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Is x^2  x > 0?
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Updated on: 30 Jan 2018, 21:50
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[GMAT math practice question] Is \(x^2x>0\)? (1) \(x>0\) (2) \(x^3+x>0\)
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Originally posted by MathRevolution on 30 Jan 2018, 02:14.
Last edited by Bunuel on 30 Jan 2018, 21:50, edited 1 time in total.
Edited the question and the OA.



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Re: Is x^2  x > 0?
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30 Jan 2018, 02:50
How is statement 1 sufficient? Given x= positive If x= 1 then x^2x = 0 and when x= 2, x^2x >0
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Re: Is x^2  x > 0?
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30 Jan 2018, 10:17
MathRevolution wrote: [GMAT math practice question]
Is \(x^2x>0?\)
\(1) x>0\) \(2) x^3+x>0\) Is x^2  x > 0 OR Is x(x1) > 0 Well, x(x1) will be greater than zero: Either when both x > 0 and x1 > 0; which will happen when x > 1 OR when both x < 0 and x1 < 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)



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Re: Is x^2  x > 0?
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30 Jan 2018, 21:41
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?
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30 Jan 2018, 21:50



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Is x^2  x > 0?
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31 Jan 2018, 05:48
MathRevolution wrote: [GMAT math practice question]
Is \(x^2x>0\)?
(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



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Re: Is x^2  x > 0?
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31 Jan 2018, 07:42
MathRevolution wrote: [GMAT math practice question]
Is \(x^2x>0\)?
(1) \(x>0\) (2) \(x^3+x>0\) Statement I: \(x = \frac{1}{2}\).. \(x^2x<0\) \(x =2\).... \(x^2x>0\).... 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?
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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: \(x^2x>0\) \(⇔ x(x1) > 0\) \(⇔ x < 0\) 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
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Re: Is x^2  x > 0?
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01 Feb 2018, 00:39
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: \(x^2x>0\) \(⇔ x(x1) > 0\) \(⇔ x < 0\) 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. Therefore, the answer is D.
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: D How first condition can be sufficient. X>0, lets assume x is 1 X^2x > 11 = 0 So when x is 1 we get the value as zero so statement A is insufficient Sent from my Redmi Note 3 using GMAT Club Forum mobile app



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Re: Is x^2  x > 0?
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01 Feb 2018, 07:24
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: \(x^2x>0\) \(⇔ x(x1) > 0\) \(⇔ x < 0\) 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. Therefore, the answer is D.
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: D 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. Please explain Stay Hungry Stay Foolish



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Is x^2x>0
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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, 18:55.
Last edited by chetan2u on 09 Aug 2018, 19:17, edited 1 time in total.
Updated topic name and OA



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Re: Is x^2x>0
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09 Aug 2018, 19:16
Afc0892 wrote: Is \(x^2\)−x>0?
a) x>0 b) \(x^3\)+x>0 \(x^2x>0........x(x1)>0\) So if x>0, x1>0, so X>1 If x<0, x1<0, so X<0 Let us see the statements a) X>0 X can be between 0 and 1, Ans is no...1^21=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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Is x^2x>0
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09 Aug 2018, 20:33
Afc0892 wrote: Is \(x^2\)−x>0?
a) x>0 b) \(x^3\)+x>0 Rephrasing question stem: \(x^2−x>0\) Or,\(x(x1)>0\) Or, \(x<0\) or \(x>1\) Rephrased question stem: Is \(x<0\) or \(x>1\) ? St1: x>0Clearly 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/isx2x258689.html?fl=similarAlready discussed in forum.
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