BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Is |x|< 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| ≠ 0

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Newbie | Next Rank: 10 Posts
Posts: 3
Joined: Tue Jun 01, 2010 10:46 am

Is |x|< 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| ≠ 0

by ankurshobhit » Tue Jun 01, 2010 11:00 am
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]
Top
Source: — Data Sufficiency |
User avatar
Legendary Member
Posts: 1460
Joined: Tue Dec 29, 2009 1:28 am
Thanked: 135 times
Followed by:7 members

by selango » Tue Jun 01, 2010 11:15 am
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
Last edited by selango on Tue Jun 01, 2010 11:33 am, edited 2 times in total.
Top
Master | Next Rank: 500 Posts
Posts: 186
Joined: Fri May 28, 2010 1:05 am
Thanked: 11 times

by jube » Tue Jun 01, 2010 11:29 am
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
Last edited by jube on Tue Jun 01, 2010 9:32 pm, edited 1 time in total.
Top
Legendary Member
Posts: 759
Joined: Mon Apr 26, 2010 10:15 am
Thanked: 85 times
Followed by:3 members

by clock60 » Tue Jun 01, 2010 12:51 pm
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
Top
Master | Next Rank: 500 Posts
Posts: 186
Joined: Fri May 28, 2010 1:05 am
Thanked: 11 times

by jube » Wed Jun 02, 2010 12:23 am
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
Top
Legendary Member
Posts: 2326
Joined: Mon Jul 28, 2008 3:54 am
Thanked: 173 times
Followed by:2 members
GMAT Score:710

by gmatmachoman » Wed Jun 02, 2010 4:11 am
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 !!
Top
GMAT Instructor
Posts: 1302
Joined: Mon Oct 19, 2009 2:13 pm
Location: Toronto
Thanked: 539 times
Followed by:164 members
GMAT Score:800

by Testluv » Thu Jun 03, 2010 2:16 am
@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.
Kaplan Teacher in Toronto
Top
Legendary Member
Posts: 2326
Joined: Mon Jul 28, 2008 3:54 am
Thanked: 173 times
Followed by:2 members
GMAT Score:710

by gmatmachoman » Sat Jun 26, 2010 10:23 pm
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!!
Top
Post new topic Post Reply
• Page 1 of 1