Is B>=0 (1) B|B|=B^2 (2) |B|+B!=2 : Quant Question Archive [LOCKED]
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# Is B>=0 (1) B|B|=B^2 (2) |B|+B!=2

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Is B>=0 (1) B|B|=B^2 (2) |B|+B!=2 [#permalink]

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13 Oct 2006, 13:02
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Is B>=0

(1) B|B|=B^2
(2) |B|+B!=2
Senior Manager
Joined: 28 Aug 2006
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13 Oct 2006, 13:18
yezz wrote:
Is B>=0

(1) B|B|=B^2
(2) |B|+B!=2

I"ll go with A

1: B|B|= B^2
Clearly B^2 is always >=0
So B|B| >=0
Since |B|>=0, B must be >=0
S0 sufficient

2: |B|+B !=2

So B could be <0,>0,=0

So not sufficient

Hence A
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13 Oct 2006, 18:37
yezz wrote:
Is B>=0

(1) B|B|=B^2
(2) |B|+B!=2

D

Statement 1:
B^2 is +ve.
|B| is +ve.
So, B = +ve SUFF

Statement 2:

Factorials are not defined for negative numbers. So, only value for B is +ve. I know the argument sounds specious.
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13 Oct 2006, 21:20
Is the second statement

|B|+B factorial =2

or

|B|+B not equal to 2
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14 Oct 2006, 08:24
cicerone wrote:
Is the second statement

|B|+B factorial =2

or

|B|+B not equal to 2

Ahhh, my mistake I guess. It must be |B| + B not equal to 2.

Now it is a sure A
14 Oct 2006, 08:24
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