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

[barlie]

Heres a question that I think is incorrectly explained.

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

(2) |x| > x

The explaination given in the answers is below
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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. 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.
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Look at the question carefully.
1. Tells us that x / |x| < x
or 1/|x| < 1
This will only hold if |x| > 0 or x E {-infinity,-1} U {1, infinity}. Eitherway, its sufficent to say that |x| > 1 which is what the question is asking

2. Tells us that |x| > x
This will only hold if x E {-infinity,0} and does no mean that |x| > 1.
eg |-0.25| = 0.25. If we assume x to be integers only (which the question does not state) then its sufficentl

In either case, (C) is incorrect. It should be A or D.



Your response ?

[StaceyKoprince]

Please remember to post the source (author and book / number / location of problem). Instructors cannot respond until we have this. Luckily, I recognize this one as one of ours.

I'm not following this:

1. Tells us that x / |x| < x
or 1/|x| < 1

The first bit is what the problem says; you rephrase it to 1/|x| < 1. How did you do this? It seems like you divided each side by x, but I'm not sure. If that is what you did, the statement is incomplete because we don't know if x is positive or negative. When we divide an inequality by a negative, we have to flip the inequality sign. So I would get two options:
if x is positive, then 1/|x| < 1
if x is negative, then 1/|x| > 1
And BOTH options now have to be tested.

You can see that this is the case more intuitively if you try some real numbers. From the question stem, x is not zero but could be anything else.
First, let's try x = -0.5. If I plug this into x / |x| < x I get -.5/.5 < -.5 or -1 < -.5 which is true, so -0.5 is a valid value for x. In this case, the absolute value of x is less than 1.
Next, let's try x = 2. If I plug this into x / |x| < x I get 2/2 < 2 or 1 < 2 which is true, so 2 is a valid value for x. In this case, the absolute value of x is greater than 1.
Contradictory results = insufficient.

Re: your point about the second statement, assume NOTHING. You can't assume that x is an integer. It is only limited to integers if the question tells you so; otherwise, decimals are possible.

[barlie]

Yes source : manhattan gmat cat test.

This is what I had said initially.

Look at the question carefully.
1. Tells us that x / |x| < x
or 1/|x| < 1
This will only hold if |x| > 0 or x E {-infinity,-1} U {1, infinity}. Eitherway, its sufficent to say that |x| > 1 which is what the question is asking

I got 1/|x| < 1 from the orginal equation by diving both sides by x and you're right. I suppose i overlooked something

x / |x| < x

=> x / |x| -x <0
=> x (1 -|x|) /|x| < 0
=> x (1 -|x|) < 0

This will only hold if either x <0 and 1-|x| > 0 => |x| < 1
OR
x > 0 and (1-|x|) < 0 => |x| > 1

[StaceyKoprince]

Yep, you got it! Which means we can't solve.

[Guest]

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:

I cant understand the logic above - can you explain?

how do we get "is -x less than 1?

[kylo]

from stat1 -
x / |x| < x
lxl > 1

hence |x| is not less than 1.

hence IMO A.



Thanks!

[JonathanSchneider]

The rephrase ("Is -x < 1?") come from the original question: "If x is not equal to 0, is |x| less than 1?"

Note that for lxl to be less than 1, x can be between -1 and 1. So, the question could originally be rephrased as: "Is x between -1 and 1?"

Here we are only examining case #2, where x is negative. If x is negative, then x < 0. Thus, the question "Is x between -1 and 1" can be limited to the following: "Is x greater than -1?" (The reason for this simplification is that, if x is negative, all that we care about is whether it is greater than -1.) In other words: "Is it true that -1 < x?" Note that the original rephrase that you asked about (listed at the top of this post) is merely the exact negative of the question just written here. (In other words, it's the same inequality, but multiplied through by -1, which flips the sign.) Why would we ask it that way? Because we are discussing absolute value. If x is negative, the absolute value bars make it positive by multiplying it by -1. We want to know if that value is less than 1.

[vinversa]

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

I agree with Answer C. This is how I solved this one...

is |x| less than 1?
indirectly means is -1<x<1 ?

GMAT given rule says
|x| = x if x>=0-----(1)
|x| = -x if x<0------(2)

Question says x is not equal to 0, therefore (1) can be eliminated because (1) needs x to be >==0. Hence (2) holds good.... |x| = -x if x<0------(2)

(1) x / |x| < x
x/-x < x
-1<x (INSUFF),

(2) |x| > x
-x>x (bring -x to the right)
0>x+x
0>2x
x<0 (INSUFF)

(1) & (2)
Gives
-1<x<0, which is SUFFICIENT to prove -1<x<1
Therefore C is good
CCCCCCCCC is good....

[mschwrtz]

That looks right vinversa. I have to admit that I don't quite follow your words, but the math part looks good. Certainly you got the right answer.