Absolute value DS

[aliensoybean]

MGMAT

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

(1) x/|x| < x

(2) |x| > x

OA is C, my answer is E.

Here is how I understand it.

Rephrasing the question, is -1<x<1 (not including 0) ?

1. Two scenarios result depending on whether X is positive or negative
a. if X is positive then...
X/X<X
1<X

b. if x is negative then...
-X/X<-X
-1<-X
1>X

okay, from this statement, if you were to graph on a number line , it would be 1<X , X<1
INSUFFICIENT

2. This tells me that X is a negative, could be a negative number or negative fraction. INSUFFICIENT


3. Together, since statement 2 tells me that X is negative, then, the following scenario is accurate:
-X/X<-X
-1<-X
1>X

Okay, X is less than 1, but is it -1<x<1 ? There is no way of telling. Hence it is E. Not C.

Please help if you think my reasoning in incorrect

[shaji]

MGMAT

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

(1) x/|x| < x

(2) |x| > x

OA is C, my answer is E.

Here is how I understand it.

Rephrasing the question, is -1<x<1 (not including 0) ?

1. Two scenarios result depending on whether X is positive or negative
a. if X is positive then...
X/X<X
1<X

b. if x is negative then...
-X/X<-X
-1<-X
1>X

okay, from this statement, if you were to graph on a number line , it would be 1<X , X<1
INSUFFICIENT

2. This tells me that X is a negative, could be a negative number or negative fraction. INSUFFICIENT


3. Together, since statement 2 tells me that X is negative, then, the following scenario is accurate:
-X/X<-X
-1<-X
1>X

Okay, X is less than 1, but is it -1<x<1 ? There is no way of telling. Hence it is E. Not C.

Please help if you think my reasoning in incorrect

The re-phrasing should have been is x a fraction? since mod(x) is always >0.
Can x be -1 04 -6; NO x has to be a fraction .

[StaceyKoprince]

There's an error in your calculation for statement 1 part b:

if x is negative then...
-X/X<-X
-1<-X
1>X

If x is negative, then when you write just X, this represents a negative number. When you write -X, this represents a positive number. So you don't write -X/X<-X. You write X/(-X) < X. The two plain X's represent negative numbers and the -X represents a positive number. Then, when you simplify, you get -1 < X.

This is confusing, I know - it might be easier to see your mistake if you pick a random value for X (say, -3) and see how that works. If X = -3, I wouldn't write -(-3) / (-3) < -(-3) (which is equivalent to what you did with your variables). I would simply write -3 / -(-3) < -3. The X substitutes for the entire statement "-3" not just for the 3.

Now, when you put the two statements together, you know that x is both negative and greater than -1, so it must be between -1 and 0.

[Guest]

MGMAT - Cat Exam Question 6

If x is not equal to 0, is |x| < 1?

1) x/|x| < x
2) |x| > x

Now, I understand how you get -1<x from statement 1, but I don't understand statement 2. Do you always convert abs value into a negative # or a positive # when converting this type of statement?

[StaceyKoprince]

For statement 2, if the absolute value of a number is greater than that number itself, it means the number is a negative number - try some real numbers again.

If x = 3, then the absolute value of 3 is not greater than 3. It's equal.
If x = -3, then the absolute value of -3 is 3, which is greater than -3. This will only be true for negative values of x.

Generally speaking, in an absolute value statement, something about the sign is going to be important, since an absolute value sign is all about taking the positive value of a number. You just have to take a look at the particular statement and see what they're trying to tell you.

Then, when you're studying, tell yourself: oh, so an alternative way for them to tell me the simple information that x is negative is to show |x| > x . I'll memorize that so that I don't have to figure it out all over again from scratch when I'm taking the test.

[Guest]

Stacey, I still don't understand this...You wrote:

"I would simply write -3 / -(-3) < -3." But with this calculation wouldn't I get -1 < -3 ? That doesn't make sense..What am I missing?

There's an error in your calculation for statement 1 part b:

if x is negative then...
-X/X<-X
-1<-X
1>X

If x is negative, then when you write just X, this represents a negative number. When you write -X, this represents a positive number. So you don't write -X/X<-X. You write X/(-X) < X. The two plain X's represent negative numbers and the -X represents a positive number. Then, when you simplify, you get -1 < X.

This is confusing, I know - it might be easier to see your mistake if you pick a random value for X (say, -3) and see how that works. If X = -3, I wouldn't write -(-3) / (-3) < -(-3) (which is equivalent to what you did with your variables). I would simply write -3 / -(-3) < -3. The X substitutes for the entire statement "-3" not just for the 3.

Now, when you put the two statements together, you know that x is both negative and greater than -1, so it must be between -1 and 0.

[StaceyKoprince]

Oh - don't use that to actually solve the problem! -3 is not actually a valid value for x in this problem (given both statements) - that's why the statement is false. I was just using that example to illustrate the specific mistake that a previous poster made.

Here's an example with a valid number: x could be -1/2, according to the final solution. So let's try that:

(-1/2) / -(-1/2) < -1/2
which simplifies to:
(-1/2) / (1/2) < -1/2
-1 < -1/2
which is true

[Guest]

Statement 1:
No matter what is value of X is (except 0 as given in the question), X/|X| is always 1 or -1.
So this statement is essentially telling us that either x>1 or x>-1 but nothing about |x|, so insufficient

Statement 2:
Says |X| > X. It is true if x is -ve. But that means |X| is +ve. But question is asking is |X|<1. So it cannot be answered using this statement alone.

Combining both Statements:
From 2 we know x is -ve. Now we know using 1 that that X > -1.
Any -ve number greater than -1 has to be a fractional number. Moreover its mode will always be less than 1 as well.

So question is answered using both, Answer C.

[StaceyKoprince]

You've got it!