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Is x greater than x^3? (1) x is negative. (2) x^2 - x^3 > 2

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Joined: 02 Sep 2009
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Is x greater than x^3? (1) x is negative. (2) x^2 - x^3 > 2  [#permalink]

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New post 19 Jan 2018, 10:57
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A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

55% (02:15) correct 45% (01:52) wrong based on 42 sessions

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Is x greater than x^3? (1) x is negative. (2) x^2 - x^3 > 2  [#permalink]

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New post 19 Jan 2018, 19:56
Bunuel wrote:
Is x greater than x^3?

(1) x is negative.
(2) x^2 - x^3 > 2



\(x>x^3.......x^3-x<0.....x(x^2-1)<0\)....
Two cases possible..
1) If x>0...
\(x^2-1<0....x^2<1\)
So 0<x<1
2) If x<0.....
\(x^2-1>0.....x^2>1\)
So x<-1

so a definite YES answer when either 0<x<1 or x<-1

Let's see statement
I...x is NEGATIVE
But if X is between 0 and -1, Ans is no
x<-1 yes
Insufficient

II..\(x^2-x^3>2\)
\(x^3-x^2<-2.....x^2(x-1)<-2\)
Two case
1) x>0..... Not possible
2) x<0... The moment X moves towards -1, the equation does not hold..
So x<-1 CONDITION II
Ans is always YES...
Suff

B
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Is x greater than x^3? (1) x is negative. (2) x^2 - x^3 > 2  [#permalink]

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New post 19 Jan 2018, 20:10
Bunuel wrote:
Is x greater than x^3?

(1) x is negative.
(2) x^2 - x^3 > 2


Statement 1: if \(x=-0.1\) then \(x^3=-0.001\) so we have \(x^3>x\).

but if \(x=-1\), then \(x^3=-1\) so we have \(x=x^3\)

and if \(x=-2\), then \(x^3=-8\), so we have \(x>x^3\). Since we have multiple values. Insufficient

Statement 2: as \(x^2-x^3\) is positive, so if \(x>1\) then \(x^3>x^2=>x^2-x^3<0\) which is not possible.

Also if \(-1≤x≤1\), then \(x^2-x^3\) will not be greater than \(2\) (you can test the boundary conditions)

Hence \(x<-1\) which implies \(x>x^3\). Sufficient

Option B
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Re: Is x greater than x^3? (1) x is negative. (2) x^2 - x^3 > 2  [#permalink]

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New post 22 Jan 2018, 23:05
Bunuel wrote:
Is x greater than x^3?

(1) x is negative.
(2) x^2 - x^3 > 2



Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the question.

We can modify the question as follows.
\(x > x^3\)
\(⇔ x^3 - x < 0\)
\(⇔ x(x^2-1) < 0\)
\(⇔ x(x+1)(x-1) < 0\)
\(⇔ x < -1\) or \(0 < x < 1\) from the graph of \(y = x(x+1)(x-1)\)

Condition 1)
\(x < 0\)
Since the range of the question does not include that of the condition 1), this is not sufficient.

Condition 2)
\(x^2 - x^3 > 2\)
\(⇔ x^3 - x^2 + 2 < 0\)
\(⇔ (x+1)(x^2 - 2x + 2 ) < 0\)
\(⇔ x + 1 < 0\) since \(x^2-2x+2 = (x-1)^2 + 1 ≥ 1 > 0\).
\(⇔ x < -1\)
Since the range of the question includes that of the condition 2), this is sufficient.

Therefore, the answer is B.

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.
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Re: Is x greater than x^3? (1) x is negative. (2) x^2 - x^3 > 2 &nbs [#permalink] 22 Jan 2018, 23:05
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