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# Is x^3 > x^2? (1) x > 0 (2) x^2 >x Isnt statement 1

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Senior Manager
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Is x^3 > x^2? (1) x > 0 (2) x^2 >x Isnt statement 1 [#permalink]  22 Sep 2011, 00:18
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Question Stats:

74% (01:34) correct 26% (00:34) wrong based on 27 sessions
Is $$x^3 > x^2?$$
(1) x > 0
(2) $$x^2 >x$$

[Reveal] Spoiler: My Doubt :
Isnt statement 1 sufficient and statement 2 sufficient?? thus shouldnt the ans be D??? but solution says c

If x > 0 then isnt $$x^3 > x^2$$since x is positive.cube is greater than square --- thus making 1 sufficient

also, same applies for 2 ... since$$x^2>x$$ $$x^3>x^2$$

Can someone clarify my doubt ??
[Reveal] Spoiler: OA
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Re: Is x^3 > x^2? [#permalink]  22 Sep 2011, 00:33
1
KUDOS
statement 1 says, x > 0 . This is not sufficient to know whether x^3 >x^2. suppose x =1/2, then x^2 = 1/4 and x^3 = 1/8 , so as 1/4 is greater than 1/8, x^2 > x ^3. but if x =2. then x^3 > X^2. so this statement is insufficient.

statement 2 says x ^ 2 > x. this means either x is a negative number or x is positive number more than 1. Essentially x can have any value except 0 to +1. so this is not enough to tell whether x ^3 is greater than x ^2 or not.

combining both statement we know that x is positive number and x is more than 1. so it sufficiently answers that x^3 > x^2. Any number more than 1 will have its value raised to power of 3 more than its value raised to power of 2.

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Re: Is x^3 > x^2? [#permalink]  22 Sep 2011, 00:39
Siddhans ,

Regarding your second doubt . your wrote " same applies for 2 ... sincex^2>x x^3>x^2"

No, X^2 > X does not necessarily means x^3 will be greater than X^2 and so on...suppose x= -2 (minus 2) then x^2 = +4 which means x^2 >x but x^3 will be -8 which is not more than +4. so you have to consider +- and fraction value while solving such questions.
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Re: Is x^3 > x^2? [#permalink]  22 Sep 2011, 01:19
1
KUDOS
Aj85 wrote:
Siddhans ,

Regarding your second doubt . your wrote " same applies for 2 ... sincex^2>x x^3>x^2"

No, X^2 > X does not necessarily means x^3 will be greater than X^2 and so on...suppose x= -2 (minus 2) then x^2 = +4 which means x^2 >x but x^3 will be -8 which is not more than +4. so you have to consider +- and fraction value while solving such questions.

Can we solve this algebrically
Question says$$x^3 > x^2$$ ?

Thus we can reduce this to $$x^3 - x^2$$ > 0?

$$X^2 (x-1)> 0?$$------ (1)

Now this equation 1 of of the form xy >0 so either x and y are both positive or both negative ...

since x^2 is always +ve this x-1 is also positive

so equation 1 become x> 0 or x> 1

thus the question becomes is x >1 ??

Now lets come to the statements

1) x>0

This just tells us x> 0 but we dont know if x > 1 ---- Insufficient

2) $$x^2 > x$$

$$X^2 - x > 0$$

thus, x (x-1) >0

x and (x-1) must have same sign either both positive or both negative

Case 1 : x and x-1 both positive

x > 0 or x-1 >0

x> 0 or x > 1

thus from case 1 we can conclude x>1

Case 2: x and x-1 both negative

x<0 or x-1<0

x<0 or x<1

thus x < 0 from case 2

combining case 1 and 2 we get :

i dont know how to proceed ahead after this... can someone help???
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Re: Is x^3 > x^2? [#permalink]  22 Sep 2011, 01:34
1
KUDOS
+1 , you have solved it correctly using algebra , can't imagine how u got it wrong first time itself.

from what you solved through equation 2. you get x >1 or x <0, so its it. on a number line x is either right of +1 or left of 0. from this we cant conclude anything. if x is more than 1, x^3 > x^2 but if x is less than 0, then x^2 > x^3. so this statement is insufficient.

But once we combine statement 1 and 2 we get to know x is positive and more than 1. positive value mentioned in statement 1 rules out one of the possibility of value of x we get from statement 2. so only 1 value remains that is x >1. sufficient.
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Re: Is x^3 > x^2? [#permalink]  22 Sep 2011, 01:54
Aj85 wrote:
+1 , you have solved it correctly using algebra , can't imagine how u got it wrong first time itself.

from what you solved through equation 2. you get x >1 or x <0, so its it. on a number line x is either right of +1 or left of 0. from this we cant conclude anything. if x is more than 1, x^3 > x^2 but if x is less than 0, then x^2 > x^3. so this statement is insufficient.

But once we combine statement 1 and 2 we get to know x is positive and more than 1. positive value mentioned in statement 1 rules out one of the possibility of value of x we get from statement 2. so only 1 value remains that is x >1. sufficient.

