Thepu,
If u go with stmt-1,
if x <1> 1, x = 3 -> Alone not SUFF
if u go with stmt-2,
x /= 3 -> Alone not SUFF.
Even if u go for both collectively, u can not reach to any conclusion.
IMO E. What's the OA? Guys am I missing something?

see..

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

if x>1 :: x+1 = 2x - 2 => x =3.

if -1 < x <1> x = 1/3.

if x<1> x = 3. This solution isn't right as x should be less thsn -1.

Now stmt 2 rules out x = 3 as a possibility.

So the correct ans is x= 1/3 and c.

Ohhh! What a silly miss! Thanks ikant 4 pointing the miss I did.

Hi,

A)
When you have Modulus on both side, I use a simpler approach

Square both sides, this removes any sign issues

(x+1)^2 = [2(x-1)]^2

x^2 + 2x + 1 = 4( x^2 - 2x + 1)

x^2 + 2x + 1 = 4 x^2 - 8x + 4

0 = 3x^2 - 10x +3

0 = 3x^2 - 9x - 1x +3
0 = 3x(x-3) -1(x -3)
0 = (x-3)( 3x-1)
=> (3x-1)= 0 or (x-3)=0
i.e x =1/3 or x = 3

not sufficient

B)|x-3| not equal to 0

means x not equal to 3

not sufficient

But comining A and B we see that x = 1/3 which is sufficient
answer is C

I got it ,thanks.

The last method is really help