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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Amortization is the process of paying off a loan (including interest) by making a series of regular, equal payments. The formula below is used to calculate payments on any amortized loan.Β
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Loan Amortization Formula
MATH 110 Sec 8-5 Lecture: Annuities β Amortization Amortization is the process of paying off a loan (including interest) by making a series of regular, equal payments. The formula below is used to calculate payments on any amortized loan.Β π (1+π) π =π
(1+π) π β1 π Loan Amortization Formula
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Loan Amortization Formula
MATH 110 Sec 8-5 Lecture: Annuities β Amortization Amortization is the process of paying off a loan (including interest) by making a series of regular, equal payments. The formula below is used to calculate payments on any amortized loan.Β This is a new formula but all of the variables are the same ones that we have used before. π (1+π) π =π
(1+π) π β1 π Loan Amortization Formula
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Loan Amortization Formula
MATH 110 Sec 8-5 Lecture: Annuities β Amortization Amortization is the process of paying off a loan (including interest) by making a series of regular, equal payments. The formula below is used to calculate payments on any amortized loan.Β This is a new formula but all of the variables are the same ones that we have used before. π‘ is the time in years (the term) π (1+π) π =π
(1+π) π β1 π Loan Amortization Formula
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Loan Amortization Formula
MATH 110 Sec 8-5 Lecture: Annuities β Amortization Amortization is the process of paying off a loan (including interest) by making a series of regular, equal payments. The formula below is used to calculate payments on any amortized loan.Β This is a new formula but all of the variables are the same ones that we have used before. π‘ is the time in years (the term) π is the annual interest rate π (1+π) π =π
(1+π) π β1 π Loan Amortization Formula
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Loan Amortization Formula
MATH 110 Sec 8-5 Lecture: Annuities β Amortization Amortization is the process of paying off a loan (including interest) by making a series of regular, equal payments. The formula below is used to calculate payments on any amortized loan.Β This is a new formula but all of the variables are the same ones that we have used before. π‘ is the time in years (the term) π is the annual interest rate π is the # of compounding periods/yr π (1+π) π =π
(1+π) π β1 π Loan Amortization Formula
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Loan Amortization Formula
MATH 110 Sec 8-5 Lecture: Annuities β Amortization Amortization is the process of paying off a loan (including interest) by making a series of regular, equal payments. The formula below is used to calculate payments on any amortized loan.Β This is a new formula but all of the variables are the same ones that we have used before. π‘ is the time in years (the term) π is the annual interest rate π is the # of compounding periods/yr π
is the amount of the periodic pmt π (1+π) π =π
(1+π) π β1 π Loan Amortization Formula
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Loan Amortization Formula
MATH 110 Sec 8-5 Lecture: Annuities β Amortization Amortization is the process of paying off a loan (including interest) by making a series of regular, equal payments. The formula below is used to calculate payments on any amortized loan.Β This is a new formula but all of the variables are the same ones that we have used before. π‘ is the time in years (the term) π is the annual interest rate π is the # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π (1+π) π =π
(1+π) π β1 π Loan Amortization Formula
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Loan Amortization Formula
MATH 110 Sec 8-5 Lecture: Annuities β Amortization Amortization is the process of paying off a loan (including interest) by making a series of regular, equal payments. The formula below is used to calculate payments on any amortized loan.Β This is a new formula but all of the variables are the same ones that we have used before. π‘ is the time in years (the term) π is the annual interest rate π is the # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= π π and π=ππ‘ π (1+π) π =π
(1+π) π β1 π Loan Amortization Formula
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Loan Amortization Formula
MATH 110 Sec 8-5 Lecture: Annuities β Amortization π‘ is the time in years (the term) π is the annual interest rate π is the # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= π π and π=ππ‘ π (1+π) π =π
(1+π) π β1 π Loan Amortization Formula Letβs illustrate the use of this formula with an example.
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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Ms. Rigsby made a $40,000 down payment on a $165,000 house and took out a 20 year mortgage at 6% compounded monthly on the balance. How much will the monthly payments be?
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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Ms. Rigsby made a $40,000 down payment on a $165,000 house and took out a 20 year mortgage at 6% compounded monthly on the balance. How much will the monthly payments be? π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= π π and π=ππ‘ π (1+π) π =π
(1+π) π β1 π
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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Ms. Rigsby made a $40,000 down payment on a $165,000 house and took out a 20 year mortgage at 6% compounded monthly on the balance. How much will the monthly payments be? π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= π π and π=ππ‘ π (1+π) π =π
(1+π) π β1 π There was a $40,000 down payment so the present value of the loan (the amount borrowed) is $125,000 ($165,000 β $40,000).
