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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 \(x>1\).

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}

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}

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.

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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

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 _________________

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 _________________

When you’re up, your friends know who you are. When you’re down, you know who your friends are.

Share some Kudos, if my posts help you. Thank you !

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. _________________

So, my final tally is in. I applied to three b schools in total this season: INSEAD – admitted MIT Sloan – admitted Wharton – waitlisted and dinged No...

HBS alum talks about effective altruism and founding and ultimately closing MBAs Across America at TED: Casey Gerald speaks at TED2016 – Dream, February 15-19, 2016, Vancouver Convention Center...