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# X/|X| <X>1 B) X>-1 C) |X|<1> 1

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Senior Manager
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X/|X| <X>1 B) X>-1 C) |X|<1> 1 [#permalink]  07 Aug 2007, 17:45
X/|X| <X>1

B) X>-1

C) |X|<1> 1
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for every person who doesn't try because he is
afraid of loosing , there is another person who
keeps making mistakes and succeeds..

GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5079
Location: Singapore
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Kudos [?]: 180 [0], given: 0

I guess you will need to post again with HTML disabled.
Senior Manager
Joined: 06 Jul 2004
Posts: 474
Location: united states
Followers: 1

Kudos [?]: 16 [0], given: 0

ywilfred wrote:
I guess you will need to post again with HTML disabled.

Thanks ywilfred. Here it is:

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

1) X > 1

2) X > -1

3) |X| < 1

4) |X| = 1

5) |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..

VP
Joined: 10 Jun 2007
Posts: 1465
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Kudos [?]: 127 [0], given: 0

shoonya wrote:
ywilfred wrote:
I guess you will need to post again with HTML disabled.

Thanks ywilfred. Here it is:

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

1) X > 1

2) X > -1

3) |X| < 1

4) |X| = 1

5) |X|^2 > 1

Only A for me.

X/|X| < X
if X<0, then X/ -X < X => X>-1 => -1<X<0
if X>0, then X>1
By definition, X will not equal to zero, so C and B are out. X cannot be equal to -1, so D is out. E cannot be true because X and be less than -1.
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5079
Location: Singapore
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Kudos [?]: 180 [0], given: 0

a) If x = 2, then x/|x| < x. If x = 1.5, then x/|x| < x. Looks good.
b) If x = 1, then x/|x| = x. If x = 2, then x/|x| <x> x. If x = -1/2, then x/|x| <x> 1. Then |x| must be an integer. But x can be positve or negative integer. If x = -4, then x/|x| > x. If x = 4, then x/|x| < x. Out.

A is best.
Senior Manager
Joined: 06 Jul 2004
Posts: 474
Location: united states
Followers: 1

Kudos [?]: 16 [0], given: 0

ywilfred wrote:
a) If x = 2, then x/|x| < x. If x = 1.5, then x/|x| < x. Looks good.
b) If x = 1, then x/|x| = x. If x = 2, then x/|x| <x> x. If x = -1/2, then x/|x| <x> 1. Then |x| must be an integer. But x can be positve or negative integer. If x = -4, then x/|x| > x. If x = 4, then x/|x| < x. Out.

A is best.

This question is from GMATCLUB's question collection 2.0 . I also got A, but the author thinks the answer is B.
_________________

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

Manager
Joined: 22 May 2006
Posts: 184
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Kudos [?]: 8 [0], given: 0

-0.5 also satisfies the equation.
SVP
Joined: 01 May 2006
Posts: 1806
Followers: 8

Kudos [?]: 97 [0], given: 0

(B) it is

X/|X| < X (Not here that X must be !=0 as the equation exists)
<=> X/|X| - X < 0
<=> X - X*|X| < 0 as |X| > 0
<=> X * (1-|X|) < 0

Implies 2 cases :

o If X > 0 then
1 - |X| < 0
<=> |X| > 1
=> X > 1 as X > 0.

o If X < 0 then
1 - |X| > 0
<=> |X| < 1
=> 0 > X > - 1 as X < 0.

So, all in all, to be sure that the equation X/|X| < X is always true, we must take an interval in the answer choice that contains both intervals above. Thus, X > -1.
SVP
Joined: 01 May 2006
Posts: 1806
Followers: 8

Kudos [?]: 97 [0], given: 0

bkk145 wrote:
Fig wrote:
(B) it is

X/|X| < X (Not here that X must be !=0 as the equation exists)
<=> X/|X| - X < 0
<=> X - X*|X| < 0 as |X| > 0
<=> X * (1-|X|) < 0

Implies 2 cases :

o If X > 0 then
1 - |X| < 0
<=> |X| > 1
=> X > 1 as X > 0.

o If X < 0 then
1 - |X| > 0
<=> |X| < 1
=> 0 > X > - 1 as X < 0.

So, all in all, to be sure that the equation X/|X| < X is always true, we must take an interval in the answer choice that contains both intervals above. Thus, X > -1.

X/|X| < X is an inequation that exists so we cannot have x = 0

So, yes.... it's voluntarily that the author asks x > -1.... because it must be true.... even if -1 < x < 0 U X > 1 is the complete solution
Senior Manager
Joined: 06 Jul 2004
Posts: 474
Location: united states
Followers: 1

Kudos [?]: 16 [0], given: 0

Fig wrote:
bkk145 wrote:
Fig wrote:
(B) it is

X/|X| < X (Not here that X must be !=0 as the equation exists)
<=> X/|X| - X < 0
<=> X - X*|X| < 0 as |X| > 0
<=> X * (1-|X|) < 0

Implies 2 cases :

o If X > 0 then
1 - |X| < 0
<=> |X| > 1
=> X > 1 as X > 0.

o If X < 0 then
1 - |X| > 0
<=> |X| < 1
=> 0 > X > - 1 as X < 0.