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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Ms. Rigsby made a $40,000 down payment on a $165,000 house and took out a 20 year mortgage at 6% compounded monthly on the balance. How much will the monthly payments be? π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= π π and π=ππ‘ π (1+π) π =π
(1+π) π β1 π There was a $40,000 down payment so the present value of the loan (the amount borrowed) is $125,000 ($165,000 β $40,000). π=
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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Ms. Rigsby made a $40,000 down payment on a $165,000 house and took out a 20 year mortgage at 6% compounded monthly on the balance. How much will the monthly payments be? π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= π π and π=ππ‘ π (1+π) π =π
(1+π) π β1 π $ (1+π) π =π
(1+π) π β1 π There was a $40,000 down payment so the present value of the loan (the amount borrowed) is $125,000 ($165,000 β $40,000). π=
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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Ms. Rigsby made a $40,000 down payment on a $165,000 house and took out a 20 year mortgage at 6% compounded monthly on the balance. How much will the monthly payments be? r= π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= π π and π=ππ‘ π (1+π) π =π
(1+π) π β1 π $ (1+π) π =π
(1+π) π β1 π There was a $40,000 down payment so the present value of the loan (the amount borrowed) is $125,000 ($165,000 β $40,000). π=
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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Ms. Rigsby made a $40,000 down payment on a $165,000 house and took out a 20 year mortgage at 6% compounded monthly on the balance. How much will the monthly payments be? r= π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= 0.06 π and π=ππ‘ π (1+π) π =π
(1+π) π β1 π $ (1+π) π =π
(1+π) π β1 π There was a $40,000 down payment so the present value of the loan (the amount borrowed) is $125,000 ($165,000 β $40,000). π=
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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Ms. Rigsby made a $40,000 down payment on a $165,000 house and took out a 20 year mortgage at 6% compounded monthly on the balance. How much will the monthly payments be? r= m=12 π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= 0.06 π and π=ππ‘ π (1+π) π =π
(1+π) π β1 π $ (1+π) π =π
(1+π) π β1 π There was a $40,000 down payment so the present value of the loan (the amount borrowed) is $125,000 ($165,000 β $40,000). π=
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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Ms. Rigsby made a $40,000 down payment on a $165,000 house and took out a 20 year mortgage at 6% compounded monthly on the balance. How much will the monthly payments be? r= m=12 π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12π‘ π (1+π) π =π
(1+π) π β1 π $ (1+π) π =π
(1+π) π β1 π There was a $40,000 down payment so the present value of the loan (the amount borrowed) is $125,000 ($165,000 β $40,000). π=
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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Ms. Rigsby made a $40,000 down payment on a $165,000 house and took out a 20 year mortgage at 6% compounded monthly on the balance. How much will the monthly payments be? t= r= m=12 π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12t π (1+π) π =π
(1+π) π β1 π $ (1+π) π =π
(1+π) π β1 π There was a $40,000 down payment so the present value of the loan (the amount borrowed) is $125,000 ($165,000 β $40,000). π=
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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Ms. Rigsby made a $40,000 down payment on a $165,000 house and took out a 20 year mortgage at 6% compounded monthly on the balance. How much will the monthly payments be? t= r= m=12 π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12(20) π (1+π) π =π
(1+π) π β1 π $ (1+π) π =π
(1+π) π β1 π There was a $40,000 down payment so the present value of the loan (the amount borrowed) is $125,000 ($165,000 β $40,000). π=
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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Ms. Rigsby made a $40,000 down payment on a $165,000 house and took out a 20 year mortgage at 6% compounded monthly on the balance. How much will the monthly payments be? t= r= m=12 π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12(20) π (1+π) π =π
(1+π) π β1 π =0.005 $ (1+π) π =π
(1+π) π β1 π =240 There was a $40,000 down payment so the present value of the loan (the amount borrowed) is $125,000 ($165,000 β $40,000). π=
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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Ms. Rigsby made a $40,000 down payment on a $165,000 house and took out a 20 year mortgage at 6% compounded monthly on the balance. How much will the monthly payments be? t= r= m=12 π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12(20) π (1+π) π =π
(1+π) π β1 π =0.005 $ (1+.005) 240 =π
(1+.005) 240 β1 .005 =240 There was a $40,000 down payment so the present value of the loan (the amount borrowed) is $125,000 ($165,000 β $40,000). π=
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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Ms. Rigsby made a $40,000 down payment on a $165,000 house and took out a 20 year mortgage at 6% compounded monthly on the balance. How much will the monthly payments be? t= r= m=12 π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12(20) π (1+π) π =π
(1+π) π β1 π =0.005 $ (1+.005) 240 =π
(1+.005) 240 β1 .005 =240 There was a $40,000 down payment so the present value of the loan (the amount borrowed) is $125,000 ($165,000 β $40,000). π=
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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Ms. Rigsby made a $40,000 down payment on a $165,000 house and took out a 20 year mortgage at 6% compounded monthly on the balance. How much will the monthly payments be? t= r= m=12 π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12(20) π (1+π) π =π
(1+π) π β1 π =0.005 $ (1.005) 240 =π
(1.005) 240 β1 .005 =240 There was a $40,000 down payment so the present value of the loan (the amount borrowed) is $125,000 ($165,000 β $40,000). π=
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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Ms. Rigsby made a $40,000 down payment on a $165,000 house and took out a 20 year mortgage at 6% compounded monthly on the balance. How much will the monthly payments be? t= r= m=12 π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12(20) π (1+π) π =π
(1+π) π β1 π =0.005 $ (1.005) 240 =π
(1.005) 240 β1 .005 =240 $125000( )=π
β1 .005
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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Ms. Rigsby made a $40,000 down payment on a $165,000 house and took out a 20 year mortgage at 6% compounded monthly on the balance. How much will the monthly payments be? t= r= m=12 π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12(20) π (1+π) π =π
(1+π) π β1 π =0.005 $ (1.005) 240 =π
(1.005) 240 β1 .005 =240 $125000( )=π
β1 .005
