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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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Is x^2 greater than x?

(1) x^2 is greater than 1
(2) x is greater than -1


[Reveal] Spoiler:
I was doing this question and did not particularly like the explanation that was given. Was my approach of rephrasing this question appropriate?

I rephrased it this way:
x^2 > x
x^2 - x > 0
x(x-1) > 0

Then from this i deduced that both of those phrases would have to be either positive or both would have to be negative. I felt that statement 1) allowed me to determine that but statement 2) did not. Was my approach correct?


OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/is-x-2-great ... 67333.html
[Reveal] Spoiler: OA

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

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New post 31 Oct 2010, 17:52
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jscott319 wrote:
Is x^2 greater than x?

1) x^2 is greater than 1

2) x is greater than -1


I was doing this question and did not particularly like the explanation that was given. Was my approach of rephrasing this question appropriate?

I rephrased it this way:
x^2 > x
x^2 - x > 0
x(x-1) > 0

Then from this i deduced that both of those phrases would have to be either positive or both would have to be negative. I felt that statement 1) allowed me to determine that but statement 2) did not. Was my approach correct?


Yes it was.

Is \(x^2 > x\)? --> is \(x(x-1)>0\)? --> is \(x\) in the following ranges: \(x<0\) or \(x>1\)?

(1) x^2 is greater than 1 --> \(x^2>1\) --> \(x<-1\) or \(x>1\). Sufficient.

(2) x is greater than -1 --> \(x>-1\). Not sufficient.

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

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New post 31 Oct 2010, 18:07
Bunuel wrote:
jscott319 wrote:
Is x^2 greater than x?

1) x^2 is greater than 1

2) x is greater than -1


I was doing this question and did not particularly like the explanation that was given. Was my approach of rephrasing this question appropriate?

I rephrased it this way:
x^2 > x
x^2 - x > 0
x(x-1) > 0

Then from this i deduced that both of those phrases would have to be either positive or both would have to be negative. I felt that statement 1) allowed me to determine that but statement 2) did not. Was my approach correct?


Yes it was.

Is \(x^2 > x\)? --> is \(x(x-1)>0\)? --> is \(x\) in the following ranges: \(x<0\) or \(x>1\)?

(1) x^2 is greater than 1 --> \(x^2>1\) --> \(x<-1\) or \(x>1\). Sufficient.

(2) x is greater than -1 --> \(x>-1\). Not sufficient.

Answer: A.


Great Thank You!

I am curious though, and maybe I was not fully aware with this in my understanding. How do you conclude to determine the ranges? \(x<0\) or \(x>1\)? Should it be \(x>0\) ?

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

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New post 31 Oct 2010, 18:24
jscott319 wrote:
Bunuel wrote:
jscott319 wrote:
Is x^2 greater than x?

1) x^2 is greater than 1

2) x is greater than -1


I was doing this question and did not particularly like the explanation that was given. Was my approach of rephrasing this question appropriate?

I rephrased it this way:
x^2 > x
x^2 - x > 0
x(x-1) > 0

Then from this i deduced that both of those phrases would have to be either positive or both would have to be negative. I felt that statement 1) allowed me to determine that but statement 2) did not. Was my approach correct?


Yes it was.

Is \(x^2 > x\)? --> is \(x(x-1)>0\)? --> is \(x\) in the following ranges: \(x<0\) or \(x>1\)?

(1) x^2 is greater than 1 --> \(x^2>1\) --> \(x<-1\) or \(x>1\). Sufficient.

(2) x is greater than -1 --> \(x>-1\). Not sufficient.

Answer: A.


Great Thank You!

I am curious though, and maybe I was not fully aware with this in my understanding. How do you conclude to determine the ranges? \(x<0\) or \(x>1\)? Should it be \(x>0\) ?


\(x(x-1)>0\) as you noted either both multiples are positive or both are negative:
\(x<0\) and \(x-1<0\), or \(x<1\) --> \(x<0\);
\(x>0\) and \(x-1>0\), or \(x>1\) --> \(x>1\);

So \(x(x-1)>0\) holds true when \(x<0\) or \(x>1\).

