hi,
decode statement 1 by taking two possible values of LHS of the eq's.

first take -x-1 = 2x-2.
which gives x = 1/3

then take x+1 = 2x-2
whih gives x = 3

now check if both these satisfy statement 1.

they do, hence statement 1 is not sufficient,

2 negates one of the values obtained above.

hence answer is c

x =1/3

bblast wrote:hi,
decode statement 1 by taking two possible values of LHS of the eq's.

first take -x-1 = 2x-2.
which gives x = 1/3

then take x+1 = 2x-2
whih gives x = 3

now check if both these satisfy statement 1.

they do, hence statement 1 is not sufficient,

2 negates one of the values obtained above.

hence answer is c

x =1/3

Bblast: can you please explain...you have taken two case for LHS i.e when |x-1| is positive and negative,y not 2 cases for RHS also??

Hi abhishek,

actually there are 4 cases :

lhs -ve rhs +ve case1
lhs -ve rhs -ve case2

lhs +ve rhs +ve case3
lhs +ve rhs -ve case4


As u can see case 2 and case 3 above are the same. similarly case 1 and case 4 are the same.
which are the actual 2 cases u need to consider.

-1 < x < 1?

Statement 1) |x + 1| = 2|x - 1| has two cases:

x + 1 = 2x - 2
x = 3

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

We always verify that both solutions satisfy the original equation:

|3 + 1| = 2|3 - 1|
|4| = 2|2|

|1/3 + 1| = 2|1/3 - 1|
|4/3| = 2|-2/3|

With these two possible values of x, statement 1 is not sufficient:

-1 < 1/3 < 1? yes
-1 < 3 < 1? no

Statement 2) |x - 3| ≠ 0

x - 3 ≠ 0
x ≠ 3

Note that no second case exists in this equation since -0 = 0.

With different possible values of x, statement 2 is not sufficient:

-1 < 0 < 1? yes
-1 < -500,000,000 < 1? no

Combined)
x = 3 OR x = 1/3
AND x ≠ 3
x = 1/3

-1 < 1/3 < 1

Sufficient.

C

we need to check if -1<x<1

a for 0.3<x<0.4 the LHS approx = RHS.
for x=3 LHS= RHS. not sufficient.

b x # 3 not sufficient.

a+b

x=0.33. sufficient.
C it is.