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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Ms. Rigsby made a $40,000 down payment on a $165,000 house and took out a 20 year mortgage at 6% compounded monthly on the balance. How much will the monthly payments be? t= r= m=12 π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12(20) π (1+π) π =π
(1+π) π β1 π =0.005 $ (1.005) 240 =π
(1.005) 240 β1 .005 =240 $125000( )=π
β1 .005
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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Ms. Rigsby made a $40,000 down payment on a $165,000 house and took out a 20 year mortgage at 6% compounded monthly on the balance. How much will the monthly payments be? t= r= m=12 π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12(20) π (1+π) π =π
(1+π) π β1 π =0.005 $ (1.005) 240 =π
(1.005) 240 β1 .005 =240 $125000( )=π
β1 .005 $125000( )=π
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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Ms. Rigsby made a $40,000 down payment on a $165,000 house and took out a 20 year mortgage at 6% compounded monthly on the balance. How much will the monthly payments be? t= r= m=12 π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12(20) π (1+π) π =π
(1+π) π β1 π =0.005 $ (1.005) 240 =π
(1.005) 240 β1 .005 =240 $125000( )=π
β1 .005 $ =π
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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Ms. Rigsby made a $40,000 down payment on a $165,000 house and took out a 20 year mortgage at 6% compounded monthly on the balance. How much will the monthly payments be? t= r= m=12 π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12(20) π (1+π) π =π
(1+π) π β1 π =0.005 $ (1.005) 240 =π
(1.005) 240 β1 .005 =240 $125000( )=π
β1 .005 $ =π
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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Ms. Rigsby made a $40,000 down payment on a $165,000 house and took out a 20 year mortgage at 6% compounded monthly on the balance. How much will the monthly payments be? t= r= m=12 π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12(20) π (1+π) π =π
(1+π) π β1 π =0.005 $ (1.005) 240 =π
(1.005) 240 β1 .005 =240 $125000( )=π
β1 .005 π
= $ $ =π
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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Ms. Rigsby made a $40,000 down payment on a $165,000 house and took out a 20 year mortgage at 6% compounded monthly on the balance. How much will the monthly payments be? t= r= m=12 π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12(20) π (1+π) π =π
(1+π) π β1 π =0.005 $ (1.005) 240 =π
(1.005) 240 β1 .005 =240 $125000( )=π
β1 .005 π
= $ $ =π
π
=$895.54
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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Ms. Rigsby made a $40,000 down payment on a $165,000 house and took out a 20 year mortgage at 6% compounded monthly on the balance. How much will the monthly payments be? Ms. Rigsbyβs monthly payments will be $
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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Ms. Rigsby made a $40,000 down payment on a $165,000 house and took out a 20 year mortgage at 6% compounded monthly on the balance. How much will the monthly payments be? Ms. Rigsbyβs monthly payments will be $ How much will Ms. Rigsby pay the mortgage company over 20 years?
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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Ms. Rigsby made a $40,000 down payment on a $165,000 house and took out a 20 year mortgage at 6% compounded monthly on the balance. How much will the monthly payments be? Ms. Rigsbyβs monthly payments will be $ How much will Ms. Rigsby pay the mortgage company over 20 years? ππππ΄πΏ ππ΄πΌπ·=πππππ»πΏπ πππ Γ 12 πππ‘π π¦π Γ20 π¦ππ
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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Ms. Rigsby made a $40,000 down payment on a $165,000 house and took out a 20 year mortgage at 6% compounded monthly on the balance. How much will the monthly payments be? Ms. Rigsbyβs monthly payments will be $ How much will Ms. Rigsby pay the mortgage company over 20 years? ππππ΄πΏ ππ΄πΌπ·=πππππ»πΏπ πππ Γ 12 πππ‘π π¦π Γ20 π¦ππ ππππ΄πΏ ππ΄πΌπ·=$895.54Γ 12 π¦π Γ20 π¦ππ
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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Ms. Rigsby made a $40,000 down payment on a $165,000 house and took out a 20 year mortgage at 6% compounded monthly on the balance. How much will the monthly payments be? Ms. Rigsbyβs monthly payments will be $ How much will Ms. Rigsby pay the mortgage company over 20 years? ππππ΄πΏ ππ΄πΌπ·=πππππ»πΏπ πππ Γ 12 πππ‘π π¦π Γ20 π¦ππ ππππ΄πΏ ππ΄πΌπ·=$895.54Γ 12 π¦π Γ20 π¦ππ ππππ΄πΏ ππ΄πΌπ·=$214,929.60
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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Ms. Rigsby made a $40,000 down payment on a $165,000 house and took out a 20 year mortgage at 6% compounded monthly on the balance. How much will the monthly payments be? Ms. Rigsbyβs monthly payments will be $ Ms. Rigsby will pay the company $214, over 20 years.
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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Ms. Rigsby made a $40,000 down payment on a $165,000 house and took out a 20 year mortgage at 6% compounded monthly on the balance. How much will the monthly payments be? Ms. Rigsbyβs monthly payments will be $ Ms. Rigsby will pay the company $214, over 20 years. How much interest will Ms. Rigsby pay the mortgage company?
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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Ms. Rigsby made a $40,000 down payment on a $165,000 house and took out a 20 year mortgage at 6% compounded monthly on the balance. How much will the monthly payments be? Ms. Rigsbyβs monthly payments will be $ Ms. Rigsby will pay the company $214, over 20 years. How much interest will Ms. Rigsby pay the mortgage company? ππππ΄πΏ πΌπππΈπ
πΈππ=ππππ΄πΏ ππ΄πΌπ· βπ΄πππππ π΅ππ
π
πππΈπ·
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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Ms. Rigsby made a $40,000 down payment on a $165,000 house and took out a 20 year mortgage at 6% compounded monthly on the balance. How much will the monthly payments be? Ms. Rigsbyβs monthly payments will be $ Ms. Rigsby will pay the company $214, over 20 years. How much interest will Ms. Rigsby pay the mortgage company? ππππ΄πΏ πΌπππΈπ
πΈππ=ππππ΄πΏ ππ΄πΌπ· βπ΄πππππ π΅ππ
π
πππΈπ· ππππ΄πΏ πΌπππΈπ
πΈππ=$214,929.60βπ΄πππππ π΅ππ
π
πππΈπ·
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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Ms. Rigsby made a $40,000 down payment on a $165,000 house and took out a 20 year mortgage at 6% compounded monthly on the balance. How much will the monthly payments be? Ms. Rigsbyβs monthly payments will be $ Ms. Rigsby will pay the company $214, over 20 years. How much interest will Ms. Rigsby pay the mortgage company? ππππ΄πΏ πΌπππΈπ
πΈππ=ππππ΄πΏ ππ΄πΌπ· βπ΄πππππ π΅ππ
π
πππΈπ· ππππ΄πΏ πΌπππΈπ
πΈππ=$214,929.60βπ΄πππππ π΅ππ
π
πππΈπ· Remember that there was a $40,000 down payment so the present value of the loan (the amount borrowed) is $125,000 ($165,000 β $40,000).
