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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
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Is x^2 - x > 0?  [#permalink]

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Difficulty:   45% (medium)

Question Stats: 70% (01:42) correct 30% (01:27) wrong based on 106 sessions

### HideShow timer Statistics [GMAT math practice question]

Is $$x^2-x>0$$?

(1) $$x>0$$
(2) $$x^3+x>0$$

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Originally posted by MathRevolution on 30 Jan 2018, 03:14.
Last edited by Bunuel on 30 Jan 2018, 22:50, edited 1 time in total.
Edited the question and the OA.
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Re: Is x^2 - x > 0?  [#permalink]

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How is statement 1 sufficient? Given x= positive

If x= 1 then x^2-x = 0
and when x= 2, x^2-x >0
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Re: Is x^2 - x > 0?  [#permalink]

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1
MathRevolution wrote:
[GMAT math practice question]

Is $$x^2-x>0?$$

$$1) x>0$$
$$2) x^3+x>0$$

Is x^2 - x > 0 OR Is x(x-1) > 0
Well, x(x-1) will be greater than zero:-
Either when both x > 0 and x-1 > 0; which will happen when x > 1
OR when both x < 0 and x-1 < 0; which will happen when x < 0
So given condition will be true either if x > 1 or if x < 0

(1) x > 0 but we dont know if x > 1 or not. Not sufficient.

(2) x^3 + x > 0 Or x*(x^2+1) > 0
Now x^2+1 will always be positive, no matter what the value of x. The above condition will be thus true only when x > 0
So in effect we are given that x > 0 but we dont know whether x > 1 or not. Not sufficient.

Combining the two statements, still we get the same information that x > 0 but we dont know whether x > 1 Or 0 < x < 1. So we cant say whether x^2 - x > 0 or not.
Not sufficient.

(can you please check the OA, its mentioned as D)
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Re: Is x^2 - x > 0?  [#permalink]

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1
Can you please check the OA? Because if x is between 0 and 1 both inequalities are not satisfied
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Re: Is x^2 - x > 0?  [#permalink]

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srija9999 wrote:
Can you please check the OA? Because if x is between 0 and 1 both inequalities are not satisfied

The correct answer is E, not D. Edited. Thank you.
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Is x^2 - x > 0?  [#permalink]

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MathRevolution wrote:
[GMAT math practice question]

Is $$x^2-x>0$$?

(1) $$x>0$$
(2) $$x^3+x>0$$

we can use critical point method

critical points are x=0 or x=1

The question is valid according to the number line

............correct zone.....0......Wrong zone...1........correct zone........

(1) $$x>0$$

If x=$$\frac{1}{2}$$....Answer is No

According to above number line it is insufficient.

(2) $$x^3+x>0$$

If x=$$\frac{1}{2}$$....Answer is No

According to above number line it is insufficient.

Combine 1 & 2 using same examples above...No conclusive answer

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Re: Is x^2 - x > 0?  [#permalink]

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MathRevolution wrote:
[GMAT math practice question]

Is $$x^2-x>0$$?

(1) $$x>0$$
(2) $$x^3+x>0$$

Statement I:

$$x = \frac{1}{2}$$.. $$x^2-x<0$$
$$x =2$$.... $$x^2-x>0$$.... So, Insufficient.

statement II:

$$x(x^2+1) > 0$$... So, its basically saying $$x > 0$$.. Same as A.
So, Insuuficient.

Combining both we get,$$x> 0$$

So, E.
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Is x^2 - x > 0?  [#permalink]

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=>

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.

In inequality questions, the law “Question is King” tells us that if the solution set of the question includes the solution set of the condition, then the condition is sufficient

Modifying the question:
$$x^2-x>0$$
$$⇔ x(x-1) > 0$$
$$⇔ x < 0$$ or $$x > 1$$ by the “LLGG” rule.

Condition 1): $$x > 0$$
Since the solution set of the question does not include the solution set of condition 1), condition 1) is not sufficient.

Condition 2):
$$x^3+x>0$$
$$⇔ x(x^2+1)>0$$
$$⇔ x>0$$, since $$x^2+1 > 0$$ is always true.
Since the solution set of the question does not include that of the condition 2) either, this is not sufficient.

