Hi,
I came across the following problem in MGMAT cat.

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

(1) x/|x|< x

(2) |x| > x

Answer for this I got is A, which I feel is valid. 'A' directly tells us that |X|>1 and hence we can sufficiently answer the question.

But, the explanation given in the CAT is as follows:

The question "Is |x| less than 1?" can be rephrased in the following way.

Case 1: If x > 0, then |x| = x. For instance, |5| = 5. So, if x > 0, then the question becomes "Is x less than 1?"

Case 2: If x < 0, then |x| = -x. For instance, |-5| = -(-5) = 5. So, if x < 0, then the question becomes "Is -x less than 1?" This can be written as follows:

-x < 1?
or, by multiplying both sides by -1, we get
x > -1?

Putting these two cases together, we get the fully rephrased question:
“Is -1 < x < 1 (and x not equal to 0)"?

Another way to achieve this rephrasing is to interpret absolute value as distance from zero on the number line. Asking "Is |x| less than 1?" can then be reinterpreted as "Is x less than 1 unit away from zero on the number line?" or "Is -1 < x < 1?" (The fact that x does not equal zero is given in the question stem.)

(1) INSUFFICIENT: If x > 0, this statement tells us that x > x/x or x > 1. If x < 0, this
statement tells us that x > x/-x or x > -1. [color=#0000FF]This is not enough to tell us if -1 < x < 1.

(2) INSUFFICIENT: When x > 0, x > x which is not true (so x < 0). When x < 0, -x > x or
x < 0. Statement (2) simply tells us that x is negative. This is not enough to tell us if -1 < x < 1.

(1) AND (2) SUFFICIENT: If we know x < 0 (statement 2), we know that x > -1 (statement 1). This means that -1 < x < 0. This means that x is definitely between -1 and 1.

The correct answer is C[/color].


I would like to differ with the explanation. While option 1 straight forward tells us that |X|>1, the approach in the explanation is complicated.Request your inputs.

Okay. Here's the problem again:

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

(1) x/|x|< x

(2) |x| > x

If I rephrase this, the only way |x| can be less than 1 (given the constraint that x cannot equal zero) is if either of the following is true:
a) 0<x<1
OR
b)-1<x<0

In other words, is x a fraction between -1 and 0 or between 0 and 1?

Let's go to the statements. I'll start with statement two because it looks easier.
(2) |x| > x

Let's evaluate this. If x is positive, these values would be equal; thus, x must be negative. However, we don't know if x is a negative fraction. x=-5 and x=-1/2 would both fit this statement. INSUFFICIENT.

Statement one:
(1) x/|x|< x

We know that |x| must be positive. Let's try to plug some numbers:
x=3 will work for this statement (1<3). Taking x=3 to answer our original question, the answer is NO.
x=1/2 will not work for this statement. Move on.
x=-1/2 will work for this statement (-1<-1/2). Taking it to answer our original question, the answer is YES.
After plugging numbers, we have a NO/YES. This statement is INSUFFICIENT.

Since we're going to need to combine the statements next, let's try a negative integer with statement one to see what happens:
x=-3 gives us -1<-3--this does not work.

So the only values that fit statement one are either negative fractions between -1 and 0 or positive integers greater than 1.

Combining the statements, (2) tells us x must be negative, and (1) tells us x must be either between -1 and 0 or greater than 1. Thus, together the statements tell us that x must be between -1 and 0. SUFFICIENT

I hope this helps.

_________________
Jamie Nelson
ManhattanGMAT Instructor

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

(1) x/|x|< x

(2) |x| > x

Thank You Nelson, but why do one like to plug the numbers in the statement 1, when equation very clearly yields |X|>1, on cancellation of X on both the sides. As Eq 1 clearly tells |X|>1, we can safely answer the question that |X| is not less than 1 and hence the statement 1 is sufficient. This was my line of analysis. If my analysis is right, answer to the question should be A.

Here's the problem. I've demonstrated that both 3 and -1/2 are valid possible values for x in statement 1. However, if I use your equation, |x|>1, -1/2 no longer works for x. Whoops! Something must have been lost when you changed x/|x|< x into |x|>1. It is clear that |x|>1 is NOT equivalent to the original statement. Please go back and review your manipulation. It's clear that you made a mistake or did something that is not mathematically allowed.

I like to pick numbers and test them precisely because I avoid these kinds of issues. It's also quick and efficient. I'd be *very* nervous about trying to simplify an inequality with unknown variables and absolute value expressions, because I could fall into the trap that you fell into.

_________________
Jamie Nelson
ManhattanGMAT Instructor