Is x^2 greater than x ? : GMAT Data Sufficiency (DS)
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# Is x^2 greater than x ?

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Is x^2 greater than x ? [#permalink]

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20 Jun 2012, 09:44
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Is x^2 greater than x ?

(1) x^2 > x^3.
(2) x^2 > x^4.

I already solved it, but I tried to solve it by also the following method, and I don't know in which part I am wrong.

(1) $$x^2 > x^3$$.
$$x^2 - x^3 > 0$$
$$x^2 * (1-x) > 0$$
So, we get: x= 0 and x = 1

-----(+)------0----(-)-------1------(+)------

Therefore, the intervals are x<0 and x>1

However, if we solve it in this way:
$$x^2 > x^3$$
We divide both sides by $$x^2$$:
$$1 > x$$
The interval is not the same.

Please, tell me in which part of my first way I am wrong.
Thanks!
[Reveal] Spoiler: OA

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Re: Is x^2 greater than x ? [#permalink]

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20 Jun 2012, 11:32
metallicafan wrote:
$$x^2 > x^3$$.
$$x^2 - x^3 > 0$$
$$x^2 * (1-x) > 0$$
So, we get: x= 0 and x = 1

-----(+)------0----(-)-------1------(+)------

Hi,

When you say,
$$x^2 (1-x) > 0$$
if $$x\neq 0$$ then,$$x^2 >0$$ for any real value of x.
thus, inequality reduces to $$1-x>0$$
or $$x < 1$$

Also, to use the number line method, you should change the coefficient of x as positive (for simplicity)
$$x^2 (x-1) < 0$$ (multiply by -1 and reverse the sign)
now, use the number line;
-----(-)-----1---(+)---------

Please go through the following post:
Solving inequalities

Regards,
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Re: Is x^2 greater than x ? [#permalink]

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20 Jun 2012, 11:41
Hi,

Is $$x^2>x$$?
or $$x^2-x>0$$?
or $$x(x-1)>0$$?
or x<0 or x>1?

Detailed solution:
Using (1),
$$x^2>x^3$$
or $$x^2(1-x)>0$$
or x < 1, for x = 0.5
$$x^2=0.25<x$$
but for x=-1,
$$x^2=1>x$$. Thus, Insufficient.

Using (2),
$$x^2>x^4$$
or $$x^2(1-x^2)>0$$
or $$x^2(1-x)(1+x)>0$$
or $$x^2(x-1)(1+x)<0$$ (multiplied by -1)
-(+)-------(-1)---(-)-----(+1)-----(+)------
or -1<x<1
Again, for x = 0.5
$$x^2=0.25<x$$
but for x=-0.5,
$$x^2=0.25>x$$. Thus, Insufficient.

Even after using (1) & (2), we have -1 < x <1, Insufficient.

Regards,
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Re: Is x^2 greater than x ? [#permalink]

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20 Jun 2012, 12:32
Thank you both!

It seems that I just memorized the method and didn't analyze it well.

I think that I get it now. And also I think that I have found another way that also can be helpful to you. Please, confirm if I am Ok. Please, Bunuel, take a look too.

(1) $$x^2 > x^3$$.
$$x^2 - x^3 > 0$$
$$x^2 * (1-x) > 0$$
So, we get: x= 0 and x = 1

If $$x>1$$ -----> $$x^2 * (1-x) < 0$$
If $$0<x<1$$ -----> $$x^2 * (1-x) > 0$$
If $$x<0$$ -----> $$x^2 * (1-x) > 0$$

So, we get:

-----(+)------0----(+)-------1------(-)------

Therefore, the solution is x < 1.

The same solution using this method:
$$x^2 > x^3$$
We divide by $$x^2$$
$$1 > x$$
$$x < 1$$.

What do you think?
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Re: Is x^2 greater than x ? [#permalink]

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20 Jun 2012, 12:43
metallicafan wrote:
If $$x>1$$ -----> $$x^2 * (1-x) < 0$$
If $$0<x<1$$ -----> $$x^2 * (1-x) > 0$$
If $$x<0$$ -----> $$x^2 * (1-x) > 0$$

Hi,

As you can see, the sign only depends on 1-x, so, even powers can be ignored here.

Regards,
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Re: Is x^2 greater than x ? [#permalink]

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21 Jun 2012, 00:03
Expert's post
1
This post was
BOOKMARKED
Is x^2 greater than x ?

Question: is $$x^2>x$$? --> is $$x(x-1)>0$$? is $$x<0$$ or $$x>1$$?

(1) x^2 > x^3. Now, from this statement we know that $$x\neq{0}$$, so $$x^2>0$$ and we can safely reduce by it to get $$1>x$$ ($$x<1$$). So, finally we have that given inequality holds true for $$x<0$$ and $$0<x<1$$ (remember we should exclude 0 from the range, since if $$x=0$$ then the given inequality doesn't hold true). Not sufficient.

(2) x^2 > x^4. Apply the same logic here: we know that $$x\neq{0}$$, so $$x^2>0$$ and we can safely reduce by it to get $$1>x^2$$ ($$x^2<1$$) --> $$-1<x<1$$. So, finally we have that given inequality holds true for $$-1<x<0$$ and $$0<x<1$$ (remember we should exclude 0 from the range, since if $$x=0$$ then the given inequality doesn't hold true). Not sufficient.Not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) gives: $$-1<x<0$$ and $$0<x<1$$, so we can still have an YES answer (consider $$x=-0.5$$) as well as a NO answer (consider $$x=0.5$$).

