Is x > 1? (1) x(|x| − 1) > 0 (2) |x| > 1

We try to find range of values of x satisfying statement 1 \(x(|x| − 1) > 0\)
Product of two numbers is positive if both numbers are positive or both numbers are negative.
If x >0 then for the product to be positive, |x|-1 must be positive.

Statement 1: \(x(|x| − 1) > 0\)
CRITICAL POINTS occur when the two sides of an inequality are EQUAL.
Herw, the two sides are equal when x=-1, x=0, or x=1.

To determine which ranges satisfy the inequality, test one value to the left and one value to the right of each critical point.
Here, we must test x<-1, -1<x<0, 0<x<1 and x>1.
If we test x=-2, x=-1/2, x=1/2 and x=2, only x=-1/2 and x=2 satisfy \(x(|x| − 1) > 0\), implying that the valid ranges are -1<x<0 and x>1.
Since the answer to the question stem is NO if -1<x<0 but YES if x>1, INSUFFICIENT.

Statement 2: |x| > 1
Here, x<-1 or x>1.
Since the answer to the question stem is NO if x<-1 but YES if x>1, INSUFFICIENT.

Statements combined:
Both statements are satisfied only by x>1.
Thus, the answer to the question stem is YES.
SUFFICIENT.

Is \(x > 1\)?

(1) \(x(|x| − 1) > 0\)
(2) \(|x| > 1\)

Solution:
from (1): insuff
x(|x| − 1) > 0
+*+ => x>0 & |x| − 1 >0 => |x|>1, so x >1 because x >0 --> answer is YES
-*- => x< 0 & |x| − 1 < 0 => |x|<1 => -1<x<0 because x <0 --> answer is NO
from (2): insuff
|x|>1
=> x < -1 --> answer is NO
and x >+1 --> answer is YES

Now combining (1) & (2), we get
x >1 --> answer is YES
Answer: C

Statement 1:

\(x(|x| − 1) > 0\), we have a product is greater than 0, for this to be true we need the two multiplying terms to be either both positive or both negative.

Case 1: \(x > 0\) and \(|x| − 1 > 0\), simplifies to \(x > 0\) and (\(x > 1\) or \(x < -1\)). Considering the "and" we get \(x > 1\) overall.

Case 2: \(x < 0\) and \(|x| − 1 < 0\), simplifies to \(x < 0\) and \(-1 < x < 1\). Combined we get \(-1 < x < 0\).

Finally combining both cases we have either \(x > 1\) or \(-1 < x < 0\) which is not always \(x > 1\), so insufficient.

Statement 2:

\(x > 1\) or \(x < -1\). Insufficient.

Combined:

We need to combine four continuous regions, look at only the overlapping parts and we find only \(x > 1\) overlapping. Then combined we have \(x > 1\), sufficient.

Ans: C

Bunuel TestPrepUnlimited GMATGuruNY chetan2u

Do questions like Is \(x > 1 ? \) warrant exclusivity?

Statement 1 suggests that x > 1 although it also suggests that -1 < x < 0 but that does not mean we cannot say x > 1 is not valid.
Since x > 1 is a valid solution range of statement 1 why shouldn't the answer be A. If feels that the whole purpose of the second statement is just to push the answer choice to C. Please share your thoughts.

From statement I, you get value of x as -0.5 or -0.7 or 2.5 or 1000 and so on.

The question is very clear: Is x>1?

x is a variable and we cannot say for sure what is its value. But here we can say that we don’t get a confirmed answer as x could be -0.8 or 19.

Statement 2 gives you another range of x.
Only the common portion of two statements can give you exact range of x

Thanks chetan2u !!
I probably read too much into this one.