Kool...thanks a lot!!! kudos to you as well for explaining
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Re: Is x^3 > x^2? [#permalink]  22 Sep 2011, 06:09
5
KUDOS
siddhans wrote:
I searched for this but couldnt find it...
Is $$x^3 > x^2?$$
(1) x > 0
(2) $$x^2 >x$$

There are many different ways to approach these questions, some of which have been discussed above. I'll just point out one useful simplification that you can often use in these questions. When we are asked if x^3 > x^2, we can rephrase this question easily. We cannot divide by x on both sides (without breaking the problem down into awkward cases), since x might be negative (and if we divide by a negative, we need to reverse the inequality). But we *can* divide by x^2 on both sides, since x^2 cannot be negative. So we can instantly rephrase the question as "Is x > 1?".

Then we can immediately see that Statement 1 is not sufficient. Further, since Statement 2 is true for any negative value of x, but is also true when x > 1, it cannot be sufficient alone either. Combining the statements, if x > 0 from Statement 1, then we can safely divide by x on both sides in Statement 2, so we get x > 1, which is what we want. So the answer is C.
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Manager
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Re: Is x^3 > x^2? [#permalink]  22 Sep 2011, 07:23
IanStewart wrote:

But we *can* divide by x^2 on both sides, since x^2 cannot be negative. So we can instantly rephrase the question as "Is x > 1?".

.

This is a new approach I learnt today. Thanks so much Ian, this approach will save lots of time.
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Re: Is x^3 > x^2? [#permalink]  22 Sep 2011, 07:35
Thanks Ian.. really nice approach

Cheers!!!

IanStewart wrote:
siddhans wrote:
I searched for this but couldnt find it...
Is $$x^3 > x^2?$$
(1) x > 0
(2) $$x^2 >x$$

There are many different ways to approach these questions, some of which have been discussed above. I'll just point out one useful simplification that you can often use in these questions. When we are asked if x^3 > x^2, we can rephrase this question easily. We cannot divide by x on both sides (without breaking the problem down into awkward cases), since x might be negative (and if we divide by a negative, we need to reverse the inequality). But we *can* divide by x^2 on both sides, since x^2 cannot be negative. So we can instantly rephrase the question as "Is x > 1?".

Then we can immediately see that Statement 1 is not sufficient. Further, since Statement 2 is true for any negative value of x, but is also true when x > 1, it cannot be sufficient alone either. Combining the statements, if x > 0 from Statement 1, then we can safely divide by x on both sides in Statement 2, so we get x > 1, which is what we want. So the answer is C.
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Re: Is x^3 > x^2? [#permalink]  22 Sep 2011, 08:40
Aj85 wrote:
IanStewart wrote:

But we *can* divide by x^2 on both sides, since x^2 cannot be negative. So we can instantly rephrase the question as "Is x > 1?".

.

This is a new approach I learnt today. Thanks so much Ian, this approach will save lots of time.

I second that!!!
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Re: Is x^3 > x^2? [#permalink]  22 Sep 2011, 22:02
I also got "C" with simple calculation , but dividing by x^2 is a very good idea and it can be beneficial in real test.

Thanks
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Re: Is x^3 > x^2? [#permalink]  26 Sep 2011, 04:55
[quote="siddhans"]I searched for this but couldnt find it...

Is $$x^3 > x^2?$$
(1) x > 0
(2) $$x^2 >x$$

I got a C in 20 seconds and I used a simple thought process:

If x^3 has to be greater than x^2, x has to be a positive integer greater than zero. In all other cases it will not hold true. Now we go to the statements and see whether we can get a YES or NO.

Statement 1 tells us that x is positive and greater than zero but tells nothing about it being an integer. Hence INSUFFICIENT

statement 2 tells us that x is an integer but says nothing about the sign of x. Hence INSUFFICIENT.

Combining both statement, we get a YES to the question. Hence SUFFICIENT
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Re: Is x^3 > x^2? [#permalink]  27 Sep 2011, 00:17
Thanks Ian & GMATMission....i liked both the explanations
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Re: Is x^3 > x^2? [#permalink]  03 Oct 2011, 01:17
alidaraei wrote:
well I thinks I don't understand this question completely!

Excuse me, alidaraei, but what is your aim of giving empty comments on different PS and DS problems?
Anyone on this forum tries to increase "common knowledge" about GMAT and to help other people.

I think this is not the best way to increase the number of your posts in order to get free access to anything which you plan.

bagrettin
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Re: Is x^3 > x^2? [#permalink]  03 Oct 2011, 01:33
bagrettin wrote:
alidaraei wrote:
well I thinks I don't understand this question completely!

Excuse me, alidaraei, but what is your aim of giving empty comments on different PS and DS problems?
Anyone on this forum tries to increase "common knowledge" about GMAT and to help other people.

I think this is not the best way to increase the number of your posts in order to get free access to anything which you plan.

bagrettin

All the posts by "alidaraei" have been deleted and the user is warned. Continuation of the user's posting attitude will get him/her banned.
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Re: Is x^3 > x^2?   [#permalink] 03 Oct 2011, 01:33
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