Condition 1) & 2):
The set satisfying both conditions together is $$x > 0$$.
Since the solution set of the question does not include that of both conditions together, they are not sufficient.

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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Originally posted by MathRevolution on 01 Feb 2018, 01:32.
Last edited by MathRevolution on 05 Feb 2018, 15:14, edited 1 time in total.
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Re: Is x^2 - x > 0?  [#permalink]

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MathRevolution wrote:
=>

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.

In inequality questions, the law “Question is King” tells us that if the solution set of the question includes the solution set of the condition, then the condition is sufficient

Modifying the question:
$$x^2-x>0$$
$$⇔ x(x-1) > 0$$
$$⇔ x < 0$$ or $$x > 1$$ by the “LLGG” rule.

Condition 1): $$x > 0$$
Since the solution set of the question includes the solution set of condition 1), condition 1) is sufficient.

Condition 2):
$$x^3+x>0$$
$$⇔ x(x^2+1)>0$$
$$⇔ x>0$$, since $$x^2+1 > 0$$ is always true.
Condition 2) is equivalent to the question, so it is sufficient.

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.

How first condition can be sufficient.

X>0, lets assume x is 1

X^2-x ---> 1-1 = 0

So when x is 1 we get the value as zero so statement A is insufficient

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Re: Is x^2 - x > 0?  [#permalink]

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MathRevolution wrote:
=>

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.

In inequality questions, the law “Question is King” tells us that if the solution set of the question includes the solution set of the condition, then the condition is sufficient

Modifying the question:
$$x^2-x>0$$
$$⇔ x(x-1) > 0$$
$$⇔ x < 0$$ or $$x > 1$$ by the “LLGG” rule.

Condition 1): $$x > 0$$
Since the solution set of the question includes the solution set of condition 1), condition 1) is sufficient.

Condition 2):
$$x^3+x>0$$
$$⇔ x(x^2+1)>0$$
$$⇔ x>0$$, since $$x^2+1 > 0$$ is always true.
Condition 2) is equivalent to the question, so it is sufficient.

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.

Not able to comprehend what you are trying to explain. In my approach I tried to fit values and found that both are insufficient hence E.

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Is $$x^2$$−x>0?

a) x>0
b) $$x^3$$+x>0
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Originally posted by Chethan92 on 09 Aug 2018, 19:55.
Last edited by chetan2u on 09 Aug 2018, 20:17, edited 1 time in total.
Updated topic name and OA
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Afc0892 wrote:
Is $$x^2$$−x>0?

a) x>0
b) $$x^3$$+x>0

$$x^2-x>0........x(x-1)>0$$
So if x>0, x-1>0, so X>1
If x<0, x-1<0, so X<0

Let us see the statements
a) X>0
X can be between 0 and 1, Ans is no...1^2-1=0 not >1
X>1 Ans is yes
Insufficient
b) x^3+X>0.......
X(x^2+1)>0
All values>0 will satisfy
If x<0, x^2+1<0, so x^2<-1 not possible
Hence x>0..
But same as above
Insufficient

Combined
0<X<1.....no
x>1.....Yes
Insufficient

E
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Afc0892 wrote:
Is $$x^2$$−x>0?

a) x>0
b) $$x^3$$+x>0

Re-phrasing question stem:-
$$x^2−x>0$$
Or,$$x(x-1)>0$$
Or, $$x<0$$ or $$x>1$$

Re-phrased question stem: Is $$x<0$$ or $$x>1$$ ?

St1:- x>0
Clearly insufficient.

St2:-$$x^3+x>0$$
Or, $$x(x^2+1)>0$$
$$(x^2+1)$$ is +ve for all real values.
So, x>0
Insufficient.

Combining, we have x>0.
Insufficient.

Ans. (E)

https://gmatclub.com/forum/is-x-2-x-258689.html?fl=similar
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Rise above the storm, you will find the sunshine Is x^2-x>0   [#permalink] 09 Aug 2018, 21:33
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