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

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Senior Manager
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X<0? 1) X^3<X^2 2) X^3<X^4 [#permalink] New post 12 Apr 2005, 15:13
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A
B
C
D
E

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(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 2 sessions
X<0?
1) X^3<X^2
2) X^3<X^4
Senior Manager
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Re: DS guys!! [#permalink] New post 12 Apr 2005, 15:21
DLMD wrote:
X<0?
1) X^3<X^2
2) X^3<X^4




(1) means either 0<x<1 or x < 0. Insufficient to answer if x<0.
(2) means either x > 1 or x < 0. Insufficient again.

Combine both. 0 < x < 1 and x > 1 can't be combined (different ranges).
Therefore, x < 0.

(C)
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 [#permalink] New post 12 Apr 2005, 16:29
one more for C


Same explanation as kaplock's
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 [#permalink] New post 13 Apr 2005, 03:04
for statement 2, is it right to simplify this way???
x^3<x^4 divide both by x^2 (it has to be positive and cant be 0)

so we have x<x^2 (which is far simpler)

awaiting your thoughts, inequalities are another weakness of mine :?
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 [#permalink] New post 13 Apr 2005, 07:27
thearch wrote:
for statement 2, is it right to simplify this way???
x^3<x^4 divide both by x^2 (it has to be positive and cant be 0)

so we have x<x^2 (which is far simpler)

awaiting your thoughts, inequalities are another weakness of mine :?



yes u can....u can do that for state 1 as well.....x^3 < x^2 ---> x < 1
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 [#permalink] New post 13 Apr 2005, 08:38
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
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 [#permalink] New post 13 Apr 2005, 10:13
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
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 [#permalink] New post 13 Apr 2005, 11:38
DLMD wrote:
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


Well for each values of x, use ½, -1/2, 2, -2, 3, -3 for trials and you will see how it is E. A little exhaustive but gives the right answer. .
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 [#permalink] New post 13 Apr 2005, 12:08
Folaa3 wrote:
DLMD wrote:
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


Well for each values of x, use ½, -1/2, 2, -2, 3, -3 for trials and you will see how it is E. A little exhaustive but gives the right 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.
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 [#permalink] New post 13 Apr 2005, 12:19
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...
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 [#permalink] New post 13 Apr 2005, 14:20
DLMD wrote:
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


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

From 1)

x^3 - x^2 <0
x^2(x-1)<0

==>> 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.
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Re: DS guys!! [#permalink] New post 13 Apr 2005, 20:47
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.
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Re: DS guys!! [#permalink] New post 14 Apr 2005, 13:38
X<0?

1) X^3<X^2
x<1
Insufficient

2) X^3<X^4
1/x<1
x<0 or x>1
Insufficient

Combined:
x<1
x<0 or x>1
=>x<0

(C)
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keep on seeking, and you will find;
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 [#permalink] New post 14 Apr 2005, 13:42
gmat2me2 wrote:
DLMD wrote:
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


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

From 1)

x^3 - x^2 <0
x^2(x-1)<0

==>> 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.


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).
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keep on seeking, and you will find;
keep on knocking, and it will be opened to you.


Last edited by HongHu on 14 Apr 2005, 20:08, edited 1 time in total.
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 [#permalink] New post 14 Apr 2005, 14:30
HongHu wrote:
gmat2me2 wrote:
DLMD wrote:
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


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

From 1)

x^3 - x^2 <0
x^2(x-1)<0

==>> 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.


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 be equal to zero either. So x must be less than zero. (C).


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
  [#permalink] 14 Apr 2005, 14:30
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