Given X/|X| < X. Which of the following must be true : Quant Question Archive [LOCKED]
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 17 Jan 2017, 03:21

### 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

# Given X/|X| < X. Which of the following must be true

Author Message
Senior Manager
Joined: 06 Jul 2004
Posts: 473
Location: united states
Followers: 1

Kudos [?]: 105 [0], given: 0

Given X/|X| < X. Which of the following must be true [#permalink]

### Show Tags

06 Aug 2006, 13:35
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Given

X/|X| < X. Which of the following must be true about X?

A) X > 1

B) X > -1

C) |X| < 1

D) |X| = 1

E) |X|^2 > 1
_________________

for every person who doesn't try because he is
afraid of loosing , there is another person who
keeps making mistakes and succeeds..

SVP
Joined: 05 Jul 2006
Posts: 1743
Followers: 6

Kudos [?]: 315 [0], given: 49

### Show Tags

06 Aug 2006, 14:32
I will go for /x/<1
Senior Manager
Joined: 06 Jul 2004
Posts: 473
Location: united states
Followers: 1

Kudos [?]: 105 [0], given: 0

### Show Tags

06 Aug 2006, 14:40
"I will go for /x/<1"

and why would you do that?
_________________

for every person who doesn't try because he is
afraid of loosing , there is another person who
keeps making mistakes and succeeds..

Senior Manager
Joined: 14 Jul 2005
Posts: 402
Followers: 1

Kudos [?]: 27 [0], given: 0

### Show Tags

06 Aug 2006, 15:34
X/|X| < X

divide by X on both sides 1/|x| < 1
multiply with |x| on both sides

1<|x| i.e. |x|>1

Hence will go with E
Director
Joined: 07 Jun 2004
Posts: 612
Location: PA
Followers: 5

Kudos [?]: 705 [0], given: 22

### Show Tags

06 Aug 2006, 15:42
I will go with A

take examples X > 1 then any value > x 1.1 or 1 milllion will be both positive and the answer will be 1 so 1 < any value > x
Manager
Joined: 22 Jul 2006
Posts: 50
Followers: 0

Kudos [?]: 3 [0], given: 0

### Show Tags

06 Aug 2006, 16:09
I will go for E

X/|X| < X implies

X/X < X or X/(-X) > X
so, x > 1 or x <-1. The only option which satisfies this condition is 'E'
Manager
Joined: 07 Aug 2005
Posts: 127
Followers: 1

Kudos [?]: 1 [0], given: 0

### Show Tags

06 Aug 2006, 20:25
shoonya wrote:
Given

X/|X| < X. Which of the following must be true about X?

A) X > 1

B) X > -1

C) |X| < 1

D) |X| = 1

E) |X|^2 > 1

rewrite as x < x * |x|
or x < x*n - where n is any +ve number(integer or fraction)
only possible solution is when n > 1

A for me.
VP
Joined: 29 Dec 2005
Posts: 1348
Followers: 10

Kudos [?]: 60 [0], given: 0

### Show Tags

06 Aug 2006, 21:15
shoonya wrote:
X/|X| < X. Which of the following must be true about X?

A) X > 1
B) X > -1
C) |X| < 1
D) |X| = 1
E) |X|^2 > 1

straight A. no matter the value of x, x > x/lxl. since x>1 and x/lxl is always 1. so its pretty clear that x>x/lxl.
Senior Manager
Joined: 22 May 2006
Posts: 371
Location: Rancho Palos Verdes
Followers: 1

Kudos [?]: 110 [0], given: 0

### Show Tags

06 Aug 2006, 22:30
X/|X| < X (X!=0 )
1. if X > 0 then 1 < X (X/X < X => 1<X)
2. if X < 0 then -1 < X <0 (X/-X < X => -1 < X)
Combine 1 and 2
-1<X<0 and X>1

Thus A.
_________________

The only thing that matters is what you believe.

