Data Sufficiency

Is |x|< 1?
(1) |x + 1| = 2|x - 1|
(2) |x - 3| &#8800; 0

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

stmt 1 - there can be two solutions for x in this case.
x+1 =2(x-1)...... x= 3.
x+1 =2(1-x)....when modulus is used both conditions of difference need to be considered....In this case x = 1/3.....Since x has two values ...Insuff

stmt 2 - x not equal to 3 is the sol.....Insuff

Both - the only possible sol is x=1/3....Suff ...........So ans is C.

can you pls confirm the answer. Thanks

Hi

Can some one pls help me whit this one?

This is the way I solved statement 1)

lx+1l = 2lx-1l

so there are two options

Option 1)

(x+1) = 2 (x-1), in which case x = 3

Option 2)

-(x+1) = -2(x-1), in which case X = 3

Both the cases X > 1, so statement 1 is sufficient

However Jess has a diff ans. Can any one confirm the right approach and explain it?

Thanks

Hi ri2007,


Your option 1 is the same as the option 2.

When we have a modulus on a side that means it can have two values +/-. When there is a modulus on both sides that means that both sides will have a +/-

|x+1| = 2|x-1|

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

which leads to the following options:

x+1 = 2(x-1) or -(x+1) = - 2(x-1) (they are same)

AND

-(x+1) = 2(x-1) or x+1 = -2(x-1) (they are same as well)

Thanks so much us. Great explaination