For alternate approach check "How to solve quadratic inequalities": x2-4x-94661.html#p731476

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

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New post 18 Mar 2011, 05:06
Let me see if I can explain this in an easy manner...
Statement tells us that x<0 or x>1 - see bunuel's explanation.
OR statement also says...x can be a positive integer or a positive fraction greater than 1. Also, x can be a negative integer or a negative fraction (Basically x<0)
Statement 1. YES - sufficient.
Statement 2. NO - Insuff. This is because x>-1 can be any number such as 0.5 (which does not solve our problem)

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

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New post 18 Mar 2011, 07:23
thesfactor wrote:
Let me see if I can explain this in an easy manner...
Statement tells us that x<0 or x>1 - see bunuel's explanation.
OR statement also says...x can be a positive integer or a positive fraction greater than 1. Also, x can be a negative integer or a negative fraction (Basically x<0)
Statement 1. YES - sufficient.
Statement 2. NO - Insuff. This is because x>-1 can be any number such as 0.5 (which does not solve our problem)


Few Tips
1. √x ≥ x for (0 ≤ x ≤ 1)
2. √x ≤ x for (1 ≤ x)
3. x³ ≤ x for (x ≤ -1) and (0 ≤ x ≤ 1)
4. x³ ≥ x for (-1 ≤ x ≤ 0) and (1 ≤ x)
5. x^2 >=x for (1 ≤ x)

So OA should be A..(Case 5)

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

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New post 29 Oct 2011, 10:14
siddhans wrote:
Is x^2 greater than x?

1) x^2 is greater than 1
2) x is greater than -1


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

1) x^2 > 1
Means;
x>1 OR x<-1; Either way it will be less than 0 or more than 1.
Sufficient.

2)x>-1
Now; x can be 2, which is greater than 1.
OR
x can be from 0 to 1.
Not Sufficient.

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

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New post 29 Oct 2011, 16:32
fluke wrote:
siddhans wrote:
Is x^2 greater than x?

1) x^2 is greater than 1
2) x is greater than -1


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

1) x^2 > 1
Means;
x>1 OR x<-1; Either way it will be less than 0 or more than 1.
Sufficient.

2)x>-1
Now; x can be 2, which is greater than 1.
OR
x can be from 0 to 1.
Not Sufficient.

Ans: "A"

Where highlighted in red, I think you mean x>0...or am I missing a property here?

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

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New post 29 Oct 2011, 17:50
martie11 wrote:
fluke wrote:
siddhans wrote:
Is x^2 greater than x?

1) x^2 is greater than 1
2) x is greater than -1


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

1) x^2 > 1
Means;
x>1 OR x<-1; Either way it will be less than 0 or more than 1.
Sufficient.

2)x>-1
Now; x can be 2, which is greater than 1.
OR
x can be from 0 to 1.
Not Sufficient.

Ans: "A"

Where highlighted in red, I think you mean x>0...or am I missing a property here?

Thx.



Fluke,



You mentioned : from st1 x>1 OR x<-1; Either way it will be less than 0 or more than 1.


So, how can we prove x^2 > x ?



does that mean x(x-1)>0
=> either both positive or both negative ?
thus x > 0 and x> 1 OR x< 0 and x <-1

Thus we have to prove x>1 or x <1?? which is what statement 1 says? hence it is A? Is this the right way ?


Also, why is 2 insufficient then? St 2 says x>-1
Is it because we cant prove x> 1 or x <1 using this statement ?

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

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New post 29 Oct 2011, 21:50
martie11 wrote:
fluke wrote:
siddhans wrote:
Is x^2 greater than x?

1) x^2 is greater than 1
2) x is greater than -1


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

1) x^2 > 1
Means;
x>1 OR x<-1; Either way it will be less than 0 or more than 1.
Sufficient.

2)x>-1
Now; x can be 2, which is greater than 1.
OR
x can be from 0 to 1.
Not Sufficient.

Ans: "A"

Where highlighted in red, I think you mean x>0...or am I missing a property here?

Thx.



Q: Is x^2>x;
Now, what values of x will satisfy this condition?

Any x that is less than 0 or more than 1 will satisfy the condition.
Say, x=-0.1; x^2=+0.01; x^2>x
x=1.1; x^2=1.21; x^2>x

But, if x is any value from 0 to 1, the expression will not hold true.
x=0; x^2=x
x=1; x^2=x
x=0.5; x^2<x

1) x^2 > 1
This tells us that x is NOT between 0 and 1.
Thus, x is number either >1 OR <0.
So, x^2>x will always be true.
Sufficient.
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Re: Is x^2 greater than x? [#permalink]

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New post 06 Dec 2011, 08:58
"
\(x(x-1)>0\) as you noted either both multiples are positive or both are negative:
\(x<0\) and \(x-1<0\), or \(x<1\) --> \(x<0\);
\(x>0\) and \(x-1>0\), or \(x>1\) --> \(x>1\);

So \(x(x-1)>0\) holds true when \(x<0\) or \(x>1\).

For alternate approach check "How to solve quadratic inequalities": x2-4x-94661.html#p731476

Hope it helps."

Hi Bunuel, sorry I did not understand how the ranges/limits change?

For x(x-1) > 0 to hold true,
either
x > 0 and x > 1
or
x < 0 and x < 1

How then does x(x-1) > 0 hold true when x < 0 and x > 1?

