I get C) OA please?

stm 1. try x=2, -2, 1/2 the only way for stm 1 to be true is x=2 ie sufficient and |x| not less than 1

stm 2. not sufficient...x=-2 and x=-1/2 both can be true

Hence a

Just saw the oa and its [spoiler]C)[/spoiler]

jayjk78 wrote:stm 1. try x=2, -2, 1/2 the only way for stm 1 to be true is x=2 ie sufficient and |x| not less than 1

stm 2. not sufficient...x=-2 and x=-1/2 both can be true

Hence a

I am not a big fan of plugging numbers but how about -1/2 ?

cramya wrote:I get C) OA please?

Could you please explain your approach ?

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

Good question

I first picked A and then stopped to check - that saved me

Question:
|x| < 1?

(1)
(x/|x|)<x
multiply both sides by |x|
x < x*|x|
here are 2 options:
a. x>0 (e.g. 5)
divide both sides by x with 1 < |x| as a result
b. x<0 (e.g. -1/2)
divide both sides by x with 1 > |x| as a result
2 options thus INSUFF

(2)
|x| > x
this means that x<0 but it not gives us anuthing about the value of x

(1)(2)
x<0 (from 2)
1 > |x| (from 1)
so, -1<x<0
SUFF

C

Wow, this question really got me!... Nice explanation 4meonly!

Logitech, where's this question from?

-BM-

logitech wrote:

cramya wrote:I get C) OA please?

Could you please explain your approach ?

One more approach logitech....
we know value of |x| will be always positive.
so,

when x>0, x/x < x or x>1
when x<0, x/-x < x or x>-1
hence insuff as we want to know if -1<x<1

from 2nd we get x<0 as 0<0 not possible when x>0
its also insuff....

combining we get
-1<x<0 hence suff...
C

hope it helps....

logitech wrote:If x is not equal to 0, is |x| less than 1?

(1) ( x / |x| ) < x


(2) |x| > x

OA C

Good discussions guys. Problem is from Manhattan GMAT.

So the question is asking if X is between - 1 and + 1 ( EXCLUDING zero )

1)

if x<0

x / |x| = -1 So, x>-1

---> -1 < x < 0

If x > 0

x / |x| = 1 So, x > 1


INSUF

2) Basically tells us X < 0

So 1&2

we know that X < 0 and -1 < x < 0


Thank you all!