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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Ms. Rigsby made a $40,000 down payment on a $165,000 house and took out a 20 year mortgage at 6% compounded monthly on the balance. How much will the monthly payments be? Ms. Rigsbyβs monthly payments will be $ Ms. Rigsby will pay the company $214, over 20 years. How much interest will Ms. Rigsby pay the mortgage company? ππππ΄πΏ πΌπππΈπ
πΈππ=ππππ΄πΏ ππ΄πΌπ· βπ΄πππππ π΅ππ
π
πππΈπ· ππππ΄πΏ πΌπππΈπ
πΈππ=$214,929.60βπ΄πππππ π΅ππ
π
πππΈπ· Remember that there was a $40,000 down payment so the present value of the loan (the amount borrowed) is $125,000 ($165,000 β $40,000).
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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Ms. Rigsby made a $40,000 down payment on a $165,000 house and took out a 20 year mortgage at 6% compounded monthly on the balance. How much will the monthly payments be? Ms. Rigsbyβs monthly payments will be $ Ms. Rigsby will pay the company $214, over 20 years. How much interest will Ms. Rigsby pay the mortgage company? ππππ΄πΏ πΌπππΈπ
πΈππ=ππππ΄πΏ ππ΄πΌπ· βπ΄πππππ π΅ππ
π
πππΈπ· ππππ΄πΏ πΌπππΈπ
πΈππ=$214,929.60β$125,000.00 Remember that there was a $40,000 down payment so the present value of the loan (the amount borrowed) is $125,000 ($165,000 β $40,000).
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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Ms. Rigsby made a $40,000 down payment on a $165,000 house and took out a 20 year mortgage at 6% compounded monthly on the balance. How much will the monthly payments be? Ms. Rigsbyβs monthly payments will be $ Ms. Rigsby will pay the company $214, over 20 years. How much interest will Ms. Rigsby pay the mortgage company? ππππ΄πΏ πΌπππΈπ
πΈππ=ππππ΄πΏ ππ΄πΌπ· βπ΄πππππ π΅ππ
π
πππΈπ· ππππ΄πΏ πΌπππΈπ
πΈππ=$214,929.60β$125,000.00 ππππ΄πΏ πΌπππΈπ
πΈππ=$89,929.60 Remember that there was a $40,000 down payment so the present value of the loan (the amount borrowed) is $125,000 ($165,000 β $40,000).
47
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Ms. Rigsby made a $40,000 down payment on a $165,000 house and took out a 20 year mortgage at 6% compounded monthly on the balance. How much will the monthly payments be? Ms. Rigsbyβs monthly payments will be $ Ms. Rigsby will pay the company $214, over 20 years. Ms. Rigsby will pay the company $89, in interest.
48
Letβs look at another problem.
MATH 110 Sec 8-5 Lecture: Annuities β Amortization Letβs look at another problem.
49
Letβs look at another problem.
MATH 110 Sec 8-5 Lecture: Annuities β Amortization Bob can afford to spend $200 each month on car payments. If he can get a 4 year auto loan at an annual rate of 12%, what is the highest selling price that can Bob afford for a car? Letβs look at another problem.
50
(Note that the selling price is just the present value of an annuity.)
MATH 110 Sec 8-5 Lecture: Annuities β Amortization Bob can afford to spend $200 each month on car payments. If he can get a 4 year auto loan at an annual rate of 12%, what is the highest selling price that can Bob afford for a car? (Note that the selling price is just the present value of an annuity.)
51
(Note that the selling price is just the present value of an annuity.)
MATH 110 Sec 8-5 Lecture: Annuities β Amortization Bob can afford to spend $200 each month on car payments. If he can get a 4 year auto loan at an annual rate of 12%, what is the highest selling price that can Bob afford for a car? (Note that the selling price is just the present value of an annuity.) This problem stills requires the use of the Loan Amortization Formula but here we know the amount of the loan payment (π
) and must find the selling priceβwhich is just the Present Value (π) of the annuity.
52
(Note that the selling price is just the present value of an annuity.)
MATH 110 Sec 8-5 Lecture: Annuities β Amortization Bob can afford to spend $200 each month on car payments. If he can get a 4 year auto loan at an annual rate of 12%, what is the highest selling price that can Bob afford for a car? (Note that the selling price is just the present value of an annuity.) π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= π π and π=ππ‘ π (1+π) π =π
(1+π) π β1 π This problem stills requires the use of the Loan Amortization Formula but here we know the amount of the loan payment (π
) and must find the selling priceβwhich is just the Present Value (π) of the annuity.
53
(Note that the selling price is just the present value of an annuity.)
