I am not sure if it is C.
Statment 1. x = 3
Statment 2. x <> 3
Is lxl< 1?
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Dream Weaver
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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.
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.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.
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Brent@GMATPrepNow
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Here's my approach:Can you explain further how you solved lx + 1l = 2lx - 1l for x? I am confused w/ the absolute value and inequality signs.
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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maihuna
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Is lxl< 1?
(1) lx + 1l = 2lx - 1l
(2) lx - 3l ≠ 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
(1) lx + 1l = 2lx - 1l
(2) lx - 3l ≠ 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