So, all in all, to be sure that the equation X/|X| < X is always true, we must take an interval in the answer choice that contains both intervals above. Thus, X > -1.

X/|X| < X is an inequation that exists so we cannot have x = 0

So, yes.... it's voluntarily that the author asks x > -1.... because it must be true.... even if -1 < x < 0 U X > 1 is the complete solution

Fig, we agree that the actual solution is (-1<x<0) U (X>1). But among the given choices, only X>1 always satisfies the inequality.

For X>-1, test X=1/2. The inequality doesn't stand. So, the answer must be A, not B.
_________________

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

Current Student
Joined: 28 Dec 2004
Posts: 3392
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 13

Kudos [?]: 180 [0], given: 2

I too dont think B can be the answer...

take x=1/2 you will 1 <1/2 which cannot be possible...
SVP
Joined: 01 May 2006
Posts: 1806
Followers: 8

Kudos [?]: 97 [0], given: 0

fresinha12 wrote:
I too dont think B can be the answer...

take x=1/2 you will 1 <1/2 which cannot be possible...

Yes.... X cannot be 1/2... but X must be > -1 to be X=2 or X=-1/2...

When we say X > -1, we do not say X = 1/2... no. That's the 'must be'

If X=6, then X > 1 or X > -1 or X > -10 000 or X must be > -1

Last edited by Fig on 08 Aug 2007, 07:23, edited 1 time in total.
SVP
Joined: 01 May 2006
Posts: 1806
Followers: 8

Kudos [?]: 97 [0], given: 0

shoonya wrote:
Fig wrote:
bkk145 wrote:
Fig wrote:
(B) it is

X/|X| < X (Not here that X must be !=0 as the equation exists)
<=> X/|X| - X < 0
<=> X - X*|X| < 0 as |X| > 0
<=> X * (1-|X|) < 0

Implies 2 cases :

o If X > 0 then
1 - |X| < 0
<=> |X| > 1
=> X > 1 as X > 0.

o If X < 0 then
1 - |X| > 0
<=> |X| < 1
=> 0 > X > - 1 as X < 0.

So, all in all, to be sure that the equation X/|X| < X is always true, we must take an interval in the answer choice that contains both intervals above. Thus, X > -1.

X/|X| < X is an inequation that exists so we cannot have x = 0

So, yes.... it's voluntarily that the author asks x > -1.... because it must be true.... even if -1 < x < 0 U X > 1 is the complete solution

Fig, we agree that the actual solution is (-1<x<0) U (X>1). But among the given choices, only X>1 always satisfies the inequality.

For X>-1, test X=1/2. The inequality doesn't stand. So, the answer must be A, not B.

'Always satisfies' is not 'must be'
Director
Joined: 03 May 2007
Posts: 894
Schools: University of Chicago, Wharton School
Followers: 6

Kudos [?]: 61 [0], given: 7

Fig wrote:
shoonya wrote:
Fig wrote:
bkk145 wrote:
Fig wrote:
(B) it is

X/|X| < X (Not here that X must be !=0 as the equation exists)
<=> X/|X| - X < 0
<=> X - X*|X| <0> 0
<=> X * (1-|X|) <0> 0 then[/b]
1 - |X| < 0
<X> 1
=> X > 1 as X > 0.

o If X <0> 0
<=> |X| <1> 0 > X > - 1 as X < 0.

So, all in all, to be sure that the equation X/|X| <X> -1.

X/|X| < X is an inequation that exists so we cannot have x = 0

So, yes.... it's voluntarily that the author asks x > -1.... because it must be true.... even if -1 < x <0> 1 is the complete solution

Fig, we agree that the actual solution is (-1<x<0>1). But among the given choices, only[b] X>1 always satisfies the inequality.

For X>-1, test X=1/2. The inequality doesn't stand. So, the answer must be A, not B.

'Always satisfies' is not 'must be'

Though i also choose A, i think it should be B as pointed by Fig. He/she has a valid point.
Senior Manager
Joined: 06 Jul 2004
Posts: 474
Location: united states
Followers: 1

Kudos [?]: 16 [0], given: 0

Darden2010 wrote:
what is with all this 'high' math stuff, GMAT is about understanding of simple concepts.

X/|X| is either +1 or -1

so we have two scenarios

+1<X

and

-1<X

put these on a line and you will see that they overlap starting from 1<X, therefore that is the solution

I think most of us know that. That's why we are discussing why the OA is other way round?
_________________

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

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