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Is x > 0?

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Is x > 0?  [#permalink]

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New post Updated on: 27 Feb 2014, 11:15
3
10
00:00
A
B
C
D
E

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  95% (hard)

Question Stats:

46% (01:14) correct 54% (01:06) wrong based on 411 sessions

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Is x > 0?

(1) x^3 < x^2

(2) x^3 < x^4

Originally posted by afife76 on 09 Nov 2004, 02:41.
Last edited by Bunuel on 27 Feb 2014, 11:15, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: DS: x>0?  [#permalink]

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New post 08 May 2005, 20:18
4
1
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^3-x^2<0
x^2*(x-1)<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^3-x^4<0
x^3*(1-x)<0
x^3<0 or 1-x<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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New post 09 Nov 2004, 03:32
1
I would answer E, X^2 or X^4 are always positive whatever X sign
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DS:x>0  [#permalink]

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New post 09 Nov 2004, 03:49
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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New post 09 Nov 2004, 05:23
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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New post 09 Nov 2004, 05:34
You are right epiphany i underestimated this one, though I am quite sure I already solved it.

Answer is C
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New post 09 Nov 2004, 08:16
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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New post 10 Nov 2004, 07:10
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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New post 01 May 2005, 15:18
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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New post 03 May 2005, 00:56
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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New post 03 May 2005, 23:37
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.

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Re: DS: x>0?  [#permalink]

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New post 04 May 2005, 00:08
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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New post 06 May 2005, 16:37
2
C

A. x^3 < x^2

=> x^2(x-1) <0
=> x-1 < 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(x-1) < 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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New post 06 May 2005, 16:52
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 :shock:
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New post 06 May 2005, 20:01
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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New post 07 May 2005, 10:10
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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New post 07 May 2005, 11:46
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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New post 31 May 2005, 21:15
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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New post 27 Feb 2014, 11:16
afife76 wrote:
Is x > 0?

(1) x^3 < x^2

(2) x^3 < x^4



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Re: Is x > 0?  [#permalink]

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New post 13 Jul 2016, 21:18
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.
Re: Is x > 0? &nbs [#permalink] 13 Jul 2016, 21:18

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