Is x 1? (1) x(|x| − 1) 0 (2) |x| 1

Question

Is  x > 1 ?

(1)  x(|x| − 1) > 0 
(2)  |x| > 1

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chetan2u · Answer

Is x > 1 ? (1) x(|x| − 1) > 0 If x>0, then |x|-1>0....that is x>1 If x<0, then |x|-1<0....that is |x|<1 or -1 1 x>1 or x<-1 Combined X cannot be <-1 as seen from statement I, so x>1 Sufficient C

riagarg07 · Answer

In (1): If x>0, then |x|-1>0....that is x>1 ? , 
How did u come to this conclusion that |x|-1>0?

Aks111 · Answer

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.

GMATGuruNY · Answer

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, -11. 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 -11. Since the answer to the question stem is NO if -11, 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. C

hiranmay · Answer

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 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

TestPrepUnlimited · Answer

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

sanjitscorps18 · Answer

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.

chetan2u · Answer

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

sanjitscorps18 · Answer

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

Ggt1234 · Answer

Can we use the wavy line method here, teaming the break points as -1,0 and 1? Then I get two areas where the equation is more than 0: x between -1 and 0 // x>1, so insufficient. The second option combined solves it.

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

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