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Is x^3 > x^2?

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Re: MGMAT inequalities Question  [#permalink]

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New post 18 Mar 2011, 17:20
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?  [#permalink]

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New post 23 Apr 2015, 07:05
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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.

Image

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.

Statement-I
St-I tells us that \(x > 0\). This does not tell us for sure if \(x >1\). Hence, the statement is insufficient.

Statement-II- Finding Range of \(x^2 -x >0\) through Wavy Line method

The expression in st-II can be simplified to \(x(x-1) > 0\). 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:

Image

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, Statement-II is insufficient.

Combining Statement-I & II
If 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 is-x-196185.html. Try solving it by Wavy Line method.

Also, an article for you on the Wavy Line method inequalities-trick-91482-80.html#p1465609


Hope it helps!

Regards
Harsh
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Re: Is x^3 > x^2?  [#permalink]

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New post 23 Apr 2015, 15:13
Hi All,

This question can be solved by TESTing VALUES and taking advantage of the built-in 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

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Re: Is x^3 > x^2?  [#permalink]

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New post 23 Apr 2015, 19:56
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?  [#permalink]

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New post 23 Apr 2015, 22:27
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 statement-II:

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, st-II 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?  [#permalink]

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New post 26 Jun 2018, 06:12
+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?  [#permalink]

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New post 26 Jun 2018, 10:14
Bunuel wrote:
seekmba wrote:
Is \(x^3 > x^2\) ?

(1) \(x>0\)
(2) \(x^2>x\)


Is \(x^3>x^2\)? --> is \(x^3-x^2>0\)? --> is \(x^2(x-1)>0\)? --> is \(x>1\)?

(1) \(x>0\). Not sufficient.

(2) \(x^2>x\) --> \(x(x-1)>0\) --> 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(x-1) > 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(-1-1) = -1*-2 which is > 0, but don't understand how you got x<0 without testing?
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Re: Is x^3 > x^2? &nbs [#permalink] 26 Jun 2018, 10:14

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