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Question from mba.com mock test

by Cybermusings » Sat Apr 14, 2007 1:26 am
The temperature of a certain cup of coffee 10 minutes after it was poured was 120 degrees Farenheit. If the temperature F of the coffee t minutes after it was poured can be determined by F = 120(2^-at)+60, where F is in Farenheit, and "a" is a constant, then the temperature of the coffee 30 minutes after it was poured was how many degrees?

1) 65
2) 75
3) 80
4) 85
5) 90

Please answer with detailed explanation. I am not getting this one right..
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Re: Question from mba.com mock test

by ajith » Sat Apr 14, 2007 3:03 am
Cybermusings wrote:The temperature of a certain cup of coffee 10 minutes after it was poured was 120 degrees Farenheit. If the temperature F of the coffee t minutes after it was poured can be determined by F = 120(2^-at)+60, where F is in Farenheit, and "a" is a constant, then the temperature of the coffee 30 minutes after it was poured was how many degrees?

1) 65
2) 75
3) 80
4) 85
5) 90

Please answer with detailed explanation. I am not getting this one right..
the temperature F of the coffee t minutes after it was poured can be determined by F = 120(2^-at)+60

Given that F aftter 10 mins =120
=> 120 = 120 (2^-10a) +60
2 = 2 (2^-10a ) +1
1= 2(2^-10a )
2^(1-10a)=2^0
=> 1-10a =0
=> a=1/10
Temp after 30 mins = 120 (2^-1/10*30)+60
= 120 (1/8) +60
= 75
Hope it is clear!
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by Cybermusings » Sat Apr 14, 2007 11:06 am
Hey thanks dude!
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by Scott@TargetTestPrep » Mon May 20, 2019 6:07 pm
Cybermusings wrote:The temperature of a certain cup of coffee 10 minutes after it was poured was 120 degrees Farenheit. If the temperature F of the coffee t minutes after it was poured can be determined by F = 120(2^-at)+60, where F is in Farenheit, and "a" is a constant, then the temperature of the coffee 30 minutes after it was poured was how many degrees?

1) 65
2) 75
3) 80
4) 85
5) 90
First we need to find the value of a using F = 120 and t = 10:

120 = 120 * 2^(-10a) + 60

60 = 120 * 2^(-10a)

1/2 = 2^(-10a)

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

With a common base, we can equate the exponents:

-1 = -10a

a = 1/10

Now we can find F using a = 1/10 and t = 30:

F = 120 * 2^(-(1/10)(30)) + 60

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

F = 120 * 1/8 + 60

F = 15 + 60

F = 75

Answer: 2/B

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by Brent@GMATPrepNow » Tue May 21, 2019 6:32 am
Cybermusings wrote:The temperature of a certain cup of coffee 10 minutes after it was poured was 120 degrees Farenheit. If the temperature F of the coffee t minutes after it was poured can be determined by F = 120(2^-at)+60, where F is in Farenheit, and "a" is a constant, then the temperature of the coffee 30 minutes after it was poured was how many degrees?

1) 65
2) 75
3) 80
4) 85
5) 90
The temperature of a certain cup of coffee 10 minutes after it was poured was 120 degrees Fahrenheit.

So, 120 = 120 * 2^[(-a)(10)] + 60
Divide both sides by 60: 2 = 2 * 2^[(-a)(10)] + 1
1 = 2 * 2^[(-a)(10)]
1/2 = 2^[(-a)(10)]
Since 2^(-1) = 1/2, we know that (-a)(10) = -1
So, a = 1/10

So, the formula is f = 120 * 2^[(-1/10)(t)] + 60

The temperature of the coffee 30 minutes after it was poured was how many degrees Fahrenheit?
f = 120 * 2^[(-1/10)(30)] + 60
= 120 * 2^[-3] + 60
= 120 * (1/8) + 60
= 15 + 60
= 75

Answer: B

Cheers,
Brent
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