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Re: Absolute Value - Could someone please explain this problem [#permalink]

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30 Dec 2008, 22:59

From stmt2: x^2 < 1, this means, -1 < x < 1 We need to compare (x-6), hence, subtract 6 from all the sides of the above inequality. Thus, (-1-6) < (x-6) < (1-6) or, -7 < (x-6) < -5

That means, absolute value of (x-6) will be between 5 and 7 i.e. 5 < |x-6| < 7

And this clearly explains that |x-6| > 5. Hence, sufficient.

Re: Absolute Value - Could someone please explain this problem [#permalink]

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30 Dec 2008, 23:06

scthakur wrote:

From stmt2: x^2 < 1, this means, -1 < x < 1 We need to compare (x-6), hence, subtract 6 from all the sides of the above inequality. Thus, (-1-6) < (x-6) < (1-6) or, -7 < (x-6) < -5

That means, absolute value of (x-6) will be between 5 and 7 i.e. 5 < |x-6| < 7

And this clearly explains that |x-6| > 5. Hence, sufficient.

so silly of me why couldn't i get this........ anyways thanks a lot

Re: Absolute Value - Could someone please explain this problem [#permalink]

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31 Dec 2008, 13:03

vksunder wrote:

Is |x-6| > 5?

1. x is an integer 2. x^2 < 1

(x^2<1) and (x<1), which is not provided in the question, are two entirely different things but both are/would be sufficient to answer the question if x<1 were also provided as supplimentary information.

(x^2<1) has limits but (x<1) has no limit. In (x^2<1), x is > -1 but < 1. In (x<1), x could have any value smaller than 1. So B is suff.
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