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Is |x|< 1?

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Source: — Data Sufficiency |
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by Morgoth » Sat Jun 27, 2009 11:24 am
|x|&#9168;< 1 ?

this simply means is -1 < x < 1

Statement I
&#9168;|x + 1|&#9168; = 2&#9168;|x - 1|&#9168;

case I
x+1 = 2(x-1)
x+1 = 2x -2
x=3

case II
-x-1 = 2(x-1)
-x-1 = 2x-2
x=1/3

x=1/3 or 3

Insufficient.

Statement II

lx - 3l&#9168; &#8800; 0

x &#8800; 3 or -3

Hence, x could be any of other value

Insufficient.


Combining I & II

x= 1/3

Hence, C.

Hope this helps.
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by avigmat11 » Sun Jun 28, 2009 1:51 am
Yes..C..nicely explained by Morgoth.
Thanks
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by aj5105 » Mon Jun 29, 2009 2:50 am
Can you please explain how you got two values for statement (2)?

Cause, I am getting just one value. x &#8800; 3.

Morgoth wrote:|x|&#9168;< 1 ?

this simply means is -1 < x < 1

Statement I
&#9168;|x + 1|&#9168; = 2&#9168;|x - 1|&#9168;

case I
x+1 = 2(x-1)
x+1 = 2x -2
x=3

case II
-x-1 = 2(x-1)
-x-1 = 2x-2
x=1/3

x=1/3 or 3

Insufficient.

Statement II

lx - 3l&#9168; &#8800; 0

x &#8800; 3 or -3

Hence, x could be any of other value

Insufficient.


Combining I & II

x= 1/3

Hence, C.

Hope this helps.
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by Morgoth » Mon Jun 29, 2009 5:29 am
aj5105 wrote:Can you please explain how you got two values for statement (2)?

Cause, I am getting just one value. x &#8800; 3.
Thanks for pointing out, statement II should only have value x &#8800; 3,
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