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

[Guest]

Original Problem:

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

(1) x/|x| < x

(2) |x| > x

I understand the rephrase and why this has to be true: -1 < x < 1 BUT, I don't understand the manipulations of statement #1. Can you please help me understand:
o (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.

[StaceyKoprince]

See thread
http://www.manhattangmat.com/forums/quantitative-question-22-on-cat-1-t504.html

[unique]

See thread
http://www.manhattangmat.com/forums/quantitative-question-22-on-cat-1-t504.html

Stacey,

Pls explain 1 in detail. I am missing the part when you consider the 2 conditions.

[StaceyKoprince]

Sure!

(1) x/|x| < x

In order to remove the absolute value sign in an equation or inequality, I have to consider what would happen if what is contained within that sign is positive and also what would happen if what is contained within that sign is negative.

If that value inside the signs is positive, I can just remove the absolute value signs: x/(x) < x or 1 < x
If the value is negative, however, I have to add that negative into the inequality : x/(-x) < x or -1 < x

For the positive result, 1 < x, or x > 1, if x is greater than 1, then it cannot be between -1 and 1. So I can answer the question definitively: no.

For the negative result, -1 < x, or x > -1, x could be be, say 1/2, in which case it is between -1 and 1. Or x could be, say, 42, in which case it is not betwen -1 and 1. I can't answer this definitively.

If I can't answer ALL of the scenarios definitively with the same answer, then the statement is insufficient.

[unique]

Sure!

(1) x/|x| < x

In order to remove the absolute value sign in an equation or inequality, I have to consider what would happen if what is contained within that sign is positive and also what would happen if what is contained within that sign is negative.

If that value inside the signs is positive, I can just remove the absolute value signs: x/(x) < x or 1 < x
If the value is negative, however, I have to add that negative into the inequality : x/(-x) < x or -1 < x

For the positive result, 1 < x, or x > 1, if x is greater than 1, then it cannot be between -1 and 1. So I can answer the question definitively: no.

For the negative result, -1 < x, or x > -1, x could be be, say 1/2, in which case it is between -1 and 1. Or x could be, say, 42, in which case it is not betwen -1 and 1. I can't answer this definitively.

If I can't answer ALL of the scenarios definitively with the same answer, then the statement is insufficient.

Stacey,

Thanks for taking the time to answer this in detail.
Here lies my confusion - let me know where I am going wrong.

If the value is negative, however, I have to add that negative into the inequality : x/(-x) < x or -1 < x

If the value is -ve which means x is -ve

-x/x < -x (bcos l-xl is positive )
-1 < -x
1 > x
x < 1

[unique]

Sure!

(1) x/|x| < x

In order to remove the absolute value sign in an equation or inequality, I have to consider what would happen if what is contained within that sign is positive and also what would happen if what is contained within that sign is negative.

If that value inside the signs is positive, I can just remove the absolute value signs: x/(x) < x or 1 < x
If the value is negative, however, I have to add that negative into the inequality : x/(-x) < x or -1 < x

For the positive result, 1 < x, or x > 1, if x is greater than 1, then it cannot be between -1 and 1. So I can answer the question definitively: no.

For the negative result, -1 < x, or x > -1, x could be be, say 1/2, in which case it is between -1 and 1. Or x could be, say, 42, in which case it is not betwen -1 and 1. I can't answer this definitively.

If I can't answer ALL of the scenarios definitively with the same answer, then the statement is insufficient.

Stacey,

Thanks for taking the time to answer this in detail.
Here lies my confusion - let me know where I am going wrong.

If the value is negative, however, I have to add that negative into the inequality : x/(-x) < x or -1 < x

If the value is -ve which means x is -ve

-x/x < -x (bcos l-xl is positive )
-1 < -x
1 > x
x < 1

Can Stacey or some MGMAT staff elaborate on my questions above. Let me know if I am making mistake when considering the negative part. I really want to know. Thanks for all the time you take to answer even the silliest of questions.

[StaceyKoprince]

Nope, not the way to solve it. It is the case that the absolute value of x is positive. But we are attempting to REMOVE the absolute value sign from the equation. The way to do this is to consider the positive and negative values for whatever is inside that absolute value sign - that's what you change in the 2nd equation, not the other representations of x.

What you've essentially done is insert new variables for x by swapping the definition: now "x" means negative x and "-x" means positive x. So x < 1 technically means -x < 1, which simplifies to x > -1.