The temperature of a certain cup of coffee 10 minutes after it was pou : Problem Solving (PS)

jakolik

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28 Jul 2010, 11:57

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Hi,

Since F=120 (2^-at)+60,
replace F=120, t=10 to find a.
Once you find a, then to find the temperature 30 min after it was poured simply replace
t=30 and value of a in the equation.

regards,
Jack

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28 Jul 2010, 13:26

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Is the question correctly written?? I am unable to find the solution.

\(F= 120* (\frac{1}{2^at}) + 60\)
Is this formula correct?

darkhorse19

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Substitute the values f=120 and t=10 in given equation, this gives value of a=0.1

now equation is f=120(2^-0.1t)+60

substitute t=30 in above equation to find f

Answer B

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Can someone explain me how
1/2 = 2^-10a becomes

2^-1 = 2^-10a

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lzielen

Can someone explain me how
1/2 = 2^-10a becomes

2^-1 = 2^-10a

Hi lzielen!

\(\frac{1}{2}\) is \(2^{-1}\).

\(x^{-n} = 1/x^n\)

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22 Mar 2018, 09:27

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Hussain15

Is the question correctly written?? I am unable to find the solution.

\(F= 120* (\frac{1}{2^at}) + 60\)
Is this formula correct?

a*t is the power of 2 it is 2^(-at)

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13 Sep 2018, 05:47

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Bunuel

kilukilam

The tempreature of a certain cup of coffee 10 minutes after it was poured was 120 degree Faranite. If the temp F of the coffee t minues after it was poured can be determinde by the formula F=120 (2^-at)+60 , where F is degree Fahrenhite and a is constant, then the tempreature of the cofffee 30 minues after it was poured was how many degree fahrenhite

a. 65
b. 75
c. 80
d. 85
e. 90

We have the formula for calculating temperature: \(F=120*2^{-at}+60\) and the value for \(t=10\) --> \(F(t=10)=120=120*2^{-10a}+60\) --> \(\frac{1}{2^{10a}}=\frac{1}{2}\) --> \(10a=1\) --> \(a=0.1\). So formula is \(F=120*2^{-0.1t}+60\)

Now for \(t=30\) the temperature in Fahrenheits will be \(F=120*2^{-0.1*30}+60=\frac{120}{2^3}+60=75\).

Answer: B.

You're solution is quite clear to me, I only have one question:

How do you conduct this step: \(\frac{1}{2^{10a}}=\frac{1}{2}\) --> \(10a=1\)

I would solve it like this:

\(\frac{1}{2^{10a}}=\frac{1}{2}\) --> 2^-1 = 2^-10a --> -1 lg(2) = -10a lg(2) --> 1 = 10a

Is there a easier possibility? Is it even right to solve the question by the use of logarithm?

Thanks in advance!

Bunuel

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13 Sep 2018, 05:50

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DominikSterk

Bunuel

kilukilam

The tempreature of a certain cup of coffee 10 minutes after it was poured was 120 degree Faranite. If the temp F of the coffee t minues after it was poured can be determinde by the formula F=120 (2^-at)+60 , where F is degree Fahrenhite and a is constant, then the tempreature of the cofffee 30 minues after it was poured was how many degree fahrenhite

a. 65
b. 75
c. 80
d. 85
e. 90

We have the formula for calculating temperature: \(F=120*2^{-at}+60\) and the value for \(t=10\) --> \(F(t=10)=120=120*2^{-10a}+60\) --> \(\frac{1}{2^{10a}}=\frac{1}{2}\) --> \(10a=1\) --> \(a=0.1\). So formula is \(F=120*2^{-0.1t}+60\)

Now for \(t=30\) the temperature in Fahrenheits will be \(F=120*2^{-0.1*30}+60=\frac{120}{2^3}+60=75\).

Answer: B.

You're solution is quite clear to me, I only have one question:

How do you conduct this step: \(\frac{1}{2^{10a}}=\frac{1}{2}\) --> \(10a=1\)

I would solve it like this:

\(\frac{1}{2^{10a}}=\frac{1}{2}\) --> 2^-1 = 2^-10a --> -1 lg(2) = -10a lg(2) --> 1 = 10a

Is there a easier possibility? Is it even right to solve the question by the use of logarithm?

Thanks in advance!

\(\frac{1}{2^{10a}}=\frac{1}{2}\);

Cross-multiply: \(2=2^{10a}\)

Equate the powers: \(10a=1\)

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Alright, so I now understand how ...
1/2 = 2^-10a
becomes
2^-1 = 2^-10a

but how does it then turn into...
-1 = -10a

to me it should be
1=-10a

any help here?

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13 Feb 2019, 06:28

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took me over 5mins to solve, is this high 600 level question ?

Bunuel

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13 Feb 2019, 06:48

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hsn81960

took me over 5mins to solve, is this high 600 level question ?

You can check difficulty level of a question in the tags above original post. For this one is 600-700. The difficulty level of a question is calculated automatically based on the timer stats from the users which attempted the question.

