

Bunuel,

I am not clear with this part

Also if x<0 then we have -x<\frac{1}{x} and now if we cross mulitply by negative x then we should flip the sign: -x^2>1 --> x^2+1<0

I am getting the below:

-x < 1/x ---> when we cross multiply by -ve x, we get (-x*-x ) x2 >( flipping) -1 ( -x/x) ----> X2+1 >0 ??

Pls help... Math Expert V
Joined: 02 Sep 2009
Posts: 59721
Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

### Show Tags

Mountain14 wrote:
Bunuel wrote:
shikhar wrote:
If x≠0, is |x| <1?

(1) x2<1
(2) |x| < 1/x

Merging similar topics.

shikhar wrote:
Explanation->
|x|<1 means x should be between -1 and 1.

from (1)
x2-1<0 => x between -1 and 1. sufficient

from (2)
if x +ve then same as 1 and sufficient
if x -ve then |x|=-x
-x2<1 or x2 > -1 not sufficient

so and A

$$x$$ cannot be negative. Refer to the solution above.

Also if $$x<0$$ then we have $$-x<\frac{1}{x}$$ and now if we cross mulitply by negative $$x$$ then we should flip the sign: $$-x^2>1$$ --> $$x^2+1<0$$ which cannot be true for any real value of $$x$$ (the sum of two positive value cannot be less than zero).

Bunuel,

I am not clear with this part

Also if x<0 then we have -x<\frac{1}{x} and now if we cross mulitply by negative x then we should flip the sign: -x^2>1 --> x^2+1<0

I am getting the below:

-x < 1/x ---> when we cross multiply by -ve x, we get (-x*-x ) x2 >( flipping) -1 ( -x/x) ----> X2+1 >0 ??

Pls help... When I say "multiply by negative x", I mean multiply by x, which is negative, so simply by x.
_________________
Retired Moderator P
Status: The best is yet to come.....
Joined: 10 Mar 2013
Posts: 481
Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

### Show Tags

If |x| <1 --> -1<x<1, why not |x| < 1/x --> -1/x<x<1/x? If so, how does it lead to x>0?
_________________
Hasan Mahmud
Intern  Joined: 20 May 2014
Posts: 34
Location: India
Schools: IIMC
GMAT 1: 700 Q51 V32 Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

### Show Tags

2
1
Hi Mahmud,

|x| <1 can be written as -1<x<1 because 1 is a constant BUT

|x| < $$\frac{1}{x}$$ cannot be written as -1/x<x<1/x because 1/x is a variable

Solving,

|x| <$$\frac{1}{x}$$

RHS has to be greater than 0 (As LHS can only be +ve or 0)
=> $$\frac{1}{x}$$ > 0
=> x>0 (x cannot be -ve or 0 ,
Because, if x is -ve then $$\frac{1}{x}$$ is -ve
if x = 0 then$$\frac{1}{x}$$ is not defined)

Rgds,
Rajat
Senior Manager  S
Joined: 08 Dec 2015
Posts: 285
GMAT 1: 600 Q44 V27 Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

### Show Tags

Hi Bunuel!

A question here.

How is St 2 sufficient? You say it yourself that 2) implies that 0<x<1 so that is just part of the range of the question stem ( is -1<x<1 ?) So statement two doesn't cover the negative area of the range. How can it be considered to be sufficient?

Do I make myself clear btw?
Math Expert V
Joined: 02 Sep 2009
Posts: 59721
Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

### Show Tags

1
iliavko wrote:
Hi Bunuel!

A question here.

How is St 2 sufficient? You say it yourself that 2) implies that 0<x<1 so that is just part of the range of the question stem ( is -1<x<1 ?) So statement two doesn't cover the negative area of the range. How can it be considered to be sufficient?

Do I make myself clear btw?

The question asks: whether x is between -1 and 1. (2) says that x is between 0 and 1, so the answer to the question is YES.
_________________
IIMA, IIMC School Moderator V
Joined: 04 Sep 2016
Posts: 1370
Location: India
WE: Engineering (Other)
Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

### Show Tags

1
Bunuel mikemcgarry IanStewart shashankism Engr2012

Quote:
Also if $$x<0$$ then we have $$-x<\frac{1}{x}$$ and now if we cross multiply by negative $$x$$ then we should flip the sign: $$-x^2>1$$ --> $$x^2+1<0$$ which cannot be true for any real value of $$x$$ (the sum of two positive value cannot be less than zero).

why am I getting $$x^2$$ > -1 for this step? I multiplied both sides of $$-x<1/x$$ with -x
_________________
It's the journey that brings us happiness not the destination.

Feeling stressed, you are not alone!!
CEO  S
Joined: 20 Mar 2014
Posts: 2560
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44 GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

### Show Tags

1
Bunuel mikemcgarry IanStewart shashankism Engr2012

Quote:
Also if $$x<0$$ then we have $$-x<\frac{1}{x}$$ and now if we cross multiply by negative $$x$$ then we should flip the sign: $$-x^2>1$$ --> $$x^2+1<0$$ which cannot be true for any real value of $$x$$ (the sum of two positive value cannot be less than zero).

why am I getting $$x^2$$ > -1 for this step? I multiplied both sides of $$-x<1/x$$ with -x

You are making an algebraic mistake. If you do assume that x<0, |x| = -x; multiplying both sides of the inequality -x<1/x by -x you should get, -x*(-x) < (1/x)*(-x) ---> x^2 < -1 and this is not possible for any real 'x'. Hence, x<0 can not be true.

Additionally, you dont need to flip the signs when you multiply by -x as we are assuming that x<0, making -x >0 and when you multiply an inequality by a positive quantity, you do not need to flip the signs.

Hope this helps.
Math Expert V
Joined: 02 Aug 2009
Posts: 8310
If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

### Show Tags

Project DS Butler: Day 18: Data Sufficiency (DS35)

If $$x\neq{0}$$, is $$|x|$$< 1?

(1) $$x^2$$ < 1

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

If $$x\neq{0}$$, is $$|x|$$< 1?
This means -1<x<1, where $$x\neq{0}$$

(1) $$x^2$$ < 1
$$x^2<1$$ means |x|<1, since only fractions with absolute value less than 1 will have their square less than 1.
Sufficient

(2) $$|x|$$ < $$1/x$$
Since |x| >0, we can write 1/|x|x>1
Since |x| and 1 are positive, x also must be positive for above to be true.
And since X is also positive, |X|=X, and X<1/X can be written as x^2<1
Therefore X is between 0 and 1
Sufficient

D
_________________
Manager  B
Joined: 07 Apr 2018
Posts: 100
Location: United States
Concentration: General Management, Marketing
GMAT 1: 600 Q45 V28 GPA: 3.8
Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

### Show Tags

the slow but robust way of doing this is solving it for x>0 and x<0.
Attachments IMG_20181125_175840.jpg [ 1.96 MiB | Viewed 4263 times ] Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x   [#permalink] 25 Nov 2018, 16:06

Go to page    1   2    Next  [ 30 posts ]

Display posts from previous: Sort by

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