Hi safina,

The question essentially asks - is -1<x<1 (excluding 0)?

Choice A: the left hand side will be 1 when x>0 and -1 when x<0. Let's look at the possibilities:
x>1 --> Statement will be TRUE because left hand side is 1 whereas x>1.
0<x<1 --> Statement will be FALSE because left hand side is 1 whereas x is a small positive number
-1<x<0 --> Statement will be TRUE because left hand side is -1 whereas x is a smaller negative number
x<-1: Statement will be FALSE because left hand side is -1 whereas x<-1

Since the statement is TRUE for x>1 and -1<x<0, it is INSUFFICIENT to determine if |x|<1.

Choice B: |x|>x only when x is negative, otherwise they'll be equal. However, this means any negative number would work. So B is INSUFFICIENT on its own.

Together: Test out both cases for choice A being true.
x>1 - makes choice A true but makes choice B false - does not work
-1<x<0 - makes choice A true and choice B true.

Therefore for A and B to be true, -1<x<0 must be true as well. The answer is C

safina wrote:Can the experts please help me with this question?

If x is not equal to 0, is |x| less than 1?
a) x/|x|<x
b) |x|>x

Thanks!!

I started with Statement 2. If the absolute value of x is greater than the actual value of x, then x must be negative; applying absolute value will make it positive, and a positive is always greater than a negative. However, x could be -1/2 (and we'd say yes to the question) or -10 (and we'd say no), so it is insufficient to say if |x| < 1.

From statement 1:

x/|x| < x

x < x|x|

If x is positive, we have x < x^2, which means x is greater than 1.

If x is negative, it must be between -1 and 0. For instance, -2 < -2(2) does not work, but -1/2 < -1/2 (1/2) does.

The only way to satisfy both statements is for x to between 0 and -1, so its absolute value is always less than 1. Thus, C

I posted an alternate approach here:

https://www.beatthegmat.com/if-x-is-not- ... 65585.html

Very nice question:


Hope that helps,
-Patrick

thanks all! I think the most important part to solve this question is the initial rephrasing..once that is done, the problem becomes easier to solve.