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

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New post 06 Dec 2011, 09:33
1. if x>1 then x^2 always greater than x
x= 2, x^2=4
2. x>-1
x=0
x^2 = x^2
x = 3
x^2>x
insufficient.
Ans. A
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Re: Is x^2 greater than x? [#permalink]

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New post 28 Apr 2014, 21:05
jscott319 wrote:
Is x^2 greater than x?

1) x^2 is greater than 1

2) x is greater than -1


I was doing this question and did not particularly like the explanation that was given. Was my approach of rephrasing this question appropriate?

I rephrased it this way:
x^2 > x
x^2 - x > 0
x(x-1) > 0

Then from this i deduced that both of those phrases would have to be either positive or both would have to be negative. I felt that statement 1) allowed me to determine that but statement 2) did not. Was my approach correct?


I think it can be answered by a bit intuitively as below:-

X^2 will be always greater than equal to 0. ( Square of a number is always positive).

Now taking the two options one by one. x^2 will be greater than 1 only when x>1 ( square of a decimal i.e 0.0~0.9 will increase the decimal place , for 0.2^2 = 0.04). Option 1 looks correct.

Exploring option 2 - For x>-1 we have to be careful that on number line , the closer one gets to zero , the bigger the number is. So extrapolating for say -0.2 ( which is greater than -1) would give value of 0.04 ( square of a negative number is positive) BUT now there is no end range. So even 1.1 is greater than -1. This makes this information insufficient.

Answer is A.


Obviously all this thought process will be done in mind so could be possibly fast compared to quadratic way.

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

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New post 28 Dec 2015, 22:44
Bunuel wrote:

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



can you explain this step please? what confuses me is the "X < 0" part.

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

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AS7x wrote:
Bunuel wrote:

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



can you explain this step please? what confuses me is the "X < 0" part.


Please check the following posts:
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Solving Quadratic Inequalities - Graphic Approach
Inequality tips

DS Inequalities Problems
PS Inequalities Problems

700+ Inequalities problems

Other Resources on Inequalities


inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html
everything-is-less-than-zero-108884.html
graphic-approach-to-problems-with-inequalities-68037.html
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Re: Is x^2 greater than x? [#permalink]

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New post 03 Jan 2016, 11:38
AS7x wrote:
Bunuel wrote:

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



can you explain this step please? what confuses me is the "X < 0" part.

This is easy one. We have to find intervals where x(x-1)>0. We have two points where this inequality turns to zero, 0 and 1. If x>1 than this inequality holds. If 0<x<1 then this is not true (negative sign). And if x<0 then inequality is true. You can check this by plugging numbers from different intervals
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Re: Is x^2 greater than x? [#permalink]

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New post 18 Jul 2016, 04:51
jscott319 wrote:
Is x^2 greater than x?

(1) x^2 is greater than 1
(2) x is greater than -1


[Reveal] Spoiler:
I was doing this question and did not particularly like the explanation that was given. Was my approach of rephrasing this question appropriate?

I rephrased it this way:
x^2 > x
x^2 - x > 0
x(x-1) > 0

Then from this i deduced that both of those phrases would have to be either positive or both would have to be negative. I felt that statement 1) allowed me to determine that but statement 2) did not. Was my approach correct?


jscott319 wrote:
Is x^2 greater than x?

(1) x^2 is greater than 1
(2) x is greater than -1


This question has more to do with DECIMAL PROPERTIES than actual inequality

(1) x^2 is greater than 1
x^2 is always greater than 1 except for two cases
i) x^2 is EQUAL to 0 When x is 0
ii) x^2 is LESS than x when x is a decimal between -1 and 1
example -0.60^2 = 0.36 (0.36 < 1)
example 0.40^2 = 0.16 (0.16 < 1)
But since here x^2 is greater than 1 we know that x is not a decimal between -1 to 1
Any other value of x (-ve or +ve) will always be smaller than its square ; therefore x^2 > x
SUFFICIENT

(2) x is greater than -1 [/quote]
x can be 0 and 0^2 is NOT > 0
x can be 4 and 4^2 > 0
INSUFFICIENT

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

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New post 09 Oct 2017, 09:55
jscott319 wrote:
Is x^2 greater than x?

(1) x^2 is greater than 1
(2) x is greater than -1


[Reveal] Spoiler:
I was doing this question and did not particularly like the explanation that was given. Was my approach of rephrasing this question appropriate?

I rephrased it this way:
x^2 > x
x^2 - x > 0
x(x-1) > 0

Then from this i deduced that both of those phrases would have to be either positive or both would have to be negative. I felt that statement 1) allowed me to determine that but statement 2) did not. Was my approach correct?




OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/is-x-2-great ... 67333.html
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Re: Is x^2 greater than x?   [#permalink] 09 Oct 2017, 09:55
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