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Is x^2 greater than x?

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Is x^2 greater than x? [#permalink]  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]  31 Oct 2010, 16:52
Expert's post
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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.

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

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]  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.

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]  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]  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]  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]  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]  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]  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]  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]  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]  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}

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]  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.

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]  30 May 2015, 01:37
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Re: Is x^2 greater than x? [#permalink]  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

[Reveal] Spoiler:
A

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Is x^2 greater than x? [#permalink]  28 Sep 2015, 10:12
It's actually an easy question:
St1: in +ve case $$x^2$$ can be only < x if 0<x<1 SUFFICIENT
St2: -ve case - x=-0,9 $$x^2$$=0,81 YES, +ve case - x=0,9 then $$x^2$$=0,8 NO... NOT SUFFICIENT
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Re: Is x^2 greater than x? [#permalink]  29 Dec 2015, 22:03
Expert's post
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution

Is x^2>x?
(1) x^2 is greater than 1
(2) x is greater than -1

In case of inequality questions, it is important to note that the conditions are sufficient if the range of the question includes the range of the conditions.
First, we need to modify the original condition and the question.
The original question asks if x^2-x>0. Then, by modifying the question, we can see that the question is asking if x(x-1)>0. This means, essentially, we need to prove if x<0 or 1<x. There is 1 variable (x), and in order to match the number of variables and the number of equations, we need 1 equation. Since the condition 1) and 2) each has 1 equation, there is high chance D is going to be the answer.
In case of the condition 1), it states x^2>1. From this, we can obtain x^2-1>0. Then, (x-1)(x+1)>0. So, x<-1 or 1<x. Since the range of the question includes the range of the condition, the condition is sufficient.
In case of the condition 2), it states x>-1. Since the range of the question includes the range of the condition, the condition is not sufficient. Hence, the correct answer is A.

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Re: Is x^2 greater than x?   [#permalink] 29 Dec 2015, 22:03
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