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Joined: 01 Feb 2011
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Re: MGMAT inequalities Question
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18 Mar 2011, 16:20
this can also be found by plugging in different numbers
1. Given x>0
x = 2 => (2^3)>(2^2) = (x^3)>(x^2)
x =0.5 => X^3<X^2
two contradicting results. so not sufficient.
2. X^2 >x x =2 => x^2>x. also X^3>x^2 x = 2 => x^2>x. but X^3>x^2
So not sufficient
together its says x^2 > x and x>0 (so no negative numbers and no fractions between 0 &1) sufficient to say x^3 > x^2. Hence answer is C.



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Re: Is x^3 > x^2?
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23 Apr 2015, 06:05
Hi Guys, The range of inequality can also be found out by using the 'Wavy Line' method. It's a powerful method of solving inequalities especially when you have trouble in finding the range algebraically. Let me illustrate briefly how to use this method. Finding Range of \(x^3  x^2 > 0\) through 'Wavy Line' method. We can simplify the expression to \(x^2 (x 1) > 0\). Now, the LHS of the inequality becomes 0 at 0 and 1. So, {0, 1} are the zero points of the inequality. Plot these points on the number line and start from the top right corner and draw a Wavy Line. Be ready to alternate the region of the wave based on how many times a a point is root to the equation. If the power of a term is odd, the wave simply passes through the point into the other region. If the power of a term is even, the wave bounces back to the same region. In this case, power of the term was 2 for the zero point 0, hence the wave bounced back. The range of the inequality will be +ve for region above the number line and ve for region below the number line. We observe from the graph that the range is +ve when \(x > 1\). So, all we need is to find from the statements I & II is if \(x >1\) or not. StatementIStI tells us that \(x > 0\). This does not tell us for sure if \(x >1\). Hence, the statement is insufficient. StatementII Finding Range of \(x^2 x >0\) through Wavy Line methodThe expression in stII can be simplified to \(x(x1) > 0\). Let's find the range by the Wavy Line method. The zero points of the expression are { 0,1}. The Wavy Line graph would be represented in the manner shown below: We see that the value of the inequality is +ve for \(x < 0\) and \(x > 1\). But the inequality in the question statement is +ve only when \(x > 1\). Hence, StatementII is insufficient. Combining StatementI & IIIf we combine both the statements, we can infer that \(x > 1\), which is what we needed. Hence, combining both the statements is sufficient to answer the question. You can practice a similar question at isx196185.html. Try solving it by Wavy Line method.
Also, an article for you on the Wavy Line method inequalitiestrick9148280.html#p1465609Hope it helps! Regards Harsh
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Re: Is x^3 > x^2?
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23 Apr 2015, 14:13
Hi All, This question can be solved by TESTing VALUES and taking advantage of the builtin Number Properties that exist in this question. We're asked if X^3 > X^2. This is a YES/NO question. Fact 1: X > 0 IF... X = 1 1^1 is NOT > 1^2 and the answer to the question is NO IF.... X = 2 2^3 IS > 2^2 and the answer to the question is YES Fact 1 is SUFFICIENT Fact 2: X^2 > X IF.... X = 2 2^3 IS > 2^2 and the answer to the question is YES IF.... X = 1 (1)^3 is NOT > (1)^2 and the answer to the question is NO Fact 2 is INSUFFICIENT Combined, we know.... X > 0 X^2 > X The combined Facts really 'limit' the possible values of X: Since X > 0, X MUST be positive. Since X^2 > X, X CANNOT be a positive fraction (between 0 and 1). X also CANNOT = 1. By extension, X MUST be > 1. In ALL of these situations, X^3 will be greater than X^2, so the answer to the question is ALWAYS YES. Combined, SUFFICIENT Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: Is x^3 > x^2?
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23 Apr 2015, 18:56
thesfactor wrote: Is x^3 > x^2?
(1) x > 0 (2) x^2 > x Statement 1: Tells us that x is bigger than 0, almost went with A... but then I remembered that x could be a fraction or 1, in which case x^3 would not be bigger than x^2 Statement 2: clears up the issue from statement 1 by telling us that x^2 is bigger than x, which means that x is not a fraction, and x is not 1 Hence why C is the answer.



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Re: Is x^3 > x^2?
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23 Apr 2015, 21:27
sabineodf wrote: thesfactor wrote: Is x^3 > x^2?
(1) x > 0 (2) x^2 > x Statement 1: Tells us that x is bigger than 0, almost went with A... but then I remembered that x could be a fraction or 1, in which case x^3 would not be bigger than x^2 Statement 2: clears up the issue from statement 1 by telling us that x^2 is bigger than x, which means that x is not a fraction, and x is not 1 Hence why C is the answer. Hi sabineodf, You were right to check about the properties of fractions, but you missed out on the properties of ve numbers. In case of a ve number, the even powers of the number would be greater than the odd powers. So, \(x^2 > x\) but \(x^3 < x^2\) For example take two cases for statementII: Case I: \(x= 2\) would give you \(x^2 > x\) and \(x^3 > x^2\) Case II: \(x = 2\) would give you \(x^2 > x\) but \(x^3 < x^2\) So,\(x^2 > x\) is possible for cases where \(x > 1\) or \(x < 0\)Hence, stII alone is not sufficient to give you a definite answer. Combining both the statements, you can definitely say that \(x\) is +ve number greater than 1. For a +ve number greater than 1, \(x^3 > x^2\). Hence the answer is C Hope this helps! Regards Harsh
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Re: Is x^3 > x^2?
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26 Jun 2018, 05:12
+1 for option C. The question is essentially asking if x>1 or not ? St 1  NS St 2  x<0 or x>1 Both together  Yes So option C it is !
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Re: Is x^3 > x^2?
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26 Jun 2018, 09:14
Bunuel wrote: seekmba wrote: Is \(x^3 > x^2\) ?
(1) \(x>0\) (2) \(x^2>x\) Is \(x^3>x^2\)? > is \(x^3x^2>0\)? > is \(x^2(x1)>0\)? > is \(x>1\)? (1) \(x>0\). Not sufficient. (2) \(x^2>x\) > \(x(x1)>0\) > either \(x<0\) or \(x>1\). Not sufficient. (1)+(2) Intersection of the ranges from (1) and (2) is the range \(x>1\). Sufficient. Answer: C. When you do x(x1) > 0 how do you get x<0 ? How can you divide 0 by x if you don't know whether it's a positive or negative? I know you could test x=1 so 1(11) = 1*2 which is > 0, but don't understand how you got x<0 without testing?




Re: Is x^3 > x^2? &nbs
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26 Jun 2018, 09:14



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