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Is x^2 greater than x? 1) x^2 is greater than 1 2) x is [#permalink]
31 Oct 2010, 16:39

2

This post was BOOKMARKED

00:00

A

B

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D

E

Difficulty:

25% (medium)

Question Stats:

71% (01:39) correct
29% (00:51) wrong based on 121 sessions

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 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.

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 \(x>0\) ?

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 \(x>1\).

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!

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)

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)

" \(x(x-1)>0\) 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 \(x>1\).

Re: Is x^2 greater than x? 1) x^2 is greater than 1 2) x is [#permalink]
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}

Re: Is x^2 greater than x? 1) x^2 is greater than 1 2) x is [#permalink]
14 Aug 2013, 05:16

Expert's post

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}

Re: Is x^2 greater than x? 1) x^2 is greater than 1 2) x is [#permalink]
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.

gmatclubot

Re: Is x^2 greater than x? 1) x^2 is greater than 1 2) x is
[#permalink]
28 Apr 2014, 20:05

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