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Manager
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Is x > 0? (1) |x + 3| = 4x 3 (2) |x 3| = |2x 3 [#permalink]
 01 Jul 2007, 09:52
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Is x > 0?
(1) |x + 3| = 4x – 3
(2) |x – 3| = |2x – 3|
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Senior Manager
Joined: 28 Feb 2007
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my vote for A.
Statement1) /x+3/>=0 so 4x-3>=0 from which x>=3/4 SUFF.
Statement2) /x-3/=/2x-3/ only x=0 or x=2 satisfy this equation: Insuff.
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Manager
Joined: 20 Mar 2005
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E
From A , X= 0 or 2 so insuff
also from B X =0 or 2 so insuff
both together dont tell anything.
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Senior Manager
Joined: 04 Jun 2007
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iamba wrote: Is x > 0?
(1) |x + 3| = 4x – 3
(2) |x – 3| = |2x – 3|
Stmt 1:
case 1: x>-3
x+3 = 4x-3
=> x = 2
case 2: x<-3
-(x+3) = 4x-3
=> x = 0
case 3: x=-3
4x-3 = 0
=> x = 3/4
Hence, insufficient.
Stmt 2:
case 1: x<3/2
-(x-3) = -(2x-3)
=> x = 0
case 2: x=3/2
-(x-3) = 0
=> x = 3
case 3: 3/2 < x < 3
-(x-3) = 2x-3
=> x = 2
case 4: x=3
2x-3 = 0
=> x = 3/2
case 5: x>3
x-3 = 2x - 3
=> x = 0
Hence, insufficient.
From both statements, x=0 or x=2. Hence, insufficient.
Therefore, E.
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Intern
Joined: 05 Apr 2007
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1) |x + 3| = 4x – 3
X can not be 0. Put 0 in (1) and you get 3 = -3
I agree with UMB.
Answer is A
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Manager
Joined: 24 Jun 2006
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---In statement 1 we have two possible value for X
we have X=2 or X=0 but if we replace the value one can see that the restriction of the inequalityonly allows for X=2
because When X equal 0 we end up with |3|=-3 which is impossible
SUFFICIENT
---Statement 2 we have also two possibilities X=0 and X=2
restrictions in the stem does not allows us to determine if x is = to zero or higher.
INSUFFICIENT
I go for A
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Senior Manager
Joined: 03 Jun 2007
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I am also with A. This is my reasoning.
From Stmt1. I took 4x - 3 >= 0 => x >= 3/4. Is this reasoning right ?
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Director
Joined: 14 Jan 2007
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Mine is 'A'
Stmt1 : x =2 So SUFF
Stmt2: for x< 3/2, x =0
for 3/2<x<3 , x = 2
So INSUFF
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Director
Joined: 16 May 2007
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Initially i thought it was E, but now i go with A since 1 results in either x=0 or x=2, if you substitute , you can easily ifnid out that x can be only 2. So this is sufficient. Therefor i go with A. (2 however is insufficient as explained earlier in this thread)
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