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# Is x^3 > x^2 ?

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Manager
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Is x^3 > x^2 ?  [#permalink]

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25 Feb 2017, 01:15
1
3
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Difficulty:

45% (medium)

Question Stats:

53% (01:00) correct 47% (00:58) wrong based on 75 sessions

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Is $$x^{3} > x^{2}$$ ?

1. x > 0
2. x < 1

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Joined: 02 Sep 2009
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Re: Is x^3 > x^2 ?  [#permalink]

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25 Feb 2017, 01:24
Is $$x^{3} > x^{2}$$ ?

Reduce by x^2 (which is non-negative): is x > 1? ($$x \neq 0$$)

(1) x > 0. Not sufficient.
(2) x < 1. Sufficient.

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Re: Is x^3 > x^2 ?  [#permalink]

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25 Feb 2017, 02:01
1
ashwink wrote:
Is $$x^{3} > x^{2}$$ ?

1. x > 0
2. x < 1

The solution given by Bunuel is THE BEST for this question but I am just mentioning a few rules for reduction for the help of readers

Rule 1: If there is something positive factor/variable on both sides of the Inequality sign then it can always be cancelled out like in this question x^2 can be cancelled from both sides as x^2 will always be positive

Rule 2: If there is something Negative factor/variable on both sides of the Inequality sign then it can always be cancelled out BUT Inequality sign Flips

Rule 3: If the sign of some factor/variable on both sides of the Inequality sign is UNKNOWN then NEVER cancel them out as one of the solutions of the variable may be "zero"

My Solution remains as mentioned by Bunuel (Saving my effort by copying the same

Reduce by x^2 (which is non-negative): is x > 1? (x≠0x≠0)

(1) x > 0. Not sufficient.
(2) x < 1. Sufficient.

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Re: Is x^3 > x^2 ?  [#permalink]

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25 Feb 2017, 03:23
Bunuel wrote:
Is $$x^{3} > x^{2}$$ ?

Reduce by x^2 (which is non-negative): is x > 1? ($$x \neq 0$$)

(1) x > 0. Not sufficient.
(2) x < 1. Sufficient.

Can we do
$$x^3-x^2>0$$
$$x^2(x-1)>0$$
and since $$x > 1$$ , also $$x^2 > 1$$
Hence, Statement 2 is sufficient.

or is it necessary to cancel out $$x^2$$ First ? because i am little confused about this concept ? please help explain in some other way, how can one judge when to cancel and when to not.
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Re: Is x^3 > x^2 ?  [#permalink]

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25 Feb 2017, 05:25
1
vabzgupta237 wrote:
Bunuel wrote:
Is $$x^{3} > x^{2}$$ ?

Reduce by x^2 (which is non-negative): is x > 1? ($$x \neq 0$$)

(1) x > 0. Not sufficient.
(2) x < 1. Sufficient.

.

Can we do
$$x^3-x^2>0$$
$$x^2(x-1)>0$$
and since $$x > 1$$ , also $$x^2 > 1$$
Hence, Statement 2 is sufficient.

or is it necessary to cancel out $$x^2$$ First ? because i am little confused about this concept ? please help explain in some other way, how can one judge when to cancel and when to not.

We can certainly do what you have suggested. In fact that's the best way to avoid any possible mistake

To know when we can cancel, please refer to the rules I have written.

If variable is positive then you can cancel the common variables from both sides.

A negative variable can be cancelled but the Inequality sign needs to change then

If nothing is mentioned about the sign of variable then never cancel it out because variable may have a solution 'zero' as well

I hope this helps!!!

Posted from my mobile device
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Re: Is x^3 > x^2 ?  [#permalink]

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25 Feb 2017, 07:06
GMATinsight wrote:
ashwink wrote:
Is $$x^{3} > x^{2}$$ ?

1. x > 0
2. x < 1

The solution given by Bunuel is THE BEST for this question but I am just mentioning a few rules for reduction for the help of readers

Rule 1: If there is something positive factor/variable on both sides of the Inequality sign then it can always be cancelled out like in this question x^2 can be cancelled from both sides as x^2 will always be positive

Rule 2: If there is something Negative factor/variable on both sides of the Inequality sign then it can always be cancelled out BUT Inequality sign Flips

Rule 3: If the sign of some factor/variable on both sides of the Inequality sign is UNKNOWN then NEVER cancel them out as one of the solutions of the variable may be "zero"

My Solution remains as mentioned by Bunuel (Saving my effort by copying the same

Reduce by x^2 (which is non-negative): is x > 1? (x≠0x≠0)

(1) x > 0. Not sufficient.
(2) x < 1. Sufficient.

Ok, Great !! So according to this, please correct me if i am wrong,

1. As $$x^2$$ is deemed to be always positive, we cancelled out $$x^2$$. Had it been $$x^5>X^4$$, we could have cancelled out $$X^4$$ also.

2. Another situtation, not this question, if it is given "x<0 or x is negative", we can cancel $$x^3$$ from $$x^6>x^3$$ and it would result in $$x^3<1$$.
and if nothing is given then $$x^3(x^3-1)>0$$ ? because both of the factors could be greater than zero.

