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# Is x^2 - x > 0?

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Math Revolution GMAT Instructor
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Is x^2 - x > 0?  [#permalink]

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Updated on: 30 Jan 2018, 22:50
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Difficulty:

55% (hard)

Question Stats:

68% (01:43) correct 33% (01:27) wrong based on 100 sessions

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[GMAT math practice question]

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

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

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"Only $99 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" 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. Manager Joined: 30 Dec 2016 Posts: 233 Schools: Schulich GMAT 1: 650 Q42 V37 GPA: 4 WE: Business Development (Other) Re: Is x^2 - x > 0? [#permalink] ### Show Tags 30 Jan 2018, 03:50 2 How is statement 1 sufficient? Given x= positive If x= 1 then x^2-x = 0 and when x= 2, x^2-x >0 _________________ Regards SandySilva ____________ Please appreciate the efforts by pressing +1 KUDOS (: DS Forum Moderator Joined: 22 Aug 2013 Posts: 1348 Location: India Re: Is x^2 - x > 0? [#permalink] ### Show Tags 30 Jan 2018, 11:17 3 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. Hence E answer (can you please check the OA, its mentioned as D) Intern Joined: 27 Jun 2017 Posts: 4 Re: Is x^2 - x > 0? [#permalink] ### Show Tags 30 Jan 2018, 22:41 1 Can you please check the OA? Because if x is between 0 and 1 both inequalities are not satisfied Math Expert Joined: 02 Sep 2009 Posts: 49999 Re: Is x^2 - x > 0? [#permalink] ### Show Tags 30 Jan 2018, 22:50 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. _________________ SVP Joined: 26 Mar 2013 Posts: 1836 Is x^2 - x > 0? [#permalink] ### Show Tags 31 Jan 2018, 06:48 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 If x=2...Answer is Yes According to above number line it is insufficient. (2) $$x^3+x>0$$ If x=$$\frac{1}{2}$$....Answer is No If x=2...Answer is Yes According to above number line it is insufficient. Combine 1 & 2 using same examples above...No conclusive answer Answer: E Senior Manager Joined: 31 Jul 2017 Posts: 477 Location: Malaysia Schools: INSEAD Jan '19 GMAT 1: 700 Q50 V33 GPA: 3.95 WE: Consulting (Energy and Utilities) Re: Is x^2 - x > 0? [#permalink] ### Show Tags 31 Jan 2018, 08:42 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. _________________ If my Post helps you in Gaining Knowledge, Help me with KUDOS.. !! Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6385 GMAT 1: 760 Q51 V42 GPA: 3.82 Is x^2 - x > 0? [#permalink] ### Show Tags Updated on: 05 Feb 2018, 15:14 => 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. Therefore, the answer is E. 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. Answer: E _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$99 for 3 month Online Course"
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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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01 Feb 2018, 01:39
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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01 Feb 2018, 08:24
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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Updated on: 09 Aug 2018, 20:17
1
Is $$x^2$$−x>0?

a) x>0
b) $$x^3$$+x>0
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Originally posted by Afc0892 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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09 Aug 2018, 20:16
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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09 Aug 2018, 21:33
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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Is x^2-x>0 &nbs [#permalink] 09 Aug 2018, 21:33
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