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# Is lxl < 1? (1) lx + 1l = 2lx - 1l (2) lx - 3l ≠ 0

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Is lxl < 1? (1) lx + 1l = 2lx - 1l (2) lx - 3l ≠ 0 [#permalink]

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02 Aug 2007, 10:05
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

1. Is lxl < 1?
(1) lx + 1l = 2lx - 1l
(2) lx - 3l ≠ 0
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cool

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02 Aug 2007, 11:17
jet1445 wrote:
1. Is lxl < 1?
(1) lx + 1l = 2lx - 1l
(2) lx - 3l ≠ 0

I get C for this one.

Stmt 1: There are 3 critical points.
x>1; -1<x<1; x<-1
Solving for each case, x=3 or 1/3. Insufficient.

Stmt 2: |x| ≠ 3. Insufficient.

Both together. x=1/3. sufficient.
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02 Aug 2007, 12:27
jet1445 wrote:
1. Is lxl < 1?
(1) lx + 1l = 2lx - 1l
(2) lx - 3l ≠ 0

Another C.

(1) Critical points are 1 and -1
if x<-1, x=3 NOT POSSIBLE
if -1<x<1, x=1/3
if x>1, x=3
Thus, x can either be 1/3 or 3, which is INSUFFICIENT.

(2) x cannot be 3 INSUFFICIENT

Together, x can only be 1/3 SUFFICIENT.
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02 Aug 2007, 12:35
1. Is lxl <1> x=3
For negative x:
-x+1=-2x+2 => x=1

In both these situations, we are able to tell that |x| is NOT less than 1

From 2:
|x-3| ≠ 0 => x is not 3 . hence Insufficient

VP
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02 Aug 2007, 12:42
pandeyrav wrote:
1. Is lxl <1> x=3
For negative x:
-x+1=-2x+2 => x=1

In both these situations, we are able to tell that |x| is NOT less than 1

From 2:
|x-3| ≠ 0 => x is not 3 . hence Insufficient

For 1, try plugging in x=1/3
This works, and it is less than 1.
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02 Aug 2007, 12:53
jet1445 wrote:
1. Is lxl < 1?
(1) lx + 1l = 2lx - 1l
(2) lx - 3l ≠ 0

I get D..here is how...

1)

|x+1| well the break point for this x>-1

|x-1| well break point for thi is that x>1

OK..so lets take x>1

if x>1

then

x+1=2(x-1) the only value of X that satisfies this equation is if x=3..therefore x=3, |x| >1...sufficient

2)

|x-3|<>0

well the break points here are if X=3, so if X>3

x-3>0; is some positive number and we know that |x|>1

however

if x<3

-(x-3) <0

-x+3<0

-x<-3

x>3 sufficient
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02 Aug 2007, 12:57
^^^ OK I was totally high when i did this...

C it is...

jeez...i hope i dont have sucha bad day when i take the gmat
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02 Aug 2007, 22:32
St2:
x could be a fraction or an integer > 1 or an integers <0. Insufficient.

St1:
If x = 3, LHS = 4, RHS = 4.
If x = 1/3, LHS = 4/3, RHS = 4/3.
Insufficient.

Using St1 and St2:
x must be 1/3. No other values work.

Sufficient.

Ans C
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03 Aug 2007, 10:11
sumande wrote:
jet1445 wrote:
1. Is lxl <1>1; -1<x<1; x<-1
Solving for each case, x=3 or 1/3. Insufficient.

Stmt 2: |x| ≠ 3. Insufficient.

Both together. x=1/3. sufficient.

i have not worked on many absolute value problems, so i please let me know what i am doing anything wrong, taking extra steps, etc.

from the question stem |x|<1 = is x<1, x<-1

from I:

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

for both positive x:

x+1 = 2x-2 .... x=3

for both negative x:

-x+1 = -2x-2 ... x=-3

Now my first question is, is there anyway to skip this step? it is an ainvalid answer I noticed no one computed it...

For +/- combinations

x+1= -2x-2 ... x=-1 another invalid answer.

-x+1 = 2x-2 ... x=1... ditto

So obviously I did something wrong because I did not come up with 1/3 for any of the answers...
Re: Absolute Value   [#permalink] 03 Aug 2007, 10:11
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