Am i right ? Thanks for the first reply though.
Also, i have a Gmat in 10Days, could use some Extra help on verbal. Current scores ranging (Q49V29)660. SC AND RC are areas to be concerned about.
Manager
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Re: Is x^3 > x^2 ?  [#permalink]

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25 Feb 2017, 08:18
B

X^2 (x-1) >0
If u draw d graph sign changes at 1
So
Statement 1
Sign chng can't say
Insufficient
Statement 2
The equation is -ve for values of x less than 0 sufficient
So
B

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Re: Is x^3 > x^2 ?  [#permalink]

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27 Feb 2017, 05:47
1
vabzgupta237 wrote:
GMATinsight wrote:
ashwink wrote:
Is $$x^{3} > x^{2}$$ ?

1. x > 0
2. x < 1

The solution given by Bunuel is THE BEST for this question but I am just mentioning a few rules for reduction for the help of readers

Rule 1: If there is something positive factor/variable on both sides of the Inequality sign then it can always be cancelled out like in this question x^2 can be cancelled from both sides as x^2 will always be positive

Rule 2: If there is something Negative factor/variable on both sides of the Inequality sign then it can always be cancelled out BUT Inequality sign Flips

Rule 3: If the sign of some factor/variable on both sides of the Inequality sign is UNKNOWN then NEVER cancel them out as one of the solutions of the variable may be "zero"

My Solution remains as mentioned by Bunuel (Saving my effort by copying the same

Reduce by x^2 (which is non-negative): is x > 1? (x≠0x≠0)

(1) x > 0. Not sufficient.
(2) x < 1. Sufficient.

Ok, Great !! So according to this, please correct me if i am wrong,

1. As $$x^2$$ is deemed to be always positive, we cancelled out $$x^2$$. Had it been $$x^5>X^4$$, we could have cancelled out $$X^4$$ also.

2. Another situtation, not this question, if it is given "x<0 or x is negative", we can cancel $$x^3$$ from $$x^6>x^3$$ and it would result in $$x^3<1$$.
and if nothing is given then $$x^3(x^3-1)>0$$ ? because both of the factors could be greater than zero.

Am i right ? Thanks for the first reply though.
Also, i have a Gmat in 10Days, could use some Extra help on verbal. Current scores ranging (Q49V29)660. SC AND RC are areas to be concerned about.

Hi vabzgupta237

About Verbal, all I can say is 'Don't experiment" in last 10 days in your strategy and don't take any guidance in these last 10 days. It may only confuse you instead of making you more confident. Take exam with your best efficiency and once the test is over then we may discuss further what next to be done if you think that there is more scope of improvement in the test score.

I hope that helps!!!
_________________

Prosper!!!
GMATinsight
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e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

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Re: Is x^3 > x^2 ?  [#permalink]

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19 Apr 2018, 07:39
+1 for option B.

x^2(x-1)>0 --> x>1 as x^2 will always be greater than zero.

St 1 - NS
St 2 - Sufficient

Hence option B.
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Re: Is x^3 > x^2 ?  [#permalink]

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19 Apr 2018, 09:28
Bunuel wrote:
Is $$x^{3} > x^{2}$$ ?

Reduce by x^2 (which is non-negative): is x > 1? ($$x \neq 0$$)

(1) x > 0. Not sufficient.
(2) x < 1. Sufficient.

Since it is not mentioned that x not = 0, hence 2nd statement is not sufficient.
The answer should be C. The question has to be modified.
Senior Manager
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Re: Is x^3 > x^2 ?  [#permalink]

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19 Apr 2018, 09:54
Hero8888 wrote:
Bunuel wrote:
Is $$x^{3} > x^{2}$$ ?

Reduce by x^2 (which is non-negative): is x > 1? ($$x \neq 0$$)

(1) x > 0. Not sufficient.
(2) x < 1. Sufficient.

Since it is not mentioned that x not = 0, hence 2nd statement is not sufficient.
The answer should be C. The question has to be modified.

Not necessarily ... as long as we know that x<1 - irrespective of whether or not the value of x is zero , The inequality will be violated. We don't care of the inequality is violated because of LHS becoming less than RHS or equal to it. The answer must remain B itself !

Hope that helps !
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Senior Manager
Joined: 29 Dec 2017
Posts: 389
Location: United States
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WE: Marketing (Telecommunications)
Re: Is x^3 > x^2 ?  [#permalink]

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19 Apr 2018, 11:41
spetznaz wrote:
Hero8888 wrote:
Bunuel wrote:
Is $$x^{3} > x^{2}$$ ?

Reduce by x^2 (which is non-negative): is x > 1? ($$x \neq 0$$)

(1) x > 0. Not sufficient.
(2) x < 1. Sufficient.

Since it is not mentioned that x not = 0, hence 2nd statement is not sufficient.
The answer should be C. The question has to be modified.

Not necessarily ... as long as we know that x<1 - irrespective of whether or not the value of x is zero , The inequality will be violated. We don't care of the inequality is violated because of LHS becoming less than RHS or equal to it. The answer must remain B itself !

Hope that helps !

Let's modify the Q stem $$x^{3} > x^{2}$$ and we get $$x^{3} - x^{2} > 0?$$
Since x<1 , then x can be -1, 0, 1/2 etc . If x=0 then (0^3-0^2) = 0 and hence is NOT > or < 0, if x =1/2 then (1/2)^3-(1/2)^2<0 - Insufficient. On a real GMAT test it's going to be a trap. Otherwise the Q should be modified with constraints x≠0
Re: Is x^3 > x^2 ? &nbs [#permalink] 19 Apr 2018, 11:41
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