Senior Manager
Joined: 19 Jul 2006
Posts: 361
Followers: 2

Kudos [?]: 6 [0], given: 0

### Show Tags

06 Aug 2006, 22:37
A

Agree with proff. explanation
SVP
Joined: 30 Mar 2006
Posts: 1737
Followers: 1

Kudos [?]: 77 [0], given: 0

### Show Tags

07 Aug 2006, 03:52
Clear winner A.

A satisfies the condition perfectly
Senior Manager
Joined: 14 Jul 2005
Posts: 402
Followers: 1

Kudos [?]: 27 [0], given: 0

### Show Tags

07 Aug 2006, 07:24
Can any of you explain if what I did was actually wrong ?

i.e.

X/|X| < X

divide by X on both sides 1/|x| < 1
multiply with |x| on both sides

1<|x| i.e. |x|>1

Hence will go with E
Senior Manager
Joined: 21 Jun 2006
Posts: 285
Followers: 1

Kudos [?]: 112 [0], given: 0

### Show Tags

07 Aug 2006, 13:18
A is the only one which works
Take X = 3
3/|3|=1<3

Take x=-3
-3||-3|=-1>X

Take X =1
Then LHS = RHS

X>1 is the only one that works
Manager
Joined: 20 Mar 2006
Posts: 200
Followers: 1

Kudos [?]: 3 [0], given: 0

### Show Tags

07 Aug 2006, 17:17
shoonya wrote:
Given

X/|X| < X. Which of the following must be true about X?

A) X > 1

B) X > -1

C) |X| < 1

D) |X| = 1

E) |X|^2 > 1

x< x|x|
when x> 0 |x| = x which implies x< x^2 or x(x-1) > 0 i.e x>1
when x<0 |x| = -x which implies x<-x^2 or x(x+1) < 0 i.e -1<x<0

Hence A

Heman
Manager
Joined: 07 Aug 2005
Posts: 127
Followers: 1

Kudos [?]: 1 [0], given: 0

### Show Tags

08 Aug 2006, 07:00
gmatornot wrote:
Can any of you explain if what I did was actually wrong ?

i.e.

X/|X| < X

divide by X on both sides 1/|x| < 1
multiply with |x| on both sides

1<|x| i.e. |x|>1

Hence will go with E

You can not divide by x, because you don't know the sign of x. If x is -ve you would need to flip the ineqality sign.
VP
Joined: 29 Dec 2005
Posts: 1348
Followers: 10

Kudos [?]: 60 [0], given: 0

### Show Tags

08 Aug 2006, 07:54
Path wrote:
gmatornot wrote:
Can any of you explain if what I did was actually wrong ?

i.e.

X/|X| < X

divide by X on both sides 1/|x| < 1
multiply with |x| on both sides

1<|x| i.e. |x|>1

Hence will go with E

You can not divide by x, because you don't know the sign of x. If x is -ve you would need to flip the ineqality sign.

agree that you cannot divide both sides by x. ok, lets suppose you are when x is +ve, but if x is a fraction x/lxl>x.
Manager
Joined: 08 Aug 2006
Posts: 82
Location: Michigan
Followers: 1

Kudos [?]: 1 [0], given: 0

### Show Tags

08 Aug 2006, 08:08
I'll try the old plug numbers routine

2/l2l<2
and so on so my answer is A
Current Student
Joined: 28 Dec 2004
Posts: 3384
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 15

Kudos [?]: 281 [0], given: 2

### Show Tags

08 Aug 2006, 11:40
actually freethinking has the best explanation...and this how u shud approach || questions...

freetheking wrote:
X/|X| < X (X!=0 )
1. if X > 0 then 1 < X (X/X < X => 1<X)
2. if X < 0 then -1 < X <0 (X/-X < X => -1 < X)
Combine 1 and 2
-1<X<0 and X>1

Thus A.
08 Aug 2006, 11:40
Display posts from previous: Sort by