MATH 110 Sec 8-5 Lecture: Annuities β Amortization Bob can afford to spend $200 each month on car payments. If he can get a 4 year auto loan at an annual rate of 12%, what is the highest selling price that can Bob afford for a car? (Note that the selling price is just the present value of an annuity.) π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= π π and π=ππ‘ ? π (1+π) π =π
(1+π) π β1 π This problem stills requires the use of the Loan Amortization Formula but here we know the amount of the loan payment (π
) and must find the selling priceβwhich is just the Present Value (π) of the annuity.
54
(Note that the selling price is just the present value of an annuity.)
MATH 110 Sec 8-5 Lecture: Annuities β Amortization Bob can afford to spend $200 each month on car payments. If he can get a 4 year auto loan at an annual rate of 12%, what is the highest selling price that can Bob afford for a car? (Note that the selling price is just the present value of an annuity.) π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= π π and π=ππ‘ ? π (1+π) π =π
(1+π) π β1 π π (1+π) π =$200 (1+π) π β1 π
55
(Note that the selling price is just the present value of an annuity.)
MATH 110 Sec 8-5 Lecture: Annuities β Amortization Bob can afford to spend $200 each month on car payments. If he can get a 4 year auto loan at an annual rate of 12%, what is the highest selling price that can Bob afford for a car? (Note that the selling price is just the present value of an annuity.) r= π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= π π and π=ππ‘ ? π (1+π) π =π
(1+π) π β1 π π (1+π) π =$200 (1+π) π β1 π
56
(Note that the selling price is just the present value of an annuity.)
MATH 110 Sec 8-5 Lecture: Annuities β Amortization m=12 Bob can afford to spend $200 each month on car payments. If he can get a 4 year auto loan at an annual rate of 12%, what is the highest selling price that can Bob afford for a car? (Note that the selling price is just the present value of an annuity.) r= π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= π π and π=ππ‘ ? π (1+π) π =π
(1+π) π β1 π π (1+π) π =$200 (1+π) π β1 π
57
(Note that the selling price is just the present value of an annuity.)
MATH 110 Sec 8-5 Lecture: Annuities β Amortization m=12 Bob can afford to spend $200 each month on car payments. If he can get a 4 year auto loan at an annual rate of 12%, what is the highest selling price that can Bob afford for a car? (Note that the selling price is just the present value of an annuity.) t= r= π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= π π and π=ππ‘ ? π (1+π) π =π
(1+π) π β1 π π (1+π) π =$200 (1+π) π β1 π
58
(Note that the selling price is just the present value of an annuity.)
MATH 110 Sec 8-5 Lecture: Annuities β Amortization m=12 Bob can afford to spend $200 each month on car payments. If he can get a 4 year auto loan at an annual rate of 12%, what is the highest selling price that can Bob afford for a car? (Note that the selling price is just the present value of an annuity.) t= r= π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= 0.12 π and π=ππ‘ ? π (1+π) π =π
(1+π) π β1 π π (1+π) π =$200 (1+π) π β1 π
59
(Note that the selling price is just the present value of an annuity.)
MATH 110 Sec 8-5 Lecture: Annuities β Amortization m=12 Bob can afford to spend $200 each month on car payments. If he can get a 4 year auto loan at an annual rate of 12%, what is the highest selling price that can Bob afford for a car? (Note that the selling price is just the present value of an annuity.) t= r= π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12π‘ ? π (1+π) π =π
(1+π) π β1 π π (1+π) π =$200 (1+π) π β1 π
60
(Note that the selling price is just the present value of an annuity.)
MATH 110 Sec 8-5 Lecture: Annuities β Amortization m=12 Bob can afford to spend $200 each month on car payments. If he can get a 4 year auto loan at an annual rate of 12%, what is the highest selling price that can Bob afford for a car? (Note that the selling price is just the present value of an annuity.) t= r= π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12(4) ? π (1+π) π =π
(1+π) π β1 π π (1+π) π =$200 (1+π) π β1 π
61
(Note that the selling price is just the present value of an annuity.)
MATH 110 Sec 8-5 Lecture: Annuities β Amortization m=12 Bob can afford to spend $200 each month on car payments. If he can get a 4 year auto loan at an annual rate of 12%, what is the highest selling price that can Bob afford for a car? (Note that the selling price is just the present value of an annuity.) t= r= π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12(4) ? π (1+π) π =π
(1+π) π β1 π π (1+π) π =$200 (1+π) π β1 π
62
(Note that the selling price is just the present value of an annuity.)
MATH 110 Sec 8-5 Lecture: Annuities β Amortization m=12 Bob can afford to spend $200 each month on car payments. If he can get a 4 year auto loan at an annual rate of 12%, what is the highest selling price that can Bob afford for a car? (Note that the selling price is just the present value of an annuity.) t= r= π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12(4) ? π (1+π) π =π
(1+π) π β1 π π (1+π) π =$200 (1+π) π β1 π π= and π=48
63
(Note that the selling price is just the present value of an annuity.)
MATH 110 Sec 8-5 Lecture: Annuities β Amortization m=12 Bob can afford to spend $200 each month on car payments. If he can get a 4 year auto loan at an annual rate of 12%, what is the highest selling price that can Bob afford for a car? (Note that the selling price is just the present value of an annuity.) t= r= π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12(4) ? π (1+π) π =π
(1+π) π β1 π π (1+.01) 48 =$200 (1+.01) 48 β1 .01 π= and π=48
64
(Note that the selling price is just the present value of an annuity.)
MATH 110 Sec 8-5 Lecture: Annuities β Amortization m=12 Bob can afford to spend $200 each month on car payments. If he can get a 4 year auto loan at an annual rate of 12%, what is the highest selling price that can Bob afford for a car? (Note that the selling price is just the present value of an annuity.) t= r= π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12(4) ? π (1+π) π =π
(1+π) π β1 π π (1+.01) 48 =$200 (1+.01) 48 β1 .01 π= and π=48
65
(Note that the selling price is just the present value of an annuity.)
