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Is x^2 greater than x?

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Is x^2 greater than x? [#permalink] New post 31 Oct 2010, 16:39
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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?
[Reveal] Spoiler: OA
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Re: Is x^2 greater than x? [#permalink] New post 31 Oct 2010, 16:52
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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] New post 31 Oct 2010, 17:07
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] New post 31 Oct 2010, 17:24
Expert's post
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] New post 31 Oct 2010, 18:39
Ah of course, I think I was just mixingup my own thinking and the 'solvinf for x' approach that is also done when solving for a quadratic. Thanks for straightening my thinking and thanks for the link!
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Re: Is x^2 greater than x? [#permalink] New post 01 Nov 2010, 07:53
Yes that is already noted as the OA

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Re: Is x^2 greater than x? [#permalink] New post 18 Mar 2011, 04:06
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] New post 18 Mar 2011, 06:23
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] New post 06 Dec 2011, 07:58
"
\(x(x-1)>0\) 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] New post 06 Dec 2011, 08:33
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] New post 14 Dec 2011, 10:44
+1 A

The numbers in the range between - 1 and 1 have a particular behavior in exponent problems.
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Re: Is x^2 greater than x? [#permalink] New post 14 Aug 2013, 05:13
Hi Bunuel, How would do this sum in your way. The official answer, which is quite confusing is below. I am unable to understand it completely. I did the sum by evaluation the combinations in regions
{ x<-1 } , { -1<x<0} , { 0<x<1} , {x,1}

Is x^2 greater than x ?

(1) x is less than -1.
(2) x^2 is greater than 1.
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Re: Is x^2 greater than x? [#permalink] New post 14 Aug 2013, 05:16
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irda wrote:
Hi Bunuel, How would do this sum in your way. The official answer, which is quite confusing is below. I am unable to understand it completely. I did the sum by evaluation the combinations in regions
{ x<-1 } , { -1<x<0} , { 0<x<1} , {x,1}

Is x^2 greater than x ?

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


Check here: is-x-2-greater-than-x-147172.html
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Re: Is x^2 greater than x? [#permalink] New post 28 Apr 2014, 20:05
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] New post 30 May 2015, 01:37
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Re: Is x^2 greater than x? [#permalink] New post 01 Jun 2015, 18:00
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Hi All,

This DS question is based on a couple of Number Property rules; if you know the rules, then you can answer this question with a more logic-based approach. If you don't know the rules, then you can still discover the patterns involved by TESTing VALUES.

We're asked if X^2 is greater than X. This is a YES/NO question.

By doing a little bit of work up-front, we can make dealing with the two Facts easier. We just have to think about what X COULD be and whether that would make X^2 greater than X (or not).

IF... X = ANY negative value....
Then X^2 = positive and X^2 > X. The answer to the question would be YES.

IF.... X = 0 or X = 1
Then X^2 = X. The answer to the question would be NO.

IF.... X = A positive FRACTION...
Then X^2 < X. The answer to the question would be NO.

IF.... X > 1
Then X^2 > X. The answer to the question would be YES.

Fact 1: X^2 is greater than 1

Here, X COULD be any negative LESS than -1 (eg. -2, -3, -4, -1.5, etc.)....and the answer to the question would be YES.
X COULD also be any positive GREATER than 1 (e.g. 2, 3, 4, 1.5, etc.)...and the answer to the question would also be YES.
The answer to the question is ALWAYS YES.
Fact 1 is SUFFICIENT.

Fact 2: X is greater than -1

X COULD be 0...and the answer to the question would be NO.
X COULD also be any positive GREATER than 1 (e.g. 2, 3, 4, 1.5, etc.)...and the answer to the question would be YES.
Fact 2 is INSUFFICIENT

Final Answer:
[Reveal] Spoiler:
A


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Re: Is x^2 greater than x? [#permalink] New post 02 Jun 2015, 08:46
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Is x^2 greater than x?

For x^2 to be greater than x, x should be greater than 1 (or less than -1) or negative if it's less than 1

(1) x^2 is greater than 1 if x^2 is greater than 1, so is x is also (or a negative integer). x = 2 or - 2, x^2 = 4, for example. Sufficient
(2) x is greater than -1 If x = 2, x^2= 4. The statement is true. But if x = 1/2 for example, x^2 = 1/4 and this is not true. Not sufficient
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Re: Is x^2 greater than x?   [#permalink] 02 Jun 2015, 08:46
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