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

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If x is not equal to 0, is |x| less than 1? x > x/|x| |x| [#permalink]  18 Dec 2007, 10:02
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If x is not equal to 0, is |x| less than 1?

x > x/|x|

|x| > x
CEO
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Re: DS abs values [#permalink]  18 Dec 2007, 10:08
young_gun wrote:
If x is not equal to 0, is |x| less than 1?

x > x/|x|

|x| > x

1. SUFF
All values over 1 make the equation hold

2. INSUFF
positive fractions do not make the equation hold and negative numbers make the equation hold.

A
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Expert's post
C

The expression "If x is not equal to 0 and |x| less than 1" is true for x e (-1,0)&(0,1)

x > x/|x| : x e (-1,0)&(1,+∞). the expression is true for x e (-1,0) and false for x e (1,+∞). INSUFF.

|x| > x: x e (-∞,0). the expression is true for x e (-1,0) and false for x e (-∞,1). INSUFF.

1&2: x e (-1,0). SUFF
Director
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Re: DS abs values [#permalink]  18 Dec 2007, 16:52
young_gun wrote:
If x is not equal to 0, is |x| less than 1?

x > x/|x|

|x| > x

A

Stmnt 1

X(|x| -1 ) > 0
means x and |x| - 1 should have same sign.
if x > 0 then |x| - 1 > 0 =>|x| > 1 . Since x>0 , x > 1
if x <0 then |x| - 1 <0> -x -1 > 0 => x > -1

So x is at least not less than 1 if we have to satisfies above equations.
Suff

Stmnt 2
obviously not sufficient.
Manager
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walker - U good =)
I got A initially ...

But could you please explain me the following:
"x > x/|x| : x e (-1,0)&(1,+∞). the expression is true for x e (-1,0) and false for x e (1,+∞). INSUFF. "

x > x/|x|: if x is -ve, then we have x > - x/x ..> x > -1, why (mathematically) do you take range (-1,0) and not (-1,+∞) excluding 0?
Senior Manager
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I think it is A.

i got C initially and then had to rethink.

The given condition can be established only if x E (-1,1)

For 1) x/|x| => x E (-1,0). So this is Sufficient.

For 2) |x| > x => x E (-∞,0). This is insufficient.

What is the OA ?
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Expert's post
Whatever wrote:
But could you please explain me the following:
"x > x/|x| : x e (-1,0)&(1,+∞). the expression is true for x e (-1,0) and false for x e (1,+∞). INSUFF. "

x > x/|x|: if x is -ve, then we have x > - x/x ..> x > -1, why (mathematically) do you take range (-1,0) and not (-1,+∞) excluding 0?

x is -ve and x > -1 give you (-1,0)
You've lost (-∞,0) condition.
Director
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Re: DS abs values [#permalink]  18 Dec 2007, 22:40
young_gun wrote:
If x is not equal to 0, is |x| less than 1?

x > x/|x|

|x| > x

Getting C.

Is |x| < 1?
If x > 0 then |x| = x and the question becomes is x < 1
If x < 0 then |x| = -x and the question becomes is x > -1
Combining the 2, the question becomes is -1 < x < 1?

Stat 1:
Since |x| is always +ve, we multiply each side by |x|
x|x| > x
x|x| - x > 0
x(|x| - 1) > 0
x > 0 or |x| - 1 > 0
x > 0 or |x| > 1
x > 0 or x > 1 & x < -1
Insuff.

Stat 2:
Tells us that x < 0. Insuff.

Together:
If x < 0 as per stat 2, then to be less than -1 as per stat 1. This means that x does not fall in the range of -1 < x < 1. The answer is a definite no. Suff.

Can someone confirm that above approach is correct? Thanks.
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Re: DS abs values [#permalink]  19 Dec 2007, 14:11
young_gun wrote:
If x is not equal to 0, is |x| less than 1?

x > x/|x|

|x| > x

Its clearly C.

1: 2>2/|2| |2| greater than 1

-1/2>-1/2|-1/2| -1/2>-1 --> |-1/2| less than 1 Insuff b/c its less than and greater than 1.

2: could be anything |-2|>-2 greater than 1 or again -1/2

together it can only be -1/2 out of our choices. Cant be 2 or -2

2>2/|2| but |2| is not greater than 2.

-2>-2|-2| -2>-1??? Nope

Only -1/2 works for both.
Re: DS abs values   [#permalink] 19 Dec 2007, 14:11
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