Data Sufficiency

Is |x|< 1?
1) |x + 1| = 2|x - 1|
(2) |x - 3| ≠ 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.

I guess the answer should be A
1) If X =+ve
X+1=2X-2; x=3
If X=-ve
-X+1=-2X-2; x=1
Both the cases X>1
so we can answer the question.
(1) is sufficient
2) does not make any sense as |x-3| can be any value
Kindly help me in finding out the correct answer [/img][/i]

From stmt1,

If X =+ve

X+1=2X-2; x=3

If X=-ve

The equation cannot be solved.[Its confusing here]

IMO A is insufficient

From stmt2 |x-3| ≠ 0

-->x cannot be equal to 3.

x can be any value like -1,0,2 except 3.

If x=0,|X|<1

If x=2,|x|>1

So B is insufficient


Answer is E

Statement 1:

1) -1< x < 1, then:
x+1 = -2x+2 or x=1/3

2) If x <- 1, then
-x-1 = -2x+2 or x=3

3) If x>1, then
x+1= 2x-2 or x=3

- insufficient


Statement 2: insufficient

Statement 1 & 2 - sufficient since it rules out x < -1 & x >1

Hence, C

---
edited to correct careless errors

my small thinking
is -1<x<1 ??
(1) |x+1|=2|x-1| here two way are possible one to square both parts and solving quadratic equation, and the second to open module step by step
_________ -1_________1 __________

|x+1|={(x+1) if x>-1. or -(x+1) if x<-1}

|x+1|={(x-1) if x>1, or -(x-1) if x<1}

the first interval
_________ -1, or x<-1

-(x+1)=-2(x-1)
x=3 but 3 does not fit in intrval x<-1
so we have no roots here

the second interval
(-1,1)

(x+1)=-2(x-1)
x=1/3 it can be root of given equation

the last interval x>1

x-3=2(x-1)
x=3 also can be root
so we have 2 answers x=3 not belong to (-1,1)
and 1/3 belong to (-1,1)
1st st insufficient

(2) x=/=3 insufficient

both 2st put off 3 as answer so we left with 1/3 within (-1,1)
the answer C

btw not sure if it would help but Manhattan GMAT's tutorial on absolute values really helped me in solving this one.

https://www.manhattangmat.com/tutorials/ ... -value.cfm

clock60 wrote:my small thinking
is -1<x<1 ??
(1) |x+1|=2|x-1| here two way are possible one to square both parts and solving quadratic equation, and the second to open module step by step
_________ -1_________1 __________

|x+1|={(x+1) if x>-1. or -(x+1) if x<-1}

|x+1|={(x-1) if x>1, or -(x-1) if x<1}

the first interval
_________ -1, or x<-1

-(x+1)=-2(x-1)
x=3 but 3 does not fit in intrval x<-1
so we have no roots here

the second interval
(-1,1)

(x+1)=-2(x-1)
x=1/3 it can be root of given equation

the last interval x>1

x-3=2(x-1)
x=3 also can be root
so we have 2 answers x=3 not belong to (-1,1)
and 1/3 belong to (-1,1)
1st st insufficient

(2) x=/=3 insufficient

both 2st put off 3 as answer so we left with 1/3 within (-1,1)
the answer C

ST 1 :1) |x + 1| = 2|x - 1|

case 1 :

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

x +1 = 2x-2

x = 3

case 2 :

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

-x-1 = -2x+2

x=3

case 3 :

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

-x-1 = 2x-2

-3x= -1

x=1/3

case 4:

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

3x =1

x= 1/3


We have 2 values x = 3 or x = 1/3. Inconsistent as 3 > 1 & 1/3 <1

Insufficient

St 2:

|x - 3| Not equal to 0

So it means X is not equal to three

Still X can be any value. Say X can be 10 or 0 or 1. Inconsistent. SO insufficient

Now combining 2 statements:

we have X is NOT equal to 3 so we have x = 1/3 only

1/3 < 1. YES

Pick C..Move on !!

@gmatmachoman: in analyzing (1), you tested 2 too many cases! Because x=y and -x=-y are identical equations, your cases 1 and 2 are identical. Because x=-y and -x=y are identical equations, your cases 3 and 4 are identical.

Testluv wrote:@gmatmachoman: in analyzing (1), you tested 2 too many cases! Because x=y and -x=-y are identical equations, your cases 1 and 2 are identical. Because x=-y and -x=y are identical equations, your cases 3 and 4 are identical.

Yeah agreed!! Thx for that!! I am all worried about probabilities..somebody on this earth plz plz help me out in some way!!