IMO 75boltach wrote:
yes it took just a little less than 60 sec. Just practice practice practice. That's the only mantra.
While practicing try doing as many steps as possible in the mind, then it will save lot of time.
user123321
Can you solve this under 2 min?
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user123321
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yes it took just a little less than 60 sec. Just practice practice practice. That's the only mantra.
While practicing try doing as many steps as possible in the mind, then it will save lot of time.
user123321
Just started my preparation 
Want to do it right the first time.
Want to do it right the first time.
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GmatKiss
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Am not getting this one!rijul007 wrote:75
F = 120*(2^-at) +160
1/2 = 2^(-10a)
-10a = -1
a = 1/10
F = 120 *2^(-3) +60 = 15+60 = 75
Could you elaborate a bit.
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shankar.ashwin
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Given,
F = 120*(2^-at) +60,
And when t=10, F =120
120 = 120*(2^-10a) + 60
Now, 120*(2^-10a) = 60
2^-10a = 1/2 = 2^-1
10a = 1
a = 1/10
Now find F for t=30
F = 120 *2^(-3) +60 = 75
F = 120*(2^-at) +60,
And when t=10, F =120
120 = 120*(2^-10a) + 60
Now, 120*(2^-10a) = 60
2^-10a = 1/2 = 2^-1
10a = 1
a = 1/10
Now find F for t=30
F = 120 *2^(-3) +60 = 75
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neelgandham
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Please find the answer to your question in green ! Hope it helps.GmatKiss wrote:Am not getting this one!rijul007 wrote:75
F = 120*(2^-at) +160
1/2 = 2^(-10a)
F = 120*(2^-at) + 60
F = 120 when t = 10
Substituting the values of F and t in the equation
120 = 120*(2^(-a*10)) + 60
120-60 = 120*(2^(-a*10))
60 = 120*(2^(-a*10)))
60/120 = (2^(-a*10))
1/2 = (2^(-a*10)) = 1/(2^(a*10))
1/(2^(a*10)) = 1/2
(2^(a*10)) = 2^1
10a = 1
a = 1/10
F = 120 *2^(-3) +60 = 15+60 = 75
Could you elaborate a bit.
Anil Gandham
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pemdas
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this q. fits right within MGMAT Number property book tips, please study the source indicated
120=120(2^-at)+120/2
120(2^0)=120(2^-at)+120*(2^-1)
Cancel 120 on both sides
2^0=2^-at + 2^-1 or 1=2^-at + ½ or ½=2^-at or 2^-1=2^-at or -1=-at, considering t=10 then 1=10a, a=1/10
F=120(2^{-1/10 * 30}) + 60 = 120(2^-3) + 60= 15+60=75
120=120(2^-at)+120/2
120(2^0)=120(2^-at)+120*(2^-1)
Cancel 120 on both sides
2^0=2^-at + 2^-1 or 1=2^-at + ½ or ½=2^-at or 2^-1=2^-at or -1=-at, considering t=10 then 1=10a, a=1/10
F=120(2^{-1/10 * 30}) + 60 = 120(2^-3) + 60= 15+60=75
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The given formula is F = 120*2^(-at) + 60.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
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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sana.noor
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that is such an easy question, put the value of temperature for 10 minutes and get the value of a. remember A is constant. putting the value of A as (-1/2) and time of 30 minutes, u will get B
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Basically, the same solution as others, but with some color that may help people understand what's what. My equation solving is a little different too.The temperature of a certain cup 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 a is a constant, then the temperature of the coffee 30 minutes after it was poured was how many degrees Fahrenheit?
A. 65
B. 75
C. 80
D. 85
E. 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 can write: 2^(-1) = 2^[(-a)(10)]
So, we know that -1 = (-a)(10)
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
= B
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