Hope it's clear.
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Re: Is x^2 greater than x ? [#permalink]

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21 Jun 2012, 00:07
Hi,

Is $$x^2>x$$?
or $$x^2-x>0$$?
or $$x(x-1)>0$$?
or x<0 or x>1?

Detailed solution:
Using (1),
$$x^2>x^3$$
or $$x^2(1-x)>0$$
or x < 1, for x = 0.5
$$x^2=0.25<x$$
but for x=-1,
$$x^2=1>x$$. Thus, Insufficient.

Using (2),
$$x^2>x^4$$
or $$x^2(1-x^2)>0$$
or $$x^2(1-x)(1+x)>0$$
or $$x^2(x-1)(1+x)<0$$ (multiplied by -1)
-(+)-------(-1)---(-)-----(+1)-----(+)------
or -1<x<1
Again, for x = 0.5
$$x^2=0.25<x$$
but for x=-0.5,
$$x^2=0.25>x$$. Thus, Insufficient.

Even after using (1) & (2), we have -1 < x <1, Insufficient.

Regards,

The red parts are not correct. You should exclude zero, from all the ranges, since if $$x=0$$, then neither of statements hold true.

Hope it's clear.
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Re: Is x^2 greater than x ? [#permalink]

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21 Jun 2012, 00:14
Hi,

Thanks for pointing it out!

Regards,
Bunuel wrote:
Hi,

Is $$x^2>x$$?
or $$x^2-x>0$$?
or $$x(x-1)>0$$?
or x<0 or x>1?

Detailed solution:
Using (1),
$$x^2>x^3$$
or $$x^2(1-x)>0$$
or x < 1, for x = 0.5
$$x^2=0.25<x$$
but for x=-1,
$$x^2=1>x$$. Thus, Insufficient.

Using (2),
$$x^2>x^4$$
or $$x^2(1-x^2)>0$$
or $$x^2(1-x)(1+x)>0$$
or $$x^2(x-1)(1+x)<0$$ (multiplied by -1)
-(+)-------(-1)---(-)-----(+1)-----(+)------
or -1<x<1
Again, for x = 0.5
$$x^2=0.25<x$$
but for x=-0.5,
$$x^2=0.25>x$$. Thus, Insufficient.

Even after using (1) & (2), we have -1 < x <1, Insufficient.

Regards,

The red parts are not correct. You should exclude zero, from all the ranges, since if $$x=0$$, then neither of statements hold true.

Hope it's clear.
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Is x² greater than x? [#permalink]

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30 Dec 2016, 05:27
Is $$x^2$$ greater than $$x$$?

(1) $$x^2$$ is greater than $$x^3$$.
(2) $$x^2$$ is greater than $$x^4$$.

OA:
[Reveal] Spoiler:
E

Last edited by paddy41 on 30 Dec 2016, 05:38, edited 1 time in total.
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Re: Is x² greater than x? [#permalink]

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30 Dec 2016, 05:36
Quote:
Explanation: $$x^2$$ is greater than x for all numbers except for those values of x between 0 and 1. Thus, we need to know whether or not x falls in that range.

Statement (1) is insucient. To simplify,divide both sides by $$x^2$$, resulting in 1 > x. If that's true, x could be between 0 and 1, but it could also be less than 0.

Statement (2) is also insucient. Again, simplify by dividing by $$x^2$$, which gives you 1 > $$x^2$$. Thus, x could be any number between -1 and 1. Again, it could be between 0 and 1, but it could also be between -1 and 0.

Taken together, it's still insucient. Both statements allow for the possibility that x is between 0 and 1, but both statements also make it possible that x is between -1 and 0. Choice (E) is correct.

Just one question about the highlighted part of the explanation: It is clear that $$x^2$$ is always positive. However, is the division by $$x^2$$ a valid inequality operation without knowing whether x is positive or negative?
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Re: Is x^2 greater than x ? [#permalink]

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30 Dec 2016, 07:18
Is $$x^2$$ greater than $$x$$?

(1) $$x^2$$ is greater than $$x^3$$.
(2) $$x^2$$ is greater than $$x^4$$.

OA:
[Reveal] Spoiler:
E

Merging topics. Please refer to the discussion above.
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Re: Is x^2 greater than x ? [#permalink]

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30 Dec 2016, 07:20
Quote:
Explanation: $$x^2$$ is greater than x for all numbers except for those values of x between 0 and 1. Thus, we need to know whether or not x falls in that range.

Statement (1) is insucient. To simplify,divide both sides by $$x^2$$, resulting in 1 > x. If that's true, x could be between 0 and 1, but it could also be less than 0.

Statement (2) is also insucient. Again, simplify by dividing by $$x^2$$, which gives you 1 > $$x^2$$. Thus, x could be any number between -1 and 1. Again, it could be between 0 and 1, but it could also be between -1 and 0.

Taken together, it's still insucient. Both statements allow for the possibility that x is between 0 and 1, but both statements also make it possible that x is between -1 and 0. Choice (E) is correct.

Just one question about the highlighted part of the explanation: It is clear that $$x^2$$ is always positive. However, is the division by $$x^2$$ a valid inequality operation without knowing whether x is positive or negative?

When dividing/multiplying an inequality by a variable we need to know its sign. If it's positive we should keep the sign and if its negative we should flip the sign. That's the rule. We know that x^2 is positive, so we can safely multiply/divide an inequality by it (so in this case it does not matter whether x itself is positive or negative).
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Re: Is x^2 greater than x ?   [#permalink] 30 Dec 2016, 07:20
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