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Is |x| < 1 ? ----------------------- (1) |x + 1| = 2|x 1| [#permalink ]
28 Jun 2007, 07:16

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Is |x| < 1 ?
-----------------------
(1) |x + 1| = 2|x – 1|
------------------------
(2) |x – 3| > 0

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B for me
Stmt1 says x=1/3 or 3 So insufficient
Stmt 2 gives |x| > 3. So B

Director

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Re: DS: Absolute Range [#permalink ]
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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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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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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A for me .1 gives x =3 in both cases (+ and -)

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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 ]
28 Jun 2007, 22:57

ArvGMAT wrote:

Is |x| < 1 ? ----------------------- (1) |x + 1| = 2|x – 1| ------------------------ (2) |x – 3| > 0

I say C.

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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(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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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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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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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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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.