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Is lxl< 1?

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Source: — Data Sufficiency |
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by keeyu2 » Tue Mar 31, 2009 8:02 am
I am not sure if it is C.

Statment 1. x = 3
Statment 2. x <> 3
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IMO E

by Dream Weaver » Tue Mar 31, 2009 11:18 am
A. Gives X=3
B. Gives X Not Equal to 3

Whats the OA?
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by ncr_10 » Wed Apr 01, 2009 12:04 am
Statement 1:
i) for x < -1, we get x = 3
ii) for -1 < x < 1 we get x = 1/3
iii) for 1 < x we get x = 3

This means x = 3 or x = 1/3. Not sufficient.

Statement 2: gives x <> 3. Not sufficient.

Both statements together we get x = 1/3. Sufficient.

Hence C.
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by apple100 » Wed Apr 08, 2009 11:46 pm
ncr_10 wrote:Statement 1:
i) for x < -1, we get x = 3
ii) for -1 < x < 1 we get x = 1/3
iii) for 1 < x we get x = 3

This means x = 3 or x = 1/3. Not sufficient.

Statement 2: gives x <> 3. Not sufficient.

Both statements together we get x = 1/3. Sufficient.

Hence C.
Can you explain further how you solved lx + 1l = 2lx - 1l for x? I am confused w/ the absolute value and inequality signs.
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by Brent@GMATPrepNow » Thu Apr 09, 2009 10:35 am
Can you explain further how you solved lx + 1l = 2lx - 1l for x? I am confused w/ the absolute value and inequality signs.
Here's my approach:
I'll begin with an aside: if |x| = 5 then x=5 or -5
In general if k = |something| then k = something or k = -(something)
If lx + 1l = 2lx - 1l then either
a) x+1 = 2(x-1) --> x=3
or
b) -(x+1) = 2(x-1) --> x= 1/3

Notice that, since we have two sets of absolutes in the original question, we could have created additional equations: -(x+1) = (2)[-(x-1)] and x+1 = (2)[-(x-1)], but these are simply versions of equations (a) and (b)

I hope that helps.
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by maihuna » Sun May 03, 2009 9:35 am
Is lxl< 1?
(1) lx + 1l = 2lx - 1l
(2) lx - 3l &#8800; 0
=================

for 1:

-------|------------------|---------------
-1 1

for x>-1, -(x+1) = -2(x-1)
=> x=3
-1<x<1: x+1 = -2(x-1)
=>3x = 1 or x=1/3
1<x x+1 = 2(x-1) or x=3
so 1 not sufficient as one possiblity is yes another no

2. for x!=3 it could be anything

combine does help as it rule out x=3 possibilities, and so is Ch
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