Is x > 0? (1) |x + 3| = 4x 3 (2) |x 3| = |2x 3 : DS Archive
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# Is x > 0? (1) |x + 3| = 4x 3 (2) |x 3| = |2x 3

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Manager
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Is x > 0? (1) |x + 3| = 4x 3 (2) |x 3| = |2x 3 [#permalink]

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01 Jul 2007, 08:52
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Is x > 0?

(1) |x + 3| = 4x – 3

(2) |x – 3| = |2x – 3|
Senior Manager
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01 Jul 2007, 09:21
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.
Manager
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03 Jul 2007, 06:56
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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Re: DS - x > 0? [#permalink]

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03 Jul 2007, 07:13
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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03 Jul 2007, 07:45
1) |x + 3| = 4x – 3

X can not be 0. Put 0 in (1) and you get 3 = -3

I agree with UMB.

Manager
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03 Jul 2007, 07:45
---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
Senior Manager
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03 Jul 2007, 11:15
I am also with A. This is my reasoning.

From Stmt1. I took 4x - 3 >= 0 => x >= 3/4. Is this reasoning right ?
Director
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04 Jul 2007, 04:17
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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04 Jul 2007, 07:45
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)
04 Jul 2007, 07:45
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