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One more hard one before I go and be merry! I will post

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One more hard one before I go and be merry! I will post [#permalink] New post 24 Dec 2006, 08:12
One more hard one before I go and be merry!

I will post answer in a couple of days. What's the best way of approaching this.
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 [#permalink] New post 24 Dec 2006, 10:38
Yes, this does look tough at first but is not very tough. Doubt I would have got it during the exam with the time pressure, although without the stress, it took me less than 2 minutes.

F = 120 (2^(-at)) + 60

Given: 120 = 120 (2^(-10a)) + 60

=> 120 (2^(-at)) = 60

=> 2^(-10a) = 1/2 = 2^(-1)

=> -10a = -1
=> a= 1/10

Thus when t = 30, F = 120(2^(-30*1/10)) + 60

= 120 (2^(-3)) + 60

= 120 * 1/8 + 60 = 75
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 [#permalink] New post 24 Dec 2006, 11:25
hsampath wrote:
Yes, this does look tough at first but is not very tough. Doubt I would have got it during the exam with the time pressure, although without the stress, it took me less than 2 minutes.

F = 120 (2^(-at)) + 60

Given: 120 = 120 (2^(-10a)) + 60

=> 120 (2^(-at)) = 60

=> 2^(-10a) = 1/2 = 2^(-1)

=> -10a = -1
=> a= 1/10

Thus when t = 30, F = 120(2^(-30*1/10)) + 60

= 120 (2^(-3)) + 60

= 120 * 1/8 + 60 = 75


Thank you for a detailed explanation :)
Manager
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 [#permalink] New post 26 Dec 2006, 04:16
OA = 75,

Thanks hsampath
  [#permalink] 26 Dec 2006, 04:16
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One more hard one before I go and be merry! I will post

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