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Exam 14: Amortization of Loans, Residential Mortgages, and Sinking Funds

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A loan of $8380.00 is repaid by equal payments made at the end of every three months for 3 years. If interest is 7% compounded quarterly, find the size of the quarterly payments and construct an amortization schedule showing the total paid and the total cost of the loan.

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8380.00 = PMT 8380.00 = PMT 8380.00 = PMT(10.7395497) $780.29 = PMT Amortization Schedule (done using Excel): 8380.00 = PMT(10.7395497)
$780.29 = PMT
Amortization Schedule (done using Excel): 8380.00 = PMT 8380.00 = PMT(10.7395497) $780.29 = PMT Amortization Schedule (done using Excel):

A factory valued at $100 000.00 is purchased for a down payment of 28% and payments of $4000.00 at the end of every three months. If interest is 9% compounded monthly, calculate the size of the final payment.

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PV = 100 000(0.72) = 72 000.00;PMT = 4000.00; i = PV = 100 000(0.72) = 72 000.00;PMT = 4000.00; i = = 0.0075; I/Y = 9; P/Y = 4;C/Y = 12; c = = 3 p = (1 + 0.0075)<sup>3</sup> - 1 = 0.022669172 72000 = 4000 0.408045094 = 1 - 1.022669172<sup>-n</sup> -n ln 1.022669172 = ln 0.591954906 n = 23.39060393 quarters Programmed solution: Present Value of final payment: PMT = 4000.00; p = 0.022669172; n = 0.39060393 PV = 4000 PV = $1538.23 Programmed solution: Interest is 1538.23(0.022669172) = $34.87 Final Payment is 1538.23 + 34.87 = $1573.10 = 0.0075; I/Y = 9; P/Y = 4;C/Y = 12; c = PV = 100 000(0.72) = 72 000.00;PMT = 4000.00; i = = 0.0075; I/Y = 9; P/Y = 4;C/Y = 12; c = = 3 p = (1 + 0.0075)<sup>3</sup> - 1 = 0.022669172 72000 = 4000 0.408045094 = 1 - 1.022669172<sup>-n</sup> -n ln 1.022669172 = ln 0.591954906 n = 23.39060393 quarters Programmed solution: Present Value of final payment: PMT = 4000.00; p = 0.022669172; n = 0.39060393 PV = 4000 PV = $1538.23 Programmed solution: Interest is 1538.23(0.022669172) = $34.87 Final Payment is 1538.23 + 34.87 = $1573.10 = 3
p = (1 + 0.0075)3 - 1 = 0.022669172
72000 = 4000 PV = 100 000(0.72) = 72 000.00;PMT = 4000.00; i = = 0.0075; I/Y = 9; P/Y = 4;C/Y = 12; c = = 3 p = (1 + 0.0075)<sup>3</sup> - 1 = 0.022669172 72000 = 4000 0.408045094 = 1 - 1.022669172<sup>-n</sup> -n ln 1.022669172 = ln 0.591954906 n = 23.39060393 quarters Programmed solution: Present Value of final payment: PMT = 4000.00; p = 0.022669172; n = 0.39060393 PV = 4000 PV = $1538.23 Programmed solution: Interest is 1538.23(0.022669172) = $34.87 Final Payment is 1538.23 + 34.87 = $1573.10 0.408045094 = 1 - 1.022669172-n
-n ln 1.022669172 = ln 0.591954906
n = 23.39060393 quarters
Programmed solution: PV = 100 000(0.72) = 72 000.00;PMT = 4000.00; i = = 0.0075; I/Y = 9; P/Y = 4;C/Y = 12; c = = 3 p = (1 + 0.0075)<sup>3</sup> - 1 = 0.022669172 72000 = 4000 0.408045094 = 1 - 1.022669172<sup>-n</sup> -n ln 1.022669172 = ln 0.591954906 n = 23.39060393 quarters Programmed solution: Present Value of final payment: PMT = 4000.00; p = 0.022669172; n = 0.39060393 PV = 4000 PV = $1538.23 Programmed solution: Interest is 1538.23(0.022669172) = $34.87 Final Payment is 1538.23 + 34.87 = $1573.10 Present Value of final payment:
PMT = 4000.00; p = 0.022669172; n = 0.39060393
PV = 4000 PV = 100 000(0.72) = 72 000.00;PMT = 4000.00; i = = 0.0075; I/Y = 9; P/Y = 4;C/Y = 12; c = = 3 p = (1 + 0.0075)<sup>3</sup> - 1 = 0.022669172 72000 = 4000 0.408045094 = 1 - 1.022669172<sup>-n</sup> -n ln 1.022669172 = ln 0.591954906 n = 23.39060393 quarters Programmed solution: Present Value of final payment: PMT = 4000.00; p = 0.022669172; n = 0.39060393 PV = 4000 PV = $1538.23 Programmed solution: Interest is 1538.23(0.022669172) = $34.87 Final Payment is 1538.23 + 34.87 = $1573.10 PV = $1538.23
Programmed solution: PV = 100 000(0.72) = 72 000.00;PMT = 4000.00; i = = 0.0075; I/Y = 9; P/Y = 4;C/Y = 12; c = = 3 p = (1 + 0.0075)<sup>3</sup> - 1 = 0.022669172 72000 = 4000 0.408045094 = 1 - 1.022669172<sup>-n</sup> -n ln 1.022669172 = ln 0.591954906 n = 23.39060393 quarters Programmed solution: Present Value of final payment: PMT = 4000.00; p = 0.022669172; n = 0.39060393 PV = 4000 PV = $1538.23 Programmed solution: Interest is 1538.23(0.022669172) = $34.87 Final Payment is 1538.23 + 34.87 = $1573.10 Interest is 1538.23(0.022669172) = $34.87
Final Payment is 1538.23 + 34.87 = $1573.10