MATH 110 Sec 8-5 Lecture: Annuities β Amortization m=12 Bob can afford to spend $200 each month on car payments. If he can get a 4 year auto loan at an annual rate of 12%, what is the highest selling price that can Bob afford for a car? (Note that the selling price is just the present value of an annuity.) t= r= π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12(4) ? π (1+π) π =π
(1+π) π β1 π 1.01= π (1+.01) 48 =$200 (1+.01) 48 β1 .01 1.01= π= and π=48
66
(Note that the selling price is just the present value of an annuity.)
MATH 110 Sec 8-5 Lecture: Annuities β Amortization m=12 Bob can afford to spend $200 each month on car payments. If he can get a 4 year auto loan at an annual rate of 12%, what is the highest selling price that can Bob afford for a car? (Note that the selling price is just the present value of an annuity.) t= r= π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12(4) ? π (1+π) π =π
(1+π) π β1 π π (1+.01) 48 =$200 (1+.01) 48 β1 .01 1.01 1.01 π= and π=48
67
(Note that the selling price is just the present value of an annuity.)
MATH 110 Sec 8-5 Lecture: Annuities β Amortization m=12 Bob can afford to spend $200 each month on car payments. If he can get a 4 year auto loan at an annual rate of 12%, what is the highest selling price that can Bob afford for a car? (Note that the selling price is just the present value of an annuity.) t= r= π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12(4) ? π (1+π) π =π
(1+π) π β1 π π (1.01) 48 =$200 (1.01) 48 β1 .01 π= and π=48
68
(Note that the selling price is just the present value of an annuity.)
MATH 110 Sec 8-5 Lecture: Annuities β Amortization m=12 Bob can afford to spend $200 each month on car payments. If he can get a 4 year auto loan at an annual rate of 12%, what is the highest selling price that can Bob afford for a car? (Note that the selling price is just the present value of an annuity.) t= r= π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12(4) ? π (1+π) π =π
(1+π) π β1 π π (1.01) 48 =$200 (1.01) 48 β1 .01 π= and π=48
69
(Note that the selling price is just the present value of an annuity.)
MATH 110 Sec 8-5 Lecture: Annuities β Amortization m=12 Bob can afford to spend $200 each month on car payments. If he can get a 4 year auto loan at an annual rate of 12%, what is the highest selling price that can Bob afford for a car? (Note that the selling price is just the present value of an annuity.) t= r= π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12(4) ? π (1+π) π =π
(1+π) π β1 π π (1.01) 48 =$200 (1.01) 48 β1 .01 = π= and π=48 =
70
(Note that the selling price is just the present value of an annuity.)
MATH 110 Sec 8-5 Lecture: Annuities β Amortization m=12 Bob can afford to spend $200 each month on car payments. If he can get a 4 year auto loan at an annual rate of 12%, what is the highest selling price that can Bob afford for a car? (Note that the selling price is just the present value of an annuity.) t= r= π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12(4) ? π (1+π) π =π
(1+π) π β1 π π ( ) =$ β1 .01 π= and π=48
71
(Note that the selling price is just the present value of an annuity.)
MATH 110 Sec 8-5 Lecture: Annuities β Amortization m=12 Bob can afford to spend $200 each month on car payments. If he can get a 4 year auto loan at an annual rate of 12%, what is the highest selling price that can Bob afford for a car? (Note that the selling price is just the present value of an annuity.) t= r= π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12(4) ? π (1+π) π =π
(1+π) π β1 π π ( ) =$ β1 .01 π= and π=48
72
(Note that the selling price is just the present value of an annuity.)
MATH 110 Sec 8-5 Lecture: Annuities β Amortization m=12 Bob can afford to spend $200 each month on car payments. If he can get a 4 year auto loan at an annual rate of 12%, what is the highest selling price that can Bob afford for a car? (Note that the selling price is just the present value of an annuity.) t= r= π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12(4) ? π (1+π) π =π
(1+π) π β1 π π ( ) =$ β1 .01 π= and π=48
73
(Note that the selling price is just the present value of an annuity.)
MATH 110 Sec 8-5 Lecture: Annuities β Amortization m=12 Bob can afford to spend $200 each month on car payments. If he can get a 4 year auto loan at an annual rate of 12%, what is the highest selling price that can Bob afford for a car? (Note that the selling price is just the present value of an annuity.) t= r= π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12(4) ? π (1+π) π =π
(1+π) π β1 π π ( ) =$ β1 .01 = π= and π=48
74
(Note that the selling price is just the present value of an annuity.)
MATH 110 Sec 8-5 Lecture: Annuities β Amortization m=12 Bob can afford to spend $200 each month on car payments. If he can get a 4 year auto loan at an annual rate of 12%, what is the highest selling price that can Bob afford for a car? (Note that the selling price is just the present value of an annuity.) t= r= π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12(4) ? π (1+π) π =π
(1+π) π β1 π π ( ) =$ π= and π=48
75
(Note that the selling price is just the present value of an annuity.)
MATH 110 Sec 8-5 Lecture: Annuities β Amortization m=12 Bob can afford to spend $200 each month on car payments. If he can get a 4 year auto loan at an annual rate of 12%, what is the highest selling price that can Bob afford for a car? (Note that the selling price is just the present value of an annuity.) t= r= π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12(4) ? π (1+π) π =π
(1+π) π β1 π π ( ) =$ π= and π=48
76
(Note that the selling price is just the present value of an annuity.)
