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looking for an alternative approach basically i am trying to

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looking for an alternative approach basically i am trying to [#permalink] New post 31 May 2006, 14:23
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

looking for an alternative approach
basically i am trying to find out the best way to deal with absolute values on DS
plz any comments or tricks welcome

IS I x-6I >5

st1 X IS AN INTEGER
St 2 X <1

THANKS
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Re: challenge 25 question 1 Absolute value [#permalink] New post 31 May 2006, 14:33
|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.
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 [#permalink] New post 31 May 2006, 19:00
Straight B.
kapslock nailed it.
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Re: challenge 25 question 1 Absolute value [#permalink] New post 11 Jun 2006, 13:32
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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 [#permalink] New post 11 Jun 2006, 21:53
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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 [#permalink] New post 12 Jun 2006, 00:49
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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 [#permalink] New post 12 Jun 2006, 01:24
As an alternative approach we may use graphs in certain cases. See post
http://www.gmatclub.com/phpbb/viewtopic ... 330#203330
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 [#permalink] New post 13 Jun 2006, 10:28
IS I x-6I >5

st1 X IS AN INTEGER
St 2 X <1

Using Option 1 , x < 1 and x > 11 will make |x-6|>5. No unique solution for x.

Using Option 2, x <1 is uniquely makes |x-6|>5

And so answer is B.
  [#permalink] 13 Jun 2006, 10:28
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