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20 May 2019, 19:13

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kilukilam

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

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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16 Feb 2020, 02:17

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Bunuel

kilukilam

The tempreature of a certain cup of coffee 10 minutes after it was poured was 120 degree Faranite. If the temp F of the coffee t minues after it was poured can be determinde by the formula F=120 (2^-at)+60 , where F is degree Fahrenhite and a is constant, then the tempreature of the cofffee 30 minues after it was poured was how many degree fahrenhite

a. 65
b. 75
c. 80
d. 85
e. 90

We have the formula for calculating temperature: \(F=120*2^{-at}+60\) and the value for \(t=10\) --> \(F(t=10)=120=120*2^{-10a}+60\) --> \(\frac{1}{2^{10a}}=\frac{1}{2}\) --> \(10a=1\) --> \(a=0.1\). So formula is \(F=120*2^{-0.1t}+60\)

Now for \(t=30\) the temperature in Fahrenheits will be \(F=120*2^{-0.1*30}+60=\frac{120}{2^3}+60=75\).

Answer: B.

Hi Bunuel

I solved it like this :

\(120 = 120*2^{-at}+60\)
\(1 = 1/2^{10a} + 1/2\)
\(2^{0}= 2^{-(10a)} + 2^{(-1)}\)
therefore comparing powers
0 = -10a - 1
1=-10a
a=-1/10

why is this wrong? Could you help pls?

Bunuel

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16 Feb 2020, 02:33

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Kritisood

Bunuel

kilukilam

The tempreature of a certain cup of coffee 10 minutes after it was poured was 120 degree Faranite. If the temp F of the coffee t minues after it was poured can be determinde by the formula F=120 (2^-at)+60 , where F is degree Fahrenhite and a is constant, then the tempreature of the cofffee 30 minues after it was poured was how many degree fahrenhite

a. 65
b. 75
c. 80
d. 85
e. 90

We have the formula for calculating temperature: \(F=120*2^{-at}+60\) and the value for \(t=10\) --> \(F(t=10)=120=120*2^{-10a}+60\) --> \(\frac{1}{2^{10a}}=\frac{1}{2}\) --> \(10a=1\) --> \(a=0.1\). So formula is \(F=120*2^{-0.1t}+60\)

Now for \(t=30\) the temperature in Fahrenheits will be \(F=120*2^{-0.1*30}+60=\frac{120}{2^3}+60=75\).

Answer: B.

Hi Bunuel

I solved it like this :

\(120 = 120*2^{-at}+60\)
\(1 = 1/2^{10a} + 1/2\)
\(2^{0}= 2^{-(10a)} + 2^{(-1)}\)
therefore comparing powers
0 = -10a - 1
1=-10a
a=-1/10

why is this wrong? Could you help pls?

Generally, if a^x = a^y + a^z, it does NOT mean that x = y + z. For example, 2^4 = 2^3 + 2^3 but 4 ≠ 3 + 3.

I think you thought that we had 2^0 = 2^(-10a -1), and if it were so then yes we could write 0 = -10a -1.

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20 Oct 2022, 03:25

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Bunuel

kilukilam

The tempreature of a certain cup of coffee 10 minutes after it was poured was 120 degree Faranite. If the temp F of the coffee t minues after it was poured can be determinde by the formula F=120 (2^-at)+60, where F is degree Fahrenhite and a is constant, then the tempreature of the cofffee 30 minues after it was poured was how many degree fahrenhite

a. 65
b. 75
c. 80
d. 85
e. 90

We have the formula for calculating temperature \(F=120*2^{-at}+60\) and the value for \(t=10\):

\(F(t=10)=120=120*2^{-10a}+60\);

\(\frac{1}{2^{10a}}=\frac{1}{2}\);

\(10a=1\);

\(a=0.1\).

So, the formula is \(F=120*2^{-0.1t}+60\)

Now, for \(t=30\) the temperature in Fahrenheits will be \(F=120*2^{-0.1*30}+60=\frac{120}{2^3}+60=75\).

Answer: B.

Bunuel, could you elaborate how did 60 > 1/2 here?
Shouldn't 1/2^{10a}[/fraction] have - sign here for moving it to other side of equaiton? Thanks

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20 Oct 2022, 03:29

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ScottTargetTestPrep

kilukilam

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

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

Great explanation ScottTargetTestPrep, If we cancel out 120 at this stage, I'm not sure how to proceed next? Could you help? Thanks

120 = 120 * 2^(-10a) + 60
1 = 2^(-10a) + 60

Bunuel

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20 Oct 2022, 07:28

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Kimberly77

Bunuel

kilukilam

The tempreature of a certain cup of coffee 10 minutes after it was poured was 120 degree Faranite. If the temp F of the coffee t minues after it was poured can be determinde by the formula F=120 (2^-at)+60, where F is degree Fahrenhite and a is constant, then the tempreature of the cofffee 30 minues after it was poured was how many degree fahrenhite

a. 65
b. 75
c. 80
d. 85
e. 90

We have the formula for calculating temperature \(F=120*2^{-at}+60\) and the value for \(t=10\):

\(F(t=10)=120=120*2^{-10a}+60\);

\(\frac{1}{2^{10a}}=\frac{1}{2}\);

\(10a=1\);

\(a=0.1\).

So, the formula is \(F=120*2^{-0.1t}+60\)

Now, for \(t=30\) the temperature in Fahrenheits will be \(F=120*2^{-0.1*30}+60=\frac{120}{2^3}+60=75\).

Answer: B.

Bunuel, could you elaborate how did 60 > 1/2 here?
Shouldn't 1/2^{10a}[/fraction] have - sign here for moving it to other side of equaiton? Thanks

We have:

\(120=120*2^{-10a}+60\).
Subtract 60 from both sides:

Divide by 120:

Rewrite 1/2 as 2^(-1):

\(2^{(-1)}=2^{-10a}\).
Equate the powers:

Hope it's clear.

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