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Is x > 0? [#permalink]
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09 Nov 2004, 01:41
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Is x > 0? (1) x^3 < x^2 (2) x^3 < x^4
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Last edited by Bunuel on 27 Feb 2014, 10:15, edited 1 time in total.
Renamed the topic, edited the question and added the OA.



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I would answer E, X^2 or X^4 are always positive whatever X sign



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ans D.
 (x^3/x^2)>1(diving by x^2 on both sides,so equality changes)
similiarly (x^4/x^3)>1..
from either of them it can e inferred if x>0.
am i correct?what is the OA?
Please correct me if i'm wrong.
P
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Re: DS: x>0? [#permalink]
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09 Nov 2004, 04:23
afife76 wrote: X>0? 1)X^3 < X^2 2)X^3 < X^4
The answer is C
[1] Tells us 0 < x < 1 in which case x > 0
Or 0 > x Insuff
[2] Tells us x > 1 in which case x > 0
Or 0 > x Insuff
Only 0 > x satisfies both statements so x is NOT greater than 0



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You are right epiphany i underestimated this one, though I am quite sure I already solved it.
Answer is C



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Epiphany is absolutely right:
1. x<1
2. x>1 OR x<0
Neither one is sufficient by itself, but if we take the intersection, we can see that x<0
Nice one... Thx afife76



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Re: DS:x>0 [#permalink]
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10 Nov 2004, 06:10
itsprem wrote: ans D.  (x^3/x^2)>1(diving by x^2 on both sides,so equality changes) similiarly (x^4/x^3)>1..
Prem the inequality does not change when you multiply or divide by a positive number. It only changes when you multiply or divide by a negative number or when you take an inverse. In this case x^2 will always be positive so the direction of inequality does not change.



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Re: DS: x>0? [#permalink]
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01 May 2005, 14:18
OA should be C.
from i, if x=5, x^3<x^2 holds.
if x=0.5, x^3<x^2 also holds.
from ii, if x=5, x^3<x^4 holds.
if x=5, x^3<x^4 also holds.
from i and ii, x=ve.



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Re: DS: x>0? [#permalink]
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02 May 2005, 23:56
afife76 wrote: X>0? 1)X^3 < X^2 2)X^3 < X^4
My answer is B.
From 1 x^3/X^2 < 1 or X < 1. This implies that X could be greater than or equal to zero or X could be negative. Hence statement 1 is not sufficient.
From 2 X^3/X^4 < 1 => 1/x < 1 ==> 1 < X or X > 1. If x > 1 then X should be greater than zero. Hence Statement 2 is alone sufficient to answer the question.
Please let me know if I am right.
Thanks



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Re: DS: x>0? [#permalink]
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03 May 2005, 22:37
MA wrote: krisrini wrote: From 2, X^3/X^4 < 1 => 1/x < 1 ==> 1 < X or X > 1. If x > 1 then X should be greater than zero. Hence Statement 2 is alone sufficient to answer the question. how about if x = ve?
MA, From statement 2, we have
1/x < 1 , then 1 < X
Or simplifying statement 2 we get X > 1 . So how can x be ve. Am I making a mistake somewhere, please let me know.
Thanks



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Re: DS: x>0? [#permalink]
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03 May 2005, 23:08
krisrini wrote: MA wrote: krisrini wrote: From 2, X^3/X^4 < 1 => 1/x < 1 ==> 1 < X or X > 1. If x > 1 then X should be greater than zero. Hence Statement 2 is alone sufficient to answer the question. how about if x = ve? MA, From statement 2, we have 1/x < 1 , then 1 < X Or simplifying statement 2 we get X > 1 . So how can x be ve. Am I making a mistake somewhere, please let me know.Thanks
ii. X^3 < X^4.
if x = +ve, there is no problem.
but x can be negative. suppose x = 5, the inequality X^3 < X^4 holds but x is not more than 0.



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C
A. x^3 < x^2
=> x^2(x1) <0
=> x1 < 0 since x^2 is always +ve
=> x< 1
x could be 5 or 0.5 so A is insufficient
B. x^4 > x^3
Simplifying we get
x(x1) < 0 since x^2 is always +ve
this means either 0<x<1 or x>1 and x<0
So B is insufficient.
C.
using both 0<x<1
because x>1 and x<1 cannot happen at the same time
C is the answer
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Re: DS: x>0? [#permalink]
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06 May 2005, 15:52
krisrini wrote: MA wrote: krisrini wrote: From 2, X^3/X^4 < 1 => 1/x < 1 ==> 1 < X or X > 1. If x > 1 then X should be greater than zero. Hence Statement 2 is alone sufficient to answer the question. how about if x = ve? MA, From statement 2, we have 1/x < 1 , then 1 < X Or simplifying statement 2 we get X > 1 . So how can x be ve. Am I making a mistake somewhere, please let me know. Thanks
I had to really think hard why you weren't getting it right.
Simplifying you got
1/x < 1
Now remember x could be +ve or ve.
The inequation will change signs if x is ve.
if x>0
1/x < 1 => 1<x
so finally
x>0 and x>1
if x<0
1/x <1 => 1>x
so finally
x<0 and x < 1
Looks like i screwed up.....its not the same as what i got earlier
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x = 1/2 , 2 , 1/2, 2
x^2= 1/4, 4, 1/4, 4
x^3 = 1/8, 8, 1/8, 8
x^4 = 1/16, 16, 1/16, 16
(1)when x = 1/2 or 2, x^3 < x^2, insufficient
(2) when x=2 or 2, x^3 < x^4, insufficient
when x=2 or 1/2, x^3 < x^2 and x^3 < x^4 at the same time
C is sufficient.



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I'll pick B
i)
X^3 < X^2
=> X < 1
doesn't say X<0 or not ... insufficient
ii) X^4 > X^3
=> X > 1 => X>0 ... sufficient
What's the OA?



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sonaketu wrote: I'll pick B
i) X^3 < X^2 => X < 1
doesn't say X<0 or not ... insufficient
ii) X^4 > X^3 => X > 1 => X>0 ... sufficient
What's the OA?
In B put x=5......and the inequality still holds true.
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Re: DS: x>0? [#permalink]
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08 May 2005, 19:18
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Remember, if you can't determine the sign, you CAN'T multiply or divide both side of an equation or ineqaulity by an algebra expression.
1)X^3 < X^2
x^3x^2<0
x^2*(x1)<0
Since x^2 can't be less than zero we get x<1. Not sufficient to determine whether x<0.
2)X^3 < X^4
x^3x^4<0
x^3*(1x)<0
x^3<0 or 1x<0
x<0 or x>1
Not sufficient to determine whether x<0.
Combined, we know that only when x<0 both inequalities are satisfied. So sufficient.
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Re: DS: x>0? [#permalink]
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31 May 2005, 20:15
Thanks ashkg and MA, I correct my mistake. Thanks much for pointing it out. I loved the solution from Honghu, her solution was very professional. Sorry for the late replies, I was on a vacation.



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Re: Is x > 0? [#permalink]
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27 Feb 2014, 10:16



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Re: Is x > 0? [#permalink]
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13 Jul 2016, 20:18
Statement 1 tells us that: a. x is negative (or) b. 0<x<1
Statement 2 tells us that: a. x is negative (or) b. x is positive AND x>1
Combining the 2 statements, x is negative. Therefore, x < 0, which is a definite answer to the problem.
Answer is C.







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