MATH 110 Sec 8-5 Lecture: Annuities β Amortization m=12 Bob can afford to spend $200 each month on car payments. If he can get a 4 year auto loan at an annual rate of 12%, what is the highest selling price that can Bob afford for a car? (Note that the selling price is just the present value of an annuity.) t= r= π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12(4) ? π (1+π) π =π
(1+π) π β1 π π ( ) =$ π= and π=48 =
77
(Note that the selling price is just the present value of an annuity.)
MATH 110 Sec 8-5 Lecture: Annuities β Amortization m=12 Bob can afford to spend $200 each month on car payments. If he can get a 4 year auto loan at an annual rate of 12%, what is the highest selling price that can Bob afford for a car? (Note that the selling price is just the present value of an annuity.) t= r= π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12(4) ? π (1+π) π =π
(1+π) π β1 π π ( ) =$ π= and π=48
78
(Note that the selling price is just the present value of an annuity.)
MATH 110 Sec 8-5 Lecture: Annuities β Amortization m=12 Bob can afford to spend $200 each month on car payments. If he can get a 4 year auto loan at an annual rate of 12%, what is the highest selling price that can Bob afford for a car? (Note that the selling price is just the present value of an annuity.) t= r= π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12(4) ? π (1+π) π =π
(1+π) π β1 π π ( ) =$ π= and π=48
79
(Note that the selling price is just the present value of an annuity.)
MATH 110 Sec 8-5 Lecture: Annuities β Amortization m=12 Bob can afford to spend $200 each month on car payments. If he can get a 4 year auto loan at an annual rate of 12%, what is the highest selling price that can Bob afford for a car? (Note that the selling price is just the present value of an annuity.) t= r= π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12(4) ? π (1+π) π =π
(1+π) π β1 π π ( ) =$ π ( ) =$ π= and π=48
80
(Note that the selling price is just the present value of an annuity.)
MATH 110 Sec 8-5 Lecture: Annuities β Amortization m=12 Bob can afford to spend $200 each month on car payments. If he can get a 4 year auto loan at an annual rate of 12%, what is the highest selling price that can Bob afford for a car? (Note that the selling price is just the present value of an annuity.) t= r= π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12(4) ? π (1+π) π =π
(1+π) π β1 π π ( ) =$ π ( ) =$ π= and π=48 π= $
81
(Note that the selling price is just the present value of an annuity.)
MATH 110 Sec 8-5 Lecture: Annuities β Amortization m=12 Bob can afford to spend $200 each month on car payments. If he can get a 4 year auto loan at an annual rate of 12%, what is the highest selling price that can Bob afford for a car? (Note that the selling price is just the present value of an annuity.) t= r= π‘ is the time in years (the term) π is the annual interest rate π is # of compounding periods/yr π
is the amount of the periodic pmt π is the Present Value of the annuity π= and π=12(4) ? π (1+π) π =π
(1+π) π β1 π π ( ) =$ π ( ) =$ π= and π=48 π= $ =$
82
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Once the monthly payment on a loan has been calculated, a table called an amortization schedule can be created.
83
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Once the monthly payment on a loan has been calculated, a table called an amortization schedule can be created. An amortization schedule provides a month-by-month breakdown of the payment, showing how much of the payment goes toward interest, how much goes toward reducing the balance owed and how much is still owed on the loan.
84
Letβs illustrate this with an example.
MATH 110 Sec 8-5 Lecture: Annuities β Amortization Once the monthly payment on a loan has been calculated, a table called an amortization schedule can be created. An amortization schedule provides a month-by-month breakdown of the payment, showing how much of the payment goes toward interest, how much goes toward reducing the balance owed and how much is still owed on the loan. Letβs illustrate this with an example.
85
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $
86
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ The monthly payment is fixed, so for every row, the amount in the βPaymentβ column will be $
87
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ The monthly payment is fixed, so for every row, the amount in the βPaymentβ column will be $
88
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $
89
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ Each row represents a month so we must use π= π π to change the 8.9% annual rate to a monthly rate.
90
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ π= π π = Each row represents a month so we must use π= π π to change the 8.9% annual rate to a monthly rate.
91
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ r= π= π π = Each row represents a month so we must use π= π π to change the 8.9% annual rate to a monthly rate.
92
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ r= π= π π = Each row represents a month so we must use π= π π to change the 8.9% annual rate to a monthly rate.
93
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ r= π= π π = Each row represents a month so we must use π= π π to change the 8.9% annual rate to a monthly rate. π=
94
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ r= π= π π = Each row represents a month so we must use π= π π to change the 8.9% annual rate to a monthly rate. π= So, this is the monthly rate.
95
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ r= π= π π = π= So, this is the monthly rate.
96
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ ? r= π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ π= So, this is the monthly rate.
97
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ ? r= π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ π= So, this is the monthly rate.
98
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ ? r= π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ ?=PREV MONTH β² S BALANCEΓπ π= So, this is the monthly rate.
99
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ ? r= π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ ?=PREV MONTH β² S BALANCEΓπ π= ?=$ Γπ So, this is the monthly rate.
100
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ ? r= π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ ?=PREV MONTH β² S BALANCEΓπ π= ?=$ Γπ So, this is the monthly rate.
101
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ ? r= π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ ?=PREV MONTH β² S BALANCEΓπ π= ?=$ Γ So, this is the monthly rate.
102
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ ? r= π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ ?=PREV MONTH β² S BALANCEΓπ π= ?=$ Γ So, this is the monthly rate.
103
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ ? r= π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ ?=PREV MONTH β² S BALANCEΓπ π= ?=$ Γ So, this is the monthly rate. ?=$39.84
104
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ $39.84 r= π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ ?=PREV MONTH β² S BALANCEΓπ π= ?=$ Γ So, this is the monthly rate. ?=$39.84
105
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ $39.84 r= π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ π= So, this is the monthly rate.
