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Is x^3 > x^2 ? (1) x > 0 (2) x < 1 25 Feb 2017, 01:15
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Is x^3 > x^2 ?

(1) x > 0
(2) x < 1
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B
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Is x^3 > x^2 ? (1) x > 0 (2) x < 1 25 Feb 2017, 01:24
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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.

Answer: B.
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Is x^3 > x^2 ? (1) x > 0 (2) x < 1 25 Feb 2017, 02:01
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Is \(x^{3} > x^{2}\) ?

1. x > 0
2. x < 1
Show more

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.

Answer: B.
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Is x^3 > x^2 ? (1) x > 0 (2) x < 1 25 Feb 2017, 03:23
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Bunuel
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.

Answer: B.
Show more


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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Is x^3 > x^2 ? (1) x > 0 (2) x < 1 25 Feb 2017, 05:25
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Bunuel
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.

Answer: B.
.


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.
Show more

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!!!

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Is x^3 > x^2 ? (1) x > 0 (2) x < 1 25 Feb 2017, 07:06
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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.

Answer: B.
Show more



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.
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Is x^3 > x^2 ? (1) x > 0 (2) x < 1 25 Feb 2017, 08:18
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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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Is x^3 > x^2 ? (1) x > 0 (2) x < 1 27 Feb 2017, 05:47
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ashwink
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.

Answer: B.



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.
Show more

Hi vabzgupta237

You are absolutely right about both your points mentioned above.

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!!! :)
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Is x^3 > x^2 ? (1) x > 0 (2) x < 1 19 Apr 2018, 07:39
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+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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Is x^3 > x^2 ? (1) x > 0 (2) x < 1 19 Apr 2018, 09:28
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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.

Answer: B.
Show more

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.
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Is x^3 > x^2 ? (1) x > 0 (2) x < 1 19 Apr 2018, 09:54
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Bunuel
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.

Answer: B.

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.
Show more

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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Is x^3 > x^2 ? (1) x > 0 (2) x < 1 19 Apr 2018, 11:41
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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.

Answer: B.

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 !
Show more

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
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Is x^3 > x^2 ? (1) x > 0 (2) x < 1 03 Dec 2019, 05:59
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Is \(x^{3} > x^{2}\) ?

1. x > 0
2. x < 1
Show more

1. x > 0 insufic

\(x=1:x^{3} > x^{2}…1=1…ans=no\)
\(x=2:x^{3} > x^{2}…8>4…ans=yes\)

2. x < 1 sufic

\(x=1/2:x^{3} > x^{2}…1/8<1/4…ans=no\)
\(x=-1/2:x^{3} > x^{2}…-1/8<1/4…ans=no\)
\(x=-2:x^{3} > x^{2}…-8<4…ans=no\)
\(x=0:x^{3} > x^{2}…0=0…ans=no\)

Ans (B)
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Is x^3 > x^2 ? (1) x > 0 (2) x < 1 14 Oct 2023, 02:12
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