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X<0? 1) X^3<X^2 2) X^3<X^4 [#permalink]
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12 Apr 2005, 16:13
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X<0?
1) X^3<X^2
2) X^3<X^4



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Re: DS guys!! [#permalink]
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12 Apr 2005, 16:21
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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one more for C
Same explanation as kaplock's



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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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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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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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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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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 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. .



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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 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. .
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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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...



Director
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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(x1)<0
==>> x^2<0 and (x1)>0 (Since both of them cannot be negative)
or x^2>0 and (x1)<0
So you cannot tell anything about x
from 2)
x^3(1x)<0
Again the same logic
x^3<0 and (1x)>0
or
x^3>0 and (1x)<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]
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13 Apr 2005, 21:47
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]
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14 Apr 2005, 14:38
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;
keep on knocking, and it will be opened to you.



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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(x1)<0 ==>> x^2<0 and (x1)>0 (Since both of them cannot be negative) or x^2>0 and (x1)<0 So you cannot tell anything about x from 2) x^3(1x)<0 Again the same logic x^3<0 and (1x)>0 or x^3>0 and (1x)<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, 21:08, edited 1 time in total.



Director
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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(x1)<0 ==>> x^2<0 and (x1)>0 (Since both of them cannot be negative) or x^2>0 and (x1)<0 So you cannot tell anything about x from 2) x^3(1x)<0 Again the same logic x^3<0 and (1x)>0 or x^3>0 and (1x)<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










