Tough one

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Source: — Data Sufficiency |

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by gmat009 » Fri Nov 14, 2008 12:50 pm
Is it A

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by schumi_gmat » Fri Nov 14, 2008 12:58 pm
1. mod(x+1) = 2 mod (x-1)

x+1>0 i.e x>-1 hence x-1>-2
so x-1 can be -ve or positive

Hence, for x>-1 and x-1>0
x+1 = 2(x-1)
x=3

For x>-1 and X-1 <0

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

for x<-1, x-1<0
-x-1 = -2(x-1)
x = 3

Hence we have 2 values and not sufficient

2) mod(x-3) not equal 0
i.e x not equal 3

Hence not sufficient

combining, we know that x not equal 3
and hence x=1/3

so mod(x) < 1

IMO C

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by gmat009 » Fri Nov 14, 2008 1:12 pm
schumi_gmat wrote:1. mod(x+1) = 2 mod (x-1)

x+1>0 i.e x>-1 hence x-1>-2
so x-1 can be -ve or positive

Hence, for x>-1 and x-1>0
x+1 = 2(x-1)
x=3

For x>-1 and X-1 <0

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

for x<-1, x-1<0
-x-1 = -2(x-1)
x = 3

Hence we have 2 values and not sufficient

2) mod(x-3) not equal 0
i.e x not equal 3

Hence not sufficient

combining, we know that x not equal 3
and hence x=1/3

so mod(x) < 1

IMO C
Yes it should be C. I missed one case

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by Tryingmybest » Fri Nov 14, 2008 1:21 pm
Here is a method if you want to avoid mistakes while plugging numbers
|x|< 1 means -1<x<1 is the question?

Statement 1.
|x + 1| = 2|x - 1|
Squaring on both sides
(x+1)^2 = 4 (x-1)^2
3x^2-10x+3 = 0
(3x-1)(x-3)=0
we get x= 3 or 1/3
It is insuficient
Statement 2:
|x - 3| &#8800; 0 => X cannot be 3
It is insufficient

Combining both X = 1/3 remains . Satisfies-1<x<1 so C :D

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by bacali » Fri Nov 14, 2008 1:44 pm
OA: C

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by logitech » Fri Nov 14, 2008 4:29 pm
Tryingmybest wrote:Combining both X = 1/3 remains . Satisfies-1<x<1 so C :D
It does not need to satisfy the -1<x<+1 - It is a YES or NO question 8)
LGTCH
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