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Coffee cup temperature

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by neeti2711 » Sun Mar 03, 2013 2:54 am
The temperature of a certain cup of coffee 10 minutes after it was poured was 120º Fahrenheit. If the temperature F of the coffee t minutes after it was poured can be determined by the equation F=120(2^-at)+60 where F is degree Fahrenheit and a is a constant then the temperature of coffee 30 minutes after it was poured was how many o Fahrenheit?

A. 65
B. 75
C. 80
D. 85
E. 90
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by aditya8062 » Sun Mar 03, 2013 3:21 am
mt take wud be B that is 75
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by GMATGuruNY » Sun Mar 03, 2013 4:12 am
The temperature of a certain cup of coffee 10 minutes it was poured was 120 degree Fahrenheit. If temperature F of the coffee t minutes after it was poured can be determined by the formula F = 120*(2^-at) + 60, where F is in degrees Fahrenheit and a is a constant, then the temperature of the coffee 30 minutes after it was poured was how many degree Fahrenheit?

A. 65
B. 75
C. 80
D. 85
E. 90
The given formula is F = 120*2^(-at) + 60.
Ignore 2^(-at).
We need to determine the MULTIPLIER: the factor by which 120 must be multiplied for every 10 minutes.

Let x = the multiplier.
The formula becomes F = 120x + 60.

Since after 10 minutes, F=120:
120 = 120x + 60
x = 1/2.

Thus, for every 10 minutes, 120 must be multiplied by 1/2.
Thus, to determine the value of F after 30 minutes, 120 must be multiplied by 1/2 three times:
F = 120(1/2)³ + 60
F = 75.

The correct answer is B.
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by Anurag@Gurome » Sun Mar 03, 2013 5:25 am
neeti2711 wrote:The temperature of a certain cup of coffee 10 minutes after it was poured was 120º Fahrenheit. If the temperature F of the coffee t minutes after it was poured can be determined by the equation F=120(2^-at)+60 where F is degree Fahrenheit and a is a constant then the temperature of coffee 30 minutes after it was poured was how many o Fahrenheit?
Temperature after t minutes = 120*[2^(-at)] + 60
Temperature after 10 minutes = 120*[2^(-10a)] + 60 = 120 --> [2^(-10a)] = 60/120 = 1/2

Temperature after 30 minutes = 120*[2^(-30a)] + 60

Now, 2^(-30a) = [2^(-10a)]^3 = (1/2)^3 = 1/8

Hence, temperature after 30 minutes = 120/8 + 60 = 15 + 60 = 75

The correct answer is B
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