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Re: looking for an alternative approach basically i am trying to [#permalink]
kapslock wrote:
|x-6| would always be +ve. The "tipping points" would be
x-6 > 5 => x > 11 OR
x - 6 < -5 => x < 1.

1. If x is an integer makes no difference. Insufficient.

2. x < 1. This is the tipping point. For x < 1, x - 6 would always be < -5 and this |x - 6| would be greater than 5.

Thus B.


Quick question, why do you reverse the inequality in the second case?
absolute(x-6)<-5

thanks
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Re: looking for an alternative approach basically i am trying to [#permalink]
Is |x-6| > 5

St1:

X is an intger. Useless. x could be 2, then |x-6| < 5 or x could be 1000, then |x-6| > 5

St:2

x < 1

All values of x that satisfies the given inequality will give |x-6| > 5.

Ans B
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Re: looking for an alternative approach basically i am trying to [#permalink]
Clear B.

1) not suff. x = 5 make |x-6| < 5 and x = -1 makes it > 5
2) for every value of x <1, |x-6| > 5
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Re: looking for an alternative approach basically i am trying to [#permalink]
As an alternative approach we may use graphs in certain cases. See post
https://www.gmatclub.com/phpbb/viewtopic ... 330#203330
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Re: looking for an alternative approach basically i am trying to [#permalink]
IS I x-6I >5

st1 X IS AN INTEGER
St 2 X 11 will make |x-6|>5. No unique solution for x.

Using Option 2, x 5

And so answer is B.



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