Rola Inc. borrowed $42 000.00 at 7% compounded semi-annually. The loan is repaid by payments of $4700.00 due at the end of every six months. a) How many payments are needed? b) How much of the principal will be repaid by the 6th payment? c) Prepare a partial amortization schedule showing the details of the last three payments and totals.

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a) n = a) n = = = b) PV<sub>n</sub> = 4700 c) Partial Amortization (rounded the nearest cent): = a) n = = = b) PV<sub>n</sub> = 4700 c) Partial Amortization (rounded the nearest cent): = a) n = = = b) PV<sub>n</sub> = 4700 c) Partial Amortization (rounded the nearest cent): a) n = = = b) PV<sub>n</sub> = 4700 c) Partial Amortization (rounded the nearest cent): b) PVn = 4700 a) n = = = b) PV<sub>n</sub> = 4700 c) Partial Amortization (rounded the nearest cent): a) n = = = b) PV<sub>n</sub> = 4700 c) Partial Amortization (rounded the nearest cent): c) Partial Amortization (rounded the nearest cent): a) n = = = b) PV<sub>n</sub> = 4700 c) Partial Amortization (rounded the nearest cent): a) n = = = b) PV<sub>n</sub> = 4700 c) Partial Amortization (rounded the nearest cent): a) n = = = b) PV<sub>n</sub> = 4700 c) Partial Amortization (rounded the nearest cent): a) n = = = b) PV<sub>n</sub> = 4700 c) Partial Amortization (rounded the nearest cent):

Karen receives pension payments of $4570.00 at the end of every six months from a retirement fund of $82 000.00. The fund earns 8.06% compounded semi-annually. a) How many payments will Karen receive? b) What is the size of the final pension payment?

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Question 4

Munawar is a first time buyer and bought a brand new townhouse in Oshawa for $212 000. He has saved enough to make a 5% down payment and will have to pay mortgage loan insurance at 2.75% of the mortgage balance. Munawar wants the insurance premium added to the maximum allowable loan balance. Determine his maximum initial mortgage balance.

