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

The question asks whether \(x<0\) or \(x>1\):So, whether x is in the blue ranges above.

(1) says that \(x<-1\) or \(x>1\). Now, let me ask you a question: is x in the blue ranges???

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:
GMAT assassins aren't born, they're made,
Rich

Solution:

We need to determine whether x^2 is greater than x. Notice that x^2 is greater than x if x > 1 or x < 0.

Statement One Alone:

Since x^2 is greater than 1, x is either greater than 1 or less than -1. Either way we have x^2 > x. Statement one alone is sufficient.

Statement Two Alone:

Statement two alone is not sufficient. For example, if x = 2, then x^2 > x. However, if x = 0, then x^2 is not greater than x.

Answer: A

BrentGMATPrepNow
Please help me understand this concept...

how does sign for x(x-1)>0 become x<0 Why does sign got changed. We didn’t multiply by negative

Posted from my mobile device

I believe you're referring to statement 1 (x² > 1)
If x² > 1, then EITHER x > 1 OR x < -1

Let's examine both cases...
case i: If x > 1, then x is positive, and (x - 1) is positive, which means x(x-1) = (positive)(positive) = positive.
In other words, if x > 1, then x(x-1) > 0
case ii: If x < -1, then x is negative, and (x - 1) is negative, which means x(x-1) = (negative)(negative) = positive.
In other words, if x < -1, then x(x-1) > 0

Since both cases yield a result in which x(x-1) > 0, we can be certain that statement 1 is sufficient

Does that help?

Bunuel
I realize that Brent discusses testing values above. But, what approach did you take to get from x(x-1)>0 to then x<0 or x>1?

"
Hi Bunuel,

for x(x-1)>0 how is x<0

If x is negative, then x(x-1) = negative*negative = positive.

Check the links below for more:

9. Inequalities

• Theory
  Inequalities Made Easy!
  Inequalities: Tips and hints
  Arithmetic with Inequalities
  Wavy Line Method Application - Complex Algebraic Inequalities
  Solving Quadratic Inequalities: Graphic Approach
  Inequalities - Quadratic Inequalities
  Graphic approach to problems with inequalities
  Inequalities trick
  INEQUATIONS(Inequalities) Part 1
  INEQUATIONS(Inequalities) Part 2
  
  • Questions
    Inequality and absolute value questions from my collection
    DS Questions
    PS Questions

­Hi guys. I'm new to Inequalities and I had a very similar doubt as many of you : How to derive the range from the equation?
Thanks to GMAT club, I now know that you use the wavy line method.

Step 1 : Rewrite the equation in the form of (x - a)(x - b) > 0.
So, x (x - 1) > 0 can be rewritten as (x - 0)(x - 1) > 0.

Step 2 : Find the points on the number line where LHS = 0. Basically, for what values of x will the LHS = 0?
So, when x = 0 : (0 - 0)(0 - 1) = 0. Therefore, LHS = 0
Or, when x = 1 : (1 - 0)(1 - 1) = 0. Therefore, LHS = 0
These two points on the number line are called critical points. (0 , 1)

Step 3 : Draw the curve on the number line.

So you can see that for all x < 0 and x > 1 : x (x - 1) > 0
and for all x such that 0 < x < 1 : x (x - 1) < 0.


This ties back to the question : is x in the following ranges: x<0 or x>1?


I hope this helps (And I also hope I'm right).
PS : I don't come from a quant background, so I need to learn from the absolute basics. I put this up for the very miniscule set of people who are in the same boat as me.
PPS : This seems like a long drawn out process, but as you can see, everyone else who has posted has pretty much calculated this in their heads in less than 2 seconds.
­

I did the same mistake !

Let me simplify :

When we solve inequality like equality like this :
x(x−1)>0 would get us x>0 and x>1 which is not WRONG , this would be correct in Equalities but not in Inequalities

Step 1: Find the Critical Points

Set the expression equal to zero(Equality method):
x(x−1)=0

The critical points are x = 0 and x = 1.

Step 2: Test Intervals

The critical points divide the number line into three intervals:

1. x<0 (e.g., x=−1)
2. 0<x<1 (e.g., x=0.5)
3. x>1 (x=2)

Now, check the sign of x(x−1)>0 in each region:

• For x<0
  Pick x=−1 then (−1)(−2)=2>0 ✅
• For 0<x<1:
  Pick x=0.5x, then (0.5)(−0.5)=−0.25<0 ❌
• For x>1:
  Pick x=2 then (2)(1)=2>0 ✅

Step 3: Write the Solution

x<0 or x>1