Absolute surprise

[levocap]

If x is not equal to 0, is |x| less than 1?

(1) (x / |x|)< x

(2) |x| > x


The OE is C. But isn't this question essentially asking if X is a fraction? Because the Abs(X) cannot be a whole number because it will always be greater than 1

[Ben Ku]

If x is not equal to 0, is |x| less than 1?

(1) (x / |x|)< x

(2) |x| > x


The OE is C. But isn't this question essentially asking if X is a fraction? Because the Abs(X) cannot be a whole number because it will always be greater than 1

Actually what is asking is, is x between -1 and 1? Remember that the question doesn't tell us whether x is positive or negative.

Statement (2) tells us that x is negative. If x were positive, then |x| = x. However, we don't know whether x is between -1 and 1. so Statement (2) is insufficient.

Statement (1) can be rephrased in this way:
If x is positive, then 1 < x. If x is negative, then -1 < x, or x > -1. With this statement, we know that -1 < x < 0 or x > 1. However we don't know if x is between -1 and 1.

Putting the two statements together, the only overlapping interval is -1 < x < 0. So therefore, we know that x is between -1 and 1. (C) is the answer.

Hope that helps.

[rlevochkin]

If x is not equal to 0, is |x| less than 1?

(1) (x / |x|)< x

(2) |x| > x


The OE is C. But isn't this question essentially asking if X is a fraction? Because the Abs(X) cannot be a whole number because it will always be greater than 1

(1) x/(x) < x

x/(x) - x < 0

x - x*(x) < 0

x(1-(x))<0, x cannot be zero, x cannot be 1.

x=2, valid and x=-2, not valid
x=1/2 not valid. So, (1) is insufficient
x=-1/2 valid
x can be -1<x<0 or x>1
(2) x < 0, Obviously Insufficient

Combined (1) and (2),
leaves us with x>1
C is correct

[RonPurewal]

you should get a takeaway from this problem, about the expressions
x / |x|
and
|x| / x
(note that these are always equal, even though they are reciprocals)
if x is POSITIVE, then both of these are equal to 1.
if x is NEGATIVE, then both of them are -1.

this makes statement #1 a lot easier to deal with, since you can just segregate the number line according to pos/neg and +/- 1, and test each zone.
If x is...
Less than -1: (x / |x|) is -1, so (x / |x|) < x is false.
Equal to -1: (x / |x|) is -1, so false.
Between -1 and 0: (x / |x|) is -1, so it's true that (x / |x|) < x.
Between 0 and 1: (x / |x|) is 1, so false.
Equal to 1: (x / |x|) is also 1, so false.
Greater than 1: (x / |x|) is 1, so it's true that (x / |x|) < x.

so statement (1) means that x is either between -1 and 0, or greater than 1.

statement (2) rules out the latter, so, (c).