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Question 5

A debt of $41 000.00 is repaid in monthly installments of $660.00. If interest is 8.4% compounded quarterly, what is the size of the final payment?

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Question 6

Li bought a house in Whitby for $305 000 5 years ago. He financed the house through Scotia Bank with a mortgage of $259 000 amortized over 25 years. Interest for the first 5 years was 4.4% compounded semi-annually and payments were made monthly. At the end of the 5 year term, Li has renewed the mortgage for another 5 years, at 3.7% compounded semi-annually. What is the monthly payment for the second term?

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Question 7

A $248 000.00 mortgage amortized by monthly payments over 35 years is renewable after five years. Interest is 8.12% compounded semi-annually. a) What is the size of the monthly payments? b) How much interest is paid during the first year? c) How much of the principal is repaid during the first five-year term? d) If the mortgage is renewed for a further five-year term at 7.16% compounded semi-annually, what will be the size of the monthly payments?

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Question 8

Zara bought a 2013 Kia Forte by taking a $19 500 loan from BMO with interest at 5.99% compounded annually and amortized by payments of $378 at the end of every month. She receives an inheritance after making 36 payments and plans to pay the balance in full. What is the payout figure just after the last regular payment made?

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Question 9

Tariq bought a commercial property valued at $130 000 for $26 000.00 down and a mortgage amortized over 25 years. He makes equal payments due at the end of every month. Interest on the mortgage is 4.1% compounded semi-annually and the mortgage is renewable after five years. What is the cost of the mortgage for the first five years?

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Question 10

What is the outstanding balance after 23 payments on a 20-year mortgage that has monthly payments of $1062.32 and an interest rate of 6.6% compounded monthly?

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Question 11

A debt of $10 000.00 with interest at 8% compounded quarterly is to be repaid by equal payments at the end of every three months for two years. a) Calculate the size of the monthly payments. b) Construct an amortization table. c) Calculate the outstanding balance after three payments.

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Question 12

What is the size of the final unequal payment of a loan that has 69.47 payments and the equal payment size is $215.64? The interest rate is 3.1% per compounding period.

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Question 13

The Korean community is planning to buy a place for their community in Markham valued at $1 697 000. They made a down payment of 10% and payments of $400 000 are required at the end of every 3 months. If interest is 9% compounded monthly, what is the size of the final payment

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Question 14

A contract valued at $47 500.00 requires payments of $6951.00 every six months. The first payment is due in four years and interest is 6.6% compounded semi-annually. a) How many payments are required? b) What is the size of the last payment? c) How much will be paid in total? d) How much of what is paid is interest?

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Question 15

Angelina has agreed to purchase her partner's share in the business by making payments of $1720.00 every three months. The agreed transfer value is $19 500.00 and interest is 9.11% compounded annually. If the first payment is due at the date of the agreement, what is the size of the final payment?

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Question 16

Olfert Inc. is repaying a loan of $52 500.00 by making payments of $4700.00 at the end of every six months. If interest is 7.5% compounded semi-annually, construct an amortization schedule showing the total paid and the total cost of the loan.

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Question 17

A bank has an interest rate of 5% compounded semi annually for a five year closed mortgage. If the mortgage has monthly payments, calculate the effective rate of interest per each term.

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Question 18

Zara bought a 2013 Kia Forte by taking a $19 500 loan from BMO with interest at 5.99% compounded annually and amortized by payments of $378 at the end of every month. She receives an inheritance after making 36 payments and plans to pay the balance in full. What is the payout figure just after the last regular payment made?

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Question 19

How much interest is paid after the first payment interval on a $100 000 mortgage? The mortgage is amortized over 25 years and the payments are monthly. The interest rate is 6% compounded semi-annually. Round your answer to the nearest dollar.

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