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The temperature of a cup of coffee 10 minutes after it was

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The temperature of a cup of coffee 10 minutes after it was [#permalink] New post 03 Oct 2006, 16:43
The temperature of a cup of coffee 10 minutes after it was poured was 120 degrees F. If the temp (F) of the cup of coffee t minutes after it was poured can be determined by F=120(2^(at)) + 60 , where a is a constant, then what was the temperature 30 minutes after it was poured?
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 [#permalink] New post 03 Oct 2006, 18:12
F=120(2^(at)) + 60

120=120(2^(a10)) + 60
1/2=2^(a10)
1/2=1/2^(-a10) ---> so we need that (-a*10)=1 so a=-1/10

For 30 min
F=120(2^((-1/10)*30)) + 60
F=120*(1/2^3)+60
F=15+60
F=75
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 [#permalink] New post 04 Oct 2006, 12:39
Got 75 also .. same working as X&Y
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 [#permalink] New post 04 Oct 2006, 12:59
Same answer.For simplifying the working process,I converted the minutes into hrs.Hence I assumed t=1/6 where t=10 and t=1/2 where t=30.This gives a=-6.Using this I get F=75.
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 [#permalink] New post 04 Oct 2006, 13:45
120 = 120(2^(10a))+60
Simplifying,
2^-1 = 2^10a
a= -1/10

After 30 minutes,

F = 120 (2^-3)+60
= (120/8)+60
F=75
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 [#permalink] New post 04 Oct 2006, 14:24
same as above, 75.

a = -1/10

after 30 mins:

120(2^(-1/10 * 30)) + 60
120(2^-3) + 60
120(1/8) + 60
15 + 60
75
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 [#permalink] New post 04 Oct 2006, 15:37
The temperature of a cup of coffee 10 minutes after it was poured was 120 degrees F. If the temp (F) of the cup of coffee t minutes after it was poured can be determined by F=120(2^(at)) + 60 , where a is a constant, then what was the temperature 30 minutes after it was poured?

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

GET A

120-60=60/120 = 1/2 = 2^-1

10A = -1 THUS A = -1/10

F= 1/8(120) +60 = 75
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 [#permalink] New post 04 Oct 2006, 16:24
75 is correct, thanks a lot guys.
  [#permalink] 04 Oct 2006, 16:24
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