106
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ $39.84 r= If the payment is $ and $39.84 of that goes to interest, then the amount paid on principal is $ β $39.84 = $ π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ π= So, this is the monthly rate.
107
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ $39.84 r= If the payment is $ and $39.84 of that goes to interest, then the amount paid on principal is $ β $39.84 = $ β = π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ π= So, this is the monthly rate.
108
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ $39.84 $317.47 r= If the payment is $ and $39.84 of that goes to interest, then the amount paid on principal is $ β $39.84 = $ β = π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ π= So, this is the monthly rate.
109
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ $39.84 $317.47 r= If the payment is $ and $39.84 of that goes to interest, then the amount paid on principal is $ β $39.84 = $ π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ π= So, this is the monthly rate.
110
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ $39.84 $317.47 r= π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ π= So, this is the monthly rate.
111
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. If $ was owed the previous month and then $ was paid on principal, then the new loan balance is $ β $ = $ π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ $39.84 $317.47 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ $39.84 $317.47 r= π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ π= So, this is the monthly rate.
112
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. If $ was owed the previous month and then $ was paid on principal, then the new loan balance is $ β $ = $ π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ $39.84 $317.47 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ $39.84 $317.47 r= β π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ π= So, this is the monthly rate.
113
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. If $ was owed the previous month and then $ was paid on principal, then the new loan balance is $ β $ = $ π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ $39.84 $317.47 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ $39.84 $317.47 $ r= β = π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ π= So, this is the monthly rate.
114
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ $39.84 $317.47 $ r= π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ π= So, this is the monthly rate.
115
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ $39.84 $317.47 $ ? r= π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ π= So, this is the monthly rate.
116
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ $39.84 $317.47 $ ? r= π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ π= π= So, this is the monthly rate.
117
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ $39.84 $317.47 $ ? r= π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ ?=PREV MONTH β² S BALANCEΓπ π= π= So, this is the monthly rate.
118
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ $39.84 $317.47 $ ? r= π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ ?=PREV MONTH β² S BALANCEΓπ π= π= ?=$ Γπ So, this is the monthly rate.
119
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ $39.84 $317.47 $ ? r= π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ ?=PREV MONTH β² S BALANCEΓπ π= π= ?=$ Γπ So, this is the monthly rate.
120
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ $39.84 $317.47 $ ? r= π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ ?=PREV MONTH β² S BALANCEΓπ π= ?=$ Γ So, this is the monthly rate.
121
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ $39.84 $317.47 $ ? r= π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ ?=PREV MONTH β² S BALANCEΓπ π= ?=$ Γ So, this is the monthly rate.
122
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ $39.84 $317.47 $ ? r= π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ ?=PREV MONTH β² S BALANCEΓπ π= ?=$ Γ So, this is the monthly rate. ?=$37.49
123
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ $39.84 $317.47 $ $37.49 r= π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ ?=PREV MONTH β² S BALANCEΓπ π= ?=$ Γ So, this is the monthly rate. ?=$37.49
124
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ $39.84 $317.47 $ $37.49 r= π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ π= So, this is the monthly rate.
125
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ $39.84 $317.47 $ $37.49 r= If the payment is $ and $37.49 of that goes to interest, then the amount paid on principal is $ β $37.49 = $ π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ π= So, this is the monthly rate.
126
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ $39.84 $317.47 $ $37.49 r= If the payment is $ and $37.49 of that goes to interest, then the amount paid on principal is $ β $37.49 = $ β = π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ π= So, this is the monthly rate.
127
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ $39.84 $317.47 $ $37.49 $319.82 r= If the payment is $ and $37.49 of that goes to interest, then the amount paid on principal is $ β $37.49 = $ β = π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ π= So, this is the monthly rate.
128
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ $39.84 $317.47 $ $37.49 $319.82 r= If the payment is $ and $37.49 of that goes to interest, then the amount paid on principal is $ β $37.49 = $ π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ π= So, this is the monthly rate.
129
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ $39.84 $317.47 $ $37.49 $319.82 r= π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ π= So, this is the monthly rate.
130
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. If $ was owed the previous month and then $ was paid on principal, then the new loan balance is $ β $ = $ π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ $39.84 $317.47 $ $37.49 $319.82 r= π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ π= So, this is the monthly rate.
131
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. If $ was owed the previous month and then $ was paid on principal, then the new loan balance is $ β $ = $ π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ $39.84 $317.47 $ $37.49 $319.82 r= π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ π= So, this is the monthly rate.
132
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. If $ was owed the previous month and then $ was paid on principal, then the new loan balance is $ β $ = $ π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ $39.84 $317.47 $ $37.49 $319.82 $ r= β = π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ π= So, this is the monthly rate.
133
MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ $39.84 $317.47 $ $37.49 $319.82 $ r= π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ π= So, this is the monthly rate.
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MATH 110 Sec 8-5 Lecture: Annuities β Amortization
Given the annual interest rate and a line of the amortization schedule for the loan (below), complete the next line of the schedule assuming monthly payments. π=12 Annual Interest Rate Payment Interest Paid Paid on Principal Balance 8.9% $357.31 $42.20 $315.11 $ $39.84 $317.47 $ $37.49 $319.82 $ r= π= π π = The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ The interest paid in a particular month is just: ππ
πΈππΌπππ ππππ π» β² π π΅π΄πΏπ΄ππΆπΈ Γπ π= And if we wanted to, we could continue producing more lines of this table. So, this is the monthly rate.
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