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Is x between 0 and 1?

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Is x between 0 and 1?  [#permalink]

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New post 06 Mar 2012, 16:03
1
4
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

55% (01:33) correct 45% (01:51) wrong based on 122 sessions

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Is x between 0 and 1?

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

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Re: Is x between 0 & 1?  [#permalink]

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New post 06 Mar 2012, 16:08
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enigma123 wrote:
Is x between 0 and 1?

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

How come statement 1 is not sufficient to answer the question guys?


Is \(0<x<1\)?

(1) -x<x^3 --> \(x^3+x>0\) --> \(x(x^2+1)>0\) --> \(x>0\) (as x^2+1 is always positive). Not sufficient.

(2) x<x^2 --> \(x^2-x>0\) --> \(x(x-1)>0\) --> either \(x>1\) or \(x<0\), so the answer is NO. Sufficient.

Answer: B.

Solving inequalities:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.
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Re: Is x between 0 & 1?  [#permalink]

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New post 06 Mar 2012, 16:13
\(x<0\)

Shouldn't be x>0 because x(x-1) > 0?
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Re: Is x between 0 & 1?  [#permalink]

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New post 06 Mar 2012, 16:19
enigma123 wrote:
\(x<0\)

Shouldn't be x>0 because x(x-1) > 0?


x(x-1) > 0 --> roots are 0 and 1 --> ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: x<0 or x>1.

Solving inequalities:
x2-4x-94661.html#p731476 (check this one first)
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.
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Re: Is x between 0 and 1 ?  [#permalink]

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New post 10 May 2012, 20:42
St1:
-x<-x^3
or x>-x^3 =>holds true for x=2 and holds true for 1/2....not suff.
st2:
x<x^2=>x(x-1)>0 =>x<0 OR x>1.....therefore x is DEFINITELY NOT between 0 and 1...suffecient

Answer : B
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Re: Is x between 0 & 1?  [#permalink]

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New post 10 Jun 2012, 07:16
Bunuel wrote:
enigma123 wrote:
\(x<0\)

Shouldn't be x>0 because x(x-1) > 0?


x(x-1) > 0 --> roots are 0 and 1 --> ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: x<0 or x>1.

Solving inequalities:
x2-4x-94661.html#p731476 (check this one first)
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.



I don't understand this either. Is there some rule that explains this?
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Re: Is x between 0 & 1?  [#permalink]

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New post 10 Jun 2012, 08:47
Stiv wrote:
Bunuel wrote:
enigma123 wrote:
\(x<0\)

Shouldn't be x>0 because x(x-1) > 0?


x(x-1) > 0 --> roots are 0 and 1 --> ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: x<0 or x>1.

Solving inequalities:
x2-4x-94661.html#p731476 (check this one first)
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.



I don't understand this either. Is there some rule that explains this?


There is indeed Stiv. That is why I gave the links to the threads explaining solution provided.

Once more, solving inequalities:
x2-4x-94661.html#p731476 (check this one first)
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863
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Re: Is x between 0 & 1?  [#permalink]

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New post 10 Jun 2012, 11:36
Hi Stiv,

This is for your reference, might come handy:
Quote:
Hi,

A better approach is to check the sign of expression on a number line.

for example: (x-1)(x-2)(x-3)(x-7) < 0

To check the intervals in which this inequality holds true, we need to pick only one value from the number line.
Lets say x = 10, then (9)(8)(7)(3) > 0, in every alternate interval the sign would be + for the above expression

---(+)-----1--(-)--2---(+)--3----(-)-------7----(+)------

Thus, inequality would hold true in the intervals:
1 < x < 2
3 < x < 7

This is the general approach which can be used when you see multiplications in inequalities.

Regards,

Stiv wrote:
Bunuel wrote:
enigma123 wrote:
\(x<0\)

Shouldn't be x>0 because x(x-1) > 0?


x(x-1) > 0 --> roots are 0 and 1 --> ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: x<0 or x>1.

Solving inequalities:
x2-4x-94661.html#p731476 (check this one first)
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.



I don't understand this either. Is there some rule that explains this?
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Re: Is x between 0 & 1?  [#permalink]

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New post 17 Jul 2012, 08:10
Bunuel

I did not understand this part

Statement 2 hovv did they both get opposite signs




Bunuel wrote:
enigma123 wrote:
Is x between 0 and 1?

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

How come statement 1 is not sufficient to answer the question guys?


Is \(0<x<1\)?

(1) -x<x^3 --> \(x^3+x>0\) --> \(x(x^2+1)>0\) --> \(x>0\) (as x^2+1 is always positive). Not sufficient.

(2) x<x^2 --> \(x^2-x>0\) --> \(x(x-1)>0\) --> either \(x>1\) or \(x<0\), so the answer is NO. Sufficient.

Answer: B.

Solving inequalities:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.
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Re: Is x between 0 & 1?  [#permalink]

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New post 17 Jul 2012, 08:16
venmic wrote:
Bunuel

I did not understand this part

Statement 2 hovv did they both get opposite signs




Bunuel wrote:
enigma123 wrote:
Is x between 0 and 1?

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

How come statement 1 is not sufficient to answer the question guys?


Is \(0<x<1\)?

(1) -x<x^3 --> \(x^3+x>0\) --> \(x(x^2+1)>0\) --> \(x>0\) (as x^2+1 is always positive). Not sufficient.

(2) x<x^2 --> \(x^2-x>0\) --> \(x(x-1)>0\) --> either \(x>1\) or \(x<0\), so the answer is NO. Sufficient.

Answer: B.

Solving inequalities:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.


Explained here:
x2-4x-94661.html#p731476 (check this one first)
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863
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Is x between 0 and 1?  [#permalink]

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New post 21 Oct 2015, 03:42
enigma123 wrote:
Is x between 0 and 1?

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


Required: Is 0 < x < 1

Statement 1: \(-x < x^3\)
On rearranging, we have
\(x^3 + x > 0\)
\(x(x^2 + 1) > 0\)
Since a perfect square is always positive, hence adding 1 to a square will also be positive
Hence we have w > 0
From this information, w can be between 0 and 1 and can be greater than 1

INSUFFICIENT

Statement 2: \(x < x^2\)
\(x^2 - x >0\)
\(x(x - 1) > 0\)

Hence the possible values of x are x < 0 or x >1
We can safely say that x does not lie between 0 and 1

SUFFICIENT

Option B
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Is x between 0 and 1?  [#permalink]

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New post 25 Oct 2015, 12:09
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is x between 0 and 1?

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

There is only 1 variable (x), and 2 equations are given so there is high chance (D) will be our answer.
If the range of the question includes that of the condition, the condition is sufficient. We can use this to solve questions quickly and accurately.
The question asks whether 0<x<1
From condition 1, x^3+x>0, x(x^2+1)>0 (as x^2+1>0). From this we get x>0. This is insufficient as the range of the question does not include this range.
From condition 2, x^2-x>0 x(x-1)>0, --> x<0 or 1<x. The condition answers the question 'no', so this is sufficient.
The answer therefore becomes (B).

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Re: Is x between 0 and 1?  [#permalink]

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Re: Is x between 0 and 1?   [#permalink] 26 Mar 2019, 19:57
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