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Is |x| < 1 ? ----------------------- (1) |x + 1| = 2|x 1|

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Senior Manager
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Is |x| < 1 ? ----------------------- (1) |x + 1| = 2|x 1| [#permalink] New post 28 Jun 2007, 07:16
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A
B
C
D
E

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Is |x| < 1 ?

-----------------------
(1) |x + 1| = 2|x – 1|

------------------------


(2) |x – 3| > 0
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 [#permalink] New post 28 Jun 2007, 09:46
B for me

Stmt1 says x=1/3 or 3 So insufficient
Stmt 2 gives |x| > 3. So B
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Re: DS: Absolute Range [#permalink] New post 28 Jun 2007, 11:09
ArvGMAT wrote:
Is |x| <1> 0


C for me.

a: if (x + 1) is +ve:
x + 1 = 2x – 2
x = 3

if (x + 1) is -ve:
- x - 1 = 2x – 2
x = 1/3. nsf.

b: x is not equal to 3. nsf.

a & b: x is not equal to 3 means it is equal to 1/3. sufficient.

I have little doubt that i missed something in a.
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 [#permalink] New post 28 Jun 2007, 18:28
Himalayan:
if (x+1) is -ve wouldnt (x-1) also be -ve,so
wouldn't it be -(x+1)=-2(x-1) ?=>x=3
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 [#permalink] New post 28 Jun 2007, 19:29
ynot wrote:
Himalayan:
if (x+1) is -ve wouldnt (x-1) also be -ve,so
wouldn't it be -(x+1)=-2(x-1) ?=>x=3


yah i know i am missing something there. therefore i mentioned my skepticism in my earlier post too.

its too much confusion when you do inequalities and modulas. i will rectify later in a fresh mood. i am too much exhusted now.
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 [#permalink] New post 28 Jun 2007, 21:30
A for me .1 gives x =3 in both cases (+ and -)
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 [#permalink] New post 28 Jun 2007, 22:27
Go with B.

Statement B tells that 3>x>2. So we know clearly that whether |x| < 1, in this case, NO.
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Re: DS: Absolute Range [#permalink] New post 28 Jun 2007, 22:57
ArvGMAT wrote:
Is |x| < 1 ?

-----------------------
(1) |x + 1| = 2|x – 1|

------------------------


(2) |x – 3| > 0


I say C. :-D

stmt 2: x <> 3. INSUFFICIENT

stmt 1:
case 1: x < -1
-(x+1) = -2*(x-1) => x=3

case 2: x > 1
(x+1) = 2*(x-1) => x=3

case 3: -1 < x < 1
(x+1) = -2*(x-1) => x=1/3

So, INSUFFICIENT.

From 1 and 2, x=1/3. So, SUFFICIENT.
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 [#permalink] New post 28 Jun 2007, 23:41
(C) for me too :)

|x| < 1 ?
<=> -1 < x < 1 ?

From 1
|x + 1| = 2|x – 1|

o If x > 1, then:
|x + 1| = 2|x – 1|
<=> x+1 = 2*(x-1) as x+1 > 0 and x-1 > 0
<=> x = 3 >>> Ok, as x > 1

o If -1 < x =< 1, then:
|x + 1| = 2|x – 1|
<=> x+1 = 2*[-(x-1)] as x+1 > 0 and x-1 =< 0
<=> x = 1/3 >>> Ok, as -1 < x =< 1

o If x =< -1, then:
|x + 1| = 2|x – 1|
<=> -[x+1] = 2*[-(x-1)] as x+1 =< 0 and x-1 < 0
<=> x = 3 >>> Out as x =< -1

So, the solutions are 3 and 1/3

INSUFF.

From 2
|x – 3| > 0
<=> x-3 != 0
<=> x != 3

INSUFF.

Both 1 & 2
We have:
o x=3 or x=1/3
or
o x != 3

Thus x = 1/3

SUFF.
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 [#permalink] New post 28 Jun 2007, 23:43
ynot wrote:
Himalayan:
if (x+1) is -ve wouldnt (x-1) also be -ve,so
wouldn't it be -(x+1)=-2(x-1) ?=>x=3


Yes... But we must remove this solution x=3 because if both x+1 < 0 and x-1 <0, we know that x must be < -1.... so x cannot be 3 if x < -1 :)
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 [#permalink] New post 29 Jun 2007, 02:09
dreamgmat1 wrote:
A for me .1 gives x =3 in both cases (+ and -)


A for me as well.

Stat 1 gives x=3 both when x<0>0

Stat 2 gives x>3 when x>0 and x<3 when x<0

Let me know if you want to see the solution.
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 [#permalink] New post 29 Jun 2007, 02:24
GK_Gmat wrote:
dreamgmat1 wrote:
A for me .1 gives x =3 in both cases (+ and -)


A for me as well.

Stat 1 gives x=3 both when x<0>0

Stat 2 gives x>3 when x>0 and x<3 when x<0

Let me know if you want to see the solution.


In stat 1, try to plug in x = 1/3 :)

o |1/3 + 1| = 4/3
o 2|x – 1| = 2*|1/3-1|= 2*|-2/3| = 2 * 2/3 = 4/3

So, we have x=3 and x=1/3 as solutions... making it insufficient to answer the question -1 < x < 1.
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 [#permalink] New post 30 Jun 2007, 03:03
Fig wrote:
(C) for me too :)

|x| < 1 ?
<=> -1 < x <1> 1, then:[/b]
|x + 1| = 2|x – 1|
<x> 0 and x-1 > 0
<x>>> Ok, as x > 1

o If -1 < x =< 1, then:
|x + 1| = 2|x – 1|
<x> 0 and x-1 =< 0
<x>>> Ok, as -1 < x =< 1

o If x =< -1, then:
|x + 1| = 2|x – 1|
<=> -[x+1] = 2*[-(x-1)] as x+1 =< 0 and x-1 < 0
<x>>> Out as x =<1> 0
<=> x-3 != 0
<=> x != 3

INSUFF.

Both 1 & 2
We have:
o x=3 or x=1/3
or
o x != 3

Thus x = 1/3

SUFF.


Perfect. I solved the same way too.
  [#permalink] 30 Jun 2007, 03:03
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