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Source: — Data Sufficiency |
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by ronniecoleman » Mon Dec 22, 2008 12:49 am
| X+1| = 2|X-1|

for x > 1

x = 3

for -1 <x < 1
x = 1/3

for x < -1

x = 3 ( discard)


2. says x != 3


so From 1 & 2 x < 1

IMO C
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Re: modulus

by tritrantran » Mon Dec 22, 2008 12:52 am
Vignesh.4384 wrote:Pls help.
Question is asking if -1<x<1 ?

(1)
x+1 = 2|x-1|, x = 3

x+1 = 2*(-x+1), x = 1/3

INSUFF

(2)
x not = 3

INSUFF

(1) and (2)

X = 1/3

Answer C
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by vittalgmat » Mon Dec 22, 2008 1:21 am
Key info regarding absolute values

|x| = x if x is +ve
AND
|x| = -x if x is -ve


Stmt 1:
|x+1| = 2|x-1|

Case 1: x is +ve
x+1 = + 2(x-1)
x+1 = 2x -2
x = 3

Case 2: x is -ve

x+1 = - 2(x-1)
x+1 = -2x +2
x = 1/3

2 values... so Insufficient.


Stmt 2:
x <> 3

Insufficient.

Combining the two stmts: x can be equal to 1/3.

So sufficient. Ans C

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by crazy4gmat » Mon Dec 22, 2008 8:43 am
I have a slight confusion here

Case 2: x is -ve

here shouldnt it be -( x+1) = - 2(x-1) instead of x+1 = - 2(x-1) ?
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by vittalgmat » Mon Dec 22, 2008 10:24 am
You are right crazy4gmat.
Pls see my solution above.

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