If anyone knows how to do either of the questions below, please let me know - thanks!
The temperature of a certain cub of coffee 10 minutes after it was poured was 120 degrees Fahrenheit. If the 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 is a constant, then the temperatre of the coffee 30 minutes after it was poured was how many degrees Fahrenheit?
(A) 65
(B) 75 (this is the answer)
(C) 80
(D) 85
(E) 90
The total of Company C's assets in 1994 was 300 percent greater than the total in 1993, which in turn was 400 percent greater than the total in 1992. If the total of Company C's assets in 1992 was N dollars, which of the following represents tht toal of Company C's assets, in dollars, in 1994?
A) 7N
B) 8N
C) 9N
D) 12N
E) 20N (this is the answer)
The answer is E, but I think it should be D.
2 GMAT Prep Questions
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First compute 'a' given the 10-min infoMonicaKhan wrote:If anyone knows how to do either of the questions below, please let me know - thanks!
The temperature of a certain cub of coffee 10 minutes after it was poured was 120 degrees Fahrenheit. If the 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 is a constant, then the temperatre of the coffee 30 minutes after it was poured was how many degrees Fahrenheit?
120 = 120 (2^(-10a)) + 60
2^-1 = 2^-10a
a = 1/10
Now, compute the 30 min info
F = 120(2^(-30/10)) + 60
= 120 * 2^-3 + 60
= 15 + 60
= 75
MonicaKhan wrote:
The total of Company C's assets in 1994 was 300 percent greater than the total in 1993, which in turn was 400 percent greater than the total in 1992. If the total of Company C's assets in 1992 was N dollars, which of the following represents tht toal of Company C's assets, in dollars, in 1994?
Lets call assets in 1993 as x and 1994 as y. We have to find y/N
x = 4N + N (x is 400% greater than N, so it is actually 5N)
x = 5N
Similarly --
y = 4x = 4*5*N = 20N
y/N = 20
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F = 120(2^-at) + 60 (f = temp of coffee "t" min after it was poured and "a" is a constant
Substituting 120 for F and 10 for t
120 = 120(2^-10a)+60
120-60 = 120(2^-10a)
60/120 = 2^-10a
1/2=2^-10a
1/2=1/2^10a
1=10a (equating the respective powers since the base is the same
Thus a =1/10
Now when t=30
F= 120[2^(-30*1/10)]+60
F=120(2^3)+60
F=120(2^-3)+60
=[120*(1/2^3)]+60
=(120/8 )+60
=75
Substituting 120 for F and 10 for t
120 = 120(2^-10a)+60
120-60 = 120(2^-10a)
60/120 = 2^-10a
1/2=2^-10a
1/2=1/2^10a
1=10a (equating the respective powers since the base is the same
Thus a =1/10
Now when t=30
F= 120[2^(-30*1/10)]+60
F=120(2^3)+60
F=120(2^-3)+60
=[120*(1/2^3)]+60
=(120/8 )+60
=75
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