X

Question

X<0?
1) X^3<X^2
2) X^3<X^4

kapslock · Answer

(1) means either 0 1 or x < 0. Insufficient again. Combine both. 0 < x < 1 and x > 1 can't be combined (different ranges). Therefore, x < 0. (C)

swath20 · Answer

one more for C


Same explanation as kaplock's

thearch · Answer

for statement 2, is it right to simplify this way??? x^3

banerjeea_98 · Answer

yes u can....u can do that for state 1 as well.....x^3 < x^2 ---> x < 1

rthothad · Answer

I would go with C

1. x can be -ve or a fraction.
2. x can be -ve or a +ve

Both put together x will be -ve

DLMD · Answer

I got C too but the answer given is E.

for these who choose E, please elaborate why E is right.

otherwise, the answer given might be wrong.

Thanks

Folaa3 · Answer

Well for each values of x, use Â½, -1/2, 2, -2, 3, -3 for trials and you will see how it is E. A little exhaustive but gives the right answer. .

banerjeea_98 · Answer

OA can't be "E"...in ur values of X all ur +ve numbers do not satisfy both statements...1/2 doesn't satisfy state 2 and 2,3 don't satisfy state 1....only x < 0 satisfies both.

FN · Answer

c for me...

1) statement 1 says x<0 or 0<x<1
two possibilites thus insuff!

2) statement 2 says
x<0 or X>1
again two possible answers...insuff

combine the two...

since both say X <0 is the only common piece...

gmat2me2 · Answer

X<0? 1) X^3> x^2<0 and (x-1)>0 (Since both of them cannot be negative) or x^2>0 and (x-1)<0 So you cannot tell anything about x from 2) x^3(1-x)<0 Again the same logic x^3<0 and (1-x)>0 or x^3>0 and (1-x)<0 which again isnt sufficient. Since(1) and (2) are giving multiple situations we cannot tell if x<0. So I believe the answer is E.

MA · Answer

from i, x = more than any negative value that is less than 1= -ve and less than 1>.
from ii, x=either -ve or +ve but more than 1.
to satisfy both, x must less than 0 but more than -1 = -1<x<0
suppose x = -0.5, 
from i, X^3<X^2
(-0.5)^3<(-0.5)^2
-0.125<0.25

from ii, X^3<X^4
(-0.5)^3<(-0.5)^4
-0.125<0.0625. therefore it is C.

HongHu · Answer

X<0? 1) X^31 Insufficient Combined: x<1 x<0 or x>1 =>x<0 (C)

HongHu · Answer

Right approach, but x^2 can never be negative. We are talking about real numbers.

Assume x>0, then you can multiple the two equations, and you get:
x^6<x^6 which is impossible. In other words it is impossible that x>0. You know that x can't be equal to zero either. So x must be less than zero. (C).

gmat2me2 · Answer

I would have cancelled x on both sides had this question came before I read your cameo on inqualities in this forum.Atleast the approach is right now.....Thanks HH

X<0? 1) X^3<X^2 2) X^3<X^4

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